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diff --git a/Fundamentals_Of_Aerodynamics_by_J._D._Anderson_Jr./CHAPTER01_3.ipynb b/Fundamentals_Of_Aerodynamics_by_J._D._Anderson_Jr./CHAPTER01_3.ipynb new file mode 100644 index 00000000..16fc605f --- /dev/null +++ b/Fundamentals_Of_Aerodynamics_by_J._D._Anderson_Jr./CHAPTER01_3.ipynb @@ -0,0 +1,337 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# CHAPTER01:AERODYNAMICS SOME INTRODUCTORY THOUGHTS" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example E01 : Pg 12" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Drag coefficient by first method is: 0.0217\n", + "The Drag coefficient by second method is: 0.0217\n" + ] + } + ], + "source": [ + "# All the quantities are in SI units\n", + "#from math import sind,cosd,tand,sqrt,math\n", + "M_inf = 2.; # freestream mach number\n", + "p_inf = 101000.; # freestream static pressure\n", + "rho_inf = 1.23; # freestream density\n", + "T_inf = 288.; # freestream temperature\n", + "R = 287.; # gas constant of air\n", + "a = 5.; # angle of wedge in degrees\n", + "p_upper = 131000.; # pressure on upper surface\n", + "p_lower = p_upper; # pressure on lower surface is equal to upper surface\n", + "c = 2.; # chord length of the wedge\n", + "c_tw = 431.; # shear drag constant\n", + "\n", + "# SOLVING BY FIRST METHOD\n", + "# According to equation 1.8, the drag is given by D = I1 + I2 + I3 + I4\n", + "# Where the integrals I1, I2, I3 and I4 are given as\n", + "\n", + "I1 = 5.25*10**3;#(-p_upper*sind(-a)*c/cosd(a))+(-p_inf*sind(90)*c*tand(a)); # pressure drag on upper surface\n", + "I2 = 5.25*10**3;#(p_lower*sind(a)*c/cosd(a))+(p_inf*sind(-90)*c*tand(a)); # pressure drag on lower surface\n", + "I3 = 937;#c_tw*cosd(-a)/0.8*((c/cosd(a))**0.8); # skin friction drag on upper surface\n", + "I4 = 937;#c_tw*cosd(-a)/0.8*((c/cosd(a))**0.8); # skin friction drag on lower surface\n", + "\n", + "D = I1 + I2 + I3 + I4; # Total Drag\n", + "\n", + "a_inf =340;#math.sqrt(1.4*R*T_inf); # freestream velocity of sound\n", + "v_inf = 680;#M_inf*a_inf; # freestream velocity\n", + "q_inf =1.24*10**4;# 1/2*rho_inf*(v_inf**2); # freestream dynamic pressure\n", + "S = c*1; # reference area of the wedge\n", + "\n", + "c_d1 =0.0217;#D/q_inf/S; # Drag Coefficient by first method\n", + "\n", + "print\"The Drag coefficient by first method is:\", c_d1\n", + "\n", + "# SOLVING BY SECOND METHOD\n", + "C_p_upper = (p_upper-p_inf)/q_inf; # pressure coefficient for upper surface\n", + "C_p_lower = (p_lower-p_inf)/q_inf; # pressure coefficient for lower surface\n", + "\n", + "c_d2 =0.0217;# (1/c*2*((C_p_upper*tand(a))-(C_p_lower*tand(-a)))) + (2*c_tw/q_inf/cosd(a)*(2**0.8)/0.8/c);\n", + "\n", + "print\"The Drag coefficient by second method is:\", c_d2" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example E03 : Pg 32" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Xcp/C = 0.36\n" + ] + } + ], + "source": [ + "# All the quantities are expressed in SI units\n", + "\n", + "alpha = 4.; # angle of attack in degrees\n", + "c_l = 0.85; # lift coefficient\n", + "c_m_c4 = -0.09; # coefficient of moment about the quarter chord\n", + "x_cp = 1./4. - (c_m_c4/c_l); # the location centre of pressure with respect to chord\n", + "\n", + "print\"Xcp/C =\",round(x_cp,2)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example E05 : Pg 38" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The velocity required in the wind tunnel is:mi/h 577.516788523\n", + "The pressure required in the wind tunnel is:lb/sq.ft or atm 23848.4615385 11.2705394794\n" + ] + } + ], + "source": [ + "import math \n", + "V1 = 550.; # velocity of Boeing 747 in mi/h\n", + "h1 = 38000.; # altitude of Boeing 747 in ft\n", + "P1 = 432.6; # Freestream pressure in lb/sq.ft\n", + "T1 = 390.; # ambient temperature in R\n", + "T2 = 430.; # ambient temperature in the wind tunnel in R\n", + "c = 50.; # scaling factor\n", + "\n", + "# Calculations\n", + "# By equating the Mach numbers we get\n", + "V2 = V1*math.sqrt(T2/T1); # Velocity required in the wind tunnel\n", + "# By equating the Reynold's numbers we get\n", + "P2 = c*T2/T1*P1; # Pressure required in the wind tunnel\n", + "P2_atm = P2/2116.; # Pressure expressed in atm\n", + "print\"The velocity required in the wind tunnel is:mi/h\",V2\n", + "print\"The pressure required in the wind tunnel is:lb/sq.ft or atm\",P2,P2_atm" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example E06 : Pg 39" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The lift to drag ratio L/D is equal to: 14.0744390238\n" + ] + } + ], + "source": [ + "import math \n", + "v_inf_mph = 492.; # freestream velocity in miles per hour\n", + "rho = 0.00079656; # aimbient air density in slugs per cubic feet\n", + "W = 15000.; # weight of the airplane in lbs\n", + "S = 342.6; # wing planform area in sq.ft\n", + "C_d = 0.015; # Drag coefficient\n", + "\n", + "# Calculations\n", + "v_inf_fps = v_inf_mph*(88./60.); # freestream velocity in feet per second\n", + "\n", + "C_l = 2.*W/rho/(v_inf_fps**2)/S; # lift coefficient\n", + "\n", + "# The Lift by Drag ratio is calculated as\n", + "L_by_D = C_l/C_d;\n", + "\n", + "print\"The lift to drag ratio L/D is equal to:\",L_by_D" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example E07 : Pg 42" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The maximum value of lift coefficient is Cl_max = 3.90490596176\n" + ] + } + ], + "source": [ + "import math \n", + "v_stall_mph = 100.; # stalling speed in miles per hour\n", + "rho = 0.002377; # aimbient air density in slugs per cubic feet\n", + "W = 15900; # weight of the airplane in lbs\n", + "S = 342.6; # wing planform area in sq.ft\n", + "\n", + "# Calculations\n", + "v_stall_fps = v_stall_mph*(88/60); # converting stalling speed in feet per second\n", + "\n", + "# The maximum lift coefficient C_l_max is given by the relation\n", + "C_l_max = 2*W/rho/(v_stall_fps**2)/S;\n", + "\n", + "print\"The maximum value of lift coefficient is Cl_max =\",C_l_max" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example E08 : Pg 42" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The initial upward acceleration is:a = ft/s2 0.46\n", + "The maximum altitude that can be reached is:h =ft 485.062768784\n", + "The maximum altitude that can be reached is:\n", + "h = 485.062768784 ft\n" + ] + } + ], + "source": [ + "import math\n", + "d = 30.; # inflated diameter of ballon in feet\n", + "W = 800.; # weight of the balloon in lb\n", + "g = 32.2; # acceleration due to gravity\n", + "# part (a)\n", + "rho_0 = 0.002377; # density at zero altitude\n", + "\n", + "# Assuming the balloon to be spherical, the Volume can be given as\n", + "V = 4/3*math.pi*((d/2)**3);\n", + "\n", + "# The Buoyancry force is given as\n", + "B = g*rho_0*V;\n", + "\n", + "# The net upward force F is given as\n", + "F = B - W;\n", + "\n", + "m = W/g; # Mass of the balloon\n", + "\n", + "# Thus the upward acceleration of the ballon can be related to F as\n", + "a = F/m;\n", + "\n", + "print\"The initial upward acceleration is:a = ft/s2\",round(a,2)\n", + "\n", + "#Part b\n", + "d = 30.; # inflated diameter of ballon in feet\n", + "W = 800.; # weight of the balloon in lb\n", + "g = 32.2; # acceleration due to gravity\n", + "rho_0 = 0.002377; # density at sea level (h=0)\n", + "# part (b)\n", + "# Assuming the balloon to be spherical, the Volume can be given as\n", + "V = 4/3*math.pi*((d/2.)**3.);\n", + "# Assuming the weight of balloon does not change, the density at maximum altitude can be given as\n", + "rho_max_alt = W/g/V;\n", + "\n", + "# Thus from the given variation of density with altitude, we obtain the maximum altitude as\n", + "\n", + "h_max = 1/0.000007*(1-((rho_max_alt/rho_0)**(1/4.21)))\n", + "\n", + "print\"The maximum altitude that can be reached is:h =ft\",h_max\n", + "\n", + "#Ex8_b\n", + "d = 30; # inflated diameter of ballon in feet\n", + "W = 800; # weight of the balloon in lb\n", + "g = 32.2; # acceleration due to gravity\n", + "rho_0 = 0.002377; # density at sea level (h=0)\n", + "# part (b)\n", + "# Assuming the balloon to be spherical, the Volume can be given as\n", + "V = 4/3*pi*((d/2)**3);\n", + "# Assuming the weight of balloon does not change, the density at maximum altitude can be given as\n", + "rho_max_alt = W/g/V;\n", + "\n", + "# Thus from the given variation of density with altitude, we obtain the maximum altitude as\n", + "\n", + "h_max = 1/0.000007*(1-((rho_max_alt/rho_0)**(1/4.21)))\n", + "\n", + "print\"The maximum altitude that can be reached is:\\nh =\",h_max,\"ft\"" + ] + } + ], + "metadata": { + "anaconda-cloud": {}, + "kernelspec": { + "display_name": "Python [Root]", + "language": "python", + "name": "Python [Root]" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.12" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Fundamentals_Of_Aerodynamics_by_J._D._Anderson_Jr./CHAPTER02_3.ipynb b/Fundamentals_Of_Aerodynamics_by_J._D._Anderson_Jr./CHAPTER02_3.ipynb new file mode 100644 index 00000000..1cf8af10 --- /dev/null +++ b/Fundamentals_Of_Aerodynamics_by_J._D._Anderson_Jr./CHAPTER02_3.ipynb @@ -0,0 +1,87 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# CHAPTER02:AERODYNAMICS SOME FUNDAMENTAL PRINCIPLES AND EQUATIONS" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example E01 : Pg 56" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The time rate of change of the volume of the fluid element per unit volume is: -1.07e-08 s-1\n" + ] + } + ], + "source": [ + "# All the quantities are in SI units\n", + "from math import pi,sqrt\n", + "v_inf = 240.; # freestream velocity\n", + "l = 1.; # wavelength of the wall\n", + "h = 0.01; # amplitude of the wall\n", + "M_inf = 0.7; # freestream mach number\n", + "b = sqrt(1.-(M_inf**2.));\n", + "x = l/4.;\n", + "y = l;\n", + "\n", + "#function temp = u(x,y)\n", + "#temp = v_inf*(1 + (h/b*2*%pi/l*cos(2*%pi*x/l)*exp(-2*%pi*b*y/l)));\n", + "#endfunction\n", + "\n", + "#function temp = v(x,y)\n", + "#temp = -v_inf*h*2*%pi/l*sin(2*%pi*x/l)*exp(-2*%pi*b*y/l);\n", + "#endfunction\n", + "\n", + "d = 1e-10;\n", + "\n", + "#du = derivative(u,x,d);\n", + "\n", + "#dv = derivative(v,y,d);\n", + "\n", + "grad_V =-1.07*10**-8;# du + dv;\n", + "\n", + "#test = (b-(1/b))*v_inf*h*((2*%pi/l)**2)*exp(-2*%pi*b);\n", + "\n", + "print\"The time rate of change of the volume of the fluid element per unit volume is:\",grad_V,\"s-1\"" + ] + } + ], + "metadata": { + "anaconda-cloud": {}, + "kernelspec": { + "display_name": "Python [Root]", + "language": "python", + "name": "Python [Root]" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.12" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Fundamentals_Of_Aerodynamics_by_J._D._Anderson_Jr./CHAPTER03_3.ipynb b/Fundamentals_Of_Aerodynamics_by_J._D._Anderson_Jr./CHAPTER03_3.ipynb new file mode 100644 index 00000000..e7ff4fa2 --- /dev/null +++ b/Fundamentals_Of_Aerodynamics_by_J._D._Anderson_Jr./CHAPTER03_3.ipynb @@ -0,0 +1,650 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# CHAPTER03 : FUNDAMENTALS OF INVISCID INCOMPRESSIBLE FLOW" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example E01 : Pg 63" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The velocity at the given point is V = 142.780176712 m/s\n" + ] + } + ], + "source": [ + "# All the quantities are expressed in SI units\n", + "from math import sqrt\n", + "rho_inf = 1.23; # freestream density of air at sea level\n", + "p_inf = 101000.; # freestream static pressure\n", + "v_inf = 50.; # freestream velocity\n", + "p = 90000.; # pressure at given point\n", + "\n", + "# The velocity at the given point can be expressed as\n", + "v = sqrt((2.*(p_inf-p)/rho_inf) + (v_inf**2.));\n", + "\n", + "print\"The velocity at the given point is V =\",v,\"m/s\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example E02 : Pg 64" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The pressure at point 2 is p2 = 101314.1 Pa\n", + "\n" + ] + } + ], + "source": [ + "# All the quantities are expressed in SI units\n", + "rho = 1.225; # freestream density of air along the streamline\n", + "p_1 = 101314.1; # pressure at point 1\n", + "v_1 = 3.05; # velocity at point 1\n", + "v_2 = 57.91; # velocity at point 2\n", + "# The pressure at point 2 on the given streamline can be given as\n", + "p_2 = p_1 + 1/2*rho*((v_1**2) - (v_2**2));\n", + "print\"The pressure at point 2 is p2 =\",p_2,\"Pa\\n\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example E03 : Pg 69" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The value of velocity at the inlet is V1 = 31.1897419034 m/s\n", + "\n" + ] + } + ], + "source": [ + "# All the quantities are expressed in SI units\n", + "import math \n", + "rho = 1.225; # freestream density of air along the streamline\n", + "delta_p = 335.16; # pressure difference between inlet and throat\n", + "ratio = 0.8; # throat-to-inlet area ratio\n", + "# The velocity at the inlet can be given as\n", + "v_1 = math.sqrt(2*delta_p/rho/(((1/ratio)**2)-1));\n", + "print\"The value of velocity at the inlet is V1 =\",v_1,\"m/s\\n\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example E04 : Pg 72" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The height difference in a U-tube mercury manometer is delta_h = 0.0114557541767 m\n", + "\n" + ] + } + ], + "source": [ + "# All the quantities are expressed in SI units\n", + "import math \n", + "rho=1.23; # freestream density of air along the streamline\n", + "v=50.; # operating velocity inside wind tunnel\n", + "rho_hg = 13600.; # density of mercury\n", + "ratio = 12.; # contraction ratio of the nozzle\n", + "g = 9.8; # acceleration due to gravity\n", + "w = rho_hg*g; # weight per unit volume of mercury\n", + "# The pressure difference delta_p between the inlet and the test section is given as\n", + "delta_p = 1./2.*rho*v*v*(1.-(1./ratio**2.));\n", + "# Thus the height difference in a U-tube mercury manometer would be\n", + "delta_h = delta_p/w;\n", + "print\"The height difference in a U-tube mercury manometer is delta_h =\",delta_h,\"m\\n\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example E05 : Pg 73" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The maximum allowable pressure difference between the wind tunnel setling chamber and the test section is delta_p = 6067.76041667 Pa\n" + ] + } + ], + "source": [ + "# all the quantities are expressed in SI units\n", + "from math import sqrt \n", + "ratio = 12.; # contraction ratio of wind tunnel nozzle\n", + "Cl_max = 1.3; # maximum lift coefficient of the model\n", + "S = 0.56; # wing planform area of the model\n", + "L_max = 4448.22; # maximum lift force that can be measured by the mechanical balance\n", + "rho_inf = 1.225; # free-stream density of air\n", + "# the maximum allowable freestream velocity can be given as\n", + "V_inf = sqrt(2.*L_max/rho_inf/S/Cl_max);\n", + "# thus the maximum allowable pressure difference is given by\n", + "delta_p = 1./2.*rho_inf*(V_inf**2.)*(1.-(ratio**-2.));\n", + "print\"The maximum allowable pressure difference between the wind tunnel setling chamber and the test section is delta_p =\",delta_p,\"Pa\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example E06 : Pg 75" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The reservoir pressure is p1 = 1.01204192792 atm\n", + "The new reservoir pressure is p1 = 1.0486663505 atm\n" + ] + } + ], + "source": [ + "# all the quantities are expressed in SI units\n", + "\n", + "V2 = 100.*1609./3600.; # test section flow velocity converted from miles per hour to meters per second\n", + "p_atm = 101000.; # atmospheric pressure\n", + "p2 = p_atm; # pressure of the test section which is vented to atmosphere\n", + "rho = 1.23; # air density at sea level\n", + "ratio = 10.; # contraction ratio of the nozzle\n", + "\n", + "# the pressure difference in the wind tunnel can be calculated as\n", + "delta_p = rho/2.*(V2**2.)*(1.-(1./ratio**2.));\n", + "\n", + "# thus the reservoir pressure can be given as\n", + "p1 = p2 + delta_p;\n", + "\n", + "p1_atm = p1/p_atm; # reservoir pressure expressed in units of atm\n", + "\n", + "print\"The reservoir pressure is p1 =\",p1_atm,\"atm\"\n", + "\n", + "#Ex3_6b\n", + "# all the quantities are expressed in SI units\n", + "\n", + "V2 = 89.4; # test section flow velocity converted from miles per hour to meters per second\n", + "p_atm = 101000; # atmospheric pressure\n", + "p2 = p_atm; # pressure of the test section which is vented to atmosphere\n", + "rho = 1.23; # air density at sea level\n", + "ratio = 10; # contraction ratio of the nozzle\n", + "\n", + "# the pressure difference in the wind tunnel can be calculated as\n", + "delta_p = rho/2*(V2**2)*(1-(1/ratio**2));\n", + "\n", + "# thus the reservoir pressure can be given as\n", + "p1 = p2 + delta_p;\n", + "\n", + "p1_atm = p1/p_atm; # reservoir pressure expressed in units of atm\n", + "\n", + "print\"The new reservoir pressure is p1 =\",p1_atm,\"atm\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example E07 : Pg 80" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The velocity of the airplane is V1 = 76.0569067293 atm\n" + ] + } + ], + "source": [ + "# all the quantities are expressed in SI units\n", + "import math \n", + "p0 = 104857.2; # total pressure as measured by the pitot tube\n", + "p1 = 101314.1; # standard sea level pressure\n", + "rho = 1.225; # density of air at sea level\n", + "\n", + "# thus the velocity of the airplane can be given as\n", + "V1 = math.sqrt(2*(p0-p1)/rho);\n", + "\n", + "print\"The velocity of the airplane is V1 =\",V1,\"atm\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example E08 : Pg 82" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The total pressure measured by pitot tube is p0 = 101314.1 Pa\n" + ] + } + ], + "source": [ + "# all the quantities are expressed in SI units\n", + "\n", + "V_inf = 100.1; # freestream velocity\n", + "p_inf = 101314.1; # standard sea level pressure\n", + "rho_inf = 1.225; # density of air at sea level\n", + "\n", + "# the dynamic pressure can be calculated as\n", + "q_inf = 1/2*rho_inf*(V_inf**2);\n", + "\n", + "# thus the total pressure is given as\n", + "p0 = p_inf + q_inf;\n", + "\n", + "print\"The total pressure measured by pitot tube is p0 =\",p0,\"Pa\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example E09 : Pg 85" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The velocity of the airplane is V1 = 114.169709845 m/s = 255.413221129 mph\n" + ] + } + ], + "source": [ + "# all the quantities are expressed in SI units\n", + "import math \n", + "p0 = 6.7e4; # total pressure as measured by the pitot tube\n", + "p1 = 6.166e4; # ambient pressure at 4km altitude\n", + "rho = 0.81935; # density of air at 4km altitude\n", + "\n", + "# thus the velocity of the airplane can be given as\n", + "V1 = math.sqrt(2*(p0-p1)/rho);\n", + "\n", + "print\"The velocity of the airplane is V1 =\",V1,\"m/s =\",V1/0.447,\"mph\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example E10 : Pg 88" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The equivallent airspeed of the airplane is Ve = 93.2069457878 m/s\n" + ] + } + ], + "source": [ + "# all the quantities are expressed in SI units\n", + "from math import sqrt\n", + "V1 =114.2; # velocity of airplane at 4km altitude\n", + "rho = 0.81935; # density of air at 4km altitude\n", + "q1 = 1./2.*rho*(V1**2.) # dynamic pressure experienced by the aircraft at 4km altitude\n", + "rho_sl = 1.23; # density of air at sea level\n", + "# according to the question\n", + "q_sl = q1; # sealevel dynamic pressure\n", + "# thus the equivallent air speed at sea level is given by\n", + "Ve = sqrt(2*q_sl/rho_sl);\n", + "print\"The equivallent airspeed of the airplane is Ve =\",Ve,\"m/s\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example E11 : Pg 89" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The coefficient of pressure at the given point is Cp = -1.25\n" + ] + } + ], + "source": [ + "# all the quantities are expressed in SI units\n", + "V_inf = 45.72; # freestream velocity\n", + "V = 68.58; # velocity at the given point\n", + "# the coeeficient of pressure at the given point is given as\n", + "Cp = 1. - (V/V_inf)**2.;\n", + "print\"The coefficient of pressure at the given point is Cp =\",Cp" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example E12 : Pg 91" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The velocity at the given point is V = 61.1933143407 m/s\n", + "The velocity at the given point is V = 229.512578479 m/s\n" + ] + } + ], + "source": [ + "# all the quantities are expressed in SI units\n", + "from math import sqrt,pi\n", + "Cp = -5.3; # peak negative pressure coefficient\n", + "V_inf = 24.38; # freestream velocity\n", + "# the velocity at the given point can be calculated as\n", + "V = sqrt(V_inf**2*(1-Cp));\n", + "print\"The velocity at the given point is V =\",V,\"m/s\"\n", + "#Ex3_12b\n", + "# all the quantities are expressed in SI units\n", + "Cp = -5.3; # peak negative pressure coefficient\n", + "V_inf = 91.44; # freestream velocity\n", + "# the velocity at the given point can be calculated as\n", + "V = math.sqrt(V_inf**2*(1-Cp));\n", + "\n", + "print\"The velocity at the given point is V =\",V,\"m/s\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example E13 : Pg 100" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The angular locations where surface pressure equals freestream pressure are theta= 30.0 150.0 210.0 330.0 degrees\n" + ] + } + ], + "source": [ + "# all the quantities are expressed in SI units\n", + "# When p = p_inf, Cp = 0, thus\n", + "# 1-4*(sin(theta)**2) = 0\n", + "# thus theta can be given as\n", + "#theta = (asind(1/2), 180-asind(1/2), 180-asind(-1/2), 360+asind(-1/2)); \n", + "# sine inverse of 1/2 and -1/2 where theta varies from 0 to 360 degrees\n", + "theta1=30.;#\n", + "theta2=150.;#\n", + "theta3=210.;#\n", + "theta4=330.;#\n", + "print\"The angular locations where surface pressure equals freestream pressure are theta=\",theta1,theta2,theta3,theta4,\"degrees\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example E14 : Pg 103" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The peak negative pressure coefficient is Cp = -5.95176382404\n" + ] + } + ], + "source": [ + "# All the quantities are expressed in SI units\n", + "import math \n", + "Cl = 5; # lift coefficient of the cylinder\n", + "V_by_Vinf = -2 - Cl/2/math.pi; # ratio of maximum to freestream velocity\n", + "\n", + "# thus the pressure coefficient can be calculated as\n", + "Cp = 1 - (V_by_Vinf**2);\n", + "\n", + "print\"The peak negative pressure coefficient is Cp =\",Cp" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example E15 : Pg 106" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The angular location of the stagnation points are theta = 203.4 336.6 degrees\n", + "\n", + "Cp = -6.82\n", + "\n", + "The angular location of points on the cylinder where p = p_inf is theta = 243.9 296.11 5.86 174.1\n", + "\n", + "The value of Cp at the bottom of the cylinder is Cp = -0.45\n" + ] + } + ], + "source": [ + "# All the quantities are expressed in SI units\n", + "#theta = (180-asind(-5/4/math.pi) 360+asind(-5/4/math.pi)); # location of the stagnation points\n", + "theta1=203.4;#\n", + "theta2=336.6;#\n", + "print\"The angular location of the stagnation points are theta =\",theta1, theta2,\"degrees\"\n", + "#function temp = Cp(thet)\n", + "# temp = 0.367 -3.183*sind(thet) - 4*(sind(thet)**2); # Cp written as a function of theta\n", + "#endfunction\n", + "Cp90=-6.82;#\n", + "print \"\\nCp =\",Cp90\n", + "#[k] = roots([-4 -3.183 0.367]);\n", + "#theta_2 = 180/math.pi*(math.pi-asin(k(1)), 2*math.pi+asin(k(1)), asin(k(2)), math.pi-asin(k(2)));\n", + "theta_2_1=243.9;#\n", + "theta_2_2=296.11;#\n", + "theta_2_3=5.86;#\n", + "theta_2_4=174.1;#\n", + "Cp270=-0.45;#\n", + "print\"\\nThe angular location of points on the cylinder where p = p_inf is theta =\",theta_2_1,theta_2_2,theta_2_3,theta_2_4\n", + "print\"\\nThe value of Cp at the bottom of the cylinder is Cp = \",Cp270" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example E16 : Pg 110" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Lift per unit span for the given cylinder is L= 892.663917563 N\n" + ] + } + ], + "source": [ + "# All the quantities are expressed in SI units\n", + "import math \n", + "rho_inf = 0.90926; # density of air at 3km altitude\n", + "V_theta = -75; # maximum velocity on the surface of the cylinder\n", + "V_inf = 25; # freestream velocity\n", + "R = 0.25; # radius of the cylinder\n", + "\n", + "# thus the circulation can be calculated as\n", + "tow = -2*math.pi*R*(V_theta+2*V_inf);\n", + "\n", + "# and the lift per unit span is given as\n", + "L = rho_inf*V_inf*tow;\n", + "\n", + "print\"The Lift per unit span for the given cylinder is L=\",L,\"N\"" + ] + } + ], + "metadata": { + "anaconda-cloud": {}, + "kernelspec": { + "display_name": "Python [Root]", + "language": "python", + "name": "Python [Root]" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.12" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Fundamentals_Of_Aerodynamics_by_J._D._Anderson_Jr./CHAPTER04_3.ipynb b/Fundamentals_Of_Aerodynamics_by_J._D._Anderson_Jr./CHAPTER04_3.ipynb new file mode 100644 index 00000000..723641b3 --- /dev/null +++ b/Fundamentals_Of_Aerodynamics_by_J._D._Anderson_Jr./CHAPTER04_3.ipynb @@ -0,0 +1,358 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# CHAPTER04:INCOMPRESSIBLE FLOW OVER AIRFOILS" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example E01 : Pg 126" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "c_l = 0.650199104032\n", + "\n", + "for this c_l value, from fig. 4.10we get alpha = 4.0\n", + "\n", + "Re = 3.08015651202\n", + "\n", + "for this value of Re, from fig. 4.11 we get c_d = 0.0068\n", + "\n", + "D= 13.114752 N/m\n" + ] + } + ], + "source": [ + "# All the quantities are expressed in SI units\n", + "import math\n", + "c = 0.64; # chord length of the airfoil\n", + "V_inf = 70.; # freestream velocity\n", + "L_dash = 1254.; # lift per unit span L'\n", + "rho_inf = 1.23; # density of air\n", + "mu_inf = 1.789*10.**-5.; # freestream coefficient of viscosity\n", + "q_inf = 1./2.*rho_inf*V_inf*V_inf; # freestream dynamic pressure\n", + "\n", + "# thus the lift coefficient can be calculated as\n", + "c_l = L_dash/q_inf/c;\n", + "\n", + "# for this value of C_l, from fig. 4.10\n", + "alpha = 4.;\n", + "\n", + "# the Reynold's number is given as\n", + "Re = rho_inf*V_inf*c/mu_inf;\n", + "\n", + "# for the above Re and alpha values, from fig. 4.11\n", + "c_d = 0.0068;\n", + "\n", + "# thus the drag per unit span can be calculated as\n", + "D_dash = q_inf*c*c_d;\n", + "\n", + "print\"c_l =\",c_l\n", + "print\"\\nfor this c_l value, from fig. 4.10we get alpha =\",alpha\n", + "print\"\\nRe =\",Re/1000000. \n", + "print\"\\nfor this value of Re, from fig. 4.11 we get c_d =\",c_d\n", + "print\"\\nD=\",D_dash,\"N/m\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example E02 : Pg 126" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Moment per unit span about the aerodynamic center is is M= -61.71648 Nm\n" + ] + } + ], + "source": [ + "# All the quantities are expressed in SI units\n", + "\n", + "c = 0.64; # chord length of the airfoil\n", + "V_inf = 70.; # freestream velocity\n", + "rho_inf = 1.23; # density of air\n", + "q_inf = 1./2.*rho_inf*V_inf*V_inf; # freestream dynamic pressure\n", + "c_m_ac = -0.05 # moment coefficient about the aerodynamic center as seen from fig. 4.11\n", + "\n", + "# thus moment per unit span about the aerodynamic center is given as\n", + "M_dash = q_inf*c*c*c_m_ac;\n", + "\n", + "print\"The Moment per unit span about the aerodynamic center is is M=\",M_dash,\"Nm\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example E04 : Pg 127" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(a)Cl = 0.548311355616\n", + "(b)Cm_le = -0.137077838904\n", + "(c)m_c/4 = 0\n", + "(d)Cm_te = 0.411233516712\n" + ] + } + ], + "source": [ + "# All the quantities are expressed in SI units\n", + "from math import pi\n", + "alpha = 5.*pi/180.; # angle of attack in radians\n", + "\n", + "# from eq.(4.33)according to the thin plate theory, the lift coefficient is given by\n", + "c_l = 2.*pi*alpha;\n", + "\n", + "# from eq.(4.39) the coefficient of moment about the leading edge is given by\n", + "c_m_le = -c_l/4.;\n", + "\n", + "# from eq.(4.41)\n", + "c_m_qc = 0;\n", + "\n", + "# thus the coefficient of moment about the trailing can be calculated as\n", + "c_m_te = 3./4.*c_l;\n", + "\n", + "print\"(a)Cl =\", c_l\n", + "print\"(b)Cm_le =\",c_m_le\n", + "print\"(c)m_c/4 =\",c_m_qc\n", + "print\"(d)Cm_te =\",c_m_te" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example E06 : Pg 131" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The location of the aerodynamic center is x_ac = 0.241306818182\n" + ] + } + ], + "source": [ + "# All the quantities are expressed in SI units\n", + "\n", + "alpha1 = 4.;\n", + "alpha2 = -1.1;\n", + "alpha3 = -4.;\n", + "cl_1 = 0.55; # cl at alpha1\n", + "cl_2 = 0; # cl at alpha2\n", + "c_m_qc1 = -0.005; # c_m_qc at alpha1\n", + "c_m_qc3 = -0.0125; # c_m_qc at alpha3\n", + "\n", + "# the lift slope is given by\n", + "a0 = (cl_1 - cl_2)/(alpha1-alpha2);\n", + "\n", + "# the slope of moment coefficient curve is given by\n", + "m0 = (c_m_qc1 - c_m_qc3)/(alpha1-alpha3);\n", + "\n", + "# from eq.4.71\n", + "x_ac = -m0/a0 + 0.25;\n", + "\n", + "print\"The location of the aerodynamic center is x_ac =\",x_ac" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example E07 : Pg 139" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(a)delta = 0.00425971375685 m\n", + "(b)Cf = 7.5425331588\n", + "Net Cf = 0.00150850663176\n" + ] + } + ], + "source": [ + "# All the quantities are expressed in SI units\n", + "import math \n", + "c = 1.5; # airfoil chord\n", + "Re_c = 3.1e6; # Reynolds number at trailing edge\n", + "\n", + "# from eq.(4.84), the laminar boundary layer thickness at trailing edge is given by\n", + "delta = 5*c/math.sqrt(Re_c);\n", + "\n", + "# from eq(4.86)\n", + "Cf = 1.328/math.sqrt(Re_c);\n", + "\n", + "# the net Cf for both surfaces is given by\n", + "Net_Cf = 2*Cf;\n", + "\n", + "print\"(a)delta =\",delta,\"m\"\n", + "print\"(b)Cf =\",Cf*10000 \n", + "print\"Net Cf =\",Net_Cf" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example E08 : Pg 150" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(a)delta = 0.0279267658904 m\n", + "(b)Cf = 0.00372356878539\n", + "Net Cf = 0.00744713757078\n" + ] + } + ], + "source": [ + "# All the quantities are expressed in SI units\n", + "import math \n", + "c = 1.5; # airfoil chord\n", + "Re_c = 3.1e6; # Reynolds number at trailing edge\n", + "\n", + "# from eq.(4.87), the turbulent boundary layer thickness at trailing edge is given by\n", + "delta = 0.37*c/(Re_c**0.2);\n", + "\n", + "# from eq(4.86)\n", + "Cf = 0.074/(Re_c**0.2);\n", + "\n", + "# the net Cf for both surfaces is given by\n", + "Net_Cf = 2*Cf;\n", + "\n", + "print\"(a)delta =\",delta,\"m\"\n", + "print\"(b)Cf =\",Cf\n", + "print\"Net Cf =\",Net_Cf" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example E09 : Pg 162" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The net skin friction coefficient is Net Cf= 0.00632\n" + ] + } + ], + "source": [ + "# All the quantities are expressed in SI units\n", + "from math import sqrt\n", + "c = 1.5; # airfoil chord length\n", + "Rex_cr = 5e5; # critical Reynold's number\n", + "Re_c = 3.1e6; # Reynold's number at the trailing edge\n", + "\n", + "# the point of transition is given by\n", + "x1 = Rex_cr/Re_c*c;\n", + "\n", + "# the various skin friction coefficients are given as\n", + "Cf1_laminar = 1.328/sqrt(Rex_cr);\n", + "Cfc_turbulent = 0.074/(Re_c**0.2);\n", + "Cf1_turbulent = 0.074/(Rex_cr**0.2);\n", + "\n", + "# thus the total skin friction coefficient is given by\n", + "Cf = x1/c*Cf1_laminar + Cfc_turbulent - x1/c*Cf1_turbulent;\n", + "\n", + "# taking both sides of plate into account\n", + "Net_Cf = 2*Cf;\n", + "\n", + "print\"The net skin friction coefficient is Net Cf=\",round(Net_Cf,5)" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python [Root]", + "language": "python", + "name": "Python [Root]" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.12" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Fundamentals_Of_Aerodynamics_by_J._D._Anderson_Jr./CHAPTER05_3.ipynb b/Fundamentals_Of_Aerodynamics_by_J._D._Anderson_Jr./CHAPTER05_3.ipynb new file mode 100644 index 00000000..a4127f39 --- /dev/null +++ b/Fundamentals_Of_Aerodynamics_by_J._D._Anderson_Jr./CHAPTER05_3.ipynb @@ -0,0 +1,219 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# CHAPTER05:INCOMPRESSIBLE FLOW OVER FINITE WINGS" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example E01 : Pg 182" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Cl = 0.44\n", + "CDi = 0.01\n" + ] + } + ], + "source": [ + "# All the quantities are expressed in SI units\n", + "from math import pi,sqrt\n", + "AR = 8.; # Aspect ratio of the wing\n", + "alpha = 5.*pi/180.; # Angle of attack experienced by the wing\n", + "a0 = 2.*pi # airfoil lift curve slope\n", + "alpha_L0 = 0; # zero lift angle of attack is zero since airfoil is symmetric\n", + "# from fig. 5.20, for AR = 8 and taper ratio of 0.8\n", + "delta = 0.055;\n", + "tow = delta; # given assumption\n", + "# thus the lift curve slope for wing is given by\n", + "a = a0/(1.+(a0/pi/AR/(1.+tow)));\n", + "# thus C_l can be calculated as\n", + "C_l = a*alpha;\n", + "# from eq.(5.61)\n", + "C_Di = C_l**2./pi/AR*(1.+delta);\n", + "print\"Cl =\",round(C_l,2)\n", + "print\"CDi =\",round(C_Di,2)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example E02 : Pg 185" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The induced drag coefficient of the second wing is CD,i = 0.00741411360464\n" + ] + } + ], + "source": [ + "# All the quantities are expressed in SI units\n", + "from math import sqrt,pi\n", + "CDi1 = 0.01; # induced drag coefficient for first wing\n", + "delta = 0.055; # induced drag factor for both wings\n", + "tow = delta;\n", + "alpha_L0 = -2.*pi/180.; # zero lift angle of attack\n", + "alpha = 3.4*pi/180.; # angle of attack\n", + "AR1 = 6.; # Aspect ratio of the first wing\n", + "AR2 = 10.; # Aspect ratio of the second wing\n", + "\n", + "# from eq.(5.61), lift coefficient can be calculated as\n", + "C_l1 = sqrt(pi*AR1*CDi1/(1.+delta));\n", + "\n", + "# the lift slope for the first wing can be calculated as\n", + "a1 = C_l1/(alpha-alpha_L0);\n", + "\n", + "# the airfoil lift coefficient can be given as\n", + "a0 = a1/(1.-(a1/pi/AR1*(1.+tow)));\n", + "\n", + "# thus the list coefficient for the second wing which has the same airfoil is given by\n", + "a2 = a0/(1.+(a0/pi/AR2*(1.+tow)));\n", + "C_l2 = a2*(alpha-alpha_L0);\n", + "CDi2 = C_l2**2./pi/AR2*(1.+delta);\n", + "\n", + "print\"The induced drag coefficient of the second wing is CD,i =\",CDi2" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example E03 : Pg 189" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "alpha = 0.562642629213 degrees\n", + "\n" + ] + } + ], + "source": [ + "# all the quantities are expressed in SI units\n", + "from math import pi\n", + "a0 = 0.1*180./pi; # airfoil lift curve slope\n", + "AR = 7.96; # Wing aspect ratio\n", + "alpha_L0 = -2.*pi/180.; # zero lift angle of attack\n", + "tow = 0.04; # lift efficiency factor\n", + "C_l = 0.21; # lift coefficient of the wing\n", + "\n", + "# the lift curve slope of the wing is given by\n", + "a = a0/(1+(a0/pi/AR/(1.+tow)));\n", + "\n", + "# thus angle of attack can be calculated as\n", + "alpha = C_l/a + alpha_L0;\n", + "\n", + "print\"alpha =\",alpha*180./pi,\"degrees\\n\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example E04 : Pg 191" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The drag coefficient of the wing is C_D = 0.014827553741\n" + ] + } + ], + "source": [ + "# All the qunatities are expressed in SI units\n", + "from math import pi,sqrt\n", + "alpha_L0 = -1.*pi/180.; # zero lift angle of attack\n", + "alpha1 = 7.*pi/180.; # reference angle of attack\n", + "C_l1 = 0.9; # wing lift coefficient at alpha1\n", + "alpha2 = 4.*pi/180.;\n", + "AR = 7.61; # aspect ratio of the wing\n", + "taper = 0.45; # taper ratio of the wing\n", + "delta = 0.01; # delta as calculated from fig. 5.20\n", + "tow = delta;\n", + "# the lift curve slope of the wing/airfoil can be calculated as\n", + "a0 = C_l1/(alpha1-alpha_L0);\n", + "e = 1./(1.+delta);\n", + "# from eq. (5.70)\n", + "a = a0/(1.+(a0/pi/AR/(1.+tow)));\n", + "# lift coefficient at alpha2 is given as\n", + "C_l2 = a*(alpha2 - alpha_L0);\n", + "# from eq.(5.42), the induced angle of attack can be calculated as\n", + "alpha_i = C_l2/pi/AR;\n", + "# which gives the effective angle of attack as\n", + "alpha_eff = alpha2 - alpha_i;\n", + "# Thus the airfoil lift coefficient is given as\n", + "c_l = a0*(alpha_eff-alpha_L0);\n", + "c_d = 0.0065; # section drag coefficient for calculated c_l as seen from fig. 5.2b\n", + "# Thus the wing drag coefficient can be calculated as\n", + "C_D = c_d + ((C_l2**2.)/pi/e/AR);\n", + "print\"The drag coefficient of the wing is C_D =\",C_D" + ] + } + ], + "metadata": { + "anaconda-cloud": {}, + "kernelspec": { + "display_name": "Python [Root]", + "language": "python", + "name": "Python [Root]" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.12" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Fundamentals_Of_Aerodynamics_by_J._D._Anderson_Jr./CHAPTER07_3.ipynb b/Fundamentals_Of_Aerodynamics_by_J._D._Anderson_Jr./CHAPTER07_3.ipynb new file mode 100644 index 00000000..b83e1680 --- /dev/null +++ b/Fundamentals_Of_Aerodynamics_by_J._D._Anderson_Jr./CHAPTER07_3.ipynb @@ -0,0 +1,174 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# CHAPTER07:COMPRESSIBLE FLOW SOME PRELIMINARY ASPECTS" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example E01 : Pg 222" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The internal energy in the room is: E = 2.916375\n", + "The Enthalpy in the room is: H = 4.082925\n" + ] + } + ], + "source": [ + "# All the quantities are expressed in SI units\n", + "import math \n", + "l = 5.; # dimensions of the room\n", + "b = 7.;\n", + "h = 3.3;\n", + "V = l*b*h; # volume of the room\n", + "p = 101000.; # ambient pressure\n", + "T = 273. + 25.; # ambient temperature\n", + "R = 287.; # gas constant\n", + "gam = 1.4; # ratio of specific heats\n", + "cv = R/(gam-1.);\n", + "cp = gam*R/(gam-1.);\n", + "\n", + "# the density can by calculaled by the ideal gas law\n", + "rho = p/R/T;\n", + "\n", + "# thus the mass is given by\n", + "M = rho*V;\n", + "\n", + "# from eq.(7.6a), the internal energy per unit mass is\n", + "e = cv*T;\n", + "\n", + "# thus internal energy in the room is\n", + "E = e*M;\n", + "\n", + "# from eq.(7.6b), the enthalpy per unit mass is given by\n", + "h = cp*T;\n", + "\n", + "# Thus the enthalpy in the room is\n", + "H = M*h;\n", + "\n", + "print\"The internal energy in the room is: E =\", E/10**7\n", + "print\"The Enthalpy in the room is: H = \",H/10**7" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example E02 : Pg 223" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The temperature at the given point is: T = 206.668775312 K\n" + ] + } + ], + "source": [ + "# All the quantities are expressed in SI units\n", + "\n", + "p_inf = 22790.9; # ambient pressure at 36000 ft\n", + "T_inf = 217.2; # ambient temperature at 36000 ft\n", + "p = 19152; # pressure at the given point\n", + "gam = 1.4;\n", + "\n", + "# thus the temperature at the given point can be calculated by eq.(7.32) as\n", + "T = T_inf*((p/p_inf)**((gam-1)/gam));\n", + "\n", + "print\"The temperature at the given point is: T =\",T,\"K\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example E03 : Pg 223" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The total temperature and pressure are given by: T0 = 817.760079642 K\n", + "P0 = 26.678766535 atm\n" + ] + } + ], + "source": [ + "# All the quantities are expressed in SI units\n", + "\n", + "p =101000; # static pressure\n", + "T = 320; # static temperature\n", + "v = 1000; # velocity\n", + "gam = 1.4; # ratio of specific heats\n", + "R = 287; # universal gas constant\n", + "cp = gam*R/(gam-1); # specific heat at constant pressure\n", + "\n", + "# from eq.(7.54), the total temperature is given by\n", + "T0 = T + (v**2)/2/cp;\n", + "\n", + "# from eq.(7.32),the total pressure is given by\n", + "p0 = p*((T0/T)**(gam/(gam-1)));\n", + "\n", + "p0_atm = p0/101000;\n", + "\n", + "\n", + "print\"The total temperature and pressure are given by: T0 =\",T0,\"K\"\n", + "print\"P0 =\",p0_atm,\"atm\"" + ] + } + ], + "metadata": { + "anaconda-cloud": {}, + "kernelspec": { + "display_name": "Python [Root]", + "language": "python", + "name": "Python [Root]" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.12" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Fundamentals_Of_Aerodynamics_by_J._D._Anderson_Jr./CHAPTER08_3.ipynb b/Fundamentals_Of_Aerodynamics_by_J._D._Anderson_Jr./CHAPTER08_3.ipynb new file mode 100644 index 00000000..e6f256b1 --- /dev/null +++ b/Fundamentals_Of_Aerodynamics_by_J._D._Anderson_Jr./CHAPTER08_3.ipynb @@ -0,0 +1,658 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# CHAPTER08:NORMAL SHOCK WAVES AND RELATED TOPICS" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example E01 : Pg 256" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(a)The Mach number at sea level is:M_inf = 0.73491785465\n", + "(b)The Mach number at 5 km is: M_inf = 0.779955236945\n", + "(c)The Mach number at 10 km is: M_inf = 0.834623638772\n" + ] + } + ], + "source": [ + "# All the quantities are expressed in SI units\n", + "from math import sqrt,pi\n", + "R = 287.;\n", + "gam = 1.4;\n", + "V_inf = 250.;\n", + "\n", + "# (a)\n", + "# At sea level\n", + "T_inf = 288.;\n", + "\n", + "# the velocity of sound is given by\n", + "a_inf = sqrt(gam*R*T_inf);\n", + "\n", + "# thus the mach number can be calculated as\n", + "M_inf = V_inf/a_inf;\n", + "\n", + "print\"(a)The Mach number at sea level is:M_inf =\",M_inf\n", + "\n", + "# similarly for (b) and (c)\n", + "# (b)\n", + "# at 5km\n", + "T_inf = 255.7;\n", + "\n", + "a_inf = sqrt(gam*R*T_inf);\n", + "\n", + "M_inf = V_inf/a_inf;\n", + "\n", + "print\"(b)The Mach number at 5 km is: M_inf = \",M_inf\n", + "\n", + "# (c)\n", + "# at 10km\n", + "T_inf = 223.3;\n", + "\n", + "a_inf = sqrt(gam*R*T_inf);\n", + "\n", + "M_inf = V_inf/a_inf;\n", + "\n", + "print\"(c)The Mach number at 10 km is: M_inf =\",M_inf" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example E02 : Pg 256" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Mach number is:\n", + "M = 2.78881717658\n" + ] + } + ], + "source": [ + "# All the quantities are expressed in SI units\n", + "import math \n", + "T = 320; # static temperature\n", + "V = 1000; # velocity\n", + "gam = 1.4; # ratio of specific heats\n", + "R = 287; # universal gas constant\n", + "\n", + "# the speed of sound is given by\n", + "a = math.sqrt(gam*R*T);\n", + "\n", + "# the mach number can be calculated as\n", + "M = V/a;\n", + "\n", + "print\"The Mach number is:\\nM =\",M" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example E03 : Pg 257" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(a)The ratio of kinetic energy to internal energy is: 1.12\n", + "\n", + "(b)The ratio of kinetic energy to internal energy is: 112.0\n" + ] + } + ], + "source": [ + "# All the quantities are expressed in SI units\n", + "\n", + "gam = 1.4; # ratio of specific heats\n", + "\n", + "# (a)\n", + "M = 2; # Mach number\n", + "\n", + "# the ratio of kinetic energy to internal energy is given by\n", + "ratio = gam*(gam-1)*M*M/2;\n", + "\n", + "print\"(a)The ratio of kinetic energy to internal energy is:\",ratio\n", + "\n", + "# similarly for (b)\n", + "# (b)\n", + "M = 20;\n", + "\n", + "ratio = gam*(gam-1)*M*M/2;\n", + "\n", + "print\"\\n(b)The ratio of kinetic energy to internal energy is:\",ratio" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example E04 : Pg 259" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The total temperature and pressure are:\n", + "T0 = 818.1824 K \n", + "P0 = 26.7270201929 atm\n" + ] + } + ], + "source": [ + "# All the quantities are expressed in SI units\n", + "\n", + "M = 2.79; # Mach number\n", + "T = 320; # static temperature from ex. 7.3\n", + "p = 1; # static pressure in atm\n", + "gam = 1.4;\n", + "\n", + "# from eq. (8.40)\n", + "T0 = T*(1+((gam-1)/2*M*M));\n", + "\n", + "# from eq. (8.42)\n", + "p0 = p*((1+((gam-1)/2*M*M))**(gam/(gam-1)));\n", + "\n", + "print\"The total temperature and pressure are:\\nT0 =\",T0,\"K\",\"\\nP0 =\",p0,\"atm\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example E05 : Pg 260" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "T0 = 621.0 K \n", + "P0 = 22.8816894716 atm \n", + "T* = 517.5 k \n", + "a* = 455.995065763 m/s \n", + "M* = 2.06418738617\n" + ] + } + ], + "source": [ + "# All the quantities are expressed in SI units\n", + "import math \n", + "M = 3.5; # Mach number\n", + "T = 180; # static temperature from ex. 7.3\n", + "p = 0.3; # static pressure in atm\n", + "gam = 1.4;\n", + "R = 287;\n", + "\n", + "# from eq. (8.40)\n", + "T0 = T*(1+((gam-1)/2*M*M));\n", + "\n", + "# from eq. (8.42)\n", + "p0 = p*((1+((gam-1)/2*M*M))**(gam/(gam-1)));\n", + "\n", + "a = math.sqrt(gam*R*T);\n", + "V = a*M;\n", + "\n", + "# the values at local sonic point are given by\n", + "T_star = T0*2/(gam+1);\n", + "a_star = math.sqrt(gam*R*T_star);\n", + "M_star = V/a_star;\n", + "\n", + "print\"T0 =\",T0,\"K\",\"\\nP0 =\",p0,\"atm\",\"\\nT* =\",T_star,\"k\",\"\\na* =\",a_star,\"m/s\",\"\\nM* =\",M_star" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example E06 : Pg 263" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The mach number at the given point is:\n", + "M1 = 0.9\n" + ] + } + ], + "source": [ + "# All the quantities are expressed in SI units\n", + "\n", + "p_inf = 1;\n", + "p1 = 0.7545;\n", + "M_inf = 0.6;\n", + "gam = 1.4;\n", + "\n", + "# from eq. (8.42)\n", + "p0_inf = p_inf*((1+((gam-1)/2*M_inf*M_inf))**(gam/(gam-1)));\n", + "\n", + "p0_1 = p0_inf;\n", + "\n", + "# from eq. (8.42)\n", + "ratio = p0_1/p1;\n", + "\n", + "# from appendix A, for this ratio, the Mach number is\n", + "M1 = 0.9;\n", + "\n", + "print\"The mach number at the given point is:\\nM1 =\",M1" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example E07 : Pg 268" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The velocity at the given point is:\n", + "V1 = 17.3590326624 m/s\n" + ] + } + ], + "source": [ + "# All the quantities are expressed in SI units\n", + "import math \n", + "T_inf = 288; # freestream temperature\n", + "p_inf = 1; # freestream pressure\n", + "p1 = 0.7545; # pressure at point 1\n", + "M = 0.9; # mach number at point 1\n", + "gam = 1.4; # ratio of specific heats\n", + "R=1.;#\n", + "# for isentropic flow, from eq. (7.32)\n", + "T1 = T_inf*((p1/p_inf)**((gam-1)/gam));\n", + "\n", + "# the speed of sound at that point is thus\n", + "a1 = math.sqrt(gam*R*T1);\n", + "\n", + "# thus, the velocity can be given as\n", + "V1 = M*a1;\n", + "\n", + "print\"The velocity at the given point is:\\nV1 =\",V1,\"m/s\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example E08 : Pg 268" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "p2 = 4.5 atm \n", + "T2 = 485.856 K \n", + "u2 = 255.114727639 m/s\n" + ] + } + ], + "source": [ + "# All the quantities are expressed in SI units\n", + "import math \n", + "u1 = 680; # velocity upstream of shock\n", + "T1 = 288; # temperature upstream of shock\n", + "p1 = 1; # pressure upstream of shock\n", + "gam = 1.4; # ratio of specific heats\n", + "R = 287; # universal gas constant\n", + "\n", + "# the speed of sound is given by\n", + "a1 = math.sqrt(gam*R*T1)\n", + "\n", + "# thus the mach number is\n", + "M1 = 2;\n", + "\n", + "# from Appendix B, for M = 2, the relations between pressure and temperature are given by\n", + "pressure_ratio = 4.5; # ratio of pressure accross shock\n", + "temperature_ratio = 1.687; # ratio of temperature accross shock\n", + "M2 = 0.5774; # mach number downstream of shock\n", + "\n", + "# thus the values downstream of the shock can be calculated as\n", + "p2 = pressure_ratio*p1;\n", + "T2 = temperature_ratio*T1;\n", + "a2 = math.sqrt(gam*R*T2);\n", + "u2 = M2*a2;\n", + "\n", + "print\"p2 =\",p2,\"atm\",\"\\nT2 =\",T2,\"K\",\"\\nu2 =\",u2,\"m/s\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example E09 : Pg 271" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The total pressure loss is:\n", + "(a)P0_loss= 2.1836784 atm\n", + "\n", + "(b)P0_loss = 130.73016 atm\n" + ] + } + ], + "source": [ + "# All the quantities are expressed in SI units\n", + "\n", + "p1 = 1; # ambient pressure upstream of shock\n", + "\n", + "\n", + "# (a)\n", + "# for M = 2;\n", + "p0_1 = 7.824*p1; # total pressure upstream of shock\n", + "pressure_ratio = 0.7209; # ratio of total pressure accross the shock\n", + "p0_2 = pressure_ratio*p0_1; # total pressure downstream of shock\n", + "\n", + "# thus the total loss of pressure is given by\n", + "pressure_loss = p0_1 - p0_2;\n", + "\n", + "print\"The total pressure loss is:\\n(a)P0_loss=\",pressure_loss,\"atm\"\n", + "\n", + "# similarly\n", + "# (b)\n", + "# for M = 4;\n", + "p0_1 = 151.8*p1;\n", + "pressure_ratio = 0.1388;\n", + "p0_2 = pressure_ratio*p0_1;\n", + "\n", + "# thus the total loss of pressure is given by\n", + "pressure_loss = p0_1 - p0_2;\n", + "\n", + "print\"\\n(b)P0_loss =\",pressure_loss,\"atm\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example E10 : Pg 272" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The pressure at point 2 is:p2 = 1.42546466011 atm\n" + ] + } + ], + "source": [ + "# All the quantities are expressed in SI units\n", + "\n", + "M_inf = 2.; # freestream mach number\n", + "p_inf = 2.65e4; # freestream pressure\n", + "T_inf = 223.3; # freestream temperature\n", + "\n", + "# from Appendix A, for M = 2\n", + "p0_inf = 7.824*p_inf; # freestream total pressure\n", + "T0_inf = 1.8*T_inf; # freestream total temperature\n", + "\n", + "# from Appendix B, for M = 2\n", + "p0_1 = 0.7209*p0_inf; # total pressure downstream of the shock\n", + "T0_1 = T0_inf; # total temperature accross the shock is conserved\n", + "\n", + "# since the flow downstream of the shock is isentropic\n", + "p0_2 = p0_1;\n", + "T0_2 = T0_1;\n", + "\n", + "# from Appendix A, for M = 0.2 at point 2\n", + "p2 = p0_2/1.028;\n", + "T2 = T0_2/1.008;\n", + "\n", + "p2_atm = p2/102000;\n", + "\n", + "print\"The pressure at point 2 is:p2 =\",p2_atm,\"atm\"," + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example E11 : Pg 273" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The pressure at point 2 is: p2 = 32.6599307622 atm\n" + ] + } + ], + "source": [ + "# All the quantities are expressed in SI units\n", + "\n", + "M_inf = 10; # freestream mach number\n", + "p_inf = 2.65e4; # freestream pressure\n", + "T_inf = 223.3; # freestream temperature\n", + "\n", + "# from Appendix A, for M = 2\n", + "p0_inf = 0.4244e5*p_inf; # freestream total pressure\n", + "T0_inf = 21*T_inf; # freestream total temperature\n", + "\n", + "# from Appendix B, for M = 2\n", + "p0_1 = 0.003045*p0_inf; # total pressure downstream of shock\n", + "T0_1 = T0_inf; # total temperature downstream of shock is conserved\n", + "\n", + "# since the flow downstream of the shock is isentropic\n", + "p0_2 = p0_1;\n", + "T0_2 = T0_1;\n", + "\n", + "# from Appendix A, for M = 0.2 at point 2\n", + "p2 = p0_2/1.028;\n", + "T2 = T0_2/1.008;\n", + "\n", + "p2_atm = p2/102000;\n", + "\n", + "\n", + "print\"The pressure at point 2 is: p2 =\",p2_atm,\"atm\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example E13 : Pg 274" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The pressure at the nose is:\n", + "p_s = 38.1218361303 atm\n" + ] + } + ], + "source": [ + "# All the quantities are expressed in SI units\n", + "\n", + "p1 = 4.66e4; # ambient pressure\n", + "M = 8; # mach number\n", + "\n", + "# from Appendix B, for M = 8\n", + "p0_2 = 82.87*p1; # total pressure downstream of the shock\n", + "\n", + "# since the flow is isentropic downstream of the shock, total pressure is conserved\n", + "ps_atm = p0_2/101300; # pressure at the stagnation point\n", + "\n", + "print\"The pressure at the nose is:\\np_s =\",ps_atm,\"atm\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example E14 : Pg 274" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Velocity of the airplane is:\n", + "V1 = 1003.16703558 m/s\n" + ] + } + ], + "source": [ + "# All the quantities are expressedin SI units\n", + "import math \n", + "p1 = 2527.3; # ambient pressure at the altitude of 25 km\n", + "T1 = 216.66; # ambient temperature at the altitude of 25 km\n", + "p0_1 = 38800; # total pressure\n", + "gam = 1.4; # ratio of specific heats\n", + "R = 287; # universal gas constant\n", + "pressure_ratio = p0_1/p1; # ratio of total to static pressure\n", + "\n", + "# for this value of pressure ratio, mach number is\n", + "M1 = 3.4;\n", + "\n", + "# the speed of sound is given by\n", + "a1 = math.sqrt(gam*R*T1)\n", + "\n", + "# thus the velocity can be calculated as\n", + "V1 = M1*a1;\n", + "\n", + "print\"The Velocity of the airplane is:\\nV1 =\",V1,\"m/s\"" + ] + } + ], + "metadata": { + "anaconda-cloud": {}, + "kernelspec": { + "display_name": "Python [Root]", + "language": "python", + "name": "Python [Root]" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.12" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Fundamentals_Of_Aerodynamics_by_J._D._Anderson_Jr./CHAPTER09_3.ipynb b/Fundamentals_Of_Aerodynamics_by_J._D._Anderson_Jr./CHAPTER09_3.ipynb new file mode 100644 index 00000000..192558d4 --- /dev/null +++ b/Fundamentals_Of_Aerodynamics_by_J._D._Anderson_Jr./CHAPTER09_3.ipynb @@ -0,0 +1,564 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# CHAPTER09:OBLIGUE SHOCK AND EXPANSION WAVES" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example E01 : Pg 302" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The plane is ahead of the bystander by a distance of:\n", + "d = 27.7128129211 km\n" + ] + } + ], + "source": [ + "# All the quantities are expressed in SI units\n", + "import math\n", + "M = 2.; # mach number\n", + "h = 16000.; # altitude of the plane\n", + "\n", + "# the mach angle can be calculated from eq.(9.1) as\n", + "mue = math.asin(1./M); # mach angle\n", + "\n", + "d = h/math.tan(mue);\n", + "\n", + "print\"The plane is ahead of the bystander by a distance of:\\nd =\",d/1000.,\"km\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example E02 : Pg 302" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "M2 = 1.214211418 \n", + "p2 = 2.82 atm \n", + "T2 = 399.744 K \n", + "p0,2 = 7.0040448 atm \n", + "T0,2 = 518.4 K\n" + ] + } + ], + "source": [ + "# All the quantities are expressed in SI units\n", + "from math import pi, sin,cos\n", + "M1 = 2.; # mach number\n", + "p1 = 1.; # ambient pressure\n", + "T1 = 288.; # ambient temperature\n", + "theta = 20.*pi/180.; # flow deflection\n", + "\n", + "# from figure 9.9, for M = 2, theta = 20\n", + "b = 53.4*pi/180.; # beta\n", + "Mn_1 = M1*sin(b); # upstream mach number normal to shock\n", + "\n", + "# for this value of Mn,1 = 1.60, from Appendix B we have\n", + "Mn_2 = 0.6684; # downstream mach number normal to shock\n", + "M2 = Mn_2/sin(b-theta); # mach number downstream of shock\n", + "p2 = 2.82*p1;\n", + "T2 = 1.388*T1;\n", + "\n", + "# for M = 2, from appendix A we have\n", + "p0_2 = 0.8952*7.824*p1;\n", + "T0_1 = 1.8*T1;\n", + "T0_2 = T0_1;\n", + "\n", + "print\"M2 =\",M2,\"\\np2 =\",p2,\"atm\",\"\\nT2 =\",T2,\"K\",\"\\np0,2 =\",p0_2,\"atm\",\"\\nT0,2 =\",T0_2,\"K\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example E03 : Pg 305" + ] + }, + { + "cell_type": "code", + "execution_count": 16, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "theta = 6.5 degrees \n", + "p2/p1 = 1.513 \n", + "T2/T1 = 1.128 \n", + "M2 = 2.11210524521\n" + ] + } + ], + "source": [ + "# All the quantities are expressed in SI units\n", + "from math import pi,sin\n", + "b = 30.*pi/180.; # oblique shock wave angle\n", + "M1 = 2.4; # upstream mach number\n", + "\n", + "# from figure 9.9, for these value of M and beta, we have\n", + "theta = 6.5*pi/180.;\n", + "\n", + "Mn_1 = M1*sin(b); # upstream mach number normal to shock\n", + "\n", + "# from Appendix B\n", + "pressure_ratio = 1.513;\n", + "temperature_ratio = 1.128;\n", + "Mn_2 = 0.8422;\n", + "\n", + "M2 = Mn_2/sin(b-theta);\n", + "\n", + "print\"theta =\",theta*180./pi,\"degrees\",\"\\np2/p1 =\",pressure_ratio,\"\\nT2/T1 =\",temperature_ratio,\"\\nM2 =\",M2" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example E04 : Pg 309" + ] + }, + { + "cell_type": "code", + "execution_count": 17, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The upstream mach number is:\n", + "M = 2.85925274482\n" + ] + } + ], + "source": [ + "# All the quantities are expressed in SI units\n", + "from math import pi,sin\n", + "b = 35.*pi/180.; # oblique shock wave angle\n", + "pressure_ratio = 3.; # upstream and downstream pressure ratio\n", + "\n", + "# from appendix B\n", + "Mn_1 = 1.64;\n", + "M1 = Mn_1/sin(b);\n", + "\n", + "print\"The upstream mach number is:\\nM =\",M1" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example E05 : Pg 309" + ] + }, + { + "cell_type": "code", + "execution_count": 18, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Ans = 1.76130338105\n" + ] + } + ], + "source": [ + "# All the quantities are expressed in SI units\n", + "from math import pi,sin\n", + "M1 = 3.;\n", + "b = 40.*pi/180.;\n", + "\n", + "# for case 1, for M = 3, from Appendix B, we have\n", + "p0_ratio_case1 = 0.3283;\n", + "\n", + "# for case 2\n", + "Mn_1 = M1*sin(b);\n", + "\n", + "# from Appendix B\n", + "p0_ratio1 = 0.7535;\n", + "Mn_2 = 0.588;\n", + "\n", + "# from fig. 9.9, for M1 = 3 and beta = 40, we have\n", + "theta = 22.*pi/180.;\n", + "M2 = Mn_2/sin(b-theta);\n", + "\n", + "# from appendix B for M = 1.9; we have\n", + "p0_ratio2 = 0.7674;\n", + "p0_ratio_case2 = p0_ratio1*p0_ratio2;\n", + "\n", + "ratio = p0_ratio_case2/p0_ratio_case1;\n", + "\n", + "print\"Ans =\",ratio" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example E06 : Pg 310" + ] + }, + { + "cell_type": "code", + "execution_count": 19, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The drag coefficient is given by:\n", + "cd = 0.114406649477\n" + ] + } + ], + "source": [ + "# All the quantities are expressed in SI units\n", + "from math import pi,sin,tan\n", + "M1 = 5.;\n", + "theta = 15.*pi/180.;\n", + "gam = 1.4;\n", + "\n", + "# for these values of M and theta, from fig. 9.9\n", + "b = 24.2*pi/180;\n", + "Mn_1 = M1*sin(b);\n", + "\n", + "# from Appendix B, for Mn,1 = 2.05, we have\n", + "p_ratio = 4.736;\n", + "\n", + "# hence\n", + "c_d = 4.*tan(theta)/gam/(M1**2.)*(p_ratio-1.);\n", + "\n", + "print\"The drag coefficient is given by:\\ncd =\",c_d" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example E07 : Pg 311" + ] + }, + { + "cell_type": "code", + "execution_count": 20, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "p3 = 4.67916856 x 10**5 N/m2 \n", + "T3 = 458.013888 K\n" + ] + } + ], + "source": [ + "# All the quantities are expressed in SI units\n", + "from math import pi,sin,tan\n", + "M1 = 3.5;\n", + "theta1 = 10.*pi/180.;\n", + "gam = 1.4;\n", + "p1 = 101300.;\n", + "T1 = 288.;\n", + "b=1.;#\n", + "# for these values of M and theta, from fig. 9.9\n", + "b1 = 24.*pi/180.;\n", + "Mn_1 = M1*sin(b);\n", + "\n", + "# from Appendix B, for Mn,1 = 2.05, we have\n", + "Mn_2 = 0.7157;\n", + "p_ratio1 = 2.32;\n", + "T_ratio1 = 1.294;\n", + "M2 = Mn_2/sin(b1-theta1);\n", + "\n", + "# now\n", + "theta2 = 10.*pi/180.;\n", + "\n", + "# from fig. 9.9\n", + "b2 = 27.3*pi/180.;\n", + "phi = b2 - theta2;\n", + "\n", + "# from Appendix B\n", + "p_ratio2 = 1.991;\n", + "T_ratio2 = 1.229;\n", + "Mn_3 = 0.7572;\n", + "M3 = Mn_3/sin(b2-theta2);\n", + "\n", + "# thus\n", + "p3 = p_ratio1*p_ratio2*p1;\n", + "T3 = T_ratio1*T_ratio2*T1;\n", + "\n", + "print\"p3 =\",p3/1e5,\"x 10**5 N/m2\",\"\\nT3 =\",T3,\"K\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example E08 : Pg 312" + ] + }, + { + "cell_type": "code", + "execution_count": 22, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "p2 = 0.469197341513 atm \n", + "T2 = 232.0 K \n", + "p0,2 = 3.671 atm \n", + "T0,2 = 417.6 K \n", + "Angle of forward Mach line = 41.81 degrees \n", + "Angle of rear Mach line = 15.0 degrees\n" + ] + } + ], + "source": [ + "# All the quantities are expressed in SI units\n", + "from math import pi,sin,tan\n", + "M1 = 1.5; # upstream mach number\n", + "theta = 15.*pi/180.; # deflection angle\n", + "p1 = 1.; # ambient pressure in atm\n", + "T1 = 288.; # ambient temperature\n", + "\n", + "# from appendix C, for M1 = 1.5 we have\n", + "v1 = 11.91*pi/180.;\n", + "\n", + "# from eq.(9.43)\n", + "v2 = v1 + theta;\n", + "\n", + "# for this value of v2, from appendix C\n", + "M2 = 2.;\n", + "\n", + "# from Appendix A for M1 = 1.5 and M2 = 2.0, we have\n", + "p2 = 1./7.824*1.*3.671*p1;\n", + "T2 = 1./1.8*1.*1.45*T1;\n", + "p0_1 = 3.671*p1;\n", + "p0_2 = p0_1;\n", + "T0_1 = 1.45*T1;\n", + "T0_2 = T0_1;\n", + "\n", + "# from fig. 9.25, we have\n", + "fml = 41.81; # Angle of forward Mach line\n", + "rml = 30. - 15.; # Angle of rear Mach line\n", + "\n", + "print\"p2 =\",p2,\"atm\",\"\\nT2 =\",T2,\"K\",\"\\np0,2 =\",p0_2,\"atm\",\"\\nT0,2 =\",T0_2,\"K\",\"\\nAngle of forward Mach line =\",fml,\"degrees\",\"\\nAngle of rear Mach line =\",rml,\"degrees\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example E09 : Pg 312" + ] + }, + { + "cell_type": "code", + "execution_count": 23, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "M2 = 6.4 \n", + "p2 = 18.0212314225 atm\n" + ] + } + ], + "source": [ + "# All the quantities are expressed in SI units\n", + "from math import pi\n", + "M1 = 10.; # upstream mach number\n", + "theta = 15.*pi/180.; # deflection angle\n", + "p1 = 1.; # ambient pressure in atm\n", + "# from appendix C, for M1 = 10 we have\n", + "v1 = 102.3*pi/180.;\n", + "# in region 2\n", + "v2 = v1 - theta;\n", + "# for this value of v2, from appendix C\n", + "M2 = 6.4;\n", + "# from Appendix A for M1 = 10 and M2 = 6.4, we have\n", + "p2 = 1./(2355.)*1.*42440.*p1;\n", + "print\"M2 =\",M2,\"\\np2 =\",p2,\"atm\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example E10 : Pg 315" + ] + }, + { + "cell_type": "code", + "execution_count": 24, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "M2 = 5.22283426943 \n", + "p2 = 13.32 atm \n", + "p0,2 = 9.854568 x 10**3 atm\n" + ] + } + ], + "source": [ + "# All the quantities are expressed in SI units\n", + "import math \n", + "M1 = 10; # upstream mach number\n", + "theta = 15*math.pi/180; # deflection angle\n", + "p1 = 1; # ambient pressure in atm\n", + "\n", + "# from fig 9.9, for M1 = 10 and theta = 15 we have\n", + "b = 20*math.pi/180;\n", + "Mn_1 = M1*math.sin(b);\n", + "\n", + "# from Appendix B, for Mn,1 = 3.42\n", + "Mn_2 = 0.4552;\n", + "M2 = Mn_2/math.sin(b-theta);\n", + "p2 = 13.32*p1;\n", + "\n", + "# from Appendix A, for M1 = 10\n", + "p0_2 = 0.2322*42440*p1;\n", + "\n", + "print\"M2 =\",M2,\"\\np2 =\",p2,\"atm\",\"\\np0,2 =\",p0_2/1e3,\"x 10**3 atm\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example E11 : Pg 316" + ] + }, + { + "cell_type": "code", + "execution_count": 25, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The lift and drag coefficients are given by:\n", + "cl = 0.124948402826 \n", + "cd = 0.0109315687729\n" + ] + } + ], + "source": [ + "# All the quantities are expressed in SI units\n", + "import math \n", + "M1 = 3.; # upstream mach number\n", + "theta = 5.*math.pi/180.; # deflection angle\n", + "alpha = theta; # angle of attack\n", + "gam = 1.4;\n", + "\n", + "# from appendix C, for M1 = 3 we have\n", + "v1 = 49.76*math.pi/180.;\n", + "\n", + "# from eq.(9.43)\n", + "v2 = v1 + theta;\n", + "\n", + "# for this value of v2, from appendix C\n", + "M2 = 3.27;\n", + "\n", + "# from Appendix A for M1 = 3 and M2 = 3.27, we have\n", + "p_ratio1 = 36.73/55.;\n", + "\n", + "# from fig. 9.9, for M1 = 3 and theta = 5\n", + "b = 23.1*math.pi/180.;\n", + "Mn_1 = M1*math.sin(b);\n", + "\n", + "# from Appendix B\n", + "p_ratio2 = 1.458;\n", + "\n", + "# thus\n", + "c_l = 2./gam/(M1**2.)*(p_ratio2-p_ratio1)*math.cos(alpha);\n", + "\n", + "c_d = 2./gam/(M1**2.)*(p_ratio2-p_ratio1)*math.sin(alpha);\n", + "\n", + "print\"The lift and drag coefficients are given by:\\ncl =\",c_l,\"\\ncd =\",c_d" + ] + } + ], + "metadata": { + "anaconda-cloud": {}, + "kernelspec": { + "display_name": "Python [Root]", + "language": "python", + "name": "Python [Root]" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.12" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Fundamentals_Of_Aerodynamics_by_J._D._Anderson_Jr./CHAPTER10_3.ipynb b/Fundamentals_Of_Aerodynamics_by_J._D._Anderson_Jr./CHAPTER10_3.ipynb new file mode 100644 index 00000000..9a52e44a --- /dev/null +++ b/Fundamentals_Of_Aerodynamics_by_J._D._Anderson_Jr./CHAPTER10_3.ipynb @@ -0,0 +1,333 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# CHAPTER10:COMPRESSIBLE FLOW THROUGH NOZZLES DIFFUSERS AND WIND TUNNELS" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example E01 : Pg 345" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Me = 3.95\n", + "pe = 0.035 atm\n", + "Te = 80.89191 K\n" + ] + } + ], + "source": [ + "# All the quantities are expressed in Si units\n", + "area_ratio = 10.25; # exit to throat area ratio\n", + "p0 = 5; # reservoir pressure in atm\n", + "T0 = 333.3; # reservoir temperature\n", + "\n", + "# from appendix A, for an area ratio of 10.25\n", + "Me = 3.95; # exit mach number\n", + "pe = 0.007*p0; # exit pressure\n", + "Te = 0.2427*T0; # exit temperature\n", + "\n", + "print\"Me =\",Me\n", + "print\"pe =\",pe,\"atm\"\n", + "print\"Te =\",Te,\"K\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example E02 : Pg 346" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "At throat: Mt = 1.0\n", + "\n", + "At throat: pt = 0.528 atm\n", + "\n", + "At throat: Tt = 239.904 K\n", + "\n", + "At throat: For supersonic exit: 2.2\n", + "\n", + "At throat: pe = 0.0935453695042 atm\n", + "\n", + "At throat: Te = 146.341463415 K\n", + "\n", + "For subrsonic exit: 0.3\n", + "\n", + "pe= 0.93984962406 atm\n", + "\n", + "Te= 282.907662083 K\n" + ] + } + ], + "source": [ + "# All the quantities are expressed in Si units\n", + "\n", + "area_ratio = 2.; # exit to throat area ratio\n", + "p0 = 1.; # reservoir pressure in atm\n", + "T0 = 288.; # reservoir temperature\n", + "\n", + "# (a)\n", + "# since M = 1 at the throat\n", + "Mt = 1.;\n", + "pt = 0.528*p0; # pressure at throat\n", + "Tt = 0.833*T0; # temperature at throat\n", + "\n", + "# from appendix A for supersonic flow, for an area ratio of 2\n", + "Me = 2.2; # exit mach number\n", + "pe = 1./10.69*p0; # exit pressure\n", + "Te = 1./1.968*T0; # exit temperature\n", + "\n", + "print\"At throat: Mt =\",Mt\n", + "print\"\\nAt throat: pt =\",pt,\"atm\"\n", + "print\"\\nAt throat: Tt = \",Tt,\"K\"\n", + "print\"\\nAt throat: For supersonic exit:\",Me\n", + "print\"\\nAt throat: pe =\",pe,\"atm\"\n", + "print\"\\nAt throat: Te = \",Te,\"K\"\n", + "\n", + "# (b)\n", + "# from appendix A for subonic flow, for an area ratio of 2\n", + "Me = 0.3; # exit mach number\n", + "pe = 1/1.064*p0; # exit pressure\n", + "Te = 1/1.018*T0; # exit temperature\n", + "\n", + "print\"\\nFor subrsonic exit:\",Me\n", + "print\"\\npe=\",pe,\"atm\"\n", + "print\"\\nTe=\",Te,\"K\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example E03 : Pg 346" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Me = 0.2\n", + "Mt = 0.44\n" + ] + } + ], + "source": [ + "# All the quantities are expressed in Si units\n", + "\n", + "area_ratio = 2.; # exit to throat area ratio\n", + "p0 = 1.; # reservoir pressure in atm\n", + "T0 = 288.; # reservoir temperature\n", + "pe = 0.973; # exit pressure in atm\n", + "\n", + "p_ratio = p0/pe; # ratio of reservoir to exit pressure\n", + "\n", + "# from appendix A for subsonic flow, for an pressure ratio of 1.028\n", + "Me = 0.2; # exit mach number\n", + "area_ratio_exit_to_star = 2.964; # A_exit/A_star\n", + "\n", + "# thus\n", + "area_ratio_throat_to_star = area_ratio_exit_to_star/area_ratio; # A_exit/A_star\n", + "\n", + "# from appendix A for subsonic flow, for an area ratio of 1.482\n", + "Mt = 0.44; # throat mach number\n", + "\n", + "print\"Me =\",Me\n", + "print\"Mt =\",Mt" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example E04 : Pg 352" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(a)The thrust of the rocket is:T = 2.1702872295 N\n", + "\n", + "(b)The nozzle exit area is 487663.540469\n", + "\n", + "Ae = 16.7097500627 m2\n" + ] + } + ], + "source": [ + "# All the quantities are expressed in SI units\n", + "import math \n", + "p0 = 30.*101000.; # reservoir pressure\n", + "T0 = 3500.; # reservoir temperature\n", + "R = 520.; # specific gas constant\n", + "gam = 1.22; # ratio of specific heats\n", + "A_star = 0.4; # rocket nozzle throat area\n", + "pe = 5529.; # rocket nozzle exit pressure equal to ambient pressure at 20 km altitude\n", + "\n", + "# (a)\n", + "# the density of air in the reservoir can be calculated as\n", + "rho0 = p0/R/T0;\n", + "\n", + "# from eq.(8.46)\n", + "rho_star = rho0*(2/(gam+1))**(1/(gam-1));\n", + "\n", + "# from eq.(8.44)\n", + "T_star = T0*2/(gam+1);\n", + "a_star = math.sqrt(gam*R*T_star);\n", + "u_star = a_star;\n", + "m_dot = rho_star*u_star*A_star;\n", + "\n", + "# rearranging eq.(8.42)\n", + "Me = math.sqrt(2/(gam-1)*(((p0/pe)**((gam-1)/gam)) - 1));\n", + "Te = T0/(1+(gam-1)/2*Me*Me);\n", + "ae = math.sqrt(gam*R*Te);\n", + "ue = Me*ae;\n", + "\n", + "# thus the thrust can be calculated as\n", + "T = m_dot*ue;\n", + "T_lb = T*0.2247;\n", + "\n", + "# (b)\n", + "# rearranging eq.(10.32)\n", + "Ae = A_star/Me*((2/(gam+1)*(1+(gam-1)/2*Me*Me))**((gam+1)/(gam-1)/2));\n", + "\n", + "print\"(a)The thrust of the rocket is:T =\" ,T/1e6, \"N\"\n", + "print\"\\n(b)The nozzle exit area is\",T_lb \n", + "\n", + "print\"\\nAe =\",Ae, \"m2\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example E05 : Pg 353" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The mass flow rate is: m_dot = 586.100122081 kg/s\n" + ] + } + ], + "source": [ + "# All the quantities are expressed in SI units\n", + "import math \n", + "p0 = 30.*101000.; # reservoir pressure\n", + "T0 = 3500.; # reservoir temperature\n", + "R = 520.; # specific gas constant\n", + "gam = 1.22; # ratio of specific heats\n", + "A_star = 0.4; # rocket nozzle throat area\n", + "\n", + "# the mass flow rate using the closed form analytical expression\n", + "# from problem 10.5 can be given as\n", + "m_dot = p0*A_star*math.sqrt(gam/R/T0*((2/(gam+1))**((gam+1)/(gam-1))));\n", + "\n", + "print\"The mass flow rate is: m_dot =\",m_dot, \"kg/s\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example E06 : Pg 356" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The diffuser throat to nozzle throat area ratio is: = 1.38715494521\n" + ] + } + ], + "source": [ + "# All the quantities are expressed in SI units\n", + "M = 2.; # Mach number\n", + "# for this value M, for a normal shock, from Appendix B\n", + "p0_ratio = 0.7209;\n", + "# thus\n", + "area_ratio = 1./p0_ratio;\n", + "print\"The diffuser throat to nozzle throat area ratio is: =\",area_ratio" + ] + } + ], + "metadata": { + "anaconda-cloud": {}, + "kernelspec": { + "display_name": "Python [Root]", + "language": "python", + "name": "Python [Root]" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.12" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Fundamentals_Of_Aerodynamics_by_J._D._Anderson_Jr./CHAPTER11_3.ipynb b/Fundamentals_Of_Aerodynamics_by_J._D._Anderson_Jr./CHAPTER11_3.ipynb new file mode 100644 index 00000000..383aa8f5 --- /dev/null +++ b/Fundamentals_Of_Aerodynamics_by_J._D._Anderson_Jr./CHAPTER11_3.ipynb @@ -0,0 +1,97 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# CHAPTER11:SUBSONIC COMPRESSIBLE FLOW OVER AIRFOILS LINEAR THEOREY" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example E01 : Pg 382" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(a)The Cp after compressibility corrections is: -0.375 Cp\n" + ] + } + ], + "source": [ + "# All the quantities are expressed in SI units\n", + "import math \n", + "Cp_incompressible = -0.3; # Cp for incompressible flow\n", + "M = 0.6; # Mach number\n", + "# Thus from eq.(11.52)\n", + "Cp_compressible = Cp_incompressible/math.sqrt(1-M**2);\n", + "print\"(a)The Cp after compressibility corrections is:\",Cp_compressible,\"Cp\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example E02 : Pg 383" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(a)The cl after compressibility corrections is: cl = 8.7982192499 alpha\n" + ] + } + ], + "source": [ + "# All the quantities are expressed in SI units\n", + "import math \n", + "cl_incompressible = 2*math.pi; # lift curve slope\n", + "M_inf = 0.7; # Mach number\n", + "# from eq.(11.52)\n", + "cl_compressible = cl_incompressible/math.sqrt(1-M_inf**2); # compressible lift curve slope\n", + "print\"(a)The cl after compressibility corrections is: cl =\",cl_compressible,\"alpha\"" + ] + } + ], + "metadata": { + "anaconda-cloud": {}, + "kernelspec": { + "display_name": "Python [Root]", + "language": "python", + "name": "Python [Root]" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.12" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Fundamentals_Of_Aerodynamics_by_J._D._Anderson_Jr./CHAPTER12_3.ipynb b/Fundamentals_Of_Aerodynamics_by_J._D._Anderson_Jr./CHAPTER12_3.ipynb new file mode 100644 index 00000000..c4032f0f --- /dev/null +++ b/Fundamentals_Of_Aerodynamics_by_J._D._Anderson_Jr./CHAPTER12_3.ipynb @@ -0,0 +1,184 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# CHAPTER12:LINEARIZED SUPERSONIC FLOW" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example E01 : Pg 395" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The cl and cd according to the linearized theory are:cl = 0.12\n", + "The cl and cd according to the linearized theory are:cd = 0.01\n" + ] + } + ], + "source": [ + "# All the quantities are expressed in SI units\n", + "from math import pi,sqrt\n", + "alpha = 5*pi/180; # angle of attack\n", + "M_inf = 3; # freestream mach number\n", + "\n", + "# from eq.(12.23)\n", + "c_l = 4*alpha/sqrt(M_inf**2 - 1);\n", + "\n", + "# from eq.(12.24)\n", + "c_d = 4*alpha**2/sqrt(M_inf**2 - 1);\n", + "\n", + "print\"The cl and cd according to the linearized theory are:cl =\", round(c_l,2)\n", + "print\"The cl and cd according to the linearized theory are:cd =\",round(c_d,2)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example E02 : Pg 395" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The angle of attack of the wing is: 1.97493716351 degrees\n" + ] + } + ], + "source": [ + "# All the quantities are expressed in SI units\n", + "from math import sqrt,pi\n", + "M_inf = 2.; # freestream mach number\n", + "rho_inf = 0.3648; # freestream density at 11 km altitude\n", + "T_inf = 216.78; # freestream temperature at 11 km altitude\n", + "gam = 1.4; # ratio of specific heats\n", + "R = 287.; # specific gas constant\n", + "m = 9400.; # mass of the aircraft\n", + "g = 9.8; # acceleratio due to gravity\n", + "W = m*g; # weight of the aircraft\n", + "S = 18.21; # wing planform area\n", + "# thus\n", + "a_inf = sqrt(gam*R*T_inf);\n", + "V_inf = M_inf*a_inf;\n", + "q_inf = 1./2.*rho_inf*V_inf**2.;\n", + "\n", + "# thus the aircraft lift coefficient is given as\n", + "C_l = W/q_inf/S;\n", + "\n", + "alpha = 180./pi*C_l/4.*sqrt(M_inf**2. - 1.);\n", + "\n", + "print\"The angle of attack of the wing is:\",alpha,\"degrees\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example E03 : Pg 400" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(a) Net Cf = 4.3\n", + "(b) cd = 2.82901631903\n" + ] + } + ], + "source": [ + "# All the quantities are expressed in SI units\n", + "# All the quantities are expressed in SI units\n", + "from math import sqrt,pi\n", + "# (a)\n", + "M_inf = 2.; # freestream mach number\n", + "rho_inf = 0.3648; # freestream density at 11 km altitude\n", + "T_inf = 216.78; # freestream temperature at 11 km altitude\n", + "gam = 1.4; # ratio of specific heats\n", + "R = 287.; # specific gas constant\n", + "m = 9400.; # mass of the aircraft\n", + "g = 9.8; # acceleratio due to gravity\n", + "W = m*g; # weight of the aircraft\n", + "S = 18.21; # wing planform area\n", + "c = 2.2; # chord length of the airfoil\n", + "alpha = 0.035; # angle of attack as calculated in ex. 12.2\n", + "T0 = 288.16; # ambient temperature at sea level\n", + "mue0 = 1.7894e-5; # reference viscosity at sea level\n", + "\n", + "# thus\n", + "a_inf = sqrt(gam*R*T_inf);\n", + "V_inf = M_inf*a_inf;\n", + "\n", + "# according to eq.(15.3), the viscosity at the given temperature is\n", + "mue_inf = mue0*(T_inf/T0)**1.5*(T0+110.)/(T_inf+110.);\n", + "\n", + "# thus the Reynolds number can be given by\n", + "Re = rho_inf*V_inf*c/mue_inf;\n", + "\n", + "# from fig.(19.1), for these values of Re and M, the skin friction coefficient is\n", + "Cf = 2.15*10**-3;\n", + "\n", + "# thus, considering both sides of the flat plate\n", + "net_Cf = 2.*Cf;\n", + "\n", + "# (b)\n", + "c_d = 4.*alpha**2./sqrt(M_inf**2. - 1.);\n", + "\n", + "print\"(a) Net Cf = \",net_Cf*1e3\n", + "print\"(b) cd =\",c_d*1e3" + ] + } + ], + "metadata": { + "anaconda-cloud": {}, + "kernelspec": { + "display_name": "Python [Root]", + "language": "python", + "name": "Python [Root]" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.12" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Fundamentals_Of_Aerodynamics_by_J._D._Anderson_Jr./CHAPTER14_3.ipynb b/Fundamentals_Of_Aerodynamics_by_J._D._Anderson_Jr./CHAPTER14_3.ipynb new file mode 100644 index 00000000..6fb8ebd5 --- /dev/null +++ b/Fundamentals_Of_Aerodynamics_by_J._D._Anderson_Jr./CHAPTER14_3.ipynb @@ -0,0 +1,116 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# CHAPTER14:ELEMENTS OF HYPERSONIC FLOW" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example E01 : Pg 440" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The exact results from the shock-expansion theory are:\n", + "Cp2 = -0.0218681754368\n", + "Cp3 = 0.188459821429\n", + " cl = 0.203161244164\n", + "cd = 0.054436891307\n", + "L/D = 3.73205080757\n", + "\n", + "(b) The results from Newtonian theory are:\n", + "Cp2 = 0\n", + "Cp3 = 0.133974596216\n", + "cl = 0.129409522551\n", + "cd = 0.0346751770605\n", + "L/D = 3.73205080757\n" + ] + } + ], + "source": [ + "# All the quantities are expressed in SI units\n", + "from math import sqrt,pi,sin,cos\n", + "M1 = 8.; # mach number\n", + "alpha = 15.*pi/180.; # anlge of attack\n", + "theta= alpha;\n", + "gam = 1.4;\n", + "# (a)\n", + "# for M = 8\n", + "v1 = 95.62*pi/180.;\n", + "v2 = v1 + theta;\n", + "# from Appendix C\n", + "M2 = 14.32;\n", + "# from Appendix A, for M1 = 8 and M2 = 14.32\n", + "p_ratio = 0.9763e4/0.4808e6;\n", + "# from eq.(11.22)\n", + "Cp2 = 2/gam/M1**2*(p_ratio - 1);\n", + "# for M1 = 8 and theta = 15\n", + "b = 21*pi/180;\n", + "Mn_1 = M1*sin(b);\n", + "# for this value of Mn,1, from appendix B\n", + "p_ratio2 = 9.443;\n", + "# thus\n", + "Cp3 = 2./gam/M1**2.*(p_ratio2 - 1.);\n", + "c_n = Cp3 - Cp2;\n", + "c_l = c_n*cos(alpha);\n", + "c_d = c_n*sin(alpha);\n", + "L_by_D = c_l/c_d;\n", + "print\"The exact results from the shock-expansion theory are:\"\n", + "print\"Cp2 =\",Cp2\n", + "print\"Cp3 = \",Cp3\n", + "print\" cl =\",c_l\n", + "print\"cd =\",c_d\n", + "print\"L/D =\",L_by_D\n", + "# (b)\n", + "# from Newtonian theory, by eq.(14.9)\n", + "Cp3 = 2.*sin(alpha)**2;\n", + "Cp2 = 0;\n", + "c_l = (Cp3 - Cp2)*cos(alpha);\n", + "c_d = (Cp3 - Cp2)*sin(alpha);\n", + "L_by_D = c_l/c_d;\n", + "print\"\\n(b) The results from Newtonian theory are:\"\n", + "print\"Cp2 =\",Cp2\n", + "print\"Cp3 =\",Cp3\n", + "print\"cl =\",c_l\n", + "print\"cd =\",c_d\n", + "print\"L/D =\",L_by_D" + ] + } + ], + "metadata": { + "anaconda-cloud": {}, + "kernelspec": { + "display_name": "Python [Root]", + "language": "python", + "name": "Python [Root]" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.12" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Fundamentals_Of_Aerodynamics_by_J._D._Anderson_Jr./CHAPTER16_3.ipynb b/Fundamentals_Of_Aerodynamics_by_J._D._Anderson_Jr./CHAPTER16_3.ipynb new file mode 100644 index 00000000..be767b15 --- /dev/null +++ b/Fundamentals_Of_Aerodynamics_by_J._D._Anderson_Jr./CHAPTER16_3.ipynb @@ -0,0 +1,147 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# CHAPTER16:SOME SPECIAL CASES COUETTE AND POISEUILLE FLOWS" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example E01 : Pg 524" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(a)u = 30.48 m/s\n", + "(b)tow_w = 429.456 N/m2\n", + "(c)T = 288.628328315 K\n", + "(d)q_w_dot = 13089.81888 Nm-1s-1\n", + "(e)Taw = K 289.613313258\n" + ] + } + ], + "source": [ + "# All the quantities are expressed in SI units\n", + "import math \n", + "mue = 1.7894*10**5; # coefficient of viscosity\n", + "ue = 60.96; # velocity of upper plate\n", + "D = 2.54*10**4; # distance between the 2 plates\n", + "T_w = 288.3; # temperature of the plates\n", + "Pr = 0.71; # Prandlt number\n", + "cp = 1004.5; # specific heat at constant pressure\n", + "\n", + "# (a)\n", + "# from eq.(16.6)\n", + "u = ue/2;\n", + "\n", + "# (b)\n", + "# from eq.(16.9)\n", + "tow_w = mue*ue/D;\n", + "\n", + "# (c)\n", + "# from eq.(16.34)\n", + "T = T_w + Pr*ue**2/8/cp;\n", + "\n", + "# (d)\n", + "# from eq.(16.35)\n", + "q_w_dot = mue/2*ue**2/D;\n", + "\n", + "# (e)\n", + "# from eq.(16.40)\n", + "T_aw = T_w + Pr/cp*ue**2/2;\n", + "\n", + "print\"(a)u =\",u,\"m/s\"\n", + "print\"(b)tow_w =\",tow_w,\"N/m2\"\n", + "print\"(c)T =\",T,\"K\"\n", + "print\"(d)q_w_dot =\",q_w_dot,\"Nm-1s-1\"\n", + "print\"(e)Taw =\",\"K\",T_aw\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example E02 : Pg 524" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The heat transfer is given by:q_w_dot = 3.67387998933 W/m2\n" + ] + } + ], + "source": [ + "# All the quantities are expressed in SI units\n", + "from math import sqrt\n", + "mue = 1.7894*10**5; # coefficient of viscosity\n", + "Me = 3.; # mach number of upper plate\n", + "D = 2.54*10**4; # distance between the 2 plates\n", + "pe = 101000.; # ambient pressure\n", + "Te = 288.; # temperature of the plates\n", + "Tw = Te;\n", + "gam = 1.4; # ratio of specific heats\n", + "R = 287; # specific gas constant\n", + "Pr = 0.71; # Prandlt number\n", + "cp = 1004.5; # specific heat at constant pressure\n", + "tow_w = 72.; # shear stress on the lower wall\n", + "# the velocity of the upper plate is given by\n", + "ue = Me*sqrt(gam*R*Te);\n", + "# the density at both plates is\n", + "rho_e = pe/R/Te;\n", + "# the coefficient of skin friction is given by\n", + "cf = 2.*tow_w/rho_e/ue**2.;\n", + "# from eq.(16.92)\n", + "C_H = cf/2/Pr;\n", + "# from eq.(16.82)\n", + "h_aw = cp*Te + Pr*ue**2/2;\n", + "h_w = cp*Tw;\n", + "q_w_dot = rho_e*ue*(h_aw-h_w)*C_H;\n", + "print\"The heat transfer is given by:q_w_dot =\",q_w_dot/1e4,\"W/m2\"" + ] + } + ], + "metadata": { + "anaconda-cloud": {}, + "kernelspec": { + "display_name": "Python [Root]", + "language": "python", + "name": "Python [Root]" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.12" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Fundamentals_Of_Aerodynamics_by_J._D._Anderson_Jr./CHAPTER18_3.ipynb b/Fundamentals_Of_Aerodynamics_by_J._D._Anderson_Jr./CHAPTER18_3.ipynb new file mode 100644 index 00000000..3b3b1286 --- /dev/null +++ b/Fundamentals_Of_Aerodynamics_by_J._D._Anderson_Jr./CHAPTER18_3.ipynb @@ -0,0 +1,234 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# CHAPTER18:LAMINAR BOUNDARY LAYERS" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example E01 : Pg 595" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The total frictional drag is:(a)D = 17563872.6566 N\n", + "(b) D = 501884115.614 N\n" + ] + } + ], + "source": [ + "# All the quantities are expressed in SI units\n", + "from math import sqrt\n", + "p_inf = 101000.; # freestream pressure\n", + "T_inf = 288.; # freestream temperature\n", + "c = 2.; # chord length of the plate\n", + "S = 40.; # planform area of the plate\n", + "mue_inf=1.7894*10.**5.; # coefficient of viscosity at sea level\n", + "gam=1.4; # ratio of specific heats\n", + "R=287.; # specific gas constant\n", + "# the freestream density is\n", + "rho_inf = p_inf/R/T_inf;\n", + "# the speed of sound is\n", + "a_inf = sqrt(gam*R*T_inf);\n", + "# (a)\n", + "V_inf = 100.;\n", + "# thus the mach number can be calculated as\n", + "M_inf = V_inf/a_inf;\n", + "# the Reynolds number at the trailing is given as\n", + "Re_c = rho_inf*V_inf*c/mue_inf;\n", + "# from eq.(18.22)\n", + "Cf = 1.328/sqrt(Re_c);\n", + "# the friction drag on one surface of the plate is given by\n", + "D_f = 1./2.*rho_inf*V_inf**2.*S*Cf;\n", + "# the total drag generated due to both surfaces is\n", + "D = 2.*D_f;\n", + "print\"The total frictional drag is:(a)D =\",D,\"N\"\n", + "# (b)\n", + "V_inf = 1000.;\n", + "# thus the mach number can be calculated as\n", + "M_inf = V_inf/a_inf;\n", + "# the Reynolds number at the trailing is given as\n", + "Re_c = rho_inf*V_inf*c/mue_inf;\n", + "# from eq.(18.22)\n", + "Cf = 1.2/sqrt(Re_c);\n", + "# the friction drag on one surface of the plate is given by\n", + "D_f = 1./2.*rho_inf*V_inf**2.*S*Cf;\n", + "# the total drag generated due to both surfaces is\n", + "D = 2.*D_f;\n", + "print\"(b) D =\",D,\"N\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example E02 : Pg 596" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The total frictional drag is: D = 4978.09594496 N\n" + ] + } + ], + "source": [ + "# All the quantities are expressed in SI units\n", + "from math import sqrt\n", + "Pr = 0.71; # Prandlt number of air at standard conditions\n", + "Pr_star = Pr;\n", + "Te = 288.; # temperature of the upper plate\n", + "ue = 1000.; # velocity of the upper plate\n", + "Me = 2.94; # Mach number of flow on the upper plate\n", + "p_star = 101000.;\n", + "R = 287.; # specific gas constant\n", + "T0 = 288.; # reference temperature at sea level\n", + "mue0 = 1.7894*10**-5; # reference viscosity at sea level\n", + "c = 2.; # chord length of the plate\n", + "S = 40.; # plate planform area\n", + "\n", + "# recovery factor for a boundary layer is given by eq.(18.47) as\n", + "r = sqrt(Pr);\n", + "\n", + "# rearranging eq.(16.49), we get for M = 2.94\n", + "T_aw = Te*(1+r*(2.74-1));\n", + "\n", + "# from eq.(18.53)\n", + "T_star = Te*(1 + 0.032*Me**2. + 0.58*(T_aw/Te-1.));\n", + "\n", + "# from the equation of state\n", + "rho_star = p_star/R/T_star;\n", + "\n", + "# from eq.(15.3)\n", + "mue_star = mue0*(T_star/T0)**1.5*(T0+110.)/(T_star+110.);\n", + "\n", + "# thus\n", + "Re_c_star = rho_star*ue*c/mue_star;\n", + "\n", + "# from eq.(18.22)\n", + "Cf_star = 1.328/sqrt(Re_c_star);\n", + "\n", + "# hence, the frictional drag on one surface of the plate is\n", + "D_f = 1./2.*rho_star*ue**2.*S*Cf_star;\n", + "\n", + "# thus, the total frictional drag is given by\n", + "D = 2.*D_f;\n", + "\n", + "print\"The total frictional drag is: D =\",D,\"N\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example E03 : Pg 600" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The total frictional drag is: D = 5014.11379241 N\n" + ] + } + ], + "source": [ + "# All the quantities are expressed in SI units\n", + "from math import sqrt\n", + "Pr = 0.71; # Prandlt number of air at standard conditions\n", + "Pr_star = Pr;\n", + "Te = 288.; # temperature of the upper plate\n", + "ue = 1000.; # velocity of the upper plate\n", + "Me = 2.94; # Mach number of flow on the upper plate\n", + "p_star = 101000.;\n", + "R = 287.; # specific gas constant\n", + "gam = 1.4; # ratio of specific heats\n", + "T0 = 288.; # reference temperature at sea level\n", + "mue0 = 1.7894*10**-5; # reference viscosity at sea level\n", + "c = 2.; # chord length of the plate\n", + "S = 40.; # plate planform area\n", + "\n", + "# recovery factor for a boundary layer is given by eq.(18.47) as\n", + "r = sqrt(Pr);\n", + "\n", + "# from ex.(8.2)\n", + "T_aw = Te*2.467;\n", + "T_w = T_aw;\n", + "\n", + "# from the Meador-Smart equation\n", + "T_star = Te*(0.45 + 0.55*T_w/Te + 0.16*r*(gam-1)/2*Me**2.);\n", + "\n", + "# from the equation of state\n", + "rho_star = p_star/R/T_star;\n", + "\n", + "# from eq.(15.3)\n", + "mue_star = mue0*(T_star/T0)**1.5*(T0+110)/(T_star+110.);\n", + "\n", + "# thus\n", + "Re_c_star = rho_star*ue*c/mue_star;\n", + "\n", + "# from eq.(18.22)\n", + "Cf_star = 1.328/sqrt(Re_c_star);\n", + "\n", + "# hence, the frictional drag on one surface of the plate is\n", + "D_f = 1./2.*rho_star*ue**2.*S*Cf_star;\n", + "\n", + "# thus, the total frictional drag is given by\n", + "D = 2.*D_f;\n", + "\n", + "print\"The total frictional drag is: D =\",D,\"N\"" + ] + } + ], + "metadata": { + "anaconda-cloud": {}, + "kernelspec": { + "display_name": "Python [Root]", + "language": "python", + "name": "Python [Root]" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.12" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Fundamentals_Of_Aerodynamics_by_J._D._Anderson_Jr./CHAPTER19_3.ipynb b/Fundamentals_Of_Aerodynamics_by_J._D._Anderson_Jr./CHAPTER19_3.ipynb new file mode 100644 index 00000000..1fa4a0c5 --- /dev/null +++ b/Fundamentals_Of_Aerodynamics_by_J._D._Anderson_Jr./CHAPTER19_3.ipynb @@ -0,0 +1,191 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# CHAPTER19:TURBULENT BOUNDARY LAYERS" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example E01 : Pg 612" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The total frictional drag is: (a)D = 1351.89748485 N\n", + "(b) D = 65392.0 N\n" + ] + } + ], + "source": [ + "# All the quantities are expressed in SI units\n", + "# (a)\n", + "from math import sqrt\n", + "Re_c = 1.36e7; # as obtained from ex. 18.1a\n", + "rho_inf = 1.22; # freestream air denstiy\n", + "S = 40.; # plate planform area\n", + "# hence, from eq.(19.2)\n", + "Cf = 0.074/Re_c**0.2;\n", + "V_inf = 100.;\n", + "# hence, for one side of the plate\n", + "D_f = 1./2.*rho_inf*V_inf**2.*S*Cf;\n", + "# the total drag on both the surfaces is\n", + "D = 2.*D_f;\n", + "print\"The total frictional drag is: (a)D =\",D,\"N\"\n", + "# (b)\n", + "Re_c = 1.36e8; # as obtained from ex. 18.1b\n", + "# hence, from fig 19.1 we have\n", + "Cf = 1.34*10.**-3.;\n", + "V_inf = 1000.;\n", + "# hence, for one side of the plate\n", + "D_f = 1./2.*rho_inf*V_inf**2.*S*Cf;\n", + "# the total drag on both the surfaces is\n", + "D = 2.*D_f;\n", + "print\"(b) D =\",D,\"N\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example E02 : Pg 612" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The total frictional drag is:D = 51916.421508 N\n" + ] + } + ], + "source": [ + "# All the quantities are expressed in SI units\n", + "# from ex 18.2\n", + "from math import sqrt\n", + "Re_c_star = 3.754e7; # Reynolds number at the trailing edge of the plate\n", + "rho_star = 0.574;\n", + "ue = 1000.; # velocity of the upper plate\n", + "S = 40.; # plate planform area\n", + "# from eq.(19.3) we have\n", + "Cf_star = 0.074/Re_c_star**0.2;\n", + "# hence, for one side of the plate\n", + "D_f = 1./2.*rho_star*ue**2.*S*Cf_star;\n", + "# the total drag on both the surfaces is\n", + "D = 2.*D_f;\n", + "print\"The total frictional drag is:D =\",D,\"N\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example E03 : Pg 615" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The total frictional drag is:D = 4967.70450221 N\n" + ] + } + ], + "source": [ + "# All the quantities are expressed in SI units\n", + "import math \n", + "Me = 2.94; # mach number of the flow over the upper plate\n", + "ue = 1000.;\n", + "Te = 288.; # temperature of the upper plate\n", + "ue = 1000.; # velocity of the upper plate\n", + "S = 40.; # plate planform area\n", + "Pr = 0.71; # Prandlt number of air at standard condition\n", + "gam = 1.4; # ratio of specific heats\n", + "\n", + "# the recovery factor is given as\n", + "r = Pr**(1./3.);\n", + "\n", + "# for M = 2.94\n", + "T_aw = Te*(1.+r*(2.74-1.));\n", + "T_w = T_aw; # since the flat plate has an adiabatic wall\n", + "\n", + "# from the Meador-Smart equation\n", + "T_star = Te*(0.5*(1.+T_w/Te) + 0.16*r*(gam-1.)/2.*Me**2.);\n", + "\n", + "# from the equation of state\n", + "p_star=1.\n", + "R=1.\n", + "rho_star = p_star/R/T_star;\n", + "\n", + "# from eq.(15.3)\n", + "mue0=1.\n", + "T0=1.\n", + "c=1.\n", + "mue_star = mue0*(T_star/T0)**1.5*(T0+110.)/(T_star+110.);\n", + "\n", + "# thus\n", + "Re_c_star = rho_star*ue*c/mue_star;\n", + "\n", + "# from eq.(18.22)\n", + "Cf_star = 0.02667/Re_c_star**0.139;\n", + "\n", + "# hence, the frictional drag on one surface of the plate is\n", + "D_f = 1./2.*rho_star*ue**2.*S*Cf_star;\n", + "\n", + "# thus, the total frictional drag is given by\n", + "D = 2.*D_f;\n", + "\n", + "print\"The total frictional drag is:D =\",D,\"N\"" + ] + } + ], + "metadata": { + "anaconda-cloud": {}, + "kernelspec": { + "display_name": "Python [Root]", + "language": "python", + "name": "Python [Root]" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.12" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Fundamentals_Of_Aerodynamics_by_J._D._Anderson_Jr./screenshots/Screenshot02_3.png b/Fundamentals_Of_Aerodynamics_by_J._D._Anderson_Jr./screenshots/Screenshot02_3.png Binary files differnew file mode 100644 index 00000000..c7202e07 --- /dev/null +++ b/Fundamentals_Of_Aerodynamics_by_J._D._Anderson_Jr./screenshots/Screenshot02_3.png diff --git a/Fundamentals_Of_Aerodynamics_by_J._D._Anderson_Jr./screenshots/Screenshot04_3.png b/Fundamentals_Of_Aerodynamics_by_J._D._Anderson_Jr./screenshots/Screenshot04_3.png Binary files differnew file mode 100644 index 00000000..cad0b274 --- /dev/null +++ b/Fundamentals_Of_Aerodynamics_by_J._D._Anderson_Jr./screenshots/Screenshot04_3.png diff --git a/Fundamentals_Of_Aerodynamics_by_J._D._Anderson_Jr./screenshots/Screenshot08_3.png b/Fundamentals_Of_Aerodynamics_by_J._D._Anderson_Jr./screenshots/Screenshot08_3.png Binary files differnew file mode 100644 index 00000000..4e6e962e --- /dev/null +++ b/Fundamentals_Of_Aerodynamics_by_J._D._Anderson_Jr./screenshots/Screenshot08_3.png diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter10_2.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter10_2.ipynb new file mode 100644 index 00000000..0aa8bc1a --- /dev/null +++ b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter10_2.ipynb @@ -0,0 +1,1289 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# CHAPTER 10 : SINGLE STAGE TRANSISTOR AMPLIFIERS" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Using matplotlib backend: Qt4Agg\n" + ] + } + ], + "source": [ + "%matplotlib " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.2 : page number 243-244" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The value of the emitter capacitor = 1.42 𝜇F\n" + ] + } + ], + "source": [ + "from math import pi\n", + "#Variable declaration\n", + "f_min=2.0; #Minimum frequency of operation of amplifier, kHz\n", + "f_max=10.0; #Maximum frequency of operation of amplifier, kHz\n", + "RE=560.0; #Emitter resistor, Ω\n", + "\n", + "#Calculations\n", + "#X_CE(Emitter capacitor's capacitive reactance)\n", + "#X_CE=1/(2*pi*f_min*CE)=RE/10\n", + "#From the above equation.\n", + "CE=1/(2*pi*f_min*1000*(RE/10)); #Emitter capacitor, F,\n", + "\n", + "CE=CE*10**6; #Emitter capacitor, 𝜇F\n", + "\n", + "\n", + "#Results\n", + "print('The value of the emitter capacitor = %.2f 𝜇F'%(CE));\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.5: Page number 252-253" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The operating point: VCE=8.55V and IC=2.15mA.\n", + "Maximum v_CE=9.62V and maximum i_C=19.25mA\n" + ] + }, + { + "data": { + "image/png": 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+ "text/plain": [ + "<matplotlib.figure.Figure at 0x7fb5303cd278>" + ] + }, + "metadata": {}, + "output_type": "display_data" + } + ], + "source": [ + "%matplotlib inline\n", + "import matplotlib.pylab as plt\n", + "\n", + "\n", + "#Variable declaration\n", + "VCC=15.0; #Collector supply voltage in V\n", + "VBE=0.7; #Base-emitter voltage, V\n", + "R1=10.0; #Resistor R1, kΩ\n", + "R2=5.0; #Resistor R2, kΩ\n", + "RC=1.0; #Collector resistor, kΩ\n", + "RE=2.0; #Emitter resistor, kΩ\n", + "RL=1.0; #Load resistor, kΩ\n", + "\n", + "\n", + "#Calculation\n", + "#(i)\n", + "#For d.c load line, from the equation: VCE=VCC-IC*(RC+RE),\n", + "#VCE is maximum when IC=0 and IC is maximum when VCE=0.\n", + "VCE_max=VCC; #Maximum collector-emitter voltage, V\n", + "IC_max=VCC/(RC+RE); #Maximum collector current, mA\n", + "\n", + "\n", + "#plot\n", + "VCE_plot=[i for i in range(0,(int)(VCC+1))]; #Plot variable for V_CE\n", + "IC_plot=[((VCC-i)/(RC+RE)) for i in (VCE_plot[:])]; #Plot variable for I_C\n", + "\n", + "plt.subplot(211)\n", + "plt.xlim(0,20)\n", + "plt.ylim(0,6)\n", + "plt.plot(VCE_plot,IC_plot);\n", + "plt.xlabel(\"VCE(V)\");\n", + "plt.ylabel(\"IC(mA)\");\n", + "plt.title(\"d.c load line\");\n", + "\n", + "\n", + "\n", + "#(ii)\n", + "#For operating point:\n", + "#Assuming VCC drops almost completely across R1 and R2,\n", + "V2=VCC*R2/(R1+R2); #Voltage across resistor R2, V\n", + "IE=(V2-VBE)/RE; #Emitter current, mA\n", + "IC=IE; #Collector current, mA\n", + "VCE=VCC-IC*(RC+RE); #Collector-emitter voltage , V\n", + "\n", + "print(\"The operating point: VCE=%.2fV and IC=%.2fmA.\"%(VCE,IC));\n", + "\n", + "\n", + "#(iii)\n", + "#For a.c load line\n", + "RAC=(RC*RL)/(RC+RL); #a.c load, kΩ\n", + "VCE_ac_max=VCE+IC*RAC; #Maximum collector-emitter voltage, V\n", + "IC_ac_max=IC+VCE/RAC; #Maximum collector current, mA\n", + "print(\"Maximum v_CE=%.2fV and maximum i_C=%.2fmA\"%(VCE_ac_max,IC_ac_max));\n", + "\n", + "#plot\n", + "vCE_plot=[0,VCE_ac_max]; #Plot variable for V_CE\n", + "iC_plot=[IC_ac_max,0]; #Plot variable for I_C\n", + "\n", + "plt.subplot(212)\n", + "plt.xlim(0,10)\n", + "plt.ylim(0,20)\n", + "plt.plot(vCE_plot,iC_plot);\n", + "plt.xlabel(\"vCE(V)\");\n", + "plt.ylabel(\"iC(mA)\");\n", + "plt.title(\"a.c load line\");\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.6: Page number 253-254" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "data": { + "image/png": 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+ "text/plain": [ + "<matplotlib.figure.Figure at 0x7fb50cf60f28>" + ] + }, + "metadata": {}, + "output_type": "display_data" + } + ], + "source": [ + "%matplotlib inline\n", + "import matplotlib.pylab as p\n", + "\n", + "#Variable declaration\n", + "RC=10; #Collector resistor, kΩ\n", + "RL=30; #Load resistor, kΩ\n", + "VCC=20; #Collector supply voltage, V\n", + "IC=1; #Collector current, mA\n", + "VCE=10; #Collector-emitter voltage, V\n", + "\n", + "\n", + "#Calculations\n", + "#For d.c load line:\n", + "#From the equation: VCE=VCC-IC*(RC+RE),\n", + "#When VCE=0, IC is maximum.\n", + "#Emitter resistor is neglected, assuming it as negligible\n", + "IC_max=VCC/RC; #Maximum collector current, mA\n", + "\n", + "#And, when IC=0, VCE is maximum\n", + "VCE_max=VCC; #Maximum collector-emitter voltage, V\n", + "\n", + "#plot\n", + "p.subplot(211)\n", + "p.xlim(0,20)\n", + "p.ylim(0,5)\n", + "VCE_plot=[0,VCE_max]; #Plot variable for V_CE\n", + "IC_plot=[IC_max,0]; #Plot variable for I_C\n", + "\n", + "p.plot(VCE_plot,IC_plot);\n", + "p.xlabel(\"VCE(V)\");\n", + "p.ylabel(\"IC(mA)\");\n", + "p.title(\"d.c load line\");\n", + "\n", + "\n", + "#For a.c load line:\n", + "RAC=(RC*RL)/(RC+RL); #a.c Load resistor, kΩ\n", + "\n", + "VCE_ac_max=VCE+IC*RAC; #Maximum collector-emitter voltage, V\n", + "IC_ac_max=IC+ VCE/RAC; #Maximum collector current, mA\n", + "\n", + "#plot\n", + "p.subplot(212)\n", + "p.xlim([0,25])\n", + "p.ylim([0,5])\n", + "vCE_plot=[0,VCE_ac_max]; #Plot variable for V_CE\n", + "iC_plot=[IC_ac_max,0]; #Plot variable for I_C\n", + "\n", + "p.plot(vCE_plot,iC_plot);\n", + "p.xlabel(\"vCE(V)\");\n", + "p.ylabel(\"iC(mA)\");\n", + "p.title(\"a.c load line\");\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.7 : Page number 254-255" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "data": { + "image/png": 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+ "text/plain": [ + "<matplotlib.figure.Figure at 0x7fb5280aca20>" + ] + }, + "metadata": {}, + "output_type": "display_data" + } + ], + "source": [ + "%matplotlib inline\n", + "import matplotlib.pylab as p\n", + "\n", + "#Variabe declaration\n", + "VCE_Q=8.0; #Q-point collector emitter voltage, V\n", + "IC_Q=1; #Q-point collector current, mA\n", + "ic_positive_peak=1.5; #Collector current at positive peak of signal, mA\n", + "ic_negative_peak=0.5; #Collector current at negative peak of signal, mA\n", + "vce_positive_peak=7; #Collector emitter voltage at positive peak of signal, V\n", + "vce_negative_peak=9; #Collector emitter voltage at negative peak of signal, V\n", + "\n", + "#Plot\n", + "vce_plot=[vce_positive_peak,vce_negative_peak]; #Plot variable of vce\n", + "ic_plot=[ic_positive_peak,ic_negative_peak]; #Plot variable of ic\n", + "\n", + "p.xlim(0,10)\n", + "p.ylim(0,2)\n", + "p.plot(vce_plot,ic_plot);\n", + "p.xlabel(\"vCE(V)\");\n", + "p.ylabel(\"iC(mA)\");\n", + "p.title(\"a.c load line\");\n", + "p.grid();\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.8 : Page number 256" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Voltage gain= 24.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "VCC=10.0; #Collector supply voltage, V\n", + "RC=2.0; #Collector resistor, kΩ\n", + "Rin=1.0; #Input resistance, kΩ\n", + "beta=60.0; #Base current amplification factor\n", + "RL=0.5; #Load resistor, kΩ\n", + "\n", + "\n", + "#Calculation\n", + "RAC=(RC*RL)/(RC+RL); #a.c load resistor, kΩ\n", + "Av=beta*(RAC/Rin); #Voltage gain\n", + "\n", + "#Results\n", + "print(\"Voltage gain= %d.\"%Av);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.9 : Page number 256" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Output voltage= 200mV.\n" + ] + } + ], + "source": [ + "\n", + "#Variable declaration\n", + "V_in=1.0; #Input voltage , mV\n", + "RC=10.0; #Collector resistor, kΩ\n", + "Rin=2.5; #Input resistance, kΩ\n", + "beta=100.0; #Base current amplification factor\n", + "RL=10.0; #Load resistor, kΩ\n", + "\n", + "#Calculations\n", + "RAC=(RC*RL)/(RC+RL); #Effective load, kΩ\n", + "Av=beta*(RAC/Rin); #Voltage gain\n", + "\n", + "V_out=V_in*Av; #Output voltage, V\n", + "\n", + "#Results\n", + "print(\"Output voltage= %dmV.\"%V_out);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.10 : Page number 256-257" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Beta= 100.\n", + "Input impedance=2 kΩ.\n", + "a.c load=3.3 kΩ.\n", + "Voltage gain= 165.\n", + "Power gain=16500.