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{
"metadata": {
"name": "",
"signature": "sha256:12bc74eaed4218a18fa109df85a443dc199f78a3b2cafffe1198babaf1458444"
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter08:Deformation of Metals"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex8.1:pg-175"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Example 8.1: critical resolved shear stress of silver\n",
" \n",
"\n",
"Ts=15;#tensile stress in Mpa\n",
"d=[1,1,0];\n",
"d1=[1,1,1];\n",
"csda=((d[0]*d1[0])+(d[1]*d1[1])+(d[2]*d1[2]))/((math.sqrt(d[0]**2+d[1]**2+d[2]**2))*math.sqrt(d1[0]**2+d1[1]**2+d1[2]**2));#angle degree\n",
"d2=[0,1,1];\n",
"csdb=((d[0]*d2[0])+(d[1]*d2[1])+(d[2]*d2[2]))/((math.sqrt(d[0]**2+d[1]**2+d[2]**2))*math.sqrt(d2[0]**2+d2[1]**2+d2[2]**2));#angle degree\n",
"t=Ts*csda*csdb;#critical resolved shear stress in MPa\n",
"print round(t,2),\"= critical resolved shear stress in MPa\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"6.12 = critical resolved shear stress in MPa\n"
]
}
],
"prompt_number": 11
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex8.2:pg-186"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Example 8.2: yield strength of material\n",
" \n",
"import numpy.linalg as lin\n",
"import math\n",
"ys1=115;# yeild strength in MN/mm**2\n",
"ys2=215;# yeild strength in MN/mm**2\n",
"d1=0.04;#diamtere in mm\n",
"d2=0.01;#diamtere in mm\n",
"A=numpy.array([[2 ,10], [1 ,10]]);\n",
"B=numpy.array([230,215]);\n",
"x=lin.solve(A,B)\n",
"si=x[0];# in MN/mm**2\n",
"k=x[1];#\n",
"d3=0.016;#in mm\n",
"sy= si +(k/math.sqrt(d3));#yeild strength for a grain size in MN/mm**2\n",
"print round(sy,1),\"=yeild strength for a grain size in MN/mm**2\"\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"173.1 =yeild strength for a grain size in MN/mm**2\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex8.3:pg-186"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Example 8.3: yield strength of material\n",
"import numpy.linalg as lin\n",
"import math\n",
"ys1=120;# yeild strength in MN/mm**2\n",
"ys2=220;# yeild strength in MN/mm**2\n",
"d1=0.04;#diamtere in mm\n",
"d2=0.01;#diamtere in mm\n",
"A=numpy.array([[2 ,10], [1 ,10]]);\n",
"B=numpy.array([240,220]);\n",
"x=lin.solve(A,B)\n",
"si=x[0];# in MN/mm**2\n",
"k=x[1];#\n",
"d3=0.025;#in mm\n",
"sy= si +(k/math.sqrt(d3));#yeild strength for a grain size in MN/mm**2\n",
"print round(sy,1),\"= yeild strength for a grain size in MN/mm**2\"\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"146.5 = yeild strength for a grain size in MN/mm**2\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex18.4:pg-193"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Example 8.4 : grain diameter\n",
"import math \n",
"\n",
"#given data :\n",
"N=9; # ASTM number\n",
"m=8*2**N; # no. of grains [er square millimetre\n",
"grain=1/math.sqrt(m);\n",
"print round(grain,4),\"=the grain diameter(mm) \"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"0.0156 =the grain diameter(mm) \n"
]
}
],
"prompt_number": 2
}
],
"metadata": {}
}
]
}
|
