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+{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:abce654f2dcdd836a8080165fb072744ef2446714ffd5c0acdbf346c961eccf3"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 1 :Simple Stresses and Strains"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Problem 1.1,page no.9"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#Given\n",
+ "#Variable declaration\n",
+ "L=150 #Length of the rod in cm\n",
+ "D=20 #Diameter of the rod in mm\n",
+ "P=20*10**3 #Axial pull in N\n",
+ "E=2.0e5 #Modulus of elasticity in N/sq.mm\n",
+ "\n",
+ "#Calculation\n",
+ "A=(math.pi/4)*(D**2) #Area in sq.mm\n",
+ " #case (i):stress\n",
+ "sigma=P/A #Stress in N/sq.mm\n",
+ " #case (ii):strain\n",
+ "e=sigma/E #Strain\n",
+ " #case (iii):elongation of the rod\n",
+ "dL=e*L #Elongation of the rod in cm\n",
+ "\n",
+ "#Result\n",
+ "print \"Stress =\",round(sigma,3),\"N/mm^2\"\n",
+ "print \"Strain =\",round(e,6)\n",
+ "print \"Elongation =\",round(dL,4),\"cm\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Stress = 63.662 N/mm^2\n",
+ "Strain = 0.000318\n",
+ "Elongation = 0.0477 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Problem 1.2,page no.10"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Given\n",
+ "#variable declaration\n",
+ "P=4000 #Load in N\n",
+ "sigma=95 #Stress in N/sq.mm\n",
+ "\n",
+ "#Calculation\n",
+ "D=round(math.sqrt(P/((math.pi/4)*(sigma))),2) #Diameter of steel wire in mm\n",
+ "\n",
+ "#Result\n",
+ "print \"Diameter of a steel wire =\",D,\"mm\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Diameter of a steel wire = 7.32 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Problem 1.3,page no.10"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Given\n",
+ "#Variable declaration\n",
+ "D=25 #Diameter of brass rod in mm\n",
+ "P=50*10**3 #Tensile load in N\n",
+ "L=250 #Length of rod in mm\n",
+ "dL=0.3 #Extension of rod in mm\n",
+ "\n",
+ "#Calculation\n",
+ "A=(math.pi/4)*(D**2) #Area of rod in sq.mm\n",
+ "sigma=round(P/A,2) #Stress in N/sq.mm\n",
+ "e=dL/L #Strain\n",
+ "E=(sigma/e) #Young's Modulus in N/sq.m\n",
+ "\n",
+ "#Result\n",
+ "print \"Young's Modulus of a rod,E =\",round(E*(10**-3),3),\"GN/m^2\" #Young's Modulus in GN/sq.m\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Young's Modulus of a rod,E = 84.883 GN/m^2\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Problem 1.4,page no.11"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Given\n",
+ "#Variable Declaration\n",
+ "D=3 #Diameter of the steel bar in cm \n",
+ "L=20 #Gauge length of the bar in cm\n",
+ "P=250 #Load at elastic limit in kN \n",
+ "dL=0.21 #Extension at a load of 150kN in mm\n",
+ "Tot_ext=60 #Total extension in mm\n",
+ "Df=2.25 #Diameter of the rod at the failure in cm\n",
+ "\n",
+ "#Calculation\n",
+ "A=round((math.pi/4)*(D**2),5) #Area of the rod in sq.m\n",
+ "\n",
+ "#case (i):Young's modulus\n",
+ "e=round((150*1000)/(7.0685),1) #stress in N/sq.m\n",
+ "sigma=dL/(L*10) #strain \n",
+ "E=round((e/sigma)*(10**-5),3) #Young's modulus in GN/sq.m\n",
+ "\n",
+ "#case (ii):stress at elastic limit\n",
+ "stress=int(round((P*1000)/A,0))*1e-2 #stress at elastic limit in MN/sq.m\n",
+ "\n",
+ "#case (iii):percentage elongation\n",
+ "Pe=(Tot_ext*1e2)/(L*10)\n",
+ "\n",
