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{
"metadata": {
"name": "",
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"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 13: Statically Indeterminate Problems"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 13.2 page number 693"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Given \n",
"#First we will solve without the reaction at middle\n",
"#Given\n",
"import numpy\n",
"l_ab = 1.0 #2L 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 0.2L 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 0.2L 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",
"Y_min = 16*min(Y)\n",
"print \"a) The maximum displacement in y direction is\",16*min(Y),\"W(l**4)/EI \"\n",
"print \"a) The maximum deflection occured at\",2*l_1[Y.index(min(Y))],\"L\"\n",
"f_bb = 2**3/48.0 #l**3/EI - flexibility coefficient\n",
"Reac = - Y_min/f_bb #WL , The reaction at the mid of the bar\n",
"print \"The reaction at the mid of the bar\",Reac ,\"WL\"\n",
"\n",
"#Graphs \n",
"Y.extend(Y) #Because of symmetry\n",
"import numpy as np\n",
"values = Y \n",
"y = np.array(values)\n",
"t = np.linspace(0,1,22)\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 minimum 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 -0.208333333333 W(l**4)/EI \n",
"a) The maximum deflection occured at 1.0 L\n",
"The reaction at the mid of the bar 1.25 WL\n"
]
},
{
"metadata": {},
"output_type": "display_data",
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LEREJpbAQEZFQCgsREQmlsBARkVAphYWZtTezYjNba2bzzaxtHe0KzKzEzNaZ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\nQmCC\n",
"text": [
"<matplotlib.figure.Figure at 0x7fad208>"
]
},
{
"output_type": "stream",
"stream": "stdout",
"text": [
"b)The above graph is beam displacement graph\n",
"b)The minimum occures in the middle from the above graph \n"
]
}
],
"prompt_number": 10
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 13.3 page number 694 "
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Given \n",
"#First we will solve without the reaction at middle\n",
"#Given\n",
"import numpy\n",
"l_ab = 1.0 #2L 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 0.2L 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 0.2L 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",
"e_1 = 16*min(Y) #WL4/EI - The maximum defection \n",
"e_2 = - F_D*((2*l_ab)**3)/24.0 #WL3/EI - The maximum angle\n",
"#Caliculating for momentum and force\n",
"f_ab = ((2*l_ab)**2)/16.0 #L2/EI \n",
"f_bb = ((2*l_ab)**3)/48.0 #L3/EI \n",
"f_aa = 2*l_ab/3.0 #L/EI\n",
"f_ba = ((l_ab)**2)/4.0 #L2/EI\n",
"#F*X = e - Matrix multiplication \n",
"#Solving for X\n",
"a = np.array([[f_aa,f_ba], [f_ba,f_bb]])\n",
"b = np.array([e_2,e_1])\n",
"x = np.linalg.solve(a, b)\n",
"print \"The reactive moment at A i.e M_A\",x[0],\"WL**2\"\n",
"print \"The reactive force at A i.e R_A\",x[1],\"WL\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The reactive moment at A i.e M_A -0.0714285714286 WL**2\n",
"The reactive force at A i.e R_A -1.14285714286 WL\n"
]
}
],
"prompt_number": 26
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Given \n"
],
"language": "python",
"metadata": {},
"outputs": []
}
],
"metadata": {}
}
]
}
|
