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diff --git a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter03_2.ipynb b/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter03_2.ipynb new file mode 100755 index 00000000..83b15053 --- /dev/null +++ b/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter03_2.ipynb @@ -0,0 +1,346 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:e94e252902947538e327888c054726e4c87fafcbcff4841a5585391ce610b765" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Chapter 03:Torsion" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.3.1, Page No:79" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable Decleration\n", + "P=20*10**3 #Power in W\n", + "f=2 #Frequency in Hz\n", + "t_max=40*10**6 #Maximum shear stress in Pa\n", + "G=83*10**9 #Bulk modulus in Pa\n", + "theta=(6*pi)/180 #Angle of twist in radians\n", + "L=3 #Length in m\n", + "\n", + "#Calculations\n", + "#Strength condition\n", + "T=P/(2*pi*f) #Torque in N.m\n", + "d1=((16*T)/(pi*t_max))**0.333 #Max allowable diameter in mm\n", + "\n", + "#Applying torque-twist relationship\n", + "d2=((32*T*L)/(G*theta*pi))**0.25 #Diameter in mm\n", + "\n", + "d=max(d1,d2)\n", + "\n", + "print \"To satisfy both strength and rigidity conditions d=\",round(d*1000,1),\"mm\"\n", + "\n", + "#NOTE:The fractional power leads to the discrepancy in the answer\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "To satisfy both strength and rigidity conditions d= 58.9 mm\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.3.2, Page No:79" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable Decleration\n", + "Ga=4*10**6 #Bulk modulus of Aluminium in psi\n", + "Gs=12*10**6 #Bulk Modulus of Steel in psi\n", + "T=10**4 #Torque in lb.in\n", + "L1=3 #Length in ft of the Steel bar\n", + "L2=6 #Length in ft of the Aluminium bar\n", + "d1=3 #Diameter of the Aluminium bar in inches\n", + "d2=2 #Diameter of the Steel bar in inches\n", + "\n", + "#Calculations\n", + "#Using Compatibility and equlibrium conditions\n", + "a=np.array([[1,1],[(L1*32)/(Gs*pi*d2**4),-((L2*32)/(Ga*d1**4*pi))]])\n", + "b=np.array([T,0])\n", + "y=np.linalg.solve(a,b)\n", + "\n", + "#Stresses\n", + "t_max_st=(16*y[0])/(pi*d2**3) #Max shear Stress in Steel in psi\n", + "t_max_al=(16*y[1])/(pi*d1**3) #Max shear stress in aluminium in psi\n", + "\n", + "print \"The maximum values of Shear Stresses are as flollows\"\n", + "print round(t_max_st),\"psi in Steel and \",round(t_max_al),\"psi in Aluminium\"\n", + "#NOTE:The shear stress for steel in the txtbook is off by 3psi" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The maximum values of Shear Stresses are as flollows\n", + "3453.0 psi in Steel and 863.0 psi in Aluminium\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.3.3, Page No:80" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable Decleration\n", + "d=2 #Diameter in ft\n", + "G=12*10**6 #Bulk Modulus in psi\n", + "#Torque in lb.ft\n", + "T1=500 #Torque 1 \n", + "T2=900 #Torque 2\n", + "T3=1000 #Torque 3\n", + "#Length in ft\n", + "L1=4 \n", + "L2=3\n", + "L3=5\n", + "\n", + "#Calculations\n", + "#Applying the sum of torques we get\n", + "Tab=T1 #Torque at section AB in lb.ft\n", + "Tbc=-T2+T1 #Torque at section BC in lb.ft\n", + "Tcd=T3-T2+T1 #Torque at Section CD in lb.ft\n", + "\n", + "#Summing the angle of twists\n", + "theta_r=(((Tab*12*L3*12)+(Tbc*12*L2*12)+(Tcd*12*L1*12))*32)/(pi*2**4*G)\n", + "theta=(theta_r*180)/pi #Angle in degrees\n", + "\n", + "print \"The angle of twist is\",round(theta,3),\"degrees\"\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The angle of twist is 1.62 degrees\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.3.4, Page No:81" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable Decleration\n", + "L=1.5 #Length of the shaft in m\n", + "t_B=200 #Torque per unit length in N.m/m\n", + "d=0.025 #Diameter