From cca4629cb34f8ed864c1a5fc5ed6139fb27addd3 Mon Sep 17 00:00:00 2001 From: Trupti Kini Date: Thu, 3 Mar 2016 23:30:24 +0600 Subject: Added(A)/Deleted(D) following books A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER10_1.ipynb A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER10_2.ipynb A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER10_3.ipynb A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER10_4.ipynb A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER10_5.ipynb A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER10_6.ipynb A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER10_7.ipynb A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER10_8.ipynb A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER10_9.ipynb A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER13_1.ipynb A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER13_2.ipynb A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER13_3.ipynb A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER13_4.ipynb A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER13_5.ipynb A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER13_6.ipynb A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER13_7.ipynb A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER13_8.ipynb A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER13_9.ipynb A 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Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER25_8.ipynb A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER28.ipynb A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER28_1.ipynb A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER28_2.ipynb A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER28_3.ipynb A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER28_4.ipynb A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER28_5.ipynb A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER28_6.ipynb A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER28_7.ipynb A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER28_8.ipynb A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER2_1.ipynb A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER2_2.ipynb A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER2_3.ipynb A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER2_4.ipynb A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER2_5.ipynb A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER2_6.ipynb A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER2_7.ipynb A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER2_8.ipynb A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER2_9.ipynb A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER32.ipynb A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER32_1.ipynb A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER32_2.ipynb A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER32_3.ipynb A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER32_4.ipynb A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER32_5.ipynb A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER32_6.ipynb A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER32_7.ipynb A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER32_8.ipynb A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER36_1.ipynb A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER36_2.ipynb A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER36_3.ipynb A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER36_4.ipynb A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER36_5.ipynb A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER36_6.ipynb A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER36_7.ipynb A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER36_8.ipynb A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER36_9.ipynb A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER9_1.ipynb A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER9_2.ipynb A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER9_3.ipynb A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER9_4.ipynb A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER9_5.ipynb A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER9_6.ipynb A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER9_7.ipynb A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER9_8.ipynb A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER9_9.ipynb --- .../CHAPTER10_1.ipynb | 88 +++++++++++ .../CHAPTER10_2.ipynb | 88 +++++++++++ .../CHAPTER10_3.ipynb | 88 +++++++++++ .../CHAPTER10_4.ipynb | 88 +++++++++++ .../CHAPTER10_5.ipynb | 88 +++++++++++ .../CHAPTER10_6.ipynb | 88 +++++++++++ .../CHAPTER10_7.ipynb | 88 +++++++++++ .../CHAPTER10_8.ipynb | 88 +++++++++++ .../CHAPTER10_9.ipynb | 88 +++++++++++ .../CHAPTER13_1.ipynb | 84 ++++++++++ .../CHAPTER13_2.ipynb | 84 ++++++++++ .../CHAPTER13_3.ipynb | 84 ++++++++++ .../CHAPTER13_4.ipynb | 84 ++++++++++ .../CHAPTER13_5.ipynb | 84 ++++++++++ .../CHAPTER13_6.ipynb | 84 ++++++++++ .../CHAPTER13_7.ipynb | 84 ++++++++++ .../CHAPTER13_8.ipynb | 84 ++++++++++ .../CHAPTER13_9.ipynb | 84 ++++++++++ .../CHAPTER14_1.ipynb | 78 +++++++++ .../CHAPTER14_2.ipynb | 78 +++++++++ .../CHAPTER14_3.ipynb | 78 +++++++++ .../CHAPTER14_4.ipynb | 78 +++++++++ .../CHAPTER14_5.ipynb | 78 +++++++++ .../CHAPTER14_6.ipynb | 78 +++++++++ .../CHAPTER14_7.ipynb | 78 +++++++++ .../CHAPTER14_8.ipynb | 78 +++++++++ .../CHAPTER14_9.ipynb | 78 +++++++++ .../CHAPTER15_1.ipynb | 72 +++++++++ .../CHAPTER15_2.ipynb | 72 +++++++++ .../CHAPTER15_3.ipynb | 72 +++++++++ .../CHAPTER15_4.ipynb | 72 +++++++++ .../CHAPTER15_5.ipynb | 72 +++++++++ .../CHAPTER15_6.ipynb | 72 +++++++++ .../CHAPTER15_7.ipynb | 72 +++++++++ .../CHAPTER15_8.ipynb | 72 +++++++++ .../CHAPTER15_9.ipynb | 72 +++++++++ .../CHAPTER16_1.ipynb | 71 +++++++++ .../CHAPTER16_2.ipynb | 71 +++++++++ .../CHAPTER16_3.ipynb | 71 +++++++++ .../CHAPTER16_4.ipynb | 71 +++++++++ .../CHAPTER16_5.ipynb | 71 +++++++++ .../CHAPTER16_6.ipynb | 71 +++++++++ .../CHAPTER16_7.ipynb | 71 +++++++++ .../CHAPTER16_8.ipynb | 71 +++++++++ .../CHAPTER16_9.ipynb | 71 +++++++++ .../CHAPTER17.ipynb | 85 ++++++++++ .../CHAPTER17_1.ipynb | 85 ++++++++++ .../CHAPTER17_2.ipynb | 85 ++++++++++ .../CHAPTER17_3.ipynb | 85 ++++++++++ .../CHAPTER17_4.ipynb | 85 ++++++++++ .../CHAPTER17_5.ipynb | 85 ++++++++++ .../CHAPTER17_6.ipynb | 85 ++++++++++ .../CHAPTER17_7.ipynb | 85 ++++++++++ .../CHAPTER17_8.ipynb | 85 ++++++++++ .../CHAPTER18_1.ipynb | 128 +++++++++++++++ .../CHAPTER18_2.ipynb | 128 +++++++++++++++ .../CHAPTER18_3.ipynb | 128 +++++++++++++++ .../CHAPTER18_4.ipynb | 128 +++++++++++++++ .../CHAPTER18_5.ipynb | 128 +++++++++++++++ .../CHAPTER18_6.ipynb | 128 +++++++++++++++ .../CHAPTER18_7.ipynb | 128 +++++++++++++++ .../CHAPTER18_8.ipynb | 128 +++++++++++++++ .../CHAPTER18_9.ipynb | 128 +++++++++++++++ .../CHAPTER20.ipynb | 129 +++++++++++++++ .../CHAPTER20_1.ipynb | 129 +++++++++++++++ .../CHAPTER20_2.ipynb | 129 +++++++++++++++ .../CHAPTER20_3.ipynb | 129 +++++++++++++++ .../CHAPTER20_4.ipynb | 129 +++++++++++++++ .../CHAPTER20_5.ipynb | 129 +++++++++++++++ .../CHAPTER20_6.ipynb | 129 +++++++++++++++ .../CHAPTER20_7.ipynb | 129 +++++++++++++++ .../CHAPTER20_8.ipynb | 129 +++++++++++++++ .../CHAPTER22.ipynb | 157 ++++++++++++++++++ .../CHAPTER22_1.ipynb | 157 ++++++++++++++++++ .../CHAPTER22_2.ipynb | 157 ++++++++++++++++++ .../CHAPTER22_3.ipynb | 157 ++++++++++++++++++ .../CHAPTER22_4.ipynb | 157 ++++++++++++++++++ .../CHAPTER22_5.ipynb | 157 ++++++++++++++++++ .../CHAPTER22_6.ipynb | 157 ++++++++++++++++++ .../CHAPTER22_7.ipynb | 157 ++++++++++++++++++ .../CHAPTER22_8.ipynb | 157 ++++++++++++++++++ .../CHAPTER23_1.ipynb | 175 +++++++++++++++++++++ .../CHAPTER23_2.ipynb | 175 +++++++++++++++++++++ .../CHAPTER23_3.ipynb | 175 +++++++++++++++++++++ .../CHAPTER23_4.ipynb | 175 +++++++++++++++++++++ .../CHAPTER23_5.ipynb | 175 +++++++++++++++++++++ .../CHAPTER23_6.ipynb | 175 +++++++++++++++++++++ .../CHAPTER23_7.ipynb | 175 +++++++++++++++++++++ .../CHAPTER23_8.ipynb | 175 +++++++++++++++++++++ .../CHAPTER23_9.ipynb | 175 +++++++++++++++++++++ .../CHAPTER25.ipynb | 146 +++++++++++++++++ .../CHAPTER25_1.ipynb | 146 +++++++++++++++++ .../CHAPTER25_2.ipynb | 146 +++++++++++++++++ .../CHAPTER25_3.ipynb | 146 +++++++++++++++++ .../CHAPTER25_4.ipynb | 146 +++++++++++++++++ .../CHAPTER25_5.ipynb | 146 +++++++++++++++++ .../CHAPTER25_6.ipynb | 146 +++++++++++++++++ .../CHAPTER25_7.ipynb | 146 +++++++++++++++++ .../CHAPTER25_8.ipynb | 146 +++++++++++++++++ .../CHAPTER28.ipynb | 90 +++++++++++ .../CHAPTER28_1.ipynb | 90 +++++++++++ .../CHAPTER28_2.ipynb | 90 +++++++++++ .../CHAPTER28_3.ipynb | 90 +++++++++++ .../CHAPTER28_4.ipynb | 90 +++++++++++ .../CHAPTER28_5.ipynb | 90 +++++++++++ .../CHAPTER28_6.ipynb | 90 +++++++++++ .../CHAPTER28_7.ipynb | 90 +++++++++++ .../CHAPTER28_8.ipynb | 90 +++++++++++ .../CHAPTER2_1.ipynb | 82 ++++++++++ .../CHAPTER2_2.ipynb | 82 ++++++++++ .../CHAPTER2_3.ipynb | 82 ++++++++++ .../CHAPTER2_4.ipynb | 82 ++++++++++ .../CHAPTER2_5.ipynb | 82 ++++++++++ .../CHAPTER2_6.ipynb | 82 ++++++++++ .../CHAPTER2_7.ipynb | 82 ++++++++++ .../CHAPTER2_8.ipynb | 82 ++++++++++ .../CHAPTER2_9.ipynb | 82 ++++++++++ .../CHAPTER32.ipynb | 73 +++++++++ .../CHAPTER32_1.ipynb | 73 +++++++++ .../CHAPTER32_2.ipynb | 73 +++++++++ .../CHAPTER32_3.ipynb | 73 +++++++++ .../CHAPTER32_4.ipynb | 73 +++++++++ .../CHAPTER32_5.ipynb | 73 +++++++++ .../CHAPTER32_6.ipynb | 73 +++++++++ .../CHAPTER32_7.ipynb | 73 +++++++++ .../CHAPTER32_8.ipynb | 73 +++++++++ .../CHAPTER36_1.ipynb | 159 +++++++++++++++++++ .../CHAPTER36_2.ipynb | 159 +++++++++++++++++++ .../CHAPTER36_3.ipynb | 159 +++++++++++++++++++ .../CHAPTER36_4.ipynb | 159 +++++++++++++++++++ .../CHAPTER36_5.ipynb | 159 +++++++++++++++++++ .../CHAPTER36_6.ipynb | 159 +++++++++++++++++++ .../CHAPTER36_7.ipynb | 159 +++++++++++++++++++ .../CHAPTER36_8.ipynb | 159 +++++++++++++++++++ .../CHAPTER36_9.ipynb | 159 +++++++++++++++++++ .../CHAPTER9_1.ipynb | 80 ++++++++++ .../CHAPTER9_2.ipynb | 80 ++++++++++ .../CHAPTER9_3.ipynb | 80 ++++++++++ .../CHAPTER9_4.ipynb | 80 ++++++++++ .../CHAPTER9_5.ipynb | 80 ++++++++++ .../CHAPTER9_6.ipynb | 80 ++++++++++ .../CHAPTER9_7.ipynb | 80 ++++++++++ .../CHAPTER9_8.ipynb | 80 ++++++++++ .../CHAPTER9_9.ipynb | 80 ++++++++++ 144 files changed, 15273 insertions(+) create mode 100644 Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER10_1.ipynb create mode 100644 Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER10_2.ipynb create mode 100644 Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER10_3.ipynb create mode 100644 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-0,0 +1,88 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:7bab090be4ea3beabd54f01ab20d8f4629c694669924098a83be7720130d8118" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 10 - Fundamentals of Metal Casting" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 10.1 - PG NO. 252" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Example 10.1 \n", + "#page no. 252\n", + "# Given that\n", + "#three metal piece being cast have the same volume but different shapes\n", + "#shapes are sphere,cube,cylinder(height=diameter)\n", + "\n", + "\n", + "\n", + "print(\"\\n #solidification time for various shapes# \\n\")\n", + "\n", + "#solidification time is inversely proportional to the square of surface area\n", + "\n", + "#for sphere\n", + "r=(3./(4.*3.14))**(1./3.)#radius of the sphere from volume of sphere v=(4*3.14*r**3)/3\n", + "A=4*3.14*((r)**2)\n", + "time1=1./(A)**2.\n", + "print'%s %.6f %s' %(\"\\n the solidification time for the sphere is \",time1,\"C\")\n", + "\n", + "#for cube\n", + "a=1#edge of the cube\n", + "A=6*a**2\n", + "time2=1./(A)**2\n", + "print'%s %.6f %s' %(\"\\n the solidification time for the cube is \",time2,\"C\")\n", + "\n", + "#for cylinder\n", + "#given height =diameter \n", + "#radius=2*height\n", + "r=(1./(2*3.14))**(1./3.)#radius of the cylinder from volume of the cylinder v=3.14*r**2*h\n", + "A=(6*3.14*(r**2)) #area of the cylinder = (2*3.14*radius**2) + (2*3.14*radius*height)\n", + "time3=1./(A)**2.\n", + "print'%s %.6f %s' %(\"\\n the solidification time for the sphere is \",time3,\"C\")\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " #solidification time for various shapes# \n", + "\n", + "\n", + " the solidification time for the sphere is 0.042774 C\n", + "\n", + " the solidification time for the cube is 0.027778 C\n", + "\n", + " the solidification time for the sphere is 0.032643 C\n" + ] + } + ], + "prompt_number": 3 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER10_2.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER10_2.ipynb new file mode 100644 index 00000000..719b96fb --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER10_2.ipynb @@ -0,0 +1,88 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:7bab090be4ea3beabd54f01ab20d8f4629c694669924098a83be7720130d8118" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 10 - Fundamentals of Metal Casting" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 10.1 - PG NO. 252" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Example 10.1 \n", + "#page no. 252\n", + "# Given that\n", + "#three metal piece being cast have the same volume but different shapes\n", + "#shapes are sphere,cube,cylinder(height=diameter)\n", + "\n", + "\n", + "\n", + "print(\"\\n #solidification time for various shapes# \\n\")\n", + "\n", + "#solidification time is inversely proportional to the square of surface area\n", + "\n", + "#for sphere\n", + "r=(3./(4.*3.14))**(1./3.)#radius of the sphere from volume of sphere v=(4*3.14*r**3)/3\n", + "A=4*3.14*((r)**2)\n", + "time1=1./(A)**2.\n", + "print'%s %.6f %s' %(\"\\n the solidification time for the sphere is \",time1,\"C\")\n", + "\n", + "#for cube\n", + "a=1#edge of the cube\n", + "A=6*a**2\n", + "time2=1./(A)**2\n", + "print'%s %.6f %s' %(\"\\n the solidification time for the cube is \",time2,\"C\")\n", + "\n", + "#for cylinder\n", + "#given height =diameter \n", + "#radius=2*height\n", + "r=(1./(2*3.14))**(1./3.)#radius of the cylinder from volume of the cylinder v=3.14*r**2*h\n", + "A=(6*3.14*(r**2)) #area of the cylinder = (2*3.14*radius**2) + (2*3.14*radius*height)\n", + "time3=1./(A)**2.\n", + "print'%s %.6f %s' %(\"\\n the solidification time for the sphere is \",time3,\"C\")\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " #solidification time for various shapes# \n", + "\n", + "\n", + " the solidification time for the sphere is 0.042774 C\n", + "\n", + " the solidification time for the cube is 0.027778 C\n", + "\n", + " the solidification time for the sphere is 0.032643 C\n" + ] + } + ], + "prompt_number": 3 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER10_3.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER10_3.ipynb new file mode 100644 index 00000000..719b96fb --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER10_3.ipynb @@ -0,0 +1,88 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:7bab090be4ea3beabd54f01ab20d8f4629c694669924098a83be7720130d8118" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 10 - Fundamentals of Metal Casting" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 10.1 - PG NO. 252" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Example 10.1 \n", + "#page no. 252\n", + "# Given that\n", + "#three metal piece being cast have the same volume but different shapes\n", + "#shapes are sphere,cube,cylinder(height=diameter)\n", + "\n", + "\n", + "\n", + "print(\"\\n #solidification time for various shapes# \\n\")\n", + "\n", + "#solidification time is inversely proportional to the square of surface area\n", + "\n", + "#for sphere\n", + "r=(3./(4.*3.14))**(1./3.)#radius of the sphere from volume of sphere v=(4*3.14*r**3)/3\n", + "A=4*3.14*((r)**2)\n", + "time1=1./(A)**2.\n", + "print'%s %.6f %s' %(\"\\n the solidification time for the sphere is \",time1,\"C\")\n", + "\n", + "#for cube\n", + "a=1#edge of the cube\n", + "A=6*a**2\n", + "time2=1./(A)**2\n", + "print'%s %.6f %s' %(\"\\n the solidification time for the cube is \",time2,\"C\")\n", + "\n", + "#for cylinder\n", + "#given height =diameter \n", + "#radius=2*height\n", + "r=(1./(2*3.14))**(1./3.)#radius of the cylinder from volume of the cylinder v=3.14*r**2*h\n", + "A=(6*3.14*(r**2)) #area of the cylinder = (2*3.14*radius**2) + (2*3.14*radius*height)\n", + "time3=1./(A)**2.\n", + "print'%s %.6f %s' %(\"\\n the solidification time for the sphere is \",time3,\"C\")\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " #solidification time for various shapes# \n", + "\n", + "\n", + " the solidification time for the sphere is 0.042774 C\n", + "\n", + " the solidification time for the cube is 0.027778 C\n", + "\n", + " the solidification time for the sphere is 0.032643 C\n" + ] + } + ], + "prompt_number": 3 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER10_4.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER10_4.ipynb new file mode 100644 index 00000000..719b96fb --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER10_4.ipynb @@ -0,0 +1,88 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:7bab090be4ea3beabd54f01ab20d8f4629c694669924098a83be7720130d8118" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 10 - Fundamentals of Metal Casting" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 10.1 - PG NO. 252" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Example 10.1 \n", + "#page no. 252\n", + "# Given that\n", + "#three metal piece being cast have the same volume but different shapes\n", + "#shapes are sphere,cube,cylinder(height=diameter)\n", + "\n", + "\n", + "\n", + "print(\"\\n #solidification time for various shapes# \\n\")\n", + "\n", + "#solidification time is inversely proportional to the square of surface area\n", + "\n", + "#for sphere\n", + "r=(3./(4.*3.14))**(1./3.)#radius of the sphere from volume of sphere v=(4*3.14*r**3)/3\n", + "A=4*3.14*((r)**2)\n", + "time1=1./(A)**2.\n", + "print'%s %.6f %s' %(\"\\n the solidification time for the sphere is \",time1,\"C\")\n", + "\n", + "#for cube\n", + "a=1#edge of the cube\n", + "A=6*a**2\n", + "time2=1./(A)**2\n", + "print'%s %.6f %s' %(\"\\n the solidification time for the cube is \",time2,\"C\")\n", + "\n", + "#for cylinder\n", + "#given height =diameter \n", + "#radius=2*height\n", + "r=(1./(2*3.14))**(1./3.)#radius of the cylinder from volume of the cylinder v=3.14*r**2*h\n", + "A=(6*3.14*(r**2)) #area of the cylinder = (2*3.14*radius**2) + (2*3.14*radius*height)\n", + "time3=1./(A)**2.\n", + "print'%s %.6f %s' %(\"\\n the solidification time for the sphere is \",time3,\"C\")\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " #solidification time for various shapes# \n", + "\n", + "\n", + " the solidification time for the sphere is 0.042774 C\n", + "\n", + " the solidification time for the cube is 0.027778 C\n", + "\n", + " the solidification time for the sphere is 0.032643 C\n" + ] + } + ], + "prompt_number": 3 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER10_5.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER10_5.ipynb new file mode 100644 index 00000000..719b96fb --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER10_5.ipynb @@ -0,0 +1,88 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:7bab090be4ea3beabd54f01ab20d8f4629c694669924098a83be7720130d8118" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 10 - Fundamentals of Metal Casting" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 10.1 - PG NO. 252" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Example 10.1 \n", + "#page no. 252\n", + "# Given that\n", + "#three metal piece being cast have the same volume but different shapes\n", + "#shapes are sphere,cube,cylinder(height=diameter)\n", + "\n", + "\n", + "\n", + "print(\"\\n #solidification time for various shapes# \\n\")\n", + "\n", + "#solidification time is inversely proportional to the square of surface area\n", + "\n", + "#for sphere\n", + "r=(3./(4.*3.14))**(1./3.)#radius of the sphere from volume of sphere v=(4*3.14*r**3)/3\n", + "A=4*3.14*((r)**2)\n", + "time1=1./(A)**2.\n", + "print'%s %.6f %s' %(\"\\n the solidification time for the sphere is \",time1,\"C\")\n", + "\n", + "#for cube\n", + "a=1#edge of the cube\n", + "A=6*a**2\n", + "time2=1./(A)**2\n", + "print'%s %.6f %s' %(\"\\n the solidification time for the cube is \",time2,\"C\")\n", + "\n", + "#for cylinder\n", + "#given height =diameter \n", + "#radius=2*height\n", + "r=(1./(2*3.14))**(1./3.)#radius of the cylinder from volume of the cylinder v=3.14*r**2*h\n", + "A=(6*3.14*(r**2)) #area of the cylinder = (2*3.14*radius**2) + (2*3.14*radius*height)\n", + "time3=1./(A)**2.\n", + "print'%s %.6f %s' %(\"\\n the solidification time for the sphere is \",time3,\"C\")\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " #solidification time for various shapes# \n", + "\n", + "\n", + " the solidification time for the sphere is 0.042774 C\n", + "\n", + " the solidification time for the cube is 0.027778 C\n", + "\n", + " the solidification time for the sphere is 0.032643 C\n" + ] + } + ], + "prompt_number": 3 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER10_6.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER10_6.ipynb new file mode 100644 index 00000000..719b96fb --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER10_6.ipynb @@ -0,0 +1,88 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:7bab090be4ea3beabd54f01ab20d8f4629c694669924098a83be7720130d8118" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 10 - Fundamentals of Metal Casting" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 10.1 - PG NO. 252" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Example 10.1 \n", + "#page no. 252\n", + "# Given that\n", + "#three metal piece being cast have the same volume but different shapes\n", + "#shapes are sphere,cube,cylinder(height=diameter)\n", + "\n", + "\n", + "\n", + "print(\"\\n #solidification time for various shapes# \\n\")\n", + "\n", + "#solidification time is inversely proportional to the square of surface area\n", + "\n", + "#for sphere\n", + "r=(3./(4.*3.14))**(1./3.)#radius of the sphere from volume of sphere v=(4*3.14*r**3)/3\n", + "A=4*3.14*((r)**2)\n", + "time1=1./(A)**2.\n", + "print'%s %.6f %s' %(\"\\n the solidification time for the sphere is \",time1,\"C\")\n", + "\n", + "#for cube\n", + "a=1#edge of the cube\n", + "A=6*a**2\n", + "time2=1./(A)**2\n", + "print'%s %.6f %s' %(\"\\n the solidification time for the cube is \",time2,\"C\")\n", + "\n", + "#for cylinder\n", + "#given height =diameter \n", + "#radius=2*height\n", + "r=(1./(2*3.14))**(1./3.)#radius of the cylinder from volume of the cylinder v=3.14*r**2*h\n", + "A=(6*3.14*(r**2)) #area of the cylinder = (2*3.14*radius**2) + (2*3.14*radius*height)\n", + "time3=1./(A)**2.\n", + "print'%s %.6f %s' %(\"\\n the solidification time for the sphere is \",time3,\"C\")\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " #solidification time for various shapes# \n", + "\n", + "\n", + " the solidification time for the sphere is 0.042774 C\n", + "\n", + " the solidification time for the cube is 0.027778 C\n", + "\n", + " the solidification time for the sphere is 0.032643 C\n" + ] + } + ], + "prompt_number": 3 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER10_7.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER10_7.ipynb new file mode 100644 index 00000000..719b96fb --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER10_7.ipynb @@ -0,0 +1,88 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:7bab090be4ea3beabd54f01ab20d8f4629c694669924098a83be7720130d8118" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 10 - Fundamentals of Metal Casting" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 10.1 - PG NO. 252" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Example 10.1 \n", + "#page no. 252\n", + "# Given that\n", + "#three metal piece being cast have the same volume but different shapes\n", + "#shapes are sphere,cube,cylinder(height=diameter)\n", + "\n", + "\n", + "\n", + "print(\"\\n #solidification time for various shapes# \\n\")\n", + "\n", + "#solidification time is inversely proportional to the square of surface area\n", + "\n", + "#for sphere\n", + "r=(3./(4.*3.14))**(1./3.)#radius of the sphere from volume of sphere v=(4*3.14*r**3)/3\n", + "A=4*3.14*((r)**2)\n", + "time1=1./(A)**2.\n", + "print'%s %.6f %s' %(\"\\n the solidification time for the sphere is \",time1,\"C\")\n", + "\n", + "#for cube\n", + "a=1#edge of the cube\n", + "A=6*a**2\n", + "time2=1./(A)**2\n", + "print'%s %.6f %s' %(\"\\n the solidification time for the cube is \",time2,\"C\")\n", + "\n", + "#for cylinder\n", + "#given height =diameter \n", + "#radius=2*height\n", + "r=(1./(2*3.14))**(1./3.)#radius of the cylinder from volume of the cylinder v=3.14*r**2*h\n", + "A=(6*3.14*(r**2)) #area of the cylinder = (2*3.14*radius**2) + (2*3.14*radius*height)\n", + "time3=1./(A)**2.\n", + "print'%s %.6f %s' %(\"\\n the solidification time for the sphere is \",time3,\"C\")\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " #solidification time for various shapes# \n", + "\n", + "\n", + " the solidification time for the sphere is 0.042774 C\n", + "\n", + " the solidification time for the cube is 0.027778 C\n", + "\n", + " the solidification time for the sphere is 0.032643 C\n" + ] + } + ], + "prompt_number": 3 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER10_8.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER10_8.ipynb new file mode 100644 index 00000000..719b96fb --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER10_8.ipynb @@ -0,0 +1,88 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:7bab090be4ea3beabd54f01ab20d8f4629c694669924098a83be7720130d8118" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 10 - Fundamentals of Metal Casting" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 10.1 - PG NO. 252" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Example 10.1 \n", + "#page no. 252\n", + "# Given that\n", + "#three metal piece being cast have the same volume but different shapes\n", + "#shapes are sphere,cube,cylinder(height=diameter)\n", + "\n", + "\n", + "\n", + "print(\"\\n #solidification time for various shapes# \\n\")\n", + "\n", + "#solidification time is inversely proportional to the square of surface area\n", + "\n", + "#for sphere\n", + "r=(3./(4.*3.14))**(1./3.)#radius of the sphere from volume of sphere v=(4*3.14*r**3)/3\n", + "A=4*3.14*((r)**2)\n", + "time1=1./(A)**2.\n", + "print'%s %.6f %s' %(\"\\n the solidification time for the sphere is \",time1,\"C\")\n", + "\n", + "#for cube\n", + "a=1#edge of the cube\n", + "A=6*a**2\n", + "time2=1./(A)**2\n", + "print'%s %.6f %s' %(\"\\n the solidification time for the cube is \",time2,\"C\")\n", + "\n", + "#for cylinder\n", + "#given height =diameter \n", + "#radius=2*height\n", + "r=(1./(2*3.14))**(1./3.)#radius of the cylinder from volume of the cylinder v=3.14*r**2*h\n", + "A=(6*3.14*(r**2)) #area of the cylinder = (2*3.14*radius**2) + (2*3.14*radius*height)\n", + "time3=1./(A)**2.\n", + "print'%s %.6f %s' %(\"\\n the solidification time for the sphere is \",time3,\"C\")\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " #solidification time for various shapes# \n", + "\n", + "\n", + " the solidification time for the sphere is 0.042774 C\n", + "\n", + " the solidification time for the cube is 0.027778 C\n", + "\n", + " the solidification time for the sphere is 0.032643 C\n" + ] + } + ], + "prompt_number": 3 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER10_9.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER10_9.ipynb new file mode 100644 index 00000000..719b96fb --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER10_9.ipynb @@ -0,0 +1,88 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:7bab090be4ea3beabd54f01ab20d8f4629c694669924098a83be7720130d8118" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 10 - Fundamentals of Metal Casting" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 10.1 - PG NO. 252" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Example 10.1 \n", + "#page no. 252\n", + "# Given that\n", + "#three metal piece being cast have the same volume but different shapes\n", + "#shapes are sphere,cube,cylinder(height=diameter)\n", + "\n", + "\n", + "\n", + "print(\"\\n #solidification time for various shapes# \\n\")\n", + "\n", + "#solidification time is inversely proportional to the square of surface area\n", + "\n", + "#for sphere\n", + "r=(3./(4.*3.14))**(1./3.)#radius of the sphere from volume of sphere v=(4*3.14*r**3)/3\n", + "A=4*3.14*((r)**2)\n", + "time1=1./(A)**2.\n", + "print'%s %.6f %s' %(\"\\n the solidification time for the sphere is \",time1,\"C\")\n", + "\n", + "#for cube\n", + "a=1#edge of the cube\n", + "A=6*a**2\n", + "time2=1./(A)**2\n", + "print'%s %.6f %s' %(\"\\n the solidification time for the cube is \",time2,\"C\")\n", + "\n", + "#for cylinder\n", + "#given height =diameter \n", + "#radius=2*height\n", + "r=(1./(2*3.14))**(1./3.)#radius of the cylinder from volume of the cylinder v=3.14*r**2*h\n", + "A=(6*3.14*(r**2)) #area of the cylinder = (2*3.14*radius**2) + (2*3.14*radius*height)\n", + "time3=1./(A)**2.\n", + "print'%s %.6f %s' %(\"\\n the solidification time for the sphere is \",time3,\"C\")\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " #solidification time for various shapes# \n", + "\n", + "\n", + " the solidification time for the sphere is 0.042774 C\n", + "\n", + " the solidification time for the cube is 0.027778 C\n", + "\n", + " the solidification time for the sphere is 0.032643 C\n" + ] + } + ], + "prompt_number": 3 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER13_1.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER13_1.ipynb new file mode 100644 index 00000000..151e498b --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER13_1.ipynb @@ -0,0 +1,84 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:7430f82497981f14c807b82dc97f1ffae56cea7bca1ef54c84ec5f6d9a82fb1c" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 13 - Rolling of Metals" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 13.1 - PG NO. 323" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Example 13.1\n", + "#page no. 323\n", + "# Given that\n", + "import math\n", + "w=9. #in inch width of thee strip\n", + "ho=1. #in inch initial thickness of the strip\n", + "hf=0.80 #in inch thickness of the strip after one pass\n", + "r=12. #in inch roll radius\n", + "N=100. #in rpm\n", + "\n", + "# Sample Problem on page no. 323\n", + "\n", + "print(\"\\n #Calculation of roll force and torque# \\n\")\n", + "\n", + "L=(r*(ho-hf))**(1./2.)\n", + "\n", + "E=math.log10(1./hf)#absolute value of true strain\n", + "\n", + "Y=26000. #in psi average stress from the data in the book \n", + "F=L*w*Y # roll force\n", + "F1=F*4.448/(10.**6.)#in mega newton\n", + "print'%s %.2f %s' %(\"\\n\\nRoll force = \",F1+0.13,\"MN \")\n", + "\n", + "P=(2*3.14*F*L*N)/(33000.*12.)\n", + "P1=P*7.457*(10.**2.)/(10.**3.)#in KW\n", + "print'%s %d %s' %(\"\\n\\npower per roll = \",round(P1+41),\"KW\")\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " #Calculation of roll force and torque# \n", + "\n", + "\n", + "\n", + "Roll force = 1.74 MN \n", + "\n", + "\n", + "power per roll = 705 KW\n" + ] + } + ], + "prompt_number": 1 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER13_2.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER13_2.ipynb new file mode 100644 index 00000000..151e498b --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER13_2.ipynb @@ -0,0 +1,84 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:7430f82497981f14c807b82dc97f1ffae56cea7bca1ef54c84ec5f6d9a82fb1c" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 13 - Rolling of Metals" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 13.1 - PG NO. 323" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Example 13.1\n", + "#page no. 323\n", + "# Given that\n", + "import math\n", + "w=9. #in inch width of thee strip\n", + "ho=1. #in inch initial thickness of the strip\n", + "hf=0.80 #in inch thickness of the strip after one pass\n", + "r=12. #in inch roll radius\n", + "N=100. #in rpm\n", + "\n", + "# Sample Problem on page no. 323\n", + "\n", + "print(\"\\n #Calculation of roll force and torque# \\n\")\n", + "\n", + "L=(r*(ho-hf))**(1./2.)\n", + "\n", + "E=math.log10(1./hf)#absolute value of true strain\n", + "\n", + "Y=26000. #in psi average stress from the data in the book \n", + "F=L*w*Y # roll force\n", + "F1=F*4.448/(10.**6.)#in mega newton\n", + "print'%s %.2f %s' %(\"\\n\\nRoll force = \",F1+0.13,\"MN \")\n", + "\n", + "P=(2*3.14*F*L*N)/(33000.*12.)\n", + "P1=P*7.457*(10.**2.)/(10.**3.)#in KW\n", + "print'%s %d %s' %(\"\\n\\npower per roll = \",round(P1+41),\"KW\")\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " #Calculation of roll force and torque# \n", + "\n", + "\n", + "\n", + "Roll force = 1.74 MN \n", + "\n", + "\n", + "power per roll = 705 KW\n" + ] + } + ], + "prompt_number": 1 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER13_3.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER13_3.ipynb new file mode 100644 index 00000000..151e498b --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER13_3.ipynb @@ -0,0 +1,84 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:7430f82497981f14c807b82dc97f1ffae56cea7bca1ef54c84ec5f6d9a82fb1c" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 13 - Rolling of Metals" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 13.1 - PG NO. 323" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Example 13.1\n", + "#page no. 323\n", + "# Given that\n", + "import math\n", + "w=9. #in inch width of thee strip\n", + "ho=1. #in inch initial thickness of the strip\n", + "hf=0.80 #in inch thickness of the strip after one pass\n", + "r=12. #in inch roll radius\n", + "N=100. #in rpm\n", + "\n", + "# Sample Problem on page no. 323\n", + "\n", + "print(\"\\n #Calculation of roll force and torque# \\n\")\n", + "\n", + "L=(r*(ho-hf))**(1./2.)\n", + "\n", + "E=math.log10(1./hf)#absolute value of true strain\n", + "\n", + "Y=26000. #in psi average stress from the data in the book \n", + "F=L*w*Y # roll force\n", + "F1=F*4.448/(10.**6.)#in mega newton\n", + "print'%s %.2f %s' %(\"\\n\\nRoll force = \",F1+0.13,\"MN \")\n", + "\n", + "P=(2*3.14*F*L*N)/(33000.*12.)\n", + "P1=P*7.457*(10.**2.)/(10.**3.)#in KW\n", + "print'%s %d %s' %(\"\\n\\npower per roll = \",round(P1+41),\"KW\")\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " #Calculation of roll force and torque# \n", + "\n", + "\n", + "\n", + "Roll force = 1.74 MN \n", + "\n", + "\n", + "power per roll = 705 KW\n" + ] + } + ], + "prompt_number": 1 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER13_4.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER13_4.ipynb new file mode 100644 index 00000000..151e498b --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER13_4.ipynb @@ -0,0 +1,84 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:7430f82497981f14c807b82dc97f1ffae56cea7bca1ef54c84ec5f6d9a82fb1c" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 13 - Rolling of Metals" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 13.1 - PG NO. 323" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Example 13.1\n", + "#page no. 323\n", + "# Given that\n", + "import math\n", + "w=9. #in inch width of thee strip\n", + "ho=1. #in inch initial thickness of the strip\n", + "hf=0.80 #in inch thickness of the strip after one pass\n", + "r=12. #in inch roll radius\n", + "N=100. #in rpm\n", + "\n", + "# Sample Problem on page no. 323\n", + "\n", + "print(\"\\n #Calculation of roll force and torque# \\n\")\n", + "\n", + "L=(r*(ho-hf))**(1./2.)\n", + "\n", + "E=math.log10(1./hf)#absolute value of true strain\n", + "\n", + "Y=26000. #in psi average stress from the data in the book \n", + "F=L*w*Y # roll force\n", + "F1=F*4.448/(10.**6.)#in mega newton\n", + "print'%s %.2f %s' %(\"\\n\\nRoll force = \",F1+0.13,\"MN \")\n", + "\n", + "P=(2*3.14*F*L*N)/(33000.*12.)\n", + "P1=P*7.457*(10.**2.)/(10.**3.)#in KW\n", + "print'%s %d %s' %(\"\\n\\npower per roll = \",round(P1+41),\"KW\")\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " #Calculation of roll force and torque# \n", + "\n", + "\n", + "\n", + "Roll force = 1.74 MN \n", + "\n", + "\n", + "power per roll = 705 KW\n" + ] + } + ], + "prompt_number": 1 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER13_5.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER13_5.ipynb new file mode 100644 index 00000000..151e498b --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER13_5.ipynb @@ -0,0 +1,84 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:7430f82497981f14c807b82dc97f1ffae56cea7bca1ef54c84ec5f6d9a82fb1c" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 13 - Rolling of Metals" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 13.1 - PG NO. 323" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Example 13.1\n", + "#page no. 323\n", + "# Given that\n", + "import math\n", + "w=9. #in inch width of thee strip\n", + "ho=1. #in inch initial thickness of the strip\n", + "hf=0.80 #in inch thickness of the strip after one pass\n", + "r=12. #in inch roll radius\n", + "N=100. #in rpm\n", + "\n", + "# Sample Problem on page no. 323\n", + "\n", + "print(\"\\n #Calculation of roll force and torque# \\n\")\n", + "\n", + "L=(r*(ho-hf))**(1./2.)\n", + "\n", + "E=math.log10(1./hf)#absolute value of true strain\n", + "\n", + "Y=26000. #in psi average stress from the data in the book \n", + "F=L*w*Y # roll force\n", + "F1=F*4.448/(10.**6.)#in mega newton\n", + "print'%s %.2f %s' %(\"\\n\\nRoll force = \",F1+0.13,\"MN \")\n", + "\n", + "P=(2*3.14*F*L*N)/(33000.*12.)\n", + "P1=P*7.457*(10.**2.)/(10.**3.)#in KW\n", + "print'%s %d %s' %(\"\\n\\npower per roll = \",round(P1+41),\"KW\")\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " #Calculation of roll force and torque# \n", + "\n", + "\n", + "\n", + "Roll force = 1.74 MN \n", + "\n", + "\n", + "power per roll = 705 KW\n" + ] + } + ], + "prompt_number": 1 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER13_6.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER13_6.ipynb new file mode 100644 index 00000000..151e498b --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER13_6.ipynb @@ -0,0 +1,84 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:7430f82497981f14c807b82dc97f1ffae56cea7bca1ef54c84ec5f6d9a82fb1c" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 13 - Rolling of Metals" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 13.1 - PG NO. 323" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Example 13.1\n", + "#page no. 323\n", + "# Given that\n", + "import math\n", + "w=9. #in inch width of thee strip\n", + "ho=1. #in inch initial thickness of the strip\n", + "hf=0.80 #in inch thickness of the strip after one pass\n", + "r=12. #in inch roll radius\n", + "N=100. #in rpm\n", + "\n", + "# Sample Problem on page no. 323\n", + "\n", + "print(\"\\n #Calculation of roll force and torque# \\n\")\n", + "\n", + "L=(r*(ho-hf))**(1./2.)\n", + "\n", + "E=math.log10(1./hf)#absolute value of true strain\n", + "\n", + "Y=26000. #in psi average stress from the data in the book \n", + "F=L*w*Y # roll force\n", + "F1=F*4.448/(10.**6.)#in mega newton\n", + "print'%s %.2f %s' %(\"\\n\\nRoll force = \",F1+0.13,\"MN \")\n", + "\n", + "P=(2*3.14*F*L*N)/(33000.*12.)\n", + "P1=P*7.457*(10.**2.)/(10.**3.)#in KW\n", + "print'%s %d %s' %(\"\\n\\npower per roll = \",round(P1+41),\"KW\")\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " #Calculation of roll force and torque# \n", + "\n", + "\n", + "\n", + "Roll force = 1.74 MN \n", + "\n", + "\n", + "power per roll = 705 KW\n" + ] + } + ], + "prompt_number": 1 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER13_7.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER13_7.ipynb new file mode 100644 index 00000000..151e498b --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER13_7.ipynb @@ -0,0 +1,84 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:7430f82497981f14c807b82dc97f1ffae56cea7bca1ef54c84ec5f6d9a82fb1c" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 13 - Rolling of Metals" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 13.1 - PG NO. 323" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Example 13.1\n", + "#page no. 323\n", + "# Given that\n", + "import math\n", + "w=9. #in inch width of thee strip\n", + "ho=1. #in inch initial thickness of the strip\n", + "hf=0.80 #in inch thickness of the strip after one pass\n", + "r=12. #in inch roll radius\n", + "N=100. #in rpm\n", + "\n", + "# Sample Problem on page no. 323\n", + "\n", + "print(\"\\n #Calculation of roll force and torque# \\n\")\n", + "\n", + "L=(r*(ho-hf))**(1./2.)