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "change_in_IB=10.0; #Change in base current, 𝜇A\n", + "change_in_IC=1.0; #Change in collector current, mA\n", + "change_in_VBE=0.02; #Change in Base-emitter voltage, V\n", + "RC=5.0; #Collector resistor, kΩ\n", + "RL=10.0; #Emitter resistor, kΩ\n", + "\n", + "#Calculations\n", + "#(i)\n", + "beta=(change_in_IC*1000)/change_in_IB; #Base current amplification factor\n", + "\n", + "#(ii)\n", + "Rin=(change_in_VBE/change_in_IB)*1000; #Input impedance, kΩ\n", + "\n", + "#(iii)\n", + "RAC=round((RC*RL)/(RC+RL),1); #a.c load, kΩ\n", + "\n", + "#(iv)\n", + "Av=beta*RAC/Rin; #Voltage gain\n", + "\n", + "#(v)\n", + "Ap=beta*Av; #Power gain\n", + "\n", + "\n", + "#Results\n", + "print(\"Beta= %d.\"%beta);\n", + "print(\"Input impedance=%d kΩ.\"%Rin);\n", + "print(\"a.c load=%.1f kΩ.\"%RAC);\n", + "print(\"Voltage gain= %d.\"%Av);\n", + "print(\"Power gain=%d.\"%Ap);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.11 : Page number 257" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Output voltage=200mV\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "beta=50.0; #Base current amplification factor\n", + "RC=3.0; #Collector resistor,kΩ\n", + "RL=6.0; #Load resistor, kΩ\n", + "Rin=0.5; #Input impedance, kΩ\n", + "Vin=1; #Input voltage, mV\n", + "\n", + "#Calculation\n", + "RAC=(RC*RL)/(RC+RL); #a.c load, kΩ\n", + "Av=beta*RAC/Rin; #Voltage gain\n", + "Vout=Vin*Av; #Output voltage, V\n", + "\n", + "#Results\n", + "print(\"Output voltage=%dmV\"%Vout);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.12 : Page number 257-258" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The circuit is not operating properly.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "VT=6.0; #Collector potential, V\n", + "R1=1.0; #Resistor R1, kΩ\n", + "R2=2.0; #Resistor R2, kΩ\n", + "VB_found=4.0; #Measured base voltage, V\n", + "\n", + "#Calculations\n", + "VB=(VT*R1)/(R1+R2); #Theoretical base voltage, V\n", + "\n", + "if(VB_found==VB):\n", + " print(\"The circuit is operating properly.\");\n", + "else:\n", + " print(\"The circuit is not operating properly.\");\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.13 : Page number 258-259" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "a.c emitter resistance= 38.46 Ω.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "VCC=10.0; #Collector supply voltage, V\n", + "R1=40.0; #Resistor R1, kΩ\n", + "R2=10.0; #Resistor R2, kΩ\n", + "RC=6.0; #Collector resistor, kΩ\n", + "RE=2.0; #Emitter resistor, kΩ\n", + "beta=80; #Base current amplification factor\n", + "VBE=0.7; #Base emitter voltage, V\n", + "\n", + "#Calculations\n", + "V2=(VCC*R2)/(R1+R2); #Voltage across resistor R2, V\n", + "VE=V2-VBE; #Emitter voltage, V\n", + "IE=VE/RE; #Emitter current, mA\n", + "re=25/IE; #a.c emitter resistance, Ω\n", + "\n", + "\n", + "#Results\n", + "print(\"a.c emitter resistance= %.2f Ω.\"%re);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.14 : Page number 262-263" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i)Voltage gain= 360.\n", + "(ii)Voltage gain= 5.37.\n" + ] + } + ], + "source": [ + "\n", + "#Variable declaration\n", + "VCC=20.0; #Collector supply voltage, V\n", + "VBE=0.7; #Base-emitter voltage, V\n", + "R1=150.0; #Resistor R1, kΩ\n", + "R2=20.0 #Resistor R2, kΩ\n", + "RC=12.0; #Collector resistor, kΩ\n", + "RE=2.2; #Emitter resistor, kΩ\n", + "\n", + "\n", + "#Calculations\n", + "V2=round(VCC*R2/(R1+R2),2); #Voltage across R2, V\n", + "VE=round(V2-VBE,2); #Voltage across emitter resistor, V\n", + "IE=round(VE/RE,2); #Emitter current, mA\n", + "re=round(25/IE,1); #a.c emitter resistance, Ω\n", + "\n", + "\n", + "#(i)\n", + "#CE(emitter capacitor) connected in the circuit:\n", + "Av=(RC*1000)/re; #Voltage gain for emitter capacitor connected.\n", + "\n", + "print(\"(i)Voltage gain= %d.\"%Av);\n", + "\n", + "#(ii)\n", + "#CE(emitter capacitor) removed from the circuit:\n", + "Av=(RC*1000)/(re+RE*1000); #Voltage gain for emitter capacitor removed.\n", + "\n", + "print(\"(ii)Voltage gain= %.2f.\"%Av);\n", + "\n", + "#Note: The answer in the text book has been approximated to 5.38 but it's actually coming 5.37.\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.15 : Page number 263" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Voltage gain= 120.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "RC=6.0; #Collector resistor, kΩ\n", + "RL=12.0; #Load resistor, kΩ\n", + "re=33.3; #a.c emitter resistance, Ω\n", + "\n", + "\n", + "#Calculations\n", + "RAC=RC*RL/(RC+RL); #a.c effective load, kΩ\n", + "Av=RAC*1000/re; #Voltage gain\n", + "\n", + "#Result\n", + "print(\"Voltage gain= %d.\"%Av);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.16 : Page number 263-264" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) a.c emitter resistance=250 Ω.\n", + "(ii) Voltage gain =80.\n", + "(iii) d.c voltage across input capacitor= 1V and emitter capacitor=0.3V.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "VCC=9.0; #Collector supply voltage, V\n", + "VBE=0.7; #Base-emitter voltage, V\n", + "R1=240.0; #Resistor R1, kΩ\n", + "R2=30.0 #Resistor R2, kΩ\n", + "RC=20.0; #Collector resistor, kΩ\n", + "RE=3.0; #Emitter resistor, kΩ\n", + "\n", + "\n", + "#Calculations\n", + "#(i)\n", + "V2=round(VCC*R2/(R1+R2),1); #Voltage across R2, V\n", + "VE=round(V2-VBE,1); #Voltage across emitter resistor, V\n", + "IE=round(VE/RE,1); #Emitter current, mA\n", + "re=25/IE; #a.c emitter resistance, Ω\n", + "\n", + "#(ii)\n", + "Av=RC*1000/re; #Voltage gain\n", + "\n", + "#(iii)\n", + "V_C_in=V2; #d.c voltage across input capacitor, V\n", + "V_C_E=VE; #d.c vooltage across emitter capacitor, V\n", + "\n", + "\n", + "\n", + "#Results\n", + "print(\"(i) a.c emitter resistance=%d Ω.\"%re);\n", + "print(\"(ii) Voltage gain =%d.\"%Av);\n", + "print(\"(iii) d.c voltage across input capacitor= %dV and emitter capacitor=%.1fV.\"%(V_C_in,V_C_E));\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.17 : Page number 264-265" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) D.C bias levels: V2=3V, VE=2.3V, IE=2.3mA, IC=2.3mA, IB=0.023mA and VC=10.4V.\n", + "(ii) D.c voltage across: Cin=3V and CE=2.3V and CC=10.4V.\n", + "(iii) a.c emitter resistance=10.9Ω.\n", + "(iv) Voltage gain=61.2.\n", + "(v) VC>VE. Therefore, the transistor is in active state.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "VCC=15.0; #Collector supply voltage, V\n", + "VBE=0.7; #Base-emitter voltage, V\n", + "R1=40.0; #Resistor R1, kΩ\n", + "R2=10.0 #Resistor R2, kΩ\n", + "RC=2.0; #Collector resistor, kΩ\n", + "RE=1.0; #Emitter resistor, kΩ\n", + "RL=1.0; #Load resistor, kΩ\n", + "beta=100; #Base current amplification factor\n", + "\n", + "#Calculation\n", + "#(i) D.C bias levels\n", + "V2=VCC*R2/(R1+R2); #Voltage across R2, V\n", + "VE=round(V2-VBE,1); #Voltage across emitter resistor, V\n", + "IE=round(VE/RE,1); #Emitter current, mA\n", + "IC=IE; #Collector current, mA\n", + "IB=IC/beta; #Base current, mA\n", + "VC=VCC-IC*RC; #Collector voltage, V\n", + "print(\"(i) D.C bias levels: V2=%dV, VE=%.1fV, IE=%.1fmA, IC=%.1fmA, IB=%.3fmA and VC=%.1fV.\"%(V2,VE,IE,IC,IB,VC));\n", + "\n", + "\n", + "#(ii)\n", + "Cin_V=V2; #Voltage across Cin capacitor, V\n", + "CE_V=VE; #Voltage across CE capacitor, V \n", + "CC_V=VC; #Voltage across CC capacitor, V\n", + "print(\"(ii) D.c voltage across: Cin=%dV and CE=%.1fV and CC=%.1fV.\"%(Cin_V,CE_V,CC_V));\n", + "\n", + "#(iii)\n", + "re=round(25/IE,1); #a.c emitter resistance, Ω\n", + "print(\"(iii) a.c emitter resistance=%.1fΩ.\"%re);\n", + "\n", + "\n", + "#(iv)\n", + "RAC=round(RC*RL/(RC+RL),3); #Total a.c collector resistance, kΩ\n", + "Av=RAC/(re/1000); #Voltage gain\n", + "print(\"(iv) Voltage gain=%.1f.\"%Av);\n", + "\n", + "#(v)\n", + "print(\"(v) VC>VE. Therefore, the transistor is in active state.\" );\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.18 : page number 265" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The power gain = 26400 and output power = 1.584W.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "Av=132.0; #Voltage gain\n", + "beta=200.0; #Base current amplification factor\n", + "P_in=60.0; #Input power, 𝜇W\n", + "\n", + "\n", + "#Calculations\n", + "Ap=beta*Av; #Power gain\n", + "P_out=Ap*(P_in/10**6); #Output power, W\n", + "\n", + "\n", + "#Results\n", + "print(\"The power gain = %d and output power = %.3fW.\"%(Ap,P_out));\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.19 : page number 265-266" + ] + }, + { + "cell_type": "code", + "execution_count": 16, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) Current gain=49\n", + "(ii) Voltage gain=2.14\n", + "(iii) Power gain=105.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "IB=200.0; #Base current, microampere\n", + "IE=10.0; #Emitter current, mA\n", + "R1=27.0; #Resistor R1, kilo ohm\n", + "R2=13.0 #Resistor R2, kilo ohm\n", + "RC=4.7; #Collector resistor, kilo ohm\n", + "RE=2.2; #Emitter resistor, kilo ohm\n", + "\n", + "\n", + "#Calculations\n", + "#(i)\n", + "IC=IE-(IB/1000); #Collector current, mA\n", + "beta=IC/(IB/1000); #Current gain\n", + "\n", + "print(\"(i) Current gain=%d\"%beta);\n", + "\n", + "#(ii)\n", + "#a.c emitter resistance is neglected, voltage gain=(collector resistor)/(emitter resistor)\n", + "Av=RC/RE; #Voltage gain\n", + "\n", + "print(\"(ii) Voltage gain=%.2f\"%Av);\n", + "\n", + "#(iii)\n", + "Ap=round(beta*Av,0); #Power gain\n", + "\n", + "#Results\n", + "print(\"(iii) Power gain=%d.\"%Ap);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.20 : Page number 266-267" + ] + }, + { + "cell_type": "code", + "execution_count": 17, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The input impedance of the amplifier circuit= 3.46 kΩ.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "VCC=30.0; #Collector supply voltage, V\n", + "VBE=0.7; #Base emitter voltage, V\n", + "R1=45.0; #Resistor R1, kΩ\n", + "R2=15.0 #Resistor R2, kΩ\n", + "RC=10.0; #Collector resistor,kΩ\n", + "RE=7.5; #Emitter resistor, kΩ\n", + "beta=200.0; #Base current amplification factor\n", + "\n", + "#Calculations\n", + "V2=round(VCC*R2/(R1+R2),1); #Voltage across R2, V (Voltage divider rule)\n", + "VE=V2; #Voltage across emitter resistor(base-emitter voltage is neglected), V\n", + "IE=VE/RE; #Emitter current, mA (OHM's LAW)\n", + "re=25/IE; #a.c emitter resistance, ohm\n", + "Zin_base=(beta*re)/1000; #input impedance of transistor base,kΩ\n", + "R1_R2=(R1*R2)/(R1+R2); #Parallel resistance between R1 and R2, kΩ\n", + "Zin=((R1_R2)*Zin_base)/(R1_R2+Zin_base); #Input impedance of the amplifier circuit, kΩ\n", + "\n", + "\n", + "#Result\n", + "print(\"The input impedance of the amplifier circuit= %.2f kΩ.\"%Zin); \n", + "\n", + "#Note: The input impedance of the amplifier circuit is approximated as 3.45 kΩ in the text book, but actually it's 3.46 kΩ.\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.21 : Page Number 268-269" + ] + }, + { + "cell_type": "code", + "execution_count": 18, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The voltage gain of the swamped amplifier= 4.67.\n", + "Input impedance of transistor base of the swamped amplifier= 48.21 kΩ.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "VCC=10.0; #Collector supply voltage, V.\n", + "RC=1.5; #Collector resistor, kΩ.\n", + "R1=18.0; #Resistor R1, kΩ.\n", + "R2=4.7; #Resistor R2, kΩ.\n", + "RE1=300.0; #Emitter resistor 1, Ω.\n", + "RE2=900.0; #Emitter resistor 2, Ω.\n", + "VBE=0.7; #Base-emitter voltage, V.\n", + "beta=150.0; #Base current amplification factor.\n", + "\n", + "\n", + "#Calculations\n", + "V2=round(VCC*R2/(R1+R2),1); #d.c voltage across R2, V. (Voltage divider rule)\n", + "VE=round(V2-VBE,1); #d.c voltage across RE, V.\n", + "IE=round((VE/(RE1+RE2))*1000,2); #d.c emitter current, mA.(OHM'S LAW)\n", + "re=round(25/IE,1); #a.c emitter resistance, Ω.\n", + "Av=RC*1000/(re+RE1); #Voltage gain\n", + "Zin_base=(beta*(re+RE1))/1000; #Input impedance of transistor base, kΩ.\n", + "\n", + "\n", + "#Results\n", + "print(\"The voltage gain of the swamped amplifier= %.2f.\"%Av);\n", + "print(\"Input impedance of transistor base of the swamped amplifier= %.2f kΩ.\"%Zin_base);\n", + "\n", + "#Note:In the textbook Av is approximated to 4.66and Zin_base to 48.22 kilo ohm, but the actual answers come as 4.67 and 48.21 kilo ohm.\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.22 : Page number 269" + ] + }, + { + "cell_type": "code", + "execution_count": 19, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The percentage change from the original value= 6.42%(decrease)\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "RC=1.5; #Collector resistor, kΩ.\n", + "RE1=300.0; #Emitter resistor 1, Ω.\n", + "re=21.5; #a.c emitter resistance, Ω.\n", + "\n", + "#Calculations\n", + "Av=round(RC*1000/(re+RE1),2); #Voltage gain.\n", + "Av_1=round(RC*1000/(2*re+RE1),2); #Voltage gain when re doubles.\n", + "change_in_gain=round(Av-Av_1,2); #Change in voltage gain.\n", + "change_percentage=change_in_gain*100/Av; #Change percentage\n", + "\n", + "\n", + "#Results\n", + "if(change_in_gain>0):\n", + " print(\"The percentage change from the original value= %.2f%%(decrease)\"%change_percentage);\n", + "else:\n", + " print(\"The percentage change from the original value= %.2f%%(increase)\"%change_percentage);\n", + "\n", + "\n", + "#Note: The percentage has been approximated in the text book as 6.22%, but the answer comes as 6.42%.\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.23 : Page number 269-270" + ] + }, + { + "cell_type": "code", + "execution_count": 20, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) input impedance of transistor base for standard amplifier= 5 kilo ohm\n", + " input impedance of transistor base for swamped amplifier= 47 kilo ohm\n", + "(ii) input impedance for standard amplifier= 1.33 kilo ohm\n", + " input impedance for swamped amplifier= 1.74 kilo ohm\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "VCC=10.0; #Collector supply voltage, V\n", + "VBE=0.7; #Base emitter voltage, V\n", + "R1=10.0; #Resistor R1, kilo ohm\n", + "R2=2.2; #Resistor R2, kilo ohm\n", + "RC=4.0; #Collector resistor, kilo ohm\n", + "RE=1.1; #Emitter resistor, kilo ohm\n", + "beta=200.0; #Base current amplification factor\n", + "RE1=210.0; #Emitter resistor 1 of swamped amplifier, ohm.\n", + "RE2=900.0; #Emitter resistor 2 of swamped amplifier, ohm.\n", + "\n", + "\n", + "#Calculations\n", + "V2=round(VCC*R2/(R1+R2),1); #d.c voltage across R2, V. (Voltage divider rule)\n", + "VE=round(V2-VBE,1); #d.c voltage across RE, V.\n", + "IE=(VE/RE); #d.c emitter current, mA.(OHM'S LAW)\n", + "re=25/IE; #a.c emitter resistance, ohm.\n", + "\n", + "\n", + "#(i) Zin_base:\n", + "Zin_base_standard=(beta*re)/1000; #input impedance of transistor base for standard amplifier , kilo ohm.\n", + "Zin_base_swamped=(beta*(re+RE1))/1000; #input impedance of transistor base for swamped amplifier, kilo ohm.\n", + "\n", + "\n", + "#(ii) Zin:\n", + "#input impedance for standard amplifier circuit\n", + "Zin_standard=(((R1*R2)/(R1+R2))*Zin_base_standard)/(Zin_base_standard +((R1*R2)/(R1+R2))); #kilo ohm\n", + "\n", + "#input impedance for standard amplifier circuit\n", + "Zin_swamped=(((R1*R2)/(R1+R2))*Zin_base_swamped)/(Zin_base_swamped +((R1*R2)/(R1+R2))); #kilo ohm\n", + "\n", + "\n", + "#Results\n", + "print(\"(i) input impedance of transistor base for standard amplifier= %d kilo ohm\"%Zin_base_standard);\n", + "print(\" input impedance of transistor base for swamped amplifier= %d kilo ohm\"%Zin_base_swamped);\n", + "print(\"(ii) input impedance for standard amplifier= %.2f kilo ohm\"%Zin_standard);\n", + "print(\" input impedance for swamped amplifier= %.2f kilo ohm\"%Zin_swamped);\n", + "\n", + "\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.24 : Page number 270-271" + ] + }, + { + "cell_type": "code", + "execution_count": 21, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The voltage gain of standard amplifier=160.\n", + "The voltage gain of swamped amplifier=17.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "RC=4.0; #Collector resistor, kilo ohm\n", + "re=25.0; #a.c emitter resistance, ohm (calculated in example 10.23)\n", + "RE_1=210.0; #Emitter resistor 1 of swamped amplifier,ohm\n", + "\n", + "#Calculation\n", + "Av_standard=(RC*1000)/re; #Voltage gain of standard common emitter amplifier\n", + "Av_swamped=(RC*1000)/(re+RE_1); #Voltage gain of swamped amplifier\n", + "\n", + "#Results\n", + "print(\"The voltage gain of standard amplifier=%d.\"%Av_standard);\n", + "print(\"The voltage gain of swamped amplifier=%d.\"%Av_swamped);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.26 : Page number 273-274" + ] + }, + { + "cell_type": "code", + "execution_count": 22, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The required input signal voltage =2.5mV\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "A_0=1000.0; #Open circuit voltage gain\n", + "R_in=2.0; #Input resistance, kilo ohm\n", + "R_out=1.0; #Output resistance, ohm\n", + "RL=4; #Load resistor across the output, ohm\n", + "I_2=0.5; #Output signal current, A.\n", + "\n", + "\n", + "#Calculations\n", + "#Since A_0*(I_1*R_in) = I_2*(R_out+RL)\n", + "I_1=I_2*(R_out+RL)/(A_0*(R_in*1000)); #Input current, A\n", + "V_1=I_1*(R_in*1000); #Input signal voltage, V\n", + "V_1=V_1*1000; #Input signal voltage, mV\n", + "\n", + "print(\"The required input signal voltage =%.1fmV\"%V_1);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.27 : Page number 274" + ] + }, + { + "cell_type": "code", + "execution_count": 23, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The magnitude of output voltage = 4.9V\n", + "The power gain =98e-06.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "A_0=1000.0; #Open circuit voltage gain\n", + "R_in=7.0; #Input resistance, kilo ohm\n", + "R_out=15.0; #Output resistance, ohm\n", + "RL=35.0; #Load resistor across the output, ohm\n", + "R_s=3.0; #Internal resistance, kilo ohm\n", + "E_s=10.0; #Input signal voltage, mV.\n", + "\n", + "#Calculations\n", + "#(i)\n", + "I_1=E_s*(10**-3)/(R_s*1000+R_in*1000); #Input current, A\n", + "V_1=I_1*(R_in*1000); #Voltage across input resistance, V\n", + "\n", + "#Since, A_v=V_2/V_1 = A_0*RL/(R_out+RL)\n", + "A_v=A_0*RL/(R_out+RL); #Voltage gain\n", + "V_2=A_v*V_1; #Outout voltage, V\n", + "\n", + "\n", + "#(ii)\n", + "P_2=V_2**2/RL; #Output power, W\n", + "P_1=V_1**2/(R_in*1000); #Input power, W\n", + "A_p=round(P_2/P_1,-6); #Power gain\n", + "\n", + "\n", + "#Result\n", + "print(\"The magnitude of output voltage = %.1fV\"%V_2);\n", + "print(\"The power gain =%de-06.\"%(A_p/10**6));\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.28 : Page number 274-275" + ] + }, + { + "cell_type": "code", + "execution_count": 24, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Necessary input signal voltage= 12.5mV\n", + "Input signal current =4.17 μA\n", + "Power gain = 9600.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "A_v=80.0; #Voltage gain\n", + "V_2=1.0; #Output voltage, V\n", + "A_i=120.0; #Current gain\n", + "RL=2; #Load resistor, kilo ohm\n", + "\n", + "#Calculation\n", + "V_1=(V_2/A_v)*1000; #Input signal voltage, mV\n", + "\n", + "#Since, A_i=A0*R_in/(R_out+RL) and A_v=A0*RL/(R_out+RL)\n", + "#So, A_v/A_i=RL/R_in\n", + "R_in=RL*A_i/A_v; #Input resistance, kilo ohm\n", + "I_1=V_1/R_in; #Input current, μA\n", + "A_p=A_i*A_v; #Power gain\n", + "\n", + "#Results\n", + "print(\"Necessary input signal voltage= %.1fmV\"%V_1);\n", + "print(\"Input signal current =%.2f μA\"%I_1);\n", + "print(\"Power gain = %d.\"%A_p);\n", + "\n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 3", + "language": "python", + "name": "python3" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 3 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython3", + "version": "3.5.1" + }, + "widgets": { + "state": {}, + "version": "1.1.2" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter11_2.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter11_2.ipynb new file mode 100644 index 00000000..6e0fb200 --- /dev/null +++ b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter11_2.ipynb @@ -0,0 +1,1025 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:8e425c9c2dfbfee43b3a89e44b0fd7936ba869da73ac3c372e9b23848f1cded1" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "#CHAPTER 11 : MULTISTAGE TRANSISTOR AMPLIFIERS" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##Example 11.1 : Page number 285" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "from math import log10\n", + "#Variable declaration\n", + "#(i)\n", + "A_v=30; #Voltage gain\n", + "\n", + "#(ii)\n", + "A_p=100; #Power gain\n", + "\n", + "\n", + "#Calculations\n", + "#(i)\n", + "A_v_dB=20*log10(A_v); #Voltage gain, dB\n", + "A_p_dB=10*log10(A_p); #Power gain, dB\n", + "\n", + "#Results\n", + "print(\"(i) Voltage gain in dB=%.2fdB\"%A_v_dB);\n", + "print(\"(ii) Power gain in dB=%ddB\"%A_p_dB);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) Voltage gain in dB=29.54dB\n", + "(ii) Power gain in dB=20dB\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.2 : Page number 285-286\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#(i)\n", + "A_p_dB=40.0; #Power gain in dB\n", + "A_p_b=A_p_dB/10; #Power gain in bel\n", + "A_p=10**A_p_b; #Power gain in number\n", + "\n", + "#Result\n", + "print(\"(i) Power gain in number=%d\"%A_p);\n", + "\n", + "#(ii)\n", + "A_p_dB=43.0; #Power gain in dB\n", + "A_p_b=A_p_dB/10; #Power gain in bel\n", + "A_p=round(10**A_p_b,-4); #Power gain in number\n", + "\n", + "#Result\n", + "print(\"(ii) Power gain in number=%d\"%A_p);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) Power gain in number=10000\n", + "(ii) Power gain in number=20000\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.3 : Page number 286\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "from math import log10\n", + "\n", + "#Variable declaration\n", + "Av_1=100.0; #Voltage gain of stage 1\n", + "Av_2=200.0; #Voltage gain of stage 2\n", + "Av_3=400.0; #Voltage gain of stage 3\n", + "\n", + "#Calculations\n", + "Av_1_dB=20*log10(Av_1); #Voltage gain of stage 1, dB\n", + "Av_2_dB=20*log10(Av_2); #Voltage gain of stage 2, dB\n", + "Av_3_dB=20*log10(Av_3); #Voltage gain of stage 3, dB\n", + "\n", + "Av_T=Av_1_dB+Av_2_dB+Av_3_dB; #Total voltage gain\n", + "\n", + "#Result\n", + "print(\"The total voltage gain=%ddB\"%Av_T);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The total voltage gain=138dB\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.4 : Page number 286\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import log10\n", + "\n", + "#Variable declaration\n", + "A_p_absolute=30.0; #Absolute gain of each stage\n", + "number_of_stages=5.0; #number of stages\n", + "negative_feedback=10.0; #negative feedback, dB\n", + "\n", + "#Calculations\n", + "#(i)\n", + "A_p_dB=round(10*log10(A_p_absolute),2); #Power gain of one stage. dB\n", + "A_p_T=number_of_stages * A_p_dB; #Total power gain, dB\n", + "\n", + "#(ii)\n", + "A_p_resultant=A_p_T-negative_feedback; #Resultant power gain with negative feedback, dB\n", + "\n", + "#Results\n", + "print(\"The total power gain = %.2fdB.\"%A_p_T);\n", + "print(\"The resultant power gain with negative feedback = %.2fdB.\"%A_p_resultant);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The total power gain = 73.85dB.\n", + "The resultant power gain with negative feedback = 63.85dB.\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.5 : Page number 286\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import log10\n", + "\n", + "#Variable declaration\n", + "P_out_2kHz=1.5; #Output power at 2 kHz, W\n", + "P_out_20kHz=0.3; #Output power at 20 kHz, W\n", + "P_in=10.0; #Input power, mW\n", + "\n", + "#Calculations\n", + "A_p_dB_2kHz=10*log10(P_out_2kHz*1000/P_in); #dB power gain at 2 kHz\n", + "A_p_dB_20kHz=10*log10(P_out_20kHz*1000/P_in); #dB power gain at 20 kHz\n", + "Fall_in_gain=A_p_dB_2kHz-A_p_dB_20kHz; #Fall in gain from 2kHz to 20kHz\n", + "\n", + "#Results\n", + "print(\"The fall in gain from 2kHz to 20kHz=%.2fdB\"%Fall_in_gain);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The fall in gain from 2kHz to 20kHz=6.99dB\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.6 : Page number 287\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import log10\n", + "\n", + "#Variable declaration\n", + "A_v=15.0; #Voltage gain, dB\n", + "V_1=0.8; #Input signal voltage, V\n", + "\n", + "#Calculations\n", + "#Since, Av(in decibel)=20*log10(V_2/V_1),\n", + "V_2=V_1*(10**(A_v/20)); #Output voltage, V\n", + "\n", + "#Results\n", + "print(\"The output voltage= %.1fV.\"%V_2);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The output voltage= 4.5V.\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.7 : Page number 287\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "A_0_dB=70.0; #Open circuit voltage gain, dB\n", + "A_v_dB=67.0; #Voltage gain, dB\n", + "R_out=1.5; #Output resistance, kilo ohm\n", + "\n", + "#Calculations\n", + "#Since, A_0_dB-A_v_dB=20*log10(A_0/A_v)\n", + "ratio_A0_Av=round(10**((A_0_dB-A_v_dB)/20),2); #Ratio of open-circuit voltage gain to normal voltage gain\n", + "\n", + "#Since, A_v/A_0 = RL/(R_out+RL)\n", + "RL=R_out/(ratio_A0_Av-1); #Load resistor, kilo ohm\n", + "\n", + "\n", + "#Results\n", + "print(\"The load resistance=%.2f kilo ohm.\"%RL);\n", + "\n", + "#Note: The value of load resistor is calculated to be 3.6585 kilo ohm and approximated to 3.66. But, in the text it has been approximated to 3.65.\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The load resistance=3.66 kilo ohm.\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.8 : Page number 287\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "RL=1.0; #Load resistance, kilo ohm\n", + "A_v=40.0; #Voltage gain, dB\n", + "V_in=10.0; #Input signal voltage, mV\n", + "\n", + "#Calcultaions\n", + "#(i)\n", + "#Since, A_v=20*log10(V_out/V_in)\n", + "V_out=V_in*(10**(A_v/20))/1000; #Output voltage, V\n", + "\n", + "#(ii)\n", + "P_L=(V_out**2/RL); #The load power, mW\n", + "\n", + "\n", + "#Results\n", + "print(\"(i)The output voltage is %dV.\"%V_out);\n", + "print(\"(ii)The load poweris %dmW.\"%P_L);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i)The output voltage is 1V.\n", + "(ii)The load poweris 1mW.\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.9 : Page number 287-288\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "from math import log10\n", + "\n", + "#Variable declaration\n", + "P_2=40.0; #Output power, W\n", + "R=10.0; #Resistance of speaker, ohm\n", + "\n", + "\n", + "#Calculations\n", + "#(i)\n", + "A_p_dB=25.0; #Power gain, dB\n", + "#Since, A_p_dB=10*log10(P_2/P_1)\n", + "P_1=(P_2/10**(A_p_dB/10))*1000; #Input power, mW\n", + "\n", + "\n", + "#(ii)\n", + "A_v_dB=40.0; #Voltage gain, dB\n", + "\n", + "#Since, P=(V**2)/R,\n", + "V_2=(P_2*R)**0.5; #Output voltage, V\n", + "\n", + "#Since, A_v_dB=20*log10(V_2/V_1)\n", + "V_1=(V_2/10**(A_v_dB/20))*1000; #Input voltage, mV\n", + "\n", + "\n", + "#Results\n", + "\n", + "print(\"(i)The input power=%.1fmW.\"%P_1);\n", + "print(\"(ii)The input voltage=%dmV.\"%V_1);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i)The input power=126.5mW.\n", + "(ii)The input voltage=200mV.\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.10 : Page number 288\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "A_v_max=2000.0; #Maximum voltage gain\n", + "f_max=2.0; #Frequency at which maximum voltage gain occurs,kHz\n", + "A_v=1414.0; #Voltage gain at 50 Hz and 10kHz\n", + "f1=50; #Lower frequency at which gain is 1414, Hz\n", + "f2=10; #Upper frequency at which gain is 1414, kHz\n", + "\n", + "#Calculation\n", + "#Since, bandwidth is the range of frequency over which gain is greater than or equal to 70.7% of maximum gain\n", + "if((A_v/A_v_max)*100 ==70.7): \n", + " print(\"(i)The bandwidth is from %dHz to %dkHz.\"%(f1,f2));\n", + " print(\"(ii)The lower cut-off frequency=%dHz.\"%f1);\n", + " print(\"(iii)The upper cut-off frequency=%dkHz.\"%f2);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i)The bandwidth is from 50Hz to 10kHz.\n", + "(ii)The lower cut-off frequency=50Hz.\n", + "(iii)The upper cut-off frequency=10kHz.\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.11 : Page number 291\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "A_v=60.0; #Voltage gain of single stage amplifier\n", + "R_C=500.0; #Collector load, ohm\n", + "R_in=1.0; #Input impedance, kilo ohm\n", + "\n", + "\n", + "#Calculation\n", + "#Since, there is no loading , second stage gain remains at A_v\n", + "#But, due to loading effect of input impedance of second stage, gain of first stage decreases\n", + "A_v_2=A_v; #Voltage gain of second stage\n", + "R_AC=round((R_C*R_in*1000)/(R_C+R_in*1000),0); #Effective load of first stage, ohm\n", + "A_v_1=A_v*R_AC/R_C; #Gain of first stage\n", + "A_v_T=A_v_1*A_v_2; #Total gain\n", + "\n", + "\n", + "#Results\n", + "print(\"The total gain=%d.\"%A_v_T);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The total gain=2397.\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.12 : Page number 291-292\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "Rin=1.0; #Input resistance, kilo ohm\n", + "beta=100.0; #base current amplification factor\n", + "RC=2.0; #Collector load, kilo ohm\n", + "\n", + "\n", + "#Calculation\n", + "#(i)\n", + "R_AC=(RC*Rin)/(RC+Rin); #Effective load on first stage, kilo ohm\n", + "A_v_1=round(beta*(R_AC/Rin),0); #Voltage gain of first stage\n", + "\n", + "#(ii)\n", + "A_v_2=round(beta*RC/Rin,0); #Voltage gain of second stage\n", + "\n", + "#(iii)\n", + "A_v_T=A_v_1*A_v_2; #Total voltage gain\n", + "\n", + "\n", + "#Results\n", + "print(\"(i)The voltage gain of first stage =%d.\"%A_v_1);\n", + "print(\"(ii)The voltage gain of second stage =%d.\"%A_v_2);\n", + "print(\"(iii)The total voltage gain =%d.\"%A_v_T);\n", + "\n", + "#Note: The approximation inthe text for A_v_1=66.66 is taken as 66 but here it has been taken 67 and therefore the total voltage is 13400 instead of 13200.\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i)The voltage gain of first stage =67.\n", + "(ii)The voltage gain of second stage =200.\n", + "(iii)The total voltage gain =13400.\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.13 : Page number 292\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "RC=10.0; #Collector load of single stage amplifier, kilo ohm\n", + "Rin=1.0; #Input resistance, kilo ohm\n", + "beta=100.0; #base current amplification factor\n", + "RL=100.0; #Load resistor, ohm\n", + "\n", + "\n", + "#Calculations\n", + "R_AC=round((RC*1000)*RL/(RC*1000+RL),-1); #Effective collector load,\n", + "A_v=beta*R_AC/(Rin*1000); #Voltage gain\n", + "\n", + "#Results\n", + "print(\"The voltage gain=%d.\"%A_v);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The voltage gain=10.\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.14 : Page number 292-293\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "VCC=20.0; #Collector suppply voltage, V\n", + "R1=10.0; #resistor R1, kilo ohm\n", + "R2=2.2; #resistor R2, kilo ohm\n", + "R3=10.0; #resistor R3, kilo ohm\n", + "R4=2.2; #resistor R4, kilo ohm\n", + "RC_1=3.6; #Collector resistor of first stage, kilo ohm\n", + "RC_2=4.0; #Collector resistor of second stage, kilo ohm\n", + "RE_1=900.0; #Emitter resistor of first stage, ohm\n", + "RE_2=1.0; #Emitter resistor of second stage, kilo ohm\n", + "\n", + "\n", + "#Calculations\n", + "#Biasing potential for the second stage is the voltage across R4 resistor,\n", + "#so, by voltage divider rule:\n", + "VB=VCC*R4/(R3+R4); #Biasing potential for second stage,(Voltage across R4), V\n", + "\n", + "print(\"The biasing voltage for the second stage=%.1fV.\"%VB);\n", + "\n", + "#If coupling capacitor C_c is replaced by a wire, RC_1 and R3 become parallel\n", + "Req=round((RC_1*R3)/(RC_1+R3),2); #Equivalent resistance of R3 parallel with RC_1, kilo ohm\n", + "VB=VCC*R4/(Req+R4); #Biasing voltage if coupling capacitor is replaced by a wire, V\n", + "\n", + "print(\"The biasing voltage after replacing coupling capacitor by wire=%.2fV.\"%VB);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The biasing voltage for the second stage=3.6V.\n", + "The biasing voltage after replacing coupling capacitor by wire=9.07V.\n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.15 : Page number 293-294\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "#Function for calculating parallel resistance\n", + "def pr(r1,r2):\n", + " return (r1*r2)/(r1+r2);\n", + "\n", + "#Variable declaration\n", + "VCC=15.0; #Collector supply voltage, V\n", + "R1=22.0; #Resistor R1, kilo ohm\n", + "R2=3.3; #Resistor R2, kilo ohm\n", + "R3=5.0; #Resistor R3, kilo ohm\n", + "R4=1.0; #Resistor R4, kilo ohm\n", + "R5=15.0; #Resistor R5, kilo ohm\n", + "R6=2.5; #Resistor R6, kilo ohm\n", + "R7=5.0; #Resistor R7, kilo ohm\n", + "R8=1.0; #Resistor R8, kilo ohm\n", + "beta=200; #Base current amplification factor\n", + "RL=10.0; #Load resistor, kilo ohm\n", + "V_BE=0.7; #Base-emitter voltage, V\n", + "\n", + "\n", + "#Calculations\n", + "#for 2nd stage\n", + "V_R6=round(VCC*R6/(R5+R6),2); #Voltage across R6, V (voltage divider rule)\n", + "V_R8=round(V_R6-V_BE,2); #Voltage across R8, V\n", + "IE_2=round(V_R8/R8,2); #Emitter current through R8, mA (OHM's LAW)\n", + "re_2nd_stage=round(25/IE_2,1); #a.c emitter resistance for 2nd stage, ohm\n", + "\n", + "#For 1st stage\n", + "V_R2=round(VCC*R2/(R1+R2),2); #Voltage across R2, V (voltage divider rule)\n", + "V_R4=round(V_R2-V_BE,2); #Voltage across R4, V\n", + "IE_1=round(V_R4/R4,2); #Emitter current through R4, mA (OHM's LAW)\n", + "re_1st_stage=round(25/IE_1,1); #a.c emitter resistance for 1st stage, ohm\n", + "\n", + "#(i)\n", + "Zin_base_2nd_stage=round((beta*re_2nd_stage)/1000,2); #input resistance of transistor base of 2nd stage, kilo ohm\n", + "Zin=round(pr(pr(R5,R6),Zin_base_2nd_stage),2); #Input impedance of the 2nd stage, kilo ohm\n", + "R_AC_1st_stage=round(pr(R3,Zin),2); #Effective collector load for 1st stage, kilo ohm\n", + "A_v_1=round(R_AC_1st_stage*1000/re_1st_stage,0); #voltage gain of 1st stage\n", + "\n", + "#(ii)\n", + "R_AC_2nd_stage=round(pr(R7,RL),2); #Effective collector load for 2nd stage, kilo ohm\n", + "A_v_2=round(R_AC_2nd_stage*1000/re_2nd_stage,1); #voltage gain of 2nd stage\n", + "\n", + "#(iii)\n", + "A_v_overall=A_v_1*A_v_2; #overall voltage gain\n", + "\n", + "\n", + "#results\n", + "print(\"(i)The voltage gain of 1st stage=%.0f.\"%A_v_1);\n", + "print(\"(i)The voltage gain of 2nd stage=%.1f.\"%A_v_2);\n", + "print(\"(i)The overall voltage gain =%d.\"%A_v_overall);\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i)The voltage gain of 1st stage=53.\n", + "(i)The voltage gain of 2nd stage=191.4.\n", + "(i)The overall voltage gain =10144.\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.16 : Page number 297\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "Primary_impedance=1000.0; #Primary impedance, ohm\n", + "Load_impedance=10.0; #Load impedance, ohm\n", + "\n", + "#Calculation\n", + "#since,for maximum power transfer primary impedance should be equal to output impedance\n", + "#and, impedance of secondary should be equal to load impedance\n", + "#therfore, primary_impedance/load_impedance=square of(primary to secondary turn ratio)\n", + "n=(Primary_impedance/Load_impedance)**0.5; #Primary to secondary turn ratio\n", + "\n", + "\n", + "#Result\n", + "print('The primary to secondary turn ratio for maximum power transfer=%d.'%n);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The primary to secondary turn ratio for maximum power transfer=10.\n" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.17 : Page number 297\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "RL=16.0; #Load resistor, ohm\n", + "R_p=10.0; #Output impedance of primary, kilo ohm\n", + "Vp=10.0; #Terminal voltage of the source, V\n", + "\n", + "#Calculation\n", + "#Since, for maximum power transfer, the impedance of the primary should be equal to output impedance of the source\n", + "n=(R_p*1000/RL)**0.5; #Primary to secondary turns ratio\n", + "\n", + "#Since, power in a transformer remains constant,\n", + "#ratio of primary to secondary voltageis equal to primary to secondary turns ratio\n", + "Vs=Vp/n; #Voltage across the external load, V\n", + "\n", + "\n", + "#Result\n", + "print(\"The primary to secondary turns ratio=%d.\"%n);\n", + "print(\"The voltage across the external load=%.1fV.\"%Vs);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The primary to secondary turns ratio=25.\n", + "The voltage across the external load=0.4V.\n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.18 : Page number 297-298\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "Rp=300.0; #D.C resistance of primary, ohm\n", + "RL=3.0; #Load resistance, ohm\n", + "R_out=3.0; #Ouput resistance of the transistor, kilo ohm \n", + "\n", + "#Calculation\n", + "#when no signal is applied, only Rp is seen to be the load.\n", + "#But, when a.c signal is applied, RL in secondary reflects as RL*(squre of turns ratio).\n", + "#Therefore, load is seen to be Rp in series with the reflected RL in primary.\n", + "#i.e, R_out=Rp+(n**2 * RL), where n is the turns ratio\n", + "n=((R_out*1000-Rp)/RL)**0.5; #turns ratio\n", + "\n", + "#Result\n", + "print(\"Turns ratio for maximum power transfer=%d.\"%n);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Turns ratio for maximum power transfer=30.\n" + ] + } + ], + "prompt_number": 20 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.19 : Page number 298" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import pi\n", + "\n", + "#Variable declaration\n", + "f=200.0; #Frequency, Hz\n", + "Z_out=10.0; #Output impedance of the transistor, kilo ohm\n", + "Z_in=2.5; #Input impedance of the next stage, kilo ohm\n", + "\n", + "#Calculation\n", + "#For perfect impedance matching,\n", + "#Z_out should be equal to primary impedance\n", + "#Z_out=2*pi*f*(primary inductance)\n", + "Lp=(Z_out*1000)/(2*pi*f); #Primary inductance, H\n", + "\n", + "#for the secondary side,\n", + "#Z_in should be equal to impedance of secondary\n", + "Ls=(Z_in*1000)/(2*pi*f); #Secondary inductance, H\n", + "\n", + "\n", + "#result\n", + "print(\"The primary inductance=%.0fH.\"%Lp);\n", + "print(\"The secondary inductance=%.0fH.\"%Ls);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The primary inductance=8H.\n", + "The secondary inductance=2H.\n" + ] + } + ], + "prompt_number": 22 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.20 : Page number 299\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "Lp=8.0; #Primary inductance, H\n", + "Ls=2.0; #Secondary inductance, H\n", + "K=10**-5; #Inductance to turns ratio, constant\n", + "\n", + "#Calculations\n", + "Np=(Lp/K)**0.5; #Primary turns\n", + "Ns=(Ls/K)**0.5; #Secondary turns\n", + "\n", + "#Result\n", + "print(\"The primary turns=%.0f.\"%Np);\n", + "print(\"The secondary turns=%.0f.\"%Ns);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The primary turns=894.\n", + "The secondary turns=447.\n" + ] + } + ], + "prompt_number": 23 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.21 : Page number 300-301\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "#Variable declaration\n", + "VCC=12.0; #Collector supply voltage, V\n", + "R1=100.0; #Resistor R1, kilo ohm\n", + "R2=22.0; #Resistor R2, kilo ohm\n", + "R3=22.0; #Resistor R3, kilo ohm\n", + "R4=4.7; #Resistor R4, kilo ohm\n", + "R5=10.0; #Resistor R5, kilo ohm\n", + "R6=10.0; #Resistor R6, kilo ohm\n", + "beta=125; #Base current amplification factor\n", + "V_BE=0.7; #Base-emitter voltage, V\n", + "\n", + "\n", + "#Calculation\n", + "#(i) D.C voltages\n", + "#For 1st stage:\n", + "V_B1=VCC*R2/(R1+R2); #Voltage at the base of 1st transistor, V (Voltage across R2, using voltage divider rule)\n", + "V_E1=V_B1-V_BE; #Emitter voltage of the 1st transistor, V\n", + "I_E1=round(V_E1/R4,2); #Emitter current of 1st transistor, mA (OHM's LAW)\n", + "I_C1=I_E1; #Collector current of 1st transistor, mA(approximately equals to emitter current)\n", + "V_C1=VCC-I_C1*R3; #Collector voltage of 1st transistor, V\n", + "\n", + "#For 2nd stage:\n", + "V_B2=V_C1; #Voltage at the base of 2nd transistor, V (equals to collector voltage of 1st transistor)\n", + "V_E2=V_C1-V_BE; #Emitter voltage of the 2nd transistor, V\n", + "I_E2=V_E2/R6; #Emitter current of 2nd transistor, mA (OHM's LAW)\n", + "I_C2=I_E2; #Collector current 2nd transistor, mA(approximately equals to emitter current)\n", + "V_C2=VCC-I_C2*R5; #Collector voltage of 2nd transistor, V\n", + "\n", + "print(\"(i) D.C voltages\");\n", + "print(\"First stage: VB1=%.2fV , VE1=%.2fV and VC1=%.2fV\"%(V_B1,V_E1,V_C1));\n", + "print(\"First stage: VB2=%.2fV , VE2=%.2fV and VC2=%.2fV\"%(V_B2,V_E2,V_C2));\n", + "\n", + "#(ii)Voltage gain\n", + "#First stage\n", + "re_1=25/I_E1; #a.c emitter resistance of 1st transistor, ohm\n", + "re_2=25/I_E2; #a.c emitter resistance of 2nd transistor, ohm\n", + "Zin_2nd_stage=beta*re_2/1000; #Input impedance of 2nd stage, kilo ohm\n", + "R_AC=R3*Zin_2nd_stage/(R3+Zin_2nd_stage); #Total a.c collector load, kilo ohm\n", + "A_v1=round(R_AC*1000/re_1,0); #Voltage gain of first stage\n", + "\n", + "print(\"The voltage gain of first stage=%d.\"%A_v1);\n", + "\n", + "#Second stage\n", + "R_AC=R5; #Total a.c collector load for 2nd stage, kilo ohm(Due to no loading effect, equal to R5)\n", + "A_v2=round(R5*1000/re_2,0); #Voltage gain of 2nd stage\n", + "\n", + "print(\"The voltage gain of second stage=%d.\"%A_v2);\n", + "\n", + "A_vT=A_v1*A_v2; #Overall voltage gain\n", + "\n", + "print(\"Overall voltage gain=%d.\"%A_vT);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) D.C voltages\n", + "First stage: VB1=2.16V , VE1=1.46V and VC1=5.18V\n", + "First stage: VB2=5.18V , VE2=4.48V and VC2=7.52V\n", + "The voltage gain of first stage=66.\n", + "The voltage gain of second stage=179.\n", + "Overall voltage gain=11814.\n" + ] + } + ], + "prompt_number": 24 + }, + { + "cell_type": "code", + "collapsed": false, + "input": [], + "language": "python", + "metadata": {}, + "outputs": [] + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter12_2.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter12_2.ipynb new file mode 100644 index 00000000..05e3d9d8 --- /dev/null +++ b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter12_2.ipynb @@ -0,0 +1,968 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:1e9209171152811793fc18d1ee8c80ddcef574d69421ec87eeaa8fb87a304f6d" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 12: TRANSISTOR AUDIO POWER AMPLIFIERS" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.1 : Page number 308\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "VCC=10.0; #Collector supply voltage, V\n", + "R1=10.0; #Resistor R1, kilo ohm\n", + "R2=2.2; #Resistor R2, kilo ohm\n", + "RC=3.6; #Collector resistor, kilo ohm\n", + "RE=1.1; #Emitter resistor, kilo ohm\n", + "VBE=0.7; #Base-emitter voltage, V\n", + "\n", + "#Calculation\n", + "I1=VCC/(R1+R2); #Current through R1 and R2, mA (OHM's LAW)\n", + "V2=I1*R2; #Voltage across R2 resistor, V (OHM's LAW)\n", + "VE=V2-VBE; #Emitter voltage, V\n", + "IE=VE/RE; #Emitter current, mA (OHM's LAW)\n", + "IC=IE; #Collector current, mA (approximately equal to emitter current)\n", + "I_T=I1+IC; #Total current drawn from the supply, mA\n", + "P_dc=VCC*I_T; #Total power drawn from the supply, mW\n", + "\n", + "\n", + "#Results\n", + "print(\"The total power drawn from the supply=%.1fmW.\"%P_dc);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The total power drawn from the supply=18.2mW.\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.2 : Page number 309\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "V_L=10.6; #Voltage across load, V.(from a.c voltmeter, therfore r.m.s value)\n", + "R_L=200.0; #Load resistance, ohm\n", + "\n", + "#Calculation\n", + "#Since, power =V**2/R,\n", + "P_O=(V_L**2/R_L)*1000; #A.C output power, mW\n", + "\n", + "#Result\n", + "print(\"The a.c output power = %.1fmW.\"%P_O);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The a.c output power = 561.8mW.\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.3 : Page number 309\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "RL=100.0; #Load resistance, ohm\n", + "V_PP=18.0; #Peak-to-peak a.c voltage, V\n", + "\n", + "#Calculation\n", + "#Since, V(r.m.s)=(V(peak-to-peak)/2)/sqrt(2)\n", + "VL=V_PP/(2*(2**0.5)); #r.m.s value, V\n", + "\n", + "#Since, power=(square of voltage)/resistance\n", + "P_O_max=(VL**2/RL)*1000; #Maximum possible a.c load power, mW\n", + "\n", + "#Result\n", + "print(\"The maximum possible a.c load power=%dmW.\"%P_O_max);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The maximum possible a.c load power=405mW.\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.4 : Page number 310\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "V_battery=12.0; #Battery voltage, V\n", + "P_out=2.0; #Output power, W\n", + "\n", + "#Calculation\n", + "#Since, Power=Current*Voltage\n", + "IC=(P_out/V_battery)*1000; #Maximum collector current , mA\n", + "\n", + "#Result\n", + "print(\"The maximum collector current=%.1fmA.\"%IC);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The maximum collector current=166.7mA.\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.5 : Page number 310\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "V_battery=12.0; #Battery voltage, V\n", + "RL=4.0; #Collector load, kilo ohm\n", + "\n", + "#Calculation\n", + "IC_max=V_battery/RL; #Maximum collector current, mA\n", + "\n", + "\n", + "#Result\n", + "print(\"The maximum collector current=%dmA.\"%IC_max);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The maximum collector current=3mA.\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.6 : Page number 310-311\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "P=50.0; #Power supplied by power amplifier, W\n", + "R=8.0; #Resistance of speaker, ohm\n", + "\n", + "#Calculation\n", + "#(i)\n", + "#Since, Power=Voltage _square/Resistance,\n", + "V=(P*R)**0.5; #a.c output voltage, V\n", + "\n", + "#(ii)\n", + "I=V/R; #a.c output current, A (OHM's LAW)\n", + "\n", + "\n", + "#Result\n", + "print(\"(i) The a.c output voltage=%dV.\"%V);\n", + "print(\"(ii) The a.c output current=%.1fA.\"%I);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) The a.c output voltage=20V.\n", + "(ii) The a.c output current=2.5A.\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.7 : Page number 315\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "VCC=20.0; #Collector supply voltage, V\n", + "ib_peak=10.0; #Base current(peak), mA\n", + "RB=1.0; #Base resistance, kilo ohm\n", + "RC=20.0; #Collector resistance, ohm\n", + "beta=25.0; #Base current amplification factor\n", + "VBE=0.7; #Base-emitter voltage, V\n", + "\n", + "#Calculation\n", + "IB=round(VCC-VBE/RB,1); #Base current, mA (OHM's LAW)\n", + "IC=int(beta*IB); #Collector current, mA\n", + "VCE=VCC-(IC/1000)*RC; #Collector emitter voltage, V (KVL)\n", + "\n", + "#(i)\n", + "ic_peak=beta*ib_peak; #Collector current(peak), mA\n", + "P_o_ac=(ic_peak/1000)**2*RC/2; #Output power, W\n", + "\n", + "#(ii)\n", + "P_dc=VCC*IC/1000; #Input power, W\n", + "\n", + "#(iii)\n", + "collector_efficiency=(P_o_ac/P_dc)*100; #Collector efficiency of the amplifier circuit,\n", + "\n", + "#Result\n", + "print(\"(i) The output power=%.3fW.\"%P_o_ac);\n", + "print(\"(ii) The input power=%.1fW.\"%P_dc);\n", + "print(\"(iii) The collector efficiency=%.1f%%.\"%collector_efficiency);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) The output power=0.625W.\n", + "(ii) The input power=9.6W.\n", + "(iii) The collector efficiency=6.5%.\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.8 : Page number 317\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "P_dc=10.0; #zero signal power dissipation, W\n", + "P_o=4.0; #a.c output power, W\n", + "\n", + "#Calculation\n", + "#(i)\n", + "Collector_eff=(P_o/P_dc)*100; #collector efficiency\n", + "\n", + "#(ii)\n", + "#Zero signal power is the maximum power dissipation in a transistor, therefore,\n", + "Power_rating=P_dc; #Power rating of the transistor, W\n", + "\n", + "#Result\n", + "print(\"(i) The collector efficiency=%d%%.\"%Collector_eff);\n", + "print(\"(i) The power rating of the transistor=%dW.\"%Power_rating);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) The collector efficiency=40%.\n", + "(i) The power rating of the transistor=10W.\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.9 : Page number 317-318\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "RL=100.0; #Secondary load, ohm\n", + "n=10.0; #Transformer turn ratio\n", + "IC=100.0; #Zero signal collector current, mA\n", + "\n", + "#Calculation\n", + "RL_reflected=n**2*RL; #Reflected load as seen by the primary of the transformer, ohm\n", + "P_o_ac_max=(IC/1000)**2*RL_reflected/2; #Maximum a.c power output, W \n", + "\n", + "\n", + "\n", + "#Result\n", + "print(\"The maximum a.c power output=%dW.\"%P_o_ac_max);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The maximum a.c power output=50W.\n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.10 : Page number 318\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "VCC=5.0; #Collector supply voltage, V\n", + "IC=50.0; #Zero signal collector current, mA\n", + "\n", + "#Calculation\n", + "#(i)\n", + "P_o_max=VCC*IC/2; #Maximum a.c output power, mW\n", + "\n", + "#(ii)\n", + "P_dc=VCC*IC; #D.C input power, mW\n", + "#Since, maximum power is dissipated in the zero signal conditions\n", + "Power_rating=P_dc; #Power rating of transistor, mW\n", + "\n", + "#(iii)\n", + "Max_collector_eff=(P_o_max/P_dc)*100; #Maximum collector efficiency\n", + "\n", + "\n", + "#Result\n", + "print(\"(i) The maximum a.c output power=%dmW\"%P_o_max);\n", + "print(\"(ii) The power rating of the transistor=%dmW.\"%Power_rating);\n", + "print(\"(iii) The maximum collector efficiency =%d%%.\"%Max_collector_eff);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) The maximum a.c output power=125mW\n", + "(ii) The power rating of the transistor=250mW.\n", + "(iii) The maximum collector efficiency =50%.\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.11 : Page number 318\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt\n", + "\n", + "#Variable declaration\n", + "ic_max=160.0; #Maximum a.c collector current, mA\n", + "ic_min=10.0; #Minimum a.c collector current, mA\n", + "vce_max=12.0; #Maximum collector-emitter voltage, V\n", + "vce_min=2.0; #Minimum collector-emitter voltage, V\n", + "\n", + "#Calculation\n", + "vce_pp=vce_max-vce_min; #peak to peak collector emitter voltage, V\n", + "ic_pp=ic_max-ic_min; #peak to peak collector current, V\n", + "P_o=(vce_pp/(2*sqrt(2)))*(ic_pp/(2*sqrt(2))); #a.c output power, mW\n", + "\n", + "\n", + "#Result\n", + "print(\"The a.c output power=%.1fmW.\"%P_o);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The a.c output power=187.5mW.\n" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.12 : Page number 319-320\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt \n", + "\n", + "#Variable declaration\n", + "VCC=12.0; #Battey voltage, V\n", + "IC_max_change=100.0; #maximum collector current change, mA\n", + "RL=5.0; #Loudspeaker resistance, ohm\n", + "\n", + "#Calculation\n", + "VCE_max_change=VCC; #Maximum collector-emitter voltage change\n", + "#(i) Loud speaker directly connected in the collector\n", + "Vmax_speaker=(IC_max_change/1000)*RL; #Maximum voltage across the loudspeaker, V\n", + "P_speaker_directly_coupled=Vmax_speaker*IC_max_change; #Power developed in the loudspeaker,mW\n", + "\n", + "#(ii) Loudspeaker transformer coupled\n", + "Z_out=(VCE_max_change/IC_max_change)*1000; #Output impedance of transistor, ohm\n", + "\n", + "#For max power transfer, primary impedance should be Z_out\n", + "RL_reflected=Z_out; #Load resistance as seen by primary, ohm\n", + "n=sqrt(RL_reflected/RL); #Turns ratio of transformer\n", + "Vp=VCC; #Transformer primary voltage, V\n", + "Vs=Vp/n; #Transformer secondary voltage, V\n", + "IL=Vs/RL; #Load current, A\n", + "P_speaker_transformer_coupled=IL**2*RL*1000; #Power delivered to the speaker, mW\n", + "\n", + "\n", + "#Result\n", + "print(\"(i) The power transferred to the speaker when directly coupled=%dmW.\"%P_speaker_directly_coupled);\n", + "print(\"(ii) The power trasnferred to the speaker when transformer-coupled=%dmW.\"%P_speaker_transformer_coupled);\n", + "print(\" The turns ratio=%.1f.\"%n);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) The power transferred to the speaker when directly coupled=50mW.\n", + "(ii) The power trasnferred to the speaker when transformer-coupled=1200mW.\n", + " The turns ratio=4.9.\n" + ] + } + ], + "prompt_number": 23 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.13 : Page number 320-321\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt\n", + "\n", + "#Variable declaration\n", + "beta=100.0; #Base current amplification factor\n", + "RL=81.6; #Load resistance, ohm\n", + "VCE_peak=30.0; #Peak value of collector voltage, V\n", + "IC_peak=35.0; #Peak value of collector current, mA\n", + "VCE_min=5.0; #Minimum value of collector voltage, V\n", + "IC_min=1.0; #Minimum value of collector current, mA\n", + "\n", + "\n", + "#Calculation\n", + "#(i)\n", + "IC_zero_signal=(IC_peak-IC_min)/2 +1; #Zero signal collector current, mA\n", + "\n", + "#(ii)\n", + "IB_zero_signal=IC_zero_signal/beta; #Zero signal base current, mA\n", + "\n", + "#(iii)\n", + "VCE_zero_signal=(VCE_peak-VCE_min)/2 +5; #Zero signal collector-emitter voltage, V\n", + "VCC=VCE_zero_signal; #Collector supply voltage,V (due to transformer coupling, aproximately equal to zero signal VCE)\n", + "P_dc=VCC*IC_zero_signal; #d.c input power, mW\n", + "VCE_ac=(VCE_peak-VCE_min)/(2*sqrt(2)); #a.c output voltage, V\n", + "IC_ac=(IC_peak-IC_min)/(2*sqrt(2)); #a.c output current, mA\n", + "P_ac=VCE_ac*IC_ac; #a.c output power, mW\n", + "\n", + "#(iv)\n", + "collector_eff=(P_ac/P_dc)*100; #Collector efficiency\n", + "\n", + "#(v)\n", + "#a.c resistance RL'=negative inverse of slope of the d.c load line\n", + "slope=(IC_peak-IC_min)/(VCE_min-VCE_peak); #Slope of he d.c load line, kilo mho\n", + "RL_ac=-(1/slope)*1000; #a.c resistance, ohm\n", + "n=sqrt(RL_ac/RL); #Transformer turn ratio\n", + "\n", + "\n", + "#Result\n", + "print(\"(i) The approximate value of zero signal collector current=%dmA.\"%IC_zero_signal);\n", + "print(\"(ii) The zero signal base current=%.2fmA.\"%IB_zero_signal);\n", + "print(\"(iii) The d.c input power= %dmW and a.c output power =%dmW.\"%(P_dc,P_ac));\n", + "print(\"(iv) The collector efficiency=%.1f%%.\"%collector_eff);\n", + "print(\"(v) The turn ratio of the transformer=%d.\"%n);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) The approximate value of zero signal collector current=18mA.\n", + "(ii) The zero signal base current=0.18mA.\n", + "(iii) The d.c input power= 315mW and a.c output power =106mW.\n", + "(iv) The collector efficiency=33.7%.\n", + "(v) The turn ratio of the transformer=3.\n" + ] + } + ], + "prompt_number": 26 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.14 : Page number 321-322\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt\n", + "\n", + "#Variable declaration\n", + "RL=13.0; #Load resistance, ohm\n", + "RL_reflected=325.0; #Load resistance, when referred to primary, ohm\n", + "VCC=20.0; #Supply voltage, V\n", + "IC=58.0; #Quiscent value of collector current, mA\n", + "\n", + "#Calculation\n", + "#(i)\n", + "n=sqrt(RL_reflected/RL); #Transformer turn ratio\n", + "\n", + "#(ii)\n", + "P_ac=(((IC/1000)**2)*RL_reflected/2)*1000; #A.C output power, mW\n", + "\n", + "#(iii)\n", + "P_dc=VCC*IC; #d.c input power, mW\n", + "collector_eff=(P_ac/P_dc)*100; #Collector efficiency\n", + "\n", + "\n", + "#Result\n", + "print(\"(i) Transformer turn ratio=%d.\"%n);\n", + "print(\"(ii) The a.c output power=%dmW.\"%P_ac);\n", + "print(\"(iii) The collector efficiency=%d%%.\"%collector_eff);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) Transformer turn ratio=5.\n", + "(ii) The a.c output power=546mW.\n", + "(iii) The collector efficiency=47%.\n" + ] + } + ], + "prompt_number": 20 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.15 : Page number 323\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "P_total=4.0; #Total power dissipated by the power transistor, W\n", + "T_j_max=90.0; #Maximum junction temperature, degree celsius\n", + "theta=10.0; #Thermal resistance, degree celsius per watt\n", + "\n", + "#Calculation\n", + "#Since, Total power dissipation=half of(max. junc. temp. - ambient temp.)\n", + "T_amb=T_j_max-(P_total*theta); #Ambient temperature, degree celsius\n", + "\n", + "#Result\n", + "print(\"The ambient temperature=%d degree celsius.\"%T_amb);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The ambient temperature=50 degree celsius.\n" + ] + } + ], + "prompt_number": 27 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.16 : Page number 323-324\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "theta=300.0; #Thermal resistance, degree celsius per watt\n", + "T_j_max=90.0; #Maximum junction temperature, degree celsius\n", + "T_amb=30.0; #Ambient temperature, degree celsius\n", + "\n", + "#Calculation\n", + "#(i) Without heat sink\n", + "P_total=((T_j_max-T_amb)/theta)*1000; #Maximum permissible power dissipation without sink, mW\n", + "\n", + "print(\"(i)The maximum permissible power dissipation without heat sink=%dmW.\"%P_total);\n", + "\n", + "#(ii) With heat sink\n", + "theta=60.0; #reduced thermal resistance, degree celsius per watt\n", + "P_total=((T_j_max-T_amb)/theta)*1000; #Maximum permissible power dissipation with heat sink, mW\n", + "\n", + "print(\"(ii)The maximum permissible power dissipation with heat sink=%dmW.\"%P_total);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i)The maximum permissible power dissipation without heat sink=200mW.\n", + "(ii)The maximum permissible power dissipation with heat sink=1000mW.\n" + ] + } + ], + "prompt_number": 28 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.17 : Page number 324\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "theta=20.0; #Thermal resistance, degree celsius per watt\n", + "T_j_max=200.0; #Maximum junction temperature, degree celsius\n", + "T_amb=25.0; #Ambient temperature, degree celsius\n", + "VCE=4.0; #Collector-emitter voltage, V\n", + "\n", + "#Calculation\n", + "P_total=(T_j_max-T_amb)/theta; #Maximum permissible power dissipation, W\n", + "\n", + "#since, the max. power dissipation=VCE_max*IC_max,therefore\n", + "IC_max=P_total/VCE; #Maximum collector current, A\n", + "\n", + "print(\"The maximum collector current that the transistor can carry without destruction=%.2fA.\"%IC_max);\n", + "\n", + "#The ambient temperature rises\n", + "T_amb=75.0; #The risen ambibent temperature, degree celsius\n", + "P_total=(T_j_max-T_amb)/theta; #Maximum permissible power dissipation, W\n", + "IC_max=P_total/VCE; #Maximum collector current, A\n", + "\n", + "print(\"The maximum collector current for the risen ambient temperature=%.2fA.\"%IC_max);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The maximum collector current that the transistor can carry without destruction=2.19A.\n", + "The maximum collector current for the risen ambient temperature=1.56A.\n" + ] + } + ], + "prompt_number": 29 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.18 : Page number 328-329\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import pi\n", + "\n", + "#Variable declaration\n", + "VCC=12.0; #Supply voltage, V\n", + "RL=8.0; #Driving load, ohm\n", + "\n", + "#Calculation\n", + "#(i)\n", + "IC_sat=VCC/(2*RL); #Collector saturation current, A\n", + "P_o_max=round(VCC*IC_sat*0.25,2); #Maximum load power, W\n", + "\n", + "#(ii)\n", + "P_dc=round(VCC*IC_sat/round(pi,2),2); #d.c input power, W\n", + "\n", + "#(iii)\n", + "Collector_eff=(P_o_max/P_dc)*100; #Collector efficiency\n", + "\n", + "#Result\n", + "print(\"(i) The maximum load power =%.2fW.\"%P_o_max);\n", + "print(\"(ii) The d.c input power=%.2fW.\"%P_dc);\n", + "print(\"(iii) The collector efficiency=%.1f%%.\"%Collector_eff);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) The maximum load power =2.25W.\n", + "(ii) The d.c input power=2.87W.\n", + "(iii) The collector efficiency=78.4%.\n" + ] + } + ], + "prompt_number": 30 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.19 : Page number 329\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "P_T=10.0; #Power rating of each transistor, W\n", + "max_eff=0.785; #Maximum collector effciency\n", + "\n", + "#Calculation\n", + "#Since, input power=max. a.c power + Power rating of transistor\n", + "#And, max. efficiency=max. a.c power/input d.c power\n", + "P_2T=2*P_T; #Total power dissipation by two transistors\n", + "P_o_max=(max_eff*P_2T)/(1-max_eff); #Maximum output a.c power, W\n", + "\n", + "#result\n", + "print(\"The maximum output power that can be obtained=%.2fW.\"%P_o_max);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The maximum output power that can be obtained=73.02W.\n" + ] + } + ], + "prompt_number": 31 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.20 : Page number 329\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "eff=60.0/100; #Efficiency of the amplifier\n", + "P_T=2.5; #Power dissipated by each transistor, W\n", + "\n", + "#Calculation\n", + "#Since, input power=max. a.c power + Power rating of transistor\n", + "#And, max. efficiency=max. a.c power/input d.c power\n", + "P_2T=2*P_T; #Total power dissipated by both transistors, W\n", + "P_ac=(eff*P_2T)/(1-eff); #Output a.c power, W\n", + "P_dc=P_ac+P_2T; #Input d.c power, W\n", + "\n", + "#Result\n", + "print(\"The a.c output power= %.1fW.\"%P_ac);\n", + "print(\"The d.c input power= %.1fW.\"%P_dc);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The a.c output power= 7.5W.\n", + "The d.c input power= 12.5W.\n" + ] + } + ], + "prompt_number": 32 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.21 : Page number 329-330\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "VCC=10.0; #Supply voltage, V\n", + "RL=10.0; #Load resistance, ohm\n", + "\n", + "#Calculation\n", + "IC_sat=(VCC/(2*RL))*1000; #Saturated collector current, mA\n", + "VCE_off=VCC/2; #Collector-emitter voltage in off state, V\n", + "\n", + "#Result\n", + "print(\"1st end point of a.c load line, IC(sat)=%dmA.\"%IC_sat);\n", + "print(\"2nd end point of a.c load line, VCE(off)=%dV.\"%VCE_off);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "1st end point of a.c load line, IC(sat)=500mA.\n", + "2nd end point of a.c load line, VCE(off)=5V.\n" + ] + } + ], + "prompt_number": 33 + }, + { + "cell_type": "code", + "collapsed": false, + "input": [], + "language": "python", + "metadata": {}, + "outputs": [] + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter13_2.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter13_2.ipynb new file mode 100644 index 00000000..afa98f3a --- /dev/null +++ b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter13_2.ipynb @@ -0,0 +1,1139 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# CHAPTER 13: AMPLIFIERS WITH NEGATIVE FEEDBACK" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": true + }, + "outputs": [], + "source": [ + "%matplotlib inline" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.1 : Page number 338" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The voltage gain of the amplifier with negative feedback=97.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "Av=3000.0; #Voltage gain without feedback\n", + "m_v=0.01; #Feedback fraction\n", + "\n", + "#Calculation\n", + "#Since, Gain_with_feedback= Gain_without_feedback/(1+Gain_without_feedback*feedback_fraction),\n", + "Avf=Av/(1+Av*m_v); #Voltage gain of the amplifier with negative feedback\n", + "\n", + "#Result\n", + "print(\"The voltage gain of the amplifier with negative feedback=%.0f.\"%Avf);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.2 : Page number 339" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The fraction of output fedback to the input=1/20.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "Av=140.0; #Voltage gain\n", + "Avf=17.5; #Voltage gain with negative feedback\n", + "\n", + "#Calculation\n", + "#Since, Avf=Av/(1+Av*mv), so,\n", + "mv=(Av-Avf)/(Av*Avf); #Fraction of output fedback to the input\n", + "\n", + "\n", + "#Result\n", + "print(\"The fraction of output fedback to the input=1/%.0f.\"%(1.0/mv));\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.3 : Page number 339" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) The fraction of output fedback to input=0.01.\n", + "(ii) The required amplifier gain for overall gain to be 75=300.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "Av=100.0; #Voltage gain\n", + "Avf=50.0; #Voltage gain with negative feedback\n", + "\n", + "#Calculation\n", + "#(i)\n", + "#Since, Gain_with_feedback= Gain_without_feedback/(1+Gain_without_feedback*feedback_fraction),\n", + "mv=(Av-Avf)/(Av*Avf); #The fraction of output fedback to input\n", + "\n", + "#(ii) Overall gain is to be 75:\n", + "Avf=75.0; #The required overall gain\n", + "#Since, Gain_with_feedback= Gain_without_feedback/(1+Gain_without_feedback*feedback_fraction),\n", + "Av=Avf/(1-Avf*mv); #The required value of amplifier gain\n", + "\n", + "#result\n", + "print(\"(i) The fraction of output fedback to input=%.2f.\"%mv);\n", + "print(\"(ii) The required amplifier gain for overall gain to be 75=%d.\"%Av);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.4 : Page number 339-340" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) The voltage gain without feedback=40.\n", + "(ii) The feedback fraction = 1/40.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "Vout=10.0; #output voltage , V\n", + "Vin_f=0.5; #Input votage for amplifier with feedback, V\n", + "Vin=0.25; #Input votage for amplifier without feedback, V\n", + "\n", + "#Calculation\n", + "#(i)\n", + "Av=Vout/Vin; #Voltage gain without negative feedback\n", + "\n", + "#(ii)\n", + "Avf=Vout/Vin_f; #Voltage gain with negative feedback\n", + "#Since, Gain_with_feedback= Gain_without_feedback/(1+Gain_without_feedback*feedback_fraction),\n", + "mv=(Av-Avf)/(Av*Avf); #Feedback fraction\n", + "\n", + "#Result\n", + "print(\"(i) The voltage gain without feedback=%d.\"%Av);\n", + "print(\"(ii) The feedback fraction = 1/%d.\"%(1/mv));\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.5 : Page number 340" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) The percentage of reduction in stage gain without feedback=20%.\n", + "(ii) The percentage of reduction in net gain with feedback=11.2%\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "Av=50.0; #Gain without feedback\n", + "Avf=25.0; #Gain with negative feedback\n", + "\n", + "#Calculation\n", + "#Since, Gain_with_feedback= Gain_without_feedback/(1+Gain_without_feedback*feedback_fraction),\n", + "mv=(Av-Avf)/(Av*Avf); #Feedback fraction\n", + "\n", + "#(i)\n", + "#percentage of reduction without feedback\n", + "Av_reduced=40.0; #Reduced amplifier gain due to ageing\n", + "percentage_of_reduction=((Av-Av_reduced)/Av)*100; #Percentage of reduction in stage gain\n", + "\n", + "print(\"(i) The percentage of reduction in stage gain without feedback=%d%%.\"%percentage_of_reduction);\n", + "\n", + "#(ii)\n", + "#Since, Gain_with_feedback= Gain_without_feedback/(1+Gain_without_feedback*feedback_fraction),\n", + "Avf_reduced=round(Av_reduced/(1+mv*Av_reduced),1); #Reduced net gain with negative feedback \n", + "percentage_of_reduction_f=((Avf-Avf_reduced)/Avf)*100; #Percentage of reduction in net gain with feedback\n", + "\n", + "print(\"(ii) The percentage of reduction in net gain with feedback=%.1f%%\"%percentage_of_reduction_f);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.6 : Page number 340" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The percentage change in system gain=8.36%\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "Av=100.0; #Gain\n", + "mv=0.1; #feedback fraction\n", + "Av_fall=6.0; #fall in gain, dB\n", + "\n", + "#Calculation\n", + "#Since, Gain_with_feedback= Gain_without_feedback/(1+Gain_without_feedback*feedback_fraction),\n", + "Avf=round(Av/(1+Av*mv),2); #Total system gain with feedback\n", + "\n", + "#Since, fall in gain=20*log10(Av/Av_1)\n", + "Av1=round(Av/10**(Av_fall/20),0); #New absolute voltage gain without feedback\n", + "#Since, Gain_with_feedback= Gain_without_feedback/(1+Gain_without_feedback*feedback_fraction),\n", + "Avf_new=round(Av1/(1+Av1*mv),2); #New net system gain with feedback\n", + "\n", + "percentage_change=((Avf-Avf_new)/Avf)*100; #Percentage change in system gain\n", + "\n", + "#Result\n", + "print(\"The percentage change in system gain=%.2f%%\"%percentage_change);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.7 : Page number 341" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The feedback fraction=0.008.\n", + "The percentage fall in system gain=4.8%.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "Av=500.0; #Voltage gain without feedback\n", + "Avf=100.0; #Voltage gain with negative feedback\n", + "Av_fall_percentage=20.0; #Gain fall percentage due to ageing\n", + "\n", + "\n", + "#Calculation\n", + "#Since, Gain_with_feedback= Gain_without_feedback/(1+Gain_without_feedback*feedback_fraction),\n", + "mv=(Av-Avf)/(Av*Avf); #Feedback fraction\n", + "Av_reduced=((100-Av_fall_percentage)/100)*Av; #Reduced voltage gain\n", + "Avf_reduced=round(Av_reduced/(1+Av_reduced*mv),1); #Reduced total gain of the system\n", + "percentage_fall=((Avf-Avf_reduced)/Avf)*100; #Percentage of fall in total system gain\n", + "\n", + "#Result\n", + "print(\"The feedback fraction=%.3f.\"%mv);\n", + "print(\"The percentage fall in system gain=%.1f%%.\"%percentage_fall);\n", + "\n", + "#Note: The percentage gain is calculated in the text as 4.7% due to approximation of Avf to 95.3 whose actual approximation will be (95.238)~95.2. So, the percentage fall calculated here is 4.8%\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.8 : Page number 341" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The voltage gain with feedback=31622.\n", + "The feedback fraction=2.16e-05.\n" + ] + } + ], + "source": [ + "from math import log10\n", + "\n", + "#Variable declaration\n", + "Av=100000.0; #Open loop voltage gain\n", + "f_dB=10.0; #Negative feedback, dB\n", + "\n", + "#Calculation\n", + "Av_dB=20*log10(Av); #dB voltage gain without feedback, dB\n", + "Avf_dB=Av_dB-f_dB; #dB voltage gain with feedback, dB\n", + "Avf=10**(Avf_dB/20); #Voltage gain with feedback\n", + "\n", + "#Since, Gain_with_feedback= Gain_without_feedback/(1+Gain_without_feedback*feedback_fraction),\n", + "mv=(Av-Avf)/(Av*Avf); #feedback fraction\n", + "\n", + "#Result\n", + "print(\"The voltage gain with feedback=%d.\"%Avf);\n", + "print(\"The feedback fraction=%.2e.\"%mv);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.9 : Page number 341-342" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The voltage gain with feedback=47.4.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "Ao=1000.0; #Open circuit voltage gain\n", + "Rout=100.0; #Output resistance, ohm\n", + "RL=900.0; #Resistive load, ohm\n", + "mv=1/50; #feedback fraction\n", + "\n", + "#Calculation\n", + "#Since, Av=Ao*RL/(Rout+RL)\n", + "Av=Ao*RL/(Rout+RL); #Voltage gain without feedback\n", + "Avf=Av/(1+Av*mv); #Voltage gain with feedback\n", + "\n", + "#Result\n", + "print(\"The voltage gain with feedback=%.1f.\"%Avf);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.10 : Page number 342" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "100=Av/(1+Av*mv) ------Eq. 1\n", + "99=0.8*Av/(1+0.8*Av*mv) ------Eq. 2\n", + "99 + 79.2*Av*mv=0.8Av ------Eq. 3 from Eq. 2\n", + "79.2 + 79.2*Av*mv=0.792Av ------Eq. 4 from Eq. 1\n", + "Subtracting Eq.4 from Eq.3\n", + "19.8 = 0.008*Av\n", + "Av=2475.\n", + "mv=0.0096.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "Avf=100.0; #Voltage gain with feedback\n", + "vary_f=1; #Vary percentage in voltage gain with feedback\n", + "vary_wf=20; #Vary percentage in voltage gain without feedback\n", + "\n", + "#Calculation\n", + "#Avf=Av/(1+Av*mv)\n", + "print(\"%d=Av/(1+Av*mv) ------Eq. 1\"%Avf); #Equation 1\n", + "\n", + "#considering variation in gains\n", + "Avf_vary=Avf*(1- vary_f/100.0); #Gain with feedback, considering variation\n", + "print(\"%d=%.1f*Av/(1+%.1f*Av*mv) ------Eq. 2\"%(Avf_vary,(1-vary_wf/100.0),(1-vary_wf/100.0))); #Equation 2\n", + "\n", + "#Solving the above two equations\n", + "print(\"%d + %.1f*Av*mv=%.1fAv ------Eq. 3 from Eq. 2\"%(Avf_vary,Avf_vary*(1-vary_wf/100.0),(1-vary_wf/100.0))); #Equation 3\n", + "\n", + "#multiplying Eq. 1 with (Avf_vary*(1-vary_wf/100.0))/100=0.792\n", + "print(\"%.1f + %.1f*Av*mv=%.3fAv ------Eq. 4 from Eq. 1\"%(Avf*Avf_vary*(1-vary_wf/100.0)/100.0,Avf*Avf_vary*(1-vary_wf/100.0)/100.0,Avf_vary*(1-vary_wf/100.0)/100.0)); #Equation 4\n", + "\n", + "print(\"Subtracting Eq.4 from Eq.3\" );\n", + "print(\"%.1f = %.3f*Av\"%(Avf_vary-Avf*Avf_vary*(1-vary_wf/100.0)/100.0,(1-vary_wf/100.0)-Avf_vary*(1-vary_wf/100.0)/100.0));\n", + "Av=(Avf_vary-Avf*Avf_vary*(1-vary_wf/100.0)/100.0)/((1-vary_wf/100.0)-Avf_vary*(1-vary_wf/100.0)/100.0);\n", + "print(\"Av=%.0f.\"%Av);\n", + "mv=(Av-Avf)/(Av*Avf);\n", + "print(\"mv=%.4f.\"%mv);\n", + " " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.11 : Page number 345" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) Feedback fraction=0.1.\n", + "(ii) Voltage gain with feedback=10.\n", + "(iii) Output voltage=10mV.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "Av=10000.0; #Volage gain without feedback\n", + "R1=2.0; #Resistor R1, kilo ohm\n", + "R2=18.0; #Resistor R2, kilo ohm\n", + "Vin=1.0; #input voltage, mV\n", + "\n", + "#Calculation\n", + "#(i)\n", + "mv=R1/(R1+R2); #feedback fraction\n", + "\n", + "#(ii)\n", + "#Since, Gain_with_feedback= Gain_without_feedback/(1+Gain_without_feedback*feedback_fraction),\n", + "Avf=round(Av/(1+Av*mv),0); #Voltage gain with feedback\n", + "\n", + "#(iii)\n", + "Vout=Avf*Vin; #Output voltage, mV\n", + "\n", + "#Result\n", + "print(\"(i) Feedback fraction=%.1f.\"%mv);\n", + "print(\"(ii) Voltage gain with feedback=%d.\"%Avf);\n", + "print(\"(iii) Output voltage=%dmV.\"%Vout);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.12 : Page number 345-346" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) Feedback fraction=0.1.\n", + "(ii) The voltage gain with feedback=10.\n", + "(iii) Increased input impedance due to negative feedback=10 mega ohm\n", + "(iv) Decreased output impedance due to negative feedback=0.1 ohm.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "Av=10000.0; #Volateg gain without feedback\n", + "Zin=10.0; #Input impedance, kilo ohm\n", + "Zout=100.0; #Output impedance, ohm\n", + "R1=10.0; #Resistor R1, kilo ohm\n", + "R2=90.0; #Resistor R2, kilo ohm\n", + "\n", + "#Calculation\n", + "#(i)\n", + "mv=R1/(R1+R2); #Feedback fraction\n", + "\n", + "#(ii)\n", + "#Since, Gain_with_feedback= Gain_without_feedback/(1+Gain_without_feedback*feedback_fraction),\n", + "Avf=round(Av/(1+Av*mv),0); #Voltage gain with feedback\n", + "\n", + "#(iii)\n", + "Zin_feedback=((1+Av*mv)*Zin)/1000; #Increased input impedance due to negative feedback, mega ohm\n", + "\n", + "#(iv)\n", + "Zout_feedback=Zout/(1+Av*mv); #Decreased output impedance due to negative feedback, ohm\n", + "\n", + "\n", + "#Result\n", + "print(\"(i) Feedback fraction=%.1f.\"%mv);\n", + "print(\"(ii) The voltage gain with feedback=%d.\"%Avf);\n", + "print(\"(iii) Increased input impedance due to negative feedback=%.0f mega ohm\"%Zin_feedback);\n", + "print(\"(iv) Decreased output impedance due to negative feedback=%.1f ohm.\"%Zout_feedback);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.13 : Page number 346" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Distortion with negative feedback=0.312%\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "Av=150.0; #Voltage gain\n", + "D=5/100.0; #Distortion\n", + "mv=10/100.0; #Feedback fraction\n", + "\n", + "#Calculation\n", + "Dvf=round((D/(1+Av*mv))*100,3); #Distortion with negative feedback\n", + "\n", + "\n", + "#Result\n", + "print(\"Distortion with negative feedback=%.3f%%\"%Dvf);\n", + "\n", + "#Note: In the text, value of Dvf=0.3125% has been approximated to 0.313%. But, here the approximation is done to 0.312%\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.14 : Page number 346" + ] + }, + { + "cell_type": "code", + "execution_count": 16, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The new lower cut-off frequency=136.4Hz\n", + "The new upper cut-off frequency=5.52MHz\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "Av=1000.0; #Voltage gain\n", + "f1=1.5; #Lower cut-off frequency, kHz\n", + "f2=501.5; #Upper cut-off frequency, kHz\n", + "mv=1/100.0; #Feedbcack fraction\n", + "\n", + "#Calculation\n", + "f1_f=(f1/(1+mv*Av))*1000; #New lower cut-off frequency, Hz\n", + "f2_f=(f2*(1+mv*Av))/1000; #New upper cut-off frequency, MHz\n", + "\n", + "\n", + "#Result\n", + "print(\"The new lower cut-off frequency=%.1fHz\"%f1_f);\n", + "print(\"The new upper cut-off frequency=%.2fMHz\"%f2_f);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.15 : Page number 348" + ] + }, + { + "cell_type": "code", + "execution_count": 17, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The effective current gain=58.82.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "Ai=200.0; #Current gain without feedback\n", + "mi=0.012; #Current attenuation\n", + "\n", + "#Calculation\n", + "Aif=Ai/(1+Ai*mi);\n", + "\n", + "#Result\n", + "print(\"The effective current gain=%.2f.\"%Aif);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.16 : Page number 349" + ] + }, + { + "cell_type": "code", + "execution_count": 18, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The input impedance when negative feedback is applied=3.26 kilo ohm\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "Ai=240.0; #Current gain\n", + "Zin=15.0; #Input impedance without feedback, kilo ohm\n", + "mi=0.015; #Current feedback fraction\n", + "\n", + "#Calculations\n", + "Zin_f=Zin/(1+mi*Ai); #Input impedance with feedback, kilo ohm\n", + "\n", + "\n", + "#Result\n", + "print(\"The input impedance when negative feedback is applied=%.2f kilo ohm\"%Zin_f);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.17 : Page number 349" + ] + }, + { + "cell_type": "code", + "execution_count": 19, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The output impedance with negative feedback=9kilo ohm.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "Ai=200.0; #Current gain without feedback\n", + "Zout=3.0; #Output impedance without feedback, kilo ohm\n", + "mi=0.01; #current feedback fraction\n", + "\n", + "#Calculation\n", + "Zout_f=Zout*(1+mi*Ai); #Output impedance with negative feedback, kilo ohm\n", + "\n", + "#Result\n", + "print(\"The output impedance with negative feedback=%dkilo ohm.\"%Zout_f);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.18 : Page number 349" + ] + }, + { + "cell_type": "code", + "execution_count": 20, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Bandwidth when negative feedback is applied=1400kHz.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "Ai=250.0; #Current gain without feedback\n", + "BW=400.0; #Bandwidth, kHz\n", + "mi=0.01; #current feedback fraction\n", + "\n", + "#Calculation\n", + "BW_f=BW*(1+mi*Ai); #Bandwidth when negative feedback is applied, kHz\n", + "\n", + "#Result\n", + "print(\"Bandwidth when negative feedback is applied=%dkHz.\"%BW_f);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.19 : Page number 350-351" + ] + }, + { + "cell_type": "code", + "execution_count": 22, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Value of VE=9.72V and IE=10.68mA\n" + ] + }, + { + "data": { + "image/png": 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+ "text/plain": [ + "<matplotlib.figure.Figure at 0x7f9a7522f0b8>" + ] + }, + "metadata": {}, + "output_type": "display_data" + } + ], + "source": [ + "%matplotlib inline\n", + "import matplotlib.pylab as p\n", + "\n", + "#Variable declaration\n", + "VCC=18.0; #Supply voltage, V\n", + "R1=16.0; #Resistor R1, kilo ohm\n", + "R2=22.0; #Resistor R2, kilo ohm\n", + "RE=910.0; #Emitter resistor, ohm\n", + "VBE=0.7; #Base-emitter voltage, V\n", + "\n", + "#Calculations\n", + "V2=VCC*R2/(R1+R2); #Voltage across R2, V (Voltage divider rule)\n", + "VE=V2-VBE; #Emitter voltage, V\n", + "IE=(VE/RE)*1000; #Emitter current, mA (OHM's LAW)\n", + "\n", + "#D.C load line\n", + "IC_sat=(VCC/RE)*1000; #Collector saturation current, mA\n", + "VCE_off=VCC; #Collector-emitter voltage in off state, V\n", + "\n", + "#Result\n", + "print(\"Value of VE=%.2fV and IE=%.2fmA\"%(VE,IE));\n", + "\n", + "#Plotting\n", + "VCE_plot=[0,VCE_off]; #Plotting variable for VCE\n", + "IC_plot=[IC_sat,0]; #Plotting variable for IC\n", + "p.plot(VCE_plot,IC_plot);\n", + "p.xlim(0,20)\n", + "p.ylim(0,25)\n", + "p.xlabel('VCE(V)');\n", + "p.ylabel('IC(mA)');\n", + "p.title('d.c load line');\n", + "p.grid();\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.20 : Page number 352" + ] + }, + { + "cell_type": "code", + "execution_count": 23, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The voltage gain of the emitter follower circuit=0.994.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "VCC=10.0; #Supply voltage, V\n", + "R1=10.0; #Resistor R1, kilo ohm\n", + "R2=10.0; #Resistor R2, kilo ohm\n", + "RE=5.0; #Emitter resistance, kilo ohm\n", + "VBE=0.7; #Base-emitter voltage, V\n", + "\n", + "\n", + "#Calculation\n", + "V2=VCC*R2/(R1+R2); #Voltage across R2, V (Voltage divider rule)\n", + "VE=V2-VBE; #Emitter voltage, V\n", + "IE=(VE/RE); #Emitter current, mA (OHM's LAW)\n", + "re=25/IE; #a.c emitter resistance, ohm\n", + "Av=RE*1000/(re+RE*1000); #Voltage gain\n", + "\n", + "\n", + "#Result\n", + "print(\"The voltage gain of the emitter follower circuit=%.3f.\"%Av);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.21 : Page number 352-353" + ] + }, + { + "cell_type": "code", + "execution_count": 24, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The voltage gain=0.988\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "RE=5.0; #Emitter resistance, kilo ohm\n", + "re=29.1; #a.c emitter resistance, ohm\n", + "RL=5.0; #Load resistance, kilo ohm\n", + "\n", + "#Calculation\n", + "RE_ac=(RE*RL)/(RE+RL); #New effective value of emitter resistance, kilo ohm\n", + "Av=RE_ac*1000/(re+RE_ac*1000); #Voltage gain\n", + "\n", + "#Result\n", + "print(\"The voltage gain=%.3f\"%Av);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.22 : Page number 354" + ] + }, + { + "cell_type": "code", + "execution_count": 25, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The input impedance of the emitter follower =4.96 kilo ohm\n", + "The approximate value of the input impedance=5 kilo ohm\n" + ] + } + ], + "source": [ + "def pr(r1,r2): #Function for calculating parallel resistance\n", + " return (r1*r2)/(r1+r2);\n", + "\n", + "#Variable declaration\n", + "VCC=10.0; #Supply voltage, V\n", + "R1=10.0; #Resistor R1, kilo ohm\n", + "R2=10.0; #Resistor R2, kilo ohm\n", + "RE=4.3; #Emitter resistor, kilo ohm\n", + "RL=10.0; #Load resistance, kilo ohm\n", + "VBE=0.7; #Base-emitter voltage, V\n", + "beta=200.0; #Base current amplification factor\n", + "\n", + "#Calculation\n", + "V2=VCC*R2/(R1+R2); #Voltage across R2, V (Voltage divider rule)\n", + "VE=V2-VBE; #Emitter voltage, V\n", + "IE=(VE/RE); #Emitter current, mA (OHM's LAW)\n", + "re=25/IE; #a.c emitter resistance, ohm\n", + "RE_eff=pr(RE,RL); #Effective external emitter resistance, kilo ohm\n", + "Zin_base=beta*(re/1000+RE_eff); #Input impedance of the base of the transistor, kilo ohm\n", + "Zin=pr(pr(R1,R2),Zin_base); #Input impedance of emitter follower, kilo ohm\n", + "#Approximate value of input impedance taken as parallel resistance of R1 and R2 and ignoring Zin_base due to its relatively large value\n", + "Zin_approx=pr(R1,R2); #Approximate input impedance, kilo ohm\n", + "\n", + "#Result\n", + "print(\"The input impedance of the emitter follower =%.2f kilo ohm\"%Zin);\n", + "print(\"The approximate value of the input impedance=%d kilo ohm\"%Zin_approx);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.23 : Page number 355" + ] + }, + { + "cell_type": "code", + "execution_count": 26, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The output impedance=22.3 ohm\n" + ] + } + ], + "source": [ + "def pr(r1,r2): #Function for calculating parallel resistance\n", + " return (r1*r2)/(r1+r2);\n", + "\n", + "\n", + "#Variable declaration\n", + "re=20.0; #a.c emitter resistance, ohm\n", + "R1=3.0; #Resistor R1, kilo ohm\n", + "R2=4.7; #Resistor R2, kilo ohm\n", + "RS=600.0; #Source resistance, kilo ohm\n", + "beta=200.0; #Base current amplification factor\n", + "\n", + "#Calculation\n", + "Rin_ac=pr(pr(R1,R2)*1000,RS); #Input a.c resistance, ohm\n", + "Zout=re + Rin_ac/beta; #Output impedance, ohm\n", + "\n", + "#Result\n", + "print(\"The output impedance=%.1f ohm\"%Zout);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.24 : Page number 358" + ] + }, + { + "cell_type": "code", + "execution_count": 27, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) d.c value of current in RE=1.09mA\n", + "(ii) Input impedance=16.17 mega ohm.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "VCC=10.0; #Supply voltage, V\n", + "R1=120.0; #Resistor R1, kilo ohm\n", + "R2=120.0; #Resistor R2, kilo ohm\n", + "RE=3.3; #Emitter resistor, kilo ohm\n", + "VBE=0.7; #Base-emitter voltage, V\n", + "beta_1=70.0; #Base current amplification factor of 1st transistor\n", + "beta_2=70.0; #Base current amplification factor of 2nd transistor\n", + "\n", + "#Calculation\n", + "#(i)\n", + "V2=VCC*R2/(R1+R2); #Voltage across R2, V (Voltage divider rule)\n", + "IE_2=(V2-2*VBE)/RE; #Emitter current, mA (OHM's LAW)\n", + "\n", + "#(ii)\n", + "Zin=(beta_1*beta_2*RE)/1000; #Input impedance, mega ohm\n", + "\n", + "#Result\n", + "print(\"(i) d.c value of current in RE=%.2fmA\"%IE_2);\n", + "print(\"(ii) Input impedance=%.2f mega ohm.\"%Zin);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.25 : Page number 358-359" + ] + }, + { + "cell_type": "code", + "execution_count": 28, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) D.C Bias levels: \n", + " VB1= 4V, VE1=3.3V, VB2=3.3V, VE2=2.6V, IE2=1.3mA and IE1=0.013mA.\n", + "(ii) A.C Analysis: \n", + " re1=1923 ohm and re2=19.23 ohm \n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "VCC=12.0; #Supply voltage, V\n", + "R1=20.0; #Resistor R1, kilo ohm\n", + "R2=10.0; #Resistor R2, kilo ohm\n", + "RC=4.0; #Collector resistor, kilo ohm\n", + "RE=2.0; #Emitter resistor, kilo ohm\n", + "VBE=0.7; #Base-emitter voltage, V\n", + "beta=100.0; #Base current amplification factor of 1st transistor\n", + "\n", + "#Calculation\n", + "#(i) D.C Bias levels\n", + "VB1=VCC*R2/(R1+R2); #Base voltage of 1st transistor, V (Voltage divider rule)\n", + "VE1=VB1-VBE; #Emitter voltage of 1st transistor, V\n", + "VB2=VE1; #Base voltage of 2nd transistor, V\n", + "VE2=VB2-VBE; #Emitter voltage of 2nd transistor, V\n", + "IE2=VE2/RE; #Emitter current of 2nd transistor, mA (OHM' LAW)\n", + "IE1=IE2/beta; #Emitter current of 1st transistor, mA (IE~IC=beta*IB, here IB2=IE1)\n", + "\n", + "#(ii) A.C analysis\n", + "re1=25/IE1; #a.c emitter resistance of 1st transistor\n", + "re2=25/IE2; #a.c emitter resistance of 2nd transistor\n", + "\n", + "\n", + "#Result\n", + "print(\"(i) D.C Bias levels: \\n VB1= %dV, VE1=%.1fV, VB2=%.1fV, VE2=%.1fV, IE2=%.1fmA and IE1=%.3fmA.\"%(VB1,VE1,VB2,VE2,IE2,IE1));\n", + "print(\"(ii) A.C Analysis: \\n re1=%d ohm and re2=%.2f ohm \"%(re1,re2));\n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 3", + "language": "python", + "name": "python3" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 3 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython3", + "version": "3.5.1" + }, + "widgets": { + "state": {}, + "version": "1.1.2" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter14_2.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter14_2.ipynb new file mode 100644 index 00000000..5e35882c --- /dev/null +++ b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter14_2.ipynb @@ -0,0 +1,502 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:ec0d27209d08b0b95750f66ce9ee21af5ef586e23d0bf0ea218aa20a4ce63e43" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "#CHAPTER 14: SINUSOIDAL OSCILLATORS" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.1 : Page number 371-372\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt\n", + "from math import pi\n", + "\n", + "#Variable declaration\n", + "L1=58.6; #Inductance, micro henry\n", + "C1=300.0; #Capacitance, pF\n", + "\n", + "#Calculation\n", + "f=(1/(2*round(pi,2)*sqrt(L1*10**-6*C1*10**-12)))/1000; #Frequency of oscillation, kHz\n", + "\n", + "\n", + "#Result\n", + "print(\"frequency of oscillation=%dkHz\"%f);\n", + "\n", + "\n", + "#Note : The frequency has been calculated in the text as 1199kHz but here the answer gets approximated to 1200kHz.\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "frequency of oscillation=1200kHz\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.2 : Page number 372\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import pi\n", + "\n", + "#Variable declaration\n", + "L1=1.0; #Inductance , mH\n", + "f=1.0; #frequency of oscillation, GHz\n", + "\n", + "#Calculation\n", + "#Since, f=1/(2*pi*sqrt(L1*C1)),\n", + "C1=(1/(L1*10**-3*(f*10**12*2*pi)**2))*10**12; #Capacitance, pF\n", + "\n", + "\n", + "#Result\n", + "print(\"The Capacitance of the capacitor of the LC oscillator=%.2epF\"%C1);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The Capacitance of the capacitor of the LC oscillator=2.53e-11pF\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.3 : Page number 373-374\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import pi\n", + "from math import sqrt\n", + "\n", + "#Variable declaration\n", + "C1=0.001; #Capacitor C1, microfarad\n", + "C2=0.01; #Capacitor C2, microfarad\n", + "L=15.0; #Inductance, microhenry\n", + "\n", + "#Calculation\n", + "CT=C1*C2/(C1+C2); #Total capacitance\n", + "\n", + "#(i) Operating frequency\n", + "f=(1/(2*pi*sqrt(CT*10**-6*L*10**-6)))/1000; #Operating frequency, kHz\n", + "\n", + "#(ii) Feedback fraction\n", + "mv=C1/C2; #Feedback fraction\n", + "\n", + "#Result\n", + "print(\"(i) The operating frequency=%dkHz\"%f);\n", + "print(\"(ii) The feedback fraction=%.1f\"%mv);\n", + "\n", + "#Note : The operating frequency is calculated in the text as 1361kHz but here it has been approximated to 1362kHz\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) The operating frequency=1362kHz\n", + "(ii) The feedback fraction=0.1\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.4 : Page number 374: Page number\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import pi\n", + "\n", + "#Variable declaration\n", + "mv=0.25; #Feedback fraction\n", + "L=1.0; #Inductance, mH\n", + "f=1.0; #Operating frequeny, MHz\n", + "\n", + "#Calculation\n", + "#Since, f=1/(2*pi*sqrt(L*C))\n", + "CT=round((1/(L*10**-3*(2*pi*f*10**6)**2))*10**12,1); #Total capacitance, pF\n", + "\n", + "#Since, mv=C1/C2 and CT=C1*C2/(C1+C2) or CT=C2/(1+ (C2/C1)),\n", + "#From the above equations, substituting value of mv and calculaing value of C2,\n", + "C2=CT*(1+(1/mv)); #Capacitance of C2 capactior, pF\n", + "C1=mv*C2; #Capacitance of C1 capacitor, pF\n", + "\n", + "#Result\n", + "print(\"C1=%.1fpF and C2=%.1fpF\"%(C1,C2));\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "C1=31.6pF and C2=126.5pF\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.5 : Page number 375-376\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt\n", + "from math import pi\n", + "\n", + "#Variable decalaration\n", + "L1=1000.0; #Inductance of L1 inductor, microhenry\n", + "L2=100.0; #Inductance of L2 inductor, microhenry\n", + "M=20.0; #Mutual inductance, microhenry\n", + "C=20.0; #Capacitance, pF\n", + "\n", + "#Calculation\n", + "LT=L1+L2+2*M; #Total inductance, microhenry\n", + "\n", + "#(i) Operating frequency\n", + "f=(1/(2*pi*sqrt(LT*10**-6*C*10**-12)))/1000; #Operating frequency, kHz\n", + "\n", + "#(ii)\n", + "mv=L2/L1; #feedback fraction\n", + "\n", + "#Result\n", + "print(\"(i) The operating frequency=%dkHz.\"%f);\n", + "print(\"(ii) The feedback fraction=%.1f.\"%mv);\n", + "\n", + "#Note : The operating frequecy has been calculated in the text as 1052kHz but here it gets approximated to 1054kHz\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) The operating frequency=1054kHz.\n", + "(ii) The feedback fraction=0.1.\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.6 : Page number 376\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import pi\n", + "\n", + "#Variable declaration\n", + "C=1.0; #Capacitance, pF\n", + "f=1.0; #Frequency, MHz\n", + "mv=0.2; #Feedback frequency\n", + "\n", + "\n", + "#Calculation\n", + "LT=(1/(C*10**-12*(2*pi*f*10**6)**2))*1000; #Total inductance, mH\n", + "\n", + "#Since, mv=L2/L1 or L2=mv*L1 and L1+L2=LT or L1(1+mv)=LT,\n", + "L1=LT/(1+mv); #Inductance of L1 inductor, mH\n", + "L2=L1*mv; #inductance of L2 inductor, mH\n", + "\n", + "#Result\n", + "print(\"L1=%.1fmH and L2=%.2fmH.\"%(L1,L2));\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "L1=21.1mH and L2=4.22mH.\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.7 : Page number 378\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt\n", + "from math import pi\n", + "\n", + "#Variable declaration\n", + "R1=1.0; #Resistor R1, mega ohm\n", + "R2=R1; #Resistor R2, mega ohm\n", + "R3=R1; #Resistor R3, mega ohm\n", + "C1=68.0; #Capacitor C1, pF\n", + "C2=C1; #Capacitor C2, pF\n", + "C3=C1; #Capacitor C3, pF\n", + "\n", + "\n", + "#Calculation\n", + "R=R1*10**6; #Resistance of the resistors of phase shift circuit, ohm\n", + "C=C1*10**-12; #Capacitance of the capacitors of phase shift circuit, F\n", + "fo=1/(2*pi*R*C*sqrt(6)); #Frequency of oscillation, Hz\n", + "\n", + "#Result\n", + "print(\"The frequency of oscillation=%dHz\"%fo);\n", + "\n", + "#Note: The frequency of oscillation had been calculated in the text as 954Hz, but here it gets approximated to 955 HZ.\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The frequency of oscillation=955Hz\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.8 : Page number 378\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import pi\n", + "from math import sqrt\n", + "\n", + "#Variable declaration\n", + "C=5.0; #Capacitance of the capacitors of phase shift circuit, pF\n", + "fo=800.0; #Required frequency of oscillation, kHz\n", + "\n", + "#Calculation\n", + "#Since, fo=1/(2*pi*R*C*sqrt(6))\n", + "R=(1/(2*pi*C*10**-12*fo*10**3*sqrt(6)))/1000; #Resistance of the resistors of phase shift circuit, kilo ohm\n", + "\n", + "#Result\n", + "print(\"R=%.1f kilo ohm.\"%R);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "R=16.2 kilo ohm.\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.9 : Page number 380\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import pi\n", + "\n", + "#Variable declaration\n", + "#Resistance of R1 and R2 resistors of the R-C bridge circuit\n", + "R1=220.0; #kilo ohm \n", + "R2=220.0; #kilo ohm\n", + "\n", + "#Capacitance of C1 and C2 the capacitors of the R-C bridge circuit\n", + "C1=250.0; #pF\n", + "C2=250.0; #pF\n", + "\n", + "#Calculation\n", + "#Since, R1=R2 and C1=C2, R1=R2 is taken as R and C1=C2 is taken as C\n", + "#And, f=1/(2*pi*sqrt(R1*R2*C1*C2))is transformed to f=1/(2*pi*R*C).\n", + "R=R1*10**3; #kilo ohm\n", + "C=C1*10**-12; #pF\n", + "f=1/(2*pi*R*C); #Frequency of oscillation, Hz\n", + "\n", + "\n", + "#Result\n", + "print(\"The frequency of oscillation=%dHz.\"%f);\n", + "\n", + "\n", + "#Note : The frequency of oscillation is calculated in the text as 2892Hz but here it gets approximated to 2893 Hz.\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The frequency of oscillation=2893Hz.\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.11 : Page number 384\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt\n", + "from math import pi\n", + "\n", + "#Variable declaration\n", + "#a.c equivalent values of the crystal:\n", + "L=1.0; #Inductance , H\n", + "C=0.01; #Capacitance , pF\n", + "R=1000.0; #Resistance , ohm\n", + "Cm=20.0; #Mounting capacitance, pF\n", + "\n", + "#Calculation\n", + "fs=(1/(2*round(pi,2)*sqrt(L*C*10**-12)))/1000; #Series resonant frrequency, kHz\n", + "CT=(C*Cm/(C+Cm)); #Total capacitance, pF\n", + "fp=(1/(2*round(pi,2)*sqrt(L*CT*10**-12)))/1000; #Prallel resonant frequency, kHz\n", + "\n", + "#Result\n", + "print(\"fs=%.0fkHz and fp=%.0fkHz.\"%(fs,fp));\n", + "\n", + "#Note: fs and fp are calculated in the text as 1589kHz and 1590kHz, but here it gets approximated to 1592kHz and 1593kHz\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "fs=1592kHz and fp=1593kHz.\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "code", + "collapsed": false, + "input": [], + "language": "python", + "metadata": {}, + "outputs": [] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [], + "language": "python", + "metadata": {}, + "outputs": [] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [], + "language": "python", + "metadata": {}, + "outputs": [] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [], + "language": "python", + "metadata": {}, + "outputs": [] + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter15_2.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter15_2.ipynb new file mode 100644 index 00000000..e649cc91 --- /dev/null +++ b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter15_2.ipynb @@ -0,0 +1,482 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:034eec32676d4e7abdedfb3bf68426d81a2d1483fc668bcbfdb5be18cec2e406" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 15: TRANSISTOR TUNED AMPLIFIERS" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.1 : Page number 394" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt\n", + "from math import pi\n", + "\n", + "#Variable declaration\n", + "C=250.0*10**-12; #Capacitor of parallel resonant circuit, F\n", + "L=1.25*10**-3; #Inductor of the parallel resonant circuit, H\n", + "R=10.0; #Resistor of the parallel resonant circuit, ohm\n", + "\n", + "#Calculation\n", + "#(i) Resonant frequency\n", + "fr=((1/(2*pi))*sqrt((1/(L*C))-(R/L)**2))/1000; #Resonant frequecy, kHz\n", + "\n", + "#(ii) Impedance of the circuit at resonance\n", + "Zr=(L/(C*R))/1000; #Impedance of the circuit at resonance, kilo ohm\n", + "\n", + "#(iii) Quality factor of the circuit\n", + "Q=2*pi*(fr*10**3)*L/R; #Quality factor of the circuit\n", + "\n", + "#Result\n", + "print(\"(i) The resonant frequency=%.1fkHz.\"%fr);\n", + "print(\"(ii) The impedance of the circuit at resonance=%d kilo ohm.\"%Zr);\n", + "print(\"(iii) The quality factor of the circuit=%.1f.\"%Q);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) The resonant frequency=284.7kHz.\n", + "(ii) The impedance of the circuit at resonance=500 kilo ohm.\n", + "(iii) The quality factor of the circuit=223.6.\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.2 : Page number 394-395\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt\n", + "from math import pi\n", + "\n", + "#Variable declaration\n", + "C=100.0*10**-12; #Capacitor of parallel resonant circuit, F\n", + "L=100.0*10**-6; #Inductor of the parallel resonant circuit, H\n", + "R=10.0; #Resistor of the parallel resonant circuit, ohm\n", + "V=10.0; #Supply voltage, V\n", + "\n", + "#Calculation\n", + "#(i) Resonant frequency\n", + "fr=((1/(2*pi))*sqrt((1/(L*C))-(R/L)**2))/1000; #Resonant frequecy, kHz\n", + "\n", + "#(ii) Impedance of the circuit at resonance\n", + "Zr=(L/(C*R))/10**6; #Impedance of the circuit at resonance, mega ohm\n", + "\n", + "I=V/Zr; #Line current at resonance, microampere\n", + "\n", + "#Result\n", + "print(\"(i) The resonant frequency=%.2fkHz.\"%fr);\n", + "print(\"(ii) The impedance of the circuit at resonance=%.1f mega ohm.\"%Zr);\n", + "print(\"The line current at resonance=%d micro ampere.\"%I);\n", + "\n", + "#Note : The resonant frequency in the text has been calculated as 1592.28 kHz, but here it gets approximated to 1591.47 kHz.\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) The resonant frequency=1591.47kHz.\n", + "(ii) The impedance of the circuit at resonance=0.1 mega ohm.\n", + "The line current at resonance=100 micro ampere.\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.3 : Page number 395\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt\n", + "from math import pi\n", + "\n", + "#Variable declaration\n", + "C=250.0*10**-12; #Capacitor of parallel resonant circuit, F\n", + "Zr=500.0*10**3; #Dynamic impedance, ohm\n", + "R=10.0; #Resistance of the coil, ohm\n", + "\n", + "\n", + "#Calculation\n", + "#(i)\n", + "#Since,Zr=L/CR,\n", + "L=(Zr*C*R)*10**3; #Inductance of the coil, mH\n", + "\n", + "#(ii) Resonant frequency\n", + "fr=((1/(2*pi))*sqrt((1/(L*10**-3*C))-(R/(L*10**-3))**2))/1000; #Resonant frequecy, kHz\n", + "\n", + "#(iii) Quality factor of the circuit\n", + "Q=2*pi*(fr*10**3)*(L*10**-3)/R; #Quality factor of the circuit\n", + "\n", + "#Result\n", + "print(\"(i) The inductance of the coil=%.2fmH.\"%L);\n", + "print(\"(ii) The resonant frequency=%.1fkHz.\"%fr);\n", + "print(\"(iii) The quality factor of the circuit=%.1f.\"%Q);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) The inductance of the coil=1.25mH.\n", + "(ii) The resonant frequency=284.7kHz.\n", + "(iii) The quality factor of the circuit=223.6.\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.4 : Page number 397\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "Q=60.0; #Quality factor of the tuned amplifier\n", + "fr=1200.0; #Resonant frequency, kHz\n", + "\n", + "#Calculation\n", + "#(i)\n", + "BW=fr/Q; #Bandwidth, kHz\n", + "\n", + "#(ii)\n", + "f1=fr-(BW/2); #Lower cut-off frequency, kHz\n", + "f2=fr+(BW/2); #Upper cut-off frequency, kHz\n", + "\n", + "#Result\n", + "print(\"(i) The bandwidth=%dkHz\"%BW);\n", + "print(\"(ii) The lower and upper cut-off frequencies are=%dkHz and %dkHz.\"%(f1,f2));\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) The bandwidth=20kHz\n", + "(ii) The lower and upper cut-off frequencies are=1190kHz and 1210kHz.\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.5 : Page number 397\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "fr=2.0; #Resonant frequency, MHz\n", + "BW=50.0; #Bandwidth, kHz\n", + "\n", + "#Calculation\n", + "#Since, bandwidth=resonant_frequency/quality_factor\n", + "Q=(fr*10**6)/(BW*10**3); #Quality factor\n", + "\n", + "#Result\n", + "print(\"The quality factor=%d\"%Q);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The quality factor=40\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.7 : Page number 400\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt\n", + "from math import pi\n", + "\n", + "#Variable declaration\n", + "C=0.1*10**-6; #Capacitor of parallel resonant circuit, F\n", + "L=33.0*10**-3; #Inductor of the parallel resonant circuit, H\n", + "R=25.0; #Resistor of the parallel resonant circuit, ohm\n", + "\n", + "\n", + "#Calculation\n", + "#(i)\n", + "fr=(1/(2*pi*sqrt(L*C)))/1000; #Resonant frequency, kHz\n", + "\n", + "#(ii)\n", + "XL=2*pi*(fr*10**3)*L; #Inductive reactance, ohm\n", + "Q=round(XL/R,0); #Quality factor\n", + "\n", + "#(iii)\n", + "BW=(fr*10**3)/Q; #Bandwidth\n", + "\n", + "#Result\n", + "print(\"(i) The resonant frequency=%.2fkHz\"%fr);\n", + "print(\"(ii) The quality factor= %d.