diff options
| author | Thomas Stephen Lee | 2015-08-28 16:53:23 +0530 |
|---|---|---|
| committer | Thomas Stephen Lee | 2015-08-28 16:53:23 +0530 |
| commit | 4a1f703f1c1808d390ebf80e80659fe161f69fab (patch) | |
| tree | 31b43ae8895599f2d13cf19395d84164463615d9 /Engineering_Mechanics_by_Tayal_A.K./chapter07_13.ipynb | |
| parent | 9d260e6fae7328d816a514130b691fbd0e9ef81d (diff) | |
| download | Python-Textbook-Companions-4a1f703f1c1808d390ebf80e80659fe161f69fab.tar.gz Python-Textbook-Companions-4a1f703f1c1808d390ebf80e80659fe161f69fab.tar.bz2 Python-Textbook-Companions-4a1f703f1c1808d390ebf80e80659fe161f69fab.zip | |
add books
Diffstat (limited to 'Engineering_Mechanics_by_Tayal_A.K./chapter07_13.ipynb')
| -rw-r--r-- | Engineering_Mechanics_by_Tayal_A.K./chapter07_13.ipynb | 551 |
1 files changed, 551 insertions, 0 deletions
diff --git a/Engineering_Mechanics_by_Tayal_A.K./chapter07_13.ipynb b/Engineering_Mechanics_by_Tayal_A.K./chapter07_13.ipynb new file mode 100644 index 00000000..aae60e42 --- /dev/null +++ b/Engineering_Mechanics_by_Tayal_A.K./chapter07_13.ipynb @@ -0,0 +1,551 @@ +{
+ "metadata": {
+ "name": "chapter7.ipynb"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 7: Application Of Friction"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.7-1,Page No:152"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "# Initilization of variables\n",
+ "\n",
+ "d1=24 # cm # diameter of larger pulley\n",
+ "d2=12 # cm # diameter of smaller pulley\n",
+ "d=30 #cm # seperation betweem 1st & the 2nd pulley\n",
+ "pie=3.14\n",
+ "\n",
+ "# Calcuations\n",
+ "\n",
+ "r1=d1*0.5 #cm # radius of 1st pulley #1/2=0.5\n",
+ "r2=d2*0.5 #cm # radius of 2nd pulley #1/2=0.5\n",
+ "theta=(arcsin((r1-r2)/d))*(180/pi) #degrees \n",
+ "\n",
+ "# Angle of lap\n",
+ "beta_1=180+(2*theta) #degree # for larger pulley\n",
+ "beta_2=180-(2*theta) #degree # for smaller pulley\n",
+ "L=pie*(r1+r2)+(2*d)+((r1-r2)**(2/d)) #cm # Length of the belt\n",
+ "\n",
+ "# Results\n",
+ "\n",
+ "print\"The angle of lap for the larger pulley is \",round(beta_1,2),\"degree\"\n",
+ "print\"he angle of lap for the smaller pulley is \",round(beta_2,2),\"degree\"\n",
+ "print\"he length of pulley required is \",round(L,2),\"cm\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The angle of lap for the larger pulley is 203.07 degree\n",
+ "he angle of lap for the smaller pulley is 156.93 degree\n",
+ "he length of pulley required is 117.52 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 22
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.7-2,Page No:153"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "# Initilization of variables\n",
+ "\n",
+ "d1=0.6 #m # diameter of larger pulley\n",
+ "d2=0.3 #m diameter of smaller pulley\n",
+ "d=3.5 #m #separation between the pulleys\n",
+ "pie=3.14\n",
+ "\n",
+ "# Calculations\n",
+ "\n",
+ "r1=d1*0.5 #m # radius of larger pulley #1/2=0.5\n",
+ "r2=d2*0.5 #m # radius of smaller pulley #1/2=0.5\n",
+ "theta=arcsin((r1+r2)/d)*(180/pi) #degree\n",
+ "\n",
+ "# Angle of lap for both the pulleys is same, i.e\n",
+ "\n",
+ "beta=180+(2*theta) # degree\n",
+ "L=((pie*(r1+r2))+(2*d)+(((r1+r2)**2)/d)) #cm # Length of the belt\n",
+ "\n",
+ "# Results\n",
+ "\n",
+ "print\"The angle of lap for the pulley is \",round(beta,1),\"degree\"\n",