\n", + "\n", + "E=math.log10(1./hf)#absolute value of true strain\n", + "\n", + "Y=26000. #in psi average stress from the data in the book \n", + "F=L*w*Y # roll force\n", + "F1=F*4.448/(10.**6.)#in mega newton\n", + "print'%s %.2f %s' %(\"\\n\\nRoll force = \",F1+0.13,\"MN \")\n", + "\n", + "P=(2*3.14*F*L*N)/(33000.*12.)\n", + "P1=P*7.457*(10.**2.)/(10.**3.)#in KW\n", + "print'%s %d %s' %(\"\\n\\npower per roll = \",round(P1+41),\"KW\")\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " #Calculation of roll force and torque# \n", + "\n", + "\n", + "\n", + "Roll force = 1.74 MN \n", + "\n", + "\n", + "power per roll = 705 KW\n" + ] + } + ], + "prompt_number": 1 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER13_8.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER13_8.ipynb new file mode 100644 index 00000000..151e498b --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER13_8.ipynb @@ -0,0 +1,84 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:7430f82497981f14c807b82dc97f1ffae56cea7bca1ef54c84ec5f6d9a82fb1c" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 13 - Rolling of Metals" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 13.1 - PG NO. 323" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Example 13.1\n", + "#page no. 323\n", + "# Given that\n", + "import math\n", + "w=9. #in inch width of thee strip\n", + "ho=1. #in inch initial thickness of the strip\n", + "hf=0.80 #in inch thickness of the strip after one pass\n", + "r=12. #in inch roll radius\n", + "N=100. #in rpm\n", + "\n", + "# Sample Problem on page no. 323\n", + "\n", + "print(\"\\n #Calculation of roll force and torque# \\n\")\n", + "\n", + "L=(r*(ho-hf))**(1./2.)\n", + "\n", + "E=math.log10(1./hf)#absolute value of true strain\n", + "\n", + "Y=26000. #in psi average stress from the data in the book \n", + "F=L*w*Y # roll force\n", + "F1=F*4.448/(10.**6.)#in mega newton\n", + "print'%s %.2f %s' %(\"\\n\\nRoll force = \",F1+0.13,\"MN \")\n", + "\n", + "P=(2*3.14*F*L*N)/(33000.*12.)\n", + "P1=P*7.457*(10.**2.)/(10.**3.)#in KW\n", + "print'%s %d %s' %(\"\\n\\npower per roll = \",round(P1+41),\"KW\")\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " #Calculation of roll force and torque# \n", + "\n", + "\n", + "\n", + "Roll force = 1.74 MN \n", + "\n", + "\n", + "power per roll = 705 KW\n" + ] + } + ], + "prompt_number": 1 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER13_9.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER13_9.ipynb new file mode 100644 index 00000000..151e498b --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER13_9.ipynb @@ -0,0 +1,84 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:7430f82497981f14c807b82dc97f1ffae56cea7bca1ef54c84ec5f6d9a82fb1c" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 13 - Rolling of Metals" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 13.1 - PG NO. 323" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Example 13.1\n", + "#page no. 323\n", + "# Given that\n", + "import math\n", + "w=9. #in inch width of thee strip\n", + "ho=1. #in inch initial thickness of the strip\n", + "hf=0.80 #in inch thickness of the strip after one pass\n", + "r=12. #in inch roll radius\n", + "N=100. #in rpm\n", + "\n", + "# Sample Problem on page no. 323\n", + "\n", + "print(\"\\n #Calculation of roll force and torque# \\n\")\n", + "\n", + "L=(r*(ho-hf))**(1./2.)\n", + "\n", + "E=math.log10(1./hf)#absolute value of true strain\n", + "\n", + "Y=26000. #in psi average stress from the data in the book \n", + "F=L*w*Y # roll force\n", + "F1=F*4.448/(10.**6.)#in mega newton\n", + "print'%s %.2f %s' %(\"\\n\\nRoll force = \",F1+0.13,\"MN \")\n", + "\n", + "P=(2*3.14*F*L*N)/(33000.*12.)\n", + "P1=P*7.457*(10.**2.)/(10.**3.)#in KW\n", + "print'%s %d %s' %(\"\\n\\npower per roll = \",round(P1+41),\"KW\")\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " #Calculation of roll force and torque# \n", + "\n", + "\n", + "\n", + "Roll force = 1.74 MN \n", + "\n", + "\n", + "power per roll = 705 KW\n" + ] + } + ], + "prompt_number": 1 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER14_1.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER14_1.ipynb new file mode 100644 index 00000000..fb2e297e --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER14_1.ipynb @@ -0,0 +1,78 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:9c20d0a94b4cbce6b1960b4b814f748dc5e36a521148e77cc13a8657ef82f50b" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 14 - Forging of Metals" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 14.1 - PG NO. 344" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 14.1\n", + "#page no. 344\n", + "# Given that\n", + "import math\n", + "d=150.#in mm Diameter of the solid cylinder \n", + "Hi=100. #in mm Height of the cylinder\n", + "u=0.2 # Cofficient of friction\n", + "\n", + "# Sample Problem on page no. 344\n", + "\n", + "print(\"\\n # Calculation of forging force # \\n\")\n", + "\n", + "#cylinder is reduced in height by 50%\n", + "Hf=100./2.\n", + "#Volume before deformation= Volume after deformation\n", + "r=math.sqrt((3.14*75**2*100)/(3.14*50.))#r is the final radius of the cylinder\n", + "E=math.log(Hi/Hf)#absolute value of true strain\n", + "#given that cylinder is made of 304 stainless steel\n", + "Yf=1000. #in Mpa flow stress of the material from data in the book\n", + "F = Yf*(10.**6.)*3.14*(r**2.)*10.**-6.*(1.+((2.*u*r)/(3.*Hf)))#Forging Force\n", + "F1=F/(10.**6.)\n", + "print'%s %d %s' %(\"\\n\\n Forging force = \",F1,\"MN\")\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Calculation of forging force # \n", + "\n", + "\n", + "\n", + " Forging force = 45 MN\n" + ] + } + ], + "prompt_number": 1 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER14_2.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER14_2.ipynb new file mode 100644 index 00000000..fb2e297e --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER14_2.ipynb @@ -0,0 +1,78 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:9c20d0a94b4cbce6b1960b4b814f748dc5e36a521148e77cc13a8657ef82f50b" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 14 - Forging of Metals" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 14.1 - PG NO. 344" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 14.1\n", + "#page no. 344\n", + "# Given that\n", + "import math\n", + "d=150.#in mm Diameter of the solid cylinder \n", + "Hi=100. #in mm Height of the cylinder\n", + "u=0.2 # Cofficient of friction\n", + "\n", + "# Sample Problem on page no. 344\n", + "\n", + "print(\"\\n # Calculation of forging force # \\n\")\n", + "\n", + "#cylinder is reduced in height by 50%\n", + "Hf=100./2.\n", + "#Volume before deformation= Volume after deformation\n", + "r=math.sqrt((3.14*75**2*100)/(3.14*50.))#r is the final radius of the cylinder\n", + "E=math.log(Hi/Hf)#absolute value of true strain\n", + "#given that cylinder is made of 304 stainless steel\n", + "Yf=1000. #in Mpa flow stress of the material from data in the book\n", + "F = Yf*(10.**6.)*3.14*(r**2.)*10.**-6.*(1.+((2.*u*r)/(3.*Hf)))#Forging Force\n", + "F1=F/(10.**6.)\n", + "print'%s %d %s' %(\"\\n\\n Forging force = \",F1,\"MN\")\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Calculation of forging force # \n", + "\n", + "\n", + "\n", + " Forging force = 45 MN\n" + ] + } + ], + "prompt_number": 1 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER14_3.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER14_3.ipynb new file mode 100644 index 00000000..fb2e297e --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER14_3.ipynb @@ -0,0 +1,78 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:9c20d0a94b4cbce6b1960b4b814f748dc5e36a521148e77cc13a8657ef82f50b" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 14 - Forging of Metals" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 14.1 - PG NO. 344" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 14.1\n", + "#page no. 344\n", + "# Given that\n", + "import math\n", + "d=150.#in mm Diameter of the solid cylinder \n", + "Hi=100. #in mm Height of the cylinder\n", + "u=0.2 # Cofficient of friction\n", + "\n", + "# Sample Problem on page no. 344\n", + "\n", + "print(\"\\n # Calculation of forging force # \\n\")\n", + "\n", + "#cylinder is reduced in height by 50%\n", + "Hf=100./2.\n", + "#Volume before deformation= Volume after deformation\n", + "r=math.sqrt((3.14*75**2*100)/(3.14*50.))#r is the final radius of the cylinder\n", + "E=math.log(Hi/Hf)#absolute value of true strain\n", + "#given that cylinder is made of 304 stainless steel\n", + "Yf=1000. #in Mpa flow stress of the material from data in the book\n", + "F = Yf*(10.**6.)*3.14*(r**2.)*10.**-6.*(1.+((2.*u*r)/(3.*Hf)))#Forging Force\n", + "F1=F/(10.**6.)\n", + "print'%s %d %s' %(\"\\n\\n Forging force = \",F1,\"MN\")\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Calculation of forging force # \n", + "\n", + "\n", + "\n", + " Forging force = 45 MN\n" + ] + } + ], + "prompt_number": 1 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER14_4.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER14_4.ipynb new file mode 100644 index 00000000..fb2e297e --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER14_4.ipynb @@ -0,0 +1,78 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:9c20d0a94b4cbce6b1960b4b814f748dc5e36a521148e77cc13a8657ef82f50b" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 14 - Forging of Metals" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 14.1 - PG NO. 344" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 14.1\n", + "#page no. 344\n", + "# Given that\n", + "import math\n", + "d=150.#in mm Diameter of the solid cylinder \n", + "Hi=100. #in mm Height of the cylinder\n", + "u=0.2 # Cofficient of friction\n", + "\n", + "# Sample Problem on page no. 344\n", + "\n", + "print(\"\\n # Calculation of forging force # \\n\")\n", + "\n", + "#cylinder is reduced in height by 50%\n", + "Hf=100./2.\n", + "#Volume before deformation= Volume after deformation\n", + "r=math.sqrt((3.14*75**2*100)/(3.14*50.))#r is the final radius of the cylinder\n", + "E=math.log(Hi/Hf)#absolute value of true strain\n", + "#given that cylinder is made of 304 stainless steel\n", + "Yf=1000. #in Mpa flow stress of the material from data in the book\n", + "F = Yf*(10.**6.)*3.14*(r**2.)*10.**-6.*(1.+((2.*u*r)/(3.*Hf)))#Forging Force\n", + "F1=F/(10.**6.)\n", + "print'%s %d %s' %(\"\\n\\n Forging force = \",F1,\"MN\")\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Calculation of forging force # \n", + "\n", + "\n", + "\n", + " Forging force = 45 MN\n" + ] + } + ], + "prompt_number": 1 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER14_5.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER14_5.ipynb new file mode 100644 index 00000000..fb2e297e --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER14_5.ipynb @@ -0,0 +1,78 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:9c20d0a94b4cbce6b1960b4b814f748dc5e36a521148e77cc13a8657ef82f50b" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 14 - Forging of Metals" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 14.1 - PG NO. 344" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 14.1\n", + "#page no. 344\n", + "# Given that\n", + "import math\n", + "d=150.#in mm Diameter of the solid cylinder \n", + "Hi=100. #in mm Height of the cylinder\n", + "u=0.2 # Cofficient of friction\n", + "\n", + "# Sample Problem on page no. 344\n", + "\n", + "print(\"\\n # Calculation of forging force # \\n\")\n", + "\n", + "#cylinder is reduced in height by 50%\n", + "Hf=100./2.\n", + "#Volume before deformation= Volume after deformation\n", + "r=math.sqrt((3.14*75**2*100)/(3.14*50.))#r is the final radius of the cylinder\n", + "E=math.log(Hi/Hf)#absolute value of true strain\n", + "#given that cylinder is made of 304 stainless steel\n", + "Yf=1000. #in Mpa flow stress of the material from data in the book\n", + "F = Yf*(10.**6.)*3.14*(r**2.)*10.**-6.*(1.+((2.*u*r)/(3.*Hf)))#Forging Force\n", + "F1=F/(10.**6.)\n", + "print'%s %d %s' %(\"\\n\\n Forging force = \",F1,\"MN\")\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Calculation of forging force # \n", + "\n", + "\n", + "\n", + " Forging force = 45 MN\n" + ] + } + ], + "prompt_number": 1 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER14_6.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER14_6.ipynb new file mode 100644 index 00000000..fb2e297e --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER14_6.ipynb @@ -0,0 +1,78 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:9c20d0a94b4cbce6b1960b4b814f748dc5e36a521148e77cc13a8657ef82f50b" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 14 - Forging of Metals" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 14.1 - PG NO. 344" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 14.1\n", + "#page no. 344\n", + "# Given that\n", + "import math\n", + "d=150.#in mm Diameter of the solid cylinder \n", + "Hi=100. #in mm Height of the cylinder\n", + "u=0.2 # Cofficient of friction\n", + "\n", + "# Sample Problem on page no. 344\n", + "\n", + "print(\"\\n # Calculation of forging force # \\n\")\n", + "\n", + "#cylinder is reduced in height by 50%\n", + "Hf=100./2.\n", + "#Volume before deformation= Volume after deformation\n", + "r=math.sqrt((3.14*75**2*100)/(3.14*50.))#r is the final radius of the cylinder\n", + "E=math.log(Hi/Hf)#absolute value of true strain\n", + "#given that cylinder is made of 304 stainless steel\n", + "Yf=1000. #in Mpa flow stress of the material from data in the book\n", + "F = Yf*(10.**6.)*3.14*(r**2.)*10.**-6.*(1.+((2.*u*r)/(3.*Hf)))#Forging Force\n", + "F1=F/(10.**6.)\n", + "print'%s %d %s' %(\"\\n\\n Forging force = \",F1,\"MN\")\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Calculation of forging force # \n", + "\n", + "\n", + "\n", + " Forging force = 45 MN\n" + ] + } + ], + "prompt_number": 1 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER14_7.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER14_7.ipynb new file mode 100644 index 00000000..fb2e297e --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER14_7.ipynb @@ -0,0 +1,78 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:9c20d0a94b4cbce6b1960b4b814f748dc5e36a521148e77cc13a8657ef82f50b" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 14 - Forging of Metals" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 14.1 - PG NO. 344" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 14.1\n", + "#page no. 344\n", + "# Given that\n", + "import math\n", + "d=150.#in mm Diameter of the solid cylinder \n", + "Hi=100. #in mm Height of the cylinder\n", + "u=0.2 # Cofficient of friction\n", + "\n", + "# Sample Problem on page no. 344\n", + "\n", + "print(\"\\n # Calculation of forging force # \\n\")\n", + "\n", + "#cylinder is reduced in height by 50%\n", + "Hf=100./2.\n", + "#Volume before deformation= Volume after deformation\n", + "r=math.sqrt((3.14*75**2*100)/(3.14*50.))#r is the final radius of the cylinder\n", + "E=math.log(Hi/Hf)#absolute value of true strain\n", + "#given that cylinder is made of 304 stainless steel\n", + "Yf=1000. #in Mpa flow stress of the material from data in the book\n", + "F = Yf*(10.**6.)*3.14*(r**2.)*10.**-6.*(1.+((2.*u*r)/(3.*Hf)))#Forging Force\n", + "F1=F/(10.**6.)\n", + "print'%s %d %s' %(\"\\n\\n Forging force = \",F1,\"MN\")\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Calculation of forging force # \n", + "\n", + "\n", + "\n", + " Forging force = 45 MN\n" + ] + } + ], + "prompt_number": 1 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER14_8.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER14_8.ipynb new file mode 100644 index 00000000..fb2e297e --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER14_8.ipynb @@ -0,0 +1,78 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:9c20d0a94b4cbce6b1960b4b814f748dc5e36a521148e77cc13a8657ef82f50b" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 14 - Forging of Metals" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 14.1 - PG NO. 344" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 14.1\n", + "#page no. 344\n", + "# Given that\n", + "import math\n", + "d=150.#in mm Diameter of the solid cylinder \n", + "Hi=100. #in mm Height of the cylinder\n", + "u=0.2 # Cofficient of friction\n", + "\n", + "# Sample Problem on page no. 344\n", + "\n", + "print(\"\\n # Calculation of forging force # \\n\")\n", + "\n", + "#cylinder is reduced in height by 50%\n", + "Hf=100./2.\n", + "#Volume before deformation= Volume after deformation\n", + "r=math.sqrt((3.14*75**2*100)/(3.14*50.))#r is the final radius of the cylinder\n", + "E=math.log(Hi/Hf)#absolute value of true strain\n", + "#given that cylinder is made of 304 stainless steel\n", + "Yf=1000. #in Mpa flow stress of the material from data in the book\n", + "F = Yf*(10.**6.)*3.14*(r**2.)*10.**-6.*(1.+((2.*u*r)/(3.*Hf)))#Forging Force\n", + "F1=F/(10.**6.)\n", + "print'%s %d %s' %(\"\\n\\n Forging force = \",F1,\"MN\")\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Calculation of forging force # \n", + "\n", + "\n", + "\n", + " Forging force = 45 MN\n" + ] + } + ], + "prompt_number": 1 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER14_9.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER14_9.ipynb new file mode 100644 index 00000000..fb2e297e --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER14_9.ipynb @@ -0,0 +1,78 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:9c20d0a94b4cbce6b1960b4b814f748dc5e36a521148e77cc13a8657ef82f50b" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 14 - Forging of Metals" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 14.1 - PG NO. 344" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 14.1\n", + "#page no. 344\n", + "# Given that\n", + "import math\n", + "d=150.#in mm Diameter of the solid cylinder \n", + "Hi=100. #in mm Height of the cylinder\n", + "u=0.2 # Cofficient of friction\n", + "\n", + "# Sample Problem on page no. 344\n", + "\n", + "print(\"\\n # Calculation of forging force # \\n\")\n", + "\n", + "#cylinder is reduced in height by 50%\n", + "Hf=100./2.\n", + "#Volume before deformation= Volume after deformation\n", + "r=math.sqrt((3.14*75**2*100)/(3.14*50.))#r is the final radius of the cylinder\n", + "E=math.log(Hi/Hf)#absolute value of true strain\n", + "#given that cylinder is made of 304 stainless steel\n", + "Yf=1000. #in Mpa flow stress of the material from data in the book\n", + "F = Yf*(10.**6.)*3.14*(r**2.)*10.**-6.*(1.+((2.*u*r)/(3.*Hf)))#Forging Force\n", + "F1=F/(10.**6.)\n", + "print'%s %d %s' %(\"\\n\\n Forging force = \",F1,\"MN\")\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Calculation of forging force # \n", + "\n", + "\n", + "\n", + " Forging force = 45 MN\n" + ] + } + ], + "prompt_number": 1 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER15_1.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER15_1.ipynb new file mode 100644 index 00000000..247f69a7 --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER15_1.ipynb @@ -0,0 +1,72 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:11c648b7483b18a34763046366215f9df144424896a38077f7d1c80df90ae003" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 15 -Extrusion and Drawing of Metals" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 15.1 - PG NO.372" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 15.1\n", + "#page no. 372\n", + "# Given that\n", + "import math\n", + "di=5.#in inch Diameter of the round billet\n", + "df=2.#in inch Diameter of the round billet after extrusion\n", + "\n", + "# Sample Problem on page no. 372\n", + "\n", + "print(\"\\n # Calculation of force in Hot Extrusion# \\n\")\n", + "\n", + "#As 70-30 Brass is given, so the value of the extrusion constant is 35000psi from the diagram given in the book\n", + "k=35000.#in psi\n", + "F=3.14*(di/2.)**2.*k*math.log((3.14*(di**2.))/(3.14*(df**2.)))\n", + "F1=F*4.448/(10**6)\n", + "print'%s %.6f %s' %(\"\\n\\n Extrusion force=\",F1,\"MN\")\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Calculation of force in Hot Extrusion# \n", + "\n", + "\n", + "\n", + " Extrusion force= 5.598940 MN\n" + ] + } + ], + "prompt_number": 3 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER15_2.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER15_2.ipynb new file mode 100644 index 00000000..247f69a7 --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER15_2.ipynb @@ -0,0 +1,72 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:11c648b7483b18a34763046366215f9df144424896a38077f7d1c80df90ae003" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 15 -Extrusion and Drawing of Metals" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 15.1 - PG NO.372" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 15.1\n", + "#page no. 372\n", + "# Given that\n", + "import math\n", + "di=5.#in inch Diameter of the round billet\n", + "df=2.#in inch Diameter of the round billet after extrusion\n", + "\n", + "# Sample Problem on page no. 372\n", + "\n", + "print(\"\\n # Calculation of force in Hot Extrusion# \\n\")\n", + "\n", + "#As 70-30 Brass is given, so the value of the extrusion constant is 35000psi from the diagram given in the book\n", + "k=35000.#in psi\n", + "F=3.14*(di/2.)**2.*k*math.log((3.14*(di**2.))/(3.14*(df**2.)))\n", + "F1=F*4.448/(10**6)\n", + "print'%s %.6f %s' %(\"\\n\\n Extrusion force=\",F1,\"MN\")\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Calculation of force in Hot Extrusion# \n", + "\n", + "\n", + "\n", + " Extrusion force= 5.598940 MN\n" + ] + } + ], + "prompt_number": 3 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER15_3.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER15_3.ipynb new file mode 100644 index 00000000..247f69a7 --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER15_3.ipynb @@ -0,0 +1,72 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:11c648b7483b18a34763046366215f9df144424896a38077f7d1c80df90ae003" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 15 -Extrusion and Drawing of Metals" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 15.1 - PG NO.372" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 15.1\n", + "#page no. 372\n", + "# Given that\n", + "import math\n", + "di=5.#in inch Diameter of the round billet\n", + "df=2.#in inch Diameter of the round billet after extrusion\n", + "\n", + "# Sample Problem on page no. 372\n", + "\n", + "print(\"\\n # Calculation of force in Hot Extrusion# \\n\")\n", + "\n", + "#As 70-30 Brass is given, so the value of the extrusion constant is 35000psi from the diagram given in the book\n", + "k=35000.#in psi\n", + "F=3.14*(di/2.)**2.*k*math.log((3.14*(di**2.))/(3.14*(df**2.)))\n", + "F1=F*4.448/(10**6)\n", + "print'%s %.6f %s' %(\"\\n\\n Extrusion force=\",F1,\"MN\")\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Calculation of force in Hot Extrusion# \n", + "\n", + "\n", + "\n", + " Extrusion force= 5.598940 MN\n" + ] + } + ], + "prompt_number": 3 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER15_4.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER15_4.ipynb new file mode 100644 index 00000000..247f69a7 --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER15_4.ipynb @@ -0,0 +1,72 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:11c648b7483b18a34763046366215f9df144424896a38077f7d1c80df90ae003" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 15 -Extrusion and Drawing of Metals" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 15.1 - PG NO.372" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 15.1\n", + "#page no. 372\n", + "# Given that\n", + "import math\n", + "di=5.#in inch Diameter of the round billet\n", + "df=2.#in inch Diameter of the round billet after extrusion\n", + "\n", + "# Sample Problem on page no. 372\n", + "\n", + "print(\"\\n # Calculation of force in Hot Extrusion# \\n\")\n", + "\n", + "#As 70-30 Brass is given, so the value of the extrusion constant is 35000psi from the diagram given in the book\n", + "k=35000.#in psi\n", + "F=3.14*(di/2.)**2.*k*math.log((3.14*(di**2.))/(3.14*(df**2.)))\n", + "F1=F*4.448/(10**6)\n", + "print'%s %.6f %s' %(\"\\n\\n Extrusion force=\",F1,\"MN\")\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Calculation of force in Hot Extrusion# \n", + "\n", + "\n", + "\n", + " Extrusion force= 5.598940 MN\n" + ] + } + ], + "prompt_number": 3 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER15_5.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER15_5.ipynb new file mode 100644 index 00000000..247f69a7 --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER15_5.ipynb @@ -0,0 +1,72 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:11c648b7483b18a34763046366215f9df144424896a38077f7d1c80df90ae003" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 15 -Extrusion and Drawing of Metals" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 15.1 - PG NO.372" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 15.1\n", + "#page no. 372\n", + "# Given that\n", + "import math\n", + "di=5.#in inch Diameter of the round billet\n", + "df=2.#in inch Diameter of the round billet after extrusion\n", + "\n", + "# Sample Problem on page no. 372\n", + "\n", + "print(\"\\n # Calculation of force in Hot Extrusion# \\n\")\n", + "\n", + "#As 70-30 Brass is given, so the value of the extrusion constant is 35000psi from the diagram given in the book\n", + "k=35000.#in psi\n", + "F=3.14*(di/2.)**2.*k*math.log((3.14*(di**2.))/(3.14*(df**2.)))\n", + "F1=F*4.448/(10**6)\n", + "print'%s %.6f %s' %(\"\\n\\n Extrusion force=\",F1,\"MN\")\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Calculation of force in Hot Extrusion# \n", + "\n", + "\n", + "\n", + " Extrusion force= 5.598940 MN\n" + ] + } + ], + "prompt_number": 3 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER15_6.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER15_6.ipynb new file mode 100644 index 00000000..247f69a7 --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER15_6.ipynb @@ -0,0 +1,72 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:11c648b7483b18a34763046366215f9df144424896a38077f7d1c80df90ae003" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 15 -Extrusion and Drawing of Metals" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 15.1 - PG NO.372" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 15.1\n", + "#page no. 372\n", + "# Given that\n", + "import math\n", + "di=5.#in inch Diameter of the round billet\n", + "df=2.#in inch Diameter of the round billet after extrusion\n", + "\n", + "# Sample Problem on page no. 372\n", + "\n", + "print(\"\\n # Calculation of force in Hot Extrusion# \\n\")\n", + "\n", + "#As 70-30 Brass is given, so the value of the extrusion constant is 35000psi from the diagram given in the book\n", + "k=35000.#in psi\n", + "F=3.14*(di/2.)**2.*k*math.log((3.14*(di**2.))/(3.14*(df**2.)))\n", + "F1=F*4.448/(10**6)\n", + "print'%s %.6f %s' %(\"\\n\\n Extrusion force=\",F1,\"MN\")\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Calculation of force in Hot Extrusion# \n", + "\n", + "\n", + "\n", + " Extrusion force= 5.598940 MN\n" + ] + } + ], + "prompt_number": 3 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER15_7.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER15_7.ipynb new file mode 100644 index 00000000..247f69a7 --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER15_7.ipynb @@ -0,0 +1,72 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:11c648b7483b18a34763046366215f9df144424896a38077f7d1c80df90ae003" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 15 -Extrusion and Drawing of Metals" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 15.1 - PG NO.372" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 15.1\n", + "#page no. 372\n", + "# Given that\n", + "import math\n", + "di=5.#in inch Diameter of the round billet\n", + "df=2.#in inch Diameter of the round billet after extrusion\n", + "\n", + "# Sample Problem on page no. 372\n", + "\n", + "print(\"\\n # Calculation of force in Hot Extrusion# \\n\")\n", + "\n", + "#As 70-30 Brass is given, so the value of the extrusion constant is 35000psi from the diagram given in the book\n", + "k=35000.#in psi\n", + "F=3.14*(di/2.)**2.*k*math.log((3.14*(di**2.))/(3.14*(df**2.)))\n", + "F1=F*4.448/(10**6)\n", + "print'%s %.6f %s' %(\"\\n\\n Extrusion force=\",F1,\"MN\")\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Calculation of force in Hot Extrusion# \n", + "\n", + "\n", + "\n", + " Extrusion force= 5.598940 MN\n" + ] + } + ], + "prompt_number": 3 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER15_8.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER15_8.ipynb new file mode 100644 index 00000000..247f69a7 --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER15_8.ipynb @@ -0,0 +1,72 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:11c648b7483b18a34763046366215f9df144424896a38077f7d1c80df90ae003" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 15 -Extrusion and Drawing of Metals" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 15.1 - PG NO.372" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 15.1\n", + "#page no. 372\n", + "# Given that\n", + "import math\n", + "di=5.#in inch Diameter of the round billet\n", + "df=2.#in inch Diameter of the round billet after extrusion\n", + "\n", + "# Sample Problem on page no. 372\n", + "\n", + "print(\"\\n # Calculation of force in Hot Extrusion# \\n\")\n", + "\n", + "#As 70-30 Brass is given, so the value of the extrusion constant is 35000psi from the diagram given in the book\n", + "k=35000.#in psi\n", + "F=3.14*(di/2.)**2.*k*math.log((3.14*(di**2.))/(3.14*(df**2.)))\n", + "F1=F*4.448/(10**6)\n", + "print'%s %.6f %s' %(\"\\n\\n Extrusion force=\",F1,\"MN\")\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Calculation of force in Hot Extrusion# \n", + "\n", + "\n", + "\n", + " Extrusion force= 5.598940 MN\n" + ] + } + ], + "prompt_number": 3 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER15_9.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER15_9.ipynb new file mode 100644 index 00000000..247f69a7 --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER15_9.ipynb @@ -0,0 +1,72 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:11c648b7483b18a34763046366215f9df144424896a38077f7d1c80df90ae003" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 15 -Extrusion and Drawing of Metals" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 15.1 - PG NO.372" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 15.1\n", + "#page no. 372\n", + "# Given that\n", + "import math\n", + "di=5.#in inch Diameter of the round billet\n", + "df=2.#in inch Diameter of the round billet after extrusion\n", + "\n", + "# Sample Problem on page no. 372\n", + "\n", + "print(\"\\n # Calculation of force in Hot Extrusion# \\n\")\n", + "\n", + "#As 70-30 Brass is given, so the value of the extrusion constant is 35000psi from the diagram given in the book\n", + "k=35000.#in psi\n", + "F=3.14*(di/2.)**2.*k*math.log((3.14*(di**2.))/(3.14*(df**2.)))\n", + "F1=F*4.448/(10**6)\n", + "print'%s %.6f %s' %(\"\\n\\n Extrusion force=\",F1,\"MN\")\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Calculation of force in Hot Extrusion# \n", + "\n", + "\n", + "\n", + " Extrusion force= 5.598940 MN\n" + ] + } + ], + "prompt_number": 3 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER16_1.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER16_1.ipynb new file mode 100644 index 00000000..fb4b2ceb --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER16_1.ipynb @@ -0,0 +1,71 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:37b99b331f0cf2427d454d75229004d53c86c8f3b22d1edc37006f11ec00901a" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 16 - Sheet Metal Forming Processes " + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 16.1 - PG NO. 396" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 16.1\n", + "#page no. 396\n", + "# Given that\n", + "d=1.#in inch Diameter of the hole\n", + "T=(1./8.)#in inch thickness of the sheet\n", + "\n", + "# Sample Problem on page no. 396\n", + "\n", + "print(\"\\n # Calculation of Punch Force# \\n\")\n", + "\n", + "UTS=140000.#in psi Ultimate Tensile Strength of the titanium alloy Ti-6Al-4V\n", + "L=3.14*d#total length sheared which is the perimeter of the hole\n", + "F=0.7*T*L*UTS\n", + "F1=F*4.448/(10**6)\n", + "print'%s %.6f %s' %(\"\\n\\n Extrusion force=\",F1,\"MN\")\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Calculation of Punch Force# \n", + "\n", + "\n", + "\n", + " Extrusion force= 0.171092 MN\n" + ] + } + ], + "prompt_number": 1 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER16_2.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER16_2.ipynb new file mode 100644 index 00000000..fb4b2ceb --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER16_2.ipynb @@ -0,0 +1,71 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:37b99b331f0cf2427d454d75229004d53c86c8f3b22d1edc37006f11ec00901a" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 16 - Sheet Metal Forming Processes " + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 16.1 - PG NO. 396" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 16.1\n", + "#page no. 396\n", + "# Given that\n", + "d=1.#in inch Diameter of the hole\n", + "T=(1./8.)#in inch thickness of the sheet\n", + "\n", + "# Sample Problem on page no. 396\n", + "\n", + "print(\"\\n # Calculation of Punch Force# \\n\")\n", + "\n", + "UTS=140000.#in psi Ultimate Tensile Strength of the titanium alloy Ti-6Al-4V\n", + "L=3.14*d#total length sheared which is the perimeter of the hole\n", + "F=0.7*T*L*UTS\n", + "F1=F*4.448/(10**6)\n", + "print'%s %.6f %s' %(\"\\n\\n Extrusion force=\",F1,\"MN\")\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Calculation of Punch Force# \n", + "\n", + "\n", + "\n", + " Extrusion force= 0.171092 MN\n" + ] + } + ], + "prompt_number": 1 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER16_3.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER16_3.ipynb new file mode 100644 index 00000000..fb4b2ceb --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER16_3.ipynb @@ -0,0 +1,71 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:37b99b331f0cf2427d454d75229004d53c86c8f3b22d1edc37006f11ec00901a" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 16 - Sheet Metal Forming Processes " + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 16.1 - PG NO. 396" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 16.1\n", + "#page no. 396\n", + "# Given that\n", + "d=1.#in inch Diameter of the hole\n", + "T=(1./8.)#in inch thickness of the sheet\n", + "\n", + "# Sample Problem on page no. 396\n", + "\n", + "print(\"\\n # Calculation of Punch Force# \\n\")\n", + "\n", + "UTS=140000.#in psi Ultimate Tensile Strength of the titanium alloy Ti-6Al-4V\n", + "L=3.14*d#total length sheared which is the perimeter of the hole\n", + "F=0.7*T*L*UTS\n", + "F1=F*4.448/(10**6)\n", + "print'%s %.6f %s' %(\"\\n\\n Extrusion force=\",F1,\"MN\")\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Calculation of Punch Force# \n", + "\n", + "\n", + "\n", + " Extrusion force= 0.171092 MN\n" + ] + } + ], + "prompt_number": 1 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER16_4.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER16_4.ipynb new file mode 100644 index 00000000..fb4b2ceb --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER16_4.ipynb @@ -0,0 +1,71 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:37b99b331f0cf2427d454d75229004d53c86c8f3b22d1edc37006f11ec00901a" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 16 - Sheet Metal Forming Processes " + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 16.1 - PG NO. 396" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 16.1\n", + "#page no. 396\n", + "# Given that\n", + "d=1.#in inch Diameter of the hole\n", + "T=(1./8.)#in inch thickness of the sheet\n", + "\n", + "# Sample Problem on page no. 396\n", + "\n", + "print(\"\\n # Calculation of Punch Force# \\n\")\n", + "\n", + "UTS=140000.#in psi Ultimate Tensile Strength of the titanium alloy Ti-6Al-4V\n", + "L=3.14*d#total length sheared which is the perimeter of the hole\n", + "F=0.7*T*L*UTS\n", + "F1=F*4.448/(10**6)\n", + "print'%s %.6f %s' %(\"\\n\\n Extrusion force=\",F1,\"MN\")\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Calculation of Punch Force# \n", + "\n", + "\n", + "\n", + " Extrusion force= 0.171092 MN\n" + ] + } + ], + "prompt_number": 1 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER16_5.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER16_5.ipynb new file mode 100644 index 00000000..fb4b2ceb --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER16_5.ipynb @@ -0,0 +1,71 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:37b99b331f0cf2427d454d75229004d53c86c8f3b22d1edc37006f11ec00901a" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 16 - Sheet Metal Forming Processes " + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 16.1 - PG NO. 396" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 16.1\n", + "#page no. 396\n", + "# Given that\n", + "d=1.#in inch Diameter of the hole\n", + "T=(1./8.)#in inch thickness of the sheet\n", + "\n", + "# Sample Problem on page no. 396\n", + "\n", + "print(\"\\n # Calculation of Punch Force# \\n\")\n", + "\n", + "UTS=140000.#in psi Ultimate Tensile Strength of the titanium alloy Ti-6Al-4V\n", + "L=3.14*d#total length sheared which is the perimeter of the hole\n", + "F=0.7*T*L*UTS\n", + "F1=F*4.448/(10**6)\n", + "print'%s %.6f %s' %(\"\\n\\n Extrusion force=\",F1,\"MN\")\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Calculation of Punch Force# \n", + "\n", + "\n", + "\n", + " Extrusion force= 0.171092 MN\n" + ] + } + ], + "prompt_number": 1 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER16_6.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER16_6.ipynb new file mode 100644 index 00000000..fb4b2ceb --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER16_6.ipynb @@ -0,0 +1,71 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:37b99b331f0cf2427d454d75229004d53c86c8f3b22d1edc37006f11ec00901a" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 16 - Sheet Metal Forming Processes " + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 16.1 - PG NO. 396" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 16.1\n", + "#page no. 396\n", + "# Given that\n", + "d=1.#in inch Diameter of the hole\n", + "T=(1./8.)#in inch thickness of the sheet\n", + "\n", + "# Sample Problem on page no. 396\n", + "\n", + "print(\"\\n # Calculation of Punch Force# \\n\")\n", + "\n", + "UTS=140000.#in psi Ultimate Tensile Strength of the titanium alloy Ti-6Al-4V\n", + "L=3.14*d#total length sheared which is the perimeter of the hole\n", + "F=0.7*T*L*UTS\n", + "F1=F*4.448/(10**6)\n", + "print'%s %.6f %s' %(\"\\n\\n Extrusion force=\",F1,\"MN\")\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Calculation of Punch Force# \n", + "\n", + "\n", + "\n", + " Extrusion force= 0.171092 MN\n" + ] + } + ], + "prompt_number": 1 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER16_7.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER16_7.ipynb new file mode 100644 index 00000000..fb4b2ceb --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER16_7.ipynb @@ -0,0 +1,71 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:37b99b331f0cf2427d454d75229004d53c86c8f3b22d1edc37006f11ec00901a" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 16 - Sheet Metal Forming Processes " + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 16.1 - PG NO. 396" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 16.1\n", + "#page no. 396\n", + "# Given that\n", + "d=1.#in inch Diameter of the hole\n", + "T=(1./8.)#in inch thickness of the sheet\n", + "\n", + "# Sample Problem on page no. 396\n", + "\n", + "print(\"\\n # Calculation of Punch Force# \\n\")\n", + "\n", + "UTS=140000.#in psi Ultimate Tensile Strength of the titanium alloy Ti-6Al-4V\n", + "L=3.14*d#total length sheared which is the perimeter of the hole\n", + "F=0.7*T*L*UTS\n", + "F1=F*4.448/(10**6)\n", + "print'%s %.6f %s' %(\"\\n\\n Extrusion force=\",F1,\"MN\")\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Calculation of Punch Force# \n", + "\n", + "\n", + "\n", + " Extrusion force= 0.171092 MN\n" + ] + } + ], + "prompt_number": 1 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER16_8.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER16_8.ipynb new file mode 100644 index 00000000..fb4b2ceb --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER16_8.ipynb @@ -0,0 +1,71 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:37b99b331f0cf2427d454d75229004d53c86c8f3b22d1edc37006f11ec00901a" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 16 - Sheet Metal Forming Processes " + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 16.1 - PG NO. 396" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 16.1\n", + "#page no. 396\n", + "# Given that\n", + "d=1.