From df60071cf1d1c18822d34f943ab8f412a8946b69 Mon Sep 17 00:00:00 2001
From: hardythe1
Date: Wed, 3 Jun 2015 15:27:17 +0530
Subject: add books
---
Testing_the_interface/chapter2_3.ipynb | 501 +++++++++++++++++++++++++++++++++
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+{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 2: Axially Loaded Members"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.1, page no. 72"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\"\"\"\n",
+ "calculating the number of revolutions for the nut\n",
+ "\"\"\"\n",
+ "\n",
+ "import math \n",
+ "\n",
+ "#initialisation\n",
+ "\n",
+ "W = 2.0 #lb\n",
+ "b = 10.5 #inch\n",
+ "c = 6.4 #inch\n",
+ "k = 4.2 #inch\n",
+ "p = 1.0/16.0 #inch\n",
+ "\n",
+ "#calculation\n",
+ "\n",
+ "n = (W*b)/(c*k*p) #inch\n",
+ "\n",
+ "#result\n",
+ "\n",
+ "print \" No. of revolution required = \", n, \"revolutions\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " No. of revolution required = 12.5 revolutions\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2, page no. 74"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\"\"\"\n",
+ "finding maximum allowable load\n",
+ "\"\"\"\n",
+ "\n",
+ "import math \n",
+ "import numpy\n",
+ "\n",
+ "#initialisation\n",
+ "\n",
+ "Fce_ = 2.0 #dummy variable\n",
+ "Fbd_ = 3.0 #dummy variable \n",
+ "Lbd = 480.0 #mm\n",
+ "Lce = 600.0 #mm\n",
+ "E = 205e6 #205Gpa\n",
+ "Abd = 1020.0 #mm\n",
+ "Ace = 520.0 #mm\n",
+ "\n",
+ "#calculation\n",
+ "Dbd_ = (Fbd_*Lbd)/(E*Abd) #dummy variable\n",
+ "Dce_ = (Fce_*Lce)/(E*Ace) #dummy variable\n",
+ "Da = 1 #limiting value\n",
+ "P = ((((450+225)/225)*(Dbd_ + Dce_) - Dce_ )**(-1)) * Da \n",
+ "Fce = 2*P # Real value in newton\n",
+ "Fbd = 3*P #real value in newton\n",
+ "Dbd = (Fbd*Lbd)/(E*Abd) #print lacement in mm\n",
+ "Dce = (Fce*Lce)/(E*Ace) # print lacement in mm\n",
+ "a = numpy.degrees(numpy.arctan(((Da+Dce)/675))) #alpha in degree\n",
+ "\n",
+ "#result\n",
+ "print \"alpha = \", round(a,2), \"degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "alpha = 0.11 degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.3, page no. 80"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\"\"\"\n",
+ "calculation if vertical displacement\n",
+ "\"\"\"\n",
+ "\n",
+ "import math \n",
+ "\n",
+ "#initialisation\n",
+ "P1 = 2100.0 #lb\n",
+ "P2 = 5600.0 #lb\n",
+ "b = 25.0 #inch\n",
+ "a = 28.0 #inch\n",
+ "A1 = 0.25 #inch^2\n",
+ "A2 = 0.15 #inch^2\n",
+ "L1 = 20.0 #inch\n",
+ "L2 = 34.8 #inch\n",
+ "E = 29e6 #29Gpa\n",
+ "\n",
+ "#Calculations\n",
+ "P3 = (P2*b)/a \n",
+ "Ra = P3-P1\n",
+ "N1 = -Ra \n",
+ "N2 = P1 \n",
+ "D = ((N1*L1)/(E*A1)) + ((N2*L2)/(E*A2)) #print lacement\n",
+ "\n",
+ "#Result\n",
+ "print \"Downward print lacement is = \", D, \"inch\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Downward print lacement is = 0.0088 inch\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.6, page no. 90"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\"\"\"\n",
+ "obtaing formula and calculating allowable load\n",
+ "\"\"\"\n",
+ "\n",
+ "import math \n",
+ "\n",
+ "#Numerical calculation of allowable load\n",
+ "\n",
+ "d1 = 4.0 #mm\n",
+ "d2 = 3.0 #mm\n",
+ "A1 = (math.pi*(d1**2))/4 #area\n",
+ "A2 = (math.pi*(d2**2))/4 #area\n",
+ "L1 = 0.4 #meter\n",
+ "L2 = 0.3 #meter\n",
+ "E1 = 72e9 #Gpa\n",
+ "E2 = 45e9 #Gpa\n",
+ "f1 = L1/(E1*A1) * 1e6 # To cpmpensate for the mm**2\n",