+ "#case (iv):percentage decrease in area\n",
+ "Pd=(D**2-Df**2)/D**2*1e2\n",
+ "\n",
+ "\n",
+ "#Result\n",
+ "print \"NOTE:The Young's Modulus found in the book is incorrect.The correct answer is,\"\n",
+ "print \"Young's modulus,E =\",E,\"GN/m^2\"\n",
+ "print \"Stress at the elastic limit,Stress =\",stress,\"MN/m^2\"\n",
+ "print \"Percentage elongation = %d%%\"%Pe\n",
+ "print \"Percentage decrease in area = %.2f%%\"%Pd\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "NOTE:The Young's Modulus found in the book is incorrect.The correct answer is,\n",
+ "Young's modulus,E = 202.104 GN/m^2\n",
+ "Stress at the elastic limit,Stress = 353.68 MN/m^2\n",
+ "Percentage elongation = 30%\n",
+ "Percentage decrease in area = 43.75%\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Problem 1.5,page no.12"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Given\n",
+ "#Variable declaration\n",
+ "sigma=125*10**6 #Safe stress in N/sq.m\n",
+ "P=2.1*10**6 #Axial load in N\n",
+ "D=0.30 #External diameter in m\n",
+ "\n",
+ "#Calculation\n",
+ " \n",
+ "d=round(math.sqrt((D**2)-P*4/(math.pi*sigma)),4)*1e2 #internal diameter in cm\n",
+ "\n",
+ "#Result\n",
+ "print \"internal diameter =\",d,\"cm\" \n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "internal diameter = 26.19 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Problem 1.6,page no.13"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Given\n",
+ "#Variable declaration\n",
+ "stress=480 #ultimate stress in N/sq.mm\n",
+ "P=1.9*10**6 #Axial load in N\n",
+ "D=200 #External diameter in mm\n",
+ "f=4 #Factor of safety\n",
+ "\n",
+ "#Calculation\n",
+ "sigma=stress/f #Working stress or Permissable stress in N/sq.mm\n",
+ "d=str(math.sqrt((D**2)-((P*4)/(math.pi*sigma))))[:6] #internal diameter in mm\n",
+ "\n",
+ "#Result\n",
+ "print \"internal diameter =\",d,\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "internal diameter = 140.85 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Problem 1.15,page no.26"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Given\n",
+ "#Variable declaration\n",
+ "D1=40 #Larger diameter in mm\n",
+ "D2=20 #Smaller diameter in mm\n",
+ "L=400 #Length of rod in mm\n",
+ "P=5000 #Axial load in N\n",
+ "E=2.1e5 #Young's modulus in N/sq.mm\n",
+ "\n",
+ "#Calculation\n",
+ "dL=float(str((4*P*L)/(math.pi*E*D1*D2))[:7]) #extension of rod in mm\n",
+ "\n",
+ "#Result\n",
+ "print \"Extension of the rod =\",dL,\"mm\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Extension of the rod = 0.01515 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Problem 1.16,page no.27"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Given\n",
+ "#Variable declaration\n",
+ "D1=30 #Larger diameter in mm\n",
+ "D2=15 #Smaller diameter in mm\n",
+ "L=350 #Length of rod in mm\n",
+ "P=5.5*10**3 #Axial load in N\n",
+ "dL=0.025 #Extension in mm\n",
+ "\n",
+ "#Calculation\n",
+ "E=int((4*P*L)/(math.pi*D1*D2*dL)) #Modulus of elasticity in N/sq.mm\n",
+ "\n",
+ "#Result\n",
+ "print \"Modulus of elasticity,E = %.5e\"%E,\"N/mm^2\" "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Modulus of elasticity,E = 2.17865e+05 N/mm^2\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Problem 1.17,page no.29"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Given\n",
+ "#Variable declaration\n",
