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+{
+ "metadata": {
+ "name": "CH7"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 7: Dislocation and Strengthening Mechanisms"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 7.1 Page no 183"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Resolved Shear Stress Computations\n",
+ "\n",
+ "#Direction for given plane\n",
+ "u1=1\n",
+ "v1=1\n",
+ "w1=0\n",
+ "u2=0\n",
+ "v2=1\n",
+ "w2=0\n",
+ "#For lamda\n",
+ "u3=-1\n",
+ "v3=1\n",
+ "w3=1\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "phi=math.acos((u1*u2+v1*v2+w1*w2)/(math.sqrt((u1**2+v1**2+w1**2)*(u2**2+v2**2+w2**2))))\n",
+ "lam=math.acos((u3*u2+v3*v2+w3*w2)/(math.sqrt((u3**2+v3**2+w3**2)*(u2**2+v2**2+w2**2))))\n",
+ "sigma=52 #in MPa, Tensile stress\n",
+ "tr=sigma*math.cos(phi)*math.cos(lam)\n",
+ "trc=30 #in MPa Critical resolved shear stress\n",
+ "sy=trc/(math.cos(phi)*math.cos(lam))\n",
+ "\n",
+ "#Result\n",
+ "print\"(a)The resolved shear stress is \",round(tr,2),\"MPa\"\n",
+ "print\"(b)Yield strength is\",round(sy,1),\"MPa\"\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a)The resolved shear stress is 21.23 MPa\n",
+ "(b)Yield strength is 73.5 MPa\n"
+ ]
+ }
+ ],
+ "prompt_number": 19
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 7.2 Page no 194"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Tensile Strength and Ductility Determinations for Cold-Worked Copper\n",
+ "\n",
+ "#Given\n",
+ "df=12.2 #Final dia in mm\n",
+ "di=15.2 #Initial dia in mm\n",
+ "\n",
+ "#Calculation\n",
+ "CW = ((di**2-df**2)/di**2)*100\n",
+ "ts=340 #in Mpa tensile strength, from fig 7.19 (b)\n",
+ "duc=7 #in % Ductility from fig 7.19 (c)\n",
+ "\n",
+ "#result\n",
+ "print\"Percent Cold Work is \",round(CW,1),\"%\"\n",
+ "print\"Tensile strength is\",ts,\"MPa\"\n",
+ "print\"Ductility is \",duc,\"%\"\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Percent Cold Work is 35.6 %\n",
+ "Tensile strength is 340 MPa\n",
+ "Ductility is 7 %\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Design Example 7.3 Page no 199"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Description of Diameter Reduction Procedure\n",
+ "\n",
+ "#Given\n",
+ "di=6.4 #Initial dia in mm\n",
+ "df=5.1 #Final dia in mm\n",
+ "\n",
+ "#Calculation\n",
+ "#Cold Work Computation\n",
+ "CW = ((di**2-df**2)/di**2)*100\n",
+ "#From Figures 7.19a and 7.19c, \n",
+ "#A yield strength of 410 MPa \n",
+ "#And a ductility of 8% EL are attained from this deformation\n",
+ "dmid = math.sqrt(df**2/(1-0.215))\n",
+ "\n",
+ "#Result\n",
+ "print\"Cold work is \",round(CW,1),\"%\"\n",
+ "print\"But required ductility and yield strength is not matched at this cold work\"\n",
+ "print\"Hence required Cold work is 21.5 %\"\n",
+ "print\"Hence original diameter for second drawing is \",round(dmid,1),\"mm\"\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Cold work is 36.5 %\n",
+ "But required ductility and yield strength is not matched at this cold work\n",
+ "Hence required Cold work is 21.5 %\n",
+ "Hence original diameter for second drawing is 5.8 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [],
+ "language": "python",
+ "metadata": {},
+ "outputs": []
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+} \ No newline at end of file