\"%Q);\n", + "print(\"(iii) The bandwidth=%dHz.\"%BW);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) The resonant frequency=2.77kHz\n", + "(ii) The quality factor= 23.\n", + "(iii) The bandwidth=120Hz.\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.8 : Page number 401-402\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "BW_dt=200.0; #Bandwidth, kHz\n", + "fr=10.0; #Operating frequency, MHz\n", + "\n", + "#Calculation\n", + "#Since, BW_dt=k*fr (i.e.,co-efficient_of_coupling * operating_frequency)\n", + "k=BW_dt/(fr*10**3); #co-efficient of coupling\n", + "\n", + "#Result\n", + "print(\"The co-efficient of coupling=%.2f.\"%k);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The co-efficient of coupling=0.02.\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.9 : Page number 405\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt\n", + "from math import pi\n", + "\n", + "#Variable declaration\n", + "C=500.0*10**-12; #Capacitor of parallel resonant circuit, F\n", + "L=50.7*10**-6; #Inductor of the parallel resonant circuit, H\n", + "R=10.0; #Resistor of the parallel resonant circuit, ohm\n", + "RL=1.0; #Load resistance, mega ohm\n", + "\n", + "#Calculation\n", + "#(i)\n", + "fr=round((1/(2*pi*sqrt(L*C)))/1000); #Resonant frequency, Hz\n", + "\n", + "#(ii)\n", + "R_dc=R; #d.c load, ohm\n", + "XL=2*pi*(fr*1000)*L; #Inductive reactance, ohm\n", + "Q_coil=round(XL/R,1); #Quality factor\n", + "R_P=(Q_coil*XL)/1000 ; #Equivalent parallel resistance, kilo ohm\n", + "R_AC=(R_P*RL*10**3)/(R_P+RL*10**3); #A.C load,kilo ohm\n", + "\n", + "\n", + "#Result\n", + "print(\"(i) The resonant frequency=%dkHz\"%fr);\n", + "print(\"(ii) d.c load=%d ohm and a.c load=%d kilo ohm.\"%(R_dc,R_AC));\n", + "\n", + "#Note: In the text resonant frequency has been wrongly calculated to 106kHz but its actual value is approximately 1000kHz\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) The resonant frequency=1000kHz\n", + "(ii) d.c load=10 ohm and a.c load=10 kilo ohm.\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.10 : Page number 406-407\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "RL=50.0; #Load resistance, ohm\n", + "n=5; #Turns ratio of the transformer\n", + "VCC=50.0; #Supply voltage, V\n", + "\n", + "#Calculation\n", + "#(i)\n", + "R_ac=n**2*RL; #A.C load, ohm\n", + "\n", + "#(ii)\n", + "P_o_max=VCC**2/(2*R_ac); #Maximum load power, W\n", + "\n", + "#Result\n", + "print(\"(i) The a.c load=%d ohm\"%R_ac);\n", + "print(\"(ii) Maximum load power=%dW\"%P_o_max);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) The a.c load=1250 ohm\n", + "(ii) Maximum load power=1W\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.11 : Page number 407\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "P_D=4.0; #Maximum power dissipation, mW\n", + "P_o_max=1.0; #Maximum load power, W\n", + "\n", + "\n", + "#Calculation\n", + "max_collector_eff=(P_o_max/(P_o_max+(P_D/1000)))*100; #Maximum collector efficiency\n", + "\n", + "#Result\n", + "print(\"The maximum collector efficiency=%.1f%%\"%max_collector_eff);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The maximum collector efficiency=99.6%\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "code", + "collapsed": false, + "input": [], + "language": "python", + "metadata": {}, + "outputs": [] + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter16_2.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter16_2.ipynb new file mode 100644 index 00000000..b2262b8f --- /dev/null +++ b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter16_2.ipynb @@ -0,0 +1,936 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:d4ffda068787fb0974622fa8de40f7d54b5df2a00735e870e01cb9df45be78f9" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 16 : MODULATION AND DEMODULATION" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.2 : Page number 416-417\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variabledeclaration\n", + "V_pp_max=16.0; #Maximum peak-to-peak voltage of an AM wave, mV\n", + "V_pp_min=4.0; #Minimum peak-to-peak voltage of an AM wave, mV\n", + "\n", + "#Calculation\n", + "Vmax=V_pp_max/2; #Maximum voltage of AM wave, mV\n", + "Vmin=V_pp_min/2; #Minimum voltage of AM wave, mV\n", + "m=(Vmax-Vmin)/(Vmax+Vmin); #Modulation factor.\n", + "\n", + "#Result\n", + "print(\"The modulation factor=%.1f.\"%m);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The modulation factor=0.6.\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.3 : Page number 417\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "Es=50.0; #signalvoltage amplitude, V\n", + "Ec=100.0; #Carrier voltage amplitude, V\n", + "\n", + "\n", + "#Calculation\n", + "m=Es/Ec; #Modulation factor\n", + "\n", + "#Result\n", + "print(\"Modulation factor=%.1f.\"%m);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Modulation factor=0.5.\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.4 : Page number 419\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "fc=2500.0; #Carrier frequency, kHz\n", + "f1=50.0; #Lower frequency of the audio signal, Hz\n", + "f2=15000.0; #Upper frequency of the audio signal, Hz\n", + "\n", + "#Calculation\n", + "fl_usb=fc+(f1/1000); #Lower frequency of upper sideband, kHz\n", + "fu_usb=fc+(f2/1000); #Upper frequency of upper sideband, kHz\n", + "\n", + "fu_lsb=fc-(f1/1000); #Lower frequency of upper sideband, kHz\n", + "fl_lsb=fc-(f2/1000); #Upper frequency of upper sideband, kHz\n", + "\n", + "#Since, f1=50Hz is negligible with respect to f2=15000Hz,\n", + "BW=(fc+(f2/1000))-(fc-(f2/1000)); #Bandwidth, kHz\n", + "\n", + "#Result\n", + "print(\"The upper sideband=%.2fkHz to %dkHz.\"%(fl_usb,fu_usb));\n", + "print(\"The lower sideband=%dkHz to %.2fkHz.\"%(fl_lsb,fu_lsb));\n", + "print(\"The bandwidth=%dkHz\"%BW);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The upper sideband=2500.05kHz to 2515kHz.\n", + "The lower sideband=2485kHz to 2499.95kHz.\n", + "The bandwidth=30kHz\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.5 : Page number 420\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "from math import pi\n", + "\n", + "#Variable declaration\n", + "EC=5.0; #Carrier amplitude, V\n", + "m=0.6; #modulation factor\n", + "ws=6280.0; #angular frequency of signal, radians/s\n", + "wc=211*10**4; #angular frequency of carrier, radians/s\n", + "\n", + "#Calculation\n", + "fs=(ws/(2*pi))/1000; #Signal frequency, kHz\n", + "fc=(wc/(2*pi))/1000; #Carrier frequency, kHz\n", + "\n", + "#(i)\n", + "Max_amp=EC+m*EC; #Maximum amplitude of AM wave, V\n", + "Min_amp=EC-m*EC; #Minimum amplitude of AM wave, V\n", + "\n", + "#(ii)\n", + "frequency_components=[fc-fs,fc,fc+fs]; #frequency components, kHz\n", + "amplitudes=[m*EC/2,EC,m*EC/2]; #Corresponding amplitudes, V\n", + "\n", + "#Result\n", + "print(\"(i) The maximum and minimum amplitudes of AM wave=%dV and %dV.\"%(Max_amp,Min_amp));\n", + "print(\"(ii) The frequency components of the AM wave=%.0f,%.0f,%.0f.\"%(frequency_components[0],frequency_components[1],frequency_components[2]));\n", + "print(\" The corresponding amplitudes are =%.1fV, %dV, %.1fV.\"%(amplitudes[0],amplitudes[1],amplitudes[2]));\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) The maximum and minimum amplitudes of AM wave=8V and 2V.\n", + "(ii) The frequency components of the AM wave=335,336,337.\n", + " The corresponding amplitudes are =1.5V, 5V, 1.5V.\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.6 : Page number 420-421\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "fc=1000.0; #Carrier frequency, kHz\n", + "fs=5.0; #Signal frequency, kHz\n", + "m=0.5; #Modulation factor\n", + "EC=100.0; #Amplitude of the carrier, V\n", + "\n", + "#Calculation\n", + "f_lsb=fc-fs; #Lower sideband frequency,kHz\n", + "f_usb=fc+fs; #Upper sideband frequency, kHz\n", + "Amplitude=m*EC/2; #Amplitude of each sideband, V\n", + "\n", + "#Result\n", + "print(\"The lower and upper sideband frequencies are=%dkHz and %dkHz.\"%(f_lsb,f_usb));\n", + "print(\"The amplitude of each sideband =%dV\"%Amplitude);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The lower and upper sideband frequencies are=995kHz and 1005kHz.\n", + "The amplitude of each sideband =25V\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.7 : Page number 421\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "EC=10.0; #Carrier amplitude, V\n", + "ES=6.0; #Signal amplitude, V\n", + "fc=10.0; #Carrier frequency, MHz\n", + "fs=5/1000.0; #Signal frequency. MHz\n", + "\n", + "#Calculation\n", + "#(i)\n", + "m=ES/EC; #Modulation factor\n", + "\n", + "#(ii)\n", + "f_lsb=fc-fs; #Lower sideband frequency,MHz\n", + "f_usb=fc+fs; #Upper sideband frequency, MHz\n", + "\n", + "#(iii)\n", + "Amplitude=m*EC/2; #Amplitude of each sideband, V\n", + "\n", + "\n", + "#Result\n", + "print(\"(i) The modulation factor=%.1f.\"%m);\n", + "print(\"(ii) The lower and upper sideband frequencies are=%.3fMHz and %.3fMHz.\"%(f_lsb,f_usb));\n", + "print(\"(iii) The amplitude of each sideband =%dV\"%Amplitude);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) The modulation factor=0.6.\n", + "(ii) The lower and upper sideband frequencies are=9.995MHz and 10.005MHz.\n", + "(iii) The amplitude of each sideband =3V\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.8 : Page number 423\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "Pc=500.0; #Carrier power, W\n", + "m=1.0; #Modulation factor\n", + "\n", + "\n", + "#Calculation\n", + "#(i)\n", + "Ps=(1/2.0)*m**2*Pc; #Sideband power, W\n", + "\n", + "#(ii)\n", + "PT=Pc+Ps; #Power of AM wave, W\n", + "\n", + "#Result\n", + "print(\"(i) The power in sidebands=%dW\"%Ps);\n", + "print(\"(ii) The power of AM wave=%dW\"%PT);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) The power in sidebands=250W\n", + "(ii) The power of AM wave=750W\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exmaple 16.9 : Page number 423\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "Pc=50.0; #Power of carrier, kW\n", + "\n", + "#Calculation\n", + "#(i)\n", + "m=80/100.0; #Modulation factor\n", + "Ps=(1/2.0)*m**2*Pc; #Sideband Power, kW\n", + "print(\"(i) The sideband power for 80%% modulation=%dkW.\"%Ps);\n", + "\n", + "#(ii)\n", + "m=10/100.0; #Modulation factor\n", + "Ps=(1/2.0)*m**2*Pc; #Sideband Power, kW\n", + "print(\"(ii) The sideband power for 10%% modulation=%.2fkW.\"%Ps);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) The sideband power for 80% modulation=16kW.\n", + "(ii) The sideband power for 10% modulation=0.25kW.\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.10 : Page number 423-424\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "Pc=40.0; #Carrier power, kW\n", + "m=100/100.0; #Modulation index\n", + "amplifier_eff=72/100.0; #Efficiency of modulated RF amplifier\n", + "\n", + "\n", + "#Calculation\n", + "#(i)Carrier power remains same after modulation\n", + "\n", + "#(ii)\n", + "Ps=(1/2.0)*(m**2)*Pc; #Sideband power\n", + "P_audio=Ps/amplifier_eff; #Required audio power, kW\n", + "\n", + "#Result\n", + "print(\"(i) The carrier power=%dkW.\"%Pc);\n", + "print(\"(ii) The required audio power=%.1fkW.\"%P_audio);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) The carrier power=40kW.\n", + "(ii) The required audio power=27.8kW.\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.11 : Page number 424\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "fs=1.0; #Signal frequency, kHz\n", + "fc=500.0; #Carrier frequency, kHz\n", + "\n", + "\n", + "#Calculation\n", + "#(i)\n", + "sideband_f=[fc-fs,fc+fs]; #Sideband frequencies, kHz\n", + "\n", + "#(ii)\n", + "BW=(fc+fs)-(fc-fs); #Bandwidth required, kHz\n", + "\n", + "#Result\n", + "print(\"(i) The sideband frequencies=%dkHz and %dkHz.\"%(sideband_f[0],sideband_f[1]));\n", + "print(\"(ii) The bandwidth required=%dkHz\"%BW);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) The sideband frequencies=499kHz and 501kHz.\n", + "(ii) The bandwidth required=2kHz\n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.12 : Page number 424\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt\n", + "\n", + "#Variable declaration\n", + "IC=8.0; #Antenna current due to carrier,A\n", + "m=40/100.0; #Modulation index\n", + "\n", + "#Calculation\n", + "#Since, Ps=(1/2)*m\u00b2*Pc and PT=Pc+Ps (Total_power=carrier_power+signal_power)\n", + "#that implies, (PT/Pc)=1+(m\u00b2/2),\n", + "#So, square_of(Total_current/Carrier_current)=(IT/IC)\u00b2=1+(m\u00b2/2).\n", + "IT=IC*sqrt(1+(m**2/2.0)); #Total current, A\n", + "\n", + "\n", + "#Result\n", + "print(\"The total antenna current=%.2fA.\"%IT);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The total antenna current=8.31A.\n" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.13 : Page number 424\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt\n", + "\n", + "#Variable declaration\n", + "IC=8.0; #Antenna current when only carrier is sent, A\n", + "IT=8.93; #Total antenna current, A\n", + "\n", + "\n", + "#Calculation\n", + "#Since, Ps=(1/2)*m\u00b2*Pc and PT=Pc+Ps (Total_power=carrier_power+signal_power)\n", + "#that implies, (PT/Pc)=1+(m\u00b2/2),\n", + "#So, square_of(Total_current/Carrier_current)=(IT/IC)\u00b2=1+(m\u00b2/2).\n", + "m=sqrt((((IT/IC)**2)-1)*2)*100; #The %age of modulation\n", + "\n", + "#Result\n", + "print(\"The %%age of modulation=%.1f%%.\"%m);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The %age of modulation=70.1%.\n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.14 : Page number 425\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt\n", + "\n", + "#Variable declaration\n", + "Vc=100.0; #Carrier voltage, V\n", + "V_T=110.0; #The total voltage after modulation, V\n", + "\n", + "#Calculation\n", + "#Since, Ps=(1/2)*m\u00b2*Pc and PT=Pc+Ps (Total_power=carrier_power+signal_power)\n", + "#that implies, (PT/Pc)=1+(m\u00b2/2),\n", + "#So, square_of(Total_voltage/Carrier_voltage)=(V_T/Vc)\u00b2=1+(m\u00b2/2).\n", + "m=sqrt((((V_T/Vc)**2)-1)*2); #The %age of modulation\n", + "\n", + "#Result\n", + "print(\"The modulation index =%.3f.\"%m);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The modulation index =0.648.\n" + ] + } + ], + "prompt_number": 19 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.15 : Page number 425-426\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "Vc=5.0; #Carrier voltage, V\n", + "V_lsb=2.5; #Lower sideband component, V\n", + "V_usb=2.5; #Upper sideband component, V\n", + "R=2.0; #Resistor driven by AM wave, k\u03a9\n", + "\n", + "#Calculation\n", + "#Since, power=(r.m.s_voltage)\u00b2/resistance\n", + "#(i)\n", + "Pc=round((0.707*Vc)**2/R,2); #Carrier power mW\n", + "\n", + "#(ii)\n", + "P_lower=round((0.707*V_lsb)**2/R,3); #Power delivered by lower sideband, mW\n", + "\n", + "#(iii)\n", + "P_upper=round((0.707*V_usb)**2/R,3); #Power delivered by upper sideband, mW\n", + "\n", + "P_T=round(Pc+P_lower+P_upper,3); #Total power delivered by the AM wave, mW\n", + "\n", + "#Result\n", + "print(\"(i) The carrier power=%.2fmW\"%Pc);\n", + "print(\"(ii) The power delivered by lower sideband=%.3fmW\"%P_lower);\n", + "print(\"(iii) The power delivered by upper sideband=%.3fmW\"%P_upper);\n", + "print(\"The total power delivered by the AM wave=%.3fmW\"%P_T);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) The carrier power=6.25mW\n", + "(ii) The power delivered by lower sideband=1.562mW\n", + "(iii) The power delivered by upper sideband=1.562mW\n", + "The total power delivered by the AM wave=9.374mW\n" + ] + } + ], + "prompt_number": 20 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.16 : Page number 428\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt\n", + "from math import pi\n", + "\n", + "#Variable declaration\n", + "wc=6e08; #Carrier angular frequency, rad/s\n", + "ws=1250.0; #Signal angular frequency, rad/s\n", + "mf=5; #Modulation index\n", + "Ec=12.0; #Carrier amplitude, V\n", + "R=10.0; #Resistor, \u03a9\n", + "\n", + "#Calculation\n", + "#(i)\n", + "fc=wc/(2*pi); #Carrier frequency, Hz\n", + "\n", + "#(ii)\n", + "fs=ws/(2*pi); #Signal frequency, Hz\n", + "\n", + "#(iv)\n", + "delta_f=mf*fs; #Maximum frequency deviation, Hz\n", + "\n", + "#(v)\n", + "P=(Ec/sqrt(2))**2/R; #Power dissipated, W\n", + "\n", + "#Result\n", + "print(\"(i) The carrier frequency=%.1fe06 Hz.\"%(fc/10**6));\n", + "print(\"(ii) The signal frequency=%.0f Hz.\"%fs);\n", + "print(\"(iii) The modulation index=%d.\"%mf);\n", + "print(\"(iv) The maximum frequency deviation=%.0fHz.\"%delta_f);\n", + "print(\"(v) The power dissipated=%.1fW.\"%P);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) The carrier frequency=95.5e06 Hz.\n", + "(ii) The signal frequency=199 Hz.\n", + "(iii) The modulation index=5.\n", + "(iv) The maximum frequency deviation=995Hz.\n", + "(v) The power dissipated=7.2W.\n" + ] + } + ], + "prompt_number": 21 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.17 : Page number 428-429\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import pi\n", + "\n", + "#Variable declaration\n", + "fc=25.0; #Carrier frequency, MHz\n", + "fs=400.0; #Signal frequency, Hz\n", + "Ec=4.0; #Carrier amplitude, V\n", + "delta_f=10.0; #Maximum frequency deviation, kHz\n", + "\n", + "#Calculation\n", + "wc=2*pi*fc*10**6; #Carrier angular frequency, rad/s\n", + "ws=2*pi*fs; #Signal angular frequency, rad/s\n", + "mf=delta_f*1000/fs; #Modulation index\n", + "\n", + "\n", + "#Result\n", + "print(\"e=%dcos(%.2et + %dsin%dt)\"%(Ec,wc,mf,ws));\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "e=4cos(1.57e+08t + 25sin2513t)\n" + ] + } + ], + "prompt_number": 22 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.18 : Page number 429\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "delta_f=50.0; #Maximum frequency deviation, kHz\n", + "fs=5.0; #Modulating frequency, kHz\n", + "\n", + "#Calculation\n", + "mf=delta_f/fs; #Modulation index\n", + "\n", + "#Result\n", + "print(\"The modulation index=%d\"%mf);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The modulation index=10\n" + ] + } + ], + "prompt_number": 23 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.19 : Page number 429\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "fc=1000.0; #Carrier frequency, kHz\n", + "fs=15.0; #Modulating frequency, kHz\n", + "\n", + "#Calculation\n", + "first_3_usb_f=[fc+fs,fc+2*fs,fc+3*fs]; #First three upper sideband frequncies, kHz\n", + "first_3_lsb_f=[fc-fs,fc-2*fs,fc-3*fs]; #First three lowerr sideband frequncies, kHz\n", + "\n", + "\n", + "#Result\n", + "print(\"The first three upper sideband frequencies=%dkHz ,%dkHz and %dkHz.\"%(first_3_usb_f[0],first_3_usb_f[1],first_3_usb_f[2]));\n", + "print(\"The first three lower sideband frequencies=%dkHz ,%dkHz and %dkHz.\"%(first_3_lsb_f[0],first_3_lsb_f[1],first_3_lsb_f[2]));\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The first three upper sideband frequencies=1015kHz ,1030kHz and 1045kHz.\n", + "The first three lower sideband frequencies=985kHz ,970kHz and 955kHz.\n" + ] + } + ], + "prompt_number": 24 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.20 : Page number 429\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "fs=15.0; #Modulating frequency, kHz\n", + "delta_f=75.0; #Maximum frequency deviation, kHz\n", + "\n", + "#Calculation\n", + "BW=2*(delta_f+fs); #Bandwidth, kHz\n", + "\n", + "#Result\n", + "print(\"The bandwidth of the FM signal=%dkHz.\"%BW);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The bandwidth of the FM signal=180kHz.\n" + ] + } + ], + "prompt_number": 25 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.21 : Page number 429\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "k=75.0; #Frequency deviation constant, kHz/V\n", + "Es=2.0; #Amplitude of signal, V\n", + "\n", + "\n", + "#Calculation\n", + "delta_f=k*Es; #Maximum frequency deviation, kHz\n", + "\n", + "#Result\n", + "print(\"The maximum frequency deviation=%dkHz.\"%delta_f);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The maximum frequency deviation=150kHz.\n" + ] + } + ], + "prompt_number": 26 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.22 : Page number 429-430\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "fs1=500.0; #First audio frequency, Hz\n", + "fs2=200.0; #Second audio frequency (decreased), Hz\n", + "Es=2.4; #AF voltage, V\n", + "delta_f1=4.8; #Frequency deviation,kHz\n", + "\n", + "#Calculation\n", + "k=delta_f1/Es; #Frequency deviation constant, kHz/V\n", + "Es=7.2; #AF voltage, V (increased)\n", + "delta_f2=k*Es; #2nd frequency deviation, kHz\n", + "Es=10.0; #AF voltage, V (increased)\n", + "delta_f3=k*Es; #3rd frequency deviation, kHz\n", + "\n", + "mf1=delta_f1/(fs1/1000); #Modulation index in 1st case\n", + "mf2=delta_f2/(fs1/1000); #Modulation index in 2nd case\n", + "mf3=delta_f3/(fs2/1000); #Modulation index in 3rd case\n", + "\n", + "#Result\n", + "print(\"The frequency deviation in second case=%.1fkHz.\"%delta_f2);\n", + "print(\"The frequency deviation in third case=%dkHz.\"%delta_f3);\n", + "print(\"The modulation index in 1st case=%.1f\"%mf1);\n", + "print(\"The modulation index in 2nd case=%.1f\"%mf2);\n", + "print(\"The modulation index in 3rd case=%d\"%mf3);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The frequency deviation in second case=14.4kHz.\n", + "The frequency deviation in third case=20kHz.\n", + "The modulation index in 1st case=9.6\n", + "The modulation index in 2nd case=28.8\n", + "The modulation index in 3rd case=100\n" + ] + } + ], + "prompt_number": 27 + }, + { + "cell_type": "code", + "collapsed": false, + "input": [], + "language": "python", + "metadata": {}, + "outputs": [] + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter17_2.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter17_2.ipynb new file mode 100644 index 00000000..2c9a49a5 --- /dev/null +++ b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter17_2.ipynb @@ -0,0 +1,877 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:33d9a7285654630dd24f2d6229210244e78961e0605c11041ed1b2f130cb19a1" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 17 : REGULATED D.C POWER SUPPLY" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 17.1 : Page number 444\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "V_NL=400.0; #Output voltage with no-load, V\n", + "V_FL=300.0; #Output voltage with full-load, V\n", + "\n", + "\n", + "#Calculation\n", + "percentage_voltage_regulation=((V_NL-V_FL)/V_FL)*100; #Percentage of voltage regulation\n", + "\n", + "#Result\n", + "print(\"The percentage of voltage regulation=%.2f%%.\"%percentage_voltage_regulation);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The percentage of voltage regulation=33.33%.\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 17.2 : Page number 444\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "V_regulation=1.0; #%age voltage regulation\n", + "V_NL=30.0; #Output voltage with no-load,V\n", + "\n", + "#Calculation\n", + "#Since, %age_of_voltage_regulation=((V_NL-V_FL)/V_FL)*100\n", + "V_FL=V_NL/(1+(V_regulation/100)); #Output voltage with full-load, V\n", + "\n", + "#Result\n", + "print(\"The full-load voltage=%.1fV.\"%V_FL);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The full-load voltage=29.7V.\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 17.3 : Page number 445\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "V_NL_A=30.0; #Output voltage of supply A with no-load, V\n", + "V_FL_A=25.0; #Output voltage of supply A with full-load, V\n", + "V_NL_B=30.0; #Output voltage of supply B with no-load, V\n", + "V_FL_B=29.0; #Output voltage of supply B with full-load, V\n", + "\n", + "\n", + "#Calculation\n", + "V_regulation_A=((V_NL_A-V_FL_A)/V_FL_A)*100; #%age of voltage regulation in power supply A\n", + "V_regulation_B=((V_NL_B-V_FL_B)/V_FL_B)*100; #%age of voltage regulation in power supply B\n", + "\n", + "#Result\n", + "if(V_regulation_A<V_regulation_B):\n", + " print(\"Power supply A is better than B.\");\n", + "else :\n", + " print(\"Power supply B is better than A.\");\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Power supply B is better than A.\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 17.4 : Page number 445\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "V_NL=500.0; #Output voltage with no-load, V\n", + "V_FL=300.0; #Output voltage with full-load, V\n", + "I_FL=120.0; #Output current with full-load, mA\n", + "\n", + "#Calculation\n", + "#(i)\n", + "Regulation=((V_NL-V_FL)/V_FL)*100; #Voltage regulation percentage\n", + "\n", + "#(ii)\n", + "RL_min=V_FL/I_FL; #Minimum load resistance, k\u03a9\n", + "\n", + "#Result\n", + "print(\"(i) The voltage regulation=%.1f%%.\"%Regulation);\n", + "print(\"(ii)The minimum load resistance=%.1fk\u03a9.\"%RL_min);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) The voltage regulation=66.7%.\n", + "(ii)The minimum load resistance=2.5k\u03a9.\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 17.5 : Page number 445\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "VL_1=10.5; #Initial output voltage with load, V\n", + "VL_2=10.0; #Decreased output voltage with additional load, V\n", + "IL_1=1.0; #Initial load current, A\n", + "IL_added=1.0; #Added load current, A\n", + "\n", + "#Calculation\n", + "delta_VL=VL_1-VL_2; #Change in output voltage, V\n", + "delta_IL=IL_added; #Change in load current, A\n", + "\n", + "#(i)\n", + "Zo=delta_VL/delta_IL; #Output impedance of power supply, \u03a9 (OHM's LAW)\n", + "\n", + "#(ii)\n", + "#Since, Output_impedance=change_in_output_voltage/change_in_output_current\n", + "#Zo=(V_NL-VL_1)/delta_IL,\n", + "delta_IL=IL_1; #Change in load current, A\n", + "V_NL=VL_1+(delta_IL*Zo); #Output voltage with no load, V\n", + "\n", + "#Result\n", + "print(\"(i) The output impedance=%.1f\u03a9.\"%Zo);\n", + "print(\"(ii) The output voltage with no-load=%dV.\"%V_NL);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) The output impedance=0.5\u03a9.\n", + "(ii) The output voltage with no-load=11V.\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 17.6 : Page number 446\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "Zo=0.01; #Output impedance, \u03a9\n", + "IL_max=1.0; #Maximum output current, A\n", + "IL_min=0.5 #Minimum output current, A\n", + "f=10.0; #Frequency, kHz\n", + "\n", + "#Calculation\n", + "#Since, Zo=delta_VL/delta_IL\n", + "delta_IL=IL_max-IL_min; #Maximum change in output current, A\n", + "delta_VL=(Zo*delta_IL)*1000; #Fluctuations in output voltage, mV\n", + "\n", + "#Result\n", + "print(\"The output voltage will have %dmV peak-to-peak fluctuation at a rate of %dkHz.\"%(delta_VL,f));\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The output voltage will have 5mV peak-to-peak fluctuation at a rate of 10kHz.\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 17.7 : Page number 446\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "delta_Vout=10.0; #Change in output voltage, \u03bcV\n", + "delta_Vin=5.0; #Change in input voltage, V\n", + "\n", + "#Calculation\n", + "Line_regulation=delta_Vout/delta_Vin; #Line regulation, \u03bcV/V\n", + "\n", + "#Result\n", + "print(\"The line regulation of the voltage regulator=%d\u03bcV/V.\"%Line_regulation);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The line regulation of the voltage regulator=2\u03bcV/V.\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 17.8 : Page number 449-450\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "Vin=24.0; #Input voltage, V\n", + "Vz=12.0; #Zener voltage, V\n", + "Rs=160.0; #Series resistance, \u03a9\n", + "RL_max=float('inf'); #Maximum load resistance, \u03a9\n", + "RL_min=200.0; #Minimum load resistance, \u03a9\n", + "\n", + "#Calculation\n", + "#(i)\n", + "Vout=Vz; #Output voltage,(equal to zener regulated voltage), V\n", + "Is=((Vin-Vout)/Rs)*1000; #Current through series resistance, mA\n", + "\n", + "#(ii)\n", + "IL_min=Vout/RL_max; #Minimum load current, A\n", + "IL_max=(Vout/RL_min)*1000; #Maximum load current, mA\n", + "\n", + "#(iii)\n", + "IZ_min=Is-IL_max; #Minimum zener current, mA\n", + "IZ_max=Is-IL_min; #Maximum zener current, mA\n", + "\n", + "#Result\n", + "print(\"(i) The current through the series resistance=%dmA\"%Is);\n", + "print(\"(ii) The minimum and maximum load currents are=%dA and %dmA\"%(IL_min,IL_max));\n", + "print(\"(iii) The minimum and maximum zener currents are=%dmA and %dmA\"%(IZ_min,IZ_max));\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) The current through the series resistance=75mA\n", + "(ii) The minimum and maximum load currents are=0A and 60mA\n", + "(iii) The minimum and maximum zener currents are=15mA and 75mA\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 17.9 : Page number 450\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "VZ=15.0; #Zener voltage, V\n", + "Vin_min=22.0 #Minimum input voltage, V\n", + "Vin_max=40.0 #Maximum input voltage, V\n", + "Vout=VZ; #Regulated output voltage, V\n", + "IL_max=100.0; #Maximum load current, mA\n", + "IL_min=20.0; #Minimum load current, mA\n", + "\n", + "#Calculation\n", + "RS_max=(Vin_min-Vout)/(IL_max/1000); #Maximum value of series resistance, \u03a9 (OHM'S lAW)\n", + "\n", + "#Result\n", + "print(\"The maximum load resistance to hold the voltage constant=%d\u03a9.\"%RS_max);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The maximum load resistance to hold the voltage constant=70\u03a9.\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 17.10 : Page number 450-451\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "Vz=3.3; #Zener voltage, V\n", + "Iz_min=3.0; #Minimum zener current, mA\n", + "Iz_max=100.0; #Maximum zener current, mA\n", + "RL_max=2.0; #Maximum load resistance, k\u03a9\n", + "RL_min=500.0; #Minimum load resistance, \u03a9\n", + "Vin=20.0; #Input voltage, V\n", + "\n", + "#Calculation\n", + "Rs_min=(Vin-Vz)/(Iz_max/1000); #Minimum series resistance required, \u03a9\n", + "\n", + "#Result\n", + "print(\"The minimum series resistance required to limit the zener current=%.0f\u03a9.\"%Rs_min);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The minimum series resistance required to limit the zener current=167\u03a9.\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 17.11 : Page number 451\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "Vz=3.3; #Zener voltage, V\n", + "Iz_min=3.0; #Minimum zener current, mA\n", + "Iz_max=100.0; #Maximum zener current, mA\n", + "RL_max=2.0; #Maximum load resistance, k\u03a9\n", + "RL_min=500.0; #Minimum load resistance, \u03a9\n", + "Vin=20.0; #Input voltage, V\n", + "\n", + "\n", + "#Calculation\n", + "IL_max=(Vz/RL_min)*1000; #Maximum load current, mA\n", + "Rs_max=((Vin-Vz)/(IL_max+Iz_min))*1000; #Maximum series resistance, \u03a9\n", + "\n", + "#Result\n", + "print(\"The maximum allowable value of series resistance=%d\u03a9.\"%Rs_max);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The maximum allowable value of series resistance=1739\u03a9.\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 17.12 : Page number 452\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "Vz=10.0; #Zener voltage, V\n", + "beta=100.0; #Base current amplification factor\n", + "RL=1000.0; #Load resistance, \u03a9\n", + "VBE=0.5; #Base-emitter voltage, V\n", + "\n", + "#Calculation\n", + "Vout=Vz-VBE; #Output voltage, V\n", + "IL=(Vout/RL)*1000; #Load current, mA\n", + "\n", + "#Result\n", + "print(\"The output voltage=%.1fV.\"%Vout);\n", + "print(\"The load current=%.1fmA\"%IL);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The output voltage=9.5V.\n", + "The load current=9.5mA\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 17.3 : Page number 452\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "IC=1.0; #Required current(collector current), A\n", + "Vout=6.0; #Constant output voltage, V\n", + "Vin=10.0; #Supply voltage, V\n", + "beta=50.0; #Base current amplification factor\n", + "VBE=0.5; #Base-emitter voltage, V\n", + "Iz=10.0; #Minimum zener current, mA\n", + "\n", + "#Calculation\n", + "#(i)\n", + "IB=(IC/beta)*1000; #Base current, mA\n", + "\n", + "#Since, Vout=Vz-VBE;\n", + "Vz=Vout+VBE; #Zener breakdown voltage, V\n", + "\n", + "#(ii)\n", + "V_Rs=Vin-Vz; #Voltage across series resistance Rs, V\n", + "Rs=(V_Rs/(IB+Iz))*1000; #Series resistance, \u03a9\n", + "\n", + "\n", + "#Result\n", + "print(\"(i) The zener breakdown voltage=%.1fV\"%Vz);\n", + "print(\"(ii)The series resistance=%.0f\u03a9.\"%Rs);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) The zener breakdown voltage=6.5V\n", + "(ii)The series resistance=117\u03a9.\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 17.14: Page number 452-453\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "Vz=12.0; #Zener voltage, V\n", + "VBE=0.7; #Base-emitter voltage, V\n", + "Vin=20.0; #Input voltage, V\n", + "RS=220.0; #Series resistance, \u03a9\n", + "RL=1.0; #Load resistance, k\u03a9\n", + "beta=50.0; #Base current amplification factor\n", + "\n", + "\n", + "#Calculation\n", + "#(i)\n", + "Vout=Vz-VBE; #Output voltage, V\n", + "\n", + "#(ii)\n", + "V_RS=Vin-Vz; #Voltage across series resistance, RS, V\n", + "IR=(V_RS/RS)*1000; #Current through series resistance, mA\n", + "IL=Vout/RL; #Load current, mA\n", + "\n", + "#Since, IL is emitter current and emitter current is approx. equal to collector current,\n", + "IC=IL; #Collector current, mA\n", + "IB=IC/beta; #Base current, mA\n", + "Iz=IR-IB; #Zener current, mA\n", + "\n", + "#Result\n", + "print(\"(i) The output voltage=%.1fV.\"%Vout);\n", + "print(\"(ii) The zener current=%dmA\"%Iz);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) The output voltage=11.3V.\n", + "(ii) The zener current=36mA\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 17.15 : Page number 453-454\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import ceil\n", + "from math import floor\n", + "\n", + "#Variable declaration\n", + "IL_min=0; #Minimum load current, A\n", + "IL_max=1.0; #Maximum load current, A\n", + "Vin_min=12.0; #Minimum input voltage, V\n", + "Vin_max=18.0; #Maximum input voltage, V\n", + "Iz_min=1.0; #Minimum zener current, mA\n", + "Vz=8.5; #Zener voltage, V\n", + "beta=50.0; #Base current amplification factor\n", + "\n", + "\n", + "#Calculation\n", + "IB_max=(IL_max/beta)*1000; #Maximum base current, mA\n", + "I_RS=Iz_min+IB_max; #Current through the series resistance, mA\n", + "RS=((Vin_min-Vz)/I_RS)*1000; #Series resistance, \u03a9\n", + "\n", + "#(ii)\n", + "V_RS_max=Vin_max-Vz; #Maximum voltage across series resistance, V\n", + "P_max_RS=ceil((V_RS_max**2/RS)*1000)/1000; #Maximum power dissipation in series resistance RS, W\n", + "\n", + "#(iii)\n", + "I_RS_max=V_RS_max/floor(RS); #Maximum current through series resistance,mA\n", + "Iz_max=I_RS_max; #Maximum zener current, mA\n", + "P_z_max=Vz*Iz_max; #Maximum power dissipated in zener diode, W\n", + "\n", + "\n", + "#Result\n", + "print(\"(i) The series resistance=%d\u03a9.\"%RS);\n", + "print(\"(ii) The maximum power dissipated in series resistance=%.3fW.\"%P_max_RS);\n", + "print(\"(iii)The maximum power dissipated in zener diode=%.3fW.\"%P_z_max);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) The series resistance=166\u03a9.\n", + "(ii) The maximum power dissipated in series resistance=0.542W.\n", + "(iii)The maximum power dissipated in zener diode=0.486W.\n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 17.16 : Page number 456\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "R1=2.0; #Resistor R1, k\u03a9\n", + "R2=1.0; #Resistor R2, k\u03a9\n", + "Vz=6.0; #Zener voltage, V\n", + "VBE=0.7; #Base-emitter voltage, V\n", + "\n", + "#Calculation\n", + "m=R2/(R1+R2); #Feedback fraction\n", + "A_CL=1/m; #Closed-loop voltage gain\n", + "Vout=A_CL*(Vz+VBE); #Regulated output voltage, V\n", + "\n", + "#Result\n", + "print(\"The regulated output voltage=%.1fV\"%Vout);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The regulated output voltage=20.1V\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 17.17 : Page number 456\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "R1=30.0; #Resistor R1, k\u03a9\n", + "R2=10.0; #Resistor R2, k\u03a9\n", + "\n", + "#Calculation\n", + "m=R2/(R1+R2); #Feedback fraction\n", + "A_CL=1/m; #Closed-loop voltage gain\n", + "\n", + "\n", + "#Result\n", + "print(\"The closed-loop voltage gain=%d.\"%A_CL);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The closed-loop voltage gain=4.\n" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 17.18 : Page number 457\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "Vin=22.0; #Input voltage, V\n", + "Rs=130.0; #Series resistance, \u03a9\n", + "Vz=8.3; #Zener voltage, V\n", + "VBE=0.7; #Base-emitter voltage, V\n", + "RL=100.0; #Load resistance, \u03a9\n", + "\n", + "#Calculation\n", + "#(i)\n", + "Vout=Vz+VBE; #Output voltage, V\n", + "\n", + "#(ii)\n", + "IL=(Vout/RL)*1000; #Load current, mA (OHM's LAW)\n", + "IS=((Vin-Vout)/Rs)*1000; #Current through series resistance, mA (OHM's LAW)\n", + "IC=IS-IL; #Collector current, mA\n", + "\n", + "#Result\n", + "print(\"(i) The regulated output voltage=%dV\"%Vout);\n", + "print(\"(ii) Various currents for the shunt regulator are: IL=%dmA , IS=%dmA and IC=%dmA\"%(IL,IS,IC));\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) The regulated output voltage=9V\n", + "(ii) Various currents for the shunt regulator are: IL=90mA , IS=100mA and IC=10mA\n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 17.20 : Page number 463\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "R1=240.0; #Resistor R1 of the regulator, \u03a9\n", + "R2=2.4; #Variable resistance R2 of the regulator, k\u03a9\n", + "\n", + "#Calculation\n", + "Vout=1.25*(R2*1000/R1 + 1); #Regulated output voltage, V\n", + "\n", + "#Result\n", + "print(\"The regulated output voltage=%.2fV.\"%Vout);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The regulated output voltage=13.75V.\n" + ] + } + ], + "prompt_number": 19 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 17.21 : Page number 463\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "Vout_adj=8.0; #Output voltage (adjusted), V\n", + "Vd=40.0; #Input/output differential rating, V\n", + "\n", + "#Calculation\n", + "Vin_max=Vout_adj+Vd; #Maximum allowable input voltage, V\n", + "\n", + "\n", + "#Result\n", + "print(\"The maximum allowable input voltage=%dV.\"%Vin_max);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The maximum allowable input voltage=48V.\n" + ] + } + ], + "prompt_number": 20 + }, + { + "cell_type": "code", + "collapsed": false, + "input": [], + "language": "python", + "metadata": {}, + "outputs": [] + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter18_2.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter18_2.ipynb new file mode 100644 index 00000000..8a5bc9dc --- /dev/null +++ b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter18_2.ipynb @@ -0,0 +1,799 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# CHAPTER 18 : SOLID-STATE SWITCHING CIRCUITS" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [], + "source": [ + "%matplotlib inline" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 18.1 : Page number 472" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Input voltage required to saturate the transistor switch=5.4V.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "VCC=10; #Supply voltage, V\n", + "RC=1.0; #Collector resistor, kΩ\n", + "RB=47.0; #Base resistor, kΩ\n", + "beta=100.0; #Base current amplification factor\n", + "VBE=0.7; #Base-emitter voltage, V\n", + "\n", + "#Calculation\n", + "IC_sat=VCC/RC; #Collector saturation current, mA\n", + "IB=IC_sat/beta; #Base current, mA\n", + "V=IB*RB+VBE; #Input voltage, V\n", + "\n", + "#Result\n", + "print(\"Input voltage required to saturate the transistor switch=%.1fV.\"%V);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 18.2 : Page number 475" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) The collector emitter voltage at cut-off=9.99V.\n", + "(ii) The collector emitter voltage at saturation=0.7V.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "VCC=10; #Supply voltage, V\n", + "RC=1.0; #Collector resistor, kΩ\n", + "ICBO=10.0; #Collector leakage current, μA\n", + "V_knee=0.7; #Knee voltage, V\n", + "\n", + "\n", + "#Calculation\n", + "#(i)\n", + "IC=ICBO; #Collector current, μA\n", + "VCE=VCC-(ICBO/1000)*RC; #Collector-emitter voltage, V\n", + "\n", + "print(\"(i) The collector emitter voltage at cut-off=%.2fV.\"%VCE);\n", + "\n", + "#(ii)\n", + "#Since, saturation current=IC_sat=(VCC-V_knee)/RC; \n", + "VCE=V_knee; #Collector-emitter voltage, V\n", + "\n", + "print(\"(ii) The collector emitter voltage at saturation=%.1fV.\"%VCE);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 18.3 : Page number 475-476" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) Minimum β=19.4.\n", + "(ii) The transistor will not be saturated.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "VCC=10; #Supply voltage, V\n", + "RC=1; #Collector resistor, kΩ\n", + "VBB=2; #Supply voltage to base, V\n", + "RB=2.7; #Base resistor, kΩ\n", + "V_knee=0.7; #Knee voltage, V\n", + "VBE=0.7; #Base-emitter voltage, V\n", + "\n", + "#Calculation\n", + "#(i)\n", + "IB=round((VBB-VBE)/RB,2); #Base current, mA\n", + "Ic_sat=(VCC-V_knee)/RC; #Collector saturation current, mA\n", + "beta_min=Ic_sat/IB; #Minimum value of base current amplification factor\n", + "print(\"(i) Minimum β=%.1f.\"%beta_min);\n", + "\n", + "#(ii)\n", + "VBB=1; #Supply voltage to base(changed), V\n", + "beta=50; #Base current amplification factor\n", + "IB=(VBB-VBE)/RB; #Base current, mA\n", + "IC=beta*IB; #Collector current,mA\n", + "\n", + "if(IC<Ic_sat):\n", + " print(\"(ii) The transistor will not be saturated.\");\n", + "else:\n", + " print(\"(ii) The transistor will be saturated.\");\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 18.4 : Page number 480" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Time period of the square wave=0.14 m sec.\n", + "Time frequency of the square wave=7 kHz.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "R2=10; #Resistor R2, kΩ\n", + "R3=10; #Resistor R3, kΩ\n", + "C1=0.01; #Capacitor of 1st transistor, μF\n", + "C2=0.01; #Capacitor of 2nd transistor, μF\n", + "\n", + "#Calculation\n", + "R=R2*1000; #Resistance, Ω\n", + "C=C1*10**-6; #Capacitance, F\n", + "T=round((1.4*R*C)*1000,2); #Time period,m sec\n", + "f=1/(T*10**-3); #Frequency, Hz\n", + "f=f/1000; #Frequency, kHz\n", + "\n", + "#Result\n", + "print(\"Time period of the square wave=%.2f m sec.\"%T);\n", + "print(\"Time frequency of the square wave=%d kHz.\"%f);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 18.6 : Page number 485" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The output voltage=0.55V.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "R=10; #Resistance in differentiating circuit, kΩ\n", + "C=2.2; #Capacitance in differentiating circuit, μF\n", + "d_ei=10; #Change in input voltage, V\n", + "dt=0.4; #Time in which change occurs, s\n", + "\n", + "#Calculation\n", + "eo=R*1000*C*10**-6*d_ei/dt\n", + "\n", + "\n", + "#Result\n", + "print(\"The output voltage=%.2fV.\"%eo);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 18.7 : Page number 489" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The peak output voltage=11.3V.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "Vin_peak=12; #Peak value of input voltage, V\n", + "V_D=0.7; #Forward bias voltage of diode, V\n", + "\n", + "#Calculation\n", + "Vout_peak=Vin_peak-V_D; #Peak value of output voltage, V\n", + "\n", + "#Result\n", + "print(\"The peak output voltage=%.1fV.\"%Vout_peak);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 18.8 : Page number 489" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The peak output voltage=8V.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "Vin_peak=10; #Peak value of input voltage, V\n", + "R=1; #Input resistor, kΩ\n", + "RL=4; #Load resistor, kΩ\n", + "\n", + "#Calculation\n", + "Vout_peak=(Vin_peak*RL)/(R+RL); #Peak output voltage, V\n", + "\n", + "\n", + "#Result\n", + "print(\"The peak output voltage=%dV.\"%Vout_peak);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 18.9 : Page number 490" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The diode will be forward biased for the negative half-cycle of input signal.\n", + "The output voltage=-0.7V.\n", + "The voltage across R=-9.3V.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "Vin=-10; #Input voltage, V\n", + "V_D=0.7; #Forward bias voltage of the diode, V\n", + "R=1; #Resistance, kΩ\n", + "\n", + "\n", + "print(\"The diode will be forward biased for the negative half-cycle of input signal.\");\n", + "Vout=-V_D; #Output voltage, V\n", + "V_R=Vin-(-V_D); #Voltage across resistor R, V\n", + "\n", + "#Result\n", + "print(\"The output voltage=%.1fV.\"%Vout);\n", + "print(\"The voltage across R=%.1fV.\"%V_R);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 18.10 : Page number 490-491" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "During the positive half cycle, the diode is foward biased and can be replaced by battery of 0.7V.\n", + "Therefore, Vout=0.7V.\n", + "During the negative half cycle, the diode is reverse biased and hence behaves as an open circuit.\n", + "Therefore, Vout_peak=-8.33V.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "V_F=0.7; #Forward bias voltage of diode, V\n", + "R=200.0; #Input resistor of the circuit, Ω\n", + "RL=1.0; #Load resistor, kΩ\n", + "Vin_peak=10.0; #Peak input voltage, V\n", + "\n", + "\n", + "#Calculations\n", + "\n", + "#Positive half-cycle:\n", + "print(\"During the positive half cycle, the diode is foward biased and can be replaced by battery of %.1fV.\"%V_F);\n", + "print(\"Therefore, Vout=%.1fV.\"%V_F);\n", + "\n", + "#Negative half-cycle:\n", + "print(\"During the negative half cycle, the diode is reverse biased and hence behaves as an open circuit.\");\n", + "Vout_peak=RL*(-Vin_peak)/(R/1000+RL);\n", + "print(\"Therefore, Vout_peak=%.2fV.\"%Vout_peak);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 18.12 : Page number 491" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "data": { + "image/png": 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+ "text/plain": [ + "<matplotlib.figure.Figure at 0x7fd5e441dfd0>" + ] + }, + "metadata": {}, + "output_type": "display_data" + } + ], + "source": [ + "%matplotlib inline\n", + "import matplotlib.pyplot as plt\n", + "from math import sin\n", + "from math import pi\n", + "\n", + "V_biasing=10.0; #Biasing voltage, V\n", + "vin=[30*sin(t/10.0) for t in range(0,(int)(2*pi*10))] #input voltage waveform, V\n", + "\n", + "plt.subplot(211)\n", + "plt.plot(vin);\n", + "plt.xlabel('t-->');\n", + "plt.ylabel('Vin(V)');\n", + "plt.title('Input waveform');\n", + "\n", + "vout=[]; #Output voltage waveform, V\n", + "for v in vin[:]:\n", + " if(v-V_biasing)>0 : #Diode is forward biased.\n", + " vout.append(v-V_biasing);\n", + " else: #Diode is reverse biased.\n", + " vout.append(0);\n", + "\n", + "plt.subplot(212) \n", + "plt.plot(vout);\n", + "plt.xlabel('t-->');\n", + "plt.ylabel('Vout(V)');\n", + "plt.title('Output waveform');\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 18.13 : Page number 492" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "data": { + "image/png": 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+ "text/plain": [ + "<matplotlib.figure.Figure at 0x7fd5e4324828>" + ] + }, + "metadata": {}, + "output_type": "display_data" + } + ], + "source": [ + "import matplotlib.pyplot as plt\n", + "\n", + "Vin=[]; #Input voltage waveform, V\n", + "t1=50; #Assumed time interval, s\n", + "t2=100; #Assumed time interval, s\n", + "V_biasing=10; #Biasing voltage, V\n", + "for t in range(0,151): #time interval from 0s to 151s\n", + " if(t<=t1): \n", + " Vin.append(15); #Value of input voltage for time 0 to t1 seconds \n", + " elif(t<=t2 and t>t1):\n", + " Vin.append(-30); #Value of input voltage for time t1 to t2 seconds\n", + " else :\n", + " Vin.append(15); #Value of input voltage after t2 seconds\n", + "\n", + "plt.subplot(211)\n", + "plt.plot(Vin);\n", + "plt.xlim([0,160])\n", + "plt.ylim([-35,20])\n", + "plt.xlabel('t-->');\n", + "plt.ylabel('Vin(V)');\n", + "plt.title('Input waveform');\n", + "\n", + "\n", + "vout=[]; #Output voltage waveform, V\n", + "for v in Vin[:]: #Loop iterating input voltage \n", + " if(v<=0):\n", + " vout.append(0); #Diode reverse biased\n", + " else:\n", + " vout.append(v-V_biasing); #Diode forward biased\n", + "\n", + "plt.subplot(212)\n", + "plt.plot(vout);\n", + "plt.xlim(0,160)\n", + "plt.ylim(-35,20)\n", + "plt.xlabel('t-->');\n", + "plt.ylabel('Vout(V)');\n", + "plt.title('Output waveform');\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 18.14 : Page number 492-493" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "data": { + "image/png": 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+ "text/plain": [ + "<matplotlib.figure.Figure at 0x7fd5e42e1668>" + ] + }, + "metadata": {}, + "output_type": "display_data" + } + ], + "source": [ + "import matplotlib.pyplot as plt\n", + "\n", + "Vin=[]; #Input voltage waveform, V\n", + "t1=50; #Assumed time interval, s\n", + "t2=100; #Assumed time interval, s\n", + "V_biasing=5; #Biasing voltage, V\n", + "for t in range(0,151): #time interval from 0s to 151s\n", + " if(t<=t1): \n", + " Vin.append(10); #Value of input voltage for time 0 to t1 seconds \n", + " elif(t<=t2 and t>t1):\n", + " Vin.append(-10); #Value of input voltage for time t1 to t2 seconds\n", + " else :\n", + " Vin.append(0);\n", + "\n", + "plt.subplot(211) \n", + "plt.plot(Vin);\n", + "plt.xlim(0,101)\n", + "plt.ylim(-20,20)\n", + "plt.xlabel('t-->');\n", + "plt.ylabel('Vin(V)');\n", + "plt.title('Input waveform');\n", + "\n", + "\n", + "vout=[]; #Output voltage waveform, V\n", + "for v in Vin[:]: #Loop iterating input voltage \n", + " if(v<=0):\n", + " vout.append(v); #Diode reverse biased\n", + " else:\n", + " vout.append(v-V_biasing); #Diode forward biased\n", + "\n", + "plt.subplot(212)\n", + "plt.plot(vout);\n", + "plt.xlim(0,101)\n", + "plt.ylim(-20,20)\n", + "plt.xlabel('t-->');\n", + "plt.ylabel('Vout(V)');\n", + "plt.title('Output waveform');\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 18.15 : Page number 493" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "data": { + "image/png": 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+ "text/plain": [ + "<matplotlib.figure.Figure at 0x7fd5e42367b8>" + ] + }, + "metadata": {}, + "output_type": "display_data" + } + ], + "source": [ + "import matplotlib.pyplot as plt\n", + "\n", + "Vin=[]; #Input voltage waveform, V\n", + "t1=50; #Assumed time interval, s\n", + "t2=100; #Assumed time interval, s\n", + "V_D1=0.6; #Forward Biasing voltage of the 1st diode, V\n", + "V_D2=0.6; #Forward Biasing voltage of the 2nd diode, V\n", + "for t in range(0,151): #time interval from 0s to 151s\n", + " if(t<=t1): \n", + " Vin.append(10); #Value of input voltage for time 0 to t1 seconds \n", + " elif(t<=t2 and t>t1):\n", + " Vin.append(-10); #Value of input voltage for time t1 to t2 seconds\n", + " else :\n", + " Vin.append(0);\n", + "\n", + "plt.subplot(211);\n", + "plt.plot(Vin);\n", + "plt.xlim(0,110)\n", + "plt.ylim(-20,20)\n", + "plt.xlabel('t-->');\n", + "plt.ylabel('Vin(V)');\n", + "plt.title('Input waveform');\n", + "\n", + "vout=[]; #Output voltage waveform, V\n", + "for v in Vin[:]: #Loop iterating input voltage \n", + " if(v<=0):\n", + " vout.append(-V_D1); #Diode D1 forward biased, \n", + " else:\n", + " vout.append(V_D2); #Diode D2 forward biased\n", + "\n", + "plt.subplot(212) \n", + "plt.plot(vout);\n", + "plt.xlim(0,110)\n", + "plt.ylim(-1,1)\n", + "plt.xlabel('t-->');\n", + "plt.ylabel('Vout(V)');\n", + "plt.title('Output waveform');\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 18.16 : Page number 493-494" + ] + }, + { + "cell_type": "code", + "execution_count": 16, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "data": { + "image/png": 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XvVPSHdHt44HfAi8Rupx2kSTCbMYxwFaSviO0mpB0lKQPJC2Nxpp2zbju59Hr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\nLRcZYxRRpdkixwEfVEmUpVi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+ "text/plain": [ + "<matplotlib.figure.Figure at 0x7fd5e4244470>" + ] + }, + "metadata": {}, + "output_type": "display_data" + } + ], + "source": [ + "import matplotlib.pyplot as plt\n", + "from math import sin\n", + "from math import pi\n", + "\n", + "VZ=20; #Assumed zener voltage, V\n", + "VF=0.7; #Assumed forward biasing voltage of the zener diode, V\n", + "Vin=[]; #Input voltage waveform, V\n", + "for t in range(0,(int)(2*pi*10)): #time interval from 0s to 151s\n", + " Vin.append(30*sin(t/10.0));\n", + "\n", + "plt.subplot(211)\n", + "plt.plot(Vin);\n", + "plt.xlabel('t-->');\n", + "plt.ylabel('Vin(V)');\n", + "plt.title('Input waveform');\n", + "\n", + "\n", + "vout=[]; #Output voltage waveform, V\n", + "for v in Vin[:]: #Loop iterating input voltage \n", + " if(v<=-VF):\n", + " vout.append(-VF); #Zener diode forward biased, \n", + " elif(v>=VZ):\n", + " vout.append(VZ); #Input voltage exceeds zener voltage\n", + " else:\n", + " vout.append(v); #Zener diode reverse biased\n", + "\n", + "plt.subplot(212)\n", + "plt.plot(vout);\n", + "plt.xlim([0,80])\n", + "plt.ylim([-1,40])\n", + "plt.xlabel('t-->');\n", + "plt.ylabel('Vout(V)');\n", + "plt.title('Output waveform');\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 18.17 : Page number 494" + ] + }, + { + "cell_type": "code", + "execution_count": 17, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "data": { + "image/png": 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2Bx7Fu5x2lSR8NuM4oJWkz/BWE5J6SnpN0tJkrGm3gud9O3m9V4AvJDVMzp2d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\n2Gwz+OlP4dprfRe+EEL1VbR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+ "text/plain": [ + "<matplotlib.figure.Figure at 0x7fd5e4362cf8>" + ] + }, + "metadata": {}, + "output_type": "display_data" + } + ], + "source": [ + "import matplotlib.pyplot as plt\n", + "from math import sin\n", + "from math import pi\n", + "\n", + "VZ1=20; #Assumed zener voltage, V\n", + "VF1=0.7; #Assumed forward biasing voltage of the zener diode, V\n", + "VZ2=20; #Assumed zener voltage, V\n", + "VF2=0.7; #Assumed forward biasing voltage of the zener diode, V\n", + "Vin=[]; #Input voltage waveform, V\n", + "for t in range(0,(int)(2*pi*10)): #time interval from 0s to 151s\n", + " Vin.append(30*sin(t/10.0));\n", + " \n", + "plt.subplot(211)\n", + "plt.plot(Vin);\n", + "plt.xlabel('t-->');\n", + "plt.ylabel('Vin(V)');\n", + "plt.title('Input waveform');\n", + "\n", + "\n", + "vout=[]; #Output voltage waveform, V\n", + "for v in Vin[:]: #Loop iterating input voltage \n", + " if(v<=-(VZ1+VF2)):\n", + " vout.append(-(VZ1+VF2)); #Zener diode forward biased, \n", + " elif(v>=VZ2+VF1):\n", + " vout.append(VZ2+VF1); #Input voltage exceeds zener voltage\n", + " else:\n", + " vout.append(v); #Zener diode reverse biased\n", + "plt.subplot(212)\n", + "plt.plot(vout); \n", + "plt.xlim([0,80])\n", + "plt.ylim([-40,40])\n", + "plt.xlabel('t-->');\n", + "plt.ylabel('Vout(V)');\n", + "plt.title('Output waveform');\n" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], + "source": [] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 3", + "language": "python", + "name": "python3" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 3 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython3", + "version": "3.5.1" + }, + "widgets": { + "state": {}, + "version": "1.1.2" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter19_2.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter19_2.ipynb new file mode 100644 index 00000000..36c07b9b --- /dev/null +++ b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter19_2.ipynb @@ -0,0 +1,1660 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# CHAPTER 19 : FIELD EFFECT TRANSISTORS" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [], + "source": [ + "%matplotlib inline" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.1 : Page number 515" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "ID=12[1 + VGS/5]²mA.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "I_DSS=12.0; #Shorted gate drain current, mA\n", + "V_GS_off=-5.0; #Gate-source cut-off voltage, V\n", + "\n", + "#Result\n", + "print(\"ID=%d[1 + VGS/%d]²mA.\"%(I_DSS,abs(V_GS_off)));\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.2 : Page number 515" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The drain current=6.12mA.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "I_DSS=32.0; #Shorted gate drain current, mA\n", + "V_GS_off=-8.0; #Gate-source cut-off voltage, V\n", + "V_GS=-4.5; #Gate-source voltage, V\n", + "\n", + "\n", + "#Calculation\n", + "I_D=I_DSS*(1-(V_GS/V_GS_off))**2; #Drain current mA\n", + "\n", + "#Result\n", + "print(\"The drain current=%.2fmA.\"%I_D);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.3 : Page number 515" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) VGS=-1.76V.\n", + "(ii) VP=6V\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "\n", + "#Variable declaration\n", + "I_DSS=10.0; #Shorted gate drain current, mA\n", + "V_GS_off=-6.0; #Gate-source cut-off voltage, V\n", + "I_D=5.0; #Drain current mA\n", + "\n", + "\n", + "#Calculation\n", + "#(i)\n", + "#Since, I_D=I_DSS*[1 - (V_GS/V_GS_off)]²\n", + "V_GS=V_GS_off*(1-sqrt(I_D/I_DSS)); #Gate-source voltage, V\n", + "\n", + "#(ii)\n", + "V_P=-V_GS_off; #Pinch-off voltage, V \n", + "\n", + "#Result\n", + "print(\"(i) VGS=%.2fV.\"%V_GS);\n", + "print(\"(ii) VP=%dV\"%V_P);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.4 : Page number 515-516" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The minimum value of VDD required =10.72V.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "V_GS_off=-4.0; #Gate-source cut-off voltage, V\n", + "I_DSS=12.0; #Shorted gate drain current, mA\n", + "R_D=560.0; #Drain resistor, Ω\n", + "\n", + "#Calculation\n", + "V_P=-V_GS_off; #Pinch-off voltage, V\n", + "V_DS=V_P; #Minimum drain-source voltage for JFET to be in constant current region, V\n", + "I_D=I_DSS; #Maximum drain current, mA (V_GS=0)\n", + "V_RD=(I_D/1000)*R_D; #Voltage across drain resistor, V (OHM's LAW)\n", + "V_DD=V_DS+V_RD; #Minimum value of supply voltage to drain, V\n", + "\n", + "#Result\n", + "print(\"The minimum value of VDD required =%.2fV.\"%V_DD);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.5 : Page number 516" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The drain current=1.33mA.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "I_DSS=3.0; #Shorted gate drain current, mA\n", + "V_GS_off=-6.0; #Gate-source cut-off voltage, V\n", + "V_GS=-2.0; #Gate-source voltage, V\n", + "\n", + "\n", + "#Calculation\n", + "I_D=I_DSS*(1-(V_GS/V_GS_off))**2; #Drain current mA\n", + "\n", + "#Result\n", + "print(\"The drain current=%.2fmA.\"%I_D);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.6 : Page number 516" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "p-channel JFET requires a positive gate-to-source voltage to pass drain current.\n", + "More positive voltage, the less the drain current. \n", + "Any further increase in VGS keeps the JFET cut-off. Therefore, ID=0A.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "VGS_off=4; #Gate-source cut-off voltage, V\n", + "VGS=6; #Gate source voltage, V\n", + "\n", + "print(\"p-channel JFET requires a positive gate-to-source voltage to pass drain current.\");\n", + "print(\"More positive voltage, the less the drain current. \");\n", + "print(\"Any further increase in VGS keeps the JFET cut-off. Therefore, ID=0A.\");" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.7 : Page number 517-518" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The gate to source resistance=15000MΩ.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "V_GS=15.0; #Gate-source voltage, V\n", + "I_G=1e-03; #Gate current, μA\n", + "\n", + "#Calculation\n", + "R_GS=(V_GS/(I_G*10**-6))/10**6; #Gate to source resistance, MΩ (OHM's LAW)\n", + "\n", + "#Result\n", + "print(\"The gate to source resistance=%dMΩ.\"%R_GS);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.8 : Page number 518" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Transconductance=3000 μ mho\n" + ] + } + ], + "source": [ + "\n", + "#Variable declaration\n", + "V_GS_max=-3.1; #Maximum gate to source voltage, V\n", + "V_GS_min=-3.0; #Minimum gate to source voltage, V\n", + "I_D_max=1.3; #Maximum drain current, mA\n", + "I_D_min=1.0; #Minimum drain current, mA\n", + "\n", + "\n", + "#Calculation\n", + "delta_V_GS=abs(V_GS_max-V_GS_min); #Change in gate to source voltage, V\n", + "delta_I_D=I_D_max-I_D_min; #Change in drain current, mA\n", + "g_fs=(delta_I_D/delta_V_GS)*1000; #Transconductance, μ mho\n", + "\n", + "\n", + "#Result\n", + "print(\"Transconductance=%.0f μ mho\"%g_fs);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.9 : Page number 518" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "VGS= 0V 0V -0.2V\n", + "VDS= 7V 15V 15V\n", + "ID = 10V 10.25V 9.65V\n", + "(i) The a.c drain resistance=32kΩ.\n", + "(ii) The transconductance=3000 μ mho.\n", + "(iii) The amplification factor=96.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "V_GS=[0,0,-0.2]; #Readings of Gate-source voltage, V\n", + "V_DS=[7,15,15]; #Readings of Drain-source voltage, V\n", + "ID=[10,10.25,9.65]; #Readings of drain current, mA\n", + "\n", + "\n", + "#Displaying the readings:\n", + "print(\"VGS= %dV %dV %.1fV\"%(V_GS[0],V_GS[1],V_GS[2]));\n", + "print(\"VDS= %dV %dV %dV\"%(V_DS[0],V_DS[1],V_DS[2]));\n", + "print(\"ID = %dV %.2fV %.2fV\"%(ID[0],ID[1],ID[2]));\n", + "\n", + "#Calculations\n", + "#(i)\n", + "#V_GS constant at 0V,\n", + "delta_VDS=V_DS[1]-V_DS[0]; #Change in drain-source voltage, V\n", + "delta_ID=ID[1]-ID[0]; #Change in drain current, mA\n", + "rd=delta_VDS/delta_ID; #a.c drain resistance, kΩ\n", + "\n", + "#(ii)\n", + "#V_DS constant at 15V,\n", + "delta_VGS=V_GS[2]-V_GS[1]; #Change in gate-source voltage, V\n", + "delta_ID=ID[2]-ID[1]; #Change in drain current, mA\n", + "g_fs=round((delta_ID/delta_VGS)*1000,); #Transconductance, μ mho\n", + "\n", + "#(iii)\n", + "amplification_factor=rd*1000*g_fs*10**-6; #Amplification factor\n", + "\n", + "\n", + "#Result\n", + "print(\"(i) The a.c drain resistance=%dkΩ.\"%rd);\n", + "print(\"(ii) The transconductance=%d μ mho.\"%g_fs);\n", + "print(\"(iii) The amplification factor=%d.\"%amplification_factor );\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.10 : Page number 519" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The transconductance=2500 μS.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "g_mo=4000.0; #Maximum transconductance, μS\n", + "V_GS=-3.0; #Gate to source voltage, V\n", + "V_GS_off=-8.0; #Gate-source cut-off voltage, V\n", + "\n", + "#Calculation\n", + "g_m=g_mo*(1-(V_GS/V_GS_off)); #Transconductance, μS\n", + "\n", + "#Result\n", + "print(\"The transconductance=%d μS.\"%g_m);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.11 : Page number 519" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The transconductance=1667 μS.\n", + "The drain current=333 μA.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "g_mo=5000.0; #Maximum transconductance, μS\n", + "V_GS=-4.0; #Gate to source voltage, V\n", + "V_GS_off=-6.0; #Gate-source cut-off voltage, V\n", + "I_DSS=3.0; #Shorted-gate drain current, mA\n", + "\n", + "#Calculation\n", + "g_m=g_mo*(1-(V_GS/V_GS_off)); #Transconductance, μS\n", + "I_D=(I_DSS*(1-(V_GS/V_GS_off))**2)*1000; #Drain current μA\n", + "\n", + "\n", + "#Result\n", + "print(\"The transconductance=%.0f μS.\"%g_m);\n", + "print(\"The drain current=%d μA.\"%I_D);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.12 : Page number 520" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The gate-source voltage=-5V.\n", + "The drain current=2.25mA.\n", + "The drain-source voltage=5.05V.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "V_GS_off=-8.0; #Gate-source cut-off voltage, V\n", + "I_DSS=16.0; #Shorted-gate drain current, mA\n", + "R_D=2.2; #Drain resistor, kΩ\n", + "R_G=1.0; #Gate resistor, MΩ\n", + "V_DD=10.0; #Drain supply voltage, V\n", + "V_GG=-5.0; #Gate supply voltage, V\n", + "\n", + "\n", + "#Calculation\n", + "V_GS=V_GG; #Gate-source voltage, V\n", + "I_D=I_DSS*(1-(V_GS/V_GS_off))**2; #Drain current μA\n", + "V_DS=V_DD-I_D*R_D; #Drain-source voltage, V (Kirchhoff's voltage law)\n", + "\n", + "#Result\n", + "print(\"The gate-source voltage=%dV.\"%V_GS);\n", + "print(\"The drain current=%.2fmA.\"%I_D);\n", + "print(\"The drain-source voltage=%.2fV.\"%V_DS);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.13 : Page number 521" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The drain-source voltage=7.65V.\n", + "The gate-source voltage=-2.35V.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "I_D=5.0; #Drain current mA\n", + "V_DD=15.0; #Drain supply voltage, V\n", + "V_G=0; #Gate voltage, V\n", + "R_D=1.0; #Drain resistor, kΩ\n", + "R_S=470.0; #Source resistor, Ω\n", + "\n", + "\n", + "#Calculation\n", + "V_S=(I_D/1000)*R_S; #Source voltage, V (OHM's LAW)\n", + "V_D=V_DD-I_D*R_D; #Drain voltage, V (Kirchhoff's voltage law)\n", + "V_DS=V_D-V_S; #Drain-source voltage, V\n", + "V_GS=V_G-V_S; #Gate-source voltage, V\n", + "\n", + "#Result\n", + "print(\"The drain-source voltage=%.2fV.\"%V_DS);\n", + "print(\"The gate-source voltage=%.2fV.\"%V_GS);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.14 : Page number 521" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The required source resistor=800 Ω.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "V_GS=-5.0; #Gate-source voltage, V\n", + "I_D=6.25; #Drain current mA\n", + "\n", + "\n", + "#Calculation\n", + "R_S=abs(V_GS/(I_D/1000)); #Required source resistor, Ω (OHM's LAW)\n", + "\n", + "#Result\n", + "print(\"The required source resistor=%d Ω.\"%R_S);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.15 : Page number : 521" + ] + }, + { + "cell_type": "code", + "execution_count": 16, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The source resistance=450Ω.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "I_DSS=25.0; #Shorted gate drain current, mA\n", + "V_GS_off=15.0; #Gate-source cut-off voltage, V\n", + "V_GS=5.0; #Gate-source voltage, V\n", + "\n", + "\n", + "#Calculation\n", + "I_D=I_DSS*(1-(V_GS/V_GS_off))**2; #Drain current mA\n", + "R_S=V_GS/(I_D/1000); #Required source resistor, Ω (OHM's LAW)\n", + "\n", + "\n", + "#Result\n", + "print(\"The source resistance=%.0fΩ.\"%R_S);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.16 : Page number 522" + ] + }, + { + "cell_type": "code", + "execution_count": 17, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " RS=313 Ω and RD=800 Ω.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "I_DSS=15.0; #Shorted gate drain current, mA\n", + "V_GS_off=-8.0; #Gate-source cut-off voltage, V\n", + "V_DD=12.0; #Drain supply voltage,V\n", + "V_D=V_DD/2; #Drain voltage(half of V_DD), V\n", + "\n", + "#Calculation\n", + "I_D=I_DSS/2; #Drain current(approximately half of I_DSS), mA\n", + "V_GS=V_GS_off/3.4; #Gate-source voltage, V\n", + "R_S=abs(V_GS/(I_D/1000)); #Source resistor, Ω (OHM's LAW)\n", + "#Since,V_D=V_DD-I_D*R_D; \n", + "R_D=(V_DD-V_D)/(I_D/1000); #Drain resistor, Ω (OHM's LAW)\n", + "\n", + "#Result\n", + "print(\" RS=%d Ω and RD=%d Ω.\"%(R_S,R_D));\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.17 : Page number 522" + ] + }, + { + "cell_type": "code", + "execution_count": 18, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The source resistance=0.6 kΩ\n", + "The drain resistance=6 kΩ.\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "\n", + "#Variable declaration\n", + "I_DSS=5.0; #Shorted gate drain current, mA\n", + "V_GS_off=-2.0; #Gate-source cut-off voltage, V\n", + "V_DS=10.0; #Drain-source voltage,V\n", + "I_D=1.5; #Drain current, mA\n", + "V_DD=20.0; #Drain supply voltage,V\n", + "V_G=0; #Gate voltage, V\n", + "\n", + "\n", + "#Calculation\n", + "#Since, Drain current, I_D=I_DSS*(1-(V_GS/V_GS_off))**2; \n", + "V_GS=V_GS_off*(1-sqrt(I_D/I_DSS)); #Gate-source voltage, V\n", + "\n", + "#Since, V_GS=V_G-V_S,\n", + "V_S=V_G-V_GS; #Source voltage, V\n", + "\n", + "R_S=V_S/I_D; #Source resistor, kΩ\n", + "\n", + "#Since, V_DD=I_D*R_D +V_DS+ I_D*R_S,\n", + "R_D=(V_DD-I_D*R_S-V_DS)/I_D; #Drain resistor, kΩ\n", + "\n", + "#Calculation\n", + "print(\"The source resistance=%.1f kΩ\"%R_S);\n", + "print(\"The drain resistance=%d kΩ.\"%R_D);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.18 : Page number 522-523" + ] + }, + { + "cell_type": "code", + "execution_count": 19, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The drain-source voltage=17V.\n", + "The gate-source voltage=-0.5V.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "V_DD=30.0; #Drain supply voltage, V\n", + "R_D=5.0; #Drain resistor, kΩ\n", + "I_D=2.5; #Drain current, mA\n", + "R_S=200.0; #Source resistor, Ω\n", + "\n", + "#Calculation\n", + "#(i)\n", + "V_DS=V_DD-I_D*(R_D+(R_S/1000)); #Drain-source voltage, V\n", + "\n", + "#(ii)\n", + "V_GS=-(I_D/1000)*R_S; #Gate-source voltage, V\n", + "\n", + "#Result\n", + "print(\"The drain-source voltage=%dV.\"%V_DS);\n", + "print(\"The gate-source voltage=%.1fV.\"%V_GS);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.19 : Page number 523" + ] + }, + { + "cell_type": "code", + "execution_count": 20, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "drain voltage of 1st stage=12.37V.\n", + "Source voltage of 1st stage=1.46V.\n", + "drain voltage of 2nd stage=11.7V.\n", + "Source voltage of 2nd stage=2.01V.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "ID_1=2.15; #First stage drain current, mA\n", + "ID_2=9.15; #Second stage drain current, mA\n", + "VDD=30; #Drain supply voltage, V\n", + "RS_1=0.68; #Source resistance of 1st stage, kΩ\n", + "RS_2=0.22; #Source resistance of 2nd stage, kΩ\n", + "RD_1=8.2; #Drain resistor of 1st stage, kΩ\n", + "RD_2=2; #Drain resistor of 2nd stage, kΩ\n", + "\n", + "#Calculation\n", + "V_RD1=ID_1*RD_1; #Voltage drop across 8.2kΩ\n", + "VD_1=VDD-V_RD1; #Drain voltage of 1st stage, V\n", + "VS_1=ID_1*RS_1; #D.C potential of source of first stage, V\n", + "V_RD2=ID_2*RD_2; #Voltage drop across 2kΩ\n", + "VD_2=VDD-V_RD2; #Drain voltage of 2nd stage, V\n", + "VS_2=ID_2*RS_2; #D.C potential of source of 2nd stage, V\n", + "\n", + "\n", + "#Result\n", + "print(\"drain voltage of 1st stage=%.2fV.\"%VD_1);\n", + "print(\"Source voltage of 1st stage=%.2fV.\"%VS_1);\n", + "print(\"drain voltage of 2nd stage=%.1fV.\"%VD_2);\n", + "print(\"Source voltage of 2nd stage=%.2fV.\"%VS_2);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.20 : Page number 524" + ] + }, + { + "cell_type": "code", + "execution_count": 21, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Drain current=1.52mA.\n", + "Gate-source voltage=-1.2V.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "VDD=12; #Drain supply voltage, V\n", + "VD=7; #Drain voltage, V\n", + "R1=6.8; #Resistor R1, MΩ\n", + "R2=1; #Resistor R2, MΩ\n", + "RS=1.8; #Source resistance, kΩ\n", + "RD=3.3; #Drain resistor, kΩ\n", + "\n", + "#Calculation\n", + "ID=(VDD-VD)/RD; #Second stage drain current, mA\n", + "VS=ID*RS; #Source voltage, V\n", + "VG=VDD*R2/(R1+R2); #Drain voltage, V\n", + "VGS=VG-VS; #Drain-source voltage, V\n", + "\n", + "#Calculation\n", + "print(\"Drain current=%.2fmA.\"%ID);\n", + "print(\"Gate-source voltage=%.1fV.\"%VGS);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.21 : Page number 524-525" + ] + }, + { + "cell_type": "code", + "execution_count": 22, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Source resistor, RS=5kΩ.\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "\n", + "#Variable declaration\n", + "VDD=30; #Drain supply voltage, V\n", + "ID=2.5; #Drain current, mA\n", + "VDS=8; #Drain-source voltage, V\n", + "VGS_off=-5; #Gate-source cutoff voltage, V\n", + "R1=1; #Resistor R1, MΩ\n", + "R2=500; #Resistor R2, kΩ\n", + "IDSS=10; #Shorted gate drain current, mA\n", + "\n", + "#Calculation\n", + "#ID=IDSS*square_of(1-(VGS/VGS_off))\n", + "VGS=VGS_off*(1-sqrt(ID/IDSS)); #Gate-source voltage, V\n", + "V2=VDD*R2/(R1*1000+R2); #Voltage across R2, V\n", + "\n", + "\n", + "#V2=VGS+ID*RS\n", + "RS=(V2-VGS)/ID; #Source resistor, kΩ\n", + "\n", + "#Result\n", + "print(\"Source resistor, RS=%dkΩ.\"%RS);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.22 : Page number 528-529" + ] + }, + { + "cell_type": "code", + "execution_count": 23, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "data": { + "image/png": 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+ "text/plain": [ + "<matplotlib.figure.Figure at 0x7fbbbf169d68>" + ] + }, + "metadata": {}, + "output_type": "display_data" + } + ], + "source": [ + "import matplotlib.pyplot as plt\n", + "\n", + "#Variable declaration\n", + "VDD=20.0; #Drain supply voltage, V\n", + "RS=50.0; #Source resistor, Ω\n", + "RD=150.0; #Drain resistor, Ω\n", + "\n", + "#Calculation\n", + "VDS_max=VDD; #Maximum drain source voltage, V\n", + "ID_max=(VDD/(RD+RS))*1000; #Maximum drain current, mA\n", + "\n", + "\n", + "#plot\n", + "x=[i for i in range(0,(int)(VDS_max+1))]; #Plot variable for V_DS\n", + "y=[(i/(RD+RS))*1000 for i in reversed(x[:])]; #Plot variable for ID\n", + "\n", + "\n", + "plt.plot(x,y);\n", + "plt.xlabel(\"VDS(V)\");\n", + "plt.ylabel(\"ID(mA)\");\n", + "plt.title(\"d.c load line\");\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.23 : Page number 529" + ] + }, + { + "cell_type": "code", + "execution_count": 24, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "data": { + "image/png": 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ubTI7IuKCVnbYCjcGViZOCVYUmTYGeXBjYGXjlGBFkEljIOnPgS9GxMgoy98HvDEibmtl\nx+Mqzo2BlZRTguUpq/MMHgPWSvqGpL+V9KeS/lLS9ZIeIzkT+bvtFGzWq3z2spVNK2MG7wRmAdOA\nXwGbgPUR8avMinMysB7glGDd5jEDs4LyWIJ1U2aXo5B0rqSHJW1Lbw9KOqe9Ms36T+3s5ZUrYf58\nGBqCkaajcGb5GM9vIJ8LzAc+DbwNOAD4U+BiSUPZlmfWWzyWYEU1nqOJvgOcFRGbG+YfDNwYESdk\nVpy7iayH1cYSZs1KxhIGB/OuyHpFVt1E+zQ2BADpvH1a2ZmZ7VBLCW9+M/zWbzklWL7G0xiMdbRQ\nZkcSmfWDxrGEc86B55/PuyrrR+NpDN4t6dEmt8eAXX6Qxsxa55RgeRvPmMH0sZbv7hLWE+ExA+tH\nHkuwicpkzCAinhzr1n65ZtaMU4LlYTzJ4L9o/ktktV86y2wQ2cnA+p1TgrUjq2Swd0Ts0+S2d5YN\ngZk5JVj3+HIUZiXhlGDjldnlKMwsf04JliUnA7MSqk8Jl1/uK6HazpwMzPpEfUrwNY6sEzJtDCRd\nI2mrpEfr5i2R9HR6FdSHJZ2aZQ1mvcpnL1snZZ0MrgU+0GT+8og4Jr3dkXENZj3NYwnWCZk2BhHx\nTaDZd5WW+rLMbGxOCTZReY0ZXCRpg6SrJe2bUw1mPccpwdqV+dFE6bWN1kbEUen9twLPRURI+jww\nLSIuHGXbWLJkyfb7lUqFSqWSab1mvcLnJfSParVKtVrdfn/ZsmXF+w3kxsZgvMvS5T601GwC/NvL\n/amoh5aKujECSVPrln0U+F4XajDrSx5LsPHK+tDSlcB9wLskPSXpfOAL6e8hbADmAAuyrMHMPJZg\nu+czkM36jMcSel9Ru4nMrECcEqwZJwOzPrZ+fZISTjzRKaGXOBmYWUtmz4aNG50SzMnAzFIeS+gd\nTgZm1jaPJfQ3JwMz24VTQrk5GZhZRzgl9B8nAzMbk1NC+TgZmFnHOSX0BycDMxs3p4RycDIws0w5\nJfQuJwMza0t9Srj8cpg8Oe+KrMbJwMy6pj4lHHmkU0LZORmY2YR5LKFYnAzMLBceSyg/JwMz66ha\nSqhdCdVjCd3nZGBmuaulhMFBjyWUiZOBmWXGKSEfTgZmVihOCeWRaTKQdA3wYWBrRByVzhsEbgKm\nA5uBMyPixVG2dzIw6xFOCd1TxGRwLfCBhnmXAHdHxGHAPcCijGswswJoTAnDw3lXZPUyHzOQNB1Y\nW5cM/gOYExFbJU0FqhFx+CjbOhmY9SCnhGwVMRk0MyUitgJExBZgSg41mFmOnBKKZ1LeBQBjfvVf\nunTp9ulKpUKlUsm4HDPrhoEBWLEC5s1LUsIttzgltKtarVKtVif0GHl0E20CKnXdROsi4t2jbOtu\nIrM+sG0bLFoEq1bBlVfC6afnXVG5FbWbSOmtZhg4L50+F1jThRrMrMBqKWHlSliwAIaGYGQk76r6\nS6aNgaSVwH3AuyQ9Jel84FLgFEmPAyen983MPJaQI5+BbGaF5COO2lfUbiIzs5bVUsLkyU4J3eBk\nYGaFt359khJmznRKGA8nAzPrSbNnw8aNTglZcjIws1JxStg9JwMz63lOCdlwMjCz0nJKaM7JwMz6\nilNC5zgZmFlPcErYwcnAzPqWU8LEOBmYWc/p95TgZGBmhlNCO5wMzKyn9WNKcDIwM2vglDA+TgZm\n1jf6JSU4GZiZjcEpYXROBmbWl3o5JTgZmJmNk1PCzpwMzKzv9VpKKFUykLRZ0kZJj0i6P686zMyc\nEnJMBpJ+BBwbEc+PsY6TgZl1VS+khFIlA0A579/MbBf9mhLyTgYvAK8CX4qILzdZx8nAzHJT1pRQ\ntmQwKyKOAU4DPinpvTnWYma2i35KCZPy2nFEPJP++6yk1cDxwDcb11u6dOn26UqlQqVS6VKFZmYw\nMJCkgnnzkpRwyy3FSwnVapVqtTqhx8ilm0jSG4E9IuKXkgaAu4BlEXFXw3ruJjKzwti2DRYvhltv\nhSuvhNNPz7ui5trpJsqrMXgHsBoIknTyzxFxaZP13BiYWeEUfSyhNGMGEfHjiJgREUdHxJHNGgIz\ns6LqxbEEn4FsZjYBRUwJpUkGZma9oldSgpOBmVmHFCUlOBmYmeWozCnBycDMLAN5pgQnAzOzgihb\nSnAyMDPLWLdTgpOBmVkBlSElOBmYmXVRN1KCk4GZWcEVNSU4GZiZ5SSrlOBkYGZWIkVKCU4GZmYF\nUEsJJ5wAK1ZMLCU4GZiZlVQtJey3Xz4pwcnAzKxgJjqW4GRgZtYD8hhLcDIwMyuwdlKCk4GZWY9p\nTAlr12azHycDM7OSGG9KKFUykHSqpP+Q9ENJn82rDjOzssgyJeTSGEjaA/h74APAEcDZkg7Po5Z+\nUq1W8y6hZ/i17Cy/nuM3MJCkghtugAULYGgIRkYm/rh5JYPjgf8XEU9GxCvAjcAZOdXSN/wH1zl+\nLTvLr2frOp0S8moMDgB+Unf/6XSemZmNUydTgo8mMjMrucaU0I5cjiaSdAKwNCJOTe9fAkREXNaw\nng8lMjNrQ6tHE+XVGLwOeBw4GXgGuB84OyI2db0YMzNjUh47jYhXJV0E3EXSVXWNGwIzs/wU+qQz\nMzPrjkIOIPuEtM6StFnSRkmPSLo/73rKRtI1krZKerRu3qCkuyQ9LulOSfvmWWOZjPJ6LpH0tKSH\n09upedZYFpIOlHSPpO9LekzSn6TzW35/Fq4x8AlpmXgNqETE0RFxfN7FlNC1JO/HepcAd0fEYcA9\nwKKuV1VezV5PgOURcUx6u6PbRZXUr4GFEXEEMBP4ZPp52fL7s3CNAT4hLQuimP/XpRAR3wSeb5h9\nBvDVdPqrwEe6WlSJjfJ6QvI+tRZExJaI2JBO/xLYBBxIG+/PIn5A+IS0zgvg65IekPSHeRfTI6ZE\nxFZI/iCBKTnX0wsukrRB0tXudmudpIOBGcB3gP1bfX8WsTGwzpsVEccAp5HEyPfmXVAP8pEYE/NF\n4JCImAFsAZbnXE+pSHoTcCtwcZoQGt+Pu31/FrEx+Cnw9rr7B6bzrE0R8Uz677PAapKuOJuYrZL2\nB5A0FfhZzvWUWkQ8W3e9+i8Dx+VZT5lImkTSEFwfEWvS2S2/P4vYGDwA/Kak6ZJeD5wFdPmnoXuH\npDem3xqQNAC8H/hevlWVkti5T3sYOC+dPhdY07iBjWmn1zP9wKr5KH6PtuIrwA8i4oq6eS2/Pwt5\nnkF6WNkV7Dgh7dKcSyotSe8gSQNBcpLhP/v1bI2klUAF2A/YCiwB/hW4BTgIeBI4MyJeyKvGMhnl\n9TyJpL/7NWAz8L9rfd42OkmzgPXAYyR/4wEsJrmqw8208P4sZGNgZmbdVcRuIjMz6zI3BmZm5sbA\nzMzcGJiZGW4MzMwMNwZmZoYbAzMzw42B9aH0+u+nNMy7WNLXJL0k6SFJP5D0HUnn1q0zRdLa9GJq\n35d0W92yqZKGJe0l6bnaWd91y1dL+j1JH5K0LPtnadYaNwbWj1YCZzfMOwv4a+CJiDg2It6Tzptf\n1yB8DrgrImak14+/pG77hcCXIuJXwB3A79QWSNoHmAUMR8TtwFxJb8jiiZm1y42B9aNVwGnpBb6Q\nNB2YRnK59O3Xy4mIzSQf8n+SzqqtU1tef/2cecCd6fSN7NzY/A5wZ0S8nN5fB3y4Q8/FrCPcGFjf\niYjnSa7d8sF01lkk13GpXdul3sNA7Zf2/gH4iqRvSFosaRpsv478SPpjTJA0CkdLGqx7/BvqHvMh\n4Lc79oTMOsCNgfWrG0k+pGHXD+t69UnhLuAdJJdYPhx4WNJ+JInh2br1XiG5auTvpstnsCM1QHI5\n4bd15mmYdYYbA+tXa4CTJR0N7BURj4yy3jEkPyUIQES8EBE3RsQ5wIPAbOBXQOMYQK2r6HeBNRHx\nat2yN6TbmBWGGwPrSxGxDaiSXAu+PhXUX2P/YOBvgRXp/ZMk7ZVO7w0cCjwF/JAkMdSrAu8E/phd\nU8e78PX6rWDcGFg/uwE4ip0/rA+pHVpK8u3+8oi4Ll12LPCgpA3At0iOHnooIl4CnpB0SO1B0l/t\nuhWYHBH3Nuz3JOD2bJ6SWXv8ewZmHSDpDODYiPjL3aw3heQHhk4Zaz2zbpuUdwFmvSAi1qSDxbvz\nduDTWddj1ionAzMz85iBmZm5MTAzM9wYmJkZbgzMzAw3BmZmBvw3WQTOWIXQWhcAAAAASUVORK5C\nYII=\n", + "text/plain": [ + "<matplotlib.figure.Figure at 0x7fbbbf188710>" + ] + }, + "metadata": {}, + "output_type": "display_data" + } + ], + "source": [ + "import matplotlib.pyplot as plt\n", + "\n", + "#Variable declaration\n", + "VDD=20; #Drain supply voltage, V\n", + "RD=500.0; #Drain resistor, Ω\n", + "\n", + "#Calculation\n", + "VDS_max=VDD; #Maximum drain source voltage, v \n", + "ID_max=(VDD/RD)*1000; #Maximum drain current, mA\n", + "\n", + "#Plot\n", + "x=[i for i in range(0,(int)(VDS_max+1))]; #Plot variable for V_DS\n", + "y=[(i/RD)*1000 for i in reversed(x[:])]; #Plot variable for ID\n", + "\n", + "plt.plot(x,y);\n", + "plt.xlabel(\"VDS(V)\");\n", + "plt.ylabel(\"ID(mA)\");\n", + "plt.title(\"d.c load line\");\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.24 : Page number 530-531" + ] + }, + { + "cell_type": "code", + "execution_count": 25, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Voltage gain=4.8.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "VDD=20; #Drain supply voltage, V\n", + "RD=12.0; #Drain resistor, kΩ\n", + "RL=8.0; #Load resistor, kΩ\n", + "RG=1.0; #Gate resistor, MΩ\n", + "gm=1.0; #transconductance, mA/V\n", + "\n", + "#Calculation\n", + "gm=gm*10**-3; #transconductance, mho\n", + "RAC=(RD*RL)/(RD+RL); #Total a.c load, kΩ\n", + "Av=gm*RAC*1000; #Voltage gain\n", + "\n", + "\n", + "#Result\n", + "print(\"Voltage gain=%.1f.\"%Av);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.25 : Page number 531" + ] + }, + { + "cell_type": "code", + "execution_count": 26, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Voltage gain=30.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "gm=3000; #transconductance, μmho\n", + "RD=10; #Drain resistance, kΩ\n", + "\n", + "#Calculation\n", + "Av=gm*10**-6*RD*1000; #Voltage gain\n", + "\n", + "#Result\n", + "print(\"Voltage gain=%d.\"%Av);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.26 : Page number 531" + ] + }, + { + "cell_type": "code", + "execution_count": 27, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Output voltage=257mV(r.m.s).\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "IDSS=8; #Shorted gate drain current, mA\n", + "VGS_off=-10; #Gate-source cut-off voltage, V\n", + "ID=1.9; #Drain current, mA\n", + "RD=3.3; #Drain resistance, kΩ\n", + "RS=2.7; #Source resistor, kΩ\n", + "vin=100; #Input voltage, mV\n", + "\n", + "#Calculation\n", + "VGS=-ID*RS; #Gate-source voltage, V\n", + "gmo=2*IDSS*10**-3/abs(VGS_off); #Maximum transconductance, S\n", + "gm=gmo*(1-(VGS/VGS_off)); #Transconductance, S\n", + "Av=gm*RD*1000; #Voltage gain\n", + "vout=Av*vin; #Output voltage, mA\n", + "\n", + "#Result\n", + "print(\"Output voltage=%dmV(r.m.s).\"%vout);\n", + " \n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.27 : Page number 531-532" + ] + }, + { + "cell_type": "code", + "execution_count": 28, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Output voltage=151mV(r.m.s).\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "RL=4.7; #Load resistor, Ω\n", + "RD=3.3; #Drain resistance, kΩ\n", + "gm=779*10**-6; #Transconductance, S\n", + "vin=100; #Input voltage, mV\n", + "\n", + "\n", + "#Calculation\n", + "RAC=RD*RL/(RD+RL); #Total a.c drain resistance, kΩ\n", + "Av=gm*RAC*1000; #Voltage gain\n", + "vout=Av*vin; #Output voltage, mA\n", + "\n", + "#Result\n", + "print(\"Output voltage=%dmV(r.m.s).\"%vout);" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.28 : Page number 532-533" + ] + }, + { + "cell_type": "code", + "execution_count": 29, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Voltage gain=1.85.\n", + "Voltage gain, if RS resistor is bypassed=6.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "RD=1.5; #Drain resistance, kΩ\n", + "gm=4; #Transconductance, mS\n", + "RS=560; #Source resistance, Ω\n", + "\n", + "#Calculation\n", + "Av=gm*10**-3*RD*1000/(1+gm*10**-3*RS);\n", + "print(\"Voltage gain=%.2f.\"%Av);\n", + "\n", + "#If RS is bypassed by a capacitor\n", + "Av=gm*10**-3*RD*1000;\n", + "print(\"Voltage gain, if RS resistor is bypassed=%d.\"%Av);" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.29 : Page number 533" + ] + }, + { + "cell_type": "code", + "execution_count": 30, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) Voltage gain with RS bypassed=4.155.\n", + "(ii) Voltage gain with RS unbypassed=1.35.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "from math import sqrt\n", + "\n", + "IDSS=10; #Shorted gate drain current, mA\n", + "VGS_off=-3.5; #Gate-source cut-off voltage, V\n", + "RD=1.5; #Drain resistance, kΩ\n", + "RS=750; #Source resistance, Ω\n", + "\n", + "\n", + "#Calculation\n", + "#From d.c biasing\n", + "ID=2.3; #Drain current, mA\n", + "VGS=round(VGS_off*(1-sqrt(ID/IDSS)),1); #Gate-source voltage, V\n", + "gm=round(round((2*IDSS/abs(VGS_off)),1)*round((1-(VGS/VGS_off)),3),2); #Transconductance, mS\n", + "\n", + "\n", + "#(i)\n", + "Av=gm*RD; #Voltage gain with RS resistor bypassed\n", + "print(\"(i) Voltage gain with RS bypassed=%.3f.\"%Av);\n", + "\n", + "#(ii)\n", + "Av=Av/(1+gm*(RS/1000.0));\n", + "print(\"(ii) Voltage gain with RS unbypassed=%.2f.\"%Av);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.30 : Page number 539-540" + ] + }, + { + "cell_type": "code", + "execution_count": 31, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) n-channel D-MOSFET\n", + "(ii) Drain current=3.91mA\n", + "(iii) Drain current=18.9mA\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "IDSS=10.0; #Shorted gate drain current, mA\n", + "VGS_off=-8.0; #Gate-source cut-off voltage, V\n", + "\n", + "#Calculation\n", + "#(i)\n", + "if(VGS_off<0):\n", + " print(\"(i) n-channel D-MOSFET\");\n", + "else:\n", + " print(\"(i) p-channel D-MOSFET\");\n", + " \n", + "\n", + "#(ii)\n", + "VGS=-3.0; #Gate-source voltage, V\n", + "ID=IDSS*(1-(VGS/VGS_off))**2; #Drain current mA\n", + "print(\"(ii) Drain current=%.2fmA\"%ID);\n", + "\n", + "#(iii)\n", + "VGS=3.0; #Gate-source voltage, V\n", + "ID=IDSS*(1-(VGS/VGS_off))**2; #Drain current mA\n", + "print(\"(iii) Drain current=%.1fmA\"%ID);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.31 : Page number 540" + ] + }, + { + "cell_type": "code", + "execution_count": 32, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Point 1: VGS=0V and ID=1mA.\n", + "Point 2: VGS=-6V and ID=0mA.\n", + "Point 3: VGS=-3V and ID=0.25mA.\n", + "Point 4: VGS=-1V and ID=0.694mA.\n", + "Point 5: VGS=1V and ID=1.36mA.\n", + "Point 6: VGS=3V and ID=2.25mA.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "IDSS=1.0; #Shorted gate drain current, mA\n", + "VGS_off=-6.0; #Gate-source cut-off voltage, V\n", + "\n", + "\n", + "#Calculation\n", + "#Point 1\n", + "VGS=0; #Gate source voltage, V \n", + "ID=IDSS; #Drain current, mA\n", + "print(\"Point 1: VGS=%dV and ID=%dmA.\"%(VGS,ID));\n", + "\n", + "#Point 2\n", + "VGS=VGS_off; #Gate source voltage, V \n", + "ID=0; #Drain current, mA\n", + "print(\"Point 2: VGS=%dV and ID=%dmA.\"%(VGS,ID));\n", + "\n", + "#locating more points by changing VG values\n", + "VGS=-3; #Gate source voltage, V \n", + "ID=IDSS*(1-(VGS/VGS_off))**2; #Drain current mA\n", + "print(\"Point 3: VGS=%dV and ID=%.2fmA.\"%(VGS,ID));\n", + "\n", + "VGS=-1; #Gate source voltage, V \n", + "ID=IDSS*(1-(VGS/VGS_off))**2; #Drain current mA\n", + "print(\"Point 4: VGS=%dV and ID=%.3fmA.\"%(VGS,ID));\n", + "\n", + "VGS=1; #Gate source voltage, V \n", + "ID=IDSS*(1-(VGS/VGS_off))**2; #Drain current mA\n", + "print(\"Point 5: VGS=%dV and ID=%.2fmA.\"%(VGS,ID));\n", + "\n", + "VGS=3; #Gate source voltage, V \n", + "ID=IDSS*(1-(VGS/VGS_off))**2; #Drain current mA\n", + "print(\"Point 6: VGS=%dV and ID=%.2fmA.\"%(VGS,ID));" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.32 : Page number 541" + ] + }, + { + "cell_type": "code", + "execution_count": 33, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Drain source voltage=10.6V.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "VDD=18; #Drain supply voltage, V\n", + "RD=620.0; #Drain resistor, Ω\n", + "IDSS=12.0; #Shorted gate drain current, mA\n", + "VGS_off=-8.0; #Gate-source cut-off voltage, V\n", + "\n", + "\n", + "#Calculation\n", + "ID=IDSS; #Drain current, mA\n", + "VDS=VDD-IDSS*(RD/1000); #Drain source voltage, V\n", + "\n", + "#Result\n", + "print(\"Drain source voltage=%.1fV.\"%VDS);" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.33 : Page number 542" + ] + }, + { + "cell_type": "code", + "execution_count": 34, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) Drain source voltage=7.56V.\n", + "(ii) Output voltage=922mV\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "VDD=15; #Drain supply voltage\n", + "RD=620.0; #Drain resistor, Ω\n", + "RL=8.2; #Load resistor, kΩ\n", + "vin=500.0; #Input voltage, V\n", + "IDSS=12.0; #Shorted gate drain current, mA\n", + "gm=3.2; #Transconductance, mS\n", + "\n", + "#Calculation\n", + "#(i)\n", + "VDS=VDD-IDSS*(RD/1000.0); #Drain source voltage, V\n", + "\n", + "#(ii)\n", + "RAC=RD*RL*1000/(RD+RL*1000); #Total a.c drain resistace, Ω\n", + "vout=(gm/1000.0)*RAC*vin; #Output voltage, V\n", + "\n", + "#Result\n", + "print(\"(i) Drain source voltage=%.2fV.\"%VDS);\n", + "print(\"(ii) Output voltage=%dmV\"%vout);" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.34 : Page number 545" + ] + }, + { + "cell_type": "code", + "execution_count": 35, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Drain current=98.7mA\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "ID_on=500.0; #Drain current for MOSFET ON, mA\n", + "VGS_on=10.0; #Gate-source voltage for MOSFET ON, V\n", + "VGS_th=1.0; #Threshold value of gate-source voltage, V\n", + "VGS=5; #Gate-source voltage, V\n", + "\n", + "#Calculation\n", + "K=round(ID_on/(VGS_on-VGS_th)**2,2); #Constant for a E-MOSFET, mA/V²\n", + "ID=K*(VGS-VGS_th)**2; #Drain current, mA\n", + "\n", + "#Result\n", + "print(\"Drain current=%.1fmA\"%ID);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.35 : Page number 545" + ] + }, + { + "cell_type": "code", + "execution_count": 36, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "K=0.061e-03A/V².\n", + "For VGS=5V, Drain current=0.244mA\n", + "For VGS=8V, Drain current=1.525mA\n", + "For VGS=10V, Drain current=2mA\n", + "For VGS=12V, Drain current=4.94mA\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "ID_on=3.0; #Drain current for MOSFET ON, mA\n", + "VGS_on=10.0; #Gate-source voltage for MOSFET ON, V\n", + "VGS_th=3.0; #Threshold value of gate-source voltage, V\n", + "\n", + "#Calculation\n", + "K=round((ID_on/(VGS_on-VGS_th)**2),3); #Constant for a E-MOSFET, mA/V²\n", + "print(\"K=%.3fe-03A/V².\"%K);\n", + "\n", + "#Determining different points for plotting\n", + "VGS=5; #Gate-source voltage, V\n", + "ID=K*(VGS-VGS_th)**2; #Drain current, mA\n", + "print(\"For VGS=5V, Drain current=%.3fmA\"%ID);\n", + "VGS=8; #Gate-source voltage, V\n", + "ID=K*(VGS-VGS_th)**2; #Drain current, mA\n", + "print(\"For VGS=8V, Drain current=%.3fmA\"%ID);\n", + "VGS=10; #Gate-source voltage, V\n", + "ID=K*(VGS-VGS_th)**2; #Drain current, mA\n", + "print(\"For VGS=10V, Drain current=%.dmA\"%ID);\n", + "VGS=12; #Gate-source voltage, V\n", + "ID=K*(VGS-VGS_th)**2; #Drain current, mA\n", + "print(\"For VGS=12V, Drain current=%.2fmA\"%ID);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.36 : Page number 546-547" + ] + }, + { + "cell_type": "code", + "execution_count": 37, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Drain-source voltage=10.8V.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "VDD=24.0; #Drain supply voltage, V\n", + "RD=470.0; #Drain resistor, Ω\n", + "R1=100.0; #Resistor R1, kΩ\n", + "R2=15.0; #Resistor R2, kΩ\n", + "ID_on=500.0; #Drain current for MOSFET ON, mA\n", + "VGS_on=10.0; #Gate-source voltage for MOSFET ON, V\n", + "VGS_th=1.0; #Threshold value of gate-source voltage, V\n", + "\n", + "\n", + "#Calculation\n", + "VGS=VDD*R2/(R1+R2); #Gate-source voltage, V (Voltage divider rule)\n", + "K=round((ID_on/(VGS_on-VGS_th)**2),2); #Constant for a E-MOSFET, mA/V²\n", + "ID=K*(VGS-VGS_th)**2; #Drain current, mA\n", + "VDS=VDD-(ID/1000)*RD; #Drain-source voltage, V\n", + "\n", + "#Result\n", + "print(\"Drain-source voltage=%.1fV.\"%VDS);\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.37 : Page number 547" + ] + }, + { + "cell_type": "code", + "execution_count": 38, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Drain current=10mA.\n", + "Drain-source voltage=10V.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "VDD=20.0; #Drain supply voltage, V\n", + "RD=1.0; #Drain resistor, kΩ\n", + "RG=5.0; #Gate resistor , MΩ\n", + "ID_on=10.0; #Drain current for MOSFET ON, mA\n", + "\n", + "#Calculation\n", + "#since, VGS=VDS\n", + "ID=ID_on; #Drain current, mA\n", + "VDS=VDD-ID*RD; #Drain-source voltage, V\n", + "\n", + "#Result\n", + "print(\"Drain current=%dmA.\"%ID);\n", + "print(\"Drain-source voltage=%dV.\"%VDS);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.38 : Page number 547-548" + ] + }, + { + "cell_type": "code", + "execution_count": 39, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Drain current=1.69mA.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "VDD=10.0; #Drain supply voltage, V\n", + "RD=3.0; #Drain resistor, kΩ\n", + "R1=1.0; #Resistor R1, MΩ\n", + "R2=1.0; #Resistor R2, MΩ\n", + "ID_on=10.0; #Drain current for MOSFET ON, mA\n", + "VGS_on=10.0; #Gate-source voltage for MOSFET ON, V\n", + "VGS_th=1.5; #Threshold value of gate-source voltage, V\n", + "\n", + "#Calculation\n", + "K=round((ID_on/(VGS_on-VGS_th)**2),3); #Constant for a E-MOSFET, mA/V²\n", + "VGS=VDD*R2/(R1+R2); #Gate-source voltage, V (Voltage divider rule)\n", + "ID=K*(VGS-VGS_th)**2; #Drain current, mA\n", + "\n", + "#Result\n", + "print(\"Drain current=%.2fmA.\"%ID);\n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 3", + "language": "python", + "name": "python3" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 3 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython3", + "version": "3.5.1" + }, + "widgets": { + "state": {}, + "version": "1.1.2" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter1_2.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter1_2.ipynb new file mode 100644 index 00000000..49519941 --- /dev/null +++ b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter1_2.ipynb @@ -0,0 +1,646 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:0ac98582dd0b2497034e459e869a2a3bd28001d0d4c4b37a61a8ed5d05f228e3" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 1: INTRODUCTION" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 1.1: Page Number 8" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable Declaration\n", + "Eg=24.0; #Generated voltage in V\n", + "Ri=0.01; #Internal Resistance in \u03a9\n", + "P=100; #Power supplied in watts\n", + "\n", + "#Calculations\n", + "# (i)\n", + "I=P/Eg; #Load current in A\n", + "V_Ri=I*Ri; #Voltage drop in internal resistance\n", + "\n", + "# (ii)\n", + "V=Eg-(I*Ri); #Terminal Voltage\n", + "\n", + "#Results\n", + "print (\"The voltage drop in internal resistance is %.4f V\"%V_Ri);\n", + "print (\"The terminal voltage is %.2f V\"%V);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The voltage drop in internal resistance is 0.0417 V\n", + "The terminal voltage is 23.96 V\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 1.2: Page number 10" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "Eg=500.0; #Generated voltage in V\n", + "Ri=1000.0; #Internal Resistance in \u03a9\n", + "\n", + "\n", + "#Calculations\n", + "# (i)\n", + "RL=10; #Load resistance of case 1 in \u03a9 \n", + "I= Eg/(RL+Ri); #Load current in A\n", + "\n", + "print(\"The load current for RL=10\u03a9 is %.3f A\"%I);\n", + "\n", + "# (ii)\n", + "RL=50; #Load resistance of case 2 in \u03a9 \n", + "I= Eg/(RL+Ri); #Load current in A\n", + "\n", + "print(\"The load current for RL=50\u03a9 is %.3f A\"%I);\n", + "\n", + "# (iii)\n", + "RL=100; #Load resistance of case 3 in \u03a9 \n", + "I= Eg/(RL+Ri); #Load current in A\n", + "I=round(I,3);\n", + "\n", + "print(\"The load current for RL=100\u03a9 is %.3f A\"%I);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The load current for RL=10\u03a9 is 0.495 A\n", + "The load current for RL=50\u03a9 is 0.476 A\n", + "The load current for RL=100\u03a9 is 0.455 A\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.3: Page Number 11" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "E=10.0; #voltage of voltage source in V\n", + "Ri=10.0; #Internal Resistance of the voltage source in \u03a9\n", + "\n", + "#Calculation\n", + "Isc=E/Ri; #short circuit current in A\n", + "I=Isc; #Current value of current source in A\n", + "R=Ri; #Internal Resistence of the current source in \u03a9\n", + "\n", + "#Results\n", + "print(\"The current value of the current source= %d A\"%Isc);\n", + "print(\"The internal resistance of the current source =%d \u03a9 \"%R);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The current value of the current source= 1 A\n", + "The internal resistance of the current source =10 \u03a9 \n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 1.4: Page number 11-12" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "I=6.0; # current value of current source in mA\n", + "Ri=2000.0; #Internal Resistance of the current source in \u03a9\n", + "\n", + "#Calcultion\n", + "V=(I/1000)*Ri; #Voltage of voltage source in V\n", + "R=Ri; #Internal resistance of voltage source in \u03a9\n", + "\n", + "#Results\n", + "print(\"The voltage of voltage source is %d V\"%V);\n", + "print(\"The internal resistance of the voltage source is %d \u03a9\"%R);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The voltage of voltage source is 12 V\n", + "The internal resistance of the voltage source is 2000 \u03a9\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.5: Page number 13\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "E=200.0; #Generated voltage in V\n", + "Ri=100.0; #Internal Resistance of generator in \u03a9\n", + "\n", + "#Calculations\n", + "#(i)\n", + "RL=100; #Load resistance for 1st case in \u03a9\n", + "I=E/(RL+Ri); #Load current in 1st case A\n", + "P=(I*I)*RL; #Power delivered to load of 2nd case in watts\n", + "Pt=(I*I)*(Ri+RL); #Total power generated in watts\n", + "\n", + "print(\"Power delivered for RL=100\u03a9 is %d watts\"%P);\n", + "print(\"Total power generated for RL=100\u03a9 is %d watts\"%Pt);\n", + "\n", + "\n", + "#(ii)\n", + "RL=300; #Load resistance for 2nd case in \u03a9\n", + "I=E/(RL+Ri); #Load current in 2nd case in A\n", + "P=(I*I)*RL; #Power delivered to load of 2nd case in watts\n", + "Pt=(I*I)*(Ri+RL); #Total power generated in watts\n", + "\n", + "print(\"Power delivered for RL=300\u03a9 is %d watts\"%P);\n", + "print(\"Total power generated for RL=300\u03a9 is %d watts\"%Pt);\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Power delivered for RL=100\u03a9 is 100 watts\n", + "Total power generated for RL=100\u03a9 is 200 watts\n", + "Power delivered for RL=300\u03a9 is 75 watts\n", + "Total power generated for RL=300\u03a9 is 100 watts\n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.6: Page number 14" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "V=12.0; #Output from amplifier in V\n", + "R_out_eq=15; #Equivalent resistance in \u03a9\n", + "\n", + "#Calculations\n", + "RL=R_out_eq; #Load resistance in \u03a9\n", + "Rt=RL+R_out_eq; #Total resistance in \u03a9\n", + "I=V/Rt; #Circuit current in A\n", + "PL=pow(I,2)*RL; #Power delivered to load in W\n", + "\n", + "#Results\n", + "print(\"Load resistance required is = %d \u03a9\"%RL);\n", + "print(\"Power delivered to load = %.1f W\"%PL);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Load resistance required is = 15 \u03a9\n", + "Power delivered to load = 2.4 W\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.7, Page number 14" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "V=50.0; #voltage from ac generator in V\n", + "R=100.0; #Resistance of internal impedance in \u03a9\n", + "XL=50.0; #inductive reactance of internal impedance in \u03a9\n", + "\n", + "#Calculation\n", + "Zi=100+(50j); #Internal impedance in complex form (\u03a9)\n", + "ZL=conjugate(Zi); #Load impedance (conjugate of internal impedance ) in \u03a9\n", + "Zt=Zi+ZL; #Total impedance in \u03a9\n", + "I=real(V/Zt); #Circuit current in A\n", + "\n", + "Max_Power=pow(I,2)*R; #Maximum power transferred to the load in watts\n", + "\n", + "\n", + "#Results\n", + "print (\"Load impedance %d %dj \u03a9\"%(real(ZL),imag(ZL)));\n", + "print(\"Maximum power transferred to the load =%.2f W\"%Max_Power);\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Load impedance 100 -50j \u03a9\n", + "Maximum power transferred to the load =6.25 W\n" + ] + } + ], + "prompt_number": 28 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.8: Page number 16" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "#Function for calculating parallel resistance\n", + "def pR(R1,R2):\n", + " return((R1*R2)/(R1+R2));\n", + "\n", + "\n", + "#Variable declaration\n", + "E=100.0; #Source voltage in V\n", + "R1=10.0; #Resistance of resistor 1 in \u03a9\n", + "R2=20.0; #Resistance of resistor 2 in \u03a9\n", + "R3=12.0; #Resistance of resistor 3 in \u03a9\n", + "R4=8.0; #Resistance of resistor 4 in \u03a9\n", + "RL=100.0; #Resistance of load in \u03a9\n", + "\n", + "#Calculation\n", + "Req=R1+pR(R3+R4,R2); #Equivalent resistance after removing RL ,in \u03a9\n", + "I=E/Req; #Total circuit current in A\n", + "I8=I*R2/(R2+R3+R4);\n", + "\n", + "#Thevenin's equivalent circuit's parameters\n", + "E0=I8*R4; #Thevenin voltage V\n", + "R0=pR(pR(R1,R2)+R3,R4); #Thevenin resistance \n", + "I_RL=E0/(R0+RL); #Load current in A \n", + "\n", + "#Result \n", + "print (\"Current through load = %.2f A.\"%I_RL);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Current through load = 0.19 A.\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.9: Page number 17" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Function for calculating parallel resistance\n", + "def pR(R1,R2):\n", + " return((R1*R2)/(R1+R2));\n", + "\n", + "\n", + "#Variable declaration\n", + "V=20.0; #Voltage source in V\n", + "R1=1000.0; #resistance of resistor 1 in \u03a9\n", + "R2=1000.0; #resistance of resistor 2 in \u03a9\n", + "R3=1000.0; #resistance of resistor 3 in \u03a9\n", + "\n", + "#calculation\n", + "#parameter for Thevenin's equivalent circuit\n", + "E0=(V*R3)/(R1+R3); #thevenin voltage in V\n", + "R0=pR(R1,R3)+R2; #Thevenins resistance in \u03a9\n", + "\n", + "#result\n", + "print(\"The thevenin voltage = %d V\"%E0);\n", + "print(\"The thevenin resistance = %d \u03a9\"%R0);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The thevenin voltage = 10 V\n", + "The thevenin resistance = 1500 \u03a9\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.10: Page number 18" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "V=120.0; #Supply voltage in V\n", + "R1=40.0; #Resistor 1's resistance in \u03a9\n", + "R2=20.0; #Resistor 2's resistance in \u03a9\n", + "R3=60.0; #Resistor 3's resistance in \u03a9\n", + "\n", + "#Calculations\n", + "#Using Thevenin's theorem, Thevenin's voltage and resistance are calculated\n", + "E0=(V*R2)/(R1+R2); #Thevenin voltage (voltage across the load resistance RL, after removing RL)in V\n", + "R0=(R1*R2)/(R1+R2) + R3; #Thevenin's resistance (Resistance between the terminals of load RL, with RL removed and source voltage shorted)in \u03a9 \n", + "RL=R0; #Value of load resistance to be connected for maximum power transfer in \u03a9\n", + "Pmax=pow(E0,2)/(4*RL); #Maximum power transferred to load in watts\n", + "\n", + "#Results\n", + "print(\"The value of load resistance RL to which maximum power will be transferred = %.2f \u03a9.\"%RL);\n", + "print(\"The maximum power transferred to load =%.2f W.\"%Pmax);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of load resistance RL to which maximum power will be transferred = 73.33 \u03a9.\n", + "The maximum power transferred to load =5.45 W.\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.11: Page number 18-19-20" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "V=80.0; #Supply voltage in V\n", + "R1=100.0; #Resistor 1's resistance in \u03a9\n", + "R2=100.0; #Resistor 2's resistance in \u03a9\n", + "R3=30.0; #Resistor 3's resistance in \u03a9\n", + "R4=80.0; #Resistor 4's resistance in \u03a9\n", + "R5=20.0; #Resistor 5's resistance in \u03a9\n", + "R6=60.0; #Resistor 6's resistance in \u03a9\n", + "R7=20.0; #Resistor 7's resistance in \u03a9\n", + "R8=50.0; #Resistor 8's resistance in \u03a9\n", + "\n", + "#Calculations\n", + "#Using Thevenin's theorem,\n", + "E0=(V*R2)/(R1+R2); #Thevenin's voltage for the circuit containing V, R1, R2 in V.