{ "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