{ "metadata": { "name": "", "signature": "sha256:ee3efcb70c7b8625c2621d2e113f06299bf0d0de2c90519d4a9ec891c7f588ab" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "Chapter3-Linear Stress-Strain Temperature Relations" ] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex1-pg88" ] }, { "cell_type": "code", "collapsed": false, "input": [ "##calculate the maximum circumferential stress\n", "## initialization of variables\n", "import math\n", "## part (a)\n", "E=72. ## G Pa\n", "v=0.33 ## Poisoon's ratio\n", "h=2. ## mm\n", "R=600. ## mm\n", "##calculations\n", "sig_cir=E*h/(2.*(1.-v**2)*R)\n", "## results\n", "print('\\n part (a) \\n')\n", "print\"%s %.2f %s\"%(' The maximum circumferential stress is ',sig_cir*10**3,' M Pa')\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "\n", " part (a) \n", "\n", " The maximum circumferential stress is 134.67 M Pa\n" ] } ], "prompt_number": 1 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex7-pg99" ] }, { "cell_type": "code", "collapsed": false, "input": [ "##calculate the stress in aluminum and steel cylinder \n", "## initialization of variables\n", "import math\n", "tR=0.02 ## t/R ration\n", "E_A=69. ##G Pa\n", "v_A=0.33 ## Poisson's ratio\n", "alpha_A=21.6*10**-6 ## /degree Celcius (Coefficient of expansion)\n", "E_S=207. ## G Pa\n", "v_S=0.280\n", "alpha_S=10.8*10**-6 ## /degree Celcius (Coefficient of expansion)\n", "## calculations\n", "## Sig_LA=a*p+b*delT+c*sig_thS\n", "## Sig_LS=v_S*Sig_thS+d*delT\n", "E_S=E_S*10**9\n", "E_A=E_A*10**9\n", "a=1./tR*E_A/E_S\n", "b=-2./3.*alpha_S*E_S\n", "c=-E_A/E_S\n", "d=-alpha_S*E_S\n", "## SigthS=e*p+f*delT\n", "## SigthA=g*p+h*delT\n", "e=37.16\n", "f=0.8639*10**6\n", "g=1./tR-e\n", "h=-f\n", "## results\n", "p=689.4 ## kPa\n", "delT=100. ## degree Celcius\n", "p=p*10**3 ## Pa\n", "SigthA=g*p+h*delT\n", "SigthS=e*p+f*delT\n", "Sig_LA=a*p+b*delT+c*SigthS\n", "Sig_LS=v_S*SigthS+d*delT\n", "print\"%s %.2f %s %.2f %s \"%('Thus, for p = ',p/10**3,' k Pa' and 'delT = ',delT,' degree celcius \\n')\n", "print\"%s %.2f %s %.2f %s \"%(' SigthA = ',SigthA/10**6,' M Pa'and 'Sig_LA = ',Sig_LA/10**6,' M Pa \\n')\n", "print\"%s %.2f %s %.2f %s \"%(' SigthS = ',SigthS/10**6,' M Pa'and 'Sig_LS = ',Sig_LS/10**6,' M Pa')\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Thus, for p = 689.40 delT = 100.00 degree celcius \n", " \n", " SigthA = -77.54 Sig_LA = -174.89 M Pa \n", " \n", " SigthS = 112.01 Sig_LS = -192.20 M Pa \n" ] } ], "prompt_number": 2 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex8-pg100" ] }, { "cell_type": "code", "collapsed": false, "input": [ "##calculate the orientation of the principal stress and deterimine stress components