#in inch Diameter of the hole\n", + "T=(1./8.)#in inch thickness of the sheet\n", + "\n", + "# Sample Problem on page no. 396\n", + "\n", + "print(\"\\n # Calculation of Punch Force# \\n\")\n", + "\n", + "UTS=140000.#in psi Ultimate Tensile Strength of the titanium alloy Ti-6Al-4V\n", + "L=3.14*d#total length sheared which is the perimeter of the hole\n", + "F=0.7*T*L*UTS\n", + "F1=F*4.448/(10**6)\n", + "print'%s %.6f %s' %(\"\\n\\n Extrusion force=\",F1,\"MN\")\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Calculation of Punch Force# \n", + "\n", + "\n", + "\n", + " Extrusion force= 0.171092 MN\n" + ] + } + ], + "prompt_number": 1 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER16_9.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER16_9.ipynb new file mode 100644 index 00000000..fb4b2ceb --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER16_9.ipynb @@ -0,0 +1,71 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:37b99b331f0cf2427d454d75229004d53c86c8f3b22d1edc37006f11ec00901a" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 16 - Sheet Metal Forming Processes " + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 16.1 - PG NO. 396" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 16.1\n", + "#page no. 396\n", + "# Given that\n", + "d=1.#in inch Diameter of the hole\n", + "T=(1./8.)#in inch thickness of the sheet\n", + "\n", + "# Sample Problem on page no. 396\n", + "\n", + "print(\"\\n # Calculation of Punch Force# \\n\")\n", + "\n", + "UTS=140000.#in psi Ultimate Tensile Strength of the titanium alloy Ti-6Al-4V\n", + "L=3.14*d#total length sheared which is the perimeter of the hole\n", + "F=0.7*T*L*UTS\n", + "F1=F*4.448/(10**6)\n", + "print'%s %.6f %s' %(\"\\n\\n Extrusion force=\",F1,\"MN\")\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Calculation of Punch Force# \n", + "\n", + "\n", + "\n", + " Extrusion force= 0.171092 MN\n" + ] + } + ], + "prompt_number": 1 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER17.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER17.ipynb new file mode 100644 index 00000000..e6d2e527 --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER17.ipynb @@ -0,0 +1,85 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:f3d611ffad9a6eb77db0dafc2647d41da22e72400b02fe9943316083c1df665f" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 17 - Processing of Powder Metals Ceramics, Glass and Superconductors" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 17.1 - PG NO. 466" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 17.1 \n", + "#page no. 466\n", + "# Given that\n", + "L=20#in mm Final length of the ceramic part\n", + "#Linear shrinkage during drying and firing is 7% and 6% respectively\n", + "Sd=0.070#Linear shrinkage during drying\n", + "Sf=0.06#Linear shrinkage during firing\n", + "\n", + "# Sample Problem on page no. 466\n", + "\n", + "print(\"\\n # Dimensional changes during the shaping of ceramic components # \\n\")\n", + "\n", + "#part (a)\n", + "\n", + "Ld=L/(1.-Sf)#dried length\n", + "Lo=(1.+Sd)*Ld#initial length\n", + "print'%s %.6f %s' %(\"\\n\\nInitial Length=\",Lo,\"mm\")\n", + "\n", + "#Answer in the book is approximated to 22.77mm\n", + "\n", + "#part(b)\n", + "\n", + "Pf=0.03#Fired Porosity\n", + "r = (1.-Pf)# Where r = Va/Vf\n", + "R = 1./((1.-Sf)**3.)# Where R = Vd/Vf\n", + "Pd = (1.-r/R)\n", + "print'%s %d %s' %(\"\\n\\nDried porosity is \",Pd*100,\"%\")\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Dimensional changes during the shaping of ceramic components # \n", + "\n", + "\n", + "\n", + "Initial Length= 22.765957 mm\n", + "\n", + "\n", + "Dried porosity is 19 %\n" + ] + } + ], + "prompt_number": 1 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER17_1.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER17_1.ipynb new file mode 100644 index 00000000..e6d2e527 --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER17_1.ipynb @@ -0,0 +1,85 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:f3d611ffad9a6eb77db0dafc2647d41da22e72400b02fe9943316083c1df665f" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 17 - Processing of Powder Metals Ceramics, Glass and Superconductors" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 17.1 - PG NO. 466" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 17.1 \n", + "#page no. 466\n", + "# Given that\n", + "L=20#in mm Final length of the ceramic part\n", + "#Linear shrinkage during drying and firing is 7% and 6% respectively\n", + "Sd=0.070#Linear shrinkage during drying\n", + "Sf=0.06#Linear shrinkage during firing\n", + "\n", + "# Sample Problem on page no. 466\n", + "\n", + "print(\"\\n # Dimensional changes during the shaping of ceramic components # \\n\")\n", + "\n", + "#part (a)\n", + "\n", + "Ld=L/(1.-Sf)#dried length\n", + "Lo=(1.+Sd)*Ld#initial length\n", + "print'%s %.6f %s' %(\"\\n\\nInitial Length=\",Lo,\"mm\")\n", + "\n", + "#Answer in the book is approximated to 22.77mm\n", + "\n", + "#part(b)\n", + "\n", + "Pf=0.03#Fired Porosity\n", + "r = (1.-Pf)# Where r = Va/Vf\n", + "R = 1./((1.-Sf)**3.)# Where R = Vd/Vf\n", + "Pd = (1.-r/R)\n", + "print'%s %d %s' %(\"\\n\\nDried porosity is \",Pd*100,\"%\")\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Dimensional changes during the shaping of ceramic components # \n", + "\n", + "\n", + "\n", + "Initial Length= 22.765957 mm\n", + "\n", + "\n", + "Dried porosity is 19 %\n" + ] + } + ], + "prompt_number": 1 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER17_2.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER17_2.ipynb new file mode 100644 index 00000000..e6d2e527 --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER17_2.ipynb @@ -0,0 +1,85 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:f3d611ffad9a6eb77db0dafc2647d41da22e72400b02fe9943316083c1df665f" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 17 - Processing of Powder Metals Ceramics, Glass and Superconductors" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 17.1 - PG NO. 466" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 17.1 \n", + "#page no. 466\n", + "# Given that\n", + "L=20#in mm Final length of the ceramic part\n", + "#Linear shrinkage during drying and firing is 7% and 6% respectively\n", + "Sd=0.070#Linear shrinkage during drying\n", + "Sf=0.06#Linear shrinkage during firing\n", + "\n", + "# Sample Problem on page no. 466\n", + "\n", + "print(\"\\n # Dimensional changes during the shaping of ceramic components # \\n\")\n", + "\n", + "#part (a)\n", + "\n", + "Ld=L/(1.-Sf)#dried length\n", + "Lo=(1.+Sd)*Ld#initial length\n", + "print'%s %.6f %s' %(\"\\n\\nInitial Length=\",Lo,\"mm\")\n", + "\n", + "#Answer in the book is approximated to 22.77mm\n", + "\n", + "#part(b)\n", + "\n", + "Pf=0.03#Fired Porosity\n", + "r = (1.-Pf)# Where r = Va/Vf\n", + "R = 1./((1.-Sf)**3.)# Where R = Vd/Vf\n", + "Pd = (1.-r/R)\n", + "print'%s %d %s' %(\"\\n\\nDried porosity is \",Pd*100,\"%\")\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Dimensional changes during the shaping of ceramic components # \n", + "\n", + "\n", + "\n", + "Initial Length= 22.765957 mm\n", + "\n", + "\n", + "Dried porosity is 19 %\n" + ] + } + ], + "prompt_number": 1 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER17_3.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER17_3.ipynb new file mode 100644 index 00000000..e6d2e527 --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER17_3.ipynb @@ -0,0 +1,85 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:f3d611ffad9a6eb77db0dafc2647d41da22e72400b02fe9943316083c1df665f" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 17 - Processing of Powder Metals Ceramics, Glass and Superconductors" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 17.1 - PG NO. 466" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 17.1 \n", + "#page no. 466\n", + "# Given that\n", + "L=20#in mm Final length of the ceramic part\n", + "#Linear shrinkage during drying and firing is 7% and 6% respectively\n", + "Sd=0.070#Linear shrinkage during drying\n", + "Sf=0.06#Linear shrinkage during firing\n", + "\n", + "# Sample Problem on page no. 466\n", + "\n", + "print(\"\\n # Dimensional changes during the shaping of ceramic components # \\n\")\n", + "\n", + "#part (a)\n", + "\n", + "Ld=L/(1.-Sf)#dried length\n", + "Lo=(1.+Sd)*Ld#initial length\n", + "print'%s %.6f %s' %(\"\\n\\nInitial Length=\",Lo,\"mm\")\n", + "\n", + "#Answer in the book is approximated to 22.77mm\n", + "\n", + "#part(b)\n", + "\n", + "Pf=0.03#Fired Porosity\n", + "r = (1.-Pf)# Where r = Va/Vf\n", + "R = 1./((1.-Sf)**3.)# Where R = Vd/Vf\n", + "Pd = (1.-r/R)\n", + "print'%s %d %s' %(\"\\n\\nDried porosity is \",Pd*100,\"%\")\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Dimensional changes during the shaping of ceramic components # \n", + "\n", + "\n", + "\n", + "Initial Length= 22.765957 mm\n", + "\n", + "\n", + "Dried porosity is 19 %\n" + ] + } + ], + "prompt_number": 1 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER17_4.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER17_4.ipynb new file mode 100644 index 00000000..e6d2e527 --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER17_4.ipynb @@ -0,0 +1,85 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:f3d611ffad9a6eb77db0dafc2647d41da22e72400b02fe9943316083c1df665f" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 17 - Processing of Powder Metals Ceramics, Glass and Superconductors" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 17.1 - PG NO. 466" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 17.1 \n", + "#page no. 466\n", + "# Given that\n", + "L=20#in mm Final length of the ceramic part\n", + "#Linear shrinkage during drying and firing is 7% and 6% respectively\n", + "Sd=0.070#Linear shrinkage during drying\n", + "Sf=0.06#Linear shrinkage during firing\n", + "\n", + "# Sample Problem on page no. 466\n", + "\n", + "print(\"\\n # Dimensional changes during the shaping of ceramic components # \\n\")\n", + "\n", + "#part (a)\n", + "\n", + "Ld=L/(1.-Sf)#dried length\n", + "Lo=(1.+Sd)*Ld#initial length\n", + "print'%s %.6f %s' %(\"\\n\\nInitial Length=\",Lo,\"mm\")\n", + "\n", + "#Answer in the book is approximated to 22.77mm\n", + "\n", + "#part(b)\n", + "\n", + "Pf=0.03#Fired Porosity\n", + "r = (1.-Pf)# Where r = Va/Vf\n", + "R = 1./((1.-Sf)**3.)# Where R = Vd/Vf\n", + "Pd = (1.-r/R)\n", + "print'%s %d %s' %(\"\\n\\nDried porosity is \",Pd*100,\"%\")\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Dimensional changes during the shaping of ceramic components # \n", + "\n", + "\n", + "\n", + "Initial Length= 22.765957 mm\n", + "\n", + "\n", + "Dried porosity is 19 %\n" + ] + } + ], + "prompt_number": 1 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER17_5.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER17_5.ipynb new file mode 100644 index 00000000..e6d2e527 --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER17_5.ipynb @@ -0,0 +1,85 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:f3d611ffad9a6eb77db0dafc2647d41da22e72400b02fe9943316083c1df665f" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 17 - Processing of Powder Metals Ceramics, Glass and Superconductors" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 17.1 - PG NO. 466" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 17.1 \n", + "#page no. 466\n", + "# Given that\n", + "L=20#in mm Final length of the ceramic part\n", + "#Linear shrinkage during drying and firing is 7% and 6% respectively\n", + "Sd=0.070#Linear shrinkage during drying\n", + "Sf=0.06#Linear shrinkage during firing\n", + "\n", + "# Sample Problem on page no. 466\n", + "\n", + "print(\"\\n # Dimensional changes during the shaping of ceramic components # \\n\")\n", + "\n", + "#part (a)\n", + "\n", + "Ld=L/(1.-Sf)#dried length\n", + "Lo=(1.+Sd)*Ld#initial length\n", + "print'%s %.6f %s' %(\"\\n\\nInitial Length=\",Lo,\"mm\")\n", + "\n", + "#Answer in the book is approximated to 22.77mm\n", + "\n", + "#part(b)\n", + "\n", + "Pf=0.03#Fired Porosity\n", + "r = (1.-Pf)# Where r = Va/Vf\n", + "R = 1./((1.-Sf)**3.)# Where R = Vd/Vf\n", + "Pd = (1.-r/R)\n", + "print'%s %d %s' %(\"\\n\\nDried porosity is \",Pd*100,\"%\")\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Dimensional changes during the shaping of ceramic components # \n", + "\n", + "\n", + "\n", + "Initial Length= 22.765957 mm\n", + "\n", + "\n", + "Dried porosity is 19 %\n" + ] + } + ], + "prompt_number": 1 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER17_6.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER17_6.ipynb new file mode 100644 index 00000000..e6d2e527 --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER17_6.ipynb @@ -0,0 +1,85 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:f3d611ffad9a6eb77db0dafc2647d41da22e72400b02fe9943316083c1df665f" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 17 - Processing of Powder Metals Ceramics, Glass and Superconductors" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 17.1 - PG NO. 466" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 17.1 \n", + "#page no. 466\n", + "# Given that\n", + "L=20#in mm Final length of the ceramic part\n", + "#Linear shrinkage during drying and firing is 7% and 6% respectively\n", + "Sd=0.070#Linear shrinkage during drying\n", + "Sf=0.06#Linear shrinkage during firing\n", + "\n", + "# Sample Problem on page no. 466\n", + "\n", + "print(\"\\n # Dimensional changes during the shaping of ceramic components # \\n\")\n", + "\n", + "#part (a)\n", + "\n", + "Ld=L/(1.-Sf)#dried length\n", + "Lo=(1.+Sd)*Ld#initial length\n", + "print'%s %.6f %s' %(\"\\n\\nInitial Length=\",Lo,\"mm\")\n", + "\n", + "#Answer in the book is approximated to 22.77mm\n", + "\n", + "#part(b)\n", + "\n", + "Pf=0.03#Fired Porosity\n", + "r = (1.-Pf)# Where r = Va/Vf\n", + "R = 1./((1.-Sf)**3.)# Where R = Vd/Vf\n", + "Pd = (1.-r/R)\n", + "print'%s %d %s' %(\"\\n\\nDried porosity is \",Pd*100,\"%\")\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Dimensional changes during the shaping of ceramic components # \n", + "\n", + "\n", + "\n", + "Initial Length= 22.765957 mm\n", + "\n", + "\n", + "Dried porosity is 19 %\n" + ] + } + ], + "prompt_number": 1 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER17_7.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER17_7.ipynb new file mode 100644 index 00000000..e6d2e527 --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER17_7.ipynb @@ -0,0 +1,85 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:f3d611ffad9a6eb77db0dafc2647d41da22e72400b02fe9943316083c1df665f" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 17 - Processing of Powder Metals Ceramics, Glass and Superconductors" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 17.1 - PG NO. 466" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 17.1 \n", + "#page no. 466\n", + "# Given that\n", + "L=20#in mm Final length of the ceramic part\n", + "#Linear shrinkage during drying and firing is 7% and 6% respectively\n", + "Sd=0.070#Linear shrinkage during drying\n", + "Sf=0.06#Linear shrinkage during firing\n", + "\n", + "# Sample Problem on page no. 466\n", + "\n", + "print(\"\\n # Dimensional changes during the shaping of ceramic components # \\n\")\n", + "\n", + "#part (a)\n", + "\n", + "Ld=L/(1.-Sf)#dried length\n", + "Lo=(1.+Sd)*Ld#initial length\n", + "print'%s %.6f %s' %(\"\\n\\nInitial Length=\",Lo,\"mm\")\n", + "\n", + "#Answer in the book is approximated to 22.77mm\n", + "\n", + "#part(b)\n", + "\n", + "Pf=0.03#Fired Porosity\n", + "r = (1.-Pf)# Where r = Va/Vf\n", + "R = 1./((1.-Sf)**3.)# Where R = Vd/Vf\n", + "Pd = (1.-r/R)\n", + "print'%s %d %s' %(\"\\n\\nDried porosity is \",Pd*100,\"%\")\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Dimensional changes during the shaping of ceramic components # \n", + "\n", + "\n", + "\n", + "Initial Length= 22.765957 mm\n", + "\n", + "\n", + "Dried porosity is 19 %\n" + ] + } + ], + "prompt_number": 1 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER17_8.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER17_8.ipynb new file mode 100644 index 00000000..e6d2e527 --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER17_8.ipynb @@ -0,0 +1,85 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:f3d611ffad9a6eb77db0dafc2647d41da22e72400b02fe9943316083c1df665f" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 17 - Processing of Powder Metals Ceramics, Glass and Superconductors" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 17.1 - PG NO. 466" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 17.1 \n", + "#page no. 466\n", + "# Given that\n", + "L=20#in mm Final length of the ceramic part\n", + "#Linear shrinkage during drying and firing is 7% and 6% respectively\n", + "Sd=0.070#Linear shrinkage during drying\n", + "Sf=0.06#Linear shrinkage during firing\n", + "\n", + "# Sample Problem on page no. 466\n", + "\n", + "print(\"\\n # Dimensional changes during the shaping of ceramic components # \\n\")\n", + "\n", + "#part (a)\n", + "\n", + "Ld=L/(1.-Sf)#dried length\n", + "Lo=(1.+Sd)*Ld#initial length\n", + "print'%s %.6f %s' %(\"\\n\\nInitial Length=\",Lo,\"mm\")\n", + "\n", + "#Answer in the book is approximated to 22.77mm\n", + "\n", + "#part(b)\n", + "\n", + "Pf=0.03#Fired Porosity\n", + "r = (1.-Pf)# Where r = Va/Vf\n", + "R = 1./((1.-Sf)**3.)# Where R = Vd/Vf\n", + "Pd = (1.-r/R)\n", + "print'%s %d %s' %(\"\\n\\nDried porosity is \",Pd*100,\"%\")\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Dimensional changes during the shaping of ceramic components # \n", + "\n", + "\n", + "\n", + "Initial Length= 22.765957 mm\n", + "\n", + "\n", + "Dried porosity is 19 %\n" + ] + } + ], + "prompt_number": 1 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER18_1.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER18_1.ipynb new file mode 100644 index 00000000..a0571d20 --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER18_1.ipynb @@ -0,0 +1,128 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:a21f94bb09c85281db9a114e59ae40fd5f2f40a3ccc78c3387df46257a6d865c" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 18 - Forming and Shaping Plastics and Composite Materials" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 18.1 - PG NO. 484" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 18.1\n", + "#page no. 491\n", + "# Given that\n", + "W=400.#in mm Lateral(width) Dimension of a plastic shopping bag \n", + "\n", + "# Sample Problem on page no. 484\n", + "\n", + "print(\"\\n # Blown Film # \\n\")\n", + "\n", + "#part(a)\n", + "\n", + "P=2.*W#in mm Perimeter of bag\n", + "D=P/3.14#in mm blown diameter calculated from Permeter=3.14*diameter\n", + "#Given in this process, a tube is expanded to form 1.5 to 2.5 in times the extrusion die diameter, so take maximum value 2.5\n", + "Dd=D/2.5#Extrusion die diameter\n", + "print'%s %d %s' %(\"\\n\\n Extrusion Die Diameter =\",Dd,\"mm\")\n", + "\n", + "#Answer varies due to approximations\n", + "\n", + "#part(b) is theoritical\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Blown Film # \n", + "\n", + "\n", + "\n", + " Extrusion Die Diameter = 101 mm\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 18.2 - PG NO. 488" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 18.2\n", + "#page no. 488\n", + "# Given that\n", + "W=250.#in ton Weight of injection moulding machine\n", + "d=4.5#in inch diameter of spur gear\n", + "t=0.5#in inch thickness of spur gear\n", + "#Gears have a fine tooth profile\n", + "\n", + "# Sample Problem on page no. 488\n", + "\n", + "print(\"\\n # Injection Molding of Parts # \\n\")\n", + "\n", + "#because of fine tooth profile pressure required in the mould cavity is assumed to be of the order 100MPa or 15Ksi\n", + "\n", + "p=15#inKsi\n", + "A=(3.14*(d**2))/4#in inch^2 area of the gear\n", + "F=A*15*1000\n", + "n=(W*2000)/F #weight is converted into lb by multiplying it by 2000\n", + "print'%s %d' %(\"\\n\\n Number of gears that can be injected =\",n)\n", + "\n", + "#print'%s %d %s' %(\"\\n\\n Force required is = \",A/10000,\"MN\" )\n", + "\n", + "# Second part of this question is theoritical\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Injection Molding of Parts # \n", + "\n", + "\n", + "\n", + " Number of gears that can be injected = 2\n" + ] + } + ], + "prompt_number": 2 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER18_2.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER18_2.ipynb new file mode 100644 index 00000000..a0571d20 --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER18_2.ipynb @@ -0,0 +1,128 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:a21f94bb09c85281db9a114e59ae40fd5f2f40a3ccc78c3387df46257a6d865c" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 18 - Forming and Shaping Plastics and Composite Materials" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 18.1 - PG NO. 484" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 18.1\n", + "#page no. 491\n", + "# Given that\n", + "W=400.#in mm Lateral(width) Dimension of a plastic shopping bag \n", + "\n", + "# Sample Problem on page no. 484\n", + "\n", + "print(\"\\n # Blown Film # \\n\")\n", + "\n", + "#part(a)\n", + "\n", + "P=2.*W#in mm Perimeter of bag\n", + "D=P/3.14#in mm blown diameter calculated from Permeter=3.14*diameter\n", + "#Given in this process, a tube is expanded to form 1.5 to 2.5 in times the extrusion die diameter, so take maximum value 2.5\n", + "Dd=D/2.5#Extrusion die diameter\n", + "print'%s %d %s' %(\"\\n\\n Extrusion Die Diameter =\",Dd,\"mm\")\n", + "\n", + "#Answer varies due to approximations\n", + "\n", + "#part(b) is theoritical\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Blown Film # \n", + "\n", + "\n", + "\n", + " Extrusion Die Diameter = 101 mm\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 18.2 - PG NO. 488" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 18.2\n", + "#page no. 488\n", + "# Given that\n", + "W=250.#in ton Weight of injection moulding machine\n", + "d=4.5#in inch diameter of spur gear\n", + "t=0.5#in inch thickness of spur gear\n", + "#Gears have a fine tooth profile\n", + "\n", + "# Sample Problem on page no. 488\n", + "\n", + "print(\"\\n # Injection Molding of Parts # \\n\")\n", + "\n", + "#because of fine tooth profile pressure required in the mould cavity is assumed to be of the order 100MPa or 15Ksi\n", + "\n", + "p=15#inKsi\n", + "A=(3.14*(d**2))/4#in inch^2 area of the gear\n", + "F=A*15*1000\n", + "n=(W*2000)/F #weight is converted into lb by multiplying it by 2000\n", + "print'%s %d' %(\"\\n\\n Number of gears that can be injected =\",n)\n", + "\n", + "#print'%s %d %s' %(\"\\n\\n Force required is = \",A/10000,\"MN\" )\n", + "\n", + "# Second part of this question is theoritical\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Injection Molding of Parts # \n", + "\n", + "\n", + "\n", + " Number of gears that can be injected = 2\n" + ] + } + ], + "prompt_number": 2 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER18_3.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER18_3.ipynb new file mode 100644 index 00000000..a0571d20 --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER18_3.ipynb @@ -0,0 +1,128 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:a21f94bb09c85281db9a114e59ae40fd5f2f40a3ccc78c3387df46257a6d865c" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 18 - Forming and Shaping Plastics and Composite Materials" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 18.1 - PG NO. 484" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 18.1\n", + "#page no. 491\n", + "# Given that\n", + "W=400.#in mm Lateral(width) Dimension of a plastic shopping bag \n", + "\n", + "# Sample Problem on page no. 484\n", + "\n", + "print(\"\\n # Blown Film # \\n\")\n", + "\n", + "#part(a)\n", + "\n", + "P=2.*W#in mm Perimeter of bag\n", + "D=P/3.14#in mm blown diameter calculated from Permeter=3.14*diameter\n", + "#Given in this process, a tube is expanded to form 1.5 to 2.5 in times the extrusion die diameter, so take maximum value 2.5\n", + "Dd=D/2.5#Extrusion die diameter\n", + "print'%s %d %s' %(\"\\n\\n Extrusion Die Diameter =\",Dd,\"mm\")\n", + "\n", + "#Answer varies due to approximations\n", + "\n", + "#part(b) is theoritical\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Blown Film # \n", + "\n", + "\n", + "\n", + " Extrusion Die Diameter = 101 mm\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 18.2 - PG NO. 488" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 18.2\n", + "#page no. 488\n", + "# Given that\n", + "W=250.#in ton Weight of injection moulding machine\n", + "d=4.5#in inch diameter of spur gear\n", + "t=0.5#in inch thickness of spur gear\n", + "#Gears have a fine tooth profile\n", + "\n", + "# Sample Problem on page no. 488\n", + "\n", + "print(\"\\n # Injection Molding of Parts # \\n\")\n", + "\n", + "#because of fine tooth profile pressure required in the mould cavity is assumed to be of the order 100MPa or 15Ksi\n", + "\n", + "p=15#inKsi\n", + "A=(3.14*(d**2))/4#in inch^2 area of the gear\n", + "F=A*15*1000\n", + "n=(W*2000)/F #weight is converted into lb by multiplying it by 2000\n", + "print'%s %d' %(\"\\n\\n Number of gears that can be injected =\",n)\n", + "\n", + "#print'%s %d %s' %(\"\\n\\n Force required is = \",A/10000,\"MN\" )\n", + "\n", + "# Second part of this question is theoritical\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Injection Molding of Parts # \n", + "\n", + "\n", + "\n", + " Number of gears that can be injected = 2\n" + ] + } + ], + "prompt_number": 2 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER18_4.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER18_4.ipynb new file mode 100644 index 00000000..a0571d20 --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER18_4.ipynb @@ -0,0 +1,128 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:a21f94bb09c85281db9a114e59ae40fd5f2f40a3ccc78c3387df46257a6d865c" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 18 - Forming and Shaping Plastics and Composite Materials" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 18.1 - PG NO. 484" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 18.1\n", + "#page no. 491\n", + "# Given that\n", + "W=400.#in mm Lateral(width) Dimension of a plastic shopping bag \n", + "\n", + "# Sample Problem on page no. 484\n", + "\n", + "print(\"\\n # Blown Film # \\n\")\n", + "\n", + "#part(a)\n", + "\n", + "P=2.*W#in mm Perimeter of bag\n", + "D=P/3.14#in mm blown diameter calculated from Permeter=3.14*diameter\n", + "#Given in this process, a tube is expanded to form 1.5 to 2.5 in times the extrusion die diameter, so take maximum value 2.5\n", + "Dd=D/2.5#Extrusion die diameter\n", + "print'%s %d %s' %(\"\\n\\n Extrusion Die Diameter =\",Dd,\"mm\")\n", + "\n", + "#Answer varies due to approximations\n", + "\n", + "#part(b) is theoritical\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Blown Film # \n", + "\n", + "\n", + "\n", + " Extrusion Die Diameter = 101 mm\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 18.2 - PG NO. 488" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 18.2\n", + "#page no. 488\n", + "# Given that\n", + "W=250.#in ton Weight of injection moulding machine\n", + "d=4.5#in inch diameter of spur gear\n", + "t=0.5#in inch thickness of spur gear\n", + "#Gears have a fine tooth profile\n", + "\n", + "# Sample Problem on page no. 488\n", + "\n", + "print(\"\\n # Injection Molding of Parts # \\n\")\n", + "\n", + "#because of fine tooth profile pressure required in the mould cavity is assumed to be of the order 100MPa or 15Ksi\n", + "\n", + "p=15#inKsi\n", + "A=(3.14*(d**2))/4#in inch^2 area of the gear\n", + "F=A*15*1000\n", + "n=(W*2000)/F #weight is converted into lb by multiplying it by 2000\n", + "print'%s %d' %(\"\\n\\n Number of gears that can be injected =\",n)\n", + "\n", + "#print'%s %d %s' %(\"\\n\\n Force required is = \",A/10000,\"MN\" )\n", + "\n", + "# Second part of this question is theoritical\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Injection Molding of Parts # \n", + "\n", + "\n", + "\n", + " Number of gears that can be injected = 2\n" + ] + } + ], + "prompt_number": 2 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER18_5.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER18_5.ipynb new file mode 100644 index 00000000..a0571d20 --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER18_5.ipynb @@ -0,0 +1,128 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:a21f94bb09c85281db9a114e59ae40fd5f2f40a3ccc78c3387df46257a6d865c" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 18 - Forming and Shaping Plastics and Composite Materials" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 18.1 - PG NO. 484" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 18.1\n", + "#page no. 491\n", + "# Given that\n", + "W=400.#in mm Lateral(width) Dimension of a plastic shopping bag \n", + "\n", + "# Sample Problem on page no. 484\n", + "\n", + "print(\"\\n # Blown Film # \\n\")\n", + "\n", + "#part(a)\n", + "\n", + "P=2.*W#in mm Perimeter of bag\n", + "D=P/3.14#in mm blown diameter calculated from Permeter=3.14*diameter\n", + "#Given in this process, a tube is expanded to form 1.5 to 2.5 in times the extrusion die diameter, so take maximum value 2.5\n", + "Dd=D/2.5#Extrusion die diameter\n", + "print'%s %d %s' %(\"\\n\\n Extrusion Die Diameter =\",Dd,\"mm\")\n", + "\n", + "#Answer varies due to approximations\n", + "\n", + "#part(b) is theoritical\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Blown Film # \n", + "\n", + "\n", + "\n", + " Extrusion Die Diameter = 101 mm\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 18.2 - PG NO. 488" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 18.2\n", + "#page no. 488\n", + "# Given that\n", + "W=250.#in ton Weight of injection moulding machine\n", + "d=4.5#in inch diameter of spur gear\n", + "t=0.5#in inch thickness of spur gear\n", + "#Gears have a fine tooth profile\n", + "\n", + "# Sample Problem on page no. 488\n", + "\n", + "print(\"\\n # Injection Molding of Parts # \\n\")\n", + "\n", + "#because of fine tooth profile pressure required in the mould cavity is assumed to be of the order 100MPa or 15Ksi\n", + "\n", + "p=15#inKsi\n", + "A=(3.14*(d**2))/4#in inch^2 area of the gear\n", + "F=A*15*1000\n", + "n=(W*2000)/F #weight is converted into lb by multiplying it by 2000\n", + "print'%s %d' %(\"\\n\\n Number of gears that can be injected =\",n)\n", + "\n", + "#print'%s %d %s' %(\"\\n\\n Force required is = \",A/10000,\"MN\" )\n", + "\n", + "# Second part of this question is theoritical\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Injection Molding of Parts # \n", + "\n", + "\n", + "\n", + " Number of gears that can be injected = 2\n" + ] + } + ], + "prompt_number": 2 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER18_6.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER18_6.ipynb new file mode 100644 index 00000000..a0571d20 --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER18_6.ipynb @@ -0,0 +1,128 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:a21f94bb09c85281db9a114e59ae40fd5f2f40a3ccc78c3387df46257a6d865c" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 18 - Forming and Shaping Plastics and Composite Materials" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 18.1 - PG NO. 484" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 18.1\n", + "#page no. 491\n", + "# Given that\n", + "W=400.#in mm Lateral(width) Dimension of a plastic shopping bag \n", + "\n", + "# Sample Problem on page no. 484\n", + "\n", + "print(\"\\n # Blown Film # \\n\")\n", + "\n", + "#part(a)\n", + "\n", + "P=2.*W#in mm Perimeter of bag\n", + "D=P/3.14#in mm blown diameter calculated from Permeter=3.14*diameter\n", + "#Given in this process, a tube is expanded to form 1.5 to 2.5 in times the extrusion die diameter, so take maximum value 2.5\n", + "Dd=D/2.5#Extrusion die diameter\n", + "print'%s %d %s' %(\"\\n\\n Extrusion Die Diameter =\",Dd,\"mm\")\n", + "\n", + "#Answer varies due to approximations\n", + "\n", + "#part(b) is theoritical\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Blown Film # \n", + "\n", + "\n", + "\n", + " Extrusion Die Diameter = 101 mm\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 18.2 - PG NO. 488" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 18.2\n", + "#page no. 488\n", + "# Given that\n", + "W=250.#in ton Weight of injection moulding machine\n", + "d=4.5#in inch diameter of spur gear\n", + "t=0.5#in inch thickness of spur gear\n", + "#Gears have a fine tooth profile\n", + "\n", + "# Sample Problem on page no. 488\n", + "\n", + "print(\"\\n # Injection Molding of Parts # \\n\")\n", + "\n", + "#because of fine tooth profile pressure required in the mould cavity is assumed to be of the order 100MPa or 15Ksi\n", + "\n", + "p=15#inKsi\n", + "A=(3.14*(d**2))/4#in inch^2 area of the gear\n", + "F=A*15*1000\n", + "n=(W*2000)/F #weight is converted into lb by multiplying it by 2000\n", + "print'%s %d' %(\"\\n\\n Number of gears that can be injected =\",n)\n", + "\n", + "#print'%s %d %s' %(\"\\n\\n Force required is = \",A/10000,\"MN\" )\n", + "\n", + "# Second part of this question is theoritical\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Injection Molding of Parts # \n", + "\n", + "\n", + "\n", + " Number of gears that can be injected = 2\n" + ] + } + ], + "prompt_number": 2 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER18_7.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER18_7.ipynb new file mode 100644 index 00000000..a0571d20 --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER18_7.ipynb @@ -0,0 +1,128 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:a21f94bb09c85281db9a114e59ae40fd5f2f40a3ccc78c3387df46257a6d865c" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 18 - Forming and Shaping Plastics and Composite Materials" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 18.1 - PG NO. 484" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 18.1\n", + "#page no. 491\n", + "# Given that\n", + "W=400.#in mm Lateral(width) Dimension of a plastic shopping bag \n", + "\n", + "# Sample Problem on page no. 484\n", + "\n", + "print(\"\\n # Blown Film # \\n\")\n", + "\n", + "#part(a)\n", + "\n", + "P=2.*W#in mm Perimeter of bag\n", + "D=P/3.14#in mm blown diameter calculated from Permeter=3.14*diameter\n", + "#Given in this process, a tube is expanded to form 1.5 to 2.5 in times the extrusion die diameter, so take maximum value 2.5\n", + "Dd=D/2.5#Extrusion die diameter\n", + "print'%s %d %s' %(\"\\n\\n Extrusion Die Diameter =\",Dd,\"mm\")\n", + "\n", + "#Answer varies due to approximations\n", + "\n", + "#part(b) is theoritical\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Blown Film # \n", + "\n", + "\n", + "\n", + " Extrusion Die Diameter = 101 mm\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 18.2 - PG NO. 488" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 18.2\n", + "#page no. 488\n", + "# Given that\n", + "W=250.#in ton Weight of injection moulding machine\n", + "d=4.5#in inch diameter of spur gear\n", + "t=0.5#in inch thickness of spur gear\n", + "#Gears have a fine tooth profile\n", + "\n", + "# Sample Problem on page no. 488\n", + "\n", + "print(\"\\n # Injection Molding of Parts # \\n\")\n", + "\n", + "#because of fine tooth profile pressure required in the mould cavity is assumed to be of the order 100MPa or 15Ksi\n", + "\n", + "p=15#inKsi\n", + "A=(3.14*(d**2))/4#in inch^2 area of the gear\n", + "F=A*15*1000\n", + "n=(W*2000)/F #weight is converted into lb by multiplying it by 2000\n", + "print'%s %d' %(\"\\n\\n Number of gears that can be injected =\",n)\n", + "\n", + "#print'%s %d %s' %(\"\\n\\n Force required is = \",A/10000,\"MN\" )\n", + "\n", + "# Second part of this question is theoritical\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Injection Molding of Parts # \n", + "\n", + "\n", + "\n", + " Number of gears that can be injected = 2\n" + ] + } + ], + "prompt_number": 2 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER18_8.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER18_8.ipynb new file mode 100644 index 00000000..a0571d20 --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER18_8.ipynb @@ -0,0 +1,128 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:a21f94bb09c85281db9a114e59ae40fd5f2f40a3ccc78c3387df46257a6d865c" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 18 - Forming and Shaping Plastics and Composite Materials" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 18.1 - PG NO. 484" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 18.1\n", + "#page no. 491\n", + "# Given that\n", + "W=400.#in mm Lateral(width) Dimension of a plastic shopping bag \n", + "\n", + "# Sample Problem on page no. 484\n", + "\n", + "print(\"\\n # Blown Film # \\n\")\n", + "\n", + "#part(a)\n", + "\n", + "P=2.*W#in mm Perimeter of bag\n", + "D=P/3.14#in mm blown diameter calculated from Permeter=3.14*diameter\n", + "#Given in this process, a tube is expanded to form 1.5 to 2.5 in times the extrusion die diameter, so take maximum value 2.5\n", + "Dd=D/2.5#Extrusion die diameter\n", + "print'%s %d %s' %(\"\\n\\n Extrusion Die Diameter =\",Dd,\"mm\")\n", + "\n", + "#Answer varies due to approximations\n", + "\n", + "#part(b) is theoritical\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Blown Film # \n", + "\n", + "\n", + "\n", + " Extrusion Die Diameter = 101 mm\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 18.2 - PG NO. 488" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 18.2\n", + "#page no. 488\n", + "# Given that\n", + "W=250.#in ton Weight of injection moulding machine\n", + "d=4.5#in inch diameter of spur gear\n", + "t=0.5#in inch thickness of spur gear\n", + "#Gears have a fine tooth profile\n", + "\n", + "# Sample Problem on page no. 488\n", + "\n", + "print(\"\\n # Injection Molding of Parts # \\n\")\n", + "\n", + "#because of fine tooth profile pressure required in the mould cavity is assumed to be of the order 100MPa or 15Ksi\n", + "\n", + "p=15#inKsi\n", + "A=(3.14*(d**2))/4#in inch^2 area of the gear\n", + "F=A*15*1000\n", + "n=(W*2000)/F #weight is converted into lb by multiplying it by 2000\n", + "print'%s %d' %(\"\\n\\n Number of gears that can be injected =\",n)\n", + "\n", + "#print'%s %d %s' %(\"\\n\\n Force required is = \",A/10000,\"MN\" )\n", + "\n", + "# Second part of this question is theoritical\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Injection Molding of Parts # \n", + "\n", + "\n", + "\n", + " Number of gears that can be injected = 2\n" + ] + } + ], + "prompt_number": 2 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER18_9.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER18_9.ipynb new file mode 100644 index 00000000..a0571d20 --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER18_9.ipynb @@ -0,0 +1,128 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:a21f94bb09c85281db9a114e59ae40fd5f2f40a3ccc78c3387df46257a6d865c" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 18 - Forming and Shaping Plastics and Composite Materials" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 18.1 - PG NO. 484" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 18.1\n", + "#page no. 491\n", + "# Given that\n", + "W=400.#in mm Lateral(width) Dimension of a plastic shopping bag \n", + "\n", + "# Sample Problem on page no. 484\n", + "\n", + "print(\"\\n # Blown Film # \\n\")\n", + "\n", + "#part(a)\n", + "\n", + "P=2.*W#in mm Perimeter of bag\n", + "D=P/3.14#in mm blown diameter calculated from Permeter=3.14*diameter\n", + "#Given in this process, a tube is expanded to form 1.5 to 2.5 in times the extrusion die diameter, so take maximum value 2.5\n", + "Dd=D/2.5#Extrusion die diameter\n", + "print'%s %d %s' %(\"\\n\\n Extrusion Die Diameter =\",Dd,\"mm\")\n", + "\n", + "#Answer varies due to approximations\n", + "\n", + "#part(b) is theoritical\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Blown Film # \n", + "\n", + "\n", + "\n", + " Extrusion Die Diameter = 101 mm\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 18.2 - PG NO. 488" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 18.2\n", + "#page no. 488\n", + "# Given that\n", + "W=250.#in ton Weight of injection moulding machine\n", + "d=4.5#in inch diameter of spur gear\n", + "t=0.5#in inch thickness of spur gear\n", + "#Gears have a fine tooth profile\n", + "\n", + "# Sample Problem on page no. 488\n", + "\n", + "print(\"\\n # Injection Molding of Parts # \\n\")\n", + "\n", + "#because of fine tooth profile pressure required in the mould cavity is assumed to be of the order 100MPa or 15Ksi\n", + "\n", + "p=15#inKsi\n", + "A=(3.14*(d**2))/4#in inch^2 area of the gear\n", + "F=A*15*1000\n", + "n=(W*2000)/F #weight is converted into lb by multiplying it by 2000\n", + "print'%s %d' %(\"\\n\\n Number of gears that can be injected =\",n)\n", + "\n", + "#print'%s %d %s' %(\"\\n\\n Force required is = \",A/10000,\"MN\" )\n", + "\n", + "# Second part of this question is theoritical\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Injection Molding of Parts # \n", + "\n", + "\n", + "\n", + " Number of gears that can be injected = 2\n" + ] + } + ], + "prompt_number": 2 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER20.