From 7876eeaf85f7c020ec1f3530963928cd2bc26a66 Mon Sep 17 00:00:00 2001 From: hardythe1 Date: Mon, 28 Jul 2014 14:01:58 +0530 Subject: adding book --- Mechanics_of_Materials/Chapter5.ipynb | 736 ++++++++++++++++++++++++++++++++++ 1 file changed, 736 insertions(+) create mode 100755 Mechanics_of_Materials/Chapter5.ipynb (limited to 'Mechanics_of_Materials/Chapter5.ipynb') diff --git a/Mechanics_of_Materials/Chapter5.ipynb b/Mechanics_of_Materials/Chapter5.ipynb new file mode 100755 index 00000000..649dd3d3 --- /dev/null +++ b/Mechanics_of_Materials/Chapter5.ipynb @@ -0,0 +1,736 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 5:Torsion" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.2 Page No 188" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Given\n", + "T1 = 4250.0 #kNmm, torque\n", + "T2 = -3000.0 #kNm\n", + "T3 = T1+T2 #kNm\n", + "r = 75.0 #mm, radius\n", + "\n", + "#Calculation\n", + "#Section Property\n", + "import math\n", + "J = (math.pi/2.0)*(r**4) #polar moment of inertia\n", + "#Shear Stress\n", + "c_a = 75 #mm\n", + "tou_a = (T3*c_a*1000)/J #tou = Tc/J\n", + "c_b = 15 #mm\n", + "tou_b = (T3*c_b*1000)/J #tou = Tc/J\n", + "\n", + "#Display\n", + "print'The shear stress developed at A = ',round(tou_a*10,1),\"ksi\"\n", + "print'The shear stress developed at B = ',round(tou_b*10,2),\"ksi\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The shear stress developed at A = 18.9 ksi\n", + "The shear stress developed at B = 3.77 ksi\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.3 Page No 189" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Given\n", + "di = 80 #mm, inside diameter\n", + "ri = 40/1000.0 #m, inside radius\n", + "d0 = 100 #mm, outside diameter\n", + "ro = d0/2000.0 #m outside radius\n", + "F = 80 #N, force\n", + "l1 = 0.2 #m, length\n", + "l2 = 0.3 #m\n", + "\n", + "#Internal Torque\n", + "T = F*(l1+l2)\n", + "#Section Property\n", + "import math\n", + "J = (math.pi/2.0)*((ro**4)-(ri**4))\n", + "#Shear Stress\n", + "c_o = 0.05#m\n", + "tou_o = (T*c_o)/(J*10**6)\n", + "c_i = 0.04 #m\n", + "tou_i = (T*c_i)/(J*10**6)\n", + "\n", + "#Display\n", + "print'The shear stress in the inner wall = ',round(tou_i,3),\"MPa\"\n", + "print'The shear stress in the outer wall = ',round(tou_o,3),\"MPa\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The shear stress in the inner wall = 0.276 MPa\n", + "The shear stress in the outer wall = 0.345 MPa\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.4 Page No 191" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Given\n", + "P = 5 #hp\n", + "N = 175 #rpm\n", + "allow_shear = 14.5 #ksi\n", + "\n", + "#Calculations\n", + "import math\n", + "P_=P*550 #ftlb/s\n", + "ang_vel = (2*math.pi*N)/60.0 # rad/s\n", + "T = P_/ang_vel #P = T*angular velocity\n", + "c = ((2*T*12)/(math.pi*allow_shear*1000))**(1/3.0)\n", + "d =2*c\n", + "\n", + "#Display\n", + "print'The required diameter of the shaft = ',round(d,3),\"inch\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The required diameter of the shaft = 0.858 inch\n" + ] + } + ], + "prompt_number": 22 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.5 Page No 205" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Given\n", + "E = 80*10**9 #N/m**2, longitudinal stress\n", + "d = 14/1000.0 #m, diameter\n", + "r = d/2.0 #m, radius\n", + "R = 100 #mm\n", + "l_ac = 0.4 #m, length\n", + "l_cd = 0.3 #m\n", + "l_de = 0.5 #m\n", + "T_c = 280 #Nm, torque\n", + "T_a = 150 #Nm\n", + "T_d = 40 #Nm\n", + "T_ac = T_a #Nm\n", + "\n", + "#Calculation\n", + "T_cd = T_ac - T_c \n", + "T_de = T_cd - T_d\n", + "#Angle of Twist\n", + "import math\n", + "J = (math.pi/2.0)*(r**4)\n", + "phiA=T_ac*l_ac/(J*E)+T_cd*l_cd/(J*E)+T_de*l_de/(J*E)\n", + "Sp=phiA*R\n", + "\n", + "#Display\n", + "print'The