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{
"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": {}
}
]
}
|
