{ "cells": [ { "cell_type": "markdown", "metadata": {}, "source": [ "# Chapter8-SIMPLE STRESSES AND STRAINS" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "# example8.1 Page number243" ] }, { "cell_type": "code", "execution_count": 1, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "sectional area= 201.06 mm^2\n", "stress= 198.94 N/mm^2\n", "strain= 0.000994718394324 N/mm^2\n", "Elongation= 0.497 mm\n" ] } ], "source": [ "from math import pi\n", "\n", "#variable declaration\n", "\n", "P=float(40000) #Load,N\n", "E=float(200000) #Modulus of elasticity for steel,N/mm^2\n", "L=500 #length of circular rod,mm\n", "d=float(16) #diameter of rod,mm\n", " \n", "A=(pi*(pow(d,2)))/4 #sectional area, mm^2\n", "p=P/A #stress, N/mm^2\n", "e=p/E #strain\n", "delta=(P*L)/(A*E) #Elongation,mm\n", "\n", "print \"sectional area=\",round(A,2),\"mm^2\"\n", "print \"stress=\",round(p,2),\"N/mm^2\"\n", "print \"strain=\",e,\"N/mm^2\"\n", "print \"Elongation=\",round(delta,3),\"mm\"\n" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "# example8.2 Page number243" ] }, { "cell_type": "code", "execution_count": 2, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "area= 11.25 mm^2\n", "Elongation= 1.6 mm\n", "Hence, if measured length is 30.0 m.\n", "Actual length is 30.0016 m\n", "Actual length of line AB= 150.008 m.\n" ] } ], "source": [ "#variable declaration\n", "\n", "P=float(120) # force applied during measurement,N\n", "E=float(200000) #Modulus of elasticity for steel,N/mm^2\n", "L=float(30) #length of Surveyor’s steel tape,mm\n", " \n", " \n", "A=15*0.75 #area, mm^2\n", "delta=((P*L*1000)/(A*E)) #Elongation,mm\n", "\n", "print \"area=\",round(A,2),\"mm^2\"\n", "print \"Elongation=\",round(delta,3),\"mm\"\n", "\n", "print \"Hence, if measured length is\", L,\"m.\"\n", "print \"Actual length is\" ,round((L+(delta/1000)),6),\"m\"\n", "\n", "print \"Actual length of line AB=\",round((150*(L+(delta/1000))/30),3),\"m.\"" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "# example 8.3 Page number 244\n" ] }, { "cell_type": "code", "execution_count": 3, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Therefore, permissible stress\n", "p= 142.857 N/mm^2\n", "Load P= 160000.0 N\n", "A= 1120.0 mm^2\n", "d= 94.32 mm\n", "t= 3.64 mm\n", "Hence, use of light section is recommended.\n" ] } ], "source": [ "from math import pi,sqrt\n", "\n", "#variable declaration\n", "\n", "Y=float(250) #Yield stress, N/mm^2\n", "FOS=float(1.75) #Factor of safety\n", "P=float(160) #Load,KN\n", "\n", "p=Y/FOS\n", "\n", "print \"Therefore, permissible stress\"\n", "print \"p=\",round(p,3), \"N/mm^2\"\n", "print \"Load P=\",P*1000,\"N\"\n", "\n", "#p=P/A\n", "\n", "A=P*1000/p #area,mm^2\n", "\n", "print \"A=\",round(A),\"mm^2\"\n", "\n", "#For hollow section of outer diameter ‘D’ and inner diameter ‘d’ A=pi*(D^2-d^2)/4\n", "D=float(101.6) #outer diameter,mm\n", "\n", "d=sqrt(pow(D,2)-(4*A/pi))\n", "\n", "print \"d=\",round(d,2),\"mm\"\n", "\n", "t=(D-d)/2\n", "print \"t=\",round(t,2),\"mm\"\n", "\n", "print \"Hence, use of light section is recommended.