angle of twist of the shaft = ',round(phiA,3),\"rad\"\n", + "print'The displacement of tooth P on gear A =',round(Sp,3),\"mm\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The angle of twist of the shaft = -0.212 rad\n", + "The displacement of tooth P on gear A = -21.212 mm\n" + ] + } + ], + "prompt_number": 28 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.6 Page No 206" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Given\n", + "T = 45\t\t #N, torque\n", + "G = 80 \t\t#GPa, pressure\n", + "d = 20/1000.0\t\t #m\n", + "r = d/2.0\t #m\n", + "l_dc = 1.5\t\t #m\n", + "l_ab = 2 \t\t#m\n", + "r1 = 75/1000.0\t\t #m\n", + "r2 = 150/1000.0\t\t #m\n", + "\n", + "#Calculation\n", + "#Internal Torque\n", + "F = T/r2\n", + "T_d_x = F*r1\n", + "#Angle of twist\n", + "import math\n", + "J = (math.pi/2)*(r**4)\n", + "phi_c = (T*l_dc)/(2*J*G*10**9)\n", + "phi_b = (phi_c*r1)/r2\n", + "phi_ab = (T*l_ab)/(J*G*10**9)\n", + "phi_a = phi_b + phi_ab\n", + "\n", + "#Display\n", + "print'The angle of twist of end A of shaft AB = ',round(phi_a,3),\"rad\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The angle of twist of end A of shaft AB = 0.085 rad\n" + ] + } + ], + "prompt_number": 31 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.7 Page No 207" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Given\n", + "d = 2 #inch, diameter\n", + "r = d/2.0 #radius\n", + "c = d/2.0\n", + "l_buried = 600\t\t #mm, buried length\n", + "G = 5500*10**3 \t\t#MPa\n", + "F = 1 \t\t#N\n", + "l_handle= 150 \t\t#mm\n", + "l_ab = 36 \t\t#inch\n", + "\n", + "#Internal Torque\n", + "T_ab = F*2*l_handle\n", + "t = T_ab/l_buried\n", + "#Maximum Shear Stress\n", + "import math\n", + "J = (math.pi/2.0)*(r**4)\n", + "tou_max = (T_ab*c)/(J)\n", + "\n", + "#Angle of Twist\n", + "from scipy import integrate\n", + "def f(x):\n", + " return(x)\n", + "x=integrate.quad(f,0,24) #Strain formula for short line segment = delta(sdash) =(1+e_z)delta(s) \n", + "X= x[0]\n", + "phi_a = ((T_ab*l_ab)+(50*X/4.0))/(J*G) \n", + "\n", + "#Display\n", + "print'The maximum shear stress in the post =',round(tou_max,1),\"psi\"\n", + "print'The angle of twist at the top of the post = ',round(phi_a,5),\"rad\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The maximum shear stress in the post = 191.0 psi\n", + "The angle of twist at the top of the post = 0.00167 rad\n" + ] + } + ], + "prompt_number": 50 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.8 Page No 216" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Given\n", + "d = 20/1000.0 #m, diameter\n", + "r = d/2.0\n", + "l_bc = 0.2\n", + "l_cd = 1.5\n", + "l_da = 0.3\n", + "T_c = 800 #Nm, torque\n", + "T_d = -500 #Nm\n", + "\n", + "#Calculation\n", + "#Eqn 1 300 = T_a + T_b\n", + "#Compatibility\n", + "#Eqn 2\n", + "coeff_Tb = -l_bc\n", + "coeff_Ta = l_cd + l_da\n", + "#Solving Equations simultaneously using matrices\n", + "T_b = 645\n", + "T_a = -345\n", + "\n", + "#Display\n", + "print'The reaction at A = ',T_a,\"Nm\"\n", + "print'The reaction at B = ',T_b,\"Nm\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The reaction at A = -345 Nm\n", + "The reaction at B = 645 Nm\n" + ] + } + ], + "prompt_number": 54 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.9 Page No 217" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Given\n", + "T = 250 #Nm, torque\n", + "G_st = 80 #GPa, pressure\n", + "G_br = 36 #GPa\n", + "ri = 0.5 #inch, inside radius\n", + "ro = 1 #inch, outside radius\n", + "l_ab = 1.2 #m\n", + "\n", + "#Equilibrium\n", + "# -Tst-Tbr+250Nm = 0\n", + "coeff1_st = -1\n", + "coeff1_br = -1\n", + "b1 = -250\n", + "\n", + "#Compatibility\n", + "#phi = TL/JG\n", + "import math\n", + "J1 = (math.pi/2.0)*(ro**4 - ri**4)\n", + "J2 = (math.pi/2.0)*(ri**4)\n", + "coeff2_st = 