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{
"metadata": {
"name": ""
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 11 : Plastic Deformation and Creep in Crystalline Materials"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 11.2, Page No 272"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#initialisation of variables\n",
"b = 2.0 \t\t# burger vector in angstrom\n",
"v = 20*b**3 \t# activation volume \n",
"tau_pn = 1000.0 # P-N stress of crystal in MNm**-2\n",
"k = 1.38e-23 \t# physical constant\n",
"t1 = 0 \t\t\t# temperature in K\n",
"t2 = 100.0\t\t# temperature in K \n",
"t3 = 300.0\t\t# temperature in K\n",
"t4 = 500.0\t\t# temperature in K\n",
"T = t1\t\n",
"\n",
"#Calculations\n",
"tau_app = tau_pn - 40.0*k*T/(v*1e-30)\n",
"print(\"The stress required to move the dislocation at temperature %dK is %d MNm**-2\" %(T,tau_app))\n",
"print(\"Part B:\")\n",
"T = t2\n",
"tau_app = tau_pn - 40*k*T/(v*1e-30*1e6)\n",
"print(\"The stress required to move the dislocation at temperature %dK is %d MNm**-2\" %(T,tau_app))\n",
"print(\"Part C:\")\n",
"T = t3\n",
"tau_app = tau_pn - 40*k*T/(v*1e-30*1e6)\n",
"if tau_app<0 :\n",
" print(\" Stress to be applied is zero\")\n",
" print(\"The stress required to move the dislocation at temperature %dK entirely overcome by thermal fluctuations\" %T)\n",
"\n",
"print(\"Part D:\")\n",
"T = t4\n",
"tau_app = tau_pn - 40*k*T/(v*1e-30*1e6)\n",
"\n",
"#Results\n",
"if tau_app<0 :\n",
" print(\"Stress to be applied is zero\")\n",
" print(\"The stress required to move the dislocation at temperature %dK entirely overcome by thermal fluctuations\" %T)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The stress required to move the dislocation at temperature 0K is 1000 MNm**-2\n",
"Part B:\n",
"The stress required to move the dislocation at temperature 100K is 655 MNm**-2\n",
"Part C:\n",
" Stress to be applied is zero\n",
"The stress required to move the dislocation at temperature 300K entirely overcome by thermal fluctuations\n",
"Part D:\n",
"Stress to be applied is zero\n",
"The stress required to move the dislocation at temperature 500K entirely overcome by thermal fluctuations\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 11.3, Page No 278"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#initialisation of variables\n",
"mu = 44.0 # shear modulus of copper in GN m^-2\n",
"b = 2.55 # burgers vector in angstrom\n",
"tau = 35.0 # shear stress in MN m^-2\n",
"\n",
"#Calculations\n",
"l = mu*1e9*b*1e-10/(tau*1e6)\n",
"rho = 1/l**2\n",
"\n",
"#Results\n",
"print(\"Dislocation density in copper is %.1e m^-2\" %rho)\n",
"# Answer in book is 1e12 m^-2\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Dislocation density in copper is 9.7e+12 m^-2\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 11.4, Page No 280"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#initialisation of variables\n",
"sigma1 = 120.0 # initial yield strength of material in MNm**-2\n",
"sigma2 = 220.0 # Final yield strength of material in MN m**-2\n",
"d1 = 0.04 # initial diameter in mm\n",
"d2 = 0.01 # final diameter in mm\n",
"n = 9.0 # astm number\n",
"\n",
"#Calculations\n",
"print(\"Example 11.4\")\n",
"k = (sigma2-sigma1)*1e6/(1/math.sqrt(d2*1e-3)-1/math.sqrt(d1*1e-3))\n",
"sigma_i = sigma1*1e6 -k/math.sqrt((d1*1e-3))\n",
"d = 1/math.sqrt(2**(n-1)*1e4/645)\n",
"sigma_y = sigma_i+k*(d*1e-3)**(-0.5)\n",
"\n",
"#Results\n",
"print(\"Yield stress for a grain size of ASTM 9 is %d MN m**-2\" %math.ceil(sigma_y/1e6))\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Example 11.4\n",
"Yield stress for a grain size of ASTM 9 is 179 MN m**-2\n"
]
}
],
"prompt_number": 6
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 11.5 Page No 283"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#initialisation of variables\n",
"n1 = 1e6 # initial number of particles\n",
"n2 = 1e3 # final number of particle\n",
"\n",
"#Calculations\n",
"k = (n1/n2)**(1.0/3)\n",
"\n",
"#Results\n",
"print(\"Yield strength would have decreased to %d percent of its initial value.\" %(100.0/k))"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Yield strength would have decreased to 10 of its initial value.\n"
]
}
],
"prompt_number": 7
}
],
"metadata": {}
}
]
}
|