\"\n" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "# example 8.4 page number 245" ] }, { "cell_type": "code", "execution_count": 4, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Area= 314.16 mm^2\n", "Stress at elastic limit= 324.68 N/mm^2\n", "Young's modulus E= 12732.4 N/mm^22\n", "Percentage elongation= 28.0 %\n", "Percentage reduction in area= 43.75 %\n", "Ultimate Tensile Stress= 0.41 N/mm^2\n" ] } ], "source": [ "from math import pi\n", "\n", "#variable declaration \n", "\n", "d=float(20) #Diameter ,mm\n", "Loadatelasticlimit=float(102) #Load at elastic limit,KN\n", "P=80 #Load for extension of o.25mm , KN\n", "delta=float(0.25) #extension in specimen of steel,mm\n", "L=200 #gauge length of specimen of steel,mm\n", "Finalextension=float(56) #total extension at fracture,mm\n", "\n", "\n", "A=(pi*pow(d,2))/4 #Area,mm^2\n", "print \"Area=\", round(A,2),\"mm^2\"\n", "\n", "Stressatelasticlimit=Loadatelasticlimit*1000/A #Stress at elastic limit,N/mm^2 \n", "print \"Stress at elastic limit=\",round(Stressatelasticlimit,2),\"N/mm^2\"\n", "\n", "E=(P*1000/A)*(delta*L) #Young’s modulus ,N/mm^2\n", "print \"Young's modulus E=\", round(E,2),\"N/mm^22\"\n", "\n", "Percentageelongation=Finalextension*100/L #percentage elongation,%\n", "print \"Percentage elongation=\", round(Percentageelongation,2),\"%\"\n", "\n", "Initialarea=(pi*pow(d,2))/4\n", "Finalarea=(pi*pow(15,2))/4 # total extension at fracture is 56 mm and diameter at neck is 15 mm.\n", "Percentagereductioninarea=(Initialarea-Finalarea)*100/Initialarea\n", "\n", "print \"Percentage reduction in area=\",round(Percentagereductioninarea,2),\"%\"\n", "\n", "UltimateLoad=130 #Maximum Load=130,kN\n", "UltimateTensileStress=UltimateLoad/A\n", "\n", "print\"Ultimate Tensile Stress=\",round(UltimateTensileStress,2),\"N/mm^2\"\n" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "# example8.5 Page number247\n" ] }, { "cell_type": "code", "execution_count": 5, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "E= 56277.19 N/mm^2\n" ] } ], "source": [ "from math import pi\n", "\n", "#variable declaration\n", "\n", "P=float(40) #Load,KN\n", "L1=150 #length of 1st portion,mm\n", "A1=pi*pow(25,2)/4 #Area of 1st portion,mm^2\n", "L2=250 #length of 2nd portion,mm\n", "A2=pi*pow(20,2)/4 #Area of 2nd portion,mm^2\n", "L3=150 #length of 3rd portion,mm\n", "A3=pi*pow(25,2)/4 #Area of 3rd portion,mm^2\n", "\n", "#E,Young's modulus ,N/mm^2\n", "\n", "#Total extension= Extension of portion 1+Extension of portion 2+Extension of portion 3\n", "\n", "#Extension=(P*1000*L)/(A*E)\n", "\n", "E=(P*1000*L1/A1)+(P*1000*L2/A2)+(P*1000*L3/A3)\n", "\n", "print \"E=\",round(E,2),\"N/mm^2\"\n", "\n" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "# example8.6 Page number247" ] }, { "cell_type": "code", "execution_count": 6, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Total extension of the bar= 0.5125 mm\n" ] } ], "source": [ "from math import pi\n", "\n", "#variable declaration\n", "\n", "P=float(30) #Load,KN\n", "L1=600 #length of 1st portion,mm\n", "A1=40*20 #Area of 1st portion,mm^2\n", "\n", "E1=200000 # material 1 Young’s modulus,N/mm^2\n", " \n", "E2=100000 # material 2 Young’s modulus,N/mm^2\n", " \n", "\n", "L2=800 #length of 2nd portion,mm\n", "A2=30*20 #Area of 2nd portion,mm^2\n", "\n", "Extensionofportion1=(P*1000*L1)/(A1*E1) #mm\n", "Extensionofportion2=(P*1000*L2)/(A2*E2) #mm\n", "\n", "Totalextensionofthebar= Extensionofportion1 + Extensionofportion2\n", "\n", "print\"Total extension of the bar=\",round(Totalextensionofthebar,4),\"mm\"\n", "\n" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "# example8.7 Page number248\n" ] }, { "cell_type": "code", "execution_count": 7, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "E= 200735.96 N/mm^2\n" ] } ], "source": [ "from math import pi\n", "\n", "#variable