+ "f2 = L2/(E2*A2) * 1e6 \n",
+ "s1 = 200e6 #stress\n",
+ "s2 = 175e6 #stress\n",
+ "\n",
+ "#Calculations\n",
+ "P1 = ( (s1*A1*(4*f1 + f2))/(3*f2) ) * 1e-6 # To cpmpensate for the mm**2\n",
+ "P2 = ( (s2*A2*(4*f1 + f2))/(6*f1) ) * 1e-6 \n",
+ "\n",
+ "#Result\n",
+ "print \"Newton Minimum allowable stress aomong the two P1 and P2 is smaller one, therefore MAS = \", P2"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Newton Minimum allowable stress aomong the two P1 and P2 is smaller one, therefore MAS = 1264.49104307\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.10, page no. 113"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\"\"\"\n",
+ "calculate stress acting on inclined section &\n",
+ "the complete state of stress\n",
+ "\"\"\"\n",
+ "\n",
+ "import math \n",
+ "\n",
+ "#initialisation\n",
+ "P = 90000.0 #newton\n",
+ "A = 1200e-6 # meter^2\n",
+ "s_x = -P/A #stress\n",
+ "t_1 = 25.0 #for the stresses on ab and cd plane\n",
+ "\n",
+ "#Calculations\n",
+ "s_1 = s_x*(math.cos(math.radians(t_1))**2)\n",
+ "T_1 = -s_x*math.cos(math.radians(t_1))*math.sin(math.radians(t_1))\n",
+ "t_2 = -65.0 #for the stresses on ad and bc plane\n",
+ "s_2 = s_x*(math.cos(math.radians(t_2))**2)\n",
+ "T_2 = -s_x*math.cos(math.radians(t_2))*math.sin(math.radians(t_2))\n",
+ "\n",
+ "#Result\n",
+ "print \"The normal and shear stresses on the plane ab and cd are\", round((T_1/1E+6),2), round((s_1/1E+6),2), \"MPa respecively\" \n",
+ "print \"respecively The normal and shear stresses on the plane ad and bc are\", round((T_2/1E+6),2), round((s_2/1E+6),2), \"MPa respecively\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The normal and shear stresses on the plane ab and cd are 28.73 -61.6 MPa respecively\n",
+ "respecively The normal and shear stresses on the plane ad and bc are -28.73 -13.4 MPa respecively\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.11, page no. 114"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\"\"\"\n",
+ "Calculate the vertical displacement of the joint\n",
+ "\"\"\"\n",
+ "\n",
+ "import math \n",
+ "\n",
+ "# Value of s_x based on allowable stresses on glued joint\n",
+ "\n",
+ "#initialisation\n",
+ "s_t = -750.0 #psi\n",
+ "t = -50.0 #degree\n",
+ "T_t = -500.0 #psi\n",
+ "\n",
+ "sg_x_1 = s_t/(math.cos(math.radians(t))**2)\n",
+ "sg_x_2 = -T_t/(math.cos(math.radians(t))*math.sin(math.radians(t)))\n",
+ "\n",
+ "# Value of s_x based on allowable stresses on plastic\n",
+ "\n",
+ "sp_x_1 = -1100.0 #psi\n",
+ "T_t_p = 600.0 #psi\n",
+ "t_p = 45.0 #degree\n",
+ "sp_x_2 = -T_t_p/(math.cos(math.radians(t_p))*math.sin(math.radians(t_p)))\n",
+ "\n",
+ "# Minimum width of bar\n",
+ "\n",
+ "P = 8000.0 #lb\n",
+ "A = P/sg_x_2\n",
+ "b_min = math.sqrt(abs(A)) #inch\n",
+ "print \"The minimum width of the bar is\", round(b_min,2), \"inch\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The minimum width of the bar is 2.81 inch\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.15, page no. 126"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\"\"\"\n",
+ "Comparison of energy-absorbing capacity with different type of bolts\n",
+ "\"\"\"\n",
+ "\n",
+ "import math \n",
+ "#Bolt with reduced shank diameter\n",
+ "\n",
+ "#initialisation\n",
+ "g = 1.50 # inch\n",
+ "d = 0.5 #inch\n",
+ "t = 0.25 #inch\n",
+ "d_r = 0.406 #inch\n",
+ "L = 13.5 #inch\n",
+ "\n",
+ "#calculation\n",