+ "print\"The length of pulley required is \",round(L,3),\"m\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The angle of lap for the pulley is 194.8 degree\n",
+ "The length of pulley required is 8.471 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 29
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.7-4,Page No:161"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "# Initilization of variables\n",
+ "\n",
+ "W1=1000 #N\n",
+ "mu=0.25 #coefficient of friction\n",
+ "pie=3.14\n",
+ "beta=pie\n",
+ "T1=W1 # Tension in the 1st belt carrying W1\n",
+ "e=2.718 #constant\n",
+ "\n",
+ "# Calculations\n",
+ "\n",
+ "T2=T1/(e**(mu*beta)) #N # Tension in the 2nd belt\n",
+ "W2=T2 #N\n",
+ "\n",
+ "# Results\n",
+ "\n",
+ "print\"The minimum weight W2 to keep W1 in equilibrium is \",round(W2),\"N\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The minimum weight W2 to keep W1 in equilibrium is 456.0 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 31
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.7-5,Page No:162"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "# Initilization of variables\n",
+ "\n",
+ "mu=0.5 # coefficient of friction between the belt and the wheel\n",
+ "W=100 #N\n",
+ "theta=45 #degree\n",
+ "e=2.718\n",
+ "Lac=0.75 #m # ength of the lever\n",
+ "Lab=0.25 #m\n",
+ "Lbc=0.50 #m\n",
+ "r=0.25 #m\n",
+ "pie=3.14 # constant\n",
+ "\n",
+ "# Calculations\n",
+ "\n",
+ "beta=((180+theta)*pie)/180 # radian # angle of lap\n",
+ "\n",
+ "# from eq'n 2\n",
+ "T1=(W*Lbc)/Lab #N \n",
+ "T2=T1/(e**(mu*beta)) #N # from eq'n 1\n",
+ "\n",
+ "# consider the F.B.D of the pulley and take moment about its center, we get Braking Moment (M)\n",
+ "M=r*(T1-T2) #N-m\n",
+ "\n",
+ "# Results\n",
+ "\n",
+ "print\"The braking moment (M) exerted by the vertical weight W is \",round(M,2),\"N-m\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The braking moment (M) exerted by the vertical weight W is 42.97 N-m\n"
+ ]
+ }
+ ],
+ "prompt_number": 32
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.7-6,Page No:163"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "# Initiization of variables\n",
+ "\n",
+ "W= 1000 #N # or 1kN\n",
+ "mu=0.3 # coefficient of friction between the rope and the cylinder\n",
+ "e=2.718 # constant\n",
+ "pie=3.14 # constant\n",
+ "alpha=90 # degree # since 2*alpha=180 egree\n",
+ "\n",
+ "# Calculations\n",
+ "\n",
+ "beta=2*pie*3 # radian # for 3 turn of the rope\n",
+ "\n",
+ "# Here T1 is the larger tension in that section of the rope which is about to slip\n",
+ "T1=W #N\n",
+ "F=W/e**(mu*(1/(sin(alpha*(pi/180))))*(beta)) #N # Here T2=F\n",
+ "\n",
+ "# Results\n",
+ "\n",
+ "print\"The force required to suport the weight of 1000 N i.e 1kN is \",round(F,1),\"N\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The force required to suport the weight of 1000 N i.e 1kN is 3.5 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 34
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.7-7,Page No:163"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "# Initilization of variables\n",
+ "\n",
+ "Pw=50 #kW\n",
+ "T_max=1500 #N\n",
+ "v=10 # m/s # velocity of rope\n",
+ "w=4 # N/m # weight of rope\n",
+ "mu=0.2 # coefficient of friction \n",
+ "g=9.81 # m/s^2 # acceleration due to gravity\n",
+ "e=2.718 # constant\n",