declaration\n", "\n", "P=float(30) #Load,KN\n", "L1=600 #length of 1st portion,mm\n", "A1=pi*pow(30,2)/4 #Area of 1st portion,mm^2\n", "L2=400 #length of 2nd portion,mm\n", "A2=pi*(pow(30,2)-pow(10,2))/4 #Area of 2nd portion,mm^2\n", "\n", "#E,Young's modulus ,N/mm^2\n", "\n", "#Total extension= Extension of portion 1+Extension of portion 2\n", "\n", "#Extension=(P*1000*L)/(A*E)\n", "\n", "T=float(0.222) #Total extension of the bar,mm\n", "\n", "E=((P*1000*L1/A1)+(P*1000*L2/A2))/T \n", "\n", "print \"E=\",round(E,2),\"N/mm^2\"\n", "\n" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "# example 8.10 Page number 251" ] }, { "cell_type": "code", "execution_count": 8, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "delta1= 0.2113 mm\n", "there is calculation mistake in book\n", "delta2= 0.48 mm^2\n", "Percentage error= 55.977 %\n", "there is calculation mistake in book\n" ] } ], "source": [ "import math\n", "#variable declaration\n", "\n", "t=10 #steel flat thickness,mm\n", "b1=float(60) #tapering from b1 to b2\n", "b2=40\n", "L=600 #steel flat length\n", "P=float(80) #Load,KN\n", "E=2*100000 #Young's Modulus,N/mm^2\n", "\n", "#Extension of the tapering bar of rectangular section\n", "\n", "delta1=(P*1000*L*math.log((b1/b2),10))/(t*E*(b1-b2))\n", "\n", "print \"delta1=\",round(delta1,4),\"mm\"\n", "print \"there is calculation mistake in book\"\n", "\n", "#If averages cross-section is considered instead of tapering cross-section, extension is given by \n", "\n", "Aav=(b1+b2)*t/2 #mm^2\n", "\n", "delta2=(P*1000*L)/(Aav*E) #mm\n", "print\"delta2=\",round(delta2,3),\"mm^2\"\n", "\n", "P= (delta2-delta1)*100/delta2\n", "\n", "print\"Percentage error=\",round(P,3),\"%\"\n", "\n", "print \"there is calculation mistake in book\"\n" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "# example8.11 page number251" ] }, { "cell_type": "code", "execution_count": 9, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "delta1= 1.194 mm\n", "delta2= 0.265 mm\n", "Total extension 1.459 mm\n" ] } ], "source": [ "from math import pi\n", "\n", "#variable declaration\n", "\n", "P=float(200) #loading,KN\n", "E=200*1000\n", "d1=40 #Young's modulus,N/mm^2\n", "A= pi*pow(d1,2)/4 #Area of uniform portion,mm^2 \n", "L1=1500 #length of uniform portion,mm \n", "d2=60 #diameter of tapered section,mm\n", "L2=500 #length of tapered section,mm\n", "#Extensions of uniform portion and tapering portion are worked out separately and then added to get extension of the given bar. \n", "\n", "#Extension of uniform portion\n", "\n", "delta1=(P*1000*L1)/(A*E)\n", "\n", "print \"delta1=\",round(delta1,3),\"mm\"\n", "\n", "delta2=(P*1000*4*L2)/(E*pi*d1*d2)\n", "\n", "print \"delta2=\",round(delta2,3),\"mm\"\n", "\n", "T=delta1 + delta2 \n", "print \"Total extension\",round(T,3),\"mm\"\n", "\n", "\n" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "# example 8.13 page number259" ] }, { "cell_type": "code", "execution_count": 10, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Poisson's ratio= 0.3\n", "E= 203718.33 N/mm^2\n", "G= 78353.2 N/mm^2\n", "K= 169765.27 N/mm^2\n" ] } ], "source": [ "from math import pi\n", "\n", "#variable declaration\n", "\n", "P=float(60) #load,KN\n", "d=float(25) #diameter,mm\n", "A=pi*pow(d,2)/4 #Area,mm^2\n", "L=float(200) #gauge length,mm\n", "\n", "delta=0.12 #extension,mm\n", "deltad=0.0045 #contraction in diameter,mm\n", "Linearstrain=delta/L\n", "Lateralstrain=deltad/d\n", "\n", "Pr=Lateralstrain/Linearstrain\n", "\n", "print \"Poisson's ratio=\",round(Pr,1)\n", "\n", "E=(P*1000*L)/(A*delta)\n", "\n", "print \"E=\",round(E,2),\"N/mm^2\"\n", "\n", "G=E/(2*(1+Pr)) #Rigidity modulus\n", "\n", "print \"G=\",round(G,1),\"N/mm^2\"\n", "\n", "K=E/(3*(1-(2*Pr))) #bulk modulus\n", "\n", "print \"K=\",round(K,2),\"N/mm^2\"\n" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "# example 8.14 page number 260" ] }, { "cell_type": "code", "execution_count": 11, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "G= 76923.1 N/mm^2\n", "K= 166666.67 N/mm^2\n", "change in volume 60.0 mm^3\n" ] } ], "source": [ "from math import pi\n", "\n", "#variable declaration\n", "\n", "E=float(2*100000) #Young's modulus,N/mm^2\n", "Pr=float(0.3) #poisson's ratio\n", "\n", "G=E/(2*(1+Pr)) #Rigidity modulus\n", "\n", "K=E/(3*(1-2*(Pr))) #Bulk modulus\n", "\n", "print \"G=\", round(G,1),\"N/mm^2\"\n", "\n", "print \"K=\", round(K,2), \"N/mm^2\"\n", "\n", "P=60 #Load,kN\n", "A=pi*pow(25,2)/4 #Area,mm^2\n", "\n", "Stress=P*1000/A #N/mm^2\n", "#Linear strain,ex\n", "\n", "ex=Stress/E\n", " \n", "#Lateralstrain,ey,ez\n", "\n", "ey=-1*Pr*ex\n", "ez=-1*Pr*ex\n", "\n", "#volumetric strain,ev=ex+ey+ez\n", "\n", "ev=ex+ey+ez\n", "\n", "v=pi*pow(25,2)*500/4\n", "Changeinvolume=ev*v\n", "\n", "print\"change in volume\",round(Changeinvolume,2),\"mm^3\"" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "# example 8.15 page number261" ] }, { "cell_type": "code", "execution_count": 12, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Change in volume= 10.8 mm^3\n" ] } ], "source": [ "#variable declaration\n", "# Let the x, y, z be the mutually perpendicular directions\n", "\n", "pr=float(0.3)\n", "PX=float(15) #Loading in x-direction,KN\n", "PY=float(80) #Loading in Y-direction(compressive),KN\n", "PZ=float(180) #Loading in Z-direction,KN\n", "\n", "#Area in X-,Y-,Z-Direction is AX,AY,AZ respectively,mm^2\n", "\n", "AX=float(10*30)\n", "AY=float(10*400)\n", "AZ=float(30*400)\n", "\n", "#stress devoloped in X-,Y-,Z- direction as px,py,pz respectively,N/mm^2\n", "\n", "px=PX*1000/AX\n", "py=PY*1000/AY\n", "pz=PZ*1000/AZ\n", "\n", "#Noting that a stress produces a strain of p/E in its own direction, the nature being same as that of stress and µ p E in lateral direction of opposite nature, and taking tensile stress as +ve, we can write expression for strains ex, ey, ez.\n", "E=2*100000 #young's modulus,N/mm^2\n", "\n", "ex=(px/E)+(pr*py/E)-(pr*pz/E)\n", "ey=(-pr*px/E)-(py/E)-(pr*pz/E)\n", "ez=(-pr*px/E)+(pr*py/E)+(pz/E)\n", "\n", "ev=ex+ey+ez #Volumetric strain\n", "\n", "volume=10*30*400\n", "\n", "Changeinvolume=ev*volume\n", "\n", "print \"Change in volume=\",round(Changeinvolume,2),\"mm^3\"\n" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "# example 8.17 page number 263" ] }, { "cell_type": "code", "execution_count": 13, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "poisson's Ratio= 0.346\n", "Bulk modulus= 227500.0 N/mm^2\n" ] } ], "source": [ "#variable declaration\n", "\n", "E=float(2.1*100000) #Young’s modulus of the material,N/mm^2\n", "G=float(0.78*100000) #modulus of rigidity,N/mm^2\n", "\n", "pr=(E/(2*G))-1\n", "\n", "print \"poisson's Ratio=\",round(pr,3)\n", "\n", "K=E/(3*(1-2*pr))\n", "\n", "print \"Bulk modulus=\",round(K,3),\"N/mm^2\" " ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "# example 8.18 page number 263" ] }, { "cell_type": "code", "execution_count": 14, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Young's