+ "L=2.8*10**3 #Length in mm\n",
+ "t=15 #Thickness in mm\n",
+ "P=40*10**3 #Axial load in N\n",
+ "a=75 #Width at bigger end in mm\n",
+ "b=30 #Width at smaller end in mm\n",
+ "E=2e5 #Young's Modulus in N/sq.mm\n",
+ "\n",
+ "#Calculation\n",
+ "dL=round((round((P*L)/(E*t*(a-b)),4)*(round(math.log(a)-math.log(b),4))),2) #extension of rod in mm\n",
+ "\n",
+ "#Result\n",
+ "print \"Extension of the rod,dL =\",dL,\"mm\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Extension of the rod,dL = 0.76 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Problem 1.18,page no.29"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Given\n",
+ "#Variable declaration\n",
+ "dL=0.21 #Extension in mm\n",
+ "L=400 #Length in mm\n",
+ "t=10 #Thickness in mm\n",
+ "a=100 #Width at bigger end in mm\n",
+ "b=50 #Width at smaller end in mm\n",
+ "E=2e5 #Young's Modulus in N/sq.mm\n",
+ "\n",
+ "#Calculation\n",
+ "P=int(dL/(round((L)/(E*t*(a-b)),6)*(round(math.log(a)-math.log(b),4))))*1e-3 #Axial load in kN\n",
+ "\n",
+ "#Result\n",
+ "print \"Axial load =\",P,\"kN\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Axial load = 75.746 kN\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Problem 1.20,page no.32"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Given\n",
+ "#Variable declaration\n",
+ "Di_s=140 #Internal diameter of steel tube in mm \n",
+ "De_s=160 #External diameter of steel tube in mm\n",
+ "Di_b=160 #Internal diameter of brass tube in mm \n",
+ "De_b=180 #External diameter of brass tube in mm\n",
+ "P=900e3 #Axial load in N\n",
+ "L=140 #Length of each tube in mm\n",
+ "Es=2e5 #Young's modulus for steel in N/sq.mm\n",
+ "Eb=1e5 #Young's modulus for brass in N/sq.mm\n",
+ "\n",
+ "#Calculation\n",
+ "As=round(math.pi/4*(De_s**2-Di_s**2),1) #Area of steel tube in sq.mm\n",
+ "Ab=round(math.pi/4*(De_b**2-Di_b**2),1) #Area of brass tube in sq.mm\n",
+ "sigmab=round(P/(2*As+Ab),2) #Stress in steel in N/sq.mm\n",
+ "sigmas=2*sigmab #Stress in brass in N/sq.mm\n",
+ "Pb=int(sigmab*Ab)*1e-3 #Load carried by brass tube in kN\n",
+ "Ps=(P*1e-3)-(Pb) #Load carried by steel tube in kN\n",
+ "dL=round(sigmab/Eb*(L),4) #Decrease in length in mm\n",
+ "\n",
+ "#Result\n",
+ "print \"Stress in brass =\",sigmab,\"N/mm^2\"\n",
+ "print \"Stress in steel =\",sigmas,\"N/mm^2\"\n",
+ "print \"Load carried by brass tube =\",Pb,\"kN\"\n",
+ "print \"Load carried by stress tube =\",Ps,\"kN\"\n",
+ "print \"Decrease in the length of the compound tube=\",dL,\"mm\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Stress in brass = 60.95 N/mm^2\n",
+ "Stress in steel = 121.9 N/mm^2\n",
+ "Load carried by brass tube = 325.515 kN\n",
+ "Load carried by stress tube = 574.485 kN\n",
+ "Decrease in the length of the compound tube= 0.0853 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Problem 1.28,page no.43"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "#Variable declaration\n",
+ "L=2*10**2 #Length of rod in cm\n",
+ "T1=10 #Initial temperature in degree celsius\n",
+ "T2=80 #Final temperature in degree celsius\n",
+ "E=1e5*10**6 #Young's Modulus in N/sq.m\n",
+ "alpha=0.000012 #Co-efficient of linear expansion \n",
+ "\n",
+ "#Calculation\n",
+ "T=T2-T1 #Rise in temperature in degree celsius\n",
+ "dL=alpha*T*L #Expansion of the rod in cm\n",
