diff options
| author | Trupti Kini | 2016-04-24 23:30:27 +0600 |
|---|---|---|
| committer | Trupti Kini | 2016-04-24 23:30:27 +0600 |
| commit | b8713def3781405ca031a9b1d91f8de72cd7d9ba (patch) | |
| tree | dd5bba760ee983d492704f743418f187721ab36b /Strength_of__Materials_by_Dr.R.K.Bansal | |
| parent | 98c79b2ca1a4a512ba5d3c491222836a83e63a2c (diff) | |
| download | Python-Textbook-Companions-b8713def3781405ca031a9b1d91f8de72cd7d9ba.tar.gz Python-Textbook-Companions-b8713def3781405ca031a9b1d91f8de72cd7d9ba.tar.bz2 Python-Textbook-Companions-b8713def3781405ca031a9b1d91f8de72cd7d9ba.zip | |
Added(A)/Deleted(D) following books
A Strength_of__Materials_by_Dr.R.K.Bansal/chapter1.ipynb
A Strength_of__Materials_by_Dr.R.K.Bansal/chapter10.ipynb
A Strength_of__Materials_by_Dr.R.K.Bansal/chapter12.ipynb
A Strength_of__Materials_by_Dr.R.K.Bansal/chapter13.ipynb
A Strength_of__Materials_by_Dr.R.K.Bansal/chapter16.ipynb
A Strength_of__Materials_by_Dr.R.K.Bansal/chapter2.ipynb
A Strength_of__Materials_by_Dr.R.K.Bansal/chapter24.ipynb
A Strength_of__Materials_by_Dr.R.K.Bansal/chapter3.ipynb
A Strength_of__Materials_by_Dr.R.K.Bansal/chapter4.ipynb
A Strength_of__Materials_by_Dr.R.K.Bansal/chapter7.ipynb
A Strength_of__Materials_by_Dr.R.K.Bansal/chapter8.ipynb
A Strength_of__Materials_by_Dr.R.K.Bansal/screenshots/Screenshot01.png
A Strength_of__Materials_by_Dr.R.K.Bansal/screenshots/Screenshot02.png
A Strength_of__Materials_by_Dr.R.K.Bansal/screenshots/Screenshot03.png
A sample_notebooks/MukteshChaudhary/ch5.ipynb
Diffstat (limited to 'Strength_of__Materials_by_Dr.R.K.Bansal')
14 files changed, 3593 insertions, 0 deletions
diff --git a/Strength_of__Materials_by_Dr.R.K.Bansal/chapter1.ipynb b/Strength_of__Materials_by_Dr.R.K.Bansal/chapter1.ipynb new file mode 100644 index 00000000..1fc731a9 --- /dev/null +++ b/Strength_of__Materials_by_Dr.R.K.Bansal/chapter1.ipynb @@ -0,0 +1,831 @@ +{
+ "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 diff --git a/Strength_of__Materials_by_Dr.R.K.Bansal/chapter10.ipynb b/Strength_of__Materials_by_Dr.R.K.Bansal/chapter10.ipynb new file mode 100644 index 00000000..5b3303b1 --- /dev/null +++ b/Strength_of__Materials_by_Dr.R.K.Bansal/chapter10.ipynb @@ -0,0 +1,86 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:eb79edf44fe3869f7971ad4e31d339c536c40824da513e054e2534b46f2f0fbe"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 10:Dams and Retaining Walls"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Problem 10.19,page no.459"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "from __future__ import division\n",
+ "import math\n",
+ "#Given\n",
+ "#Variable declaration\n",
+ "h=20 #height in m\n",
+ "D=4 #External diameter in m\n",
+ "d=2 #Internal diameter in m\n",
+ "p=1 #Horizontal wind pressure in kN/sq.m\n",
+ "w=22 #specific weight in kN/m^3 \n",
+ "K=2/3 #Co-efficient of wind resistance\n",
+ "\n",
+ "#Calculation\n",
+ "A1=(math.pi/4)*(D**2-d**2) #Area of cross-section\n",
+ "W=w*A1*h #Weigth of the chimney in kN\n",
+ "sigma0=W/A1 #Direct stress in kN/sq.mm\n",
+ "A2=D*h #Projected area of the surface exposed to wind in sq.m\n",
+ "F=K*p*A2 #Wind Force in kN\n",
+ "M=F*h/2 #Bending moment in kNm\n",
+ "I=(math.pi/64)*(D**4-d**4) #Moment of inertia\n",
+ "y=D/2 #Distance between C.G. of the base section and extreme edge of the base\n",
+ "Z=I/y #Section modulus\n",
+ "sigmab=M/Z #Bending stress \n",
+ "sigma_max=round(sigma0+sigmab,2) #Maximum stress in kN/sq.m\n",
+ "sigma_min=round(sigma0-sigmab,2) #Minimum stress in kN/sq.m\n",
+ "\n",
+ "#Result\n",
+ "print\"Maximum stress =\",sigma_max,\"kN/m^2\"\n",
+ "print\"Minimum stress =\",sigma_min,\"kN/m^2\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Maximum stress = 530.54 kN/m^2\n",
+ "Minimum stress = 349.46 kN/m^2\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [],
+ "language": "python",
+ "metadata": {},
+ "outputs": []
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Strength_of__Materials_by_Dr.R.K.Bansal/chapter12.ipynb b/Strength_of__Materials_by_Dr.R.K.Bansal/chapter12.ipynb new file mode 100644 index 00000000..cf8facd0 --- /dev/null +++ b/Strength_of__Materials_by_Dr.R.K.Bansal/chapter12.ipynb @@ -0,0 +1,380 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:ced402d1abd8ee6656b1d5779752e65300ae4ab42f06bdc7c60dba927fb54a5a"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 12:Deflection of Beams"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Problem 12.1,page no.518"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Given\n",
+ "#Variable declaration\n",
+ "L=6*1000 #Length in mm\n",
+ "W=50*1000 #Point load in N\n",
+ "I=78e6 #Moment of Inertia in mm^4\n",
+ "E=2.1e5 #Young's modulus in N/sq.mm\n",
+ "\n",
+ "#Calculation\n",
+ "yc=round((W*L**3)/(48*E*I),3) #The deflection at the centre in mm\n",
+ "thetaB=round(math.degrees((W*L**2)/(16*E*I)),4) #The slope at the supports\n",
+ "\n",
+ "#Result\n",
+ "print \"Deflection at the centre =\",yc,\"mm\"\n",
+ "print \"NOTE:The answer given for slope at the support is wrong.The correct answer is,\"\n",
+ "print \"Slope at the support =\",thetaB,\"degree\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Deflection at the centre = 13.736 mm\n",
+ "NOTE:The answer given for slope at the support is wrong.The correct answer is,\n",
+ "Slope at the support = 0.3935 degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Problem 12.2,page no.518"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Given\n",
+ "#Variable declaration\n",
+ "L=4*1000 #Length in mm\n",
+ "\n",
+ "#Calculation\n",
+ "thetaA=round(math.radians(1),5) #Slope at the ends in radians\n",
+ "yc=float(str(thetaA*(L/3))[:5]) #Deflection at the centre in mm\n",
+ "\n",
+ "#Result\n",
+ "print \"Deflection at the centre =\",yc,\"mm\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Deflection at the centre = 23.26 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Problem 12.3,page no.519"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Given\n",
+ "#Variable declaration\n",
+ "L=3*1000 #Length in mm\n",
+ "\n",
+ "#Calculation\n",
+ "thetaA=round(math.radians(1),5) #Slope at the ends in radians\n",
+ "yc=float(str(thetaA*(L/3))[:5]) #Deflection at the centre in mm\n",
+ "\n",
+ "#Result\n",
+ "print \"Deflection at the centre =\",yc,\"mm\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Deflection at the centre = 17.45 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Problem 12.4,page no.526"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Given\n",
+ "#Variable declaration\n",
+ "L=5*1000 #Length in mm\n",
+ "W=5*1000 #Point load in N\n",
+ "a=3*1000 #Distance between point load and left end in mm\n",
+ "E=2e5 #Young's modulus in N/sq.mm\n",
+ "I=1e8 #Moment of Inertia in mm^4\n",
+ "\n",
+ "#Calculation\n",
+ "b=L-a #Width in mm\n",
+ "#case(i):The slope at the left support\n",
+ "thetaA=-(W*a*b)/(6*E*I*L)*(a+2*b)\n",
+ "#case(iii): The deflection under the load\n",
+ "yc=(W*a**2*b**2)/(3*E*I*L)\n",
+ "#case(iii):The maximum deflection\n",
+ "y_max=round((W*b)/(9*math.sqrt(3)*E*I*L)*(((a**2)+(2*a*b))**(3/2)),4)\n",
+ "\n",
+ "#Result\n",
+ "print\"slope at the left support =\",thetaA,\"radians\"\n",