modulus= 102857.143 N\n", "Poisson's Ratio 0.2857\n" ] } ], "source": [ "#variable declaration\n", "\n", "G=float(0.4*100000) #modulus of rigidity of material,N/mm^2\n", "K=float(0.8*100000) #bulk modulus,N/mm^2\n", "\n", "E=(9*G*K)/(3*K+G)\n", "\n", "\n", "print \"Young's modulus=\",round(E,3),\"N\"\n", "\n", "pr=(E/(2*G))-1\n", "\n", "print \"Poisson's Ratio\",round(pr,4)\n" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "# example 8.19 page number 264" ] }, { "cell_type": "code", "execution_count": 15, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Stress in aluminium strip= 23.08 N/mm^2\n", "Stress in steel strip= 46.15 N/mm^2\n", "Extension of the compound bar= 0.138 mm\n" ] } ], "source": [ "#variable declaration\n", "\n", "L=float(600) #compound bar of length,mm\n", "P=float(60) #compound bar when axial tensile force ,KN\n", "\n", "Aa=float(40*20) #area of aluminium strip,mm^2\n", "As=float(60*15) #area of steel strip,mm^2\n", "\n", "Ea=1*100000 # elastic modulus of aluminium,N/mm^2\n", "Es=2*100000 # elastic modulus of steel,N/mm^2\n", "\n", "#load shared by aluminium strip be Pa and that shared by steel be Ps. Then from equilibrium condition Pa+Ps=P\n", "#From compatibility condition, deltaAL=deltaS\n", "Pa=(P*1000)/(1+((As*Es)/(Aa*Ea)))\n", "Ps=Pa*((As*Es)/(Aa*Ea))\n", "\n", "Sias=Pa/Aa\n", "print \"Stress in aluminium strip=\",round(Sias,2),\"N/mm^2\"\n", "Siss=Ps/As\n", "print \"Stress in steel strip=\",round(Siss,2),\"N/mm^2\"\n", "\n", "L=600\n", "#Extension of the compound bar \n", "deltal=(Pa*L)/(Aa*Ea)\n", "print\"Extension of the compound bar=\",round(deltal,3),\"mm\"" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "# example 8.20 page number 265" ] }, { "cell_type": "code", "execution_count": 16, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "stress in Copper= 75.76 N/mm^2\n", "stress in Steel= 126.27 N/mm^2\n" ] } ], "source": [ "from math import pi\n", "\n", "#variable declaration\n", "\n", "Es=float(2*100000) #Young's modulus of steel rod ,N/mm^2\n", "Ec=float(1.2*100000) #Young's modulus of copper tube,N/mm^2\n", "\n", "di=float(25) #internal diameter,mm\n", "de=float(40) #external diameter,mm\n", "\n", "As=pi*pow(di,2)/4 #Area of steel rod,mm^2\n", "Ac=pi*(pow(de,2)-pow(di,2))/4 #Area of copper tube,mm^2\n", "P=120 #load, KN\n", "#From equation of equilibrium, Ps+Pc=P,where Ps is the load shared by steel rod and Pc is the load shared by the copper tube.\n", "#From compatibility condition,deltaS=deltaC\n", "\n", "Pc=(P*1000)/(1+((As*Es)/(Ac*Ec)))\n", "Ps=Pc*((As*Es)/(Ac*Ec))\n", "\n", "SIC=Pc/Ac #stress in copper, N/mm^2\n", "SIS=Ps/As #stress in steel,N/mm^2\n", "\n", "print \"stress in Copper=\",round(SIC,2),\"N/mm^2\"\n", "print \"stress in Steel=\",round(SIS,2),\"N/mm^2\"\n" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "# example 8.21 page number 266" ] }, { "cell_type": "code", "execution_count": 17, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "stress in Concrete= 4.51 N/mm^2\n", "stress in Steel= 81.2 N/mm^2\n" ] } ], "source": [ "from math import pi\n", "\n", "#variable declaration\n", "#Es/Ec=18(given)\n", "Er=float(18) #young modulus ratio Er=Es/Ec\n", "d=float(16) #steel bar diameter,mm\n", "#8 steel bars\n", "As=8*pi*pow(d,2)/4 #Area of steel bar,mm^2\n", "Ac=(300*500)-As #Area of concrete,mm^2\n", "\n", "P=800 #Compressive force, KN\n", "#From equation of equilibrium, Ps+Pc=P,where Ps is the load shared by steel