+ "sigma=int((alpha*T*E)*1e-6) #Thermal stress in N/sq.mm\n",
+ "\n",
+ "#Result\n",
+ "print \"Expansion of the rod =\",dL,\"cm\"\n",
+ "print \"Thermal stress =\",sigma,\"N/mm^2\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Expansion of the rod = 0.168 cm\n",
+ "Thermal stress = 84 N/mm^2\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Problem 1.29,page no.43"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Given\n",
+ "#Variable declaration\n",
+ "d=3*10 #Diameter of the rod in mm\n",
+ "L=5*10**3 #Area of the rod in sq.mm\n",
+ "T1=95 #Initial temperature in degree celsius\n",
+ "T2=30 #Final temperature in degree celsius\n",
+ "E=2e5*10**6 #Young's Modulus in N/sq.m\n",
+ "alpha=12e-6 #Co-efficient of linear expansion in per degree celsius\n",
+ "\n",
+ "#Calculation\n",
+ "A=math.pi/4*(d**2) #Area of the rod\n",
+ "T=T1-T2 #Fall in temperature in degree celsius\n",
+ "\n",
+ "#case(i) When the ends do not yield \n",
+ "stress1=int(alpha*T*E*1e-6) #Stress in N/sq.mm\n",
+ "Pull1=round(stress1*A,1) #Pull in the rod in N\n",
+ "\n",
+ "#case(ii) When the ends yield by 0.12cm\n",
+ "delL=0.12*10\n",
+ "stress2=int((alpha*T*L-delL)*E/L*1e-6) #Stress in N/sq.mm\n",
+ "Pull2=round(stress2*A,1) #Pull in the rod in N\n",
+ "\n",
+ "#Result\n",
+ "print \"Stress when the ends do not yield =\",stress1,\"N/mm^2\"\n",
+ "print \"Pull in the rod when the ends do not yield =\",Pull1,\"N\"\n",
+ "print \"Stress when the ends yield =\",stress2,\"N/mm^2\"\n",
+ "print \"Pull in the rod when the ends yield =\",Pull2,\"N\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Stress when the ends do not yield = 156 N/mm^2\n",
+ "Pull in the rod when the ends do not yield = 110269.9 N\n",
+ "Stress when the ends yield = 108 N/mm^2\n",
+ "Pull in the rod when the ends yield = 76340.7 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Problem 1.30,page no.45"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "from __future__ import division\n",
+ "import math\n",
+ "#Given\n",
+ "#Variable declaration\n",
+ "Ds=20 #Diameter of steel rod in mm\n",
+ "Di_c=40 #Internal diameter of copper tube in mm\n",
+ "De_c=50 #External diameter of copper tube in mm\n",
+ "Es=200*10**3 #Young's modulus of steel in N/sq.mm\n",
+ "Ec=100*10**3 #Young's modulus of copper in N/sq.mm\n",
+ "alpha_s=12e-6 #Co-efficient of linear expansion of steel in per degree celsius\n",
+ "alpha_c=18e-6 #Co-efficient of linear expansion of copper in per degree celsius\n",
+ "T=50 #Rise of temperature in degree celsius\n",
+ "\n",
+ "#Calculation\n",
+ "As=(math.pi/4)*(Ds**2) #Area of steel rod in sq.mm\n",
+ "Ac=(math.pi/4)*(De_c**2-Di_c**2) #Area of copper tube in sq.mm\n",
+ "sigmac=float(str(((alpha_c-alpha_s)*T)/(((Ac/As)/Es)+(1/Ec)))[:6]) #Compressive stress in copper \n",
+ "sigmas=round(sigmac*(Ac/As),2) #Tensile stress in steel \n",
+ "\n",
+ "#Result\n",
+ "print \"Stress in copper =\",sigmac,\"N/mm^2\"\n",
+ "print \"Stress in steel =\",sigmas,\"N/mm^2\"\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Stress in copper = 14.117 N/mm^2\n",
+ "Stress in steel = 31.76 N/mm^2\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Problem 1.31,page no.47"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Given\n",
+ "#Variable declaration\n",
+ "Dc=15 #Diameter of copper rod in mm\n",
+ "Di_s=20 #Internal diameter of steel in mm\n",