+ "print\"Deflection under the load =\",yc,\"mm\"\n",
+ "print\"Maximum deflection =\",y_max,\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "slope at the left support = -0.00035 radians\n",
+ "Deflection under the load = 0.6 mm\n",
+ "Maximum deflection = 0.6173 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Problem 12.5,page no,529"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "from __future__ import division\n",
+ "import math\n",
+ "\n",
+ "#Given\n",
+ "#Variable declaration\n",
+ "b=200 #Width in mm\n",
+ "d=300 #Depth in mm\n",
+ "L=5*1000 #Span in mm\n",
+ "L_star=5 #Span in m\n",
+ "w=9*1000 #Uniformly distributed load in N/m \n",
+ "E=1e4 #Young's modulus in N/sq.mm\n",
+ "\n",
+ "#Calculation\n",
+ "W=w*L_star #Total load in N\n",
+ "I=b*d**3/12 #Moment of Inertia in mm^4\n",
+ "\n",
+ "#case(i):the slope at the support\n",
+ "thetaA=round(-(W*(L**2))/(24*E*I),4)\n",
+ "\n",
+ "#case(ii):maximum deflection\n",
+ "yc=str((W*L**3)/(E*I)*(5/384))[:5]\n",
+ "\n",
+ "#Result\n",
+ "print\"Slope at the support =\",-thetaA,\"radians\"\n",
+ "print\"Maximum deflection =\",yc,\"mm\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Slope at the support = 0.0104 radians\n",
+ "Maximum deflection = 16.27 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Problem 12.6,page no.529"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "#Variable declaration\n",
+ "L=5*1000 #Length in mm\n",
+ "L_star=5 #Length in m\n",
+ "w=9 #Uniformly distributed load in kN/m\n",
+ "f=7 #Bending stress in N/sq.mm\n",
+ "E=1e4 #Young's modulus in N/sq.mm\n",
+ "yc=10 #Central deflection in mm\n",
+ "\n",
+ "#Calculation\n",
+ "W=w*L_star*1e3 #Total load in N \n",
+ "bd3=((W*(L**3)*12*5)/(E*yc*384)) #width X depth^3 in mm^4\n",
+ "M=(W*L/8) #Maximum bending moment in Nmm \n",
+ "bd2=round(M*12/(f*2),2) #width X depth^2 in mm^3\n",
+ "d=round(bd3/bd2,2) #Depth of beam in mm\n",
+ "b=str(M*12/(f*2)/d**2)[:6] #Width of beam in mm\n",
+ "\n",
+ "#Result\n",
+ "print \"Depth of beam =\",d,\"mm\"\n",
+ "print \"Width of beam =\",b,\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Depth of beam = 364.58 mm\n",
+ "Width of beam = 181.36 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Problem 12.7,page no.531"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given \n",
+ "#Variable declaration\n",
+ "L=5*1000 #Length in mm\n",
+ "f=8 #Bending stress in N/sq.mm\n",
+ "yc=10 #Central deflection in mm\n",
+ "E=1.2e4 #Young's modulus in N/sq.mm\n",
+ "\n",
+ "#Calculation\n",
+ "d=round((5*L**2*(f*2*8))/(E*384*yc)*1e-1,2) #Depth of beam in cm\n",
+ "\n",
+ "#Result\n",
+ "print \"Depth of beam =\",d,\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Depth of beam = 34.72 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Problem 12.8,page no.534"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "#Variable declaration\n",
+ "L=6*1000 #Length in mm\n",
+ "W=40*1000 #Point load in N\n",
+ "a=4*1000 #Distance of point load from left support in mm\n",
+ "I=7.33e7 #Moment of Inertia in mm^4\n",
+ "E=2e5 #Young's modulus in sq.mm\n",
+ "\n",
+ "#Calculation\n",
+ "b=L-a #Width of beam in mm\n",
+ "yc=round(-(W*a**2*b**2)/(3*E*I*L),1) #Deflection under the load in mm\n",
+ "\n",
+ "#Result\n",
+ "print \"Deflection under the load =\",yc,\"mm\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Deflection under the load = -9.7 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [],
+ "language": "python",
+ "metadata": {},
+ "outputs": []
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Strength_of__Materials_by_Dr.R.K.Bansal/chapter13.ipynb b/Strength_of__Materials_by_Dr.R.K.Bansal/chapter13.ipynb new file mode 100644 index 00000000..f1b03206 --- /dev/null +++ b/Strength_of__Materials_by_Dr.R.K.Bansal/chapter13.ipynb @@ -0,0 +1,341 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:645699e2826749157978cc3139b5b4362e98aa81aba9d727772115e7ecd189e3"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 13:Deflection of Cantilevers"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Problem 13.1,page no.557"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Given\n",
+ "#Variable declaration\n",
+ "L=3*1000 #Length in mm\n",
+ "W=25*1000 #Point load in N\n",
+ "I=1e8 #Moment of Inertia in mm^4\n",
+ "E=2.1e5 #Young's modulus in N/sq.mm\n",
+ "\n",
+ "#Calculation\n",
+ "#case(i):Slope of the cantilever at the free end\n",
+ "thetaB=round((W*(L**2))/(2*E*I),6)\n",
+ "#case(ii):Deflection at the free end\n",
+ "yB=round((W*L**3)/(E*I*3),2)\n",
+ "\n",
+ "#Result\n",
+ "print \"Slope at the free end =\",thetaB,\"rad\"\n",
+ "print \"Deflection at the free end =\",yB,\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Slope at the free end = 0.005357 rad\n",
+ "Deflection at the free end = 10.71 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Problem 13.2,page no.557"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Given\n",
+ "#Variable declaration\n",
+ "L=3*1000 #Length in mm\n",
+ "W=50*1000 #Point load in N\n",
+ "a=2*1000 #Distance between the load and fixed end in mm\n",
+ "I=1e8 #Moment of Inertia in mm^4\n",
+ "E=2e5 #Young's modulus in N/sq.mm\n",
+ "\n",
+ "#Calculation\n",
+ "#case(i):Slope at the free end\n",
+ "thetaB=(W*(a**2))/(2*E*I)\n",
+ "#case(ii):Deflection at the free end\n",
+ "yB=round(((W*a**3)/(E*I*3))+((W*(a**2))/(2*E*I)*(L-a)),2)\n",
+ "\n",
+ "#Result\n",
+ "print \"Slope at the free end =\",thetaB,\"rad\"\n",
+ "print \"Deflection at the free end =\",yB,\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Slope at the free end = 0.005 rad\n",
+ "Deflection at the free end = 11.67 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Problem 13.3,page no.559"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Given\n",
+ "#Variable declaration\n",
+ "L=2.5*1000 #Length in mm\n",
+ "w=16.4 #Uniformly distributed load in kN/m \n",
+ "I=7.95e7 #Moment of Inertia in mm^4\n",
+ "E=2e5 #Young's modulus in N/sq.mm\n",
+ "\n",
+ "#Calculation\n",
+ "W=w*L #Total load in N\n",
+ "yB=round((W*L**3)/(E*I*8),3) #Deflection at the free end in mm\n",
+ "\n",
+ "#Result\n",
+ "print \"Deflection at the free end =\",yB,\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Deflection at the free end = 5.036 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Problem 13.4(A),page no.560"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "#Variable declaration\n",
+ "b=120 #Width in mm\n",
+ "d=200 #Depth in mm\n",
+ "L_star=2.5 #Length in m\n",
+ "L=2.5*1000 #Length in mm\n",
+ "yB=5 #Deflection at free end in mm\n",
+ "E=2e5 #Young's modulus in N/sq.mm\n",
+ "\n",
+ "#Calculation\n",
+ "I=(b*d**3)/12 #Moment of Inertia in mm^4\n",
+ "w=(yB*8*E*I)/(L**3*L_star)/1e3 #Uniformly distributed load in N/m\n",
+ "\n",
+ "#Result\n",
+ "print \"Uniformly distributed load =\",w,\"kN/m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Uniformly distributed load = 16.384 kN/m\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Problem 13.5,page no.563"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Given\n",
+ "#Variable declaration\n",
+ "L=3*1000 #Length in mm\n",
+ "w=10 #Uniformly distributed load in N/mm\n",
