{ "metadata": { "name": "" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "Chapter 12 : Fracture" ] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 12.1, Page No 302" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "#initialisation of variables\n", "c = 2.0 \t# crack of half length in micro meter\n", "Y = 70.0 \t# young\u2019s modulus in GN m^-2\n", "Gamma = 1.0\t # specific surface energy \n", "\n", "#Calculations\n", "sigma_f = math.sqrt(2*Gamma*Y*1e9/(math.pi*c*1e-6))/1e6\n", "r = Y*1e3/sigma_f\n", "\n", "#Results\n", "print(\" Fracture strength of %d MN m^-2 is 1/%d th of young\u2019s modulus. \"%(math.ceil(sigma_f),math.ceil(r/100.0)*100.0))\n", "print(\" Thus Griffiths criterion bridges the gap between the observed and ideal strengths of brittle material\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " Fracture strength of 150 MN m^-2 is 1/500 th of young\u2019s modulus. \n", " Thus Griffiths criterion bridges the gap between the observed and ideal strengths of brittle material\n" ] } ], "prompt_number": 1 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 12.2, Page No 304" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "#initialisation of variables\n", "Gamma = 1.5\t\t# specific surface energy in J/m^2\n", "Y = 200.0\t \t# Young\u2019s modulus in GN/m^2\n", "c = 2.0\t\t # half length of crack\n", "\n", "#Calculations\n", "sigma_f = math.sqrt(2*Gamma*Y*1e9/(math.pi*c*1e-6))\n", "\n", "#Results\n", "print(\" Brittle fracture strength at low temperatures is %d MNm^-2 \" %(sigma_f/1e6))# answer in book is 310MNm^-2\n", "\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " Brittle fracture strength at low temperatures is 309 MNm^-2 \n" ] } ], "prompt_number": 3 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 12.3, Page No 306" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "#initialisation of variables\n", "Gamma = 2.0\t\t # specific surface energy in J/m**2\n", "Y = 350.0\t\t # Young\u2019s modulus in GN/m**2\n", "c = 2.0\t\t\t # half length of crack\n", "de_dt1 = 1e-2 \t\t# strain rate\n", "de_dt2 = 1e-5 \t# strain rate\n", "\n", "#Calculations\n", "print(\" Part A:\")\n", "sigma_f = math.sqrt(2*Gamma*Y*1e9/(math.pi*c*1e-6))\n", "sigma_y = sigma_f/1e6\n", "T = 173600/(sigma_y-20.6-61.3*math.log(de_dt1,10))\t# temperature calculation\n", "\n", "print(\" Transition temperature for strain rate %.0e s**-1 is %d K\" %(de_dt1,T)) # answer in book is 300 K\n", "print(\" Part B:\")\n", "\n", "T = 173600/(sigma_y-20.6-61.3*math.log(de_dt2,10))\t# temperature calculation\n", "\n", "\n", "#Results\n", "print(\" Transition temperature for strain rate %.0e s**-1 is %d K\" %(de_dt2,T)) # answer in book is 230 K\n", "# Solution in book for two parts is divided into three parts\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " Part A:\n", " Transition temperature for strain rate 1e-02 s**-1 is 302 K\n", " Part B:\n", " Transition temperature for strain rate 1e-05 s**-1 is 229 K\n" ] } ], "prompt_number": 5 } ], "metadata": {} } ] }