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diff --git a/Solid_Mechanics_by_S._M._A._Kazimi/Chapter5.ipynb b/Solid_Mechanics_by_S._M._A._Kazimi/Chapter5.ipynb new file mode 100755 index 00000000..ce308344 --- /dev/null +++ b/Solid_Mechanics_by_S._M._A._Kazimi/Chapter5.ipynb @@ -0,0 +1,558 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:768e5620d31d4b5603faf8e19db8db3178570776ab6e28491caaf3d036876dc4"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter5-Uniaxial Deformations"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1-pg139"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "##initialization of variables\n",
+ "#find the depth of clay bed \n",
+ "l=20. ##cm\n",
+ "dL=1. ##m\n",
+ "dl=0.004 ##cm\n",
+ "##calculations\n",
+ "L=l*dL/dl ##m\n",
+ "##results\n",
+ "print'%s %.2f %s'%('The depth of the clay bed is ',L,' m')\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The depth of the clay bed is 5000.00 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2-pg140"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "##initialization of variables\n",
+ "#find the total extension of the rod and draw the force and extension diagrams\n",
+ "A=1. ##unit area\n",
+ "E=2.*10**6 ##kg/cm^2\n",
+ "## calculations\n",
+ "db=3000.*90./(A*E)\n",
+ "dc=db+5000.*60./(A*E)\n",
+ "dd=dc+4000.*30./(A*E)\n",
+ "##results\n",
+ "print'%s %.2e %s %.2e %s %.2e %s '%('The extension of the rod in part AB is ',db,' cm'and'in part BC is ',dc,' cm'and' \\n and in part CD is ',dd,' cm')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The extension of the rod in part AB is 1.35e-01 in part BC is 2.85e-01 \n",
+ " and in part CD is 3.45e-01 cm \n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3-pg141"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "##initialization of variables\n",
+ "#findthe extension under its own weight\n",
+ "A=3. ##cm^2\n",
+ "L=18. ##m\n",
+ "E= 2*10**6 ##kg/cm^2\n",
+ "r=7833. ##kg/m^3\n",
+ "##calculations\n",
+ "e=r*(L*100)**2./(2*E*10**6)\n",
+ "## results\n",
+ "print'%s %.4f %s'%('The elongation is ',e,' cm')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The elongation is 0.0063 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4-pg142"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "##initialization of variables\n",
+ "#find the extension under a concertrated load of 3 tonne at the bottom\n",
+ "## linked to 5_3\n",
+ "P=3 ##tonne\n",
+ "E=2*10**6 ##kg/cm^2\n",
+ "d_0= 1. ##cm\n",
+ "d_l=2.8 ##cm\n",
+ "## calculations\n",
+ "e=4*P*1000.*d_l*10**3/(d_l**2*math.pi*E*(1-((d_l-d_0)/d_l)))\n",
+ "##results\n",
+ "print'%s %.2f %s'%('The total elongation is ',e,' cm')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The total elongation is 1.91 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6-145"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "##initialization of variables\n",
+ "import numpy\n",
+ "from numpy import linalg\n",
+ "import scipy as Sci\n",
+ "from scipy import linalg\n",
+ "P=10 ##tonne\n",
+ "import numpy as np\n",
+ "E=2*10**6 ##kg/cm^2\n",
+ "## calculations\n",
+ "## We have to solve linear system Ax=B\n",
+ "A=numpy.matrix([[1, 1, 1, 0], [3, 1, -3, 0],[-2, 2, 0, -E],[0, -1, 2, -E]])\n",
+ "B=numpy.matrix([[P*10**3],[0],[0],[0]])\n",
+ "x=numpy.dot(np.linalg.inv(A),B)\n",
+ "W1=x[0,0]/1000.\n",
+ "W2=x[1,0]/1000.\n",
+ "W3=x[2,0]/1000.\n",
+ "th=x[3,0]\n",
+ "##results\n",
+ "print'%s %.2f %s %.2f %s %.2f %s '%('The load taken by each rod is',W1,' tonne'and'',W2,' tonne'and'',W3,'tonne')\n",
