{ "metadata": { "name": "", "signature": "sha256:2781545e4a86c16cb74b3bfc3908d3f7649eaff2f44e3b1448ebf17923520f6f" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "Chapter12-Plastic Behaviour of Solids" ] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex1-pg345" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "#calculate plastic yeilding and max allowable stress and axial stress and transverse stress and total elonganation for stability in meter\n", "alpha=45.\n", "sigmayp=35.*10**6 ##Pa\n", "k=840.##MPa\n", "n=0.2\n", "L0=3 ##m\n", "Aad=10.*10**-5 ##m**2\n", "Acd=10.*10**-5 ##m**2\n", "Abd=15.*10**-5 ##m**2\n", "\n", "P=sigmayp*Abd+2.*sigmayp*Aad*math.cos(45/57.3)\n", "print'%s %.2f %s'%(\"plastic yeilding in N is= \",P,\"\")\n", "sigma=k*n**n\n", "print'%s %.2f %s'%(\"maxi allowable stress in MPa is= \",sigma,\"\")\n", "epsilon1=n\n", "print'%s %.2f %s'%(\"axial stress is= \",epsilon1,\"\")\n", "epsilon2=-0.1\n", "print'%s %.2f %s'%(\"transverse stress is= \",epsilon2,\"\")\n", "epsilon3=-0.1\n", "print'%s %.2f %s'%(\"transverse stress is= \",epsilon3,\"\")\n", "z=3*n\n", "print'%s %.2f %s'%(\"total elongation for stability in meter is= \",z,\"\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "plastic yeilding in N is= 10200.03 \n", "maxi allowable stress in MPa is= 608.81 \n", "axial stress is= 0.20 \n", "transverse stress is= -0.10 \n", "transverse stress is= -0.10 \n", "total elongation for stability in meter is= 0.60 \n" ] } ], "prompt_number": 1 } ], "metadata": {} } ] }