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diff --git a/Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter12.ipynb b/Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter12.ipynb deleted file mode 100755 index 13fb52c7..00000000 --- a/Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter12.ipynb +++ /dev/null @@ -1,405 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:aaf5a5f62a391d916c40aa2d720de6a3e7681d1c9c64ec2fdbeff148819b3c75"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Chapter 12:Energy and Virtual-work Methods"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.1 page number 645 "
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "#Virtual loading\n",
- "p_ab = -0.833 #lb The recorded virtual loading\n",
- "p_bc = + 0.833 #lb The recorded virtual loading\n",
- "F_ab = 2500 #lb\n",
- "F_bc = -2500 #lb\n",
- "l_ab = 60 #in - The length of the rod\n",
- "l_bc = 60 #in - The length of the rod\n",
- "A_ab = 0.15 #in2 the areaof ab\n",
- "A_bc = 0.25 #in2 the areaof bc\n",
- "E = 30*(10**6) #psi The youngs modulus of the material\n",
- "#Part_a\n",
- "e_a =p_ab*l_ab*F_ab/(A_ab*E) + p_bc*l_bc*F_bc/(A_bc*E) #in the deflection\n",
- "if e_a<0:\n",
- " print \"a) The deflection is downwards\",round(-e_a,3),\"in\"\n",
- "else:\n",
- " print \"a) The deflection is upwards\",round(e_a,3),\"in\"\n",
- "#part-b\n",
- "x = 0.125 #Shortening of member Ab\n",
- "e_b = p_ab*(-x) + p_bc*0 #in - in\n",
- "if e_b<0:\n",
- " print \"b) The deflection is downwards\",round(-e_b,3),\"in\"\n",
- "else:\n",
- " print \"b) The deflection is upwards\",round(e_b,3),\"in\"\n",
- "#Part-c\n",
- "S = 6.5*(10**-6) #Thermal specific heat\n",
- "T = 120 #F - The cahnge in temperature\n",
- "e_t = -S*T*l_ab #in - The change in length of member\n",
- "e_c = p_bc*e_t #in the deflection\n",
- "if e_c<0:\n",
- " print \"c) The deflection is downwards\",round(-e_c,3),\"in\"\n",
- "else:\n",
- " print \"c) The deflection is upwards\",round(e_c,3),\"in\"\n",
- "\n",
- "\n",
- " \n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "a) The deflection is downwards 0.044 in\n",
- "b) The deflection is upwards 0.104 in\n",
- "c) The deflection is downwards 0.039 in\n"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.3 page number 648"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "#Virtual loading\n",
- "#Two parts \n",
- "#Part -1 \n",
- "p_ab = 5 #KN The recorded virtual loading\n",
- "p_bc = -4 #KN The recorded virtual loading\n",
- "F_ab = 10 #KN\n",
- "F_bc = -8 #KN\n",
- "l_ab = 2.5 #mt - The length of the rod\n",
- "l_bc = 2 #mt - The length of the rod\n",
- "A_ab = 5*(10**-4) #mt2 the areaof ab\n",
- "A_bc = 5*(10**-3) #mt2 the areaof bc\n",
- "E = 70 #Gpa The youngs modulus of the material\n",
- "e_a =(p_ab*l_ab*F_ab/(A_ab*E) + p_bc*l_bc*F_bc/(A_bc*E))*(10**-6) #KN-m\n",
- "#Part -2 due to flexure\n",
- "I = 60*10**6 #mm4 - the moment of inertia \n",
- "#After solving the integration \n",
- "e_b = 0.01525 #KN-m\n",
- "#Total\n",
- "e = (e_a+e_b)*1 #m\n",
- "print \"The point C deflects\",round(e,3),\"mt down\"\n",
- "\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The point C deflects 0.019 mt\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.5 page number 651"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "#Virtual loading Without f_d\n",
- "p_ab = -0.833 #lb The recorded virtual loading\n",