+ "print'%s %.3e %s'%('\\n and the slope is theta = ',th,' radians') \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The load taken by each rod is 2.33 4.00 3.67 tonne \n",
+ "\n",
+ " and the slope is theta = 1.667e-03 radians\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8-pg147"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "## initialization of variables\n",
+ "#calculate the safe load taken by the column\n",
+ "b=30. ## cm\n",
+ "h=30. ##cm\n",
+ "n=6.\n",
+ "A=36. ##cm^2\n",
+ "ss_s=1500. ##kg/cm^2\n",
+ "ss_c=60. ##kg/cm^2\n",
+ "Er=15. ## Elasticity ratio\n",
+ "## calculations\n",
+ "L=A*Er*ss_c+(b*h-A)*ss_c\n",
+ "## results\n",
+ "print'%s %.2f %s'%('The safe load is ',L,'.kg')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The safe load is 84240.00 .kg\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex9-pg148"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#find the stress in steel and concerete after redisturbution of stress in steel and cncerete\n",
+ "## initiaization of variables\n",
+ "import math\n",
+ "gs_b=10. ##cm\n",
+ "gs_h=10. ##cm\n",
+ "d_b=2. ##cm\n",
+ "d_h=2. ##cm\n",
+ "As= 1. ##cm^2\n",
+ "s=10000. ##kg/cm^2\n",
+ "## part (a)\n",
+ "Es=2*10**6 ##kg/cm^2\n",
+ "Ec=2*10**5 ##kg/cm^2\n",
+ "## calculations\n",
+ "e=s/Es\n",
+ "Ac=gs_b*gs_h-(d_b*d_h)\n",
+ "e_c=e*Es*As/(Ec*Ac+Es*As)\n",
+ "s_c=Ec*e_c\n",
+ "e_s=e-e_c\n",
+ "s_s=Es*e_s\n",
+ "## results\n",
+ "print'%s %.2f %s %.2f %s '%('part (a) \\n The stress in steel and concrete are respectively ',s_s,''and '',s_c,' kg/cm^2')\n",
+ "## part(b)\n",
+ "P=8000. ##kg\n",
+ "## calculations\n",
+ "e_c=(e*Es*As-P)/(Ec*Ac+Es*As)\n",
+ "e_s=e-e_c\n",
+ "s_c=Ec*e_c\n",
+ "s_s=Es*e_s\n",
+ "## results\n",
+ "print'%s %.2f %s %.2f %s'%('\\n part (b) \\n The stress in steel and concrete are respectively ',s_s,''and '',s_c,'kg/cm^2')\n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "part (a) \n",
+ " The stress in steel and concrete are respectively 9056.60 94.34 kg/cm^2 \n",
+ "\n",
+ " part (b) \n",
+ " The stress in steel and concrete are respectively 9811.32 18.87 kg/cm^2\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex10-pg151"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##calculate temperature which sleeve must be heated if the room temperature is 10c\n",
+ "#and pressure and find axial force necessary to separate the two room temperature and the temperature at which sleeve will easily come off\n",
+ "## initialization\n",
+ "import math\n",
+ "d=10 ##cm\n",
+ "D=9.99 ##cm\n",
+ "t=3 ##mm\n",
+ "E=1.0*10**6 ##kg/cm^2\n",
+ "a=2.02*10**-5 ## degree/celcius\n",
+ "## part(a)\n",
+ "Tr=10. ##degree C\n",
+ "T=(d-D)/D*1/a\n",
+ "print'%s %.2f %s'%('part(a) \\n The sleeve must be heated to ',T+Tr,' degree C or more for this purpose')\n",
+ "\n",
+ "##part(b)\n",
+ "s_th=a*T*E\n",
+ "p=s_th*t*2./(d*10.)\n",
+ "print'%s %.2f %s'%('\\n part(b) \\n The pressure developed between the rod and sleeve is',p,' kg/cm^2')\n",
+ "\n",
+ "## part(c)\n",
+ "f=0.2\n",
+ "o=10. ## overlap: cm\n",
+ "A=math.pi*d*o\n",
+ "F=f*p*A\n",
+ "print'%s %.2f %s'%('\\n part (c) \\n The axial force required is ',F,' kg')\n",
+ "\n",
+ "##part (d)\n",
+ "## linked to part c\n",
+ "T2=20. ##degree C\n",
+ "a2=1.17*10**-5 ## /degree C\n",
+ "Ts=(a-a2)*(T2-Tr)*E\n",
+ "Ts=s_th-Ts\n",
+ "p2=p*Ts/s_th\n",
+ "F2=F*Ts/s_th\n",
+ "print'%s %.2f %s'%('\\n part(d)\\n The pressure developed between the rod and sleeve is',p2,' kg/cm^2')\n",