+ "a=2*1000 #Length of Uniformly distributed load from fixed end in mm\n",
+ "I=1e8 #Moment of Inertia in mm^4\n",
+ "E=2e5 #Young's modulus in N/sq.mm\n",
+ "\n",
+ "#Calculation\n",
+ "#case(i):Slope at the free end\n",
+ "thetaB=str((w*(a**3))/(6*E*I))[:7]\n",
+ "#case(ii):Deflection at the free end\n",
+ "yB=round(((w*a**4)/(E*I*8))+((w*(a**3))/(6*E*I)*(L-a)),2)\n",
+ "\n",
+ "#Result\n",
+ "print \"Slope at the free end =\",thetaB,\"rad\"\n",
+ "print \"Deflection at the free end =\",yB,\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Slope at the free end = 0.00066 rad\n",
+ "Deflection at the free end = 1.67 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Problem 13.6,page no.563"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Given\n",
+ "#Variable declaration\n",
+ "L=3*1000 #Length in mm\n",
+ "w=10 #Uniformly distributed load in N/mm\n",
+ "a=2*1000 #Length of Uniformly distributed load from fixed end in mm\n",
+ "I=1e8 #Moment of Inertia in mm^4\n",
+ "E=2e5 #Young's modulus in N/sq.mm\n",
+ "\n",
+ "#Calculation\n",
+ "#case(i):Slope at the free end\n",
+ "thetaB=round(((w*(L**3))/(6*E*I))-((w*((L-a)**3))/(6*E*I)),6)\n",
+ "#case(ii):Deflection at the free end\n",
+ "yB=round(((w*L**4)/(E*I*8))-(((w*(L-a)**4)/(8*E*I))+((w*(L-a)**3)/(6*E*I)*a)),4)\n",
+ "\n",
+ "#Result\n",
+ "print \"Slope at the free end =\",thetaB,\"rad\"\n",
+ "print \"Deflection at the free end =\",yB,\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Slope at the free end = 0.002167 rad\n",
+ "Deflection at the free end = 4.8333 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Problem 13.10,page no.569"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Given\n",
+ "#Variable declaration\n",
+ "L=4*1000 #Length in mm\n",
+ "w=50 #load at fixed end in N/mm\n",
+ "I=1e8 #Moment of Inertia in mm^4\n",
+ "E=2e5 #Young's modulus in N/sq.mm\n",
+ "\n",
+ "#Calculation\n",
+ "#case(i):Slope at the free end\n",
+ "thetaB=round(-(w*(L**3))/(24*E*I),5)\n",
+ "#case(ii):Deflection at the free end\n",
+ "yB=round((w*L**4)/(E*I*30),2)\n",
+ "\n",
+ "#Result\n",
+ "print \"Slope at the free end =\",-thetaB,\"rad\"\n",
+ "print \"Deflection at the free end =\",yB,\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Slope at the free end = 0.00667 rad\n",
+ "Deflection at the free end = 21.33 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [],
+ "language": "python",
+ "metadata": {},
+ "outputs": []
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Strength_of__Materials_by_Dr.R.K.Bansal/chapter16.ipynb b/Strength_of__Materials_by_Dr.R.K.Bansal/chapter16.ipynb new file mode 100644 index 00000000..05d4efe7 --- /dev/null +++ b/Strength_of__Materials_by_Dr.R.K.Bansal/chapter16.ipynb @@ -0,0 +1,319 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:5150ceeff814b37f327edfa239de19ec011af3bd135e00c577336409efc3a800"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 16:Torsion of Shafts and Springs"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Problem 16.1,page no.675"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#Given\n",
+ "#Variable declaration\n",
+ "D=150 #Diameter of the shaft in mm\n",
+ "tau=45 #Maximum shear stress in N/sq.mm\n",
+ "\n",
+ "#Calculation\n",
+ "T=int(math.pi/16*tau*D**3)*1e-3 #Maximum torque transmitted by the shaft in N-m\n",
+ "\n",
+ "#Result\n",
+ "print \"Maximum torque =\",T,\"N-m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Maximum torque = 29820.586 N-m\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Problem 16.3,page no.677"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Given\n",
+ "#Variable declaration\n",
+ "Do=200 #Outer diameter in mm\n",
+ "Di=100 #Inner diameter in mm\n",
+ "tau=40 #Maximum shear stress in N/sq.mm\n",
+ "\n",
+ "#Calculation\n",
+ "T=int(round((math.pi)/16*tau*((Do**4-Di**4)/Do),-1))*1e-3 #Maximum torque transmitted by the shaft in Nm\n",
+ "\n",
+ "#Result\n",
+ "print \"Maximum torque transmitted by the shaft =\",T,\"Nm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Maximum torque transmitted by the shaft = 58904.86 Nm\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Problem 16.7,page no.682"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "from __future__ import division\n",
+ "import math\n",
+ "\n",
+ "#Given\n",
+ "#Variable declaration\n",
+ "Do=120 #External diameter in mm\n",
+ "P=300*1000 #Power in W\n",
+ "N=200 #Speed in r.p.m\n",
+ "tau=60 #Maximum shear stress in N/sq.mm\n",
+ "\n",
+ "#Calculation\n",
+ "T=round((P*60)/(2*math.pi*N),1)*1e3 #Torque transmitted in Nmm \n",
+ "Di=round(((Do**4)-((T*16*Do)/(math.pi*tau)))**(1/4),1) #Maximum internal diameter in mm\n",
+ "\n",
+ "#Result\n",
+ "print \"Maximum Internal diameter =\",Di,\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Maximum Internal diameter = 88.5 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Problem 16.8,page no.683"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#Given\n",
+ "#Variable declaration\n",
+ "D=15*10 #Diameter of shaft in mm\n",
+ "P=150*1e3 #Power transmitted in W\n",
+ "N=180 #Speed of shaft in r.p.m\n",
+ "\n",
+ "#Calculation\n",
+ "T=(P*60)/(2*math.pi*N)*1e3 #Torque transmitted in Nmm\n",
+ "tau=int((16*T)/(math.pi*D**3)) #Maximum shear stress in N/sq.mm\n",
+ "\n",
+ "#Result\n",
+ "print \"Maximum shear stress =\",tau,\"N/mm^2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Maximum shear stress = 12 N/mm^2\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Problem 16.9,page no.683"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "from __future__ import division\n",
+ "import math\n",
+ "#Given\n",
+ "#Variable declaration\n",
+ "P=300*1000 #Power in W\n",
+ "N=100 #Speed in r.p.m\n",
+ "tau=80 #Maximum shear stress in N/sq.mm\n",
+ "\n",
+ "#Calculation\n",
+ "#case(a):\n",
+ "T=(P*60)/(2*math.pi*N)*1e3 #Torque transmitted in Nmm\n",
+ "D=round(((16*T)/(math.pi*tau))**(1/3),0) #Diameter of solid shaft in mm\n",
+ "#case(b):\n",
+ "Do=round(((T*16)/(math.pi*tau*(1-0.6**4)))**(1/3),0) #External diameter of hollow shaft in mm\n",
+ "Di=0.6*Do #Internal diameter of hollow shaft in mm\n",
+ "Per=(D**2-(Do**2-Di**2))/(D**2)*100 #Percentage saving in weight\n",
+ "\n",
+ "#Result\n",
+ "print \"Diameter of solid shaft =\",D,\"mm\"\n",
+ "print \"Percentage saving in weight = %.2f%%\"%Per"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Diameter of solid shaft = 122.0 mm\n",
+ "Percentage saving in weight = 29.55%\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Problem 16.10,page no.685"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#Given\n",
+ "#Variable declaration\n",
+ "P=75e3 #Power transmitted in W\n",
+ "N=200 #R.P.M of the shaft \n",
+ "tau=70 #Shear stress in N/sq.mm\n",
+ "\n",
+ "#Calculation\n",
+ "T=P*60/(math.pi*2*N)*1e3 #Mean Torque transmitted in Nmm\n",
+ "Tmax=1.3*T #Maximum Torque transmitted in Nmm\n",
+ "D=round((16*Tmax/(math.pi*tau))**(1/3),0) #Suitable diameter of the shaft in mm\n",
+ "\n",
+ "#Result\n",
+ "print \"Diameter of the shaft = %d mm\"%D"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Diameter of the shaft = 70 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Problem 16.11,page no.685"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#Given\n",
