From 73b0bffc1781c6cf1eec459813767508da507c33 Mon Sep 17 00:00:00 2001 From: hardythe1 Date: Mon, 28 Jul 2014 14:01:58 +0530 Subject: adding book --- Mechanics_of_Materials/Chapter7.ipynb | 390 ++++++++++++++++++++++++++++++++++ 1 file changed, 390 insertions(+) create mode 100755 Mechanics_of_Materials/Chapter7.ipynb (limited to 'Mechanics_of_Materials/Chapter7.ipynb') diff --git a/Mechanics_of_Materials/Chapter7.ipynb b/Mechanics_of_Materials/Chapter7.ipynb new file mode 100755 index 00000000..2461bf06 --- /dev/null +++ b/Mechanics_of_Materials/Chapter7.ipynb @@ -0,0 +1,390 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 7:Transverse Shear" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.1 Page No 369" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Given\n", + "V = 4\t\t #kN, load\n", + "co = 0.05\t\t#mm, outside radius\n", + "ci = 0.02\t\t #mm, inside radius\n", + "t1 = 0.1 \t#mm, thickness\n", + "t2=0.06\n", + "\n", + "#Part (a)\n", + "#Section Properties\n", + "import math\n", + "Isolid=1/4.0*(math.pi)*co**4\n", + "Itube = 1/4.0*(math.pi)*(co**4-ci**4)\n", + "Qsolid=4*co/(3*math.pi)*(math.pi*co**2/2.0)\n", + "Qtube=4*co/(3*math.pi)*(math.pi*co**2/2.0)-4*ci/(3*math.pi)*(math.pi*ci**2/2.0)\n", + "Tsolid=V*10**3*Qsolid/(Isolid*t1)\n", + "Ttube=V*10**3*Qsolid/(Itube*t2)\n", + "\n", + "#Display\n", + "print\"The shear stress in solid = \",round(Tsolid/1000,1),\"KPa\"\n", + "print'The shear stress in tube = ',round(Ttube/1000000,2),\"MPa\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The shear stress in solid = 679.1 KPa\n", + "The shear stress in tube = 1.16 MPa\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.3 Page No 370" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Given\n", + "V = 80.0 \t\t\t#kN, load\n", + "thick_1 = 20/1000.0 \t#m, thickness\n", + "thick_2 = 15/1000.0 \t#m\n", + "l = 300/1000.0 \t\t#m, length\n", + "y = 100/1000.0 \t\t#m\n", + "h = 2*y\n", + "y_dash = y +thick_1/2.0\n", + "\n", + "#Part(a)\n", + "I1 = (thick_2*(h**3))/12.0\n", + "I2 = (l*(thick_1**3))/12.0\n", + "I3 = (l*thick_1*(y_dash)**2)\n", + "I = I1+2*(I2+I3) #Moment of inertia\n", + "Q_b = y_dash*l*thick_1\n", + "#At B'\n", + "tou_b_dash = (V*Q_b)/(I*l*1000)\n", + "#At B\n", + "tou_b = (V*Q_b)/(I*thick_2*1000)\n", + "\n", + "#At C\n", + "Q_c = (y_dash*l*thick_1)+(y*thick_2*y/2.0)\n", + "tou_c = (V*Q_c)/(I*thick_2*1000)\n", + "\n", + "#Display\n", + "print\"The shear stress at B dash = \",round(tou_b_dash,1),\"MPa\"\n", + "print\"The shear stress at B = \",round(tou_b,1),\"MPa\"\n", + "print\"The shear stress at C = \",round(tou_c,1),\"MPa\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The shear stress at B dash = 1.1 MPa\n", + "The shear stress at B = 22.6 MPa\n", + "The shear stress at C = 25.2 MPa\n" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.4 Page No 372" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Given\n", + "udl = 6.5\t\t #kN, force\n", + "l_bc = 8 \t\t#m, length\n", + "l = 150/1000.0\t\t#m\n", + "t = 30/1000.0\t\t#m, thickness\n", + "\n", + "#Calculation\n", + "#Internal Shear\n", + "w = udl*l_bc/2.0\n", + "l_wc = l_bc/4.0\n", + "l_bw = l_bc - l_wc\n", + "V = (w*l_bw)/l_bc\n", + "R_b = w - V\n", + "\n", + "#Section Properties\n", + "y1= l/2.0\n", + "A = (l*t)\n", + "y2= l+(t/2.0)\n", + "y_dash = (y1*A + y2*A)/(2*A)\n", + "I1 = (t*l**3)/12.0\n", + "I2 = (A*(y_dash-y1)**2)\n", + "I3 = (l*t**3)/12.0\n", + "I4 = (A*(y2 - y_dash)**2)\n", + "I = I1+I2+I3+I4\n", + "Q = ((l+t)-(t/2.0)-y_dash)*A\n", + "#Shear Stress\n", + "tou_max = (V*Q)/(I*t*1000)\n", + "\n", + "#Display\n", + "print\"The maximum shear stress in the glue necessary to hold the boards together\",round(tou_max,2),\"MPa\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The maximum shear stress in the glue necessary to hold the boards together 4.88 MPa\n" + ] + } + ], + "prompt_number": 20 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.5 Page No 380" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Given\n", + "V = 850 #kN, force\n", + "#The given dimension are\n", + "l1 =250/1000.0 #m.