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diff --git a/Mechanics_of_Materials/Chapter12.ipynb b/Mechanics_of_Materials/Chapter12.ipynb new file mode 100755 index 00000000..ac815672 --- /dev/null +++ b/Mechanics_of_Materials/Chapter12.ipynb @@ -0,0 +1,437 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 12:Deflection of Beams and Shaft"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.10 Page No 610"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E = 200*10**6 #kN/m**2, stress\n",
+ "I = 17*10**-6 #mm**4, moment of inertia\n",
+ "\n",
+ "#The given dimension are\n",
+ "l_ac = 2 #m\n",
+ "l_cF = 4 #m\n",
+ "l_Fb = 2 #m\n",
+ "l_cb = 6 #m\n",
+ "l_aF = 6 #m\n",
+ "l_ab = 8 #m\n",
+ "F = 16 #kN\n",
+ "R_b = (F*l_cb)/l_ab\n",
+ "R_a = F - R_b\n",
+ "\n",
+ "#Calculation\n",
+ "mc = R_a*l_ac\n",
+ "mf = R_b*l_Fb\n",
+ "theta_ca = (0.5*l_ac*mc)/(E*I)\n",
+ "A1 = 0.5*l_aF*mf\n",
+ "t1_ba = (l_Fb + l_aF/3)*(A1)\n",
+ "A2 = 0.5*l_Fb*mf\n",
+ "t2_ba = (l_Fb*2*A2)/3\n",
+ "t_ba = (t1_ba+t2_ba)/(E*I)\n",
+ "theta_c = (t_ba/l_ab)-(theta_ca)\n",
+ "\n",
+ "#Display\n",
+ "print\"The slope at point C of the steel beam = \",round(theta_c,4),\"rad\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The slope at point C of the steel beam = 0.0094 rad\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.12 Page No 612"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E = 29*10**3 #ksi, stress\n",
+ "I = 125 #inch**4, moment of inertia\n",
+ "l_ab = 12 #ft, dimension\n",
+ "l_bc =12 #ft\n",
+ "l_ac = l_ab+l_bc\n",
+ "R_a = -5 #kip, normal force\n",
+ "R_b = 10 #kip\n",
+ "R_c = 5 #kip\n",
+ "\n",
+ "#calculation\n",
+ "mb = R_a*l_ab\n",
+ "#Moment-Area Theorem\n",
+ "t_ca = (l_ab*0.5*l_ac*mb)/(E*I)\n",
+ "t_ba = ((1/3.0)*l_ab*0.5*l_ab*mb)/(E*I)\n",
+ "del_c = (t_ca - 2*t_ba)*1728\n",
+ "\n",
+ "#Display\n",
+ "print\"The displacement at point C for the steel overhanging beam =\",round(del_c,2),\"inch\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The displacement at point C for the steel overhanging beam = -2.75 inch\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.13 Page No 620"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "w = 2 #kN/m, load\n",
+ "L = 8 #m, length\n",
+ "P = 8 #kN\n",
+ "\n",
+ "#Calculations\n",
+ "EI_theta_A1 = (3*w*L**3)/(128) #ThetaA1 = (3wL**3)/(128EI)\n",
+ "EI_nu_C1 = (5*w*L**4)/(768) #NuC1 = (5wL**4)/(768EI)\n",
+ "EI_theta_A2 = (P*L**2)/(16) #theta_A2 = (PL**2)/(16EI)\n",
+ "EI_nu_C2 = (P*L**3)/(48) #nu_C2 = (PL**3)/(48EI)\n",
+ "theta_A = EI_theta_A1 + EI_theta_A2\n",
+ "nu_C = EI_nu_C1 + EI_nu_C2\n",
+ "\n",
+ "#Display\n",
+ "print'The slope at A in terms of EI =',theta_A,\"kNm**2\"\n",
+ "print'The displacement at point C in terms of EI =',nu_C,\"kNm**2\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The slope at A in terms of EI = 56 kNm**2\n",
+ "The displacement at point C in terms of EI = 138 kNm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.14 Page No 621"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "w = 5 #kN/m, force per unit length\n",
+ "l_ab = 4 #m, length\n",
+ "l_bc = 2 #m\n",
+ "P = 10 #kN, load\n",
+ "M = w*l_ab #kNm\n",
+ "\n",
+ "#Calculations\n",
+ "EI_theta_B1 = (w*l_ab**3)/(24) #ThetaB1 = (wL**3)/(24EI)\n",
+ "EI_nu_C1 = l_bc*EI_theta_B1\n",
+ "\n",
+ "EI_theta_B2 = (M*l_ab)/(3) #\n",
+ "EI_nu_C2 = l_bc*EI_theta_B2\n",
+ "EI_nu_C3 = (P*l_bc**3)/(3) #nuC3 = (PL**3)/(24EI)\n",
+ "nu_C = -EI_nu_C1 + EI_nu_C2 + EI_nu_C3\n",
+ "\n",
+ "#Display\n",
+ "print'The displacement at end C of the overhanging beam, in terms of EI = ',nu_C,\"kNm**3\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The displacement at end C of the overhanging beam, in terms of EI = 52 kNm**3\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.15 Page No 622"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "w = 4 #kN/m, force per unit length\n",