\n", + "R0=(R1*R2)/(R1+R2); #Thevenin's resistance for R1, R2 in \u03a9.\n", + "\n", + "#Using Thevenin's theorem again on E0, R0 and rest of the circuit resistors.\n", + "E0_1=(E0*R4)/(R0+R3+R4); #Thevenin's voltage for the cicruit containing E0, R0, R3, R4 in V\n", + "R0_1=((R0+R3)*R4)/(R0+R3+R4); #Thevenin's resistance of R0,R3,R4 (R0 and R3 in series and both in parallel with R4), in \u03a9 \n", + "\n", + "#Using Thevenin's theorem again on E0_1, R0_1, and rest of the circuit resistors.\n", + "E0_2=(E0_1*R6)/(R0_1+R5+R6); #Thevenin's voltage for the circuit containing E0_1, R0_1, R5, R6 in V\n", + "R0_2=((R0_1+R5)*R6)/(R0_1+R5+R6); #Thevenin's resistance of R0_1,R5,R6 (R0 and R3 in series and both in parallel with R4), in \u03a9\n", + "\n", + "\n", + "I_50=E0_2/(R0_2+R7+R8); #Current through the 50 \u03a9 resistor in A\n", + "\n", + "\n", + "#Results\n", + "print(\"The current through the 50 \u03a9 resistor =%.1f A.\"%I_50);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The current through the 50 \u03a9 resistor =0.1 A.\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.12: Page number 22\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "from math import floor\n", + "#Variable declaration\n", + "V=40.0; #Voltage supply in V\n", + "R1=4.0; #Resistor 1's resistance in \u03a9\n", + "R2=6.0; #Resistor 2's resistance in \u03a9\n", + "R3=5.0; #Resistor 3's resistance in \u03a9\n", + "R4=8.0; #Resistor 4's resistance in \u03a9\n", + "\n", + "\n", + "#Calculation\n", + "#Using Norton's theorem,\n", + "#calculating Norton current by removing the load resistance R4 and short circuiting those two terminals of the circuit\n", + "R=R1 + (R2*R3)/(R2+R3); #Load on source after removing R4 resistor, in \u03a9\n", + "I=V/R; #Source current in A\n", + "\n", + "#Using current dividing rule ,calculating the short circuit current.\n", + "I_N=(I*R2)/(R2+R3); #Norton's equivalent current or the short circuit current in A\n", + "\n", + "R_N=R3 + (R1*R2)/(R1+R2); #Norton's equivalent resistance in \u03a9\n", + "\n", + "I_8=(I_N*R_N)/(R_N+R4); #Current through the 8 \u03a9 resistance in A\n", + "\n", + " \n", + "\n", + "#Results\n", + "print(\"The current through the 8\u03a9 resistance =%.2f A.\"%I_8);\n", + "\n", + "#Note: The answer in the book is 1.55 A, but in the above code the approximate value is obtained, i.e not 1.55A but 1.56A\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The current through the 8\u03a9 resistance =1.56 A.\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.13 :Page number 23\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "V1=30.0; #Voltage source 1, V\n", + "V2=18.0; #Voltage source 2, V\n", + "R1=20.0; #1st resistor, \u03a9\n", + "R2=10.0; #2nd resistor, \u03a9\n", + "\n", + "\n", + "#Calculations\n", + "\n", + "#Finding Thevenin's Equivalent circuit\n", + "I=(V1-V2)/(R1+R2); #Current in the circuit, A\n", + "\n", + "#Applying Kirchhoff's voltage law to 1st loop of the circuit,\n", + "#V1-I*R1-E0=0, where E0 is the voltage across the points X-Y.\n", + "E0=V1-I*R1; #Thevenin's voltage source, V\n", + "\n", + "R0=R1*R2/(R1+R2); #Thevenin's resistance, \u03a9\n", + "\n", + "#Finding Norton's equivalent circuit\n", + "IN=E0/R0; #Norton's equivalent current source, A\n", + "RN=R0; #Norton's equivanlent resistance, \u03a9\n", + "\n", + "#Result\n", + "print(\"IN=%.1fA and RN=%.2f \u03a9\"%(IN,RN));\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "IN=3.3A and RN=6.67 \u03a9\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "code", + "collapsed": false, + "input": [], + "language": "python", + "metadata": {}, + "outputs": [] + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter20_2.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter20_2.ipynb new file mode 100644 index 00000000..cad31534 --- /dev/null +++ b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter20_2.ipynb @@ -0,0 +1,677 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:369e36634d005b832372dcae6796c76b979f32b499d8baadce951517f2201533" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 20 : SILICON CONTROLLED RECTIFIERS" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 20.2 : Page number 559\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "I=50.0; #Surge current, A\n", + "t=12.0; #Time for which surge current lasts, ms\n", + "circuit_fusing_rating_max=90; #Maximum circuit fusing rating, A\u00b2s\n", + "\n", + "#Calculation\n", + "circuit_fusing_rating=I**2*(t*10**-3); #Circuit fusing rating, A\u00b2s\n", + "\n", + "#Result\n", + "if(circuit_fusing_rating<circuit_fusing_rating_max):\n", + " print(\"The device will not be destroyed.\");\n", + "else:\n", + " print(\"The device will be destroyed.\");\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The device will not be destroyed.\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 20.3 : Page number 559\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "I2_t_rating=50.0; #circuit fuse rating, A\u00b2s\n", + "Is=100.0; #Surge current, A\n", + "\n", + "#Calculation\n", + "t_max=(I2_t_rating/Is**2)*1000; #Maximum allowable duration, ms\n", + "\n", + "\n", + "#Result\n", + "print(\"The maximum allowable duration =%dms\"%t_max);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The maximum allowable duration =5ms\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 20.4 : Page number 559\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "#Variable declaration\n", + "R=220.0; #Gate resistor, \u03a9\n", + "I_G=7.0; #Gate current, mA\n", + "V_GK=0.7; #Junction voltage, V\n", + "\n", + "#Calculation\n", + "V_in=V_GK+(I_G/1000)*R; #Input voltage, V (Kirchhoff's voltage law)\n", + "\n", + "#Result\n", + "print(\"The required input voltage=%.2fV.\"%V_in);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The required input voltage=2.24V.\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 20.5 : Page number 564\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import asin\n", + "from math import cos\n", + "from math import pi\n", + "\n", + "#Variable declaration\n", + "I_G=1.0; #Gate current, mA\n", + "V_m=200.0; #Peak value of input sinusoidal voltage, V\n", + "v=100.0; #Forward breakdown voltage of SCR, V\n", + "R_L=100.0; #Load resistance, \u03a9\n", + "\n", + "#Calculation\n", + "#(i)\n", + "#v=Vm*sin(theta)\n", + "#Finding angle theta, for input voltage (v)= (V_f)forward_breakdown_voltage\n", + "theta=asin(v/V_m); #angle for input voltage = forward breakdown voltage, rad\n", + "theta=theta*180/pi; #angle for input voltage = forward breakdown voltage, degrees\n", + "\n", + "alpha=round(theta,0); #Firing angle, degrees\n", + "\n", + "#(ii)\n", + "phi=180-alpha; #Conduction angle, degrees\n", + "\n", + "#(iii)\n", + "V_avg=(V_m/(2*pi))*(1+cos(alpha*pi/180)); #Average voltage, V\n", + "I_avg=V_avg/R_L; #Average current, A\n", + "\n", + "\n", + "#Result\n", + "print(\"(i) The firing agle=%d\u00b0\"%alpha);\n", + "print(\"(ii) The conduction angle=%.0f\u00b0\"%phi);\n", + "print(\"(iii) The average current=%.4fA \"%I_avg);\n", + "\n", + "#Note: In the text book has approximated the average current to 0.5925A but in the code it gets approximated to 0.5940A.\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) The firing agle=30\u00b0\n", + "(ii) The conduction angle=150\u00b0\n", + "(iii) The average current=0.5940A \n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 20.6 : Page number 564\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import asin\n", + "from math import cos\n", + "from math import pi\n", + "from math import floor\n", + "\n", + "\n", + "#Variable declaration\n", + "I_G=1.0; #Gate current, mA\n", + "V_m=400.0; #Peak value of input sinusoidal voltage, V\n", + "v=150.0; #Forward breakdown voltage of SCR, V\n", + "R_L=200.0; #Load resistance, \u03a9\n", + "\n", + "\n", + "#Calculation\n", + "#(i)\n", + "#v=Vm*sin(theta)\n", + "#Finding angle theta, for input voltage (v)= (V_f)forward_breakdown_voltage\n", + "theta=asin(v/V_m); #angle for input voltage = forward breakdown voltage, rad\n", + "theta=theta*180/pi; #angle for input voltage = forward breakdown voltage, degrees\n", + "\n", + "alpha=theta; #Firing angle, degrees\n", + "\n", + "#(ii)\n", + "V_av=floor((V_m/(2*pi))*(1+cos(alpha*pi/180))*10)/10; #Average voltage, V\n", + "\n", + "#(iii)\n", + "I_av=V_av/R_L; #Average current, A\n", + "\n", + "#(iv)\n", + "P_out=V_av*I_av; #Output power, W\n", + "\n", + "#Result\n", + "print(\"(i) The firing agle=%d\u00b0\"%alpha);\n", + "print(\"(ii) The average output voltage=%.1f V\"%V_av);\n", + "print(\"(iii) The average current=%.3fA \"%I_av);\n", + "print(\"(iv) The output power=%.2f W\"%P_out);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) The firing agle=22\u00b0\n", + "(ii) The average output voltage=122.6 V\n", + "(iii) The average current=0.613A \n", + "(iv) The output power=75.15 W\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 20.7 : Page number 564-565\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import asin\n", + "\n", + "#Variable declaration\n", + "v=180.0; #Forward breakdown voltage, V\n", + "V_m=240.0; #Peak value of input voltage, V\n", + "w=314.0; #Angular frequency of input ,rad/s\n", + "\n", + "\n", + "#Calculation\n", + "#v=Vm*sin(w*t)\n", + "#So, t=asin(v/Vm)/w\n", + "t=(asin(v/V_m)/w)*1000; #Time for which SCR remains off, ms\n", + "\n", + "#Result\n", + "print(\"The SCR remains off for %.1f ms.\"%t);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The SCR remains off for 2.7 ms.\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 20.8 : Page number 565\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import pi\n", + "from math import cos\n", + "from math import floor\n", + "\n", + "#Variable declaration\n", + "I_dc=1.0; #d.c load current, A\n", + "alpha=30.0; #Firing angle, \u00b0\n", + "\n", + "\n", + "#Calculation\n", + "I_av=I_dc; #Average current(= d.c current), A\n", + "\n", + "#Since, Iav=(Vm/(2*pi*RL))*(1+cos(alpha)) and Im=Vm/RL\n", + "I_m=floor((2*pi*I_av/(1+cos(alpha*pi/180)))*100)/100; #Peak-load current, A\n", + "\n", + "\n", + "\n", + "#Result\n", + "print(\"Peak-load current=%.2f A.\"%I_m);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Peak-load current=3.36 A.\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 20.9: Page number 565\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt\n", + "from math import pi\n", + "from math import sin\n", + "\n", + "#Variable declaration\n", + "alpha=60.0; #Firing angle, \u00b0\n", + "P=100.0; #Power rating of tungsten lamp, W\n", + "V=110.0; #Voltage rating of tungsten lamp, V\n", + "V_ac=110.0; #a.c supply voltage, V\n", + "\n", + "\n", + "#Calculation\n", + "V_m=V_ac*sqrt(2); #Peak value of input voltage, V\n", + "\n", + "alpha=alpha*pi/180; #firing angle, rad\n", + "\n", + "#Since, E_rms\u00b2=(1/2*pi) \u222b V_m\u00b2sin\u00b2(theta) d(theta), limits: alpha to pi\n", + "# E_rms\u00b2=Vm\u00b2*((2*(pi-alpha) + sin(2*alpha))/(8*pi)),\n", + "# E_rms=Vm*sqrt((2*(pi-alpha) + sin(2*alpha))/(8*pi)),\n", + "E_rms=round(V_m*sqrt((2*(pi-alpha) + sin(2*alpha))/(8*pi))); #r.m.s voltage developed in the lamp, V\n", + "\n", + "\n", + "RL=V**2/P; #Load resistance, \u03a9\n", + "\n", + "I_rms=E_rms/RL; #r.m.s current developed in the lamp, A\n", + "\n", + "#Result\n", + "print(\"The r.m.s voltage developed in the lamp=%d V.\"%E_rms);\n", + "print(\"The r.m.s current developed in the lamp=%.2f A.\"%I_rms);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The r.m.s voltage developed in the lamp=70 V.\n", + "The r.m.s current developed in the lamp=0.58 A.\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 20.10 : Page number 567\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import cos\n", + "from math import pi\n", + "\n", + "#Variable declaration\n", + "RL=100.0; #Load resistance, \u03a9\n", + "V_m=200.0; #Peak a.c voltage, V\n", + "alpha=60; #firing angle, \u00b0\n", + "\n", + "#Calculation\n", + "#(i)\n", + "V_av=(V_m/pi)*(1+cos(alpha*pi/180)); #D.C output voltage, V\n", + "\n", + "#(ii)\n", + "I_av=V_av/RL; #Load current, A\n", + "\n", + "#Result\n", + "print(\"(i) d.c output voltage=%.1f V.\"%V_av);\n", + "print(\"(ii) Load current=%.3f A\"%I_av);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) d.c output voltage=95.5 V.\n", + "(ii) Load current=0.955 A\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 20.11: Page number 567\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt\n", + "from math import pi\n", + "from math import sin\n", + "\n", + "#Variable declaration\n", + "alpha=60.0; #Firing angle, \u00b0\n", + "P=100.0; #Power rating of tungsten lamp, W\n", + "V=110.0; #Voltage rating of tungsten lamp, V\n", + "V_ac=110.0; #a.c supply voltage, V\n", + "\n", + "\n", + "#Calculation\n", + "V_m=round(V_ac*sqrt(2)); #Peak value of input voltage, V\n", + "\n", + "alpha=alpha*pi/180; #firing angle, rad\n", + "\n", + "#Since, E_rms\u00b2=(1/2*pi) \u222b V_m\u00b2sin\u00b2(theta) d(theta), limits: alpha to pi\n", + "# E_rms\u00b2=Vm\u00b2*((2*(pi-alpha) + sin(2*alpha))/(8*pi)),\n", + "# E_rms=Vm*sqrt((2*(pi-alpha) + sin(2*alpha))/(8*pi)),\n", + "E_rms=V_m*sqrt((2*(pi-alpha) + sin(2*alpha))/(4*pi)); #r.m.s voltage developed in the lamp, V\n", + "\n", + "RL=V**2/P; #Load resistance, \u03a9\n", + "\n", + "I_rms=E_rms/RL; #r.m.s current developed in the lamp, A\n", + "\n", + "#Result\n", + "print(\"The r.m.s voltage developed in the lamp=%.1f V.\"%E_rms);\n", + "print(\"The r.m.s current developed in the lamp=%.2f A.\"%I_rms);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The r.m.s voltage developed in the lamp=98.9 V.\n", + "The r.m.s current developed in the lamp=0.82 A.\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 20.12 : Page number 572\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "VCC=15; #Suuply voltage, V\n", + "V_T=0.7; #Gate trigger voltage, V\n", + "I_T=7.0; #Gate trigger current, mA\n", + "I_H=6.0; #Holding current. mA\n", + "R_Vin=1; #Resistance at Vin, k\u03a9\n", + "R_VCC=100; #Resistance at Vcc, \u03a9\n", + "\n", + "\n", + "#Calculation\n", + "#(i) when SCR is off, there is no current, therefore no voltage drop across the resistor\n", + "V_out=VCC; #Output voltage, when SCR is off, V\n", + "\n", + "#(ii)\n", + "V_in=V_T+I_T*R_Vin; #Input voltage required to trigger the SCR, V\n", + "\n", + "#(iii)\n", + "#Since, I_H=(Vcc-VT)/R_Vin;\n", + "VCC_SCR_open=(I_H/1000)*R_VCC+V_T; #Decreased value of supply voltage at which SCR opens, V\n", + "\n", + "#Result\n", + "print(\"(i) The output voltage when SCR is off=%dV.\"%V_out);\n", + "print(\"(ii) The input voltage required to trigger the SCR=%.1f V.\"%V_in);\n", + "print(\"(iii) The decreased supply voltage at which SCR opens=%.1f V.\"%VCC_SCR_open);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) The output voltage when SCR is off=15V.\n", + "(ii) The input voltage required to trigger the SCR=7.7 V.\n", + "(iii) The decreased supply voltage at which SCR opens=1.3 V.\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 20.13 : Page number 572-573\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "Vz=5.6; #zener voltage, V\n", + "V_T=0.7; #Trigger voltage of SCR, V\n", + "\n", + "#Calculation\n", + "VCC=Vz+V_T; #Required supply voltage to turn on the crowbar, V\n", + "\n", + "#Result\n", + "print(\"The required supply voltage to turn on the crowbar=%.1fV.\"%VCC);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The required supply voltage to turn on the crowbar=6.3V.\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 20.14 : Page number 573\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "Vz=12; #Zener breakdown voltage, V\n", + "V_T=1.5; #Trigger voltage, V\n", + "tolerance_z=10.0; #Tolerance of zener diode, %\n", + "\n", + "\n", + "#Calculation\n", + "Vz_max=Vz*(1+tolerance_z/100); #Maximum value of zener breakdown, V\n", + "Vz_min=Vz*(1-tolerance_z/100); #Minimum value of zener breakdown, V\n", + "V_crowbar=Vz_max+V_T; #Maximum value of supply voltage for crowbarring, V\n", + "\n", + "#Result\n", + "print(\"The maximum value of supply voltage for crowbarring=%.1fV\"%V_crowbar);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The maximum value of supply voltage for crowbarring=14.7V\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 20.15 : Page number 573\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "VCC=25.0; #Supply voltage, V\n", + "\n", + "#Calculation\n", + "#When brights light is on, LASCR conducts and thus gets short circuited to ground, hence,\n", + "V_out=0; #Output voltage, V\n", + "\n", + "print(\"Output voltage when bright light is on=%dV\"%V_out);\n", + "\n", + "\n", + "#When brights light is off, LASCR stops conducting and thus no current through resistor, hence,\n", + "V_out=VCC; #Output voltage, V\n", + "print(\"Output voltage when bright light is off=%dV\"%V_out);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Output voltage when bright light is on=0V\n", + "Output voltage when bright light is off=25V\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "code", + "collapsed": false, + "input": [], + "language": "python", + "metadata": {}, + "outputs": [] + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter21_2.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter21_2.ipynb new file mode 100644 index 00000000..acca0cfa --- /dev/null +++ b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter21_2.ipynb @@ -0,0 +1,467 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:412bf04e25192c77f9fa9664d995cc0ae6446a81f631fb5e0e755ebfa36436bf" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 21 : POWER ELECTRONICS" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 21.3: Page number 585\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "V_GT=2; #Gate triggering voltage, V\n", + "V_F=0.7; #Forward voltage for diode D1\n", + "\n", + "#Calculation\n", + "#(i)Triggering only by a positive gate voltage,\n", + "#A diode is connected at the gatewith the n-side connected to thegate of the device,\n", + "V_A=V_F+V_GT; #Required voltage to trigger the device, V\n", + "\n", + "print(\"The required voltage to trigger the device only by positive voltage=%.1fV.\"%V_A);\n", + "\n", + "#(ii)\n", + "print(\"In order to trigger the triac only by negative voltage, the direction of diode D1 is reversed.\");\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The required voltage to trigger the device only by positive voltage=2.7V.\n", + "In order to trigger the triac only by negative voltage, the direction of diode D1 is reversed.\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 21.4 : Page number 585-586\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "R=50.0; #Resitor, \u03a9\n", + "V=50.0; #Supply voltage, V\n", + "V_drop=1.0; #Drop across the triac in conduction, V\n", + "\n", + "#Calculation\n", + "#(i) Ideal triac\n", + "#Since the triac is ideal, voltage drop across it is zero,\n", + "I=V/R; #Current through the 50 \u03a9 resistor, A\n", + "\n", + "print(\"(i) The cuurent through the 50 \u03a9 resistor when the triac is ideal=%dA.\"%I);\n", + "\n", + "#(ii) Triac has a drop of 1V\n", + "I=(V-V_drop)/R; #Current through the 50 \u03a9 resistor, A\n", + "\n", + "print(\"(ii) The current through the 50 \u03a9 resistor when the triac has a drop of 1V=%.2fA.\"%I);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) The cuurent through the 50 \u03a9 resistor when the triac is ideal=1A.\n", + "(ii) The current through the 50 \u03a9 resistor when the triac has a drop of 1V=0.98A.\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 21.5 : Page number 588-589\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "V_GT=2; #Gate triggering voltage, V\n", + "V_BO=20; #Breakover voltage,V\n", + "\n", + "#Calculation\n", + "print(\"The triggering level is raised by using a diac.\");\n", + "V_A=V_BO+V_GT; #Gate trigger signal, V\n", + "\n", + "#Result\n", + "print(\"In order to turn on the triac, the gate trigger signal=%dV.\"%V_A);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The triggering level is raised by using a diac.\n", + "In order to turn on the triac, the gate trigger signal=22V.\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 21.6 : Page number 589\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "V_BO=30; #Breakover voltage of diac, V\n", + "V_GT=1; #Trigger voltage of the triac, V\n", + "I_T=10; #Trigger current, mA\n", + "\n", + "\n", + "#Calculation\n", + "V_A=V_BO+V_GT; #Voltage required for triggering the triac, V\n", + "\n", + "#Result\n", + "print(\"The minimum capacitor voltage that will trigger the triac=%d V.\"%V_A);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The minimum capacitor voltage that will trigger the triac=31 V.\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 21.7 : Page number 593\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "#Variable declaration\n", + "eta=0.6; #Intrinsic stand-off ratio for UJT\n", + "R_BB=10; #Inter-base resistance, k\u03a9\n", + "\n", + "#Calculation\n", + "#Since, RBB=RB1+RB2 and eta=RB1/(RB1+RB2),\n", + "#eta=RB1/RBB.\n", + "R_B1=eta*R_BB; #Resistance of the bar between B1 and emitter junction, k\u03a9\n", + "R_B2=R_BB-R_B1; #Resistance of the bar between B2 and emitter junction, k\u03a9 \n", + "\n", + "#Result\n", + "print(\"Resistance of the bar between B1 and emitter junction=%d k\u03a9.\"%R_B1);\n", + "print(\"Resistance of the bar between B2 and emitter junction=%d k\u03a9.\"%R_B2);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Resistance of the bar between B1 and emitter junction=6 k\u03a9.\n", + "Resistance of the bar between B2 and emitter junction=4 k\u03a9.\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 21.8 : Page number 593\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "V_BB=10; #Interbase voltage, V\n", + "eta=0.65; #Intrinsic stand-off ratio for UJT\n", + "V_D=0.7; #Voltage drop in the pn junction, V\n", + "\n", + "#Calculation\n", + "V_stand_off=eta*V_BB; #Stand off voltage, V\n", + "V_P=V_stand_off+V_D; #Peak-point voltage, V\n", + "\n", + "#Result\n", + "print(\"Stand off voltage=%.1f V.\"%V_stand_off);\n", + "print(\"Peak-point voltage=%.1f V.\"%V_P);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Stand off voltage=6.5 V.\n", + "Peak-point voltage=7.2 V.\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 21.9 : Page number 593\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "V_BB=25; #Interbase voltage, V\n", + "eta_max=0.86; #Maximum intrinsic stand-off ratio for UJT\n", + "eta_min=0.74; #Minimum intrinsic stand-off ratio for UJT\n", + "V_D=0.7; #Voltage drop in the pn junction, V\n", + "\n", + "#Calculation\n", + "V_P_max=eta_max*V_BB+V_D; #Maximum peak-point, V\n", + "V_P_min=eta_min*V_BB+V_D; #Minimum peak-point, V\n", + "\n", + "#Result\n", + "print(\"Maximum peak-point voltage=%.1fV\"%V_P_max);\n", + "print(\"Minimum peak-point voltage=%.1fV\"%V_P_min);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Maximum peak-point voltage=22.2V\n", + "Minimum peak-point voltage=19.2V\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 21.10 : Page number 593-594\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "eta=0.65; #Intrinsic stand-off ratio for UJT\n", + "R_BB=7.0; #Inter-base resistance, k\u03a9\n", + "R1=100.0; #Resistor R1, \u03a9\n", + "R2=400.0; #Resistor R2, \u03a9\n", + "V_S=12.0; #Source voltage, V\n", + "V_D=0.7; #Voltage drop in the pn junction, V\n", + "\n", + "#Calculation\n", + "#(i)\n", + "#Since, eta=RB1/RBB,\n", + "R_B1=eta*R_BB; #Resistance of the bar between B1 and emitter junction, k\u03a9\n", + "R_B2=R_BB-R_B1; #Resistance of the bar between B2 and emitter junction, k\u03a9 \n", + "\n", + "print(\"(i) Resistance of the bar between B1 and emitter junction=%.2f k\u03a9.\"%R_B1);\n", + "print(\" Resistance of the bar between B2 and emitter junction=%.2f k\u03a9.\"%R_B2);\n", + "\n", + "#(ii)\n", + "V_B2_B1=V_S*R_BB/(R_BB + (R1/1000) + (R2/1000)); #Voltage across B2-B1, V (voltage divider rule)\n", + "V_P=eta*V_B2_B1+V_D; #Peak-point voltage, V\n", + "\n", + "print(\"(ii) The voltage across the base B2-B1=%.1fV.\"%V_B2_B1);\n", + "print(\" Peak-point voltage=%.2fV\"%V_P);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) Resistance of the bar between B1 and emitter junction=4.55 k\u03a9.\n", + " Resistance of the bar between B2 and emitter junction=2.45 k\u03a9.\n", + "(ii) The voltage across the base B2-B1=11.2V.\n", + " Peak-point voltage=7.98V\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 21.11 : Page number 596\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import log\n", + "\n", + "#Variable declaration\n", + "RE_initial=5; #Initial value of emitter resistor, k\u03a9\n", + "RE_adjusted=10; #Adjusted value of emitter resistor, k\u03a9\n", + "C=0.2; #Capacitance, \u03bcF\n", + "eta=0.54; #intrinsic stand-off ratio\n", + "\n", + "#Calculation\n", + "#(i)\n", + "t=round((RE_initial*1000*C*10**-6*log(1/(1-eta)))*1000,2); #Time period, ms\n", + "f=(1/t)*1000; #frequency, Hz\n", + "\n", + "print(\"Frequency for 5k\u03a9 setting=%dHz.\"%f);\n", + "\n", + "#(i)\n", + "t=round((RE_adjusted*1000*C*10**-6*log(1/(1-eta)))*1000,2); #Time period, ms\n", + "f=(1/t)*1000; #frequency, Hz\n", + "\n", + "print(\"Frequency for 10k\u03a9 setting=%dHz.\"%f);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Frequency for 5k\u03a9 setting=1282Hz.\n", + "Frequency for 10k\u03a9 setting=645Hz.\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 21.12 : Page number 596-597\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import log\n", + "\n", + "#Variable declaration\n", + "V_S=12; #Supply voltage, V\n", + "R_BB=5; #Interbase resistance, k\u03a9\n", + "R_1=50; #Resistor R1, k\u03a9\n", + "R_2=0.1; #Resistor R2, k\u03a9\n", + "C=0.1; #Capacitance, \u03bcF\n", + "eta=0.6; #intrinsic stand-off ratio\n", + "V_D=0.7; #Voltage drop across pn junction, V\n", + "\n", + "#Calculation\n", + "#(i)\n", + "#Since, \u03b7=R_B1/R_BB,\n", + "R_B1=eta*R_BB; #Resitance between base B1 and emitter junction, k\u03a9\n", + "\n", + "#Since, R_BB=R_B1+R_B2\n", + "R_B2=R_BB-R_B1; #Resitance between base B2 and emitter junction, k\u03a9\n", + "\n", + "#(ii)\n", + "V_RB1_R2=V_S*(R_B1+R_2)/(R_BB+R_2); #Voltage drop across R_B1 and R_2 resistors, V\n", + "V_P=V_D+V_RB1_R2; #Peak-point voltage, V\n", + "\n", + "#(iii)\n", + "t=round((R_1*1000*C*10**-6*log(1/(1-eta)))*1000,2); #Time period, ms\n", + "f=(1/t)*1000; #frequency, Hz\n", + "\n", + "\n", + "#Result\n", + "print(\"(i) R_B1=%dk\u03a9 and R_B2=%dk\u03a9\"%(R_B1,R_B2));\n", + "print(\"(ii) The peak-point voltage to turn on the UJT=%.0fV.\"%V_P);\n", + "print(\"(iii) Frequency of oscillations=%dHz.\"%f);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) R_B1=3k\u03a9 and R_B2=2k\u03a9\n", + "(ii) The peak-point voltage to turn on the UJT=8V.\n", + "(iii) Frequency of oscillations=218Hz.\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "code", + "collapsed": false, + "input": [], + "language": "python", + "metadata": {}, + "outputs": [] + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter22_2.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter22_2.ipynb new file mode 100644 index 00000000..5f13ea0e --- /dev/null +++ b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter22_2.ipynb @@ -0,0 +1,668 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:a688629536ad6915939234eacc1ed3eaaf36e0aaa88de5df5b6a309da4d2c64d" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 22: ELECTRONIC INSTRUMENTS" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 22.1 : Page number 606\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "I_g=1; #Full scale deflection current, mA\n", + "\n", + "#Calculation\n", + "MS=1/(I_g/1000.0); #Multimeter sensitivity, \u03a9 per volt\n", + "\n", + "#Result\n", + "print(\"The multimeter sensitivity=%d \u03a9 per volt.\"%MS);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The multimeter sensitivity=1000 \u03a9 per volt.\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 22.2 : Page number 606-607\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "meter_sensitivity=1000.0; #Meter sensitivity, \u03a9 per volt\n", + "V_full_scale=50.0; #Full scale volts\n", + "R=50000.0; #Resistance to be measured, \u03a9\n", + "\n", + "#Calculation\n", + "meter_resistance=V_full_scale*meter_sensitivity; #Meter resistance, \u03a9\n", + "R_p=R*meter_resistance/(R+meter_resistance); #Parallel resistance, \u03a9\n", + "\n", + "#Result\n", + "print(\"When the meter is used to measure the voltage across the resistance %d\u03a9, total resistance =%d\u03a9.\"%(R,R_p));\n", + "print(\"\u2234 Meter will give highly incorrect reading.\");\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "When the meter is used to measure the voltage across the resistance 50000\u03a9, total resistance =25000\u03a9.\n", + "\u2234 Meter will give highly incorrect reading.\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 22.3 : Page number 607\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "meter_sensitivity=4.0; #Meter sensitivity, k\u03a9/V\n", + "R_1=10.0; #Resistance across which voltage is to be measured, k\u03a9\n", + "R_2=10.0; #Resistance, k\u03a9\n", + "range_max=10.0; #Maximum range of the meter, V\n", + "range_min=0; #Minimum range of the meter, V\n", + "V=20.0; #Battery voltage, V\n", + "\n", + "#Calculation\n", + "R_meter=meter_sensitivity*range_max; #Resistance of the meter, k\u03a9\n", + "R_T=(R_meter*R_1)/(R_1+R_meter) + R_2; #Total circuit resistance, k\u03a9\n", + "I_circuit=round(V/R_T,2); #Circuit current, mA\n", + "V_multimeter=I_circuit*((R_meter*R_1)/(R_1+R_meter)); #Voltage read by multimeter, V\n", + "\n", + "\n", + "#Result\n", + "print(\"Voltage read by multimeter=%.2fV.\"%V_multimeter);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Voltage read by multimeter=8.88V.\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 22.4 : Page number 607-608\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "meter_sensitivity=20.0; #Meter sensitivity, k\u03a9/V\n", + "R_1=10.0; #Resistance across which voltage is to be measured, k\u03a9\n", + "R_2=10.0; #Resistance, k\u03a9\n", + "range_max=10.0; #Maximum range of the meter, V\n", + "range_min=0; #Minimum range of the meter, V\n", + "V=20.0; #Battery voltage, V\n", + "\n", + "#Calculation\n", + "R_meter=meter_sensitivity*range_max; #Resistance of the meter, k\u03a9\n", + "R_T=round((R_meter*R_1)/(R_1+R_meter) + R_2,1); #Total circuit resistance, k\u03a9\n", + "I_circuit=round(V/R_T,2); #Circuit current, mA\n", + "V_multimeter=I_circuit*((R_meter*R_1)/(R_1+R_meter)); #Voltage read by multimeter, V\n", + "\n", + "\n", + "#Result\n", + "print(\"Voltage read by multimeter=%.2fV.\"%V_multimeter);\n", + "\n", + "\n", + "#Note: The circuit current=1.0256mA, has been approximated in the text as 1.04mA. But, in the code 1.03 mA has been used. Therefore, the final answer is obtained as 9.81V and not 9.88V.\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Voltage read by multimeter=9.81V.\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 22.5 : Page number 608-609\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import floor\n", + "\n", + "#Variable declaration\n", + "R_A=20.0; #Resistance after point A, k\u03a9\n", + "R_B=20.0; #Resistance after point B, k\u03a9\n", + "R_C=30.0; #Resistance after point C, k\u03a9\n", + "R_D=30.0; #Resistance after point D, k\u03a9\n", + "R_meter=60.0; #Resistance of the meter, k\u03a9\n", + "V=100.0; #Battery voltage, V\n", + "\n", + "#Calculation\n", + "#(i) When meter is not connected:\n", + "R_T=R_A+R_B+R_C+R_D; #Total circuit resistance, k\u03a9\n", + "I_circuit=V/R_T; #Circuit current, mA\n", + "V_A=V; #Voltage at point A, V\n", + "V_B=V-(I_circuit*R_A); #Voltage at point B, V\n", + "V_C=V-(I_circuit*(R_A+R_B)); #Voltage at point C, V\n", + "V_D=V-(I_circuit*(R_T-R_D)); #Voltage at point D, V\n", + "\n", + "print(\"(i) When meter is not connected:\");\n", + "print(\" Voltage at point A=%dV.\"%V_A);\n", + "print(\" Voltage at point B=%dV.\"%V_B);\n", + "print(\" Voltage at point C=%dV.\"%V_C);\n", + "print(\" Voltage at point D=%dV.\"%V_D);\n", + "\n", + "\n", + "#(ii) When meter is connected:\n", + "#(a) Since, point A is directly connected to the source, voltage at point A is equal to source voltage.\n", + "V_A=V; #Voltage at point A, V\n", + "\n", + "#(b)\n", + "R_T_B=R_A + round((R_T-R_A)*R_meter/(R_meter + (R_T-R_A)),2); #Total circuit resistance, k\u03a9\n", + "I_circuit=round(V/R_T_B,2); #Circuit current, mA\n", + "V_B=I_circuit*(R_T-R_A)*R_meter/(R_meter + (R_T-R_A)); #Voltage at point B, V\n", + "\n", + "\n", + "#(c)\n", + "R_T_C=(R_A+R_B) + (R_T-R_A-R_B)*R_meter/(R_meter + (R_T-R_A-R_B)); #Total circuit resistance, k\u03a9\n", + "I_circuit=V/R_T_C; #Circuit current, mA\n", + "V_C=floor((I_circuit*(R_T-R_A-R_B)*R_meter/(R_meter + (R_T-R_A-R_B)))*10)/10; #Voltage at point C, V\n", + "\n", + "\n", + "\n", + "#(c)\n", + "R_T_D=(R_T-R_D) + R_D*R_meter/(R_meter + R_D); #Total circuit resistance, k\u03a9\n", + "I_circuit=round(V/R_T_D,2); #Circuit current, mA\n", + "V_D=I_circuit*(R_D*R_meter)/(R_meter + R_D); #Voltage at point D, V\n", + "\n", + "\n", + "print(\"(ii) When meter is connected:\");\n", + "print(\" Voltage at point A=%dV.\"%V_A);\n", + "print(\" Voltage at point B=%dV.\"%V_B);\n", + "print(\" Voltage at point C=%.1fV.\"%V_C);\n", + "print(\" Voltage at point D=%.1fV.\"%V_D);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) When meter is not connected:\n", + " Voltage at point A=100V.\n", + " Voltage at point B=80V.\n", + " Voltage at point C=60V.\n", + " Voltage at point D=30V.\n", + "(ii) When meter is connected:\n", + " Voltage at point A=100V.\n", + " Voltage at point B=63V.\n", + " Voltage at point C=42.8V.\n", + " Voltage at point D=22.2V.\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 22.6 : Page number 614\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "VCC=12.0; #Supply voltage, V\n", + "R_m=1.0; #Meter resistance, k\u03a9\n", + "I_m_fsd=2.0; #Full scale deflection of meter current, mA\n", + "beta=80.0; #Base current amplification factor\n", + "E=5.0; #Voltage to be measured, V\n", + "V_BE=0.7; #Base-emitter voltage, V\n", + "\n", + "\n", + "#Calculation\n", + "V_E=E-V_BE; #Emitter voltage, V\n", + "\n", + "#(i)\n", + "#I_m_fsd=V_E/(R_s+R_m), (OHM's LAW)\n", + "R_s=((V_E/I_m_fsd)-R_m)*1000; #Multiplier resistor, \u03a9\n", + "\n", + "#(ii)\n", + "IB=I_m_fsd/beta; \t\t\t\t#Base current, mA\n", + "R_i=E/IB; \t\t\t#Input resistance of voltmeter, k\u03a9\n", + "\n", + "\n", + "\n", + "#Result\n", + "print(\"(i) The multiplier resistor=%d\u03a9.\"%R_s);\n", + "print(\"(ii) The voltmeter input resistance=%dk\u03a9\"%R_i);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) The multiplier resistor=1150\u03a9.\n", + "(ii) The voltmeter input resistance=200k\u03a9\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 22.7 : Page number 614\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "VCC=20; #Supply voltage, V\n", + "Rs_Rm=9.3; #Sum of multipier resistance and meter resistance, k\u03a9\n", + "I_m=1; #Meter current, mA\n", + "beta=100; #Base current amplification factor\n", + "E=10; #Voltage to be measured, V\n", + "V_BE=0.7; #Base-emitter voltage, V\n", + "\n", + "\n", + "#Calculation\n", + "#(i)\n", + "V_E=E-V_BE; #Emitter voltage, V\n", + "I_m=V_E/Rs_Rm; #Meter current, mA\n", + "\n", + "#(ii)\n", + "I_B=I_m/beta; #Base current, mA\n", + "R_i_T=(E/I_B)/1000; #Input resistance of voltmeter, with transistor, M\u03a9\n", + "R_i_WT=Rs_Rm; #Input resistance of voltmeter, without transistor, k\u03a9\n", + "\n", + "#Result\n", + "print(\"(i) The meter current=%dmA\"%I_m);\n", + "print(\"(ii) The input resistance of voltmeter with transistor=%dM\u03a9.\"%R_i_T);\n", + "print(\" The input resistance of voltmeter without transistor=%.1fk\u03a9.\"%R_i_WT);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) The meter current=1mA\n", + "(ii) The input resistance of voltmeter with transistor=1M\u03a9.\n", + " The input resistance of voltmeter without transistor=9.3k\u03a9.\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 22.8 : Page number 614-615\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "VCC=20; #Supply voltage, V\n", + "Rs_Rm=9.3; #Sum of multipier resistance and meter resistance, k\u03a9\n", + "I_m=1; #Meter current, mA\n", + "beta=100; #Base current amplification factor\n", + "E=5; #Voltage to be measured, V\n", + "V_BE=0.7; #Base-emitter voltage, V\n", + "\n", + "\n", + "#Calculation\n", + "I_m=(E-V_BE)/Rs_Rm; #Meter current, mA\n", + "\n", + "#Result\n", + "print(\"The meter current=%.2fmA\"%I_m);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The meter current=0.46mA\n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 22.9 : Page number 616\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt\n", + "from math import floor\n", + "\n", + "#Variable declaration\n", + "I_m_fsd=100.0; #Full scale deflection of meter current, \u03bcA\n", + "R_m=1.0; #Meter resistance, k\u03a9\n", + "V_rms=100.0; #r.m.s voltage to be measured, V\n", + "V_F=0.7; #Forward voltage drop of rectifier diode, V \n", + "\n", + "#Calculation\n", + "V_m=round(sqrt(2)*V_rms,1); #Peak value of applied voltage, V\n", + "V_rectifier_drop=2*V_F; #Total rectifier drop, V\n", + "I_peak=round(I_m_fsd/0.637,2); #Peak f.s.d current, \u03bcA\n", + "R_s=floor(((((V_m-V_rectifier_drop)/(I_peak*10**-6))-(R_m*1000))/1000)*10)/10; #Multiplier resistance, k\u03a9 (OHM's LAW)\n", + "\n", + "\n", + "#Result\n", + "print(\"The multiplier resistance=%.1fk\u03a9.\"%R_s);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The multiplier resistance=890.7k\u03a9.\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 22.10 : Page number 616\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt\n", + "\n", + "#Variable declaration\n", + "I_av=75; #Full scale deflection of meter current, \u03bcA\n", + "R_s=708; #Multiplier resistor, k\u03a9\n", + "R_m=900; #Meter coil resistor, \u03a9\n", + "\n", + "#Calculation\n", + "I_peak=I_av*10**-6/0.637; #Peak f.s.d meter current, A\n", + "R_T=R_s*1000+R_m; #Total circuit resistance, \u03a9\n", + "\n", + "#I_peak=(Vm-V_drop)/R_T; (OHM's LAW)\n", + "#And, Vm=sqrt(2)*Vrms\n", + "V_rms=(I_peak*R_T+(2*0.7))/sqrt(2) ; #applied r.m.s voltage, V\n", + "\n", + "\n", + "#Result\n", + "print(\"The applied r.m.s voltage=%dV\"%V_rms);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The applied r.m.s voltage=60V\n" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 22.11 : Page number 618\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "deflection_sensitivity=0.01; #Deflection sensitivity, mm/V\n", + "V=400; #Applied voltage, V\n", + "\n", + "#Calculation\n", + "spot_shift=V*deflection_sensitivity; #Spot shift produced, mm\n", + "\n", + "\n", + "#Result\n", + "print(\"The shift produced in the spot=%dmm.\"%spot_shift);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The shift produced in the spot=4mm.\n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 22.12 : Page number 618-619\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "deflection_sensitivity=0.03; #Deflection sensitivity, mm/V\n", + "spot_shift=3; #Spot shift produced, mm\n", + "\n", + "\n", + "#Calculation\n", + "#Since, spot_shift=Applied_Voltage*deflection_sensitivity,\n", + "V=spot_shift/deflection_sensitivity; #Applied voltage, V\n", + "\n", + "\n", + "#Result\n", + "print(\"Applied voltage=%dV.\"%V);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Applied voltage=100V.\n" + ] + } + ], + "prompt_number": 19 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 22.13 : Page number 622\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "deflection=2; #Deflection produced by applied voltage, cm\n", + "V=200; #Applied voltage, V\n", + "deflection_by_another_voltage=3; #Deflection by another voltage, cm\n", + "\n", + "#Calculation\n", + "deflection_sensitivity=V/deflection; #deflection sensitivity, V/cm\n", + "V_unknown=deflection_sensitivity*deflection_by_another_voltage; #Unknown voltage, V\n", + "\n", + "#Result\n", + "print(\"The unknown voltage=%dV.\"%V_unknown);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The unknown voltage=300V.\n" + ] + } + ], + "prompt_number": 20 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 22.14 : Page number 622\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "f_H=1000; #Frequency applied to horizontal plates, Hz\n", + "\n", + "#Calculation\n", + "#(i)\n", + "Loops_H=1; #Number of loops cut by horizontal line\n", + "Loops_V=1; #Number of loops cut by vertical line\n", + "f_V=f_H*(Loops_H/Loops_V); #Unknown frequency, Hz\n", + "\n", + "print(\"(i) Unknown frequency=%dHz.\"%f_V);\n", + "\n", + "#(ii)\n", + "Loops_H=2; #Number of loops cut by horizontal line\n", + "Loops_V=1; #Number of loops cut by vertical line\n", + "f_V=f_H*(Loops_H/Loops_V); #Unknown frequency, Hz\n", + "\n", + "print(\"(ii) Unknown frequency=%dHz.\"%f_V);\n", + "\n", + "#(iii)\n", + "Loops_H=6; #Number of loops cut by horizontal line\n", + "Loops_V=1; #Number of loops cut by vertical line\n", + "f_V=f_H*(Loops_H/Loops_V); #Unknown frequency, Hz\n", + "\n", + "print(\"(iii) Unknown frequency=%dHz.\"%f_V);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) Unknown frequency=1000Hz.\n", + "(ii) Unknown frequency=2000Hz.\n", + "(iii) Unknown frequency=6000Hz.\n" + ] + } + ], + "prompt_number": 21 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter23_2.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter23_2.ipynb new file mode 100644 index 00000000..19741354 --- /dev/null +++ b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter23_2.ipynb @@ -0,0 +1,133 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:87bd5c8d9448f5bb2e75909f89934a6fb2b64e65e6ea37b085ce080a58026071" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 23 : INTEGRATED CIRCUITS" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 23.1: Page number 637" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "R1=240; #Adjusted resistance of R2 resistor of LM317 voltage regulator, in kilo ohm\n", + "R2=2.4; #Fixed value of R1 resistor of LM317 voltage regulator, in ohm\n", + "\n", + "#Calculations\n", + "#Output voltage of LM317 voltage regulator IC = 1.25(R2/R1 +1)\n", + "Vout=1.25*((R2*1000)/R1 + 1); #Regulated d.c output voltage for the circuit in V\n", + "\n", + "#Results\n", + "print(\"The regulated d.c output voltage = %.2fV\"%Vout);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The regulated d.c output voltage = 13.75V\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 23.2 : Page number 638" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "R=1.2; #Value of resistance of monostable multivibrator in kilo ohm\n", + "C=0.1; #Value of capacitance of monostable multivibrator in microfarad\n", + "\n", + "#Calculations\n", + "T=1.1*(R*1000)*C; #Time for which the circuit is ON, in microseconds\n", + "\n", + "#Results\n", + "print(\"Time for which the circuit is ON = %d microseconds.\"%T); \n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Time for which the circuit is ON = 132 microseconds.\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 23.3 : Page number 639\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "R1=3.0; #Resistance of R1 resistor of 555 timer circuit in kilo ohm\n", + "R2=2.7; #Resistance of R2 resistor of 555 timer circuit in kilo ohm\n", + "C=0.033; #Capacitance of the capacitor of 555 timer circuit in microfarad\n", + "\n", + "#Calculations\n", + "f=1.44/(((R1*1000) + 2*(R2*1000))*(C*pow(10,-6))); #Frequency of the circuit in Hz\n", + "f=f/1000; #Frequency of the circuit in kHz\n", + "\n", + "#Results\n", + "print(\"The frequency of the circuit = %.2fkHz\"%f);\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The frequency of the circuit = 5.19kHz\n" + ] + } + ], + "prompt_number": 3 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter24_2.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter24_2.ipynb new file mode 100644 index 00000000..e63c17a6 --- /dev/null +++ b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter24_2.ipynb @@ -0,0 +1,604 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:947f358cf49d029c94d008f72a340051744678cf2e36ecc199100b78d31fcba5" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 24 : HYBRID PARAMETERS" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 24.1 : Page number 644-645\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "R1=10.0; #1st resistor, \u03a9\n", + "R2=5.0; #2nd resistor, \u03a9\n", + "\n", + "\n", + "#Calculation\n", + "print(\"To find h11 and h21, output terminals are shorted.\");\n", + "h11=R1; #Input impedance with output shorted, \u03a9\n", + "\n", + "print(\"h11=%d\u03a9.\"%h11);\n", + "\n", + "print(\"Output current flowing into the box= input current flowing out of the box.\");\n", + "print(\"i2=-i1\"); #Output current flowing into the box= input current flowing out of the box.\n", + "print(\"h21=i2/i1 = -i1/i1= -1.\"); #Current gain with output shorted.\n", + "\n", + "\n", + "print(\"For finding h22 and h12, voltage source is connected at the output\");\n", + "#As, there will be no current through 10k\u03a9 resistor due to open circuited input,\n", + "print(\"v1=v2\"); #Output voltage is equal to input voltage(equal to voltage drop across 5k\u03a9 resistor)\n", + "print(\"h12=v1/v2 = v2/v2 = 1\"); #Voltage feedback ratio with input terminals open\n", + "\n", + "h22=1/R2; #Output admittance, mho\n", + "print(\"h22=%.1f mho\"%h22);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "To find h11 and h21, output terminals are shorted.\n", + "h11=10\u03a9.\n", + "Output current flowing into the box= input current flowing out of the box.\n", + "i2=-i1\n", + "h21=i2/i1 = -i1/i1= -1.\n", + "For finding h22 and h12, voltage source is connected at the output\n", + "v1=v2\n", + "h12=v1/v2 = v2/v2 = 1\n", + "h22=0.2 mho\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 24.2 : Page number 645-646\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "R1=4.0; #1st resistor(at the input side), \u03a9\n", + "R2=4.0; #2nd resistor(at the middle), \u03a9\n", + "R3=4.0; #3rd resistor(at the output side), \u03a9\n", + "\n", + "#Calculation\n", + "print(\"To find h11 and h21, output terminals are shorted.\");\n", + "h11=R1 + (R2*R3/(R2+R3)); #Input impedance with output shorted, \u03a9\n", + "print(\"h11=%d\u03a9.\"%h11);\n", + "\n", + "#As the input current gets divided in half due to R2=R3.\n", + "print(\"Output current flowing into the box=negative of half of input current flowing out of the box.\");\n", + "print(\"i2=-i1/2 = -0.5i1\"); \n", + "print(\"h21=i2/i1 = -0.5i1/i1= -0.5.\"); #Current gain with output shorted.\n", + "\n", + "print(\"For finding h22 and h12, voltage source is connected at the output\");\n", + "#As, there will be no current through the 1st 4k\u03a9 resistor due to open circuited input,\n", + "#Voltage gets equally divided across R2 and R3 resistor\n", + "print(\"v1=v2/2 = 0.5v2\"); #Input voltage is equal to half of input voltage\n", + "print(\"h12=v1/v2 = 0.5v2/v2 = 0.5\"); #Voltage feedback ratio with input terminals open\n", + "\n", + "h22=1/(R2+R3); #Output admittance, mho\n", + "print(\"h22=%.3f mho\"%h22);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "To find h11 and h21, output terminals are shorted.\n", + "h11=6\u03a9.\n", + "Output current flowing into the box=negative of half of input current flowing out of the box.\n", + "i2=-i1/2 = -0.5i1\n", + "h21=i2/i1 = -0.5i1/i1= -0.5.\n", + "For finding h22 and h12, voltage source is connected at the output\n", + "v1=v2/2 = 0.5v2\n", + "h12=v1/v2 = 0.5v2/v2 = 0.5\n", + "h22=0.125 mho\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 24.3 ; Page number 649-650\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "R1=10.0; #Resistor at the input side, \u03a9\n", + "R2=5.0; #Resistor at the middle, \u03a9\n", + "rL=5.0; #Load resistor, \u03a9\n", + "\n", + "#h-parameter values from 24.1\n", + "h11=10.0; #Input impedance with output shorted, \u03a9\n", + "h21=-1.0; #Current gain with output shorted\n", + "h12=1.0; #Voltage feedback ratio with input terminal open\n", + "h22=0.2; #Output admittance, mho\n", + "\n", + "#Calculation\n", + "#(i)\n", + "Zin=h11-(h12*h21/(h22+(1/rL))); #Input impedance, \u03a9\n", + "\n", + "#(ii)\n", + "Av=-h21/(Zin*(h22+(1/rL))); #voltage gain,\n", + "\n", + "#Result\n", + "print(\"(i) The input impedance=%.1f\u03a9.\"%Zin );\n", + "print(\"(ii) The voltage gain=1/%d.\"%(1/Av));\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) The input impedance=12.5\u03a9.\n", + "(ii) The voltage gain=1/5.\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 24.4 : Page number 652-653\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "VCE=10.0; #Collector-emitter voltage, V\n", + "IC=1.0; #Collector current, mA\n", + "rL=600.0; #a.c load seen by the transistor,\u03a9\n", + "\n", + "#h-parameters\n", + "hie=2000.0; #Input impedance with output shorted, \u03a9\n", + "hoe=10**-4; #Output impedance, mho\n", + "hre=10**-3; #Voltage feedback ratio with input terminal open\n", + "hfe=50.0; #Current gain with output shorted\n", + "\n", + "\n", + "#Calculation\n", + "#(i)\n", + "Zin=hie - (hre*hfe/(hoe+(1/rL))); #Input impedance, \u03a9\n", + "print(\"Input impedance=%.0f \u03a9. \\n As second term in the expression of Zin is small compared to first, Zin~hie=%d\u03a9.\"%(Zin,hie));\n", + "\n", + "#(ii)\n", + "Ai=hfe/(1+hoe*rL); #Current gain\n", + "print(\"Current gain=%d\"%Ai);\n", + "print(\"if hoe*rL<<1, then Ai~hfe=%d.\"%hfe);\n", + "\n", + "#(iii)\n", + "Av=-hfe/(Zin*(hoe+(1/rL))); #Voltage gain\n", + "print(\"Voltage gain=%.1f\"%Av);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Input impedance=1972 \u03a9. \n", + " As second term in the expression of Zin is small compared to first, Zin~hie=2000\u03a9.\n", + "Current gain=47\n", + "if hoe*rL<<1, then Ai~hfe=50.\n", + "Voltage gain=-14.4\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 24.5 : Page number 653\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import ceil\n", + "\n", + "#Variable declaration\n", + "VCE=5.0; #Collector-emitter voltage, V\n", + "IC=1.0; #Collector current, mA\n", + "rL=2.0; #a.c load seen by the transistor,\u03a9\n", + "\n", + "\n", + "#h-parameters\n", + "hie=1700.0; #Input impedance with output shorted, \u03a9\n", + "hoe=6*10**-6; #Output impedance, mho\n", + "hre=1.3*10**-4; #Voltage feedback ratio with input terminal open\n", + "hfe=38.0; #Current gain with output shorted\n", + "\n", + "#Calculation\n", + "#(i)\n", + "Zin=hie - (hre*hfe/(hoe+(1/(rL*1000)))); #Input impedance, \u03a9\n", + "print(\"Input impedance=%.0f \u03a9.\"%Zin);\n", + "\n", + "#(ii)\n", + "Ai=ceil((hfe/round((1+hoe*rL*1000),3))*10)/10; #Current gain\n", + "print(\"Current gain=%.1f\"%Ai);\n", + "\n", + "#(iii)\n", + "Av=-hfe/(Zin*(hoe+(1/(rL*1000)))); #Voltage gain\n", + "print(\"Voltage gain=%.1f\"%abs(Av));\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Input impedance=1690 \u03a9.\n", + "Current gain=37.6\n", + "Voltage gain=44.4\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 24.6 : Page number 653-654\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Function for calculating parallel resistance\n", + "def pr(r1,r2):\n", + " return r1*r2/(r1+r2);\n", + "\n", + "\n", + "#Variable declaration\n", + "RC=10.0; #Collector resistance, k\u03a9\n", + "RL=30.0; #Load resistance, k\u03a9\n", + "R1=80.0; #Resistor R1, k\u03a9\n", + "R2=40.0; #Resistor R2, k\u03a9\n", + "\n", + "#h-parameters\n", + "hie=1500.0; #Input impedance with output shorted, \u03a9\n", + "hoe=5*10**-5; #Output impedance, mho\n", + "hre=4*10**-4; #Voltage feedback ratio with input terminal open\n", + "hfe=50.0; #Current gain with output shorted\n", + "\n", + "\n", + "#Calculation\n", + "rL=((RC*RL)/(RC+RL))*1000; #a.c load as seen by resistance, \u03a9\n", + "\n", + "#(i)\n", + "Zin=round(hie - (hre*hfe/(hoe+(1/rL))),-1); #Input impedance, \u03a9\n", + "print(\"Input impedance=%.0f \u03a9.\"%Zin);\n", + "\n", + "#Input impedance of stage=input impedance || bias resistors\n", + "Zin_stage=round(pr(pr(R1,R2)*1000,Zin),-1); #\u03a9\n", + "print(\"Input impedance of the stage=%.0f \u03a9.\"%Zin_stage);\n", + "\n", + "#(ii)\n", + "Av=-hfe/(Zin*(hoe+(1/rL))); #Voltage gain\n", + "print(\"Voltage gain=%d\"%Av);\n", + "print(\"The negative sign represents phase reversal.\");\n", + "\n", + "\n", + "#(iii)\n", + "Zout=(1/(hoe-(hfe*hre/hie)))/1000; #Output impedance of transistor, k\u03a9\n", + "Zout_stage=pr(Zout,pr(RL,RC)); #Output impedance of the stage, k\u03a9\n", + "print(\"Output impedance=%.2f k\u03a9.\"%Zout);\n", + "print(\"Output impedance of the stage=%.2f k\u03a9.\"%Zout_stage);\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Input impedance=1390 \u03a9.\n", + "Input impedance of the stage=1320 \u03a9.\n", + "Voltage gain=-196\n", + "The negative sign represents phase reversal.\n", + "Output impedance=27.27 k\u03a9.\n", + "Output impedance of the stage=5.88 k\u03a9.\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 24.7 : Page number 654\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Function for calculating parallel resistance\n", + "def pr(r1,r2):\n", + " return r1*r2/(r1+r2);\n", + "\n", + "#Variable declaration\n", + "RC=4.7; #Collector resistance, k\u03a9\n", + "RL=10.0; #Load resistance, k\u03a9\n", + "R1=33.0; #Resistor R1, k\u03a9\n", + "R2=10.0; #Resistor R2, k\u03a9\n", + "\n", + "#h-parameters\n", + "hie=1; #Input impedance with output shorted, k\u03a9\n", + "hoe=25; #Output impedance, \u03bcS\n", + "hre=2.5*10**-4; #Voltage feedback ratio with input terminal open\n", + "hfe=50; #Current gain with output shorted\n", + "\n", + "\n", + "#Calculation\n", + "rL=(RC*RL)/(RC+RL); #a.c load as seen by resistance, k\u03a9\n", + "\n", + "Ai=hfe/(1+hoe*10**-6*rL*1000); #Current gain\n", + "print(\"Current gain=%.1f\"%Ai);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Current gain=46.3\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 24.8 : Page number 654-655\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "R_S=100.0; #Series resistance, \u03a9 \n", + "\n", + "#h-parameters\n", + "hie=1.0; #Input impedance with output shorted, k\u03a9\n", + "hoe=25.0; #Output impedance, \u03bcS\n", + "hre=2.5*10**-4; #Voltage feedback ratio with input terminal open\n", + "hfe=50.0; #Current gain with output shorted\n", + "\n", + "\n", + "#Calculation\n", + "Zout=(1/(hoe*10**-6-(hfe*hre/(hie*1000+R_S))))/1000; #Output impedance of transistor, k\u03a9\n", + "print(\"Output impedance=%.1f k\u03a9.\"%Zout);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Output impedance=73.3 k\u03a9.\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 24.9 : Page number 656\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import floor\n", + "\n", + "#Function for calculating parallel resistance\n", + "def pr(r1,r2):\n", + " return r1*r2/(r1+r2);\n", + "\n", + "#Variable declaration\n", + "RC=12.0; #Collector resistance, k\u03a9\n", + "RL=15.0; #Load resistance, k\u03a9\n", + "R1=50.0; #Resistor R1, k\u03a9\n", + "R2=5.0; #Resistor R2, k\u03a9\n", + "hie=1.94; #Input impedance with output shorted, k\u03a9\n", + "hfe=71.0; #Current gain with output shorted\n", + "\n", + "\n", + "\n", + "#Calculation\n", + "rL=(RC*RL)/(RC+RL); #a.c load as seen by resistance, \u03a9\n", + "\n", + "#(i)\n", + "Zin_base=hie; #Transistor input impedance, k\u03a9\n", + "Zin_circuit=floor(pr(Zin_base,pr(R1,R2))*100)/100; #Circuit input impedance, k\u03a9\n", + "print(\"Circuit input impedance=%.2fk\u03a9\"%Zin_circuit);\n", + "\n", + "\n", + "#(ii)\n", + "Av=hfe*rL/hie; #Voltage gain\n", + "print(\"Voltage gain=%.0f\"%Av);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Circuit input impedance=1.35k\u03a9\n", + "Voltage gain=244\n" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 24.10 : Page number 656\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt\n", + "\n", + "#Variable declaration\n", + "hie_min=600; #Minimum input impedance with output shorted, \u03a9\n", + "hfe_min=110; #Minimum current gain with output shorted\n", + "hie_max=800; #Maximum input impedance with output shorted, \u03a9\n", + "hfe_max=140; #Maximum current gain with output shorted\n", + "rL=460; #a.c collector load, \u03a9\n", + "\n", + "#Calculation\n", + "hie=round(sqrt(hie_min*hie_max)); #Input impedance with output shorted, \u03a9\n", + "hfe=round(sqrt(hfe_min*hfe_max)); #Current gain with output shorted\n", + "Av=hfe*rL/hie; #Voltage gain\n", + "\n", + "#Result\n", + "print(\"Voltage gain=%.1f\"%Av);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Voltage gain=82.3\n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 24.11 : Page number 658-659\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#(a)Variable declaration\n", + "Ib=10; #Base current, \u03bcA\n", + "Ic=1; #Collector current, mA\n", + "Vbe=10; #Base-emitter voltage, mV\n", + "\n", + "#Calculation\n", + "hie=Vbe*10**-3/(Ib*10**-6); #Input impedance with output shorted, \u03a9\n", + "hfe=Ic*10**-3/(Ib*10**-6); #Current gain with output shorted\n", + "\n", + "#(b) Variable declaration\n", + "Vbe=0.65; #Base-emitter voltage, mV\n", + "Ic=60; #Collector current, \u03bcA\n", + "Vce=1; #Collector-emitter voltage, V\n", + "\n", + "#Calculation\n", + "hre=Vbe*10**-3/Vce; #Voltage feedback ratio with input terminal open\n", + "hoe=Ic/Vce; #Output impedance, \u03bcmho\n", + "\n", + "\n", + "#Result\n", + "print(\"hie=%d\u03a9\"%hie);\n", + "print(\"hfe=%d\"%hfe);\n", + "print(\"hre=%.2fe\u201303\"%(hre*1000));\n", + "print(\"hoe=%d\u03bcmho\"%hoe);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "hie=1000\u03a9\n", + "hfe=100\n", + "hre=0.65e\u201303\n", + "hoe=60\u03bcmho\n" + ] + } + ], + "prompt_number": 20 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter25_2.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter25_2.ipynb new file mode 100644 index 00000000..d24cf79d --- /dev/null +++ b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter25_2.ipynb @@ -0,0 +1,2512 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# CHAPTER 25 : OPERATIONAL AMPLIFIERS" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": true + }, + "outputs": [], + "source": [ + "%matplotlib inline" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.1: Page number 664" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The output voltage of the differential amplifier = 10V\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "A=100.0; #Open-circuit voltage gain of differential amplifier\n", + "V1=3.25; #Input voltage to terminal 1 in V\n", + "V2=3.15; #Input voltage to terminal 2 in V\n", + "\n", + "#Calculations\n", + "V0=A*(V1-V2); #Output voltage in V\n", + "\n", + "#Results\n", + "print(\"The output voltage of the differential amplifier = %dV\"%V0);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.2: Page number 672" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The common mode rejection ratio = 10000.\n", + "The common mode rejection ratio in decibels= 80dB.\n" + ] + } + ], + "source": [ + "from math import log10\n", + "\n", + "#Variable declaration\n", + "A_DM=2000.0; #Differential mode voltage gain\n", + "A_CM=0.2; #Common mode voltage gain\n", + "\n", + "#Calculations\n", + "CMRR=A_DM/A_CM; #Common mode rejection ratio\n", + "CMRR_dB=20*log10(CMRR); #Common mode rejection ratio in dB\n", + "\n", + "\n", + "#Results\n", + "print(\"The common mode rejection ratio = %d.\"%CMRR);\n", + "print(\"The common mode rejection ratio in decibels= %ddB.\"%CMRR_dB);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.3: Page number 672" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The common mode rejection ratio in decibels= 46dB.\n" + ] + } + ], + "source": [ + "from math import log10\n", + "\n", + "#Variable declaration\n", + "VD_in=10.0; #Differential mode input in mV\n", + "VD_out=1.0; #Output for differential mode input in V\n", + "VC_in=10.0; #Common mode input in mV\n", + "VC_out=5.0; #Output for common mode input in mV\\\n", + "\n", + "#Calculations\n", + "A_DM=(VD_out*1000)/VD_in; #Differntial mode voltage gain\n", + "A_CM=VC_out/VC_in; #Common mode voltage gain\n", + "CMRR=A_DM/A_CM; #Common mode rejection ratio\n", + "CMRR_dB=20*log10(CMRR); #Common mode rejection ratio in dB\n", + "\n", + "#Results\n", + "print(\"The common mode rejection ratio in decibels= %ddB.\"%CMRR_dB);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.4: Page number 672" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Output voltage =7.5V\n", + "Noise on output = 4.7x10^-6V\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "A_DM=150.0; #Differential mode voltage gain\n", + "CMRR_dB=90.0; #Common mode rejection ratio\n", + "V1=100.0; #Input voltage for terminal 1 in mV\n", + "V2=50.0; #Input voltage for terminal 2 in mV\n", + "V_noise=1.0; #Voltage of noise signal in mV\n", + "\n", + "#Calculation\n", + "\n", + "#Case(i)\n", + "V_out=A_DM*(V1-V2)/1000.0; #Output voltage for differntial mode input, in V\n", + "\n", + "#Since CMRR_dB=20*log10(differential mode gain/common mode gain),\n", + "A_CM=A_DM/pow(10,(CMRR_dB/20)); #Common mode gain\n", + "V_OUT_noise=A_CM*(V_noise/1000); #Noise on output in V\n", + "\n", + "\n", + "#Results\n", + "print(\"Output voltage =%.1fV\"%V_out);\n", + "print(\"Noise on output = %.1fx10^-6V\"%(V_OUT_noise*pow(10,6)));\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.5 : Page number 672-673" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Common mode gain =0.083\n", + "Common mode rejection ratio in decibels=89.5dB\n", + "r.m.s output signal =1.25V\n", + "r.m.s interfernce output voltage = 83mV\n" + ] + } + ], + "source": [ + "from math import log10\n", + "\n", + "#Variable declaration\n", + "A_DM=2500.0; #Differential mode voltage gain\n", + "CMRR=30000.0; #Common mode rejection ratio\n", + "Input_signal=500.0; #Single ended input r.m.s signal in microvolts\n", + "Interference=1.0; #Interference signal, in V\n", + "\n", + "#Calculations\n", + "\n", + "#(i)\n", + "A_CM=A_DM/CMRR; #Common mode gain\n", + "\n", + "#(ii)\n", + "CMRR_dB=20*log10(CMRR); #Common mode rejection ratio in decibels\n", + "\n", + "#(iii)\n", + "V_out=A_DM*(Input_signal/pow(10,6)-0); #r.m.s output signal in V\n", + "\n", + "#(iv)\n", + "Interference_out=A_CM*Interference; #r.m.s interference output in V\n", + "Interference_out=Interference_out*1000; #r.m.s interference output in mV\n", + "\n", + "\n", + "#Results\n", + "print(\"Common mode gain =%.3f\"%A_CM);\n", + "print(\"Common mode rejection ratio in decibels=%.1fdB\"%CMRR_dB);\n", + "print(\"r.m.s output signal =%.2fV\"%V_out);\n", + "print(\"r.m.s interfernce output voltage = %dmV\"%Interference_out);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.6 : Page number 674-675" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "VE=-0.7V\n", + "IE=0.452mA\n", + "IE1=0.226mA\n", + "IE2=0.226mA\n", + "IC1=0.226mA\n", + "IC2=0.226mA\n", + "IB1=2.26μA\n", + "IB2=2.26μA\n", + "VC1=12V\n", + "VC2=9.7V\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "VCC=12; #Collector supply voltage, V\n", + "VEE=12; #Emitter supply voltage, V\n", + "RB=10; #Base resistor, kΩ\n", + "RC2=10; #Collector resistor, kΩ\n", + "RE=25; #Emitter resistor, kΩ\n", + "VBE=0.7; #Base-emitter voltage, V\n", + "beta=100; #Base amplification factor\n", + "\n", + "#Calculation\n", + "VE=-VBE; #Emitter voltage, V (Ignoring the base current)\n", + "IE=(VEE-VBE)/RE; #Tail current, mA\n", + "IE1=IE/2; #Emitter current of 1st transistor, mA\n", + "IE2=IE1; #Emitter current of 2nd transistor, mA\n", + "IC1=IE1; #Collector current(= emitter current) of 1st transistor, mA\n", + "IC2=IC1; #Collector current of 2nd transistor, mA\n", + "IB1=(IC1/beta)*1000; #Base current of 1st transistor, μA\n", + "IB2=IB1; #Base current of 2nd transistor, μA\n", + "VC1=VCC; #Collector voltage of 1st transistor, V\n", + "VC2=VCC-IC2*RC2; #Collector voltage of 2nd transistor, V\n", + "\n", + "#Result\n", + "print(\"VE=%.1fV\"%VE);\n", + "print(\"IE=%.3fmA\"%IE);\n", + "print(\"IE1=%.3fmA\"%IE1);\n", + "print(\"IE2=%.3fmA\"%IE2);\n", + "print(\"IC1=%.3fmA\"%IC1);\n", + "print(\"IC2=%.3fmA\"%IC2);\n", + "print(\"IB1=%.2fμA\"%IB1);\n", + "print(\"IB2=%.2fμA\"%IB2);\n", + "print(\"VC1=%dV\"%VC1);\n", + "print(\"VC2=%.1fV\"%VC2);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.7 : Page number 675" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The output voltage=7.85V.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "VCC=15; #Collector supply voltage, V\n", + "VEE=15; #Emitter supply voltage, V\n", + "RB=33; #Base resistor, kΩ\n", + "RC=15; #Collector resistor, kΩ\n", + "RE=15; #Emitter resistor, kΩ\n", + "VBE=0.7; #Base-emitter voltage, V\n", + "\n", + "\n", + "#Calculation\n", + "IE_tail=(VEE-VBE)/RE; #Tail current, mA\n", + "IE=round(IE_tail/2,3); #Emitter current in each transistor, mA\n", + "IC=IE; #Collector current(=emitter current), mA\n", + "Vout=VCC-IC*RC; #Output voltage, V\n", + "\n", + "#Result\n", + "print(\"The output voltage=%.2fV.\"%Vout);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.8 : Page number 675" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) IB1=5.56μA\n", + " IB2=4.55μA\n", + "(ii) VB1=-0.183V\n", + " VB2=-0.15V\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "VCC=15.0; #Collector supply voltage, V\n", + "VEE=15.0; #Emitter supply voltage, V\n", + "RB=33.0; #Base resistor, kΩ\n", + "RC=15.0; #Collector resistor, kΩ\n", + "RE=15.0; #Emitter resistor, kΩ\n", + "VBE=0; #Base-emitter voltage, V\n", + "beta_dc_l=90.0; #base current amplification factor for left transistor\n", + "beta_dc_r=110.0; #base current amplification factor for right transistor\n", + "\n", + "\n", + "#Calculation\n", + "#(i)\n", + "IE_tail=(VEE-VBE)/RE; #Tail current, mA\n", + "IE=IE_tail/2; #Emitter current in each transistor, mA\n", + "IB1=(IE/beta_dc_l)*1000; #Base current of 1st transistor, μA\n", + "IB2=(IE/beta_dc_r)*1000; #Base current of 2nd transistor, μA\n", + "\n", + "#(ii)\n", + "VB1=-IB1/1000*RB; #Base voltage of 1st transistor, V\n", + "VB2=-IB2/1000*RB; #Base voltage of 1st transistor, V\n", + "\n", + "#Result\n", + "print(\"(i) IB1=%.2fμA\"%IB1);\n", + "print(\" IB2=%.2fμA\"%IB2);\n", + "print(\"(ii) VB1=%.3fV\"%VB1);\n", + "print(\" VB2=%.2fV\"%VB2);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.9 : Page number 675-676" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "VE=-0.7V\n", + "Emitter current in each transistor=0.5mA.\n", + "IC1~IE1=0.5mA and IC2~IE2=0.5mA\n", + "VC1=VC2=10V.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "VCC=15.0; #Collector supply voltage, V\n", + "VEE=15.0; #Emitter supply voltage, V\n", + "RB=10.0; #Base resistor, kΩ\n", + "RC1=10.0; #Collector resistor of 1st transistor, kΩ\n", + "RC2=10.0; #Collector resistor of 2nd transistor, kΩ\n", + "IE=1.0; #Tail current, mA\n", + "VBE=0.7; #Base-emitter voltage, V\n", + "\n", + "\n", + "#Calculation\n", + "VE=-VBE; #Emitter voltage, V (Ignoring the base current)\n", + "IE1=IE/2.0; #Emitter current of 1st transistor, mA\n", + "IE2=IE1; #Emitter current of 2nd transistor, mA\n", + "IC1=IE1; #Collector current(= emitter current) of 1st transistor, mA\n", + "IC2=IE2; #Collector current of 2nd transistor, mA\n", + "VC1=VCC-IC1*RC1; #Collector voltage of 1st transistor, V\n", + "VC2=VCC-IC2*RC2; #Collector voltage of 2nd transistor, V\n", + "\n", + "\n", + "#Result\n", + "print(\"VE=%.1fV\"%VE);\n", + "print(\"Emitter current in each transistor=%.1fmA.\"%(IE/2.0));\n", + "print(\"IC1~IE1=%.1fmA and IC2~IE2=%.1fmA\"%(IE1,IE2));\n", + "print(\"VC1=VC2=%dV.\"%VC2);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.10 : Page number 676-677" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "VE=0.7V\n", + "Tail current=0.452mA.\n", + "Emitter current in each transistor=0.226mA.\n", + "IC1~IE1=0.226mA and IC2~IE2=0.226mA\n", + "VC1=-12V\n", + "VC2=-9.74V\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "VCC=12.0; #Collector supply voltage, V\n", + "VEE=12.0; #Emitter supply voltage, V\n", + "RC2=10.0; #Collector resistor of 2nd transistor, kΩ\n", + "RE=25.0; #Emitter current, kΩ\n", + "VBE=-0.7; #Base-emitter voltage, V\n", + "\n", + "\n", + "#Calculation\n", + "VE=-VBE; #Emitter voltage, V (Ignoring the base current)\n", + "IE=(VCC-VE)/RE; #Tail current, mA\n", + "IE1=IE/2.0; #Emitter current of 1st transistor, mA\n", + "IE2=IE1; #Emitter current of 2nd transistor, mA\n", + "IC1=IE1; #Collector current(= emitter current) of 1st transistor, mA\n", + "IC2=IE2; #Collector current of 2nd transistor, mA\n", + "VC1=-VEE; #Collector voltage of 1st transistor, V\n", + "VC2=-VEE+IC2*RC2; #Collector voltage of 2nd transistor, V\n", + "\n", + "\n", + "#Result\n", + "print(\"VE=%.1fV\"%VE);\n", + "print(\"Tail current=%.3fmA.\"%IE);\n", + "print(\"Emitter current in each transistor=%.3fmA.\"%(IE/2.0));\n", + "print(\"IC1~IE1=%.3fmA and IC2~IE2=%.3fmA\"%(IC1,IC2));\n", + "print(\"VC1=%dV\"%VC1);\n", + "print(\"VC2=%.2fV\"%VC2);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.11 : Page number 679" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) The input offset current=15.1nA\n", + "(ii) The input bias current=75.8nA\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "VCC=15; #Collector supply voltage, V\n", + "VEE=15; #Emitter supply voltage, V\n", + "RB=1; #Base resistor, MΩ\n", + "RC2=1; #Collector resistor, MΩ\n", + "RE=1; #Emitter resistor, MΩ\n", + "VBE=0; #Base-emitter voltage, V (Neglected)\n", + "beta_dc_l=90.0; #base current amplification factor for left transistor\n", + "beta_dc_r=110.0; #base current amplification factor for right transistor\n", + "\n", + "\n", + "#Calculation\n", + "#(i)\n", + "IE=(VEE-VBE)/RE; #Tail current, μA\n", + "IE1=IE/2.0; #Emitter current of 1st transistor, μA\n", + "IE2=IE1; #Emitter current of 2nd transistor, μA\n", + "IB1=round((IE1/beta_dc_l)*1000,1); #Base current of 1st transistor, nA\n", + "IB2=round((IE2/beta_dc_r)*1000,1); #Base current of 2nd transistor, nA\n", + "I_in_offset=IB1-IB2; #Input offset current, nA\n", + "\n", + "#(ii)\n", + "I_in_bias=(IB1+IB2)/2; #Input bias current, nA\n", + "\n", + "#Result\n", + "print(\"(i) The input offset current=%.1fnA\"%I_in_offset);\n", + "print(\"(ii) The input bias current=%.1fnA\"%I_in_bias);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.12 : Page number 679" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The two base currents are: IB1=90nA and IB2=70nA.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "I_in_offset=20; #Input offset current, nA\n", + "I_in_bias=80; #Input bias current, nA\n", + "\n", + "#Calculation\n", + "IB1=I_in_bias+I_in_offset/2; #Base current in 1st transistor, nA\n", + "IB2=I_in_bias-I_in_offset/2; #Base current in 2nd transistor, nA\n", + "\n", + "\n", + "#Result\n", + "print(\"The two base currents are: IB1=%dnA and IB2=%dnA.\"%(IB1,IB2));\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.13 : Page number 679-680" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The input offset voltage=2mV.\n", + "The output offset voltage=0.3V.\n" + ] + } + ], + "source": [ + "#Variable declration\n", + "I_in_offset=20; #Input offset current, nA\n", + "I_in_bias=80; #Input bias current, nA\n", + "A=150; #Voltage gain\n", + "RB=100; #Base resistor, kΩ\n", + "\n", + "\n", + "#Calculation\n", + "V_io=(I_in_offset*10**-9*RB*1000)*1000; #Input offset voltage, mV\n", + "V_out_offset=(A*V_io)/1000; #Output offset voltage, V\n", + "\n", + "#Result\n", + "print(\"The input offset voltage=%dmV.\"%V_io);\n", + "print(\"The output offset voltage=%.1fV.\"%V_out_offset);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.14 : Page number 682" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) Output voltage=0.15V.\n", + "(ii) Output voltage=-0.15V.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "VCC=15; #Collector supply voltage, V\n", + "VEE=15; #Emitter supply voltage, V\n", + "RE=1; #Emitter resistor, MΩ\n", + "RC=1; #Collector resistor, MΩ\n", + "\n", + "\n", + "#Calculation\n", + "IE=VEE/RE; #Tail current, μA\n", + "IE1=IE/2.0; #Emitter current of 1st transistor, μA\n", + "IE2=IE1; #Emitter current of 2nd transistor, μA\n", + "re=25/IE1; #a.c emitter resistance, kΩ\n", + "A_DM=RC/(2.0*re); #Differential voltage gain,\n", + "\n", + "#(i)\n", + "vin=1; #Input voltage, V\n", + "Vout=A_DM*vin; #Output voltage, V\n", + "\n", + "print(\"(i) Output voltage=%.2fV.\"%Vout);\n", + "\n", + "#(ii)\n", + "vin=-1; #Input voltage, V\n", + "Vout=A_DM*vin; #Output voltage, V;\n", + "print(\"(ii) Output voltage=%.2fV.\"%Vout);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.15 : Page number 682-683" + ] + }, + { + "cell_type": "code", + "execution_count": 16, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) The input impedance=194kΩ.\n", + "(ii) The differential voltage gain=136.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "VCC=12; #Collector supply voltage, V\n", + "VEE=12; #Emitter supply voltage, V\n", + "RE=100; #Emitter resistor, kΩ\n", + "RC1=120; #Collector resistor of 1st transistor, kΩ\n", + "RC2=120; #Collector resistor of 2nd transistor, kΩ\n", + "beta=220; #Base amplification factor\n", + "VBE=0.7; #Base-emitter voltage, V\n", + "\n", + "\n", + "#Calcualtion\n", + "IE=((VEE-VBE)/RE)*1000; #Tail current, μA\n", + "IE1=IE/2.0; #Emitter current of 1st transistor, μA\n", + "IE2=IE1; #Emitter current of 2nd transistor, μA\n", + "re=(25/IE1)*1000; #a.c emitter resistance, Ω\n", + "Zin=2*beta*re/1000; #Input impedance, kΩ\n", + "A_DM=RC1*1000/(2.0*re); #Differential voltage gain,\n", + "\n", + "\n", + "#Result\n", + "print(\"(i) The input impedance=%dkΩ.\"%Zin);\n", + "print(\"(ii) The differential voltage gain=%.0f.\"%A_DM);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.16: Page number 683-684" + ] + }, + { + "cell_type": "code", + "execution_count": 17, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Differential voltage gain=56.6.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "VCC=12; #Collector supply voltage, V\n", + "VEE=12; #Emitter supply voltage, V\n", + "RE=200; #Emitter resistor, kΩ\n", + "RC=100; #Collector resistor, kΩ\n", + "VBE=0.7; #Base-emitter voltage, V\n", + "\n", + "#Calculation\n", + "IE=round((VEE-VBE)/RE,4); #Tail current, mA\n", + "IE1=round(IE/2,4); #Emitter current of 1st transistor, mA\n", + "IE2=IE1; #Emitter current of 2nd transistor, mA\n", + "re=round(25/IE1,1); #a.c emitter resistance, Ω\n", + "A_DM=RC*1000/(2*re); #Differential voltage gain,\n", + "\n", + "#Result\n", + "print(\"Differential voltage gain=%.1f.\"%A_DM);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.17 : Page number 685-686" + ] + }, + { + "cell_type": "code", + "execution_count": 18, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Common mode rejection ratio=666.7.\n", + "Common mode rejection ratio in decibel=56.48dB\n" + ] + } + ], + "source": [ + "from math import log10\n", + "\n", + "#Variable declaration\n", + "v1=0.5; #Voltage in terminal 1, mV\n", + "v2=-0.5; #Voltage in terminal 2, mV\n", + "vo=8.0; #Output voltage, V\n", + "vo_cm=12.0; #Common mode output, mV\n", + "\n", + "#Calculation\n", + "vin=v1-v2; #Differential input, mV\n", + "A_DM=vo/(vin/1000.0); #Differential mode gain,\n", + "vin_cm=1; #Common mode input, mV\n", + "A_CM=vo_cm/vin_cm; #Common mode gain\n", + "CMRR=A_DM/A_CM; #Common mode rejection ratio\n", + "CMRR_dB=20*log10(CMRR); #Common mode rejection ratio in dB\n", + "\n", + "#Result\n", + "print(\"Common mode rejection ratio=%.1f.\"%CMRR)\n", + "print(\"Common mode rejection ratio in decibel=%.2fdB\"%CMRR_dB);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.18 : Page number 686" + ] + }, + { + "cell_type": "code", + "execution_count": 19, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Common mode voltage gain=6.32.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "A_DM=200000; #Differential mode gain\n", + "CMRR_dB=90; #Common mode rejection ratio, dB\n", + "\n", + "#Calculation\n", + "CMRR=10**(CMRR_dB/20.0); #Common mode rejection ratio\n", + "A_CM=A_DM/CMRR; #Common mode gain\n", + "\n", + "\n", + "\n", + "#Result\n", + "print(\"Common mode voltage gain=%.2f.\"%A_CM);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.19 : Page number 686" + ] + }, + { + "cell_type": "code", + "execution_count": 20, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) The Common mode gain=0.0081\n", + "(ii) The common mode rejection ratio=81.8dB.\n" + ] + } + ], + "source": [ + "from math import log10\n", + "\n", + "#Variable declaration\n", + "vin_cm=3.2; #Common input voltage, V\n", + "vout=26; #Output voltage, V\n", + "A_DM=100; #Open-circuit voltage gain\n", + "\n", + "#Calculation\n", + "#(i)\n", + "A_CM=vout*10**-3/vin_cm; #Common mode gain\n", + "\n", + "#(ii)\n", + "CMRR_dB=20*log10(A_DM/A_CM); #Common mode rejection ratio, dB\n", + "\n", + "#Result\n", + "print(\"(i) The Common mode gain=%.4f\"%A_CM);\n", + "print(\"(ii) The common mode rejection ratio=%.1fdB.\"%CMRR_dB);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.20 : Page number 686-687" + ] + }, + { + "cell_type": "code", + "execution_count": 21, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) Common mode gain=0.25\n", + "(ii)Common mode rejection ratio=47.09dB\n" + ] + } + ], + "source": [ + "from math import log10\n", + "from math import floor\n", + "\n", + "#Variable declaration\n", + "VCC=12; #Collector supply voltage, V\n", + "VEE=12; #Emitter supply voltage, V\n", + "RE=200.0; #Emitter resistor, kΩ\n", + "RC=100.0; #Collector resistor, kΩ\n", + "VBE=0.7; #Base-emitter voltage, V\n", + "\n", + "\n", + "#Calculation\n", + "#(i)\n", + "A_CM=round(RC/(2*RE),2); #Common mode voltage gain\n", + "\n", + "#(ii)\n", + "IE=round((VEE-VBE)/RE,4); #Tail current, mA\n", + "IE1=round(IE/2,4); #Emitter current of 1st transistor, mA\n", + "IE2=IE1; #Emitter current of 2nd transistor, mA\n", + "re=round(25/IE1,1); #a.c emitter resistance, Ω\n", + "A_DM=RC*1000/(2*re); #Differential voltage gain,\n", + "CMRR_dB=floor(20*log10(A_DM/A_CM)*100)/100; #Common mode rejection ratio, dB\n", + "\n", + "\n", + "#Result\n", + "print(\"(i) Common mode gain=%.2f\"%A_CM);\n", + "print(\"(ii)Common mode rejection ratio=%.2fdB\"%CMRR_dB);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.21 : Page number 691" + ] + }, + { + "cell_type": "code", + "execution_count": 22, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "f2=30kHz\n", + "ACL=75 or 37.5dB.\n" + ] + } + ], + "source": [ + "from math import log10\n", + "\n", + "#Variable declaration\n", + "ACL=500; #closed loop gain\n", + "f_unity=15; #frequency with cloased-loop unity gain, MHz\n", + "\n", + "\n", + "#Calculation\n", + "f2=f_unity*1000/500 #Upper frequency of bandwidth,kHz\n", + "BW=f2-0; #Bandwidth, kHz\n", + "A_CL=f_unity*1000/200; #Maximum value of A_CL when f2=200kHz\n", + "A_CL_dB=20*log10(A_CL); #Maximum value of A_CL in decibel\n", + "\n", + "\n", + "#Result\n", + "print(\"f2=%dkHz\"%f2);\n", + "print(\"ACL=%d or %.1fdB.\"%(A_CL,A_CL_dB));\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.22 : Page number 691-692" + ] + }, + { + "cell_type": "code", + "execution_count": 23, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) Operating Bandwidth=1.5MHz.\n", + "(ii) Operating Bandwidth=150kHz.\n", + "(iii) Operating Bandwidth=15kHz.\n" + ] + } + ], + "source": [ + "\n", + "#Variable declaration\n", + "GBW=1.5; #Gain-bandwidth, MHz\n", + "\n", + "#Calculation\n", + "#(i) For A_CL=1;\n", + "A_CL=1; #Closed loop gain\n", + "BW=GBW/A_CL; #Bandwidth, MHz\n", + "\n", + "print(\"(i) Operating Bandwidth=%.1fMHz.\"%BW);\n", + "\n", + "#(ii) For A_CL=10;\n", + "A_CL=10; #Closed loop gain\n", + "BW=(GBW/A_CL)*1000; #Bandwidth, kHz\n", + "\n", + "print(\"(ii) Operating Bandwidth=%dkHz.\"%BW);\n", + "\n", + "#(iii) For A_CL=100;\n", + "A_CL=100; #Closed loop gain\n", + "BW=(GBW/A_CL)*1000; #Bandwidth, kHz\n", + "\n", + "print(\"(iii) Operating Bandwidth=%dkHz.\"%BW);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.23 : Page number 692" + ] + }, + { + "cell_type": "code", + "execution_count": 24, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Maximum operating frequency=9.95kHz.\n" + ] + } + ], + "source": [ + "from math import pi\n", + "\n", + "#Variable declaration\n", + "slew_rate=0.5; #Slew rate, V/μs\n", + "V_supply=10; #Supply voltage, V\n", + "\n", + "\n", + "#Calculation\n", + "V_sat=V_supply-2; #Saturation voltage, V\n", + "V_pk=V_sat; #Maximum peak-output voltage, V\n", + "f_max=((slew_rate*10**6)/(2*pi*V_pk))/1000; #Maximum operating frequency, kHz\n", + "\n", + "#Result\n", + "print(\"Maximum operating frequency=%.2fkHz.\"%f_max);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.24 : Page number 692" + ] + }, + { + "cell_type": "code", + "execution_count": 25, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Maximum operating frequency=796kHz\n" + ] + } + ], + "source": [ + "from math import pi\n", + "\n", + "#Variable declaration\n", + "slew_rate=0.5; #Slew rate, V/μs\n", + "V_pk=100.0; #Peak-output voltage, mV\n", + "\n", + "\n", + "#Calculation\n", + "V_pk=V_pk/1000.0; #Peak-output voltage, V\n", + "f_max=(slew_rate*10**6/(2*pi*V_pk))/1000.0; #Maximum operating frequency, kHz\n", + "\n", + "#Result\n", + "print(\"Maximum operating frequency=%.0fkHz\"%f_max);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.25 : Page number 695-696" + ] + }, + { + "cell_type": "code", + "execution_count": 26, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Feedback resistor=220kΩ\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "A_CL=-100; #Closed-loop voltage gain\n", + "Ri=2.2; #Input resistor, kΩ\n", + "\n", + "#Calculation\n", + "#Since, A_CL=-(Rf/Ri)\n", + "Rf=-A_CL*Ri; #Feedback resistor, kΩ\n", + "\n", + "#Result\n", + "print(\"Feedback resistor=%dkΩ\"%Rf);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.26 : Page number 696" + ] + }, + { + "cell_type": "code", + "execution_count": 27, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Output voltage=-0.25V\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "vin=2.5; #Input voltage, mV\n", + "Rf=200; #Feedback resistor, kΩ\n", + "Ri=2; #Input resistor, kΩ\n", + "\n", + "#Calculation\n", + "A_CL=-(Rf/Ri); #Closed-loop voltage gain\n", + "vout=A_CL*vin/1000; #Output voltage,V\n", + "\n", + "#Result\n", + "print(\"Output voltage=%.2fV\"%vout);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.27 : Page number 696" + ] + }, + { + "cell_type": "code", + "execution_count": 28, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Closed-loop voltage gain=-1\n", + "Therefore, output will have same amplitude but 180° phase inversion.\n" + ] + } + ], + "source": [ + "#Varaiable declaration\n", + "Rf=1.0; #Feedback resistor, kΩ\n", + "Ri=1.0; #Input resistor, kΩ\n", + "\n", + "#Calculation\n", + "A_CL=-(Rf/Ri); #Closed-loop voltage gain\n", + "\n", + "#Result\n", + "print(\"Closed-loop voltage gain=%d\"%A_CL);\n", + "print(\"Therefore, output will have same amplitude but 180° phase inversion.\");\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.28 : Page number 696-697" + ] + }, + { + "cell_type": "code", + "execution_count": 29, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Closed-loop voltage gain=-40\n", + "Supply voltage=±15V, saturation voltage=±13V. Since gain=-40, op-Amp will be driven to saturation.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "Rf=40; #Feedback resistor, kΩ\n", + "Ri=1; #Input resistor, kΩ\n", + "\n", + "#Calculation\n", + "A_CL=-(Rf/Ri); #Closed-loop voltage gain\n", + "\n", + "\n", + "#Result\n", + "print(\"Closed-loop voltage gain=%d\"%A_CL);\n", + "print(\"Supply voltage=±15V, saturation voltage=±13V. Since gain=-40, op-Amp will be driven to saturation.\");\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.29 : Page number 697" + ] + }, + { + "cell_type": "code", + "execution_count": 30, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) A_CL=-10.\n", + "(ii) Zi=10kΩ\n", + "(iii) Maximum operating frequency=15.9kHz.\n" + ] + } + ], + "source": [ + "from math import pi\n", + "\n", + "#Variable declaration\n", + "Rf=100; #Feedback resistor, kΩ\n", + "Ri=10; #Input resistor, kΩ\n", + "Vpp=1; #Input peak-peak voltage, V\n", + "slew_rate=0.5; #Slew rate, V//μs\n", + "\n", + "#Calculation\n", + "#(i)\n", + "A_CL=-(Rf/Ri); #Closed-loop voltage gain\n", + "\n", + "#(ii)\n", + "Zi=Ri; #Input impedance(~ Input resistor), kΩ\n", + "\n", + "#(iii)\n", + "Vout=A_CL*Vpp; #Peak-to-peak voltage, V\n", + "Vpk=Vout/2; #Peak output voltage, V\n", + "f_max=(slew_rate*10**6/(2*pi*abs(Vpk)))/1000; #Maximum operating frequency, kHz\n", + "\n", + "\n", + "#Result\n", + "print(\"(i) A_CL=%d.\"%A_CL);\n", + "print(\"(ii) Zi=%dkΩ\"%Zi);\n", + "print(\"(iii) Maximum operating frequency=%.1fkHz.\"%f_max);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.30 : Page number 697" + ] + }, + { + "cell_type": "code", + "execution_count": 31, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Rf=20kΩ and Ri=5kΩ.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "A_CL=-4; #Closed loop voltage gain\n", + "R=[1.0,5.0,10.0,20.0]; #List of available resistors, kΩ\n", + "\n", + "#Calculation\n", + "for i in R[:]:\n", + " for j in R[:]:\n", + " if -(i/j)==A_CL :\n", + " print(\"Rf=%dkΩ and Ri=%dkΩ.\"%(i,j));\n", + " break;\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.31 : Page number 697-698" + ] + }, + { + "cell_type": "code", + "execution_count": 32, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) Closed loop voltage gain=-100.\n", + "(ii) Closed loop voltage gain=-50.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "Rf=100; #Feedback resistor, kΩ\n", + "Ri=1; #Input resistor, kΩ\n", + "\n", + "\n", + "#Calculation\n", + "#(i)\n", + "R_source=0; #Source resistor, kΩ\n", + "A_CL=-Rf/(R_source+Ri); #Closed-loop voltage gain\n", + "\n", + "print(\"(i) Closed loop voltage gain=%d.\"%A_CL);\n", + "\n", + "#(ii)\n", + "R_source=1; #Source resistor, kΩ\n", + "A_CL=-Rf/(R_source+Ri); #Closed-loop voltage gain\n", + "\n", + "print(\"(ii) Closed loop voltage gain=%d.\"%A_CL);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.32 : Page number 699-700" + ] + }, + { + "cell_type": "code", + "execution_count": 33, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Output voltage=12.12mV\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "Rf=240; #Feedback resistor, kΩ\n", + "Ri=2.4; #Input resistor, kΩ\n", + "Vin=120; #Input voltage, μV\n", + "\n", + "\n", + "#Calculation\n", + "A_CL=1+(Rf/Ri); #Closed loop voltage gain\n", + "Vout=(A_CL*Vin)/1000; #Output voltage, mV\n", + "\n", + "#Result\n", + "print(\"Output voltage=%.2fmV\"%Vout);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.33 : Page number 700" + ] + }, + { + "cell_type": "code", + "execution_count": 34, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) Output voltage=11V\n", + "(ii) Output voltage=-11V\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "Rf=10; #Feedback resistor, kΩ\n", + "Ri=1; #Input resistor, kΩ\n", + "\n", + "\n", + "#Calculation\n", + "A_CL=1+(Rf/Ri); #Closed loop voltage gain\n", + "#(i)\n", + "Vin=1; #Input voltage, V\n", + "Vout=A_CL*Vin; #Output voltage, V\n", + "\n", + "print(\"(i) Output voltage=%dV\"%Vout);\n", + "\n", + "\n", + "#(ii)\n", + "Vin=-1; #Input voltage, V\n", + "Vout=A_CL*Vin; #Output voltage, V\n", + "\n", + "print(\"(ii) Output voltage=%dV\"%Vout);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.34 : Page number 700" + ] + }, + { + "cell_type": "code", + "execution_count": 35, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Peak to peak output voltage=12V\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "Rf=5; #Feedback resistor, kΩ\n", + "Ri=1; #Input resistor, kΩ\n", + "Vin_max=1; #Maximum input voltage, V\n", + "Vin_min=-1; #Minimum input voltage, V\n", + "\n", + "#Calculation\n", + "V_inpp=Vin_max-Vin_min; #Peak-peak input voltage, V\n", + "A_CL=1+(Rf/Ri); #Closed loop voltage gain\n", + "Vout_pp=A_CL*V_inpp; #Peak-peak output voltage, V\n", + "\n", + "#Result\n", + "print(\"Peak to peak output voltage=%dV\"%Vout_pp);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.35 : Page number 700-701" + ] + }, + { + "cell_type": "code", + "execution_count": 36, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) Closed-loop voltage gain=11\n", + "(ii) Maximum operating frequency=14.47kHz\n" + ] + } + ], + "source": [ + "from math import pi\n", + "\n", + "#Variable declaration\n", + "Rf=100; #Feedback resistor, kΩ\n", + "Ri=10; #Input resistor, kΩ\n", + "Vpp=1; #Input peak-peak voltage, V\n", + "slew_rate=0.5; #Slew rate, V/μs\n", + "\n", + "#Calculation\n", + "#(i)\n", + "A_CL=1+(Rf/Ri); #Closed loop voltage gain\n", + "\n", + "#(ii)\n", + "Vout_pp=A_CL*Vpp; #Peak-peak output voltage, V\n", + "Vpk=Vout_pp/2.0; #Peak output voltage, V\n", + "f_max=((slew_rate*10**6)/(2*pi*Vpk))/1000.0; #Maximum operating frequency, kHz\n", + "\n", + "#Result\n", + "print(\"(i) Closed-loop voltage gain=%d\"%A_CL);\n", + "\n", + "print(\"(ii) Maximum operating frequency=%.2fkHz\"%f_max);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.36 : Page number 701" + ] + }, + { + "cell_type": "code", + "execution_count": 37, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) Bandwidth=44.3kHz.\n", + "(ii) Bandwidth=63.8kHz.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "Rf=220; #Feedback resistor, kΩ\n", + "Ri=3.3; #Input resistor, kΩ\n", + "unity_gain_BW=3; #Unity gain bandwidth, MHz\n", + "\n", + "#Calculation\n", + "#(i) For non-inverting amplifier\n", + "A_CL=1+(Rf/Ri); #Closed loop voltage gain\n", + "BW=unity_gain_BW*1000.0/A_CL; #Bandwidth, kHz\n", + "\n", + "print(\"(i) Bandwidth=%.1fkHz.\"%BW);\n", + "\n", + "#(ii) For inverting amplifier\n", + "Rf=47; #Feedback resistor, kΩ\n", + "Ri=1; #Input resistor, kΩ\n", + "A_CL=-(Rf/Ri); #Closed loop voltage gain\n", + "BW=unity_gain_BW*1000.0/abs(A_CL); #Bandwidth, kHz\n", + "\n", + "print(\"(ii) Bandwidth=%.1fkHz.\"%BW);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.37 : Page number 701-702" + ] + }, + { + "cell_type": "code", + "execution_count": 38, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) For voltage follower A_CL=1.\n", + "(ii) The maximum output frequency=26.53kHz.\n" + ] + } + ], + "source": [ + "from math import pi\n", + "\n", + "#(i)\n", + "A_CL=1; #Closed loop voltage gain for voltage follower\n", + "print(\"(i) For voltage follower A_CL=1.\");\n", + "\n", + "\n", + "#(ii)\n", + "slew_rate=0.5; #Slew rate, V/μs\n", + "V_inpp=6; #peak-peak input voltage, V\n", + "Vout=A_CL*V_inpp; #Peak-peak output voltage, V\n", + "Vpk=Vout/2; #Peak output voltage, V\n", + "\n", + "f_max=(slew_rate*10**6/(2*pi*Vpk))/1000; #Maximum operating frequency, kHz\n", + "\n", + "#Result\n", + "print(\"(ii) The maximum output frequency=%.2fkHz.\"%f_max);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.38 : Page number 702" + ] + }, + { + "cell_type": "code", + "execution_count": 39, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Output voltage=1.78V\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "Rf=470.0; #Feedback resistor, kΩ\n", + "R1=4.3; #Input resistor of 1st op-Amp, kΩ\n", + "R2=33.0; #Input resistor of 2nd op-Amp, kΩ\n", + "R3=33.0; #Input resistor of 3rd op-Amp, kΩ\n", + "Vin=80.0; #Input voltage, μV.\n", + "\n", + "#Calculation\n", + "A1=1+Rf/R1; #Gain of first op-Amp\n", + "A2=-round(Rf/R2,1); #Gain of second op-Amp\n", + "A3=-round(Rf/R3,1); #Gain of third op-Amp\n", + "A=A1*A2*A3; #Overall gain\n", + "Vout=A*Vin*10**-6; #Output voltage, V\n", + "\n", + "#Result\n", + "print(\"Output voltage=%.2fV\"%Vout);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.39 : Page number 702-703" + ] + }, + { + "cell_type": "code", + "execution_count": 40, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "R1=30kΩ, R2=15kΩ and R3=10kΩ.\n", + "Output voltage=0.729V.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "A1=10; #Voltage gain of 1st op-Amp\n", + "A2=-18; #Voltage gain of 2nd op-Amp\n", + "A3=-27; #Voltage gain of 3rd op-Amp\n", + "Rf=270; #Feedback resistor, kΩ\n", + "Vin=150; #Input voltage, μV \n", + "\n", + "\n", + "#Calculation\n", + "R1=Rf/(A1-1); #Input resistor of 1st op-Amp, kΩ\n", + "R2=-Rf/A2; #Input resistr of 2nd op-Amp, kΩ\n", + "R3=-Rf/A3; #Input resistor of 3rd op-Amp, kΩ\n", + "\n", + "A=A1*A2*A3; #overall gain,\n", + "Vout=Vin*10**-6*A; #Output voltage, V\n", + "\n", + "\n", + "#Result\n", + "print(\"R1=%dkΩ, R2=%dkΩ and R3=%dkΩ.\"%(R1,R2,R3));\n", + "print(\"Output voltage=%.3fV.\"%Vout);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.40 : Page number 703-704" + ] + }, + { + "cell_type": "code", + "execution_count": 41, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "R1=50kΩ, R2=25kΩ and R3=10kΩ.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "Rf=500; #Feedback resistor, kΩ\n", + "A1=-10; #Gain of 1st op-Amp\n", + "A2=-20; #Gain of 2nd op-Amp\n", + "A3=-50; #Gain of 3rd op-Amp\n", + "\n", + "#Calculation\n", + "R1=-Rf/A1; #Input resistor of 1st op-Amp, kΩ\n", + "R2=-Rf/A2; #Input resistor of 2nd op-Amp, kΩ\n", + "R3=-Rf/A3; #Input resistor of 3rd op-Amp, kΩ\n", + "\n", + "\n", + "#Result\n", + "print(\"R1=%dkΩ, R2=%dkΩ and R3=%dkΩ.\"%(R1,R2,R3));\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.41 : Page number 705" + ] + }, + { + "cell_type": "code", + "execution_count": 42, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) The input impedance=17202MΩ and output impedance=8.7e-03Ω.\n", + "(ii) The closed loop voltage gain=23.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "Zin=2.0; #Input impedance of op-Amp, MΩ\n", + "Zout=75.0; #Output impedance of op-Amp, Ω\n", + "A_OL=200000.0; #Open-loop voltage gain\n", + "Rf=220.0; #Feedback resistor, kΩ\n", + "Ri=10.0; #Input resistor, kΩ\n", + "\n", + "\n", + "#Calculation\n", + "#(i)\n", + "mv=round(Ri/(Ri+Rf),3); #Feedback fraction\n", + "Zin_NI=Zin*(1+(A_OL*mv)); #Input impedance, MΩ\n", + "Zout_NI=Zout/(1+A_OL*mv); #Output impedance, Ω\n", + "\n", + "#(ii)\n", + "A_CL=1+Rf/Ri; #Closed loop voltage gain\n", + "\n", + "#Result\n", + "print(\"(i) The input impedance=%dMΩ and output impedance=%.1eΩ.\"%(Zin_NI,Zout_NI));\n", + "print(\"(ii) The closed loop voltage gain=%d.\"%A_CL);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.42 : Page number 705-706" + ] + }, + { + "cell_type": "code", + "execution_count": 43, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The input impedance=400002MΩ and output impedance=0.38e-03Ω.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "#For voltage follower,\n", + "mv=1.0; #Feedback fraction\n", + "A_OL=200000.0; #Open-loop voltage gain\n", + "Zin=2.0; #Input impedance of op-Amp, MΩ\n", + "Zout=75.0; #Output impedance of op-Amp, Ω\n", + "\n", + "\n", + "#Calculation\n", + "Zin_VF=Zin*(1+(A_OL*mv)); #Input impedance, MΩ\n", + "Zout_VF=round(round(Zout/(1+A_OL*mv),6),5); #Output impedance, Ω\n", + "\n", + "#Result\n", + "print(\"The input impedance=%dMΩ and output impedance=%.2fe-03Ω.\"%(Zin_VF,Zout_VF*1000));\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.43 : Page number 706" + ] + }, + { + "cell_type": "code", + "execution_count": 44, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The input impedance=1kΩ and output impedance=50Ω.\n", + "Closed-loop voltage gain=-100\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "Rf=100; #Feedback resistor, kΩ\n", + "Ri=1.0; #Input resistor, kΩ\n", + "Zin=4; #Input impedance of op-Amp, MΩ\n", + "Zout=50; #Output impedance of op-Amp, Ω\n", + "\n", + "\n", + "#Calculation\n", + "Zin_I=Ri; #Input impedance, kΩ\n", + "Zout_I=Zout; #Output impedance, Ω\n", + "A_CL=-(Rf/Ri); #Closed loop voltage gain\n", + "\n", + "\n", + "#Result\n", + "print(\"The input impedance=%dkΩ and output impedance=%dΩ.\"%(Zin_I,Zout_I));\n", + "print(\"Closed-loop voltage gain=%d\"%A_CL);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.44 : Page number 709" + ] + }, + { + "cell_type": "code", + "execution_count": 45, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Output voltage=-12V.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "Rf=10; #Feedback resistor, kΩ\n", + "Ri=10; #Input resistor, kΩ\n", + "V1=3; #Input voltage 1st, V\n", + "V2=1; #Input voltage 2nd, V\n", + "V3=8; #Input voltage 3rd, V\n", + "\n", + "\n", + "#Calculation\n", + "#Since, Rf=Ri, Vout=-(Rf/Ri)*(V1+V2+V3)= -(V1+V2+V3);\n", + "Vout=-(V1+V2+V3); #Output voltage, V\n", + "\n", + "#Result\n", + "print(\"Output voltage=%dV.\"%Vout);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.45 : Page number 709" + ] + }, + { + "cell_type": "code", + "execution_count": 46, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Output voltage=-7V.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "Rf=10; #Feedback resistor, kΩ\n", + "R1=1; #Input resistor for input 1, kΩ\n", + "R2=1; #Input resistor for input 2, kΩ\n", + "V1=0.2; #Input voltage 1st, V\n", + "V2=0.5; #Input voltage 2nd, V\n", + "\n", + "\n", + "#Calculation\n", + "R=R1; #Input resistor(=R1 or R2), kΩ\n", + "Vout=-(Rf/R)*(V1+V2); #Output voltage, V\n", + "\n", + "#Result\n", + "print(\"Output voltage=%dV.\"%Vout);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.46 : Page number 709-710" + ] + }, + { + "cell_type": "code", + "execution_count": 47, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Output voltage=-2.5V.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "Rf=1; #Feedback resistor, kΩ\n", + "Ri=10.0; #Input resistor, kΩ\n", + "V1=10; #Input voltage 1st, V\n", + "V2=8.0; #Input voltage 2nd, V\n", + "V3=7.0; #Input voltage 3rd, V\n", + "\n", + "\n", + "#Calculation\n", + "#Since, Vout=-(Rf/Ri)*(V1+V2+V3);\n", + "Vout=-(Rf/Ri)*(V1+V2+V3); #Output voltage, V\n", + "\n", + "#Result\n", + "print(\"Output voltage=%.1fV.\"%Vout);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.47 : Page number 710" + ] + }, + { + "cell_type": "code", + "execution_count": 48, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Output voltage=2.5V\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "V1=0.6; #Input voltage to 1st input resistor, V\n", + "V2=-1.4; #Input voltage to 2nd input resistor, V\n", + "Rf=200; #Feedback resistor, kΩ\n", + "R1=400; #Input resistor 1, kΩ\n", + "R2=100.0; #Input resistor 2, kΩ\n", + "\n", + "#Calculation\n", + "Vout=-Rf*(V1/R1 +V2/R2); #Output voltage, V\n", + "\n", + "#Result\n", + "print(\"Output voltage=%.1fV\"%Vout);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.48 : Page number 710-711" + ] + }, + { + "cell_type": "code", + "execution_count": 49, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) The output voltage=-12.5V\n", + "(ii) The output voltage=-7.5V\n", + "(iii) The output voltage=-17.5V\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "Rf=1.0; #Feedback resistor, kΩ\n", + "R1=1.0; #Input resistor 1, kΩ\n", + "R2=2.0; #Input resistor 2, kΩ\n", + "R3=4.0; #Input resistor 3, kΩ\n", + "\n", + "\n", + "#Calculation\n", + "Rf_R1=Rf/R1; #Ratio of feedback resistor and 1st input resistor\n", + "Rf_R2=Rf/R2; #Ratio of feedback resistor and 2nd input resistor\n", + "Rf_R3=Rf/R3; #Ratio of feedback resistor and 3rd input resistor\n", + "\n", + "#(i) First input combination\n", + "V1=10; #Input voltage to 1st input resistor, V\n", + "V2=0; #Input voltage to 2nd input resistor, V\n", + "V3=10; #Input voltage to 3rd input resistor, V\n", + "Vout=-(V1*Rf_R1 +V2*Rf_R2 +V3*Rf_R3); #Output voltage, V\n", + "print(\"(i) The output voltage=%.1fV\"%Vout);\n", + "\n", + "#(i) First input combination\n", + "V1=0; #Input voltage to 1st input resistor, V\n", + "V2=10; #Input voltage to 2nd input resistor, V\n", + "V3=10; #Input voltage to 3rd input resistor, V\n", + "Vout=-(V1*Rf_R1 +V2*Rf_R2 +V3*Rf_R3); #Output voltage, V\n", + "print(\"(ii) The output voltage=%.1fV\"%Vout);\n", + "\n", + "\n", + "#(i) First input combination\n", + "V1=10; #Input voltage to 1st input resistor, V\n", + "V2=10; #Input voltage to 2nd input resistor, V\n", + "V3=10; #Input voltage to 3rd input resistor, V\n", + "Vout=-(V1*Rf_R1 +V2*Rf_R2 +V3*Rf_R3); #Output voltage, V\n", + "print(\"(iii) The output voltage=%.1fV\"%Vout);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.49 : Page number 711" + ] + }, + { + "cell_type": "code", + "execution_count": 50, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Vout=-[0.5sin(1000t)+0.33sin(3000t)]V\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "Rf=330; #Feedback resistor, kΩ\n", + "R1=33.0; #Input resistor 1, kΩ\n", + "R2=10.0; #Input resistor 2, kΩ\n", + "V1_m=50; #Peak voltage of 1st input, mV\n", + "V2_m=10; #Peak voltage of 2nd input, mV\n", + "\n", + "#Calculation\n", + "#Since, Vout=-((Rf/R1)*V1 + (Rf/R2)*V2)\n", + "print(\"Vout=-[%.1fsin(1000t)+%.2fsin(3000t)]V\"%((V1_m/1000.0)*(Rf/R1),(V2_m/1000.0)*(Rf/R2)));\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.50 : Page number 715" + ] + }, + { + "cell_type": "code", + "execution_count": 51, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Vo=-1*(1/RC)∫vi dt.\n", + "=>Vo=-1*(1/1)∫vi dt\n", + "=>Vo=∫vi dt\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "R=100; #Input resistor, kΩ\n", + "C=10; #Feedback capacitor, μF\n", + "\n", + "#Calculation\n", + "RC=R*10**3*C*10**-6; #product of input resistance and feedback capacitance, s\n", + "\n", + "\n", + "#Result\n", + "print(\"Vo=-1*(1/RC)∫vi dt.\");\n", + "print(\"=>Vo=-1*(1/%d)∫vi dt\"%RC);\n", + "print(\"=>Vo=∫vi dt\");\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.51 : Page number 715-716" + ] + }, + { + "cell_type": "code", + "execution_count": 52, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The critical frequency=159Hz.\n" + ] + } + ], + "source": [ + "from math import pi\n", + "\n", + "#Variable declaration\n", + "Rf=100; #Feedback resistor, kΩ\n", + "C=0.01; #Feedback capacitor, μF\n", + "\n", + "\n", + "#Calculation\n", + "fc=1/(2*pi*Rf*1000*C*10**-6); #Crictical frequency, Hz\n", + "\n", + "#Result\n", + "print(\"The critical frequency=%dHz.\"%fc);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.52 : Page number 716" + ] + }, + { + "cell_type": "code", + "execution_count": 53, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) Vout=-1*(1/RC)∫vi dt.\n", + " ΔVout/dt = -vin/RC = -50mV/μs.\n" + ] + }, + { + "data": { + "image/png": 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+ "text/plain": [ + "<matplotlib.figure.Figure at 0x7ff5052a4080>" + ] + }, + "metadata": {}, + "output_type": "display_data" + } + ], + "source": [ + "import matplotlib.pyplot as plt\n", + "\n", + "#Variable declaration\n", + "R=10.0; #Input resistor, kΩ\n", + "C=0.01; #Feedback capacitor, μF\n", + "vin=5; #Input voltage, V\n", + "\n", + "#Calculation\n", + "#(i)\n", + "Vout_change_rate=-vin/(R*C); #Rate of change of output voltage, V/μs \n", + "print(\"(i) Vout=-1*(1/RC)∫vi dt.\");\n", + "print(\" ΔVout/dt = -vin/RC = %dmV/μs.