and orientation of the principal axes of strain\n", "## initialization of variables\n", "import math\n", "import numpy\n", "## Material constants\n", "Ex=14700. ## M Pa\n", "Ey=1000. ## M Pa\n", "Ez=735. ## M Pa\n", "Gxy=941. ## M Pa\n", "Gxz=1147. ## M Pa\n", "Gyz=103. ## M pa\n", "Vxy=0.292\n", "Vxz=0.449\n", "Vyz=0.39\n", "## Stresses at a point\n", "Sxx=7. ## M pa\n", "Syy=2.1 ## M Pa\n", "Szz=-2.8 ##M Pa\n", "Sxy=1.4 ## M Pa\n", "Sxz=0. ##M Pa\n", "Syz=0. ## M Pa\n", "## part (a)\n", "th=1/2.*math.atan(2.*Sxy/(Sxx-Syy))*180./math.pi\n", "I1=Sxx+Syy+Szz\n", "I2=Sxx*Syy-Sxy**2+Szz*Sxx-Sxz**2+Szz*Syy-Syz**2\n", "M=numpy.matrix([[Sxx, Sxy, Sxz],\n", " [Sxy, Syy, Syz],\n", " [Sxz, Syz, Szz]])\n", "I3=numpy.linalg.det(M) \n", "p=([1, -I1, I2 ,-I3])\n", "S=numpy.roots(p)\n", "## results\n", "print('Part (a) \\n')\n", "print\"%s %.2f %s\"%('The maximum principal stress is S1 = ',S[0],' M Pa')\n", "print\"%s %.2f %s\"%('\\n and occurs in direction th = ',th,' degrees')\n", "print\"%s %.2f %s %.2f %s \"%('\\n and the intermediate principal stress S2 = ',S[2],' M Pa'and 'occurs in the direction th = ',th+90,' degrees \\n')\n", "print\"%s %.2f %s\"%(' The minimum principal stress is S3 = Szz = ',S[1],' M Pa')\n", "Ex=Ex*10**6\n", "Ey=Ey*10**6\n", "Ez=Ez*10**6\n", "Gxy=Gxy*10**6\n", "Gxz=Gxz*10**6\n", "Gyz=Gyz*10**6\n", "## part (b) is to find strains\n", "Exx=Sxx/Ex-Vxy*Syy/Ey-Vxz*Szz/Ez\n", "Eyy=-Vxy*Sxx/Ex+Syy/Ey-Vyz*Szz/Ez\n", "Ezz=-Vxz*Sxx/Ex-Vyz*Syy/Ey+Szz/Ez\n", "Exy=Sxy/Gxy\n", "Exz=Sxz/Gxz\n", "Eyz=Syz/Gyz\n", "print('\\n Part (b)')\n", "print('\\n The srains are')\n", "print\"%s %.2e %s %.2e %s %.2e %s\"%('\\n Exx = ',Exx,'' and'Eyy =',Eyy,''and 'Ezz = ',Ezz,'')\n", "print\"%s %.2e %s %.2d %s %.2d %s\"%('\\n Exy = ',Exy,''and ' Exz = ',Exz,''and 'Eyz = ',Eyz,'')\n", "## Wrong Exx value in the textbook\n", "th=1/2.*math.atan(Exy/(Exx-Eyy))\n", "th=th*180./math.pi\n", "th2=th+90.\n", "print('\\n part (c)')\n", "print\"%s %.2f %s %.2f %s \"%('\\n theta = ',th,'' and 'or theta=',th2,'degrees')\n", "## Wrong theta too since Example given in textbook is wrong\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Part (a) \n", "\n", "The maximum principal stress is S1 = 7.37 M Pa\n", "\n", " and occurs in direction th = 14.87 degrees\n", "\n", " and the intermediate principal stress S2 = 1.73 occurs in the direction th = 104.87 degrees \n", " \n", " The minimum principal stress is S3 = Szz = -2.80 M Pa\n", "\n", " Part (b)\n", "\n", " The srains are\n", "\n", " Exx = 1.57e-09 3.45e-09 -4.84e-09 \n", "\n", " Exy = 1.49e-09 00 00 \n", "\n", " part (c)\n", "\n", " theta = -19.23 70.77 degrees \n" ] } ], "prompt_number": 3 } ], "metadata": {} } ] }