{ "metadata": { "name": "", "signature": "sha256:42031753a466b3ba5ee8c11468e375015d1a3d943c879daacc89b28eec81ab67" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "Chapter 10:Deflections of beams " ] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 10.1 page number 501" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Given \n", "dia = 400 #mm - The diameter of a pulley\n", "E = 200 #Gpa - Youngs modulus\n", "t = 0.6 #mm - The thickness of band\n", "c = t/2 #mm - The maximum stress is seen \n", "#Caliculations\n", "\n", "stress_max = E*c*(10**3)/(dia/2) #Mpa - The maximum stress on the crossection occurs at the ends\n", "print \"The maximum bending stress developed in the saw \",stress_max,\"Mpa\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The maximum bending stress developed in the saw 300.0 Mpa\n" ] } ], "prompt_number": 3 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 10.3 page number 512" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Given\n", "import numpy\n", "l_ab = 1.0 #L in - The length of the beam\n", "F_D = 1.0 #W lb/in - The force distribution \n", "F = F_D*l_ab #WL - The force applied\n", "#Beause of symmetry the moment caliculations can be neglected\n", "#F_Y = 0\n", "R_A = F/2 #wl - The reactive force at A\n", "R_B = F/2 #wl - The reactive force at B\n", "#EI - The flxure rigidity is constant and 1/EI =1 # k\n", "\n", "#part - A\n", "#section 1--1\n", "l_1 = [0,0.1,0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9,1] #L taking each section at 0.1L distance \n", "M_1 = [0,0,0,0,0,0,0,0,0,0,0]\n", "v = [0,0,0,0,0,0,0,0,0,0,0]\n", "for i in range(10):\n", " v[i] = R_A - F_D*l_1[i] \n", " M_1[i] = R_A*l_1[i] - F_D*(l_1[i]**2)/2\n", "# (EI)y'' = M_1[i] we will integrate M_1[i] twice where variable is l_1[i]\n", "#(EI)y'- \n", "\n", "M_1_intg1 = R_A*(l_1[i]**2)/4 - F_D*(l_1[i]**3)/6 - F_D*(l_ab**3)*l_1[i]/24 #integration of x**n = x**n+1/n+1\n", "#(EI)y- Using end conditions for caliculating constants \n", "\n", "M_1_intg2 = R_A*(l_1[i]**3)/12.0 - F_D*(l_1[i]**4)/24.0 + F_D*(l_ab**3)*l_1[i]/24.0 \n", "#Equations \n", "\n", "l_1 = [0,0.1,0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9,1] #L taking each section at 0.1L distance \n", "M_1_intg2 = [0,0,0,0,0,0,0,0,0,0,0]\n", "Y = [0,0,0,0,0,0,0,0,0,0,0]\n", "for i in range(10):\n", " M_1_intg2[i] = (l_1[i]**3)/12.0 - (l_1[i]**4)/24.0 - l_1[i]/24.0 # discluding every term for ruling out float values\n", " Y[i] = M_1_intg2[i] #W(l**4)/EI k = 1/EI\n", "#The precision is very less while caliculating through this equation because the least count in X direction is 0.1\n", "print \"a) The maximum displacement in y direction is\",min(Y),\"W(l**4)/EI \"\n", "print \"a) The maximum deflection occured at\",l_1[Y.index(min(Y))],\"L\"\n", "\n", "#Part - B\n", "#Graphs\n", "import numpy as np\n", "values = M_1\n", "y = np.array(values)\n", "t = np.linspace(0,1,11)\n", "poly_coeff = np.polyfit(t, y, 2)\n", "import matplotlib.pyplot as plt\n", "plt.plot(t, y, 'o')\n", "plt.plot(t, np.poly1d(poly_coeff)(t), '-')\n", "plt.show()\n", "print \"b) The above graph is bending moment graph\"\n", "import numpy as np\n", "values = Y \n", "y = np.array(values)\n", "t = np.linspace(0,1,11)\n", "poly_coeff = np.polyfit(t, y, 