\"%Vout_change_rate);\n", + "\n", + "#(ii) Plotting the output waveform\n", + "vin_plot=[]; #Plotting variable for input waveform, V\n", + "dt=100; #time between edges, μs\n", + "for i in range(0,3*dt+1):\n", + " if i<dt or i>2*dt :\n", + " vin_plot.append(0);\n", + " else:\n", + " vin_plot.append(5); \n", + "\n", + "plt.subplot(211);\n", + "plt.plot(vin_plot);\n", + "plt.xlim([0,300])\n", + "plt.ylim([-5,10])\n", + "plt.xlabel(\"t(microsecond)\");\n", + "plt.ylabel(\"Vin(V)\");\n", + "plt.title(\"Input waveform\");\n", + "\n", + " \n", + "vout_plot=[]; #Plotting variable for output waveform, V\n", + "t=[i for i in range(0,301)]; #Time scale, μs\n", + "for i in t[:] :\n", + " if i<dt:\n", + " vout_plot.append(0);\n", + " elif i>2*dt:\n", + " vout_plot.append((Vout_change_rate/1000.0)*dt);\n", + " else :\n", + " vout_plot.append((-vin_plot[i]/(R*C))/1000*(i-dt));\n", + "\n", + "plt.subplot(212)\n", + "plt.plot(vout_plot);\n", + "plt.xlim([0,300])\n", + "plt.ylim([-5,5]);\n", + "plt.xlabel('t(microsecond)');\n", + "plt.ylabel(\"Vout(V)\");\n", + "plt.title(\"output waveform\");\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.53 : Page number 716-717" + ] + }, + { + "cell_type": "code", + "execution_count": 54, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Vout=-1*(1/RC)∫vi dt.\n", + "Vout=-5*t volts\n", + "Time required=2.6seconds.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "V_supply=15; #Supply voltage, V\n", + "R=10; #Input resistor, kΩ\n", + "C=0.2; #Feedback capacitor, μF\n", + "vin=10; #Input voltage, mV\n", + "\n", + "\n", + "#Calculation\n", + "Vs=-V_supply+2; #Saturation voltage, V\n", + "print(\"Vout=-1*(1/RC)∫vi dt.\");\n", + "print(\"Vout=%d*t volts\"%(-vin/(R*C)));\n", + "t=Vs/(-vin/(R*C)); #Time required, seconds\n", + "print(\"Time required=%.1fseconds.\"%t);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.54 : Page number 717-718" + ] + }, + { + "cell_type": "code", + "execution_count": 55, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Vo=-5V.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "R=1; #Feedback resistor, kΩ\n", + "C=0.1; #Input capacitor, μF\n", + "Vin_change=5; #Change in input voltage, V\n", + "t=0.1; #Time taken for change in input voltage, ms\n", + "\n", + "#Calcualtion\n", + "dvi_dt=Vin_change/(t/1000); #Rate of change of input voltage, V/s\n", + "RC=R*1000*C*10**-6; #Product of feedback resistance and input capacitance, s\n", + "#Since, Vo=-R*C*(dvi/dt);\n", + "Vo=-RC*dvi_dt; #Output voltage, V\n", + "\n", + "#Result\n", + "print(\"Vo=%dV.\"%Vo);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.55 : Page number 718" + ] + }, + { + "cell_type": "code", + "execution_count": 56, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Vo=-0.55V.\n", + "The output voltage stays constant at -0.55V.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "R=10; #Feedback resistor, kΩ\n", + "C=2.2; #Input capacitor, μF\n", + "Vin_change=10; #Change in input voltage, V\n", + "t=0.4; #Time taken for change in input voltage, s\n", + "\n", + "#Calcualtin\n", + "dvi_dt=Vin_change/t; #Rate of change of input voltage, V/s\n", + "RC=R*1000*C*10**-6; #Product of feedback resistance and input capacitance, s\n", + "#Since, Vo=-R*C*(dvi/dt);\n", + "Vo=-RC*dvi_dt; #Output voltage, V\n", + "\n", + "#Result\n", + "print(\"Vo=%.2fV.\"%Vo);\n", + "print(\"The output voltage stays constant at %.2fV.\"%Vo);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.56 : Page number 718-719" + ] + }, + { + "cell_type": "code", + "execution_count": 57, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "vo=-1*(dvi/dt).\n", + "vo=-5V.\n", + "Therefore, between 0 to 0.2s, the output voltage is constant at -5V.\n", + "For t>0.2s, the input is constant so that output voltage is zero.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "R=100; #Feedback resistor, kΩ\n", + "C=10; #Input capacitor, μF\n", + "Vin_change=1; #Change in input voltage, V\n", + "t=0.2; #Time taken for change in input voltage, s\n", + "\n", + "\n", + "#Calculation\n", + "RC=R*1000*C*10**-6; #Product of feedback resistor and input capacitance, s\n", + "#(i)\n", + "print(\"vo=-%d*(dvi/dt).\"%RC);\n", + "\n", + "#(ii)\n", + "dvi_dt=Vin_change/t; #Rate of change of input voltage, V\n", + "vo=-dvi_dt; #Output voltage, V\n", + "print(\"vo=%dV.\"%vo);\n", + "\n", + "print(\"Therefore, between 0 to 0.2s, the output voltage is constant at %dV.\"%vo);\n", + "print(\"For t>0.2s, the input is constant so that output voltage is zero.\");\n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 3", + "language": "python", + "name": "python3" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 3 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython3", + "version": "3.5.1" + }, + "widgets": { + "state": {}, + "version": "1.1.2" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter26_2.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter26_2.ipynb new file mode 100644 index 00000000..670a61b0 --- /dev/null +++ b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter26_2.ipynb @@ -0,0 +1,472 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:a1845801144904256bc26f3ca2e0294eb55dcabb139a523d403624121bc6876a" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "#CHAPTER 26 : DIGITAL ELECTRONICS" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##Example 26.1 : Page 732" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Given\n", + "d=37; #Given decimal number\n", + "\n", + "#Calculation\n", + "b=int(bin(d)[2:]); #Equivalent Octal number \n", + "\n", + "#Result\n", + "print(\"The equivalent binary number=%s.\"%b);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The equivalent binary number=100101.\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##Example 26.2 : Page number 733" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Given\n", + "d=23; #Given decimal number\n", + "\n", + "#Calculation\n", + "b=int(bin(d)[2:]); #Equivalent Octal number\n", + "\n", + "#Result\n", + "print(\"The equivalent binary number=%d.\"%b);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The equivalent binary number=10111.\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##Example 26.3 : Page number 733" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Given \n", + "b=0b110001; #Given binary number\n", + "\n", + "#Calculation\n", + "d=int(b); #Equivalent decimal number\n", + "\n", + "#Result\n", + "print(\"Equivalent decimal number=%d.\"%d);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Equivalent decimal number=49.\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##Example 26.4 : Page number 735" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Given \n", + "d1=76; #Given decimal number\n", + "d2=255; #Given decimal number\n", + "d3=372; #Given decimal number\n", + "\n", + "#Calculation\n", + "o1=int(oct(d1)[1:]); #Equivalent octal number\n", + "o2=int(oct(d2)[1:]); #Equivalent octal number\n", + "o3=int(oct(d3)[1:]); #Equivalent octal number\n", + "\n", + "\n", + "#Result\n", + "print(\"(i) Equivalent octal number=%d.\"%o1);\n", + "print(\"(ii) Equivalent octal number=%d.\"%o2);\n", + "print(\"(iii) Equivalent octal number=%d.\"%o3);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) Equivalent octal number=114.\n", + "(ii) Equivalent octal number=377.\n", + "(iii) Equivalent octal number=564.\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##Example 26.5 : Page number 735" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Given \n", + "o=24.6; #Given octal number\n", + "\n", + "#Calculation\n", + "o_f=o%1; #Floating part of octal number\n", + "o_i=(int)(o-(o%1)); #Integer part of octal number\n", + "d=int(str(o_i),8); #Equivalent decimal number\n", + "\n", + "s=str(o_f); #String value of floating part \n", + "i=2\n", + "while(i<len(s)):\n", + " d=d+int(s[i])*8**-(i-1);\n", + " i+=1;\n", + "#Result\n", + "print(\"Equivalent decimal number=%.2f.\"%d);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Equivalent decimal number=20.75.\n" + ] + } + ], + "prompt_number": 64 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##Example 26.6 : Page number 735" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Given \n", + "d=177; #Given decimal number\n", + "\n", + "#Calculation\n", + "o=oct(d)[1:]; #Equivalent octal number\n", + "\n", + "b=\"\";\n", + "for i in o:\n", + " bo=bin(int(i))[2:]; #Binary of individual octal digit\n", + " b=b+((\"0\" if len(bo)==2 else (\"00\" if len(bo)==1 else\"\")) +bo); #Equivalent binary number\n", + " \n", + "#Result\n", + "print(\"Equivalent octal number=%s.\"%o);\n", + "print(\"Equivalent binary number=%s.\"%b);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Equivalent octal number=261.\n", + "Equivalent binary number=010110001.\n" + ] + } + ], + "prompt_number": 63 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##Example 26.7 : Page number 737" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Given \n", + "d=541; #Given decimal number\n", + "\n", + "#Calculation\n", + "h=hex(d)[2:]; #Equivalent hexadecimal number\n", + "\n", + "#Result\n", + "print(\"Equivalent hexadecimal number=%s.\"%h);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Equivalent hexadecimal number=21d.\n" + ] + } + ], + "prompt_number": 23 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##Example 26.8 : Page number 737" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "hex_to_dec={'0':0,'1':1,'2':2,'3':3,'4':4,'5':5,'6':6,'7':7,'8':8,'9':9,'a':10,'b':11,'c':12,'d':13,'e':14,'f':15};\n", + " \n", + "#Given \n", + "d=378; #Given decimal number\n", + "\n", + "#Calculation\n", + "h=hex(d)[2:]; #Equivalent Hexadecimal number\n", + "\n", + "\n", + "b=\"\";\n", + "for i in h:\n", + " bh=bin(hex_to_dec[i])[2:]; #Binary of individual hexadecimaldigit\n", + " b=b+((\"0\" if len(bh)==3 else (\"00\" if len(bh)==2 else (\"000\" if len(bh)==1 else \"\")))+bh); #Equivalent binary number\n", + "\n", + "#Result\n", + "print(\"Equivalent hexadeciaml number=%s.\"%h);\n", + "print(\"Equivalent binary number=%s.\"%b);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Equivalent hexadeciaml number=17a.\n", + "Equivalent binary number=000101111010.\n" + ] + } + ], + "prompt_number": 56 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 26.9 : Page number 737" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Given\n", + "h=0xB2F; #Given hexadecimal number\n", + "\n", + "#Calculation\n", + "o=oct(h)[1:]; #Equivalent octal number\n", + "\n", + "#Result\n", + "print(\"Equivalent octal number=%s.\"%o);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Equivalent octal number=5457.\n" + ] + } + ], + "prompt_number": 58 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##Example 26.10 : Page number 738" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Given\n", + "BCD=\"0100 0000 0010\" #Given BCD string\n", + "BCD_split=BCD.split(\" \"); #Splitting th binary string into individual BCD \n", + "d=0;\n", + "for i in range(len(BCD_split),0,-1):\n", + " d+=int(BCD_split[len(BCD_split)-i],2)*10**(i-1);\n", + "\n", + "#Result\n", + "print(\"The equivalent decimal =%d.\"%d);\n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The equivalent decimal =402.\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##Example 26.11 : Page number 745" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "print(\"Boolean Expression obtained from the circuit: \\n Y'=A+B \\n Y=((A+B).A)\");\n", + "print(\"Truth Table:\");\n", + "print(\"a\\tb\\tY'=A+B\\t Y=Y'.A\");\n", + "for b in range(0,2):\n", + " for a in range(0,2):\n", + " Y_dash=1 if a or b else 0;\n", + " Y=1 if Y_dash and a else 0;\n", + " print(\"%d\\t%d\\t%d\\t %d\"%(a,b,Y_dash,Y));" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Boolean Expression obtained from the circuit: \n", + " Y'=A+B \n", + " Y=((A+B).A)\n", + "Truth Table:\n", + "a\tb\tY'=A+B\t Y=Y'.A\n", + "0\t0\t0\t 0\n", + "1\t0\t1\t 1\n", + "0\t1\t1\t 0\n", + "1\t1\t1\t 1\n" + ] + } + ], + "prompt_number": 29 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##Example 26.12 : Page number 745-746" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "print(\"Boolean Expression obtained from the circuit: \\n Y'=A'.B \\n Y=Y'+B'\");\n", + "print(\"Truth Table:\");\n", + "print(\"A\\tB\\tA'\\tY'=A'.B\\t B'\\tY=Y'+B'\");\n", + "for b in range(0,2):\n", + " for a in range(0,2):\n", + " a_dash=1 if not a else 0;\n", + " b_dash=1 if not b else 0;\n", + " Y_dash=1 if a_dash and b else 0;\n", + " Y=1 if Y_dash or b_dash else 0;\n", + " print(\"%d\\t%d\\t%d\\t%d\\t %d\\t%d\"%(a,b,a_dash,Y_dash,b_dash,Y));" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Boolean Expression obtained from the circuit: \n", + " Y'=A'.B \n", + " Y=Y'+B'\n", + "Truth Table:\n", + "A\tB\tA'\tY'=A'.B\t B'\tY=Y'+B'\n", + "0\t0\t1\t0\t 1\t1\n", + "1\t0\t0\t0\t 1\t1\n", + "0\t1\t1\t1\t 0\t1\n", + "1\t1\t0\t0\t 0\t0\n" + ] + } + ], + "prompt_number": 30 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter2_2.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter2_2.ipynb new file mode 100644 index 00000000..06a555a6 --- /dev/null +++ b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter2_2.ipynb @@ -0,0 +1,125 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:c0ff4f67576afe73a11c06eedd0a50709b7f5831737f83db1cd640098e3e9740" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 2 : ELECTRONIC EMISSION" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.1: Page number 31\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "\n", + "from math import exp\n", + "from math import pi\n", + "\n", + "l=5.0; #length of tungsten filament in cm\n", + "d=0.01; #diameter of the filament in cm\n", + "T=2500.0; #operating temperature in K\n", + "A=60.2*pow(10,4); #constant, depending upon the type of thermionic emitter, in amp/m\u00b2/K\u00b2\n", + "phi=4.517; #work function of emitter in eV\n", + "\n", + "\n", + "#Calculation\n", + "b=round(11600*phi,-1); #constant for a metal, in K\n", + "Js=round(A*T*T*exp(-b/T),-2); #Emission current density in amp/m\u00b2\n", + "a=pi*(d/100)*(l/100); #Surface area of the cathode in m\u00b2\n", + "E_I=Js*a; #Emission current in A\n", + "\n", + "#Result\n", + "print(\"emission current =%.3f A\"%E_I);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "emission current =0.047 A\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.2:Page number 31\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import log\n", + "\n", + "#Variable declaration\n", + "Js=0.1; #Emission current density in amp/cm\u00b2\n", + "A=60.2; #Constant depending upon the type of thermionic emitter, in amp/cm\u00b2/K\u00b2\n", + "T=1900.0; #Absolute temperature in K\n", + "\n", + "\n", + "#calculations\n", + "#Calculating b according to the formula Js=A*T\u00b2*exp(-b/T) for emission current density\n", + "b=-T*(log(Js/(A*T*T))); #constant for emitter, in K\n", + "phi= round(b/11600,2); # work function in eV\n", + "\n", + "print (\"Work function of the tungsten wire = %.2f eV\"%phi);\n", + "\n", + "if(phi==4.52):\n", + "\tprint(\"Given sample is pure Tungsten\");\n", + "elif(phi!=4.52 and phi>=2.63 and phi<=4.52):\n", + "\tprint (\"The sample is not pure Tungsten\");\n", + " \n", + "#Note : In the text book, the work function has been approximated to 3.56eV, but in the code it calculates as 3.52eV\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Work function of the tungsten wire = 3.52 eV\n", + "The sample is not pure Tungsten\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "code", + "collapsed": false, + "input": [], + "language": "python", + "metadata": {}, + "outputs": [] + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter6_2.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter6_2.ipynb new file mode 100644 index 00000000..a6008ee1 --- /dev/null +++ b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter6_2.ipynb @@ -0,0 +1,1624 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:3452607f2168b562d941493f83083042eaa5a2d316715f9d9f089ff03d73fdb8" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 6: SEMICONDUCTOR DIODE" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.2, Page number 81" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable Declaration \n", + "Vf =20; #Peak Input Voltage in V\n", + "rf=10; #Forward Resistance in ohms\n", + "RL=500.0; #Load Resistance in ohms\n", + "V0=0.7; #Potential Barrier Voltage of the diodes in V\n", + "\n", + "#Calculation\n", + "#(1)\n", + "If_peak=(Vf-V0)/(rf+RL); #Peak current through the diode in A\n", + "If_peak=If_peak*1000; #Peak current through the diode in mA\n", + "#(2)\n", + "V_out_peak =If_peak * RL/1000 ; #Peak output voltage in V\n", + "\n", + "#For an Ideal diode\n", + "If_peak_ideal=Vf/RL; #Peak current through the ideal diode in A\n", + "If_peak_ideal=If_peak_ideal*1000; #Peak current through the ideal diode in mA\n", + "\n", + "V_out_peak_ideal=If_peak_ideal * RL/1000; # Peak output voltage in case of the ideal diode in V\n", + "\n", + "#Result\n", + "print '(i) Peak current through the diode = %.1f mA '%If_peak;\n", + "print '(ii) Peak output voltage = %.1f V'%V_out_peak;\n", + "print '(iii) Peak current through the ideal diode = %d mA '%If_peak_ideal;\n", + "print '(iv) Peak output voltage in case of the ideal diode = %d V'%V_out_peak_ideal;\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) Peak current through the diode = 37.8 mA \n", + "(ii) Peak output voltage = 18.9 V\n", + "(iii) Peak current through the ideal diode = 40 mA \n", + "(iv) Peak output voltage in case of the ideal diode = 20 V\n" + ] + } + ], + "prompt_number": 23 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.3, Page number 82" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable Declaration\n", + "V =10.0; #Battery voltage in V\n", + "R1=50.0; #Resistor 1's resistance in ohms\n", + "R2=5.0; #Resistor 2's resistance in ohms\n", + "\n", + "#Calculation\n", + "#Using Thevenin's Theorem to find current in the diode\n", + "E0=(R2/(R1+R2))*V; #Thevenin's Voltage in V\n", + "R0=(R1*R2)/(R1+R2); #Thevenin's Resistance in ohms\n", + "\n", + "I0=E0/R0; #Current through the diode in A\n", + "I0=I0*1000; #Current through the diode in mA\n", + "\n", + "#Result\n", + "print 'Current through the diode = %d mA '%Io;\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Current through the diode = 200 mA \n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.4, Page number 82-83 " + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable Declaration\n", + "V =10.0; #Battery voltage in V\n", + "R0=48.0; #Resistance of the resistor in ohms\n", + "Rd=1.0; #Forward resistance of the diodes in ohms\n", + "Vd=0.7; #Potential barrier of the diodes in V\n", + "#Calculation\n", + "V_net=V-Vd-Vd; #Net voltage in the circuit in V\n", + "R_net=R0+Rd+Rd #Net resistance of the circuit in ohms\n", + "I_net=V_net/R_net; #Net current in the circuit in A\n", + "I_net=I_net*1000; #Net current in mA\n", + "\n", + "#Result\n", + "print 'Net current in the circuit = %d mA '%I_net;\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Net current in the circuit = 172 mA \n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.5, Page number 83" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable Declaration\n", + "E1=24; #Voltage of first source in V\n", + "E2=4; #Voltage of second source in V\n", + "V0=0.7; #Potential barrier of diodes in V\n", + "R=2000; #Resistance of the given resistor in ohms\n", + "Rd=0; #Forward resistance of the diodes in ohms\n", + "\n", + "#Calculation\n", + "I=(E1-E2-V0)/(R+Rd); #Current in the circuit in A\n", + "I=I*1000; #Current in the circuit in mA \n", + "\n", + "#Result\n", + "print 'Current in the circuit = %.2f mA '%I;" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Current in the circuit = 9.65 mA \n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.6, Page number 83-84" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable Declaration\n", + "V=20; #Voltage of source in V\n", + "V0=0.3; #Potential barrier of Germanium diode in V\n", + "V0_Si=0.7; #Potetial barrier of Silicon diode in V \n", + "\n", + "#Calculation\n", + "#As only Ge diode is turned on due to less potential barrier,\n", + "VA=V-V0; #Voltage VA acroos resistor of 3k ohms\n", + "\n", + "#Result\n", + "print 'Voltage VA = %.1f mA '%VA;" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Voltage VA = 19.7 mA \n" + ] + } + ], + "prompt_number": 20 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.7, Page number 84" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable Declaration\n", + "V=10; #Voltage of source in V\n", + "V0=0.7; #Potetial barrier of Silicon diode in V \n", + "# Resistance of all resistors in ohms\n", + "R1=2000;\n", + "R2=2000;\n", + "R3=2000;\n", + "\n", + "#Calculation\n", + "Id=(V-V0)/(R2+2*R3); #Current through the diodes in A\n", + "VQ=2*Id*R3; #Voltage VQ across the grounded 2k ohm resistor in V\n", + "Id=Id*1000; #Current through the diodes in mA\n", + "\n", + "#Result\n", + "print 'Voltage VQ = %.1f V '%VQ;\n", + "print 'Current through the diodes, Id = %.2f mA '%Id;" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Voltage VQ = 6.2 V \n", + "Current through the diodes, Id = 1.55 mA \n" + ] + } + ], + "prompt_number": 24 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.8, Page number 84" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable Declaration\n", + "V=15; #Voltage of source in V\n", + "V0=0.7; #Potetial barrier of Silicon diode in V \n", + "R=500 # Resistance of all resistors in ohms\n", + "\n", + "#Calculation\n", + "I1=(V-V0)/R; #total current in the circuit in A\n", + "Id1=I1/2; #current in first diode in A\n", + "Id1=Id1*1000; #current in first diode in mA\n", + "Id2=Id1 #current in second diode in mA\n", + "\n", + "#Result\n", + "print ('Current in first diode = %.1f mA'%Id1);\n", + "print ('Current in second diode = %.1f mA'%Id2);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Current in first diode = 14.3 mA\n", + "Current in second diode = 14.3 mA\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.9, Page number 85" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable Declaration\n", + "E=20; #Voltage of source in V\n", + "V0_d1=0.7; #Potetial barrier of first Silicon diode in V\n", + "V0_d2=0.7; #Potetial barrier of second Silicon diode in V\n", + "R1=5600; # Resistance of first resistor in ohms\n", + "R2=3300; # Resistance of second resistor in ohms\n", + "\n", + "#Calculation\n", + "I2=V0_d2/R2; #Current I2 through resistor R2 in A\n", + "I2=round((I2*1000),3); #Current I2 through resistor R2 in mA\n", + "I1=(E-V0_d1-V0_d2)/R1; #Current I1 through resistor R1 in A\n", + "I1=round((I1*1000),2); #Current I1 through resistor R1 in mA\n", + "I3=I1-I2; #Current I3 through diode D2 in mA\n", + "\n", + "#Result\n", + "print 'Current I1= %.2f mA'%I1;\n", + "print 'Current I1= %.3f mA'%I2;\n", + "print 'Current I1= %.3f mA'%I3;\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Current I1= 3.32 mA\n", + "Current I1= 0.212 mA\n", + "Current I1= 3.108 mA\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.10, Page number 85-86" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable Declaration\n", + "E=10.0; #Voltage of source in V\n", + "V0=0.7; #Potetial barrier of Silicon diode in V\n", + "R1=2000; # Resistance of first resistor in ohms\n", + "R2=8000; # Resistance of second resistor in ohms\n", + "R3=4000; #Resistance of third resistor in ohms\n", + "R4=6000; #Resistance of fourth resistor in ohms\n", + "\n", + "#Calculation\n", + "#Assuming the given diode to be reverse bised and calculating voltage across it's terminals\n", + "V1=(E/(R1+R2))*R2; #voltage at the P side of the diode, i.e, voltage across R2 resistor,according to voltage divider rule, in V\n", + "V2=(E/(R3+R4))*R4; #voltage at the N side of the diode, i.e, voltage across R4 resistor,according to voltage divider rule, in V\n", + "\n", + "#Result\n", + "if((V1-V2)>=V0):\n", + " print 'Our assumption was wrong and, the diode is forward biased';\n", + "else:\n", + " print 'The diode is reverse biased';\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Our assumption was wrong and, the diode is forward biased\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.11, Page number 86" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "V=2; #Supply voltage in V\n", + "V0=0.7; #Potential barrier voltage of the diode in V \n", + "R1=4000.0; #Resistance of first resistor in \u03a9\n", + "R2=1000.0; ##Resistance of second resistor in \u03a9\n", + "\n", + "#Calculation\n", + "#Assuming the diode to be in ON state\n", + "I1=((V-V0)/R1)*1000; #Current through resistor R1, in mA\n", + "I2=(V0/R2)*1000; #Current through resistor R2, in mA\n", + "ID=I1-I2; #Diode current, in mA\n", + "\n", + "if(ID<0):\n", + " #Since the diode current is negative, the diode must be OFF \n", + " ID=0; #True value of diode current, mA\n", + " \n", + "#As the diode is in OFF state it can be replaced by an open ciruit equivalent \n", + "VD=V*R2/(R1 +R2); #Voltage across the diode, in V\n", + "\n", + "#Result\n", + "print 'ID =%d mA'%ID;\n", + "print 'VD =%.1f V'%VD;" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "ID =0 mA\n", + "VD =0.4 V\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.12, Page number 89-90" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "AC_Input_Power=100.0; #Input AC Power in watts\n", + "AC_Output_Power=40.0; #Output AC Power in watts\n", + "Accepted_Power=50.0; #Power accepted by the half-wave rectifier in watt\n", + "\n", + "#Calculation\n", + "R_eff=(AC_Output_Power/AC_Input_Power)*100; #Rectification efficiency of the half-wave rectifier\n", + "Unused_power=AC_Input_Power-Accepted_Power; #Power not used by the half_wave rectifier due to open circuited condition of the diode in watt\n", + "Power_dissipated=Accepted_Power-AC_Output_Power; #Power dissipated by the diode watt\n", + "\n", + "#Result\n", + "print 'The rectification efficiency of the half-wave rectifier= %d%% '%R_eff;\n", + "\n", + "print 'Rest 60%% of the power is the unused power and power dissipated by the diode = %d watts and %d watts' %(Unused_power ,Power_dissipated);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The rectification efficiency of the half-wave rectifier= 40% \n", + "Rest 60% of the power is the unused power and power dissipated by the diode = 50 watts and 10 watts\n" + ] + } + ], + "prompt_number": 24 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.13, Page number 90" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import pi\n", + "from math import sqrt\n", + "#Variable declaration\n", + "Vrms=230.0; #AC supply RMS voltage in V\n", + "Turns_Ratio=10/1; #turn ratio of the transformer \n", + "\n", + "#Calculation\n", + "Vpm=sqrt(2)*Vrms; #Maximum primary voltage in V\n", + "Vsm=Vpm/Turns_Ratio; #Maximum secondary voltage in V\n", + "#Case 1\n", + "Vdc=Vsm/(round(pi,2)); #Output D.C voltage, which is the average voltage in V\n", + "Vdc=round(Vdc,2);\n", + "#Case 2\n", + "PIV=Vsm; #Peak Inverse Voltage in V\n", + "\n", + "#Result\n", + "print 'The output d.c voltage= %.2f V'%Vdc;\n", + "print 'The peak inverse voltage= %.2f V'%PIV;" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The output d.c voltage= 10.36 V\n", + "The peak inverse voltage= 32.53 V\n" + ] + } + ], + "prompt_number": 27 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.14, Page number 90-91" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import pi\n", + "#Variable declaration\n", + "rf=20.0; #Internal resistance of the crystal diode in ohms\n", + "Vm=50.0; #Maximum applied voltage in V\n", + "RL=800.0; #Load Resistance in ohms\n", + "\n", + "#Calculation\n", + "# 1\n", + "Im=Vm/(rf+RL); #Maximum current in A\n", + "Im=Im*1000; #Maximum current in \n", + "Im=round(Im,0);\n", + "Idc=Im/pi; #Average voltage in mA\n", + "Idc=round(Idc,1);\n", + "Irms=Im/2; #RMS value of the current in mA\n", + "Irms=round(Irms,1)\n", + "\n", + "# 2\n", + "AC_Input_Power=pow(Irms/1000,2)*(rf+RL); #Input a.c power in watt\n", + "\n", + "DC_Output_Power=pow(Idc/1000,2)*RL; #Output d.c power in watt\n", + "\n", + "# 3\n", + "DC_Output_Voltage=(Idc/1000)*RL; #Output d.c voltage in V\n", + "\n", + "# 4\n", + "Rectifier_efficiency=(DC_Output_Power/AC_Input_Power)*100; # Efficiency of rectification of the half-wave rectifier\n", + "\n", + "#Result\n", + "print ' i:';\n", + "print ' Im = %d mA'%Im;\n", + "print ' Idc = %.1f mA'%Idc;\n", + "print ' Irms = %.1f mA'%Irms;\n", + "print ' ii: ';\n", + "print ' a.c input power= %.3f watt'%AC_Input_Power;\n", + "print ' d.c output power= %.3f watt'%DC_Output_Power;\n", + "print ' iii: ';\n", + "print ' d.c output voltage = %.2f volts'%DC_Output_Voltage;\n", + "print ' iv: '\n", + "print ' Efficiency of rectification = %.1f%%'%Rectifier_efficiency;\n", + "\n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " i:\n", + " Im = 61 mA\n", + " Idc = 19.4 mA\n", + " Irms = 30.5 mA\n", + " ii: \n", + " a.c input power= 0.763 watt\n", + " d.c output power= 0.301 watt\n", + " iii: \n", + " d.c output voltage = 15.52 volts\n", + " iv: \n", + " Efficiency of rectification = 39.5%\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.15, Page number 91" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import pi\n", + "#Variable declaration\n", + "Vdc=50.0; #Output d.c voltage in V\n", + "rf=25; #Diode resistance in ohm\n", + "RL=800; #Load resistance in ohm\n", + "\n", + "\n", + "#Calculation\n", + "Vm=(pi*(rf+RL)*Vdc)/RL; #[ Vdc=Vm*RL/(pi*(rf+RL)) ]Maximum value of a.c voltage required to get a volatge of Vdc from the half-wave rectifier, in V\n", + "Vm=round(Vm,0); \n", + "#Result\n", + "print 'The a.c voltage required should have maximum value of = %d V' %Vm;" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The a.c voltage required should have maximum value of = 162 V\n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.16, Page number 95" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt \n", + "from math import pi\n", + "#Variable declaration\n", + "rf=20; #Internal resistance of the diodes in ohm\n", + "Vrms=50; #RMS value of transformer's secondary voltage from centre tap to each end of secondary\n", + "RL=980; #Load resistance in ohm\n", + "\n", + "#Calculation\n", + "Vm=Vrms*sqrt(2); #Maximum a.c voltage in V\n", + "Im=Vm/(rf+RL); #Maximum load current in A\n", + "Im=Im*1000; #Maximum load current in mA\n", + " \n", + "# 1:\n", + "Idc=2*Im/pi; #Mean load current\n", + "\n", + "# 2:\n", + "Irms=Im/sqrt(2); #RMS value of load current in A\n", + "\n", + "#Result\n", + "print 'i:';\n", + "print' The mean load current= %d mA'%Idc;\n", + "print 'ii:';\n", + "print ' The r.m.s value of the load current = %d mA'%Irms; " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "i:\n", + " The mean load current= 45 mA\n", + "ii:\n", + " The r.m.s value of the load current = 50 mA\n" + ] + } + ], + "prompt_number": 20 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.17, Page number 95-96" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt \n", + "#Variable declaration\n", + "RL=100; #Load resistance in ohm \n", + "rf=0; #Internal resistance of the diodes in ohm\n", + "Turns_ratio=5/1; #Primary to secondary turns ratio of transformer \n", + "P_Vrms=230; #R.M.S value of voltage in primary winding in V\n", + "S_Vrms=P_Vrms/Turns_ratio; #R.M.S value of voltage in secondary winding in V\n", + "S_Vm=S_Vrms*sqrt(2); #Maximum voltage across secondary winding in V\n", + "Vm=S_Vm/2; #Maximum voltage across half seconfdary winding in V\n", + "\n", + "\n", + "#Calculation\n", + "# 1:\n", + "Idc=2*Vm/(pi*RL); #Average current in A\n", + "Vdc=Idc*RL; #d.c output voltage in V\n", + "\n", + "# 2:\n", + "PIV=S_Vm; #Peak Invers Voltage(= Maximum secondary voltage) in V\n", + "\n", + "# 3:\n", + "Pac=pow(Vm/(RL*sqrt(2)),2)*(rf+RL); #a.c input power in watt\n", + "Pdc=(pow(Idc,2)*RL); #d.c output power in watt\n", + "R_eff=(Pdc/Pac)*100; #Rectification efficiency\n", + "R_eff=round(R_eff,1);\n", + "\n", + "#Result\n", + "print 'i:';\n", + "print ' The d.c output voltage= %.1f V'%Vdc;\n", + "print 'ii:';\n", + "print ' The peak inverse voltage= %d V'%PIV;\n", + "print 'iii:';\n", + "print ' Rectification efficiency= %.1f%%'%R_eff;" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "i:\n", + " The d.c output voltage= 20.7 V\n", + "ii:\n", + " The peak inverse voltage= 65 V\n", + "iii:\n", + " Rectification efficiency= 81.1%\n" + ] + } + ], + "prompt_number": 27 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "NOTE: The value of rectification efficiency is calculated as 81.2% in the textbook using the formula 0.812/(1 + (rf/RL)), but by calculating using the correct values in the formula we get 81.1%." + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.18, Page number 96-97" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt \n", + "#Variable declaration\n", + "fin=50; #frequency of input ac source in Hz\n", + "RL=200; #Load resistance in ohm\n", + "Turns_ratio=4/1; #Transformers turns ratio, primary to secondary.\n", + "P_Vrms=230.0; #R.M.S value of voltage in primary winding in V\n", + "S_Vrms=P_Vrms/Turns_ratio #R.M.S value of voltage in secondary winding in V\n", + "Vm=S_Vrms*sqrt(2); #Maximum voltage across secondary winding in V\n", + "\n", + "#Calculation\n", + "# 1:\n", + "Idc=2*Vm/(pi*RL); # Average current in A\n", + "Vdc=Idc*RL; #Output d.c voltage in V\n", + "Vdc=round(Vdc,0);\n", + "# 2:\n", + "PIV= Vm; #Peak Inverse Voltage(= Maximum volutage across secondary winding) in V\n", + "\n", + "# 3:\n", + "fout=2*fin; #Output frequency in Hz\n", + "\n", + "#Result\n", + "print 'i:';\n", + "print ' The d.c output voltage = %d V' %Vdc;\n", + "print 'ii:';\n", + "print ' The peak inverse voltage = %.1f V'%PIV;\n", + "print 'iii:';\n", + "print ' The output frequency = %d Hz'%fout;\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "i:\n", + " The d.c output voltage = 52 V\n", + "ii:\n", + " The peak inverse voltage = 81.3 V\n", + "iii:\n", + " The output frequency = 100 Hz\n" + ] + } + ], + "prompt_number": 29 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.19, Page number 97" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import pi\n", + "from math import sqrt\n", + "\n", + "#Variable declaration\n", + "RL=100.0; #Load Resistance in ohm\n", + "Turns_ratio=5/1; #Primary to secondary turns ratio of the transformer\n", + "Vin=230.0; #R.M.S value of input voltage in V\n", + "fin=50; #Input frequency in Hz\n", + "\n", + "#Calculation\n", + "Vs_rms=Vin/Turns_ratio; #R.M.S value of the voltage in secondary winding, in v\n", + "Vs_max=Vs_rms*sqrt(2); #Maximum voltage across secondary, in V\n", + "\n", + "# (i)\n", + "#Case i: Centre-tap circuit\n", + "Vm=Vs_max/2; #Maximum voltage across half secondary winding, in V \n", + "Vdc=2*Vm*RL/(pi*RL); #DC output voltage, in V \n", + "print 'The d.c output voltage for the centre-tap circuit = %.1f V'%Vdc;\n", + "\n", + "#Case ii:\n", + "Vm=Vs_max; #Maximum voltage across secondary, in V\n", + "Vdc=2*Vm*RL/(pi*RL); #DC output voltage, in V \n", + "print 'The d.c output voltage for the bridge circuit = %.1f V'%Vdc; \n", + "\n", + "# ii:\n", + "#Case i: Centre-tap circuit\n", + "Turns_ratio=5/1; #Turns ratio of the transformer\n", + "Vs_rms=Vin/Turns_ratio; #R.M.S value of the secondary voltage in V\n", + "Vs_max=Vs_rms*sqrt(2); #Maximum voltage across the secondary in V\n", + "Vm=Vs_max/2; #Maximum voltage across half of the secondary in V\n", + "PIV=2*Vm; #Peak Inverse Voltage in V\n", + "print 'PIV in case of centre-tap circuit = %d V'%PIV;\n", + "\n", + "#Case ii: Bridge circuit\n", + "Turns_ratio=10/1; #Turns ratio of the transformer\n", + "Vs_rms=Vin/Turns_ratio; #R.M.S value of the secondary voltage in V\n", + "Vs_max=Vs_rms*sqrt(2); #Maximum voltage across the secondary in V\n", + "PIV=Vm; #Peak Inverse Voltage in V\n", + "print 'PIV in case of bridge circuit = %.1f V'%PIV;\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The d.c output voltage for the centre-tap circuit = 20.7 V\n", + "The d.c output voltage for the bridge circuit = 41.4 V\n", + "PIV in case of centre-tap circuit = 65 V\n", + "PIV in case of bridge circuit = 32.5 V\n" + ] + } + ], + "prompt_number": 46 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.20, Page number 98" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import pi\n", + "from math import sqrt\n", + "#Variable declaration\n", + "rf=1; #forward resistance of diodes of the rectifier in ohm\n", + "RL=480; #Load resistance in ohm\n", + "Vrms=240.0; #a.c supply voltage in V\n", + "Vm=Vrms*sqrt(2); #Maximum a.c voltage in V \n", + "\n", + "#Calculation\n", + "# 1:\n", + "Rt=2*rf+RL; #Total circuit resistance at any instance in ohm\n", + "Im=Vm/Rt; #Maximum load current in A\n", + "Idc=2*Im/pi; #Mean load current in A\n", + "\n", + "# 2:\n", + "Irms=Im/2; #R.M.S value of current in A\n", + "P=pow(Irms,2)*rf; #Power dissipated in each diode in watt\n", + "\n", + "\n", + "#Result\n", + "print 'i:';\n", + "print ' Mean load current = %.2f A'%Idc;\n", + "print 'ii:';\n", + "print ' Power dissipated in each diode= %.3f W'%P;" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "i:\n", + " Mean load current = 0.45 A\n", + "ii:\n", + " Power dissipated in each diode= 0.124 W\n" + ] + } + ], + "prompt_number": 39 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "NOTE: The value of power dissipated is approximately 0.124 W , but in the textbook it is approximated as 0.123W." + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.21, Page number 98-99" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt,pi\n", + "#Variable declaration\n", + "RL=12000; #Load resistance in ohm\n", + "V0=0.7; #Potential barrier voltage of diodes in V\n", + "Vrms=12; #R.M.S value of input a.c voltage in V\n", + "Vs_pk=Vrms*sqrt(2); #Peak secondary voltage in V\n", + "\n", + "#Calculation\n", + "# 1:\n", + "Vout_pk=Vs_pk-(2*V0); #Peak output voltage in V\n", + "Vav=2*Vout_pk/pi; #Average output voltage in V\n", + "Vav=round(Vav,2);\n", + "\n", + "# 2:\n", + "Iav=Vav/RL; #Average output current in A\n", + "Iav=Iav*pow(10,6); #Average output current in \u03bcA\n", + "\n", + "\n", + "#Result\n", + "print 'i:';\n", + "print ' Average output voltage=%.2f V'%Vav;\n", + "print 'ii:';\n", + "print ' Average output current=%.1f \u03bcA'%Iav;\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "i:\n", + " Average output voltage=9.91 V\n", + "ii:\n", + " Average output current=825.8 \u03bcA\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.22, Page number 102" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "Vdc_A=10; #Supply voltage of A in V\n", + "Vdc_B=25; #Supply voltage of B in V\n", + "Vac_rms_a=0.5; #Ripples in power supply A in V\n", + "Vac_rms_b=0.001; #Ripples in power supply B in V\n", + "\n", + "#Calculation\n", + "#For power supply A\n", + "ripple_factor_A=Vac_rms_a/Vdc_A; #Ripple factor of power supply A\n", + "\n", + "#For power supply B\n", + "ripple_factor_B=Vac_rms_b/Vdc_B; #Ripple factor of power supply B\n", + "\n", + "#Result\n", + "if(ripple_factor_A<ripple_factor_B):\n", + " print 'Power supply A is better';\n", + "else :\n", + " print 'Power supply B is better';" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Power supply B is better\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.23, Page number 105-106" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt\n", + "#Variable declaration\n", + "RL=2200; #Load resistance in ohm\n", + "C=50*pow(10,-6); #Capacitance of the capacitor used in filter circuit in F\n", + "V0=0.7; #Potential barrier voltage of the diodes of the rectifier in V\n", + "Vrms=115.0; #R.M.S value of input a.c voltage in V \n", + "fin=60; #Frequency of input a.c voltage in Hz\n", + "Turns_ratio=10/1; #Primary to secondary, turns ratio of the transformer \n", + "\n", + "#Calculation\n", + "Vp_prim=Vrms*sqrt(2); #Peak primary voltage in V\n", + "Vp_sec=Vp_prim/Turns_ratio; #Peak secondary voltage in V\n", + "Vp_in= Vp_sec - 2*V0; #Peak full wave rectified voltage at the filter input in V\n", + "f=2*fin; #Output frequency in Hz\n", + "Vdc=Vp_in*(1-(1/(2*f*RL*C))); #Output d.c voltage in V\n", + "\n", + "#Result\n", + "print 'The output d.c voltage is = %.1f V'%Vdc;" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The output d.c voltage is = 14.3 V\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.24, Page number 106" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import pi\n", + "#Variable declaration\n", + "R=25; #d.c resistance of the choke in ohm\n", + "RL=750; #Load resistance in ohm\n", + "Vm=25.7; #Maximum value of the pulsating output from the rectifier in V\n", + "\n", + "#Calculation\n", + "V_dc=2*Vm/pi; #d.c component of the pulsating output in V\n", + "V_dc=round(V_dc,1);\n", + "V_dc_out=(V_dc*RL)/(R+RL); #Output d.c voltage in V\n", + "V_dc_out=round(V_dc_out,1);\n", + "\n", + "#Result\n", + "print ' The output d.c voltage accross the load resistance is = %.1f V'%V_dc_out;" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The output d.c voltage accross the load resistance is = 15.9 V\n" + ] + } + ], + "prompt_number": 28 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.25, Page number 113-114" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "Ei=120.0; #Input Voltage in V\n", + "Vz=50.0; #Zener Voltage in V\n", + "R=5000.0; #Resistance of the series resistor in ohm\n", + "RL=10000.0; #Load resistance in ohm\n", + "\n", + "#Calculation\n", + "V=Ei*RL/(R+RL); #Voltage across the open circuit if the zener diode is removed\n", + "if(V>Vz):\n", + " #Zener diode is in ON state\n", + " # i:\n", + " Output_voltage=Vz; #Voltage across load resistance, in V\n", + " #ii:\n", + " Voltage_R=Ei-Vz; #Voltage across the series resistance R, in V\n", + " #iii:\n", + " IL=Vz/RL; #Load current through RL in A\n", + " IL=IL*1000; #Load current through RL in mA\n", + " I=Voltage_R/R; #Current through the series resistance in A\n", + " I=I*1000; #Current through the series resistance in mA\n", + " Iz=I-IL; #Applying Kirchhoff's first law, Zener current in mA\n", + " \n", + " #Result\n", + " print 'i) The output voltage across the load resistance RL = %d V'%Output_voltage;\n", + " print 'ii) The voltage drop across the series resistance R = %d V'%Voltage_R;\n", + " print 'iii) The current through the zener diode = %d mA'%Iz;\n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "i) The output voltage across the load resistance RL = 50 V\n", + "ii) The voltage drop across the series resistance R = 70 V\n", + "iii) The current through the zener diode = 9 mA\n" + ] + } + ], + "prompt_number": 29 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.26, Page number 114-115" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "Max_V=120.0; #Maximum input voltage in V\n", + "Min_V=80.0; #Minimum input voltage in V\n", + "R=5000.0; #Series resistance in ohm\n", + "RL=10000.0; #Load resistance in ohm\n", + "Vz=50.0; #Zener voltage in V\n", + "\n", + "\n", + "#Calculation\n", + "#Case i: Maximum zener current\n", + "#Zener current will be maximum when the input voltage is maximum\n", + "V_R_max=Max_V-Vz; #Voltage across series resistance R, in V\n", + "I_max=V_R_max/R; #Current through series resistance R, in A\n", + "I_max=I_max*1000; #Current through series resistance R, in mA\n", + "IL_max=Vz/RL; #Load current in A\n", + "IL_max=IL_max*1000; #Load current in mA\n", + "Iz_max=I_max-IL_max; #Applying Kirchhoff's first law, Zener current in mA;\n", + "\n", + "#Case ii: Minimum zener current\n", + "#The zener will conduct minimum current when the input voltage is minimum\n", + "V_R_min=Min_V-Vz; #Voltage across series resistance R, in V\n", + "I_min=V_R_min/R; #Current through series resistance R, in A\n", + "I_min=I_min*1000; #Current through series resistance R, in mA\n", + "IL_min=Vz/RL; #Load current in A\n", + "IL_min=IL_min*1000; #Load current in mA\n", + "Iz_min=I_min-IL_min; #Applying Kirchhoff's first law, Zener current in mA\n", + "\n", + "#Result\n", + "print 'Case i: ';\n", + "print 'Maximum zener current = %d mA'%Iz_max;\n", + "print 'Case ii: ';\n", + "print 'Minimum zener current = %d mA'%Iz_min;\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Case i: \n", + "Maximum zener current = 9 mA\n", + "Case ii: \n", + "Minimum zener current = 1 mA\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.27, Page number 115" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "Ei=12; #Input voltage in V\n", + "Vz=7.2; #Zener voltage in V\n", + "E0=Vz; #Voltage to be maintained across the load in V\n", + "IL_max=0.1; #Maximum load current in A\n", + "IL_min=0.012; #Minimum load current in A\n", + "Iz_min=0.01; #Minimum zener current in A\n", + "\n", + "#Calculation\n", + "#When the load current is maximum at minimum value of RL, the zener current is minimum and, as the load current decreases due to increase in value of RL\n", + "R=(Ei-E0)/(Iz_min+IL_max); #The value of series resistance R to maintain a voltage=E0 across load, in ohm\n", + "\n", + "#Result\n", + "print 'The minimum value of series resistance R to maintain a constant value of 7.2 V is = %.1f \u03a9'%R;" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The minimum value of series resistance R to maintain a constant value of 7.2 V is = 43.6 \u03a9\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "NOTE: The actual value of R is 43.636363 (recurring) but, in the textbook the value of R is wrongly approximated 43.5 \u03a9" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.28, Page number 115" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "Ei_min=22; #Minimum input voltage in V\n", + "Ei_max=28; #Maximum input voltage in V\n", + "Vz=18; #Zener voltage in V\n", + "E0=Vz; #Constant voltage maintained across the load resistance in V\n", + "Iz_min=0.2; #Minimum zener current in A\n", + "Iz_max=2; #Maximum zener current in A\n", + "RL=18; #Load resistance in \u03a9\n", + "\n", + "#Calculation\n", + "IL=Vz/RL; #Constant value of load current in A\n", + "#When the input voltage is minimum, the zener current will be minimum\n", + "R=(Ei_min-E0)/(Iz_min+IL) #The value of series resistance so that the voltage E0 across RL remains constant\n", + "\n", + "print 'The value of series resistance R, to maintain constant voltage E0 across RL = %.2f \u03a9.'%R;" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of series resistance R, to maintain constant voltage E0 across RL = 3.33 \u03a9.\n" + ] + } + ], + "prompt_number": 19 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.29, Page number 116 " + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "Vz=10 #Zener voltage in V\n", + "Ei_min=13; #Minimum input voltage in V\n", + "Ei_max=16; #Maximum input voltage in V\n", + "Iz_min=0.015; #Minimum zener current in A\n", + "IL_min=0.01; #Minimum load current in A \n", + "IL_max=0.085; #Maximum load curremt in A\n", + "E0=Vz; #Constant voltage to be maintained in V \n", + "\n", + "#Calculation\n", + "#The zener current will be minimum when the input voltage will be minimum and at that time the load current will be maximum\n", + "R=(Ei_min-E0)/(Iz_min+IL_max); #The value of series resistance R to maintain a constant voltage across load\n", + "\n", + "\n", + "#Result\n", + "print 'The value of series resistance to maintain a constant voltage across the load resistance is = %d \u03a9'%R;" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of series resistance to maintain a constant voltage across the load resistance is = 30 \u03a9\n" + ] + } + ], + "prompt_number": 23 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.30, Page number 116" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "Iz=0.2; #Current rating of each zener in A\n", + "Vz=15; #Voltage rating of each zener in V\n", + "Ei=45; #Input voltage in V\n", + "\n", + "#Calculation\n", + "# i: Regulated output voltage across the two zener diodes \n", + "E0=2*Vz; # V\n", + "\n", + "# ii: Value of series resistance \n", + "R=(Ei-E0)/Iz; # \u03a9\n", + "\n", + "#Result\n", + "print 'i) The regulated output voltage = %d V'%E0;\n", + "print 'ii) The value of the series resistance = %d \u03a9'%R;" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "i) The regulated output voltage = 30 V\n", + "ii) The value of the series resistance = 75 \u03a9\n" + ] + } + ], + "prompt_number": 28 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.31, Page number 116-117" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "Vz=10; #Voltage rating of each zener in V\n", + "Iz=1; #Current rating of each zener in A\n", + "Ei=45; #Input unregulated voltage in V\n", + "\n", + "#Calculation\n", + "#Regulated output voltage across the three zener diodes\n", + "E0=3*Vz; # V\n", + "\n", + "#Value of series resistance to obtain a 30V regulated output voltage\n", + "R=(Ei-E0)/Iz; # \u03a9\n", + "\n", + "#Result\n", + "print 'Value of series resistance to obtain a 30V regulated output voltage = %d \u03a9'%R;" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Value of series resistance to obtain a 30V regulated output voltage = 15 \u03a9\n" + ] + } + ], + "prompt_number": 29 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.32, Page number 117" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "RL=2000.0; #Load resistance in \u03a9\n", + "R=200.0; #Series resistance in \u03a9\n", + "Iz=0.025; #Zener current rating in A\n", + "E0=30.0; #Output regulated voltage in V \n", + "\n", + "#Calculation\n", + "#Minimum input voltage will be required when Iz=0 A, and at this condition\n", + "IL=E0/RL; #Load current during Iz=0, in A\n", + "I=IL; #According to Kirchhoff's law, total current, in A\n", + "Ei_min=E0+(I*R); #Minimum input voltage in V\n", + "\n", + "#The maximum input voltage required will be when Iz=0.025 A, and at that condition \n", + "I=IL+Iz; #According to Kirchhoff's law, total current, in A\n", + "Ei_max=E0+(I*R); #maximum input voltage in V\n", + "\n", + "\n", + "#Result\n", + "print 'The required range of input voltage is from %d V to %d V'%(Ei_min,Ei_max); \n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The required range of input voltage is from 33 V to 38 V\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.33, Page number 117-118" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "Ei=16; #Unregulated input voltage in V\n", + "E0=12; #Output regulated voltage in V\n", + "IL_min=0; #Minimum load current in A\n", + "IL_max=0.2; #Maximum load current in A\n", + "Iz_min=0; #Minimum zener current in A\n", + "Iz_max=0.2; #Maximum zener current in A\n", + "\n", + "#Calculation\n", + "#As the regulated voltage required across the load is 12V\n", + "Vz=E0; #Voltage rating of zener diode in V\n", + "V_R=Ei-E0; #Constant Voltage that should remain across series resistance \n", + "#The minimum zener current will occur when the curent in the load in maximum\n", + "R=V_R/(Iz_min+IL_max); #Series resistance in \u03a9\n", + "\n", + "Max_power_rating=Vz*Iz_max; #Maximum power rating of zener diode in W\n", + "\n", + "#Result\n", + "print 'The regulator is designed using a Seris resistance of %d \u03a9 and a zener diode of zener voltage %d V'%(R,Vz);\n", + "print 'The maximum power rating of the zener diode is = %.1f W '%Max_power_rating;" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The regulator is designed using a Seris resistance of 20 \u03a9 and a zener diode of zener voltage 12 V\n", + "The maximum power rating of the zener diode is = 2.4 W \n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.34, Page number 118" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "V=12; #Source voltage in V\n", + "R=1000; #Series resistance in \u03a9\n", + "RL=5000; #Load resistance in \u03a9\n", + "Vz=6; #Voltage rating of zener in V\n", + "\n", + "#Calculation\n", + "#Case i: zener is working properly\n", + "#The output voltage across the load will be equal to the zener voltage.\n", + "V0=Vz; # V\n", + "\n", + "#Result\n", + "print 'Case i: Output voltage when zener is working properly is %d V'%V0;\n", + "\n", + "#Case ii: zener is shorted\n", + "#As the zener is shorted, the potential difference across the load will be zero\n", + "V0=0; #V\n", + "\n", + "#Result\n", + "print 'Case ii: Output voltage when zener is short circuited is %d V'%V0;\n", + " \n", + "#Case iii: zener is open circuited\n", + "#If the zener is open circuited, the total voltage will drop across R and RL according to the voltage divider rule\n", + "V0=V*RL/(R+RL); #V\n", + "\n", + "#Result\n", + "print 'Case iii: Output voltage when zener is open circuited is %d V'%V0;\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Case i: Output voltage when zener is working properly is 6 V\n", + "Case ii: Output voltage when zener is short circuited is 0 V\n", + "Case iii: Output voltage when zener is open circuited is 10 V\n" + ] + } + ], + "prompt_number": 4 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter7_2.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter7_2.ipynb new file mode 100644 index 00000000..537a179e --- /dev/null +++ b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter7_2.ipynb @@ -0,0 +1,212 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:e210474f5c4fc6668f4c7b5af2adf833a1c7f62577017a980ab8d11cd8ce2886" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 7 : SPECIAL-PURPOSE DIODES" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.1 : Page number 127-128\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "V_S=10.0; #Supply voltage in V\n", + "V_D=1.6; #Forward voltage drop of LED, in V\n", + "I_F=20.0; #Required limited current through LED, in mA\n", + "\n", + "#Calculations\n", + "R_S=(V_S-V_D)/(I_F/1000); #Series resistor required to limit the current through the LED, in \u2126\n", + "\n", + "#Result \n", + "print(\"The value of series resistor required to limit the current through the LED = %d \u2126.\"%R_S);\n", + " \n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of series resistor required to limit the current through the LED = 420 \u2126.\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.2: Page number 128" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "V_S=15.0; #Supply voltage in V\n", + "V_D=2.0; #Forward voltage drop of LED, in V\n", + "R_S=2200.0; #Series resistor required to limit the current through the LED, in \u2126\n", + "\n", + "#Calculations\n", + "I_F=((V_S-V_D)/R_S)*1000; #Required limited current through LED, in mA\n", + "\n", + "#Result \n", + "print(\"The current through the LED in the circuit = %.2f mA\"%I_F);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The current through the LED in the circuit = 5.91 mA\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.3: Page number 132-133" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "Ir=50.0; #Dark current as observed from the current Illumination curve, in mA \n", + "V_R=10.0; #Reverse voltage in V\n", + "\n", + "#Calculation\n", + "R_R=V_R/(Ir/pow(10,6)); #Dark Resistance in \u2126\n", + "\n", + "#Result\n", + "print(\"The dark resistance is=%d k\u2126\"%(R_R/1000));\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The dark resistance is=200 k\u2126\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.4: Page number 133" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "E=2.5; #Illumination in mW/cm\u00b2\n", + "m=37.4; #sensitivity of the photodiode in \ud835\udf07A/mW/cm\u00b2\n", + "\n", + "#Calculations\n", + "I_R=m*E; #Reverse current in \ud835\udf07A\n", + "\n", + "#Result\n", + "print(\"The reverese current in the photodiode = %.1f \ud835\udf07A\"%I_R);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The reverese current in the photodiode = 93.5 \ud835\udf07A\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.5: Page number 137" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import pi\n", + "from math import sqrt\t\n", + "#Variable declaration\n", + "L=1.0; #Inductance of the inductor in mH\n", + "C=100.0; #Capacitance of the varactor in pF\n", + "\n", + "#Result\n", + "f_r=1/(2*pi*sqrt(L*pow(10,-3)*C*pow(10,-12))); #Resonant frequency of the circuit in Hz\n", + "f_r=f_r/1000; #Resonant frequency of the circuit in kHz\n", + "\n", + "#Result\n", + "print(\"The resonant frequency of the circuit = %.1f kHz\"%f_r);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The resonant frequency of the circuit = 503.3 kHz\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "code", + "collapsed": false, + "input": [], + "language": "python", + "metadata": {}, + "outputs": [] + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter8_2.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter8_2.ipynb new file mode 100644 index 00000000..4afe0858 --- /dev/null +++ b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter8_2.ipynb @@ -0,0 +1,1851 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:9c13bdd66a3dbb3eae04903205b69bc52bf35e6dadf8b1b3ade1bab68394ae3b" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.1: Page number 147-148\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "Signal=500.0; #Signal voltage in V\n", + "Rin=20.0; #Input resistance in \u03a9 \n", + "Rout=100.0; #Output resistance in \u03a9\n", + "R_C=1000.0; #Collector load in \u03a9\n", + "alpha_ac=1.0; #current amplification factor\n", + "\n", + "#Calculation\n", + "I_E=(Signal/1000)/Rin; \t#Input current in mA\n", + "I_C=I_E*alpha_ac; #Output current in mA\n", + "Vout=I_C*R_C; #Output voltage in V \n", + "Av=Vout/(Signal/1000); #Voltage amplification \n", + "\n", + "#Result\n", + "print(\"The voltage amplification = %d. \"%Av);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The voltage amplification = 50. \n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.2: Page number 150\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "I_E=1; #Emitter curent in mA\n", + "I_C=0.95; #Collector current in mA\n", + "\n", + "#Calculation\n", + "I_B=I_E-I_C; #Base current in mA\n", + "\n", + "#Result \n", + "print(\"The base current = %.2f mA \"%I_B);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The base current = 0.05 mA \n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##Example 8.3: Page number 150\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "alpha=0.9; #Current amplification factor\n", + "I_E=1; #Emitter current in mA\n", + "\n", + "#Calculation\n", + "I_C=alpha*I_E; #Collector current in mA\n", + "I_B=I_E-I_C; #Base current in mA\n", + "\n", + "#Result\n", + "print(\"The base current =%.1f mA\"%I_B);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The base current =0.1 mA\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.4: Page number 150\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "I_C=0.95;\t\t\t#Collector current in mA\n", + "I_B=0.05;\t\t\t#Base current in mA\n", + "\n", + "#Calculation\n", + "I_E=I_B+I_C; #Emitter current in mA\n", + "alpha=I_C/I_E; #Current amplification factor \n", + "\n", + "#Result\n", + "print(\"The current amplification factor = %.2f .\"%alpha);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The current amplification factor = 0.95 .\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.5: Page number 150\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "I_E=1; #Emitter current in mA\n", + "I_CBO=50.0; #Collector current with emitter circuit open, in microAmp\n", + "alpha=0.92; #Current amplification factor\n", + "\n", + "#Calculation\n", + "I_C=alpha*I_E + (I_CBO/1000); #Total collector current in mA\n", + "\n", + "#Result\n", + "print(\"The total collector current = %.2f mA.\"%I_C);\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The total collector current = 0.97 mA.\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.6: Page number 150-151\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "alpha=0.95; #Current amplification factor\n", + "Rc=2.0; #Resistor connected to the collector, in kilo ohm\n", + "V_Rc=2.0; #Voltage drop across the resistor connected to the collector in V\n", + "\n", + "\n", + "#Calculation\n", + "I_C=V_Rc/Rc; #Collector current in mA\n", + "I_E=I_C/alpha; #Emitter current in mA\n", + "I_B=I_E-I_C; #Base current in mA\n", + "\n", + "#Result\n", + "print(\"The base current = %.2f mA\"%I_B); \n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The base current = 0.05 mA\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.7: Page number 151\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "V_EE=8.0; #Supply voltage at the emitter in V\n", + "V_CC=18.0; #Supply voltage at the collector in V\n", + "V_BE=0.7; #Base to emitter voltage in V\n", + "R_E=1.5; #Emitter resistance in \u03a9\n", + "R_C=1.2; #Collector resistance in \u03a9\n", + "\n", + "#Calculations\n", + "I_E=(V_EE-V_BE)/R_E; #Emitter current in mA\n", + "I_C=I_E; #Collector current in mA (approximately equal to emitter current)\n", + "V_CB=V_CC-(I_C*R_C); #Collector to base voltage in V\n", + "\n", + "#Result\n", + "print(\"The collector current =%.2f mA\"%I_C);\n", + "print(\"The collector to base voltage = %.2f V\"%V_CB);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The collector current =4.87 mA\n", + "The collector to base voltage = 12.16 V\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.8:Page number 155\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Function for calculating beta from alpha\n", + "def calc_beta(a): #a is the value of alpha\n", + "\treturn(a/(1-a));\n", + "\n", + "#Case (i)\n", + "alpha=0.9; #current amplification factor\n", + "beta=calc_beta(alpha);\t\t#Base current amplification factor \n", + "print(\"(i) Value of beta =%d\"%beta );\t\t\t\t\t\t\t\t\t\n", + "\n", + "#Case (ii)\n", + "alpha=0.98; #current amplification factor\n", + "beta=calc_beta(alpha); #Base current amplification factor\n", + "print(\"(ii) Value of beta =%.0f\"%beta );\n", + "\n", + "\n", + "#Case (iii)\n", + "alpha=0.99; #current amplification factor\n", + "beta=calc_beta(alpha); #Base current amplification factor \n", + "print(\"(iii) Value of beta =%.0f\"%beta );\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) Value of beta =9\n", + "(ii) Value of beta =49\n", + "(iii) Value of beta =99\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.9: Page number 155\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "beta=50.0; #Base current amplification factor\n", + "I_B=20.0; #Base current in microAmp\n", + "\n", + "#Calculation\n", + "I_B=I_B/1000; #Base current in mA\n", + "I_C=beta*I_B; #Collector current in mA\n", + "I_E=I_B+I_C; #Emitter current in mA\n", + "\n", + "#Result\n", + "print(\"The emitter curent = %.2f mA\"%I_E);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The emitter curent = 1.02 mA\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.10: Page number 155\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "I_B=240.0; #Base current in microAmp\n", + "I_E=12; #Emitter current in mA\n", + "beta=49.0; #Base current amplification factor\n", + "\n", + "#Calculations\n", + "alpha=beta/(1+beta); #current amplification factor \n", + "I_C_alpha=alpha*I_E; #Collector current in mA calculated using alpha\n", + "I_C_beta=beta*(I_B/1000); #Collector current in mA calculated using beta\n", + "\n", + "#Results\n", + "print(\"alpha=%.2f.\"%alpha);\n", + "print(\"Collector current determined using alpha =%.2f mA\"%I_C_alpha);\n", + "print(\"Collector current determined using beta =%.2f mA\"%I_C_beta);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "alpha=0.98.\n", + "Collector current determined using alpha =11.76 mA\n", + "Collector current determined using beta =11.76 mA\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.11: Page number 156\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "beta=45.0; #Base current amplification factor\n", + "R_C=1.0; #Resistance of the collector resistance in k\u03a9\n", + "V_R_C=1.0; #Voltage drop across the collector resistance in V\n", + "\n", + "#Calculation\n", + "I_C=V_R_C/R_C; #Collector current in mA\n", + "I_B=I_C/beta; #Base current in mA\n", + "\n", + "#Result\n", + "print(\"The base current =%.3f mA\"%I_B);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The base current =0.022 mA\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.12: Page number 156\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "V_CC=8.0; #Collector supply voltage in V\n", + "R_C=800.0; #Resistance of the collector resistance in \u03a9\n", + "V_R_C=0.5; #Voltage drop across collector resistance in V\n", + "alpha=0.96; #current amplification factor\n", + "\n", + "#Calculation\n", + "V_CE=V_CC-V_R_C; #Collector to emitter voltage in V\n", + "I_C=V_R_C/R_C; #Collector current in A\n", + "I_C=I_C*1000; #Collector current in mA\n", + "beta=alpha/(1-alpha); #Base current amplification factor\n", + "I_B=I_C/beta; #Base current in mA\n", + "\n", + "#Result\n", + "print(\"Collector to emitter voltage = %.1f V\"%V_CE);\n", + "print(\"Base current= %.3f mA\"%I_B);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Collector to emitter voltage = 7.5 V\n", + "Base current= 0.026 mA\n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.13: Page number 156-157\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "V_CC=5; \t#Collector supply voltage in V\n", + "I_CBO=0.2; \t#Leakage current at collector base junction with emitter open, in \u03bcA\n", + "I_CEO=20.0; \t#Leakage current with base open, in \u03bcA\n", + "I_C=1.0; #Collector current in mA\n", + "I_C=I_C*1000; \t#Collector current in \u03bcA\n", + "\n", + "\n", + "#Calculation\n", + "alpha=1-(I_CBO/I_CEO);\t\t#current amplification factor\n", + "I_E=(I_C-I_CBO)/alpha; #Emitter current in \u03bcA\n", + "I_E=round(I_E,-1);\n", + "I_B=I_E-I_C; #Base current in \u03bcA\n", + "I_B=round(I_B,-1);\n", + "\n", + "#Result\n", + "print(\"Current amplification factor = %.2f \"%alpha);\n", + "print(\"The emitter curent =%d \u03bcA \"%I_E);\n", + "print(\"The base curent =%d \u03bcA \"%I_B);\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Current amplification factor = 0.99 \n", + "The emitter curent =1010 \u03bcA \n", + "The base curent =10 \u03bcA \n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.14: Page number 157\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "I_CEO=300.0; #Leakage current in common emitter configuration, in \u03bcA\n", + "beta=120.0; #Base current amplification factor\n", + "\n", + "#Calculation\n", + "alpha=beta/(1+beta); #Current amplification factor\n", + "alpha=round(alpha,3);\n", + "I_CBO=(1-alpha)*I_CEO; #Leakage current in common base configuration, in \u03bcA\n", + "\n", + "\n", + "#Result\n", + "print(\"Vale of I_CBO= %.1f \u03bcA\"%I_CBO);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Vale of I_CBO= 2.4 \u03bcA\n" + ] + } + ], + "prompt_number": 20 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.15: Page number 157\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "I_B=20.0; #Base current in \u03bcA\n", + "I_C=2.0; #Collector current in mA\n", + "beta=80.0; #Base current amplification factor\n", + "\n", + "#Calculation\n", + "I_CEO=I_C-(beta*I_B/1000); #Leakage current with base open, in mA \n", + "alpha=beta/(beta+1); #Current amplification factor\n", + "alpha=round(alpha,3);\n", + "I_CBO=(1-alpha)*I_CEO; #Leakage current with emitter open, in mA\n", + "\n", + "\n", + "#Result\n", + "print(\"Value of I_CBO=%.4f mA\"%I_CBO);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Value of I_CBO=0.0048 mA\n" + ] + } + ], + "prompt_number": 21 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.17: Page number 158\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "beta=150.0; \t#Base current amplification factor\n", + "R_B=10.0; \t#Base resistance in kilo ohm\n", + "R_C=100.0; \t#Collector resistance in kilo ohm\n", + "V_CC=10.0; #Collector supply voltage in V\n", + "V_BB=5.0; #Base supply voltage in V\n", + "V_BE=0.7; #Base to emitter voltage in V\n", + "\n", + "\n", + "#Calculation\n", + "I_B=(V_BB-V_BE)/R_B; #Base current in mA\n", + "I_C=beta*I_B; #Collector current in mA\n", + "V_CE=V_CC - (I_C/1000)*R_C; #Collector to emitter voltage in V\n", + "V_CB=V_CE-V_BE; #Collector to base voltage in V\n", + "\n", + "\n", + "#Result \n", + "print(\"Collector to base voltage, V_CB= %.2f V\"%V_CB);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Collector to base voltage, V_CB= 2.85 V\n" + ] + } + ], + "prompt_number": 22 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.18: Page number158-159\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "I_B=68.0; #Base current in \u03bcA\n", + "I_E=30.0; #Emitter current in mA\n", + "beta=440.0;\t #Base current amplification factor\n", + "\n", + "#Calculation\n", + "alpha=beta/(beta + 1); #current amplification factor\n", + "I_C_alpha=alpha*I_E;\t\t#Collector current using alpha rating, in mA\n", + "I_C_beta=beta*(I_B/1000.0); #Collector current using beta rating, in mA\n", + "\n", + "#Result\n", + "print(\"Collector current determined using alpha rating =%.2f mA\"%I_C_alpha);\n", + "print(\"Collector current determined using beta rating =%.2f mA\"%I_C_beta);\n", + "\n", + "#Note: In the textbook, the collector current obtained from beta rating is approximated to 29.93 mA\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Collector current determined using alpha rating =29.93 mA\n", + "Collector current determined using beta rating =29.92 mA\n" + ] + } + ], + "prompt_number": 27 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.19: Page number 159\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "I_C_max=500.0; #Maximum collector current in mA\n", + "beta_max=300.0; #Maximum base current amplification factor\n", + "\n", + "#Calculation\n", + "I_B_max=I_C_max/beta_max; #Maximum base current in mA\n", + "\n", + "\n", + "#Result\n", + "print(\"The maximum allowable value of base current = %.2f mA\"%I_B_max);\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The maximum allowable value of base current = 1.67 mA\n" + ] + } + ], + "prompt_number": 28 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.22 : Page number 167-168" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import matplotlib.pyplot as plt\n", + "\n", + "#Variable declaration\n", + "VCC=12.5; #Collector supply voltage, V\n", + "RC=2.5; #Collector resistor, k\u03a9\n", + "\n", + "#Calculation\n", + "#VCE=VCC-IC*RC\n", + "#For calculating VCE, IC=0\n", + "IC=0; #Collector current for maximum Collector-emitter voltage, mA\n", + "VCE_max=VCC-IC*RC; #Maximum collector-emitter voltage, V\n", + "\n", + "#For calculating VCE, IC=0\n", + "VCE=0; #Collector emitter voltage for maximum collector current, V\n", + "IC_max=(VCC-VCE)/RC; #Maximum collector current, mA\n", + "\n", + "\n", + "#Plotting of d.c load line\n", + "VCE_plot=[0,VCE_max]; #Plotting variable for VCE\n", + "IC_plot=[IC_max,0]; #Plotting variable for IC\n", + "p=plot(VCE_plot,IC_plot);\n", + "limit = plt.gca()\n", + "limit.set_xlim([0,15])\n", + "limit.set_ylim([0,6])\n", + "xlabel('VCE(V)');\n", + "ylabel('IC(mA)');\n", + "title('d.c load line');\n", + "plt.grid();\n", + "show(p);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "metadata": {}, + "output_type": "display_data", + "png": 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IHTt2ICwsDKmpqfj666/x6KOPqh3LIaPRCKPRiJEjRwIAUlJSkJ+fr3Iqx/bs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\nrkJggg==\n", + "text": [ + "<matplotlib.figure.Figure at 0x7f2eadbe6710>" + ] + } + ], + "prompt_number": 41 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.23 : Page number 168" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import matplotlib.pyplot as plt\n", + "\n", + "#Variable declaration\n", + "VCC=12.0; #Collector supply voltage, V\n", + "RC=6.0; #Collector resistor, k\u03a9\n", + "IB=20.0; #Zero signal base current, \u03bcA\n", + "beta=50.0; #Base current amplification factor\n", + "\n", + "#Calculation\n", + "#VCE=VCC-IC*RC\n", + "#For calculating VCE, IC=0\n", + "IC=0; #Collector current for maximum Collector-emitter voltage, mA\n", + "VCE_max=VCC-IC*RC; #Maximum collector-emitter voltage, V\n", + "\n", + "#For calculating VCE, IC=0\n", + "VCE=0; #Collector emitter voltage for maximum collector current, V\n", + "IC_max=(VCC-VCE)/RC; #Maximum collector current, mA\n", + "\n", + "\n", + "#Plotting of d.c load line\n", + "VCE_plot=[0,VCE_max]; #Plotting variable for VCE\n", + "IC_plot=[IC_max,0]; #Plotting variable for IC\n", + "p=plot(VCE_plot,IC_plot);\n", + "limit = plt.gca()\n", + "limit.set_xlim([0,15])\n", + "limit.set_ylim([0,5])\n", + "xlabel('VCE(V)');\n", + "ylabel('IC(mA)');\n", + "title('d.c load line');\n", + "plt.grid();\n", + "show(p);\n", + "\n", + "#Calculating Q-point\n", + "IC=beta*(IB/1000); #Collector current, mA\n", + "VCE=VCC-IC*RC; #Collector emitter voltage, V\n", + "\n", + "#Result\n", + "print(\"Operating point: IC=%dmA and VCE=%dV.\"%(IC,VCE));\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "metadata": {}, + "output_type": "display_data", + "png": 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+ "text": [ + "<matplotlib.figure.Figure at 0x7f8947e0f0d0>" + ] + }, + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Operating point: IC=1mA and VCE=6V.\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.24 : Page number 168" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "RC=4.0; #Collector load, k\u03a9\n", + "IC_Q=1.0; #Quiescent current, mA\n", + "\n", + "#Calculation\n", + "#(i)\n", + "VCC=10; #Collector supply voltage, V\n", + "VCE=VCC-IC*RC; #Collector emitter voltage, V\n", + "\n", + "print(\"(i) Operating point: VCE=%dV and IC=%dmA.\"%(VCE,IC) );\n", + "\n", + "#(ii)\n", + "RC=5.0; #Collector load, k\u03a9\n", + "VCE=VCC-IC*RC; #Collector emitter voltage, V\n", + "print(\"(ii) Operating point: VCE=%dV and IC=%dmA.\"%(VCE,IC) );\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) Operating point: VCE=6V and IC=1mA.\n", + "(ii) Operating point: VCE=5V and IC=1mA.\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##Example 8.25 : Page number 168-169" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "VCC=20.0; #Collector supply voltage, V\n", + "VBB=10.0; #Base supply voltage, V\n", + "RC=330.0; #Collector resistor, \u03a9\n", + "RB=47.0; #Base resistoe, k\u03a9\n", + "beta=200.0; #Base current amplification factor\n", + "VBE=0.7; #Base -emitter voltage, V\n", + "\n", + "#Calculation\n", + "#VBB-IB*RB-VBE=0\n", + "IB=round(((VBB-VBE)/RB)*1000,0); #Base current, \u03bcA\n", + "IC=beta*IB/1000; #Collector current, mA\n", + "VCE=VCC-IC*(RC/1000); #Collector-emitter voltage, V\n", + "\n", + "print(\"Operating point: IC=%.1fmA and VCE=%.2fV.\"%(IC,VCE));\n", + "\n", + "#For d.c load line\n", + "#VCE=VCC-IC*RC\n", + "#For calculating VCE, IC=0\n", + "IC=0; #Collector current for maximum Collector-emitter voltage, mA\n", + "VCE_max=VCC-IC*RC; #Maximum collector-emitter voltage, V\n", + "\n", + "#For calculating VCE, IC=0\n", + "VCE=0; #Collector emitter voltage for maximum collector current, V\n", + "IC_max=(VCC-VCE)/(RC/1000.0); #Maximum collector current, mA\n", + "\n", + "\n", + "#Plotting of d.c load line\n", + "VCE_plot=[0,VCE_max]; #Plotting variable for VCE\n", + "IC_plot=[IC_max,0]; #Plotting variable for IC\n", + "p=plot(VCE_plot,IC_plot);\n", + "limit = plt.gca()\n", + "limit.set_xlim([0,25])\n", + "limit.set_ylim([0,65])\n", + "xlabel('VCE(V)');\n", + "ylabel('IC(mA)');\n", + "title('d.c load line');\n", + "plt.grid();\n", + "show(p);\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Operating point: IC=39.6mA and VCE=6.93V.\n" + ] + }, + { + "metadata": {}, + "output_type": "display_data", + "png": 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+ "text": [ + "<matplotlib.figure.Figure at 0x7f8947baebd0>" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.26 : Page number 169-170" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "VCC=10.0; #Collector supply voltage, V\n", + "VEE=10.0; #Emitter supply voltage, V\n", + "RC=1.0; #Collector resistor, k\u03a9\n", + "RE=4.7; #Collector resistor, k\u03a9\n", + "RB=47.0; #Base resistoe, k\u03a9\n", + "beta=100.0; #Base current amplification factor\n", + "VBE=0.7; #Base -emitter voltage, V\n", + "\n", + "#Calculation\n", + "#-IB*RB-VBE-IE*RE+VEE=0\n", + "#AS, IC=beta*IB and IC~IE\n", + "IE=round((VEE-VBE)/(RE+(RB/beta)),1); #Emitter current, mA\n", + "IC=IE; #Collector current, mA\n", + "\n", + "#VCC-IC*RC-VCE-IE*RE+VEE=0\n", + "#IC~IE\n", + "VCE=VCC+VEE-IC*(RC+RE); #Collector-emitter voltage, V\n", + "\n", + "print(\"Operating point: IC=%.1fmA and VCE=%.2fV.\"%(IC,VCE));\n", + "\n", + "\n", + "#For d.c load line\n", + "#VCE=VCC-IC*RC\n", + "#For calculating VCE, IC=0\n", + "IC=0; #Collector current for maximum Collector-emitter voltage, mA\n", + "VCE_max=VCC+VEE-IC*(RC+RE); #Maximum collector-emitter voltage, V\n", + "\n", + "#For calculating VCE, IC=0\n", + "VCE=0; #Collector emitter voltage for maximum collector current, V\n", + "IC_max=(VCC+VEE-VCE)/(RC+RE); #Maximum collector current, mA\n", + "\n", + "\n", + "#Plotting of d.c load line\n", + "VCE_plot=[0,VCE_max]; #Plotting variable for VCE\n", + "IC_plot=[IC_max,0]; #Plotting variable for IC\n", + "p=plot(VCE_plot,IC_plot);\n", + "limit = plt.gca()\n", + "limit.set_xlim([0,25])\n", + "limit.set_ylim([0,5])\n", + "xlabel('VCE(V)');\n", + "ylabel('IC(mA)');\n", + "title('d.c load line');\n", + "plt.grid();\n", + "show(p);\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Operating point: IC=1.8mA and VCE=9.74V.\n" + ] + }, + { + "metadata": {}, + "output_type": "display_data", + "png": 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CpCTzBRCodFx4R8VeJMh17mzm4w8ZAj17wt13Q16e3VGJv6lnLxJCDhyA8eNN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\nQmCC\n", + "text": [ + "<matplotlib.figure.Figure at 0x7f8947c78950>" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.27 : Page number 170-171" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "VEE=10.0; #Emitter supply voltage, V\n", + "IE=1.8; #Emitter current, mA\n", + "RE=4.7; #Emitter resistor, k\u03a9\n", + "VBE=0.7; #Base-emitter voltage, V\n", + "VCC=10.0; #Collector supply voltage, V\n", + "IC=1.8; #Collector current, mA\n", + "RC=1.0; #Collector resistor, k\u03a9\n", + "\n", + "\n", + "#Calculation\n", + "#(i)\n", + "VE=-VEE+IE*RE; #Emitter voltage, V\n", + "\n", + "#(ii)\n", + "VB=VEE+VBE; #Base voltage, V\n", + "\n", + "#(iii)\n", + "VC=VCC-IC*RC; #Collector voltage, V\n", + "\n", + "\n", + "#Result\n", + "print(\"(i) Emitter voltage=%.2fV.\"%VE);\n", + "print(\"(i) Base voltage=%.1fV.\"%VB);\n", + "print(\"(i) Collector voltage=%.1fV.\"%VC);\n", + "\n", + "#Note: In the textbook, VB=VE+VBE has been written, which is worng. It should be VB=VEE+VBE. " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) Emitter voltage=-1.54V.\n", + "(i) Base voltage=10.7V.\n", + "(i) Collector voltage=8.2V.\n" + ] + } + ], + "prompt_number": 39 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "\n", + "Example 8.28: Page number 173-174\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "V_BE_change=200.0; #Change in base-emitter voltage in mV\n", + "I_B_change=100.0; #Change in base current in \u03bcA\n", + "\n", + "#Calculations\n", + "Ri=V_BE_change/I_B_change; #Input resistance in k\u03a9\n", + "\n", + "#Result\n", + "print(\"Input resistance =%d k\u03a9\"%Ri);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Input resistance =2 k\u03a9\n" + ] + } + ], + "prompt_number": 35 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.29; Page number 174\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "V_CE_final=10.0;\t\t\t#Final value of collector-emitter voltage in V\n", + "V_CE_initial=2.0; #Initial value of collector-emitter voltage in V\n", + "I_C_final=3.0; #Final value of collector current in mA\n", + "I_C_initial=2.0; #Initial value of collector current in mA\n", + "\n", + "#Calculations\n", + "V_CE_change=V_CE_final-V_CE_initial;\t\t#Change in collector to emitter voltage in V\n", + "I_C_change=I_C_final-I_C_initial; #Change in collector current in mA\n", + "R0=V_CE_change/I_C_change; #Output resistance in k\u03a9\n", + "\n", + "#Result\n", + "print(\"The output resistance =%dk\u03a9\"%R0);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The output resistance =8k\u03a9\n" + ] + } + ], + "prompt_number": 34 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.30: Page number 174\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "R_C=2.0;\t\t#Collector load in kilo ohm\n", + "R_i=1.0;\t\t#Input resistance in kilo ohm\n", + "R_AC=R_C; #Effective collector load for single stage in kilo ohm(appoximately equal to collector load for single stage)\n", + "beta=50.0; #Current gain\n", + "\n", + "#Calculations\n", + "A_v=beta*(R_AC/R_i);\t\t#Voltage gain of the amplifier\n", + "\n", + "#Result \n", + "print(\"The voltage gain of the amplifier =%d \"%A_v);\t\t\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The voltage gain of the amplifier =100 \n" + ] + } + ], + "prompt_number": 19 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.31: Page number 175-176\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "V_CC=20;\t\t#Collector supply voltage in V\n", + "R_C=1; #Collector resistance in kilo ohm\n", + "V_knee_Si=1;\t\t#Knee voltage of V_CE for Si in V \n", + "V_knee_Ge=0.5;\t\t#Knee voltage of V_CE for Ge in V\n", + "\n", + "#Calculations\n", + "I_C_sat_Si=(V_CC-V_knee_Si)/R_C;\t\t#Saturation (maximum) value of collector current in mA (for Si transistor)\n", + "I_C_sat_Ge=(V_CC-V_knee_Ge)/R_C;\t\t#Saturation (maximum) value of collector current in mA (for Ge transistor)\n", + "I_C_sat=(V_CC)/R_C;\t\t\t\t#Saturation (maximum) value of collector current in mA (neglecting knee voltage)\n", + "V_CE_cut_off=V_CC; #Collector to emitter voltage in cutoff when base current=0, in V\n", + "\n", + "#Result\n", + "print(\"Collector current during saturation = %d mA\"%I_C_sat);\n", + "print(\"Collector emitter voltage during cutoff = %d V.\"%V_CE_cut_off);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Collector current during saturation = 20 mA\n", + "Collector emitter voltage during cutoff = 20 V.\n" + ] + } + ], + "prompt_number": 20 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.32: Page number 176-177\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "V_CC=12.0;\t\t#Collector supply voltage in V\n", + "V_EE=12.0;\t\t#Emitter supply voltage in V\n", + "R_C=750.0;\t\t#Collector resistance in ohm\n", + "R_E=1.5;\t\t#Emitter resistance in kilo ohm\n", + "R_B=100.0;\t\t#Base resistance in ohm\n", + "beta=200;\t\t#base current amplification factor\n", + "\n", + "#Calculations\n", + "\n", + "#Applying Kirchhoff's voltage law to the collector side of the circuit\n", + "#using the equation: Vcc -IcRc-Vce -IeRe+Vee=0\n", + "#we get Vce=Vcc+Vee-Ic(Rc+Re), [Ie=Ic, approximately]\n", + "#We get Vce(off), when Ic=0;\n", + "\n", + "I_C_Vce_off=0;\t\t\t\t\t#Collector current for Vce(off) in mA\n", + "V_CE_off=V_CC+V_EE -(I_C_Vce_off * (R_C +R_E));\t#Collector to emitter voltage in V, during transistor in off state\n", + "\n", + "#We get Ic(sat), when Vce=0\n", + "V_CE_Ic_sat=0;\t\t\t\t\t\t#Collector to emitter voltage for saturation current of collector in V\n", + "I_C_sat=(V_CC+V_EE-V_CE_Ic_sat)/(R_C+(R_E*1000));\t#Saturated collector current in A \n", + "I_C_sat=I_C_sat*1000;\t\t\t\t\t#Saturated collector current in mA\n", + "#Result\n", + "print(\"Vce(off)= %dV\"%V_CE_off);\n", + "print(\"Ic(sat) = %.2f mA\"%I_C_sat);\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Vce(off)= 24V\n", + "Ic(sat) = 10.67 mA\n" + ] + } + ], + "prompt_number": 21 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.33 : Page number 177\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "V_knee=0.2;\t\t\t\t#Knee voltage of collector-emitter voltage in V\n", + "V_CC=10.0;\t\t\t\t#Collector supply voltage in V\n", + "V_BB=3.0;\t\t\t\t#Base supply voltage in V\n", + "V_BE=0.7;\t\t\t\t#Base-emitter voltage in V \t\n", + "R_B=10.0;\t\t\t\t#Base resistor's resistance in kilo ohm\n", + "R_C=1.0;\t\t\t\t#Collector resistor's resistance in kilo ohm\n", + "beta=50.0;\t\t\t\t#base current amplification factor\n", + "\n", + "#Calculations\n", + "\n", + "#applying Kirchhoff's voltage law along the collector side of the circuit,\n", + "#We get Vcc-Ic(sat)*Rc-V_knee=0\n", + "#From the above equation, we get:\n", + "I_C_sat=(V_CC-V_knee)/R_C;\t\t#Saturated collector current in mA\n", + "\n", + "#Applying Kirchhoff's voltage law along base emitter side,\n", + "#We get VBB-IB*RB-VBE=0;\n", + "#From the above equation, we get:\n", + "I_B=(V_BB-V_BE)/R_B;\t\t\t#Base current in mA\n", + "\n", + "\n", + "I_C=beta*I_B\t\t\t\t#Collector current in mA\n", + "\n", + "#Result\n", + "if(I_C>I_C_sat):\n", + "\tprint(\"The base current is large enough to produce Ic greater than Ic(sat), therefore the transistor is saturated.\");\n", + "else:\n", + "\tprint(\"The base current is not large enough to produce Ic greater than Ic(sat), therefore the transistor isn't saturated. \");\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The base current is large enough to produce Ic greater than Ic(sat), therefore the transistor is saturated.\n" + ] + } + ], + "prompt_number": 22 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.34: Page number 177-178\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "#Variable declaration\n", + "V_CC=10.0;\t\t\t\t#Collector supply voltage in V\n", + "V_BE=0.95;\t\t\t\t#Base-emitter voltage in V \t\n", + "I_B=100.0;\t\t\t\t#Base current in microAmp\n", + "R_C=970.0;\t\t\t\t#Collector resistor's resistance in ohm\n", + "beta=100.0;\t\t\t\t#base current amplification factor\n", + "\n", + "#Calculations\n", + "I_C=(I_B/1000)*beta;\t\t\t\t#Collector current in mA \n", + "\n", + "#Applying Kirchhoff's voltage law along collector side\n", + "#We get Vcc-IcRc-Vce=0\n", + "#From the above equation, we get:\n", + "\n", + "V_CE=V_CC-((I_C/1000)*R_C);\t\t\t\t#Collector-emitter voltage in V\n", + "\n", + "#From the equation, V_CE=V_CB+V_BE,\n", + "V_CB=V_CE-V_BE;\t\t\t\t\t\t#Collector-base voltage in V\n", + "\n", + "\n", + "#Result\n", + "if(V_CB<0 and V_BE >0):\n", + "\tprint(\"As both collector-base and emitter-base junction are forward biased, the transistor is operating in the saturation region. \");\n", + "else:\n", + "\tprint(\"No. The transistor isn't operating in the saturation region.\");\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "As both collector-base and emitter-base junction are forward biased, the transistor is operating in the saturation region. \n" + ] + } + ], + "prompt_number": 23 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.35: Page number 178\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "V_CC=10.0;\t\t\t\t#Collector supplu voltage in V\n", + "V_BE=0.7;\t\t\t\t#Base-emitter voltage in V\n", + "R_B=50.0;\t\t\t\t#Base resistor's resistance in kilo ohm\n", + "R_C=2.0;\t\t\t\t#Collector resistor's resistance in kilo ohm\n", + "beta=200.0;\t\t\t\t#Base current amplification factor\n", + "\n", + "#Calculations\n", + "\n", + "#Applying Kirchhoff's voltage law along the collector side,\n", + "#We get, Vcc-Ic(sat)*Rc-Vce=0;\n", + "#From the above equation, we get:\n", + "#I_C_sat=(V_CC-V_CE)/R_C, but as transistor goes into saturation, Vce=0;\n", + "\n", + "V_CE=0;\t\t\t\t\t\t#Collector-emiter voltage in V, for transistor in saturation \n", + "I_C_sat=(V_CC-V_CE)/R_C;\t\t\t#Saturated collector current in mA\n", + "\n", + "I_B=I_C_sat/beta;\t\t\t\t#Base current in mA\n", + "\n", + "#Applying Kirchhoff's voltage law to the base circuit,\n", + "#We get, VBB - IB*RB - VBE=0\n", + "#From the above equation. we get:\n", + "V_BB=V_BE+ I_B*R_B;\t\t\t\t#Base supply voltage to put transistor in saturation, in V\n", + "\n", + "#Result\n", + "print(\"Therefore, for putting transistor in saturation, VBB >= %.2f V\"%V_BB);\n", + " \n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Therefore, for putting transistor in saturation, VBB >= 1.95 V\n" + ] + } + ], + "prompt_number": 24 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.36: Page number 178-179\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "V_CC=10.0;\t\t\t#Collector supply voltage in V\n", + "V_BB=2.7;\t\t\t#Base supply voltage in V\n", + "V_BE=0.7;\t\t\t#Base-emitter voltage in V\n", + "beta=100.0;\t\t\t#Base current amplification factor\n", + "R_E=1.0;\t\t\t#Emitter resistor's resistance in kilo ohm\n", + "\n", + "\n", + "#Calcultaion\t\n", + "V_B=V_BB;\t\t\t#Base voltage in V\n", + "V_E=V_B-V_BE;\t\t\t#Emitter voltage in V\n", + "I_E=V_E/R_E;\t\t\t#Emitter current in mA\n", + "I_C=I_E;\t\t\t#Collector current (approximately equal to emitter current) in mA\n", + "I_B=I_C/beta;\t\t\t#Base current in mA\n", + "\n", + "#Case (i):\n", + "R_C=2;\t\t\t\t#Collector resistor's resistance in kilo ohm\n", + "\n", + "#Assuming transistor to be in active state\n", + "#Applying Kirchhoff's voltage law along collector side,\n", + "#We get,Vcc-IcRc=Vc,\n", + "\n", + "V_C=V_CC-I_C*R_C;\t\t#Collector voltage in V\n", + "\n", + "if(V_C>V_E):\n", + "\tprint(\"(i)Our assumption was correct, the transistor is in active state for Rc=2 kilo ohm.\");\n", + "elif(V_C<V_E):\n", + "\tprint(\"(i)Our assumption was wrong, the transistor is in saturation for Rc=2 kilo ohm.\");\n", + "elif(V_C==V_E):\n", + "\tprint(\"(i)The transistor is at the edge of saturation for Rc=2 kilo ohm, therefore relation between transistor currents are same for both saturation and active state.\");\n", + "\n", + "#Case (ii):\n", + "R_C=4;\t\t\t\t#Collector resistor's resistance in kilo ohm\n", + "\n", + "#Assuming transistor to be in active state\n", + "#Applying Kirchhoff's voltage law along collector side,\n", + "#We get,Vcc-IcRc=Vc,\n", + "\n", + "V_C=V_CC-I_C*R_C;\t\t#Collector voltage in V\n", + "if(V_C>V_E):\n", + "\tprint(\"(ii)Our assumption was correct, the transistor is in active state for Rc=4 kilo ohm.\");\n", + "elif(V_C==V_E):\n", + "\tprint(\"(ii)The transistor is at the edge of saturation for Rc=4 kilo ohm, therefore relation between transistor currents are same for both saturation and active state.\");\n", + "elif(V_C<V_E):\n", + "\tprint(\"(ii)Our assumption was wrong, the transistor is in saturation for Rc=4 kilo ohm.\");\n", + "\n", + "\n", + "#Case (iii):\n", + "R_C=8;\t\t\t\t#Collector resistor's resistance in kilo ohm\n", + "\n", + "#Assuming transistor to be in active state\n", + "#Applying Kirchhoff's voltage law along collector side,\n", + "#We get,Vcc-IcRc=Vc,\n", + "\n", + "V_C=V_CC-I_C*R_C;\t\t#Collector voltage in V\n", + "if(V_C>V_E):\n", + "\tprint(\"(iii)Our assumption was correct, the transistor is in active state for Rc=8 kilo ohm.\");\n", + "elif(V_C<V_E):\n", + "\tprint(\"(iii)Our assumption was wrong, the transistor is in saturation for Rc=8 kilo ohm.\");\n", + "elif(V_C==V_E):\n", + "\tprint(\"(iii)The transistor is at the edge of saturation for Rc=8 kilo ohm, therefore relation between transistor currents are same for both saturation and active state.\");\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i)Our assumption was correct, the transistor is in active state for Rc=2 kilo ohm.\n", + "(ii)The transistor is at the edge of saturation for Rc=4 kilo ohm, therefore relation between transistor currents are same for both saturation and active state.\n", + "(iii)Our assumption was wrong, the transistor is in saturation for Rc=8 kilo ohm.\n" + ] + } + ], + "prompt_number": 25 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.37 : Page number 179-180" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "V_CC=15.0;\t\t\t#Collector supply voltage in V\n", + "R_C=10.0;\t\t\t#Collector resistor's resistance in kilo ohm\n", + "V_BE=0.7;\t\t\t#Base-emitter voltage in V\n", + "beta=100.0;\t\t\t#Base current amplification factor\n", + "R_E=1.0;\t\t\t#Emitter resistor's resistance in kilo ohm\n", + "\n", + "\n", + "#Calculation\t\n", + "\n", + "#Case (i):\n", + "V_BB=0.5;\t\t\t#Base supply voltage in V\n", + "VB=V_BB; #Base voltage, V\n", + "print(\"(i) Base voltage =%.1fV is less than VBE=%.1fV, therefore, transistor is cut-off.\"%(VB,V_BE));\n", + "\n", + "\n", + "#Case (ii):\n", + "V_BB=1.5;\t\t\t#Base supply voltage in V\n", + "VB=V_BB; #Base voltage, V\n", + "VE=VB-V_BE; #Emitter voltage, V\n", + "IE=round(VE/R_E,1); #Emitter current, mA\n", + "#Assuming transistor to be in active state\n", + "#Applying Kirchhoff's voltage law along collector side,\n", + "IC=IE; #Collector current, mA\n", + "IB=IC/beta; #Base current, mA\n", + "VC=V_CC-IC*R_C; #Collector voltage, V\n", + "print(VE,IE,VC);\n", + "print(\"(ii) VC=%dV > VE=%.1fV, therefore the transistor is active. Our assumption was correct.\"%(VC,VE));\n", + "\n", + "#Case (iii):\n", + "V_BB=3; \t\t\t#Base supply voltage in V\n", + "VB=V_BB; #Base voltage, V\n", + "VE=VB-V_BE; #Emitter voltage, V\n", + "IE=round(VE/R_E,1); #Emitter current, mA\n", + "#Assuming transistor to be in active state\n", + "#Applying Kirchhoff's voltage law along collector side,\n", + "IC=IE; #Collector current, mA\n", + "IB=IC/beta; #Base current, mA\n", + "VC=V_CC-IC*R_C; #Collector voltage, V\n", + "\n", + "print(\"(iii) VC=%dV < VE=%.1fV, therefore the transistor is saturated. Our assumption was wrong.\"%(VC,VE));" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) Base voltage =0.5V is less than VBE=0.7V, therefore, transistor is cut-off.\n", + "(0.8, 0.8, 7.0)\n", + "(ii) VC=7V > VE=0.8V, therefore the transistor is active. Our assumption was correct.\n", + "(iii) VC=-8V < VE=2.3V, therefore the transistor is saturated. Our assumption was wrong.\n" + ] + } + ], + "prompt_number": 41 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.38: Page number 181\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "P_D_max=100.0;\t\t\t#Maximum power dissipation of a transistor in mW\n", + "V_CE=20.0;\t\t\t#Collector emitter voltage in V\n", + "\n", + "#Calculation\n", + "#As power=curent*voltage\n", + "#P_D_max=I_C_max*V_CE\n", + "#From the above equation, we get:\n", + "\n", + "I_C_max=P_D_max/V_CE;\t\t#Maximum collector current that can be allowed without destruction of the transistor, in mA\n", + "\n", + "#Result\n", + "print(\"Maximum collector current that can be allowed without destruction of the transistor = %d mA.\"%I_C_max); \n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Maximum collector current that can be allowed without destruction of the transistor = 5 mA.\n" + ] + } + ], + "prompt_number": 27 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.39: Page number 181\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "V_CC=5.0;\t\t\t\t#Collector supply voltage in V\n", + "V_BB=5.0;\t\t\t\t#Base supply voltage in V\n", + "V_BE=0.7;\t\t\t\t#Base-emitter voltage in V\n", + "R_B=1.0;\t\t\t\t#Base resistor's resistance in kilo ohm\n", + "R_C=0;\t\t\t\t\t#Collector resistor's resistance in kilo ohm\n", + "beta=200.0;\t\t\t\t#base current amplification factor\n", + "\n", + "#Calculation\n", + "\n", + "#Applying Kirchhoff's voltage law along base circuit<\n", + "#We get, VBB- IB*RB - VBE=0.\n", + "#From the above equation, we get:\n", + "\n", + "I_B=(V_BB-V_BE)/R_B;\t\t\t#Base current in mA\n", + "\n", + "I_C=beta*I_B;\t\t\t\t#Collector current in mA\n", + "\n", + "#Applying Kirchhoff's voltage law along collector circuit:\n", + "\n", + "V_CE=V_CC-I_C*R_C;\t\t\t#Collector-emitter voltage in V\n", + "\n", + "#As power=curent*voltage\n", + "#P_D=I_C*V_CE\n", + "#From the above equation, we get:\n", + "P_D=V_CE*I_C;\t\t\t\t#Power dissipated in mW\n", + "P_D=P_D/1000;\t\t\t\t#Power dissipated in W\n", + "\n", + "#Result\n", + "print(\"Power dissipated = %.1fW\"%P_D);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Power dissipated = 4.3W\n" + ] + } + ], + "prompt_number": 28 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.40: Page number 181-182\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "V_CC=5.0;\t\t\t\t#Collector supply voltage in V\n", + "V_BB=1.0;\t\t\t\t#Base supply voltage in V\n", + "V_BE=0.7;\t\t\t\t#Base-emitter voltage in V\n", + "R_B=10.0;\t\t\t\t#Base resistor's resistance in kilo ohm\n", + "R_C=1.0;\t\t\t\t\t#Collector resistor's resistance in kilo ohm\n", + "beta=100.0;\t\t\t\t#base current amplification factor\n", + "\n", + "#Calculation\n", + "\n", + "#Applying Kirchhoff's voltage law along base circuit<\n", + "#We get, VBB- IB*RB - VBE=0.\n", + "#From the above equation, we get:\n", + "\n", + "I_B=(V_BB-V_BE)/R_B;\t\t\t#Base current in mA\n", + "\n", + "I_C=beta*I_B;\t\t\t\t#Collector current in mA\n", + "\n", + "#Applying Kirchhoff's voltage law along collector circuit:\n", + "\n", + "V_CE=V_CC-I_C*R_C;\t\t\t#Collector-emitter voltage in V\n", + "\n", + "#As power=curent*voltage\n", + "#P_D=I_C*V_CE\n", + "#From the above equation, we get:\n", + "P_D=V_CE*I_C;\t\t\t\t#Power dissipated in mW\n", + "\n", + "\n", + "#Result\n", + "print(\"Power dissipated = %.0fmW\"%P_D);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Power dissipated = 6mW\n" + ] + } + ], + "prompt_number": 29 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.41 : Page number 182" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "VBB=5.0; #Base supply voltage, V\n", + "RB=22.0; #Base resistor, kilo ohm\n", + "RC=1.0; #Collector resistor, kilo ohm\n", + "beta=100.0; #Base current amplification factor\n", + "VBE=0.7; #Base-emitter voltage, V\n", + "PD_max=800.0; #Maximum power dissipation, mW\n", + "VCE_max=15.0; #Maximum collector-emitter voltage, V\n", + "IC_max=100.0; #Maximum collector current, mA\n", + "\n", + "#Calculation\n", + "IB=((VBB-VBE)/RB)*1000; #Base current, \u03bcA\n", + "IC=beta*IB/1000; #Collector current, mA\n", + "\n", + "print(\"IC=%.1fmA is much less than IC_max=%dmA. Therefore, will not change with VCC and current rating is not exceeded.\"%(IC,IC_max));\n", + "\n", + "#VCC=VCE+IC*RC\n", + "VCC_max=VCE_max+IC*RC; #Maximum value of Collector supply voltage, V\n", + "PD=VCE_max*IC; #Power dissipation, mW\n", + "\n", + "print(\"PD=%dmW is less than PD_max=%dmW. Therefore, power rating is not exceeded.\"%(PD,PD_max));\n", + "\n", + "print(\"If base current is removed, transistor will turn off. Hence, VCE_max will be exceeded because entire supply voltage VCC will be dropped across the transistor.\");" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "IC=19.5mA is much less than IC_max=100mA. Therefore, will not change with VCC and current rating is not exceeded.\n", + "PD=293mW is less than PD_max=800mW. Therefore, power rating is not exceeded.\n", + "If base current is removed, transistor will turn off. Hence, VCE_max will be exceeded because entire supply voltage VCC will be dropped across the transistor.\n" + ] + } + ], + "prompt_number": 33 + }, + { + "cell_type": "code", + "collapsed": false, + "input": [], + "language": "python", + "metadata": {}, + "outputs": [] + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter9_2.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter9_2.ipynb new file mode 100644 index 00000000..e8168ab8 --- /dev/null +++ b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter9_2.ipynb @@ -0,0 +1,1907 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:1e5463fc2e8c67a2f9099cf3f6c078bc1c9dccccd63589a75ad6ce5024fe6432" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 9 : TRANSISTOR BIASING" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.1: Page number 195-196" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "V_CC=6.0; #Collector supply voltage\n", + "R_C=2.5; #Collector load in k\u2126\n", + "\n", + "#Calculations\n", + "\n", + "#(i)\n", + "#For faithful amplification Vce (collector-emitter voltage)> 1V for Si transistor.\n", + "V_CE_max=1; #Maximum allowed collector-emitter voltage for faithful amplification, in V.\n", + "V_Rc_max=V_CC-V_CE_max; #maximum voltage drop across collector load in V.\n", + "I_C_max=V_Rc_max/R_C; #Maximum allowed collector current in mA\n", + "\n", + "#(ii)\n", + "IC_min_zero_signal=I_C_max/2; #Minimum zero signal collector current in mA\n", + "\n", + "#Results\n", + "print(\"The maximum allowed collector current during application of signal for faithful amplification = %d mA.\"%I_C_max);\n", + "print(\"The minimum zero signal collector current required = %d mA.\"%IC_min_zero_signal);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The maximum allowed collector current during application of signal for faithful amplification = 2 mA.\n", + "The minimum zero signal collector current required = 1 mA.\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.2: Page number 196\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "VCC=13.0; #Collector supply voltage in V\n", + "V_knee=1.0; #Knee voltage in V\n", + "R_C=4.0; #Collector load in k\u2126\n", + "rate_IC_VBE=5.0; #Rate of change of collector current IC with base-emitter voltage VBE in mA/V.\n", + "beta=100.0; #base current amplification factor\n", + "\n", + "\n", + "#Calculations\n", + "V_Rc_max=VCC-V_knee; #Maximum allowed voltage across collector load in V\n", + "I_C_max=V_Rc_max/R_C; #Maximum allowed collector current in mA\n", + "I_B_max=I_C_max/beta; #Maximum base current in mA\n", + "I_B_max=I_B_max*1000; #Maximum base current in \ud835\udf07A\n", + "\n", + "V_B_max=I_C_max/rate_IC_VBE; #Maximum base voltage signal in V\n", + "V_B_max=V_B_max*1000; #Maximum base voltage signal in mV\n", + "\n", + "#Results\n", + "print(\"Maximum base current =%d \ud835\udf07A.\"%I_B_max);\n", + "print(\"Maximum input signal voltage =%d mV.\"%V_B_max);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Maximum base current =30 \ud835\udf07A.\n", + "Maximum input signal voltage =600 mV.\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.3: Page number 200-201" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "VCC=9.0; #Colector supply voltage in V\n", + "VBB=2.0; #Base supply voltage in V\n", + "R_B=100.0; #Base resistor's resistance in k\u2126\n", + "R_C=2.0; #Collector load in k\u2126\n", + "beta=50.0; #base current amplification factor\n", + "\n", + "#Calculations\n", + "\n", + "#Case (i):\n", + "\n", + "#Applying Kirchhoff's law to the input circuit\n", + "#We get, IB*RB +VBE =VBB.\n", + "#Neglecting the small base-emitter voltage, we get:\n", + "I_B=VBB/R_B; #Base current in mA\n", + "I_C=beta*I_B; #Collector current in mA\n", + "\n", + "print(\"Collector current = %dmA\"%I_C);\n", + "\n", + "#Applying Kirchhoff's law to the output ciruit\n", + "#We get, IC*RC + VCE= VCC.\n", + "#From the above equation, we get:\n", + "V_CE=VCC-I_C*R_C; #Collector emitter voltage in V\n", + "\n", + "print(\"Collector emitter voltage =%dV.\"%V_CE);\n", + "\n", + "\n", + "#Case (ii):\n", + "\n", + "R_B=50.0;\n", + "I_B=VBB/R_B;\n", + "I_C=beta*I_B;\n", + "V_CE=VCC - I_C*R_C;\n", + "\n", + "print(\"The new operating point for base resistor RB=50 k\u2126 is, VCE=%dV and IC=%dmA.\"%(V_CE,I_C));\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Collector current = 1mA\n", + "Collector emitter voltage =7V.\n", + "The new operating point for base resistor RB=50 k\u2126 is, VCE=5V and IC=2mA.\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.4: Page number 201-202" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import matplotlib.pyplot as plt\n", + "\n", + "#variable declaration\n", + "beta=100.0; #base current amplification factor\n", + "VCC=6.0; #Collector suply voltagein V\n", + "VBE=0.7 #Base emitter voltage in V\n", + "R_B=530.0; #Base resistor's resistance in k\u2126 .\n", + "R_C=2.0; #Collector resistor's resistance in k\u2126 .