{
"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": {}
}
]
}