2)\n", "import matplotlib.pyplot as plt\n", "plt.plot(t, y, 'o')\n", "plt.plot(t, np.poly1d(poly_coeff)(t), '-')\n", "plt.show()\n", "print \"b)The above graph is beam displacement graph\"\n", "print \"b)The maximum occures in the middle from the above graph \"\n", "\n", "\n", " \n", "\n", "\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "a) The maximum displacement in y direction is -0.0130208333333 W(l**4)/EI \n", "a) The maximum deflection occured at 0.5 L\n" ] }, { "metadata": {}, "output_type": 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R6pNEAAAAAElFTkSuQmCC\n", "text": [ "" ] }, { "output_type": "stream", "stream": "stdout", "text": [ "b) The above graph is bending moment graph\n" ] }, { "metadata": {}, "output_type": "display_data", "png": 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"text": [ "" ] }, { "output_type": "stream", "stream": "stdout", "text": [ "b)The above graph is beam displacement graph\n", "b)The maximum occures in the middle from the above graph \n" ] } ], "prompt_number": 4 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 10.4 page number 514" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Given\n", "import numpy\n", "l_ab = 1.0 #L in - The length of the beam\n", "F_D = 1.0 #W lb/in - The force distribution \n", "F = F_D*l_ab #WL - The force applied\n", "#Beause of symmetry the moment caliculations can be neglected\n", "#F_Y = 0\n", "R_A = F/2 #wl - The reactive force at A\n", "R_B = F/2 #wl - The reactive force at B\n", "#EI - The flxure rigidity is constant and 1/EI =1 # k\n", "#M_A and M_B are applied at the ends\n", "\n", "#part - A\n", "#section 1--1\n", "l_1 = [0,0.1,0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9,1] #L taking each section at 0.1L distance \n", "M = [0,0,0,0,0,0,0,0,0,0,0]\n", "for i in range(10):\n", " M[i] = l_1[i]/2.0 - (l_1[i]**2)/2.0 -1.0/12.0 #The moment euation at 1--1 section\n", "# M_1 = R_A*l_1[i]/2.0 - F_D*(l_1[i]**2)/2.0 -F_D*(l_ab**2)/12.0 #The moment euation at 1--1 section \n", "# (EI)y'' = M_1[i] we will integrate M_1[i] twice where variable is l_1[i]\n", "#(EI)y'\n", "M_1_intg1 = R_A*(l_1[i]**2)/4 - F_D*(l_1[i]**3)/6 - F_D*(l_ab**2)*l_1[i]/12.0 #integration of x**n = x**n+1/n+1\n", "#(EI)y\n", "M_1_intg2[i] = R_A*(l_1[i]**3)/12.0 - F_D*(l_1[i]**4)/24.0 + F_D*(l_ab**2)*(l_1[i]**2)/24.0 \n", "\n", "l_1 = [0,0.1,0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9,1] #L taking each section at 0.1L distance \n", "M_1_intg2 = [0,0,0,0,0,0,0,0,0,0,0]\n", "Y = [0,0,0,0,0,0,0,0,0,0,0]\n", "for i in range(10):\n", " M_1_intg2[i] = (l_1[i]**3)/12.0 - (l_1[i]**4)/24.0 - (l_1[i]**2)/24.0 # discluding every term for ruling out float values\n", " Y[i] = M_1_intg2[i] #W(l**4)/EI k = 1/EI\n", " \n", "#Part - B\n", "#Graphs\n", "import numpy as np\n", "values = M\n", "y = np.array(values)\n", "t = np.linspace(0,1,11)\n", "poly_coeff = np.polyfit(t, y, 2)\n", "import matplotlib.pyplot as plt\n", "plt.plot(t, y, 'o')\n", "plt.plot(t, np.poly1d(poly_coeff)(t), '-')\n", "plt.show()\n", "print \"b) The above graph is bending moment graph\"\n", "import numpy as np\n", "values = Y \n", "y = np.array(values)\n", "t = np.linspace(0,1,11)\n", "poly_coeff = np.polyfit(t, y, 2)\n", "import matplotlib.pyplot as plt\n", "plt.plot(t, y, 