+ "#Variable declaration\n",
+ "P=300e3 #Power transmitted in W\n",
+ "N=80 #speed of the shaft in r.p.m\n",
+ "tau=60 #Maximum shear stress in N/sq.mm\n",
+ "\n",
+ "#Calculation\n",
+ "T=P*60/(math.pi*2*N)*1e3 #Mean Torque transmitted in Nmm\n",
+ "Tmax=1.4*T #Maximum Torque transmitted in Nmm\n",
+ "D=round((16*Tmax/(math.pi*tau))**(1/3),0) #Suitable diameter of the shaft in mm\n",
+ "Do=round(((Tmax*16)/(math.pi*tau*(1-0.6**4)))**(1/3),0) #External diameter of hollow shaft in mm\n",
+ "Di=0.6*Do #Internal diameter of hollow shaft in mm\n",
+ "\n",
+ "#Result\n",
+ "print \"External diameter of hollow shaft = %d mm\"%Do\n",
+ "print \"Internal diameter of hollow shaft = %d mm\"%Di"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "External diameter of hollow shaft = 170 mm\n",
+ "Internal diameter of hollow shaft = 102 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [],
+ "language": "python",
+ "metadata": {},
+ "outputs": []
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Strength_of__Materials_by_Dr.R.K.Bansal/chapter2.ipynb b/Strength_of__Materials_by_Dr.R.K.Bansal/chapter2.ipynb new file mode 100644 index 00000000..44097fc8 --- /dev/null +++ b/Strength_of__Materials_by_Dr.R.K.Bansal/chapter2.ipynb @@ -0,0 +1,354 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:a847652c7729f38097f73bfa8bb0c1fa136b92fa8a1c23926acab13f8bc56911"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 2:Elastic Constants"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Problem 2.1,page no.60"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "from __future__ import division\n",
+ "import math\n",
+ "\n",
+ "#Given\n",
+ "#Variable declaration\n",
+ "L=4*(10**3) #Length of the bar in mm\n",
+ "b=30 #Breadth of the bar in mm\n",
+ "t=20 #Thickness of the bar in mm\n",
+ "P=30*(10**3) #Axial pull in N\n",
+ "E=2e5 #Young's modulus in N/sq.mm\n",
+ "mu=0.3 #Poisson's ratio\n",
+ "\n",
+ "#Calculation\n",
+ "A=b*t #Area of cross-section in sq.mm\n",
+ "long_strain=P/(A*E) #Longitudinal strain \n",
+ "delL=long_strain*L #Change in length in mm\n",
+ "lat_strain=mu*long_strain #Lateral strain\n",
+ "delb=b*lat_strain #Change in breadth in mm\n",
+ "delt=t*lat_strain #Change in thickness in mm\n",
+ "\n",
+ "#Result\n",
+ "print \"change in length =\",delL,\"mm\"\n",
+ "print \"change in breadth =\",delb,\"mm\"\n",
+ "print \"change in thickness =\",delt,\"mm\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "change in length = 1.0 mm\n",
+ "change in breadth = 0.00225 mm\n",
+ "change in thickness = 0.0015 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Problem 2.2,page no.61"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "#Variable declaration\n",
+ "L=30 #Length in cm\n",
+ "b=4 #Breadth in cm\n",
+ "d=4 #Depth in cm\n",
+ "P=400*(10**3) #Axial compressive load in N\n",
+ "delL=0.075 #Decrease in length in cm\n",
+ "delb=0.003 #Increase in breadth in cm\n",
+ "\n",
+ "#Calculation\n",
+ "A=(b*d)*1e2 #Area of cross-section in sq.mm\n",
+ "long_strain=delL/L #Longitudinal strain\n",
+ "lat_strain=delb/b #Lateral strain\n",
+ "mu=lat_strain/long_strain #Poisson's ratio\n",
+ "E=int((P)/(A*long_strain)) #Young's modulus\n",
+ "\n",
+ "#Result\n",
+ "print \"Poisson's ratio =\",mu\n",
+ "print \"Young's modulus = %.e N/mm^2\"%E\n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Poisson's ratio = 0.3\n",
+ "Young's modulus = 1e+05 N/mm^2\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Problem 2.3,page no.63"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "#Variable declaration\n",
+ "L=4000 #Length of the bar in mm\n",
+ "b=30 #Breadth of the bar in mm\n",
+ "t=20 #Thickness of the bar in mm\n",
+ "mu=0.3 #Poisson's ratio\n",
+ "delL=1.0 #delL from problem 2.1\n",
+ "\n",
+ "#Calculation\n",
+ "ev=(delL/L)*(1-2*mu) #Volumetric strain \n",
+ "V=L*b*t #Original volume in mm^3\n",
+ "delV=ev*V #Change in volume in mm^3\n",
+ "F=int(V+delV) #Final volume in mm^3\n",
+ "\n",
+ "#Result\n",
+ "print \"Volumetric strain =\",ev\n",
+ "print \"Final volume =\",F,\"mm^3\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Volumetric strain = 0.0001\n",
+ "Final volume = 2400240 mm^3\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Problem 2.4,page no.63"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "from __future__ import division\n",
+ "#Given\n",
+ "#Variable declaration\n",
+ "L=300 #Length in mm\n",
+ "b=50 #Width in mm\n",
+ "t=40 #Thickness in mm\n",
+ "P=300*10**3 #Pull in N\n",
+ "E=2*10**5 #Young's modulus in N/sq.mm\n",
+ "mu=0.25 #Poisson's ratio\n",
+ "\n",
+ "#Calculation\n",
+ "V=L*b*t #Original volume in mm^3\n",
+ "Area=b*t #Area in sq.mm \n",
+ "stress=P/Area #Stress in N/sq.mm \n",
+ "ev=(stress/E)*(1-2*mu) #Volumetric strain \n",
+ "delV=int(ev*V) #Change in volume in mm^3 \n",
+ "\n",
+ "#Result\n",
+ "print \"Change in volume =\",delV,\"mm^3\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " Change in volume = 225 mm^3\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Problem 2.7,page no.69"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Given\n",
+ "#Variable declaration\n",
+ "L=5*10**3 #Length in mm\n",
+ "d=30 #Diameter in mm\n",
+ "P=50*10**3 #Tensile load in N\n",
+ "E=2e5 #Young's modulus in N/sq.mm\n",
+ "mu=0.25 #Poisson's ratio\n",
+ "\n",
+ "#Calculation\n",
+ "V=int(round((math.pi*d**2*L)/4,-2)) #Volume in mm^3 \n",
+ "e=P*4/(math.pi*(d**2)*E) #Strain of length\n",
+ "delL=round(e*L,3) #Change in length in mm\n",
+ "lat_strain=round(mu*round(e,7),7) #Lateral strain \n",
+ "deld=lat_strain*d #Change in diameter in mm\n",
+ "delV=round(V*(0.0003536-(2*lat_strain)),2) #Change in volume in mm^3\n",
+ "\n",
+ "#Result\n",
+ "print \"Change in length =\",delL,\"mm\"\n",
+ "print \"Change in diameter =\",deld,\"mm\"\n",
+ "print \"Change in volume =\",delV,\"mm^3\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Change in length = 1.768 mm\n",
+ "Change in diameter = 0.002652 mm\n",
+ "Change in volume = 624.86 mm^3\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Problem 2.10,page no.79"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "#Variable declaration\n",
+ "E=1.2e5 #Young's modulus in N/sq.mm\n",
+ "C=4.8e4 #Modulus of rigidity in N/sq.mm\n",
+ "\n",
+ "#Calculation\n",
+ "mu=(E/(2*C))-1 #Poisson's ratio \n",
+ "K=int(E/(3*(1-2*mu))) #Bulk modulus in N/sq.mm\n",
+ "\n",
+ "#Result\n",
+ "print \"Poisson's ratio =\",mu\n",
+ "print \"Bulk modulus = %.0e N/mm^2\"%K\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Poisson's ratio = 0.25\n",
+ "Bulk modulus = 8e+04 N/mm^2\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Problem 2.11,page no.79"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "#Variable declaration\n",
+ "A=8*8 #Area of section in sq.mm\n",
+ "P=7000 #Axial pull in N\n",
+ "Ldo=8 #Original Lateral dimension in mm\n",
+ "Ldc=7.9985 #Changed Lateral dimension in mm\n",
+ "C=0.8e5 #modulus of rigidity in N/sq.mm\n",
+ "\n",
+ "#Calculation\n",
+ "lat_strain=(Ldo-Ldc)/Ldo #Lateral strain\n",
+ "sigma=P/A #Axial stress in N/sq.mm\n",
+ "mu=round(1/((sigma/lat_strain)/(2*C)-1),3) #Poisson's ratio\n",
