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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": {}
+  }
+ ]
+}
+\ No newline at end of file