+ "pie=3.14 # constant\n",
+ "alpha=30 # degree # since 2*alpha=60 \n",
+ "\n",
+ "# Calcuations\n",
+ "\n",
+ "beta=pie # radian # angle of lap\n",
+ "T_e=(w*v**2)/g # N # where T_e is the centrifugal tension\n",
+ "T1=(T_max)-(T_e) #N\n",
+ "T2=T1/(e**(mu*(1/sin(alpha*(pi/180)))*(beta))) #N # From eq'n T1/T2=e^(mu*cosec(alpha)*beta)\n",
+ "P=(T1-T2)*v*(10**-3) #kW # power transmitted by a single rope\n",
+ "N=Pw/P # Number of ropes required\n",
+ "\n",
+ "# Results\n",
+ "\n",
+ "print\"The number of ropes required to transmit 50 kW is \",round(N),\"Nos\"\n",
+ "# approx no of ropes is 5\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The number of ropes required to transmit 50 kW is 5.0 Nos\n"
+ ]
+ }
+ ],
+ "prompt_number": 36
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.7-8,Page No:164"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "# Initilization of variables\n",
+ "\n",
+ "d1=0.45 #m # diameter of larger pulley\n",
+ "d2=0.20 #m # diameter of smaller pulley\n",
+ "d=1.95 #m # separation between the pulley's\n",
+ "T_max=1000 #N # or 1kN which is the maximum permissible tension\n",
+ "mu=0.20 # coefficient of friction\n",
+ "N=100 # r.p.m # speed of larger pulley\n",
+ "e=2.718 # constant\n",
+ "pie=3.14 # constant\n",
+ "T_e=0 #N # as the data for calculating T_e is not given we assume T_e=0\n",
+ "\n",
+ "# Calculations\n",
+ "\n",
+ "r1=d1*0.5 #m # radius of larger pulley #1/2=0.5\n",
+ "r2=d2*0.5 #m # radius of smaller pulley #1/2=0.5\n",
+ "theta=arcsin((r1+r2)/d)*(180/pi) # degree\n",
+ "\n",
+ "# for cross drive the angle of lap for both the pulleys is same\n",
+ "beta=(180+(2*(theta)))*(pi/180) #radian\n",
+ "T1=T_max-T_e #N\n",
+ "T2=T1/(e**(mu*(beta))) #N # from formulae, T1/T2=e^(mu*beta)\n",
+ "v=((2*pie)*N*r1)/60 # m/s # where v=velocity of belt which is given as, v=wr=2*pie*N*r/60\n",
+ "P=(T1-T2)*v*(10**-3) #kW # Power\n",
+ "\n",
+ "# Results\n",
+ "\n",
+ "print\"The power transmitted by the cross belt drive is \",round(P,1),\"kW\"\n",
+ "# the approx answer is 1.3 kW The answer given in the book (i.e 1.81kW) is wrong.\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The power transmitted by the cross belt drive is 1.2 kW\n"
+ ]
+ }
+ ],
+ "prompt_number": 48
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.7-9,Page No:165"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "# Initilization of variabes\n",
+ "\n",
+ "b=0.1 #m #width of the belt\n",
+ "t=0.008 #m #thickness of the belt\n",
+ "v=26.67 # m/s # belt speed\n",
+ "pie=3.14 # constant\n",
+ "beta=165 # radian # angle of lap for the smaller belt\n",
+ "mu=0.3 # coefficient of friction\n",
+ "sigma_max=2 # MN/m^2 # maximum permissible stress in the belt\n",
+ "m=0.9 # kg/m # mass of the belt\n",
+ "g=9.81 # m/s^2\n",
+ "e=2.718 # constant\n",
+ "\n",
+ "# Calculations\n",
+ "\n",
+ "A=b*t # m^2 # cross-sectional area of the belt\n",
+ "T_e=m*v**2 # N # where T_e is the Centrifugal tension\n",
+ "T_max=(sigma_max)*(A)*(10**6) # N # maximum tension in the belt\n",
+ "T1=(T_max)-(T_e) # N \n",
+ "T2=T1/(e**((mu*pie*beta)/180)) #N # from formulae T1/T2=e^(mu*beta)\n",
+ "P=(T1-T2)*v*(10**-3) #kW # Power transmitted\n",
+ "T_o=(T1+T2)/2 # N # Initial tension\n",
+ "\n",
+ "# Now calculations to transmit maximum power\n",
+ "\n",