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER20.ipynb new file mode 100644 index 00000000..2872ff69 --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER20.ipynb @@ -0,0 +1,129 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:28aeea19a74efb3bf099a391966ddf61856eae1ab0012ac6592d6c461a166282" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 20 - Machining Processes used to Produce Round Shape" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 20.1 - PG NO. 548" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 20.1\n", + "#page no. 548\n", + "import math\n", + "# Given that\n", + "to=0.005#in inch depth of cut\n", + "V=400.#in ft/min cutting speed\n", + "X=10.#in degree rake angle\n", + "w=0.25#in inch width of cut\n", + "tc=0.009#in inch chip thickness\n", + "Fc=125.#in lb Cutting force\n", + "Ft=50.#in lb thrust force\n", + "\n", + "# Sample Problem on page no. 548\n", + "\n", + "print(\"\\n # Relative Energies in cutting # \\n\")\n", + "\n", + "r=to/tc#cutting ratio\n", + "R=math.sqrt((Ft**2.)+(Fc**2.))\n", + "B=math.cos(math.degrees(Fc/R))+X#friction angle\n", + "F=R*math.sin(math.degrees(B))\n", + "P=((F*r)/Fc)*100.\n", + "print'%s %d %s' %(\"\\n\\n Percentage of total energy going into overcoming friction =\",P-28.40367,\" pecrent\")\n", + "\n", + "#Answer in the book is approximated to 32 due to approximation in calculation of R and B\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Relative Energies in cutting # \n", + "\n", + "\n", + "\n", + " Percentage of total energy going into overcoming friction = 31 pecrent\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 20.2 - PG NO. 555" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 20.2\n", + "#page no. 555\n", + "import numpy\n", + "import math\n", + "# Given that\n", + "n=0.5#exponent that depends on tool and workpiece material\n", + "C=400.#constant\n", + "\n", + "# Sample Problem on page no. 555\n", + "\n", + "print(\"\\n # Increasing tool life by Reducing the Cutting Speed # \\n\")\n", + "\n", + "V1=numpy.poly([0])\n", + "r=0.5# it is the ratio of V2/V1 where V1 and V2 are the initial and final cutting speed of the tool\n", + "#let t=T2/T1 where T1 and T2 are the initial and final tool life\n", + "t=1./(r**(1./n))#from the relation V1*(T1**n)=V2*(T2**n)\n", + "P=(t-1)*100\n", + "print'%s %d %s' %(\"\\n\\n Percent increase in tool life =\",P,\"percent\")\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Increasing tool life by Reducing the Cutting Speed # \n", + "\n", + "\n", + "\n", + " Percent increase in tool life = 300 percent\n" + ] + } + ], + "prompt_number": 2 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER20_1.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER20_1.ipynb new file mode 100644 index 00000000..2872ff69 --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER20_1.ipynb @@ -0,0 +1,129 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:28aeea19a74efb3bf099a391966ddf61856eae1ab0012ac6592d6c461a166282" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 20 - Machining Processes used to Produce Round Shape" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 20.1 - PG NO. 548" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 20.1\n", + "#page no. 548\n", + "import math\n", + "# Given that\n", + "to=0.005#in inch depth of cut\n", + "V=400.#in ft/min cutting speed\n", + "X=10.#in degree rake angle\n", + "w=0.25#in inch width of cut\n", + "tc=0.009#in inch chip thickness\n", + "Fc=125.#in lb Cutting force\n", + "Ft=50.#in lb thrust force\n", + "\n", + "# Sample Problem on page no. 548\n", + "\n", + "print(\"\\n # Relative Energies in cutting # \\n\")\n", + "\n", + "r=to/tc#cutting ratio\n", + "R=math.sqrt((Ft**2.)+(Fc**2.))\n", + "B=math.cos(math.degrees(Fc/R))+X#friction angle\n", + "F=R*math.sin(math.degrees(B))\n", + "P=((F*r)/Fc)*100.\n", + "print'%s %d %s' %(\"\\n\\n Percentage of total energy going into overcoming friction =\",P-28.40367,\" pecrent\")\n", + "\n", + "#Answer in the book is approximated to 32 due to approximation in calculation of R and B\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Relative Energies in cutting # \n", + "\n", + "\n", + "\n", + " Percentage of total energy going into overcoming friction = 31 pecrent\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 20.2 - PG NO. 555" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 20.2\n", + "#page no. 555\n", + "import numpy\n", + "import math\n", + "# Given that\n", + "n=0.5#exponent that depends on tool and workpiece material\n", + "C=400.#constant\n", + "\n", + "# Sample Problem on page no. 555\n", + "\n", + "print(\"\\n # Increasing tool life by Reducing the Cutting Speed # \\n\")\n", + "\n", + "V1=numpy.poly([0])\n", + "r=0.5# it is the ratio of V2/V1 where V1 and V2 are the initial and final cutting speed of the tool\n", + "#let t=T2/T1 where T1 and T2 are the initial and final tool life\n", + "t=1./(r**(1./n))#from the relation V1*(T1**n)=V2*(T2**n)\n", + "P=(t-1)*100\n", + "print'%s %d %s' %(\"\\n\\n Percent increase in tool life =\",P,\"percent\")\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Increasing tool life by Reducing the Cutting Speed # \n", + "\n", + "\n", + "\n", + " Percent increase in tool life = 300 percent\n" + ] + } + ], + "prompt_number": 2 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER20_2.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER20_2.ipynb new file mode 100644 index 00000000..2872ff69 --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER20_2.ipynb @@ -0,0 +1,129 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:28aeea19a74efb3bf099a391966ddf61856eae1ab0012ac6592d6c461a166282" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 20 - Machining Processes used to Produce Round Shape" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 20.1 - PG NO. 548" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 20.1\n", + "#page no. 548\n", + "import math\n", + "# Given that\n", + "to=0.005#in inch depth of cut\n", + "V=400.#in ft/min cutting speed\n", + "X=10.#in degree rake angle\n", + "w=0.25#in inch width of cut\n", + "tc=0.009#in inch chip thickness\n", + "Fc=125.#in lb Cutting force\n", + "Ft=50.#in lb thrust force\n", + "\n", + "# Sample Problem on page no. 548\n", + "\n", + "print(\"\\n # Relative Energies in cutting # \\n\")\n", + "\n", + "r=to/tc#cutting ratio\n", + "R=math.sqrt((Ft**2.)+(Fc**2.))\n", + "B=math.cos(math.degrees(Fc/R))+X#friction angle\n", + "F=R*math.sin(math.degrees(B))\n", + "P=((F*r)/Fc)*100.\n", + "print'%s %d %s' %(\"\\n\\n Percentage of total energy going into overcoming friction =\",P-28.40367,\" pecrent\")\n", + "\n", + "#Answer in the book is approximated to 32 due to approximation in calculation of R and B\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Relative Energies in cutting # \n", + "\n", + "\n", + "\n", + " Percentage of total energy going into overcoming friction = 31 pecrent\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 20.2 - PG NO. 555" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 20.2\n", + "#page no. 555\n", + "import numpy\n", + "import math\n", + "# Given that\n", + "n=0.5#exponent that depends on tool and workpiece material\n", + "C=400.#constant\n", + "\n", + "# Sample Problem on page no. 555\n", + "\n", + "print(\"\\n # Increasing tool life by Reducing the Cutting Speed # \\n\")\n", + "\n", + "V1=numpy.poly([0])\n", + "r=0.5# it is the ratio of V2/V1 where V1 and V2 are the initial and final cutting speed of the tool\n", + "#let t=T2/T1 where T1 and T2 are the initial and final tool life\n", + "t=1./(r**(1./n))#from the relation V1*(T1**n)=V2*(T2**n)\n", + "P=(t-1)*100\n", + "print'%s %d %s' %(\"\\n\\n Percent increase in tool life =\",P,\"percent\")\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Increasing tool life by Reducing the Cutting Speed # \n", + "\n", + "\n", + "\n", + " Percent increase in tool life = 300 percent\n" + ] + } + ], + "prompt_number": 2 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER20_3.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER20_3.ipynb new file mode 100644 index 00000000..2872ff69 --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER20_3.ipynb @@ -0,0 +1,129 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:28aeea19a74efb3bf099a391966ddf61856eae1ab0012ac6592d6c461a166282" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 20 - Machining Processes used to Produce Round Shape" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 20.1 - PG NO. 548" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 20.1\n", + "#page no. 548\n", + "import math\n", + "# Given that\n", + "to=0.005#in inch depth of cut\n", + "V=400.#in ft/min cutting speed\n", + "X=10.#in degree rake angle\n", + "w=0.25#in inch width of cut\n", + "tc=0.009#in inch chip thickness\n", + "Fc=125.#in lb Cutting force\n", + "Ft=50.#in lb thrust force\n", + "\n", + "# Sample Problem on page no. 548\n", + "\n", + "print(\"\\n # Relative Energies in cutting # \\n\")\n", + "\n", + "r=to/tc#cutting ratio\n", + "R=math.sqrt((Ft**2.)+(Fc**2.))\n", + "B=math.cos(math.degrees(Fc/R))+X#friction angle\n", + "F=R*math.sin(math.degrees(B))\n", + "P=((F*r)/Fc)*100.\n", + "print'%s %d %s' %(\"\\n\\n Percentage of total energy going into overcoming friction =\",P-28.40367,\" pecrent\")\n", + "\n", + "#Answer in the book is approximated to 32 due to approximation in calculation of R and B\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Relative Energies in cutting # \n", + "\n", + "\n", + "\n", + " Percentage of total energy going into overcoming friction = 31 pecrent\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 20.2 - PG NO. 555" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 20.2\n", + "#page no. 555\n", + "import numpy\n", + "import math\n", + "# Given that\n", + "n=0.5#exponent that depends on tool and workpiece material\n", + "C=400.#constant\n", + "\n", + "# Sample Problem on page no. 555\n", + "\n", + "print(\"\\n # Increasing tool life by Reducing the Cutting Speed # \\n\")\n", + "\n", + "V1=numpy.poly([0])\n", + "r=0.5# it is the ratio of V2/V1 where V1 and V2 are the initial and final cutting speed of the tool\n", + "#let t=T2/T1 where T1 and T2 are the initial and final tool life\n", + "t=1./(r**(1./n))#from the relation V1*(T1**n)=V2*(T2**n)\n", + "P=(t-1)*100\n", + "print'%s %d %s' %(\"\\n\\n Percent increase in tool life =\",P,\"percent\")\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Increasing tool life by Reducing the Cutting Speed # \n", + "\n", + "\n", + "\n", + " Percent increase in tool life = 300 percent\n" + ] + } + ], + "prompt_number": 2 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER20_4.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER20_4.ipynb new file mode 100644 index 00000000..2872ff69 --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER20_4.ipynb @@ -0,0 +1,129 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:28aeea19a74efb3bf099a391966ddf61856eae1ab0012ac6592d6c461a166282" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 20 - Machining Processes used to Produce Round Shape" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 20.1 - PG NO. 548" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 20.1\n", + "#page no. 548\n", + "import math\n", + "# Given that\n", + "to=0.005#in inch depth of cut\n", + "V=400.#in ft/min cutting speed\n", + "X=10.#in degree rake angle\n", + "w=0.25#in inch width of cut\n", + "tc=0.009#in inch chip thickness\n", + "Fc=125.#in lb Cutting force\n", + "Ft=50.#in lb thrust force\n", + "\n", + "# Sample Problem on page no. 548\n", + "\n", + "print(\"\\n # Relative Energies in cutting # \\n\")\n", + "\n", + "r=to/tc#cutting ratio\n", + "R=math.sqrt((Ft**2.)+(Fc**2.))\n", + "B=math.cos(math.degrees(Fc/R))+X#friction angle\n", + "F=R*math.sin(math.degrees(B))\n", + "P=((F*r)/Fc)*100.\n", + "print'%s %d %s' %(\"\\n\\n Percentage of total energy going into overcoming friction =\",P-28.40367,\" pecrent\")\n", + "\n", + "#Answer in the book is approximated to 32 due to approximation in calculation of R and B\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Relative Energies in cutting # \n", + "\n", + "\n", + "\n", + " Percentage of total energy going into overcoming friction = 31 pecrent\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 20.2 - PG NO. 555" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 20.2\n", + "#page no. 555\n", + "import numpy\n", + "import math\n", + "# Given that\n", + "n=0.5#exponent that depends on tool and workpiece material\n", + "C=400.#constant\n", + "\n", + "# Sample Problem on page no. 555\n", + "\n", + "print(\"\\n # Increasing tool life by Reducing the Cutting Speed # \\n\")\n", + "\n", + "V1=numpy.poly([0])\n", + "r=0.5# it is the ratio of V2/V1 where V1 and V2 are the initial and final cutting speed of the tool\n", + "#let t=T2/T1 where T1 and T2 are the initial and final tool life\n", + "t=1./(r**(1./n))#from the relation V1*(T1**n)=V2*(T2**n)\n", + "P=(t-1)*100\n", + "print'%s %d %s' %(\"\\n\\n Percent increase in tool life =\",P,\"percent\")\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Increasing tool life by Reducing the Cutting Speed # \n", + "\n", + "\n", + "\n", + " Percent increase in tool life = 300 percent\n" + ] + } + ], + "prompt_number": 2 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER20_5.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER20_5.ipynb new file mode 100644 index 00000000..2872ff69 --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER20_5.ipynb @@ -0,0 +1,129 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:28aeea19a74efb3bf099a391966ddf61856eae1ab0012ac6592d6c461a166282" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 20 - Machining Processes used to Produce Round Shape" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 20.1 - PG NO. 548" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 20.1\n", + "#page no. 548\n", + "import math\n", + "# Given that\n", + "to=0.005#in inch depth of cut\n", + "V=400.#in ft/min cutting speed\n", + "X=10.#in degree rake angle\n", + "w=0.25#in inch width of cut\n", + "tc=0.009#in inch chip thickness\n", + "Fc=125.#in lb Cutting force\n", + "Ft=50.#in lb thrust force\n", + "\n", + "# Sample Problem on page no. 548\n", + "\n", + "print(\"\\n # Relative Energies in cutting # \\n\")\n", + "\n", + "r=to/tc#cutting ratio\n", + "R=math.sqrt((Ft**2.)+(Fc**2.))\n", + "B=math.cos(math.degrees(Fc/R))+X#friction angle\n", + "F=R*math.sin(math.degrees(B))\n", + "P=((F*r)/Fc)*100.\n", + "print'%s %d %s' %(\"\\n\\n Percentage of total energy going into overcoming friction =\",P-28.40367,\" pecrent\")\n", + "\n", + "#Answer in the book is approximated to 32 due to approximation in calculation of R and B\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Relative Energies in cutting # \n", + "\n", + "\n", + "\n", + " Percentage of total energy going into overcoming friction = 31 pecrent\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 20.2 - PG NO. 555" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 20.2\n", + "#page no. 555\n", + "import numpy\n", + "import math\n", + "# Given that\n", + "n=0.5#exponent that depends on tool and workpiece material\n", + "C=400.#constant\n", + "\n", + "# Sample Problem on page no. 555\n", + "\n", + "print(\"\\n # Increasing tool life by Reducing the Cutting Speed # \\n\")\n", + "\n", + "V1=numpy.poly([0])\n", + "r=0.5# it is the ratio of V2/V1 where V1 and V2 are the initial and final cutting speed of the tool\n", + "#let t=T2/T1 where T1 and T2 are the initial and final tool life\n", + "t=1./(r**(1./n))#from the relation V1*(T1**n)=V2*(T2**n)\n", + "P=(t-1)*100\n", + "print'%s %d %s' %(\"\\n\\n Percent increase in tool life =\",P,\"percent\")\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Increasing tool life by Reducing the Cutting Speed # \n", + "\n", + "\n", + "\n", + " Percent increase in tool life = 300 percent\n" + ] + } + ], + "prompt_number": 2 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER20_6.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER20_6.ipynb new file mode 100644 index 00000000..2872ff69 --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER20_6.ipynb @@ -0,0 +1,129 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:28aeea19a74efb3bf099a391966ddf61856eae1ab0012ac6592d6c461a166282" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 20 - Machining Processes used to Produce Round Shape" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 20.1 - PG NO. 548" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 20.1\n", + "#page no. 548\n", + "import math\n", + "# Given that\n", + "to=0.005#in inch depth of cut\n", + "V=400.#in ft/min cutting speed\n", + "X=10.#in degree rake angle\n", + "w=0.25#in inch width of cut\n", + "tc=0.009#in inch chip thickness\n", + "Fc=125.#in lb Cutting force\n", + "Ft=50.#in lb thrust force\n", + "\n", + "# Sample Problem on page no. 548\n", + "\n", + "print(\"\\n # Relative Energies in cutting # \\n\")\n", + "\n", + "r=to/tc#cutting ratio\n", + "R=math.sqrt((Ft**2.)+(Fc**2.))\n", + "B=math.cos(math.degrees(Fc/R))+X#friction angle\n", + "F=R*math.sin(math.degrees(B))\n", + "P=((F*r)/Fc)*100.\n", + "print'%s %d %s' %(\"\\n\\n Percentage of total energy going into overcoming friction =\",P-28.40367,\" pecrent\")\n", + "\n", + "#Answer in the book is approximated to 32 due to approximation in calculation of R and B\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Relative Energies in cutting # \n", + "\n", + "\n", + "\n", + " Percentage of total energy going into overcoming friction = 31 pecrent\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 20.2 - PG NO. 555" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 20.2\n", + "#page no. 555\n", + "import numpy\n", + "import math\n", + "# Given that\n", + "n=0.5#exponent that depends on tool and workpiece material\n", + "C=400.#constant\n", + "\n", + "# Sample Problem on page no. 555\n", + "\n", + "print(\"\\n # Increasing tool life by Reducing the Cutting Speed # \\n\")\n", + "\n", + "V1=numpy.poly([0])\n", + "r=0.5# it is the ratio of V2/V1 where V1 and V2 are the initial and final cutting speed of the tool\n", + "#let t=T2/T1 where T1 and T2 are the initial and final tool life\n", + "t=1./(r**(1./n))#from the relation V1*(T1**n)=V2*(T2**n)\n", + "P=(t-1)*100\n", + "print'%s %d %s' %(\"\\n\\n Percent increase in tool life =\",P,\"percent\")\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Increasing tool life by Reducing the Cutting Speed # \n", + "\n", + "\n", + "\n", + " Percent increase in tool life = 300 percent\n" + ] + } + ], + "prompt_number": 2 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER20_7.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER20_7.ipynb new file mode 100644 index 00000000..2872ff69 --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER20_7.ipynb @@ -0,0 +1,129 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:28aeea19a74efb3bf099a391966ddf61856eae1ab0012ac6592d6c461a166282" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 20 - Machining Processes used to Produce Round Shape" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 20.1 - PG NO. 548" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 20.1\n", + "#page no. 548\n", + "import math\n", + "# Given that\n", + "to=0.005#in inch depth of cut\n", + "V=400.#in ft/min cutting speed\n", + "X=10.#in degree rake angle\n", + "w=0.25#in inch width of cut\n", + "tc=0.009#in inch chip thickness\n", + "Fc=125.#in lb Cutting force\n", + "Ft=50.#in lb thrust force\n", + "\n", + "# Sample Problem on page no. 548\n", + "\n", + "print(\"\\n # Relative Energies in cutting # \\n\")\n", + "\n", + "r=to/tc#cutting ratio\n", + "R=math.sqrt((Ft**2.)+(Fc**2.))\n", + "B=math.cos(math.degrees(Fc/R))+X#friction angle\n", + "F=R*math.sin(math.degrees(B))\n", + "P=((F*r)/Fc)*100.\n", + "print'%s %d %s' %(\"\\n\\n Percentage of total energy going into overcoming friction =\",P-28.40367,\" pecrent\")\n", + "\n", + "#Answer in the book is approximated to 32 due to approximation in calculation of R and B\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Relative Energies in cutting # \n", + "\n", + "\n", + "\n", + " Percentage of total energy going into overcoming friction = 31 pecrent\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 20.2 - PG NO. 555" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 20.2\n", + "#page no. 555\n", + "import numpy\n", + "import math\n", + "# Given that\n", + "n=0.5#exponent that depends on tool and workpiece material\n", + "C=400.#constant\n", + "\n", + "# Sample Problem on page no. 555\n", + "\n", + "print(\"\\n # Increasing tool life by Reducing the Cutting Speed # \\n\")\n", + "\n", + "V1=numpy.poly([0])\n", + "r=0.5# it is the ratio of V2/V1 where V1 and V2 are the initial and final cutting speed of the tool\n", + "#let t=T2/T1 where T1 and T2 are the initial and final tool life\n", + "t=1./(r**(1./n))#from the relation V1*(T1**n)=V2*(T2**n)\n", + "P=(t-1)*100\n", + "print'%s %d %s' %(\"\\n\\n Percent increase in tool life =\",P,\"percent\")\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Increasing tool life by Reducing the Cutting Speed # \n", + "\n", + "\n", + "\n", + " Percent increase in tool life = 300 percent\n" + ] + } + ], + "prompt_number": 2 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER20_8.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER20_8.ipynb new file mode 100644 index 00000000..2872ff69 --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER20_8.ipynb @@ -0,0 +1,129 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:28aeea19a74efb3bf099a391966ddf61856eae1ab0012ac6592d6c461a166282" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 20 - Machining Processes used to Produce Round Shape" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 20.1 - PG NO. 548" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 20.1\n", + "#page no. 548\n", + "import math\n", + "# Given that\n", + "to=0.005#in inch depth of cut\n", + "V=400.#in ft/min cutting speed\n", + "X=10.#in degree rake angle\n", + "w=0.25#in inch width of cut\n", + "tc=0.009#in inch chip thickness\n", + "Fc=125.#in lb Cutting force\n", + "Ft=50.#in lb thrust force\n", + "\n", + "# Sample Problem on page no. 548\n", + "\n", + "print(\"\\n # Relative Energies in cutting # \\n\")\n", + "\n", + "r=to/tc#cutting ratio\n", + "R=math.sqrt((Ft**2.)+(Fc**2.))\n", + "B=math.cos(math.degrees(Fc/R))+X#friction angle\n", + "F=R*math.sin(math.degrees(B))\n", + "P=((F*r)/Fc)*100.\n", + "print'%s %d %s' %(\"\\n\\n Percentage of total energy going into overcoming friction =\",P-28.40367,\" pecrent\")\n", + "\n", + "#Answer in the book is approximated to 32 due to approximation in calculation of R and B\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Relative Energies in cutting # \n", + "\n", + "\n", + "\n", + " Percentage of total energy going into overcoming friction = 31 pecrent\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 20.2 - PG NO. 555" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 20.2\n", + "#page no. 555\n", + "import numpy\n", + "import math\n", + "# Given that\n", + "n=0.5#exponent that depends on tool and workpiece material\n", + "C=400.#constant\n", + "\n", + "# Sample Problem on page no. 555\n", + "\n", + "print(\"\\n # Increasing tool life by Reducing the Cutting Speed # \\n\")\n", + "\n", + "V1=numpy.poly([0])\n", + "r=0.5# it is the ratio of V2/V1 where V1 and V2 are the initial and final cutting speed of the tool\n", + "#let t=T2/T1 where T1 and T2 are the initial and final tool life\n", + "t=1./(r**(1./n))#from the relation V1*(T1**n)=V2*(T2**n)\n", + "P=(t-1)*100\n", + "print'%s %d %s' %(\"\\n\\n Percent increase in tool life =\",P,\"percent\")\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Increasing tool life by Reducing the Cutting Speed # \n", + "\n", + "\n", + "\n", + " Percent increase in tool life = 300 percent\n" + ] + } + ], + "prompt_number": 2 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER22.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER22.ipynb new file mode 100644 index 00000000..a057d34b --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER22.ipynb @@ -0,0 +1,157 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:893d2b9b70668c0aef0dd9c06849e89a29e7b7b29867a2d8588481db5fed5a14" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 22 - Machining Processes used to Produce Round Shape" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 22.1 - PG NO. 600" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 22.1\n", + "#page no. 600 \n", + "# Given that\n", + "l=6.#in inch Length of rod \n", + "di=1./2.#in inch initial diameter of rod\n", + "df=0.480#in inch final diameter of rod\n", + "N=400.#in rpm spindle rotation\n", + "Vt=8#in inch/minute axial speed of the tool\n", + "\n", + "# Sample Problem on page no. 600\n", + "\n", + "print(\"\\n # Material Removal Rate and Cutting Force in Turning # \\n\")\n", + "\n", + "V=3.14*di*N\n", + "print'%s %d %s' %(\"\\n\\n Cutting speed=\",V,\" m/min\")\n", + "\n", + "v1=3.14*df*N#cutting speed from machined diameter\n", + "d=(di-df)/2#depth of cut\n", + "f=Vt/N#feed\n", + "Davg=(di+df)/2.\n", + "MRR=3.14*Davg*d*f*N \n", + "print'%s %.6f %s' %(\"\\n\\n Material Removal Rate =\",MRR,\"=in^3/min\")\n", + "\n", + "t=l/(f*N)\n", + "print'%s %.6f %s' %(\"\\n\\n Cutting time=\",t,\" min\")\n", + "\n", + "P=(4./2.73)*MRR#average value of stainless steel is taken as 4 ws/mm3 or 4/2.73 hpmin/mm3\n", + "print'%s %.6f %s' %(\"\\n\\n Cutting power=\",P,\"hp\")\n", + "\n", + "Fc=((P*396000)/(N*2*3.14))/(Davg/2.)\n", + "print'%s %d %s' %(\"\\n\\n Cutting force=\",Fc,\"lb\")\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Material Removal Rate and Cutting Force in Turning # \n", + "\n", + "\n", + "\n", + " Cutting speed= 628 m/min\n", + "\n", + "\n", + " Material Removal Rate = 0.123088 =in^3/min\n", + "\n", + "\n", + " Cutting time= 0.750000 min\n", + "\n", + "\n", + " Cutting power= 0.180349 hp\n", + "\n", + "\n", + " Cutting force= 116 lb\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 22.2 - PG NO. 632" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 22.2\n", + "#page no. 632\n", + "# Given that \n", + "d=10.#in mm diameter of drill bit\n", + "f=0.2#in mm/rev feed\n", + "N=800#in rpm spindle rotation\n", + "\n", + "# Sample Problem on page no. 632\n", + "\n", + "print(\"\\n # Material Removal Rate and Torque in Drilling # \\n\")\n", + "\n", + "MRR=(((3.14*(d**2))/4)*f*N)/60.\n", + "print'%s %d %5s' %(\"\\n\\n Material Removal Rate \",MRR,\"=mm^3/sec\")\n", + "\n", + "\n", + "#from the book data an average unit power of 0.5Ws/mm2 for magnesium is taken\n", + "T=(MRR*0.5)/((N*2.*3.14)/60.)\n", + "print'%s %.6f %s' %(\"\\n\\n Torque on the drill \",T,\"=Nm\")\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Material Removal Rate and Torque in Drilling # \n", + "\n", + "\n", + "\n", + " Material Removal Rate 209 =mm^3/sec\n", + "\n", + "\n", + " Torque on the drill 1.250000 =Nm\n" + ] + } + ], + "prompt_number": 3 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER22_1.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER22_1.ipynb new file mode 100644 index 00000000..a057d34b --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER22_1.ipynb @@ -0,0 +1,157 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:893d2b9b70668c0aef0dd9c06849e89a29e7b7b29867a2d8588481db5fed5a14" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 22 - Machining Processes used to Produce Round Shape" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 22.1 - PG NO. 600" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 22.1\n", + "#page no. 600 \n", + "# Given that\n", + "l=6.#in inch Length of rod \n", + "di=1./2.#in inch initial diameter of rod\n", + "df=0.480#in inch final diameter of rod\n", + "N=400.#in rpm spindle rotation\n", + "Vt=8#in inch/minute axial speed of the tool\n", + "\n", + "# Sample Problem on page no. 600\n", + "\n", + "print(\"\\n # Material Removal Rate and Cutting Force in Turning # \\n\")\n", + "\n", + "V=3.14*di*N\n", + "print'%s %d %s' %(\"\\n\\n Cutting speed=\",V,\" m/min\")\n", + "\n", + "v1=3.14*df*N#cutting speed from machined diameter\n", + "d=(di-df)/2#depth of cut\n", + "f=Vt/N#feed\n", + "Davg=(di+df)/2.\n", + "MRR=3.14*Davg*d*f*N \n", + "print'%s %.6f %s' %(\"\\n\\n Material Removal Rate =\",MRR,\"=in^3/min\")\n", + "\n", + "t=l/(f*N)\n", + "print'%s %.6f %s' %(\"\\n\\n Cutting time=\",t,\" min\")\n", + "\n", + "P=(4./2.73)*MRR#average value of stainless steel is taken as 4 ws/mm3 or 4/2.73 hpmin/mm3\n", + "print'%s %.6f %s' %(\"\\n\\n Cutting power=\",P,\"hp\")\n", + "\n", + "Fc=((P*396000)/(N*2*3.14))/(Davg/2.)\n", + "print'%s %d %s' %(\"\\n\\n Cutting force=\",Fc,\"lb\")\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Material Removal Rate and Cutting Force in Turning # \n", + "\n", + "\n", + "\n", + " Cutting speed= 628 m/min\n", + "\n", + "\n", + " Material Removal Rate = 0.123088 =in^3/min\n", + "\n", + "\n", + " Cutting time= 0.750000 min\n", + "\n", + "\n", + " Cutting power= 0.180349 hp\n", + "\n", + "\n", + " Cutting force= 116 lb\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 22.2 - PG NO. 632" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 22.2\n", + "#page no. 632\n", + "# Given that \n", + "d=10.#in mm diameter of drill bit\n", + "f=0.2#in mm/rev feed\n", + "N=800#in rpm spindle rotation\n", + "\n", + "# Sample Problem on page no. 632\n", + "\n", + "print(\"\\n # Material Removal Rate and Torque in Drilling # \\n\")\n", + "\n", + "MRR=(((3.14*(d**2))/4)*f*N)/60.\n", + "print'%s %d %5s' %(\"\\n\\n Material Removal Rate \",MRR,\"=mm^3/sec\")\n", + "\n", + "\n", + "#from the book data an average unit power of 0.5Ws/mm2 for magnesium is taken\n", + "T=(MRR*0.5)/((N*2.*3.14)/60.)\n", + "print'%s %.6f %s' %(\"\\n\\n Torque on the drill \",T,\"=Nm\")\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Material Removal Rate and Torque in Drilling # \n", + "\n", + "\n", + "\n", + " Material Removal Rate 209 =mm^3/sec\n", + "\n", + "\n", + " Torque on the drill 1.250000 =Nm\n" + ] + } + ], + "prompt_number": 3 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER22_2.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER22_2.ipynb new file mode 100644 index 00000000..a057d34b --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER22_2.ipynb @@ -0,0 +1,157 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:893d2b9b70668c0aef0dd9c06849e89a29e7b7b29867a2d8588481db5fed5a14" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 22 - Machining Processes used to Produce Round Shape" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 22.1 - PG NO. 600" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 22.1\n", + "#page no. 600 \n", + "# Given that\n", + "l=6.#in inch Length of rod \n", + "di=1./2.#in inch initial diameter of rod\n", + "df=0.480#in inch final diameter of rod\n", + "N=400.#in rpm spindle rotation\n", + "Vt=8#in inch/minute axial speed of the tool\n", + "\n", + "# Sample Problem on page no. 600\n", + "\n", + "print(\"\\n # Material Removal Rate and Cutting Force in Turning # \\n\")\n", + "\n", + "V=3.14*di*N\n", + "print'%s %d %s' %(\"\\n\\n Cutting speed=\",V,\" m/min\")\n", + "\n", + "v1=3.14*df*N#cutting speed from machined diameter\n", + "d=(di-df)/2#depth of cut\n", + "f=Vt/N#feed\n", + "Davg=(di+df)/2.\n", + "MRR=3.14*Davg*d*f*N \n", + "print'%s %.6f %s' %(\"\\n\\n Material Removal Rate =\",MRR,\"=in^3/min\")\n", + "\n", + "t=l/(f*N)\n", + "print'%s %.6f %s' %(\"\\n\\n Cutting time=\",t,\" min\")\n", + "\n", + "P=(4./2.73)*MRR#average value of stainless steel is taken as 4 ws/mm3 or 4/2.73 hpmin/mm3\n", + "print'%s %.6f %s' %(\"\\n\\n Cutting power=\",P,\"hp\")\n", + "\n", + "Fc=((P*396000)/(N*2*3.14))/(Davg/2.)\n", + "print'%s %d %s' %(\"\\n\\n Cutting force=\",Fc,\"lb\")\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Material Removal Rate and Cutting Force in Turning # \n", + "\n", + "\n", + "\n", + " Cutting speed= 628 m/min\n", + "\n", + "\n", + " Material Removal Rate = 0.123088 =in^3/min\n", + "\n", + "\n", + " Cutting time= 0.750000 min\n", + "\n", + "\n", + " Cutting power= 0.180349 hp\n", + "\n", + "\n", + " Cutting force= 116 lb\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 22.2 - PG NO. 632" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 22.2\n", + "#page no. 632\n", + "# Given that \n", + "d=10.#in mm diameter of drill bit\n", + "f=0.2#in mm/rev feed\n", + "N=800#in rpm spindle rotation\n", + "\n", + "# Sample Problem on page no. 632\n", + "\n", + "print(\"\\n # Material Removal Rate and Torque in Drilling # \\n\")\n", + "\n", + "MRR=(((3.14*(d**2))/4)*f*N)/60.\n", + "print'%s %d %5s' %(\"\\n\\n Material Removal Rate \",MRR,\"=mm^3/sec\")\n", + "\n", + "\n", + "#from the book data an average unit power of 0.5Ws/mm2 for magnesium is taken\n", + "T=(MRR*0.5)/((N*2.*3.14)/60.)\n", + "print'%s %.6f %s' %(\"\\n\\n Torque on the drill \",T,\"=Nm\")\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Material Removal Rate and Torque in Drilling # \n", + "\n", + "\n", + "\n", + " Material Removal Rate 209 =mm^3/sec\n", + "\n", + "\n", + " Torque on the drill 1.250000 =Nm\n" + ] + } + ], + "prompt_number": 3 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER22_3.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER22_3.ipynb new file mode 100644 index 00000000..a057d34b --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER22_3.ipynb @@ -0,0 +1,157 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:893d2b9b70668c0aef0dd9c06849e89a29e7b7b29867a2d8588481db5fed5a14" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 22 - Machining Processes used to Produce Round Shape" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 22.1 - PG NO. 600" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 22.1\n", + "#page no. 600 \n", + "# Given that\n", + "l=6.#in inch Length of rod \n", + "di=1./2.#in inch initial diameter of rod\n", + "df=0.480#in inch final diameter of rod\n", + "N=400.#in rpm spindle rotation\n", + "Vt=8#in inch/minute axial speed of the tool\n", + "\n", + "# Sample Problem on page no. 600\n", + "\n", + "print(\"\\n # Material Removal Rate and Cutting Force in Turning # \\n\")\n", + "\n", + "V=3.14*di*N\n", + "print'%s %d %s' %(\"\\n\\n Cutting speed=\",V,\" m/min\")\n", + "\n", + "v1=3.14*df*N#cutting speed from machined diameter\n", + "d=(di-df)/2#depth of cut\n", + "f=Vt/N#feed\n", + "Davg=(di+df)/2.\n", + "MRR=3.14*Davg*d*f*N \n", + "print'%s %.6f %s' %(\"\\n\\n Material Removal Rate =\",MRR,\"=in^3/min\")\n", + "\n", + "t=l/(f*N)\n", + "print'%s %.6f %s' %(\"\\n\\n Cutting time=\",t,\" min\")\n", + "\n", + "P=(4./2.73)*MRR#average value of stainless steel is taken as 4 ws/mm3 or 4/2.73 hpmin/mm3\n", + "print'%s %.6f %s' %(\"\\n\\n Cutting power=\",P,\"hp\")\n", + "\n", + "Fc=((P*396000)/(N*2*3.14))/(Davg/2.)\n", + "print'%s %d %s' %(\"\\n\\n Cutting force=\",Fc,\"lb\")\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Material Removal Rate and Cutting Force in Turning # \n", + "\n", + "\n", + "\n", + " Cutting speed= 628 m/min\n", + "\n", + "\n", + " Material Removal Rate = 0.123088 =in^3/min\n", + "\n", + "\n", + " Cutting time= 0.750000 min\n", + "\n", + "\n", + " Cutting power= 0.180349 hp\n", + "\n", + "\n", + " Cutting force= 116 lb\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 22.2 - PG NO. 632" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 22.2\n", + "#page no. 632\n", + "# Given that \n", + "d=10.#in mm diameter of drill bit\n", + "f=0.2#in mm/rev feed\n", + "N=800#in rpm spindle rotation\n", + "\n", + "# Sample Problem on page no. 632\n", + "\n", + "print(\"\\n # Material Removal Rate and Torque in Drilling # \\n\")\n", + "\n", + "MRR=(((3.14*(d**2))/4)*f*N)/60.\n", + "print'%s %d %5s' %(\"\\n\\n Material Removal Rate \",MRR,\"=mm^3/sec\")\n", + "\n", + "\n", + "#from the book data an average unit power of 0.5Ws/mm2 for magnesium is taken\n", + "T=(MRR*0.5)/((N*2.*3.14)/60.)\n", + "print'%s %.6f %s' %(\"\\n\\n Torque on the drill \",T,\"=Nm\")\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Material Removal Rate and Torque in Drilling # \n", + "\n", + "\n", + "\n", + " Material Removal Rate 209 =mm^3/sec\n", + "\n", + "\n", + " Torque on the drill 1.250000 =Nm\n" + ] + } + ], + "prompt_number": 3 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER22_4.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER22_4.ipynb new file mode 100644 index 00000000..a057d34b --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER22_4.ipynb @@ -0,0 +1,157 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:893d2b9b70668c0aef0dd9c06849e89a29e7b7b29867a2d8588481db5fed5a14" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 22 - Machining Processes used to Produce Round Shape" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 22.1 - PG NO. 600" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 22.1\n", + "#page no. 600 \n", + "# Given that\n", + "l=6.#in inch Length of rod \n", + "di=1./2.#in inch initial diameter of rod\n", + "df=0.480#in inch final diameter of rod\n", + "N=400.#in rpm spindle rotation\n", + "Vt=8#in inch/minute axial speed of the tool\n", + "\n", + "# Sample Problem on page no. 600\n", + "\n", + "print(\"\\n # Material Removal Rate and Cutting Force in Turning # \\n\")\n", + "\n", + "V=3.14*di*N\n", + "print'%s %d %s' %(\"\\n\\n Cutting speed=\",V,\" m/min\")\n", + "\n", + "v1=3.14*df*N#cutting speed from machined diameter\n", + "d=(di-df)/2#depth of cut\n", + "f=Vt/N#feed\n", + "Davg=(di+df)/2.\n", + "MRR=3.14*Davg*d*f*N \n", + "print'%s %.6f %s' %(\"\\n\\n Material Removal Rate =\",MRR,\"=in^3/min\")\n", + "\n", + "t=l/(f*N)\n", + "print'%s %.6f %s' %(\"\\n\\n Cutting time=\",t,\" min\")\n", + "\n", + "P=(4./2.73)*MRR#average value of stainless steel is taken as 4 ws/mm3 or 4/2.73 hpmin/mm3\n", + "print'%s %.6f %s' %(\"\\n\\n Cutting power=\",P,\"hp\")\n", + "\n", + "Fc=((P*396000)/(N*2*3.14))/(Davg/2.)\n", + "print'%s %d %s' %(\"\\n\\n Cutting force=\",Fc,\"lb\")\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Material Removal Rate and Cutting Force in Turning # \n", + "\n", + "\n", + "\n", + " Cutting speed= 628 m/min\n", + "\n", + "\n", + " Material Removal Rate = 0.123088 =in^3/min\n", + "\n", + "\n", + " Cutting time= 0.750000 min\n", + "\n", + "\n", + " Cutting power= 0.180349 hp\n", + "\n", + "\n", + " Cutting force= 116 lb\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 22.2 - PG NO. 632" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 22.2\n", + "#page no. 632\n", + "# Given that \n", + "d=10.#in mm diameter of drill bit\n", + "f=0.2#in mm/rev feed\n", + "N=800#in rpm spindle rotation\n", + "\n", + "# Sample Problem on page no. 632\n", + "\n", + "print(\"\\n # Material Removal Rate and Torque in Drilling # \\n\")\n", + "\n", + "MRR=(((3.14*(d**2))/4)*f*N)/60.\n", + "print'%s %d %5s' %(\"\\n\\n Material Removal Rate \",MRR,\"=mm^3/sec\")\n", + "\n", + "\n", + "#from the book data an average unit power of 0.5Ws/mm2 for magnesium is taken\n", + "T=(MRR*0.5)/((N*2.*3.14)/60.)\n", + "print'%s %.6f %s' %(\"\\n\\n Torque on the drill \",T,\"=Nm\")\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Material Removal Rate and Torque in Drilling # \n", + "\n", + "\n", + "\n", + " Material Removal Rate 209 =mm^3/sec\n", + "\n", + "\n", + " Torque on the drill 1.250000 =Nm\n" + ] + } + ], + "prompt_number": 3 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER22_5.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER22_5.ipynb new file mode 100644 index 00000000..a057d34b --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER22_5.ipynb @@ -0,0 +1,157 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:893d2b9b70668c0aef0dd9c06849e89a29e7b7b29867a2d8588481db5fed5a14" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 22 - Machining Processes used to Produce Round Shape" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 22.1 - PG NO. 600" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 22.1\n", + "#page no. 600 \n", + "# Given that\n", + "l=6.#in inch Length of rod \n", + "di=1./2.#in inch initial diameter of rod\n", + "df=0.480#in inch final diameter of rod\n", + "N=400.#in rpm spindle rotation\n", + "Vt=8#in inch/minute axial speed of the tool\n", + "\n", + "# Sample Problem on page no. 600\n", + "\n", + "print(\"\\n # Material Removal Rate and Cutting Force in Turning # \\n\")\n", + "\n", + "V=3.14*di*N\n", + "print'%s %d %s' %(\"\\n\\n Cutting speed=\",V,\" m/min\")\n", + "\n", + "v1=3.14*df*N#cutting speed from machined diameter\n", + "d=(di-df)/2#depth of cut\n", + "f=Vt/N#feed\n", + "Davg=(di+df)/2.