+ "E=round((sigma/lat_strain)/((sigma/lat_strain)/(2*C)-1),-1) #Modulus of elasticity in N/sq.mm\n",
+ "\n",
+ "#Result\n",
+ "print \"Poisson's ratio =\",mu\n",
+ "print \"Modulus of elasticity = %.4e N/mm^2\"%E\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Poisson's ratio = 0.378\n",
+ "Modulus of elasticity = 2.2047e+05 N/mm^2\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [],
+ "language": "python",
+ "metadata": {},
+ "outputs": []
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Strength_of__Materials_by_Dr.R.K.Bansal/chapter24.ipynb b/Strength_of__Materials_by_Dr.R.K.Bansal/chapter24.ipynb new file mode 100644 index 00000000..6c709b70 --- /dev/null +++ b/Strength_of__Materials_by_Dr.R.K.Bansal/chapter24.ipynb @@ -0,0 +1,132 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:087506e84c2c9ece60ca0c23e6e5e2f5e309357e67157657330423d5e748517d"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 24 :Theories of Failure"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Problem 24.10,page no.1019"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "from __future__ import division\n",
+ "import math\n",
+ "\n",
+ "#Given\n",
+ "#Variable declaration\n",
+ "P=9*1000 #Axial pull in N\n",
+ "F=4.5*1000 #Shear force in N \n",
+ "sigmat_star=225 #Elastic limit in tension in N/sq.mm\n",
+ "Sf=3 #Factor of safety \n",
+ "mu=0.3 #Poisson's ratio \n",
+ "sigma3=0 #third principle stress\n",
+ "\n",
+ "#Calculation\n",
+ "sigmat=sigmat_star/Sf \n",
+ "sigma=(P/(math.pi/4))\n",
+ "tau=float(str(F/(math.pi/4))[:6])\n",
+ "sigma1=float(str((tau)+int(round(math.sqrt((sigma/2)**2+tau**2),0)))[:7])\n",
+ "sigma2=float(str((tau)-int(round(math.sqrt((sigma/2)**2+tau**2),0)))[:8])\n",
+ "d=round(((((sigma1-sigma2)**2+(sigma2-sigma3)**2+(sigma3-sigma1)**2)/(2*(sigmat**2)))**(1/4)),3) \n",
+ "\n",
+ "#Result\n",
+ "print\"Diameter of the bolt =\",d,\"mm\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Diameter of the bolt = 14.217 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Problem 24.12,page no.1027"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "from __future__ import division\n",
+ "import math\n",
+ "#Given\n",
+ "#Variable declaration\n",
+ "d=1.2 #Diameter in m\n",
+ "p=1.5 #Internal pressure in MN/sq.m\n",
+ "sigmat_star=200 #Yield stress in MN/sq.m\n",
+ "Sf=3 #Factor of safety\n",
+ "\n",
+ "#Calculation\n",
+ "sigmat=sigmat_star/Sf #Permissible stress in simple tension in MN/sq.m\n",
+ "\n",
+ "#case(i):Thickness on the basis of Maximum principal stress theory\n",
+ "t1=((p*d)/2)/sigmat*1e3\n",
+ "\n",
+ "#case(ii):Thickness on the basis of Maximum shear stress theory\n",
+ "t2=((p*d)/2)/sigmat*1e3\n",
+ "\n",
+ "#case(iii):Thickness on the basis of Maximum shear strain energy theory\n",
+ "t3=round(math.sqrt((((p*d/2)**2)+((p*d/4)**2)-((p*d/2)*(p*d/4)))/(sigmat**2)),4)\n",
+ "\n",
+ "#Result\n",
+ "print \"Thickness of plate on the basis of maximum principal stress theory =\",\"%.1fmm\"%t1\n",
+ "print \"Thickness of plate on the basis of maximum shear stress theory =\",\"%.1fmm\"%t2\n",
+ "print \"Thickness of plate on the basis of maximum shear strain energy theory =\",\"%.4fmm\"%t3"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Thickness of plate on the basis of maximum principal stress theory = 13.5mm\n",
+ "Thickness of plate on the basis of maximum shear stress theory = 13.5mm\n",
+ "Thickness of plate on the basis of maximum shear strain energy theory = 0.0117mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [],
+ "language": "python",
+ "metadata": {},
+ "outputs": []
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Strength_of__Materials_by_Dr.R.K.Bansal/chapter3.ipynb b/Strength_of__Materials_by_Dr.R.K.Bansal/chapter3.ipynb new file mode 100644 index 00000000..43fe6e11 --- /dev/null +++ b/Strength_of__Materials_by_Dr.R.K.Bansal/chapter3.ipynb @@ -0,0 +1,220 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:1d0b1f24c9d10ade871ede95388991dbd93c7a64d0a0c3e5beb35a6a045b6a61"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 3:Principal Stresses and Strains"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Problem 3.8,page no.98"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Given\n",
+ "#Variable declaration\n",
+ "sigma1=100 #Major principal stress in N/sq.mm\n",
+ "sigma2=-60 #Minor principal stress in N/sq.mm\n",
+ "theta=90-50 #Angle of inclination in degrees\n",
+ "\n",
+ "#Calculation\n",
+ "sigman=round(((sigma1+sigma2)/2)+(((sigma1-sigma2)/2)*math.cos(math.radians(2*theta))),2)\n",
+ "sigmat=round((sigma1-sigma2)/2*(math.sin(math.radians(2*theta))),3)\n",
+ "sigmaR=round(math.sqrt(sigman**2+sigmat**2),3)\n",
+ "sigmat_max=int((sigma1-sigma2)/2)\n",
+ "\n",
+ "#Result\n",
+ "print \"Normal stress =\",sigman,\"N/mm^2\"\n",
+ "print \"Shear stress =\",sigmat,\"N/mm^2\"\n",
+ "print \"Resultant stress =\",sigmaR,\"N/mm^2\"\n",
+ "print \"Maximum shear stress =\",sigmat_max,\"N/mm^2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Normal stress = 33.89 N/mm^2\n",
+ "Shear stress = 78.785 N/mm^2\n",
+ "Resultant stress = 85.765 N/mm^2\n",
+ "Maximum shear stress = 80 N/mm^2\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Problem 3.9,page no.99"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Given\n",
+ "#Variable declaration\n",
+ "sigma1=100 #Major principal stress in N/sq.mm\n",
+ "sigma2=-40 #Minor principal stress in N/sq.mm\n",
+ "theta=90-60 #Angle of inclination in degrees\n",
+ "\n",
+ "#Calculation\n",
+ "sigman=((sigma1+sigma2)/2)+(((sigma1-sigma2)/2)*math.cos(math.radians(2*theta)))\n",
+ "sigmat=round((sigma1-sigma2)/2*(math.sin(math.radians(2*theta))),2)\n",
+ "sigmaR=round(math.sqrt(sigman**2+sigmat**2),1)\n",
+ "sigmat_max=int((sigma1-sigma2)/2)\n",
+ "phi=int(math.degrees(math.atan(sigmat/sigman)))\n",
+ "\n",
+ "#Result\n",
+ "print \"Resultant stress in magnitude =\",sigmaR,\"N/mm^2\"\n",
+ "print \"Direction of resultant stress =\",phi,\"degrees\"\n",
+ "print \"Maximum shear stress =\",sigmat_max,\"N/mm^2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Resultant stress in magnitude = 88.9 N/mm^2\n",
+ "Direction of resultant stress = 43 degrees\n",
+ "Maximum shear stress = 70 N/mm^2\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Problem 3.13,page no.111"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Given\n",
+ "#Variable declaration\n",
+ "sigma1=120 #Major tensile stress in N/sq.mm\n",
+ "sigma2=-90 #Minor compressive stress in N/sq.mm\n",
+ "sigma_gp=150 #Greatest principal stress in N/sq.mm\n",
+ "\n",
+ "#Calculation\n",
+ " #case(a):Magnitude of the shearing stresses on the two planes\n",
+ "tau=round(math.sqrt(((sigma_gp-((sigma1+sigma2)/2))**2)-(((sigma1-sigma2)/2)**2)),3)\n",
+ " #case(b):Maximum shear stress at the point\n",
+ "sigmat_max=int((math.sqrt((sigma1-sigma2)**2+(4*tau**2)))/2)\n",
+ "\n",
+ "#Result\n",
+ "print \"Shear stress on the two planes =\",tau,\"N/mm^2\"\n",
+ "print \"Maximum shear stress at the point =\",sigmat_max,\"N/mm^2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Shear stress on the two planes = 84.853 N/mm^2\n",
+ "Maximum shear stress at the point = 135 N/mm^2\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Problem 3.16,page no.115"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Given\n",