of the shaft in m\n", + "G=80*10**9 #Bulk Modulus for steel in Pa\n", + "\n", + "\n", + "#Calculations\n", + "#Part(1)\n", + "#After carrying out the variable integration\n", + "T_A=0.5*t_B*L #Torque about point A in N.m\n", + "#Using equation of max stress\n", + "tau_Max=(16*T_A)*(pi*d**3)**-1 #Maximum stress in the shaft in Pa\n", + "\n", + "#Part(2)\n", + "J=(pi*d**4)*32**-1 #Polar moment of inertia in m^4\n", + "#After carrying out the computation for angle of twist we get\n", + "theta_r=(t_B*L**2)*(3*G*J)**-1 #Angle of twist in radians\n", + "theta=theta_r*(180*pi**-1) #Angle of twist in degrees\n", + "\n", + "#Result\n", + "print \"Result for part (1)\"\n", + "print \"Maximum Shear Stress in the shaft is\",round(tau_Max/10**6,1),\"MPa\"\n", + "print \"Result for part (2)\"\n", + "print \"The angle of twist in the shaft is\",round(theta,2),\"degrees\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Result for part (1)\n", + "Maximum Shear Stress in the shaft is 48.9 MPa\n", + "Result for part (2)\n", + "The angle of twist in the shaft is 2.8 degrees\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.3.5, Page No:91" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable Decleration\n", + "L=6 #Length of the tube in ft\n", + "t=3*8**-1 #Constant wall thickness in inches\n", + "G=12*10**6 #Bulk modulus of the tube in psi\n", + "w1=6 #Width on the top in inches\n", + "w2=4 #Width at the bottom in inches\n", + "h=5 #Height in inches\n", + "theta=0.5 #Angle of twist in radians\n", + "\n", + "#Calculations\n", + "#Part(1)\n", + "Ao=(w1+w2)*2**-1*h #Area enclosed by the median line in sq.in\n", + "S=w1+w2+2*(sqrt(1**2+h**2)) #Length of the median line in inches\n", + "#Using the torsional stifness formula we get\n", + "k=4*G*Ao**2*t*(L*12*S)**-1*(pi*180**-1) #tortional Stiffness in lb.in/rad\n", + "\n", + "#Part(2)\n", + "T=k*theta #Torque required to produce an angle of twist of theta in lb.in\n", + "q=T*(2*Ao)**-1 #Shear flow in lb/in\n", + "tau=q/t #Shear stress in the wall in psi\n", + "\n", + "\n", + "#Result \n", + "print \"Part(1) results\"\n", + "print \"Torsional stiffness is\",round(k),\"lb.in/deg\"\n", + "print \"Part(2) results\"\n", + "print \"The shear stress in the wall is\", round(tau),\"psi\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Part(1) results\n", + "Torsional stiffness is 135017.0 lb.in/deg\n", + "Part(2) results\n", + "The shear stress in the wall is 3600.0 psi\n" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.3.6, Page No:92" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable Decleration\n", + "L=1.2 #Length of the tube in m\n", + "tau=40*10**6 #MAximum shear stress in MPa\n", + "t=0.002 #Thickness in m\n", + "r=0.025 #Radius of the semicircle in m\n", + "G=28*10**9 #Bulk Modulus in Pa\n", + "t1=2 #Thickness in mm\n", + "t2=3 #thickness in mm\n", + "\n", + "#Calculations\n", + "#Part(1)\n", + "q=tau*t #Shear flow causing the stress in N/m\n", + "Ao=pi*r**2*0.5 #Area of the semi-circle in m^2\n", + "T=2*Ao*q #Torque causing the shear stress in N.m\n", + "\n", + "#Part(2)\n", + "#After computing the median lines integration we get\n", + "S=(pi*25*t1**-1)+(2*25*t2**-1) #Length of median line \n", + "theta_r=T*L*S*(4*G*Ao**2)**-1 #Angle of twist in radians\n", + "theta=theta_r*(180*pi**-1) #Angle of twist in degrees\n", + "\n", + "#Result\n", + "print \"Result for part(1)\"\n", + "print \"The torque causing the stress of 40MPa is\",round(T,2),\"N.m\"\n", + "print \"Result for part (2)\"\n", + "print \"The angle of twist is\",round(theta,1),\"degrees\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Result for part(1)\n", + "The torque causing the stress of 40MPa is 157.08 N.m\n", + "Result for part (2)\n", + "The angle of twist is 5.6 degrees\n" + ] + } + ], + "prompt_number": 21 + } + ], + "metadata": {} + } + ] +}
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