bar and Pc is the load shared by the Concrete\n", "#From compatibility condition,deltaS=deltaC\n", "\n", "Pc=(P*1000)/(1+((As*Er)/(Ac)))\n", "Ps=Pc*((As*Er)/(Ac))\n", "\n", "SIC=Pc/Ac #stress in Concrete, N/mm^2\n", "SIS=Ps/As #stress in steel,N/mm^2\n", "\n", "print \"stress in Concrete=\",round(SIC,2),\"N/mm^2\"\n", "print \"stress in Steel=\",round(SIS,2),\"N/mm^2\"" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "# example 8.22 page number 267" ] }, { "cell_type": "code", "execution_count": 18, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "stress in Aluminium= 66.96 N/mm^2\n", "stress in Steel= 89.29 N/mm^2\n" ] } ], "source": [ "#variable declaration\n", "\n", "Es=float(2*100000) #Young's modulus of steel ,N/mm^2\n", "Ea=float(1*100000) #Young's modulus of aluminium,N/mm^2\n", "Ls=240 #length of steel,mm\n", "La=160 #length of aluminium,mm\n", "Aa=1200 #Area of aluminium,mm^2\n", "As=1000 #Area of steel,mm^2\n", "P=250 #load, KN\n", "#From equation of equilibrium, Ps+2Pa=P,et force shared by each aluminium pillar be Pa and that shared by steel pillar be Ps. \n", "#From compatibility condition,deltaS=deltaC\n", "\n", "Pa=(P*1000)/(2+((As*Es*La)/(Aa*Ea*Ls)))\n", "Ps=Pa*((As*Es*La)/(Aa*Ea*Ls))\n", "\n", "SIA=Pa/Aa #stress in aluminium, N/mm^2\n", "SIS=Ps/As #stress in steel,N/mm^2\n", "\n", "print \"stress in Aluminium=\",round(SIA,2),\"N/mm^2\"\n", "print \"stress in Steel=\",round(SIS,2),\"N/mm^2\"" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "# example 8.23 page number 268\n" ] }, { "cell_type": "code", "execution_count": 19, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "ps= 91.73 N/mm^2\n", "pc= 44.96 N/mm^2\n" ] } ], "source": [ "from math import pi\n", "\n", "#variable declaration\n", "\n", "# Let the force shared by bolt be Ps and that by tube be Pc. Since there is no external force, static equilibrium condition gives Ps + Pc = 0 or Ps = – Pc i.e., the two forces are equal in magnitude but opposite in nature. Obviously bolt is in tension and tube is in compression.\n", "#Let the magnitude of force be P. Due to quarter turn of the nut\n", "\n", "#[Note. Pitch means advancement of nut in one full turn] \n", "\n", "Ls=float(600) #length of whole assembly,mm\n", "Lc=float(600) #length of whole assembly,mm\n", "delta=float(0.5)\n", "ds=float(20) #diameter,mm\n", "di=float(28) #internal diameter,mm\n", "de=float(40) #external diameter,mm\n", "Es=float(2*100000) #Young's modulus, N/mm^2\n", "Ec=float(1.2*100000)\n", "As=pi*pow(ds,2)/4 #area of steel bolt,mm^2\n", "Ac=pi*(pow(de,2)-pow(di,2))/4 #area of copper tube,mm^2\n", "\n", "P= (delta*(1/Ls))/((1/(As*Es))+(1/(Ac*Ec))) #Load,N\n", "\n", "ps=P/As #stress,N/mm^2\n", "pc=P/Ac #copper,N/mm^2\n", "\n", "print \"ps=\",round(ps,2),\"N/mm^2\"\n", "print \"pc=\",round(pc,2),\"N/mm^2\"" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "# example 8.24 page number 271" ] }, { "cell_type": "code", "execution_count": 20, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "(a) p= 52.8 \tN/mm^2\n", "(b) p= 27.8 \tN/mm^2\n", " (iii) delta= 1.968 mm\n" ] } ], "source": [ "#variable declaration\n", "E=float(2*100000) #Young's modulus,N/mm^2\n", "alpha=float(0.000012) #expansion coeffecient,/°c\n", "L=float(12) #length,m\n", "t=float(40-18) #temperature difference,°c\n", "\n", "delta=alpha*t*L*1000 #free expansion of the rails,mm \n", "# Provide a minimum gap of 3.168 mm between the rails, so that temperature stresses do not develop\n", " \n", "# a) If no expansion joint is provided, free expansion prevented is equal to 3.168 mm.