+ "#Variable declaration\n",
+ "sigma1=600 #Major tensile stress in N/sq.mm\n",
+ "sigma2=300 #Minor tensile stress in N/sq.mm\n",
+ "tau=450 #Shear stress in N/sq.mm\n",
+ "theta1=45 #Angle of inclination in degrees\n",
+ "theta2=135 #Angle of inclination in degrees\n",
+ "\n",
+ "#Calculation\n",
+ "sigman1=int(((sigma1+sigma2)/2)+(((sigma1-sigma2)/2)*math.cos(math.radians(2*theta1)))+(tau*math.sin(math.radians(2*theta1)))) \n",
+ "sigman2=int(((sigma1+sigma2)/2)+(((sigma1-sigma2)/2)*math.cos(math.radians(2*theta2)))+(tau*math.sin(math.radians(2*theta2)))) \n",
+ "sigmat1=int(round((sigma1-sigma2)/2*(math.sin(math.radians(2*theta1)))-(tau*math.cos(math.radians(2*theta1))),0))\n",
+ "sigmat2=int(round((sigma1-sigma2)/2*(math.sin(math.radians(2*theta2)))-(tau*math.cos(math.radians(2*theta2))),0)) \n",
+ "\n",
+ "#Result\n",
+ "print \"Normal stress(when theta is 45 degrees)=\",sigman1,\"N/mm^2\"\n",
+ "print \"Normal stress(when theta is 135 degrees)=\",sigman2,\"N/mm^2\" \n",
+ "print \"Tangential stress(when theta is 45 degrees)=\",sigmat1,\"N/mm^2\"\n",
+ "print \"Tangential stress(when theta is 135 degrees)=\",sigmat2,\"N/mm^2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Normal stress(when theta is 45 degrees)= 900 N/mm^2\n",
+ "Normal stress(when theta is 135 degrees)= 0 N/mm^2\n",
+ "Tangential stress(when theta is 45 degrees)= 150 N/mm^2\n",
+ "Tangential stress(when theta is 135 degrees)= -150 N/mm^2\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [],
+ "language": "python",
+ "metadata": {},
+ "outputs": []
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Strength_of__Materials_by_Dr.R.K.Bansal/chapter4.ipynb b/Strength_of__Materials_by_Dr.R.K.Bansal/chapter4.ipynb new file mode 100644 index 00000000..1bf2c1be --- /dev/null +++ b/Strength_of__Materials_by_Dr.R.K.Bansal/chapter4.ipynb @@ -0,0 +1,675 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:08a19915ea28ff1c8eacb09b86820e50e5a6bce0905af69520d3a90c0261cd74"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 4:Strain Energy and Impact Loading"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Problem 4.1,page no.145"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Given\n",
+ "#Variable declaration\n",
+ "P=60*10**3 #Load in N\n",
+ "d=4*10 #diameter in mm\n",
+ "L=5*10**3 #Length of rod in mm\n",
+ "E=2e5 #Young's Modulus in N/sq.mm\n",
+ "\n",
+ "\n",
+ "#Calculation\n",
+ "A=(math.pi/4)*d**2 #Area in sq.mm\n",
+ "V=int(A*L) #Volume of rod in cubic.mm\n",
+ "#case (ii):stress in the rod\n",
+ "sigma=round(P/A,3) #stress in N/sq.mm\n",
+ "\n",
+ "#case (i):stretch in the rod\n",
+ "x=round((sigma/E)*L,2) #stretch or extension in mm\n",
+ "\n",
+ "#case (iii):strain energy absorbed by the rod\n",
+ "U=round((sigma**2/(2*E)*V),-1)*1e-3 #strain energy absorbed by the rod in Nm\n",
+ "\n",
+ "\n",
+ "#Result\n",
+ "print \"stress in the rod =\",sigma,\"N/mm^2\"\n",
+ "print \"stretch in the rod =\",x,\"mm\"\n",
+ "print \"strain energy absorbed by the rod =\",U,\"N-m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "stress in the rod = 47.746 N/mm^2\n",
+ "stretch in the rod = 1.19 mm\n",
+ "strain energy absorbed by the rod = 35.81 N-m\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Problem 4.3,page no.146"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "#Variable declaration\n",
+ "A=10*10**2 #Area of bar in sq.mm\n",
+ "L=3*10**3 #Length of bar in mm\n",
+ "x=1.5 #Extension due to suddenly applied load in mm\n",
+ "E=2e5 #Young's Modulus in N/sq.mm\n",
+ "\n",
+ "#Calculation\n",
+ "sigma=int(x*E/L) #Instantaneous stress due to sudden load in N/sq.mm \n",
+ "P=int((sigma*A)/2*1e-3) #Suddenly applied load in kN\n",
+ "\n",
+ "#Result\n",
+ "print \"Instantaneous stress produced by a sudden load =\",sigma,\"N/mm^2\"\n",
+ "print \"Suddenly applied load =\",P,\"kN\" "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Instantaneous stress produced by a sudden load = 100 N/mm^2\n",
+ "Suddenly applied load = 50 kN\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Problem 4.4,page no.147"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Given\n",
+ "#Variable declaration\n",
+ "L=2*10**3 #Length in mm\n",
+ "d=50 #Diameter in mm\n",
+ "P=100*10**3 #Suddenly applied load in N\n",
+ "E=200e3 #Young's Modulus in N/sq.mm\n",
+ "\n",
+ "#Calculation\n",
+ "A=(math.pi/4)*d**2 #Area in sq.mm\n",
+ "sigma=round(2*P/A,2) #Instantaneous stress induced in N/sq.mm\n",
+ "dL=(sigma*L)/E #Elongation in mm\n",
+ "\n",
+ "#Result\n",
+ "print \"Instantaneous stress induced =\",sigma,\"N/mm^2\"\n",
+ "print \"Instantaneous elongation =\",dL,\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Instantaneous stress induced = 101.86 N/mm^2\n",
+ "Instantaneous elongation = 1.0186 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Problem 4.5,page no.147"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "#Variable declaration\n",
+ "A=700 #Area in sq.mm\n",
+ "L=1.5*10**3 #Length of a metal bar in mm\n",
+ "sigma=160 #Stress at elastic limit in N/sq.mm\n",
+ "E=2e5 #Young's Modulus in N/sq.mm\n",
+ "\n",
+ "\n",
+ "#Calculation\n",
+ "V=A*L #Volume of bar in sq.mm\n",
+ "Pr=(sigma**2/(2*E)*V)*1e-3 #Proof resilience in N-m\n",
+ "P=int(sigma*A/2*1e-3) #Suddenly applied load in kN\n",
+ "P1=int(sigma*A*1e-3) #gradually applied load in kN\n",
+ "\n",
+ "#Result\n",
+ "print \"Proof resilience =\",Pr,\"N-m\"\n",
+ "print \"Suddenly applied load =\",P,\"kN\"\n",
+ "print \"Gradually applied load =\",P1,\"kN\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Proof resilience = 67.2 N-m\n",
+ "Suddenly applied load = 56 kN\n",
+ "Gradually applied load = 112 kN\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Problem 4.9,page no.154"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Given\n",
+ "#Variable declaration\n",
+ "P=10*10**3 #Falling weight in N\n",
+ "h=30 #Falling height in mm\n",
+ "L=4*10**3 #Length of bar in mm\n",
+ "A=1000 #Area of bar in sq.m\n",
+ "E=2.1e5 #Young's modulus in N/sq.mm\n",
+ "\n",
+ "#Calculation\n",
+ "sigma=float(str((P/A)*(1+(math.sqrt(1+((2*E*A*h)/(P*L))))))[:5])\n",
+ "delL=round(sigma*L/E,3)\n",
+ "\n",
+ "#Result \n",
+ "print \"Instantaneous elongation due to falling weight =\",delL,\"mm\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Instantaneous elongation due to falling weight = 3.575 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Problem 4.10,page no.155"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#Given\n",
+ "#Variable declaration\n",
+ "P=100 #Impact load in N\n",
+ "h=2*10 #Height in mm\n",
+ "L=1.5*1000 #Length of bar in mm\n",
+ "A=1.5*100 #Area of bar in sq.mm\n",
+ "E=2e5 #Modulus of elasticity in N/sq.mm\n",
+ "\n",
+ "#Calculation\n",
+ "V=A*L #Volume in mm^3\n",
+ "#case(i):Maximum instantaneous stress induced in the vertical bar\n",
+ "sigma=round((P/A)*(1+(math.sqrt(1+((2*E*A*h)/(P*L))))),2)\n",
+ "#case(ii):Maximum instantaneous elongation\n",
+ "delL=round(sigma*L/E,3)\n",
+ "#case(iii):Strain energy stored in the vertical rod\n",
+ "U=round(sigma**2/(2*E)*V*1e-3,3)\n",
+ "\n",
+ "#Result\n",
