{ "metadata": { "name": "", "signature": "sha256:cfde82eb2eac726301e92e1b27c27d3628db6f9b0e8b5d7b0cd15017ca0006f7" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "Chapter 9:Elastic stress analysis and design" ] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 9.4 pagenumber 465" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Given \n", "b = 40.0 #mm - The width of the beam crossection\n", "h = 300.0 #mm - The length of the beam crossection \n", "V = 40.0 #KN - The shear stress in teh crossection\n", "M = 10.0 #KN-m - The bending moment on K----K crossection \n", "c = h/2 #mm -The position at which maximum stress occurs on the crossection\n", "I = b*(h**3)/12 #mmm4 - the moment of inertia \n", "#Caliculations \n", "\n", "stress_max_1 = M*c*(10**6)/I #The maximum stress occurs at the end\n", "stress_max_2 = -M*c*(10**6)/I #The maximum stress occurs at the end\n", "y = 140 #mm The point of interest, the distance of element from com\n", "n = y/(c) # The ratio of the distances from nuetral axis to the elements\n", "stress_L_1 = n*stress_max_1 #The normal stress on elements L--L\n", "stress_L_2 = -n*stress_max_1 #The normal stress on elements L--L\n", "x = 10 #mm The length of the element\n", "A = b*x #mm3 The area of the element \n", "y_1 = y+x/2 # the com of element from com of whole system\n", "stress_xy = V*A*y_1*(10**3)/(I*b) #Mpa - The shear stress on the element \n", "#stresses acting in plane 30 degrees \n", "o = 60 #degrees - the plane angle\n", "stress_theta = stress_L_1/2 + stress_L_1*(math.cos(math.radians(o)))/2 - stress_xy*(math.sin(math.radians(o))) #Mpa by direct application of equations\n", "stress_shear = -stress_L_1*(math.sin(math.radians(o)))/2 - stress_xy*(math.cos(math.radians(o))) #Mpa Shear stress\n", " \n", "print \"a)The principle stresses are \",round(stress_max_1,2),\"MPa,\",round(stress_max_2,2),\"Mpa\"\n", "print \"b)The stresses on inclines plane \",round(stress_theta,2),\"Mpa noraml, \",round(stress_shear,2),\"Mpa shear \"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "a)The principle stresses are 16.67 MPa, -16.67 Mpa\n", "b)The stresses on inclines plane 11.11 Mpa noraml, -7.06 Mpa shear \n" ] } ], "prompt_number": 20 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 9.5 page number 476" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Given\n", "M = 10 #KN-m moment\n", "v = 8.0 #KN - shear Stress \n", "stress_allow = 8 #MPa - The maximum allowable stress\n", "shear_allow_per = 1.4 #Mpa - The allowable stress perpendicular to grain\n", "stress_allow_shear = 0.7 #MPa - The maximum allowable shear stress\n", "#Caliculations \n", "\n", "S = M*(10**6)/stress_allow #mm3 \n", "#lets arbitarly assume h = 2b\n", "#S = b*(h**2)/6\n", "h = pow(12*S,0.333) #The depth of the beam\n", "b = h/2 #mm The width of the beam\n", "A = h*b #mm2 The area of the crossection , assumption\n", "stress_shear = 3*v*(10**3)/(2*A) #Mpa The strear stress \n", "if stress_shear<stress_allow_shear:\n", " print \"The stress developed \",round(stress_shear,2),\" is in allowable ranges for \",round(A,2),\"mm2 area\"\n", "else:\n", " print \"The stress developed\",stress_shear,\" is in non allowable ranges\",A,\"area\"\n", "Area_allow = v*(10**3)/shear_allow_per #mm - the allowable area\n", "print \"The minimum area is \",Area_allow ,\"mm2\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The stress developed 0.4 is in allowable ranges for 30077.85 mm2 area\n", "The minimum area is 5714.28571429 mm2\n" ] }, { "metadata": {}, "output_type": "pyout", "prompt_number": 36, "text": [ "8.0" ] } ], "prompt_number": 36 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 9.6 page number 478" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Given\n", "stress_allow = 24 #ksi - The maximum allowable stress\n", "stress_allow_shear = 14.5 #ksi- The maximum allowable shear stress\n", "M_max = 36 #k-ft The maximum moment\n", "l = 16 #in-The length of the rod\n", "w = 2 #k/ft - The force distribution on the rod\n", "R_A = 6.4 #k - The reaction at A\n", "R_B = 25.6 #k - the reaction at B\n", "v_max = R_B-l*w #kips the maximum stress, from diagram\n", "#W8x24 is used from the appendix table 3 and 4 \n", "l =0.245 #in - W8x24 crossesction length\n", "#Caliculations \n", "\n", "stress_xy = v_max/A #ksi the approximate shear stress \n", "if stress_xy < stress_allow_shear:\n", " print \"W8x24 gives the allowable ranges of shear stress\"\n", "else:\n", " print \"W8x24 doesnot gives the allowable ranges of shear stress\"\n", "k = 7.0/8 #in the distance from the outer face of the flange to the webfillet\n", "#at+kt should not exceed 0.75 of yeild stress\n", "#a1t+2kt should not exceed 0.75 of yeild stress\n", "Stress_yp = 36 #Ksi - The yeild stress\n", "t = 0.245 #in thickness of the web\n", "#support a \n", "a = R_A/(0.75*Stress_yp*t)-k #in lengths of the bearings\n", "#support b\n", "a_1 = R_B/(0.75*Stress_yp*t)-2*k #in lengths of the bearings\n", "print \"lengths of the bearing at A \",round(a,3),\"in\"\n", "print \"lengths of the bearing at B\",round(a_1,3),\"in\"\n", " \n", "\n", "\n", "\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "W8x24 gives the allowable ranges of shear stress\n", "lengths of the bearing at A 0.092 in\n", "lengths of the bearing at B 2.12 in\n" ] }, { "metadata": {}, "output_type": "pyout", "prompt_number": 44, "text": [ "0.875" ] } ], "prompt_number": 44 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 9.8 page number 483" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#given \n", "hp = 63000 #horse power\n", "T = hp*20*(10**-3)/63 #k-in the torsion implies due to horse power\n", "stress_allow_shear = 6 #ksi- The maximum allowable shear stress\n", "M_ver = 6.72/2 #k-in the vertical component of the moment \n", "M_hor = 9.10 #k-in the horizantal component of the moment \n", "#Caliculations \n", "\n", "M = pow(((M_ver**2)+(M_hor**2)),0.5) #K-in The resultant \n", "d = pow((16*(((M**2)+(T**2))**0.5)/(stress_allow_shear*3.14)),0.333) #in, The suggested diameter from derivation\n", "print \"The suggested diameter is\",round(d,2),\"in\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The suggested diameter is 2.66 in\n" ] } ], "prompt_number": 49 }, { "cell_type": "code", "collapsed": false, "input": [], "language": "python", "metadata": {}, "outputs": [] } ], "metadata": {} } ] }