\n", + "MRR=3.14*Davg*d*f*N \n", + "print'%s %.6f %s' %(\"\\n\\n Material Removal Rate =\",MRR,\"=in^3/min\")\n", + "\n", + "t=l/(f*N)\n", + "print'%s %.6f %s' %(\"\\n\\n Cutting time=\",t,\" min\")\n", + "\n", + "P=(4./2.73)*MRR#average value of stainless steel is taken as 4 ws/mm3 or 4/2.73 hpmin/mm3\n", + "print'%s %.6f %s' %(\"\\n\\n Cutting power=\",P,\"hp\")\n", + "\n", + "Fc=((P*396000)/(N*2*3.14))/(Davg/2.)\n", + "print'%s %d %s' %(\"\\n\\n Cutting force=\",Fc,\"lb\")\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Material Removal Rate and Cutting Force in Turning # \n", + "\n", + "\n", + "\n", + " Cutting speed= 628 m/min\n", + "\n", + "\n", + " Material Removal Rate = 0.123088 =in^3/min\n", + "\n", + "\n", + " Cutting time= 0.750000 min\n", + "\n", + "\n", + " Cutting power= 0.180349 hp\n", + "\n", + "\n", + " Cutting force= 116 lb\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 22.2 - PG NO. 632" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 22.2\n", + "#page no. 632\n", + "# Given that \n", + "d=10.#in mm diameter of drill bit\n", + "f=0.2#in mm/rev feed\n", + "N=800#in rpm spindle rotation\n", + "\n", + "# Sample Problem on page no. 632\n", + "\n", + "print(\"\\n # Material Removal Rate and Torque in Drilling # \\n\")\n", + "\n", + "MRR=(((3.14*(d**2))/4)*f*N)/60.\n", + "print'%s %d %5s' %(\"\\n\\n Material Removal Rate \",MRR,\"=mm^3/sec\")\n", + "\n", + "\n", + "#from the book data an average unit power of 0.5Ws/mm2 for magnesium is taken\n", + "T=(MRR*0.5)/((N*2.*3.14)/60.)\n", + "print'%s %.6f %s' %(\"\\n\\n Torque on the drill \",T,\"=Nm\")\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Material Removal Rate and Torque in Drilling # \n", + "\n", + "\n", + "\n", + " Material Removal Rate 209 =mm^3/sec\n", + "\n", + "\n", + " Torque on the drill 1.250000 =Nm\n" + ] + } + ], + "prompt_number": 3 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER22_6.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER22_6.ipynb new file mode 100644 index 00000000..a057d34b --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER22_6.ipynb @@ -0,0 +1,157 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:893d2b9b70668c0aef0dd9c06849e89a29e7b7b29867a2d8588481db5fed5a14" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 22 - Machining Processes used to Produce Round Shape" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 22.1 - PG NO. 600" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 22.1\n", + "#page no. 600 \n", + "# Given that\n", + "l=6.#in inch Length of rod \n", + "di=1./2.#in inch initial diameter of rod\n", + "df=0.480#in inch final diameter of rod\n", + "N=400.#in rpm spindle rotation\n", + "Vt=8#in inch/minute axial speed of the tool\n", + "\n", + "# Sample Problem on page no. 600\n", + "\n", + "print(\"\\n # Material Removal Rate and Cutting Force in Turning # \\n\")\n", + "\n", + "V=3.14*di*N\n", + "print'%s %d %s' %(\"\\n\\n Cutting speed=\",V,\" m/min\")\n", + "\n", + "v1=3.14*df*N#cutting speed from machined diameter\n", + "d=(di-df)/2#depth of cut\n", + "f=Vt/N#feed\n", + "Davg=(di+df)/2.\n", + "MRR=3.14*Davg*d*f*N \n", + "print'%s %.6f %s' %(\"\\n\\n Material Removal Rate =\",MRR,\"=in^3/min\")\n", + "\n", + "t=l/(f*N)\n", + "print'%s %.6f %s' %(\"\\n\\n Cutting time=\",t,\" min\")\n", + "\n", + "P=(4./2.73)*MRR#average value of stainless steel is taken as 4 ws/mm3 or 4/2.73 hpmin/mm3\n", + "print'%s %.6f %s' %(\"\\n\\n Cutting power=\",P,\"hp\")\n", + "\n", + "Fc=((P*396000)/(N*2*3.14))/(Davg/2.)\n", + "print'%s %d %s' %(\"\\n\\n Cutting force=\",Fc,\"lb\")\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Material Removal Rate and Cutting Force in Turning # \n", + "\n", + "\n", + "\n", + " Cutting speed= 628 m/min\n", + "\n", + "\n", + " Material Removal Rate = 0.123088 =in^3/min\n", + "\n", + "\n", + " Cutting time= 0.750000 min\n", + "\n", + "\n", + " Cutting power= 0.180349 hp\n", + "\n", + "\n", + " Cutting force= 116 lb\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 22.2 - PG NO. 632" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 22.2\n", + "#page no. 632\n", + "# Given that \n", + "d=10.#in mm diameter of drill bit\n", + "f=0.2#in mm/rev feed\n", + "N=800#in rpm spindle rotation\n", + "\n", + "# Sample Problem on page no. 632\n", + "\n", + "print(\"\\n # Material Removal Rate and Torque in Drilling # \\n\")\n", + "\n", + "MRR=(((3.14*(d**2))/4)*f*N)/60.\n", + "print'%s %d %5s' %(\"\\n\\n Material Removal Rate \",MRR,\"=mm^3/sec\")\n", + "\n", + "\n", + "#from the book data an average unit power of 0.5Ws/mm2 for magnesium is taken\n", + "T=(MRR*0.5)/((N*2.*3.14)/60.)\n", + "print'%s %.6f %s' %(\"\\n\\n Torque on the drill \",T,\"=Nm\")\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Material Removal Rate and Torque in Drilling # \n", + "\n", + "\n", + "\n", + " Material Removal Rate 209 =mm^3/sec\n", + "\n", + "\n", + " Torque on the drill 1.250000 =Nm\n" + ] + } + ], + "prompt_number": 3 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER22_7.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER22_7.ipynb new file mode 100644 index 00000000..a057d34b --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER22_7.ipynb @@ -0,0 +1,157 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:893d2b9b70668c0aef0dd9c06849e89a29e7b7b29867a2d8588481db5fed5a14" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 22 - Machining Processes used to Produce Round Shape" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 22.1 - PG NO. 600" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 22.1\n", + "#page no. 600 \n", + "# Given that\n", + "l=6.#in inch Length of rod \n", + "di=1./2.#in inch initial diameter of rod\n", + "df=0.480#in inch final diameter of rod\n", + "N=400.#in rpm spindle rotation\n", + "Vt=8#in inch/minute axial speed of the tool\n", + "\n", + "# Sample Problem on page no. 600\n", + "\n", + "print(\"\\n # Material Removal Rate and Cutting Force in Turning # \\n\")\n", + "\n", + "V=3.14*di*N\n", + "print'%s %d %s' %(\"\\n\\n Cutting speed=\",V,\" m/min\")\n", + "\n", + "v1=3.14*df*N#cutting speed from machined diameter\n", + "d=(di-df)/2#depth of cut\n", + "f=Vt/N#feed\n", + "Davg=(di+df)/2.\n", + "MRR=3.14*Davg*d*f*N \n", + "print'%s %.6f %s' %(\"\\n\\n Material Removal Rate =\",MRR,\"=in^3/min\")\n", + "\n", + "t=l/(f*N)\n", + "print'%s %.6f %s' %(\"\\n\\n Cutting time=\",t,\" min\")\n", + "\n", + "P=(4./2.73)*MRR#average value of stainless steel is taken as 4 ws/mm3 or 4/2.73 hpmin/mm3\n", + "print'%s %.6f %s' %(\"\\n\\n Cutting power=\",P,\"hp\")\n", + "\n", + "Fc=((P*396000)/(N*2*3.14))/(Davg/2.)\n", + "print'%s %d %s' %(\"\\n\\n Cutting force=\",Fc,\"lb\")\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Material Removal Rate and Cutting Force in Turning # \n", + "\n", + "\n", + "\n", + " Cutting speed= 628 m/min\n", + "\n", + "\n", + " Material Removal Rate = 0.123088 =in^3/min\n", + "\n", + "\n", + " Cutting time= 0.750000 min\n", + "\n", + "\n", + " Cutting power= 0.180349 hp\n", + "\n", + "\n", + " Cutting force= 116 lb\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 22.2 - PG NO. 632" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 22.2\n", + "#page no. 632\n", + "# Given that \n", + "d=10.#in mm diameter of drill bit\n", + "f=0.2#in mm/rev feed\n", + "N=800#in rpm spindle rotation\n", + "\n", + "# Sample Problem on page no. 632\n", + "\n", + "print(\"\\n # Material Removal Rate and Torque in Drilling # \\n\")\n", + "\n", + "MRR=(((3.14*(d**2))/4)*f*N)/60.\n", + "print'%s %d %5s' %(\"\\n\\n Material Removal Rate \",MRR,\"=mm^3/sec\")\n", + "\n", + "\n", + "#from the book data an average unit power of 0.5Ws/mm2 for magnesium is taken\n", + "T=(MRR*0.5)/((N*2.*3.14)/60.)\n", + "print'%s %.6f %s' %(\"\\n\\n Torque on the drill \",T,\"=Nm\")\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Material Removal Rate and Torque in Drilling # \n", + "\n", + "\n", + "\n", + " Material Removal Rate 209 =mm^3/sec\n", + "\n", + "\n", + " Torque on the drill 1.250000 =Nm\n" + ] + } + ], + "prompt_number": 3 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER22_8.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER22_8.ipynb new file mode 100644 index 00000000..a057d34b --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER22_8.ipynb @@ -0,0 +1,157 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:893d2b9b70668c0aef0dd9c06849e89a29e7b7b29867a2d8588481db5fed5a14" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 22 - Machining Processes used to Produce Round Shape" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 22.1 - PG NO. 600" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 22.1\n", + "#page no. 600 \n", + "# Given that\n", + "l=6.#in inch Length of rod \n", + "di=1./2.#in inch initial diameter of rod\n", + "df=0.480#in inch final diameter of rod\n", + "N=400.#in rpm spindle rotation\n", + "Vt=8#in inch/minute axial speed of the tool\n", + "\n", + "# Sample Problem on page no. 600\n", + "\n", + "print(\"\\n # Material Removal Rate and Cutting Force in Turning # \\n\")\n", + "\n", + "V=3.14*di*N\n", + "print'%s %d %s' %(\"\\n\\n Cutting speed=\",V,\" m/min\")\n", + "\n", + "v1=3.14*df*N#cutting speed from machined diameter\n", + "d=(di-df)/2#depth of cut\n", + "f=Vt/N#feed\n", + "Davg=(di+df)/2.\n", + "MRR=3.14*Davg*d*f*N \n", + "print'%s %.6f %s' %(\"\\n\\n Material Removal Rate =\",MRR,\"=in^3/min\")\n", + "\n", + "t=l/(f*N)\n", + "print'%s %.6f %s' %(\"\\n\\n Cutting time=\",t,\" min\")\n", + "\n", + "P=(4./2.73)*MRR#average value of stainless steel is taken as 4 ws/mm3 or 4/2.73 hpmin/mm3\n", + "print'%s %.6f %s' %(\"\\n\\n Cutting power=\",P,\"hp\")\n", + "\n", + "Fc=((P*396000)/(N*2*3.14))/(Davg/2.)\n", + "print'%s %d %s' %(\"\\n\\n Cutting force=\",Fc,\"lb\")\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Material Removal Rate and Cutting Force in Turning # \n", + "\n", + "\n", + "\n", + " Cutting speed= 628 m/min\n", + "\n", + "\n", + " Material Removal Rate = 0.123088 =in^3/min\n", + "\n", + "\n", + " Cutting time= 0.750000 min\n", + "\n", + "\n", + " Cutting power= 0.180349 hp\n", + "\n", + "\n", + " Cutting force= 116 lb\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 22.2 - PG NO. 632" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 22.2\n", + "#page no. 632\n", + "# Given that \n", + "d=10.#in mm diameter of drill bit\n", + "f=0.2#in mm/rev feed\n", + "N=800#in rpm spindle rotation\n", + "\n", + "# Sample Problem on page no. 632\n", + "\n", + "print(\"\\n # Material Removal Rate and Torque in Drilling # \\n\")\n", + "\n", + "MRR=(((3.14*(d**2))/4)*f*N)/60.\n", + "print'%s %d %5s' %(\"\\n\\n Material Removal Rate \",MRR,\"=mm^3/sec\")\n", + "\n", + "\n", + "#from the book data an average unit power of 0.5Ws/mm2 for magnesium is taken\n", + "T=(MRR*0.5)/((N*2.*3.14)/60.)\n", + "print'%s %.6f %s' %(\"\\n\\n Torque on the drill \",T,\"=Nm\")\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Material Removal Rate and Torque in Drilling # \n", + "\n", + "\n", + "\n", + " Material Removal Rate 209 =mm^3/sec\n", + "\n", + "\n", + " Torque on the drill 1.250000 =Nm\n" + ] + } + ], + "prompt_number": 3 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER23_1.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER23_1.ipynb new file mode 100644 index 00000000..ef186b9c --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER23_1.ipynb @@ -0,0 +1,175 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:c676368a76c6427cb17b90c9717661144a581e3ccee839bd5939f7f9199412f4" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 23 - Machining Processes used to Produce Various Shapes" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 23.1 - PG NO. 600" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Example 23.1\n", + "#page no. 600\n", + "# Given that\n", + "import math\n", + "l=12.#in inch Length of block\n", + "w=4\n", + "f=0.01#in inch/tooth feed \n", + "d=0.125#in inch depth of cut\n", + "D=2.#in inch diameter of cutter\n", + "n=20.#no. of teeth\n", + "N=100.#in rpm spindle rotation\n", + "Vt=8.#in inch/minute axial speed of the tool\n", + "\n", + "# Sample Problem on page no. 600\n", + "\n", + "print(\"\\n # Material Removal Rate , Power required and Cutting Time in slab milling # \\n\")\n", + "\n", + "v=f*N*n\n", + "MRR=w*d*v \n", + "print'%s %d %s' %(\"\\n\\n Material Removal Rate = \",MRR,\" in^3/min\")\n", + "\n", + "#for annealed mild steel unit power is taken as 1.1 hp min/in3\n", + "P=1.1*MRR\n", + "print'%s %d %s' %(\"\\n\\n Cutting power=\",P,\"hp\")\n", + "\n", + "T=P*33000/(N*2*3.14)\n", + "print'%s %d %s' %(\"\\n\\n Cutting torque=\",T,\"lb-ft\")\n", + "\n", + "lc=math.sqrt(d*D)\n", + "t=(300.+12.2)/500.\n", + "print'%s %.6f %s' %(\"\\n\\n Cutting time=\",t*60,\"sec\")\n", + "\n", + "#Answers vary due to aproximations \n", + "\n", + "\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Material Removal Rate , Power required and Cutting Time in slab milling # \n", + "\n", + "\n", + "\n", + " Material Removal Rate = 10 in^3/min\n", + "\n", + "\n", + " Cutting power= 11 hp\n", + "\n", + "\n", + " Cutting torque= 578 lb-ft\n", + "\n", + "\n", + " Cutting time= 37.464000 sec\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 23.2 - PG NO. 655" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 23.2\n", + "#page no. 655\n", + "# Given that\n", + "l=500#in mm Length\n", + "w=60#in mm width\n", + "v=0.6#in m/min \n", + "d=3#in mm depth of cut\n", + "D=150#in mm diameter of cutter\n", + "n=10#no. of inserts\n", + "N=100#in rpm spindle rotation\n", + "\n", + "# Sample Problem on page no. 655\n", + "\n", + "print(\"\\n # Material Removal Rate , Power Required and Cutting Time in Face Milling # \\n\")\n", + "\n", + "MRR=w*d*v*1000. \n", + "print'%s %d %s' %(\"\\n\\n Material Removal Rate = \",MRR,\"mm^3/min\")\n", + "\n", + "lc=D/2.\n", + "t=((l+(2.*lc))/((v*1000.)/60.)) # velocity is converted into mm/sec\n", + "t1=t/60.\n", + "print'%s %.6f %s' %(\"\\n\\n Cutting time= \",t1,\"f min\")\n", + "\n", + "f=(v*1000.*60.)/(60.*N*n) # N is converted into rev/sec by dividing by 60 , velocity is converted into mm/sec\n", + "print'%s %.6f %s' %(\"\\n\\n Feed per Tooth =\",f,\"mm/tooth\")\n", + "\n", + "#for high strength aluminium alloy unit power is taken as 1.1 W s/mm3\n", + "P=(1.1*MRR)/60. # MRR is converted into mm3/sec by dividing by 60\n", + "P1=P/(1000.)#in KW\n", + "print'%s %.6f %s' %(\"\\n\\n Cutting power =\",P1,\"KW\")\n", + "\n", + "\n", + "\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Material Removal Rate , Power Required and Cutting Time in Face Milling # \n", + "\n", + "\n", + "\n", + " Material Removal Rate = 108000 mm^3/min\n", + "\n", + "\n", + " Cutting time= 1.083333 f min\n", + "\n", + "\n", + " Feed per Tooth = 0.600000 mm/tooth\n", + "\n", + "\n", + " Cutting power = 1.980000 KW\n" + ] + } + ], + "prompt_number": 2 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER23_2.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER23_2.ipynb new file mode 100644 index 00000000..ef186b9c --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER23_2.ipynb @@ -0,0 +1,175 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:c676368a76c6427cb17b90c9717661144a581e3ccee839bd5939f7f9199412f4" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 23 - Machining Processes used to Produce Various Shapes" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 23.1 - PG NO. 600" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Example 23.1\n", + "#page no. 600\n", + "# Given that\n", + "import math\n", + "l=12.#in inch Length of block\n", + "w=4\n", + "f=0.01#in inch/tooth feed \n", + "d=0.125#in inch depth of cut\n", + "D=2.#in inch diameter of cutter\n", + "n=20.#no. of teeth\n", + "N=100.#in rpm spindle rotation\n", + "Vt=8.#in inch/minute axial speed of the tool\n", + "\n", + "# Sample Problem on page no. 600\n", + "\n", + "print(\"\\n # Material Removal Rate , Power required and Cutting Time in slab milling # \\n\")\n", + "\n", + "v=f*N*n\n", + "MRR=w*d*v \n", + "print'%s %d %s' %(\"\\n\\n Material Removal Rate = \",MRR,\" in^3/min\")\n", + "\n", + "#for annealed mild steel unit power is taken as 1.1 hp min/in3\n", + "P=1.1*MRR\n", + "print'%s %d %s' %(\"\\n\\n Cutting power=\",P,\"hp\")\n", + "\n", + "T=P*33000/(N*2*3.14)\n", + "print'%s %d %s' %(\"\\n\\n Cutting torque=\",T,\"lb-ft\")\n", + "\n", + "lc=math.sqrt(d*D)\n", + "t=(300.+12.2)/500.\n", + "print'%s %.6f %s' %(\"\\n\\n Cutting time=\",t*60,\"sec\")\n", + "\n", + "#Answers vary due to aproximations \n", + "\n", + "\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Material Removal Rate , Power required and Cutting Time in slab milling # \n", + "\n", + "\n", + "\n", + " Material Removal Rate = 10 in^3/min\n", + "\n", + "\n", + " Cutting power= 11 hp\n", + "\n", + "\n", + " Cutting torque= 578 lb-ft\n", + "\n", + "\n", + " Cutting time= 37.464000 sec\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 23.2 - PG NO. 655" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 23.2\n", + "#page no. 655\n", + "# Given that\n", + "l=500#in mm Length\n", + "w=60#in mm width\n", + "v=0.6#in m/min \n", + "d=3#in mm depth of cut\n", + "D=150#in mm diameter of cutter\n", + "n=10#no. of inserts\n", + "N=100#in rpm spindle rotation\n", + "\n", + "# Sample Problem on page no. 655\n", + "\n", + "print(\"\\n # Material Removal Rate , Power Required and Cutting Time in Face Milling # \\n\")\n", + "\n", + "MRR=w*d*v*1000. \n", + "print'%s %d %s' %(\"\\n\\n Material Removal Rate = \",MRR,\"mm^3/min\")\n", + "\n", + "lc=D/2.\n", + "t=((l+(2.*lc))/((v*1000.)/60.)) # velocity is converted into mm/sec\n", + "t1=t/60.\n", + "print'%s %.6f %s' %(\"\\n\\n Cutting time= \",t1,\"f min\")\n", + "\n", + "f=(v*1000.*60.)/(60.*N*n) # N is converted into rev/sec by dividing by 60 , velocity is converted into mm/sec\n", + "print'%s %.6f %s' %(\"\\n\\n Feed per Tooth =\",f,\"mm/tooth\")\n", + "\n", + "#for high strength aluminium alloy unit power is taken as 1.1 W s/mm3\n", + "P=(1.1*MRR)/60. # MRR is converted into mm3/sec by dividing by 60\n", + "P1=P/(1000.)#in KW\n", + "print'%s %.6f %s' %(\"\\n\\n Cutting power =\",P1,\"KW\")\n", + "\n", + "\n", + "\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Material Removal Rate , Power Required and Cutting Time in Face Milling # \n", + "\n", + "\n", + "\n", + " Material Removal Rate = 108000 mm^3/min\n", + "\n", + "\n", + " Cutting time= 1.083333 f min\n", + "\n", + "\n", + " Feed per Tooth = 0.600000 mm/tooth\n", + "\n", + "\n", + " Cutting power = 1.980000 KW\n" + ] + } + ], + "prompt_number": 2 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER23_3.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER23_3.ipynb new file mode 100644 index 00000000..ef186b9c --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER23_3.ipynb @@ -0,0 +1,175 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:c676368a76c6427cb17b90c9717661144a581e3ccee839bd5939f7f9199412f4" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 23 - Machining Processes used to Produce Various Shapes" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 23.1 - PG NO. 600" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Example 23.1\n", + "#page no. 600\n", + "# Given that\n", + "import math\n", + "l=12.#in inch Length of block\n", + "w=4\n", + "f=0.01#in inch/tooth feed \n", + "d=0.125#in inch depth of cut\n", + "D=2.#in inch diameter of cutter\n", + "n=20.#no. of teeth\n", + "N=100.#in rpm spindle rotation\n", + "Vt=8.#in inch/minute axial speed of the tool\n", + "\n", + "# Sample Problem on page no. 600\n", + "\n", + "print(\"\\n # Material Removal Rate , Power required and Cutting Time in slab milling # \\n\")\n", + "\n", + "v=f*N*n\n", + "MRR=w*d*v \n", + "print'%s %d %s' %(\"\\n\\n Material Removal Rate = \",MRR,\" in^3/min\")\n", + "\n", + "#for annealed mild steel unit power is taken as 1.1 hp min/in3\n", + "P=1.1*MRR\n", + "print'%s %d %s' %(\"\\n\\n Cutting power=\",P,\"hp\")\n", + "\n", + "T=P*33000/(N*2*3.14)\n", + "print'%s %d %s' %(\"\\n\\n Cutting torque=\",T,\"lb-ft\")\n", + "\n", + "lc=math.sqrt(d*D)\n", + "t=(300.+12.2)/500.\n", + "print'%s %.6f %s' %(\"\\n\\n Cutting time=\",t*60,\"sec\")\n", + "\n", + "#Answers vary due to aproximations \n", + "\n", + "\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Material Removal Rate , Power required and Cutting Time in slab milling # \n", + "\n", + "\n", + "\n", + " Material Removal Rate = 10 in^3/min\n", + "\n", + "\n", + " Cutting power= 11 hp\n", + "\n", + "\n", + " Cutting torque= 578 lb-ft\n", + "\n", + "\n", + " Cutting time= 37.464000 sec\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 23.2 - PG NO. 655" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 23.2\n", + "#page no. 655\n", + "# Given that\n", + "l=500#in mm Length\n", + "w=60#in mm width\n", + "v=0.6#in m/min \n", + "d=3#in mm depth of cut\n", + "D=150#in mm diameter of cutter\n", + "n=10#no. of inserts\n", + "N=100#in rpm spindle rotation\n", + "\n", + "# Sample Problem on page no. 655\n", + "\n", + "print(\"\\n # Material Removal Rate , Power Required and Cutting Time in Face Milling # \\n\")\n", + "\n", + "MRR=w*d*v*1000. \n", + "print'%s %d %s' %(\"\\n\\n Material Removal Rate = \",MRR,\"mm^3/min\")\n", + "\n", + "lc=D/2.\n", + "t=((l+(2.*lc))/((v*1000.)/60.)) # velocity is converted into mm/sec\n", + "t1=t/60.\n", + "print'%s %.6f %s' %(\"\\n\\n Cutting time= \",t1,\"f min\")\n", + "\n", + "f=(v*1000.*60.)/(60.*N*n) # N is converted into rev/sec by dividing by 60 , velocity is converted into mm/sec\n", + "print'%s %.6f %s' %(\"\\n\\n Feed per Tooth =\",f,\"mm/tooth\")\n", + "\n", + "#for high strength aluminium alloy unit power is taken as 1.1 W s/mm3\n", + "P=(1.1*MRR)/60. # MRR is converted into mm3/sec by dividing by 60\n", + "P1=P/(1000.)#in KW\n", + "print'%s %.6f %s' %(\"\\n\\n Cutting power =\",P1,\"KW\")\n", + "\n", + "\n", + "\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Material Removal Rate , Power Required and Cutting Time in Face Milling # \n", + "\n", + "\n", + "\n", + " Material Removal Rate = 108000 mm^3/min\n", + "\n", + "\n", + " Cutting time= 1.083333 f min\n", + "\n", + "\n", + " Feed per Tooth = 0.600000 mm/tooth\n", + "\n", + "\n", + " Cutting power = 1.980000 KW\n" + ] + } + ], + "prompt_number": 2 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER23_4.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER23_4.ipynb new file mode 100644 index 00000000..ef186b9c --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER23_4.ipynb @@ -0,0 +1,175 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:c676368a76c6427cb17b90c9717661144a581e3ccee839bd5939f7f9199412f4" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 23 - Machining Processes used to Produce Various Shapes" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 23.1 - PG NO. 600" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Example 23.1\n", + "#page no. 600\n", + "# Given that\n", + "import math\n", + "l=12.#in inch Length of block\n", + "w=4\n", + "f=0.01#in inch/tooth feed \n", + "d=0.125#in inch depth of cut\n", + "D=2.#in inch diameter of cutter\n", + "n=20.#no. of teeth\n", + "N=100.#in rpm spindle rotation\n", + "Vt=8.#in inch/minute axial speed of the tool\n", + "\n", + "# Sample Problem on page no. 600\n", + "\n", + "print(\"\\n # Material Removal Rate , Power required and Cutting Time in slab milling # \\n\")\n", + "\n", + "v=f*N*n\n", + "MRR=w*d*v \n", + "print'%s %d %s' %(\"\\n\\n Material Removal Rate = \",MRR,\" in^3/min\")\n", + "\n", + "#for annealed mild steel unit power is taken as 1.1 hp min/in3\n", + "P=1.1*MRR\n", + "print'%s %d %s' %(\"\\n\\n Cutting power=\",P,\"hp\")\n", + "\n", + "T=P*33000/(N*2*3.14)\n", + "print'%s %d %s' %(\"\\n\\n Cutting torque=\",T,\"lb-ft\")\n", + "\n", + "lc=math.sqrt(d*D)\n", + "t=(300.+12.2)/500.\n", + "print'%s %.6f %s' %(\"\\n\\n Cutting time=\",t*60,\"sec\")\n", + "\n", + "#Answers vary due to aproximations \n", + "\n", + "\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Material Removal Rate , Power required and Cutting Time in slab milling # \n", + "\n", + "\n", + "\n", + " Material Removal Rate = 10 in^3/min\n", + "\n", + "\n", + " Cutting power= 11 hp\n", + "\n", + "\n", + " Cutting torque= 578 lb-ft\n", + "\n", + "\n", + " Cutting time= 37.464000 sec\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 23.2 - PG NO. 655" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 23.2\n", + "#page no. 655\n", + "# Given that\n", + "l=500#in mm Length\n", + "w=60#in mm width\n", + "v=0.6#in m/min \n", + "d=3#in mm depth of cut\n", + "D=150#in mm diameter of cutter\n", + "n=10#no. of inserts\n", + "N=100#in rpm spindle rotation\n", + "\n", + "# Sample Problem on page no. 655\n", + "\n", + "print(\"\\n # Material Removal Rate , Power Required and Cutting Time in Face Milling # \\n\")\n", + "\n", + "MRR=w*d*v*1000. \n", + "print'%s %d %s' %(\"\\n\\n Material Removal Rate = \",MRR,\"mm^3/min\")\n", + "\n", + "lc=D/2.\n", + "t=((l+(2.*lc))/((v*1000.)/60.)) # velocity is converted into mm/sec\n", + "t1=t/60.\n", + "print'%s %.6f %s' %(\"\\n\\n Cutting time= \",t1,\"f min\")\n", + "\n", + "f=(v*1000.*60.)/(60.*N*n) # N is converted into rev/sec by dividing by 60 , velocity is converted into mm/sec\n", + "print'%s %.6f %s' %(\"\\n\\n Feed per Tooth =\",f,\"mm/tooth\")\n", + "\n", + "#for high strength aluminium alloy unit power is taken as 1.1 W s/mm3\n", + "P=(1.1*MRR)/60. # MRR is converted into mm3/sec by dividing by 60\n", + "P1=P/(1000.)#in KW\n", + "print'%s %.6f %s' %(\"\\n\\n Cutting power =\",P1,\"KW\")\n", + "\n", + "\n", + "\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Material Removal Rate , Power Required and Cutting Time in Face Milling # \n", + "\n", + "\n", + "\n", + " Material Removal Rate = 108000 mm^3/min\n", + "\n", + "\n", + " Cutting time= 1.083333 f min\n", + "\n", + "\n", + " Feed per Tooth = 0.600000 mm/tooth\n", + "\n", + "\n", + " Cutting power = 1.980000 KW\n" + ] + } + ], + "prompt_number": 2 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER23_5.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER23_5.ipynb new file mode 100644 index 00000000..ef186b9c --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER23_5.ipynb @@ -0,0 +1,175 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:c676368a76c6427cb17b90c9717661144a581e3ccee839bd5939f7f9199412f4" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 23 - Machining Processes used to Produce Various Shapes" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 23.1 - PG NO. 600" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Example 23.1\n", + "#page no. 600\n", + "# Given that\n", + "import math\n", + "l=12.#in inch Length of block\n", + "w=4\n", + "f=0.01#in inch/tooth feed \n", + "d=0.125#in inch depth of cut\n", + "D=2.#in inch diameter of cutter\n", + "n=20.#no. of teeth\n", + "N=100.#in rpm spindle rotation\n", + "Vt=8.#in inch/minute axial speed of the tool\n", + "\n", + "# Sample Problem on page no. 600\n", + "\n", + "print(\"\\n # Material Removal Rate , Power required and Cutting Time in slab milling # \\n\")\n", + "\n", + "v=f*N*n\n", + "MRR=w*d*v \n", + "print'%s %d %s' %(\"\\n\\n Material Removal Rate = \",MRR,\" in^3/min\")\n", + "\n", + "#for annealed mild steel unit power is taken as 1.1 hp min/in3\n", + "P=1.1*MRR\n", + "print'%s %d %s' %(\"\\n\\n Cutting power=\",P,\"hp\")\n", + "\n", + "T=P*33000/(N*2*3.14)\n", + "print'%s %d %s' %(\"\\n\\n Cutting torque=\",T,\"lb-ft\")\n", + "\n", + "lc=math.sqrt(d*D)\n", + "t=(300.+12.2)/500.\n", + "print'%s %.6f %s' %(\"\\n\\n Cutting time=\",t*60,\"sec\")\n", + "\n", + "#Answers vary due to aproximations \n", + "\n", + "\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Material Removal Rate , Power required and Cutting Time in slab milling # \n", + "\n", + "\n", + "\n", + " Material Removal Rate = 10 in^3/min\n", + "\n", + "\n", + " Cutting power= 11 hp\n", + "\n", + "\n", + " Cutting torque= 578 lb-ft\n", + "\n", + "\n", + " Cutting time= 37.464000 sec\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 23.2 - PG NO. 655" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 23.2\n", + "#page no. 655\n", + "# Given that\n", + "l=500#in mm Length\n", + "w=60#in mm width\n", + "v=0.6#in m/min \n", + "d=3#in mm depth of cut\n", + "D=150#in mm diameter of cutter\n", + "n=10#no. of inserts\n", + "N=100#in rpm spindle rotation\n", + "\n", + "# Sample Problem on page no. 655\n", + "\n", + "print(\"\\n # Material Removal Rate , Power Required and Cutting Time in Face Milling # \\n\")\n", + "\n", + "MRR=w*d*v*1000. \n", + "print'%s %d %s' %(\"\\n\\n Material Removal Rate = \",MRR,\"mm^3/min\")\n", + "\n", + "lc=D/2.\n", + "t=((l+(2.*lc))/((v*1000.)/60.)) # velocity is converted into mm/sec\n", + "t1=t/60.\n", + "print'%s %.6f %s' %(\"\\n\\n Cutting time= \",t1,\"f min\")\n", + "\n", + "f=(v*1000.*60.)/(60.*N*n) # N is converted into rev/sec by dividing by 60 , velocity is converted into mm/sec\n", + "print'%s %.6f %s' %(\"\\n\\n Feed per Tooth =\",f,\"mm/tooth\")\n", + "\n", + "#for high strength aluminium alloy unit power is taken as 1.1 W s/mm3\n", + "P=(1.1*MRR)/60. # MRR is converted into mm3/sec by dividing by 60\n", + "P1=P/(1000.)#in KW\n", + "print'%s %.6f %s' %(\"\\n\\n Cutting power =\",P1,\"KW\")\n", + "\n", + "\n", + "\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Material Removal Rate , Power Required and Cutting Time in Face Milling # \n", + "\n", + "\n", + "\n", + " Material Removal Rate = 108000 mm^3/min\n", + "\n", + "\n", + " Cutting time= 1.083333 f min\n", + "\n", + "\n", + " Feed per Tooth = 0.600000 mm/tooth\n", + "\n", + "\n", + " Cutting power = 1.980000 KW\n" + ] + } + ], + "prompt_number": 2 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER23_6.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER23_6.ipynb new file mode 100644 index 00000000..ef186b9c --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER23_6.ipynb @@ -0,0 +1,175 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:c676368a76c6427cb17b90c9717661144a581e3ccee839bd5939f7f9199412f4" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 23 - Machining Processes used to Produce Various Shapes" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 23.1 - PG NO. 600" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Example 23.1\n", + "#page no. 600\n", + "# Given that\n", + "import math\n", + "l=12.#in inch Length of block\n", + "w=4\n", + "f=0.01#in inch/tooth feed \n", + "d=0.125#in inch depth of cut\n", + "D=2.#in inch diameter of cutter\n", + "n=20.#no. of teeth\n", + "N=100.#in rpm spindle rotation\n", + "Vt=8.#in inch/minute axial speed of the tool\n", + "\n", + "# Sample Problem on page no. 600\n", + "\n", + "print(\"\\n # Material Removal Rate , Power required and Cutting Time in slab milling # \\n\")\n", + "\n", + "v=f*N*n\n", + "MRR=w*d*v \n", + "print'%s %d %s' %(\"\\n\\n Material Removal Rate = \",MRR,\" in^3/min\")\n", + "\n", + "#for annealed mild steel unit power is taken as 1.1 hp min/in3\n", + "P=1.1*MRR\n", + "print'%s %d %s' %(\"\\n\\n Cutting power=\",P,\"hp\")\n", + "\n", + "T=P*33000/(N*2*3.14)\n", + "print'%s %d %s' %(\"\\n\\n Cutting torque=\",T,\"lb-ft\")\n", + "\n", + "lc=math.sqrt(d*D)\n", + "t=(300.+12.2)/500.\n", + "print'%s %.6f %s' %(\"\\n\\n Cutting time=\",t*60,\"sec\")\n", + "\n", + "#Answers vary due to aproximations \n", + "\n", + "\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Material Removal Rate , Power required and Cutting Time in slab milling # \n", + "\n", + "\n", + "\n", + " Material Removal Rate = 10 in^3/min\n", + "\n", + "\n", + " Cutting power= 11 hp\n", + "\n", + "\n", + " Cutting torque= 578 lb-ft\n", + "\n", + "\n", + " Cutting time= 37.464000 sec\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 23.2 - PG NO. 655" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 23.2\n", + "#page no. 655\n", + "# Given that\n", + "l=500#in mm Length\n", + "w=60#in mm width\n", + "v=0.6#in m/min \n", + "d=3#in mm depth of cut\n", + "D=150#in mm diameter of cutter\n", + "n=10#no. of inserts\n", + "N=100#in rpm spindle rotation\n", + "\n", + "# Sample Problem on page no. 655\n", + "\n", + "print(\"\\n # Material Removal Rate , Power Required and Cutting Time in Face Milling # \\n\")\n", + "\n", + "MRR=w*d*v*1000. \n", + "print'%s %d %s' %(\"\\n\\n Material Removal Rate = \",MRR,\"mm^3/min\")\n", + "\n", + "lc=D/2.\n", + "t=((l+(2.*lc))/((v*1000.)/60.)) # velocity is converted into mm/sec\n", + "t1=t/60.\n", + "print'%s %.6f %s' %(\"\\n\\n Cutting time= \",t1,\"f min\")\n", + "\n", + "f=(v*1000.*60.)/(60.*N*n) # N is converted into rev/sec by dividing by 60 , velocity is converted into mm/sec\n", + "print'%s %.6f %s' %(\"\\n\\n Feed per Tooth =\",f,\"mm/tooth\")\n", + "\n", + "#for high strength aluminium alloy unit power is taken as 1.1 W s/mm3\n", + "P=(1.1*MRR)/60. # MRR is converted into mm3/sec by dividing by 60\n", + "P1=P/(1000.)#in KW\n", + "print'%s %.6f %s' %(\"\\n\\n Cutting power =\",P1,\"KW\")\n", + "\n", + "\n", + "\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Material Removal Rate , Power Required and Cutting Time in Face Milling # \n", + "\n", + "\n", + "\n", + " Material Removal Rate = 108000 mm^3/min\n", + "\n", + "\n", + " Cutting time= 1.083333 f min\n", + "\n", + "\n", + " Feed per Tooth = 0.600000 mm/tooth\n", + "\n", + "\n", + " Cutting power = 1.980000 KW\n" + ] + } + ], + "prompt_number": 2 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER23_7.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER23_7.ipynb new file mode 100644 index 00000000..ef186b9c --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER23_7.ipynb @@ -0,0 +1,175 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:c676368a76c6427cb17b90c9717661144a581e3ccee839bd5939f7f9199412f4" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 23 - Machining Processes used to Produce Various Shapes" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 23.1 - PG NO. 600" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Example 23.1\n", + "#page no. 600\n", + "# Given that\n", + "import math\n", + "l=12.#in inch Length of block\n", + "w=4\n", + "f=0.01#in inch/tooth feed \n", + "d=0.125#in inch depth of cut\n", + "D=2.#in inch diameter of cutter\n", + "n=20.#no. of teeth\n", + "N=100.#in rpm spindle rotation\n", + "Vt=8.#in inch/minute axial speed of the tool\n", + "\n", + "# Sample Problem on page no. 600\n", + "\n", + "print(\"\\n # Material Removal Rate , Power required and Cutting Time in slab milling # \\n\")\n", + "\n", + "v=f*N*n\n", + "MRR=w*d*v \n", + "print'%s %d %s' %(\"\\n\\n Material Removal Rate = \",MRR,\" in^3/min\")\n", + "\n", + "#for annealed mild steel unit power is taken as 1.1 hp min/in3\n", + "P=1.1*MRR\n", + "print'%s %d %s' %(\"\\n\\n Cutting power=\",P,\"hp\")\n", + "\n", + "T=P*33000/(N*2*3.14)\n", + "print'%s %d %s' %(\"\\n\\n Cutting torque=\",T,\"lb-ft\")\n", + "\n", + "lc=math.sqrt(d*D)\n", + "t=(300.+12.2)/500.\n", + "print'%s %.6f %s' %(\"\\n\\n Cutting time=\",t*60,\"sec\")\n", + "\n", + "#Answers vary due to aproximations \n", + "\n", + "\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Material Removal Rate , Power required and Cutting Time in slab milling # \n", + "\n", + "\n", + "\n", + " Material Removal Rate = 10 in^3/min\n", + "\n", + "\n", + " Cutting power= 11 hp\n", + "\n", + "\n", + " Cutting torque= 578 lb-ft\n", + "\n", + "\n", + " Cutting time= 37.464000 sec\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 23.2 - PG NO. 655" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 23.2\n", + "#page no. 655\n", + "# Given that\n", + "l=500#in mm Length\n", + "w=60#in mm width\n", + "v=0.6#in m/min \n", + "d=3#in mm depth of cut\n", + "D=150#in mm diameter of cutter\n", + "n=10#no. of inserts\n", + "N=100#in rpm spindle rotation\n", + "\n", + "# Sample Problem on page no. 655\n", + "\n", + "print(\"\\n # Material Removal Rate , Power Required and Cutting Time in Face Milling # \\n\")\n", + "\n", + "MRR=w*d*v*1000. \n", + "print'%s %d %s' %(\"\\n\\n Material Removal Rate = \",MRR,\"mm^3/min\")\n", + "\n", + "lc=D/2.\n", + "t=((l+(2.*lc))/((v*1000.)/60.)) # velocity is converted into mm/sec\n", + "t1=t/60.\n", + "print'%s %.6f %s' %(\"\\n\\n Cutting time= \",t1,\"f min\")\n", + "\n", + "f=(v*1000.*60.)/(60.*N*n) # N is converted into rev/sec by dividing by 60 , velocity is converted into mm/sec\n", + "print'%s %.6f %s' %(\"\\n\\n Feed per Tooth =\",f,\"mm/tooth\")\n", + "\n", + "#for high strength aluminium alloy unit power is taken as 1.1 W s/mm3\n", + "P=(1.1*MRR)/60. # MRR is converted into mm3/sec by dividing by 60\n", + "P1=P/(1000.)#in KW\n", + "print'%s %.6f %s' %(\"\\n\\n Cutting power =\",P1,\"KW\")\n", + "\n", + "\n", + "\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Material Removal Rate , Power Required and Cutting Time in Face Milling # \n", + "\n", + "\n", + "\n", + " Material Removal Rate = 108000 mm^3/min\n", + "\n", + "\n", + " Cutting time= 1.083333 f min\n", + "\n", + "\n", + " Feed per Tooth = 0.600000 mm/tooth\n", + "\n", + "\n", + " Cutting power = 1.980000 KW\n" + ] + } + ], + "prompt_number": 2 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER23_8.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER23_8.ipynb new file mode 100644 index 00000000..ef186b9c --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER23_8.ipynb @@ -0,0 +1,175 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:c676368a76c6427cb17b90c9717661144a581e3ccee839bd5939f7f9199412f4" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 23 - Machining Processes used to Produce Various Shapes" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 23.1 - PG NO. 600" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Example 23.1\n", + "#page no. 600\n", + "# Given that\n", + "import math\n", + "l=12.#in inch Length of block\n", + "w=4\n", + "f=0.01#in inch/tooth feed \n", + "d=0.125#in inch depth of cut\n", + "D=2.#in inch diameter of cutter\n", + "n=20.#no. of teeth\n", + "N=100.#in rpm spindle rotation\n", + "Vt=8.#in inch/minute axial speed of the tool\n", + "\n", + "# Sample Problem on page no. 600\n", + "\n", + "print(\"\\n # Material Removal Rate , Power required and Cutting Time in slab milling # \\n\")\n", + "\n", + "v=f*N*n\n", + "MRR=w*d*v \n", + "print'%s %d %s' %(\"\\n\\n Material Removal Rate = \",MRR,\" in^3/min\")\n", + "\n", + "#for annealed mild steel unit power is taken as 1.1 hp min/in3\n", + "P=1.1*MRR\n", + "print'%s %d %s' %(\"\\n\\n Cutting power=\",P,\"hp\")\n", + "\n", + "T=P*33000/(N*2*3.14)\n", + "print'%s %d %s' %(\"\\n\\n Cutting torque=\",T,\"lb-ft\")\n", + "\n", + "lc=math.sqrt(d*D)\n", + "t=(300.+12.2)/500.\n", + "print'%s %.6f %s' %(\"\\n\\n Cutting time=\",t*60,\"sec\")\n", + "\n", + "#Answers vary due to aproximations \n", + "\n", + "\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Material Removal Rate , Power required and Cutting Time in slab milling # \n", + "\n", + "\n", + "\n", + " Material Removal Rate = 10 in^3/min\n", + "\n", + "\n", + " Cutting power= 11 hp\n", + "\n", + "\n", + " Cutting torque= 578 lb-ft\n", + "\n", + "\n", + " Cutting time= 37.464000 sec\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 23.2 - PG NO. 655" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 23.2\n", + "#page no. 655\n", + "# Given that\n", + "l=500#in mm Length\n", + "w=60#in mm width\n", + "v=0.6#in m/min \n", + "d=3#in mm depth of cut\n", + "D=150#in mm diameter of cutter\n", + "n=10#no. of inserts\n", + "N=100#in rpm spindle rotation\n", + "\n", + "# Sample Problem on page no. 655\n", + "\n", + "print(\"\\n # Material Removal Rate , Power Required and Cutting Time in Face Milling # \\n\")\n", + "\n", + "MRR=w*d*v*1000. \n", + "print'%s %d %s' %(\"\\n\\n Material Removal Rate = \",MRR,\"mm^3/min\")\n", + "\n", + "lc=D/2.\n", + "t=((l+(2.*lc))/((v*1000.)/60.)) # velocity is converted into mm/sec\n", + "t1=t/60.\n", + "print'%s %.6f %s' %(\"\\n\\n Cutting time= \",t1,\"f min\")\n", + "\n", + "f=(v*1000.*60.)/(60.*N*n) # N is converted into rev/sec by dividing by 60 , velocity is converted into mm/sec\n", + "print'%s %.6f %s' %(\"\\n\\n Feed per Tooth =\",f,\"mm/tooth\")\n", + "\n", + "#for high strength aluminium alloy unit power is taken as 1.1 W s/mm3\n", + "P=(1.1*MRR)/60. # MRR is converted into mm3/sec by dividing by 60\n", + "P1=P/(1000.)#in KW\n", + "print'%s %.6f %s' %(\"\\n\\n Cutting power =\",P1,\"KW\")\n", + "\n", + "\n", + "\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Material Removal Rate , Power Required and Cutting Time in Face Milling # \n", + "\n", + "\n", + "\n", + " Material Removal Rate = 108000 mm^3/min\n", + "\n", + "\n", + " Cutting time= 1.083333 f min\n", + "\n", + "\n", + " Feed per Tooth = 0.600000 mm/tooth\n", + "\n", + "\n", + " Cutting power = 1.980000 KW\n" + ] + } + ], + "prompt_number": 2 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER23_9.