- "p_bc = + 0.833 #lb The recorded virtual loading\n",
- "F_ab = 2500 #lb\n",
- "F_bc = -2500 #lb\n",
- "l_ab = 60 #in - The length of the rod\n",
- "l_bc = 60 #in - The length of the rod\n",
- "A_ab = 0.15 #in2 the areaof ab\n",
- "A_bc = 0.25 #in2 the areaof bc\n",
- "E = 30*(10**6) #psi The youngs modulus of the material\n",
- "#Part_a\n",
- "e_a =p_ab*l_ab*F_ab/(A_ab*E) + p_bc*l_bc*F_bc/(A_bc*E) #lb-in the deflection\n",
- "#With f_d\n",
- "p_bd = 1 #lb The recorded virtual loading \n",
- "F_bd = 1 #lb\n",
- "l_bd = 40 #in - The length of the rod\n",
- "A_bd = 0.1 #in2 the areaof ab\n",
- "e_a_1 =p_ab*p_ab*l_ab/(A_ab*E) + p_bc*p_bc*l_bc/(A_bc*E) +p_bd*p_bd*l_bd/(A_bd*E) #lb-in the deflection\n",
- "#Since the produced defelection should compensate the other one\n",
- "x_d = e_a/e_a_1\n",
- "print \"The reaction force at D is\",round(-x_d,2),\"lb\"\n",
- "\n",
- "#Part - B\n",
- "e_b = -x_d*l_bd/(A_bd*E ) #in - The deflection of nodal point B\n",
- "print\"The deflection of nodal point B\",round(e_b,4),\"in\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The reaction force at D is 1578.98 lb\n",
- "The deflection of nodal point B 0.0211 in\n"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.6 page number 655"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "#Virtual loading\n",
- "p_ab = -0.833 #lb The recorded virtual loading\n",
- "p_bc = + 0.833 #lb The recorded virtual loading \n",
- "l_ab = 60 #in - The length of the rod\n",
- "l_bc = 60 #in - The length of the rod\n",
- "A_ab = 0.15 #in2 the areaof ab\n",
- "A_bc = 0.25 #in2 the areaof bc\n",
- "E = 30*(10**6) #psi The youngs modulus of the material\n",
- "K_1 = A_ab*E/l_ab #k/in - Stiffness\n",
- "K_2 = A_bc*E/l_bc #k/in - Stiffness\n",
- "#soving for e_1 and e_2 gives a liner euations to solve\n",
- "# 128*e_1 + 24*e_2 = 0\n",
- "#24*e_1 + 72*e_2 = -3\n",
- "#Solving for e_1,e_2\n",
- "a = np.array([[128,24], [24,72]])\n",
- "b = np.array([0,-3])\n",
- "x = np.linalg.solve(a, b)\n",
- "e_1 = x[0] #in\n",
- "e_2 = x[1] #in\n",
- "u_1 = 0.8*e_1 - 0.6*e_2 #Taking each components\n",
- "F_1 = K_1*u_1*(10**-3) #k The reaction at A Force = stiffness x dislacement \n",
- "u_2 = 0.8*e_1 + 0.6*e_2 #Taking each components\n",
- "F_2 = K_2*u_2*(10**-3) #k The reaction at B Force\n",
- "print \"The reaction at A \",F_1,\"k\"\n",
- "print \"The reaction at B \",F_2,\"k\"\n",
- "\n",
- "\n",
- "\n",
- "\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The reaction at A 2.5 k\n",
- "The reaction at B -2.5 k\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.7 page number 655"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Virtual loading\n",
- "p_ab = -0.833 #lb The recorded virtual loading\n",
- "p_bc = + 0.833 #lb The recorded virtual loading \n",
- "l_ab = 60 #in - The length of the rod\n",
- "l_bc = 60 #in - The length of the rod\n",
- "A_ab = 0.15 #in2 the areaof ab\n",
- "A_bc = 0.25 #in2 the areaof bc\n",
- "E = 30*(10**6) #psi The youngs modulus of the material\n",
- "K_1 = A_ab*E/l_ab #k/in - Stiffness\n",
- "K_2 = A_bc*E/l_bc #k/in - Stiffness\n",
- "p_bd = 1 #lb The recorded virtual loading \n",
- "F_bd = 1 #lb\n",
- "l_bd = 40 #in - The length of the rod\n",
- "A_bd = 0.1 #in2 the areaof ab\n",
- "K_3 = A_ab*E/l_ab #k/in - Stiffness\n",
- "#soving for e_1 and e_2 gives a liner euations to solve\n",
- "# 128*e_1 + 24*e_2 = 0\n",