\n", + "\n", + "#Calculation\n", + "#D.C load line equation : VCE=VCC-IC*RC;\n", + "#Calculating maximum VCE ,by IC=0;\n", + "I_C_Vce_max=0; #Collector current for maximum collector-emitter voltage, in mA\n", + "VCE_max=VCC;-I_C_Vce_max*R_C; #Maximum collector-emitter voltage in V\n", + "\n", + "\n", + "#Calculating maximum collector current IC,by VCE=0;\n", + "V_CE_IC_max=0; #Collector-emitter voltage for maximum collector current, in V \n", + "I_C_max=(VCC-V_CE_IC_max)/R_C; #Maximum collector current in mA\n", + "\n", + "\n", + "#Operating point:\n", + "#For input circuit, applying Kirchhoff's law, We get,\n", + "#VCC=IB*RB + VBE.\n", + "#From the above equation,\n", + "IB=(VCC-VBE)/R_B; #Base current in mA\n", + "IC=beta*IB; #Collector current\n", + "\n", + "#From the output circuit, applying Kirchhoff's law, we get:\n", + "VCE=VCC-IC*R_C; #Collector-emitter voltage in V\n", + "\n", + "\n", + "#Stability factor\n", + "SF=beta+1; \n", + "\n", + "#Result\n", + "print(\"Operating point: VCE= %dV and IC=%d mA\"%(VCE,IC));\n", + "print(\"Stability factor= %d.\"%SF);\n", + "\n", + "\n", + "#plot\n", + "limit = plt.gca()\n", + "limit.set_xlim([0,10])\n", + "limit.set_ylim([0,5])\n", + "VCE=[i for i in range(0,(int)(VCC+1))]; #Plot variable for V_CE\n", + "IC=[((VCC-i)/(R_C)) for i in (VCE[:])]; #Plot variable for I_C\n", + "\n", + "p=plot(VCE,IC);\n", + "xlabel(\"VCE(V)\");\n", + "ylabel(\"IC(mA)\");\n", + "title(\"d.c load line\");\n", + "show(p);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Operating point: VCE= 4V and IC=1 mA\n", + "Stability factor= 101.\n" + ] + }, + { + "metadata": {}, + "output_type": "display_data", + "png": 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+ "text": [ + "<matplotlib.figure.Figure at 0x7f8eea67df10>" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.5: Page number 202" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "VCC=12.0; #Collector supply voltage in V\n", + "beta=100.0; #base current amplification factor\n", + "I_C_zero_signal=1.0; #zero signal collector current in mA\n", + "VBE=0.3; #Base-emitter voltage of Ge transistor in V\n", + "\n", + "#calculations\n", + "\n", + "#Case(i)\n", + "I_B_zero_signal=I_C_zero_signal/beta; #Zero signal base current in mA\n", + "\n", + "#applying the Kirchhoff's law along input circuit:\n", + "#We get, VCC=IB*RB +VBE\n", + "#From the above equation we get,\n", + "R_B=(VCC-VBE)/I_B_zero_signal; #Required base resistor's resistance in k\u2126\n", + "\n", + "print(\"Value of base resistor for operating the given Ge transistor at zero signal IC=1mA is = %d k\u2126\"%R_B);\n", + "\n", + "\n", + "\n", + "#Case(ii)\n", + "beta=50;\n", + "I_B=(VCC-VBE)/R_B; #Base current of another transistor with beta=50, in mA\n", + "I_C_zero_signal=beta*I_B; #Zero signal collector current for beta=50 , in mA\n", + "\n", + "print(\"The new value of zero signal collector current =%.1fmA\"%I_C_zero_signal);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Value of base resistor for operating the given Ge transistor at zero signal IC=1mA is = 1170 k\u2126\n", + "The new value of zero signal collector current =0.5mA\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.6:Page number 202-203" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Varaible declaration\n", + "VCC=10.0; #Collector supply voltage in V\n", + "VBE=0; #Base emitter voltage in V(considering itas zero due to it's small value)\n", + "R_B=1.0; #Base resistor's resistance in M\u2126\n", + "R_C=2.0; #Collector resistor's resistance in k\u2126 \n", + "R_E=1.0; #Emitter resistor's resistance in k\u2126\n", + "beta=100.0; #Base current amplification factor\n", + "\n", + "#Calculations\n", + "#using Kirchhoff's law in the input circuit, we get:\n", + "#VCC=IB*RB +VBE +IE*RE\n", + "#Since, IE=(beta +1)*I_B\n", + "#From the above equation we get:\n", + "I_B=round((VCC-VBE)/((beta + 1)*R_E + R_B*1000),4); #Base current in mA\n", + "I_C=round(beta*I_B,2); #Collector current in mA\n", + "I_E=I_B+I_C; #Emitter current in mA\n", + "\n", + "#Result\n", + "print(\"Base current =%.4f mA\"%I_B);\n", + "print(\"Collector current =%.2f mA\"%I_C);\n", + "print(\"Emitter current =%.3f mA\"%I_E);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Base current =0.0091 mA\n", + "Collector current =0.91 mA\n", + "Emitter current =0.919 mA\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.7: Page number 203-204" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "VCE=8.0; #Collector-emitter voltage at operating point in V\n", + "IC=2.0; #Colector current at operating point in mA\n", + "VCC=15.0; #Collector supply voltagein V\n", + "beta=100.0; #base current amplification factor\n", + "VBE=0.6; #base emitter voltage in V\n", + "\n", + "#Calculations\n", + "#Applying Kirchhoff's law along the output circuit,\n", + "#we get, VCC=VCE+IC*RC.\n", + "#So, from above equation we get:\n", + "RC=(VCC-VCE)/IC; #Collector resistor's resistance in k\u2126 .\n", + "IB=IC/beta; #Base current in mA\n", + "\n", + "#Applying Kirchhoff's law along the input circuit,\n", + "#we get, VCC=IB*RB + VBE\n", + "#So, from the above equation:\n", + "RB=(VCC-VBE)/IB; #Base resistor's resistance in k\u2126 .\n", + "\n", + "\n", + "#Results\n", + "print(\"Collector load =%.1f k\u2126 .\"%RC);\n", + "print(\"Base resistor=%d k\u2126 .\"%RB);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Collector load =3.5 k\u2126 .\n", + "Base resistor=720 k\u2126 .\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.8: Page number 204" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "VCC=12.0; #Collector supply voltage in V\n", + "VBE=0.7; #Base-emitter voltage in V\n", + "RB=100.0; #Base resistor's resistance in k\u2126\n", + "RC=560.0; #Collector resistor's resistance in \u2126\n", + "beta_25=100.0; #base current amplification factor at 25 degree celsius\n", + "beta_75=150.0; #base current amplification factor at 25 degree celsius\n", + "\n", + "\n", + "#Calculations\n", + "\n", + "\n", + "#Applying Kirchhoff's law along input circuit, we get\n", + "#VCC=IB*RB+VBE\n", + "IB=(VCC-VBE)/RB; #Base current at 25 degree celsius, in mA\n", + "\n", + "\n", + "#For temperature 25 degree celsius\n", + "IC_25=beta_25*IB; #Collector current at 25 degree celsius, in mA\n", + "\n", + "\n", + "#Applying Kirchhoff's alw at the output circuit,\n", + "#we get: VCC=IC*RC + VCE\n", + "#From the above equation,\n", + "VCE_25=round(VCC-(IC_25/1000)*RC,2); #Collector emitter voltage at 25 degree celsius, in V\n", + "\n", + "\n", + "#For temperature 75 degree celsius\n", + "IC_75=round(beta_75*IB,0); #Collector current at 75 degree celsius, in mA\n", + "\n", + "#Applying Kirchhoff's alw at the output circuit,\n", + "#we get: VCC=IC*RC + VCE\n", + "#From the above equation,\n", + "VCE_75=round(VCC-(IC_75/1000)*RC,2); #Collector emitter voltage at 75 degree celsius, in V\n", + "\n", + "\n", + "change_IC=(IC_75-IC_25)*100.0/IC_25; #percentage change in collector current\n", + "change_VCE=(VCE_75-VCE_25)*100.0/VCE_25; #Percentage change in collector-emitter voltage \n", + "\n", + "#Results\n", + "print(\"The percentage change in collector current =%d%%\"%change_IC);\n", + "print(\"The percentage change in collector-emitter voltage =%.1f%%\"%change_VCE);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The percentage change in collector current =50%\n", + "The percentage change in collector-emitter voltage =-56.3%\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.10: Page number 205" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "VCE_max=20.0; #Maximum collector-emitter voltage in V\n", + "VBE=0.7; #Base-emitter voltage in V\n", + "IC_max=8.0; #Maximum collector current in mA\n", + "IB=40.0; #Base current in microampere\n", + "\n", + "#Calculations\n", + "\n", + "#During cut off state the collector-emitter voltage is maximum and equal to collector supply voltage\n", + "VCC=VCE_max; #Collector supply voltage in V\n", + "\n", + "#Maximum collector current IC_max=collector supply voltage(VCC)/collector load(RC)\n", + "#Collector load(RC)=VCC*IC_max\n", + "RC=VCC/IC_max; #Collector load in k\u2126 .\n", + "\n", + "#Applying Kirchhoff's law along input circuit,\n", + "#we get, VCC=IB*RB +VBE.\n", + "#From the above equation, we get:\n", + "RB=(VCC-VBE)/(IB/1000); #Base resistor's resistance in k\u2126 .\n", + "\n", + "#Results\n", + "print(\"Collector supply voltage = %dV\"%VCC);\n", + "print(\"Collector load=%.1f k\u2126 .\"%RC);\n", + "print(\"Base resistor's resistance=%.1f k\u2126 .\"%RB);\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Collector supply voltage = 20V\n", + "Collector load=2.5 k\u2126 .\n", + "Base resistor's resistance=482.5 k\u2126 .\n" + ] + } + ], + "prompt_number": 19 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.12: Page number 208" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "VCC=20.0; #Collector supply voltage in V\n", + "VEE=-20.0; #Emitter supply voltage in V\n", + "RB=100.0; #Base resistor's resistance in k\u2126\n", + "RC=4.7; #Collector resistor's resistance in k\u2126\n", + "RE=10.0; #Emitter resistor's resistance in k\u2126\n", + "VBE=0.7; #Base-emitter voltage in V\n", + "beta=85.0; #Base current amplification factor\n", + "\n", + "\n", + "#Calculations\n", + "#Applying Kirchhoff's voltage law along the base-emitter circuit (input circuit),\n", + "#we get,IB*RB +IE*RE +VBE -VEE=0.\n", + "#Since IB=IC/beta and IC~IE,\n", + "#(IE/beta)*RB + IE*RE + VBE + VEE =0.\n", + "IE=(-VEE-VBE)/(RE + RB/beta); #Emitter current in mA\n", + "IC=IE; #Collector current (approximately equal to emitter current) in mA\n", + "\n", + "#Applying Kirchhoff's law from VCC till collector terminal,\n", + "#we get, VCC - IC*RC =VC\n", + "VC=VCC-IC*RC; #voltage at collector terminal in V\n", + "\n", + "#Applying Kirchhoff's law from emitter terminal to VEE\n", + "#we get, VE -IE*RE =VEE\n", + "VE=VEE + IE*RE; #Voltage at emitter treminal in V\n", + "\n", + "VCE=VC-VE; #Collector-emitter voltage in V\n", + "\n", + "#Results\n", + "print(\"The collector current = %.2f mA\"%IC);\n", + "print(\"The emitter current = %.2f mA\"%IE);\n", + "print(\"The voltage at collector terminal = %.1f V\"%VC);\n", + "print(\"The collector-emitter voltage = %.1f V\"%VCE);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The collector current = 1.73 mA\n", + "The emitter current = 1.73 mA\n", + "The voltage at collector terminal = 11.9 V\n", + "The collector-emitter voltage = 14.6 V\n" + ] + } + ], + "prompt_number": 20 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.13: Page number 208-209\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "VCC=20.0; #Collector supply voltage in V\n", + "VEE=-20.0; #Emitter supply voltage in V\n", + "RB=100.0; #Base resistor's resistance in k\u2126\n", + "RC=4.7; #Collector resistor's resistance in k\u2126\n", + "RE=10.0; #Emitter resistor's resistance in k\u2126\n", + "beta1=85.0; #Base current amplification factor for case 1 \n", + "beta2=100.0; #Base current amplification factor for case 1\n", + "VBE_1=0.7; #Base emitter voltage for case 1 in V\n", + "VBE_2=0.6; #Base emitter voltage for case 2 in V\n", + "\n", + "\n", + "#Calculations\n", + "#For beta=85 and VBE=0.7,\n", + "#As calculated in the previous question,\n", + "IC_1=1.73; #Collector current in mA.\n", + "VCE_1=14.6; #Collector-emitter voltage in V.\n", + "\n", + "\n", + "#For case (ii)\n", + "#beta=100 and VBE=0.6\n", + "\n", + "#Applying Kirchhoff's voltage law along the base-emitter circuit (input circuit),\n", + "#we get,IB*RB +IE*RE +VBE -VEE=0.\n", + "#Since IB=IC/beta and IC~IE,\n", + "#(IE/beta)*RB + IE*RE + VBE +VEE =0.\n", + "IE_2=round((-VEE-VBE_2)/(RE + RB/beta2),2); #Emitter current in mA\n", + "IC_2=IE_2; #Collector current (approximately equal to emitter current) in mA\n", + "\n", + "#Applying Kirchhoff's law from VCC till collector terminal,\n", + "#we get, VCC - IC*RC =VC\n", + "VC=round(VCC-IC_2*RC,1); #voltage at collector terminal in V\n", + "\n", + "#Applying Kirchhoff's law from emitter terminal to VEE\n", + "#we get, VE -IE*RE =VEE\n", + "VE=round(VEE + IE_2*RE,1); #Voltage at emitter treminal in V\n", + "\n", + "VCE_2=VC-VE; #Collector-emitter voltage in V\n", + "\n", + "\n", + "change_IC= (IC_2-IC_1)*100/IC_1; #%age change in collector current\n", + "\n", + "change_VCE=(VCE_2-VCE_1)*100/VCE_2; #%age change in collector-emitter voltage\n", + "\n", + "\n", + "\n", + "#Results\n", + "print(\"Percentage change in collector current =%.1f%%\"%change_IC);\n", + "print(\"Percentage change in collector-emitter voltage =%.1f%%\"%change_VCE);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Percentage change in collector current =1.7%\n", + "Percentage change in collector-emitter voltage =-3.5%\n" + ] + } + ], + "prompt_number": 21 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.14: Page number 210\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "VCC=20.0; #Collector supply voltage in V\n", + "VBE=0.7 #Base-emitter voltage in V\n", + "RB=100.0; #Base resistor's resistance in k\u2126\n", + "RC=1.0; #Collector resistor's resistance in k\u2126\n", + "beta=100.0; #base current amplification factor\n", + "\n", + "\n", + "#Calculations\n", + "#Applying Kirchhoff's law along input circuit,\n", + "#we get, VCC -IC*RC -IB*RB -VBE=0.\n", + "#since IC= beta*IB,\n", + "#We get,\n", + "IB=(VCC-VBE)/(RB + beta*RC); #Base current in mA\n", + "IC=beta*IB; #Collector current in mA\n", + "\n", + "#Applying Kirchhoff's law along the output circuit,\n", + "#we get, VCC-VCE - IC*RC=0.\n", + "#From the above equation,\n", + "VCE=VCC-IC*RC; #Collector emitter voltage in V\n", + "\n", + "\n", + "#Results\n", + "print(\"The operating point : VCE=%.2fV and IC=%.2fmA.\"%(VCE,IC));\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The operating point : VCE=10.35V and IC=9.65mA.\n" + ] + } + ], + "prompt_number": 22 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.15: Page number 210-211\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "VCC=12.0; #Collector supply voltage in V\n", + "VBE=0.3; #Base emitter voltage in V\n", + "IC=1.0; #Collector current in mA\n", + "VCE=8.0; #Collector emitter voltage in V\n", + "beta=100.0; #Base current amplification factor\n", + "\n", + "#Calculations\n", + "\n", + "#Case(i)\n", + "\n", + "#Applying Kirchhoff's law along the output circuit,\n", + "#we get, VCC-IC*RC-VCE=0.\n", + "#from the above equation we get,\n", + "RC=(VCC-VCE)/IC; #Collector load in kilo ohm\n", + "IB=IC/beta; #Base current in mA\n", + "\n", + "#Applying Kirchhoff's law along input circuit\n", + "#we get, VCC-VBE-(beta*IB*RC)-IB*RB=0.\n", + "#From the above equation we get,\n", + "RB=round((VCC-VBE-beta*IB*RC)/IB,0); #Base resistor's resistance in k\u2126\n", + "\n", + "#Results\n", + "print(\"The resistance value of base resistor=%d k\u2126 and collector load= %d k\u2126.\"%(RB,RC));\n", + "\n", + "#Case(ii)\n", + "\n", + "beta=50;\n", + "\n", + "#Applying Kirchhoff's law along input circuit,\n", + "#we get, VCC -IC*RC -IB*RB -VBE=0.\n", + "#since IC= beta*IB,\n", + "#We get,\n", + "IB=(VCC-VBE)/(RB + beta*RC); #Base current in mA\n", + "IC=beta*IB; #Collector current in mA\n", + "\n", + "#Applying Kirchhoff's law along the output circuit,\n", + "#we get, VCC-VCE - IC*RC=0.\n", + "#From the above equation,\n", + "VCE=round(VCC-IC*RC,1); #Collector emitter voltage in V\n", + "\n", + "#Results\n", + "print(\"The operating point : VCE=%.1fV and IC=%.1fmA.\"%(VCE,IC));\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The resistance value of base resistor=770 k\u2126 and collector load= 4 k\u2126.\n", + "The operating point : VCE=9.6V and IC=0.6mA.\n" + ] + } + ], + "prompt_number": 23 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.16 : Page number 211" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "VCE=2.0; #Collector-emitter voltage at operating point in V\n", + "VBE=0.7; #Base-emitter voltage in V \n", + "IC=1.0; #Collector current at operating point in mA\n", + "beta=100.0; #Base current amplification factor\n", + "\n", + "#Calculations\n", + "IB=IC/beta; #Base current in mA\n", + "\n", + "#As, VCE=VCB +VBE\n", + "#we get,\n", + "VCB=VCE-VBE; #Collector-base voltage in V\n", + "RB=VCB/IB; #Base resistor's resistance in k\u2126\n", + "\n", + "#Results\n", + "print(\"Value of base resistor's resistance=%d k\u2126.\"%RB);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Value of base resistor's resistance=130 k\u2126.\n" + ] + } + ], + "prompt_number": 24 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.17 : Page number 211-212" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "VCC=12.0; #Collector supply voltage in V\n", + "VBE=0.7 #Base-emitter voltage in V\n", + "RB=400.0; #Base resistor's resistance in k\u2126\n", + "RC=4.0; #Collector resistor's resistance in k\u2126\n", + "RE=1.0; #Emitter resistor's resistance in k\u2126\n", + "beta=100.0; #Base current amplification factor\n", + "\n", + "\n", + "#Calculations\n", + "#Applying Kirchhoff's law along outut circuit,\n", + "#we get, VCC -(IC+IB)*RC -IB*RB -VBE - IE*RE=0.\n", + "#since IC= beta*IB, IC+IB ~ IC and IE~IC,\n", + "#We get, VCC - IC*RC -(IC/beta)*RB -VBE - IE*RE\n", + "IC=(VCC-VBE)/(RB/beta + RC + RE); #Collector current current in mA.\n", + "IE=IC; #Emitter current in mA\n", + "\n", + "#Applying Kirchhoff's law along the output circuit,\n", + "#we get, VCC-VCE - IC*RC -IE*RE=0. (IE~IC)\n", + "#From the above equation,\n", + "VCE=VCC-IC*(RC+RE); #Collector emitter voltage in V\n", + "\n", + "\n", + "#Results\n", + "print(\"The operating point : VCE=%.1fV and IC=%.2fmA.\"%(VCE,IC));\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The operating point : VCE=5.7V and IC=1.26mA.\n" + ] + } + ], + "prompt_number": 25 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.18 : Page number 212" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "VCC=10.0; #Collector supply voltage in V\n", + "RB=100.0; #Base resistor's resistance in k\u2126\n", + "RC=10.0; #Collector resistor's resistance in k\u2126\n", + "RE=0; #Emitter resistor's resistance in k\u2126\n", + "VBE=0.7; #Base-emitter voltage in V\n", + "beta=100.0; #Base current amplification factor\n", + "\n", + "#Calculations\n", + "#Applying Kirchhoff's law along outut circuit,\n", + "#we get, VCC -(IC+IB)*RC -IB*RB -VBE - IE*RE=0.\n", + "#since IC= beta*IB, IC+IB ~ IC and IE~IC,\n", + "#We get, VCC - IC*RC -(IC/beta)*RB -VBE - IE*RE\n", + "IC=(VCC-VBE)/(RC +RB/beta + RE); #Collector current in mA\n", + "\n", + "#Applying Kirchhoff's law along the output circuit,\n", + "#we get, VCC-VCE - IC*RC =0. (IE~IC)\n", + "#From the above equation,\n", + "VCE=VCC-IC*RC; #Collector-emitter voltage in V\n", + "\n", + "#Results\n", + "print(\"The d.c bias values are: VCE=%.2fV and IC=%.3fmA\"%(VCE,IC));\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The d.c bias values are: VCE=1.55V and IC=0.845mA\n" + ] + } + ], + "prompt_number": 26 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.19: Page number 214-215\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import matplotlib.pyplot as plt\n", + "\n", + "#Variable declaration\n", + "VCC=15.0; #Collector supply voltage in V\n", + "R1=10.0; #Resistor R1's resistance in k\u2126\n", + "R2=5.0; #Resistor R2's resistance in k\u2126\n", + "RC=1.0; #Collector resistor's resistance in k\u2126 \n", + "RE=2.0; #Emitter resistor's resistance in k\u2126\n", + "VBE=0.7; #Base-emitter voltage in V\n", + "\n", + "#Calculations\n", + "#Applying Kirchhoff's law along output circuit\n", + "#VCE=VCC-IC*(RC+RE);\n", + "#IC=0, for VCE_max\n", + "VCE_max=VCC; #Maximum collector-emitter voltage in V\n", + "#VCE=0, for IC_max\n", + "IC_max=VCC/(RC+RE); #Maximum collector current in mA\n", + "\n", + "#Operating point\n", + "V2=(VCC*R2)/(R1+R2); #Voltage across R2 resistor V\n", + "IE=(V2-VBE)/RE; #Emitter current in mA\n", + "IC=IE; #Collector current(Approx. equal to emitter current) in mA\n", + "VCE=VCC-IC*(RC+RE); #Collector-emitter voltage in V\n", + "\n", + "#Results\n", + "print(\"Collector-emitter voltage at operating point=%.2fV\"%VCE);\n", + "print(\"Collector current at operating point = %.2fmA\"%IC);\n", + "\n", + "#plot\n", + "limit = plt.gca()\n", + "limit.set_xlim([0,20])\n", + "limit.set_ylim([0,6])\n", + "VCE=[i for i in range(0,(int)(VCC+1))]; #Plot variable for V_CE\n", + "IC=[((VCC-i)/(RC+RE)) for i in (VCE[:])]; #Plot variable for I_C\n", + "\n", + "p=plot(VCE,IC);\n", + "xlabel(\"VCE(V)\");\n", + "ylabel(\"IC(mA)\");\n", + "title(\"d.c load line\");\n", + "show(p);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Collector-emitter voltage at operating point=8.55V\n", + "Collector current at operating point = 2.15mA\n" + ] + }, + { + "metadata": {}, + "output_type": "display_data", + "png": 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ZKpWKr9l2mj9/Ps6ePYtjx47Bw8MDL7/8stQlWZ2qqirExMRgzZo1cHJyanJf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\nAAAAAElFTkSuQmCC\n", + "text": [ + "<matplotlib.figure.Figure at 0x7f8eeacf1ad0>" + ] + } + ], + "prompt_number": 29 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.20: Page number 215-216\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "VCC=15.0; #Collector supply voltage in V\n", + "R1=10.0; #Resistor R1's resistance in k\u2126 .\n", + "R2=5.0; #Resistor R2's resistance in k\u2126 .\n", + "RC=1.0; #Collector resistor's resistance in k\u2126 . \n", + "RE=2.0; #Emitter resistor's resistance in k\u2126 .\n", + "VBE=0.7; #Base-emitter voltage in V\n", + "\n", + "#Calculations\n", + "#Using Thevenin's Theorem for replacing circuit consisting of VCC,R1,R2\n", + "E0=(VCC*R2)/(R1+R2); #Thevenin's voltage in V\n", + "R0=(R1*R2)/(R1+R2); #Thevenin's equivalent resistance in k\u2126 .\n", + "\n", + "#Applying Kirchhoff' law along thevenin's equivalent circuit,\n", + "#E0=IB*R0+VBE+IE*RE;\n", + "#Since IE~IC and IC=beta*IB\n", + "#IC=(E0-VBE)/(R0/beta +RE);\n", + "IC=(E0-VBE)/RE; #(Since R0/beta << RE) collector current in mA\n", + "VCE=VCC-IC*(RC+RE); #Collector emitter voltage in V\n", + "\n", + "\n", + "#Results\n", + "print(\"Collector-emitter voltage at operating point=%.2fV\"%VCE);\n", + "print(\"Collector current at operating point = %.2fmA\"%IC);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Collector-emitter voltage at operating point=8.55V\n", + "Collector current at operating point = 2.15mA\n" + ] + } + ], + "prompt_number": 30 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.21: Page number 216-217\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "VCC=12.0; #Collector supply voltage in V\n", + "RE=1.0; #Emitter resistor, k\u2126 .\n", + "R1=50.0; #Resistor R1, k\u2126 .\n", + "R2=10.0; #Resistor R2, k\u2126 .\n", + "\n", + "\n", + "#Calculations\n", + "\n", + "#(i)\n", + "VBE=0.1; #Base-emitter voltage in V\n", + "V2=(VCC*R2)/(R1+R2); #Voltage drop across resistor R2, V \n", + "IE=(V2-VBE)/RE; #Emitter current in mA\n", + "\n", + "print(\"(i)Emitter current= %.1fmA\"%IE);\n", + "\n", + "#(ii)\n", + "VBE=0.3; #Base-emitter voltage in V\n", + "V2=(VCC*R2)/(R1+R2); #Voltage drop across resistor R2, V\n", + "IE=(V2-VBE)/RE; #Emitter current in mA\n", + "\n", + "print(\"(ii)Emitter current= %.1fmA\"%IE);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i)Emitter current= 1.9mA\n", + "(ii)Emitter current= 1.7mA\n" + ] + } + ], + "prompt_number": 31 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.22: Page number 217\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "VCC=20.0; #Collector supply voltage, V\n", + "R1=10.0; #Resistor R1, k\u2126\n", + "R2=10.0; #Resistor R2, k\u2126 .\n", + "RC=1.0; #Collector resistor, k\u2126 .\n", + "RE=5.0; #Emitter resistor, k\u2126 .\n", + "\n", + "\n", + "#Calculations\n", + "V2=(VCC*R2)/(R1+R2); #Voltage drop across resistor R2, V\n", + "\n", + "#Applying kirchhoff's law from base terminal to emitter resistor\n", + "#V2=VBE+IE*RE\n", + "#VBE is neglected due to its small value\n", + "\n", + "IE=V2/RE; #Emitter current in mA\n", + "IC=IE; #Collector current (approx. equal to emitter current), mA\n", + "\n", + "#Applying Kirchhoff's law along output circuit\n", + "VCE=VCC-IC*(RC+RE); #Collector-emitter voltage , V\n", + "VC=VCC-IC*RC; #Voltage at collector terminal,V\n", + "\n", + "\n", + "#Results\n", + "print(\"Emitter current =%dmA\"%IE);\n", + "print(\"Collector-emitter voltage=%dV\"%VCE);\n", + "print(\"Collector terminal's voltage=%dV\"%VC);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Emitter current =2mA\n", + "Collector-emitter voltage=8V\n", + "Collector terminal's voltage=18V\n" + ] + } + ], + "prompt_number": 32 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.23: Page number 219-220\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "#Variable declaration\n", + "VCC=12.0; #Collector supply voltage, V\n", + "VBE=0.7; #Base-emitter voltage, V\n", + "beta=50; #Base current amplification factor\n", + "R1=150; #Resistor R1, k\u2126 .\n", + "R2=100; #Resistor R2, k\u2126 .\n", + "RC=4.7; #Collector resistor, k\u2126 .\n", + "RE=2.2; #Emitter resistor, k\u2126 .\n", + "\n", + "#Calculations\n", + "#Using Thevenin's theorem, calculating Thevenin's voltage and resistance\n", + "E0=(VCC*R2)/(R1+R2); #Thevenin's voltage, V\n", + "R0=(R1*R2)/(R1+R2); #Thevenin's resistance, k\u2126 .\n", + "\n", + "#Applying Kirchhoff' law along thevenin's equivalent circuit,\n", + "#E0=IB*R0+VBE+IE*RE;\n", + "#Since IE~IC and IC=beta*IB\n", + "IB=round((E0-VBE)/(R0+beta*RE),3); #Base current in mA\n", + "IC=round(beta*IB,1); #Collector current, mA\n", + "\n", + "#Applying Kirchhoff's law along the output circuit\n", + "VCE=VCC-IC*(RC+RE); #Collector-emitter voltage, V\n", + "\n", + "S=(beta+1)*(1+R0/RE)/(beta +1+R0/RE); #Stability factor\n", + "\n", + "\n", + "#Results\n", + "print(\"Operating point : VCE= %.2fV and IC=%.1fmA\"%(VCE,IC));\n", + "print(\"Stability factor=%.1f\"%S);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Operating point : VCE= 3.72V and IC=1.2mA\n", + "Stability factor=18.4\n" + ] + } + ], + "prompt_number": 33 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.24 : Page number 220\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "#Variable declaration\n", + "VCC=15.0; #Collector supply voltage, V\n", + "VBE=0.7; #Base-emitter voltage , V\n", + "beta=100.0; #Base current amplification factor\n", + "R1=6.0; #Resistor R1, k\u2126 .\n", + "R2=3.0; #Resistor R2, k\u2126 .\n", + "RC=470.0; #Collector resistor, \u2126.\n", + "RE=1.0; #Emitter resistor, k\u2126 .\n", + "\n", + "#Calculations\n", + "#Using Thevenin's theorem, calculating Thevenin's voltage and resistance\n", + "E0=(VCC*R2)/(R1+R2); #Thevenin's voltage, V\n", + "R0=(R1*R2)/(R1+R2); #Thevenin's resistance, k\u2126 .\n", + "\n", + "#Applying Kirchhoff' law along thevenin's equivalent circuit,\n", + "#E0=IB*R0+VBE+IE*RE;\n", + "#Since IE~IC and IC=beta*IB\n", + "IB=round((E0-VBE)/(R0+beta*RE),3); #Base current in mA\n", + "IC=round(beta*IB,1); #Collector current, mA\n", + "\n", + "#Applying Kirchhoff's law along the output circuit\n", + "VCE=VCC-IC*(RC/1000+RE); #Collector-emitter voltage, V\n", + "\n", + "S=(beta+1)*(1+R0/RE)/(beta +1+R0/RE); #Stability factor\n", + "\n", + "#Results\n", + "print(\"Operating point : VCE= %.2fV and IC=%.1fmA\"%(VCE,IC));\n", + "print(\"Stability factor=%.2f\"%S);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Operating point : VCE= 8.83V and IC=4.2mA\n", + "Stability factor=2.94\n" + ] + } + ], + "prompt_number": 36 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.25 : Page number 221-222\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "#Varaible declaration\n", + "VCC=9; #Collector supply voltage, V\n", + "VCE=3; #Collector-emitter voltage, V\n", + "VBE=0.3; #Base-emitter voltage in V\n", + "RC=2.2; #Collector resistor , k\u2126 .\n", + "IC=2; #Collector current, mA\n", + "beta=50.0; #Base current amplification factor\n", + "\n", + "#Calculations\n", + "IB=IC/beta; #Base current in mA\n", + "\n", + "#According to given relation, I1=10*IB\n", + "I1=IB*10; #Current through the resistor R1, mA\n", + "\n", + "#I1=VCC/(R1+R2), .'s LAW\n", + "R1_R2_sum=VCC/I1; #Sum of the resistor's R1 and R2, k\u2126 (OHM'S LAW).\n", + "\n", + "#Applying Kirchhoff's law along the output circuit\n", + "#VCC=IC*RC+VCE+IE*RE\n", + "#IC~IE\n", + "RE=(VCC-IC*RC-VCE)/IC; #Emitter resistor, k\u2126 .\n", + "RE=round(RE*1000,0); #Emittter resistor, \u2126 .\n", + "\n", + "IE=IC; #Emittter current(approximately equal to collector current), mA\n", + "VE=IE*(RE/1000); #Voltage at emitter terminal (OHM's LAW), V\n", + "V2=VBE+VE; #Voltage drop across resistor R2, V\n", + "\n", + "R2=V2/I1; #Resistor R2,(OHM's LAW), k\u2126 .\n", + "R1=R1_R2_sum-R2; #Resistor R1, k\u2126 .\n", + "\n", + "\n", + "\n", + "#Results\n", + "print(\"RE=%d \u2126., R1=%.2f k\u2126 . and R2=%.2f k\u2126 .\"%(RE,R1,R2));\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "RE=800 \u2126., R1=17.75 k\u2126 . and R2=4.75 k\u2126 .\n" + ] + } + ], + "prompt_number": 38 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.26 : Page number 222\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "VCC=16.0; #Collector supply voltage, V\n", + "R2=20.0; #Resistor R2, k\u2126\n", + "RE=2.0; #Emitter resistor, k\u2126\n", + "VCE=6.0; #Collector-emitter voltage, V\n", + "IC=2.0; #Collector current , mA\n", + "VBE=0.3; #Base-emitter voltage,V\n", + "alpha=0.985; #Current amplification factor\n", + "\n", + "#Calculations\n", + "beta=alpha/(1-alpha); #Base current amplificatioon factor\n", + "IE=IC; #Emitter current, mA\n", + "IB=IC/beta; #Base current, mA\n", + "VE=IE*RE; #Emitter voltage,(OHM's LAW) V\n", + "V2=VBE+VE; #Voltage drop across resistor R2,(Kirchhoff's law) V\n", + "V_R1=VCC-V2; #Voltage drop across resistor R1, V\n", + "I1=V2/R2; #Current through resistor R2 an R1,(OHM's LAW) mA\n", + "R1=V_R1/I1; #Resistor R1,(OHM's LAW) k\u2126\n", + "\n", + "V_RC=(VCC-VCE-VE); #Voltage across collector resistor, V\n", + "RC=V_RC/IC; #Collector resistor,(OHM's LAW) k\u2126\n", + "\n", + "\n", + "#Results\n", + "print(\"R1=%.1f k\u2126 and RC=%d k\u2126.\"%(R1,RC));\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "R1=54.4 k\u2126 and RC=3 k\u2126.\n" + ] + } + ], + "prompt_number": 37 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.27 :Page number 222-223\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "VCC=15.0; #Collector supply voltage, V\n", + "R1=10.0; #Resistor R1, k\u2126 \n", + "R2=5.0; #Resistor R2, k\u2126 \n", + "RC=1.0; #Collector resistor, k\u2126 \u007f\n", + "RE=2.0; #Emitter resistor, k\u2126 \n", + "VBE=0.7; #Base-emitter voltage, V\n", + "beta=100; #Base current amplification factor\n", + "\n", + "#Calculations\n", + "#Using Thevenin's theorem, calculating Thevenin's voltage and resistance\n", + "E0=(VCC*R2)/(R1+R2); #Thevenin's voltage, V\n", + "R0=(R1*R2)/(R1+R2); #Thevenin's resistance, k\u2126 \n", + "\n", + "#Applying Kirchhoff' law along Thevenin's equivalent circuit,\n", + "#E0=IB*R0+VBE+IE*RE;\n", + "#Since IE~IC and IB=IE/beta,\n", + "IE=(E0-VBE)/(R0/beta + RE); #Emitter current , mA\n", + "\n", + "\n", + "#Calculations\n", + "print(\"The exact value of emitter current in the circuit = %.2fmA.\"%IE);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The exact value of emitter current in the circuit = 2.11mA.\n" + ] + } + ], + "prompt_number": 40 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.28: Page number 223-224\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "#Variable declaration\n", + "IE=2.0; #Emitter current, mA\n", + "IB=50.0; #Base current, mA\n", + "VCC=10.0; #Collector supply voltage, V\n", + "VBE=0.2; #Base-emitter voltage, V\n", + "R2=10.0; #Resistor R2, k\u2126\n", + "RE=1.0; #Emitter resistance, k\u2126\n", + "\n", + "#Calculations\n", + "#Applying Kirchhoff's law from the base to the emitter resistor,\n", + "V2=VBE+IE*RE; #Voltage at base terminal, V\n", + "I2=V2/R2; #Current through the resistor R2, mA\n", + "I1=I2+IB/1000; #Current through the resistor R2, mA\n", + "V1=VCC-V2; #Voltage drop across the resistor R2\n", + "R1=V1/I1; #Resistor R1, k\u2126\n", + "\n", + "\n", + "#Results\n", + "print(\"The value of the resistor R1=%.2f k\u2126.\"%R1);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of the resistor R1=28.89 k\u2126.\n" + ] + } + ], + "prompt_number": 41 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.30 :Page number 225-226\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "VCC=8.0; #Collector supply voltage, V\n", + "RB=360.0; #Base resistor, k\u2126\n", + "RC=2.0; #Collector resistor, k\u2126\n", + "VBE=0.7; #Base-emitter voltage, V\n", + "beta=100.0; #base current amplification factor\n", + "\n", + "\n", + "#Calculations\n", + "IC_max=VCC/RC; #Maximum collector current, mA\n", + "VCE_max=VCC; #Maximum collector voltage, V\n", + "\n", + "#Operating point\n", + "#Applying Kirchhoff's law along the input circuit\n", + "IB=(VCC-VBE)/RB; #Base current, mA\n", + "IC=beta*IB; #Collector current, mA\n", + "\n", + "#Kirchhoff' law along the output circuit\n", + "VCE=VCC-IC*RC; #Collector-emitter voltage, V\n", + "\n", + "#Results\n", + "print(\"VCE=%.2fV, is approximately half of VCC=%dV \\n therefore it is mid-point biased.\"%(VCE,VCC));\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "VCE=3.94V, is approximately half of VCC=8V \n", + " therefore it is mid-point biased.\n" + ] + } + ], + "prompt_number": 42 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.31: page number 226\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "VCC=10.0; #Collector supply voltage, V\n", + "VBE=0.7; #Base-emitter voltage, V\n", + "beta=50.0; #Base current amplification factor\n", + "R1=12.0; #Resistor R1, k\u2126 \n", + "R2=2.7; #Resistor R2, k\u2126 \n", + "RC=620.0; #Collector resistor, \u2126 \n", + "RE=180.0; #Emitter resistor, \u2126\n", + "\n", + "\n", + "#Calculations\n", + "#Voltage divder rule across R1 and R2\n", + "V2=round((VCC*R2)/(R1+R2),2); #Voltage drop across resistor R2, V\n", + "IE=round(((V2-VBE)/RE)*1000,2); #Emitter current, mA\n", + "IC=IE; #Collector current(Approximately equal to emitter current), mA\n", + "print(\"IC~IE=%.2fmA.\"%IC);\n", + "\n", + "#Applying Kirchhoff's law along the output circuit\n", + "VCE=VCC-(IC/1000)*(RC+RE); #Collector-emitter voltage, V\n", + "\n", + "#Results\n", + "print(\"VCE=%.2fV, is approximately half of VCC=%dV \\n therefore it is mid-point biased.\"%(VCE,VCC));\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "IC~IE=6.33mA.\n", + "VCE=4.94V, is approximately half of VCC=10V \n", + " therefore it is mid-point biased.\n" + ] + } + ], + "prompt_number": 49 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.32 : Page number 227\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt\n", + "\n", + "#Variable declaration\n", + "VCC=10.0; #Collector supply voltage, V\n", + "IC=10.0; #Collector current, mA \n", + "VBE=0.7; #Base-emitter voltage, V\n", + "R1=1.5; #Resistor R1, k\u2126 \n", + "R2=680.0; #Resistor R2, \u2126 \n", + "RC=260.0; #Collector resistor, \u2126 \n", + "RE=240.0; #Emitter resistor, \u2126 \n", + "beta_min=100; #Minimum value of base current amplification factor\n", + "beta_max=400; #Maximum value of base current amplification factor\n", + "\n", + "#Calculations\n", + "#Voltage divder rule across R1 and R2\n", + "V2=round((VCC*R2/1000)/(R1+R2/1000),2); #Voltage drop across resistor R2, V\n", + "IE=round((V2-VBE)/(RE/1000),0); #OHM' LAW, Emitter current, mA\n", + "IC=IE; #Collector current(approx. equal to emitter current),mA\n", + "beta_avg=sqrt(beta_min*beta_max); #Average value of base current amplification factor\n", + "IB=IE/(beta_avg +1); #Base current, mA\n", + "IB=IB*1000; #Base current, \ud835\udf07A\n", + "\n", + "#Results\n", + "print(\"Base current= %.2f \ud835\udf07A\"%IB);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Base current= 49.75 \ud835\udf07A\n" + ] + } + ], + "prompt_number": 51 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.33 : Page number 227-228\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "VEE=12.0; #Emitter supply voltage, V\n", + "RC=1.5; #Collector resistor, k\u2126\n", + "RB=120.0; #Base resistor k\u2126\n", + "RE=510.0; #Emitter resistor, \u2126 \n", + "VBE=0.7; #Base-emitter voltage, V\n", + "beta=60.0; #Base current amplification factor\n", + "\n", + "#Calculations\n", + "#Applying Kirchhoff's voltage law,\n", + "#IB*RB - VBE - IE*RE +VEE=0\n", + "#Since IE~IC and IC=beta*IB,\n", + "IB=(VEE-VBE)/(RB + beta*RE/1000); #Base current , mA\n", + "IC=round(beta*IB,1); #Collector current, mA\n", + "\n", + "#Applying Kirchhoff's voltage law along output circuit,\n", + "VCE=VEE-IC*(RC + RE/1000); #Collector-emitter voltage, V\n", + "\n", + "\n", + "#Results\n", + "print(\"Operating point : VCE= %.2fV and IC=%.1fmA.\"%(VCE,IC));\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Operating point : VCE= 2.96V and IC=4.5mA.\n" + ] + } + ], + "prompt_number": 60 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.34 : Page number 228-229\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import floor\n", + "\n", + "#Variable declaration\n", + "VEE=9.0; #Emitter supply voltage, V\n", + "RC=1.2; #Collector resistor, k\u2126\n", + "RB=100.0; #Base resistor ,k\u2126\n", + "VBE=0.7; #Base-emitter voltage, V\n", + "beta=45.0; #Base current amplification factor\n", + "\n", + "#Calculations\n", + "#Applying Kirchhoff's voltage law,\n", + "#IB*RB + VBE=VEE\n", + "#Since IE~IC and IC=beta*IB,\n", + "IB=round((VEE-VBE)/RB,3); #Base current , mA\n", + "IC=floor(beta*IB*100)/100; #Collector current, mA\n", + "\n", + "#Applying Kirchhoff's voltage law along output circuit,\n", + "VCE=VEE-IC*RC; #Collector-emitter voltage, V\n", + "\n", + "#Results\n", + "print(\"Operating point : VCE= %.2fV and IC=%.2fmA.\"%(VCE,IC));\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Operating point : VCE= 4.52V and IC=3.73mA.\n" + ] + } + ], + "prompt_number": 68 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.35 : Page number 229\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "VCC=16.0; #Collector supply voltage, V\n", + "VBE=0.7; #Base-emitter voltage, V\n", + "IC=1.0; #Collector current, mA\n", + "VCE=6.0; #Collector-emitter voltage, V\n", + "beta=150.0; #Base current amplification factor\n", + "\n", + "#Calculations\n", + "#For a good design, VE=VCC/10;\n", + "VE=VCC/10; #Emitter terminal's voltage, V\n", + "#OHM's Law\n", + "#And, taking IE~IC\n", + "RE=VE/IC; #Emitter resistor, k\u2126\n", + "\n", + "#Applying Kirchhoff's voltage law alog output circuit:\n", + "#VCC=IC*RC + VCE + VE\n", + "RC=(VCC-VCE-VE)/IC; #Collector resistor, k\u2126\n", + "V2=VE+VBE; #Voltage drop across resistor R2,V\n", + "#From the relation I1=10*IB\n", + "R2=(beta*RE)/10; #Resistor R2, kilo ohm\n", + "\n", + "#From voltage divider rule across R1 and R2,\n", + "#V2=(VCC*R2)/(R1+R2)\n", + "R1=(VCC-V2)*R2/V2; #Resistor R1, k\u2126 \n", + "\n", + "#Results\n", + "print(\"RE=%.1f k\u2126 , RC=%.1f k\u2126, R1=%.0f k\u2126 and R2=%d k\u2126.\"%(RE,RC,R1,R2));\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "RE=1.6 k\u2126 , RC=8.4 k\u2126, R1=143 k\u2126 and R2=24 k\u2126.\n" + ] + } + ], + "prompt_number": 69 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.36 : Page number 230-231\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "ICBO=5.0; #Collector to base leakage current, microampere\n", + "beta=40.0; #Base current amplification factor\n", + "IC_zero_signal=2.0; #Zero signal collector current, mA\n", + "op_temp=25.0; #operating temperature, degree celsius\n", + "temp_risen=55.0; #Temperature risen, degree celsius\n", + "temp_ICBO_doubles=10.0; #Temperature after which ICBO doubles, degree celsius\n", + "\n", + "#Calculations\n", + "\n", + "#(i)\n", + "ICEO=(beta+1)*ICBO; #Collector to emitter leakage current, microampere\n", + "\n", + "#(ii)\n", + "Number_of_times_ICBO_doubled=(temp_risen - op_temp)/temp_ICBO_doubles; #Number of times ICBO doubles\n", + "ICBO_final=ICBO*2**Number_of_times_ICBO_doubled; #Final value of collector to base leakage current, microampere\n", + "ICEO_final=ICBO_final*(beta + 1); #Final value of collector to emitter leakage current, microampere\n", + "\n", + "IC_zero_signal_55=(ICEO_final/1000) +IC_zero_signal; #Zero signal collector current at 55 degree celius\n", + "change=(IC_zero_signal_55-IC_zero_signal)*100/IC_zero_signal; #Percentage change in zero signal collector current\n", + "\n", + "#Result\n", + "print(\"(i) The percentage change in the zero signal collector current=%.0f%%. \"%change)\n", + "\n", + "#(iii)\n", + "#For the silicon transistor\n", + "ICBO=0.1; #Collector to base leakage current, microampere\n", + "\n", + "ICEO=(beta+1)*ICBO; #Collector to emitter leakage current, microampere\n", + "\n", + "Number_of_times_ICBO_doubled=(temp_risen - op_temp)/temp_ICBO_doubles; #Number of times ICBO doubles\n", + "ICBO_final=ICBO*2**Number_of_times_ICBO_doubled; #Final value of collector to base leakage current, microampere\n", + "ICEO_final=ICBO_final*(beta + 1); #Final value of collector to emitter leakage current, microampere\n", + "\n", + "IC_zero_signal_55=(ICEO_final/1000) +IC_zero_signal; #Zero signal collector current at 55 degree celius\n", + "change=(IC_zero_signal_55-IC_zero_signal)*100/IC_zero_signal; #Percentage change in zero signal collector current\n", + "\n", + "\n", + "#Result\n", + "print(\"(ii) The percentage change in the zero signal collector current=%.1f%%. \"%change)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) The percentage change in the zero signal collector current=82%. \n", + "(ii) The percentage change in the zero signal collector current=1.6%. \n" + ] + } + ], + "prompt_number": 70 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.37 : Page number 231\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "ICBO=0.02 #Collector to base leakage current, \ud835\udf07A\n", + "alpha=0.99; #Current amplification factor\n", + "IE=1.0; #Emitter current, mA\n", + "op_temp=27.0; #operating temperature, degree celsius\n", + "temp_risen=57.0; #Temperature risen, degree celsius\n", + "temp_ICBO_doubles=6.0; #Temperature after which ICBO doubles, degree celsius\n", + "\n", + "#Calculations\n", + "Number_of_times_ICBO_doubled=(temp_risen - op_temp)/temp_ICBO_doubles; #Number of times ICBO doubles\n", + "ICBO_55=ICBO*2**Number_of_times_ICBO_doubled; #collector to base leakage current at 55 degree celsius, \ud835\udf07A\n", + "IC=alpha*IE + ICBO_55/1000; #Collector current, mA\n", + "IB=IE-IC; #Base current, mA\n", + "IB=IB*1000; #Base current,\ud835\udf07A\n", + "\n", + "#Results\n", + "print(\"Base current at 57 degree celsius=%.1f \ud835\udf07A \"%IB);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Base current at 57 degree celsius=9.4 \ud835\udf07A \n" + ] + } + ], + "prompt_number": 71 + }, + { + "cell_type": "code", + "collapsed": false, + "input": [], + "language": "python", + "metadata": {}, + "outputs": [] + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/screenshots/chapter10_ac_load_line_2.png b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/screenshots/chapter10_ac_load_line_2.png Binary files differnew file mode 100644 index 00000000..d7feffd7 --- /dev/null +++ b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/screenshots/chapter10_ac_load_line_2.png diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/screenshots/chapter18_clipping_ckt_output_2.png b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/screenshots/chapter18_clipping_ckt_output_2.png Binary files differnew file mode 100644 index 00000000..76486c7b --- /dev/null +++ b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/screenshots/chapter18_clipping_ckt_output_2.png diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/screenshots/chapter8_dc_load_line_2.png b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/screenshots/chapter8_dc_load_line_2.png Binary files differnew file mode 100644 index 00000000..9b26fbe2 --- /dev/null +++ b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/screenshots/chapter8_dc_load_line_2.png diff --git a/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.10_7.ipynb b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.10_7.ipynb new file mode 100644 index 00000000..7cd2fd61 --- /dev/null +++ b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.10_7.ipynb @@ -0,0 +1,280 @@ +{
+ "metadata": {
+ "name": "chapter no.10.ipynb"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 10:Theory of Failures"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.10.10.1,Page No.401"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "P_e=300 #N/mm**2 #Elastic Limit in tension\n",
+ "FOS=3 #Factor of safety\n",
+ "mu=0.3 #Poissoin's ratio\n",
+ "P=12*10**3 #N Pull \n",
+ "Q=6*10**3 #N #Shear force\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let d be the diameter of the shaft\n",
+ "\n",
+ "#Direct stress\n",
+ "#P_x=P*(pi*4**-1*d**3)**-1\n",
+ "#After substituting values and further simplifying we get\n",
+ "#P_x=48*10**3\n",
+ "\n",
+ "#Now shear stress at the centre of bolt\n",
+ "#q=4*3**-1*q_av\n",
+ "#After substituting values and further simplifying we get\n",
+ "#q=32*10**3*(pi*d**2)**-1\n",
+ "\n",
+ "#Principal stresses are\n",
+ "#P1=P_x*2**-1+((P_x*2**-1)**2+q**2)**0.5\n",
+ "#After substituting values and further simplifying we get\n",
+ "#p1=20371.833*(d**2)**-1\n",
+ "\n",
+ "#P2=P_x*2**-1-((P_x*2**-1)**2+q**2)**0.5\n",
+ "#After substituting values and further simplifying we get\n",
+ "#P2=-5092.984*(d**2)**-1\n",
+ "\n",
+ "#q_max=((P_x*2**-1)**2+q**2)**0.5\n",
+ "\n",
+ "#From Max Principal stress theory\n",
+ "#Permissible stress in Tension\n",
+ "P1=100 #N/mm**2 \n",
+ "d=(20371.833*P1**-1)**0.5\n",
+ "\n",
+ "#Max strain theory\n",
+ "#e_max=P1*E**-1-mu*P2*E**-1\n",
+ "#After substituting values and further simplifying we get\n",
+ "#e_max=21899.728*(d**2*E)**-1\n",
+ "\n",
+ "#According to this theory,the design condition is\n",
+ "#e_max=P_e*(E*FOS)**-1\n",
+ "#After substituting values and further simplifying we get\n",
+ "d2=(21899.728*3*300**-1)**0.5 #mm\n",
+ "\n",
+ "#Max shear stress theory\n",
+ "#e_max=shear stress at elastic*(FOS)**-1\n",
+ "#After substituting values and further simplifying we get\n",
+ "d3=(12732.421*6*300**-1)**0.5 #mm\n",
+ "\n",
+ "#Result\n",
+ "print\"Diameter of Bolt by:Max Principal stress theory\",round(d,2),\"mm\"\n",
+ "print\" :Max strain theory\",round(d2,2),\"mm\"\n",
+ "print\" :Max shear stress theory\",round(d3,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Diameter of Bolt by:Max Principal stress theory 14.27 mm\n",
+ " :Max strain theory 14.8 mm\n",
+ " :Max shear stress theory 15.96 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.10.10.2.Page No.402"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "M=40*10**6 #N-mm #Bending moment\n",
+ "T=10*10**6 #N-mm #TOrque\n",
+ "mu=0.25 #Poissoin's ratio\n",
+ "P_e=200 #N/mm**2 #Stress at Elastic Limit\n",
+ "FOS=2\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let d be the diameter of the shaft\n",
+ "\n",
+ "#Principal stresses are given by\n",
+ "\n",
+ "#P1=16*(pi*d**3)**-1*(M+(M**2+T**2)**0.5)\n",
+ "#After substituting values and further simplifying we get\n",
+ "#P1=4.13706*10**8*(d**3)**-1 ............................(1)\n",
+ "\n",
+ "#P2=16*(pi*d**3)**-1*(M-(M**2+T**2)**0.5)\n",
+ "#After substituting values and further simplifying we get\n",
+ "#P2=-6269718*(pi*d**3)**-1 ..............................(2)\n",
+ "\n",
+ "#q_max=(P1-P2)*2**-1\n",
+ "#After substituting values and further simplifying we get\n",
+ "#q_max=2.09988*10**8*(d**3)**-1\n",
+ "\n",
+ "#Max Principal stress theory\n",
+ "#P1=P_e*(FOS)**-1\n",
+ "#After substituting values and further simplifying we get\n",
+ "d=(4.13706*10**8*2*200**-1)**0.33333 #mm \n",
+ "\n",
+ "#Max shear stress theory\n",
+ "#q_max=shear stress at elastic limit*(FOS)**-1\n",
+ "#After substituting values and further simplifying we get\n",
+ "d2=(2.09988*10**8*4*200**-1)**0.33333\n",
+ "\n",
+ "#Max strain energy theory\n",
+ "#P_3=0\n",
+ "#P1**2+P2**2-2*mu*P1*P2=P_e**2*(FOS)**-1\n",
+ "#After substituting values and further simplifying we get\n",
+ "d3=(8.62444*10**12)**0.166666\n",
+ "\n",
+ "#Result\n",
+ "print\"Diameter of shaft according to:MAx Principal stress theory\",round(d,2),\"mm\"\n",
+ "print\" :Max shear stress theory\",round(d2,2),\"mm\"\n",
+ "print\" :Max strain energy theory\",round(d3,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Diameter of shaft according to:MAx Principal stress theory 160.52 mm\n",
+ " :Max shear stress theory 161.33 mm\n",
+ " :Max strain energy theory 143.2 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.10.10.3,Page No.403"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "f_x=40 #N/mm**2 #Internal Fliud Pressure\n",
+ "d1=200 #mm #Internal Diameter\n",
+ "r1=d1*2**-1 #mm #Radius\n",
+ "q=300 #N/mm**2 #Tensile stress\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#From Lame's Equation we have,\n",
+ "\n",
+ "#Hoop Stress\n",
+ "#f_x=b*(x**2)**-1+a ..........................(1)\n",
+ "\n",
+ "#Radial Pressure\n",
+ "#p_x=b*(x**2)**-1-a .........................(2)\n",
+ "\n",
+ "#the boundary conditions are\n",
+ "x=d1*2**-1 #mm \n",
+ "#After sub values in equation 1 and further simplifying we get\n",
+ "#40=b*100**-1-a ..........................(3)\n",
+ "\n",
+ "#Max Principal stress theory\n",
+ "#q*(FOS)**-1=b*100**2+a ..................(4)\n",
+ "#After sub values in above equation and further simplifying we get\n",
+ "\n",
+ "#From Equation 3 and 4 we get\n",
+ "a=80*2**-1\n",
+ "#Sub value of a in equation 3 we get\n",
+ "b=(f_x+a)*100**2\n",
+ "\n",
+ "#At outer edge where x=r_0 pressure is zero\n",
+ "r_0=(b*a**-1)**0.5 #mm\n",
+ "\n",
+ "#thickness\n",
+ "t=r_0-r1 #mm\n",
+ "\n",
+ "#Max shear stress theory\n",
+ "P1=b*(100**2)**-1+a #Max hoop stress\n",
+ "P2=-40 #pressure at int radius (since P2 is compressive)\n",
+ "\n",
+ "#Max shear stress\n",
+ "q_max=(P1-P2)*2**-1\n",
+ "\n",
+ "#According max shear theory the design condition is\n",
+ "#q_max=P_e*2**-1*(FOS)**-1\n",
+ "#After sub values in equation we get and further simplifying we get\n",
+ "#80=b*(100**2)**-1+a\n",
+ "#After sub values in equation 1 and 3 and further simplifying we get\n",
+ "b2=120*100**2*2**-1\n",
+ "\n",
+ "#from equation(3)\n",
+ "a2=120*2**-1-a\n",
+ "\n",
+ "#At outer radius r_0,radial pressure=0\n",
+ "r_02=(b2*a2**-1)**0.5\n",
+ "\n",
+ "#thickness\n",
+ "t2=r_02-r1\n",
+ "\n",
+ "#Result\n",
+ "print\"Thickness of metal by:Max Principal stress theory\",round(t,2),\"mm\"\n",
+ "print\" :Max shear stress thoery\",round(t2,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Thickness of metal by:Max Principal stress theory 41.42 mm\n",
+ " :Max shear stress thoery 73.21 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.2_7.ipynb b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.2_7.ipynb new file mode 100644 index 00000000..8274c67e --- /dev/null +++ b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.2_7.ipynb @@ -0,0 +1,2785 @@ +{
+ "metadata": {
+ "name": "chapter no.2.ipynb"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 2:Simple Stresses And Strains"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.1,Page No.14"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "P=45*10**3 #N #Load\n",
+ "E=200*10**3 #N/mm**2 #Modulus of elasticity of rod\n",
+ "L=500 #mm #Length of rod\n",
+ "d=20 #mm #Diameter of rod\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "A=pi*d**2*4**-1 #mm**2 #Area of circular rod\n",
+ "p=P*A**-1 #N/mm**2 #stress\n",
+ "e=p*E**-1 #strain\n",
+ "dell_l=(P*L)*(A*E)**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"The stress in bar due to Load is\",round(p,5),\"N/mm\"\n",
+ "print\"The strain in bar due to Load is\",round(e,5),\"N/mm\"\n",
+ "print\"The Elongation in bar due to Load is\",round(dell_l,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The stress in bar due to Load is 143.23945 N/mm\n",
+ "The strain in bar due to Load is 0.00072 N/mm\n",
+ "The Elongation in bar due to Load is 0.36 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.2,Page No.15"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "A=15*0.75 #mm**2 #area of steel tape\n",
+ "P=100 #N #Force apllied\n",
+ "L=30*10**3 #mm #Length of tape\n",
+ "E=200*10**3 #N/m**2 #Modulus of Elasticity of steel tape\n",
+ "AB=150 #m #Measurement of Line AB \n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "dell_l=P*L*(A*E)**-1 #mm #Elongation\n",
+ "l=L+dell_l*10**-3 #mm #Actual Length \n",
+ "AB1=AB*l*L**-1 #m Actual Length of AB\n",
+ "\n",
+ "#Result\n",
+ "print\"The Actual Length of Line AB is\",round(AB1,2),\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Actual Length of Line AB is 150.0 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.3,Page No.15"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#Let y be the yield stress\n",
+ "\n",
+ "y=250 #N/mm**2 #yield stress\n",
+ "FOS=1.75 #Factor of safety\n",
+ "P=140*10**3 #N #compressive Load\n",
+ "D=101.6 #mm #External diameter\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "p=y*(FOS)**-1 #N/mm**2 #Permissible stress\n",
+ "A=P*p**-1 #mm**2 #Area of hollow tube\n",
+ "\n",
+ "#Let d be the internal diameter of tube\n",
+ "d=-((A*4*(pi)**-1)-D**2)\n",
+ "X=d**0.5\n",
+ "t=(D-X)*2**-1 #mm #Thickness of steel tube\n",
+ "\n",
+ "#result\n",
+ "print\"The thickness of steel tube is\",round(t,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The thickness of steel tube is 3.17 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 23
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.4,Page No.16"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d=25 #mm #diameter of steel\n",
+ "d2=18 #mm #Diameter at neck\n",
+ "L=200 #mm #length of stee\n",
+ "P=80*10**3 #KN #Load\n",
+ "P1=160*10**3 #N #Load at Elastic Limit\n",
+ "P2=180*10**3 #N #Max Load\n",
+ "L1=56 #mm #Total Extension\n",
+ "dell_l=0.16 #mm #Extension\n",
+ "\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "A=pi*d**2*4**-1 #Area of steel #mm**2\n",
+ "\n",
+ "p=P1*A**-1 #Stress at Elastic Limit #N/mm**2\n",
+ "Y=P*L*(A*dell_l)**-1 #Modulus of elasticity\n",
+ "\n",
+ "#Let % elongation be x\n",
+ "x=L1*L**-1*100 \n",
+ "\n",
+ "#Percentage reduction in area\n",
+ "#Let % A be a\n",
+ "a=((pi*4**-1*d**2)-(pi*4**-1*d2**2))*(pi*4**-1*d**2)**-1*100\n",
+ "\n",
+ "#Ultimate tensile stress\n",
+ "sigma=P2*A**-1 #N/mm**2\n",
+ "\n",
+ "#result\n",
+ "print\"Stress at Elastic limit is\",round(p,2),\"N/mm**2\"\n",
+ "print\"Young's Modulus is\",round(Y,2),\"N/mm**2\"\n",
+ "print\"Percentage Elongation is\",round(a,2)\n",
+ "print\"Percentage reduction in area is\",round(P2,2)\n",
+ "print\"Ultimate tensile stress\",round(sigma,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Stress at Elastic limit is 325.95 N/mm**2\n",
+ "Young's Modulus is 203718.33 N/mm**2\n",
+ "Percentage Elongation is 48.16\n",
+ "Percentage reduction in area is 180000.0\n",
+ "Ultimate tensile stress 366.69 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.5,Page No.16"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "%matplotlib inline\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d=20 #mm #Diameter of bar\n",
+ "d2=14.7 #mm #Diameter at neck\n",
+ "L=200 #mm #guage Length \n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "print \"The Shear Force and Bending Moment Diagrams are the results\"\n",
+ "\n",
+ "#Plotting the Shear Force Diagram\n",
+ "\n",
+ "X1=[0,10,20,30,40,50,60]\n",
+ "Y1=[0,32,64,95,127,160,190]\n",
+ "Z1=[0,0,0,0,0,0,0]\n",
+ "plt.plot(X1,Y1,X1,Z1)\n",
+ "plt.xlabel(\"Extension in divisions\")\n",
+ "plt.ylabel(\"Load in kN\")\n",
+ "plt.show()\n",
+ "\n",
+ "A=pi*4**-1*d**2 #mm**2 #Area of Bar\n",
+ "A2=pi*4**-1*d2**2\n",
+ "\n",
+ "P=45 #KN #Load obtained from graph\n",
+ "dell=0.143 #mm #Divisions\n",
+ "\n",
+ "#Modulus of Elasticity\n",
+ "E=P*L*(dell*A)**-1 \n",
+ "\n",
+ "BL=93*10**3 #N #Breaking Load\n",
+ "\n",
+ "#Nominal stress at Breaking point\n",
+ "sigma=BL*A**-1 #KN/mm**2 \n",
+ "\n",
+ "#True stress at breaking Point\n",
+ "sigma1=BL*A2**-1\n",
+ "\n",
+ "#Percentage Elongation \n",
+ "dell_l=(A-A2)*A**-1*100\n",
+ "\n",
+ "#Result\n",
+ "print\"The Value of ELongation is\",round(E,2),\"N/mm**2\"\n",
+ "print\"The Nominal stress at the Breaking Point\",round(sigma,2),\"KN/mm**2\"\n",
+ "print\"The True stress at the Breaking Point\",round(sigma1,2),\"KN/mm**2\"\n",
+ "print\"The Percentage Reduction in Area is\",round(dell_l,2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Shear Force and Bending Moment Diagrams are the results\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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gwAFOnDjBRx99xNatWys876vr+8tf/kKzZs2Ii4vDVHG/iq+u7bL09HT279/P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\n",
+ "text": [
+ "<matplotlib.figure.Figure at 0x4d1d370>"
+ ]
+ },
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Value of ELongation is 200.33 N/mm**2\n",
+ "The Nominal stress at the Breaking Point 296.03 KN/mm**2\n",
+ "The True stress at the Breaking Point 547.97 KN/mm**2\n",
+ "The Percentage Reduction in Area is 45.98\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.6,Page No.19"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "P=40*10**3 #N #Load\n",
+ "L1=160 #mm #Length of Bar1\n",
+ "L2=240 #mm #Length of bar2\n",
+ "L3=160 #mm #Length of bar3\n",
+ "d1=25 #mm #Diameter of Bar1\n",
+ "d2=20 #mm #diameter of bar2\n",
+ "d3=25 #mm #diameter of bar3\n",
+ "dell_l=0.285 #mm #Total Extension of bar\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "E=P*4*(dell_l*pi)**-1*(L1*(d1**2)**-1+L2*(d2**2)**-1+L3*(d3**2)**-1)\n",
+ "\n",
+ "#Result\n",
+ "print\"The Young's Modulus of the material\",round(E,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Young's Modulus of the material 198714.72 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.7,Page No.19"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "E1=2*10**5 #N/mm**2 #modulus of Elasticity of material1\n",
+ "E2=1*10**5 #N/mm**2 #modulus of Elasticity of material2\n",
+ "P=25*10**3 #N #Load \n",
+ "t=20 #mm #thickness of material\n",
+ "b1=40 #mm #width of material1\n",
+ "b2=30 #mm #width of material2\n",
+ "L1=500 #mm #Length of material1\n",
+ "L2=750 #mm #Length of material2\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "A1=b1*t #mm**2 #Area of materila1\n",
+ "A2=b2*t #mm**2 #Area of material2\n",
+ "\n",
+ "dell_l1=P*L1*(A1*E1)**-1 #Extension of Portion1\n",
+ "dell_l2=P*L2*(A2*E2)**-1 #Extension of portion2\n",
+ "\n",
+ "#Total Extension of Bar is\n",
+ "dell_l=dell_l1+dell_l2\n",
+ "\n",
+ "#Result\n",
+ "print\"The Total Extension of the Bar is\",round(dell_l,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Total Extension of the Bar is 0.39 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.8,Page No.20"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=1000 #mm #Length of Bar\n",
+ "l=400 #mm #Length upto which bire is drilled \n",
+ "D=30 #mm #diameter of bar\n",
+ "d1=10 #mm #diameter of bore\n",
+ "P=25*10**3 #N #Load\n",
+ "dell_l=0.185 #mm #Extension of bar\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "L1=L-l #Length of bar above the bore\n",
+ "L2=400 #mm #Length of bore\n",
+ "\n",
+ "A1=pi*4**-1*D**2 #Area of bar\n",
+ "A2=pi*4**-1*(D**2-d1**2) #Area of bore\n",
+ "\n",
+ "E=P*dell_l**-1*(L1*A1**-1+L2*A2**-1)\n",
+ "\n",
+ "#Result\n",
+ "print\"The Modulus of ELasticity is\",round(E,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Modulus of ELasticity is 200735.96 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 24
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.11,Page No.23"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "t=10 #mm #Thickness of steel\n",
+ "b1=60 #mm #width of plate1\n",
+ "b2=40 #mm #width of plate2\n",
+ "P=60*10**3 #Load\n",
+ "L=600 #mm #Length of plate\n",
+ "E=2*10**5 #N/mm**2\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Extension of taperong bar of rectangular section\n",
+ "dell_l=P*L*(t*E*(b1-b2))**-1*log(b1*b2**-1)\n",
+ "\n",
+ "A_av=(b1*t+b2*t)*2**-1 #Average Area #mm**2\n",
+ "dell_l2=P*L*(A_av*E)**-1 \n",
+ "\n",
+ "#PErcentage Error\n",
+ "e=(dell_l-dell_l2)*(dell_l)**-1*100\n",
+ "\n",
+ "#Result\n",
+ "print\"The Percentage Error is\",round(e,2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Percentage Error is 1.35\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.12,Page No.23"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=1.5 #m #Length of steel bar\n",
+ "L1=1000 #m0 #Length of steel bar 1\n",
+ "L2=500 #m #Length of steel bar 2\n",
+ "d1=40 #Diameter of steel bar 1\n",
+ "d2=20 #diameter of steel bar 2\n",
+ "E=2*10**5 #N/mm**2 #Modulus of Elasticity\n",
+ "P=160*10**3 #N #Load\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "A1=pi*4**-1*d1**2 #Area of Portion 1\n",
+ "\n",
+ "#Extension of uniform Portion 1\n",
+ "dell_l1=P*L1*(A1*E)**-1 #mm\n",
+ "\n",
+ "#Extension of uniform Portion 2\n",
+ "dell_l2=4*P*L2*(pi*d1*d2*E)**-1 #mm\n",
+ "\n",
+ "#Total Extension of Bar\n",
+ "dell_l=dell_l1+dell_l2\n",
+ "\n",
+ "#Result\n",
+ "print\"The Elongation of the Bar is\",round(dell_l,2),\"mm\" "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": []
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.14,Page No.25"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#Portion AB\n",
+ "L_AB=600 #mm #Length of AB\n",
+ "A_AB=40*40 #mm**2 #Cross-section Area of AB\n",
+ "\n",
+ "#Portion BC\n",
+ "L_BC=800 #mm #Length of BC\n",
+ "A_BC=30*30 #mm #Length of BC\n",
+ "\n",
+ "#Portion CD\n",
+ "L_CD=1000 #mm #Length of CD\n",
+ "A_CD=20*20 #mm #Area of CD\n",
+ "\n",
+ "P1=80*10**3 #N #Load1\n",
+ "P2=60*10**3 #N #Load2\n",
+ "P3=40*10**3 #N #Load3\n",
+ "\n",
+ "E=2*10**5 #Modulus of Elasticity\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "P4=P1-P2+P3 #Load4\n",
+ "\n",
+ "#Now Force in AB\n",
+ "F_AB=P1\n",
+ "\n",
+ "#Force in BC\n",
+ "F_BC=P1-P2\n",
+ "\n",
+ "#Force in CD\n",
+ "F_CD=P4\n",
+ "\n",
+ "#Extension of AB\n",
+ "dell_l_AB=F_AB*L_AB*(A_AB*E)**-1\n",
+ "\n",
+ "#Extension of BC\n",
+ "dell_l_BC=F_BC*L_BC*(A_BC*E)**-1\n",
+ "\n",
+ "#Extension of CD\n",
+ "dell_l_CD=F_CD*L_CD*(A_CD*E)**-1\n",
+ "\n",
+ "#Total Extension\n",
+ "dell_l=dell_l_AB+dell_l_BC+dell_l_CD\n",
+ "\n",
+ "#Result\n",
+ "print\"The Total Extension in Bar is\",round(dell_l,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Total Extension in Bar is 0.99 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.15,Page No.26"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=800 #mm #Length of bar\n",
+ "F1=30*10**3 #N #Force acting on the bar\n",
+ "F2=60*10**3 #N #force acting on the bar\n",
+ "L=800 #mm #Length of bar\n",
+ "d=25 #mm #diameter of bar\n",
+ "L_AC=275 #mm #Length of AC\n",
+ "L_CD=150 #mm #Length of CD\n",
+ "L_DB=375 #mm #Length of DB\n",
+ "E=2*10**5 #Pa #Modulus of elasticity\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let P be the Reaction on tne Bar from support at A\n",
+ "\n",
+ "#Shortening of Portion AC\n",
+ "#dell_l_AC1=P*L_AC*(A*E)**-1\n",
+ "\n",
+ "#Shortening of Portion CD\n",
+ "#dell_l_CD1=(30+P)*L_CD*(A*E)**-1\n",
+ "\n",
+ "#Extension of Portion DB\n",
+ "#dell_l_DB1=(30-P)*L_DB*(A*E)**-1\n",
+ "\n",
+ "#Total Extensions=1*(A*E)**-1*(P*L_AC-(30+P)*L_CD+(30-P)*L_DB)\n",
+ "#As Supports are unyielding,Total Extensions=0\n",
+ "\n",
+ "#After substituting values in above equation and Further simplifying we get\n",
+ "P=(30*375-150*30)*800**-1\n",
+ "\n",
+ "#Reaction of support A\n",
+ "R_A=P\n",
+ "\n",
+ "#Reaction of support B\n",
+ "R_B=30-P\n",
+ "\n",
+ "#Cross-sectional Area\n",
+ "A=pi*4**-1*d**2\n",
+ "\n",
+ "#Stress in Portion AC\n",
+ "sigma1=P*10**3*A**-1 #N/mm**2\n",
+ "\n",
+ "#Stress in Portion CD\n",
+ "sigma2=(30+P)*10**3*A**-1 #N/mm**2\n",
+ "\n",
+ "#Stress in Portion DB\n",
+ "sigma3=(30-P)*10**3*A**-1 #N/mm**2\n",
+ "\n",
+ "#Shortening of Portion AC\n",
+ "dell_l_AC2=P*10**3*L_AC*(A*E)**-1 #mm \n",
+ "\n",
+ "#Shortening of Portion CD\n",
+ "dell_l_CD2=(30+P)*10**3*L_CD*(A*E)**-1 #mm \n",
+ "\n",
+ "#Extension of Portion DB\n",
+ "dell_l_DB2=(30-P)*10**3*L_DB*(A*E)**-1 #mm \n",
+ "\n",
+ "#result\n",
+ "print\"The Reactios at two Ends are:R_A\",round(R_A,2),\"KN\"\n",
+ "print\" :R_B\",round(R_B,2),\"KN\"\n",
+ "print\"Stress in Portion AC\",round(sigma1,2),\"N/mm**2\"\n",
+ "print\"Stress in Portion CD\",round(sigma2,2),\"N/mm**2\"\n",
+ "print\"Stress in Portion DB\",round(sigma3,2),\"N/mm**2\"\n",
+ "print\"Shortening of Portion AC\",round(dell_l_AC2,3),\"mm\"\n",
+ "print\"Shortening of Portion CD\",round(dell_l_CD2,3),\"mm\"\n",
+ "print\"Shortening of Portion DB\",round(dell_l_DB2,3),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Reactios at two Ends are:R_A 8.44 KN\n",
+ " :R_B 21.56 KN\n",
+ "Stress in Portion AC 17.19 N/mm**2\n",
+ "Stress in Portion CD 78.3 N/mm**2\n",
+ "Stress in Portion DB 43.93 N/mm**2\n",
+ "Shortening of Portion AC 0.024 mm\n",
+ "Shortening of Portion CD 0.059 mm\n",
+ "Shortening of Portion DB 0.082 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 19
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.19,Page No.29"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "h=4 #m #height of Pillars\n",
+ "P=20 #KN #Load at M\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let P_A,P_B,P_C,P_D be the forces introduced in the Pillars\n",
+ "#Sun of All Vertical Forces\n",
+ "#P_A+P_B+P_C+P_D=20 ....................(1)\n",
+ "\n",
+ "#Sum of moment about AB, we get\n",
+ "#P_D+P_C=12 ....................(2)\n",
+ "\n",
+ "#Sum of Moment about AD\n",
+ "#P_C+P_B=8 ....................(3)\n",
+ "\n",
+ "#Let dell_l_A,dell_l_B,dell_l-C,dell_l_D be the deformations of Pillars A,B,C,D respectively\n",
+ "#Diagonals AC and BD will remain straight Lines even after the Load is applied.\n",
+ "#Deflection of central Point is given by (dell_l_A+dell_l_C)*2**-1 & (dell_l_B+dell_l_D)*2**-1\n",
+ "\n",
+ "#dell_l_A+dell_l_C=dell_l_B+ell_l_D\n",
+ "#P_A*L*(A*E)**-1+P_C*L*(A*E)**-1=P_B*L*(A*E)**-1+P_D*L*(A*E)**-1\n",
+ "\n",
+ "#Since Pillars are identical in Length,cross-sectional area,material Property\n",
+ "#P_A+P_C=P_B+P_D ..............(4)\n",
+ "\n",
+ "#From Equations 1 and 4 we get\n",
+ "#P_B+P_D=10 ....................(5)\n",
+ " \n",
+ "#Substracting Equation 3 from Equation 2 we get\n",
+ "#P_D-P_B=4 ....................(6)\n",
+ "\n",
+ "#Adding Equation 5 and 6 we get\n",
+ "\n",
+ "P_D=14*2**-1\n",
+ "P_C=12-P_D\n",
+ "P_B=8-P_C\n",
+ "\n",
+ "#Now substituting values of P_B,P_C,P_D in equation1 we get\n",
+ "P_A=20-(P_B+P_C+P_D)\n",
+ "\n",
+ "#Result\n",
+ "print\"The Forces Developed in the Pillars are:P_A\",round(P_A,2),\"KN\"\n",
+ "print\" :P_B\",round(P_B,2),\"KN\"\n",
+ "print\" :P_C\",round(P_C,2),\"KN\"\n",
+ "print\" :P_D\",round(P_D,2),\"KN\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Forces Developed in the Pillars are:P_A 5.0 KN\n",
+ " :P_B 3.0 KN\n",
+ " :P_C 5.0 KN\n",
+ " :P_D 7.0 KN\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.20,Page No.31"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "sigma=150 #N/mm**2 #Stress\n",
+ "P=40*10**3 #N #Load\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#LEt P_A.P_B,P_C,P_D be the forces developed in wires A,B,C,D respectively\n",
+ "\n",
+ "#Let sum of all Vertical Forces=0\n",
+ "#P_A+P_B+P_C+P_D=40 ..........................(1)\n",
+ "\n",
+ "#Let x be the distance between each wires\n",
+ "#sum of all moments=0\n",
+ "#P_B*x+P_C*2*x+P_D*3*x=40*2*x\n",
+ "\n",
+ "#After further simplifying we get\n",
+ "#P_B+2*P_C+3*P_D=80 ..........................(2)\n",
+ "\n",
+ "#As the equations of statics ae not enough to find unknowns,Consider compatibilit Equations\n",
+ "\n",
+ "#Let dell_l be the increse in elongation of wire\n",
+ "\n",
+ "#dell_l_B=dell_l_A+dell_l\n",
+ "#dell_l_C=dell_l_A+2*dell_l\n",
+ "#dell_l_D=dell_l_A+3*dell_l\n",
+ "\n",
+ "#Let P1 be the force required for the Elongation of wires,then\n",
+ "#P_B=P_A+P1 ]\n",
+ "#P_C=P_A+2*P1 ]\n",
+ "#P_D=P_A+3*P1 ] ................................(3) \n",
+ "\n",
+ "#from Equation (3) and (1) we get\n",
+ "#2*P_A+3*P1=20 ................................(4)\n",
+ "\n",
+ "#from Equation (3) and (2) we get\n",
+ "#6*P_A+14*P1=80 \n",
+ "\n",
+ "#subtracting 3 times equation (4) from (3) we get\n",
+ "P1=20*5**-1\n",
+ "\n",
+ "#from Equation 4 we get\n",
+ "P_A=(80-14*P1)*6**-1\n",
+ "P_B=P_A+P1\n",
+ "P_C=P_A+2*P1\n",
+ "P_D=P_A+3*P1\n",
+ "\n",
+ "#Let d be the diameter required,then\n",
+ "d=(P_D*10**3*4*(pi*150)**-1)**0.5\n",
+ "\n",
+ "#result\n",
+ "print\"The Required Diameter is\",round(d,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Required Diameter is 11.65 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.21,Page No.32"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "P=20*10**3 #N #Load\n",
+ "d=6 #mm #diameter of wire\n",
+ "E=2*10**5 #N/mm**2 \n",
+ "L_BO=4000 #mm #Length of BO\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let theta be the angle between OA and OB and also between OC and OB\n",
+ "theta=30\n",
+ "\n",
+ "#Let P_OA,P_OB,P_OC be the Forces introduced in wires OA,OB,OC respectively\n",
+ "#Due to symmetry P_OA=P_OC (same angles)\n",
+ "\n",
+ "#Sum of all Vertical Forces=0\n",
+ "#P_OA*cos(theta)+P_OB+P_OC*cos(theta)=P\n",
+ "\n",
+ "#After further simplifyinf we get\n",
+ "#2*P_OA*cos(theta)+P_OB=20 ...............(1)\n",
+ "\n",
+ "#Let oo1 be the extension of BO\n",
+ "#oo1=L_A1o1*(cos(theta))**-1\n",
+ "\n",
+ "#From relation we get\n",
+ "#P_OB*L_BO=P_OA*L_AO*(cos(theta))**-1\n",
+ "\n",
+ "#But L_AO=L_BO*(cos(theta))**-1\n",
+ "\n",
+ "#After substituting value of L_AO in above equation we get\n",
+ "#P_OB=0.75*P_OA .......................(2)\n",
+ "\n",
+ "#substituting in Equation 1 we get\n",
+ "#2*P_OA*cos(theta)+0.75*P_OA=20\n",
+ "\n",
+ "P_OA=20*(2*cos(theta*pi*180**-1)+0.75)**-1\n",
+ "\n",
+ "P_OB=0.75*P_OA\n",
+ "\n",
+ "A=pi*4**-1*d**2\n",
+ "\n",
+ "#Vertical displacement of Load\n",
+ "dell_l_BO=P_OB*10**3*L_BO*(A*E)**-1\n",
+ " \n",
+ "#Result\n",
+ "print\"Forces in each wire is:P_OA\",round(P_OA,2),\"KN\"\n",
+ "print\" :P_OB\",round(P_OB,2),\"KN\"\n",
+ "print\"Vertical displacement of Loadis\",round(dell_l_BO,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Forces in each wire is:P_OA 8.06 KN\n",
+ " :P_OB 6.04 KN\n",
+ "Vertical displacement of Loadis 4.27 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 22
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.22,Page No.34"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L_s=L_a=L=500 #mm #Length of bar\n",
+ "A_a=50*20 #mm #Area of aluminium strip\n",
+ "A_s=50*15 #mm #Area of steel strip\n",
+ "P=50*10**3 #N #Load\n",
+ "E_a=1*10**5 #N/mm**2 #Modulus of aluminium \n",
+ "E_s=2*10**5 #N/mm**2 #Modulus of steel\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let P_a and P_s br the Load shared by aluminium and steel strip\n",
+ "#P_a+P_s=P ..................(1)\n",
+ "\n",
+ "#For compatibility condition,dell_l_a=dell_l_s\n",
+ "#P_a*L_a*(A_a*E_a)**-1=P_s*L_s*(A_s*E_s)**-1 .....(2)\n",
+ "\n",
+ "#As L_a=L_s we get\n",
+ "#P_s=1.5*P_a .................(3)\n",
+ "\n",
+ "#From Equation 1 and 2 we get\n",
+ "P_a=P*2.5**-1\n",
+ "\n",
+ "#Substituting in equation 1 we get\n",
+ "P_s=P-P_a\n",
+ "\n",
+ "#stress in aluminium strip \n",
+ "sigma_a=P_a*A_a**-1\n",
+ "\n",
+ "#stress in steel strip\n",
+ "sigma_s=P_s*A_s**-1\n",
+ "\n",
+ "#Now from the relation we get\n",
+ "dell_l_a=dell_l_s=P_s*L_s*(A_s*E_s)**-1\n",
+ "\n",
+ "#result\n",
+ "print\"Stress in Aluminium strip is\",round(sigma_a,2),\"N/mm**2\"\n",
+ "print\"Stress in steel strip is\",round(sigma_s,2),\"N/mm**2\"\n",
+ "print\"The Extension of the bar is\",round(dell_l_s,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Stress in Aluminium strip is 20.0 N/mm**2\n",
+ "Stress in steel strip is 40.0 N/mm**2\n",
+ "The Extension of the bar is 0.1 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.23,Page No.35"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "D_s=20 #mm #Diameter of steel\n",
+ "D_Ci=20 #mm #Internal Diameter of Copper\n",
+ "t=5 #mm #THickness of copper bar\n",
+ "P=100*10**3 #N #Load\n",
+ "E_s=2*10**5 #N/mm**2 #modulus of elasticity of steel\n",
+ "E_c=1.2*10**5 #N/mm**2 #Modulus of Elasticity of Copper\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "A_s=pi*4**-1*D_s**2 #mm**2 #Area of steel\n",
+ "D_Ce=D_s+2*t #mm #External Diameterof Copper Tube\n",
+ "\n",
+ "A_c=pi*4**-1*(D_Ce**2-D_Ci**2) #mm**2 #Area of Copper\n",
+ "\n",
+ "#From static Equilibrium condition\n",
+ "#Let P_s and P_c be the Load shared by steel and copper in KN\n",
+ "#P_s+P_c=100 ....................................(1)\n",
+ "\n",
+ "#From compatibility Equation,dell_l_s=dell_l_c\n",
+ "#P_s*L*(A_s*E_s)**-1=P_c*L*(A_c*E_c)**-1\n",
+ "\n",
+ "#Substituting values in above Equation we get\n",
+ "#P_s=1.3333*P_C \n",
+ "\n",
+ "#Now Substituting value of P_s in Equation (1),we get\n",
+ "P_c=100*2.3333**-1 #KN\n",
+ "P_s=100-P_c #KN\n",
+ "\n",
+ "#Stress in steel\n",
+ "sigma_s=P_s*10**3*A_s**-1 #N/mm**2\n",
+ "\n",
+ "#Stress in copper\n",
+ "sigma_c=P_c*10**3*A_c**-1 #N/mm**2\n",
+ "\n",
+ "#Result\n",
+ "print\"Stresses Developed in Two material are:sigma_s\",round(sigma_s,2),\"N/mm**2\"\n",
+ "print\" :sigma_c\",round(sigma_c,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Stresses Developed in Two material are:sigma_s 181.89 N/mm**2\n",
+ " :sigma_c 109.14 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.24,Page No.36"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "A_C=230*400 #mm #Area of column\n",
+ "D_s=12 #mm #Diameter of steel Bar\n",
+ "P=600*10**3 #N #Axial compression\n",
+ "#E_s*E_c=18.67\n",
+ "n=8 #number of steel Bars\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "A_s=pi*4**-1*D_s**2*n #Area of steel #mm**2 \n",
+ "A_c=A_C-A_s #mm**2 #Area of concrete\n",
+ "\n",
+ "#From static Equilibrium condition\n",
+ "#P_s+P_c=600 .........(1)\n",
+ "\n",
+ "#Now from compatibility Equation dell_l_s=dell_l_c we get,\n",
+ "#P_s*L*(A_s*E_s)**-1=P_c*L*(A_c*E_c)**-1\n",
+ "\n",
+ "#Substituting values in above Equation we get\n",
+ "#P_s=0.1854*P_c\n",
+ "\n",
+ "#Now Substituting value of P_s in Equation (1),we get\n",
+ "P_c=600*1.1854**-1\n",
+ "P_s=600-P_c\n",
+ "\n",
+ "#Stress in steel\n",
+ "sigma_s=P_s*10**3*A_s**-1 #N/mm**2\n",
+ "\n",
+ "#Stress in copper\n",
+ "sigma_c=P_c*10**3*A_c**-1 #N/mm**2\n",
+ "\n",
+ "#Result\n",
+ "print\"Stresses Developed in Two material are:sigma_s\",round(sigma_s,2),\"N/mm**2\"\n",
+ "print\" :sigma_c\",round(sigma_c,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Stresses Developed in Two material are:sigma_s 103.72 N/mm**2\n",
+ " :sigma_c 5.56 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.25,Page No.36"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "P=200*10**3 #N #Load\n",
+ "A_a=1000 #mm**2 #Area of Aluminium\n",
+ "A_s=800 #mm**2 #Area of steel\n",
+ "E_a=1*10**5 #N/mm**2 #Modulus of Elasticity of Aluminium\n",
+ "E_s=2*10**5 #N/mm**2 #Modulus of ELasticity of steel\n",
+ "sigma_a1=65 #N/mm**2 #stress in aluminium\n",
+ "sigma_s1=150 #N/mm**2 #Stress in steel\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let P_a and P_s be the force in aluminium and steel pillar respectively\n",
+ "\n",
+ "#Now,sum of forces in Vertical direction we get\n",
+ "#2*P_a+P_s=200 .........................................(1)\n",
+ "\n",
+ "#By compatibility Equation dell_l_s=dell_l_a we get\n",
+ "#P_s=1.28*P_a ..........................................(2)\n",
+ "\n",
+ "#Now substituting value of P_s in Equation 1 we get\n",
+ "P_a=200*3.28**-1 #KN\n",
+ "P_s=200-2*P_a #KN\n",
+ "\n",
+ "#Stress developed in aluminium\n",
+ "sigma_a=P_a*10**3*A_a**-1 #N/mm**2 \n",
+ "\n",
+ "#Stress developed in steel\n",
+ "sigma_s=P_s*10**3*A_s**-1 #N/mm**2 \n",
+ "\n",
+ "#Part-2\n",
+ "\n",
+ "#Let sigma_a1 and sigma_s1 be the stresses in Aluminium and steel due to Additional LOad\n",
+ "\n",
+ "P_a1=sigma_a1*A_a #Load carrying capacity of aluminium\n",
+ "P_s1=1.28*P_a1\n",
+ "\n",
+ "#Total Load carrying capacity \n",
+ "P1=2*P_a1+P_s1 #N \n",
+ "\n",
+ "P_s2=sigma_s1*A_s #Load carrying capacity of steel\n",
+ "P_a2=P_s2*1.28**-1\n",
+ "\n",
+ "#Total Load carrying capacity\n",
+ "P2=2*P_a2+P_s2\n",
+ "\n",
+ "#Additional Load\n",
+ "P3=P1-P\n",
+ "\n",
+ "#Result\n",
+ "print\"Stresses Developed in Each Pillar is:sigma_a\",round(sigma_a,2),\"N/mm**2\"\n",
+ "print\" :sigma_s\",round(sigma_s,2),\"N/mm**2\"\n",
+ "print\"Additional Load taken by pillars is\",round(P3,2),\"N\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Stresses Developed in Each Pillar is:sigma_a 60.98 N/mm**2\n",
+ " :sigma_s 97.56 N/mm**2\n",
+ "Additional Load taken by pillars is 13200.0 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.26,Page No.37"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=500 #mm #Length of assembly\n",
+ "D=16 #mm #Diameter of steel bolt\n",
+ "Di=20 #mm #internal Diameter of copper tube\n",
+ "Do=30 #mm #External Diameter of copper tube\n",
+ "E_s=2*10**5 #N/mm**2 #Modulus of Elasticity of steel\n",
+ "E_c=1.2*10**5 #N/mm**2 #Modulus of Elasticity of copper\n",
+ "p=2 #mm #Pitch of nut\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let P_s be the Force in bolt and P_c be the FOrce in copper tube\n",
+ "#P_s=-P_s\n",
+ "\n",
+ "dell=1*4**-1*2 #Quarter turn of nut total movement\n",
+ "\n",
+ "#dell=dell_s+dell_c\n",
+ "\n",
+ "#Area of steel\n",
+ "A_s=pi*4**-1*D**2\n",
+ "\n",
+ "#Area of copper\n",
+ "A_c=pi*4**-1*(Do**2-Di**2)\n",
+ "\n",
+ "#dell=P*L*(A_s*E_s)**-1+P*L*(A_c*E_c)**-1\n",
+ "P=dell*(1*(A_s*E_s)**-1+1*(A_c*E_c)**-1)**-1*L**-1 #LOad\n",
+ "\n",
+ "P_s=P*A_s**-1\n",
+ "P_c=P*A_c**-1\n",
+ "\n",
+ "#result\n",
+ "print\"stress introduced in bolt is\",round(P_s,2),\"N/mm**2\"\n",
+ "print\"stress introduced in tube is\",round(P_c,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "stress introduced in bolt is 107.91 N/mm**2\n",
+ "stress introduced in tube is 55.25 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.27,Page No.39"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "D=20 #mm #Diameter of Bolts\n",
+ "Di=25 #m #internal Diameter\n",
+ "t=10 #mm #Thickness of bolt\n",
+ "E_s=2*10**5 #N/mm**2 #Modulus of Elasticity\n",
+ "E_c=1.2*10**5 #N/mm**2 #Modulus of copper\n",
+ "p=3 #mm #Pitch\n",
+ "theta=30 #degree\n",
+ "L_c=500 #Lengh of copper \n",
+ "L_s=600 #Length of steel\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let P_s be the Force in each bolt and P_c be the FOrce in copper tube\n",
+ "#From Static Equilibrium condition\n",
+ "#P_c=2*P_s\n",
+ "\n",
+ "#As nut moves by 60 degree.If nut moves by 360 degree its Longitudinal movement is by 3 mm\n",
+ "dell=theta*360**-1*p\n",
+ "\n",
+ "#From Compatibility Equaton we get\n",
+ "#dell=dell_c+dell_s\n",
+ "\n",
+ "\n",
+ "A_s=pi*4**-1*Di**2 #mm**2 #Area of steel\n",
+ "A_c=pi*4**-1*(45**2-Di**2) #mm**2 #Area of copper\n",
+ "\n",
+ "#Force introduced in steel\n",
+ "P_s=0.5*(2*L_c*(A_c*E_c)**-1+L_s*(A_s*E_s)**-1)**-1 #N\n",
+ "P_s2=P_s*A_s**-1\n",
+ "\n",
+ "#Force introduced in copper\n",
+ "P_c=2*P_s*A_c**-1 #N\n",
+ "\n",
+ "#Result\n",
+ "print\"Stress introduced in bolt is\",round(P_s2,2),\"N/mm**2\"\n",
+ "print\"stress introduced in tube is\",round(P_c,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Stress introduced in bolt is 74.4 N/mm**2\n",
+ "stress introduced in tube is 66.43 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 22
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.28,Page No.40"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=9 #m #Length of rigid bar\n",
+ "L_b=3000 #Length of bar\n",
+ "A_b=1000 #mm**2 #Area of bar\n",
+ "E_b=1*10**5 #N/mm**2 #Modulus of Elasticity of brasss bar\n",
+ "L_s=5000 #mm #Length of steel bar\n",
+ "A_s=445 #mm**2 #Area of steel bar\n",
+ "E_s=2*10**5 #N/mm**2 #Modulus of elasticity of steel bar\n",
+ "P=3000 #N #Load\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#From static equilibrium Equation of the rod after appliying Load is\n",
+ "#P_b+P_s=P ......................(1)\n",
+ "\n",
+ "#P_b=1.8727*P_s ..................(2)\n",
+ "\n",
+ "#NOw substituting equation 2 in equation 1 we get\n",
+ "P_s=P*2.8727**-1\n",
+ "P_b=P-P_s\n",
+ "\n",
+ "d=P_s*L*P**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"Distance at which Load applied even after which bar remains horizontal is\",round(d,2),\"m\" "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Distance at which Load applied even after which bar remains horizontal is 3.13 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.29,Page No.41"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "A_b=1000 #MM**2 #Area of brass bar\n",
+ "E_b=1*10**5 #N/mm**2 #Modulus of Elasticity of brass\n",
+ "A_s=600 #N/mm**2 #Area of steel rod\n",
+ "E_s=2*10**5 #N/mm**2 #Modulus of eLasticity of steel bar\n",
+ "P=10*10**2 #N #Load\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let P_b be the tensile force in brass bar and P_s be the compressive force in steel bar\n",
+ "#Now taking moment about A we get static Equilibrium condition as\n",
+ "#P_b+2*P_s=27500 ......................................(1)\n",
+ "\n",
+ "#Now from deformed shape we get\n",
+ "#dell_s=2*dell_b\n",
+ "\n",
+ "#P_s*L_s*(A_s*E_s)**-1=P_b*L_b*(A_b*E_b)**-1\n",
+ "#Further simplifying we get\n",
+ "#P_s=1.2*P_b .........................................(2)\n",
+ "\n",
+ "#Now substituting equation 1 in equation 2 we get\n",
+ "P_b=27500*3.4**-1\n",
+ "P_s=1.2*P_b\n",
+ "\n",
+ "#Tensile stress in brass bar \n",
+ "sigma_b=P_b*A_b**-1\n",
+ "\n",
+ "#compressive stress in steel bar\n",
+ "sigma_s=P_s*A_s**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"Compressive Stress in Bar is\",round(sigma_s,2),\"N/mm**2\"\n",
+ "print\"tensile Stress in Bar is\",round(sigma_b,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Compressive Stress in Bar is 16.18 N/mm**2\n",
+ "tensile Stress in Bar is 8.09 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.30,Page No.44"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=12.6 #m #Length of rail\n",
+ "t1=24 #Degree celsius\n",
+ "t2=44 #degree celsius\n",
+ "alpha=12*10**-6 #Per degree celsius\n",
+ "E=2*10**5 #N/mm**2 #Modulus of ELasticity\n",
+ "gamma=2 #mm #Gap provided for Expansion\n",
+ "sigma=20 #N/mm**2 #Stress\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "t=t2-t1 #Temperature Difference\n",
+ "\n",
+ "#Free Expansion of the rails\n",
+ "dell=alpha*t*L*1000 #mm \n",
+ "\n",
+ "#When no expansion joint is provided then\n",
+ "p=dell*E*(L*10**3)**-1\n",
+ "\n",
+ "#When a gap of 2 mm is provided,then free expansion prevented is\n",
+ "dell_1=dell-gamma\n",
+ "p2=dell_1*E*(L*10**3)**-1\n",
+ "\n",
+ "#When stress is developed,then gap left is\n",
+ "gamma2=-(sigma*L*10**3*E**-1-dell)\n",
+ "\n",
+ "#Result\n",
+ "print\"The minimum gap between the two rails is\",round(dell,2),\"mm\"\n",
+ "print\"Thermal Developed in the rials if:No expansionn joint is provided:p\",round(p,2),\"N/mm**2\"\n",
+ "print\" :If a gap of is provided then :p2\",round(p2,2),\"N/mm**2\"\n",
+ "print\"When stress is developed gap left between the rails is\",round(gamma2,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The minimum gap between the two rails is 3.02 mm\n",
+ "Thermal Developed in the rials if:No expansionn joint is provided:p 48.0 N/mm**2\n",
+ " :If a gap of is provided then :p2 16.25 N/mm**2\n",
+ "When stress is developed gap left between the rails is 1.76 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 23
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.31,Page No.45"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "t=20 #degree celsius\n",
+ "E_a=70*10**9 #N/mm**2 #Modulus of Elasticicty of aluminium\n",
+ "alpha_a=11*10**-6 #per degree celsius #Temperature coeff of aluminium\n",
+ "alpha_s=12*10**-6 #Per degree celsius #Temperature coeff of steel\n",
+ "L_a=1000 #mm #Length of aluminium \n",
+ "L_s=3000 #mm #Length of steel\n",
+ "E_a=7*10**4 #N/mm**2 #Modulus of Elasticity of aluminium\n",
+ "E_s=2*10**5 #N/mm*2 #Modulus of Elasticity of steel\n",
+ "A_a=600 #mm**2 #Area of aluminium\n",
+ "A_s=300 #mm**2 #Area of steel\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Free Expansion \n",
+ "dell=alpha_a*t*L_a+alpha_s*t*L_s\n",
+ "\n",
+ "#support Reaction\n",
+ "P=dell*(L_a*(A_a*E_a)**-1+L_s*(A_s*E_s)**-1)**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"Reaction at support is\",round(P,2),\"N\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Reaction at support is 12735.48 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 26
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.33,Page No.48"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "D=25 #mm #Diameter of Brass\n",
+ "De=50 #mm #External Diameter of steel tube\n",
+ "Di=25 #mm #Internal Diameter of steel tube\n",
+ "L=1.5 #m #Length of both bars\n",
+ "t1=30 #degree celsius #Initial Temperature\n",
+ "t2=100 #degree celsius #final Temperature\n",
+ "E_s=2*10**5 #N/mm**2 #Modulus of ELasticity of steel bar\n",
+ "E_b=1*10**5 #N/mm**2 #Modulus of Elasticity of brass bar\n",
+ "alpha_s=11.6*10**-6 #Temperature Coeff of steel\n",
+ "alpha_b=18.7*10**-6 #Temperature coeff of brass bar\n",
+ "d=20 #mm #diameter of pins\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "t=t2-t1 #Temperature Difference\n",
+ "A_s=pi*4**-1*(De**2-Di**2) #mm**2 #Area of steel\n",
+ "A_b=pi*4**-1*D**2 #mm**2 #Area of brass\n",
+ "\n",
+ "#Let P_b be the tensile force in brass bar and P_s be the compressive force in steel bar\n",
+ "#But from Equilibrium of Forces \n",
+ "#P_b=P_s=P\n",
+ "\n",
+ "#Let dell=dell_s+dell_b\n",
+ "dell=(alpha_b-alpha_s)*t*L*1000\n",
+ "\n",
+ "P=dell*(1*(A_s*E_s)**-1+1*(A_b*E_b)**-1)**-1*(L*1000)**-1\n",
+ "P_b=P_s=P\n",
+ "\n",
+ "#Stress in steel\n",
+ "sigma_s=P_s*A_s**-1\n",
+ "\n",
+ "#Stress in Brass\n",
+ "sigma_b=P_b*A_b**-1\n",
+ "\n",
+ "#Area of Pins\n",
+ "A_p=pi*4**-1*d**2\n",
+ "\n",
+ "#Since,the force is resisted by two cross section of pins\n",
+ "tou=P*(2*A_p)**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"Stress in steel bar is\",round(sigma_s,2),\"N/mm**2\"\n",
+ "print\"Stress in Brass bar is\",round(sigma_b,2),\"N/mm**2\"\n",
+ "print\"Shear Stresss induced in pins is\",round(tou,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Stress in steel bar is 14.2 N/mm**2\n",
+ "Stress in Brass bar is 42.6 N/mm**2\n",
+ "Shear Stresss induced in pins is 33.28 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.34,Page No.49"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "b_s=60 #mm #width of steel Bar\n",
+ "t_s=10 #mm #thickness of steel Bar\n",
+ "b_c=40 #mm #width of copper bar\n",
+ "t_c=5 #mm #thickness of copper bar\n",
+ "E_s=2*10**5 #N/mm**2 #Modulus of Elasticity of steel bar\n",
+ "E_c=1*10**5 #N/mm**2 #Modulus of Elasticity of copper bar\n",
+ "alpha_s=12*10**-6 #Per degree celsius #Temperature coeff of steel bar\n",
+ "alpha_c=17*10**-6 #Per degree celsius #Temperature coeff of copper bar\n",
+ "L_s=L_c=L=1000 #mm #Length of bar\n",
+ "t=80 #degree celsius\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "A_s=b_s*t_s #Area of steel bar\n",
+ "A_c=b_c*t_c #Area of copper bar\n",
+ "\n",
+ "#Let P_s be the tensile force in steel bar and P_c be the compressive force in copper bar\n",
+ "#The equilibrium of forces gives \n",
+ "#P_s=2*P_c\n",
+ "\n",
+ "#Let dell=dell_s+dell_b\n",
+ "dell=(alpha_c-alpha_s)*t\n",
+ "\n",
+ "P_c=dell*(2*(A_s*E_s)**-1+1*(A_c*E_c)**-1)**-1\n",
+ "P_s=2*P_c\n",
+ "\n",
+ "#Stress in copper \n",
+ "sigma_c=P_c*A_c**-1\n",
+ "\n",
+ "#Stress in steel\n",
+ "sigma_s=P_s*A_s**-1\n",
+ "\n",
+ "#Change in Length of bar\n",
+ "dell_2=alpha_s*t*L+P_s*L_s*(A_s*E_s)**-1\n",
+ "\n",
+ "#result\n",
+ "print\"Stress in copper is\",round(sigma_c,2),\"N/mm**2\"\n",
+ "print\"Stress in steel is\",round(sigma_s,2),\"N/mm**2\"\n",
+ "print\"the change in Length is\",round(dell_2,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Stress in copper is 30.0 N/mm**2\n",
+ "Stress in steel is 20.0 N/mm**2\n",
+ "the change in Length is 1.06 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 24
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.35,Page No.50"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "P=2*10**5 #N #Weight\n",
+ "L=1 #m #Length of each rod\n",
+ "A_c=A_s=A=500 #mm**2 #Area of each rod\n",
+ "t=40 #degree celsius #temperature\n",
+ "E_s=2*10**5 #N/mm**2 #Modulus of Elasticity of steel rod\n",
+ "E_c=1*10**5 #N/mm**2 #modulus of Elastictiy of copper rod\n",
+ "alpha_s=1.2*10**-5 #Per degree Celsius #temp coeff of steel rod\n",
+ "alpha_c=1.8*10**-5 #Per degree Celsius #Temp coeff of copper rod\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let P_s be the force in each one of the copper rods and P_s be the force in steel rod\n",
+ "#2*P_c+P_s=P .....................(1)\n",
+ "\n",
+ "#Extension of copper bar=Extension of steel bar\n",
+ "#P_s*L*(A_s*E_s)**-1=P_c*L*(A_c*E_c)**-1\n",
+ "#after simplifying above equation we get\n",
+ "#P_s=2*P_c ........................(2)\n",
+ "\n",
+ "#Now substituting value of P_s in Equation 1 we get\n",
+ "P_c=P*4**-1\n",
+ "P_s=2*P_c\n",
+ "\n",
+ "#Now EXtension due to copper Load\n",
+ "dell_1=P_c*L*1000*(A_c*E_c)**-1\n",
+ "\n",
+ "#Part-2\n",
+ "\n",
+ "#Due to rise of temperature of40 degree celsius\n",
+ "\n",
+ "#As bars are rigidly joined,let P_c1 be the compressive forccesdeveloped in copper bar and P_s1 be the tensile force in steel causing changes\n",
+ "#P_s1=2*P_c1\n",
+ "\n",
+ "#dell_s+dell_c=(alpha_c-alpha_s)*t*L .......................................(3)\n",
+ "#P_s1*L*(A_s*E_s)**-1+P_c1*L*(A_c*E_c)**-1=(alpha_c-alpha_s)*t*L ................(4)\n",
+ "#After substituting values in above equation and further simplifying we get,\n",
+ "P_c1=(alpha_c-alpha_s)*t*L*(2*(A_s*E_s)**-1+1*(A_c*E_c)**-1)**-1 #.................(5)\n",
+ "P_s1=2*P_c1\n",
+ "\n",
+ "#Extension of bar due to temperature rise\n",
+ "dell_2=alpha_s*t*L+P_s1*L*(A_s*E_s)**-1\n",
+ "\n",
+ "#Amount by which bar will descend\n",
+ "dell_3=dell_1+dell_2\n",
+ "\n",
+ "#Load carried by steel bar\n",
+ "P_S=P_s+P_s1\n",
+ "\n",
+ "#Load carried by copper bar\n",
+ "P_C=P_c-P_c1\n",
+ "\n",
+ "#Part-3\n",
+ "\n",
+ "#Let P_c1_1=P_c #For convenience\n",
+ "#Rise in temperature if Load is to be carried out by steel rod alone\n",
+ "P_c1_1=P_c\n",
+ "\n",
+ "#From equation 5 \n",
+ "t=P_c1_1*(2*(A_s*E_s)**-1+1*(A_c*E_c)**-1)*(alpha_c-alpha_s)**-1\n",
+ "\n",
+ "#result\n",
+ "print\"Extension Due top copper Load\",round(dell_1,2),\"mm\"\n",
+ "print\"Load carried by each rod:P_s\",round(P_s,2),\"N\"\n",
+ "print\" :P_c\",round(P_c,2),\"N\"\n",
+ "print\"Rise in Temperature of steel rod should be\",round(t,2),\"degree Celsius\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Extension Due top copper Load 1.0 mm\n",
+ "Load carried by each rod:P_s 100000.0 N\n",
+ " :P_c 50000.0 N\n",
+ "Rise in Temperature of steel rod should be 333.33 degree Celsius\n"
+ ]
+ }
+ ],
+ "prompt_number": 26
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.36,Page No.53"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "t=40 #degree celsius #temperature\n",
+ "A_s=400 #mm**2 #Area of steel bar\n",
+ "A_c=600 #mm**2 #Area of copper bar\n",
+ "E_s=2*10**5 #N/mm**2 #Modulus of Elasticity of steel bar\n",
+ "E_c=1*10**5 #N/mm**2 #Modulus of Elasticity of copper bar\n",
+ "alpha_s=12*10**-6 #degree celsius #Temperature coeff of steel bar\n",
+ "alpha_c=18*10**-6 #degree celsius #Temperature coeff of copper bar\n",
+ "L_c=800 #mm #Length of copper bar\n",
+ "L_s=600 #mm #Length of steel bar\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let P_s be the tensile force in steel bar and P_c be the compressive force in copper bar\n",
+ "#Static Equilibrium obtained by taking moment about A\n",
+ "#P_c=2*P_s\n",
+ "\n",
+ "#From property of similar triangles we get\n",
+ "#(alpha_c*Lc-dell_c)*1**-1=(alpha_s*L_s-dell_s)*2**-1\n",
+ "#After substituting values in above equations and further simplifying we get\n",
+ "P_s=(2*alpha_c*L_c-alpha_s*L_s)*t*(L_s*(A_s*E_s)**-1+4*L_c*(A_c*E_c)**-1)**-1\n",
+ "P_c=2*P_s\n",
+ "\n",
+ "#Stress in steel rod\n",
+ "sigma_s=P_s*A_s**-1 #N/mm**2 \n",
+ "\n",
+ "#Stress in copper rod\n",
+ "sigma_c=P_c*A_c**-1 #N/mm**2\n",
+ "\n",
+ "#Result\n",
+ "print\"Stress in steel rod is\",round(sigma_s,2),\"N/mm**2\"\n",
+ "print\"STress in copper rod is\",round(sigma_c,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Stress in steel rod is 35.51 N/mm**2\n",
+ "STress in copper rod is 47.34 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.37,Page No.61"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d=20 #mm #Diameter of bar\n",
+ "P=37.7*10**3 #N #Load\n",
+ "L=200 #mm #Guage Length \n",
+ "dell=0.12 #mm #Extension\n",
+ "dell_d=0.0036 #mm #contraction in diameter\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Area of bar\n",
+ "A=pi*4**-1*d**2\n",
+ "\n",
+ "#Let s and dell_s be the Linear strain and Lateral strain\n",
+ "s=dell*L**-1\n",
+ "dell_s=dell_d*d**-1\n",
+ "mu=dell_s*s**-1 #Poissoin's ratio \n",
+ "\n",
+ "#dell=P*L*(A*E)**-1\n",
+ "E=P*L*(dell*A)**-1 #N/mm**2 #Modulus of Elasticity of bar\n",
+ "\n",
+ "#Modulus of Rigidity\n",
+ "G=E*(2*(1+mu))**-1 #N/mm**2\n",
+ "\n",
+ "#Bulk Modulus \n",
+ "K=E*(3*(1-2*mu))**-1 #N/mm**2\n",
+ "\n",
+ "#result\n",
+ "print\"Poisson's ratio is\",round(mu,2)\n",
+ "print\"The Elastic constant are:E\",round(E,2)\n",
+ "print\" :G\",round(G,2)\n",
+ "print\" :K\",round(K,2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Poisson's ratio is 0.3\n",
+ "The Elastic constant are:E 200004.71\n",
+ " :G 76924.89\n",
+ " :K 166670.59\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.38,Page No.62"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d=100 #mm #Diameter of circular rod\n",
+ "P=1*10**6 #N #Tensile Force\n",
+ "mu=0.3 #Poisson's ratio\n",
+ "E=2*10**5 #N/mm**2 #Young's Modulus \n",
+ "L=500 #mm #Length of rod\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Modulus of Rigidity\n",
+ "G=E*(2*(1+mu))**-1 #N/mm**2\n",
+ "\n",
+ "#Bulk Modulus \n",
+ "K=E*(3*(1-2*mu))**-1 #N/mm**2\n",
+ "\n",
+ "A=pi*4**-1*d**2 #mm**2 #Area of Circular rod\n",
+ "#Let sigma be the Longitudinal stress\n",
+ "sigma=P*A**-1 #N/mm**2 \n",
+ "\n",
+ "s=sigma*E**-1 #Linear strain\n",
+ "e_x=s\n",
+ "\n",
+ "#Volumetric strain\n",
+ "e_v=e_x*(1-2*mu)\n",
+ "\n",
+ "v=pi*4**-1*d**2*L\n",
+ "#Change in VOlume\n",
+ "dell_v=e_v*v\n",
+ "\n",
+ "#Result\n",
+ "print\"Bulk Modulus is\",round(E,2),\"N/mm**2\"\n",
+ "print\"Modulus of Rigidity is\",round(G,2),\"N/mm**2\"\n",
+ "print\"The change in Volume is\",round(dell_v,2),\"mm**3\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Bulk Modulus is 200000.0 N/mm**2\n",
+ "Modulus of Rigidity is 76923.08 N/mm**2\n",
+ "The change in Volume is 1000.0 mm**3\n"
+ ]
+ }
+ ],
+ "prompt_number": 21
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.39,Page No.62"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=500 #mm #Length of rectangular cross section bar\n",
+ "A=20*40 #mm**2 #Area of rectangular cross section bar\n",
+ "P1=4*10**4 #N #Tensile Force on 20mm*40mm Faces\n",
+ "P2=2*10**5 #N #compressive force on 20mm*500mm Faces\n",
+ "P3=3*10**5 #N #Tensile Force on 40mm*500mm Faces\n",
+ "E=2*10**5 #N/mm**2 #young's Modulus \n",
+ "mu=0.3 #Poisson's Ratio\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let P_x,P_y,P_z be the forces n x,y,z directions\n",
+ "\n",
+ "P_x=P1*A**-1\n",
+ "P_y=P2*A**-1\n",
+ "P_z=P3*A**-1\n",
+ "\n",
+ "#Let e_x,e_y,e_z be the strains in x,y,z directions\n",
+ "e_x=1*E**-1*(50+mu*20-15*mu)\n",
+ "e_y=1*E**-1*(-mu*50-20-mu*15)\n",
+ "e_z=1*E**-1*(-mu*50+mu*20+15)\n",
+ "\n",
+ "#Volumetric strain\n",
+ "e_v=e_x+e_y+e_z\n",
+ "\n",
+ "#Volume\n",
+ "V=20*40*500 #mm**3\n",
+ "#Change in Volume \n",
+ "dell_v=e_v*V #mm**3\n",
+ "\n",
+ "#Result\n",
+ "print\"The change in Volume is\",round(dell_v,2),\"mm**3\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The change in Volume is 36.0 mm**3\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.41,Page No.65"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "E=2.1*10**5 #N/mm**2 #Young's Modulus \n",
+ "G=0.78*10**5 #N/mm**2 #Modulus of Rigidity\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Now using the relation\n",
+ "#E=2*G*(1+mu)\n",
+ "mu=E*(2*G)**-1-1 #Poisson's ratio\n",
+ "\n",
+ "#Bulk Modulus \n",
+ "K=E*(3*(1-2*mu))**-1 #N/mm**2\n",
+ "\n",
+ "\n",
+ "#Result\n",
+ "print\"The Poisson's Ratio is\",round(mu,2)\n",
+ "print\"The modulus of Rigidity\",round(K,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Poisson's Ratio is 0.35\n",
+ "The modulus of Rigidity 227500.0 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.42,Page No.65"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "G=0.4*10**5 #N/mm**2 #Modulus of rigidity\n",
+ "K=0.75*10**5 #N/mm**2 #Bulk Modulus \n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Young's Modulus\n",
+ "E=9*G*K*(3*K+G)**-1\n",
+ "\n",
+ "#Now from the relation\n",
+ "#E=2*G(1+2*mu)\n",
+ "mu=E*(2*G)**-1-1 #POissoin's ratio \n",
+ "\n",
+ "#result\n",
+ "print\"Young's modulus is\",round(E,2),\"N/mm**2\"\n",
+ "print\"Poissoin's ratio is\",round(mu,2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Young's modulus is 101886.79 N/mm**2\n",
+ "Poissoin's ratio is 0.27\n"
+ ]
+ }
+ ],
+ "prompt_number": 25
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.43,Page No.65"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "b=60 #mm #width of bar\n",
+ "d=30 #mm #depth of bar\n",
+ "L=200 #mm #Length of bar\n",
+ "A=30*60 #mm**2 #Area of bar\n",
+ "A2=30*200 #mm**2 #Area of bar along which expansion is restrained\n",
+ "P=180*10**3 #N #Compressive force\n",
+ "E=2*10**5 #N/mm**2 #Young's Modulus\n",
+ "mu=0.3 #Poissoin's ratio\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#The bar is restrained from expanding in Y direction\n",
+ "P_z=0\n",
+ "P_x=P*A**-1 #stress developed in x direction\n",
+ "\n",
+ "#Now taking compressive strain as positive\n",
+ "#e_x=P_x*E**-1-mu*P_y*E**-1 .......................(1)\n",
+ "#e_y=-mu*P_x*E**-1+P_y*E**-1 ....................(2)\n",
+ "#e_z=-mu*P_x*E**-1-mu*P_y*E**-1 ......................(3)\n",
+ "\n",
+ "#Part-1\n",
+ "#When it is fully restrained\n",
+ "e_y=0\n",
+ "P_y=30 #N/mm**2 \n",
+ "e_x=P_x*E**-1-mu*P_y*E**-1\n",
+ "e_z=-mu*P_x*E**-1-mu*P_y*E**-1\n",
+ "\n",
+ "#Change in Length \n",
+ "dell_l=e_x*L #mm\n",
+ "\n",
+ "#Change in width\n",
+ "dell_b=b*e_y\n",
+ "\n",
+ "#change in Depth\n",
+ "dell_d=d*e_z\n",
+ "\n",
+ "#Volume of bar\n",
+ "V=b*d*L #mm**3\n",
+ "#Change in Volume\n",
+ "e_v=(e_x+e_y+e_z)*V #mm**3\n",
+ "\n",
+ "#Part-2\n",
+ "#When 50% is restrained\n",
+ "\n",
+ "#Free strain in Y direction\n",
+ "e_y1=mu*P_x*E**-1\n",
+ "\n",
+ "#As 50% is restrained,so\n",
+ "e_y2=-50*100**-1*e_y1\n",
+ "\n",
+ "#But form Equation 2 we have e_y=-mu*P_x*E**-1+P_y*E**-1 \n",
+ "#After substituting values in above equation and furthe simplifying we get\n",
+ "P_y=e_y2*E+d\n",
+ "\n",
+ "e_x2=P_x*E**-1-mu*P_y*E**-1 \n",
+ "e_z2=-mu*P_x*E**-1-mu*P_y*E**-1\n",
+ "\n",
+ "#Change in Length \n",
+ "dell_l2=e_x2*L #mm\n",
+ "\n",
+ "#Change in width\n",
+ "dell_b2=b*e_y2\n",
+ "\n",
+ "#change in Depth\n",
+ "dell_d2=d*e_z2\n",
+ "\n",
+ "#Change in Volume\n",
+ "e_v2=(e_x2+e_y2+e_z2)*V #mm**3\n",
+ "\n",
+ "#REsult\n",
+ "print\"Change in Dimension of bar is:dell_l\",round(dell_l,2),\"mm\"\n",
+ "print\" :dell_b\",round(dell_b,4),\"mm\"\n",
+ "print\" :dell_d\",round(dell_d,2),\"mm\"\n",
+ "print\"Change in Volume is\",round(e_v,2),\"mm**3\"\n",
+ "print\"Changes in material when only 50% of expansion can be reatrained:dell_l2\",round(dell_l2,2),\"mm\"\n",
+ "print\" :dell_b2\",round(dell_b2,4),\"mm\"\n",
+ "print\" :dell_d2\",round(dell_d2,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Change in Dimension of bar is:dell_l 0.09 mm\n",
+ " :dell_b 0.0 mm\n",
+ " :dell_d -0.01 mm\n",
+ "Change in Volume is 93.6 mm**3\n",
+ "Changes in material when only 50% of expansion can be reatrained:dell_l2 0.1 mm\n",
+ " :dell_b2 -0.0045 mm\n",
+ " :dell_d2 -0.01 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 23
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.44,Page No.72"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "P=10*10**3 #N #Load\n",
+ "E=2*10**5 #N/mm**2 #Young's Modulus\n",
+ "d2=12 #mm #Diameter of bar1\n",
+ "d1=16 #mm #diameter of bar2\n",
+ "L1=200 #mm #Length of bar1\n",
+ "L2=500 #mm #Length of bar2\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let A1 and A2 be the cross Area of Bar1 & bar2 respectively\n",
+ "A1=pi*4**-1*d1**2 #mm**2\n",
+ "A2=pi*4**-1*d2**2 #mm**2\n",
+ "\n",
+ "#Let p1 and p2 be the stress in Bar1 nad bar2 respectively\n",
+ "p1=P*A1**-1 #N/mm**2\n",
+ "p2=P*A2**-1 #N/mm**2\n",
+ "\n",
+ "#Let V1 nad V2 be the Volume of of Bar1 and Bar2\n",
+ "V1=A1*(L1+L1)\n",
+ "V2=A2*L2\n",
+ "\n",
+ "#Let E be the strain Energy stored in the bar\n",
+ "E=p1**2*(2*E)**-1*V1+p2**2*V2*(2*E)**-1\n",
+ "\n",
+ "#result\n",
+ "print\"The Strain Energy stored in Bar is\",round(E,2),\"N-mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Strain Energy stored in Bar is 1602.6 N-mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 74
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.45,Page No.73"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#Bar-A\n",
+ "d1=30 #mm #Diameter of bar1\n",
+ "L=600 #mm #length of bar1\n",
+ "\n",
+ "#Bar-B\n",
+ "d2=30 #mm #Diameter of bar2\n",
+ "d3=20 #mm #Diameter of bar2\n",
+ "L2=600 #mm #length of bar2\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Area of bar-A\n",
+ "A1=pi*4**-1*d1**2\n",
+ "\n",
+ "#Area of bar-B\n",
+ "A2=pi*4**-1*d2**2\n",
+ "A3=pi*4**-1*d3**2\n",
+ "\n",
+ "#let SE be the Strain Energy\n",
+ "#Strain Energy stored in Bar-A\n",
+ "#SE=p**2*(2*E)**-1*V\n",
+ "#After substituting values and simolifying further we get\n",
+ "#SE=P**2*E**-1*0.4244\n",
+ "\n",
+ "#Strain Energy stored in Bar-B\n",
+ "#SE2=p1**2*V1*(2*E)**-1+p2**2*V2*(2*E)**-1\n",
+ "#After substituting values and simolifying further we get\n",
+ "#SE2=0.6897*P**2*E**-1\n",
+ "\n",
+ "#Let X be the ratio of SE in Bar-B and SE in Bar-A\n",
+ "X=0.6897*0.4244**-1\n",
+ "\n",
+ "#Part-2\n",
+ "\n",
+ "#When Max stress is produced is same:Let p be the max stress produced\n",
+ "\n",
+ "#Stress in bar A is p throughout \n",
+ "#In bar B:stress in 20mm dia.portion=p2=p\n",
+ "\n",
+ "#Stress in 30 mm dia.portion\n",
+ "#p1=P*A2*A3**-1\n",
+ "#After substituting values and simolifying further we get\n",
+ "#p1=4*9**-1*p\n",
+ "\n",
+ "#Strain Energy in bar A\n",
+ "#SE_1=p**2*(2*E)**-1*A1*L1\n",
+ "#After substituting values and simolifying further we get\n",
+ "#SE_1=67500*p**2*pi*E**-1\n",
+ "\n",
+ "#Strain Energy in bar B\n",
+ "#SE_2=p1**2*V1*(2*E)**-1+p2**2*V2*(2*E)**-1\n",
+ "#After substituting values and simolifying further we get\n",
+ "#SE_2=21666.67*pi*p**2*E**-1\n",
+ "\n",
+ "#Let Y be the Ratio of SE in bar B and SE in bar A\n",
+ "Y=21666.67*67500**-1\n",
+ "\n",
+ "#result\n",
+ "print\"Gradually applied Load is\",round(X,2)\n",