+ "ratio = ((g*(d**2))/(((g-t)*(d_r**2))+(t*(d**2)))) #U2/U1\n",
+ "\n",
+ "print \"The energy absorbing capacity of the bolts with reduced shank diameter\", round(ratio,2)\n",
+ "ratio_1 = ( (((L-t)*(d_r**2))+(t*(d**2))) / ((2*(g-t)*(d_r**2))+2*(t*(d**2))) ) #U3/2U1\n",
+ "print \"The energy absorbing capacity of the long bolts\", round(ratio_1,2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The energy absorbing capacity of the bolts with reduced shank diameter 1.4\n",
+ "The energy absorbing capacity of the long bolts 4.18\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ " Example 2.16, page no. 133"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\"\"\"\n",
+ "Determine the maximum elongation and tensile stress\n",
+ "\"\"\"\n",
+ "\n",
+ "import math \n",
+ "\n",
+ "#initialisation\n",
+ "# Maximum elongation\n",
+ "M = 20 #kg\n",
+ "g = 9.81 #m/s^2\n",
+ "L = 2 #meter\n",
+ "E = 210e9 #210Gpa\n",
+ "h = 0.15 #meter\n",
+ "diameter = 0.015 #milimeter\n",
+ "\n",
+ "#Calculations & Result\n",
+ "A = (math.pi/4)*(diameter**2) #area\n",
+ "D_st = ((M*g*L)/(E*A)) \n",
+ "D_max = D_st*(1+(1+(2*h/D_st))**0.5) \n",
+ "D_max_1 = math.sqrt(2*h*D_st) # another approach to find D_max\n",
+ "i = D_max / D_st # Impact factor\n",
+ "print \"Maximum elongation is\",round((D_max/1E-3),2), \"mm\" # Maximum tensile stress\n",
+ "s_max = (E*D_max)/L #Maximum tensile stress\n",
+ "s_st = (M*g)/A #static stress\n",
+ "i_1 = s_max / s_st #Impact factor \n",
+ "print \"Maximum tensile stress is \", round((s_max/1E+6),2), \"MPa\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Maximum elongation is 1.79 mm\n",
+ "Maximum tensile stress is 188.13 MPa\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.18, page no. 148"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\"\"\"\n",
+ "determine displacement at the lower end of bar in various conditions\n",
+ "\"\"\"\n",
+ "\n",
+ "import math \n",
+ "\n",
+ "\n",
+ "#initialisation\n",
+ "P1 = 108000.0 #Newton\n",
+ "P2 = 27000.0 #Newton\n",
+ "L = 2.2 #meter\n",
+ "A = 480.0 #mm^2\n",
+ "\n",
+ "\n",
+ "#calculations\n",
+ "\n",
+ "# Displacement due to load P1 acting alone\n",
+ "s = (P1/A) #stress in MPa\n",
+ "e = (s/70000) + (1/628.2)*((s/260)**10) #strain\n",
+ "D_b = e*L*1e3 #elongation in mm\n",
+ "print \"elongation when only P1 load acting is = \", round(D_b,2), \" mm\"\n",
+ "\n",
+ "# Displacement due to load P2 acting alone\n",
+ "s_1 = (P2/A) #stress in MPa\n",
+ "e_1 = (s_1/70000) + (1/628.2)*((s_1/260)**10) #strain\n",
+ "D_b_1 = e_1*(L/2)*1e3 #elongation in mm (no elongation in lower half)\n",
+ "print \"elongation when only P2 load acting is = \", round(D_b_1,2), \" mm\"\n",
+ "\n",
+ "# Displacement due to both load acting simonmath.taneously\n",
+ "#upper half\n",
+ "s_2 = (P1/A) #stress in MPa\n",
+ "e_2 = (s_2/70000) + (1/628.2)*((s_2/260)**10) #strain\n",
+ "\n",
+ "#lower half\n",
+ "s_3 = (P1+P2)/A #stress in MPa\n",
+ "e_3 = (s_3/70000) + (1/628.2)*((s_3/260)**10) #strain\n",
+ "D_b_2 = ((e_2*L)/2 + (e_3*L)/2) * 1e3 # elongation in mm\n",
+ "print \"elongation when P1 and P2 both loads are acting is = \", round(D_b_2,2), \" mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "elongation when only P1 load acting is = 7.9 mm\n",
+ "elongation when only P2 load acting is = 0.88 mm\n",
+ "elongation when P1 and P2 both loads are acting is = 12.21 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file
--
cgit