- "#24*e_1 + 72*e_2 = -3\n",
- "#Solving for e_1,e_2\n",
- "a = np.array([[128,24], [24,147]])\n",
- "b = np.array([0,-3])\n",
- "x = np.linalg.solve(a, b)\n",
- "e_1 = x[0] #in\n",
- "e_2 = x[1] #in\n",
- "u_1 = 0.8*e_1 - 0.6*e_2 #Taking each components\n",
- "F_1 = K_1*u_1*(10**-3) #k The reaction at A Force = stiffness x dislacement \n",
- "u_2 = 0.8*e_1 + 0.6*e_2 #Taking each components\n",
- "F_2 = K_2*u_2*(10**-3) #k The reaction at B Force\n",
- "u_3 = e_2 #Taking each components\n",
- "F_3 = K_3*u_3*(10**-3) #k The reaction at D Force\n",
- "print \"The reaction at A \",round(F_1,2),\"k\"\n",
- "print \"The reaction at B \",round(F_2,2),\"k\"\n",
- "print \"The reaction at D \",round(F_3,2),\"k\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The reaction at A 1.18 k\n",
- "The reaction at B -1.18 k\n",
- "The reaction at D -1.58 k\n"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.8 page number 659"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "#First we will solve part B\n",
- "u_1 =5 #L/AE, elastic elongation\n",
- "u_2 =25 #L/AE,elastic elongation\n",
- "f_1 = u_1#, Units got neutralized , Constitutive relation for elastic bars\n",
- "f_2 = u_2# Units got neutralized\n",
- "#u_1 = 0.8*e_1 - 0.6*e_2\n",
- "#u_2 = 0.8*e_1 + 0.6*e_2\n",
- "#u = A*e Matric multiplication \n",
- "A = np.array([[0.8,-0.6],[0.8,0.6]]) #The matrix form of A\n",
- "F = np.array([[f_1],[f_2]])\n",
- "P = np.dot((A.T),F) #Nodal forces matrix\n",
- "print \"b) The vertical component of the nodal force is\",P[1],\"\"\n",
- "print \"b) The vertical component of the nodal force is\",P[0],\"\"\n",
- "#Part A\n",
- "#F_1 = (5/8.0)*P_1 - (5/6.0)*p_2 , From statics\n",
- "#F_1 = (5/8.0)*P_1 + (5/6.0)*p_2\n",
- "#F = BP ,Matric multiplication \n",
- "B = np.array([[(5/8.0),-(5/6.0)],[(5/8.0),(5/6.0)]]) #The matrix form of A\n",
- "U = np.array([[u_1],[u_2]])\n",
- "e = P = np.dot((B.T),U) #L/AE, Nodal forces matrix\n",
- "print \"a) The components of displacement of point B are\",round(e[0],2),\"L/AE and\",round(e[1],2),\"L/AE\" \n",
- "\n",
- "\n",
- "\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "b) The vertical component of the nodal force is [ 12.] \n",
- "b) The vertical component of the nodal force is [ 24.] \n",
- "a) The components of displacement of point B are 18.75 L/AE and 16.67 L/AE\n"
- ]
- }
- ],
- "prompt_number": 26
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.10 page number 667"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "A_1 = 0.125 #in2 , The area of the crossection of AB\n",
- "A_2 = 0.219 #in2 , The area of the crossection of BC\n",
- "l_1 = 3*(5**0.5) #in , The length of AB\n",
- "l_2 = 6*(2**0.5) #in , The length of BC\n",
- "p = 3 #k , Force acting on the system \n",
- "E = 10.6*(10**3) #Ksi - youngs modulus of the material\n",
- "p_1 = (5**0.5)*p/3 #P, The component of p on AB\n",
- "p_2 = -2*(2**0.5)*p/3 #P, The component of p on AB\n",
- "\n",
- "e = p_1*l_1*p_1/(p*E*A_1) + p_2*l_2*p_2/(p*E*A_2) #in, By virtual deflection method \n",
- "print \"The deflection is\",round(e,3),\"in\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The deflection is 0.018 in\n"
- ]
- }
- ],
- "prompt_number": 32
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [],
- "language": "python",
- "metadata": {},
- "outputs": []
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file |