{ "metadata": { "name": "chapter 12 som.ipynb" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "Chapter 12:Propped Cantilever" ] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Problem no 12.1,Page No.286" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#Initilization of variables\n", "\n", "L=6 #m #Length of Beam\n", "L_1=4 #m #Length of Beam with udl Load\n", "w=10 #KN/m #u.d.l\n", "\n", "#Calculation\n", "\n", "#Deflection of cantileverat C due to udl on AB\n", "y_c=w*L_1**4*8**-1+w*L_1**3*6**-1*(L-L_1) \n", "\n", "#Deflection of cantileverat C due to prop reaction alone\n", "#y_c_2=R_c*L**3*3**-1\n", "\n", "#Since both Deflection are Equal\n", "#y_c=y_c_2\n", "\n", "R_c=y_c*(6**3)**-1*3 #Reaction at C\n", "\n", "#Result\n", "print\"The Reaction at End C is\",round(R_c,3),\"KN\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The Reaction at End C is 7.407 KN\n" ] } ], "prompt_number": 8 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Problem no 12.2,Page No.286" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#Initilization of variables\n", "\n", "L=10 #m #Length\n", "b=15 #cm #Width\n", "d=40 #cm #Depth\n", "y_c=1.5*10**-2 #m #Deflection\n", "E=120*10**9 \n", "y=0.2\n", "sigma=10*10**6 #Bending stress\n", "\n", "#Calculations\n", "\n", "I=b*d**3*12**-1*10**-8 #cm #M.I\n", "\n", "#From Deflection at the centre of cantilever we get\n", "w=y_c*192*E*I*(L**4)**-1*10**-3 #udl distributed over the cantilever\n", "\n", "#From Bending Moment Equation we get\n", "w_2=sigma*I*y**-1*16*(L**2)**-1*10**-3 #udl distributed over the cantilever\n", "\n", "#Result\n", "print\"The safe uniformly Distributed Load is\",round(w_2,2),\"KN/m\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The safe uniformly Distributed Load is 6.4 KN/m\n" ] } ], "prompt_number": 23 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Problem no 12.3,Page No.287" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#Initilization of variables\n", "\n", "L=6 #m #span of beam\n", "w=30*10**3 #KN/m #u.d.l\n", "P=160*10**3 #N #concentrated Load\n", "\n", "#Calculations\n", "\n", "#Consider a section at a distance x from the fixed end A and B.M at x\n", "#M_x=R_b*(6-x)-30*2**-1*(6-x)**2-160*(3-x)\n", "\n", "#E*I*d**2y*(dx**2)**-1=-M_x=-R_b*(6-x)+15*(6-x)+160*(3-x)\n", "\n", "#Now Integrating above term we get\n", "#E*I*dy*(dx)**-1=R_b*2**-1*(6-x)**2-5*(6-x)**3-80*(3-x)**2+C_1 (Equation 1)\n", "\n", "#Now on Integrating we get\n", "#E*I*y=-R_b*6**-1*(6-x)**3+5*4**-1*(6-x)**2+80*3**-1*(3-x)**3+C_1*x+C_2 (Equation 2)\n", "\n", "#At x=0,dy*dx**-1=0\n", "#substituting in equation 1 we get\n", "#C_1=1800-R_b\n", "\n", "#At x=0,y=0\n", "#substituting in equation 2 we get\n", "#C_2=36*R_b-2340\n", "\n", "#At x=6,y=0\n", "R_b=72**-1*(10800-2340)\n", "\n", "#At x=0\n", "x=0\n", "M_x=R_b*(6-x)-30*2**-1*(6-x)**2-160*(3-x)\n", "\n", "#Result\n", "print\"Bending Moment at A is\",round(M_x,2),\"KNm\"\n", "print\"The Reaction at B\",round(R_b,2),\"KN\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Bending Moment at A is -315.0 KNm\n", "The Reaction at B 117.5 KN\n" ] } ], "prompt_number": 5 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Problem no 12.4,Page No.288" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#Initilization of variables\n", "\n", "L=4 #span of beam\n", "w_1=20*10**3 #Nm #u.d.l\n", "w_2=30*10**3 #Nm #u.d.l\n", "\n", "#Calculations\n", "\n", "#consider a section at a distance x from A and B.M at this section is \n", "#M_x=R_b*(3-x)-10*x**2+90*x-195\n", "\n", "#Now integrating above equation we get\n", "#E*I*dy*(dx)**-1=-R_b(3*x-x**2*2**-1)+10*x**3*3**-1-45*x**2+195*x+C_1\n", "\n", "#again