\n", + "l2 = 300/1000.0 #m\n", + "l3 = 125/1000.0 #m\n", + "t = 10/1000.0 #m\n", + "h = 200/1000.0 #m\n", + "\n", + "#Calculation\n", + "A1 = l1*t\n", + "A2 = l2*t\n", + "A3 = l3*t\n", + "y1 = l2+(t/2.0)\n", + "y2 = l2/2.0\n", + "y3 = h+(t/2.0)\n", + "y_dash = (2*y2*A2 + A1*y1 + A3*y3)/(2*A2 + A1 + A3)\n", + "I1 = ((l1*t**3)/12.0) +(A1 * (l2+(t/2.0)-y_dash)**2)\n", + "I2 = ((t*l2**3)/12.0) +(A2 * (y_dash - (l2/2.0))**2)\n", + "I3 = ((l3*t**3)/12.0) +(A1 * (h+(t/2.0)-y_dash)**2)\n", + "I = 2*I2 + I1 + I3\n", + "Q_b = (l2+(t/2.0) - y_dash)*A1 #Q = y'A'\n", + "Q_c = (h+(t/2.0) - y_dash)*A3 #Q = y'A'\n", + "\n", + "#Shear Flow\n", + "q_b = (V*Q_b)/I\n", + "q_c = (V*Q_c)/I\n", + "q_b = q_b/(2*1000)\n", + "q_c = q_c/(2*1000)\n", + "\n", + "#Display\n", + "print\"The shear flow at B, resisted by the glue is \",round(q_b,2),\"MN/m\"\n", + "print\"The shear flow at C, resisted by the glue is \",round(q_c,4),\"MN/m\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The shear flow at B, resisted by the glue is 1.31 MN/m\n", + "The shear flow at C, resisted by the glue is 0.0498 MN/m\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.6 Page No 381" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Given\n", + "V = 80 #lb, load\n", + "#The given dimension are\n", + "t = 1.5 #inch\n", + "a = 7.5 #inch\n", + "b = a-2*t #inch\n", + "F_nail= 30 #lb\n", + "\n", + "\n", + "#Calculation\n", + "#Section Properties\n", + "I = (a*a**3 - b*b**3 )/12.0\n", + "Q_b = (((a-2*t)/2.0)+(t/2.0))*a*t #Q = y'A'\n", + "Q_c = (((a-2*t)/2.0)+(t/2.0))*(a-2*t)*t #Q = y'A'\n", + "\n", + "#Shear Flow\n", + "q_b = (V*Q_b)/I\n", + "q_c = (V*Q_c)/I\n", + "s_b = F_nail/(q_b/2.0)\n", + "s_c = F_nail/(q_c/2.0)\n", + "\n", + "#Display\n", + "print\"The maximum spacing of nails required at B is =\",s_b,\"inch\"\n", + "print\"The maximum spacing of nails required at C is =\",s_c,\"inch\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The maximum spacing of nails required at B is = 5.1 inch\n", + "The maximum spacing of nails required at C is = 8.5 inch\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.7 Page No 382" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Given\n", + "F = 40 #lb, force\n", + "#The other dimension are\n", + "s = 9.0 #inch\n", + "h = 5 #inch\n", + "t = 0.5 #inch\n", + "w = 3 #inch\n", + "w_3 = w/3.0 #inch\n", + "\n", + "#Calculations\n", + "I = (w*h**3)/12.0 - (2*w_3*(h - 2*t)**3)/12.0\n", + "#Case 1\n", + "Q1 = ((h-t)/2.0)*(w*t)\n", + "V1 =((F/s)*I)/Q1 #q = VQ/I\n", + "\n", + "#Case2\n", + "Q2 = ((h-t)/2.0)*(w_3*t)\n", + "V2 =((F/s)*I)/Q2 #q = VQ/I\n", + "\n", + "#Display\n", + "print\"The largest vertical shear that can be supported in Case 1 = \",round(V1,1),\"lb\"\n", + "print\"The largest vertical shear that can be supported in Case 2 = \",round(V2,1),\"lb\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The largest vertical shear that can be supported in Case 1 = 27.1 lb\n", + "The largest vertical shear that can be supported in Case 2 = 81.3 lb\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.8 Page No 381" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Given\n", + "V = 10 #lb, load\n", + "b1 = 6 #inch\n", + "h1 = 8 #inch\n", + "t = 1 #inch\n", + "b2 = b1-2*t\n", + "h2 = h1-2*t #inch\n", + "b3 = 4 #inch\n", + "\n", + "#Calculations\n", + "I = ((b2/2.0*(b1+t)**3))/12.0 +2*((b3+t)*t*((h1-t)/2.0)**2)\n", + "q_b = 0\n", + "Q_c = (h1-t)/2.0*(b3+t)*t \n", + "q_c = 1/2.0*(V*Q_c)/I\n", + "Q_d = 2*((h1-t)/4.0)*(h1-t)/2.0*t+(h1-t)/2.0*(b3+t)*t\n", + "q_d = 1/2.0*(V*Q_d)/I #Q = VQ/I\n", + "\n", + "#Display\n", + "print\"Variation of shear flow at B = \",q_b,\"kip/inch\"\n", + "print\"Variation of shear flow at C = \",round(q_c,3),\"kip/inch\"\n", + "print'Variation of shear flow at D = ',round(q_d,3),\"kip/inch\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Variation of shear flow at B = 0 kip/inch\n", + "Variation of shear flow at C = 0.487 kip/inch\n", + "Variation of shear flow at D = 0.828 kip/inch\n" + ] + } + ], + "prompt_number": 30 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file -- cgit