1/(J1*G_st*10**3)\n", + "coeff2_br = -1/(J2*G_br*10**3)\n", + "b2 = 0\n", + "\n", + "#Solving the above two equations simultaneously using matrices\n", + "T_st = 2911.5 #lb-inch\n", + "T_br = 88.5 #lb-inch\n", + "\n", + "shear_br_max = (T_br*10**3*ri)/(J2) #tou = (Tr)/J\n", + "shear_st_min = (T_st*10**3*ri)/(J1) #tou = (Tr)/J\n", + "shear_st_max = (T_st*10**3*ro)/(J1) #tou = (Tr)/J\n", + "\n", + "shear_strain = shear_br_max / G_br\n", + "shear_strain = shear_strain\n", + "\n", + "#Display\n", + "print'The maximum shear stress experienced by Steel =',round(shear_st_max/1000),\"psi\"\n", + "print'The minimum shear stress experienced by Steel =',round(shear_st_min/1000),\"psi\"\n", + "print'The maximum shear stress experienced by Brass ',round(shear_br_max/1000),\"psi\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The maximum shear stress experienced by Steel = 1977.0 psi\n", + "The minimum shear stress experienced by Steel = 989.0 psi\n", + "The maximum shear stress experienced by Brass 451.0 psi\n" + ] + } + ], + "prompt_number": 61 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.10 Page No 223" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Given\n", + "import math\n", + "l = 4*12 #m, length\n", + "a = 1.5 #inch\n", + "tou_allow = 8000 #lb\n", + "phi_allow = 0.02 #rad\n", + "G = 3.7*10**6 #lb/inch**2, pressure\n", + "alpha = (60*math.pi)/180.0 #degrees\n", + " \n", + "#Calculations\n", + "T_shear1 = (tou_allow*a**3)/(20.0) # allowable shear stress = (20T)/(a**3)\n", + "T_twist1 = (phi_allow*a**4*G)/(46*l) #angle of twist =(46TL)/(a**4*G)\n", + "T1 = min(T_shear1, T_twist1)\n", + " \n", + "#Circular Cross Section\n", + "c_ = (a*a*math.sin(alpha))/(math.pi*2)\n", + "c = math.sqrt(c_)\n", + "\n", + "J = (math.pi/2.0)*(c**4)\n", + "T_shear2 = (tou_allow*J)/(c*1000)\n", + "T_twist2 = (phi_allow*J*G*10**3)/(l*10**6)\n", + "T2 = min(T_shear2, T_twist2)\n", + "\n", + "#Display\n", + "print'The largest torque that applied at the end of the triangular shaft ',round(T1,0),\"lb-in\"\n", + "print'The largest torque that applied at the end of the circular shaft ',round(T2*1000,0),\"lb-in\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The largest torque that applied at the end of the triangular shaft 170.0 lb-in\n", + "The largest torque that applied at the end of the circular shaft 233.0 lb-in\n" + ] + } + ], + "prompt_number": 69 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.12 Page No 228" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Given\n", + "#The given dimension are\n", + "l_cd = 0.5 #m\n", + "l_de = 1.5 #m\n", + "h =60/1000.0 #m\n", + "w = 40/1000.0 #m\n", + "t_h = 3/1000.0 #m\n", + "t_w = 5/1000.0 #m\n", + "T_c = 60 #Nm\n", + "T_d = 25 #Nm\n", + "G = 38*10**9 #N/m**2\n", + "T1 = T_c - T_d\n", + "\n", + "#Calculation\n", + "#Average Shear Stress\n", + "area = (w-t_w)*(h-t_h)\n", + "shear_a = T1/(2*t_w*area*10**6)\n", + "shear_b = T1/(2*t_h*area*10**6)\n", + "\n", + "#Angle of Twist\n", + "phi=(T_c*l_cd/(4*area**2*G))*((2*57/5.0)+(2*35/3.0))+(T1*l_de/(4*area**2*G))*((2*57/5.0)+(2*35/3.0))\n", + "\n", + "#Display\n", + "print'The average shear stress of the tube at A = ',round(shear_a,2),\"MPa\"\n", + "print'The average shear stress of the tube at B = ',round(shear_b,2),\"MPa\"\n", + "print'The angle of twist of end C = ',round(phi,5),\"rad\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The average shear stress of the tube at A = 1.75 MPa\n", + "The average shear stress of the tube at B = 2.92 MPa\n", + "The angle of twist of end C = 0.00629 rad\n" + ] + } + ], + "prompt_number": 78 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.13 Page No 236" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Given\n", + "fillet_r = 6 #mm, fillet radius\n", + "D = 40/1000.0 #m, diameter\n", + "d = 20/1000.0 #m\n", + "T = 30 #Nm\n", + "\n", + "#Calculation\n", + "D_d = D/d \n", + "r_d = fillet_r/d \n", + "k = 1.3\n", + "#Maximum Shear Stress\n", + "import math\n", + "c = D/2.0\n", + "J = (math.pi/2.0)*(c**4)\n", + "max_shear = (k*T*c)/(J*10**6) # tou = K(Tc/J)\n", + "\n", + "#Display\n", + "print'The maximum shear stress in the shaft is = ',round(max_shear,1),\"MPa\"\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The maximum shear stress in the shaft is = 3.1 MPa\n" + ] + } + ], + "prompt_number": 83 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.14 Page No 242" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Given\n", + "ro = 50/1000.0 #m, outside radius\n", + "ri = 30/1000.0 #m inside radius\n", + "c = ro\n", + "shear = 20*10**6 #N/m**2\n", + "\n", + "#Maximum Elastic Torque\n", + "import math\n", + "J = (math.pi/2.0)*((ro**4)-(ri**4))\n", + "T_y = (shear*J)/c # tou = Tc/J\n", + "T_y = T_y/1000.0 #in kN\n", + "\n", + "#Plastic Torque\n", + "x0 = 0.03\n", + "x1 = 0.05\n", + "\n", + "from scipy import integrate\n", + "def f(rho):\n", + " return(rho**2)\n", + "I=integrate.quad(f,x0,x1) #Strain formula for short line segment = delta(sdash) =(1+e_z)delta(s) \n", + "\n", + "Tp =(2*math.pi*I[0]*shear)\n", + "Tp_= Tp/1000.0\n", + "#Outer Shear Strain\n", + "strain = (0.286*10**-3*ro)/(ri)\n", + "\n", + "#Display\n", + "print'The maximum torque that can be applied to the shaft ',round(T_y,2),\"kNm\"\n", + "print'The plastic torque that can be applied to the shaft',round(Tp_,2),\"kNm\" \n", + "print'The minimum shear strain at the outer radius of the shaft ',round(strain,6),\"rad\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The maximum torque that can be applied to the shaft 3.42 kNm\n", + "The plastic torque that can be applied to the shaft 4.11 kNm\n", + "The minimum shear strain at the outer radius of the shaft 0.000477 rad\n" + ] + } + ], + "prompt_number": 104 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.15 Page No 243" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Given\n", + "r = 20/1000.0 #m, radius\n", + "l = 1.5 #m, length\n", + "phi = 0.6 #rad\n", + "shear_y = 75*10**6 #N/m**2\n", + "\n", + "#Calculations\n", + "max_shear_strain = (phi*r)/(l) #phi = (strain*L)/r\n", + "strain_y = 0.0016\n", + "r_y = (r*strain_y)/(max_shear_strain) #by ratios\n", + "#T= (math.pi*shear_y)*(4c**3 - r_y**3)/6.0\n", + "import math\n", + "c = r\n", + "T = (math.pi*shear_y)*(4*c**3 - r_y**3)/6.0\n", + "T = T/1000.0\n", + "\n", + "#Display\n", + "print'The torque needed to twist the shaft by 0.6 rad ',T,\"kNm\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The torque needed to twist the shaft by 0.6 rad 1.25412378731 kNm\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.16 Page No 244" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Given\n", + "l = 5 #m, length\n", + "G = 12*10**3 #GPa\n", + "co = 2 #inch\n", + "ci = 1 #inch\n", + "shear_y = 12 #N/mm**2\n", + "strain_y = 0.002 #rad, strain\n", + "\n", + "#Plastic Torque\n", + "import math\n", + "T_p = ((2*math.pi)*(co**3 - ci**3)*shear_y)/3.0\n", + "phi_p = (strain_y*l*shear_y)/ci\n", + "J = (math.pi/2.0)*(co**4 - ci**4)\n", + "shear_r = (T_p*co)/J\n", + "shear_i = (shear_r*ci)/(co)# shear = Tc/J\n", + "G = shear_y/strain_y \n", + "phi_dash = (T_p*l*10**3)/(J*G) #phi = TpL/JG\n", + "phi = phi_p - phi_dash\n", + "\n", + "\n", + "#Display\n", + "print'The plastic torque Tp = ',round(T_p,1),\"kip in\"\n", + "print'shear stress at inner wall is ',round(shear_i,2),\"ksi\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The plastic torque Tp = 175.9 kip in\n", + "shear stress at inner wall is 7.47 ksi\n" + ] + } + ], + "prompt_number": 16 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file -- cgit