+ "l = 6 #m, length\n",
+ "l_bc =2 #m\n",
+ "\n",
+ "#Calculations\n",
+ "EI_theta_B = (w*l**3)/(24.0) #ThetaB1 = (wL**3)/(24EI)\n",
+ "EI_nu_B = (w*l**4)/(30.0) #nuB = (wL**4)/(30EI)\n",
+ "nu_C = EI_nu_B + (EI_theta_B*l_bc)\n",
+ "\n",
+ "#Display\n",
+ "print'The displacement at end C of the cantilever beam, in terms of EI = ',nu_C,\"kNm**3\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The displacement at end C of the cantilever beam, in terms of EI = 244.8 kNm**3\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.16 Page No 623"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "k = 15.0 #kip/ft, force per unit length\n",
+ "F = 3 #kip, force\n",
+ "E = 29*10**3 #ksi, stress\n",
+ "l_ab = 3.0 #ft, length\n",
+ "l_ac = 1 #ft\n",
+ "l_cb = 2 #ft\n",
+ "I = 12 #in**4, moment of inertia\n",
+ "R_a = (F*l_cb)/(l_ab)\n",
+ "R_b = F-R_a\n",
+ "\n",
+ "#Calculations\n",
+ "mu_a = (R_a)/k\n",
+ "mu_b = (R_b)/k\n",
+ "mu_c1 = mu_b + (l_cb/l_ab)*(mu_a - mu_b)\n",
+ "#From fig b\n",
+ "a=3 #ft\n",
+ "b=6 #ft\n",
+ "L=9 #ft\n",
+ "mu_c2 = ((F*a*b)*(L**2 - a**2 - b**2))/(6*E*144*I*(1/20736.0)*L)\n",
+ "mu_c = mu_c1 + mu_c2\n",
+ "\n",
+ "#Display\n",
+ "print'The vertical displacement of the force at C = ',round(mu_c,3),\"ft\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The vertical displacement of the force at C = 0.126 ft\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.21 Page No 643"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "l = 10 #ft, length\n",
+ "P = 8 #kip, load\n",
+ "w = 2 #kip/ft, force per unit length\n",
+ "\n",
+ "#Calculation\n",
+ "#Compatibility Equation\n",
+ "EI_nu_b1 = (w*l**4)/8.0 + (5*P*l**3)/48.0 #nu_b = (wl**4)/8EI + (5Pl**3)/48EI\n",
+ "EI_nu_b2 = (l**3)/3.0\n",
+ "B_y = EI_nu_b1 / EI_nu_b2\n",
+ "\n",
+ "#Display\n",
+ "print\"The reactions at roller support B = \",B_y,\"kip\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The reactions at roller support B = 10.0 kip\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.22 Page No 644"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given:\n",
+ "l = 8 #ft, length\n",
+ "l_ab = 5 #ft\n",
+ "l_bc = 5 #ft\n",
+ "l_af = 1/2.0 #inch\n",
+ "b = 12/1000.0 #m\n",
+ "w = 8 #kip, force per unit length\n",
+ "E = 29*10**3 #Ksi\n",
+ "I = 475.0 # inch**4, moment of inertia\n",
+ "\n",
+ "#Compatibility Equation:\n",
+ "import math\n",
+ "A=math.pi/4.0*(l_af**2)\n",
+ "muB__byFbc=l*12/(A*E)\n",
+ "L=l_ab+l_bc\n",
+ "muB=5*w*L**3*12/(48*E*I)\n",
+ "muB_byFbc=L**3*12/(3*E*I)\n",
+ "#From equation muB__=muB-muB_\n",
+ "Fbc=muB/(muB_byFbc+muB__byFbc)\n",
+ "\n",
+ "#Display:\n",
+ "print \"Force developed in the rod is\",round(Fbc,3),\"kip\"\n",
+ "print\"In the book: Calculation mistake\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Force developed in the rod is 0.042 kip\n",
+ "In the book: Calculation mistake\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.23 Page No: 646"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given:\n",
+ "L = 12 #ft\n",
+ "#E and I are constant say\n",
+ "E = 29*10**3 #Ksi\n",
+ "I = 475 # inch**4\n",
+ "w = 3 #kip/ft\n",
+ "\n",
+ "#calculation\n",
+ "thetab=w*L**3/(48*E*I)\n",
+ "mub=7*w*L**4/(384*E*I)\n",
+ "theta_bbyBy=L**2/(2*E*I)\n",
+ "mu_bbyBy=L**3/(3*E*I)\n",
+ "theta__byMb=L/(E*I)\n",
+ "muB__byMb=L**2/(2*E*I)\n",
+ "\n",
+ "#From eq 1 and 2 (solving by matrix)\n",
+ "#72By+12Mb=-108\n",
+ "#576By+72Mb=-1134\n",
+ "M = array([[576, 72], [72, 12]])\n",
+ "N=([-108,-1134])\n",
+ "X=inv(M)*N #Inverse matrix\n",
+ "a=X[0,0]\n",
+ "b=X[1,0]*2.5\n",
+ "\n",
+ "#Display:\n",
+ "print\"Moment at B is\",b,\"kip-ft\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Moment at B is 11.25 kip-ft\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file |