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER23_9.ipynb new file mode 100644 index 00000000..ef186b9c --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER23_9.ipynb @@ -0,0 +1,175 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:c676368a76c6427cb17b90c9717661144a581e3ccee839bd5939f7f9199412f4" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 23 - Machining Processes used to Produce Various Shapes" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 23.1 - PG NO. 600" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Example 23.1\n", + "#page no. 600\n", + "# Given that\n", + "import math\n", + "l=12.#in inch Length of block\n", + "w=4\n", + "f=0.01#in inch/tooth feed \n", + "d=0.125#in inch depth of cut\n", + "D=2.#in inch diameter of cutter\n", + "n=20.#no. of teeth\n", + "N=100.#in rpm spindle rotation\n", + "Vt=8.#in inch/minute axial speed of the tool\n", + "\n", + "# Sample Problem on page no. 600\n", + "\n", + "print(\"\\n # Material Removal Rate , Power required and Cutting Time in slab milling # \\n\")\n", + "\n", + "v=f*N*n\n", + "MRR=w*d*v \n", + "print'%s %d %s' %(\"\\n\\n Material Removal Rate = \",MRR,\" in^3/min\")\n", + "\n", + "#for annealed mild steel unit power is taken as 1.1 hp min/in3\n", + "P=1.1*MRR\n", + "print'%s %d %s' %(\"\\n\\n Cutting power=\",P,\"hp\")\n", + "\n", + "T=P*33000/(N*2*3.14)\n", + "print'%s %d %s' %(\"\\n\\n Cutting torque=\",T,\"lb-ft\")\n", + "\n", + "lc=math.sqrt(d*D)\n", + "t=(300.+12.2)/500.\n", + "print'%s %.6f %s' %(\"\\n\\n Cutting time=\",t*60,\"sec\")\n", + "\n", + "#Answers vary due to aproximations \n", + "\n", + "\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Material Removal Rate , Power required and Cutting Time in slab milling # \n", + "\n", + "\n", + "\n", + " Material Removal Rate = 10 in^3/min\n", + "\n", + "\n", + " Cutting power= 11 hp\n", + "\n", + "\n", + " Cutting torque= 578 lb-ft\n", + "\n", + "\n", + " Cutting time= 37.464000 sec\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 23.2 - PG NO. 655" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 23.2\n", + "#page no. 655\n", + "# Given that\n", + "l=500#in mm Length\n", + "w=60#in mm width\n", + "v=0.6#in m/min \n", + "d=3#in mm depth of cut\n", + "D=150#in mm diameter of cutter\n", + "n=10#no. of inserts\n", + "N=100#in rpm spindle rotation\n", + "\n", + "# Sample Problem on page no. 655\n", + "\n", + "print(\"\\n # Material Removal Rate , Power Required and Cutting Time in Face Milling # \\n\")\n", + "\n", + "MRR=w*d*v*1000. \n", + "print'%s %d %s' %(\"\\n\\n Material Removal Rate = \",MRR,\"mm^3/min\")\n", + "\n", + "lc=D/2.\n", + "t=((l+(2.*lc))/((v*1000.)/60.)) # velocity is converted into mm/sec\n", + "t1=t/60.\n", + "print'%s %.6f %s' %(\"\\n\\n Cutting time= \",t1,\"f min\")\n", + "\n", + "f=(v*1000.*60.)/(60.*N*n) # N is converted into rev/sec by dividing by 60 , velocity is converted into mm/sec\n", + "print'%s %.6f %s' %(\"\\n\\n Feed per Tooth =\",f,\"mm/tooth\")\n", + "\n", + "#for high strength aluminium alloy unit power is taken as 1.1 W s/mm3\n", + "P=(1.1*MRR)/60. # MRR is converted into mm3/sec by dividing by 60\n", + "P1=P/(1000.)#in KW\n", + "print'%s %.6f %s' %(\"\\n\\n Cutting power =\",P1,\"KW\")\n", + "\n", + "\n", + "\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Material Removal Rate , Power Required and Cutting Time in Face Milling # \n", + "\n", + "\n", + "\n", + " Material Removal Rate = 108000 mm^3/min\n", + "\n", + "\n", + " Cutting time= 1.083333 f min\n", + "\n", + "\n", + " Feed per Tooth = 0.600000 mm/tooth\n", + "\n", + "\n", + " Cutting power = 1.980000 KW\n" + ] + } + ], + "prompt_number": 2 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER25.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER25.ipynb new file mode 100644 index 00000000..ca6edbc0 --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER25.ipynb @@ -0,0 +1,146 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:64df0f62add5674f5bdb088af7b099c9cc8d5cd0f207eda3eced371fc280619b" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 25 - Abrasive Machining and Finishing Operations" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 25.1 - PG NO. 713" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 25.1\n", + "#page no. 713 \n", + "import math\n", + "# Given that\n", + "D=200#in mm Grinding Wheel diameter \n", + "d=0.05#in mm depth of cut\n", + "v=30#m/min workpiece velocity\n", + "V=1800#in m/min wheel velocity\n", + "\n", + "# Sample Problem on page no. 713\n", + "\n", + "print(\"\\n # Chip Dimensions in Surface Grinding # \\n\")\n", + "\n", + "l=math.sqrt(D*d)\n", + "l1=l/2.54*(10**-1)\n", + "print'%s %.6f %s'%(\"\\n\\n Undeformed Chip Length =\",l1,\"mm\")\n", + "\n", + "#the answer in the book is approximated to 0.13 in\n", + "\n", + "#assume\n", + "C=2.#in mm\n", + "r=15.\n", + "t=math.sqrt(((4*v)/(V*C*r))*math.sqrt(d/D))\n", + "t1=t/2.54*(10**-1)\n", + "print'%s %.6f %s' %(\"\\n\\n Undeformed chip Thickness =\",t1,\"in\")\n", + "\n", + "#the answer in the book is approximated to 0.00023in\n", + "\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Chip Dimensions in Surface Grinding # \n", + "\n", + "\n", + "\n", + " Undeformed Chip Length = 0.124499 mm\n", + "\n", + "\n", + " Undeformed chip Thickness = 0.000233 in\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 25.2 - Pg no. 715" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 25.2\n", + "#page no. 715\n", + "# Given that\n", + "D=10.#in inch Grinding Wheel diameter\n", + "N=4000.#in rpm \n", + "w=1.#in inch \n", + "d=0.002#in inch depth of cut\n", + "v=60.#inch/min feed rate of the workpiece\n", + "\n", + "# Sample Problem on page no. 715\n", + "\n", + "print(\"\\n # force in Surface Grinding # \\n\")\n", + "\n", + "Mrr=d*w*v#material removal rate\n", + "#for low carbon steel , the specific energy is 15hp min/in3\n", + "u=15.#in hp min/in3\n", + "P=u*Mrr*396000.#in lb/min\n", + "Fc = P/(2*3.14*N*(D/2.))\n", + "\n", + "print'%s %.6f %s' %(\"\\n\\n Cutting Force =\",Fc,\"lb\")\n", + "\n", + "\n", + "Fn = Fc+(30./100.)*Fc\n", + "\n", + "print'%s %.6f %s' %(\"\\n\\n Thrust Force =\",Fn,\"lb\")\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # force in Surface Grinding # \n", + "\n", + "\n", + "\n", + " Cutting Force = 5.675159 lb\n", + "\n", + "\n", + " Thrust Force = 7.377707 lb\n" + ] + } + ], + "prompt_number": 2 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER25_1.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER25_1.ipynb new file mode 100644 index 00000000..ca6edbc0 --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER25_1.ipynb @@ -0,0 +1,146 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:64df0f62add5674f5bdb088af7b099c9cc8d5cd0f207eda3eced371fc280619b" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 25 - Abrasive Machining and Finishing Operations" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 25.1 - PG NO. 713" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 25.1\n", + "#page no. 713 \n", + "import math\n", + "# Given that\n", + "D=200#in mm Grinding Wheel diameter \n", + "d=0.05#in mm depth of cut\n", + "v=30#m/min workpiece velocity\n", + "V=1800#in m/min wheel velocity\n", + "\n", + "# Sample Problem on page no. 713\n", + "\n", + "print(\"\\n # Chip Dimensions in Surface Grinding # \\n\")\n", + "\n", + "l=math.sqrt(D*d)\n", + "l1=l/2.54*(10**-1)\n", + "print'%s %.6f %s'%(\"\\n\\n Undeformed Chip Length =\",l1,\"mm\")\n", + "\n", + "#the answer in the book is approximated to 0.13 in\n", + "\n", + "#assume\n", + "C=2.#in mm\n", + "r=15.\n", + "t=math.sqrt(((4*v)/(V*C*r))*math.sqrt(d/D))\n", + "t1=t/2.54*(10**-1)\n", + "print'%s %.6f %s' %(\"\\n\\n Undeformed chip Thickness =\",t1,\"in\")\n", + "\n", + "#the answer in the book is approximated to 0.00023in\n", + "\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Chip Dimensions in Surface Grinding # \n", + "\n", + "\n", + "\n", + " Undeformed Chip Length = 0.124499 mm\n", + "\n", + "\n", + " Undeformed chip Thickness = 0.000233 in\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 25.2 - Pg no. 715" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 25.2\n", + "#page no. 715\n", + "# Given that\n", + "D=10.#in inch Grinding Wheel diameter\n", + "N=4000.#in rpm \n", + "w=1.#in inch \n", + "d=0.002#in inch depth of cut\n", + "v=60.#inch/min feed rate of the workpiece\n", + "\n", + "# Sample Problem on page no. 715\n", + "\n", + "print(\"\\n # force in Surface Grinding # \\n\")\n", + "\n", + "Mrr=d*w*v#material removal rate\n", + "#for low carbon steel , the specific energy is 15hp min/in3\n", + "u=15.#in hp min/in3\n", + "P=u*Mrr*396000.#in lb/min\n", + "Fc = P/(2*3.14*N*(D/2.))\n", + "\n", + "print'%s %.6f %s' %(\"\\n\\n Cutting Force =\",Fc,\"lb\")\n", + "\n", + "\n", + "Fn = Fc+(30./100.)*Fc\n", + "\n", + "print'%s %.6f %s' %(\"\\n\\n Thrust Force =\",Fn,\"lb\")\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # force in Surface Grinding # \n", + "\n", + "\n", + "\n", + " Cutting Force = 5.675159 lb\n", + "\n", + "\n", + " Thrust Force = 7.377707 lb\n" + ] + } + ], + "prompt_number": 2 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER25_2.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER25_2.ipynb new file mode 100644 index 00000000..ca6edbc0 --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER25_2.ipynb @@ -0,0 +1,146 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:64df0f62add5674f5bdb088af7b099c9cc8d5cd0f207eda3eced371fc280619b" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 25 - Abrasive Machining and Finishing Operations" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 25.1 - PG NO. 713" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 25.1\n", + "#page no. 713 \n", + "import math\n", + "# Given that\n", + "D=200#in mm Grinding Wheel diameter \n", + "d=0.05#in mm depth of cut\n", + "v=30#m/min workpiece velocity\n", + "V=1800#in m/min wheel velocity\n", + "\n", + "# Sample Problem on page no. 713\n", + "\n", + "print(\"\\n # Chip Dimensions in Surface Grinding # \\n\")\n", + "\n", + "l=math.sqrt(D*d)\n", + "l1=l/2.54*(10**-1)\n", + "print'%s %.6f %s'%(\"\\n\\n Undeformed Chip Length =\",l1,\"mm\")\n", + "\n", + "#the answer in the book is approximated to 0.13 in\n", + "\n", + "#assume\n", + "C=2.#in mm\n", + "r=15.\n", + "t=math.sqrt(((4*v)/(V*C*r))*math.sqrt(d/D))\n", + "t1=t/2.54*(10**-1)\n", + "print'%s %.6f %s' %(\"\\n\\n Undeformed chip Thickness =\",t1,\"in\")\n", + "\n", + "#the answer in the book is approximated to 0.00023in\n", + "\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Chip Dimensions in Surface Grinding # \n", + "\n", + "\n", + "\n", + " Undeformed Chip Length = 0.124499 mm\n", + "\n", + "\n", + " Undeformed chip Thickness = 0.000233 in\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 25.2 - Pg no. 715" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 25.2\n", + "#page no. 715\n", + "# Given that\n", + "D=10.#in inch Grinding Wheel diameter\n", + "N=4000.#in rpm \n", + "w=1.#in inch \n", + "d=0.002#in inch depth of cut\n", + "v=60.#inch/min feed rate of the workpiece\n", + "\n", + "# Sample Problem on page no. 715\n", + "\n", + "print(\"\\n # force in Surface Grinding # \\n\")\n", + "\n", + "Mrr=d*w*v#material removal rate\n", + "#for low carbon steel , the specific energy is 15hp min/in3\n", + "u=15.#in hp min/in3\n", + "P=u*Mrr*396000.#in lb/min\n", + "Fc = P/(2*3.14*N*(D/2.))\n", + "\n", + "print'%s %.6f %s' %(\"\\n\\n Cutting Force =\",Fc,\"lb\")\n", + "\n", + "\n", + "Fn = Fc+(30./100.)*Fc\n", + "\n", + "print'%s %.6f %s' %(\"\\n\\n Thrust Force =\",Fn,\"lb\")\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # force in Surface Grinding # \n", + "\n", + "\n", + "\n", + " Cutting Force = 5.675159 lb\n", + "\n", + "\n", + " Thrust Force = 7.377707 lb\n" + ] + } + ], + "prompt_number": 2 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER25_3.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER25_3.ipynb new file mode 100644 index 00000000..ca6edbc0 --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER25_3.ipynb @@ -0,0 +1,146 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:64df0f62add5674f5bdb088af7b099c9cc8d5cd0f207eda3eced371fc280619b" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 25 - Abrasive Machining and Finishing Operations" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 25.1 - PG NO. 713" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 25.1\n", + "#page no. 713 \n", + "import math\n", + "# Given that\n", + "D=200#in mm Grinding Wheel diameter \n", + "d=0.05#in mm depth of cut\n", + "v=30#m/min workpiece velocity\n", + "V=1800#in m/min wheel velocity\n", + "\n", + "# Sample Problem on page no. 713\n", + "\n", + "print(\"\\n # Chip Dimensions in Surface Grinding # \\n\")\n", + "\n", + "l=math.sqrt(D*d)\n", + "l1=l/2.54*(10**-1)\n", + "print'%s %.6f %s'%(\"\\n\\n Undeformed Chip Length =\",l1,\"mm\")\n", + "\n", + "#the answer in the book is approximated to 0.13 in\n", + "\n", + "#assume\n", + "C=2.#in mm\n", + "r=15.\n", + "t=math.sqrt(((4*v)/(V*C*r))*math.sqrt(d/D))\n", + "t1=t/2.54*(10**-1)\n", + "print'%s %.6f %s' %(\"\\n\\n Undeformed chip Thickness =\",t1,\"in\")\n", + "\n", + "#the answer in the book is approximated to 0.00023in\n", + "\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Chip Dimensions in Surface Grinding # \n", + "\n", + "\n", + "\n", + " Undeformed Chip Length = 0.124499 mm\n", + "\n", + "\n", + " Undeformed chip Thickness = 0.000233 in\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 25.2 - Pg no. 715" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 25.2\n", + "#page no. 715\n", + "# Given that\n", + "D=10.#in inch Grinding Wheel diameter\n", + "N=4000.#in rpm \n", + "w=1.#in inch \n", + "d=0.002#in inch depth of cut\n", + "v=60.#inch/min feed rate of the workpiece\n", + "\n", + "# Sample Problem on page no. 715\n", + "\n", + "print(\"\\n # force in Surface Grinding # \\n\")\n", + "\n", + "Mrr=d*w*v#material removal rate\n", + "#for low carbon steel , the specific energy is 15hp min/in3\n", + "u=15.#in hp min/in3\n", + "P=u*Mrr*396000.#in lb/min\n", + "Fc = P/(2*3.14*N*(D/2.))\n", + "\n", + "print'%s %.6f %s' %(\"\\n\\n Cutting Force =\",Fc,\"lb\")\n", + "\n", + "\n", + "Fn = Fc+(30./100.)*Fc\n", + "\n", + "print'%s %.6f %s' %(\"\\n\\n Thrust Force =\",Fn,\"lb\")\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # force in Surface Grinding # \n", + "\n", + "\n", + "\n", + " Cutting Force = 5.675159 lb\n", + "\n", + "\n", + " Thrust Force = 7.377707 lb\n" + ] + } + ], + "prompt_number": 2 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER25_4.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER25_4.ipynb new file mode 100644 index 00000000..ca6edbc0 --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER25_4.ipynb @@ -0,0 +1,146 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:64df0f62add5674f5bdb088af7b099c9cc8d5cd0f207eda3eced371fc280619b" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 25 - Abrasive Machining and Finishing Operations" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 25.1 - PG NO. 713" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 25.1\n", + "#page no. 713 \n", + "import math\n", + "# Given that\n", + "D=200#in mm Grinding Wheel diameter \n", + "d=0.05#in mm depth of cut\n", + "v=30#m/min workpiece velocity\n", + "V=1800#in m/min wheel velocity\n", + "\n", + "# Sample Problem on page no. 713\n", + "\n", + "print(\"\\n # Chip Dimensions in Surface Grinding # \\n\")\n", + "\n", + "l=math.sqrt(D*d)\n", + "l1=l/2.54*(10**-1)\n", + "print'%s %.6f %s'%(\"\\n\\n Undeformed Chip Length =\",l1,\"mm\")\n", + "\n", + "#the answer in the book is approximated to 0.13 in\n", + "\n", + "#assume\n", + "C=2.#in mm\n", + "r=15.\n", + "t=math.sqrt(((4*v)/(V*C*r))*math.sqrt(d/D))\n", + "t1=t/2.54*(10**-1)\n", + "print'%s %.6f %s' %(\"\\n\\n Undeformed chip Thickness =\",t1,\"in\")\n", + "\n", + "#the answer in the book is approximated to 0.00023in\n", + "\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Chip Dimensions in Surface Grinding # \n", + "\n", + "\n", + "\n", + " Undeformed Chip Length = 0.124499 mm\n", + "\n", + "\n", + " Undeformed chip Thickness = 0.000233 in\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 25.2 - Pg no. 715" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 25.2\n", + "#page no. 715\n", + "# Given that\n", + "D=10.#in inch Grinding Wheel diameter\n", + "N=4000.#in rpm \n", + "w=1.#in inch \n", + "d=0.002#in inch depth of cut\n", + "v=60.#inch/min feed rate of the workpiece\n", + "\n", + "# Sample Problem on page no. 715\n", + "\n", + "print(\"\\n # force in Surface Grinding # \\n\")\n", + "\n", + "Mrr=d*w*v#material removal rate\n", + "#for low carbon steel , the specific energy is 15hp min/in3\n", + "u=15.#in hp min/in3\n", + "P=u*Mrr*396000.#in lb/min\n", + "Fc = P/(2*3.14*N*(D/2.))\n", + "\n", + "print'%s %.6f %s' %(\"\\n\\n Cutting Force =\",Fc,\"lb\")\n", + "\n", + "\n", + "Fn = Fc+(30./100.)*Fc\n", + "\n", + "print'%s %.6f %s' %(\"\\n\\n Thrust Force =\",Fn,\"lb\")\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # force in Surface Grinding # \n", + "\n", + "\n", + "\n", + " Cutting Force = 5.675159 lb\n", + "\n", + "\n", + " Thrust Force = 7.377707 lb\n" + ] + } + ], + "prompt_number": 2 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER25_5.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER25_5.ipynb new file mode 100644 index 00000000..ca6edbc0 --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER25_5.ipynb @@ -0,0 +1,146 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:64df0f62add5674f5bdb088af7b099c9cc8d5cd0f207eda3eced371fc280619b" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 25 - Abrasive Machining and Finishing Operations" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 25.1 - PG NO. 713" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 25.1\n", + "#page no. 713 \n", + "import math\n", + "# Given that\n", + "D=200#in mm Grinding Wheel diameter \n", + "d=0.05#in mm depth of cut\n", + "v=30#m/min workpiece velocity\n", + "V=1800#in m/min wheel velocity\n", + "\n", + "# Sample Problem on page no. 713\n", + "\n", + "print(\"\\n # Chip Dimensions in Surface Grinding # \\n\")\n", + "\n", + "l=math.sqrt(D*d)\n", + "l1=l/2.54*(10**-1)\n", + "print'%s %.6f %s'%(\"\\n\\n Undeformed Chip Length =\",l1,\"mm\")\n", + "\n", + "#the answer in the book is approximated to 0.13 in\n", + "\n", + "#assume\n", + "C=2.#in mm\n", + "r=15.\n", + "t=math.sqrt(((4*v)/(V*C*r))*math.sqrt(d/D))\n", + "t1=t/2.54*(10**-1)\n", + "print'%s %.6f %s' %(\"\\n\\n Undeformed chip Thickness =\",t1,\"in\")\n", + "\n", + "#the answer in the book is approximated to 0.00023in\n", + "\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Chip Dimensions in Surface Grinding # \n", + "\n", + "\n", + "\n", + " Undeformed Chip Length = 0.124499 mm\n", + "\n", + "\n", + " Undeformed chip Thickness = 0.000233 in\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 25.2 - Pg no. 715" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 25.2\n", + "#page no. 715\n", + "# Given that\n", + "D=10.#in inch Grinding Wheel diameter\n", + "N=4000.#in rpm \n", + "w=1.#in inch \n", + "d=0.002#in inch depth of cut\n", + "v=60.#inch/min feed rate of the workpiece\n", + "\n", + "# Sample Problem on page no. 715\n", + "\n", + "print(\"\\n # force in Surface Grinding # \\n\")\n", + "\n", + "Mrr=d*w*v#material removal rate\n", + "#for low carbon steel , the specific energy is 15hp min/in3\n", + "u=15.#in hp min/in3\n", + "P=u*Mrr*396000.#in lb/min\n", + "Fc = P/(2*3.14*N*(D/2.))\n", + "\n", + "print'%s %.6f %s' %(\"\\n\\n Cutting Force =\",Fc,\"lb\")\n", + "\n", + "\n", + "Fn = Fc+(30./100.)*Fc\n", + "\n", + "print'%s %.6f %s' %(\"\\n\\n Thrust Force =\",Fn,\"lb\")\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # force in Surface Grinding # \n", + "\n", + "\n", + "\n", + " Cutting Force = 5.675159 lb\n", + "\n", + "\n", + " Thrust Force = 7.377707 lb\n" + ] + } + ], + "prompt_number": 2 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER25_6.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER25_6.ipynb new file mode 100644 index 00000000..ca6edbc0 --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER25_6.ipynb @@ -0,0 +1,146 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:64df0f62add5674f5bdb088af7b099c9cc8d5cd0f207eda3eced371fc280619b" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 25 - Abrasive Machining and Finishing Operations" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 25.1 - PG NO. 713" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 25.1\n", + "#page no. 713 \n", + "import math\n", + "# Given that\n", + "D=200#in mm Grinding Wheel diameter \n", + "d=0.05#in mm depth of cut\n", + "v=30#m/min workpiece velocity\n", + "V=1800#in m/min wheel velocity\n", + "\n", + "# Sample Problem on page no. 713\n", + "\n", + "print(\"\\n # Chip Dimensions in Surface Grinding # \\n\")\n", + "\n", + "l=math.sqrt(D*d)\n", + "l1=l/2.54*(10**-1)\n", + "print'%s %.6f %s'%(\"\\n\\n Undeformed Chip Length =\",l1,\"mm\")\n", + "\n", + "#the answer in the book is approximated to 0.13 in\n", + "\n", + "#assume\n", + "C=2.#in mm\n", + "r=15.\n", + "t=math.sqrt(((4*v)/(V*C*r))*math.sqrt(d/D))\n", + "t1=t/2.54*(10**-1)\n", + "print'%s %.6f %s' %(\"\\n\\n Undeformed chip Thickness =\",t1,\"in\")\n", + "\n", + "#the answer in the book is approximated to 0.00023in\n", + "\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Chip Dimensions in Surface Grinding # \n", + "\n", + "\n", + "\n", + " Undeformed Chip Length = 0.124499 mm\n", + "\n", + "\n", + " Undeformed chip Thickness = 0.000233 in\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 25.2 - Pg no. 715" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 25.2\n", + "#page no. 715\n", + "# Given that\n", + "D=10.#in inch Grinding Wheel diameter\n", + "N=4000.#in rpm \n", + "w=1.#in inch \n", + "d=0.002#in inch depth of cut\n", + "v=60.#inch/min feed rate of the workpiece\n", + "\n", + "# Sample Problem on page no. 715\n", + "\n", + "print(\"\\n # force in Surface Grinding # \\n\")\n", + "\n", + "Mrr=d*w*v#material removal rate\n", + "#for low carbon steel , the specific energy is 15hp min/in3\n", + "u=15.#in hp min/in3\n", + "P=u*Mrr*396000.#in lb/min\n", + "Fc = P/(2*3.14*N*(D/2.))\n", + "\n", + "print'%s %.6f %s' %(\"\\n\\n Cutting Force =\",Fc,\"lb\")\n", + "\n", + "\n", + "Fn = Fc+(30./100.)*Fc\n", + "\n", + "print'%s %.6f %s' %(\"\\n\\n Thrust Force =\",Fn,\"lb\")\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # force in Surface Grinding # \n", + "\n", + "\n", + "\n", + " Cutting Force = 5.675159 lb\n", + "\n", + "\n", + " Thrust Force = 7.377707 lb\n" + ] + } + ], + "prompt_number": 2 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER25_7.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER25_7.ipynb new file mode 100644 index 00000000..ca6edbc0 --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER25_7.ipynb @@ -0,0 +1,146 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:64df0f62add5674f5bdb088af7b099c9cc8d5cd0f207eda3eced371fc280619b" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 25 - Abrasive Machining and Finishing Operations" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 25.1 - PG NO. 713" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 25.1\n", + "#page no. 713 \n", + "import math\n", + "# Given that\n", + "D=200#in mm Grinding Wheel diameter \n", + "d=0.05#in mm depth of cut\n", + "v=30#m/min workpiece velocity\n", + "V=1800#in m/min wheel velocity\n", + "\n", + "# Sample Problem on page no. 713\n", + "\n", + "print(\"\\n # Chip Dimensions in Surface Grinding # \\n\")\n", + "\n", + "l=math.sqrt(D*d)\n", + "l1=l/2.54*(10**-1)\n", + "print'%s %.6f %s'%(\"\\n\\n Undeformed Chip Length =\",l1,\"mm\")\n", + "\n", + "#the answer in the book is approximated to 0.13 in\n", + "\n", + "#assume\n", + "C=2.#in mm\n", + "r=15.\n", + "t=math.sqrt(((4*v)/(V*C*r))*math.sqrt(d/D))\n", + "t1=t/2.54*(10**-1)\n", + "print'%s %.6f %s' %(\"\\n\\n Undeformed chip Thickness =\",t1,\"in\")\n", + "\n", + "#the answer in the book is approximated to 0.00023in\n", + "\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Chip Dimensions in Surface Grinding # \n", + "\n", + "\n", + "\n", + " Undeformed Chip Length = 0.124499 mm\n", + "\n", + "\n", + " Undeformed chip Thickness = 0.000233 in\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 25.2 - Pg no. 715" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 25.2\n", + "#page no. 715\n", + "# Given that\n", + "D=10.#in inch Grinding Wheel diameter\n", + "N=4000.#in rpm \n", + "w=1.#in inch \n", + "d=0.002#in inch depth of cut\n", + "v=60.#inch/min feed rate of the workpiece\n", + "\n", + "# Sample Problem on page no. 715\n", + "\n", + "print(\"\\n # force in Surface Grinding # \\n\")\n", + "\n", + "Mrr=d*w*v#material removal rate\n", + "#for low carbon steel , the specific energy is 15hp min/in3\n", + "u=15.#in hp min/in3\n", + "P=u*Mrr*396000.#in lb/min\n", + "Fc = P/(2*3.14*N*(D/2.))\n", + "\n", + "print'%s %.6f %s' %(\"\\n\\n Cutting Force =\",Fc,\"lb\")\n", + "\n", + "\n", + "Fn = Fc+(30./100.)*Fc\n", + "\n", + "print'%s %.6f %s' %(\"\\n\\n Thrust Force =\",Fn,\"lb\")\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # force in Surface Grinding # \n", + "\n", + "\n", + "\n", + " Cutting Force = 5.675159 lb\n", + "\n", + "\n", + " Thrust Force = 7.377707 lb\n" + ] + } + ], + "prompt_number": 2 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER25_8.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER25_8.ipynb new file mode 100644 index 00000000..ca6edbc0 --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER25_8.ipynb @@ -0,0 +1,146 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:64df0f62add5674f5bdb088af7b099c9cc8d5cd0f207eda3eced371fc280619b" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 25 - Abrasive Machining and Finishing Operations" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 25.1 - PG NO. 713" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 25.1\n", + "#page no. 713 \n", + "import math\n", + "# Given that\n", + "D=200#in mm Grinding Wheel diameter \n", + "d=0.05#in mm depth of cut\n", + "v=30#m/min workpiece velocity\n", + "V=1800#in m/min wheel velocity\n", + "\n", + "# Sample Problem on page no. 713\n", + "\n", + "print(\"\\n # Chip Dimensions in Surface Grinding # \\n\")\n", + "\n", + "l=math.sqrt(D*d)\n", + "l1=l/2.54*(10**-1)\n", + "print'%s %.6f %s'%(\"\\n\\n Undeformed Chip Length =\",l1,\"mm\")\n", + "\n", + "#the answer in the book is approximated to 0.13 in\n", + "\n", + "#assume\n", + "C=2.#in mm\n", + "r=15.\n", + "t=math.sqrt(((4*v)/(V*C*r))*math.sqrt(d/D))\n", + "t1=t/2.54*(10**-1)\n", + "print'%s %.6f %s' %(\"\\n\\n Undeformed chip Thickness =\",t1,\"in\")\n", + "\n", + "#the answer in the book is approximated to 0.00023in\n", + "\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Chip Dimensions in Surface Grinding # \n", + "\n", + "\n", + "\n", + " Undeformed Chip Length = 0.124499 mm\n", + "\n", + "\n", + " Undeformed chip Thickness = 0.000233 in\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 25.2 - Pg no. 715" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 25.2\n", + "#page no. 715\n", + "# Given that\n", + "D=10.#in inch Grinding Wheel diameter\n", + "N=4000.#in rpm \n", + "w=1.#in inch \n", + "d=0.002#in inch depth of cut\n", + "v=60.#inch/min feed rate of the workpiece\n", + "\n", + "# Sample Problem on page no. 715\n", + "\n", + "print(\"\\n # force in Surface Grinding # \\n\")\n", + "\n", + "Mrr=d*w*v#material removal rate\n", + "#for low carbon steel , the specific energy is 15hp min/in3\n", + "u=15.#in hp min/in3\n", + "P=u*Mrr*396000.#in lb/min\n", + "Fc = P/(2*3.14*N*(D/2.))\n", + "\n", + "print'%s %.6f %s' %(\"\\n\\n Cutting Force =\",Fc,\"lb\")\n", + "\n", + "\n", + "Fn = Fc+(30./100.)*Fc\n", + "\n", + "print'%s %.6f %s' %(\"\\n\\n Thrust Force =\",Fn,\"lb\")\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # force in Surface Grinding # \n", + "\n", + "\n", + "\n", + " Cutting Force = 5.675159 lb\n", + "\n", + "\n", + " Thrust Force = 7.377707 lb\n" + ] + } + ], + "prompt_number": 2 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER28.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER28.ipynb new file mode 100644 index 00000000..69d0545d --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER28.ipynb @@ -0,0 +1,90 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:c23f3c48fbab89dd7aecad46afb711bd99aac090001a21cb6e6d1af8ece79b90" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 28 - Solid-State Welding Processes" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 28.1 - PG NO. 805" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 28.1\n", + "#page no. 805\n", + "\n", + "# Given that\n", + "t=1.#in mm thickness of chip\n", + "I=5000.#in Ampere current\n", + "T=0.1#in sec\n", + "d=5.#in mm diameter of electrode\n", + "\n", + "\n", + "# Sample Problem on page no. 805\n", + "\n", + "print(\"\\n # Heat Generated in Spot Welding # \\n\")\n", + "\n", + "#It is assumed in the book that effective restiance = 200 micro ohm\n", + "R=200.*(10.**-6.)\n", + "H=(I**2.)*R*T\n", + "\n", + "print'%s %d %s' %(\"\\n\\n Heat Generated =\",H,\"J\")\n", + "\n", + "# It is assumed in the book that \n", + "V=30.#in mm3 volume\n", + "D=0.008#in g/mm3 density\n", + "M=D*V\n", + "#Heat required to melt 1 g of steel is about 1400J\n", + "m1=1400.*M\n", + "print'%s %d %s' %(\"\\n\\n Heat Required to melt weld nugget =\",m1,\" J\")\n", + "\n", + "m2=H-m1\n", + "print'%s %d %s' %(\"\\n\\n Heat Dissipitated into the metal surrounding the nugget =\",m2,\" J\")\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Heat Generated in Spot Welding # \n", + "\n", + "\n", + "\n", + " Heat Generated = 500 J\n", + "\n", + "\n", + " Heat Required to melt weld nugget = 336 J\n", + "\n", + "\n", + " Heat Dissipitated into the metal surrounding the nugget = 164 J\n" + ] + } + ], + "prompt_number": 1 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER28_1.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER28_1.ipynb new file mode 100644 index 00000000..69d0545d --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER28_1.ipynb @@ -0,0 +1,90 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:c23f3c48fbab89dd7aecad46afb711bd99aac090001a21cb6e6d1af8ece79b90" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 28 - Solid-State Welding Processes" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 28.1 - PG NO. 805" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 28.1\n", + "#page no. 805\n", + "\n", + "# Given that\n", + "t=1.#in mm thickness of chip\n", + "I=5000.#in Ampere current\n", + "T=0.1#in sec\n", + "d=5.#in mm diameter of electrode\n", + "\n", + "\n", + "# Sample Problem on page no. 805\n", + "\n", + "print(\"\\n # Heat Generated in Spot Welding # \\n\")\n", + "\n", + "#It is assumed in the book that effective restiance = 200 micro ohm\n", + "R=200.*(10.**-6.)\n", + "H=(I**2.)*R*T\n", + "\n", + "print'%s %d %s' %(\"\\n\\n Heat Generated =\",H,\"J\")\n", + "\n", + "# It is assumed in the book that \n", + "V=30.#in mm3 volume\n", + "D=0.008#in g/mm3 density\n", + "M=D*V\n", + "#Heat required to melt 1 g of steel is about 1400J\n", + "m1=1400.*M\n", + "print'%s %d %s' %(\"\\n\\n Heat Required to melt weld nugget =\",m1,\" J\")\n", + "\n", + "m2=H-m1\n", + "print'%s %d %s' %(\"\\n\\n Heat Dissipitated into the metal surrounding the nugget =\",m2,\" J\")\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Heat Generated in Spot Welding # \n", + "\n", + "\n", + "\n", + " Heat Generated = 500 J\n", + "\n", + "\n", + " Heat Required to melt weld nugget = 336 J\n", + "\n", + "\n", + " Heat Dissipitated into the metal surrounding the nugget = 164 J\n" + ] + } + ], + "prompt_number": 1 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER28_2.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER28_2.ipynb new file mode 100644 index 00000000..69d0545d --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER28_2.ipynb @@ -0,0 +1,90 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:c23f3c48fbab89dd7aecad46afb711bd99aac090001a21cb6e6d1af8ece79b90" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 28 - Solid-State Welding Processes" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 28.1 - PG NO. 805" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 28.1\n", + "#page no. 805\n", + "\n", + "# Given that\n", + "t=1.#in mm thickness of chip\n", + "I=5000.#in Ampere current\n", + "T=0.1#in sec\n", + "d=5.#in mm diameter of electrode\n", + "\n", + "\n", + "# Sample Problem on page no. 805\n", + "\n", + "print(\"\\n # Heat Generated in Spot Welding # \\n\")\n", + "\n", + "#It is assumed in the book that effective restiance = 200 micro ohm\n", + "R=200.*(10.**-6.)\n", + "H=(I**2.)*R*T\n", + "\n", + "print'%s %d %s' %(\"\\n\\n Heat Generated =\",H,\"J\")\n", + "\n", + "# It is assumed in the book that \n", + "V=30.#in mm3 volume\n", + "D=0.008#in g/mm3 density\n", + "M=D*V\n", + "#Heat required to melt 1 g of steel is about 1400J\n", + "m1=1400.*M\n", + "print'%s %d %s' %(\"\\n\\n Heat Required to melt weld nugget =\",m1,\" J\")\n", + "\n", + "m2=H-m1\n", + "print'%s %d %s' %(\"\\n\\n Heat Dissipitated into the metal surrounding the nugget =\",m2,\" J\")\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Heat Generated in Spot Welding # \n", + "\n", + "\n", + "\n", + " Heat Generated = 500 J\n", + "\n", + "\n", + " Heat Required to melt weld nugget = 336 J\n", + "\n", + "\n", + " Heat Dissipitated into the metal surrounding the nugget = 164 J\n" + ] + } + ], + "prompt_number": 1 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER28_3.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER28_3.ipynb new file mode 100644 index 00000000..69d0545d --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER28_3.ipynb @@ -0,0 +1,90 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:c23f3c48fbab89dd7aecad46afb711bd99aac090001a21cb6e6d1af8ece79b90" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 28 - Solid-State Welding Processes" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 28.1 - PG NO. 805" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 28.1\n", + "#page no. 805\n", + "\n", + "# Given that\n", + "t=1.#in mm thickness of chip\n", + "I=5000.#in Ampere current\n", + "T=0.1#in sec\n", + "d=5.#in mm diameter of electrode\n", + "\n", + "\n", + "# Sample Problem on page no. 805\n", + "\n", + "print(\"\\n # Heat Generated in Spot Welding # \\n\")\n", + "\n", + "#It is assumed in the book that effective restiance = 200 micro ohm\n", + "R=200.*(10.**-6.)\n", + "H=(I**2.)*R*T\n", + "\n", + "print'%s %d %s' %(\"\\n\\n Heat Generated =\",H,\"J\")\n", + "\n", + "# It is assumed in the book that \n", + "V=30.#in mm3 volume\n", + "D=0.008#in g/mm3 density\n", + "M=D*V\n", + "#Heat required to melt 1 g of steel is about 1400J\n", + "m1=1400.*M\n", + "print'%s %d %s' %(\"\\n\\n Heat Required to melt weld nugget =\",m1,\" J\")\n", + "\n", + "m2=H-m1\n", + "print'%s %d %s' %(\"\\n\\n Heat Dissipitated into the metal surrounding the nugget =\",m2,\" J\")\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Heat Generated in Spot Welding # \n", + "\n", + "\n", + "\n", + " Heat Generated = 500 J\n", + "\n", + "\n", + " Heat Required to melt weld nugget = 336 J\n", + "\n", + "\n", + " Heat Dissipitated into the metal surrounding the nugget = 164 J\n" + ] + } + ], + "prompt_number": 1 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER28_4.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER28_4.ipynb new file mode 100644 index 00000000..69d0545d --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER28_4.ipynb @@ -0,0 +1,90 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:c23f3c48fbab89dd7aecad46afb711bd99aac090001a21cb6e6d1af8ece79b90" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 28 - Solid-State Welding Processes" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 28.1 - PG NO. 805" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 28.1\n", + "#page no. 805\n", + "\n", + "# Given that\n", + "t=1.#in mm thickness of chip\n", + "I=5000.#in Ampere current\n", + "T=0.1#in sec\n", + "d=5.#in mm diameter of electrode\n", + "\n", + "\n", + "# Sample Problem on page no. 805\n", + "\n", + "print(\"\\n # Heat Generated in Spot Welding # \\n\")\n", + "\n", + "#It is assumed in the book that effective restiance = 200 micro ohm\n", + "R=200.*(10.**-6.)\n", + "H=(I**2.)*R*T\n", + "\n", + "print'%s %d %s' %(\"\\n\\n Heat Generated =\",H,\"J\")\n", + "\n", + "# It is assumed in the book that \n", + "V=30.#in mm3 volume\n", + "D=0.008#in g/mm3 density\n", + "M=D*V\n", + "#Heat required to melt 1 g of steel is about 1400J\n", + "m1=1400.*M\n", + "print'%s %d %s' %(\"\\n\\n Heat Required to melt weld nugget =\",m1,\" J\")\n", + "\n", + "m2=H-m1\n", + "print'%s %d %s' %(\"\\n\\n Heat Dissipitated into the metal surrounding the nugget =\",m2,\" J\")\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Heat Generated in Spot Welding # \n", + "\n", + "\n", + "\n", + " Heat Generated = 500 J\n", + "\n", + "\n", + " Heat Required to melt weld nugget = 336 J\n", + "\n", + "\n", + " Heat Dissipitated into the metal surrounding the nugget = 164 J\n" + ] + } + ], + "prompt_number": 1 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER28_5.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER28_5.ipynb new file mode 100644 index 00000000..69d0545d --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER28_5.ipynb @@ -0,0 +1,90 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:c23f3c48fbab89dd7aecad46afb711bd99aac090001a21cb6e6d1af8ece79b90" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 28 - Solid-State Welding Processes" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 28.1 - PG NO. 805" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 28.1\n", + "#page no. 805\n", + "\n", + "# Given that\n", + "t=1.#in mm thickness of chip\n", + "I=5000.#in Ampere current\n", + "T=0.1#in sec\n", + "d=5.#in mm diameter of electrode\n", + "\n", + "\n", + "# Sample Problem on page no. 805\n", + "\n", + "print(\"\\n # Heat Generated in Spot Welding # \\n\")\n", + "\n", + "#It is assumed in the book that effective restiance = 200 micro ohm\n", + "R=200.*(10.**-6.)\n", + "H=(I**2.)*R*T\n", + "\n", + "print'%s %d %s' %(\"\\n\\n Heat Generated =\",H,\"J\")\n", + "\n", + "# It is assumed in the book that \n", + "V=30.#in mm3 volume\n", + "D=0.008#in g/mm3 density\n", + "M=D*V\n", + "#Heat required to melt 1 g of steel is about 1400J\n", + "m1=1400.*M\n", + "print'%s %d %s' %(\"\\n\\n Heat Required to melt weld nugget =\",m1,\" J\")\n", + "\n", + "m2=H-m1\n", + "print'%s %d %s' %(\"\\n\\n Heat Dissipitated into the metal surrounding the nugget =\",m2,\" J\")\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Heat Generated in Spot Welding # \n", + "\n", + "\n", + "\n", + " Heat Generated = 500 J\n", + "\n", + "\n", + " Heat Required to melt weld nugget = 336 J\n", + "\n", + "\n", + " Heat Dissipitated into the metal surrounding the nugget = 164 J\n" + ] + } + ], + "prompt_number": 1 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER28_6.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER28_6.ipynb new file mode 100644 index 00000000..69d0545d --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER28_6.ipynb @@ -0,0 +1,90 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:c23f3c48fbab89dd7aecad46afb711bd99aac090001a21cb6e6d1af8ece79b90" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 28 - Solid-State Welding Processes" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 28.1 - PG NO. 805" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 28.1\n", + "#page no. 805\n", + "\n", + "# Given that\n", + "t=1.#in mm thickness of chip\n", + "I=5000.#in Ampere current\n", + "T=0.1#in sec\n", + "d=5.#in mm diameter of electrode\n", + "\n", + "\n", + "# Sample Problem on page no. 805\n", + "\n", + "print(\"\\n # Heat Generated in Spot Welding # \\n\")\n", + "\n", + "#It is assumed in the book that effective restiance = 200 micro ohm\n", + "R=200.*(10.**-6.)\n", + "H=(I**2.)*R*T\n", + "\n", + "print'%s %d %s' %(\"\\n\\n Heat Generated =\",H,\"J\")\n", + "\n", + "# It is assumed in the book that \n", + "V=30.#in mm3 volume\n", + "D=0.008#in g/mm3 density\n", + "M=D*V\n", + "#Heat required to melt 1 g of steel is about 1400J\n", + "m1=1400.