\n", "\n", "#delta=(P*L)/(A*E) & p=P/A where p is stress, P,A is load,area \n", "\n", "p1=(delta*E)/(L*1000) #stress developed , N/mm^2\n", "\n", "print \"(a) p=\", round(p1,1),\"\tN/mm^2\"\n", "\n", "#(b) If a gap of 1.5 mm is provided, free expansion prevented delta2 = 3.168 – 1.5 = 1.668 mm.\n", "\n", "delta2=1.668 #mm\n", "#delta2=(P*L)/(A*E) & p=P/A where p is stress, P,A is load,area \n", "\n", "p2=(delta2*E)/(L*1000) #stress developed , N/mm^2\n", "\n", "print \"(b) p=\", round(p2,1),\"\tN/mm^2\"\n", "\n", "# If the stress developed is 20 N/mm2, then p = P/ A\n", "p3=20 #stress developed,N/mm^2\n", "delta3=delta-(p3*L*1000/E)\n", "\n", "print \" (iii) delta=\",round(delta3,3),\"mm\"\n" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "# example 8.25 page number 272\n" ] }, { "cell_type": "code", "execution_count": 21, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "stress p= 360.0 N/mm^2\n" ] } ], "source": [ "from math import pi\n", "\n", "#variable declaration\n", "\n", "# Let D be the diameter of ring after heating and ‘d’ be its diameter before heating\n", "D=float(1.2*1000) #mm\n", "\n", "#Circumference of ring after heating Ca= pi*D & Circumference of ring before heating Cb= pi*d\n", "\n", "Ca=pi*D\n", "Cb=pi*d\n", "alphas=float(0.000012) #coefficient of expansion,/°C\n", "t=150 #temperature change,°C\n", "Es=2*100000 #young's modulus,N/mm^2\n", "d=(Ca-Cb)/(alphas*t*pi)\n", "\n", "#when it cools expansion prevented\n", "#delta=pi*(D-d)\n", "delta=alphas*t*pi*d\n", "\n", "p=(delta*Es)/(pi*d) #stress,N/mm^2\n", "\n", "print \"stress p=\",round(p,2),\"N/mm^2\"\n" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "# example 8.26 page number 272\n" ] }, { "cell_type": "code", "execution_count": 22, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "P= 12907.3 N\n" ] } ], "source": [ "#variable declaration\n", "\n", "Ea=70*1000 #Young's modulus of aluminium,N/mm^2\n", "Es=200*1000 #Young's modulus of steel,N/mm^2\n", "\n", "alphaa=float(0.000011) #expansion coefficient,/°C\n", "alphas=float(0.000012) #expansion coefficient,/°C\n", "\n", "Aa=600 #Area of aluminium portion,mm^2\n", "As=400 #Area of steel, mm^2\n", "La=float(1.5) #length of aluminium portion,m\n", "Ls=float(3.0) #length of steel portion,m\n", "t=18 #temperature,°C\n", "\n", "delta=(alphaa*t*La*1000)+(alphas*t*Ls*1000) #mm\n", "\n", "P=(delta)/(((La*1000)/(Aa*Ea))+((Ls*1000)/(As*Es)))\n", "\n", "print \"P=\" ,round(P,1),\"N\"\n", "\n" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "# example8.27 page number 273" ] }, { "cell_type": "code", "execution_count": 23, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Corresponding maximum stress = 120.0 N/mm^2\n" ] } ], "source": [ "from math import pi\n", "\n", "#variable declaration\n", "\n", "d1=float(25) # variation linearly in diameter from 25 mm to 50 mm \n", "d2=float(50)\n", "L=float(500) #length,mm\n", "alpha=float(0.000012) #expansion coeffecient,/°C\n", "t=25 #rise in temperture,°C\n", "E=2*100000 #Young's modulus,N/mm^2\n", "\n", "delta=alpha*t*L\n", "\n", "#If P is the force developed by supports, then it can cause a contraction of 4*P*L/(pi*d1*d2*E)\n", "\n", "P=(delta*pi*d1*d2*E)/(4*L)\n", "Am=pi*pow(d1,2)/4\n", "Ms=P/Am\n", "\n", "print \"Corresponding maximum stress = \",round(Ms,1),\"N/mm^2\"\n" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "# example 