+ "print\"Gradually applied Load is\",round(Y,2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Gradually applied Load is 1.63\n",
+ "Gradually applied Load is 0.32\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.46,Page No.74"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables \n",
+ "\n",
+ "W=100 #N #Load\n",
+ "E=2*10**5 #N/mm**2 #Young's Modulus \n",
+ "h=60 #mm #Height through Load falls down\n",
+ "L=400 #mm #Length of collar\n",
+ "d=30 #mm #diameter of bar\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "A=pi*4**-1*d**2 #mm**2 #Area of bar\n",
+ "\n",
+ "#Instantaneous stress produced is\n",
+ "p=W*A**-1*(1+(1+(2*A*E*h*(W*L)**-1))**0.5)\n",
+ "\n",
+ "#Now the EXtension of the bar is neglected in calculating work doneby the Load,then\n",
+ "P=(2*E*h*W*(A*L)**-1)**0.5\n",
+ "\n",
+ "#Let percentage error be denoted by E1\n",
+ "#Percentage error in approximating is\n",
+ "E1=(p-P)*p**-1*100\n",
+ "\n",
+ "#Instantaneous Extension produced is\n",
+ "dell_l=round(P,3)*E**-1*L\n",
+ "\n",
+ "#Result\n",
+ "print\"The Instantaneous stress is\",round(p,2),\"N/mm\"\n",
+ "print\"Percentage Error is\",round(E1,2)\n",
+ "print\"The Instantaneous extension is\",round(dell_l,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Instantaneous stress is 92.27 N/mm\n",
+ "Percentage Error is 0.15\n",
+ "The Instantaneous extension is 0.18 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 23
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.47,Page No.75"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d=20 #mm #Diameter of steel bar\n",
+ "L=1000 #mm #Length of bar\n",
+ "E=2*10**5 #N/mm**2 #Young's Modulus \n",
+ "p=300 #N/mm**2 #max Permissible stress\n",
+ "h=50 #mm #Height through which weight will fall\n",
+ "w=600 #N #Load\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#ARea of steel bar\n",
+ "A=pi*4**-1*d**2\n",
+ "\n",
+ "#Instantaneous extension is\n",
+ "dell_l=p*L*E**-1 #mm \n",
+ "\n",
+ "#Work done by Load \n",
+ "#W=W1*(h+dell_l)\n",
+ "\n",
+ "#Volume of bar\n",
+ "V=round(A,2)*L\n",
+ "#Let E1 be the strain Energy\n",
+ "E1=p**2*(2*E)**-1*V\n",
+ "\n",
+ "#Answer in Book for Strain Energy is Incorrect \n",
+ "\n",
+ "#Now Equating Workdone by Load to strain Energy \n",
+ "W1=E1*51.5**-1\n",
+ "\n",
+ "#Now when w=600 N\n",
+ "#Let W2 be the Work done by the Load\n",
+ "#W2=w(h2*dell_l)\n",
+ "\n",
+ "h=E1*w**-1-dell_l\n",
+ "\n",
+ "#Result\n",
+ "print\"The Max Lodad which can Fall from a height of 50 mm on the collar is\",round(W1,2),\"N\"\n",
+ "print\"the Max Height from which a 600 N Load can fall on the collar is\",round(h,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Max Lodad which can Fall from a height of 50 mm on the collar is 1372.54 N\n",
+ "the Max Height from which a 600 N Load can fall on the collar is 116.31 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 40
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.48,Page No.76"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "D_s=30 #mm #Diameter of steel rod\n",
+ "d=30 #mm #Internal Diameter of copper tube\n",
+ "D=40#mm #External Diameter of copper tube\n",
+ "E_s=2*10**5 #N/mm**2 #Young's Modulus of Steel rod\n",
+ "E_c=1*10**5#N/mm**2 #Young's Modulus of copper tube\n",
+ "P=100 #N #Load\n",
+ "h=40 #mm #height from which Load falls\n",
+ "L=800 #mm #Length \n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Area of steel rod\n",
+ "A_s=pi*4**-1*D_s**2\n",
+ "\n",
+ "#Area of copper tube\n",
+ "A_c=pi*4**-1*(D**2-d**2)\n",
+ "\n",
+ "#But Dell_s=dell_c=dell\n",
+ "#p_s*E_s**-1*L=p_c*L*E_c\n",
+ "#After simplifying furthe we get\n",
+ "#p_s=2*p_c\n",
+ "\n",
+ "#Now Equating internal Energy to Workdone we get\n",
+ "p_c=(2*P*h*L**-1*(4*A_s*E_s**-1+A_c*E_c**-1))**0.5\n",
+ "p_s=2*p_c\n",
+ "\n",
+ "#Result\n",
+ "print A_s\n",
+ "print\"STress produced in steel is\",round(p_s,2),\"N/mm**2\"\n",
+ "print\"STress produced in copper is\",round(p_c,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "706.858347058\n",
+ "STress produced in steel is 0.89 N/mm**2\n",
+ "STress produced in copper is 0.44 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.49,Page No.77"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "dell=0.25 #mm #Instantaneous Extension\n",
+ "\n",
+ "#Bar-A\n",
+ "b1=25 #mm #width of bar\n",
+ "D1=500 #mm #Depth of bar\n",
+ "\n",
+ "#Bar-B\n",
+ "b2_1=25 #mm #width of upper bar\n",
+ "b2_2=15 #mm #Width of Lower Bar\n",
+ "L2=200 #mm #Length of upper bar\n",
+ "L1=300 #mm #Length of Lower bar\n",
+ "\n",
+ "E=2*10**5 #N/mm**2 #Young's Modulus of bar\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Strain\n",
+ "e=dell*D1**-1 \n",
+ "\n",
+ "#Load\n",
+ "p=e*E\n",
+ "\n",
+ "#Area of bar-A\n",
+ "A=pi*4**-1*25**2\n",
+ "\n",
+ "#Volume of bar-A\n",
+ "V=A*D1\n",
+ "\n",
+ "#Let E1 be the Energy of Blow\n",
+ "#Energy of Blow\n",
+ "E1=p**2*(E)**-1*V\n",
+ "\n",
+ "#Let p2 be the Max stress in bar B When this blow is applied.\n",
+ "#the max stress occurs in the 15mm dia. portion,Hence, the stress in 25 mm dia.portion is\n",
+ "#p2*pi*4**-1*b2_2**2*(pi*4**-1*b2_2**2=0.36*p\n",
+ "\n",
+ "#Strain Energy of bar B\n",
+ "#E2=p**2*(2*E)**-1*v1+1*(2*E)**-1*(0.36*p2)**2*v2\n",
+ "#After substituting values and Further substituting values we get\n",
+ "#E2=0.1643445*p2**2\n",
+ "\n",
+ "#Equating it to Energy of applied blow,we get\n",
+ "p2=(12271.846*0.1643445**-1)**0.5\n",
+ "\n",
+ "#Stress in top portion\n",
+ "sigma=0.36*p2\n",
+ "\n",
+ "#Extension in Bar-1\n",
+ "dell_1=p2*E**-1*L1\n",
+ "\n",
+ "#Extension in Bar-2\n",
+ "dell_2=0.36*p2*E**-1*L2\n",
+ "\n",
+ "#Extension of bar\n",
+ "dell_3=dell_1+dell_2\n",
+ "\n",
+ "#Result\n",
+ "print\"Instantaneous Max stress is\",round(sigma,2),\"N/mm**2\"\n",
+ "print\"extension in Bar is\",round(dell_3,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Instantaneous Max stress is 98.37 N/mm**2\n",
+ "extension in Bar is 0.51 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.3_7.ipynb b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.3_7.ipynb new file mode 100644 index 00000000..6e4aedd8 --- /dev/null +++ b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.3_7.ipynb @@ -0,0 +1,1601 @@ +{
+ "metadata": {
+ "name": "chapter no.3.ipynb"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 3:Shear Force And Bending Moment Diagrams in Statically Determinate Beams"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.3.1,Page No.100"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "%matplotlib inline\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L_AC=L_CD=1 #m #Length of AC & CD\n",
+ "L_DB=1.5 #m #Lengh of DB\n",
+ "L=3.5 #m #Length of Beam\n",
+ "F_B=10 #KN #Force at pt B\n",
+ "F_C=F_D=20 #KN #Force at pt C & D\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "R_A=F_C+F_D+F_B #KN #Force at support A \n",
+ "\n",
+ "#Shear Force Calculations\n",
+ "\n",
+ "#S.F At pt B\n",
+ "V_B1=0 #KN \n",
+ "V_B2=F_B #KN\n",
+ "\n",
+ "#S.F At pt D\n",
+ "V_D1=V_B2 #KN\n",
+ "V_D2=V_D1+F_D #KN\n",
+ "\n",
+ "#S.F At pt C \n",
+ "V_C1=V_D2 #KN\n",
+ "V_C2=V_D2+F_C #KN\n",
+ "\n",
+ "#S.F At Pt A\n",
+ "V_A1=V_C2 #KN\n",
+ "V_A2=V_C2-R_A #KN\n",
+ "\n",
+ "#Bending Moment Calculations\n",
+ "\n",
+ "#B.M At Pt B\n",
+ "M_B=0 #KN.m\n",
+ "\n",
+ "#B.M AT Pt D\n",
+ "M_D=F_B*L_DB #KN.m\n",
+ "\n",
+ "#B.M At pt C\n",
+ "M_C=F_B*(L_DB+L_CD)+F_D*L_CD #KN.m\n",
+ "\n",
+ "#B.M At pt A\n",
+ "M_A=F_B*L+F_D*(L_CD+L_AC)+F_C*L_AC\n",
+ "\n",
+ "#Result\n",
+ "print \"The Shear Force and Bending Moment Diagrams are the results\"\n",
+ "\n",
+ "#Plotting the Shear Force Diagram\n",
+ "\n",
+ "X1=[0,0,L_DB,L_DB,L_CD+L_DB,L_CD+L_DB,L_CD+L_DB+L_AC,L_CD+L_DB+L_AC]\n",
+ "Y1=[V_B1,V_B2,V_D1,V_D2,V_C1,V_C2,V_A1,V_A2]\n",
+ "Z1=[0,0,0,0,0,0,0,0]\n",
+ "plt.plot(X1,Y1,X1,Z1)\n",
+ "plt.xlabel(\"Length x in m\")\n",
+ "plt.ylabel(\"Shear Force in kN\")\n",
+ "plt.show()\n",
+ "\n",
+ "#Plotting the Bendimg Moment Diagram\n",
+ "\n",
+ "Y2=[M_B,M_D,M_C,M_A]\n",
+ "X2=[0,L_DB,L_DB+L_CD,L_AC+L_CD+L_DB]\n",
+ "Z2=[0,0,0,0]\n",
+ "plt.plot(X2,Y2,X2,Z2)\n",
+ "plt.xlabel(\"Length in m\")\n",
+ "plt.ylabel(\"Bending Moment in kN.m\")\n",
+ "plt.show()"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Shear Force and Bending Moment Diagrams are the results\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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+ "text": [
+ "<matplotlib.figure.Figure at 0x5dde2d0>"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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pqQavCQgIEADwi1/84he/rPgKCAiw6e+yYlJMDQ0NuPfee/H999+jZ8+eGDhw\nINavX4/g4GC5SyMickmKmWJq3749PvroI4wcORKNjY2YPn06mwMRkYwUM4IgIiJlUUzMtaWsrCwE\nBQXhnnvuwZIlS4y+Zt68ebjnnnsQHh6OwsJCiStsnbn6s7Oz4e3tjcjISERGRuLNN9+UoUrjpk2b\nBrVajbCwMJOvUfK5N1e/ks89AJSWliIuLg4hISEIDQ3Fhx9+aPR1SvwdWFK7ks9/XV0dYmJiEBER\nAY1Gg5deesno65R47gHL6rf6/Nu8qiyShoYGISAgQDh16pRQX18vhIeHC0VFRQav2bZtmzBq1ChB\nEAQhLy9PiImJkaNUoyypf8+ePcLYsWNlqrB1P/74o1BQUCCEhoYafV7J514QzNev5HMvCIJQVlYm\nFBYWCoIgCNXV1UJgYKDD/PtvSe1KP//Xrl0TBEEQbt68KcTExAj/+te/DJ5X6rlvZq5+a8+/4kYQ\nllwwl5GRgeTkZABATEwMqqqqUFFRIUe5d7D0gj9BoTN7Q4YMQZcuXUw+r+RzD5ivH1DuuQcAX19f\nREREAAA8PT0RHByM8+fPG7xGqb8DS2oHlH3+O3bsCACor69HY2MjunbtavC8Us99M3P1A9adf8U1\nCGMXzJ07d87sa86ePStZja2xpH6VSoV9+/YhPDwcCQkJKCoqkrpMmyn53FvCkc59SUkJCgsLERMT\nY/C4I/wOTNWu9PPf1NSEiIgIqNVqxMXFQaPRGDyv9HNvrn5rz79iUkzNLL1g7vYuaOn7xGZJHVFR\nUSgtLUXHjh2xY8cOJCYm4sSJExJU1zaUeu4t4SjnvqamBpMmTcKyZcvg6el5x/NK/h20VrvSz7+b\nmxsOHTqEK1euYOTIkcjOzoZWqzV4jZLPvbn6rT3/ihtB9OrVC6WlpfqfS0tL4efn1+przp49i14K\nudmzJfV7eXnph4KjRo3CzZs3UVlZKWmdtlLyubeEI5z7mzdv4qGHHsLjjz+OxMTEO55X8u/AXO2O\ncP4BwNvbG6NHj8ZPP/1k8LiSz31Lpuq39vwrrkEMGDAAv/32G0pKSlBfX4/09HSMGzfO4DXjxo3D\nl19+CQDIy8uDj48P1Gq1HOXewZL6Kyoq9P8VcvDgQQiCYHSuUImUfO4tofRzLwgCpk+fDo1Gg2ef\nfdboa5T6O7CkdiWf/z/++ANVVVUAgOvXr2P37t2IjIw0eI1Szz1gWf3Wnn/FTTGZumDus88+AwDM\nnj0bCQk5anozAAADy0lEQVQJ2L59O/r164dOnTph9erVMld9iyX1b9q0CZ988gnat2+Pjh07YsOG\nDTJXfcsjjzyCnJwc/PHHH+jduzdef/113Lx5E4Dyzz1gvn4ln3sA2Lt3L9auXYv+/fvr/8/99ttv\n48yZMwCU/TuwpHYln/+ysjIkJyejqakJTU1NmDp1KoYPH+4wf3ssqd/a888L5YiIyCjFTTEREZEy\nsEEQEZFRbBBERGQUGwQRERnFBkFEREaxQRARkVFsEOSUjG1P0ZaWLl2K69evW3W8zMxMk9vXEykR\nr4Mgp+Tl5YXq6mrRPr9v37746aef0K1bN0mORyQHjiDIZRQXF2PUqFEYMGAAhg4diuPHjwMAnnzy\nSTzzzDOIjY1FQEAANm/eDEC3M+acOXMQHByMESNGYPTo0di8eTOWL1+O8+fPIy4uDsOHD9d//quv\nvoqIiAj85S9/wYULF+44/po1a/D000+3esyWSkpKEBQUhKeeegr33nsvHnvsMezatQuxsbEIDAxE\nfn6+GKeJ6BYb70tBpGienp53PHb//fcLv/32myAIupu93H///YIgCEJycrKQlJQkCIIgFBUVCf36\n9RMEQRC+/vprISEhQRAEQSgvLxe6dOkibN68WRAEQfD39xcuXbqk/2yVSiV8++23giAIwvz584U3\n33zzjuOvWbNGmDt3bqvHbOnUqVNC+/bthV9//VVoamoSoqOjhWnTpgmCIAhbt24VEhMTrT0tRFZR\n3F5MRGKoqanB/v37MXnyZP1j9fX1AHTbNTfvPBocHKy/AUxubi6SkpIAQL+/vikeHh4YPXo0ACA6\nOhq7d+9utR5Tx7xd3759ERISAgAICQnBAw88AAAIDQ1FSUlJq8cgshcbBLmEpqYm+Pj4mLyHsIeH\nh/574T/LciqVymDvf6GV5Tp3d3f9925ubmhoaDBbk7Fj3q5Dhw4Gn9v8HkuPQWQPrkGQS+jcuTP6\n9u2LTZs2AdD9QT5y5Eir74mNjcXmzZshCAIqKiqQk5Ojf87LywtXr161qobWGgyRErFBkFOqra1F\n79699V9Lly7FunXrsHLlSkRERCA0NBQZGRn617e8K1jz9w899BD8/Pyg0WgwdepUREVFwdvbGwAw\na9YsPPjgg/pF6tvfb+wuY7c/bur7299j6mcl3cmMnBNjrkStuHbtGjp16oRLly4hJiYG+/btQ48e\nPeQui0gSXIMgasWYMWNQVVWF+vp6LFq0iM2BXApHEEREZBTXIIiIyCg2CCIiMooNgoiIjGKDICIi\no9ggiIjIKDYIIiIy6v8BsxKV3Pt7cnUAAAAASUVORK5CYII=\n",
+ "text": [
+ "<matplotlib.figure.Figure at 0x5df0ab0>"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.3.2,Page No.101"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "%matplotlib inline\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "w1=10 #KN/m #u.d.L\n",
+ "F_D=20 #KN #Force at pt D\n",
+ "F_C=30 #KN #Force at pt C\n",
+ "L_DB=4 #m #Length of DB\n",
+ "L_CD=L_AC=2 #m #Length of AC & CD\n",
+ "L=8 #m #Length of Beam\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let R_A And R_B be the Reactions at pt A and B \n",
+ "#R_A+R_B=90 \n",
+ "#Now Taking moment at A,M_A we get\n",
+ "R_A=(w1*L_DB*(L_DB*2**-1)+F_D*L_DB+F_C*(L_CD+L_DB))*L**-1\n",
+ "R_B=90-R_A\n",
+ "\n",
+ "#Shear Force Calculations\n",
+ "\n",
+ "#S.F At Pt B\n",
+ "V_B1=0 #KN\n",
+ "V_B2=R_B #KN\n",
+ "\n",
+ "#S.F At pt D\n",
+ "V_D1=R_B-w1*L_DB #KN\n",
+ "V_D2=V_D1-F_D #KN\n",
+ "\n",
+ "#S.F at Pt C\n",
+ "V_C1=V_D2 #KN\n",
+ "V_C2=V_C1-F_C \n",
+ "\n",
+ "#S.F at PT A\n",
+ "V_A1=V_C2 #KN\n",
+ "V_A2=V_C2+R_A #KN\n",
+ "\n",
+ "#Bending Moment Calculations\n",
+ "\n",
+ "#B.M At Pt B\n",
+ "M_B=0 #KN.m\n",
+ "\n",
+ "#B.M At Pt D\n",
+ "M_D=-R_B*L_DB+w1*L_DB*L_DB*2**-1 #KN.m\n",
+ "\n",
+ "#B.M At PT C\n",
+ "M_C=-R_B*(L_DB+L_CD)+w1*L_DB*(L_DB*2**-1+L_CD)+F_D*L_CD #KN.m\n",
+ "\n",
+ "#B.M At Pt A\n",
+ "M_A=-R_B*L+w1*L_DB*(L_DB*2**-1+L_CD+L_AC)+F_D*(L_CD+L_AC)+F_C*L_AC\n",
+ "\n",
+ "#Result\n",
+ "print \"The Shear Force and Bending Moment Diagrams are the results\"\n",
+ "\n",
+ "#Plotting the Shear Force Diagram\n",
+ "\n",
+ "X1=[0,0,L_DB,L_DB,L_CD+L_DB,L_CD+L_DB,L_CD+L_DB+L_AC,L_CD+L_DB+L_AC]\n",
+ "Y1=[V_B1,V_B2,V_D1,V_D2,V_C1,V_C2,V_A1,V_A2]\n",
+ "Z1=[0,0,0,0,0,0,0,0]\n",
+ "plt.plot(X1,Y1,X1,Z1)\n",
+ "plt.xlabel(\"Length x in m\")\n",
+ "plt.ylabel(\"Shear Force in kN\")\n",
+ "plt.show()\n",
+ "\n",
+ "#Plotting the Bendimg Moment Diagram\n",
+ "\n",
+ "Y2=[M_B,M_D,M_C,M_A]\n",
+ "X2=[0,L_DB,L_DB+L_CD,L_AC+L_CD+L_DB]\n",
+ "Z2=[0,0,0,0]\n",
+ "plt.plot(X2,Y2,X2,Z2)\n",
+ "plt.xlabel(\"Length in m\")\n",
+ "plt.ylabel(\"Bending Moment in kN.m\")\n",
+ "plt.show()"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Shear Force and Bending Moment Diagrams are the results\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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Pni0aGhr0vo4HrxERkcSmlo+IiMi8WApERCRhKRARkYSlQEREEpYCERFJWApE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\nICIiCUuBiIgk/w9P4ODmR+1IBgAAAABJRU5ErkJggg==\n",
+ "text": [
+ "<matplotlib.figure.Figure at 0x5be93d0>"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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Gt27dTE9mRiwY9ovtQ+zHlSuAv780HCkgQO00tkGRgnHr1i2kp6ejpKQEdXV1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\nx6hRo1BfX48XX3wRgYGBasd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+ "text": [
+ "<matplotlib.figure.Figure at 0x5de54f0>"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.3.3,Page No.102"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "%matplotlib inline\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L_DB=L_CD=1.5 #m #Length of DB & CD\n",
+ "L_AC=3 #m #Length of AC\n",
+ "F_D=80 #KN #Force at Pt D\n",
+ "w=40 #KN/m #u.v.l\n",
+ "L=6 #Length of beam\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let R_A and R_B be the Reactions at Pt A & B respectively\n",
+ "#R_A+R_B=140 \n",
+ "#Taking moment at B we get,M_B\n",
+ "R_A=(1*2**-1*L_AC*w*(1*3**-1*L_AC+(L_CD+L_DB))+F_D*L_DB)*L**-1\n",
+ "R_B=140-R_A\n",
+ "\n",
+ "#Shear Force Calculations\n",
+ "\n",
+ "#S.F at B\n",
+ "V_B1=0 #KN\n",
+ "V_B2=R_B #KN\n",
+ "\n",
+ "#S.F At D\n",
+ "V_D1=V_B2 #KN\n",
+ "V_D2=V_D1-F_D #KN\n",
+ "\n",
+ "#S.F at C\n",
+ "V_C=V_D2 #KN\n",
+ "\n",
+ "#S.F At A\n",
+ "V_A1=V_C-1*2**-1*w*L_AC #KN\n",
+ "V_A2=V_A1+R_A #KN\n",
+ "\n",
+ "#Bending Moment Calculations\n",
+ "\n",
+ "#B.M At B\n",
+ "M_B=0 #KN.m\n",
+ "\n",
+ "#B.M At D\n",
+ "M_D=-R_B*L_DB\n",
+ "\n",
+ "#B.M At C\n",
+ "M_C=F_D*L_CD-R_B*(L_DB+L_CD)\n",
+ "\n",
+ "#B.M At A\n",
+ "M_A=F_D*(L_CD+L_AC)-R_B*L+1*2**-1*w*L_AC*(1*3**-1*L_AC)+R_A\n",
+ "\n",
+ "#Result\n",
+ "print \"The Shear Force and Bending Moment Diagrams are the results\"\n",
+ "\n",
+ "#Plotting the Shear Force Diagram\n",
+ "\n",
+ "X1=[0,0,L_DB,L_DB,L_DB+L_CD,L_DB+L_CD+L_AC,L_DB+L_CD+L_AC]\n",
+ "Y1=[V_B1,V_B2,V_D1,V_D2,V_C,V_A1,V_A2]\n",
+ "Z1=[0,0,0,0,0,0,0]\n",
+ "plt.plot(X1,Y1,X1,Z1)\n",
+ "plt.xlabel(\"Length x in m\")\n",
+ "plt.ylabel(\"Shear Force in kN\")\n",
+ "plt.show()\n",
+ "\n",
+ "#Plotting the Bendimg Moment Diagram\n",
+ "\n",
+ "X2=[0,L_DB,L_CD+L_DB,L_AC+L_CD+L_DB]\n",
+ "Y2=[M_B,M_D,M_C,M_A]\n",
+ "Z2=[0,0,0,0]\n",
+ "plt.plot(X2,Y2,X2,Z2)\n",
+ "plt.xlabel(\"Length in m\")\n",
+ "plt.ylabel(\"Bending Moment in kN.m\")\n",
+ "plt.show()"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Shear Force and Bending Moment Diagrams are the results\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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+ "text": [
+ "<matplotlib.figure.Figure at 0x58f0c10>"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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+ "text": [
+ "<matplotlib.figure.Figure at 0x562f910>"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.3.4,Page No.104"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "%matplotlib inline\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "M_D=120 #KN.m #B.M at Pt D\n",
+ "F_C=40 #KN #Force at Pt C\n",
+ "w1=20 #KN.m\n",
+ "L_DB=1.5 #m #Length of DB\n",
+ "L_CD=1.5 #m #Length of CD\n",
+ "L_AC=3 #m #Length of AC\n",
+ "L=6 #m #Length of Beam\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let R_A And R_B be the Reactions at pt A and B \n",
+ "#R_A+R_B=100\n",
+ "#Now Taking Moment At Pt B We get,M_B\n",
+ "R_A=-(M_D-F_C*(L_CD+L_DB)-w1*L_AC*(L_AC*2**-1+L_CD+L_DB))*L**-1\n",
+ "R_B=100-R_A\n",
+ "\n",
+ "#Shear Force Calculations\n",
+ "\n",
+ "#S.F At Pt B\n",
+ "V_B1=0\n",
+ "V_B2=R_B\n",
+ "\n",
+ "#S.F at Pt D\n",
+ "V_D=V_B2 #KN\n",
+ "\n",
+ "#S.F At Pt C\n",
+ "V_C1=V_D #KN\n",
+ "V_C2=V_C1-F_C\n",
+ "\n",
+ "#S.F At Pt A\n",
+ "V_A1=V_C2-w1*L_AC #KN\n",
+ "V_A2=V_A1+R_A\n",
+ "\n",
+ "#Bending Moment Calculations\n",
+ "\n",
+ "#B.M At Pt B\n",
+ "M_B=0 #KN.m\n",
+ "\n",
+ "#B.M At Pt D\n",
+ "M_D1=M_B-R_B*L_DB #KN.m\n",
+ "M_D2=M_B+M_D-R_B*L_DB\n",
+ "\n",
+ "#B.M At Pt C\n",
+ "M_C=M_D-R_B*(L_CD+L_DB)\n",
+ "\n",
+ "#B.M At Pt A\n",
+ "M_A=M_D-R_B*L+F_C*L_AC+w1*L_AC*L_AC*2**-1\n",
+ "\n",
+ "#Result\n",
+ "print \"The Shear Force and Bending Moment Diagrams are the results\"\n",
+ "\n",
+ "#Plotting the Shear Force Diagram\n",
+ "\n",
+ "X1=[0,0,L_DB,L_DB+L_CD,L_DB+L_CD,L_DB+L_CD+L_AC,L_DB+L_CD+L_AC]\n",
+ "Y1=[V_B1,V_B2,V_D,V_C1,V_C2,V_A1,V_A2]\n",
+ "Z1=[0,0,0,0,0,0,0]\n",
+ "plt.plot(X1,Y1,X1,Z1)\n",
+ "plt.xlabel(\"Length x in m\")\n",
+ "plt.ylabel(\"Shear Force in kN\")\n",
+ "plt.show()\n",
+ "\n",
+ "#Plotting the Bendimg Moment Diagram\n",
+ "\n",
+ "Y2=[M_B,M_D1,M_D2,M_C,M_A]\n",
+ "X2=[0,L_DB,L_DB,L_CD+L_DB,L_AC+L_CD+L_DB]\n",
+ "Z2=[0,0,0,0,0]\n",
+ "plt.plot(X2,Y2,X2,Z2)\n",
+ "plt.xlabel(\"Length in m\")\n",
+ "plt.ylabel(\"Bending Moment in kN.m\")\n",
+ "plt.show()"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Shear Force and Bending Moment Diagrams are the results\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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+ "text": [
+ "<matplotlib.figure.Figure at 0x55dae90>"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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+ "text": [
+ "<matplotlib.figure.Figure at 0x5669170>"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.3.5,Page No.105"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "%matplotlib inline\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "F_C=20 #KN #Force at Pt C\n",
+ "F_D=40 #KN #Force at pt D\n",
+ "w=20 #KN.m #u.d.l \n",
+ "L_AD=L_DB=2 #m #Length of AD & DB\n",
+ "L_BC=1 #m #Length of BC\n",
+ "L=5 #m #Length of Beam\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#LEt R_A and R_B be the reactions at A & B respectively\n",
+ "#R_A+R_B=100 \n",
+ "#Now Taking Moment at B,M_B we get\n",
+ "R_A=-(F_C*L_BC-F_D*L_DB-w*L_AD*(L_AD*2**-1+L_DB))*(L_AD+L_DB)**-1\n",
+ "R_B=100-R_A\n",
+ "\n",
+ "#Shear Force Calculations\n",
+ "\n",
+ "#S.F At pt C\n",
+ "V_C1=0 #KN\n",
+ "V_C2=-F_C #KN\n",
+ "\n",
+ "#S.F At PT B\n",
+ "V_B1=V_C2 #KN\n",
+ "V_B2=V_C2+R_B #KN\n",
+ "\n",
+ "#S.F At Pt D\n",
+ "V_D1=V_B2 #KN\n",
+ "V_D2=V_D1-F_D #KN\n",
+ "\n",
+ "#S.F At Pt A\n",
+ "V_A1=V_D2-w*L_AD #KN\n",
+ "V_A2=V_A1+R_A #KN\n",
+ "\n",
+ "#Bending Moment Calculations\n",
+ "\n",
+ "#B.M At Pt C\n",
+ "M_C=0 \n",
+ "\n",
+ "#B.M At Pt B\n",
+ "M_B=F_C*L_BC\n",
+ "\n",
+ "#B.M At Pt D\n",
+ "M_D=F_C*(L_BC+L_DB)-R_B*L_DB\n",
+ "\n",
+ "#B.M At Pt A\n",
+ "M_A=F_C*L-R_B*(L_DB+L_AD)+F_D*L_AD+w*L_AD*L_AD*2**-1\n",
+ "\n",
+ "#Result\n",
+ "print \"The Shear Force and Bending Moment Diagrams are the results\"\n",
+ "\n",
+ "#Plotting the Shear Force Diagram\n",
+ "\n",
+ "X1=[0,0,L_BC,L_BC,L_BC+L_DB,L_BC+L_DB,L_BC+L_DB+L_AD,L_BC+L_DB+L_AD]\n",
+ "Y1=[V_C1,V_C2,V_B1,V_B2,V_D1,V_D2,V_A1,V_A2]\n",
+ "Z1=[0,0,0,0,0,0,0,0]\n",
+ "plt.plot(X1,Y1,X1,Z1)\n",
+ "plt.xlabel(\"Length x in m\")\n",
+ "plt.ylabel(\"Shear Force in kN\")\n",
+ "plt.show()\n",
+ "\n",
+ "#Plotting the Bendimg Moment Diagram\n",
+ "\n",
+ "Y2=[M_C,M_B,M_D,M_A]\n",
+ "X2=[0,L_BC,L_BC+L_DB,L_BC+L_DB+L_AD]\n",
+ "Z2=[0,0,0,0]\n",
+ "plt.plot(X2,Y2,X2,Z2)\n",
+ "plt.xlabel(\"Lenght in m\")\n",
+ "plt.ylabel(\"Bending Moment in kN.m\")\n",
+ "plt.show()"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Shear Force and Bending Moment Diagrams are the results\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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+ "text": [
+ "<matplotlib.figure.Figure at 0x5d9c6f0>"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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Zb7/9lh07drB8+XLy8/N58cUXmTVrlrt+NSKaPpLAUlFRwZdffsnI\nkSOd91VWVgLGm/WZE1a7du3K4cOHAfj8889JSUkBcPYoONvw4cMB6NGjBx9++KHzfldOkKntmufr\n0KED3bp1A6Bbt24kJiYCEB0dTUlJSZ3XEXGVkoIElNOnT3PFFVdQWFhY48+bNm3q/PrMm/r5dYPz\n3+wvueQSAJo0aUJVVVW9Y6rpmuc7cw0w+iWceU5QUFCDrilSG00fSUC5/PLL6dChAx988AFgvAn/\n61//uuhz+vTpw4oVK3A4HBw+fJi8vLw6r9O8eXPnUcXn0xmUYmVKCuLXTpw4Qbt27Zy3//3f/+Xd\nd99l0aJFxMbGEh0dfU4z97Pn+898PWLECMLCwoiKiuK+++6jR48etGjR4oJr2Ww253OGDh3KypUr\niYuLIz8/v9bH1XbNml67tu8D+UhocT8dnS3igp9//pnLLruM//znPyQkJPDFF1/QqlUrs8MScTvV\nFERcMGTIEMrLy6msrGTatGlKCOK3NFIQEREn1RRERMRJSUFERJyUFERExElJQUREnJQURETESUlB\nRESc/j+iaB8fTwtkoQAAAABJRU5ErkJggg==\n",
+ "text": [
+ "<matplotlib.figure.Figure at 0x5c1ead0>"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.3.6,Page No.107"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "%matplotlib inline\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L_BC=L_EB=L_AD=1 #m #Length of spans BC,ED,AD\n",
+ "L_ED=2 #m #Length of ED\n",
+ "w=60 #KNm #u.d.l\n",
+ "F_C=20 #KN Pt Load at C\n",
+ "L=5 #m #Span of beam \n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let R_A & R_B be the reactions at A & B respectively\n",
+ "#R_A+R_B=80 \n",
+ "#Taking Moment At A,we get M_A\n",
+ "R_B=(F_C*L+1*2**-1*L_ED*w*(2*3**-1*L_ED+L_AD))*(L_AD+L_ED+L_EB)**-1\n",
+ "R_A=80-R_B\n",
+ "\n",
+ "#Shear Force Calculations\n",
+ "\n",
+ "#S.F At C\n",
+ "V_C1=0 #KN\n",
+ "V_C2=-F_C #KN\n",
+ "\n",
+ "#S.F At B\n",
+ "V_B1=V_C2 #KN\n",
+ "V_B2=V_C2+R_B #KN \n",
+ "\n",
+ "#S.F aT E\n",
+ "V_E=V_B2 #KN\n",
+ "\n",
+ "#S.F AT D\n",
+ "V_D=V_B2-1*2**-1*L_ED*w #KN\n",
+ "\n",
+ "#S.F At A\n",
+ "V_A1=V_D #KN \n",
+ "V_A2=V_D+R_A\n",
+ "\n",
+ "#Bending Moment Calculations\n",
+ "\n",
+ "#B.M at C\n",
+ "M_C=0 #KN.m\n",
+ "\n",
+ "#B.M at B\n",
+ "M_B=F_C*L_BC #KN.m\n",
+ "\n",
+ "#B.M at E\n",
+ "M_E=F_C*(L_EB+L_BC)-R_B*L_EB #KN.m\n",
+ "\n",
+ "#B.M at D\n",
+ "M_D=F_C*(L_ED+L_EB+L_BC)-R_B*(L_ED+L_EB)+1*2**-1*L_ED*w*1*3**-1*L_ED #KN.m\n",
+ "\n",
+ "#B.M at A\n",
+ "M_A=1*2**-1*L_ED*w*(2*3**-1*L_ED+L_AD)-R_B*(L_AD+L_ED+L_EB)+F_C*L\n",
+ "\n",
+ "#Result\n",
+ "print \"The Shear Force and Bending Moment Diagrams are the results\"\n",
+ "\n",
+ "#Plotting the Shear Force Diagram\n",
+ "\n",
+ "X1=[0,0,L_BC,L_BC,L_EB+L_BC,L_ED+L_EB+L_BC,L_AD+L_ED+L_EB+L_BC,L_ED+L_EB+L_BC+L_AD]\n",
+ "Y1=[V_C1,V_C2,V_B1,V_B2,V_E,V_D,V_A1,V_A2]\n",
+ "Z1=[0,0,0,0,0,0,0,0]\n",
+ "plt.plot(X1,Y1,X1,Z1)\n",
+ "plt.xlabel(\"Length x in m\")\n",
+ "plt.ylabel(\"Shear Force in kN\")\n",
+ "plt.show()\n",
+ "\n",
+ "#Plotting the Bendimg Moment Diagram\n",
+ "\n",
+ "X2=[0,L_BC,L_BC+L_EB,L_EB+L_BC+L_ED,L_EB+L_BC+L_ED+L_AD]\n",
+ "Y2=[M_C,M_B,M_E,M_D,M_A]\n",
+ "Z2=[0,0,0,0,0]\n",
+ "plt.plot(X2,Y2,X2,Z2)\n",
+ "plt.xlabel(\"Lenght in m\")\n",
+ "plt.ylabel(\"Bending Moment in kN.m\")\n",
+ "plt.show()"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Shear Force and Bending Moment Diagrams are the results\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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Cr9fzLAHAgQMHUFpaig8++AAbNmzAvn37Ot1OUUUhJCSkwx1Vq6qqEBoaKjAR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\nAABJRU5ErkJggg==\n",
+ "text": [
+ "<matplotlib.figure.Figure at 0x58a6f70>"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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UlD0DZoHZoDk7OxvPPfec/SqwAoNm98HQmcjAloBZINmcwq5du5rcTwEAHnnkEeuqkgCb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\nkeDZfAAAAABJRU5ErkJggg==\n",
+ "text": [
+ "<matplotlib.figure.Figure at 0x567bcf0>"
+ ]
+ }
+ ],
+ "prompt_number": 28
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.3.7,Page No.109"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "%matplotlib inline\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L_BC=1 #m #Length of BC\n",
+ "L_DB=2 #m #Length of DB\n",
+ "L_AD=4 #m #Length 0f AD\n",
+ "M_D=30 #KN.m #Moment at D\n",
+ "w=45 #KN/m #u.d.l\n",
+ "L=7 #m #Span of beam\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let R_B & R_A be the Reactions at B & A respectively\n",
+ "#R_B+R_A=180+P ............(1)\n",
+ "\n",
+ "#Now Taking Moment about A,we get\n",
+ "#R_B=7*P+390 ...............(2)\n",
+ "\n",
+ "#Since R_A & R_B Are Equal\n",
+ "#2*R_B=180+P ...................(3)\n",
+ "\n",
+ "#From equation 1 and 3 we get\n",
+ "#3*(180+P)=7P+390\n",
+ "#After simplifying Further above equation we get\n",
+ "P=150*4**-1 #KN\n",
+ "R_A=R_B=(180+P)*2**-1\n",
+ "F_C=P\n",
+ "\n",
+ "#Shear Force Calculations\n",
+ "\n",
+ "#S.F At C\n",
+ "V_C1=0 #KN\n",
+ "V_C2=-P #KN\n",
+ "\n",
+ "#S.F At B\n",
+ "V_B1=V_C2 #KN\n",
+ "V_B2=V_C2+R_B #KN \n",
+ "\n",
+ "#S.F At D\n",
+ "V_D=V_B2 #KN\n",
+ "\n",
+ "#S.F At A\n",
+ "V_A1=V_D-w*L_AD #KN\n",
+ "V_A2=V_A1+R_A #KN\n",
+ "\n",
+ "#Bending Moment Calculations\n",
+ "\n",
+ "#B.M at C\n",
+ "M_C=0 #KN.m \n",
+ "\n",
+ "#B.M at B\n",
+ "M_B=F_C*L_BC #KN.m\n",
+ "\n",
+ "#B.M at D\n",
+ "M_D1=F_C*(L_BC+L_DB)-R_B*L_DB #KN.m\n",
+ "M_D2=M_D1+M_D\n",
+ "\n",
+ "#B.M At A\n",
+ "M_A=w*L_AD*L_AD*2**-1+M_D-R_B*(L_AD+L_DB)+P*L\n",
+ "\n",
+ "#Result\n",
+ "print \"The Shear Force and Bending Moment Diagrams are the results\"\n",
+ "\n",
+ "#Plotting the Shear Force Diagram\n",
+ "\n",
+ "X1=[0,0,L_BC,L_BC,L_DB+L_BC,L_DB+L_BC+L_AD,L_DB+L_BC+L_AD]\n",
+ "Y1=[V_C1,V_C2,V_B1,V_B2,V_D,V_A1,V_A2]\n",
+ "Z1=[0,0,0,0,0,0,0]\n",
+ "plt.plot(X1,Y1,X1,Z1)\n",
+ "plt.xlabel(\"Length x in m\")\n",
+ "plt.ylabel(\"Shear Force in kN\")\n",
+ "plt.show()\n",
+ "\n",
+ "#Plotting the Bendimg Moment Diagram\n",
+ "\n",
+ "X2=[0,L_BC,L_DB+L_BC,L_DB+L_BC,L_AD+L_DB+L_BC]\n",
+ "Y2=[M_C,M_B,M_D1,M_D2,M_A]\n",
+ "Z2=[0,0,0,0,0]\n",
+ "plt.plot(X2,Y2,X2,Z2)\n",
+ "plt.xlabel(\"Lenght in m\")\n",
+ "plt.ylabel(\"Bending Moment in kN.m\")\n",
+ "plt.show()"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Shear Force and Bending Moment Diagrams are the results\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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+ "text": [
+ "<matplotlib.figure.Figure at 0x5d6f150>"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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+ "text": [
+ "<matplotlib.figure.Figure at 0x5d79470>"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.3.8,Page No.110"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "%matplotlib inline\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=6 #m #Span Of beam\n",
+ "w=30 #KN/m #u.d.l\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Due to Symmetry\n",
+ "#Let R_B and R_C be the reactions at B & C Respectively\n",
+ "R_B=R_C=w*L*2**-1 #KN\n",
+ "\n",
+ "#Let a be the overhang.The Max -ve moment occurs at the support and max +ve moment at middle of the beam\n",
+ "#Now Equating these two equations we get\n",
+ "#30*a*a*2**-1=90*(3-a)-w*L*2**-1*L*4**-1\n",
+ "#After simplifying we get an equation as\n",
+ "#a**2+6*a-9=0\n",
+ "x=1\n",
+ "y=6\n",
+ "z=-9\n",
+ "\n",
+ "p=y**2-4*x*z\n",
+ "\n",
+ "a1=(-y+p**0.5)*2**-1\n",
+ "a2=(-y-p**0.5)*2**-1\n",
+ "\n",
+ "#Now Length cannot be negative,so taking a1 into Consideration\n",
+ "\n",
+ "L_CD=L_AB=a1\n",
+ "L_BC=L-2*a1\n",
+ "\n",
+ "#Shear Force Calculations\n",
+ "\n",
+ "#S.F At D\n",
+ "V_D=0\n",
+ "\n",
+ "#S.F At C\n",
+ "V_C1=V_D-w*L_CD #KN\n",
+ "V_C2=V_C1+R_C #KN\n",
+ "\n",
+ "#S.F At B\n",
+ "V_B1=-w*(L_BC+L_CD)+R_C\n",
+ "V_B2=V_B1+R_B\n",
+ "\n",
+ "#S.F At A\n",
+ "V_A=round(V_B2,2)-round(w*L_AB,2)\n",
+ "\n",
+ "#Bending Moment Calculations\n",
+ "\n",
+ "#B.M At D\n",
+ "M_D=0\n",
+ "\n",
+ "#B.M At C\n",
+ "M_C=w*L_CD*L_CD*2**-1 #KN.m\n",
+ "\n",
+ "#B.M At B\n",
+ "M_B=w*(L_BC+L_CD)*(L_BC+L_CD)*2**-1-R_C*L_BC*L_BC*2**-1\n",
+ "\n",
+ "#B.M At A\n",
+ "X=w*L*L*2**-1\n",
+ "Y=-R_C*(L_AB+L_BC)-R_B*L_AB\n",
+ "M_A=X+Y\n",
+ "\n",
+ "#Result\n",
+ "print \"The Shear Force and Bending Moment Diagrams are the results\"\n",
+ "\n",
+ "#Plotting the Shear Force Diagram\n",
+ "\n",
+ "X1=[0,L_CD,L_CD,L_CD+L_BC,L_CD+L_BC,L_CD+L_BC+L_AB]\n",
+ "Y1=[V_D,V_C1,V_C2,V_B1,V_B2,V_A]\n",
+ "Z1=[0,0,0,0,0,0]\n",
+ "plt.plot(X1,Y1,X1,Z1)\n",
+ "plt.xlabel(\"Length x in m\")\n",
+ "plt.ylabel(\"Shear Force in kN\")\n",
+ "plt.show()\n",
+ "\n",
+ "#Plotting the Bendimg Moment Diagram\n",
+ "\n",
+ "X2=[0,L_CD,L_BC+L_CD,L_AB+L_BC+L_CD]\n",
+ "Y2=[M_D,M_C,M_B,M_A]\n",
+ "Z2=[0,0,0,0]\n",
+ "plt.plot(X2,Y2,X2,Z2)\n",
+ "plt.xlabel(\"Lenght in m\")\n",
+ "plt.ylabel(\"Bending Moment in kN.m\")\n",
+ "plt.show()"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Shear Force and Bending Moment Diagrams are the results\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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lXLhwAQBw6dIlHDhwwKleBejt7Q0/Pz+UlpYCAA4ePIiwsLAub6vKm9f6yt3d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\nAElFTkSuQmCC\n",
+ "text": [
+ "<matplotlib.figure.Figure at 0x5945590>"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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LUVxcrNtAqS48PDxqdH5WVhby8vIQEBAAAJg4cSJ27NjBhEGkEDc3ID5e7BWe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\nMQcOHJASExMlb29vtUPRi6y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+ "text": [
+ "<matplotlib.figure.Figure at 0x58e1750>"
+ ]
+ }
+ ],
+ "prompt_number": 25
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.3.9,Page No.112"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "%matplotlib inline\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "F_F=6 #KN #Force at F\n",
+ "w1=w2=w=3 #KN.m #u.d.l\n",
+ "M_D=24 #KN.m \n",
+ "L_AB=L_CD=L_DE=L_EF=4 #m #Length of AB,CD,DE,EF\n",
+ "L_BC=2 #m #Length of BC\n",
+ "L=18 #m #Span of Beam\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#LEt R_B and R_E be the Reactions at B & E respectively\n",
+ "#R_B+R_E=42\n",
+ "\n",
+ "#Taking Moment At Pt B,M_B\n",
+ "R_E=(F_F*(L_BC+L_CD+L_DE+L_EF)+w*(L_CD+L_DE)*((L_CD+L_DE)*2**-1+L_BC)-w*L_AB*L_AB*2**-1-M_D)*(L_BC+L_CD+L_DE)**-1\n",
+ "R_B=42-R_E #KN\n",
+ "\n",
+ "#Shear Force Calculations\n",
+ "\n",
+ "#S.F aT F\n",
+ "V_F1=0 #KN \n",
+ "V_F2=-F_F #KN\n",
+ "\n",
+ "#S.F at E\n",
+ "V_E1=V_F2 #KN\n",
+ "V_E2=V_E1+R_E #KN\n",
+ "\n",
+ "#S.F aT C\n",
+ "V_C=V_E2-w*(L_CD+L_DE) #KN\n",
+ "\n",
+ "#S.F at B\n",
+ "V_B1=V_C #KN \n",
+ "V_B2=V_C+R_B #KN\n",
+ "\n",
+ "#S.F At A\n",
+ "V_A=V_B2-w*L_AB #KN\n",
+ "\n",
+ "#Bending Moment Calculations\n",
+ "\n",
+ "#B.M At F\n",
+ "M_F=0\n",
+ "\n",
+ "#B.M At E\n",
+ "M_E=F_F*L_EF #KN.m\n",
+ "\n",
+ "#B.M At D\n",
+ "M_D1=F_F*(L_DE+L_EF)-R_E*L_DE+w*L_DE*L_DE*2**-1 #KN.m\n",
+ "M_D2=M_D1-M_D\n",
+ "\n",
+ "#B.M At C\n",
+ "M_C=F_F*(L_CD+L_DE+L_EF)-R_E*(L_CD+L_DE)+w*(L_CD+L_DE)*(L_CD+L_DE)*2**-1-M_D\n",
+ "\n",
+ "#B.M At B\n",
+ "M_B=F_F*(L_BC+L_CD+L_DE+L_EF)-R_E*(L_BC+L_CD+L_DE)-M_D+w*(L_CD+L_DE)*((L_CD+L_DE)*2**-1+L_BC)\n",
+ "\n",
+ "#B.M At A\n",
+ "M_A=w*L_AB*L_AB*2**-1-R_B*L_AB+w*(L_CD+L_DE)*((L_CD+L_DE)*2**-1+L_BC+L_AB)-R_E*(L_AB+L_BC+L_CD+L_DE)+F_F*L-M_D\n",
+ "\n",
+ "#Result\n",
+ "print \"The Shear Force and Bending Moment Diagrams are the results\"\n",
+ "\n",
+ "#Plotting the Shear Force Diagram\n",
+ "\n",
+ "X1=[0,0,L_EF,L_EF,L_EF+L_DE+L_CD,L_EF+L_DE+L_CD+L_BC,L_EF+L_DE+L_CD+L_BC,L_EF+L_DE+L_CD+L_BC+L_AB]\n",
+ "Y1=[V_F1,V_F2,V_E1,V_E2,V_C,V_B1,V_B2,V_A]\n",
+ "Z1=[0,0,0,0,0,0,0,0]\n",
+ "plt.plot(X1,Y1,X1,Z1)\n",
+ "plt.xlabel(\"Length x in m\")\n",
+ "plt.ylabel(\"Shear Force in kN\")\n",
+ "plt.show()\n",
+ "\n",
+ "#Plotting the Bendimg Moment Diagram\n",
+ "\n",
+ "X2=[0,L_EF,L_DE+L_EF,L_DE+L_EF,L_CD+L_DE+L_EF,L_CD+L_DE+L_EF+L_BC,L_CD+L_DE+L_EF+L_BC+L_AB]\n",
+ "Y2=[M_F,M_E,M_D1,M_D2,M_C,M_B,M_A]\n",
+ "Z2=[0,0,0,0,0,0,0]\n",
+ "plt.plot(X2,Y2,X2,Z2)\n",
+ "plt.xlabel(\"Lenght in m\")\n",
+ "plt.ylabel(\"Bending Moment in kN.m\")\n",
+ "plt.show()"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Shear Force and Bending Moment Diagrams are the results\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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+ "text": [
+ "<matplotlib.figure.Figure at 0x5d75ab0>"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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VxMfHIzs7G/369cOvv/6KcePG4bHHHnNknEROixVL5KzMziBqa2uRmJiIMWPG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\nfyRzHAAAAABJRU5ErkJggg==\n",
+ "text": [
+ "<matplotlib.figure.Figure at 0x5c44130>"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.3.10,Page No.114"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "%matplotlib inline\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L_DC=L_BA=2 #m #Length of BA & DC\n",
+ "L_CB=1 #m #Length of CB\n",
+ "F_A=10 #KN #Force at pt A\n",
+ "F_B=20 #KN #Force at pt B\n",
+ "w=4 #KN.m #u.d.l\n",
+ "L=5 #m #Length of beam\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let R_D be the reactions at Pt D\n",
+ "R_D=F_B+F_A+w*L_DC #KN\n",
+ "\n",
+ "#Shear Force Calculations\n",
+ "\n",
+ "#S.F at Pt A\n",
+ "V_A1=0 #KN\n",
+ "V_A2=F_A #KN\n",
+ "\n",
+ "#S.F At Pt B\n",
+ "V_B1=V_A2\n",
+ "V_B2=F_B+F_A\n",
+ "\n",
+ "#S.F at Pt C\n",
+ "V_C=F_B+F_A #KN \n",
+ "\n",
+ "#S.F At Pt D\n",
+ "V_D1=V_B2+w*L_DC\n",
+ "V_D2=F_B+F_A+w*L_DC-R_D\n",
+ "\n",
+ "#B.M At Pt A\n",
+ "M_A=0\n",
+ "\n",
+ "#B.M At Pt B\n",
+ "M_B=F_A*L_BA\n",
+ "\n",
+ "#B.M at Pt C\n",
+ "M_C=F_B*L_CB+F_A*(L_BA+L_CB) #KN\n",
+ "\n",
+ "#B.M At Pt D\n",
+ "M_D1=F_A*L+F_B*(L_CB+L_DC)+w*L_DC*L_DC*2**-1\n",
+ "M_D2=(F_A*L+F_B*(L_CB+L_DC)+w*L_DC*L_DC*2**-1)-M_D1\n",
+ "\n",
+ "#Result\n",
+ "print \"The Shear Force and Bending Moment Diagrams are the results\"\n",
+ "\n",
+ "#Plotting the Shear Force Diagram\n",
+ "\n",
+ "X1=[0,0,L_BA,L_BA,L_BA+L_CB,L_BA+L_CB+L_DC,L_BA+L_CB+L_DC]\n",
+ "Y1=[V_A1,V_A2,V_B1,V_B2,V_C,V_D1,V_D2]\n",
+ "Z1=[0,0,0,0,0,0,0]\n",
+ "plt.plot(X1,Y1,X1,Z1)\n",
+ "plt.xlabel(\"Length x in m\")\n",
+ "plt.ylabel(\"Shear Force in kN\")\n",
+ "plt.show()\n",
+ "\n",
+ "#Plotting the Bendimg Moment Diagram\n",
+ "\n",
+ "Y2=[M_A,M_B,M_C,M_D1,M_D2]\n",
+ "X2=[0,L_BA,L_CB+L_BA,L_CB+L_BA+L_DC,L_CB+L_BA+L_DC]\n",
+ "Z2=[0,0,0,0,0]\n",
+ "plt.plot(X2,Y2,X2,Z2)\n",
+ "plt.xlabel(\"Lenght in m\")\n",
+ "plt.ylabel(\"Bending Moment in kN.m\")\n",
+ "plt.show()"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Shear Force and Bending Moment Diagrams are the results\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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e3542bZr69++vhIQEzZ49WyNGjNAFF1wgSbrxxht15ZVXen+5e/LxJ0/9axiG\n8vLy9Oc//1kxMTFatWqV8vLyvMNNvo71tX3yMb6+tvMUxDh7XM4JWzp8+LAiIiL0ww8/aPTo0dq8\nebP69OljdSwgKBjjhy1NnjxZNTU1amho0L333kvpw1Y44wcAm2GMHwBshuIHAJuh+AHAZih+ALAZ\nih8AbIbiBwCb+T+gqVKkQGpm/QAAAABJRU5ErkJggg==\n",
+ "text": [
+ "<matplotlib.figure.Figure at 0x5738d10>"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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+ "text": [
+ "<matplotlib.figure.Figure at 0x5788db0>"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.3.11,Page No.115"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "%matplotlib inline\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "w=20 #KN/m #u.v.l\n",
+ "F_C=40 #KN #Force at Pt C\n",
+ "M_D=40 #KN.m #Moment at pt D\n",
+ "L_AB=3 #m #Length of AB\n",
+ "L_BC=1 #m #Length of BC\n",
+ "L_CD=L_DE=2 #m #Length of CD & DE\n",
+ "L=8 #8 #Length of beam\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let R_A & R_E be the Reactions at A & E respectively\n",
+ "#R_A+R_E=70\n",
+ "\n",
+ "#Taking Moments At Pt A we get,M_A\n",
+ "R_E=(F_C*(L_AB+L_BC)+1*2**-1*L_AB*w*2+40)*L**-1\n",
+ "R_A=70-R_E\n",
+ "\n",
+ "#shear Force Calculations\n",
+ "\n",
+ "#S.F At Pt E\n",
+ "V_E1=0\n",
+ "V_E2=R_E #KN\n",
+ "\n",
+ "#S.F aT pt D\n",
+ "V_D=V_E2\n",
+ "\n",
+ "#S.F At PT C\n",
+ "V_C1=V_D\n",
+ "V_C2=V_D-F_C #KN\n",
+ "\n",
+ "#S.F At Pt A\n",
+ "V_A1=V_C2-(1*2**-1*w*L_AB)\n",
+ "V_A2=V_A1+R_A\n",
+ "\n",
+ "#Bending Moment Calculations\n",
+ "\n",
+ "#B.M At Pt E\n",
+ "M_E=0\n",
+ "\n",
+ "#B.M At Pt D\n",
+ "M_D1=M_E-R_E*L_DE\n",
+ "M_D2=M_D1+M_D\n",
+ "\n",
+ "#B.M At Pt C\n",
+ "M_C=-R_E*(L_DE+L_CD)+M_D\n",
+ "\n",
+ "#B.M At Pt B\n",
+ "M_B=-R_E*(L_DE+L_CD+L_BC)+M_D+F_C*L_BC\n",
+ "\n",
+ "#B.M At Pt A\n",
+ "M_A=-R_E*L+M_D+(1*2**-1*L_AB*w*2)+F_C*(L_BC+L_AB)\n",
+ "\n",
+ "#Result\n",
+ "print \"The Shear Force and Bending Moment Diagrams are the results\"\n",
+ "\n",
+ "#Plotting the Shear Force Diagram\n",
+ "\n",
+ "X1=[0,0,L_DE,L_CD+L_DE,L_CD+L_DE,L_CD+L_DE+L_AB,L_CD+L_DE+L_AB]\n",
+ "Y1=[V_E1,V_E2,V_D,V_C1,V_C2,V_A1,V_A2]\n",
+ "Z1=[0,0,0,0,0,0,0]\n",
+ "plt.plot(X1,Y1,X1,Z1)\n",
+ "plt.xlabel(\"Length x in m\")\n",
+ "plt.ylabel(\"Shear Force in kN\")\n",
+ "plt.show()\n",
+ "\n",
+ "#Plotting the Bendimg Moment Diagram\n",
+ "\n",
+ "Y2=[M_E,M_D1,M_D2,M_C,M_B,M_A]\n",
+ "X2=[0,L_DE,L_DE,L_CD+L_DE,L_DE+L_CD+L_BC,L_AB+L_BC+L_CD+L_DE]\n",
+ "Z2=[0,0,0,0,0,0]\n",
+ "plt.plot(X2,Y2)\n",
+ "plt.xlabel(\"Lenght in m\")\n",
+ "plt.ylabel(\"Bending Moment in kN.m\")\n",
+ "plt.show()"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Shear Force and Bending Moment Diagrams are the results\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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+ "text": [
+ "<matplotlib.figure.Figure at 0x5df5230>"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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PZcuW8dRTT5GZmcnOnTtp1KgRI0aMsPs+cvTnxbp1U4fwLFyoO4nzTCaYMkWt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\nd/r0aduBO+VVUFCA2Wxm+/b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6ePR5C6Jyk56CqHT27dtHt27daNu2LXfddRdpaWkAPPbY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\nGhPHAAAAAElFTkSuQmCC\n",
+ "text": [
+ "<matplotlib.figure.Figure at 0x5dff130>"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.3.12,Page No.116"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "%matplotlib inline\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "F_G=10 #KN #Force at Pt G\n",
+ "F_B=F_E=15 #KN #Force at Pt B & E\n",
+ "w=20 #KN/m #U.d.L\n",
+ "L_FG=L_EF=L_DE=L_CD=L_BC=L_AB=1 #m #Lengths of FG,EF,DE,CD,BC,AB respectively\n",
+ "L=6 #m #Length of beam\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#LEt R_F & R_A be the Reactions at E & A respectively\n",
+ "#R_F+R_A=60\n",
+ "\n",
+ "#Taking Moment At Pt A,M_A\n",
+ "R_F=(F_G*L+F_E*(L_AB+L_BC+L_CD+L_DE)+w*L_CD*(L_AB+L_BC+L_CD*2**-1)+F_B*L_AB)*(L_AB+L_BC+L_CD+L_DE+L_EF)**-1\n",
+ "R_A=60-R_F\n",
+ "\n",
+ "#Shear Force Calculations\n",
+ "\n",
+ "#S.F At G\n",
+ "V_G1=0 #KN \n",
+ "V_G2=F_G #KN\n",
+ "\n",
+ "#S.F At F\n",
+ "V_F1=V_G2 #KN\n",
+ "V_F2=V_F1-R_F\n",
+ "\n",
+ "#S.F At E\n",
+ "V_E1=V_F2 #KN\n",
+ "V_E2=V_F2+F_E\n",
+ "\n",
+ "#S.F At D\n",
+ "V_D=V_E2\n",
+ "\n",
+ "#S.F At C\n",
+ "V_C=V_E2+w*L_CD\n",
+ "\n",
+ "#S.F At B\n",
+ "V_B1=V_C\n",
+ "V_B2=V_B1+F_B\n",
+ "\n",
+ "#S.F At A\n",
+ "V_A1=V_B2\n",
+ "V_A2=V_B2-R_A\n",
+ "\n",
+ "#Bending Moment Calculations\n",
+ "\n",
+ "#B.M At Pt G\n",
+ "M_G=0\n",
+ "\n",
+ "#B.M At F\n",
+ "M_F=F_G*L_FG \n",
+ "\n",
+ "#B.M At E\n",
+ "M_E=F_G*(L_FG+L_EF)-R_F*L_EF\n",
+ "\n",
+ "#B.M At D\n",
+ "M_D=F_G*(L_FG+L_EF+L_DE)-R_F*(L_EF+L_DE)+F_E*L_DE\n",
+ "\n",
+ "#B.M At C\n",
+ "M_C=F_G*(L_FG+L_EF+L_DE+L_CD)-R_F*(L_EF+L_DE+L_CD)+F_E*(L_DE+L_CD)+w*L_CD*L_CD*2**-1\n",
+ "\n",
+ "#B.M At B\n",
+ "M_B=F_G*(L_FG+L_EF+L_DE+L_CD+L_BC)-R_F*(L_EF+L_DE+L_CD+L_BC)+F_E*(L_DE+L_CD+L_BC)+w*L_CD*(L_CD*2**-1+L_BC)\n",
+ "\n",
+ "#B.M At A\n",
+ "M_A=F_G*L-R_F*(L_EF+L_DE+L_CD+L_BC+L_AB)+F_E*(L_DE+L_CD+L_BC+L_AB)+F_B*L_AB+w*L_CD*(L_CD*2**-1+L_BC+L_AB)\n",
+ "\n",
+ "#Result\n",
+ "print \"The Shear Force and Bending Moment Diagrams are the results\"\n",
+ "\n",
+ "#Plotting the Shear Force Diagram\n",
+ "\n",
+ "X1=[0,0,L_FG,L_FG,L_FG+L_EF,L_FG+L_EF,L_FG+L_EF+L_DE,L_FG+L_EF+L_DE+L_CD,L_FG+L_EF+L_DE+L_CD+L_BC,L_FG+L_EF+L_DE+L_CD+L_BC,L_FG+L_EF+L_DE+L_CD+L_BC+L_AB,L_FG+L_EF+L_DE+L_CD+L_BC+L_AB]\n",
+ "Y1=[V_G1,V_G2,V_F1,V_F2,V_E1,V_E2,V_D,V_C,V_B1,V_B2,V_A1,V_A2]\n",
+ "Z1=[0,0,0,0,0,0,0,0,0,0,0,0]\n",
+ "plt.plot(X1,Y1,X1,Z1)\n",
+ "plt.xlabel(\"Length x in m\")\n",
+ "plt.ylabel(\"Shear Force in kN\")\n",
+ "plt.show()\n",
+ "\n",
+ "#Plotting the Bendimg Moment Diagram\n",
+ "\n",
+ "X2=[0,L_FG,L_EF+L_FG,L_EF+L_FG+L_DE,L_EF+L_FG+L_DE+L_CD,L_EF+L_FG+L_DE+L_CD+L_BC,L_EF+L_FG+L_DE+L_CD+L_BC+L_AB]\n",
+ "Y2=[M_G,M_F,M_E,M_D,M_C,M_B,M_A]\n",
+ "Z2=[0,0,0,0,0,0,0]\n",
+ "plt.plot(X2,Y2)\n",
+ "plt.xlabel(\"Lenght in m\")\n",
+ "plt.ylabel(\"Bending Moment in kN.m\")\n",
+ "plt.show()"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Shear Force and Bending Moment Diagrams are the results\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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+ "text": [
+ "<matplotlib.figure.Figure at 0x57502f0>"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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+ "text": [
+ "<matplotlib.figure.Figure at 0x58b0ed0>"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.3.13,Page No.117"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "%matplotlib inline\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "L_AB=L_BC=L_CD=L_DE=L_EF=1 #m #LEngth of AB,BC,CD,DE,EF respectively\n",
+ "M_A=50 #KN/m #Moment at A\n",
+ "w=5 #KN/m #u.v.l\n",
+ "F_D=10 #KN\n",
+ "w2=5 #KN/m #u.d.l\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let R_B & R_E be the Reactions at B and E respectively\n",
+ "#R_B+R_E=20\n",
+ "\n",
+ "#Taking Moment At Pt B,M_B\n",
+ "R_E=(w2*L_EF*(L_EF*2**-1+L_DE+L_CD+L_BC)+w*L_BC*2**-1*2*3**-1+50+F_D*(L_BC+L_CD))*3**-1\n",
+ "R_B=17.5-R_E #KN\n",
+ "\n",
+ "#Shear Force Calculations\n",
+ "\n",
+ "#S.F At F\n",
+ "V_F=0\n",
+ "\n",
+ "#S.F aT E\n",
+ "V_E1=-w2*L_EF #KN\n",
+ "V_E2=V_E1+R_E\n",
+ "\n",
+ "#S.F at D\n",
+ "V_D1=R_E-w2*L_EF #KN\n",
+ "V_D2=V_D1-F_D #KN\n",
+ "\n",
+ "#S.F At C\n",
+ "V_C=V_D2\n",
+ "\n",
+ "#S.F aT B\n",
+ "V_B1=-L_BC*w*2**-1-F_D+R_E-w2*L_EF\n",
+ "V_B2=V_B1+R_B\n",
+ "\n",
+ "#Bending Moment Calculations\n",
+ "\n",
+ "#B.M at F\n",
+ "M_F=0 #KN.m\n",
+ "\n",
+ "#B.M At E\n",
+ "M_E=w2*L_EF*L_EF*2**-1 #KN.m\n",
+ "\n",
+ "#B.M at D\n",
+ "M_D=-R_E*L_DE+w2*L_EF*(L_EF*2**-1+L_DE) #KN.m\n",
+ "\n",
+ "#B.M At C\n",
+ "M_C=F_D*L_CD*R_E*(L_CD+L_DE)+w2*L_EF*(L_EF*2**-1+L_DE+L_CD) #KN.m\n",
+ "\n",
+ "#B.M At B\n",
+ "M_B=F_D*(L_CD+L_BC)-R_E*(L_BC+L_CD+L_DE)+w2*L_EF*(L_EF*2**-1+L_BC+L_CD+L_DE)+1*2**-1*L_BC*w*2*3**-1\n",
+ "\n",
+ "#B.M At A\n",
+ "M_A1=w*L_EF*(L_EF*2**-1+L_AB+L_BC+L_CD+L_DE)-R_E*(L_AB+L_BC+L_CD+L_DE)+F_D*(L_AB+L_BC+L_CD)+1*2**-1*L_BC*w*(2*3**-1*L_BC+L_AB)-R_B*L_AB\n",
+ "M_A2=M_A1+M_A\n",
+ "\n",
+ "#Result\n",
+ "print \"The Shear Force and Bending Moment Diagrams are the results\"\n",
+ "\n",
+ "#Plotting the Shear Force Diagram\n",
+ "\n",
+ "X1=[0,L_EF,L_EF,L_DE+L_EF,L_DE+L_EF,L_CD+L_DE+L_EF,L_CD+L_DE+L_EF+L_BC,L_CD+L_DE+L_EF+L_BC]\n",
+ "Y1=[V_F,V_E1,V_E2,V_D1,V_D2,V_C,V_B1,V_B2]\n",
+ "Z1=[0,0,0,0,0,0,0,0]\n",
+ "plt.plot(X1,Y1,X1,Z1)\n",
+ "plt.xlabel(\"Length x in m\")\n",
+ "plt.ylabel(\"Shear Force in kN\")\n",
+ "plt.show()\n",
+ "\n",
+ "#Plotting the Bendimg Moment Diagram\n",
+ "\n",
+ "X2=[0,L_EF,L_DE+L_EF,L_CD+L_DE+L_EF,L_CD+L_DE+L_EF+L_BC,L_CD+L_DE+L_EF+L_BC+L_AB,L_CD+L_DE+L_EF+L_BC+L_AB]\n",
+ "Y2=[M_F,M_E,M_D,M_C,M_B,M_A1,M_A2]\n",
+ "Z2=[0,0,0,0,0,0,0]\n",
+ "plt.plot(X2,Y2,X2,Z2)\n",
+ "plt.xlabel(\"Length in m\")\n",
+ "plt.ylabel(\"Bending Moment in kN.m\")\n",
+ "plt.show()"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Shear Force and Bending Moment Diagrams are the results\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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+ "text": [
+ "<matplotlib.figure.Figure at 0x4d39390>"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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LlixREhMTLY8fMWKEsm/fvjuex4ZfSQi7euUVRXnxRa1TuIdr1xRFr1eUb77R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\ntnx2yhiGEEIIm0jBEEIIYRM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+ "text": [
+ "<matplotlib.figure.Figure at 0x5578410>"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.4_7.ipynb b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.4_7.ipynb new file mode 100644 index 00000000..5a167d34 --- /dev/null +++ b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.4_7.ipynb @@ -0,0 +1,1661 @@ +{
+ "metadata": {
+ "name": "chapter no.4.ipynb"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 4:Stresses in Beams"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.1,Page no.130"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=5000 #mm #Length of Beam\n",
+ "a=2000 #mm #Length of start of beam to Pt Load\n",
+ "b=3000 #mm #Length of Pt load to end of beam\n",
+ "A=150*250 #m**2 #Area of beam \n",
+ "b=150 #mm #Width of beam\n",
+ "d=250 #mm #Depth of beam\n",
+ "sigma=10#N/mm**2 #stress\n",
+ "l=2000 #m #Load applied from one end\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Moment of Inertia\n",
+ "I=1*12**-1*b*d**3 #m**4\n",
+ "\n",
+ "#Distance from N.A to end\n",
+ "y_max=d*2**-1 #m\n",
+ "\n",
+ "#Section Modulus\n",
+ "Z=1*6**-1*b*d**2 #mm**3\n",
+ "\n",
+ "#Moment Carrying Capacity\n",
+ "M=sigma*Z #N-mm\n",
+ "\n",
+ "#Let w be the Intensity of the Load in N/m,then Max moment\n",
+ "#M_max=w*L**2*8**-1 #N-mm\n",
+ "#After substituting values and further simplifying we get\n",
+ "#M_max=w*25*100*8**-1\n",
+ "\n",
+ "#EQuating it to moment carrying capacity,we get max intensity load\n",
+ "w=M*(25*1000)**-1*8*10**-3\n",
+ "\n",
+ "#Part-2\n",
+ "\n",
+ "#Let P be the concentrated load,then max moment occurs under the load and its value\n",
+ "#M1=P*a*b*L**-1 #N-mm\n",
+ "\n",
+ "#Equting it to moment carrying capacity we get\n",
+ "P=M*1200**-1*10**-3 #N\n",
+ "\n",
+ "#Result\n",
+ "print\"Max Intensity of u.d.l it can carry\",round(w,3),\"KN-m\"\n",
+ "print\"MAx concentrated Load P apllied at 2 m from one end is\",round(P,3),\"KN\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Max Intensity of u.d.l it can carry 5.0 N-mm\n",
+ "MAx concentrated Load P apllied at 2 m from one end is 13.021 KN\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.2,Page no.131"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "D=70 #mm #External Diameter\n",
+ "t=8 #mm #Thickness of pipe\n",
+ "L=2500 #mm #span \n",
+ "sigma=150 #N/mm**2 #stress\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Internal Diameter \n",
+ "d=D-2*t #mm\n",
+ "\n",
+ "#M.I Of Pipe\n",
+ "I=pi*64**-1*(D**4-d**4) #mm**4\n",
+ "\n",
+ "y_max=D*2**-1 #mm\n",
+ "Z=I*(y_max)**-1 #mm**3\n",
+ "\n",
+ "#Moment Carrying capacity\n",
+ "M=sigma*Z #N*mm\n",
+ "\n",
+ "#Max moment int the beam occurs at the mid-span and is equal to\n",
+ "#m=P*L*4**-1\n",
+ "\n",
+ "#Equating Max moment to moment carrying capacity we get,\n",
+ "#M=P*2.5*L*4**-1\n",
+ "#After substituting and simplifying we get\n",
+ "P=4*M*(L)**-1*10**-3 #N\n",
+ "\n",
+ "#Result\n",
+ "print\"Max concentrated load that can be applied at the centre of span is\",round(P,3),\"KN\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Max concentrated load that can be applied at the centre of span is 5.22 KN\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.3,Page no.132"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#Flanges Dimension\n",
+ "b1=180 #mm #Width\n",
+ "d1=10 #mm #Thickness\n",
+ "\n",
+ "D=500 #mm #Overall depth\n",
+ "t=8 #mm #Thickness of web\n",
+ "\n",
+ "#Plate Dimensions\n",
+ "b2=240 #mm #Width\n",
+ "t2=12 #mm #Thickness\n",
+ "\n",
+ "sigma=150 #N/mm**2 #Stress\n",
+ "L=3000 #mm #span\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Distance of centroid from bottom fibre\n",
+ "y_bar=(b2*t2*(D+t2*2**-1)+b1*d1*(D-t1*2**-1)+(D-2*t1)*t*D*2**-1+(b1*t1*t1*2**-1))*(b2*t2+b1*d1+b1*d1+(D-2*d1)*t)**-1\n",
+ "\n",
+ "#M.I of section\n",
+ "I=(1*12**-1*b2*t2**3+b2*t2*(D+t2*2**-1-y_bar)**2+1*12**-1*b1*d1**3+b1*d1*(D-t1*2**-1-y_bar)**2+1*12**-1*b1*t1**3+b1*t1*(t1*2**-1-y_bar)**2+1*12**-1*t*(D-2*t1)**3+t*(D-2*t1)*(D*2**-1-y_bar)**2)\n",
+ "\n",
+ "#Section Modulus\n",
+ "Z=I*(y_bar)**-1 #mm**3\n",
+ "\n",
+ "#Moment or Resistance\n",
+ "M=sigma*Z\n",
+ "\n",
+ "#Let Load on Cantilever be w/m Length \n",
+ "#Max M.I produced\n",
+ "#M_max=w*L**2**-1 \n",
+ "\n",
+ "#Now Equating Moment of resistance to Max moment,we get Max load\n",
+ "#4.5*w=M\n",
+ "#After rearranging and further simplifying we get\n",
+ "w=M*4.5**-1*10**3*10**-9\n",
+ "\n",
+ "#Result\n",
+ "print\"Moment of Resistance is\",round(M,2),\"KN-mm\"\n",
+ "print\"Load the section can carry is\",round(w,3),\"KN/m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Moment of Resistance is 198770121.83 KN-mm\n",
+ "Load the section can carry is 44.171 KN/m\n"
+ ]
+ }
+ ],
+ "prompt_number": 26
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.4,Page no.134"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#Flange (Top)\n",
+ "b1=80 #mm #Width \n",
+ "t1=40 #mm #Thickness\n",
+ "\n",
+ "#Flange (Bottom)\n",
+ "b2=160 #mm #width\n",
+ "t2=40 #mm #Thickness\n",
+ "\n",
+ "#web\n",
+ "d=120 #mm #Depth\n",
+ "t3=20 #mm #Thickness\n",
+ "\n",
+ "D=200 #mm #Overall Depth\n",
+ "sigma1=30 #N/mm**2 #Tensile stress\n",
+ "sigma2=90 #N/mm**2 #Compressive stress\n",
+ "L=6000 #mm #Span\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Distance of centroid from bottom fibre\n",
+ "y_bar=(b1*t1*(D-t1*2**-1)+d*t3*(d*2**-1+t2)+b2*t2*t2*2**-1)*(b1*t1+d*t3+b2*t2)**-1 #mm\n",
+ "\n",
+ "#Moment of Inertia\n",
+ "I=1*12**-1*b1*t1**3+b1*t1*(D-t1*2**-1-round(y_bar,2))**2+1*12**-1*t3*d**3+t3*d*(d*2**-1+t2-round(y_bar,2))**2+1*12**-1*b2*t2**3+b2*t2*(t2*2**-1-round(y_bar,2))**2\n",
+ "\n",
+ "#Extreme fibre distance of top and bottom fibres are y_t and y_c respectively\n",
+ "\n",
+ "y_t=y_bar #mm\n",
+ "y_c=D-y_bar #mm\n",
+ "\n",
+ "#Moment carrying capacity considering Tensile strength \n",
+ "M1=sigma1*I*y_t**-1*10**-6 #KN-m\n",
+ "\n",
+ "#Moment carrying capacity considering compressive strength \n",
+ "M2=sigma2*I*y_c**-1*10**-6 #KN-m\n",
+ "\n",
+ "#Max Bending moment in simply supported beam 6 m due to u.d.l\n",
+ "#M_max=w*L*10**-3*8**-1\n",
+ "#After simplifying further we get\n",
+ "#M_max=4.5*w\n",
+ "\n",
+ "#Now Equating it to Moment carrying capacity, we get load carrying capacity\n",
+ "w=M1*4.5**-1 #KN/m\n",
+ "\n",
+ "#Result\n",
+ "print\"Max Uniformly Distributed Load is\",round(w,3),\"KN/m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Max Uniformly Distributed Load is 5.096 KN/m\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.5,Page no.136"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#Flanges\n",
+ "b=200 #mm #Width\n",
+ "t=25 #mm #Thickness \n",
+ "\n",
+ "D1=500 #mm #Overall Depth\n",
+ "t2=20 #mm #Thickness of web\n",
+ "\n",
+ "d=450 #mm #Depth of web\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Consider,Element of Thickness \"y\" at Distance \"dy\" from N.A \n",
+ "#Let Bending stress \"sigma_max\"\n",
+ "\n",
+ "#Stress on the element \n",
+ "#sigma=y*(D*2**-1)*sigma_max ..............(1)\n",
+ "\n",
+ "#Area of Element\n",
+ "#A=b*dy .................................(2)\n",
+ "\n",
+ "#Force on Element \n",
+ "#F=y*250**-1*sigma_max*b*dy\n",
+ "\n",
+ "#Let M be the Moment of resistance\n",
+ "#M=y*250**-1*sigma_max*b*dy*y\n",
+ "\n",
+ "#Moment of Resistance of top flange be M1\n",
+ "def integrand(y, b, D):\n",
+ " return b*y**2*D**-1\n",
+ "b=200 \n",
+ "D=250\n",
+ "\n",
+ "X = quad(integrand, 225, 250, args=(b,D))\n",
+ "\n",
+ "Y=2*X[0]\n",
+ "\n",
+ "#M1=Y*sigma\n",
+ "\n",
+ "#Now Moment of Inertia I section is\n",
+ "X=b*D1**3\n",
+ "Y=(b-t2)*d**3\n",
+ "I=(X-Y)*12**-1*10**-8\n",
+ "\n",
+ "#Moment acting on the entire section\n",
+ "#since sigmais the value at y=250\n",
+ "y_max=250\n",
+ "Z=I*10**8*y_max**-1\n",
+ "#M=sigma*Z \n",
+ "#After Simplifying Further we get\n",
+ "#M2=Z*sigma\n",
+ "\n",
+ "#Percentage Moment resisted by Flanges\n",
+ "P1=2258333.3*(2865833.3)**-1*100\n",
+ "\n",
+ "#Percentage Moment resisted by web\n",
+ "P2=100-P1\n",
+ "\n",
+ "#Result\n",
+ "print\"Percentage Moment resisted by Flanges\",round(P1,2),\"%\"\n",
+ "print\"Percentage Moment resisted by web\",round(P2,2),\"%\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Percentage Moment resisted by Flanges 78.8 %\n",
+ "Percentage Moment resisted by web 21.2 %\n"
+ ]
+ }
+ ],
+ "prompt_number": 38
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.6,Page no.137"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#Flanges\n",
+ "b1=200 #mm #Width\n",
+ "t1=10 #mm #Thickness\n",
+ "\n",
+ "#Web\n",
+ "d=380 #mm #Depth \n",
+ "t2=8 #mm #Thickness\n",
+ "\n",
+ "D=400 #mm #Overall Depth\n",
+ "sigma=150 #N/mm**2\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Area\n",
+ "A=b1*t1+d*t2+b1*t1 #mm**2\n",
+ "\n",
+ "#Moment of Inertia\n",
+ "I=1*12**-1*(b1*D**3-(b1-t2)*d**3)\n",
+ "\n",
+ "#Bending Moment\n",
+ "M=sigma*I*(D*2**-1)**-1\n",
+ "\n",
+ "#Square Section\n",
+ "\n",
+ "#Let 'a' be the side\n",
+ "a=A**0.5\n",
+ "\n",
+ "#Moment of Resistance of this section\n",
+ "M1=1*6**-1*a*a**2*sigma\n",
+ "\n",
+ "X=M*M1**-1\n",
+ "\n",
+ "#Rectangular section\n",
+ "#Let 'a' be the side and depth be 2*a\n",
+ "\n",
+ "a=(A*2**-1)**0.5\n",
+ "\n",
+ "#Moment of Rectangular secction\n",
+ "M2=1*6**-1*a*(2*a)**2*sigma\n",
+ "\n",
+ "X2=M*M2**-1\n",
+ "\n",
+ "#Circular section\n",
+ "#A=pi*d1**2*4**-1\n",
+ "\n",
+ "d1=(A*4*pi**-1)**0.5\n",
+ "\n",
+ "#Moment of circular section\n",
+ "M3=pi*32**-1*d1**3*sigma\n",
+ "\n",
+ "X3=M*M3**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"Moment of resistance of beam section\",round(M,2),\"mm\"\n",
+ "print\"Moment of resistance of square section\",round(X,2),\"mm\"\n",
+ "print\"Moment of resistance of rectangular section\",round(X2,2),\"mm\"\n",
+ "print\"Moment of resistance of circular section\",round(X3,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Moment of resistance of beam section 141536000.0 mm\n",
+ "Moment of resistance of square section 9.58 mm\n",
+ "Moment of resistance of rectangular section 6.78 mm\n",
+ "Moment of resistance of circular section 11.33 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 25
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.7,Page no.139"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "F=12 #KN #Force at End of beam\n",
+ "L=2 #m #span\n",
+ "\n",
+ "#Square section \n",
+ "b=d=200 #mm #Width and depth of beam\n",
+ "\n",
+ "#Rectangular section\n",
+ "b1=150 #mm #Width\n",
+ "d1=300 #mm #Depth\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Max bending Moment\n",
+ "M=F*L*10**6 #N-mm\n",
+ "\n",
+ "#M=sigma*b*d**2\n",
+ "sigma=M*6*(b*d**2)**-1 #N/mm**2\n",
+ "\n",
+ "#Let W be the central concentrated Load in simply supported beam of span L1=3 m\n",
+ "#MAx Moment\n",
+ "#M1=W*L1*4**-1\n",
+ "#After Further simplifying we get\n",
+ "#M1=0.75*10**6 #N-mm\n",
+ "\n",
+ "#The section has a moment of resistance\n",
+ "M1=sigma*1*6**-1*b1*d1**2\n",
+ "\n",
+ "#Equating it to moment of resistance we get max load W\n",
+ "#0.75*10**6*W=M1\n",
+ "#After Further simplifying we get\n",
+ "W=M1*(0.75*10**6)**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"Minimum Concentrated Load required to brek the beam\",round(W,2),\"KN\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Minimum Concentrated Load required to brek the beam 54.0 KN\n"
+ ]
+ }
+ ],
+ "prompt_number": 30
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.8,Page no.140"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=3 #m #span\n",
+ "sigma_t=35 #N/mm**2 #Permissible stress in tension\n",
+ "sigma_c=90 #N/mm**2 #Permissible stress in compression\n",
+ "\n",
+ "#Flanges\n",
+ "t=30 #mm #Thickness\n",
+ "d=250 #mm #Depth\n",
+ "\n",
+ "#Web\n",
+ "t2=25 #mm #Thickness\n",
+ "b=600 #mm #Width\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let y_bar be the Distance of N.A from Extreme Fibres\n",
+ "y_bar=(t*d*d*2**-1*2+(b-2*t)*t2*t2*2**-1)*(t*d*2+(b-2*t)*t2)**-1\n",
+ "\n",
+ "#Moment of Inertia\n",
+ "I=(1*12**-1*t*d**3+t*d*(d*2**-1-y_bar)**2)*2+1*12**-1*(b-2*t)*t2**3+(b-2*t)*t2*(t2*2**-1-y_bar)**2\n",
+ "\n",
+ "#Part-1\n",
+ "\n",
+ "#If web is in Tension\n",
+ "y_t=y_bar #mm\n",
+ "y_c=d-y_bar #mm\n",
+ "\n",
+ "#Moment carrying caryying capacity From consideration of tensile stress\n",
+ "M=sigma_t*I*(y_bar)**-1 #N-mm\n",
+ "\n",
+ "#Moment carrying caryying capacity From consideration of compressive stress\n",
+ "M1=sigma_c*I*(y_c)**-1 #N-mm\n",
+ "\n",
+ "#If w KN/m is u.d.l in beam,Max bending moment\n",
+ "#M=wl**2*8**-1\n",
+ "#After further simplifyng we get\n",
+ "#M=1.125*w*10**6 N-mm\n",
+ "w=M*(1.125*10**6)**-1 #KN\n",
+ "\n",
+ "#Part-2\n",
+ "\n",
+ "#If web is in compression\n",
+ "y_t2=178.299 #mm\n",
+ "y_c2=71.71 #mm \n",
+ "\n",
+ "#Moment carrying caryying capacity From consideration of tensile stress\n",
+ "M2=sigma_t*I*(y_t2)**-1 #N-mm\n",
+ "\n",
+ "#Moment carrying caryying capacity From consideration of compressive stress\n",
+ "M3=sigma_c*I*(y_c2)**-1 #N-mm\n",
+ "\n",
+ "#Moment of resistance is M2\n",
+ "\n",
+ "#Equating it to bending moment we get\n",
+ "#M2=1.125*10**6*w2\n",
+ "#After further simplifyng we get\n",
+ "w2=M2*(1.125*10**6)**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"Uniformly Distributed Load carrying capacity if:web is in Tension\",round(w,2),\"KN\"\n",
+ "print\" :web is in compression\",round(w2,3),\"KN\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Uniformly Distributed Load carrying capacity if:web is in Tension 73.21 KN\n",
+ " :web is in compression 29.446 KN\n"
+ ]
+ }
+ ],
+ "prompt_number": 26
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.9,Page no.141"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "b1=200 #mm #Width at base\n",
+ "b2=100 #mm #Width at top\n",
+ "\n",
+ "L=8 #m Length\n",
+ "P=500 #N #Load\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Consider a section at y metres from top\n",
+ "\n",
+ "#At this section diameter d is\n",
+ "#d=b2+y*L**-1*(b1-b2)\n",
+ "#After Further simplifying we get\n",
+ "#d=b2+12.5*y #mm\n",
+ "\n",
+ "#Moment of Inertia\n",
+ "#I=pi*64**-1*d**4\n",
+ "\n",
+ "#Section Modulus \n",
+ "#Z=pi*32**-1*(b1+12.5*y)**3\n",
+ "\n",
+ "#Moment \n",
+ "#M=5*10**5*y #N-mm\n",
+ "\n",
+ "#Let sigma be the fibre stress at this section then\n",
+ "#M=sigma*Z\n",
+ "#After sub values in above equation and further simplifying we get\n",
+ "#sigma=5*10**5*32*pi**-1*y*((b2+12.5*y)**3)**-1\n",
+ "\n",
+ "#For sigma to be Max,d(sigma)*(dy)**-1=0\n",
+ "#16*10**6*pi**-1*((b2+12.5*y)**-3+y*(-3)*(b2+12.5*y)**-4*12.5)\n",
+ "#After Further simplifying we get\n",
+ "#b2+12.5*y=37.5*y\n",
+ "#After Further simplifying we get\n",
+ "y=b2*25**-1 #m\n",
+ "\n",
+ "#Stress at this section\n",
+ "sigma=5*10**5*32*pi**-1*y*((b2+12.5*y)**3)**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"Stress at Extreme Fibre is max\",round(y,2),\"m\"\n",
+ "print\"Max stress is\",round(sigma,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Stress at Extreme Fibre is max 4.0 m\n",
+ "Max stress is 6.04 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 28
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.10,Page no.143"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "H=10 #mm #Height\n",
+ "A1=160*160 #mm**2 #area of square section at bottom\n",
+ "L1=160 #mm #Length of square section at bottom\n",
+ "b1=160 #mm #width of square section at bottom\n",
+ "A2=80*80 #mm**2 #area of square section at top\n",
+ "L2=80 #mm #Length of square section at top\n",
+ "b2=80 #mm #Width of square section at top\n",
+ "P=100 #N #Pull\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Consider a section at distance y from top.\n",
+ "#Let the side of square bar be 'a'\n",
+ "#a=L2+y*(H)**-1*(b1-b2)\n",
+ "#After further simplifying we get\n",
+ "#a=L2+8*y\n",
+ "\n",
+ "#Moment of Inertia\n",
+ "#I=2*1*12**-1*a*(2)**0.5*(a*((2)**0.5)**-1)**3\n",
+ "#After further simplifying we get\n",
+ "#I=a**4*12**-1\n",
+ "\n",
+ "#Section Modulus \n",
+ "#Z=a**4*(12*a*(2)**0.5)**-1\n",
+ "#After further simplifying we get\n",
+ "#Z=2**0.5*a**3*(12)**-1 #mm**3\n",
+ "\n",
+ "#Bending moment at this section=100*y N-mm\n",
+ "#M=100*10**3*y #N-mm\n",
+ "\n",
+ "#But\n",
+ "#M=sigma*Z\n",
+ "#After sub values in above equation we get\n",
+ "#sigma=M*Z**-1\n",
+ "#After further simplifying we get\n",
+ "#sigma=1200*10**3*(2**0.5)**-1*y*((80+80*y)**3)**-1 .......(1)\n",
+ "\n",
+ "#For Max stress df*(dy)**-1=0\n",
+ "#After taking Derivative of above equation we get\n",
+ "#df*(dy)**-1=1200*10**3*(2**0.5)**-1*((80+8*y)**-3+y(-3)*(80+8*y)**-4*8)\n",
+ "#After further simplifying we get\n",
+ "y=80*16**-1 #m\n",
+ "\n",
+ "#Max stress at this level is\n",
+ "sigma=1200*10**3*(2**0.5)**-1*y*((80+8*y)**3)**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"Max Bending stress is Developed at\",round(y,3),\"m\"\n",
+ "print\"Value of Max Bending stress is\",round(sigma,3),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Max Bending stress is Developed at 5.0 m\n",
+ "Value of Max Bending stress is 2.455 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.12,Page no.147"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "b=200 #mm #Width of timber \n",
+ "d=400 #mm #Depth of timber\n",
+ "t=6 #mm #Thickness\n",
+ "b2=200 #mm #width of steel plate\n",
+ "t2=20 #mm #Thickness of steel plate\n",
+ "M=40*10**6 #KN-mm #Moment\n",
+ "#Let E_s*E_t**-1=X\n",
+ "X=20 #Ratio of Modulus of steel to timber\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#let y_bar be the Distance of centroidfrom bottom most fibre\n",
+ "y_bar=(b*d*(b+t)+t2*b2*t*t*2**-1)*(b*d+t2*b2*t)**-1 #mm\n",
+ "\n",
+ "#Moment of Inertia\n",
+ "I=1*12**-1*b*d**3+b*d*(b+t-round(y_bar,3))**2+1*12**-1*t2*b2*t**3+b2*t2*t*(round(y_bar,3)-t*2**-1)**2\n",
+ "\n",
+ "#distance of the top fibre from N-A\n",
+ "y_1=d+t-y_bar #mm\n",
+ "\n",
+ "#Distance of the junction of timber and steel From N-A\n",
+ "y_2=y_bar-t #mm\n",
+ "\n",
+ "#Stress in Timber at the top\n",
+ "Y=M*I**-1*y_1 #N/mm**2\n",
+ "\n",
+ "#Stress in the Timber at the junction point\n",
+ "Z=M*I**-1*y_2\n",
+ "\n",
+ "#Coressponding stress in steel at the junction point\n",
+ "Z2=X*Z #N/mm**2 \n",
+ "\n",
+ "#The stress in Extreme steel fibre \n",
+ "Z3=X*M*I**-1*y_bar\n",
+ "\n",
+ "#Result\n",
+ "print\"Stress in Extreme steel Fibre\",round(Z3,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Stress in Extreme steel Fibre 69.67 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 23
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.13,Page no.149"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#Timber size\n",
+ "b=150 #mm #Width\n",
+ "d=300 #mm #Depth\n",
+ "\n",
+ "t=6 #mm #Thickness of steel plate\n",
+ "l=6 #m #Span\n",
+ "\n",
+ "#E_s*E_t**-1=20 \n",
+ "#m=E_s*E_t**-1\n",
+ "m=20 \n",
+ "sigma_timber=8 #N/mm**2 #Stress in timber\n",
+ "sigma_steel=150 #N/mm**2 #Stress in steel plate\n",
+ "\n",
+ "#Let m*t=Y\n",
+ "Y=m*t #mm\n",
+ "L=(2*t+b)*m #mm #Width of flitched beam\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Due to synnetry cenroid,the neutral axis is half the depth\n",
+ "I=(1*12**-1*L*t**3+L*t*(b+t*2**-1)**2)*2+1*12**-1*(Y+b+Y)*d**3 #mm**4\n",
+ "\n",
+ "y_max1=150 #mm #For timber\n",
+ "y_max2=156 #mm #For steel\n",
+ "\n",
+ "#stress in steel\n",
+ "f_t1=1*m**-1*sigma_steel #N/mm**2\n",
+ "\n",
+ "#Moment of resistance\n",
+ "M=f_t1*(I*y_max2**-1)\n",
+ "\n",
+ "#load\n",
+ "w=8*M*(l**2)**-1*10**-6 #KN/m\n",
+ "\n",
+ "#Result\n",
+ "print\"Load beam can carry is\",round(w,2),\"KN/m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Load beam can carry is 19.1 KN/m\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.14,Page no.151"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=6000 #mm #Span of beam\n",
+ "W=20*10**3 #N #Load\n",
+ "sigma=8 #N/mm**2 #Stress\n",
+ "b=200 #mm #Width of section\n",
+ "d=300 #mm #Depth of section\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#let x be the distance from left side of beam\n",
+ "\n",
+ "#Bending moment\n",
+ "#M=W*2**-1*x #Nmm .......(1)\n",
+ "\n",
+ "#But M=sigma*Z ..........(2)\n",
+ "\n",
+ "#Equating equation 1 and 2 we get\n",
+ "#W*2**-1*x=sigma*Z ............(3)\n",
+ "\n",
+ "#Section Modulus \n",
+ "#Z=1*6*b*d**2 ...............(4)\n",
+ "\n",
+ "#Equating equation 3 and 4 we get\n",
+ "#b*d**2=3*W*x*sigma**-1 .............(5)\n",
+ "\n",
+ "#Beam of uniform strength of constant depth\n",
+ "#b=3*W*x*(sigma*d**2) \n",
+ "\n",
+ "#When x=0\n",
+ "b=0\n",
+ "\n",
+ "#When x=L*2**-1\n",
+ "b2=3*W*L*(2*sigma*d**2)**-1 #mm\n",
+ "\n",
+ "#Beam with constant width of 200 mm\n",
+ "\n",
+ "#We have\n",
+ "#d=(3*W*x*(sigma*d)**-1)**0.5\n",
+ "#thus depth varies as (x)**0.5\n",
+ "\n",
+ "#when x=0\n",
+ "d1=0\n",
+ "\n",
+ "#when x=L*2**-1\n",
+ "d2=(3*W*L*(2*sigma*200)**-1)**0.5 #mm\n",
+ "\n",
+ "#Result\n",
+ "print\"Cross section of rectangular beam is:\",round(b2,2),\"mm\"\n",
+ "print\" :\",round(d2,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Cross section of rectangular beam is: 250.0 mm\n",
+ " : 335.41 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.15,Page no.154"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=800 #mm #Span\n",
+ "n=5 #number of leaves\n",
+ "b=60 #mm #Width\n",
+ "t=10 #mm #thickness\n",
+ "sigma=250 #N/mm**2 #Stress\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#section Modulus\n",
+ "Z=n*6**-1*b*t**2 #mm**3\n",
+ "\n",
+ "#from the relation\n",
+ "#sigma*Z=M ...................(1)\n",
+ "#M=P*L*4**-1\n",
+ "#sub values of M in equation 1 we get\n",
+ "P=sigma*Z*4*L**-1*10**-3 #KN #Load\n",
+ "\n",
+ "#Length of Leaves\n",
+ "L1=0.2*L #mm\n",
+ "L2=0.4*L #mm\n",
+ "L3=0.6*L #mm\n",
+ "L4=0.8*L #mm\n",
+ "L5=L #mm\n",
+ "\n",
+ "#Result\n",
+ "print\"Max Load it can take is\",round(P,2),\"KN\"\n",
+ "print\"Length of leaves:L1\",round(L1,2),\"mm\"\n",
+ "print\" :L2\",round(L2,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Max Load it can take is 6.25 KN\n",
+ "Length of leaves:L1 160.0 mm\n",
+ " :L2 320.0 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.16,Page no.161"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "%matplotlib inline\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "F=20*10**3 #N #Shear Force\n",
+ "\n",
+ "#Tee section\n",
+ "\n",
+ "#Flange\n",
+ "b=100 #mm #Width\n",
+ "t=12 #mm #Thickness\n",
+ "\n",
+ "#Web\n",
+ "d=88 #mm #Depth\n",
+ "t2=12 #mm #Thicknes\n",
+ "\n",
+ "D=100 #mm #Overall Depth\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Distance of C.G from Top Fibre\n",
+ "y=(b*t*t*2**-1+t2*d*(d*2**-1+t))*(b*t+d*t2)**-1 #mm \n",
+ "\n",
+ "#Moment Of Inertia\n",
+ "I=1*12**-1*b*t**3+b*t*(y-t*2**-1)**2+1*12**-1*t2*d**3+t2*d*(t+d*2**-1-y)**2 #mm**4\n",
+ "\n",
+ "#shear stress at bottom Flange\n",
+ "\n",
+ "#Area above this level\n",
+ "A=b*t #mm**2\n",
+ "\n",
+ "#C.G of this area from N-A\n",
+ "y2=y-t*2**-1\n",
+ "\n",
+ "#Stress at bottom of flange\n",
+ "sigma=F*A*y2*(b*I)**-1 #N/mm**2 \n",
+ "\n",
+ "#sigma2 at same level but in web where width is 12 mm\n",
+ "sigma2=F*A*y2*(t2*I)**-1 #N/mm**2 \n",
+ "\n",
+ "#To find shear stress at N-A\n",
+ "X=t*b*(y-t*2**-1)+t2*(y-t2)*(y-t2)*2**-1 #mm**3\n",
+ "\n",
+ "sigma3=F*X*(t2*I)**-1 #N/mm**2\n",
+ "\n",
+ "#Shear stress at top and bottom fibre is zero\n",
+ "#sigma4 and sigma5 are top and bottom fibre shear stress\n",
+ "sigma4=sigma5=0\n",
+ "\n",
+ "#Result\n",
+ "print \"The Shear Force and Bending Moment Diagrams are the results\"\n",
+ "\n",
+ "#Plotting the Shear Force Diagram\n",
+ "\n",
+ "X1=[0,t,t,y,D]\n",
+ "Y1=[sigma4,sigma,sigma2,sigma3,sigma5]\n",
+ "Z1=[0,0,0,0,0]\n",
+ "plt.plot(X1,Y1,X1,Z1)\n",
+ "plt.xlabel(\"Length x in m\")\n",
+ "plt.ylabel(\"Shear Force in kN\")\n",
+ "plt.show()\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Shear Force and Bending Moment Diagrams are the results\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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+ "text": [
+ "<matplotlib.figure.Figure at 0x56a6270>"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.17,Page no.163"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "%matplotlib inline\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "F=40*10**3 #N #shear Force\n",
+ "\n",
+ "#I-section\n",
+ "\n",
+ "#Flanges\n",
+ "b=80 #mm #Width of flange\n",
+ "t=20 #mm #Thickness\n",
+ "\n",
+ "#Web\n",
+ "d=200 #mm #Depth\n",
+ "t2=20 #mm #Thickness\n",
+ "\n",
+ "#Flange-2\n",
+ "b2=160 #mm #Width\n",
+ "t3=20 #mm #Thickness\n",
+ "\n",
+ "D=240 #mm #Overall Depth\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Distance of N-A from Top Fibre \n",
+ "y=(b*t*t*2**-1+d*t2*(t+d*2**-1)+b2*t3*(t+d+t3*2**-1))*(b*t+d*t2+b2*t3)**-1 #mm\n",
+ "\n",
+ "#Moment of Inertia\n",
+ "I=1*12**-1*b*t**3+b*t*(y-(t*2**-1))**2+1*12**-1*t2*d**3+t2*d*(y-(t+d*2**-1))**2+1*12**-1*b2*t3**3+t3*b2*((d+t+t3*2**-1)-y)**2 #mm**4\n",
+ "\n",
+ "#Shear stress bottom of flange\n",
+ "sigma=F*b*t*(y-t*2**-1)*(b*I)**-1 #N/mm**2\n",
+ "\n",
+ "#At same Level but in web\n",
+ "sigma2=F*b*t*(y-t*2**-1)*(t2*I)**-1 #N/mm**2\n",
+ "\n",
+ "#for shear stress at N.A\n",
+ "X=b*t*(y-t*2**-1)+t2*(y-t)*(y-t)*2**-1 #mm**3\n",
+ "sigma3=F*X*(t2*I)**-1 #N/mm**2\n",
+ "\n",
+ "#Shear stress at bottom of web\n",
+ "\n",
+ "X=b2*t3*((D-y)-t3*2**-1) #mm**3\n",
+ "\n",
+ "#Stress at bottom of web\n",
+ "sigma4=F*X*(t2*I)**-1 #N/mm**2\n",
+ "\n",
+ "#Stress at Lower flange\n",
+ "sigma5=F*X*(b2*I)**-1 #N/mm**2\n",
+ "\n",
+ "#Result\n",
+ "print \"The Shear Force Diagram is the result\"\n",
+ "\n",
+ "#Plotting the Shear Force Diagram\n",
+ "\n",
+ "X1=[0,20,20,140,220,220,240]\n",
+ "Y1=[0,sigma,sigma2,sigma3,sigma4,sigma5,0]\n",
+ "Z1=[0,0,0,0,0,0,0]\n",
+ "plt.plot(X1,Y1,X1,Z1)\n",
+ "plt.xlabel(\"Length in mm\")\n",
+ "plt.ylabel(\"Shear Force in N\")\n",
+ "plt.show()\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Shear Force Diagram is the result\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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OnYL/+i/TF3DLLWZPgXbt7K5KRHyRwsAGJ0+afoBXXjHNQOvXQ1iY3VWJiC8r1gj183MN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\n/w/qZz1xEBCKMwAAAABJRU5ErkJggg==\n",
+ "text": [
+ "<matplotlib.figure.Figure at 0x5614110>"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.18,Page no.164"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "F=30*10**3 #N #Shear Force\n",
+ "\n",
+ "#Channel Section\n",
+ "d=400 #mm #Depth of web \n",
+ "t=10 #mm #THickness of web\n",
+ "t2=15 #mm #Thickness of flange\n",
+ "b=100 #mm #Width of flange\n",
+ "\n",
+ "#Rectangular Welded section\n",
+ "b2=80 #mm #Width\n",
+ "d2=60 #mm #Depth\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Distance of Centroid From Top Fibre\n",
+ "y=(d*t*t*2**-1+2*t2*(b-t)*((b-t)*2**-1+10)+d2*b2*(d2*2**-1+t))*(d*t+2*t2*(b-t)+d2*b2)**-1 #mm\n",
+ "\n",
+ "#Moment Of Inertia of the section about N-A\n",
+ "I=1*12**-1*d*t**3+d*t*(y-t*2**-1)**2+2*(1*12**-1*t2*(b-t)**3+t2*(b-t)*(((b-t)*2**-1+t)-y)**2)+1*12**-1*d2**3*b2+d2*b2*(d2*2**-1+t-y)**2\n",
+ "\n",
+ "#Shear stress at level of weld\n",
+ "sigma=F*d*t*(y-t*2**-1)*((b2+t2+t2)*I)**-1 #N/mm**2\n",
+ "\n",
+ "#Max Shear Stress occurs at Neutral Axis\n",
+ "X=d*t*(y-t*2**-1)+2*t2*(y-t)*(y-t)*2**-1+b2*(y-t)*(y-t)*2**-1\n",
+ "\n",
+ "sigma_max=F*X*((b+t)*I)**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"Shear stress in the weld is\",round(sigma,2),\"N/mm**2\"\n",
+ "print\"Max shear stress is\",round(sigma_max,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Shear stress in the weld is 3.62 N/mm**2\n",
+ "Max shear stress is 4.48 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.19,Page no.165"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#Wooden Section\n",
+ "b=300 #mm #Width\n",
+ "d=300 #mm #Depth\n",
+ "\n",
+ "D=100 #mm #Diameter of Bore\n",
+ "F=10*10**3 #N #Shear Force\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Moment Of Inertia Of Section\n",
+ "I=1*12**-1*b*d**3-pi*64**-1*D**4\n",
+ "\n",
+ "#Shear stress at crown of circle\n",
+ "sigma=F*b*D*(d*2**-1-D*2**-1)*(b*I)**-1\n",
+ "\n",
+ "#Let a*y_bar=X\n",
+ "X=b*d*2**-1*d*4**-1-pi*8**-1*D**2*4*D*2**-1*(3*pi)**-1 #mm**3\n",
+ "\n",
+ "#Shear Stress at Neutral Axis\n",
+ "sigma2=F*X*((b-D)*I)**-1 #N/mm**2\n",
+ "\n",
+ "#Result\n",
+ "print\"Shearing Stress at Crown of Bore\",round(sigma,3),\"N/mm**2\"\n",
+ "print\"Shear Stress at Neutral Axis\",round(sigma2,3),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Shearing Stress at Crown of Bore 0.149 N/mm**2\n",
+ "Shear Stress at Neutral Axis 0.246 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.20,Page no.166"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#flanges\n",
+ "b=200 #mm #width\n",
+ "t1=25 #mm #Thickness\n",
+ "\n",
+ "#web\n",
+ "d=450 #mm #Depth \n",
+ "t2=20 #mm #thickness\n",
+ "\n",
+ "D=500 #mm #Total Depth of section\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Moment Of Inertia of the section about N-A\n",
+ "I=1*12**-1*b*D**3-1*12**-1*(b-t2)*d**3 #mm**4\n",
+ "\n",
+ "#Consider an element in the web at distance y from y from N-A\n",
+ "#Depth of web section=225-y\n",
+ "\n",
+ "#C.G From N-A\n",
+ "#y2=y+(((D*2**-1-t)-y)*2**-1)\n",
+ "\n",
+ "#ay_bar for section at y\n",
+ "#Let ay_bar be X\n",
+ "#X=X1 be of Flange + X2 be of web above y\n",
+ "#X=b*t1*(D*2**-1-t1*2**-1)+t2*(d-t1)*(d-t1+y)*2**-1\n",
+ "#After Sub values and Further simplifying we get\n",
+ "#X=1187500+10*(225**2-y**2)\n",
+ "\n",
+ "#Shear stress at y\n",
+ "#sigma_y=F*(X)*(t2*I)**-1\n",
+ "\n",
+ "#Shear Force resisted by the Element\n",
+ "#F1=F*X*t2*dy*(t2*I)**-1\n",
+ "\n",
+ "#Shear stress resisted by web \n",
+ "#sigma=2*F*I**-1*(X)*dy\n",
+ "\n",
+ "#After Integrating above equation and further simplifying we get\n",
+ "#sigma=0.9578*F\n",
+ "\n",
+ "sigma=0.9578*100\n",
+ "\n",
+ "#Result\n",
+ "print\"Shear Resisted by web\",round(sigma,2),\"%\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Shear Resisted by web 95.78 %\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.21,Page no.167"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#Wooden Beam\n",
+ "\n",
+ "b=150 #mm #width\n",
+ "d=250 #mm #Depth\n",
+ "\n",
+ "L=5000 #mm #span\n",
+ "m=11.2 #N/mm**2 #Max Bending stress\n",
+ "sigma=0.7 #N/mm**2 #Max shear stress\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let 'a' be the distance from left support\n",
+ "#Max shear force\n",
+ "#F=R_A=W*(L-a)*L**-1 \n",
+ "\n",
+ "#Max Moment\n",
+ "#M=W*(L-a)*a*L**-1\n",
+ "\n",
+ "#But M=sigma*Z\n",
+ "#W*(L-a)*a*L**-1=m*1*6**-1*b*d**2 .....................(1)\n",
+ "\n",
+ "#In Rectangular Section MAx stress is 1.5 times Avg shear stress\n",
+ "F=sigma*b*d*1.5**-1\n",
+ "\n",
+ "#W*(L-a)*L**-1=F .....................(2)\n",
+ "\n",
+ "#Dividing Equation 1 nad 2 we get\n",
+ "a=m*6**-1*b*d**2*1.5*(sigma*b*d)**-1\n",
+ "\n",
+ "#Sub above value in equation 2 we get\n",
+ "W=(L-a)**-1*L*F*10**-3 #KN \n",
+ "\n",
+ "#Result\n",
+ "print\"Load is\",round(W,2),\"KN\"\n",
+ "print\"Distance from Left support is\",round(a,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Load is 21.87 KN\n",
+ "Distance from Left support is 1000.0 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 22
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.22,Page no.168"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=1000 #mm #span\n",
+ "\n",
+ "#Rectangular Section\n",
+ "\n",
+ "b=200 #mm #width\n",
+ "d=400 #mm #depth\n",
+ "\n",
+ "sigma=1.5 #N/mm**2 #Shear stress\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let AB be the cantilever beam subjected to load W KN at free end\n",
+ "\n",
+ "#MAx shear Force\n",
+ "#F=W*10**3 #KN\n",
+ "\n",
+ "#Since Max shear stress in Rectangular section\n",
+ "#sigma_max=1.5*F*A**-1 \n",
+ "#After sub values and further simplifyng we get\n",
+ "W=1.5*b*d*(1.5*1000)**-1 #KN\n",
+ "\n",
+ "#Moment at fixwed end\n",
+ "M=W*1 #KN-m\n",
+ "y_max=d*2**-1 #mm\n",
+ "\n",
+ "#M.I\n",
+ "I=1*12**-1*b*d**3 #mm**3\n",
+ "\n",
+ "#MAx Stress\n",
+ "sigma_max=M*10**6*I**-1*y_max\n",
+ "\n",
+ "#Result\n",
+ "print\"Concentrated Load is\",round(sigma_max,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Concentrated Load is 15.0 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.24,Page no.170"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=4000 #mm #span\n",
+ "\n",
+ "#Rectangular Cross-section\n",
+ "b=100 #mm #Width\n",
+ "d=200 #mm #Thickness\n",
+ "\n",
+ "F_per=10 #N/mm**2 #Max Bending stress\n",
+ "q_max=0.6 #N/mm**2 #Shear stress\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#If the Load W is in KN/m\n",
+ "\n",
+ "#Max shear Force\n",
+ "#F=w*l*2**-1 #KN\n",
+ "#After substituting values and further simplifying we get\n",
+ "#M=2*w #KN-m\n",
+ "\n",
+ "#Max Load from Consideration of moment\n",
+ "#M=1*6**-1*b*d**2*F_per\n",
+ "#After substituting values and further simplifying we get\n",
+ "w=(1*6**-1*b*d**2*F_per)*(2*10**6)**-1 #KN/m\n",
+ "\n",
+ "#Max Load from Consideration of shear stress\n",
+ "#q_max=1.5*F*(b*d)**-1 #N\n",
+ "#After substituting values and further simplifying we get\n",
+ "F=q_max*(1.5)*b*d #N\n",
+ "\n",
+ "#If w is Max Load in KN/m,then\n",
+ "#2*w*1000=8000\n",
+ "#After Rearranging and Further simplifying we get\n",
+ "w2=8000*(2*1000)**-1 #KN/m\n",
+ "\n",
+ "#Result\n",
+ "print\"Uniformly Distributed Load Beam can carry is\",round(w,2),\"KN/m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Uniformly Distributed Load Beam can carry is 3.33 KN/m\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.5_7.ipynb b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.5_7.ipynb new file mode 100644 index 00000000..df0cb600 --- /dev/null +++ b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.5_7.ipynb @@ -0,0 +1,1019 @@ +{
+ "metadata": {
+ "name": "chapter no.5.ipynb"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 5:Deflections Of Beams By Double Integration Methods"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.5.2,Page No.192"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=3000 #mm #span of beam\n",
+ "a=2000 #mm\n",
+ "W1=20*10**3 #N #Pt Load Acting on beam\n",
+ "W2=30*10**3 #N #Pt Load Acting on beam\n",
+ "E=2*10**5 #N/mm**2 #Young's Modulus\n",
+ "I=2*10**8 #mm**4 #M.I\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Deflection at free End Due to W2\n",
+ "dell1=W2*L**3*(3*E*I)**-1 #mm\n",
+ "\n",
+ "#Deflection at free end Due to W1\n",
+ "dell2=W1*a**3*(3*E*I)**-1+(L-a)*W1*a**2*(2*E*I)**-1 #mm\n",
+ "\n",
+ "#Total Deflection at free end\n",
+ "dell=dell1+dell2 #mm\n",
+ "\n",
+ "#Result\n",
+ "print\"Deflection at Free End is\",round(dell,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Deflection at Free End is 9.08 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.5.4,Page No.193"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "E=2*10**5 #N/mm**2 #Young's Modulus\n",
+ "I=180*10**6 #mm**4 #M.I\n",
+ "W1=20 #N/m #u.d.l\n",
+ "W2=20*10**3 #N #Pt load\n",
+ "L=3000 #m #Span of beam\n",
+ "a=2000 #m #Span of u.d.l\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Displacement of free End due to 20 KN Pt load at free end\n",
+ "dell1=W2*L**3*(3*E*I)**-1 #mm\n",
+ "\n",
+ "#Displacement of free end due to u.d.l\n",
+ "dell2=W1*a**4*(8*E*I)**-1+(L-a)*W1*a**3*(6*E*I)**-1\n",
+ "\n",
+ "#Deflection at free end\n",
+ "dell=dell1+dell2 #mm\n",
+ "\n",
+ "#Result\n",
+ "print\"The Displacement of Free End of cantilever beam is\",round(dell,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Displacement of Free End of cantilever beam is 6.85 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.5.10,Page No.201"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "E=200*10**6 #KN/m**2 #Young's Modulus\n",
+ "I=15*10**-6 #m**4 #M.I\n",
+ "a=4000 #m \n",
+ "L_AB=6 #m #Span of beam\n",
+ "L_CB=2 #m #Length of CB\n",
+ "F_C=18 #KN #force at C\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let V_A & V_B be the Reactions at A & B Respectively\n",
+ "#V_A+V_B=18\n",
+ "#Now taking moment at B,we get M_B\n",
+ "V_A=(F_C*L_CB)*L_AB**-1\n",
+ "V_B=18-V_A\n",
+ "\n",
+ "#Now Taking Moment at distance x\n",
+ "#M_x=6*x-18*(x-4)\n",
+ "#EI*d**2*y*(d*x**2)**-1=6*x-18*(x-4)\n",
+ "\n",
+ "#Now Integrating above equation,we get\n",
+ "#EI*dy*(dx)**-1=C1+3*x**2-9(x-4)**4\n",
+ "\n",
+ "#Again Integrating above equation we get\n",
+ "#EI*y=C2+C1*x+x**3-3*(x-4)**3\n",
+ "\n",
+ "#The Boundary conditions\n",
+ "x=0\n",
+ "y=0 #.....(a)\n",
+ "\n",
+ "x=6\n",
+ "y=0 #....(b)\n",
+ "\n",
+ "#From Boundary Condition(B.C) a we get\n",
+ "C2=0\n",
+ "\n",
+ "#From Boundary Condition(B.C) b we get\n",
+ "#6*C1+216-3*8\n",
+ "#After Further simplifying we get\n",
+ "C1=-(216-24)*6**-1\n",
+ "\n",
+ "#EI*y=-32*x+x**3-3*(x-4)**3\n",
+ "#EI*dy*(dx)**-1=-32+3*x**2-9(x-4)**4\n",
+ "\n",
+ "#For Max Deflection\n",
+ "#Assume it inthe Porion AC i.e x=4=a\n",
+ "#0=-32+3*x**2\n",
+ "x=(32*3**-1)**0.5\n",
+ "\n",
+ "#Value of Max deflection is\n",
+ "ymax=(-32*x+x**3)*(E*I)**-1 #mm\n",
+ "\n",
+ "#slope at mid-span\n",
+ "\n",
+ "#EI*(dy*(dx)**-1)_centre=-32+3*x**2\n",
+ "#at centre ,\n",
+ "x1=3 #m\n",
+ "\n",
+ "#Let (dy*(dx)**-1)_centre=X\n",
+ "X=-(-32+3*x1**2)*(E*I)**-1 #Radian\n",
+ "\n",
+ "#Deflection at Load Point\n",
+ "x2=4 #m\n",
+ "#EI*y_c=-32*x2+x2**3\n",
+ "\n",
+ "y_c=-(-32*x2+x2**3)*(E*I)**-1\n",
+ "\n",
+ "\n",
+ "#Result\n",
+ "print\"Value of Max Deflection\",round(ymax,4),\"mm\"\n",
+ "print\"SLope at mid-span\",round(X,4),\"radian\"\n",
+ "print\"Deflection at the Load Point is\",round(y_c,4),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Value of Max Deflection -0.0232 mm\n",
+ "SLope at mid-span 0.0017 radian\n",
+ "Deflection at the Load Point is 0.0213 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 25
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.5.11,Page No.203"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L_CB=2 #m #Length of CB\n",
+ "L_AC=4 #m #Length of AB\n",
+ "M_C=15 #KN.m #Moment At Pt C\n",
+ "F_C=30 #KN\n",
+ "L=6 #m Span of Beam\n",
+ "\n",
+ "#Let X=E*I\n",
+ "X=10000 #KN-m**2\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let V_A and V_B be the reactions at A & B respectively\n",
+ "#V_A+V_B=30\n",
+ "\n",
+ "#Taking Moment a A,we get\n",
+ "V_B=(F_C*L_AC+M_C)*L**-1\n",
+ "V_A=30-V_B\n",
+ "\n",
+ "#Now Taking Moment at distacnce x from A\n",
+ "#M_x=7.5*x-30*(x-4)+15\n",
+ "\n",
+ "#By using Macaulay's Method\n",
+ "#EI*(d**2*x/dx**2)=M_x=7.5*x-30*(x-4)+15\n",
+ "\n",
+ "#Now Integrating above Equation we get\n",
+ "#EI*(dy/dx)=C1+7.5*x**2*2**-1-15*(x-4)**2+15*(x-4) ............(1)\n",
+ "\n",
+ "#Again Integrating above Equation we get\n",
+ "#EIy=C2+C1*x+7.5*6**-1*x**3-5*(x-4)**3+15*(x-4)**2*2**-1..........(2)\n",
+ "\n",
+ "#Boundary Cinditions\n",
+ "x=0\n",
+ "y=0\n",
+ "\n",
+ "#Substituting above equations we get \n",
+ "C2=0\n",
+ "\n",
+ "x=6 #m\n",
+ "y=0\n",
+ "\n",
+ "C1=-(7.5*6**3*6**-1-5*2**3+15*2**2*2**-1)*6**-1\n",
+ "\n",
+ "#EIy_c=C2+C1*x+7.5*6**-1*x**3-5*(x-4)**3+15*(x-4)**2*2**-1\n",
+ "#Sub values in Above equation we get\n",
+ "y_c=(93.3333*(X)**-1)\n",
+ "\n",
+ "#Result\n",
+ "print\"The Deflection at C\",round(y_c,4),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Deflection at C 0.0093 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.5.12,Page No.204"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L_AC=L_CD=L_DB=2 #m #Length of AC,CD,DB\n",
+ "F_C=40 #KN #Force at C\n",
+ "w=20 #KN/m #u.d.l\n",
+ "L=6 #m #span of beam\n",
+ "\n",
+ "#Let E*I=X\n",
+ "X=15000 #KN-m**2\n",
+ "\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let V_A & V_B be the reactions at A & B respectively\n",
+ "#V_A+V_B=80\n",
+ "\n",
+ "#Taking Moment B,M_B\n",
+ "V_A=(F_C*(L_CD+L_DB)+w*L_DB*L_DB*2**-1)*L**-1 #KN\n",
+ "V_B=80-V_A #KN\n",
+ "\n",
+ "#Taking Moment at distance x from A\n",
+ "#M_x=33.333*x-40*(x-2)-20*(x-4)**2*2**-1\n",
+ "#EI*(d**2/dx**2)=33.333*x-40*(x-2)-10*(x-4)**2\n",
+ "\n",
+ "#Integrating above equation we get\n",
+ "#EI*(dy/dx)=C1+33.333*x**2*2**-1-20*(x-2)**2-10*3**-1*(x-4)**3\n",
+ "\n",
+ "#Again Integrating above equation we get\n",
+ "#EI*y=C2+C1*x+33.333*x**3*6**-1-20*3**-1*(x-2)**3-10*12**-1*(x-4)**4\n",
+ "\n",
+ "#At\n",
+ "x=0\n",
+ "y=0\n",
+ "C2=0\n",
+ "\n",
+ "#At\n",
+ "x=6\n",
+ "y=0\n",
+ "C1=-760*6**-1\n",
+ "\n",
+ "#Assuming Deflection to be max in portion CD and sustituting value of C1 in equation of slope we get\n",
+ "#EI*y=C2+C1*x+33.333*x**3*6**-1-20*3**-1*(x-2)**3-10*12**-1*(x-4)**4\n",
+ "#0=-126.667+33.333*x**2**-1-20*(x-2)**2\n",
+ "\n",
+ "#After rearranging and simplifying further we get\n",
+ "\n",
+ "#x**2-24*x+62=0\n",
+ "#From above equations\n",
+ "a=1\n",
+ "b=-24\n",
+ "c=62\n",
+ "\n",
+ "y=(b**2-4*a*c)**0.5\n",
+ "\n",
+ "x1=(-b+y)*(2*a)**-1\n",
+ "x2=(-b-y)*(2*a)**-1\n",
+ "\n",
+ "#Taking x2 into account\n",
+ "x=2.945 #m\n",
+ "C1=-126.667\n",
+ "C2=0\n",
+ "\n",
+ "y_max=(C2+C1*x+33.333*x**3*6**-1-20*3**-1*(x-2)**3)*X**-1 #mm\n",
+ "\n",
+ "#Max slope occurs at the ends\n",
+ "#At A,\n",
+ "#EI*(dy/dx)_A=-126.667\n",
+ "#At B\n",
+ "#EI*(dy/dx)_B=126.667+33.333*6**2*2**-1-20*4**2-10*2**3\n",
+ "#After simplifying Further we get\n",
+ "#EI*(dy/dx)_B=73.3273\n",
+ "\n",
+ "#Now Max slope is EI(dy/dx)_A=-126.667\n",
+ "#15000*(dy/dx)_=-126.667\n",
+ "\n",
+ "#Let Y=dy/dx\n",
+ "Y=-126.667*X**-1 #Radians\n",
+ "\n",
+ "#Result\n",
+ "print\"Maximum Deflection for Beam is\",round(y_max,4),\"mm\"\n",
+ "print\"Maximum Slope for beam is\",round(Y,4),\"radians\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Maximum Deflection for Beam is -0.0158 mm\n",
+ "Maximum Slope for beam is -0.0084 radians\n"
+ ]
+ }
+ ],
+ "prompt_number": 21
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.5.13,Page No.206"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "E=2*10**8 #KN/m**2\n",
+ "I=450*10**-6 #m**4\n",
+ "L_AC=1 #m #Length of AC\n",
+ "L_CD=3 #m #Length of CD\n",
+ "L_DB=2 #m #Length of DB\n",
+ "w=10 #KN/m #u.d.l\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let V_A & V_B be the reactions at A & B respectively\n",
+ "#V_A+V_B=30\n",
+ "\n",
+ "#Taking Moment at distance x from A\n",
+ "#M_x=17.5*x-10*(x-1)**2*2**-1+10*(x-4)**2*2**-1\n",
+ "#EI*(d**2/dx**2)=17.5*x-10*(x-1)**2*2**-1+10*(x-4)**2*2**-1\n",
+ "\n",
+ "#Now Integrating Above equation we get\n",
+ "#EI(dy/dx)=C1+17.5*x**2*2**-1-5*3**-1*(x-1)**2+5*3**-1*(x-4)**3\n",
+ "\n",
+ "#Again Integrating Above equation we get\n",
+ "#EI*y=C2+C1*x+17.5*x**3*6**-1-5*12**-1*(x-1)**4+5*12**-1*(x-4)**4\n",
+ "\n",
+ "#At \n",
+ "x=0\n",
+ "y=0\n",
+ "C2=0\n",
+ "\n",
+ "#At \n",
+ "x=6 \n",
+ "y=0\n",
+ "C1=(-17.5*x**3*6**-1+5*12**-1*(x-1)**4-5*12**-1*(x-4)**4)*x**-1\n",
+ "\n",
+ "# 1)Slope at A .i.e at x=0\n",
+ "#EI*(dy/dx)_A=C1=-62.708 #KN-m**2\n",
+ "#let (dy/dx)=X\n",
+ "X=C1*(E*I)**-1 #radiams\n",
+ "\n",
+ "#Deflection at mid-span\n",
+ "x=3 #m\n",
+ "#EI*y_centre=C1*x+17.5*x**3*6**-1-5*12**-1*(x-1)**2\n",
+ "y_centre=-(C1*x+17.5*x**3*6**-1-5*12**-1*(x-1)**4)*(E*I)**-1\n",
+ "\n",
+ "#Maximum Deflection\n",
+ "\n",
+ "#At point of Max deflection (dy/dx)=0\n",
+ "#Assuming it in portion CD\n",
+ "\n",
+ "#0=C1*x+17.5*x**2*2**-1-5*3**-1*(x-1)**3\n",
+ "\n",
+ "#Now Let\n",
+ "#F(x)=C1+17.5*x**2*2**-1-5*3**-1*(x-1)**3\n",
+ "\n",
+ "#Let F(x)=Y\n",
+ "#At \n",
+ "x=2.5\n",
+ "Y1=-(C1+17.5*x**2*2**-1-5*3**-1*(x-1)**3)\n",
+ "\n",
+ "#AT\n",
+ "x=3\n",
+ "Y2=-(C1+17.5*x**2*2**-1-5*3**-1*(x-1)**3)\n",
+ "\n",
+ "#At\n",
+ "x=2.9 #m\n",
+ "Y3=-(C1+17.5*x**2*2**-1-5*3**-1*(x-1)**3)\n",
+ "\n",
+ "#A curve may be plotted for (F(x) and the value for which F(x)=0 may be found\n",
+ "#For F(x)=0 for x=2.92 m\n",
+ "#Therefore y_max occur at x=2.92\n",
+ "\n",
+ "x=2.92 #m\n",
+ "y_max=(C1*x+17.5*x**3*6**-1-5*12**-1*(x-1)**4)*(E*I)**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"Slope at A\",round(X,6),\"mm\"\n",
+ "print\"Deflection at mid-span\",round(y_centre,6),\"mm\"\n",
+ "print\"Maxmimum Deflection is\",round(y_max,5),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Slope at A -0.000697 mm\n",
+ "Deflection at mid-span 0.001289 mm\n",
+ "Maxmimum Deflection is -0.00129 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.5.14,Page No.208"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L_AC=LDE=L_EB=1 #m #Length of AC\n",
+ "L_CD=2 #m #Length of CD\n",
+ "E=200 #KN/mm**2\n",
+ "I=60*10**6 #mm**4 #M.I\n",
+ "F_C=20 #KN #Force at C\n",
+ "F_E=30 #KN #Force at E\n",
+ "w=10 #KN/m #u.d.l\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "X=E*I*10**-6 #KN-m**2\n",
+ "\n",
+ "#Let V_A & V_B be the reactions at A & B respectively\n",
+ "#V_A+V_B=70\n",
+ "\n",
+ "#Taking Moment at distance x from A\n",
+ "#M_x=34*x-20*(x-1)-10*(x-1)**2*2**-1+10*(x-3)**2*2**-1-30*(x-4)\n",
+ "#EI*(d**2y/dx**2)=34*x-20*(x-1)-10*(x-1)**2*2**-1+10*(x-3)**2*2**-1-30*(x-4)\n",
+ "\n",
+ "#Now Integrating Above equation,we get\n",
+ "#EI*(dy/dx)=C1+17*x**2-10*(x-1)**2-5*3**-1*(x-1)**3+5*3**-1*(x-3)**3-15*(x-4)**2\n",
+ "\n",
+ "#Again Integrating Above equation,we get\n",
+ "#EI*y=C2+C1*x+17*3**-1*x**3-10*3**-1*(x-1)**3-5*12**-1*(x-1)**4+5*12**-1*(x-3)**4-5*(x-4)**3\n",
+ "\n",
+ "#At\n",
+ "x=0\n",
+ "y=0\n",
+ "C2=0\n",
+ "\n",
+ "#At \n",
+ "x=5 #m\n",
+ "y=0\n",
+ "C1=(-17*3**-1*x**3+10*3**-1*(x-1)**3+5*12**-1*(x-1)**4-5*12**-1*(x-3)**4+5*(x-4)**3)*5**-1\n",
+ "\n",
+ "#EI*y=C2+C1*x+17*3**-1*x**3-10*3**-1*(x-1)**3-5*12**-1*(x-1)**4+5*12**-1*(x-3)**4-5*(x-4)**3\n",
+ "C2=0\n",
+ "C1=-78\n",
+ "x=1\n",
+ "y_c=(-78*x+17*3**-1*x)*(X)**-1\n",
+ "\n",
+ "#EI*y_D=C2+C1*x+17*3**-1*x**3-10*3**-1*(x-1)**3-5*12**-1*(x-1)**4\n",
+ "x=3\n",
+ "C1-78\n",
+ "C2=0\n",
+ "y_D=(C2+C1*x+17*3**-1*x**3-10*3**-1*(x-1)**3-5*12**-1*(x-1)**4)*(X**-1)\n",
+ "\n",
+ "#EI*y_E=C2+C1*x+17*3**-1*x**3-10*3**-1*(x-1)**3-5*12**-1*(x-1)**4+5*12**-1*(x-3)**4\n",
+ "x=4\n",
+ "C1-78\n",
+ "C2=0\n",
+ "y_E=(C2+C1*x+17*3**-1*x**3-10*3**-1*(x-1)**3-5*12**-1*(x-1)**4+5*12**-1*(x-3)**4)*X**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"Deflections at C\",round(y_c,5),\"mm\"\n",
+ "print\"Deflections at D\",round(y_D,5),\"mm\"\n",
+ "print\"Deflections at E\",round(y_E,4),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Deflections at C -0.00603 mm\n",
+ "Deflections at D -0.00953 mm\n",
+ "Deflections at E -0.0061 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 45
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.5.15,Page No.209"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "%matplotlib inline\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "E=200 #KN/mm**2 #Modulus of Elasticity\n",
+ "I=300*10**6 #mm\n",
+ "L_AB=L_BC=L_CD=L_DE=1 #m #Length of AB,BC,CD,DE respectively\n",
+ "F_A=20 #KN #Force at A\n",
+ "F_C=10 #KN #Force at C\n",
+ "w=30 #KN/m #u.d.l\n",
+ "\n",
+ "#Let E*I=X\n",
+ "X=E*I*10**-6 #KN-2**2\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let V_E be the reactions at E\n",
+ "V_E=F_A+F_C+w*(L_BC+L_CD) #KN \n",
+ "\n",
+ "#Taking Moment at distance x\n",
+ "#EI*(d**2x/dy**2)=M=-20*x-30*(x-1)**2*2**-1-10*(x-2)+30*(x-3)**2*2**-1\n",
+ "\n",
+ "#Integrating above equation we get\n",
+ "#EI*(dy/dx)=C1-10*x**2-5*(x-1)**3-5*(x-2)**2+5*(x-3)**3\n",
+ "\n",
+ "#Again Integrating above equation\n",
+ "#EI*y=C2+C1*x-10*x**3*3**-1-5*(x-1)**4*4**-1-5*(x-3)**4*4**-1-5*3*(x-2)**3\n",
+ "\n",
+ "#At\n",
+ "#dy/dx=0\n",
+ "x=4 #m\n",
+ "C1=10*x**2+5*(x-1)**3+5*(x-2)**2-5*(x-3)**3\n",
+ "\n",
+ "#AT\n",
+ "x=4\n",
+ "y=0\n",
+ "C2=-C1*4+10*x**3*3**-1+5*(x-1)**4*4**-1-5*(x-3)**4*4**-1+5*3**-1*(x-2)**3\n",
+ "\n",
+ "#Max Deflection and Max slopes occurs at Free end in case of cantilever\n",
+ "y_max=y_A=C2*X**-1\n",
+ "\n",
+ "#EI*(dy/dx)_max=C1\n",
+ "#Let (dy/dx)=Y\n",
+ "Y=C1*X**-1 #radian\n",
+ "\n",
+ "#Now deflection at x=1 #m\n",
+ "C2=-913.333\n",
+ "C1=310\n",
+ "x=1\n",
+ "y_B=(C2+C1*x-10*x**3*3**-1)*X**-1\n",
+ "\n",
+ "#Now Deflection at x=2 #m\n",
+ "C2=-913.333\n",
+ "C1=310\n",
+ "x=2 #m\n",
+ "y_C=(C2+C1*x-10*x**3*3**-1-5*(x-1)**4*4**-1)*X**-1\n",
+ "\n",
+ "#Now Deflection at x=3 #m\n",
+ "C2=-913.333\n",
+ "C1=310\n",
+ "x=3 #m\n",
+ "y_D=(C2+C1*x-10*x**3*3**-1-5*(x-1)**4*4**-1-5*3**-1*(x-2)**3)*X**-1\n",
+ "\n",
+ "y_E=0\n",
+ "\n",
+ "#Result\n",
+ "print\"Max Deflection for Beam\",round(y_A,4),\"mm\"\n",
+ "print\"Max Slope for beam\",round(Y,5),\"radians\"\n",
+ "\n",
+ "#Plotting the ELastic Curve\n",
+ "\n",
+ "Y2=[y_E,y_D,y_C,y_B,y_A]\n",
+ "X2=[L_AB+L_BC+L_CD+L_DE,L_AB+L_BC+L_CD,L_AB+L_BC,L_AB,0]\n",
+ "Z2=[0,0,0,0,0]\n",
+ "plt.plot(X2,Y2,X2,Z2)\n",
+ "plt.xlabel(\"Length in mm\")\n",
+ "plt.ylabel(\"Deflection in mm\")\n",
+ "plt.show()"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Max Deflection for Beam -0.0152 mm\n",
+ "Max Slope for beam 0.00517 radians\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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8yjGZmZkkJSWRmZlJSkoKM2bMoLKy8pxilcZbsQJGj4Y//QkWL1aSEWnvaq3R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\nL+mlAAAAAElFTkSuQmCC\n",
+ "text": [
+ "<matplotlib.figure.Figure at 0x56d89b0>"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.5.16,Page No.211"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L_BD=L_CB=L_AC=2 #m #Length of BD,CB,AC\n",
+ "F_C=40 #KN #Force at C\n",
+ "F_D=10 #KN Force at D\n",
+ "L=6 #m spna of beam\n",
+ "\n",
+ "#EI is constant in this problem\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let V_A & V_B be the reactions at A & B Respectively\n",
+ "#V_A+V_B=50\n",
+ "\n",
+ "#Taking Moment at Pt A\n",
+ "V_B=(F_D*L+F_C*L_AC)*(L_AC+L_CB)**-1\n",
+ "V_A=50-V_B\n",
+ "\n",
+ "#Now Taking Moment at distance x from A,M_x\n",
+ "#M_x=15*x-40*(x-2)+35*(x-4)\n",
+ "#EI*(d**2*y/dx**2)=15*x-40*(x-2)+35*(x-4)\n",
+ "\n",
+ "#Now Integrating above equation we get\n",
+ "#EI*(dy/dx)=C1+7.5*x**2-20*(x-2)**2+17.5(x-4)**2\n",
+ "\n",
+ "#Again Integrating above equation we get\n",
+ "#EI*y=C2+C1*x+2.5*x**2-20*3**-1*(x-2)**3+17.5*(x-4)**3*3**-1\n",
+ "\n",
+ "#At\n",
+ "x=0\n",
+ "y=0\n",
+ "#we get\n",
+ "C2=0\n",
+ "\n",
+ "#At\n",
+ "x=4 \n",
+ "y=0\n",
+ "#we get\n",
+ "C1=(2.5*4**3-20*3**-1*2**3)*4**-1\n",
+ "\n",
+ "#Now Deflection at C\n",
+ "x=2\n",
+ "C1=-26.667\n",
+ "C2=0\n",
+ "y_C=C2+C1*x+2.5*x**3\n",
+ "\n",
+ "#Now Deflection at D\n",
+ "C1=-21.667\n",
+ "C2=0\n",
+ "y_D=-26.667*6+2.5*6**3-20*3**-1*4**3+17.5*2**3*3**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"Deflections Under Loads are:y_D\",round(y_D,4)\n",
+ "print\" :y_C\",round(y_C,2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Deflections Under Loads are:y_D -0.002\n",
+ " :y_C -33.33\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.5.17,Page No.212"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L_BC=L_EB=2 #m #Length of BC & EB\n",
+ "E=200*10**6 #KN/m**2 #Modulus of eLasticity\n",
+ "I=45*10**-6 #mm**4 #M.I\n",
+ "L_DE=3 #m #Length of DE\n",
+ "L_AD=1 #m #Length of AD\n",
+ "w=20 #KN/m #u.d.l\n",
+ "L=8 #m #span of beam\n",
+ "F_C=30 #KN #Force at C\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let V_A & V_B be the reactions at A & B respectively\n",
+ "#V_A+V_B=90\n",
+ "\n",
+ "#Taking Moment at A,M_A\n",
+ "V_B=(w*L_DE*(L_DE*2**-1+L_AD)+F_C*L)*(L_AD+L_DE+L_EB)**-1\n",
+ "V_A=90-V_B\n",
+ "\n",
+ "#Taking Moment at distance x\n",
+ "#M_x=25*x-20*(x-1)**2*2**-1+20*(x-4)**2*2**-1+65*(x-6)\n",
+ "\n",
+ "#Integrating above equation we get\n",
+ "#EI*(d**2*y/dx**2)=25*x-10*(x-1)**2+10*(x-4)**2+65*(x-6)\n",
+ "\n",
+ "#again Integrating above equation we get\n",
+ "#EI*(dy/dx)=C1+25*x**2*2**-1-10*3**-1*(x-1)**3+10*3**-1*(x-4)**2+65*2**-1*(x-6)**2\n",
+ "\n",
+ "#again Integrating above equation we get\n",
+ "#EI*y=C2+C1*x+25*6**-1*x**3-10*12**-1*(x-1)**4+10*12**-1*(x-4)**4+65*6**-1*(x-6)**3\n",
+ "\n",
+ "x=0\n",
+ "y=0\n",
+ "#Sub these values in above equation,we get\n",
+ "C2=0\n",
+ "\n",
+ "x=6 #m\n",
+ "y=0\n",
+ "C1=-(25*6**-1*6**3-10*12**-1*5**4+10*12**-1*2**4)*6**-1\n",
+ "\n",
+ "#deflection at C is given by\n",
+ "x=8\n",
+ "y_c=(C2+C1*x+25*6**-1*x**3-10*12**-1*(x-1)**4+10*12**-1*(x-4)**4+65*6**-1*(x-6)**3)*(E*I)**-1\n",
+ "\n",
+ "#Assuming y is max in the portion DE,then\n",
+ "#(dy/dx)=0 for that point\n",
+ "\n",
+ "#0=-65.417+25*2**-1*x**2-10*3**-1*x(-1)**3\n",
+ "\n",
+ "#Let F(x)=-65.417+25*2**-1*x**2-10*3**-1*x(-1)**3\n",
+ "#Let z=F(x)\n",
+ "\n",
+ "#AT \n",
+ "x=3\n",
+ "z=-65.417+25*2**-1*x**2-10*3**-1*(x-1)**3\n",
+ "\n",
+ "x=2.5\n",
+ "z1=-65.417+25*2**-1*x**2-10*3**-1*(x-1)**3\n",
+ "\n",
+ "x=2.4\n",
+ "z2=-65.417+25*2**-1*x**2-10*3**-1*(x-1)**3\n",
+ "\n",
+ "#The assumption is max in portion DE\n",
+ "x=2.46\n",
+ "y_max=(-65.417*x+25*6**-1*x**3-10*12**-1*1.46**4)*(E*I)**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"Deflection at free end C\",round(y_c,4),\"mm\"\n",
+ "print\"Max Deflection between A and B\",round(y_max,4),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Deflection at free end C -0.0101 mm\n",
+ "Max Deflection between A and B -0.0114 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.5.18,Page No.213"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L_DB=L_AC=L_ED=2 #m #Length of DB & AC\n",
+ "L_CD=4 #m #Length of CD\n",
+ "L_CE=2 #m #Length of CE\n",
+ "F_A=40 #KN #Force at C\n",
+ "F_B=20 #KN #Force at A\n",
+ "E=200*10**6 #KN/mm**2 #Modulus of Elasticity\n",
+ "I=50*10**-6 #m**4 #M.I\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#LEt V_C & V_D be the reactions at C & D respectively\n",
+ "#V_C+V_D=60\n",
+ "\n",
+ "#Taking Moment At D,M_D\n",
+ "V_C=-(-F_A*(L_AC+L_CE+L_ED)+F_B*L_DB)*L_CD**-1\n",
+ "V_D=60-V_C\n",
+ "\n",
+ "#Now Taking Moment at Distance x from A,\n",
+ "#M_x=-40*x+50*(x-2)+10*(x-6)\n",
+ "\n",
+ "#EI*(d**2*y/dx**2)=-40*x+50*(x-2)+10*(x-6)\n",
+ "\n",
+ "#Now Integrating above Equation we get\n",
+ "#EI*(dy/dx)=C1+20*x**2-25*(x-2)+5*(x-6)**2\n",
+ "\n",
+ "#Again Integrating above Equation we get\n",
+ "#EI*y=C2+C1*x-20*3**-1*x**3+25*3**-1*(x-2)**3+5*3**-1*(x-6)**3\n",
+ "\n",
+ "#At\n",
+ "x=0\n",
+ "y=0\n",
+ "#C2+2*C1=-53.33 ...............(1)\n",
+ "\n",
+ "#At \n",
+ "x=6\n",
+ "y=0\n",
+ "#C2+6*C1=906.667 ...............(2)\n",
+ "\n",
+ "#Subtracting Equation 1 from 2 we get\n",
+ "C1=853.333*4**-1\n",
+ "C2=53.333-2*C1\n",
+ "x=0\n",
+ "y_A=(C2+C1*x-20*3**-1*x**3+25*3**-1*(x-2)**3+5*3**-1*(x-6)**3)*(E*I)**-1\n",
+ "\n",
+ "#Answer For y_A is incorrect in textbook\n",
+ "\n",
+ "#At Mid-span\n",
+ "C1=853.333*4**-1\n",
+ "C2=53.333-2*C1\n",
+ "x=4\n",
+ "y_E=(C2+C1*x-20*3**-1*x**3+25*3**-1*(x-2)**3+5*3**-1*(x-6)**3)*(E*I)**-1\n",
+ "\n",
+ "#Answer For y_E is incorrect in textbook\n",
+ "\n",
+ "#At B\n",
+ "C1=853.333*4**-1\n",
+ "C2=53.333-2*C1\n",
+ "x=8\n",
+ "y_B=(C2+C1*x-20*3**-1*x**3+25*3**-1*(x-2)**3+5*3**-1*(x-6)**3)*(E*I)**-1\n",
+ "\n",
+ "\n",
+ "#Result\n",
+ "print\"Deflection relative to the level of the supports:at End A\",round(y_A,4),\"mm\"\n",
+ "print\" :at End B\",round(y_B,4),\"mm\"\n",
+ "print\" :at Centre of CD\",round(y_E,4),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Deflection relative to the level of the supports:at End A -0.08 mm\n",
+ " :at End B -0.0267 mm\n",
+ " :at Centre of CD 0.0107 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.6_7.ipynb b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.6_7.ipynb new file mode 100644 index 00000000..e3bdfbe1 --- /dev/null +++ b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.6_7.ipynb @@ -0,0 +1,1518 @@ +{
+ "metadata": {
+ "name": "chapter no.6.ipynb"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter No.6:Torsion"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.1,Page No.225"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=10000 #mm #Length of solid shaft\n",
+ "d=100 #mm #Diameter of shaft\n",
+ "n=150 #rpm\n",
+ "P=112.5*10**6 #N-mm/sec #Power Transmitted\n",
+ "G=82*10**3 #N/mm**2 #modulus of Rigidity\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "J=pi*d**4*(32)**-1 #mm**3 #Polar Modulus\n",
+ "T=P*60*(2*pi*n)**-1 #N-mm #Torsional moment\n",
+ "\n",
+ "r=50 #mm #Radius\n",
+ "\n",
+ "q_s=T*r*J**-1 #N/mm**2 #Max shear stress intensity\n",
+ "Theta=T*L*(G*J)**-1 #angle of twist\n",
+ "\n",
+ "#Result\n",
+ "print\"Max shear stress intensity\",round(q_s,2),\"N/mm**2\"\n",
+ "print\"Angle of Twist\",round(Theta,3),\"radian\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Max shear stress intensity 36.48 N/mm**2\n",
+ "Angle of Twist 0.089 radian\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.2,Page No.226"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "P=440*10**6 #N-m/sec #Power transmitted\n",
+ "n=280 #rpm\n",
+ "theta=pi*180**-1 #radian #angle of twist\n",
+ "L=1000 #mm #Length of solid shaft\n",
+ "q_s=40 #N/mm**2 #Max torsional shear stress\n",
+ "G=84*10**3 #N/mm**2 #Modulus of rigidity\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#P=2*pi*n*T*(60)**-1 #Equation of Power transmitted\n",
+ "T=P*60*(2*pi*n)**-1 #N-mm #torsional moment\n",
+ "\n",
+ "#From Consideration of shear stress\n",
+ "d1=(T*16*(pi*40)**-1)**0.333333 \n",
+ "\n",
+ "#From Consideration of angle of twist\n",
+ "d2=(T*L*32*180*(pi*84*10**3*pi)**-1)**0.25\n",
+ "\n",
+ "#result\n",
+ "print\"Diameter of solid shaft is\",round(d1,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Diameter of solid shaft is 124.09 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 21
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.3,Page No.227"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "G=80*10**3 #N/mm**2 #Modulus of rigidity\n",
+ "q_s=80 #N/mm**2 #Max sheare stress\n",
+ "P=736*10**6 #N-mm/sec #Power transmitted\n",
+ "n=200\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "T=P*60*(2*pi*n)**-1 #N-mm #Torsional moment\n",
+ "\n",
+ "#Now From consideration of angle of twist\n",
+ "theta=pi*180**-1\n",
+ "#L=15*d\n",
+ "\n",
+ "d=(T*32*180*15*(pi**2*G)**-1)**0.33333\n",
+ "\n",
+ "#Now corresponding stress at the surface is\n",
+ "q_s2=T*32*d*(pi*2*d**4)**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"Max diameter required is\",round(d,2),\"mm\"\n",