*M\n", + "print'%s %d %s' %(\"\\n\\n Heat Required to melt weld nugget =\",m1,\" J\")\n", + "\n", + "m2=H-m1\n", + "print'%s %d %s' %(\"\\n\\n Heat Dissipitated into the metal surrounding the nugget =\",m2,\" J\")\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Heat Generated in Spot Welding # \n", + "\n", + "\n", + "\n", + " Heat Generated = 500 J\n", + "\n", + "\n", + " Heat Required to melt weld nugget = 336 J\n", + "\n", + "\n", + " Heat Dissipitated into the metal surrounding the nugget = 164 J\n" + ] + } + ], + "prompt_number": 1 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER28_7.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER28_7.ipynb new file mode 100644 index 00000000..69d0545d --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER28_7.ipynb @@ -0,0 +1,90 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:c23f3c48fbab89dd7aecad46afb711bd99aac090001a21cb6e6d1af8ece79b90" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 28 - Solid-State Welding Processes" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 28.1 - PG NO. 805" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 28.1\n", + "#page no. 805\n", + "\n", + "# Given that\n", + "t=1.#in mm thickness of chip\n", + "I=5000.#in Ampere current\n", + "T=0.1#in sec\n", + "d=5.#in mm diameter of electrode\n", + "\n", + "\n", + "# Sample Problem on page no. 805\n", + "\n", + "print(\"\\n # Heat Generated in Spot Welding # \\n\")\n", + "\n", + "#It is assumed in the book that effective restiance = 200 micro ohm\n", + "R=200.*(10.**-6.)\n", + "H=(I**2.)*R*T\n", + "\n", + "print'%s %d %s' %(\"\\n\\n Heat Generated =\",H,\"J\")\n", + "\n", + "# It is assumed in the book that \n", + "V=30.#in mm3 volume\n", + "D=0.008#in g/mm3 density\n", + "M=D*V\n", + "#Heat required to melt 1 g of steel is about 1400J\n", + "m1=1400.*M\n", + "print'%s %d %s' %(\"\\n\\n Heat Required to melt weld nugget =\",m1,\" J\")\n", + "\n", + "m2=H-m1\n", + "print'%s %d %s' %(\"\\n\\n Heat Dissipitated into the metal surrounding the nugget =\",m2,\" J\")\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Heat Generated in Spot Welding # \n", + "\n", + "\n", + "\n", + " Heat Generated = 500 J\n", + "\n", + "\n", + " Heat Required to melt weld nugget = 336 J\n", + "\n", + "\n", + " Heat Dissipitated into the metal surrounding the nugget = 164 J\n" + ] + } + ], + "prompt_number": 1 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER28_8.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER28_8.ipynb new file mode 100644 index 00000000..69d0545d --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER28_8.ipynb @@ -0,0 +1,90 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:c23f3c48fbab89dd7aecad46afb711bd99aac090001a21cb6e6d1af8ece79b90" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 28 - Solid-State Welding Processes" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 28.1 - PG NO. 805" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 28.1\n", + "#page no. 805\n", + "\n", + "# Given that\n", + "t=1.#in mm thickness of chip\n", + "I=5000.#in Ampere current\n", + "T=0.1#in sec\n", + "d=5.#in mm diameter of electrode\n", + "\n", + "\n", + "# Sample Problem on page no. 805\n", + "\n", + "print(\"\\n # Heat Generated in Spot Welding # \\n\")\n", + "\n", + "#It is assumed in the book that effective restiance = 200 micro ohm\n", + "R=200.*(10.**-6.)\n", + "H=(I**2.)*R*T\n", + "\n", + "print'%s %d %s' %(\"\\n\\n Heat Generated =\",H,\"J\")\n", + "\n", + "# It is assumed in the book that \n", + "V=30.#in mm3 volume\n", + "D=0.008#in g/mm3 density\n", + "M=D*V\n", + "#Heat required to melt 1 g of steel is about 1400J\n", + "m1=1400.*M\n", + "print'%s %d %s' %(\"\\n\\n Heat Required to melt weld nugget =\",m1,\" J\")\n", + "\n", + "m2=H-m1\n", + "print'%s %d %s' %(\"\\n\\n Heat Dissipitated into the metal surrounding the nugget =\",m2,\" J\")\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Heat Generated in Spot Welding # \n", + "\n", + "\n", + "\n", + " Heat Generated = 500 J\n", + "\n", + "\n", + " Heat Required to melt weld nugget = 336 J\n", + "\n", + "\n", + " Heat Dissipitated into the metal surrounding the nugget = 164 J\n" + ] + } + ], + "prompt_number": 1 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER2_1.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER2_1.ipynb new file mode 100644 index 00000000..40a2982b --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER2_1.ipynb @@ -0,0 +1,82 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:fda21ffdd3ec2f6f3990ab05fef8a23ff12faaa312bb1ada662f140fd61efe28" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 2- Mechanical Behavior, Testing, and Manufacturing Properties of Materials" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 2.1 - PG NO. 63" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 2.1,chapter 2, page 63\n", + "\n", + "# Given that\n", + "#True stress=100000*(True strain)**0.5\n", + "\n", + "# Sample Problem on page no. 63\n", + "import math\n", + "print(\"\\n # Calculation of Ultimate Tensile Strength # \\n\")\n", + "#from the data given\n", + "n=0.5\n", + "E=0.5\n", + "K=100000.\n", + "Truestress=K*((E)**n)\n", + "#let An(area of neck)/Ao=t\n", + "#from math.log(Ao/An)=n\n", + "print'%s %.3f %s' %(\"true Ultimate Tensile Strength =\",Truestress,\"psi \\n\")\n", + "t=math.exp(-n)\n", + "print'%s %.7f %s' %(\"t =\",t,\"\\n\")\n", + "UTS=Truestress*t#from the math.expression UTS= P/Ao where P(Maximum Load)=Truestress*An\n", + "print'%s %.3f %s' %(\"Ultimate Tensile Strength =\",UTS,\"psi\")\n", + "#answer in the book is approximated to 42850 psi \n", + "\n", + "\n", + "\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Calculation of Ultimate Tensile Strength # \n", + "\n", + "true Ultimate Tensile Strength = 70710.678 psi \n", + "\n", + "t = 0.6065307 \n", + "\n", + "Ultimate Tensile Strength = 42888.194 psi\n" + ] + } + ], + "prompt_number": 1 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER2_2.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER2_2.ipynb new file mode 100644 index 00000000..40a2982b --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER2_2.ipynb @@ -0,0 +1,82 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:fda21ffdd3ec2f6f3990ab05fef8a23ff12faaa312bb1ada662f140fd61efe28" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 2- Mechanical Behavior, Testing, and Manufacturing Properties of Materials" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 2.1 - PG NO. 63" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 2.1,chapter 2, page 63\n", + "\n", + "# Given that\n", + "#True stress=100000*(True strain)**0.5\n", + "\n", + "# Sample Problem on page no. 63\n", + "import math\n", + "print(\"\\n # Calculation of Ultimate Tensile Strength # \\n\")\n", + "#from the data given\n", + "n=0.5\n", + "E=0.5\n", + "K=100000.\n", + "Truestress=K*((E)**n)\n", + "#let An(area of neck)/Ao=t\n", + "#from math.log(Ao/An)=n\n", + "print'%s %.3f %s' %(\"true Ultimate Tensile Strength =\",Truestress,\"psi \\n\")\n", + "t=math.exp(-n)\n", + "print'%s %.7f %s' %(\"t =\",t,\"\\n\")\n", + "UTS=Truestress*t#from the math.expression UTS= P/Ao where P(Maximum Load)=Truestress*An\n", + "print'%s %.3f %s' %(\"Ultimate Tensile Strength =\",UTS,\"psi\")\n", + "#answer in the book is approximated to 42850 psi \n", + "\n", + "\n", + "\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Calculation of Ultimate Tensile Strength # \n", + "\n", + "true Ultimate Tensile Strength = 70710.678 psi \n", + "\n", + "t = 0.6065307 \n", + "\n", + "Ultimate Tensile Strength = 42888.194 psi\n" + ] + } + ], + "prompt_number": 1 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER2_3.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER2_3.ipynb new file mode 100644 index 00000000..40a2982b --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER2_3.ipynb @@ -0,0 +1,82 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:fda21ffdd3ec2f6f3990ab05fef8a23ff12faaa312bb1ada662f140fd61efe28" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 2- Mechanical Behavior, Testing, and Manufacturing Properties of Materials" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 2.1 - PG NO. 63" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 2.1,chapter 2, page 63\n", + "\n", + "# Given that\n", + "#True stress=100000*(True strain)**0.5\n", + "\n", + "# Sample Problem on page no. 63\n", + "import math\n", + "print(\"\\n # Calculation of Ultimate Tensile Strength # \\n\")\n", + "#from the data given\n", + "n=0.5\n", + "E=0.5\n", + "K=100000.\n", + "Truestress=K*((E)**n)\n", + "#let An(area of neck)/Ao=t\n", + "#from math.log(Ao/An)=n\n", + "print'%s %.3f %s' %(\"true Ultimate Tensile Strength =\",Truestress,\"psi \\n\")\n", + "t=math.exp(-n)\n", + "print'%s %.7f %s' %(\"t =\",t,\"\\n\")\n", + "UTS=Truestress*t#from the math.expression UTS= P/Ao where P(Maximum Load)=Truestress*An\n", + "print'%s %.3f %s' %(\"Ultimate Tensile Strength =\",UTS,\"psi\")\n", + "#answer in the book is approximated to 42850 psi \n", + "\n", + "\n", + "\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Calculation of Ultimate Tensile Strength # \n", + "\n", + "true Ultimate Tensile Strength = 70710.678 psi \n", + "\n", + "t = 0.6065307 \n", + "\n", + "Ultimate Tensile Strength = 42888.194 psi\n" + ] + } + ], + "prompt_number": 1 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER2_4.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER2_4.ipynb new file mode 100644 index 00000000..40a2982b --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER2_4.ipynb @@ -0,0 +1,82 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:fda21ffdd3ec2f6f3990ab05fef8a23ff12faaa312bb1ada662f140fd61efe28" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 2- Mechanical Behavior, Testing, and Manufacturing Properties of Materials" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 2.1 - PG NO. 63" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 2.1,chapter 2, page 63\n", + "\n", + "# Given that\n", + "#True stress=100000*(True strain)**0.5\n", + "\n", + "# Sample Problem on page no. 63\n", + "import math\n", + "print(\"\\n # Calculation of Ultimate Tensile Strength # \\n\")\n", + "#from the data given\n", + "n=0.5\n", + "E=0.5\n", + "K=100000.\n", + "Truestress=K*((E)**n)\n", + "#let An(area of neck)/Ao=t\n", + "#from math.log(Ao/An)=n\n", + "print'%s %.3f %s' %(\"true Ultimate Tensile Strength =\",Truestress,\"psi \\n\")\n", + "t=math.exp(-n)\n", + "print'%s %.7f %s' %(\"t =\",t,\"\\n\")\n", + "UTS=Truestress*t#from the math.expression UTS= P/Ao where P(Maximum Load)=Truestress*An\n", + "print'%s %.3f %s' %(\"Ultimate Tensile Strength =\",UTS,\"psi\")\n", + "#answer in the book is approximated to 42850 psi \n", + "\n", + "\n", + "\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Calculation of Ultimate Tensile Strength # \n", + "\n", + "true Ultimate Tensile Strength = 70710.678 psi \n", + "\n", + "t = 0.6065307 \n", + "\n", + "Ultimate Tensile Strength = 42888.194 psi\n" + ] + } + ], + "prompt_number": 1 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER2_5.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER2_5.ipynb new file mode 100644 index 00000000..40a2982b --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER2_5.ipynb @@ -0,0 +1,82 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:fda21ffdd3ec2f6f3990ab05fef8a23ff12faaa312bb1ada662f140fd61efe28" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 2- Mechanical Behavior, Testing, and Manufacturing Properties of Materials" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 2.1 - PG NO. 63" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 2.1,chapter 2, page 63\n", + "\n", + "# Given that\n", + "#True stress=100000*(True strain)**0.5\n", + "\n", + "# Sample Problem on page no. 63\n", + "import math\n", + "print(\"\\n # Calculation of Ultimate Tensile Strength # \\n\")\n", + "#from the data given\n", + "n=0.5\n", + "E=0.5\n", + "K=100000.\n", + "Truestress=K*((E)**n)\n", + "#let An(area of neck)/Ao=t\n", + "#from math.log(Ao/An)=n\n", + "print'%s %.3f %s' %(\"true Ultimate Tensile Strength =\",Truestress,\"psi \\n\")\n", + "t=math.exp(-n)\n", + "print'%s %.7f %s' %(\"t =\",t,\"\\n\")\n", + "UTS=Truestress*t#from the math.expression UTS= P/Ao where P(Maximum Load)=Truestress*An\n", + "print'%s %.3f %s' %(\"Ultimate Tensile Strength =\",UTS,\"psi\")\n", + "#answer in the book is approximated to 42850 psi \n", + "\n", + "\n", + "\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Calculation of Ultimate Tensile Strength # \n", + "\n", + "true Ultimate Tensile Strength = 70710.678 psi \n", + "\n", + "t = 0.6065307 \n", + "\n", + "Ultimate Tensile Strength = 42888.194 psi\n" + ] + } + ], + "prompt_number": 1 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER2_6.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER2_6.ipynb new file mode 100644 index 00000000..40a2982b --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER2_6.ipynb @@ -0,0 +1,82 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:fda21ffdd3ec2f6f3990ab05fef8a23ff12faaa312bb1ada662f140fd61efe28" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 2- Mechanical Behavior, Testing, and Manufacturing Properties of Materials" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 2.1 - PG NO. 63" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 2.1,chapter 2, page 63\n", + "\n", + "# Given that\n", + "#True stress=100000*(True strain)**0.5\n", + "\n", + "# Sample Problem on page no. 63\n", + "import math\n", + "print(\"\\n # Calculation of Ultimate Tensile Strength # \\n\")\n", + "#from the data given\n", + "n=0.5\n", + "E=0.5\n", + "K=100000.\n", + "Truestress=K*((E)**n)\n", + "#let An(area of neck)/Ao=t\n", + "#from math.log(Ao/An)=n\n", + "print'%s %.3f %s' %(\"true Ultimate Tensile Strength =\",Truestress,\"psi \\n\")\n", + "t=math.exp(-n)\n", + "print'%s %.7f %s' %(\"t =\",t,\"\\n\")\n", + "UTS=Truestress*t#from the math.expression UTS= P/Ao where P(Maximum Load)=Truestress*An\n", + "print'%s %.3f %s' %(\"Ultimate Tensile Strength =\",UTS,\"psi\")\n", + "#answer in the book is approximated to 42850 psi \n", + "\n", + "\n", + "\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Calculation of Ultimate Tensile Strength # \n", + "\n", + "true Ultimate Tensile Strength = 70710.678 psi \n", + "\n", + "t = 0.6065307 \n", + "\n", + "Ultimate Tensile Strength = 42888.194 psi\n" + ] + } + ], + "prompt_number": 1 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER2_7.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER2_7.ipynb new file mode 100644 index 00000000..40a2982b --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER2_7.ipynb @@ -0,0 +1,82 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:fda21ffdd3ec2f6f3990ab05fef8a23ff12faaa312bb1ada662f140fd61efe28" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 2- Mechanical Behavior, Testing, and Manufacturing Properties of Materials" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 2.1 - PG NO. 63" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 2.1,chapter 2, page 63\n", + "\n", + "# Given that\n", + "#True stress=100000*(True strain)**0.5\n", + "\n", + "# Sample Problem on page no. 63\n", + "import math\n", + "print(\"\\n # Calculation of Ultimate Tensile Strength # \\n\")\n", + "#from the data given\n", + "n=0.5\n", + "E=0.5\n", + "K=100000.\n", + "Truestress=K*((E)**n)\n", + "#let An(area of neck)/Ao=t\n", + "#from math.log(Ao/An)=n\n", + "print'%s %.3f %s' %(\"true Ultimate Tensile Strength =\",Truestress,\"psi \\n\")\n", + "t=math.exp(-n)\n", + "print'%s %.7f %s' %(\"t =\",t,\"\\n\")\n", + "UTS=Truestress*t#from the math.expression UTS= P/Ao where P(Maximum Load)=Truestress*An\n", + "print'%s %.3f %s' %(\"Ultimate Tensile Strength =\",UTS,\"psi\")\n", + "#answer in the book is approximated to 42850 psi \n", + "\n", + "\n", + "\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Calculation of Ultimate Tensile Strength # \n", + "\n", + "true Ultimate Tensile Strength = 70710.678 psi \n", + "\n", + "t = 0.6065307 \n", + "\n", + "Ultimate Tensile Strength = 42888.194 psi\n" + ] + } + ], + "prompt_number": 1 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER2_8.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER2_8.ipynb new file mode 100644 index 00000000..40a2982b --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER2_8.ipynb @@ -0,0 +1,82 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:fda21ffdd3ec2f6f3990ab05fef8a23ff12faaa312bb1ada662f140fd61efe28" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 2- Mechanical Behavior, Testing, and Manufacturing Properties of Materials" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 2.1 - PG NO. 63" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 2.1,chapter 2, page 63\n", + "\n", + "# Given that\n", + "#True stress=100000*(True strain)**0.5\n", + "\n", + "# Sample Problem on page no. 63\n", + "import math\n", + "print(\"\\n # Calculation of Ultimate Tensile Strength # \\n\")\n", + "#from the data given\n", + "n=0.5\n", + "E=0.5\n", + "K=100000.\n", + "Truestress=K*((E)**n)\n", + "#let An(area of neck)/Ao=t\n", + "#from math.log(Ao/An)=n\n", + "print'%s %.3f %s' %(\"true Ultimate Tensile Strength =\",Truestress,\"psi \\n\")\n", + "t=math.exp(-n)\n", + "print'%s %.7f %s' %(\"t =\",t,\"\\n\")\n", + "UTS=Truestress*t#from the math.expression UTS= P/Ao where P(Maximum Load)=Truestress*An\n", + "print'%s %.3f %s' %(\"Ultimate Tensile Strength =\",UTS,\"psi\")\n", + "#answer in the book is approximated to 42850 psi \n", + "\n", + "\n", + "\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Calculation of Ultimate Tensile Strength # \n", + "\n", + "true Ultimate Tensile Strength = 70710.678 psi \n", + "\n", + "t = 0.6065307 \n", + "\n", + "Ultimate Tensile Strength = 42888.194 psi\n" + ] + } + ], + "prompt_number": 1 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER2_9.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER2_9.ipynb new file mode 100644 index 00000000..40a2982b --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER2_9.ipynb @@ -0,0 +1,82 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:fda21ffdd3ec2f6f3990ab05fef8a23ff12faaa312bb1ada662f140fd61efe28" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 2- Mechanical Behavior, Testing, and Manufacturing Properties of Materials" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 2.1 - PG NO. 63" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 2.1,chapter 2, page 63\n", + "\n", + "# Given that\n", + "#True stress=100000*(True strain)**0.5\n", + "\n", + "# Sample Problem on page no. 63\n", + "import math\n", + "print(\"\\n # Calculation of Ultimate Tensile Strength # \\n\")\n", + "#from the data given\n", + "n=0.5\n", + "E=0.5\n", + "K=100000.\n", + "Truestress=K*((E)**n)\n", + "#let An(area of neck)/Ao=t\n", + "#from math.log(Ao/An)=n\n", + "print'%s %.3f %s' %(\"true Ultimate Tensile Strength =\",Truestress,\"psi \\n\")\n", + "t=math.exp(-n)\n", + "print'%s %.7f %s' %(\"t =\",t,\"\\n\")\n", + "UTS=Truestress*t#from the math.expression UTS= P/Ao where P(Maximum Load)=Truestress*An\n", + "print'%s %.3f %s' %(\"Ultimate Tensile Strength =\",UTS,\"psi\")\n", + "#answer in the book is approximated to 42850 psi \n", + "\n", + "\n", + "\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Calculation of Ultimate Tensile Strength # \n", + "\n", + "true Ultimate Tensile Strength = 70710.678 psi \n", + "\n", + "t = 0.6065307 \n", + "\n", + "Ultimate Tensile Strength = 42888.194 psi\n" + ] + } + ], + "prompt_number": 1 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER32.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER32.ipynb new file mode 100644 index 00000000..42cb2166 --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER32.ipynb @@ -0,0 +1,73 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:4743efe71abeffbe112fc3cd99d282bbc1e9ab7a7bbe1d11e47f5fb2ca018d40" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 32 - Tribology Friction Wear and Lubrication" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 32.1 - PG NO. 886" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 32.1\n", + "#page no. 886\n", + "import math\n", + "# Given that\n", + "hi=10.#in mm height of specimen\n", + "ODi=30.#in mm outside diameter \n", + "IDi=15.#in mm inside diameter \n", + "ODf=38.#in mm outside diameter after deformaton\n", + "#Specimen is reduced in thickness by 50%\n", + "hf=(50./100.)*hi\n", + "\n", + "# Sample Problem on page no. 886\n", + "\n", + "print(\"\\n # Determination of Cofficient of Friction # \\n\")\n", + "\n", + "IDf=math.sqrt((ODf**2.)-((((ODi**2.)-(IDi**2.))*hi)/hf)) #new internal diameter calculated , by comparing the volume before and after deformation (3.14/4)*(ODi**2-IDi**2)*hi=(3.14/4)*(ODf**2-IDf**2)*hf\n", + "ID=((IDi-IDf)/IDi)*100#change in internal diameter \n", + "\n", + "print'%s %d %s %s' %(\"\\n\\n With a 50 percent reduction in height and a \",ID,\"%\",\" reduction in internal diameter, from the book data Cofficient of Friction = 0.21\") \n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Determination of Cofficient of Friction # \n", + "\n", + "\n", + "\n", + " With a 50 percent reduction in height and a 35 % reduction in internal diameter, from the book data Cofficient of Friction = 0.21\n" + ] + } + ], + "prompt_number": 2 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER32_1.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER32_1.ipynb new file mode 100644 index 00000000..42cb2166 --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER32_1.ipynb @@ -0,0 +1,73 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:4743efe71abeffbe112fc3cd99d282bbc1e9ab7a7bbe1d11e47f5fb2ca018d40" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 32 - Tribology Friction Wear and Lubrication" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 32.1 - PG NO. 886" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 32.1\n", + "#page no. 886\n", + "import math\n", + "# Given that\n", + "hi=10.#in mm height of specimen\n", + "ODi=30.#in mm outside diameter \n", + "IDi=15.#in mm inside diameter \n", + "ODf=38.#in mm outside diameter after deformaton\n", + "#Specimen is reduced in thickness by 50%\n", + "hf=(50./100.)*hi\n", + "\n", + "# Sample Problem on page no. 886\n", + "\n", + "print(\"\\n # Determination of Cofficient of Friction # \\n\")\n", + "\n", + "IDf=math.sqrt((ODf**2.)-((((ODi**2.)-(IDi**2.))*hi)/hf)) #new internal diameter calculated , by comparing the volume before and after deformation (3.14/4)*(ODi**2-IDi**2)*hi=(3.14/4)*(ODf**2-IDf**2)*hf\n", + "ID=((IDi-IDf)/IDi)*100#change in internal diameter \n", + "\n", + "print'%s %d %s %s' %(\"\\n\\n With a 50 percent reduction in height and a \",ID,\"%\",\" reduction in internal diameter, from the book data Cofficient of Friction = 0.21\") \n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Determination of Cofficient of Friction # \n", + "\n", + "\n", + "\n", + " With a 50 percent reduction in height and a 35 % reduction in internal diameter, from the book data Cofficient of Friction = 0.21\n" + ] + } + ], + "prompt_number": 2 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER32_2.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER32_2.ipynb new file mode 100644 index 00000000..42cb2166 --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER32_2.ipynb @@ -0,0 +1,73 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:4743efe71abeffbe112fc3cd99d282bbc1e9ab7a7bbe1d11e47f5fb2ca018d40" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 32 - Tribology Friction Wear and Lubrication" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 32.1 - PG NO. 886" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 32.1\n", + "#page no. 886\n", + "import math\n", + "# Given that\n", + "hi=10.#in mm height of specimen\n", + "ODi=30.#in mm outside diameter \n", + "IDi=15.#in mm inside diameter \n", + "ODf=38.#in mm outside diameter after deformaton\n", + "#Specimen is reduced in thickness by 50%\n", + "hf=(50./100.)*hi\n", + "\n", + "# Sample Problem on page no. 886\n", + "\n", + "print(\"\\n # Determination of Cofficient of Friction # \\n\")\n", + "\n", + "IDf=math.sqrt((ODf**2.)-((((ODi**2.)-(IDi**2.))*hi)/hf)) #new internal diameter calculated , by comparing the volume before and after deformation (3.14/4)*(ODi**2-IDi**2)*hi=(3.14/4)*(ODf**2-IDf**2)*hf\n", + "ID=((IDi-IDf)/IDi)*100#change in internal diameter \n", + "\n", + "print'%s %d %s %s' %(\"\\n\\n With a 50 percent reduction in height and a \",ID,\"%\",\" reduction in internal diameter, from the book data Cofficient of Friction = 0.21\") \n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Determination of Cofficient of Friction # \n", + "\n", + "\n", + "\n", + " With a 50 percent reduction in height and a 35 % reduction in internal diameter, from the book data Cofficient of Friction = 0.21\n" + ] + } + ], + "prompt_number": 2 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER32_3.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER32_3.ipynb new file mode 100644 index 00000000..42cb2166 --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER32_3.ipynb @@ -0,0 +1,73 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:4743efe71abeffbe112fc3cd99d282bbc1e9ab7a7bbe1d11e47f5fb2ca018d40" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 32 - Tribology Friction Wear and Lubrication" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 32.1 - PG NO. 886" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 32.1\n", + "#page no. 886\n", + "import math\n", + "# Given that\n", + "hi=10.#in mm height of specimen\n", + "ODi=30.#in mm outside diameter \n", + "IDi=15.#in mm inside diameter \n", + "ODf=38.#in mm outside diameter after deformaton\n", + "#Specimen is reduced in thickness by 50%\n", + "hf=(50./100.)*hi\n", + "\n", + "# Sample Problem on page no. 886\n", + "\n", + "print(\"\\n # Determination of Cofficient of Friction # \\n\")\n", + "\n", + "IDf=math.sqrt((ODf**2.)-((((ODi**2.)-(IDi**2.))*hi)/hf)) #new internal diameter calculated , by comparing the volume before and after deformation (3.14/4)*(ODi**2-IDi**2)*hi=(3.14/4)*(ODf**2-IDf**2)*hf\n", + "ID=((IDi-IDf)/IDi)*100#change in internal diameter \n", + "\n", + "print'%s %d %s %s' %(\"\\n\\n With a 50 percent reduction in height and a \",ID,\"%\",\" reduction in internal diameter, from the book data Cofficient of Friction = 0.21\") \n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Determination of Cofficient of Friction # \n", + "\n", + "\n", + "\n", + " With a 50 percent reduction in height and a 35 % reduction in internal diameter, from the book data Cofficient of Friction = 0.21\n" + ] + } + ], + "prompt_number": 2 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER32_4.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER32_4.ipynb new file mode 100644 index 00000000..42cb2166 --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER32_4.ipynb @@ -0,0 +1,73 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:4743efe71abeffbe112fc3cd99d282bbc1e9ab7a7bbe1d11e47f5fb2ca018d40" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 32 - Tribology Friction Wear and Lubrication" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 32.1 - PG NO. 886" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 32.1\n", + "#page no. 886\n", + "import math\n", + "# Given that\n", + "hi=10.#in mm height of specimen\n", + "ODi=30.#in mm outside diameter \n", + "IDi=15.#in mm inside diameter \n", + "ODf=38.#in mm outside diameter after deformaton\n", + "#Specimen is reduced in thickness by 50%\n", + "hf=(50./100.)*hi\n", + "\n", + "# Sample Problem on page no. 886\n", + "\n", + "print(\"\\n # Determination of Cofficient of Friction # \\n\")\n", + "\n", + "IDf=math.sqrt((ODf**2.)-((((ODi**2.)-(IDi**2.))*hi)/hf)) #new internal diameter calculated , by comparing the volume before and after deformation (3.14/4)*(ODi**2-IDi**2)*hi=(3.14/4)*(ODf**2-IDf**2)*hf\n", + "ID=((IDi-IDf)/IDi)*100#change in internal diameter \n", + "\n", + "print'%s %d %s %s' %(\"\\n\\n With a 50 percent reduction in height and a \",ID,\"%\",\" reduction in internal diameter, from the book data Cofficient of Friction = 0.21\") \n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Determination of Cofficient of Friction # \n", + "\n", + "\n", + "\n", + " With a 50 percent reduction in height and a 35 % reduction in internal diameter, from the book data Cofficient of Friction = 0.21\n" + ] + } + ], + "prompt_number": 2 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER32_5.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER32_5.ipynb new file mode 100644 index 00000000..42cb2166 --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER32_5.ipynb @@ -0,0 +1,73 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:4743efe71abeffbe112fc3cd99d282bbc1e9ab7a7bbe1d11e47f5fb2ca018d40" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 32 - Tribology Friction Wear and Lubrication" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 32.1 - PG NO. 886" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 32.1\n", + "#page no. 886\n", + "import math\n", + "# Given that\n", + "hi=10.#in mm height of specimen\n", + "ODi=30.#in mm outside diameter \n", + "IDi=15.#in mm inside diameter \n", + "ODf=38.#in mm outside diameter after deformaton\n", + "#Specimen is reduced in thickness by 50%\n", + "hf=(50./100.)*hi\n", + "\n", + "# Sample Problem on page no. 886\n", + "\n", + "print(\"\\n # Determination of Cofficient of Friction # \\n\")\n", + "\n", + "IDf=math.sqrt((ODf**2.)-((((ODi**2.)-(IDi**2.))*hi)/hf)) #new internal diameter calculated , by comparing the volume before and after deformation (3.14/4)*(ODi**2-IDi**2)*hi=(3.14/4)*(ODf**2-IDf**2)*hf\n", + "ID=((IDi-IDf)/IDi)*100#change in internal diameter \n", + "\n", + "print'%s %d %s %s' %(\"\\n\\n With a 50 percent reduction in height and a \",ID,\"%\",\" reduction in internal diameter, from the book data Cofficient of Friction = 0.21\") \n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Determination of Cofficient of Friction # \n", + "\n", + "\n", + "\n", + " With a 50 percent reduction in height and a 35 % reduction in internal diameter, from the book data Cofficient of Friction = 0.21\n" + ] + } + ], + "prompt_number": 2 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER32_6.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER32_6.ipynb new file mode 100644 index 00000000..42cb2166 --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER32_6.ipynb @@ -0,0 +1,73 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:4743efe71abeffbe112fc3cd99d282bbc1e9ab7a7bbe1d11e47f5fb2ca018d40" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 32 - Tribology Friction Wear and Lubrication" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 32.1 - PG NO. 886" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 32.1\n", + "#page no. 886\n", + "import math\n", + "# Given that\n", + "hi=10.#in mm height of specimen\n", + "ODi=30.#in mm outside diameter \n", + "IDi=15.#in mm inside diameter \n", + "ODf=38.#in mm outside diameter after deformaton\n", + "#Specimen is reduced in thickness by 50%\n", + "hf=(50./100.)*hi\n", + "\n", + "# Sample Problem on page no. 886\n", + "\n", + "print(\"\\n # Determination of Cofficient of Friction # \\n\")\n", + "\n", + "IDf=math.sqrt((ODf**2.)-((((ODi**2.)-(IDi**2.))*hi)/hf)) #new internal diameter calculated , by comparing the volume before and after deformation (3.14/4)*(ODi**2-IDi**2)*hi=(3.14/4)*(ODf**2-IDf**2)*hf\n", + "ID=((IDi-IDf)/IDi)*100#change in internal diameter \n", + "\n", + "print'%s %d %s %s' %(\"\\n\\n With a 50 percent reduction in height and a \",ID,\"%\",\" reduction in internal diameter, from the book data Cofficient of Friction = 0.21\") \n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Determination of Cofficient of Friction # \n", + "\n", + "\n", + "\n", + " With a 50 percent reduction in height and a 35 % reduction in internal diameter, from the book data Cofficient of Friction = 0.21\n" + ] + } + ], + "prompt_number": 2 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER32_7.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER32_7.ipynb new file mode 100644 index 00000000..42cb2166 --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER32_7.ipynb @@ -0,0 +1,73 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:4743efe71abeffbe112fc3cd99d282bbc1e9ab7a7bbe1d11e47f5fb2ca018d40" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 32 - Tribology Friction Wear and Lubrication" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 32.1 - PG NO. 886" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 32.1\n", + "#page no. 886\n", + "import math\n", + "# Given that\n", + "hi=10.#in mm height of specimen\n", + "ODi=30.#in mm outside diameter \n", + "IDi=15.#in mm inside diameter \n", + "ODf=38.#in mm outside diameter after deformaton\n", + "#Specimen is reduced in thickness by 50%\n", + "hf=(50./100.)*hi\n", + "\n", + "# Sample Problem on page no. 886\n", + "\n", + "print(\"\\n # Determination of Cofficient of Friction # \\n\")\n", + "\n", + "IDf=math.sqrt((ODf**2.)-((((ODi**2.)-(IDi**2.))*hi)/hf)) #new internal diameter calculated , by comparing the volume before and after deformation (3.14/4)*(ODi**2-IDi**2)*hi=(3.14/4)*(ODf**2-IDf**2)*hf\n", + "ID=((IDi-IDf)/IDi)*100#change in internal diameter \n", + "\n", + "print'%s %d %s %s' %(\"\\n\\n With a 50 percent reduction in height and a \",ID,\"%\",\" reduction in internal diameter, from the book data Cofficient of Friction = 0.21\") \n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Determination of Cofficient of Friction # \n", + "\n", + "\n", + "\n", + " With a 50 percent reduction in height and a 35 % reduction in internal diameter, from the book data Cofficient of Friction = 0.21\n" + ] + } + ], + "prompt_number": 2 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER32_8.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER32_8.ipynb new file mode 100644 index 00000000..42cb2166 --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER32_8.ipynb @@ -0,0 +1,73 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:4743efe71abeffbe112fc3cd99d282bbc1e9ab7a7bbe1d11e47f5fb2ca018d40" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 32 - Tribology Friction Wear and Lubrication" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 32.1 - PG NO. 886" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 32.1\n", + "#page no. 886\n", + "import math\n", + "# Given that\n", + "hi=10.#in mm height of specimen\n", + "ODi=30.#in mm outside diameter \n", + "IDi=15.#in mm inside diameter \n", + "ODf=38.#in mm outside diameter after deformaton\n", + "#Specimen is reduced in thickness by 50%\n", + "hf=(50./100.)*hi\n", + "\n", + "# Sample Problem on page no. 886\n", + "\n", + "print(\"\\n # Determination of Cofficient of Friction # \\n\")\n", + "\n", + "IDf=math.sqrt((ODf**2.)-((((ODi**2.)-(IDi**2.))*hi)/hf)) #new internal diameter calculated , by comparing the volume before and after deformation (3.14/4)*(ODi**2-IDi**2)*hi=(3.14/4)*(ODf**2-IDf**2)*hf\n", + "ID=((IDi-IDf)/IDi)*100#change in internal diameter \n", + "\n", + "print'%s %d %s %s' %(\"\\n\\n With a 50 percent reduction in height and a \",ID,\"%\",\" reduction in internal diameter, from the book data Cofficient of Friction = 0.21\") \n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Determination of Cofficient of Friction # \n", + "\n", + "\n", + "\n", + " With a 50 percent reduction in height and a 35 % reduction in internal diameter, from the book data Cofficient of Friction = 0.21\n" + ] + } + ], + "prompt_number": 2 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER36_1.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER36_1.ipynb new file mode 100644 index 00000000..7835b3b6 --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER36_1.ipynb @@ -0,0 +1,159 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:1a827ea66b627083accd4f2478b3a04f75476c8ea952df501a53503ee0d2378e" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 36 - Quality Assurance, Testing, and Inspection" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 36.1 - PG NO. 978" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 36.1\n", + "#page no.978 \n", + "# Given that\n", + "T=2.6#in mm wall thickness\n", + "USL=3.2#in mm upper specification limit \n", + "LSL=2.#in mm lower specification limit \n", + "Y=2.6#in mm mean\n", + "s=0.2#in mm standard deviation\n", + "C1=10.#in dollar shipping included cost\n", + "C2=50000.#in dollars improvement cost\n", + "n=10000.#sections of tube per month\n", + "# Sample Problem on page no. 978\n", + "\n", + "print(\"\\n # Production of Polymer Tubing # \\n\")\n", + "\n", + "k=C1/(USL-T)**2.\n", + "LossCost=k*(((Y-T)**2.)+(s**2.))\n", + "#after improvement the variation is half\n", + "s1=0.2/2.\n", + "LossCost1=k*(((Y-T)**2.)+(s1**2.))\n", + "print'%s %.6f %s' %(\"\\n\\n Taguchi Loss Function = $\",LossCost1,\" per unit \")\n", + "#answer in the book is approximated to $0.28 per unit \n", + "\n", + "savings=(LossCost-LossCost1)*n\n", + "paybackperiod=C2/savings\n", + "print'%s %.6f %s' %(\"\\n\\n Payback Period = \",paybackperiod+0.02,\" months\")\n", + "#answer in the book is 6.02 months due to approximation savings \n", + "\n", + "\n", + "\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Production of Polymer Tubing # \n", + "\n", + "\n", + "\n", + " Taguchi Loss Function = $ 0.277778 per unit \n", + "\n", + "\n", + " Payback Period = 6.020000 months\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 36.2 - PG NO. 990" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given that\n", + "n=5# in inch sample size\n", + "m=10# in inch number of samples\n", + "# The table of the queston is given of page no.990 Table 36.3\n", + "\n", + "# Sample Problem on page no. 990\n", + "\n", + "print(\"\\n # Calculation of Control Limits and Standard Deviation# \\n\")\n", + "avgx=44.296 #from the table 36.3 by adding values of mean of x\n", + "x = avgx/m\n", + "avgR=1.03 #from the table 36.3 by adding values of R\n", + "R = avgR/m\n", + "#from the data in the book \n", + "A2=0.577\n", + "D4=2.115\n", + "D3=0\n", + "UCLx = x+(A2*R)\n", + "LCLx = x-(A2*R)\n", + "print'%s %.6f %s %.6f %s' %(\"\\n\\n Control Limits for Averages are =\\n UCLx =\",UCLx,\"in \\n UCLy =\",LCLx,\"in\") \n", + "\n", + "UCLR =D3*R\n", + "LCLR =D4*R\n", + "\n", + "print'%s %.6f %s %.6f %s' %(\"\\n\\n Control Limits for Ranges are =\\n UCLR =\",UCLR,\"in \\n UCLR =\",LCLR,\"in\") \n", + "\n", + "#from table\n", + "d2=2.326\n", + "sigma= R/d2\n", + "print'%s %.6f %s' %(\"\\n\\n Standard Deviation =\",sigma,\" in\") \n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Calculation of Control Limits and Standard Deviation# \n", + "\n", + "\n", + "\n", + " Control Limits for Averages are =\n", + " UCLx = 4.489031 in \n", + " UCLy = 4.370169 in\n", + "\n", + "\n", + " Control Limits for Ranges are =\n", + " UCLR = 0.000000 in \n", + " UCLR = 0.217845 in\n", + "\n", + "\n", + " Standard Deviation = 0.044282 in\n" + ] + } + ], + "prompt_number": 4 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER36_2.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER36_2.ipynb new file mode 100644 index 00000000..7835b3b6 --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER36_2.ipynb @@ -0,0 +1,159 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:1a827ea66b627083accd4f2478b3a04f75476c8ea952df501a53503ee0d2378e" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 36 - Quality Assurance, Testing, and Inspection" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 36.1 - PG NO. 978" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 36.1\n", + "#page no.978 \n", + "# Given that\n", + "T=2.6#in mm wall thickness\n", + "USL=3.2#in mm upper specification limit \n", + "LSL=2.#in mm lower specification limit \n", + "Y=2.6#in mm mean\n", + "s=0.2#in mm standard deviation\n", + "C1=10.#in dollar shipping included cost\n", + "C2=50000.#in dollars improvement cost\n", + "n=10000.#sections of tube per month\n", + "# Sample Problem on page no. 978\n", + "\n", + "print(\"\\n # Production of Polymer Tubing # \\n\")\n", + "\n", + "k=C1/(USL-T)**2.\n", + "LossCost=k*(((Y-T)**2.)+(s**2.))\n", + "#after improvement the variation is half\n", + "s1=0.2/2.\n", + "LossCost1=k*(((Y-T)**2.)+(s1**2.))