8.28 page number 275" ] }, { "cell_type": "code", "execution_count": 24, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "stress in steel= 12.17 N/mm^2\n", "Stress in brass= 36.51 N/mm^2\n", "Shear stress in pin 18.26 N/mm^2\n" ] } ], "source": [ "from math import pi\n", "\n", "#variable declaration\n", "\n", "Db=float(20) #diameter of brass rod,mm\n", "Dse=float(40) #external diameter of steel tube,mm\n", "Dsi=float(20) #internal diameter of steel tube,mm\n", "Es=float(2*100000 ) #Young's modulus steel, N/mm^2\n", "Eb=float(1*100000 ) #Young's modulus brass, N/mm^2\n", "alphas=float(0.0000116) #coeffcient of expansion of steel,/°C\n", "alphab=float(0.0000187) #coeffcient of expansion of brass,/°C\n", "t=60 #raise in temperature, °C\n", "As=pi*(pow(Dse,2)-pow(Dsi,2))/4 #Area of steel tube, mm^2\n", "Ab=pi*(pow(Db,2))/4 #Area of brass rod,mm^2\n", "L=1200 #length,mm\n", "#Since free expansion of brass is more than free expansion of steel , compressive force Pb develops in brass and tensile force Ps develops in steel to keep the final position at CC \n", "\n", "#Horizontal equilibrium condition gives Pb = Ps, say P. \n", "\n", "P=((alphab-alphas)*t*L)/((L/(As*Es))+(L/(Ab*Eb)))\n", "\n", "ps=P/As\n", "pb=P/Ab\n", "\n", "print \"stress in steel=\",round(ps,2),\"N/mm^2\"\n", "print \"Stress in brass=\",round(pb,2),\"N/mm^2\"\n", "\n", "#the pin resist the force P at the two cross- sections at junction of two bars.\n", "\n", "Shearstress=P/(2*Ab)\n", "print \"Shear stress in pin\",round(Shearstress,2),\"N/mm^2\"\n", "\n" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "# example 8.29 page number 276" ] }, { "cell_type": "code", "execution_count": 25, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Change in length= 1.07 mm\n", "Hoop stress f= 83.33 N/mm^2\n" ] } ], "source": [ "#variable declaration\n", "\n", "L=float(1000) #length of the bar at normal temperature,mm\n", "As=float(50*10) #Area of steel,mm^2\n", "Ac=float(40*5) #Area of copper,mm^2\n", "#Ac = Free expansion of copper is greater than free expansion of steel . To bring them to the same position, tensile force Ps acts on steel plate and compressive force Pc acts on each copper plate. \n", "alphas=float(0.000012) #Expansion of coeffcient of steel,/°C\n", "alphac=float(0.000017 ) #Expansion of coeffcient of copper,/°C\n", "t=80 #raise by temperature, °C\n", "Es=2*100000 #Young's modulus of steel,N/mm^2\n", "Ec=1*100000 #Young's modulus of copper,N/mm^2\n", "Pc=((alphac-alphas)*t*L)/((2*L/(As*Es)) +(L/(Ac*Ec)))\n", "Ps=2*Pc\n", "\n", "pc=Pc/Ac #Stress in copper,N/mm^2\n", "ps=Ps/As #Stress in steel, N/mm^2\n", "\n", "Changeinlength=alphas*t*L+(Ps*L/(As*Es))\n", "\n", "\n", "print\"Change in length=\",round(Changeinlength,2),\"mm\"\n", "\n", "##example 8.30 page number 278\n", "\n", "#variable declaration\n", "\n", "p=float(2) #internal pressure, N/mm^2\n", "t=12 #thickness of thin cylinder,mm\n", "D=float(1000) #internal diameter,mm\n", "\n", "f=(p*D)/(2*t) #Hoop stress,N/mm^2\n", "\n", "print \"Hoop stress f=\",round(f,2),\"N/mm^2\"\n", "\n", "\n" ] } ], "metadata": { "kernelspec": { "display_name": "Python [Root]", "language": "python", "name": "Python [Root]" }, "language_info": { "codemirror_mode": { "name": "ipython", "version": 2 }, "file_extension": ".py", "mimetype": "text/x-python", "name": "python", "nbconvert_exporter": "python", "pygments_lexer": "ipython2", "version": "2.7.12" } }, "nbformat": 4, "nbformat_minor": 0 }