'o')\n", "plt.plot(t, np.poly1d(poly_coeff)(t), '-')\n", "plt.show()\n", "\n", "\n", " \n" ], "language": "python", "metadata": {}, "outputs": [ { "metadata": {}, "output_type": "display_data", "png": 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"text": [ "" ] }, { "output_type": "stream", "stream": "stdout", "text": [ "b) The above graph is bending moment graph\n" ] }, { "metadata": {}, "output_type": "display_data", "png": 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"text": [ "" ] } ], "prompt_number": 68 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 10.5 page number 517" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Given \n", "#because of symmetry the problem can be solved by considering first half\n", "#Given\n", "import numpy\n", "l_ab = 1.0 #L in - The length of the beam\n", "F_D = 1.0 #W lb/in - The force distribution \n", "F = F_D*l_ab #WL - The force applied\n", "#Beause of symmetry the moment caliculations can be neglected\n", "#EI - The flxure rigidity is constant and 1/EI =1 # k\n", "\n", "#part - A\n", "#section 1--1\n", "l_1 = [0,0.1,0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9,1] #L taking each section at 0.1L distance \n", "M_1 = [0,0,0,0,0,0,0,0,0,0,0]\n", "v = [0,0,0,0,0,0,0,0,0,0,0]\n", "for i in range(10):\n", " v[i] = R_A - F_D*l_1[i] \n", " M_1[i] = R_A*l_1[i] - F_D*(l_1[i]**2)/2\n", "# (EI)y'' = M_1[i] we will integrate M_1[i] twice where variable is l_1[i]\n", "#(EI)y'\n", "M_1_intg1 = R_A*(l_1[i]**2)/2 - F_D*(l_1[i]**3)/6 - F_D*(l_ab**3)*l_1[i]/48 #integration of x**n = x**n+1/n+1\n", "#(EI)y\n", "M_1_intg2[i] = R_A*(l_1[i]**3)/6 - F_D*(l_1[i]**4)/24.0 - F_D*(l_ab**3)*l_1[i]/48.0 \n", "#Equations \n", "#R_A = #wl Unknown- The reactive force at A\n", "#R_B = #wl - The reactive force at B\n", "\n", "# M_1_intg2[10] = 0, the displacement at the end of rod is 0 since its rigid \n", "R_A = (F_D*(l_1[10]**4)/24.0 + F_D*(l_ab**3)*l_1[10]/48.0)/((l_1[10]**3)/6.0)\n", "R_C = R_A #WL - symmetry\n", "R_B = 1-R_A+R_B # WL - F_Y = 0, the equilibrium in Y direction\n", "print \"The reaction at A is\",R_A ,\"WL\"\n", "print \"The reaction at B is\",R_B ,\"WL\"\n", "print \"The reaction at C is\",R_C ,\"WL\"\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " The reaction at A is 0.375 WL\n", "The reaction at B is 2.375 Wl\n", "The reaction at C is 0.375 Wl\n" ] } ], "prompt_number": 82 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 10.7 page number 521 " ] }, { "cell_type": "code", "collapsed": false, "input": [ "l_ac = 5 #m - The length of the beam \n", "l_ab = 4 #m - The length of ac on beam \n", "l_bc = 1 #m - The length of bc on beam \n", "F = 20 #N - force applied on beam at 'b'\n", "I_ab = 4 #I The moment of inertia of part AB \n", "I_bc = 1 #I - The momemt of inertia of part BC\n", "R_A = F*(l_bc/l_ac) #N- The reaction at joint A\n", "R_B = F*(l_ab/l_ac) #N- The reaction at joint B\n", "E = 1 #E youngs modulus\n", "\n", "#0" ] }, { "output_type": "stream", "stream": "stdout", "text": [ "b) The shape from x belongs to 0" ] }, { "output_type": "stream", "stream": "stdout", "text": [ "b) The shape from x belongs to 4" ] }, { "metadata": {}, "output_type": "display_data", "png": 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