+ "De_s=30 #External diameter of steel in mm\n",
+ "T1=10 #Initial temperature in degree celsius\n",
+ "T2=200 #Raised temperature in degree celsius\n",
+ "Es=2.1e5 #Young's modulus of steel in N/sq.mm\n",
+ "Ec=1e5 #Young's modulus of copper in N/sq.mm\n",
+ "alpha_s=11e-6 #Co-efficient of linear expansion of steel in per degree celsius\n",
+ "alpha_c=18e-6 #Co-efficient of linear expansion of copper in per degree celsius\n",
+ "\n",
+ "#Calculation\n",
+ "Ac=(math.pi/4)*Dc**2 #Area of copper tube in sq.mm\n",
+ "As=(math.pi/4)*(De_s**2-Di_s**2) #Area of steel rod in sq.mm\n",
+ "T=T2-T1 #Rise of temperature in degree celsius\n",
+ "sigmas=round(((alpha_c-alpha_s)*T)/((round(As/Ac,2)/Ec)+(1/Es)),3)\n",
+ "sigmac=round(sigmas*round(As/Ac,2),2)\n",
+ "\n",
+ "#Result\n",
+ "print \"NOTE: The answers in the book for stresses are wrong.The correct answers are,\"\n",
+ "print \"Stress in steel =\",sigmas,\"N/mm^2\"\n",
+ "print \"Stress in copper =\",sigmac,\"N/mm^2\"\n",
+ "\n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "NOTE: The answers in the book for stresses are wrong.The correct answers are,\n",
+ "Stress in steel = 49.329 N/mm^2\n",
+ "Stress in copper = 109.51 N/mm^2\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Problem 1.32,page no.48"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#Given\n",
+ "#Variable declaration\n",
+ "Dg=20 #Diameter of gun metal rod in mm\n",
+ "Di_s=25 #Internal diameter of steel in mm\n",
+ "De_s=30 #External diameter of steel in mm\n",
+ "T1=30 #Temperature in degree celsius\n",
+ "T2=140 #Temperature in degree celsius\n",
+ "Es=2.1e5 #Young's modulus of steel in N/sq.mm\n",
+ "Eg=1e5 #Young's modulus of gun metal in N/sq.mm\n",
+ "alpha_s=12e-6 #Co-efficient of linear expansion of steel in per degree celsius\n",
+ "alpha_g=20e-6 #Co-efficient of linear expansion of gun metal in per degree celsius\n",
+ "\n",
+ "#Calculation\n",
+ "Ag=(math.pi/4)*Dg**2 #Area of gun metal in sq.mm\n",
+ "As=(math.pi/4)*(De_s**2-Di_s**2) #Area of steel in sq.mm\n",
+ "T=T2-T1 #Fall in temperature in degree celsius\n",
+ "sigmag=round(((alpha_g-alpha_s)*T)/(((Ag/As)/Es)+(1/Eg)),2)\n",
+ "sigmas=round(sigmag*(Ag/As),2)\n",
+ "\n",
+ "#Result\n",
+ "print \"Stress in gun metal rod =\",sigmag,\"N/mm^2\"\n",
+ "print \"Stress in steel =\",sigmas,\"N/mm^2\"\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Stress in gun metal rod = 51.99 N/mm^2\n",
+ "Stress in steel = 75.62 N/mm^2\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Problem 1.33,page no.52"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Given\n",
+ "#Variable declaration\n",
+ "P=600e3 #Axial load in N\n",
+ "L=20e3 #Length in mm\n",
+ "w=0.00008 #Weight per unit volume in N/sq.mm\n",
+ "A2=400 #Area of bar at lower end in sq.mm\n",
+ "\n",
+ "#Calculation\n",
+ "sigma=int(P/A2) #Uniform stress on the bar in N/sq.mm\n",
+ "A1=round(A2*round(math.exp(round(w*L/sigma,7)),5),3)\n",
+ "\n",
+ "#Result\n",
+ "print \"Area of the bar at the upper end =\",A1,\"mm^2\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Area of the bar at the upper end = 400.428 mm^2\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [],
+ "language": "python",
+ "metadata": {},
+ "outputs": []
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+} \ No newline at end of file