+ "Te=T_max/3 # N # max tension\n",
+ "u=(T_max/(3*m))**0.5 # m/s # belt speed for max power\n",
+ "T_1=T_max-Te # N # T1 for case 2\n",
+ "T_2=T_1/(e**((mu*pie*beta)/180)) # N \n",
+ "P_max=(T_1-T_2)*u*(10**-3) # kW # Max power transmitted\n",
+ "\n",
+ "# Results\n",
+ "\n",
+ "print\"The initial power transmitted is \",round(P,2),\"kW\"\n",
+ "print\"The initial tension in the belt is \",round(T_o,1),\"N\"\n",
+ "print\"The maximum power that can be transmitted is \",round(P_max,2),\"kW\" # the answer is approx 15.017kW wheres the answer 14.99kW given in book is wrong \n",
+ "print\"The maximum power is transmitted at a belt speed of \",round(u,2),\"m/s\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The initial power transmitted is 14.8 kW\n",
+ "The initial tension in the belt is 682.3 N\n",
+ "The maximum power that can be transmitted is 15.02 kW\n",
+ "The maximum power is transmitted at a belt speed of 24.34 m/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 54
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.7-10,Page No:169"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "# Initilization of variables\n",
+ "\n",
+ "p=0.0125 # m # pitch of screw\n",
+ "d=0.1 #m # diameter of the screw\n",
+ "r=0.05 #m # radius of the screw\n",
+ "l=0.5 #m # length of the lever\n",
+ "W=50 #kN # load on the lever\n",
+ "mu=0.20 # coefficient of friction \n",
+ "pie=3.14 #constant\n",
+ "\n",
+ "# Calculations\n",
+ "\n",
+ "theta=arctan(p/(2*pie*r))*(180/pi) #degree # theta is the Helix angle\n",
+ "phi=arctan(mu)*(180/pi) # degree # phi is the angle of friction\n",
+ "\n",
+ "# Taking the leverage due to handle into account,force F1 required is,\n",
+ "F1=(W*(tan(theta*(pi/180)+phi*(pi/180))))*(r/l) #kN\n",
+ "\n",
+ "# To lower the load\n",
+ "F2=(W*(tan(theta*(pi/180)-phi*(pi/180))))*(r/l) #kN # -ve sign of F2 indicates force required is in opposite sense\n",
+ "E=(tan(theta*(pi/180))/tan(theta*(pi/180)+phi*(pi/180)))*100 # % # here E=eata=efficiency in %\n",
+ "\n",
+ "# Results\n",
+ "\n",
+ "print\"The force required (i.e F1) to raise the weight is \",round(F1,3),\"kN\" #due to decimal variance answer varies by 0.004kN \n",
+ "print\"The force required (i.e F2) to lower the weight is \",round(F2,3),\"kN\"\n",
+ "print\"The efficiency of the jack is \",round(E,2),\"percent\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The force required (i.e F1) to raise the weight is 1.209 kN\n",
+ "The force required (i.e F2) to lower the weight is -0.795 kN\n",
+ "The efficiency of the jack is 16.47 percent\n"
+ ]
+ }
+ ],
+ "prompt_number": 58
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.7-11,Page No:172"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "# Initilization of variabes\n",
+ "\n",
+ "P=20000 #N #Weight of the shaft\n",
+ "D=0.30 #m #diameter of the shaft\n",
+ "R=0.15 #m #radius of the shaft\n",
+ "mu=0.12 # coefficient of friction\n",
+ "\n",
+ "# Calculations\n",
+ "\n",
+ "# Friction torque T is given by formulae,\n",
+ "T=(0.666)*P*R*mu #N-m #2/3=0.666\n",
+ "M=T #N-m\n",
+ "\n",
+ "# Results\n",
+ "\n",
+ "print\"The frictional torque is \",round(M),\"N-m\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The frictional torque is 240.0 N-m\n"
+ ]
+ }
+ ],
+ "prompt_number": 61
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file |