+ "print'%s %.2f %s'%('\\n The axial force required is ',F2,' kg')\n",
+ "##part(e)\n",
+ "T3=Tr+(s_th/((a-a2)*10**6))\n",
+ "print'%s %.2f %s'%('\\n part(e) \\n The temperature at which the sleeve comes off easily is ',T3,' C')\n",
+ "\n",
+ "print('calculations in the text: rounding off errors')\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "part(a) \n",
+ " The sleeve must be heated to 59.55 degree C or more for this purpose\n",
+ "\n",
+ " part(b) \n",
+ " The pressure developed between the rod and sleeve is 60.06 kg/cm^2\n",
+ "\n",
+ " part (c) \n",
+ " The axial force required is 3773.68 kg\n",
+ "\n",
+ " part(d)\n",
+ " The pressure developed between the rod and sleeve is 54.96 kg/cm^2\n",
+ "\n",
+ " The axial force required is 3453.24 kg\n",
+ "\n",
+ " part(e) \n",
+ " The temperature at which the sleeve comes off easily is 127.76 C\n",
+ "calculations in the text: rounding off errors\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex11-pg154"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "##initialization of variables\n",
+ "#calculate the radius of curvature of this strip at temperautre of 93.3\n",
+ "T1=37.8 ## degre C\n",
+ "t=0.355 ##mm\n",
+ "T2=93.3 ## degree C\n",
+ "L=2 ##cm\n",
+ "m=1\n",
+ "n=1.53\n",
+ "a=1.86*10**-5\n",
+ "##calculations\n",
+ "R=2*t*(3*(1+m)**2.+(1+m*n)*(m**2+(m*n)**-1))\n",
+ "R=R/(6.*a*(T2-T1)*(1+m**2)) ## mm\n",
+ "R=R/10.\n",
+ "D=L**2./(8.*R)\n",
+ "## results\n",
+ "print'%s %.2f %s'%('The radius of curvature is ',R,' cm')\n",
+ "print'%s %.4f %s'%('\\n The deflection is',D ,' cm')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The radius of curvature is 92.76 cm\n",
+ "\n",
+ " The deflection is 0.0054 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex12-pg155"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "## initialization of variables\n",
+ "#find the energy stored in bolt\n",
+ "L=5. ##cm\n",
+ "D=1.8 ##cm\n",
+ "l=2.5 ##cm\n",
+ "d=1.5 ##cm\n",
+ "F=1 ##tonne\n",
+ "E=2.1*10**6 ##kg/cm^2\n",
+ "## calculations\n",
+ "s1=F*1000.*4./(D**2*math.pi)\n",
+ "s2=F*1000.*4./(d**2*math.pi)\n",
+ "U1=1/2.*s1**2./E\n",
+ "U1=U1*L*D**2*math.pi/4.\n",
+ "U2=1/2.*s2**2./E\n",
+ "U2=U2*l*d**2*math.pi/4.\n",
+ "U=U1+U2\n",
+ "## results\n",
+ "print'%s %.1f %s'%('The energy stored in the bolt is ',U,' kg-cm')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The energy stored in the bolt is 0.8 kg-cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex13-pg159"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "## initialization of variables\n",
+ "#calculate which of the two ways will give stronger joint\n",
+ "t=16. ##mm\n",
+ "Pt=1500. ##kg/cm^2\n",
+ "Ps=1025. ##kg/cm^2\n",
+ "Pb=2360. ##kg/cm^2\n",
+ "\n",
+ "##part (a)\n",
+ "p=6. ##cm\n",
+ "r=24. ##mm\n",
+ "d=r/10.+0.15\n",
+ "Ft=t*(p-d)*Pt/10.\n",
+ "Fs=math.pi*d**2*Ps/4.\n",
+ "Fb=d*t*Pb\n",
+ "x=min(Ft,Fs,Fb)\n",
+ "effA=x*100./(p*t/10.*Pt)\n",
+ "\n",
+ "##part (b)\n",
+ "p=9. ##cm\n",
+ "r=30. ##mm\n",
+ "d=r/10.+0.2\n",
+ "Ft=t*(p-d)*Pt/10.\n",
+ "Fs=math.pi*d**2*Ps/4.\n",
+ "Fb=d*t*Pb\n",
+ "x=min(Ft,Fs,Fb)\n",
+ "effB=x*100./(p*t/10.*Pt)\n",
+ "\n",
+ "## results\n",
+ "print'%s %.2f %s %.2f %s '%('The efficiencies corresponding to cases a and b are ',effA,'' and '',effB,'')\n",
+ "print('\\n Hence part b is better than part a')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The efficiencies corresponding to cases a and b are 36.35 38.16 \n",
+ "\n",
+ " Hence part b is better than part a\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file |