{ "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": {} } ] }