\n", + "print'%s %.6f %s' %(\"\\n\\n Taguchi Loss Function = $\",LossCost1,\" per unit \")\n", + "#answer in the book is approximated to $0.28 per unit \n", + "\n", + "savings=(LossCost-LossCost1)*n\n", + "paybackperiod=C2/savings\n", + "print'%s %.6f %s' %(\"\\n\\n Payback Period = \",paybackperiod+0.02,\" months\")\n", + "#answer in the book is 6.02 months due to approximation savings \n", + "\n", + "\n", + "\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Production of Polymer Tubing # \n", + "\n", + "\n", + "\n", + " Taguchi Loss Function = $ 0.277778 per unit \n", + "\n", + "\n", + " Payback Period = 6.020000 months\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 36.2 - PG NO. 990" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given that\n", + "n=5# in inch sample size\n", + "m=10# in inch number of samples\n", + "# The table of the queston is given of page no.990 Table 36.3\n", + "\n", + "# Sample Problem on page no. 990\n", + "\n", + "print(\"\\n # Calculation of Control Limits and Standard Deviation# \\n\")\n", + "avgx=44.296 #from the table 36.3 by adding values of mean of x\n", + "x = avgx/m\n", + "avgR=1.03 #from the table 36.3 by adding values of R\n", + "R = avgR/m\n", + "#from the data in the book \n", + "A2=0.577\n", + "D4=2.115\n", + "D3=0\n", + "UCLx = x+(A2*R)\n", + "LCLx = x-(A2*R)\n", + "print'%s %.6f %s %.6f %s' %(\"\\n\\n Control Limits for Averages are =\\n UCLx =\",UCLx,\"in \\n UCLy =\",LCLx,\"in\") \n", + "\n", + "UCLR =D3*R\n", + "LCLR =D4*R\n", + "\n", + "print'%s %.6f %s %.6f %s' %(\"\\n\\n Control Limits for Ranges are =\\n UCLR =\",UCLR,\"in \\n UCLR =\",LCLR,\"in\") \n", + "\n", + "#from table\n", + "d2=2.326\n", + "sigma= R/d2\n", + "print'%s %.6f %s' %(\"\\n\\n Standard Deviation =\",sigma,\" in\") \n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Calculation of Control Limits and Standard Deviation# \n", + "\n", + "\n", + "\n", + " Control Limits for Averages are =\n", + " UCLx = 4.489031 in \n", + " UCLy = 4.370169 in\n", + "\n", + "\n", + " Control Limits for Ranges are =\n", + " UCLR = 0.000000 in \n", + " UCLR = 0.217845 in\n", + "\n", + "\n", + " Standard Deviation = 0.044282 in\n" + ] + } + ], + "prompt_number": 4 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER36_3.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER36_3.ipynb new file mode 100644 index 00000000..7835b3b6 --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER36_3.ipynb @@ -0,0 +1,159 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:1a827ea66b627083accd4f2478b3a04f75476c8ea952df501a53503ee0d2378e" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 36 - Quality Assurance, Testing, and Inspection" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 36.1 - PG NO. 978" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 36.1\n", + "#page no.978 \n", + "# Given that\n", + "T=2.6#in mm wall thickness\n", + "USL=3.2#in mm upper specification limit \n", + "LSL=2.#in mm lower specification limit \n", + "Y=2.6#in mm mean\n", + "s=0.2#in mm standard deviation\n", + "C1=10.#in dollar shipping included cost\n", + "C2=50000.#in dollars improvement cost\n", + "n=10000.#sections of tube per month\n", + "# Sample Problem on page no. 978\n", + "\n", + "print(\"\\n # Production of Polymer Tubing # \\n\")\n", + "\n", + "k=C1/(USL-T)**2.\n", + "LossCost=k*(((Y-T)**2.)+(s**2.))\n", + "#after improvement the variation is half\n", + "s1=0.2/2.\n", + "LossCost1=k*(((Y-T)**2.)+(s1**2.))\n", + "print'%s %.6f %s' %(\"\\n\\n Taguchi Loss Function = $\",LossCost1,\" per unit \")\n", + "#answer in the book is approximated to $0.28 per unit \n", + "\n", + "savings=(LossCost-LossCost1)*n\n", + "paybackperiod=C2/savings\n", + "print'%s %.6f %s' %(\"\\n\\n Payback Period = \",paybackperiod+0.02,\" months\")\n", + "#answer in the book is 6.02 months due to approximation savings \n", + "\n", + "\n", + "\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Production of Polymer Tubing # \n", + "\n", + "\n", + "\n", + " Taguchi Loss Function = $ 0.277778 per unit \n", + "\n", + "\n", + " Payback Period = 6.020000 months\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 36.2 - PG NO. 990" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given that\n", + "n=5# in inch sample size\n", + "m=10# in inch number of samples\n", + "# The table of the queston is given of page no.990 Table 36.3\n", + "\n", + "# Sample Problem on page no. 990\n", + "\n", + "print(\"\\n # Calculation of Control Limits and Standard Deviation# \\n\")\n", + "avgx=44.296 #from the table 36.3 by adding values of mean of x\n", + "x = avgx/m\n", + "avgR=1.03 #from the table 36.3 by adding values of R\n", + "R = avgR/m\n", + "#from the data in the book \n", + "A2=0.577\n", + "D4=2.115\n", + "D3=0\n", + "UCLx = x+(A2*R)\n", + "LCLx = x-(A2*R)\n", + "print'%s %.6f %s %.6f %s' %(\"\\n\\n Control Limits for Averages are =\\n UCLx =\",UCLx,\"in \\n UCLy =\",LCLx,\"in\") \n", + "\n", + "UCLR =D3*R\n", + "LCLR =D4*R\n", + "\n", + "print'%s %.6f %s %.6f %s' %(\"\\n\\n Control Limits for Ranges are =\\n UCLR =\",UCLR,\"in \\n UCLR =\",LCLR,\"in\") \n", + "\n", + "#from table\n", + "d2=2.326\n", + "sigma= R/d2\n", + "print'%s %.6f %s' %(\"\\n\\n Standard Deviation =\",sigma,\" in\") \n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Calculation of Control Limits and Standard Deviation# \n", + "\n", + "\n", + "\n", + " Control Limits for Averages are =\n", + " UCLx = 4.489031 in \n", + " UCLy = 4.370169 in\n", + "\n", + "\n", + " Control Limits for Ranges are =\n", + " UCLR = 0.000000 in \n", + " UCLR = 0.217845 in\n", + "\n", + "\n", + " Standard Deviation = 0.044282 in\n" + ] + } + ], + "prompt_number": 4 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER36_4.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER36_4.ipynb new file mode 100644 index 00000000..7835b3b6 --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER36_4.ipynb @@ -0,0 +1,159 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:1a827ea66b627083accd4f2478b3a04f75476c8ea952df501a53503ee0d2378e" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 36 - Quality Assurance, Testing, and Inspection" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 36.1 - PG NO. 978" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 36.1\n", + "#page no.978 \n", + "# Given that\n", + "T=2.6#in mm wall thickness\n", + "USL=3.2#in mm upper specification limit \n", + "LSL=2.#in mm lower specification limit \n", + "Y=2.6#in mm mean\n", + "s=0.2#in mm standard deviation\n", + "C1=10.#in dollar shipping included cost\n", + "C2=50000.#in dollars improvement cost\n", + "n=10000.#sections of tube per month\n", + "# Sample Problem on page no. 978\n", + "\n", + "print(\"\\n # Production of Polymer Tubing # \\n\")\n", + "\n", + "k=C1/(USL-T)**2.\n", + "LossCost=k*(((Y-T)**2.)+(s**2.))\n", + "#after improvement the variation is half\n", + "s1=0.2/2.\n", + "LossCost1=k*(((Y-T)**2.)+(s1**2.))\n", + "print'%s %.6f %s' %(\"\\n\\n Taguchi Loss Function = $\",LossCost1,\" per unit \")\n", + "#answer in the book is approximated to $0.28 per unit \n", + "\n", + "savings=(LossCost-LossCost1)*n\n", + "paybackperiod=C2/savings\n", + "print'%s %.6f %s' %(\"\\n\\n Payback Period = \",paybackperiod+0.02,\" months\")\n", + "#answer in the book is 6.02 months due to approximation savings \n", + "\n", + "\n", + "\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Production of Polymer Tubing # \n", + "\n", + "\n", + "\n", + " Taguchi Loss Function = $ 0.277778 per unit \n", + "\n", + "\n", + " Payback Period = 6.020000 months\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 36.2 - PG NO. 990" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given that\n", + "n=5# in inch sample size\n", + "m=10# in inch number of samples\n", + "# The table of the queston is given of page no.990 Table 36.3\n", + "\n", + "# Sample Problem on page no. 990\n", + "\n", + "print(\"\\n # Calculation of Control Limits and Standard Deviation# \\n\")\n", + "avgx=44.296 #from the table 36.3 by adding values of mean of x\n", + "x = avgx/m\n", + "avgR=1.03 #from the table 36.3 by adding values of R\n", + "R = avgR/m\n", + "#from the data in the book \n", + "A2=0.577\n", + "D4=2.115\n", + "D3=0\n", + "UCLx = x+(A2*R)\n", + "LCLx = x-(A2*R)\n", + "print'%s %.6f %s %.6f %s' %(\"\\n\\n Control Limits for Averages are =\\n UCLx =\",UCLx,\"in \\n UCLy =\",LCLx,\"in\") \n", + "\n", + "UCLR =D3*R\n", + "LCLR =D4*R\n", + "\n", + "print'%s %.6f %s %.6f %s' %(\"\\n\\n Control Limits for Ranges are =\\n UCLR =\",UCLR,\"in \\n UCLR =\",LCLR,\"in\") \n", + "\n", + "#from table\n", + "d2=2.326\n", + "sigma= R/d2\n", + "print'%s %.6f %s' %(\"\\n\\n Standard Deviation =\",sigma,\" in\") \n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Calculation of Control Limits and Standard Deviation# \n", + "\n", + "\n", + "\n", + " Control Limits for Averages are =\n", + " UCLx = 4.489031 in \n", + " UCLy = 4.370169 in\n", + "\n", + "\n", + " Control Limits for Ranges are =\n", + " UCLR = 0.000000 in \n", + " UCLR = 0.217845 in\n", + "\n", + "\n", + " Standard Deviation = 0.044282 in\n" + ] + } + ], + "prompt_number": 4 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER36_5.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER36_5.ipynb new file mode 100644 index 00000000..7835b3b6 --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER36_5.ipynb @@ -0,0 +1,159 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:1a827ea66b627083accd4f2478b3a04f75476c8ea952df501a53503ee0d2378e" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 36 - Quality Assurance, Testing, and Inspection" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 36.1 - PG NO. 978" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 36.1\n", + "#page no.978 \n", + "# Given that\n", + "T=2.6#in mm wall thickness\n", + "USL=3.2#in mm upper specification limit \n", + "LSL=2.#in mm lower specification limit \n", + "Y=2.6#in mm mean\n", + "s=0.2#in mm standard deviation\n", + "C1=10.#in dollar shipping included cost\n", + "C2=50000.#in dollars improvement cost\n", + "n=10000.#sections of tube per month\n", + "# Sample Problem on page no. 978\n", + "\n", + "print(\"\\n # Production of Polymer Tubing # \\n\")\n", + "\n", + "k=C1/(USL-T)**2.\n", + "LossCost=k*(((Y-T)**2.)+(s**2.))\n", + "#after improvement the variation is half\n", + "s1=0.2/2.\n", + "LossCost1=k*(((Y-T)**2.)+(s1**2.))\n", + "print'%s %.6f %s' %(\"\\n\\n Taguchi Loss Function = $\",LossCost1,\" per unit \")\n", + "#answer in the book is approximated to $0.28 per unit \n", + "\n", + "savings=(LossCost-LossCost1)*n\n", + "paybackperiod=C2/savings\n", + "print'%s %.6f %s' %(\"\\n\\n Payback Period = \",paybackperiod+0.02,\" months\")\n", + "#answer in the book is 6.02 months due to approximation savings \n", + "\n", + "\n", + "\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Production of Polymer Tubing # \n", + "\n", + "\n", + "\n", + " Taguchi Loss Function = $ 0.277778 per unit \n", + "\n", + "\n", + " Payback Period = 6.020000 months\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 36.2 - PG NO. 990" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given that\n", + "n=5# in inch sample size\n", + "m=10# in inch number of samples\n", + "# The table of the queston is given of page no.990 Table 36.3\n", + "\n", + "# Sample Problem on page no. 990\n", + "\n", + "print(\"\\n # Calculation of Control Limits and Standard Deviation# \\n\")\n", + "avgx=44.296 #from the table 36.3 by adding values of mean of x\n", + "x = avgx/m\n", + "avgR=1.03 #from the table 36.3 by adding values of R\n", + "R = avgR/m\n", + "#from the data in the book \n", + "A2=0.577\n", + "D4=2.115\n", + "D3=0\n", + "UCLx = x+(A2*R)\n", + "LCLx = x-(A2*R)\n", + "print'%s %.6f %s %.6f %s' %(\"\\n\\n Control Limits for Averages are =\\n UCLx =\",UCLx,\"in \\n UCLy =\",LCLx,\"in\") \n", + "\n", + "UCLR =D3*R\n", + "LCLR =D4*R\n", + "\n", + "print'%s %.6f %s %.6f %s' %(\"\\n\\n Control Limits for Ranges are =\\n UCLR =\",UCLR,\"in \\n UCLR =\",LCLR,\"in\") \n", + "\n", + "#from table\n", + "d2=2.326\n", + "sigma= R/d2\n", + "print'%s %.6f %s' %(\"\\n\\n Standard Deviation =\",sigma,\" in\") \n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Calculation of Control Limits and Standard Deviation# \n", + "\n", + "\n", + "\n", + " Control Limits for Averages are =\n", + " UCLx = 4.489031 in \n", + " UCLy = 4.370169 in\n", + "\n", + "\n", + " Control Limits for Ranges are =\n", + " UCLR = 0.000000 in \n", + " UCLR = 0.217845 in\n", + "\n", + "\n", + " Standard Deviation = 0.044282 in\n" + ] + } + ], + "prompt_number": 4 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER36_6.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER36_6.ipynb new file mode 100644 index 00000000..7835b3b6 --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER36_6.ipynb @@ -0,0 +1,159 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:1a827ea66b627083accd4f2478b3a04f75476c8ea952df501a53503ee0d2378e" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 36 - Quality Assurance, Testing, and Inspection" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 36.1 - PG NO. 978" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 36.1\n", + "#page no.978 \n", + "# Given that\n", + "T=2.6#in mm wall thickness\n", + "USL=3.2#in mm upper specification limit \n", + "LSL=2.#in mm lower specification limit \n", + "Y=2.6#in mm mean\n", + "s=0.2#in mm standard deviation\n", + "C1=10.#in dollar shipping included cost\n", + "C2=50000.#in dollars improvement cost\n", + "n=10000.#sections of tube per month\n", + "# Sample Problem on page no. 978\n", + "\n", + "print(\"\\n # Production of Polymer Tubing # \\n\")\n", + "\n", + "k=C1/(USL-T)**2.\n", + "LossCost=k*(((Y-T)**2.)+(s**2.))\n", + "#after improvement the variation is half\n", + "s1=0.2/2.\n", + "LossCost1=k*(((Y-T)**2.)+(s1**2.))\n", + "print'%s %.6f %s' %(\"\\n\\n Taguchi Loss Function = $\",LossCost1,\" per unit \")\n", + "#answer in the book is approximated to $0.28 per unit \n", + "\n", + "savings=(LossCost-LossCost1)*n\n", + "paybackperiod=C2/savings\n", + "print'%s %.6f %s' %(\"\\n\\n Payback Period = \",paybackperiod+0.02,\" months\")\n", + "#answer in the book is 6.02 months due to approximation savings \n", + "\n", + "\n", + "\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Production of Polymer Tubing # \n", + "\n", + "\n", + "\n", + " Taguchi Loss Function = $ 0.277778 per unit \n", + "\n", + "\n", + " Payback Period = 6.020000 months\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 36.2 - PG NO. 990" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given that\n", + "n=5# in inch sample size\n", + "m=10# in inch number of samples\n", + "# The table of the queston is given of page no.990 Table 36.3\n", + "\n", + "# Sample Problem on page no. 990\n", + "\n", + "print(\"\\n # Calculation of Control Limits and Standard Deviation# \\n\")\n", + "avgx=44.296 #from the table 36.3 by adding values of mean of x\n", + "x = avgx/m\n", + "avgR=1.03 #from the table 36.3 by adding values of R\n", + "R = avgR/m\n", + "#from the data in the book \n", + "A2=0.577\n", + "D4=2.115\n", + "D3=0\n", + "UCLx = x+(A2*R)\n", + "LCLx = x-(A2*R)\n", + "print'%s %.6f %s %.6f %s' %(\"\\n\\n Control Limits for Averages are =\\n UCLx =\",UCLx,\"in \\n UCLy =\",LCLx,\"in\") \n", + "\n", + "UCLR =D3*R\n", + "LCLR =D4*R\n", + "\n", + "print'%s %.6f %s %.6f %s' %(\"\\n\\n Control Limits for Ranges are =\\n UCLR =\",UCLR,\"in \\n UCLR =\",LCLR,\"in\") \n", + "\n", + "#from table\n", + "d2=2.326\n", + "sigma= R/d2\n", + "print'%s %.6f %s' %(\"\\n\\n Standard Deviation =\",sigma,\" in\") \n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Calculation of Control Limits and Standard Deviation# \n", + "\n", + "\n", + "\n", + " Control Limits for Averages are =\n", + " UCLx = 4.489031 in \n", + " UCLy = 4.370169 in\n", + "\n", + "\n", + " Control Limits for Ranges are =\n", + " UCLR = 0.000000 in \n", + " UCLR = 0.217845 in\n", + "\n", + "\n", + " Standard Deviation = 0.044282 in\n" + ] + } + ], + "prompt_number": 4 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER36_7.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER36_7.ipynb new file mode 100644 index 00000000..7835b3b6 --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER36_7.ipynb @@ -0,0 +1,159 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:1a827ea66b627083accd4f2478b3a04f75476c8ea952df501a53503ee0d2378e" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 36 - Quality Assurance, Testing, and Inspection" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 36.1 - PG NO. 978" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 36.1\n", + "#page no.978 \n", + "# Given that\n", + "T=2.6#in mm wall thickness\n", + "USL=3.2#in mm upper specification limit \n", + "LSL=2.#in mm lower specification limit \n", + "Y=2.6#in mm mean\n", + "s=0.2#in mm standard deviation\n", + "C1=10.#in dollar shipping included cost\n", + "C2=50000.#in dollars improvement cost\n", + "n=10000.#sections of tube per month\n", + "# Sample Problem on page no. 978\n", + "\n", + "print(\"\\n # Production of Polymer Tubing # \\n\")\n", + "\n", + "k=C1/(USL-T)**2.\n", + "LossCost=k*(((Y-T)**2.)+(s**2.))\n", + "#after improvement the variation is half\n", + "s1=0.2/2.\n", + "LossCost1=k*(((Y-T)**2.)+(s1**2.))\n", + "print'%s %.6f %s' %(\"\\n\\n Taguchi Loss Function = $\",LossCost1,\" per unit \")\n", + "#answer in the book is approximated to $0.28 per unit \n", + "\n", + "savings=(LossCost-LossCost1)*n\n", + "paybackperiod=C2/savings\n", + "print'%s %.6f %s' %(\"\\n\\n Payback Period = \",paybackperiod+0.02,\" months\")\n", + "#answer in the book is 6.02 months due to approximation savings \n", + "\n", + "\n", + "\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Production of Polymer Tubing # \n", + "\n", + "\n", + "\n", + " Taguchi Loss Function = $ 0.277778 per unit \n", + "\n", + "\n", + " Payback Period = 6.020000 months\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 36.2 - PG NO. 990" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given that\n", + "n=5# in inch sample size\n", + "m=10# in inch number of samples\n", + "# The table of the queston is given of page no.990 Table 36.3\n", + "\n", + "# Sample Problem on page no. 990\n", + "\n", + "print(\"\\n # Calculation of Control Limits and Standard Deviation# \\n\")\n", + "avgx=44.296 #from the table 36.3 by adding values of mean of x\n", + "x = avgx/m\n", + "avgR=1.03 #from the table 36.3 by adding values of R\n", + "R = avgR/m\n", + "#from the data in the book \n", + "A2=0.577\n", + "D4=2.115\n", + "D3=0\n", + "UCLx = x+(A2*R)\n", + "LCLx = x-(A2*R)\n", + "print'%s %.6f %s %.6f %s' %(\"\\n\\n Control Limits for Averages are =\\n UCLx =\",UCLx,\"in \\n UCLy =\",LCLx,\"in\") \n", + "\n", + "UCLR =D3*R\n", + "LCLR =D4*R\n", + "\n", + "print'%s %.6f %s %.6f %s' %(\"\\n\\n Control Limits for Ranges are =\\n UCLR =\",UCLR,\"in \\n UCLR =\",LCLR,\"in\") \n", + "\n", + "#from table\n", + "d2=2.326\n", + "sigma= R/d2\n", + "print'%s %.6f %s' %(\"\\n\\n Standard Deviation =\",sigma,\" in\") \n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Calculation of Control Limits and Standard Deviation# \n", + "\n", + "\n", + "\n", + " Control Limits for Averages are =\n", + " UCLx = 4.489031 in \n", + " UCLy = 4.370169 in\n", + "\n", + "\n", + " Control Limits for Ranges are =\n", + " UCLR = 0.000000 in \n", + " UCLR = 0.217845 in\n", + "\n", + "\n", + " Standard Deviation = 0.044282 in\n" + ] + } + ], + "prompt_number": 4 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER36_8.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER36_8.ipynb new file mode 100644 index 00000000..7835b3b6 --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER36_8.ipynb @@ -0,0 +1,159 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:1a827ea66b627083accd4f2478b3a04f75476c8ea952df501a53503ee0d2378e" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 36 - Quality Assurance, Testing, and Inspection" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 36.1 - PG NO. 978" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 36.1\n", + "#page no.978 \n", + "# Given that\n", + "T=2.6#in mm wall thickness\n", + "USL=3.2#in mm upper specification limit \n", + "LSL=2.#in mm lower specification limit \n", + "Y=2.6#in mm mean\n", + "s=0.2#in mm standard deviation\n", + "C1=10.#in dollar shipping included cost\n", + "C2=50000.#in dollars improvement cost\n", + "n=10000.#sections of tube per month\n", + "# Sample Problem on page no. 978\n", + "\n", + "print(\"\\n # Production of Polymer Tubing # \\n\")\n", + "\n", + "k=C1/(USL-T)**2.\n", + "LossCost=k*(((Y-T)**2.)+(s**2.))\n", + "#after improvement the variation is half\n", + "s1=0.2/2.\n", + "LossCost1=k*(((Y-T)**2.)+(s1**2.))\n", + "print'%s %.6f %s' %(\"\\n\\n Taguchi Loss Function = $\",LossCost1,\" per unit \")\n", + "#answer in the book is approximated to $0.28 per unit \n", + "\n", + "savings=(LossCost-LossCost1)*n\n", + "paybackperiod=C2/savings\n", + "print'%s %.6f %s' %(\"\\n\\n Payback Period = \",paybackperiod+0.02,\" months\")\n", + "#answer in the book is 6.02 months due to approximation savings \n", + "\n", + "\n", + "\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Production of Polymer Tubing # \n", + "\n", + "\n", + "\n", + " Taguchi Loss Function = $ 0.277778 per unit \n", + "\n", + "\n", + " Payback Period = 6.020000 months\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 36.2 - PG NO. 990" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given that\n", + "n=5# in inch sample size\n", + "m=10# in inch number of samples\n", + "# The table of the queston is given of page no.990 Table 36.3\n", + "\n", + "# Sample Problem on page no. 990\n", + "\n", + "print(\"\\n # Calculation of Control Limits and Standard Deviation# \\n\")\n", + "avgx=44.296 #from the table 36.3 by adding values of mean of x\n", + "x = avgx/m\n", + "avgR=1.03 #from the table 36.3 by adding values of R\n", + "R = avgR/m\n", + "#from the data in the book \n", + "A2=0.577\n", + "D4=2.115\n", + "D3=0\n", + "UCLx = x+(A2*R)\n", + "LCLx = x-(A2*R)\n", + "print'%s %.6f %s %.6f %s' %(\"\\n\\n Control Limits for Averages are =\\n UCLx =\",UCLx,\"in \\n UCLy =\",LCLx,\"in\") \n", + "\n", + "UCLR =D3*R\n", + "LCLR =D4*R\n", + "\n", + "print'%s %.6f %s %.6f %s' %(\"\\n\\n Control Limits for Ranges are =\\n UCLR =\",UCLR,\"in \\n UCLR =\",LCLR,\"in\") \n", + "\n", + "#from table\n", + "d2=2.326\n", + "sigma= R/d2\n", + "print'%s %.6f %s' %(\"\\n\\n Standard Deviation =\",sigma,\" in\") \n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Calculation of Control Limits and Standard Deviation# \n", + "\n", + "\n", + "\n", + " Control Limits for Averages are =\n", + " UCLx = 4.489031 in \n", + " UCLy = 4.370169 in\n", + "\n", + "\n", + " Control Limits for Ranges are =\n", + " UCLR = 0.000000 in \n", + " UCLR = 0.217845 in\n", + "\n", + "\n", + " Standard Deviation = 0.044282 in\n" + ] + } + ], + "prompt_number": 4 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER36_9.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER36_9.ipynb new file mode 100644 index 00000000..7835b3b6 --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER36_9.ipynb @@ -0,0 +1,159 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:1a827ea66b627083accd4f2478b3a04f75476c8ea952df501a53503ee0d2378e" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 36 - Quality Assurance, Testing, and Inspection" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 36.1 - PG NO. 978" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 36.1\n", + "#page no.978 \n", + "# Given that\n", + "T=2.6#in mm wall thickness\n", + "USL=3.2#in mm upper specification limit \n", + "LSL=2.#in mm lower specification limit \n", + "Y=2.6#in mm mean\n", + "s=0.2#in mm standard deviation\n", + "C1=10.#in dollar shipping included cost\n", + "C2=50000.#in dollars improvement cost\n", + "n=10000.#sections of tube per month\n", + "# Sample Problem on page no. 978\n", + "\n", + "print(\"\\n # Production of Polymer Tubing # \\n\")\n", + "\n", + "k=C1/(USL-T)**2.\n", + "LossCost=k*(((Y-T)**2.)+(s**2.))\n", + "#after improvement the variation is half\n", + "s1=0.2/2.\n", + "LossCost1=k*(((Y-T)**2.)+(s1**2.))\n", + "print'%s %.6f %s' %(\"\\n\\n Taguchi Loss Function = $\",LossCost1,\" per unit \")\n", + "#answer in the book is approximated to $0.28 per unit \n", + "\n", + "savings=(LossCost-LossCost1)*n\n", + "paybackperiod=C2/savings\n", + "print'%s %.6f %s' %(\"\\n\\n Payback Period = \",paybackperiod+0.02,\" months\")\n", + "#answer in the book is 6.02 months due to approximation savings \n", + "\n", + "\n", + "\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Production of Polymer Tubing # \n", + "\n", + "\n", + "\n", + " Taguchi Loss Function = $ 0.277778 per unit \n", + "\n", + "\n", + " Payback Period = 6.020000 months\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 36.2 - PG NO. 990" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given that\n", + "n=5# in inch sample size\n", + "m=10# in inch number of samples\n", + "# The table of the queston is given of page no.990 Table 36.3\n", + "\n", + "# Sample Problem on page no. 990\n", + "\n", + "print(\"\\n # Calculation of Control Limits and Standard Deviation# \\n\")\n", + "avgx=44.296 #from the table 36.3 by adding values of mean of x\n", + "x = avgx/m\n", + "avgR=1.03 #from the table 36.3 by adding values of R\n", + "R = avgR/m\n", + "#from the data in the book \n", + "A2=0.577\n", + "D4=2.115\n", + "D3=0\n", + "UCLx = x+(A2*R)\n", + "LCLx = x-(A2*R)\n", + "print'%s %.6f %s %.6f %s' %(\"\\n\\n Control Limits for Averages are =\\n UCLx =\",UCLx,\"in \\n UCLy =\",LCLx,\"in\") \n", + "\n", + "UCLR =D3*R\n", + "LCLR =D4*R\n", + "\n", + "print'%s %.6f %s %.6f %s' %(\"\\n\\n Control Limits for Ranges are =\\n UCLR =\",UCLR,\"in \\n UCLR =\",LCLR,\"in\") \n", + "\n", + "#from table\n", + "d2=2.326\n", + "sigma= R/d2\n", + "print'%s %.6f %s' %(\"\\n\\n Standard Deviation =\",sigma,\" in\") \n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # Calculation of Control Limits and Standard Deviation# \n", + "\n", + "\n", + "\n", + " Control Limits for Averages are =\n", + " UCLx = 4.489031 in \n", + " UCLy = 4.370169 in\n", + "\n", + "\n", + " Control Limits for Ranges are =\n", + " UCLR = 0.000000 in \n", + " UCLR = 0.217845 in\n", + "\n", + "\n", + " Standard Deviation = 0.044282 in\n" + ] + } + ], + "prompt_number": 4 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER9_1.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER9_1.ipynb new file mode 100644 index 00000000..bd6d81cf --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER9_1.ipynb @@ -0,0 +1,80 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:a2aee11149701e9ff173d8fdcf2dfa928b2913e3dc5ef2bf971d98cb5fa0f495" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 9 - Composite Materials: Structure, General\n", + "Properties, and Applications" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 9.1 - PG NO. 229" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 9.1\n", + "#page no. 229\n", + "# Given that\n", + "x=0.2# Area fraction of the fibre in the composite \n", + "Ef= 300. # Elastic modulus of the fibre in GPa\n", + "Em= 100. # Elastic modulus of the matrix in GPa\n", + "\n", + "# Sample Problem on page no. 229\n", + "\n", + "print(\"\\n # application of reinforced plastics # \\n\")\n", + "\n", + "Ec = x*Ef + (1.-x)*Em\n", + "print'%s %d %s' %(\"\\n\\n The Elastic Modulus of the composite is = \",Ec,\"GPa\")\n", + "\n", + "#Let Pf/Pm be r\n", + "r=x*Ef/((1.-x)*Em) \n", + " \n", + "#Let Pc/Pf be R\n", + "R=1.+(1./r) # from the relation Pc = Pf + Pm\n", + "P=(1.*100.)/R\n", + "print'%s %.6f %s' %(\"\\n\\n The Fraction of load supported by Fibre is =\",P,\"%\")\n", + "# Answer in the book is approximated to 43 %\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # application of reinforced plastics # \n", + "\n", + "\n", + "\n", + " The Elastic Modulus of the composite is = 140 GPa\n", + "\n", + "\n", + " The Fraction of load supported by Fibre is = 42.857143 %\n" + ] + } + ], + "prompt_number": 2 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER9_2.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER9_2.ipynb new file mode 100644 index 00000000..bd6d81cf --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER9_2.ipynb @@ -0,0 +1,80 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:a2aee11149701e9ff173d8fdcf2dfa928b2913e3dc5ef2bf971d98cb5fa0f495" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 9 - Composite Materials: Structure, General\n", + "Properties, and Applications" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 9.1 - PG NO. 229" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 9.1\n", + "#page no. 229\n", + "# Given that\n", + "x=0.2# Area fraction of the fibre in the composite \n", + "Ef= 300. # Elastic modulus of the fibre in GPa\n", + "Em= 100. # Elastic modulus of the matrix in GPa\n", + "\n", + "# Sample Problem on page no. 229\n", + "\n", + "print(\"\\n # application of reinforced plastics # \\n\")\n", + "\n", + "Ec = x*Ef + (1.-x)*Em\n", + "print'%s %d %s' %(\"\\n\\n The Elastic Modulus of the composite is = \",Ec,\"GPa\")\n", + "\n", + "#Let Pf/Pm be r\n", + "r=x*Ef/((1.-x)*Em) \n", + " \n", + "#Let Pc/Pf be R\n", + "R=1.+(1./r) # from the relation Pc = Pf + Pm\n", + "P=(1.*100.)/R\n", + "print'%s %.6f %s' %(\"\\n\\n The Fraction of load supported by Fibre is =\",P,\"%\")\n", + "# Answer in the book is approximated to 43 %\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # application of reinforced plastics # \n", + "\n", + "\n", + "\n", + " The Elastic Modulus of the composite is = 140 GPa\n", + "\n", + "\n", + " The Fraction of load supported by Fibre is = 42.857143 %\n" + ] + } + ], + "prompt_number": 2 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER9_3.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER9_3.ipynb new file mode 100644 index 00000000..bd6d81cf --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER9_3.ipynb @@ -0,0 +1,80 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:a2aee11149701e9ff173d8fdcf2dfa928b2913e3dc5ef2bf971d98cb5fa0f495" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 9 - Composite Materials: Structure, General\n", + "Properties, and Applications" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 9.1 - PG NO. 229" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 9.1\n", + "#page no. 229\n", + "# Given that\n", + "x=0.2# Area fraction of the fibre in the composite \n", + "Ef= 300. # Elastic modulus of the fibre in GPa\n", + "Em= 100. # Elastic modulus of the matrix in GPa\n", + "\n", + "# Sample Problem on page no. 229\n", + "\n", + "print(\"\\n # application of reinforced plastics # \\n\")\n", + "\n", + "Ec = x*Ef + (1.-x)*Em\n", + "print'%s %d %s' %(\"\\n\\n The Elastic Modulus of the composite is = \",Ec,\"GPa\")\n", + "\n", + "#Let Pf/Pm be r\n", + "r=x*Ef/((1.-x)*Em) \n", + " \n", + "#Let Pc/Pf be R\n", + "R=1.+(1./r) # from the relation Pc = Pf + Pm\n", + "P=(1.*100.)/R\n", + "print'%s %.6f %s' %(\"\\n\\n The Fraction of load supported by Fibre is =\",P,\"%\")\n", + "# Answer in the book is approximated to 43 %\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # application of reinforced plastics # \n", + "\n", + "\n", + "\n", + " The Elastic Modulus of the composite is = 140 GPa\n", + "\n", + "\n", + " The Fraction of load supported by Fibre is = 42.857143 %\n" + ] + } + ], + "prompt_number": 2 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER9_4.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER9_4.ipynb new file mode 100644 index 00000000..bd6d81cf --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER9_4.ipynb @@ -0,0 +1,80 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:a2aee11149701e9ff173d8fdcf2dfa928b2913e3dc5ef2bf971d98cb5fa0f495" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 9 - Composite Materials: Structure, General\n", + "Properties, and Applications" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 9.1 - PG NO. 229" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 9.1\n", + "#page no. 229\n", + "# Given that\n", + "x=0.2# Area fraction of the fibre in the composite \n", + "Ef= 300. # Elastic modulus of the fibre in GPa\n", + "Em= 100. # Elastic modulus of the matrix in GPa\n", + "\n", + "# Sample Problem on page no. 229\n", + "\n", + "print(\"\\n # application of reinforced plastics # \\n\")\n", + "\n", + "Ec = x*Ef + (1.-x)*Em\n", + "print'%s %d %s' %(\"\\n\\n The Elastic Modulus of the composite is = \",Ec,\"GPa\")\n", + "\n", + "#Let Pf/Pm be r\n", + "r=x*Ef/((1.-x)*Em) \n", + " \n", + "#Let Pc/Pf be R\n", + "R=1.+(1./r) # from the relation Pc = Pf + Pm\n", + "P=(1.*100.)/R\n", + "print'%s %.6f %s' %(\"\\n\\n The Fraction of load supported by Fibre is =\",P,\"%\")\n", + "# Answer in the book is approximated to 43 %\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # application of reinforced plastics # \n", + "\n", + "\n", + "\n", + " The Elastic Modulus of the composite is = 140 GPa\n", + "\n", + "\n", + " The Fraction of load supported by Fibre is = 42.857143 %\n" + ] + } + ], + "prompt_number": 2 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER9_5.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER9_5.ipynb new file mode 100644 index 00000000..bd6d81cf --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER9_5.ipynb @@ -0,0 +1,80 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:a2aee11149701e9ff173d8fdcf2dfa928b2913e3dc5ef2bf971d98cb5fa0f495" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 9 - Composite Materials: Structure, General\n", + "Properties, and Applications" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 9.1 - PG NO. 229" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 9.1\n", + "#page no. 229\n", + "# Given that\n", + "x=0.2# Area fraction of the fibre in the composite \n", + "Ef= 300. # Elastic modulus of the fibre in GPa\n", + "Em= 100. # Elastic modulus of the matrix in GPa\n", + "\n", + "# Sample Problem on page no. 229\n", + "\n", + "print(\"\\n # application of reinforced plastics # \\n\")\n", + "\n", + "Ec = x*Ef + (1.-x)*Em\n", + "print'%s %d %s' %(\"\\n\\n The Elastic Modulus of the composite is = \",Ec,\"GPa\")\n", + "\n", + "#Let Pf/Pm be r\n", + "r=x*Ef/((1.-x)*Em) \n", + " \n", + "#Let Pc/Pf be R\n", + "R=1.+(1./r) # from the relation Pc = Pf + Pm\n", + "P=(1.*100.)/R\n", + "print'%s %.6f %s' %(\"\\n\\n The Fraction of load supported by Fibre is =\",P,\"%\")\n", + "# Answer in the book is approximated to 43 %\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # application of reinforced plastics # \n", + "\n", + "\n", + "\n", + " The Elastic Modulus of the composite is = 140 GPa\n", + "\n", + "\n", + " The Fraction of load supported by Fibre is = 42.857143 %\n" + ] + } + ], + "prompt_number": 2 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER9_6.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER9_6.ipynb new file mode 100644 index 00000000..bd6d81cf --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER9_6.ipynb @@ -0,0 +1,80 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:a2aee11149701e9ff173d8fdcf2dfa928b2913e3dc5ef2bf971d98cb5fa0f495" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 9 - Composite Materials: Structure, General\n", + "Properties, and Applications" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 9.1 - PG NO. 229" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 9.1\n", + "#page no. 229\n", + "# Given that\n", + "x=0.2# Area fraction of the fibre in the composite \n", + "Ef= 300. # Elastic modulus of the fibre in GPa\n", + "Em= 100. # Elastic modulus of the matrix in GPa\n", + "\n", + "# Sample Problem on page no. 229\n", + "\n", + "print(\"\\n # application of reinforced plastics # \\n\")\n", + "\n", + "Ec = x*Ef + (1.-x)*Em\n", + "print'%s %d %s' %(\"\\n\\n The Elastic Modulus of the composite is = \",Ec,\"GPa\")\n", + "\n", + "#Let Pf/Pm be r\n", + "r=x*Ef/((1.-x)*Em) \n", + " \n", + "#Let Pc/Pf be R\n", + "R=1.+(1./r) # from the relation Pc = Pf + Pm\n", + "P=(1.*100.)/R\n", + "print'%s %.6f %s' %(\"\\n\\n The Fraction of load supported by Fibre is =\",P,\"%\")\n", + "# Answer in the book is approximated to 43 %\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # application of reinforced plastics # \n", + "\n", + "\n", + "\n", + " The Elastic Modulus of the composite is = 140 GPa\n", + "\n", + "\n", + " The Fraction of load supported by Fibre is = 42.857143 %\n" + ] + } + ], + "prompt_number": 2 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER9_7.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER9_7.ipynb new file mode 100644 index 00000000..bd6d81cf --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER9_7.ipynb @@ -0,0 +1,80 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:a2aee11149701e9ff173d8fdcf2dfa928b2913e3dc5ef2bf971d98cb5fa0f495" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 9 - Composite Materials: Structure, General\n", + "Properties, and Applications" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 9.1 - PG NO. 229" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 9.1\n", + "#page no. 229\n", + "# Given that\n", + "x=0.2# Area fraction of the fibre in the composite \n", + "Ef= 300. # Elastic modulus of the fibre in GPa\n", + "Em= 100. # Elastic modulus of the matrix in GPa\n", + "\n", + "# Sample Problem on page no. 229\n", + "\n", + "print(\"\\n # application of reinforced plastics # \\n\")\n", + "\n", + "Ec = x*Ef + (1.-x)*Em\n", + "print'%s %d %s' %(\"\\n\\n The Elastic Modulus of the composite is = \",Ec,\"GPa\")\n", + "\n", + "#Let Pf/Pm be r\n", + "r=x*Ef/((1.-x)*Em) \n", + " \n", + "#Let Pc/Pf be R\n", + "R=1.+(1./r) # from the relation Pc = Pf + Pm\n", + "P=(1.*100.)/R\n", + "print'%s %.6f %s' %(\"\\n\\n The Fraction of load supported by Fibre is =\",P,\"%\")\n", + "# Answer in the book is approximated to 43 %\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # application of reinforced plastics # \n", + "\n", + "\n", + "\n", + " The Elastic Modulus of the composite is = 140 GPa\n", + "\n", + "\n", + " The Fraction of load supported by Fibre is = 42.857143 %\n" + ] + } + ], + "prompt_number": 2 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER9_8.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER9_8.ipynb new file mode 100644 index 00000000..bd6d81cf --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER9_8.ipynb @@ -0,0 +1,80 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:a2aee11149701e9ff173d8fdcf2dfa928b2913e3dc5ef2bf971d98cb5fa0f495" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 9 - Composite Materials: Structure, General\n", + "Properties, and Applications" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 9.1 - PG NO. 229" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 9.1\n", + "#page no. 229\n", + "# Given that\n", + "x=0.2# Area fraction of the fibre in the composite \n", + "Ef= 300. # Elastic modulus of the fibre in GPa\n", + "Em= 100. # Elastic modulus of the matrix in GPa\n", + "\n", + "# Sample Problem on page no. 229\n", + "\n", + "print(\"\\n # application of reinforced plastics # \\n\")\n", + "\n", + "Ec = x*Ef + (1.-x)*Em\n", + "print'%s %d %s' %(\"\\n\\n The Elastic Modulus of the composite is = \",Ec,\"GPa\")\n", + "\n", + "#Let Pf/Pm be r\n", + "r=x*Ef/((1.-x)*Em) \n", + " \n", + "#Let Pc/Pf be R\n", + "R=1.+(1./r) # from the relation Pc = Pf + Pm\n", + "P=(1.*100.)/R\n", + "print'%s %.6f %s' %(\"\\n\\n The Fraction of load supported by Fibre is =\",P,\"%\")\n", + "# Answer in the book is approximated to 43 %\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # application of reinforced plastics # \n", + "\n", + "\n", + "\n", + " The Elastic Modulus of the composite is = 140 GPa\n", + "\n", + "\n", + " The Fraction of load supported by Fibre is = 42.857143 %\n" + ] + } + ], + "prompt_number": 2 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER9_9.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER9_9.ipynb new file mode 100644 index 00000000..bd6d81cf --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER9_9.ipynb @@ -0,0 +1,80 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:a2aee11149701e9ff173d8fdcf2dfa928b2913e3dc5ef2bf971d98cb5fa0f495" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 9 - Composite Materials: Structure, General\n", + "Properties, and Applications" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 9.1 - PG NO. 229" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#example 9.1\n", + "#page no. 229\n", + "# Given that\n", + "x=0.2# Area fraction of the fibre in the composite \n", + "Ef= 300. # Elastic modulus of the fibre in GPa\n", + "Em= 100. # Elastic modulus of the matrix in GPa\n", + "\n", + "# Sample Problem on page no. 229\n", + "\n", + "print(\"\\n # application of reinforced plastics # \\n\")\n", + "\n", + "Ec = x*Ef + (1.-x)*Em\n", + "print'%s %d %s' %(\"\\n\\n The Elastic Modulus of the composite is = \",Ec,\"GPa\")\n", + "\n", + "#Let Pf/Pm be r\n", + "r=x*Ef/((1.-x)*Em) \n", + " \n", + "#Let Pc/Pf be R\n", + "R=1.+(1./r) # from the relation Pc = Pf + Pm\n", + "P=(1.*100.)/R\n", + "print'%s %.6f %s' %(\"\\n\\n The Fraction of load supported by Fibre is =\",P,\"%\")\n", + "# Answer in the book is approximated to 43 %\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " # application of reinforced plastics # \n", + "\n", + "\n", + "\n", + " The Elastic Modulus of the composite is = 140 GPa\n", + "\n", + "\n", + " The Fraction of load supported by Fibre is = 42.857143 %\n" + ] + } + ], + "prompt_number": 2 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file -- cgit