on Integrating we get\n", "#E*I*y=-R_b*(3*x**2*2**-1-x**3*6**-1)+10*x**4*12**-1-15*x**3+195*x**2*2**-1+C_1*x+C_2\n", "\n", "#At x=0,dy*(dx)**-1=0 Therefore C_1=0\n", "\n", "#At x=0,y=0 Therefore C_2=0\n", "\n", "#At x=3m, y=0\n", "x=3\n", "C_1=0\n", "C_2=0\n", "R_b=-(-10*x**4*12**-1+15*x**3-195*x**2*2**-1-C_1*x-C_2)*(3*x**2*2**-1-x**3*6**-1)**-1\n", "\n", "#result\n", "print\"Load taken by prop is\",round(R_b,2),\"KN\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Load taken by prop is 60.0 KN\n" ] } ], "prompt_number": 3 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Problem no 12.5,Page No.289" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#Initilization of variables\n", "\n", "L=2 #m #Span of beam\n", "w=10 #KN/m #u.d.l\n", "\n", "#Calculations\n", "\n", "#Downward deflection at B(of Beam AB) due to u.d.l of 10 KN/m is\n", "Y_B_1=w*L**4*8**-1 \n", "\n", "#Upward deflection at B due to reaction at C alone is\n", "#Y_B_2=R_c*8*3**-1\n", "\n", "#Net downward deflection of cantilever at AB at B\n", "#Y_B=Y_B_1-Y_B_2\n", "\n", "#Downward Deflection of Beam CD at C due to the reaction\n", "#R_c=R_c*(3*E*I)**-1\n", "\n", "#since both deflection at C and B are equal\n", "R_c=20*(1*3**-1+8*3**-1)**-1\n", "\n", "#Result\n", "print\"Reaction at C is\",round(R_c,2),\"KN\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Reaction at C is 6.67 KN\n" ] } ], "prompt_number": 7 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Problem no 12.8,Page No.292" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "%matplotlib inline\n", "\n", "#Initilization of variables\n", "\n", "L=8 #m #span\n", "W=24*10**3 #N/m #U.D.L\n", "y=2*10**-2 #m #deflection\n", "E=20*10**9\n", "I=10**-5 #m**4\n", "\n", "#Calculations\n", "\n", "#The Downward deflection at C Due to u.d.l\n", "#Y_c=5*W*L**3*(384*E*I)**-1\n", "\n", "#The Upward Deflection at C due to prop Reaction P \n", "#Y_c_1=P*L**3*(48*E*I)**-1\n", "\n", "#Since the prop is at the same level as end supports\n", "#Y_c_1=Y_c\n", "P_1=5*W*8**-1*10**-3 #KN\n", "\n", "#The reaction at A and B is equal\n", "R_a=R_b=(24-15)*2**-1\n", "\n", "#Shear Force at B\n", "V_B=4.5 #KN\n", "\n", "#Shear Force at C\n", "V_C1=4.5-24*2**-1\n", "V_C2=4.5-24*2**-1+15\n", "\n", "#Shea rForce at A\n", "V_A=-4.5 #KN\n", "\n", "#B.M at C due to u.d.l\n", "M_C1=W*L*8**-1*10**-3 #KN*m\n", "\n", "#B.M due to only prop reaction P=15 KN\n", "P=15\n", "M_C2=-P*L*4**-1 #KN*m\n", "\n", "#B.M at D\n", "M_D=4.5*1.5-24*8**-1*1.5**2*2**-1\n", "\n", "#In second case prop sinks by 2 cm\n", "#Y_c-Y_c_1=2 \n", "\n", "#So Further simplifying and sunstituting values in above equation we get\n", "P=-(2*100**-1-(5*W*L**3*(384*E*I)**-1))*(L**3*(48*E*I)**-1)**-1\n", "\n", "#Let Each end reaction be X\n", "X=(24-14.625)*2**-1\n", "\n", "#Result\n", "print\"prop reaction is\",round(P_1,2),\"KN\"\n", "print\"The End Reaction is\",round(X,2),\"KN\"\n", "\n", "#Plotting the SHear Force Diagram\n", "\n", "X1=[0,4,4,8,8]\n", "Y1=[V_B,V_C1,V_C2,V_A,0]\n", "Z1=[0,0,0,0,0]\n", "plt.plot(X1,Y1,X1,Z1)\n", "plt.xlabel(\"Length x in m\")\n", "plt.ylabel(\"Shear Force in kN\")\n", "plt.show()\n", "\n", "#Plotting the Bendimg Moment Diagram\n", "\n", "X2=[0,4,4]\n", "Y2=[0,M_C1,0]\n", "Z2=[0,0,0]\n", "plt.plot(X2,Y2)\n", "plt.xlabel(\"Lenght in m\")\n", "plt.ylabel(\"Bending Moment in kN.m\")\n", "plt.show()" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "prop reaction is 15.0 KN\n", "The End Reaction is 4.69 KN\n" ] }, { "metadata": {}, "output_type": "display_data", "png": 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