+ "print \"NOTE:The answer in the book for instantaneous stress is incorrect.The correct answer is,\"\n",
+ "print \"Maximum instantaneous stress =\",sigma,\"N/mm^2\"\n",
+ "print \"Maximum instantaneous elongation =\",delL,\"mm\"\n",
+ "print \"Strain energy =\",U,\"N-m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "NOTE:The answer in the book for instantaneous stress is incorrect.The correct answer is,\n",
+ "Maximum instantaneous stress = 60.3 N/mm^2\n",
+ "Maximum instantaneous elongation = 0.452 mm\n",
+ "Strain energy = 2.045 N-m\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Problem 4.11,page no.155"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#Given\n",
+ "#Variable declaration\n",
+ "delL=2.1 #Instantaneous extension in mm\n",
+ "L=3*10**3 #Length of bar in mm\n",
+ "A=5*100 #Area of bar in mm\n",
+ "h=4*10 #Height in mm\n",
+ "E=2e5 #Modulus of elasticity in N/sq.mm\n",
+ "\n",
+ "#Calculation\n",
+ "V=A*L #Volume of bar in mm^3\n",
+ "\n",
+ "#case(i):Instantaneous stress induced in the vertical bar\n",
+ "sigma=int(E*delL/L) \n",
+ "\n",
+ "#case(ii):Unknown weight\n",
+ "P=round(((sigma**2)/(2*E)*V)/(h+delL),1) \n",
+ "\n",
+ "#Result\n",
+ "print\"Instantaneous stress =\",sigma,\"N/mm^2\"\n",
+ "print\"Unknown weight =\",P,\"N\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Instantaneous stress = 140 N/mm^2\n",
+ "Unknown weight = 1745.8 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Problem 4.13,page no.157"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#Given\n",
+ "#Variable declaration\n",
+ "d=12 #Diameter of bar in mm \n",
+ "delL=3 #Increase in length in mm\n",
+ "W=8000 #Steady load in N\n",
+ "P=800 #Falling weight in N\n",
+ "h=8*10 #Vertical distance in mm\n",
+ "E=2e5 #Young's modulus in N/sq.mm\n",
+ "\n",
+ "#Calculation\n",
+ "A=round((math.pi/4)*d**2,1) #Area of bar in sq.mm\n",
+ "L=round(E*A*delL/W,1) #Length of the bar in mm\n",
+ "sigma=round(round(P/A,4)*float(str(1+(math.sqrt(1+round((2*E*A*h)/(P*L),2))))[:7]),4) \n",
+ "sigma=float(str(sigma)[:7]) #Stress produced by the falling weight in N/sq.mm\n",
+ "\n",
+ "#Result\n",
+ "print \"Stress produced by the falling weight =\",sigma,\"N/mm^2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Stress produced by the falling weight = 170.578 N/mm^2\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Problem 4.14,page no.158"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#Given\n",
+ "#Variable declaration\n",
+ "d=12.5 #Diameter of the rod in mm\n",
+ "delL=3.2 #Increase in length in mm\n",
+ "W=10*1000 #Steady load in N\n",
+ "P=700 #Falling load in N\n",
+ "h=75 #Falling height in mm\n",
+ "E=2.1e5 #Young's modulus in N/sq.mm\n",
+ "\n",
+ "#Calculation\n",
+ "A=round((math.pi/4)*d**2,2) #Area of rod in sq.mm \n",
+ "L=round(E*A*delL/W,1) #Length of the rod in mm\n",
+ "sigma=round((P/A)*(1+(math.sqrt(1+((2*E*A*h)/(P*L))))),2) #Stress produced by the falling weight in N/mm^2\n",
+ "\n",
+ "#Result\n",
+ "print \"NOTE:The given answer for stress is wrong.The correct answer is,\"\n",
+ "print \"Stress = %.2f N/mm^2\"%sigma"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "NOTE:The given answer for stress is wrong.The correct answer is,\n",
+ "Stress = 153.42 N/mm^2\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Problem 4.15,page no.159"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Given\n",
+ "#Variable declaration\n",
+ "L=1.82*1000 #Length of rod in mm\n",
+ "h1=30 #Height through which load falls in mm\n",
+ "h2=47.5 #Fallen height in mm\n",
+ "sigma=157 #Maximum stress induced in N/sq.mm\n",
+ "E=2.1e5 #Young's modulus in N/sq.mm\n",
+ "\n",
+ "#Calculation\n",
+ "U=sigma**2/(2*E) #Strain energy stored in the rod in N-m\n",
+ "delL=sigma*L/E #Extension of the rod in mm\n",
+ "Tot_dist=h1+delL #Total distance in mm\n",
+ "\n",
+ "#case(i):Stress induced in the rod if the load is applied gradually\n",
+ "sigma1=round((U/Tot_dist)*L,1)\n",
+ "\n",
+ "#case(ii):Maximum stress if the load had fallen from a height of 47.5 mm\n",
+ "sigma2=round((sigma1)*(1+(math.sqrt(1+((2*E*h2)/(sigma1*L))))),2)\n",
+ "\n",
+ "#Result\n",
+ "print \"Stress induced in the rod = %.1f N/mm^2\"%sigma1\n",
+ "print \"NOTE:The given answer for stress(2nd case) in the book is wrong.The correct answer is,\"\n",
+ "print \"Maximum stress if the load has fallen = %.2f N/mm^2\"%sigma2"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Stress induced in the rod = 3.4 N/mm^2\n",
+ "NOTE:The given answer for stress(2nd case) in the book is wrong.The correct answer is,\n",
+ "Maximum stress if the load has fallen = 196.48 N/mm^2\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Problem 4.17,page no.162"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Given\n",
+ "#Variable declaration\n",
+ "L=4*10**3 #Length of bar in mm\n",
+ "A=2000 #Area of bar in sq.mm\n",
+ "P1=3000 #Falling weight in N(for 1st case)\n",
+ "h1=20*10 #Height in mm(for 1st case)\n",
+ "P2=30*1000 #Falling weight in N(for 2nd case)\n",
+ "h2=2*10 #Height in mm(for 2nd case)\n",
+ "E=2e5 #Young's modulus in N/sq.mm\n",
+ "\n",
+ "#Calculation\n",
+ "V=A*L #Volume of bar in mm^3\n",
+ "\n",
+ "#case(i):Maximum stress when a 3000N weight falls through a height of 20cm\n",
+ "sigma1=round(((math.sqrt((2*E*P1*h1)/(A*L)))),1)\n",
+ "\n",
+ "#case(ii):Maximum stress when a 30kN weight falls through a height of 2cm\n",
+ "sigma2=round((P2/A)*(1+(math.sqrt(1+((2*E*A*h2)/(P2*L))))),2)\n",
+ "\n",
+ "#Result\n",
+ "print\"Maximum stress induced(when a weight of 3000N falls through a height of 20cm)=\",sigma1,\"N/mm^2\"\n",
+ "print\"Maximum stress induced(when a weight of 30kN falls through a height of 2cm)=\",sigma2,\"N/mm^2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Maximum stress induced(when a weight of 3000N falls through a height of 20cm)= 173.2 N/mm^2\n",
+ "Maximum stress induced(when a weight of 30kN falls through a height of 2cm)= 188.85 N/mm^2\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Problem 4.18,page no.163"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "from __future__ import division\n",
+ "import math\n",
+ "\n",
+ "#Given \n",
+ "#Variable declaration\n",
+ "A=6.25*100 #Area in sq.mm\n",
+ "W=10*10**3 #Load in N\n",
+ "V=(40/60) #Velocity in m/s\n",
+ "L=10000 #Length of chain unwound in mm\n",
+ "E=2.1e5 #Young's modulus in N/sq.mm\n",
+ "g=9.81 #acceleration due to gravity\n",
+ "\n",
+ "#Calculation\n",
+ "K_E=round(((W/g)*(V**2))/2,1)*1e3 #K.E of the crane in N mm\n",
+ "sigma=round(math.sqrt(K_E*E*2/(A*L)),2) #Stress induced in the chain in N/sq.mm\n",
+ "\n",
+ "#Result\n",
+ "print \"Stress induced in the chain due to sudden stoppage =\",sigma,\"N/mm^2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Stress induced in the chain due to sudden stoppage = 123.37 N/mm^2\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Problem 4.19,page no.164"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Given\n",
+ "#Variable declaration\n",
+ "W=60*10**3 #Weight in N\n",
+ "V=1 #Velocity in m/s\n",
+ "L=15*10**3 #Free length in mm\n",
+ "A=25*100 #Area in sq.mm\n",
+ "E=2e5 #Young's modulus in N/sq.mm\n",
+ "g=9.81 #acceleration due to gravity\n",
+ "\n",
+ "#Calculation\n",