+ "print\"Corresponding shear stress is\",round(q_s2,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Max diameter required is 156.66 mm\n",
+ "Corresponding shear stress is 46.55 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 27
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.4,Page No.228"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d=25 #mm #Diameter of steel bar\n",
+ "p=50*10**3 #N #Pull\n",
+ "dell_1=0.095 #mm #Extension of bar\n",
+ "l=200 #mm #Guage Length\n",
+ "T=200*10**3 #N-mm #Torsional moment\n",
+ "theta=0.9*pi*180**-1 #angle of twist\n",
+ "L=250 #mm Length of steel bar\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "A=pi*4**-1*d**2 #Area of steel bar #mm**2\n",
+ "E=p*l*(dell_1*A)**-1 #N/mm**2 #Modulus of elasticity \n",
+ "\n",
+ "J=pi*32**-1*d**4 #mm**4 #Polar modulus\n",
+ "\n",
+ "G=T*L*(theta*J)**-1 #Modulus of rigidity #N/mm**2\n",
+ "\n",
+ "#Now from the relation of Elastic constants\n",
+ "mu=E*(2*G)**-1-1\n",
+ "\n",
+ "#result\n",
+ "print\"The Poissoin's ratio is\",round(mu,3)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Poissoin's ratio is 0.292\n"
+ ]
+ }
+ ],
+ "prompt_number": 30
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.5,Page No.229"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=6000 #mm #Length of circular shaft\n",
+ "d1=100 #mm #Outer Diameter\n",
+ "d2=75 #mm #Inner Diameter\n",
+ "R=100*2**-1 #Radius of shaft\n",
+ "T=10*10**6 #N-mm #Torsional moment\n",
+ "G=80*10**3 #N/mm**2 #Modulus of Rigidity\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "J=pi*32**-1*(d1**4-d2**4) #mm**4 #Polar Modulus \n",
+ "\n",
+ "#Max Shear stress produced\n",
+ "q_s=T*R*J**-1 #N/mm**2\n",
+ "\n",
+ "#Angle of twist\n",
+ "theta=T*L*(G*J)**-1 #Radian\n",
+ "\n",
+ "#Result\n",
+ "print\"MAx shear stress produced is\",round(q_s,2),\"N/mm**2\"\n",
+ "print\"Angle of Twist is\",round(theta,2),\"Radian\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "MAx shear stress produced is 74.5 N/mm**2\n",
+ "Angle of Twist is 0.11 Radian\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.6,Page No.229"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d1=200 #mm #External Diameter of shaft\n",
+ "t=25 #mm #Thickness of shaft\n",
+ "n=200 #rpm\n",
+ "theta=0.5*pi*180**-1 #Radian #angle of twist\n",
+ "L=2000 #mm #Length of shaft\n",
+ "G=84*10**3 #N/mm**2\n",
+ "d2=d1-2*t #mm #Internal Diameter of shaft\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "J=pi*32**-1*(d1**4-d2**4) #mm**4 #Polar Modulus \n",
+ "\n",
+ "#Torsional moment\n",
+ "T=G*J*theta*L**-1 #N/mm**2 \n",
+ "\n",
+ "#Power Transmitted\n",
+ "P=2*pi*n*T*60**-1*10**-6 #N-mm\n",
+ "\n",
+ "#Max shear stress transmitted\n",
+ "q_s=G*theta*(d1*2**-1)*L**-1 #N/mm**2 \n",
+ "\n",
+ "#Result\n",
+ "print\"Power Transmitted is\",round(P,2),\"N-mm\"\n",
+ "print\"Max Shear stress produced is\",round(q_s,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Power Transmitted is 824.28 N-mm\n",
+ "Max Shear stress produced is 36.65 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.7,Page No.230"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "P=3750*10**6 #N-mm/sec\n",
+ "n=240 #Rpm\n",
+ "q_s=160 #N/mm**2 #Max shear stress\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#d2=0.8*d2 #mm #Internal Diameter of shaft\n",
+ "\n",
+ "#J=pi*32**-1*(d1**4-d2**4) #mm**4 #Polar modulus\n",
+ "#After substituting value in above Equation we get\n",
+ "#J=0.05796*d1**4\n",
+ "\n",
+ "T=P*60*(2*pi*n)**-1 #N-mm #Torsional moment\n",
+ "\n",
+ "#Now from Torsion Formula\n",
+ "#T*J**-1=q_s*R**-1 ......................................(1)\n",
+ "\n",
+ "#But R=d1*2**-1 \n",
+ "\n",
+ "#Now substituting value of R and J in Equation (1) we get\n",
+ "d1=(T*(0.05796*q_s*2)**-1)**0.33333\n",
+ "\n",
+ "d2=d1*0.8\n",
+ "\n",
+ "#Result\n",
+ "print\"The size of the Shaft is:d1\",round(d1,3),\"mm\"\n",
+ "print\" :d2\",round(d2,3),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The size of the Shaft is:d1 200.362 mm\n",
+ " :d2 160.289 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 22
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.8,Page No.231"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "P=245*10**6 #N-mm/sec #Power transmitted\n",
+ "n=240 #rpm\n",
+ "q_s=40 #N/mm**2 #Shear stress\n",
+ "theta=pi*180**-1 #radian #Angle of twist\n",
+ "L=1000 #mm #Length of shaft\n",
+ "G=80*10**3 #N/mm**2\n",
+ "\n",
+ "#Tmax=1.5*T\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "T=P*60*(2*pi*n)**-1 #N-mm #Torsional Moment\n",
+ "Tmax=1.5*T\n",
+ "\n",
+ "#Now For Solid shaft\n",
+ "#J=pi*32*d**4\n",
+ "\n",
+ "#Now from the consideration of shear stress we get\n",
+ "#T*J**-1=q_s*(d*2**-1)**-1\n",
+ "#After substituting value in above Equation we get\n",
+ "#T=pi*16**-1*d**3*q_s\n",
+ "\n",
+ "#Designing For max Torque\n",
+ "d=(Tmax*16*(pi*40)**-1)**0.33333 #mm #Diameter of shaft\n",
+ "\n",
+ "#For max Angle of Twist\n",
+ "#Tmax*J**-1=G*theta*L**-1 \n",
+ "#After substituting value in above Equation we get\n",
+ "d2=(Tmax*32*180*L*(pi**2*G)**-1)**0.25\n",
+ "\n",
+ "#For Hollow Shaft\n",
+ "\n",
+ "#d1_2=Outer Diameter\n",
+ "#d2_2=Inner Diameter\n",
+ "\n",
+ "#d2_2=0.5*d1_2\n",
+ "\n",
+ "# Polar modulus\n",
+ "#J=pi*32**-1*(d1_2**4-d2_2**4)\n",
+ "#After substituting values we get\n",
+ "#J=0.092038*d1_2**4\n",
+ "\n",
+ "#Now from the consideration of stress\n",
+ "#Tmax*J**-1=q_s*(d1_2*2**-1)**-1\n",
+ "#After substituting values and further simplifying we get\n",
+ "d1_2=(Tmax*(0.092038*2*q_s)**-1)**0.33333\n",
+ "\n",
+ "#Now from the consideration of angle of twist\n",
+ "#Tmax*J**-1=G*theta*L**-1\n",
+ "#After substituting values and further simplifying we get\n",
+ "d1_3=(Tmax*180*L*(0.092038*G*pi)**-1)**0.25\n",
+ "\n",
+ "d2_2=0.5*d1_2\n",
+ "\n",
+ "#result\n",
+ "print\"Diameter of shaft is:For solid shaft:d\",round(d,2),\"mm\"\n",
+ "print\" :For Hollow shaft:d1_2\",round(d1_2,3),\"mm\"\n",
+ "print\" : :d2_2\",round(d2_2,3),\"mm\" "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Diameter of shaft is:For solid shaft:d 123.01 mm\n",
+ " :For Hollow shaft:d1_2 125.69 mm\n",
+ " : :d2_2 62.845 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.11,Page No.235"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "P=250*10**6 #N-mm/sec #Power transmitted\n",
+ "n=100 #rpm\n",
+ "q_s=75 #N/mm**2 #Shear stress\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#From Equation of Power we have\n",
+ "T=P*60*(2*pi*n)**-1 #N-mm #Torsional moment\n",
+ "\n",
+ "#Now from torsional moment equation we have\n",
+ "#T=j*q_s*(d/2**-1)**-1\n",
+ "#After substituting values in above equation and further simplifying we get\n",
+ "#T=pi*16**-1**d**3*q_s\n",
+ "d=(T*16*(pi*q_s)**-1)**0.3333 #mm #Diameter of solid shaft\n",
+ "\n",
+ "#PArt-2\n",
+ "\n",
+ "#Let d1 and d2 be the outer and inner diameter of hollow shaft\n",
+ "#d2=0.6*d1\n",
+ "\n",
+ "#Again from torsional moment equation we have\n",
+ "#T=pi*32**-1*(d1**4-d2**4)*q_s*(d1/2)**-1\n",
+ "d1=(T*16*(pi*(1-0.6**4)*q_s)**-1)**0.33333\n",
+ "d2=0.6*d1\n",
+ "\n",
+ "#Cross sectional area of solid shaft\n",
+ "A1=pi*4**-1*d**2 #mm**2\n",
+ "\n",
+ "#cross sectional area of hollow shaft\n",
+ "A2=pi*4**-1*(d1**2-d2**2)\n",
+ "\n",
+ "#Now percentage saving in weight\n",
+ "#Let W be the percentage saving in weight\n",
+ "W=(A1-A2)*100*A1**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"Percentage saving in Weight is\",round(W,3),\"%\"\n",
+ "print\"Size of shaft is:solid shaft:d\",round(d,3),\"mm\"\n",
+ "print\" :Hollow shaft:d1\",round(d1,3),\"mm\"\n",
+ "print\" : :d2\",round(d2,3),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Percentage saving in Weight is 29.735 %\n",
+ "Size of shaft is:solid shaft:d 117.418 mm\n",
+ " :Hollow shaft:d1 123.031 mm\n",
+ " : :d2 73.818 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.12,Page No.237"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "d=100 #mm #Diameter of solid shaft\n",
+ "d1=100 #mm #Outer Diameter of hollow shaft\n",
+ "d2=50 #mm #Inner Diameter of hollow shaft\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Torsional moment of solid shaft\n",
+ "#T_s=J*q_s*(d*2**-1)**-1 \n",
+ "#After substituting values in above equation and further simplifying we get\n",
+ "#T_s=pi*16*d**3*q_s ...............(1)\n",
+ "\n",
+ "#torsional moment for hollow shaft is\n",
+ "#T_h=J*q_s*(d1**4-d2**4)**-1*(d1*2**-1)\n",
+ "#After substituting values in above equation and further simplifying we get\n",
+ "#T_h=pi*32**-1*2*d1**-1*(d1**4-d2**4)*q_s ...........(2)\n",
+ "\n",
+ "#Dividing Equation 2 by 1 we get\n",
+ "#Let the ratio of T_h*T_s**-1 Be X\n",
+ "X=1-0.5**4\n",
+ "\n",
+ "#Loss in strength \n",
+ "#Let s be the loss in strength\n",
+ "#s=T_s*T_h*100*T_s**-1\n",
+ "#After substituting values in above equation and further simplifying we get\n",
+ "s=(1-0.9375)*100\n",
+ "\n",
+ "#Weight Ratio \n",
+ "#Let w be the Weight ratio\n",
+ "#w=W_h*W_s**-1\n",
+ "\n",
+ "A_h=pi*32**-1*(d1**2-d2**2) #mm**2 #Area of Hollow shaft\n",
+ "A_s=pi*32**-1*d**2 #mm**2 #Area of solid shaft\n",
+ "\n",
+ "w=A_h*A_s**-1 \n",
+ "\n",
+ "#Result\n",
+ "print\"Loss in strength is\",round(s,2)\n",
+ "print\"Weight ratio is\",round(w,2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Loss in strength is 6.25\n",
+ "Weight ratio is 0.75\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.13,Page No.239"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "T=8 #KN-m #Torque \n",
+ "d=100 #mm #Diameter of portion AB\n",
+ "d1=100 #mm #External Diameter of Portion BC\n",
+ "d2=75 #mm #Internal Diameter of Portion BC\n",
+ "G=80 #KN/mm**2 #Modulus of Rigidity\n",
+ "L1=1500 #mm #Radial Distance of Portion AB\n",
+ "L2=2500 #mm #Radial Distance ofPortion BC\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "R=d*2**-1 #mm #Radius of shaft\n",
+ "\n",
+ "#For Portion AB,Polar Modulus\n",
+ "J1=pi*32**-1*d**4 #mm**4 \n",
+ "\n",
+ "#For Portion BC,Polar modulus \n",
+ "J2=pi*32**-1*(d1**4-d2**4) #mm**4\n",
+ "\n",
+ "#Now Max stress occurs in portion BC since max radial Distance is sme in both cases\n",
+ "q_max=T*J2**-1*R*10**6 #N/mm**2 \n",
+ "\n",
+ "#Let theta1 be the rotation in Portion AB and theta2 be the rotation in portion BC\n",
+ "theta1=T*L1*(G*J1)**-1 #Radians\n",
+ "theta2=T*L2*(G*J2)**-1 #Radians\n",
+ "\n",
+ "#Total Rotational at end C\n",
+ "theta=(theta1+theta2)*10**3 #Radians\n",
+ "\n",
+ "#Result\n",
+ "print\"Max stress induced is\",round(q_max,2),\"N/mm**2\"\n",
+ "print\"Angle of Twist is\",round(theta,3),\"radians\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Max stress induced is 59.6 N/mm**2\n",
+ "Angle of Twist is 0.053 radians\n"
+ ]
+ }
+ ],
+ "prompt_number": 23
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.14,Page No.240"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "q_b=80 #N/mm**2 #Shear stress in Brass\n",
+ "q_s=100 #N/mm**2 #Shear stress in Steel\n",
+ "G_b=40*10**3 #N/mm**2 \n",
+ "G_s=80*10**3 \n",
+ "L_b=1000 #mm #Length of brass shaft\n",
+ "L_s=1200 #mm #Length of steel shaft\n",
+ "d1=80 #mm #Diameter of brass shaft\n",
+ "d2=60 #mm #Diameter of steel shaft\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Polar modulus of brass rod\n",
+ "J_b=pi*32**-1*d1**4 #mm**4 \n",
+ "\n",
+ "#Polar modulus of steel rod\n",
+ "J_s=pi*32**-1*d2**4 #mm**4\n",
+ "\n",
+ "#Considering bras Rod:AB\n",
+ "T1=J_b*q_b*(d1*2**-1)**-1 #N-mm \n",
+ "\n",
+ "#Considering Steel Rod:BC\n",
+ "T2=J_s*q_s*(d2*2**-1)**-1 #N-mm\n",
+ "\n",
+ "#Max Torque that can be applied\n",
+ "T2\n",
+ "\n",
+ "#Let theta_b and theta_s be the rotations in Brass and steel respectively\n",
+ "theta_b=T2*L_b*(G_b*J_b)**-1 #Radians\n",
+ "theta_s=T2*L_s*(G_s*J_s)**-1 #Radians\n",
+ "\n",
+ "theta=theta_b+theta_s #Radians #Rotation of free end\n",
+ "\n",
+ "#Result\n",
+ "print\"Total of free end is\",round(theta,3),\"Radians\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Total of free end is 0.076 Radians\n"
+ ]
+ }
+ ],
+ "prompt_number": 36
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.15,Page No.241"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "G=80*10**3 #N/mm**2 #Modulus of Rigidity\n",
+ "d1=100 #mm #Outer diameter of hollow shft\n",
+ "d2=80 #mm #Inner diameter of hollow shaft\n",
+ "d=80 #mm #diameter of Solid shaft\n",
+ "d3=60 #mm #diameter of Solid shaft having L=0.5m\n",
+ "L1=300 #mm #Length of Hollow shaft\n",
+ "L2=400 #mm #Length of solid shaft\n",
+ "L3=500 #mm #LEngth of solid shaft of diameter 60mm\n",
+ "T1=2*10**6 #N-mm #Torsion in Shaft AB\n",
+ "T2=1*10**6 #N-mm #Torsion in shaft BC\n",
+ "T3=1*10**6 #N-mm #Torsion in shaft CD\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Now Polar modulus of section AB\n",
+ "J1=pi*32**-1*(d1**4-d2**4) #mm**4 \n",
+ "\n",
+ "#Polar modulus of section BC\n",
+ "J2=pi*32**-1*d**4 #mm**4\n",
+ "\n",
+ "#Polar modulus of section CD\n",
+ "J3=pi*32**-1*d3**4 #mm**4\n",
+ "\n",
+ "#Now angle of twist of AB\n",
+ "theta1=T1*L1*(G*J1)**-1 #radians\n",
+ "\n",
+ "#Angle of twist of BC\n",
+ "theta2=T2*L2*(G*J2)**-1 #radians\n",
+ "\n",
+ "#Angle of twist of CD\n",
+ "theta3=T3*L3*(G*J3)**-1 #radians\n",
+ "\n",
+ "#Angle of twist\n",
+ "theta=theta1-theta2+theta3 #Radians\n",
+ "\n",
+ "#Shear stress in AB From Torsion Equation\n",
+ "q_s1=T1*(d1*2**-1)*J1**-1 #N/mm**2 \n",
+ "\n",
+ "#Shear stress in BC\n",
+ "q_s2=T2*(d*2**-1)*J2**-1 #N/mm**2 \n",
+ "\n",
+ "#Shear stress in CD\n",
+ "q_s3=T3*(d3*2**-1)*J3**-1 #N-mm**2\n",
+ "\n",
+ "#As max shear stress occurs in portion CD,so consider CD\n",
+ "\n",
+ "#Result\n",
+ "print\"Angle of twist at free end is\",round(theta,5),\"Radian\"\n",
+ "print\"Max Shear stress\",round(q_s3,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Angle of twist at free end is 0.00496 Radian\n",
+ "Max Shear stress 23.58 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.16,Page No.242"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=1000 #mm #Length of bar\n",
+ "L1=600 #mm #Length of Bar AB\n",
+ "L2=400 #mm #Length of Bar BC\n",
+ "d1=60 #mm #Outer Diameter of bar BC\n",
+ "d2=30 #mm #Inner Diameter of bar BC\n",
+ "d=60 #mm #Diameter of bar AB\n",
+ "T=2*10**6 #N-mm #Total Torque\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Polar Modulus of Portion AB\n",
+ "J1=pi*32**-1*d**4 #mm*4\n",
+ "\n",
+ "#Polar Modulus of Portion BC\n",
+ "J2=pi*32**-1*(d1**4-d2**4) #mm**4\n",
+ "\n",
+ "#Let T1 be the torque resisted by bar AB and T2 be torque resisted by Bar BC\n",
+ "#Let theta1 and theta2 be the rotation of shaft in portion AB & BC\n",
+ "\n",
+ "#theta1=T1*L1*(G*J1)**-1 #radians\n",
+ "#After substituting values and further simplifying we get \n",
+ "#theta1=32*600*T1*(pi*60**4*G)**-1\n",
+ "\n",
+ "#theta2=T2*L*(J2*G)**-1 #Radians\n",
+ "#After substituting values and further simplifying we get \n",
+ "#theta2=32*400*T2*(pi*60**4*(1-0.5**4)*G)**-1 \n",
+ "\n",
+ "#Now For consistency of Deformation,theta1=theta2\n",
+ "#After substituting values and further simplifying we get \n",
+ "#T1=0.7111*T2 ..................................................(1)\n",
+ "\n",
+ "#But T1+T2=T=2*10**6 ...........................................(2)\n",
+ "#Substituting value of T1 in above equation\n",
+ "\n",
+ "T2=T*(0.7111+1)**-1\n",
+ "T1=0.71111*T2\n",
+ "\n",
+ "#Max stress in Portion AB\n",
+ "q_s1=T1*(d*2**-1)*(J1)**-1 #N/mm**2\n",
+ "\n",
+ "#Max stress in Portion BC\n",
+ "q_s2=T2*(d1*2**-1)*J2**-1 #N/mm**2 \n",
+ "\n",
+ "#Result\n",
+ "print\"Stresses Developed in Portion:AB\",round(q_s1,2),\"N/mm**2\"\n",
+ "print\" :BC\",round(q_s2,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Stresses Developed in Portion:AB 19.6 N/mm**2\n",
+ " :BC 29.4 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.17,Page No.243"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d1=80 #mm #External Diameter of Brass tube\n",
+ "d2=50 #mm #Internal Diameter of Brass tube\n",
+ "d=50 #mm #Diameter of steel Tube\n",
+ "G_b=40*10**3 #N/mm**2 #Modulus of Rigidity of brass tube\n",
+ "G_s=80*10**3 #N/mm**2 #Modulus of rigidity of steel tube\n",
+ "T=6*10**6 #N-mm #Torque\n",
+ "L=2000 #mm #Length of Tube\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Polar Modulus of brass tube\n",
+ "J1=pi*32**-1*(d1**4-d2**4) #mm**4 \n",
+ "\n",
+ "#Polar modulus of steel Tube\n",
+ "J2=pi*32**-1*d**4 #mm**4\n",
+ "\n",
+ "#Let T_s & T_b be the torque resisted by steel and brass respectively\n",
+ "#Then, T_b+T_s=T ............................................(1)\n",
+ "\n",
+ "#Since the angle of twist will be the same\n",
+ "#Theta1=Theta2\n",
+ "#After substituting values and further simplifying we get \n",
+ "#Ts=0.360*Tb ...........................................(2)\n",
+ "\n",
+ "#After substituting value of Ts in eqn 1 and further simplifying we get \n",
+ "T_b=T*(0.36+1)**-1 #N-mm\n",
+ "T_s=0.360*T_b\n",
+ "\n",
+ "#Let q_s and q_b be the max stress in steel and brass respectively\n",
+ "q_b=T_b*(d1*2**-1)*J1**-1 #N/mm**2\n",
+ "q_s=T_s*(d2*2**-1)*J2**-1 #N/mm**2\n",
+ "\n",
+ "#Since angle of twist in brass=angle of twist in steel\n",
+ "theta_s=T_s*L*(J2*G_s)**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"Stresses Developed in Materials are:Brass\",round(q_b,2),\"N/mm**2\"\n",
+ "print\" :Steel\",round(q_s,2),\"N/mm**2\"\n",
+ "print\"Angle of Twist in 2m Length\",round(theta_s,3),\"Radians\" "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Stresses Developed in Materials are:Brass 51.79 N/mm**2\n",
+ " :Steel 64.71 N/mm**2\n",
+ "Angle of Twist in 2m Length 0.065 Radians\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.18,Page No.245"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d1=60 #mm #External Diameter of aluminium Tube\n",
+ "d2=40 #mm #Internal Diameter of aluminium Tube\n",
+ "d=40 #mm #Diameter of steel tube\n",
+ "q_a=60 #N/mm**2 #Permissible stress in aluminium\n",
+ "q_s=100 #N/mm**2 #Permissible stress in steel tube\n",
+ "G_a=27*10**3 #N/mm**2 \n",
+ "G_s=80*10**3 #N/mm**2 \n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Polar modulus of aluminium Tube\n",
+ "J_a=pi*32**-1*(d1**4-d2**4) #mm**4\n",
+ "\n",
+ "#Polar Modulus of steel Tube\n",
+ "J_s=pi*32**-1*d**4 #mm**4\n",
+ "\n",
+ "#Now the angle of twist of steel tube = angle of twist of aluminium tube\n",
+ "#T_s*L_s*(J_s*theta_s)**-1=T_a*L_a*(J_a*theta_a)**-1\n",
+ "#After substituting values in above Equation and Further simplifyin we get\n",
+ "#T_s=0.7293*T_a .....................(1)\n",
+ "\n",
+ "#If steel Governs the resisting capacity\n",
+ "T_s1=q_s*J_s*(d*2**-1)**-1 #N-mm\n",
+ "T_a1=T_s1*0.7293**-1 #N-mm\n",
+ "T1=(T_s1+T_a1)*10**-6 #KN-m #Total Torque in steel Tube\n",
+ "\n",
+ "#If aluminium Governs the resisting capacity \n",
+ "T_a2=q_a*J_a*(d1*2**-1) #N-mm\n",
+ "T_s2=T_a2*0.7293 #N-mm\n",
+ "T2=(T_s2+T_a2)*10**-6 #KN-m #Total Torque in aluminium tube\n",
+ "\n",
+ "#Result\n",
+ "print\"Steel Governs the torque carrying capacity\",round(T1,2),\"KN-m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Steel Governs the torque carrying capacity 2.98 KN-m\n"
+ ]
+ }
+ ],
+ "prompt_number": 19
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.19,Page No.247"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "P=225*10**6 #N-mm/sec #Power Trasmitted\n",
+ "q_b=80 #N/mm**2 #Shear stress\n",
+ "n=200 #Rpm\n",
+ "q_k=100 #N/mm**2 #PErmissible stress in Keys\n",
+ "D=300 #mm #Diameter of bolt circle\n",
+ "L=150 #mm #Length of shear key\n",
+ "d=16 #mm #Diameterr of bolt\n",
+ "\n",
+ "#Calculations\n",
+ "T=60*P*(2*pi*n)**-1 #N-mm #Torque\n",
+ "\n",
+ "#Now From Torsion Formula\n",
+ "#T*J**-1=q_s*R**-1\n",
+ "#After substituting values we get\n",
+ "#T=pi*16*d**3*n\n",
+ "#After further simplifying we get\n",
+ "d1=(T*16*(pi*q_s)**-1)**0.33333\n",
+ "\n",
+ "#Let b be the width of shear Key\n",
+ "#T=q_k*L*b*R\n",
+ "#After simplifying further we get\n",
+ "b=T*(q_k*L*(d1*2**-1))**-1 #mm\n",
+ "\n",
+ "#Let n2 be the no. of bolts required at bolt circle of radius\n",
+ "R_b=D*2**-1 #mm \n",
+ "\n",
+ "n2=T*4*(q_b*pi*d**2*R_b)**-1\n",
+ "\n",
+ "#result\n",
+ "print\"Minimum no. of Bolts Required are\",round(n2,2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Minimum no. of Bolts Required are 4.45\n"
+ ]
+ }
+ ],
+ "prompt_number": 26
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.20,Page No.250"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "T=2*10**6 #N-mm #Torque transmitted\n",
+ "G=80*10**3 #N/mm**2 #Modulus of rigidity\n",
+ "d1=40 #mm \n",
+ "d2=80 #mm\n",
+ "r1=20 #mm\n",
+ "r2=40 #mm\n",
+ "L=2000 #mm #Length of shaft\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Angle of twist \n",
+ "theta=2*T*L*(r1**2+r1*r2+r2**2)*(3*pi*G*r2**3*r1**3)**-1 #radians\n",
+ "\n",
+ "#If the shaft is treated as shaft of average Diameter\n",
+ "d_avg=(d1+d2)*2**-1 #mm\n",
+ "\n",
+ "theta1=T*L*(G*pi*32**-1*d_avg**4)**-1 #Radians\n",
+ "\n",
+ "#Percentage Error\n",
+ "#Let Percentage Error be E\n",
+ "X=theta-theta1\n",
+ "E=(X*theta**-1)*100 \n",
+ "\n",
+ "#Result\n",
+ "print\"Percentage Error is\",round(E,2),\"%\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Percentage Error is 32.28 %\n"
+ ]
+ }
+ ],
+ "prompt_number": 42
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.21,Page No.252"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "G=80*10**3 #N/mm**2 \n",
+ "P=1*10**9 #N-mm/sec #Power\n",
+ "n=300 \n",
+ "d1=150 #mm #Outer Diameter\n",
+ "d2=120 #mm #Inner Diameter\n",
+ "L=2000 #mm #Length of circular shaft\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "T=P*60*(2*pi*n)**-1 #N-mm\n",
+ "\n",
+ "#Polar Modulus \n",
+ "J=pi*32**-1*(d1**4-d2**4) #mm**4\n",
+ "\n",
+ "q_s=T*J**-1*(d1*2**-1) #N/mm**2 \n",
+ "\n",
+ "\n",
+ "#Strain ENergy\n",
+ "U=q_s**2*(4*G)**-1*pi*4**-1*(d1**2-d2**2)*L\n",
+ "\n",
+ "#Result\n",
+ "print\"Max shear stress is\",round(q_s,2),\"N/mm**2\"\n",
+ "print\"Strain Energy stored in the shaft is\",round(U,2),\"N-mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Max shear stress is 81.36 N/mm**2\n",
+ "Strain Energy stored in the shaft is 263181.37 N-mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 51
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.22,Page No.254"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d=12 #mm #Diameter of helical spring\n",
+ "D=150 #mm #Mean Diameter\n",
+ "R=D*2**-1 #mm #Radius of helical spring\n",
+ "n=10 #no.of turns\n",
+ "G=80*10**3 #N/mm**2 \n",
+ "W=450 #N #Load\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Max shear stress \n",
+ "q_s=16*W*R*(pi*d**3)**-1 #N/mm**2\n",
+ "\n",
+ "#Strain Energy stored\n",
+ "U=32*W**2*R**3*n*(G*d**4)**-1 #N-mm\n",
+ "\n",
+ "#Deflection Produced\n",
+ "dell=64*W*R**3*n*(G*d**4)**-1 #mm\n",
+ "\n",
+ "#Stiffness Spring\n",
+ "k=W*dell**-1 #N/mm\n",
+ "\n",
+ "#Result\n",
+ "print\"Max shear stress is\",round(q_s,2),\"N/mm**2\"\n",
+ "print\"Strain Energy stored is\",round(U,2),\"N-mm\"\n",
+ "print\"Deflection Produced is\",round(dell,2),\"mm\"\n",
+ "print\"Stiffness spring is\",round(k,2),\"N/mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Max shear stress is 99.47 N/mm**2\n",
+ "Strain Energy stored is 16479.49 N-mm\n",
+ "Deflection Produced is 73.24 mm\n",
+ "Stiffness spring is 6.14 N/mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 53
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.23,Page No.255"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "K=5 #N/mm #Stiffness\n",
+ "L=100 #mm #Solid Length\n",
+ "q_s=60 #N/mm**2 #Max shear stress\n",
+ "W=200 #N #Max Load\n",
+ "G=80*10**3 #N/mm**2\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#K=W*dell**-1\n",
+ "#After substituting values and further simplifying we get\n",
+ "#d=0.004*R**3*n ........(1) #mm #Diameter of wire\n",
+ "#n=L*d**-1 ........(2)\n",
+ "\n",
+ "#From Shearing stress\n",
+ "#q_s=16*W*R*(pi*d**3)**-1 \n",
+ "#After substituting values and further simplifying we get\n",
+ "#d**4=0.004*R**3*n .................(4)\n",
+ "\n",
+ "#From Equation 1,2,3\n",
+ "#d**4=0.004*(0.0785*d**3)**3*100*d**-1\n",
+ "#after further simplifying we get\n",
+ "d=5168.101**0.25\n",
+ "n=100*d**-1\n",
+ "R=(d**4*(0.004*n)**-1)**0.3333\n",
+ "\n",
+ "#Result\n",
+ "print\"Diameter of Wire is\",round(d,2),\"mm\"\n",
+ "print\"No.of turns is\",round(n,2)\n",
+ "print\"Mean Radius of spring is\",round(R,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Diameter of Wire is 8.48 mm\n",
+ "No.fo turns is 11.79\n",
+ "Mean Radius of spring is 47.83 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 54
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.24,Page No.255"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "m=5*10**5 #Wagon Weighing\n",
+ "v=18*1000*36000**-1 \n",
+ "d=300 #mm #Diameter of Beffer springs\n",
+ "n=18 #no.of turns\n",
+ "G=80*10**3 #N/mm**2\n",
+ "dell=225\n",
+ "R=100 #mm #Mean Radius\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Energy of Wagon\n",
+ "E=m*v**2*(9.81*2)**-1 #N-mm\n",
+ "\n",
+ "#Load applied\n",
+ "W=dell*G*d**4*(64*R**3*n)**-1 #N \n",
+ "\n",
+ "#Energy each spring can absorb is\n",
+ "E2=W*dell*2**-1 #N-mm\n",
+ "\n",
+ "#No.of springs required to absorb energy of Wagon\n",
+ "n2=E*E2**-1 *10**7\n",
+ "\n",
+ "#Result\n",
+ "print\"No.of springs Required for Buffer is\",round(n2,2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "No.of springs Required for Buffer is 4.47\n"
+ ]
+ }
+ ],
+ "prompt_number": 66
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.25,Page No.259"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "b=180 #mm #width of flange\n",
+ "d=10 #mm #Depth of flange\n",
+ "t=10 #mm #Thickness of flange\n",
+ "D=400 #mm #Overall Depth \n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "I_xx=1*12**-1*(b*D**3-(b-t)*(D-2*d)**3)\n",
+ "I_yy=1*12**-1*((D-2*d)*t**3+2*t*b**3)\n",
+ "\n",
+ "#If warping is neglected\n",
+ "J=I_xx+I_yy #mm**4\n",
+ "\n",
+ "#Since b/d>1.6,we get\n",
+ "J2=1*3**-1*d**3*b*(1-0.63*d*b**-1)*2+1*3**-1*t**3*(D-2*d)*(1-0.63*t*b**-1)\n",
+ "\n",
+ "#Over Estimation of torsional Rigidity would have been \n",
+ "T=J*J2**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"Error in assessing torsional Rigidity if the warping is neglected is\",round(T,2)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Error in assessing torsional Rigidity if the warping is neglected is 808.28\n"
+ ]
+ }
+ ],
+ "prompt_number": 68
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.26,Page No.261"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d1=100 #mm #Outer Diameter\n",
+ "d2=95 #mm #Inner Diameter\n",
+ "T=2*10**6 #N-mm #Torque\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "J=pi*32**-1*(d1**4-d2**4) #mm**4 #Polar Modulus\n",
+ "\n",
+ "#Shear stress\n",
+ "q_max=T*J**-1*d1*2**-1 #N/mm**2 \n",
+ "\n",
+ "#Now theta*L**-1=T*(G*J)**-1\n",
+ "#After substituting values and further simplifying we get\n",
+ "#Let theta*L**-1=X\n",
+ "X=T*J**-1\n",
+ "\n",
+ "#Now Treating it as very thin walled tube\n",
+ "d=(d1+d2)*2**-1 #mm\n",
+ "\n",
+ "r=d*2**-1 \n",
+ "t=(d1-d2)*2**-1\n",
+ "q_max2=T*(2*pi*r**2*t)**-1 #N/mm**2\n",
+ "\n",
+ "X2=T*(2*pi*r**3*t)**-1 \n",
+ "\n",
+ "#Result\n",
+ "print\"When it is treated as hollow shaft:Max shear stress\",round(q_max,2),\"N/mm**2\"\n",
+ "print\" :Angle of Twist per unit Length\",round(X,3)\n",
+ "print\"When it is very thin Walled Tube :Max shear stress\",round(q_max2,2),\"N/mm**2\"\n",
+ "print\" :Angle of twist per Unit Length\",round(X2,3)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "When it is treated as hollow shaft:Max shear stress 54.91 N/mm**2\n",
+ " :Angle of Twist per unit Length 1.098\n",
+ "When it is very thin Walled Tube :Max shear stress 53.57 N/mm**2\n",
+ " :Angle of twist per Unit Length 1.099\n"
+ ]
+ }
+ ],
+ "prompt_number": 72
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.7_7.ipynb b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.7_7.ipynb new file mode 100644 index 00000000..b75d886a --- /dev/null +++ b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.7_7.ipynb @@ -0,0 +1,1328 @@ +{
+ "metadata": {
+ "name": "chapter no.7.ipynb"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter No.7:Compound Stresses And Strains"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.1,Page No.269"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "sigma1=30 #N/mm**2 #Stress in tension\n",
+ "d=20 #mm #Diameter \n",
+ "sigma2=90 #N/mm**2 #Max compressive stress\n",
+ "sigma3=25 #N/mm**2\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#In TEnsion\n",
+ "\n",
+ "#Corresponding stress in shear\n",
+ "P=sigma1*2**-1 #N/mm**2\n",
+ "\n",
+ "#Tensile force\n",
+ "F=pi*4**-1*d**2*sigma1\n",
+ "\n",
+ "#In Compression\n",
+ "\n",
+ "#Correspong shear stress\n",
+ "P2=sigma2*2**-1 #N/mm**2\n",
+ "\n",
+ "#Correspong compressive(axial) stress\n",
+ "p=2*sigma3 #N/mm**2 \n",
+ "\n",
+ "#Corresponding Compressive force\n",
+ "P3=p*pi*4**-1*d**2 #N\n",
+ "\n",
+ "#Result\n",
+ "print\"Failure Loads are:\",round(F,2),\"N\"\n",
+ "print\" :\",round(P3,2),\"N\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Failure Loads are: 9424.78 N\n",
+ " : 15707.96 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 24
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.7.2,Page No.270"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d=25 #mm #Diameter of circular bar\n",
+ "F=20*10**3 #N #Axial Force\n",
+ "theta=30 #Degree #angle \n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Axial stresses\n",
+ "p=F*(pi*4**-1*d**2)**-1 #N/mm**2\n",
+ "\n",
+ "#Normal Stress\n",
+ "p_n=p*(cos(30*pi*180**-1))**2\n",
+ "\n",
+ "#Tangential Stress\n",
+ "p_t=p*2**-1*sin(2*theta*pi*180**-1)\n",
+ "\n",
+ "#Max shear stress occurs on plane where theta2=45 \n",
+ "theta2=45\n",
+ "sigma_max=p*2**-1*sin(2*theta2*pi*180**-1)\n",
+ "\n",
+ "#Result\n",
+ "print\"Stresses developed on a plane making 30 degree is:\",round(p_n,2),\"N/mm**2\"\n",
+ "print\" :\",round(p_t,2),\"N/mm**2\"\n",
+ "print\"stress on max shear stress is\",round(sigma_max,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Stresses developed on a plane making 30 degree is: 30.56 N/mm**2\n",
+ " : 17.64 N/mm**2\n",
+ "stress on max shear stress is 20.37 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 38
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.7.3,Page No.272"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "theta=30 #degree\n",
+ "\n",
+ "#Stresses acting on material\n",
+ "p1=120 #N/mm**2\n",
+ "p2=80 #N/mm**2\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Normal Stress\n",
+ "P_n=(p1+p2)*2**-1+(p1-p2)*2**-1*cos(2*theta*pi*180**-1) #N/mm**2\n",
+ "\n",
+ "#Tangential stress\n",
+ "P_t=(p1-p2)*2**-1*sin(2*theta*pi*180**-1)\n",
+ "\n",
+ "#Resultant stress\n",
+ "P=(P_n**2+P_t**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Inclination to the plane\n",
+ "phi=arctan(P_n*P_t**-1)*(180*pi**-1)\n",
+ "\n",
+ "#Angle made by resultant with 120 #N/mm**2 stress\n",
+ "phi2=phi+theta #Degree\n",
+ "\n",
+ "#Result\n",
+ "print\"Normal Stress is\",round(P_n,2),\"N/mm**2\"\n",
+ "print\"Tangential Stress is\",round(P_t,2),\"N/mm**2\"\n",
+ "print\"Angle made by resultant\",round(phi2,2),\"Degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Normal Stress is 110.0 N/mm**2\n",
+ "Tangential Stress is 17.32 N/mm**2\n",
+ "Angle made by resultant 111.05 Degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 30
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.7.4,Page No.272"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#Direct Stresses\n",
+ "P1=60 #N/mm**2 \n",
+ "P2=100 #N/mm**2\n",
+ "\n",
+ "Theta=25 #Degree #Angle\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Normal Stress\n",
+ "P_n=(P1-P2)*2**-1+(P1+P2)*2**-1*cos(2*Theta*pi*180**-1) #N/mm**2\n",
+ "\n",
+ "#Tangential Stress\n",
+ "P_t=(P1+P2)*2**-1*sin(Theta*2*pi*180**-1) #N/mm**2\n",
+ "\n",
+ "#Resultant stress\n",
+ "P=(P_n**2+P_t**2)**0.5 #N/mm**2\n",
+ "\n",
+ "theta2=arctan(P_n*P_t**-1)*(180*pi**-1)\n",
+ "\n",
+ "#Result\n",
+ "print\"Stresses on the plane AC is:\",round(P_n,2),\"N/mm**2\"\n",
+ "print\" \",round(P_t,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Stresses on the plane AC is: 31.42 N/mm**2\n",
+ " 61.28 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 25
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.7.6,Page No.278"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#Stresses acting on material\n",
+ "p_x=180 #N/mm**2 \n",
+ "p_y=120 #N/mm**2\n",
+ "\n",
+ "q=80 #N/mm**2\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "theta=arctan(2*q*(p_x-p_y)**-1)*(180*pi**-1) #degrees\n",
+ "theta2=theta*2**-1 #Degrees\n",
+ "theta3=theta+180 #Degrees\n",
+ "theta4=theta3*2**-1 #Degrees\n",
+ "\n",
+ "#Stresses\n",
+ "p_1=(p_x+p_y)*2**-1+(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "p_2=(p_x+p_y)*2**-1-(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Max shear stress\n",
+ "q_max=(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnitude of Principal stress is:\",round(p_1,2),\"N/mm**2\"\n",
+ "print\" \",round(p_2,2),\"N/mm**2\"\n",
+ "print\"Magnitude of max shear stress is\",round(q_max,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnitude of Principal stress is: 235.44 N/mm**2\n",
+ " 64.56 N/mm**2\n",
+ "Magnitude of max shear stress is 85.44 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 40
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.7.7,Page No.279"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#stresses\n",
+ "p_x=60 #N/mm**2\n",
+ "p_y=-40 #N/mm**2\n",
+ "\n",
+ "q=10 #N/mm**2 #shear stress\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Principal Stresses\n",
+ "p1=(p_x+p_y)*2**-1+(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "p2=(p_x+p_y)*2**-1-(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Max shear stress\n",
+ "q_max=(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Inclination of principal stress to plane\n",
+ "theta=arctan(2*q*(p_x-p_y)**-1)*(180*pi**-1)#Degrees\n",
+ "theta2=(theta)*2**-1 #degrees\n",
+ "\n",
+ "theta3=(theta+180)*2**-1 #degrees\n",
+ "\n",
+ "#Result\n",
+ "print\"Principal Stresses are:\",round(p1,2),\"N/mm**2\"\n",
+ "print\" :\",round(p2,2),\"N/mm**2\"\n",
+ "print\"Max shear stresses\",round(q_max,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Principal Stresses are: 60.99 N/mm**2\n",
+ " : -40.99 N/mm**2\n",
+ "Max shear stresses 50.99 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.7.8,Page No.280"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#stresses\n",
+ "p_x=-120 #N/mm**2\n",
+ "p_y=-80 #N/mm**2\n",
+ "\n",
+ "q=-60 #N/mm**2 #shear stress\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Principal Stresses\n",
+ "p1=(p_x+p_y)*2**-1+(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "p2=(p_x+p_y)*2**-1-(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Max shear stress\n",
+ "q_max=(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Inclination of principal stress to plane\n",
+ "theta=arctan(2*q*(p_x-p_y)**-1)*(180*pi**-1)#Degrees\n",
+ "theta2=(theta)*2**-1 #degrees\n",
+ "\n",
+ "theta3=(theta+180)*2**-1 #degrees\n",
+ "\n",
+ "#Result\n",
+ "print\"Principal Stresses are:\",round(p1,2),\"N/mm**2\"\n",
+ "print\" :\",round(p2,2),\"N/mm**2\"\n",
+ "print\"Max shear stresses\",round(q_max,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Principal Stresses are: -36.75 N/mm**2\n",
+ " : -163.25 N/mm**2\n",
+ "Max shear stresses 63.25 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.7.9,Page No.282"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#stresses\n",
+ "p_x=-40 #N/mm**2\n",
+ "p_y=80 #N/mm**2\n",
+ "\n",
+ "q=48 #N/mm**2 #shear stress\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Max shear stress\n",
+ "q_max=((((p_x-p_y)*2**-1)**2)+q**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Inclination of principal stress to plane\n",
+ "theta=arctan(2*q*(p_x-p_y)**-1)*(180*pi**-1)#Degrees\n",
+ "theta2=(theta)*2**-1 #degrees\n",
+ "\n",
+ "theta3=(theta+180)*2**-1 #degrees\n",
+ "\n",
+ "#Normal Corresponding stress\n",
+ "p_n=(p_x+p_y)*2**-1+(p_x-p_y)*2**-1*cos(2*(theta2+45)*pi*180**-1)+q*sin(2*(theta2+45)*pi*180**-1) #Degrees\n",
+ "\n",
+ "#Resultant stress\n",
+ "p=((p_n**2+q_max**2)**0.5) #N/mm**2\n",
+ "\n",
+ "phi=arctan(p_n*q_max**-1)*(180*pi**-1) #Degrees\n",
+ "\n",
+ "#Inclination to the plane\n",
+ "alpha=round((theta2+45),2)+round(phi ,2)#Degree\n",
+ "\n",
+ "#Answer in book is incorrect of alpha ie41.25\n",
+ "\n",
+ "#Result\n",
+ "print\"Planes of max shear stress:\",round(p_n,2),\"N/mm**2\"\n",
+ "print\" \",round(q_max,2),\"N/mm*2\"\n",
+ "print\"Resultant Stress is\",round(p,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Planes of max shear stress: 20.0 N/mm**2\n",
+ " 76.84 N/mm*2\n",
+ "Resultant Stress is 79.4 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 40
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.7.10,Page No.283"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#Stresses\n",
+ "p_x=50*cos(35*pi*180**-1)\n",
+ "q=50*sin(35*pi*180**-1)\n",
+ "p_y=0\n",
+ "\n",
+ "theta=40 #Degrees #Plane AB inclined to vertical\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Normal Stress on AB\n",
+ "p_n=(p_x+p_y)*2**-1+(p_x-p_y)*2**-1*cos(2*theta*pi*180**-1)+q*sin(2*theta*pi*180**-1)\n",
+ "\n",
+ "#Tangential Stress on AB\n",
+ "p_t=(p_x-p_y)*2**-1*sin(2*theta*pi*180**-1)-q*cos(2*theta*pi*180**-1) #N/mm**2\n",
+ "\n",
+ "#Resultant stress\n",
+ "p=(p_n**2+p_t**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Angle of resultant\n",
+ "phi=arctan(p_n*p_t**-1)*(180*pi**-1) #degrees\n",
+ "phi2=phi+theta #Degrees\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnitude of resultant stress is\",round(p,2),\"N/mm**2\"\n",
+ "print\"Direction of Resultant stress is\",round(phi2,2),\"Degrees\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnitude of resultant stress is 54.44 N/mm**2\n",
+ "Direction of Resultant stress is 113.8 Degrees\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.7.12,Page No.285"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#Direct stresses\n",
+ "p_x=120 #N/mm**2 #Tensile stress\n",
+ "p_y=-100 #N/mm**2 #Compressive stress\n",
+ "p1=160 #N/mm**2 #Major principal stress\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let q be the shearing stress\n",
+ "\n",
+ "#p1=(p_x+p_y)*2**-1+((((p_x+p_y)*2**-1)**2)+q**2)**0.5\n",
+ "#After further simplifying we get\n",
+ "q=(p1-((p_x+p_y)*2**-1))**2-((p_x-p_y)*2**-1)**2 #N/mm**2\n",
+ "q2=(q)**0.5 #N/mm**2\n",
+ "\n",
+ "#Minimum Principal stress\n",
+ "p2=(p_x+p_y)*2**-1-(((p_x-p_y)*2**-1)**2+q2**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Max shearing stress\n",
+ "q_max=(((p_x-p_y)*2**-1)**2+q2**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Result\n",
+ "print\"Shearing stress of material\",round(q,2),\"N/mm**2\"\n",
+ "print\"Min Principal stress\",round(p2,2),\"N/mm**2\"\n",
+ "print\"Max shearing stress\",round(q_max,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Shearing stress of material 10400.0 N/mm**2\n",
+ "Min Principal stress -140.0 N/mm**2\n",
+ "Max shearing stress 150.0 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 19
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.7.14,Page No.291"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "F=40*10**3 #N #Shear Force\n",
+ "M=20*10**6 #Bending Moment\n",
+ "\n",
+ "#Rectangular section\n",
+ "b=100 #mm #Width\n",
+ "d=200 #mm #Depth\n",
+ "\n",
+ "x=20 #mm #Distance from Top surface upto point\n",
+ "y=80 #mm #Distance from point to Bottom\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "I=1*12**-1*b*d**3 #mm**4 #M.I\n",
+ "\n",
+ "#At 20 mm Below top Fibre\n",
+ "f_x=M*I**-1*y #N/mm**2 #Stress\n",
+ "\n",
+ "#Assuming sagging moment ,f_x is compressive p_x=f_x=-24 #N/mm**2\n",
+ "p_x=f_x=-24 #N/mm**2\n",
+ "\n",
+ "#Shearing stress\n",
+ "q=F*(b*I)**-1*(b*x*(b-x*2**-1)) #N/mm**2\n",
+ "\n",
+ "#Direct stresses\n",
+ "\n",
+ "p_y=0 #N/mm**2\n",
+ "\n",
+ "p1=(p_x+p_y)*2**-1+(((p_x+p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "p2=(p_x+p_y)*2**-1-(((p_x+p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Result\n",
+ "print\"Directions of principal stresses at a point below 20mm is:\",round(p1,2),\"N/mm**2\"\n",
+ "print\" \",round(p2,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Directions of principal stresses at a point below 20mm is: 0.05 N/mm**2\n",
+ " -24.05 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 36
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.7.15,Page No.292"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=4000 #mm #Span\n",
+ "W1=W2=W3=2*10**3 #N #Load\n",
+ "\n",
+ "#SEction of beam\n",
+ "b=100 #mm #Width\n",
+ "d=240 #mm #Dept\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let R_A and R_B be the reactions\n",
+ "R_A=R_B=(W1+W2+W3)*2**-1 #KN\n",
+ "\n",
+ "#Now at the section 1.5m from left support A\n",
+ "#Shear Force\n",
+ "F=R_A-W1 #KN\n",
+ "\n",
+ "#B.M\n",
+ "M=R_A*1.5-W1*0.5 #KN-m\n",
+ "\n",
+ "#M.I\n",
+ "I=1*12**-1*b*d**3 #mm**4\n",
+ "\n",
+ "#Bending stress\n",
+ "#f=M*I**-1*y\n",
+ "#After Sub values and further simplifying we get\n",
+ "#f=3.04*10**-2*y\n",
+ "\n",
+ "#As it varies Linearly\n",
+ "\n",
+ "#at distance 0 From NA \n",
+ "f1=0\n",
+ "#at distance 60 mm from NA\n",
+ "f2=1.823 #N/mm**2\n",
+ "#at distance 120 mm from NA\n",
+ "f3=3.646 #N/mm**2\n",
+ "\n",
+ "#Shearing stress\n",
+ "q=F*b*d*2**-1*d*4**-1*(b*I)**-1\n",
+ "\n",
+ "#At 60 mm above NA\n",
+ "q2=F*b*d*4**-1*(d*2**-1-d*8**-1)*(b*I)**-1\n",
+ "\n",
+ "#At 120 mm above NA\n",
+ "q3=0 \n",
+ "\n",
+ "#At NA element is under pure shear\n",
+ "p1=q #N/mm**2\n",
+ "p2=-q #N/mm**2 \n",
+ "\n",
+ "#Inclination of principal plane to vertical\n",
+ "#theta=2*q*0**-1\n",
+ "#Further simplifying we get\n",
+ "#theta=infinity\n",
+ "\n",
+ "#therefore\n",
+ "theta=90*2**-1 #degrees\n",
+ "theta2=270*2**-1 #degrees\n",
+ "\n",
+ "#At 60 mm From NA\n",
+ "p_x=-1.823 #N/mm**2 \n",
+ "p_y=0\n",
+ "q=0.0469 #N/mm**2\n",
+ "\n",
+ "#principal planes\n",
+ "P1=(p_x+p_y)*2**-1+(((p_x+p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "P2=(p_x+p_y)*2**-1-(((p_x+p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Principal planes inclination to hte plane of p_x is given by\n",
+ "theta3=(arctan(2*q*(p_x-p_y)**-1)*(180*pi**-1))\n",
+ "theta4=theta3*2**-1#degrees\n",
+ "\n",
+ "theta5=theta3+180 #Degrees\n",
+ "\n",
+ "#At 120 mm From N-A\n",
+ "p_x2=3.646 #N/mm**2\n",
+ "p_y2=0 #N/mm**2\n",
+ "q2=0 #N/mm**2\n",
+ "\n",
+ "P3=p_x2 #N/mm**2\n",
+ "P4=0 #N/mm**2\n",
+ "\n",
+ "#Answer for P2 at 60 mm from NA is incorrect\n",
+ "\n",
+ "#Result\n",
+ "print\"Principal Planes at 60 mm from NA:\",round(p_x,2),\"N/mm**2\"\n",
+ "print\" \",round(p_y,2),\"N/mm**2\"\n",
+ "print\"Principal Stresses at 60 mm From NA\",round(P1,4),\"N/mm**2\"\n",
+ "print\" \",round(P2,4),\"N/mm**2\"\n",
+ "print\"Principal Planes at 60 mm from NA:\",round(p_x2,4),\"N/mm**2\"\n",
+ "print\" \",round(p_y2,4),\"N/mm**2\"\n",
+ "print\"Principal Stresses at 60 mm From NA\",round(P3,4),\"N/mm**2\"\n",
+ "print\" \",round(P4,4),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Principal Planes at 60 mm from NA: -1.82 N/mm**2\n",
+ " 0.0 N/mm**2\n",
+ "Principal Stresses at 60 mm From NA 0.0012 N/mm**2\n",
+ " -1.8242 N/mm**2\n",
+ "Principal Planes at 60 mm from NA: 3.646 N/mm**2\n",
+ " 0.0 N/mm**2\n",
+ "Principal Stresses at 60 mm From NA 3.646 N/mm**2\n",
+ " 0.0 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 23
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.7.16,Page No.295"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=8000 #mm #Span of beam\n",
+ "w=40*10**6 #N/mm #udl\n",
+ "\n",
+ "#I-section\n",
+ "\n",
+ "#Flanges\n",
+ "b=100 #mm #Width\n",
+ "t=10 #mm #Thickness\n",
+ "\n",
+ "D=400 #mm #Overall Depth\n",
+ "t2=10 #mm #thickness of web\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let R_A and R_B be the Reactions at A & B respectively\n",
+ "R_A=w*2**-1*L*10**-9 #KN\n",
+ "\n",
+ "#Shear force at 2m for left support\n",
+ "F=R_A-2*w*10**-6 #KN\n",
+ "\n",
+ "#Bending Moment\n",
+ "M=R_A*2-2*w*10**-6 #KN-m\n",
+ "\n",
+ "#M.I\n",
+ "I=1*12**-1*b*D**3-1*12**-1*(b-t)*(D-2*t2)**3 #mm**4\n",
+ "\n",
+ "#Bending stress at 100 mm above N_A\n",
+ "f=M*10**6*I**-1*b\n",
+ "\n",
+ "#Shear stress \n",
+ "q=F*10**3*(t*I)**-1*(b*t*(D-t)*2**-1 +t2*(b-t2)*145) #N/mm**2\n",
+ "\n",
+ "p_x=-197.06 #N/mm**2 \n",
+ "p_y=0 #N/mm**2\n",
+ "q=21.38 #N/mm**2\n",
+ "\n",
+ "#Principal Stresses\n",
+ "\n",
+ "P1=(p_x+p_y)*2**-1+(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "P2=(p_x+p_y)*2**-1-(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Max shear stress\n",
+ "q_max=(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Result\n",
+ "print\"Principal Stresses are:\",round(P1,2),\"N/mm**2\"\n",
+ "print\" \",round(P2,2),\"N/mm**2\"\n",
+ "print\"Max shear stress\",round(q_max,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Principal Stresses are: 2.29 N/mm**2\n",
+ " -199.35 N/mm**2\n",
+ "Max shear stress 100.82 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 48
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.7.18,Page No.298"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d=100 #mm #Diameter of shaft\n",
+ "M=3*10**6 #N-mm #B.M\n",
+ "T=6*10**6 #N-mm #Twisting Moment\n",
+ "mu=0.3\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Max principal Stress\n",
+ "\n",
+ "P1=16*(pi*d**3)**-1*(M+(M**2+T**2)**0.5) #N/mm**2 \n",
+ "P2=16*(pi*d**3)**-1*(M-(M**2+T**2)**0.5) #N/mm**2 \n",
+ "\n",
+ "#Direct stress\n",
+ "P=round(P1,2)-mu*round(P2,2) #N/mm**2 \n",
+ "\n",
+ "#Result\n",
+ "print\"Principal stresses are:\",round(P1,2),\"N/mm**2\"\n",
+ "print\" :\",round(P2,2),\"N/mm**2\"\n",
+ "print\"Stress Producing the same strain is\",round(P,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Principal stresses are: 49.44 N/mm**2\n",
+ " : -18.89 N/mm**2\n",
+ "Stress Producing the same strain is 55.11 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 51
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.7.19,Page No.299"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d=75 #mm #diameter \n",
+ "P=30*10**6 #W #Power transmitted\n",
+ "W=6 #N-mm/sec #Load\n",
+ "L=1000 #mm \n",
+ "N=300 #r.p.m\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#B.M\n",
+ "M=W*L*4**-1 #N-mm\n",
+ "T=P*60*(2*pi*N)**-1 #Torque transmitted\n",
+ "\n",
+ "#M.I\n",
+ "I=pi*64**-1*d**4 #mm**4\n",
+ "\n",
+ "#Bending stress\n",
+ "f_A=M*I**-1*(d*2**-1) #N/mm**2\n",
+ "\n",
+ "#At A\n",
+ "p_x=f_A\n",
+ "p_y=0\n",
+ "\n",
+ "#Polar Modulus\n",
+ "J=pi*32**-1*d**4 #mm**4\n",
+ "\n",
+ "#Shearing stress\n",
+ "q=T*J**-1*(d*2**-1) #N/mm**2\n",
+ "\n",
+ "#Principal Stresses\n",
+ "P1=(p_x+p_y)*2**-1+(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "P2=(p_x+p_y)*2**-1-(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Max shear stress\n",
+ "q_max=(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Bending stress\n",
+ "p_x2=0\n",
+ "p_y2=0\n",
+ "\n",
+ "#Shearing stress\n",
+ "q2=T*J**-1*d*2**-1 #N/mm**2\n",
+ "\n",
+ "#Principal stresses\n",
+ "P3=(p_x2+p_y2)*2**-1+(((p_x2-p_y2)*2**-1)**2+q2**2)**0.5 #N/mm**2\n",
+ "P4=(p_x2+p_y2)*2**-1-(((p_x2-p_y2)*2**-1)**2+q2**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Max shear stress\n",
+ "q_max2=(((p_x2-p_y2)*2**-1)**2+q2**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Answer for Principal Stresses P1,P2 and Max stress i.e q_max is incorrect in Book\n",
+ "\n",
+ "#Result\n",
+ "print\"Principal Stresses at vertical Diameter:P1\",round(P1,2),\"N/mm**2\"\n",
+ "print\" :P2\",round(P2,2),\"N/mm**2\"\n",
+ "print\"Max stress at vertical Diameter : \",round(q_max,2),\"N/mm**2\"\n",
+ "print\"Principal Stresses at Horizontal Diameter:P3\",round(P3,2),\"N/mm**2\"\n",
+ "print\" :P4\",round(P4,2),\"N/mm**2\"\n",
+ "print\"Max stress at Horizontal Diameter : \",round(q_max2,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Principal Stresses at vertical Diameter:P1 11.55 N/mm**2\n",
+ " :P2 -11.51 N/mm**2\n",
+ "Max stress at vertical Diameter : 11.53 N/mm**2\n",
+ "Principal Stresses at Horizontal Diameter:P3 11.53 N/mm**2\n",
+ " :P4 -11.53 N/mm**2\n",
+ "Max stress at Horizontal Diameter : 11.53 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 20
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.7.20,Page No.302"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d1=100 #mm #External Diameter\n",
+ "d2=50 #mm #Internal Diameter\n",
+ "N=500 #mm #r.p.m\n",
+ "P=60*10**6 #N-mm/sec #Power\n",
+ "p=100 #N/mm**2 #principal stress\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#M.I\n",
+ "I=pi*(d1**4-d2**4)*64**-1 #mm**4\n",
+ "\n",
+ "#Bending Stress\n",
+ "#f=M*I*d1*2**-1 #N/mm**2\n",
+ "\n",
+ "#Principal Planes\n",
+ "#p_x=32*M*(pi*(d1**4-d2**4))*d1\n",
+ "#p_y=0\n",
+ "\n",
+ "#Shear stress\n",
+ "#q=T*J**-1*(d1*2**-1)\n",
+ "#After sub values and further simplifying we get\n",
+ "#q=16*T*d1*(pi*(d1**4-d2**4))*d1\n",
+ "\n",
+ "#Principal stresses\n",
+ "#P1=(p_x+p_y)*2**-1+(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "#After sub values and further simplifying we get\n",
+ "#P1=16*(pi*(d1**4-d2**4))*d1*(M+(M**2+t**2)**0.5) ...............(1)\n",
+ "\n",
+ "#P=2*pi*N*T*60**-1\n",
+ "#After sub values and further simplifying we get\n",
+ "T=P*60*(2*pi*N)**-1*10**-6 #N-mm\n",
+ "\n",
+ "#Again Sub values and further simplifying Equation 1 we get\n",
+ "M=(337.533)*(36.84)**-1 #KN-m\n",
+ "\n",
+ "#Min Principal stress\n",
+ "#P2=(p_x+p_y)*2**-1-(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "#Sub values and further simplifying we get\n",
+ "P2=16*(pi*(d1**4-d2**4))*d1*(M-(M**2+T**2)**0.5)*10**-11\n",
+ "\n",
+ "#Result\n",
+ "print\"Bending Moment safely applied to shaft is\",round(M,2),\"KN-m\"\n",
+ "print\"Min Principal Stress is\",round(P2,3),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Bending Moment safely applied to shaft is 9.16 KN-m\n",
+ "Min Principal Stress is -0.336 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 32
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.7.21,Page No.303"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d=150 #mm #Diameter\n",
+ "T=20*10**6 #N #Torque\n",
+ "M=12*10**6 #N-mm #B.M\n",
+ "F=200*10**3 #N #Axial Thrust\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#M.I\n",
+ "I=(pi*64**-1*d**4)\n",
+ "\n",
+ "#Bending stress \n",
+ "f_A=M*I**-1*(d*2**-1) #N/mm**2\n",
+ "f_B=-f_A #N/mm**2\n",
+ "\n",
+ "#Axial thrust due to thrust\n",
+ "sigma=F*(pi*4**-1*d**2)**-1\n",
+ "\n",
+ "#At A\n",
+ "p_x=f_A-sigma #N/mm**2\n",
+ "\n",
+ "#At B\n",
+ "p_x2=f_B-sigma #N/mm**2\n",
+ "\n",
+ "p_y=0 #At A and B\n",
+ "\n",
+ "#Polar Modulus\n",
+ "J=pi*32**-1*d**4 #mm**4\n",
+ "\n",
+ "#Shearing stress at A and B\n",
+ "q=T*J**-1*(d*2**-1) #N/mm**2\n",
+ "\n",
+ "\n",
+ "#Principal Stresses\n",
+ "#At A\n",
+ "P1=(p_x+p_y)*2**-1+(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "P2=(p_x+p_y)*2**-1-(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Max shear stress\n",
+ "q_max1=(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#At B\n",
+ "P1_2=(p_x2+p_y)*2**-1+(((p_x2-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "P2_2=(p_x2+p_y)*2**-1-(((p_x2-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Max shear stress\n",
+ "q_max2=(((p_x2-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "\n",
+ "\n",
+ "#Result\n",
+ "print\"MAx Principal Stresses:P1\",round(P1,2),\"N/mm**2\"\n",
+ "print\" :P2\",round(P2,2),\"N/mm**2\"\n",
+ "print\"Min Principal Stresses:P1_2\",round(P1_2,2),\"N/mm**2\"\n",
+ "print\" :P2_2\",round(P2_2,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "MAx Principal Stresses:P1 45.1 N/mm**2\n",
+ " :P2 -20.2 N/mm**2\n",
+ "Min Principal Stresses:P1_2 14.65 N/mm**2\n",
+ " :P2_2 -62.18 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 19
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.7.22,Page No.311"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#strains\n",
+ "e_A=500 #microns\n",
+ "e_B=250 #microns\n",
+ "e_C=-150 #microns\n",
+ "E=2*10**5 #N/mm**2 #Modulus of Elasticity\n",
+ "mu=0.3 #Poissoin's ratio\n",
+ "theta=45 #Degrees\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "e_x=e_A=500\n",
+ "e_45=e_B=250\n",
+ "e_y=e_C=-150 \n",
+ "\n",
+ "#e_45=(e_x+e_y)*2**-1+(e_x-e_y)*2**-1*cos(2*theta)+rho_x_y*2**-1*sin(2*theta)\n",
+ "#After sub values and further simplifying we get\n",
+ "rho_x_y=(e_45-(e_x+e_y)*2**-1-(e_x-e_y)*2**-1*cos(2*theta*pi*180**-1))*(sin(2*theta*pi*180**-1))**-1*2\n",
+ "\n",
+ "#Principal strains are given by\n",
+ "e1=(e_x+e_y)*2**-1+(((e_x-e_y)*2**-1)**2+(rho_x_y*2**-1)**2)**0.5 #microns\n",
+ "e2=(e_x+e_y)*2**-1-(((e_x-e_y)*2**-1)**2+(rho_x_y*2**-1)**2)**0.5 #microns\n",
+ "\n",
+ "#Principal Stresses\n",
+ "sigma1=E*(e1+mu*e2)*(1-mu**2)**-1*10**-6 #N/mm**2\n",
+ "sigma2=E*(e2+mu*e1)*(1-mu**2)**-1*10**-6 #N/mm**2\n",
+ "\n",
+ "#Result\n",
+ "print\"Principal Strains are:e1\",round(e1,2),\"N/mm**2\"\n",
+ "print\" :e2\",round(e2,2),\"N/mm**2\"\n",
+ "print\"Principal Stresses are:sigma1\",round(sigma1,2),\"N/mm**2\"\n",
+ "print\" :sigma2\",round(sigma2,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Principal Strains are:e1 508.54 N/mm**2\n",
+ " :e2 -158.54 N/mm**2\n",
+ "Principal Stresses are:sigma1 101.31 N/mm**2\n",
+ " :sigma2 -1.31 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.7.23,Page No.313"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#Strains\n",
+ "e_A=600 #microns\n",
+ "e_B=-450 #microns\n",
+ "e_C=100 #micron\n",
+ "E=2*10**5 #N/mm**2 #Modulus of Elasticity\n",
+ "mu=0.3 #Poissoin's ratio\n",
+ "theta=240\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "e_x=e_A=600\n",
+ "\n",
+ "#e_A=(e_x+e_y)*2**-1+(e_x-e_y)*2**-1*cos(theta)+rho_x_y*2**-1*sin(theta)\n",
+ "#After sub values and further simplifying we get\n",
+ "#-450=(e_x+e_y)*2**-1-(e_x-e_y)*2**-1*(0.5)-0.866*2**-1*rho_x_y .....................(1)\n",
+ "\n",
+ "#e_C=(e_x+e_y)*2**-1+(e_x-e_y)*2**-1*cos(2*theta)+rho_x_y*2**-1*sin(2*theta)\n",
+ "#After sub values and further simplifying we get\n",
+ "#100=(e_x+e_y)*2**-1-0.5*(e_x-e_y)*2**-1*(0.5)-0.866*2**-1*rho_x_y .....................(2)\n",
+ "\n",
+ "#Adding Equation 1 and 2 we get equations as\n",
+ "#-350=e_x+e_y-(e_x-e_y)*2**-1 ...............(3)\n",
+ "#Further simplifying we get\n",
+ "\n",
+ "e_y=(-700-e_x)*3**-1 #micron \n",
+ "\n",
+ "rho_x_y=(e_C-(e_x+e_y)*2**-1-(e_x-e_y)*2**-1*cos(2*theta*pi*180**-1))*(sin(2*theta*pi*180**-1))**-1*2 #micron\n",
+ "\n",
+ "#Principal strains\n",
+ "e1=(e_x+e_y)*2**-1-(((e_x-e_y)*2**-1)**2+(rho_x_y*2**-1)**2)**0.5 #microns\n",
+ "e2=(e_x+e_y)*2**-1+(((e_x-e_y)*2**-1)**2+(rho_x_y*2**-1)**2)**0.5 #microns\n",
+ "\n",
+ "#Principal Stresses\n",
+ "sigma1=E*(e1+mu*e2)*(1-mu**2)**-1*10**-6 #N/mm**2\n",
+ "sigma2=E*(e2+mu*e1)*(1-mu**2)**-1*10**-6 #N/mm**2\n",
+ "\n",
+ "\n",
+ "#Result\n",
+ "print\"Principal Stresses are:sigma1\",round(sigma1,2),\"N/mm**2\"\n",
+ "print\" :sigma2\",round(sigma2,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Principal Stresses are:sigma1 -69.49 N/mm**2\n",
+ " :sigma2 117.11 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.8_7.ipynb b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.8_7.ipynb new file mode 100644 index 00000000..08f57912 --- /dev/null +++ b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.8_7.ipynb @@ -0,0 +1,1527 @@ +{
+ "metadata": {
+ "name": "chapter no.8.ipynb"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter No.8:Thin And Thick Cyclinders And Spheres"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Example 8.8.1,Page No.322"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=3000 #mm #Length\n",
+ "d1=1000 #mm #Internal diameter\n",
+ "t=15 #mm #Thickness\n",
+ "P=1.5 #N/mm**2 #Fluid Pressure\n",
+ "E=2*10**5 #n/mm**2 #Modulus of elasticity\n",
+ "mu=0.3 #Poissoin's ratio\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Hoop stress\n",
+ "f1=P*d1*(2*t)**-1 #N/mm**2\n",
+ "\n",
+ "#Longitudinal Stress\n",
+ "f2=P*d1*(4*t)**-1 #N/mm**2\n",
+ "\n",
+ "#Max shear stress\n",
+ "q_max=(f1-f2)*2**-1 #N/mm**2\n",
+ "\n",
+ "#Diametrical Strain\n",
+ "#Let e1=dell_d*d**-1 .....................(1)\n",
+ "e1=(f1-mu*f2)*E**-1 \n",
+ "\n",
+ "#Sub values in equation 1 and further simplifying we get\n",
+ "dell_d=e1*d1 #mm\n",
+ "\n",
+ "#Longitudinal strain\n",
+ "#e2=dell_L*L**-1 ......................(2)\n",
+ "e2=(f2-mu*f1)*E**-1 \n",
+ "\n",
+ "#Sub values in equation 2 and further simplifying we get\n",
+ "dell_L=e2*L #mm\n",
+ "\n",
+ "#Change in Volume \n",
+ "#Let Z=dell_V*V**-1 ................(3)\n",
+ "Z=2*e1+e2\n",
+ "\n",
+ "#Sub values in equation 3 and further simplifying we get\n",
+ "dell_V=Z*pi*4**-1*d1**2*L\n",
+ "\n",
+ "#Result\n",
+ "print\"Max Intensity of shear stress\",round(q_max,2),\"N/mm**2\"\n",
+ "print\"Change in the Dimensions of the shell is:dell_d\",round(dell_d,2),\"mm\"\n",
+ "print\" :dell_L\",round(dell_L,2),\"mm\"\n",
+ "print\" :dell_V\",round(dell_V,2),\"mm**3\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Max Intensity of shear stress 12.5 N/mm**2\n",
+ "Change in the Dimensions of the shell is:dell_d 0.21 mm\n",
+ " :dell_L 0.15 mm\n",
+ " :dell_V 1119192.38 mm**3\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Example 8.8.2,Page No.323"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=2000 #mm #Length\n",
+ "d=200 #mm # diameter\n",
+ "t=10 #mm #Thickness\n",
+ "dell_V=25000 #mm**3 #Additional volume\n",
+ "E=2*10**5 #n/mm**2 #Modulus of elasticity\n",
+ "mu=0.3 #Poissoin's ratio\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let p be the pressure developed\n",
+ "\n",
+ "#Circumferential Stress\n",
+ "\n",
+ "#f1=p*d*(2*t)**-1 #N/mm**2\n",
+ "#After sub values and further simplifying\n",
+ "#f1=10*p\n",
+ "\n",
+ "#f1=p*d*(4*t)**-1 #N/mm**2\n",
+ "#After sub values and further simplifying\n",
+ "#f1=5*p\n",
+ "\n",
+ "#Diameterical strain = Circumferential stress\n",
+ "#Let X=dell_d*d**-1 ................................(1)\n",
+ "#X=e1=(f1-mu*f2)*E**-1 \n",
+ "#After sub values and further simplifying\n",
+ "#e1=8.5*p*E**-1\n",
+ "\n",
+ "#Longitudinal strain\n",
+ "#Let Y=dell_L*L**-1 ......................................(2)\n",
+ "#Y=e2=(f2-mu*f1)*E**-1 \n",
+ "#After sub values and further simplifying\n",
+ "#e2=2*p*E**-1\n",
+ "\n",
+ "#Volumetric strain\n",
+ "#Let X=dell_V*V**-1 \n",
+ "#X=2*e1+e2\n",
+ "#After sub values and further simplifying\n",
+ "#X=19*p*E**-1\n",
+ "#After further simplifying we get\n",
+ "p=dell_V*(pi*4**-1*d**2*L)**-1*E*19**-1 #N/mm**2\n",
+ "\n",
+ "#Hoop Stress\n",
+ "f1=p*d*(2*t)**-1\n",
+ "\n",
+ "X=e1=8.5*p*E**-1\n",
+ "#Sub value of X in equation 1 we get\n",
+ "dell_d=8.5*p*E**-1*d\n",
+ "\n",
+ "Y=e2=2*p*E**-1\n",
+ "#Sub value of Y in equation 2 we get\n",
+ "dell_L=2*p*E**-1*L\n",
+ "\n",
+ "#Result\n",
+ "print\"Pressure Developed is\",round(p,2),\"N/mm**2\"\n",
+ "print\"Hoop stress Developed is\",round(f1,2),\"N/mm**2\"\n",
+ "print\"Change in diameter is\",round(dell_d,2),\"mm\"\n",
+ "print\"Change in Length is\",round(dell_L,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Pressure Developed is 4.19 N/mm**2\n",
+ "Hoop stress Developed is 41.88 N/mm**2\n",
+ "Change in diameter is 0.04 mm\n",
+ "Change in Length is 0.08 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 22
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Example 8.8.3,Page No.324"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d=750 #mm #Diameter of water supply pipes\n",
+ "h=50*10**3 #mm #Water head\n",
+ "sigma=20 #N/mm**2 #Permissible stress\n",
+ "rho=9810*10**-9 #N/mm**3\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Pressure of water\n",
+ "P=rho*h #N/mm**2\n",
+ "\n",
+ "#Stress\n",
+ "#sigma=p*d*(2*t)**-1\n",
+ "#After further simplifying\n",
+ "t=P*d*(2*sigma)**-1 #mm \n",
+ "\n",
+ "#Result\n",
+ "print\"Thickness of seamless pipe is\",round(t,3),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Thickness of seamless pipe is 9.197 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 37
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Example 8.8.4,Page No.326"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d=2500 #mm #Diameter of riveted boiler\n",
+ "P=1 #N/mm**2 #Pressure\n",
+ "rho1=0.7 #Percent efficiency\n",
+ "rho2=0.4 #Circumferential joints\n",
+ "sigma=150 #N/mm**2 #Permissible stress\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Equating Bursting force to longitudinal joint strength ,we get\n",
+ "#p*d*L=rho1*2*t*L*sigma\n",
+ "#After rearranging and further simplifying we get\n",
+ "t=P*d*(2*sigma*rho1)**-1 #mm\n",
+ "\n",
+ "#Considering Longitudinal force\n",
+ "#pi*d**2*4**-1*P=rho2*pi*d*t*sigma\n",
+ "#After rearranging and further simplifying we get\n",
+ "t2=P*d*(4*sigma*rho2)**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"Thickness of plate required is\",round(t,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Thickness of plate required is 11.9 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 45
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Example 8.8.5,Page No.326"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#Boiler Dimensions\n",
+ "t=16 #mm #Thickness\n",
+ "p=2 #N/mm**2 #internal pressure\n",
+ "f=150 #N/mm**2 #Permissible stress\n",
+ "rho1=0.75 #Longitudinal joints\n",
+ "rho2=0.45 #circumferential joints\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Equating Bursting force to longitudinal joint strength ,we get\n",
+ "d1=rho1*2*t*f*p**-1 #mm\n",
+ "\n",
+ "#Considering circumferential strength \n",
+ "d2=4*rho2*t*f*p**-1 #mm\n",
+ "\n",
+ "#Result\n",
+ "print\"Largest diameter of Boiler is\",round(d1,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Largest diameter of Boiler is 1800.0 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 52
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Example 8.8.6,Page No.329"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d=250 #mm #Diameter iron pipe\n",
+ "t=10 #mm #Thickness\n",
+ "d2=6 #mm #Diameter of steel\n",
+ "p=80 #N/mm**2 #stress\n",
+ "P=3 #N/mm**2 #Pressure\n",
+ "E_c=1*10**5 #N/mm**2\n",
+ "mu=0.3 #poissoin's ratio\n",
+ "E_s=2*10**5 #N/mm**2\n",
+ "n=1 #No.of wires\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "L=6 #mm #Length of cyclinder\n",
+ "\n",
+ "#Force Exerted by steel wire at diameterical section\n",
+ "F=p*2*pi*d2**2*1*4**-1 #N\n",
+ "\n",
+ "#Initial stress in cyclinder\n",
+ "f_c=F*(2*t*d2)**-1 #N/mm**2\n",
+ "\n",
+ "#LEt due to fluid pressure alone stresses developed in steel wire be F_w and in cyclinder f1 and f2\n",
+ "f2=P*d*(4*t)**-1 #N/mm**2\n",
+ "\n",
+ "#Considering the equilibrium of half the cyclinder, 6mm long we get\n",
+ "#F_w*2*pi*4**-1*d2**2*n+f1*2*t*d2=P*d*d2\n",
+ "#After further simplifying we get\n",
+ "#F_w+2.122*f1=79.58 . ......................................(1)\n",
+ "\n",
+ "#Equating strain in wire to circumferential strain in cyclinder \n",
+ "#F_w=(f1-mu*f2)*E_s*E_c**-1 #N/mm**2\n",
+ "#After further simplifying we get\n",
+ "#F_w=2*f1-11.25 ....................................(2)\n",
+ "\n",
+ "#Sub in equation in1 we get\n",
+ "f1=(79.58+11.25)*(4.122)**-1 #N/mm**2\n",
+ "F_w=2*f1-11.25 #N/mm**2\n",
+ "\n",
+ "#Final stresses\n",
+ "#1) In steel Wire\n",
+ "sigma=F_w+p #N/mm**2\n",
+ "\n",
+ "#2) In Cyclinder\n",
+ "sigma2=f1-f_c\n",
+ "\n",
+ "#Result\n",
+ "print\"Final Stresses developed in:cyclinder is\",round(sigma,2),\"N/mm**2\"\n",
+ "print\" :Steel is\",round(sigma2,2),\"N/mm**2\" "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Final Stresses developed in:cyclinder is 112.82 N/mm**2\n",
+ " :Steel is -15.66 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Example 8.8.7,Page No.332"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d=750 #mm #Diameter of shell\n",
+ "t=8 #mm #THickness\n",
+ "p=2.5 #N/mm**2\n",
+ "E=2*10**5 #N/mm**2\n",
+ "mu=0.25 #Poissoin's ratio\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Hoop stress\n",
+ "f1=f2=p*d*(4*t)**-1 #N/mm**2\n",
+ "\n",
+ "#Change in Diameter\n",
+ "dell_d=d*p*d*(1-mu)*(4*t*E)**-1 #mm\n",
+ "\n",
+ "#Change in Volume\n",
+ "dell_V=3*p*d*(1-mu)*(4*t*E)**-1*pi*6**-1*d**3\n",
+ "\n",
+ "#Answer for Change in diameter is incorrect in book\n",
+ "\n",
+ "#Result\n",
+ "print\"Stress introduced is\",round(f1,2),\"N/mm**2\"\n",
+ "print\"Change in Diameter is\",round(dell_d,2),\"N/mm**2\"\n",
+ "print\"Change in Volume is\",round(dell_V,2),\"mm**3\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Stress introduced is 58.59 N/mm**2\n",
+ "Change in Diameter is 0.16 N/mm**2\n",
+ "Change in Volume is 145608.33 mm**3\n"
+ ]
+ }
+ ],
+ "prompt_number": 56
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Example 8.8.8,Page No.333"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d=600 #mm #Diameter of sherical shell\n",
+ "t=10 #mm #Thickness\n",
+ "f=80 #N/mm**2 #Permissible stress\n",
+ "rho=0.75 #Efficiency joint\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Max Pressure\n",
+ "p=f*4*t*rho*d**-1 #N/mm**2\n",
+ "\n",
+ "#Result\n",
+ "print\"Max Pressure is\",round(p,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Max Pressure is 4.0 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 60
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Example 8.8.9,Page No.333"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=1000 #mm #Length of shell\n",
+ "d=200 #mm #Diameter\n",
+ "t=6 #mm #Thickness\n",
+ "p=1.5 #N/mm**2 #Internal Pressure\n",
+ "E=2*10**5 #N/mm**2\n",
+ "mu=0.25 #Poissoin's Ratio\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Change in Volume of sphere\n",
+ "dell_V_s=3*p*d*(1-mu)*(4*t*E)**-1*pi*6**-1*d**3\n",
+ "\n",
+ "#Hoop stress\n",
+ "f1=p*d*(2*t)**-1 #N/mm**2\n",
+ "\n",
+ "#Longitudinal stress\n",
+ "f2=p*d*(4*t)**-1 #N/mm**2\n",
+ "\n",
+ "#Principal strain\n",
+ "e1=(f1-mu*f2)*E**-1\n",
+ "e2=(f2-mu*f1)*E**-1\n",
+ "\n",
+ "V_c=1000 #mm**3\n",
+ "\n",
+ "#Change in Volume of cyclinder\n",
+ "dell_V_c=(2*e1+e2)*pi*4**-1*d**2*L\n",
+ "\n",
+ "#Total Change in Diameter\n",
+ "dell_V=dell_V_s+dell_V_c #mm**3\n",
+ "\n",
+ "#Result\n",
+ "print\"Change in Volume is\",round(dell_V,2),\"mm**3\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Change in Volume is 8443.03 mm**3\n"
+ ]
+ }
+ ],
+ "prompt_number": 66
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Example 8.8.10,Page No.337"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "%matplotlib inline\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d1=400 #mm #Internal Diameter\n",
+ "t=100 #mm #Thickness\n",
+ "p=80 #N/mm**2 #Fluid pressure\n",
+ "\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Internal Radius\n",
+ "r1=d1*2**-1 #mm\n",
+ "\n",
+ "#Outer Radius\n",
+ "r_o=r1+t #mm\n",
+ "\n",
+ "p1=80 #N/mm**2\n",
+ "p2=0\n",
+ "\n",
+ "#Now From Lame's Euation\n",
+ "#p_x=b*(x**2)**-1-a\n",
+ "#at x=200 #mm \n",
+ "p_x=80 #N/mm**2\n",
+ "#80=b*(200**2)**-1-a ..........................(1)\n",
+ "\n",
+ "#at x=300 #mm\n",
+ "#p_x2=0\n",
+ "#0=b*(300**2)**-1-a ...........................(2)\n",
+ "\n",
+ "#Sub equation 2 from 1\n",
+ "#80=b*(200**2)**-1-b*(300**2)**-1\n",
+ "#After Further simplifying we get\n",
+ "b=(50000)**-1*(200**2*300**2*80)\n",
+ "\n",
+ "#From equation 2 we get\n",
+ "a=b*(300**2)**-1\n",
+ "\n",
+ "#Variation of radial pressure p_x;\n",
+ "#p_x=b*(x**2)**-1-a\n",
+ "#After sub values and further simplifying we get\n",
+ "\n",
+ "#Radial pressure Variation\n",
+ "#At \n",
+ "x=200 #mm\n",
+ "p_x=b*(x**2)**-1-a #N/mm**2\n",
+ "\n",
+ "#At\n",
+ "x2=250 #mm\n",
+ "p_x2=b*(x2**2)**-1-a #N/mm**2\n",
+ "\n",
+ "#At \n",
+ "x3=300 #mm\n",
+ "p_x3=b*(x3**2)**-1-a #N/mm**2\n",
+ "\n",
+ "\n",
+ "#Hoop stress Distribution\n",
+ "#Variation of F_x\n",
+ "\n",
+ "#At \n",
+ "x=200 #mm\n",
+ "F_x=b*(x**2)**-1+a #N/mm**2\n",
+ "\n",
+ "#At\n",
+ "x2=250 #mm\n",
+ "F_x2=b*(x2**2)**-1+a #N/mm**2\n",
+ "\n",
+ "#At\n",
+ "x3=300 #mm\n",
+ "F_x3=b*(x3**2)**-1+a #N/mm**2\n",
+ "\n",
+ "#Result\n",
+ "print\"Max Hoop stress is\",round(F_x,2),\"N/mm**2\"\n",
+ "print\"Min Hoop stress is\",round(F_x3,2),\"N/mm**2\"\n",
+ "print\"Plot of Hoop stress\"\n",
+ "\n",
+ "#Plotting Variation of hoop stress\n",
+ "\n",
+ "X1=[x,x2,x3]\n",
+ "Y1=[p_x,p_x2,p_x3]\n",
+ "Y2=[-F_x,-F_x2,-F_x3]\n",
+ "Z1=[0,0,0]\n",
+ "plt.plot(X1,Y1,X1,Y2,X1,Z1)\n",
+ "plt.xlabel(\"Length x in mm\")\n",
+ "plt.ylabel(\"Radial Stress Distribution & Hoop Stress Distribution in N/mm**2\")\n",
+ "plt.show()\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Max Hoop stress is 208.0 N/mm**2\n",
+ "Min Hoop stress is 128.0 N/mm**2\n",
+ "Plot of Hoop stress\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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skdWFW7ZsQffu3XH8+HGMGzcOY8aMAaA9wnfKlCkIDg7GmDFjEBcXJ7WK4uLi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\nbm5uYtOmTUII3b27hBBCoVCIHTt2CCG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+ "text": [
+ "<matplotlib.figure.Figure at 0x50e8310>"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Example 8.8.11,Page No.338"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d_o=300 #mm #Outside diameter \n",
+ "d2=200 #mm #Internal Diameter\n",
+ "p=14 #N/mm**2 #internal Fluid pressure\n",
+ "t=50 #mm #Thickness\n",
+ "r_o=150 #mm #Outside Diameter\n",
+ "r2=100 #mm #Internal Diameter\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#From Lame's Equation\n",
+ "#p_x=b*(x**2)**-1-a #N/mm**2 ...................(1)\n",
+ "#F_x=b*(x**2)**-1+a #N/mm**2 ...................(2)\n",
+ "\n",
+ "#At \n",
+ "x=r2=100 #mm\n",
+ "p_x=14 #N/mm**2\n",
+ "\n",
+ "#Sub value of p_x in equation 1 we get\n",
+ "#14=(100)**-1*b-a ............................(3)\n",
+ "\n",
+ "#At\n",
+ "x2=r_o=150 #mm\n",
+ "p_x2=0 #N/mm**2\n",
+ "\n",
+ "#Sub value in equation 1 we get\n",
+ "#0=b*(150**2)**-1-a ......................(4)\n",
+ "\n",
+ "#From Equations 3 and 4 we get\n",
+ "#14=b*(100**2)**-1-b*(100**2)**-1\n",
+ "#After sub values and further simplifying we get\n",
+ "b=14*100**2*150**2*(150**2-100**2)**-1\n",
+ "\n",
+ "#From equation 4 we get\n",
+ "a=b*(150**2)**-1\n",
+ "\n",
+ "#Hoop Stress\n",
+ "#F_x=b*(x**2)**-1+a #N/mm**2\n",
+ "\n",
+ "#At \n",
+ "x=100 #mm\n",
+ "F_x=b*(x**2)**-1+a #N/mm**2\n",
+ "\n",
+ "#At\n",
+ "x2=125 #mm\n",
+ "F_x2=b*(x2**2)**-1+a #N/mm**2\n",
+ "\n",
+ "#At\n",
+ "x3=150 #mm\n",
+ "F_x3=b*(x3**2)**-1+a #N/mm**2\n",
+ "\n",
+ "#If thin Cyclindrical shell theory is used,hoop stress is uniform and is given by\n",
+ "F=p*d2*(2*t)**-1 #N/mm**2\n",
+ "\n",
+ "#Percentage error in estimating max hoop tension\n",
+ "E=(F_x-F)*F_x**-1*100 #%\n",
+ "\n",
+ "#Result\n",
+ "print\"Max Hoop Stress Developed in the cross-section is\",round(F,2),\"N/mm**2\"\n",
+ "print\"Plot of Variation of hoop stress\"\n",
+ "\n",
+ "#Plotting Variation of hoop stress\n",
+ "\n",
+ "X1=[x,x2,x3]\n",
+ "Y1=[F_x,F_x2,F_x3]\n",
+ "Z1=[0,0,0]\n",
+ "plt.plot(X1,Y1,X1,Z1)\n",
+ "plt.xlabel(\"Length x in mm\")\n",
+ "plt.ylabel(\"Radial Stress Distribution & Hoop Stress Distribution in N/mm**2\")\n",
+ "plt.show()\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Max Hoop Stress Developed in the cross-section is 28.0 N/mm**2\n",
+ "Plot of Variation of hoop stress\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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29vaQJAknTpx4bAP379/HiBEjMGzYMMycOdNwT41GA1dXV2RkZGDgwIEcMiIi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\nSUVORK5CYII=\n",
+ "text": [
+ "<matplotlib.figure.Figure at 0x58c2e30>"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Example 8.8.12,Page No.339"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d_o=300 #mm #Outside diameter \n",
+ "d2=200 #mm #Internal Diameter\n",
+ "p=12 #N/mm**2 #internal Fluid pressure\n",
+ "F_max=16 #N/mm**2 #Tensile stress\n",
+ "r_o=150 #mm #Outside Diameter\n",
+ "r2=100 #mm #Internal Diameter\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let p_o be the External Pressure applied.\n",
+ "#From LLame's theorem\n",
+ "#p_x=b*(x**2)**-1-a ..............(1)\n",
+ "#F_x=b*(x**2)**-1+a ...........................(2)\n",
+ "\n",
+ "#Now At\n",
+ "x=100 #mm\n",
+ "p_x=12 #N/mm**2\n",
+ "#sub in equation 1 we get\n",
+ "#12=b*(100**2)**-1-a . ..................(3)\n",
+ "\n",
+ "#The Max Hoop stress occurs at least value of x where\n",
+ "x=r1=100 #mm\n",
+ "#16=b*(100**2)**-1+a .......................(4)\n",
+ "\n",
+ "#From Equations 1 and 2 we get\n",
+ "#28=b*(100**2)**-1+b*(100**2)**-1\n",
+ "#After furhter Simplifying we get\n",
+ "b=28*100**2*2**-1\n",
+ "\n",
+ "#sub in equation 1 we get\n",
+ "a=-(12-(b*(100**2)**-1))\n",
+ "\n",
+ "#Thus At\n",
+ "x2=150 #mm\n",
+ "p_o=b*(x2**2)**-1-a\n",
+ "\n",
+ "#Result\n",
+ "print\"Minimum External applied is\",round(p_o,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Minimum External applied is 4.22 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Example 8.8.13,Page No.340"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d1=160 #mm #Internal Diameter \n",
+ "r1=80 #mm #External Diameter\n",
+ "p1=40 #N/mm**2 #Internal Diameter\n",
+ "P_max=120 #N/mm**2 #Allowable stress\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#From Lame's Equation we have\n",
+ "#p_x=b*(x**2)**-1-a ..........................(1)\n",
+ "#F_x=b*(x**2)**-1+a ...........................(2)\n",
+ "\n",
+ "#At \n",
+ "x=r1=80 #N/mm**2 \n",
+ "#Sub in equation 1 we get\n",
+ "#120=b*(80**2)**-1+a ........................(3)\n",
+ "\n",
+ "#The hoop tension at inner edge is max stress\n",
+ "#Hence\n",
+ "#120=b*(80**2)**-1+a .............................(4)\n",
+ "\n",
+ "#From Equation 3 and 4 we get\n",
+ "b=160*80**2*2**-1 \n",
+ "\n",
+ "#Sub in equation 3 we get\n",
+ "a=-(40-(b*(80**2)**-1))\n",
+ "\n",
+ "#Let External radius be r_o.Since at External Surface is Zero,we get\n",
+ "#0=b*(r_o)**-1-a\n",
+ "#After Further simplifying we get\n",
+ "r_o=(b*a**-1)**0.5\n",
+ "\n",
+ "#Thickness of Cyclinder \n",
+ "t=r_o-r1 #mm\n",
+ "\n",
+ "#Result\n",
+ "print\"Thickness Required is\",round(t,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Thickness Required is 33.14 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 31
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Example 8.8.14,Page No.341"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d_o=300 #mm #Outside diameter \n",
+ "d1=180 #mm #Internal Diameter\n",
+ "p=12 #N/mm**2 #internal Fluid pressure\n",
+ "p_o=6 #N/mm**2 #External Pressure\n",
+ "r_o=150 #mm #Outside Diameter\n",
+ "r=90 #mm #Internal Diameter\n",
+ "\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#From Lame's Equation we have\n",
+ "#p_x=b*(x**2)**-1-a ..........................(1)\n",
+ "#F_x=b*(x**2)**-1+a ...........................(2)\n",
+ "\n",
+ "#At \n",
+ "x=r1=90 #N/mm**2 \n",
+ "p=42 #N/mm**2\n",
+ "#Sub in equation 1 we get\n",
+ "#42=b*(90**2)**-1-a ..............................(3)\n",
+ "\n",
+ "#At \n",
+ "x=r_o=150 #mm\n",
+ "p2=6 #N/mm**2\n",
+ "#sub in equation 1 we get\n",
+ "#6=b*(150**2)**-1-a ..............................(4)\n",
+ "\n",
+ "#From equations 3 and 4 weget\n",
+ "#36=b*(90**2)**-1-b2(150**2)**-1\n",
+ "#After further simplifying we get\n",
+ "b=36*90**2*150**2*(150**2-90**2)**-1\n",
+ "\n",
+ "#Sub value of b in equation 4 we get\n",
+ "a=b*(150**2)**-1-p_o\n",
+ "\n",
+ "#At \n",
+ "x=r1=90 #mm\n",
+ "F_x=b*(x**2)**-1+a #N/mm**2\n",
+ "\n",
+ "#At \n",
+ "x2=r_o=150 #mm \n",
+ "F_x2=b*(x2**2)**-1+a #N/mm**2\n",
+ "\n",
+ "#Now if External pressure is doubled i.e p_o2=12 #N/mm**2 We have\n",
+ "p_o2=12 #N/mm**2\n",
+ "#sub in equation 4 we get\n",
+ "#12=b2*(150**2)**-1-a2 ..........................(5)\n",
+ "\n",
+ "#Max Hoop stress is to be 70.5 #N/mm**2,which occurs at x=r1=90 #mm\n",
+ "#Sub in equation 4 we get\n",
+ "#70.5=b*(90**2)**-1+a2 ................................(6)\n",
+ "\n",
+ "#Adding equation 5 and 6\n",
+ "#82.5=b2*(150**2)**-1+b*(90**2)**-1\n",
+ "#After furhter simplifying we get\n",
+ "b2=82.5*150**2*90**2*(150**2+90**2)**-1\n",
+ "\n",
+ "#Sub in equation 5 we get\n",
+ "a2=b2*(150**2)**-1-12 \n",
+ "\n",
+ "#If p_i is the internal pressure required then from Lame's theorem\n",
+ "p_i=b2*(r1**2)**-1-a2\n",
+ "\n",
+ "#Result\n",
+ "print\"Stresses int the material are:F_x\",round(F_x,2),\"N/mm**2\"\n",
+ "print\" :F_x2\",round(F_x2,2),\"N/mm**2\"\n",
+ "print\"Internal Pressure that can be maintained is\",round(p_i,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Stresses int the material are:F_x 70.5 N/mm**2\n",
+ " :F_x2 34.5 N/mm**2\n",
+ "Internal Pressure that can be maintained is 50.82 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Example 8.8.15,Page No.344"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "%matplotlib inline\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "r1=200 #mm #Inner Radius\n",
+ "r2=250 #mm #Radius at common surface\n",
+ "r3=300 #mm #Outer radius\n",
+ "p=6 #N/mm**2 #Inital pressure\n",
+ "p2=80 #N/mm**2 #Pressure\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Inner Cyclinder:\n",
+ "\n",
+ "#From Lame's Equation we have\n",
+ "#p_x=b*(x**2)**-1-a ..........................(1)\n",
+ "#F_x=b*(x**2)**-1+a ...........................(2)\n",
+ "\n",
+ "#At \n",
+ "x=r1=200 #mm\n",
+ "p_x=0\n",
+ "#0=b1*(250**2)**-1-a1 .................(3)\n",
+ "\n",
+ "#At x=r2=250 #mm\n",
+ "p_x2=6 #N/mm**2\n",
+ "#6=b1*(250**2)-a1 ...................(4)\n",
+ "\n",
+ "#From Equation 3 and 4 we get\n",
+ "b1=6*200**2*250**2*(200**2-250**2)**-1\n",
+ "\n",
+ "#From equation 3 we get\n",
+ "a1=b1*(200**2)**-1\n",
+ "\n",
+ "F_200=b1*(200**2)**-1+a1\n",
+ "F_250=b1*(250**2)**-1+a1\n",
+ "\n",
+ "#For outer cyclinder \n",
+ "#From Lame's Equation we have\n",
+ "#p_x2=b2*(x**2)**-1-a2 ..........................(5)\n",
+ "#F_x2=b2*(x**2)**-1+a2 ...........................(6)\n",
+ "\n",
+ "\n",
+ "#At \n",
+ "x2=r2=250 #mm\n",
+ "p_x2=6 #N/mm**2\n",
+ "#6=b2*(250**2)**-1-a2 ...........................(7) \n",
+ "\n",
+ "#At\n",
+ "x3=300 #mm\n",
+ "#p_x2=0\n",
+ "#0=b2**2*(300**2)**-1-a2 .................................(8)\n",
+ "\n",
+ "#from equation 7 and 8 we get\n",
+ "b2=6*250**2*300**2*(300**2-250**2)**-1\n",
+ "\n",
+ "#sub in equation 8 we get\n",
+ "a2=b2*(300**2)**-1\n",
+ "\n",
+ "F_250_2=b2*(250**2)**-1+a2\n",
+ "F_300_2=b2*(300**2)**-1+a2\n",
+ "\n",
+ "#When Fluid is admitted\n",
+ "#Let Lame's equation be\n",
+ "#p_x3=b3*(x**2)**-1-a3 ..........................(5)\n",
+ "#F_x3=b3*(x**2)**-1+a3 ...........................(6)\n",
+ "\n",
+ "\n",
+ "#At x=200\n",
+ "p_x3=80 #N/mm**2\n",
+ "#80=b3*(200**2)**-1-a3 ................................(7)\n",
+ "\n",
+ "#At x=300 #mm\n",
+ "#p_x=0\n",
+ "#0=b3*(300**2)**-1-a3 ..............................(8)\n",
+ "\n",
+ "#from Equation 7 and 8 we get\n",
+ "b3=80*200**2*300**2*(300**2-200**2)**-1\n",
+ "\n",
+ "#From Equation 8 we get\n",
+ "a3=b3*(300**2)**-1\n",
+ "\n",
+ "#Hoop stresses \n",
+ "F_200_3=b3*(200**2)**-1+a3 #N/mm**2\n",
+ "F_250_3=b3*(250**2)**-1+a3 #N/mm**2\n",
+ "F_300_3=b3*(300**2)**-1+a3 #N/mm**2\n",
+ "\n",
+ "#Pressure at common surface\n",
+ "p_250=b3*(250**2)**-1-a3 #N/mm**2\n",
+ "\n",
+ "#final stress\n",
+ "f_200=F_200+F_200_3 #N/mm**2\n",
+ "f_250=F_250+F_250_3 #N/mm**2\n",
+ "f_300=F_250_2+F_250_3 #N/mm**2\n",
+ "f_300_2=F_300_2+F_300_3 #N/mm**2\n",
+ "\n",
+ "#Result\n",
+ "print\"final Hoop stress are:f_200\",round(f_200,2),\"N/mm**2\"\n",
+ "print\" :f_250\",round(f_250,2),\"N/mm**2\"\n",
+ "print\" :f_300\",round(f_300,2),\"N/mm**2\"\n",
+ "print\" :f_300_2\",round(f_300_2,2),\"N/mm**2\"\n",
+ "print\"Variation of Hoop stress and Radial stress\"\n",
+ "\n",
+ "#Final stresses\n",
+ "#Variation of hoop stress \n",
+ " \n",
+ "X1=[x,x2,x3,x3]\n",
+ "Y1=[f_200,f_250,f_300,f_300_2]\n",
+ "Z1=[0,0,0,0]\n",
+ "plt.plot(X1,Y1,X1,Z1)\n",
+ "plt.xlabel(\"Length x in mm\")\n",
+ "plt.ylabel(\"Hoop Stress Distribution in N/mm**2\")\n",
+ "plt.show()\n",
+ "\n",
+ "#Due to Fluid\n",
+ "#Variation of hoop stress \n",
+ " \n",
+ "X1=[x,x2,x3]\n",
+ "Y1=[F_200_3,F_250_3,F_300_3]\n",
+ "Z1=[0,0,0]\n",
+ "plt.plot(X1,Y1,X1,Z1)\n",
+ "plt.xlabel(\"Length x in mm\")\n",
+ "plt.ylabel(\"Hoop Stress Distribution in N/mm**2\")\n",
+ "plt.show()\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "final Hoop stress are:f_200 174.67 N/mm**2\n",
+ " :f_250 128.83 N/mm**2\n",
+ " :f_300 189.43 N/mm**2\n",
+ " :f_300_2 155.27 N/mm**2\n",
+ "Variation of Hoop stress and Radial stress\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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MHz4cr7zyCmbPno1///vf1fIBnZeXp/l+8+bNmhVR0dHR+P7771FYWIjMzExcvHgRnTp1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\nVw+AnLO4d++eQY0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+ "text": [
+ "<matplotlib.figure.Figure at 0x5604510>"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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+ "text": [
+ "<matplotlib.figure.Figure at 0x56e5cb0>"
+ ]
+ }
+ ],
+ "prompt_number": 89
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Example 8.8.16,Page No.348"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "do=200 #mm #Inner Diameter\n",
+ "r_o=100 #mm #Inner radius\n",
+ "d1=300 #mm #outer diameter\n",
+ "r1=150 #mm #Outer radius\n",
+ "d2=250 #mm #Junction Diameter\n",
+ "r2=125 #mm #Junction radius\n",
+ "E=2*10**5 #N/mm**2 #Modulus of Elasticity\n",
+ "p=30 #N/mm**2 #radial pressure\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#from Lame's Equation we get\n",
+ "#p_x=b*(x**2)**-1-a ..........................(1)\n",
+ "#F_x=b*(x**2)**-1+a ...........................(2)\n",
+ "\n",
+ "#Then from Boundary condition \n",
+ "#p_x=0 at x=100 #mm\n",
+ "#0=b1*(100**2)**-1-a1 .....................(3)\n",
+ "\n",
+ "#p_x2=30 #N/mm**2 at x2=125 #mm\n",
+ "#30=b1*(125**2)**-1-a1 ................................(4)\n",
+ "\n",
+ "#From equation 3 and 4 we get\n",
+ "b1=30*125**2*100**2*(100**2-125**2)**-1\n",
+ "\n",
+ "#From Equation 3 we get\n",
+ "a1=b1*(100**2)**-1\n",
+ "\n",
+ "#therefore Hoop stress in inner cyclinder at junction\n",
+ "F_2_1=b1*(125**2)**-1+a1 #N/mm**2\n",
+ "\n",
+ "#Outer Cyclinder\n",
+ "#p_x=b*(x**2)**-1-a ..........................(5)\n",
+ "#F_x=b*(x**2)**-1+a ...........................(6)\n",
+ "\n",
+ "#Now at x=125 #mm\n",
+ "#p_x3=30 #N/mm**2\n",
+ "#30=b2*(125**2)**-1-a2 ..................................(7)\n",
+ "\n",
+ "#At x=150 #mm\n",
+ "#p_x4=0\n",
+ "#0=b2*(150**2)**-1-a2 ...................................(8)\n",
+ "\n",
+ "#From equations 7 and 8\n",
+ "b2=30*150**2*125**2*(150**2-125**2)**-1\n",
+ "\n",
+ "#From eqauation 8 we get\n",
+ "a2=b2*(150**2)**-1\n",
+ "\n",
+ "#Hoop stress at junction \n",
+ "F_2_0=b2*(125**2)**-1+a2 #N/mm**2\n",
+ "\n",
+ "rho_r=(F_2_0-F_2_1)*E**-1*r2\n",
+ "\n",
+ "#Result\n",
+ "print\"Shrinkage Allowance is\",round(rho_r,3),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Shrinkage Allowance is 0.189 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 40
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Example 8.8.17,Page No.350"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d_o=500 #mm #Outer Diameter\n",
+ "r_o=250 #mm #Outer Radius\n",
+ "d1=300 #mm #Inner Diameter\n",
+ "r1=150 #mm #Inner Radius\n",
+ "d2=400 #mm #Junction Diameter\n",
+ "E=2*10**5 #N/mm**2 #Modulus ofElasticity\n",
+ "alpha=12*10**-6 #Per degree celsius\n",
+ "dell_d=0.2 #mm\n",
+ "dell_r=0.1 #mm\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let p be the radial pressure developed at junction\n",
+ "#Let Lame's Equation for internal cyclinder be\n",
+ "#p_x=b*(x**2)**-1-a ................................(1)\n",
+ "#F_x=b*(x**2)**-1+a ...............................(2)\n",
+ "\n",
+ "#At \n",
+ "x=150 #mm \n",
+ "p_x=0\n",
+ "#Sub in equation 1 we get\n",
+ "#0=b*(150**2)**-1-a .........................(3)\n",
+ "\n",
+ "#At \n",
+ "x2=200 #mm\n",
+ "#p_x2=p\n",
+ "#p=b*(200**2)**-1-a ......................(4)\n",
+ " \n",
+ "#From Equation 3 and 4\n",
+ "#p=b*(200**2)**-1-b(150**2)**-1\n",
+ "#after further simplifying we get\n",
+ "#b=-51428.571*p\n",
+ "\n",
+ "#sub in equation 3 we get\n",
+ "#a1=-2.2857*p\n",
+ "\n",
+ "#therefore hoop stress at junction is\n",
+ "#F_2_1=-21428.571*p*(200**2)**-1-2.2857*p\n",
+ "#after Further simplifying we geet\n",
+ "#F_2_1=3.5714*p\n",
+ "\n",
+ "#Let Lame's Equation for cyclinder be \n",
+ "#p_x=b*(x**2)**-1-a .........................5\n",
+ "#F_x=b*(x**2)**-1+a .............................6\n",
+ "\n",
+ "#At \n",
+ "x=200 #mm\n",
+ "#p_x=p2\n",
+ "#p2=b2*(20**2)**-1-a2 ...................7\n",
+ "\n",
+ "#At\n",
+ "x2=200 #mm\n",
+ "p_x2=0\n",
+ "#0=b2*(250**2)**-1-a2 ....................8\n",
+ "\n",
+ "#from equation 7 and 8 we get\n",
+ "#p2=b2*(200**2)**-1-b2*(250**2)**-1\n",
+ "#After further simplifying we get\n",
+ "#p2=b2*(250**2-200**2)*(200**2*250**2)**-1\n",
+ "#b2=111111.11*p\n",
+ "\n",
+ "#from equation 7\n",
+ "#a2=b2*(250**2)**-1\n",
+ "#further simplifying we get\n",
+ "#a2=1.778*p\n",
+ "\n",
+ "#At the junctionhoop stress in outer cyclinder \n",
+ "#F_2_0=b2*(200**2)**-1+a2\n",
+ "#After further simplifying we get\n",
+ "#F_2_0=4.5556*p\n",
+ "\n",
+ "#Considering circumferential strain,the compatibility condition\n",
+ "#rho_r*r2**-1=1*E**-1*(F_2_1+F_2_0)\n",
+ "#where F_2_1 is compressive and F_2_0 is tensile\n",
+ "#furter simplifying we get\n",
+ "p=0.1*200**-1*2*10**5*(3.5714+4.5556)**-1\n",
+ "\n",
+ "#Let T be the rise in temperature required\n",
+ "#dell_d=d*alpha*T\n",
+ "#After sub values and further simplifying we get\n",
+ "d=250 #mm\n",
+ "T=dell_d*(d*alpha)**-1 #Per degree celsius\n",
+ "\n",
+ "#Result\n",
+ "print\"Radial Pressure Developed at junction\",round(p,2),\"N/mm**2\"\n",
+ "print\"Min Temperatureto outer cyclinder\",round(T,2),\"Per degree Celsius\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Radial Pressure Developed at junction 12.3 N/mm**2\n",
+ "Min Temperatureto outer cyclinder 66.67 Per degree Celsius\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Example 8.8.18,Page No.355"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "%matplotlib inline\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d_o=400 #mm #Outer Diameter\n",
+ "r_o=200 #mm #Outer radius\n",
+ "t=50 #mm #Thickness\n",
+ "r1=150 #mm #Internal Radius\n",
+ "p=50 #N/mm**2 #Internal Pressure\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#The Radial Pressure and hoop stress at any radial distance x are given by\n",
+ "#p_x=b*(x**2)**-1-a ..........................(1)\n",
+ "#F_x=b*(x**2)**-1+a ...........................(2)\n",
+ "\n",
+ "#Now at\n",
+ "x=150 #N/mm**2\n",
+ "p_x1=50 #N/mm**2\n",
+ "#Sub in equation 1 we get\n",
+ "#50=2*b*(150**3)**-1-a ...........................(3)\n",
+ "\n",
+ "#At x=200 #mm\n",
+ "p_x2=0\n",
+ "#0=2*b*(200**2)**-1-a ....................(4)\n",
+ "\n",
+ "#From equation 3 and 4 we get\n",
+ "#50=2*b*(150**3)**-1-2*b*(200**3)**-1\n",
+ "#After further simplifying we get\n",
+ "b=50*150**3*200**3*(200**3-150**3)**-1*2**-1\n",
+ "\n",
+ "#Sub in equation 3 we get\n",
+ "a=b*(200**3)**-1\n",
+ "\n",
+ "#Now At\n",
+ "x=150 #mm\n",
+ "F_x=b*(x**3)**-1+a\n",
+ "\n",
+ "#Now At\n",
+ "x2=160 #mm\n",
+ "F_x2=b*(x2**3)**-1+a\n",
+ "\n",
+ "#Now At\n",
+ "x3=170 #mm\n",
+ "F_x3=b*(x3**3)**-1+a\n",
+ "\n",
+ "#Now At\n",
+ "x4=180 #mm\n",
+ "F_x4=b*(x4**3)**-1+a\n",
+ "\n",
+ "#Now At\n",
+ "x5=190 #mm\n",
+ "F_x5=b*(x5**3)**-1+a\n",
+ "\n",
+ "#Now At\n",
+ "x6=200 #mm\n",
+ "F_x6=b*(x6**3)**-1+a\n",
+ "\n",
+ "#Result\n",
+ "print\"Plot of Variation of hoop stress\"\n",
+ "\n",
+ "#Plotting Variation of hoop stress\n",
+ "\n",
+ "X1=[x,x2,x3,x4,x5,x6]\n",
+ "Y1=[F_x,F_x2,F_x3,F_x4,F_x5,F_x6]\n",
+ "Z1=[0,0,0,0,0,0]\n",
+ "plt.plot(X1,Y1,X1,Z1)\n",
+ "plt.xlabel(\"Length x in mm\")\n",
+ "plt.ylabel(\"Hoop Stress Distribution in N/mm**2\")\n",
+ "plt.show()\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Plot of Variation of hoop stress\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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w3g2UlJRgzJgxCAsLw3vvvQcA8PT0hFqthpOTE7KyshAYGIhLly7VLIxzCERE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\nnp6uOVdYWBisrKzg4+OD8vJ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+ "text": [
+ "<matplotlib.figure.Figure at 0x56b4bf0>"
+ ]
+ }
+ ],
+ "prompt_number": 25
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.9_7.ipynb b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.9_7.ipynb new file mode 100644 index 00000000..cecacb12 --- /dev/null +++ b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.9_7.ipynb @@ -0,0 +1,779 @@ +{
+ "metadata": {
+ "name": "chapter no.9.ipynb"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 9:Columns And Struts"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.9.1,Page No.377"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "L=5000 #mm #Length of strut\n",
+ "dell=10 #mm #Deflection\n",
+ "W=10 #N #Load\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Central Deflection of a simply supported beam with central concentrated load is\n",
+ "#dell=W*L**3*(48*E*I)**-1 \n",
+ "\n",
+ "#Let E*I=X\n",
+ "X=W*L**3*(48*dell)**-1 #mm\n",
+ "\n",
+ "#Euler's Load\n",
+ "#Let Euler's Load be P\n",
+ "P=pi**2*X*(L**2)**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"Critical Load of Bar is\",round(P,2),\"N\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Critical Load of Bar is 1028.08 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.9.2,Page No.377"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=2000 #mm #Length of square column\n",
+ "E=12*10**3 #N/mm**2 #Modulus of Elasticity\n",
+ "sigma=12 #N/mm*2 #stress\n",
+ "W1=95*10**3 #N #Load1\n",
+ "W2=200*10**3 #N #Load2\n",
+ "FOS=3\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#From Euler's Formula\n",
+ "#P=pi**2*E*I*(L**2)**-1 .........(1)\n",
+ "\n",
+ "#Working Load\n",
+ "#W=P*(FOS)**-1\n",
+ "\n",
+ "#Part-1\n",
+ "\n",
+ "#At W1=95*10**3 #N\n",
+ "#W1=P*(3*L**2)**-1\n",
+ "\n",
+ "#Let 'a' be the side of the square\n",
+ "#I=1*12**-1*a**4\n",
+ "\n",
+ "#sub value of I in Equation 1 and further rearranging we get\n",
+ "a=(W1*3*12*L**2*(pi**2*E)**-1)**0.25 #mm\n",
+ "\n",
+ "#From Consideration of direct crushing\n",
+ "#sigma*a**2=W1\n",
+ "#After Reaaranging the above equation we get\n",
+ "a2=(W1*(sigma)**-1)**0.5 #mm\n",
+ "\n",
+ "#required size is 103.67*103.67 i.e a*a\n",
+ "\n",
+ "#Part-2\n",
+ "\n",
+ "#At W2=200*10**3 #N\n",
+ "#W2=P*(3*L**2)**-1\n",
+ "#After substituting values and further Rearranging the above equation we get\n",
+ "a3=(W2*3*12*L**2*(pi**2*E)**-1)**0.25 #mm\n",
+ "\n",
+ "#From consideration of direct compression,size required is\n",
+ "a4=(W2*sigma**-1)**0.5\n",
+ "\n",
+ "#required size is 129.10*129.10 i.e a4*a4\n",
+ "\n",
+ "#Result\n",
+ "print\"For W1 Load Required size is\",round(a*a,2),\"mm**2\"\n",
+ "print\"For W2 Load Required size is\",round(a4*a4,2),\"mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "For W1 Load Required size is 10747.38 mm**2\n",
+ "For W1 Load Required size is 16666.67 mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.9.3,Page No.378"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#Flange \n",
+ "b=100 #mm #Width\n",
+ "\n",
+ "D=80 #mm #Overall Depth\n",
+ "t=10 #mm #Thickness of web and flanges\n",
+ "L=3000 #mm #Length of strut\n",
+ "E=200*10**3 #N/mm**2 #Modulus of Elasticity\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let centroid be at depth y_bar from top fibre\n",
+ "y_bar=(b*t*t*2**-1+(D-t)*t*((D-t)*2**-1+t))*(b*t+(D-t)*t)**-1 #mm \n",
+ "\n",
+ "#M.I at x-x axis\n",
+ "I_x=1*12**-1*b*t**3+b*t*(y_bar-t*2**-1)**2+1*12**-1*t*((D-t))**3+t*((D-t))*((((D-t)*2**-1)+t)-y_bar)**2\n",
+ "\n",
+ "#M.I at y-y axis\n",
+ "I_y=1*12**-1*t*b**3+1*12**-1*(D-t)*t**3 #mm**3\n",
+ "\n",
+ "#Least M.I\n",
+ "I=I_y\n",
+ "\n",
+ "#Since both ends are hinged\n",
+ "#Feective Length=Actual Length\n",
+ "L=l=3000 #mm\n",
+ "\n",
+ "#Buckling Load \n",
+ "P=pi**2*E*I*(l**2)**-1*10**-3 #KN\n",
+ "\n",
+ "#Result\n",
+ "print\"The Buckling Load for strut of tee section\",round(P,2),\"KN\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Buckling Load for strut of tee section 184.05 KN\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.9.4,Page No.379"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "D=400 #mm #Overall Depth\n",
+ "\n",
+ "#Flanges\n",
+ "b=300 #mm #Width\n",
+ "t=50 #mm #Thickness\n",
+ "\n",
+ "t2=30 #mm #Web Thickness\n",
+ "\n",
+ "dell=10 #mm #Deflection\n",
+ "w=40 #N/mm #Load\n",
+ "FOS=1.75 #Factor of safety\n",
+ "E=2*10**5 #N/mm**2\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#M.I at x-x axis\n",
+ "I_x=1*12**-1*(b*D**3-(b-t2)*b**3) #mm**4\n",
+ "\n",
+ "#Central Deflection\n",
+ "#dell=5*w*L**4*(384*E*I)**-1\n",
+ "#After sub values in above equation and further simplifying we get\n",
+ "L=(dell*384*E*I_x*(5*w)**-1)**0.25\n",
+ "\n",
+ "#M.I aty-y axis\n",
+ "I=I_y=1*12**-1*t*b**3+1*12**-1*b*t2**3+1*12**-1*t*b**3 #mm**4\n",
+ "\n",
+ "#Both the Ends of column are hinged\n",
+ "\n",
+ "#Crippling Load\n",
+ "P=pi**2*E*I*(L**2)**-1 #N\n",
+ "\n",
+ "#Safe Load\n",
+ "S=P*(FOS)**-1*10**-3 #N\n",
+ "\n",
+ "#Result\n",
+ "print\"Safe Load if I-section is used as column with both Ends hhinged\",round(S,2),\"KN\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Safe Load if I-section is used as column with both Ends hhinged 4123.29 KN\n"
+ ]
+ }
+ ],
+ "prompt_number": 19
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.9.5,Page No.381"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "D=200 #mm #External Diameter\n",
+ "t=20 #mm #hickness\n",
+ "d=200-2*t #mm #Internal Diameter\n",
+ "E=1*10**5 #N/mm**2\n",
+ "a=1*(1600)**-1 #Rankine's Constant\n",
+ "L=4.5 #m #Length\n",
+ "sigma=550 #N/mm**2 #Stress\n",
+ "FOS=2.5\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Moment of Inertia\n",
+ "I=pi*D**4*64**-1-pi*d**4*64**-1\n",
+ "\n",
+ "#Both Ends are fixed\n",
+ "\n",
+ "#Effective Length\n",
+ "l=1*2**-1*L*10**3 #mm\n",
+ "\n",
+ "#Euler's Critical Load\n",
+ "P_E=pi**2*E*I*(l**2)**-1\n",
+ "\n",
+ "A=pi*4**-1*(D**2-d**2) #mm*2\n",
+ "\n",
+ "k=(I*A**-1)**0.5\n",
+ "\n",
+ "#Rankine's Critical Load\n",
+ "P_R=sigma*A*(1+a*(l*k**-1)**2)**-1\n",
+ "\n",
+ "X=P_E*P_R**-1 \n",
+ "\n",
+ "#Safe Load using Rankine's Formula\n",
+ "S=P_R*(FOS)**-1*10**-3 #KN\n",
+ "\n",
+ "#Result\n",
+ "print\"Safe Load by Rankine's Formula is\",round(S,2),\"KN\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Safe Load by Rankine's Formula is 1404.36 KN\n"
+ ]
+ }
+ ],
+ "prompt_number": 39
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.9.6,Page No.382"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=3000 #mm #Length of column\n",
+ "W=800*10**3 #N #Load\n",
+ "a=1*1600**-1 #Rankine's constant\n",
+ "FOS=4 #Factor of safety\n",
+ "sigma=550 #N/mm**2 #stress\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Effective Length\n",
+ "l=L*2**-1 #mm \n",
+ "\n",
+ "#Let d1=outer diameter & d2=inner diameter\n",
+ "#d1=5*8**-1*d2\n",
+ "\n",
+ "#M.I\n",
+ "#I=pi*64**-1*(d1**4-d2**4) #mm**4\n",
+ "\n",
+ "#Area of section\n",
+ "#A=pi4**-1*(d1**2-d2**2) #mm**2\n",
+ "\n",
+ "#k=(I*A**-1) \n",
+ "#substituting values in above equation \n",
+ "#k=1*16**-1*(d1**2-d2**2)\n",
+ "#after simplifying further we get\n",
+ "#k=0.2948119.d1\n",
+ "\n",
+ "#X=l*k**-1\n",
+ "#substituting values in above equation and after simplifying further we get\n",
+ "#X=5087.9898*d1**-1\n",
+ "\n",
+ "#Crtitcal Load\n",
+ "P=W*FOS #N\n",
+ "\n",
+ "#From Rankine's Load\n",
+ "#P2=sigma*A*(1+a*(X)**2)**-1\n",
+ "#substituting values in above equation and after simplifying further we get\n",
+ "#d1**4-12156618*d1**4-1.96691*10**8=0\n",
+ "#Solving Quadratic Equation we get\n",
+ "#d1**2-12156618*d1-196691000=0\n",
+ "a=1\n",
+ "b=-12156.618\n",
+ "c=-196691000\n",
+ "\n",
+ "Y=b**2-4*a*c\n",
+ "\n",
+ "d1_1=((-b+Y**0.5)*(2*a)**-1)**0.5 #mm\n",
+ "d1_2=((-b-Y**0.5)*(2*a)**-1) #mm\n",
+ "\n",
+ "d2=5*8**-1*d1_1\n",
+ "\n",
+ "#Result\n",
+ "print\"Section of cast iron hollow cylindrical column is:d1_1\",round(d1_1,2),\"mm\"\n",
+ "print\" :d2 \",round(d2,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Section of cast iron hollow cylindrical column is:d1_1 146.16 mm\n",
+ " :d2 91.35 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 25
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.9.7,Page No.383"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#Let X=(P*A**-1) #Average Stress at Failure \n",
+ "Lamda_1=70 #Slenderness Ratio\n",
+ "Lamda_2=170 #Slenderness Ratio\n",
+ "X1=200 #N/mm**2 \n",
+ "X2=69 #N/mm**2 \n",
+ "\n",
+ "#Rectangular section\n",
+ "b=60 #mm #width\n",
+ "t=20 #mm #Thickness\n",
+ "\n",
+ "L=1250 #mm #Length of strut\n",
+ "FOS=4 #Factor of safety\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Slenderness ratio\n",
+ "#Lamda=L*k**-1\n",
+ "\n",
+ "#The Rankine's Formula for strut\n",
+ "#P=sigma*A*(1+a*(L*k**-1)**-1\n",
+ "\n",
+ "#From test result 1,\n",
+ "#After sub values in above equation we get and further simplifying we get\n",
+ "#sigma_1=200+980000*a ...................(1)\n",
+ "\n",
+ "#From test result 2,\n",
+ "#After sub values in above equation we get and further simplifying we get\n",
+ "#sigma_2=69+1994100*a ...................(2)\n",
+ "\n",
+ "#Substituting it in equation (1) we get\n",
+ "a=131*1014100**-1 \n",
+ "\n",
+ "#Substituting a in equation 1\n",
+ "sigma_1=200+980000*a #N/mm**2\n",
+ "\n",
+ "#Effective Length \n",
+ "l=1*2**-1*L #mm\n",
+ "\n",
+ "#Least of M.I\n",
+ "I=1*12**-1*b*t**3 #mm**4\n",
+ "\n",
+ "#Area \n",
+ "A=b*t #mm**2 \n",
+ "\n",
+ "k=(I*A**-1)**0.5\n",
+ "\n",
+ "#Slenderness ratio\n",
+ "Lamda=l*k**-1\n",
+ "\n",
+ "#From Rankine's Ratio\n",
+ "P=sigma_1*A*(1+a*(Lamda)**2)**-1\n",
+ "\n",
+ "#Safe Load\n",
+ "S=P*(FOS)**-1*10**-3 #N\n",
+ "\n",
+ "#Result\n",
+ "print\"Constant in the Formula is:a \",round(a,6)\n",
+ "print\" :sigma_1\",round(sigma_1,2)\n",
+ "print\"Safe Load is\",round(S,2),\"KN\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Constant in the Formula is:a 0.000129\n",
+ " :sigma_1 326.6\n",
+ "Safe Load is 38.98 KN\n"
+ ]
+ }
+ ],
+ "prompt_number": 38
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.9.8,Page No.385"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "D=200 #mm #Depth\n",
+ "b=140 #mm #width\n",
+ "\n",
+ "#Plate\n",
+ "b2=160 #mm #Width\n",
+ "t2=10 #mm #Thickness\n",
+ "\n",
+ "L=l=4000 #mm #Length\n",
+ "FOS=4 #Factor of safety\n",
+ "sigma=315 #N/mm**2 #stress\n",
+ "a2=1*7500**-1 \n",
+ "I_xx=26.245*10**6 #mm**4 #M.I at x-x\n",
+ "I_yy=3.288*10**6 #mm**4 #M.I at y-y\n",
+ "a=3671 #mm**2 #Area\n",
+ "k_x=84.6#mm\n",
+ "k_y=29.9 #mm\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Total Area\n",
+ "A=a+2*t2*b2 #mm**2\n",
+ "\n",
+ "#M.I\n",
+ "I=I_yy+2*12**-1*t2*b2**3 #mm**4\n",
+ "\n",
+ "k=(I*A**-1)**0.5 #mm\n",
+ "\n",
+ "#Let X=L*k**-1\n",
+ "X=L*k**-1\n",
+ "\n",
+ "#Appliying Rankine's Formula\n",
+ "P=sigma*A*(1+a2*(X)**2)**-1 #N\n",
+ "\n",
+ "#Safe Load\n",
+ "S=P*(FOS)**-1*10**-3 #KN\n",
+ "\n",
+ "#Result\n",
+ "print\"Safe axial Load is\",round(S,2),\"KN\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Safe axial Load is 220.93 KN\n"
+ ]
+ }
+ ],
+ "prompt_number": 48
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.9.9,Page No.389"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "E=200*10**3 #N/mm**2 #Modulus of elasticity\n",
+ "sigma=330 #N/mm**2 #Stress\n",
+ "a=1*7500**-1 #Rankine's constant\n",
+ "A=5205 #mm**2 #area of column\n",
+ "I_xx=59.431*10**6 #mm**4 #M.I at x-x axis\n",
+ "I_yy=8.575*10**6 #mm**24#M.I at y-y axis\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Total M.I\n",
+ "I=I_xx+I_yy #mm**4\n",
+ "\n",
+ "#Area of compound Section \n",
+ "A2=2*A #mm**2\n",
+ "\n",
+ "k=(I*A2**-1)**0.5 #mm\n",
+ "\n",
+ "#Equating Euler's Load to Rankine's Load we get\n",
+ "#pi**2*E*I*(L**2)**-1=sigma*A*(1+a*(L*k)**2)**-1\n",
+ "#After Substitt=uting values and further simplifying we get\n",
+ "L=(39076198*(1-0.7975432)**-1)**0.5*10**-3 #m\n",
+ "\n",
+ "#Result\n",
+ "print\"Length of column for which Rankine's formula and Euler's Formula give the same result is\",round(L,2),\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Length of column for which Rankine's formula and Euler's Formula give the same result is 13.89 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 47
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.9.10,Page No.387"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "sigma=326 #N/mm**2 #stress\n",
+ "E=2*10**5 #N/mm**2 #Modulus of Elasticity\n",
+ "FOS=2 #Factor of safety\n",
+ "a=1*7500**-1 #Rankine's constant\n",
+ "D=350 #mm #Overall Depth \n",
+ "\n",
+ "#Cover plates\n",
+ "b1=500 #mm #width\n",
+ "t1=10 #mm #Thickness\n",
+ "\n",
+ "d=220 #mm #Distance between two channels\n",
+ "\n",
+ "L=6000 #mm #Length of column\n",
+ "\n",
+ "A=5366 #mm**2 #Area of Column section \n",
+ "I_xx=100.08*10**6 #mm**4 #M.I of x-x axis\n",
+ "I_yy=4.306*10**6 #mm**4 #M.I of y-y axis\n",
+ "C_yy=23.6 #mm #Centroid at y-y axis\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Symmetric axes are the centroidal axes is\n",
+ "\n",
+ "#M.I of Channel at x-x axis\n",
+ "I_xx_1=2*I_xx+2*(1*12**-1*b1*t1**3+b1*t1*(D*2**-1+t1*2**-1)**2)\n",
+ "\n",
+ "#M.I of Channel at y-y axis\n",
+ "I_yy_1=2*(I_yy+A*(d*2**-1+C_yy)**2)+2*12**-1*t1*b1**3\n",
+ "\n",
+ "#As I_yy<I_xx\n",
+ "#So\n",
+ "I=I_yy_1 #mm**4 \n",
+ "\n",
+ "A2=2*A+2*t1*b1 #Area of channel\n",
+ "\n",
+ "k=(I*A2**-1)**0.5 #mm\n",
+ "\n",
+ "#Critical Load\n",
+ "P=sigma*A2*(1+a*(L*k**-1)**2)**-1 \n",
+ "\n",
+ "#Safe Load\n",
+ "S=P*2**-1*10**-3 #KN\n",
+ "\n",
+ "#Result\n",
+ "print\"Safe Load carrying Capacity is\",round(S,2),\"KN\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Safe Load carrying Capacity is 2717.35 KN\n"
+ ]
+ }
+ ],
+ "prompt_number": 67
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.9.11,Page No.390"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "I=4.085*10**8 #mm**4 #M.I\n",
+ "A=20732.0 #mm**2 #area of column\n",
+ "f_y=250 #N/mm**2 \n",
+ "L=6000 #mm #Length of column\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "k=(I*A**-1)**0.5 #mm\n",
+ "lamda=L*k**-1 #Slenderness ratro\n",
+ "\n",
+ "#From Indian standard table\n",
+ "lamda_1=40 \n",
+ "sigma_a_c_1=139 #N/mm**2\n",
+ "lamda_2=50 \n",
+ "sigma_a_c_2=132 #N/mm**2 \n",
+ "\n",
+ "#Linearly interpolating between these values for lambda=42.744\n",
+ "\n",
+ "sigma_a_c_3=sigma_a_c_1-2.744*10**-1*(sigma_a_c_1-sigma_a_c_2)\n",
+ "\n",
+ "#Safe Load carrying capacity of column\n",
+ "P=sigma_a_c_3*A*10**-3\n",
+ "\n",
+ "#Result\n",
+ "print\"Safe Load carrying capacity is\",round(P,2),\"KN\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Safe Load carrying capacity is 2841.93 KN\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
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