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-{
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "# Chapter 15: Fundamentals of Metalworking"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "### Example 15.1, Mechanics of Metal Working, Page No. 506"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 9,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "\n",
- "Enginering Strain = 1\n",
- "True Strain = 0.693147\n",
- "Reduction = 1\n",
- "\n",
- "\n",
- "Enginering Strain = -0.5\n",
- "True Strain = -0.693147\n",
- "Reduction = -1\n"
- ]
- }
- ],
- "source": [
- "from math import log\n",
- "\n",
- "#For Bar which is double in length\n",
- "#variable declaration 1\n",
- "L2=2;\n",
- "L1=1;\n",
- "\n",
- "#calculation 1\n",
- "e=(L2-L1)/L1;\n",
- "e1=log(L2/L1);\n",
- "r=1-L1/L2;\n",
- "\n",
- "#result 1\n",
- "print('\\nEnginering Strain = %g\\nTrue Strain = %g\\nReduction = %g')%(e,e1,r);\n",
- "\n",
- "\n",
- "\n",
- "#For bar which is halved in length\n",
- "#variable declaration 2\n",
- "L1=1;\n",
- "L2=0.5;\n",
- "\n",
- "#calculation 2\n",
- "e=(L2-L1)/L1;\n",
- "e1=log(L2/L1);\n",
- "r=1-L1/L2;\n",
- "\n",
- "#result 2\n",
- "print('\\n\\nEnginering Strain = %g\\nTrue Strain = %g\\nReduction = %g')%(e,e1,r);\n"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {
- "collapsed": true
- },
- "source": [
- "### Example 15.2, Mechanics of Metal Working, Page No. 511"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 10,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "\n",
- "Plastic work done in 1st step = 39752.1 lb/in^2\n",
- "Plastic work done in 2nd step = 97934.8 lb/in^2\n",
- "\n"
- ]
- }
- ],
- "source": [
- "\n",
- "from scipy.integrate import quad\n",
- "from math import log\n",
- "\n",
- "#variable declaration\n",
- "D0=25.0;\n",
- "D1=20.0;\n",
- "D2=15.0;\n",
- "def integrand(e):\n",
- " return 200000*e**0.5\n",
- "\n",
- "#calculation\n",
- "ep1=log((D0/D1)**2);\n",
- "U1,U1_err=quad(integrand,0,ep1);\n",
- "ep2=log((D1/D2)**2);\n",
- "U2,U2_err=quad(integrand,ep1,ep1+ep2);\n",
- "\n",
- "#result\n",
- "print('\\nPlastic work done in 1st step = %g lb/in^2\\nPlastic work done in 2nd step = %g lb/in^2\\n')%(U1,U2);"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "### Example 15.3, Hodography, Page No. 517"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 11,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Pressure = 2.88675\n"
- ]
- }
- ],
- "source": [
- "\n",
- "from math import sin\n",
- "from math import radians\n",
- "\n",
- "#variable declaration\n",
- "alpha=60;\n",
- "\n",
- "#calculation\n",
- "r=radians(alpha);\n",
- "mu=1/sin(r);\n",
- "p_2k=mu*5/2;\n",
- "\n",
- "#result\n",
- "print('Pressure = %g')%(p_2k);"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "### Example 15.4, Temperature in Metalworking, Page No. 526"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 12,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "\n",
- "Temperature Rise for aluminium = 78.4808 C\n",
- "Temperature Rise for titanium = 162.686 C\n",
- "\n"
- ]
- }
- ],
- "source": [
- "\n",
- "\n",
- "#variable declaration\n",
- "Al_s=200;\n",
- "Al_e=1;\n",
- "Al_p=2.69;\n",
- "Al_c=0.215;\n",
- "Ti_s=400;\n",
- "Ti_e=1;\n",
- "Ti_p=4.5;\n",
- "Ti_c=0.124;\n",
- "J=4.186;\n",
- "b=0.95;\n",
- "\n",
- "#calculation\n",
- "Al_Td=Al_s*Al_e*b/(Al_p*Al_c*J);\n",
- "Ti_Td=Ti_s*Ti_e*b/(Ti_p*Ti_c*J);\n",
- "\n",
- "#result\n",
- "print('\\nTemperature Rise for aluminium = %g C\\nTemperature Rise for titanium = %g C\\n')%(Al_Td,Ti_Td);\n"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "### Example 15.5, Friction and Lubrication, Page No. 546"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 13,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "\n",
- "For OD after deformation being 70 mm, Di = 22.3607 mm\n",
- "Precent change in inside diameter = 25.4644 percent\n",
- "Peak pressure = 1.93531\n",
- "\n",
- "\n",
- "\n",
- "\n",
- "For OD after deformation being 81.4 mm, Di = 35.0137 mm\n",
- "Precent change in inside diameter = -16.7124 percent\n",
- "Peak pressure = 1.17321\n"
- ]
- }
- ],
- "source": [
- "\n",
- "\n",
- "from math import sqrt\n",
- "\n",
- "#variable declaration\n",
- "Do=60;\n",
- "Di=30;\n",
- "def1=70;\n",
- "def2=81.4;\n",
- "h=10;\n",
- "a=30;\n",
- "\n",
- "#calculation1\n",
- "di=sqrt((Do**2-Di**2)*2-def1**2);\n",
- "pr=(Di-di)/Di*100;\n",
- "m=0.27;\n",
- "p_s=1+2*m*a/(sqrt(3)*h);\n",
- "\n",
- "#result 1\n",
- "print('\\nFor OD after deformation being 70 mm, Di = %g mm\\nPrecent change in inside diameter = %g percent\\nPeak pressure = %g')%(di,pr,p_s);\n",
- "\n",
- "#calculation 2\n",
- "di=sqrt(def2**2-(Do**2-Di**2)*2);\n",
- "pr=(Di-di)/Di*100;\n",
- "m=0.05;\n",
- "p_s=1+2*m*a/(sqrt(3)*h);\n",
- "\n",
- "#result 2\n",
- "print('\\n\\n\\n\\nFor OD after deformation being 81.4 mm, Di = %g mm\\nPrecent change in inside diameter = %g percent\\nPeak pressure = %g')%(di,pr,p_s);\n"
- ]
- }
- ],
- "metadata": {
- "kernelspec": {
- "display_name": "Python 2",
- "language": "python",
- "name": "python2"
- },
- "language_info": {
- "codemirror_mode": {
- "name": "ipython",
- "version": 2
- },
- "file_extension": ".py",
- "mimetype": "text/x-python",
- "name": "python",
- "nbconvert_exporter": "python",
- "pygments_lexer": "ipython2",
- "version": "2.7.9"
- }
- },
- "nbformat": 4,
- "nbformat_minor": 0
-}