+ "K_E=((W/g)*(V**2))/2*1e3 #Kinetic Energy of the cage in N mm\n",
+ "sigma=round(math.sqrt(K_E*E*2/(A*L)),2) #Maximum stress in N/sq.mm\n",
+ "\n",
+ "#Result\n",
+ "print\"Maximum stress produced in the rope =\",sigma,\"N/mm^2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Maximum stress produced in the rope = 180.61 N/mm^2\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Problem 4.20,page no.166"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given \n",
+ "#Variable declaration\n",
+ "tau=50 #Shear stress in N/sq.mm\n",
+ "C=8e4 #Modulus of rigidity in N/sq.mm\n",
+ "\n",
+ "#Calculation\n",
+ "ste=(tau**2)/(2*C) #Strain energy per unit volume in N/sq.mm\n",
+ "\n",
+ "#Result\n",
+ "print\"Strain energy per unit volume =\",ste,\"N/mm^2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Strain energy per unit volume = 0.015625 N/mm^2\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [],
+ "language": "python",
+ "metadata": {},
+ "outputs": []
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Strength_of__Materials_by_Dr.R.K.Bansal/chapter7.ipynb b/Strength_of__Materials_by_Dr.R.K.Bansal/chapter7.ipynb new file mode 100644 index 00000000..eb76c96c --- /dev/null +++ b/Strength_of__Materials_by_Dr.R.K.Bansal/chapter7.ipynb @@ -0,0 +1,130 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:9a486aadfd6b9085cbabcb01b6c4b7704df2013ab12ac6b342be3bf11e26b98d"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 7:Bending Stresses in Beams"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Problem 7.1,page no.298"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "#Variable declaration\n",
+ "b=120 #Width of plate in mm\n",
+ "t=20 #Thickness of plate in mm\n",
+ "R=10*10**3 #Radius of curvature in mm\n",
+ "E=2e5 #Young's modulus in N/sq.mm\n",
+ "\n",
+ "#Calculation\n",
+ "I=b*t**3/12 #Moment of inertia in mm^4\n",
+ "y_max=t/2 #Maximum distance in mm\n",
+ "sigma_max=int((E/R)*y_max) #Maximum stress in N/sq.mm\n",
+ "M=round((E/R*I)*(10**-6),1) #Bending moment in kNm\n",
+ "\n",
+ "#Result\n",
+ "print \"Maximum stress =\",sigma_max,\"N/mm^2\"\n",
+ "print \"Bending moment =\",M,\"kNm\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Maximum stress = 200 N/mm^2\n",
+ "Bending moment = 1.6 kNm\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Problem 7.8,page no.305"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "from __future__ import division\n",
+ "import math\n",
+ "\n",
+ "#Given\n",
+ "#Variable declaration\n",
+ "W=20*1000 #Total load in N\n",
+ "L=3.6 #Span in m\n",
+ "sigma_max=7 #Maximum stress in N/sq.mm\n",
+ "\n",
+ "#Calculation\n",
+ "M1=W*L/8*1e3 #Maximum Bending moment in Nmm\n",
+ "b1=round((M1*3/(sigma_max*2))**(1/3),1) #Breadth of the beam in mm\n",
+ "d1=int(round(2*b1,0)) #depth of the beam in mm\n",
+ "M2=W*L/4*1e3 #Maximum Bending moment in Nmm\n",
+ "b2=float(str(round((M2*3/(sigma_max*2))**(1/3),4))[:6]) #Breadth of the beam in mm\n",
+ "d2=2*b2 #depth of the beam in mm\n",
+ "\n",
+ "#Result\n",
+ "print \"Dimensions of the cross-section:\"\n",
+ "print \"Breadth of beam =\",b1,\"mm\"\n",
+ "print \"Depth of beam\",d1,\"mm\"\n",
+ "\n",
+ "print \"Dimensions of the cross-section when the beam carries a point load at the centre:\"\n",
+ "print \"Breadth of beam =\",b2,\"mm\"\n",
+ "print \"Depth of beam\",d2,\"mm\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Dimensions of the cross-section:\n",
+ "Breadth of beam = 124.5 mm\n",
+ "Depth of beam 249 mm\n",
+ "Dimensions of the cross-section when the beam carries a point load at the centre:\n",
+ "Breadth of beam = 156.82 mm\n",
+ "Depth of beam 313.64 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [],
+ "language": "python",
+ "metadata": {},
+ "outputs": []
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Strength_of__Materials_by_Dr.R.K.Bansal/chapter8.ipynb b/Strength_of__Materials_by_Dr.R.K.Bansal/chapter8.ipynb new file mode 100644 index 00000000..934ee9d5 --- /dev/null +++ b/Strength_of__Materials_by_Dr.R.K.Bansal/chapter8.ipynb @@ -0,0 +1,125 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:7d21feb46ef90fdbbfbaad4563a2c375000b58700df6f4805ecd9fd8506686a5"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 8:Shear Stresses in Beams"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Problem 8.6,page no.355"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "from __future__ import division\n",
+ "import math\n",
+ "#Given\n",
+ "#Variable declaration\n",
+ "D=100 #Diameter in mm\n",
+ "R=D/2 #Radius in mm\n",
+ "F=5*10**3 #Shear force in N\n",
+ "y=40 #given distance from N.A. in mm\n",
+ "\n",
+ "#Calculation\n",
+ "#case(i):Average shear stress \n",
+ "A=math.pi*R**2\n",
+ "tau_avg=round(F/A,4)\n",
+ "#case(ii):Maximum shear stress for a circular section\n",
+ "tau_max=round(4/3*tau_avg,4)\n",
+ "#case(iii):Shear stress at a distance 40mm from N.A.\n",
+ "I=math.pi/64*D**4\n",
+ "tau=float(str(F/(3*I)*(R**2-y**2))[:6])\n",
+ "\n",
+ "#Result\n",
+ "print \"Average shear stress =\",tau_avg,\"N/mm^2\"\n",
+ "print \"Maximum shear stress =\",tau_max,\"N/mm^2\"\n",
+ "print \"Shear stress at a distance 40mm from N.A. =\",tau,\"N/mm^2\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Average shear stress = 0.6366 N/mm^2\n",
+ "Maximum shear stress = 0.8488 N/mm^2\n",
+ "Shear stress at a distance 40mm from N.A. = 0.3055 N/mm^2\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Problem 8.12,page no.369"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "from __future__ import division\n",
+ "import math\n",
+ "\n",
+ "#Given\n",
+ "#Variable declaration\n",
+ "F=50*10**3 #Shear force in N\n",
+ "b=250 #Base width in mm\n",
+ "h=200 #height in mm\n",
+ "\n",
+ "#Calculation\n",
+ "tau_max=int((3*F)/(b*h)) #Maximum shear stress in N/sq.mm\n",
+ "tau=round((8*F)/(3*b*h),2) #Shear stress at N.A. in N/sq.mm\n",
+ "\n",
+ "#Result\n",
+ "print \"Maximum shear stress =\",tau_max,\"N/mm^2\"\n",
+ "print \"Shear stress at N.A. =\",tau,\"N/mm^2\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Maximum shear stress = 3 N/mm^2\n",
+ "Shear stress at N.A. = 2.67 N/mm^2\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [],
+ "language": "python",
+ "metadata": {},
+ "outputs": []
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Strength_of__Materials_by_Dr.R.K.Bansal/screenshots/Screenshot01.png b/Strength_of__Materials_by_Dr.R.K.Bansal/screenshots/Screenshot01.png Binary files differnew file mode 100644 index 00000000..4e56b1e6 --- /dev/null +++ b/Strength_of__Materials_by_Dr.R.K.Bansal/screenshots/Screenshot01.png diff --git a/Strength_of__Materials_by_Dr.R.K.Bansal/screenshots/Screenshot02.png b/Strength_of__Materials_by_Dr.R.K.Bansal/screenshots/Screenshot02.png Binary files differnew file mode 100644 index 00000000..c509ba38 --- /dev/null +++ b/Strength_of__Materials_by_Dr.R.K.Bansal/screenshots/Screenshot02.png diff --git a/Strength_of__Materials_by_Dr.R.K.Bansal/screenshots/Screenshot03.png b/Strength_of__Materials_by_Dr.R.K.Bansal/screenshots/Screenshot03.png Binary files differnew file mode 100644 index 00000000..92a8f897 --- /dev/null +++ b/Strength_of__Materials_by_Dr.R.K.Bansal/screenshots/Screenshot03.png |
