From 5d4ad197775046773493a55d4ae90b7e7a8cd51e Mon Sep 17 00:00:00 2001 From: Trupti Kini Date: Tue, 22 Dec 2015 23:30:15 +0600 Subject: Added(A)/Deleted(D) following books A Design_With_Operational_Amplifiers_And_Analog_Integrated_Circuits_by_Sergio_Franco/chapter10_1.ipynb A Design_With_Operational_Amplifiers_And_Analog_Integrated_Circuits_by_Sergio_Franco/chapter11_1.ipynb A Design_With_Operational_Amplifiers_And_Analog_Integrated_Circuits_by_Sergio_Franco/chapter12_1.ipynb A Design_With_Operational_Amplifiers_And_Analog_Integrated_Circuits_by_Sergio_Franco/chapter13_1.ipynb A Design_With_Operational_Amplifiers_And_Analog_Integrated_Circuits_by_Sergio_Franco/chapter1_1.ipynb A Design_With_Operational_Amplifiers_And_Analog_Integrated_Circuits_by_Sergio_Franco/chapter2_1.ipynb A Design_With_Operational_Amplifiers_And_Analog_Integrated_Circuits_by_Sergio_Franco/chapter3_1.ipynb A Design_With_Operational_Amplifiers_And_Analog_Integrated_Circuits_by_Sergio_Franco/chapter4_1.ipynb A Design_With_Operational_Amplifiers_And_Analog_Integrated_Circuits_by_Sergio_Franco/chapter5_1.ipynb A Design_With_Operational_Amplifiers_And_Analog_Integrated_Circuits_by_Sergio_Franco/chapter6_1.ipynb A Design_With_Operational_Amplifiers_And_Analog_Integrated_Circuits_by_Sergio_Franco/chapter7_1.ipynb A Design_With_Operational_Amplifiers_And_Analog_Integrated_Circuits_by_Sergio_Franco/chapter8_1.ipynb A Design_With_Operational_Amplifiers_And_Analog_Integrated_Circuits_by_Sergio_Franco/chapter9_1.ipynb A Design_With_Operational_Amplifiers_And_Analog_Integrated_Circuits_by_Sergio_Franco/screenshots/Frequency_1.png A Design_With_Operational_Amplifiers_And_Analog_Integrated_Circuits_by_Sergio_Franco/screenshots/Saturation_1.png A Design_With_Operational_Amplifiers_And_Analog_Integrated_Circuits_by_Sergio_Franco/screenshots/Step_1.png A Electronic_Devices_by_S._Sharma/README.txt A Manufacturing_Science_by_A._Ghosh_And_A._K._Mallik/README.txt A Optoelectronics:_An_Introduction_by_John_Wilson_&_John_Hawkes/Chapter10_2.ipynb A Optoelectronics:_An_Introduction_by_John_Wilson_&_John_Hawkes/Chapter1_2.ipynb A Optoelectronics:_An_Introduction_by_John_Wilson_&_John_Hawkes/Chapter2_2.ipynb A Optoelectronics:_An_Introduction_by_John_Wilson_&_John_Hawkes/Chapter3_2.ipynb A Optoelectronics:_An_Introduction_by_John_Wilson_&_John_Hawkes/Chapter4_2.ipynb A Optoelectronics:_An_Introduction_by_John_Wilson_&_John_Hawkes/Chapter5_2.ipynb A Optoelectronics:_An_Introduction_by_John_Wilson_&_John_Hawkes/Chapter6_2.ipynb A Optoelectronics:_An_Introduction_by_John_Wilson_&_John_Hawkes/Chapter7_2.ipynb A Optoelectronics:_An_Introduction_by_John_Wilson_&_John_Hawkes/Chapter8_2.ipynb A Optoelectronics:_An_Introduction_by_John_Wilson_&_John_Hawkes/Chapter9_2.ipynb A Optoelectronics:_An_Introduction_by_John_Wilson_&_John_Hawkes/screenshots/CHAPTER9_1.png A Optoelectronics:_An_Introduction_by_John_Wilson_&_John_Hawkes/screenshots/Fig_1.24_2.png A Optoelectronics:_An_Introduction_by_John_Wilson_&_John_Hawkes/screenshots/Fig_8.26_2.png A "sample_notebooks/KAVANA B/chapter3.ipynb" A sample_notebooks/RohitPhadtare/chapter_no.6.ipynb A "sample_notebooks/S PRASHANTHS PRASHANTH/Chapter_1_4.ipynb" --- sample_notebooks/RohitPhadtare/chapter_no.6.ipynb | 1494 +++++++++++++++++++++ 1 file changed, 1494 insertions(+) create mode 100644 sample_notebooks/RohitPhadtare/chapter_no.6.ipynb (limited to 'sample_notebooks/RohitPhadtare') diff --git a/sample_notebooks/RohitPhadtare/chapter_no.6.ipynb b/sample_notebooks/RohitPhadtare/chapter_no.6.ipynb new file mode 100644 index 00000000..4657aa0b --- /dev/null +++ b/sample_notebooks/RohitPhadtare/chapter_no.6.ipynb @@ -0,0 +1,1494 @@ +{ + "metadata": { + "name": "chapter no.6.ipynb" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter No.6:Torsion" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.6.1,Page No.225" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Initilization of Variables\n", + "\n", + "L=10000 #mm #Length of solid shaft\n", + "d=100 #mm #Diameter of shaft\n", + "n=150 #rpm\n", + "P=112.5*10**6 #N-mm/sec #Power Transmitted\n", + "G=82*10**3 #N/mm**2 #modulus of Rigidity\n", + "\n", + "#Calculations\n", + "\n", + "J=pi*d**4*(32)**-1 #mm**3 #Polar Modulus\n", + "T=P*60*(2*pi*n)**-1 #N-mm #Torsional moment\n", + "\n", + "r=50 #mm #Radius\n", + "\n", + "q_s=T*r*J**-1 #N/mm**2 #Max shear stress intensity\n", + "Theta=T*L*(G*J)**-1 #angle of twist\n", + "\n", + "#Result\n", + "print\"Max shear stress intensity\",round(q_s,2),\"N/mm**2\"\n", + "print\"Angle of Twist\",round(Theta,3),\"radian\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Max shear stress intensity 36.48 N/mm**2\n", + "Angle of Twist 0.089 radian\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.6.2,Page No.226" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Initilization of Variables\n", + "\n", + "P=440*10**6 #N-m/sec #Power transmitted\n", + "n=280 #rpm\n", + "theta=pi*180**-1 #radian #angle of twist\n", + "L=1000 #mm #Length of solid shaft\n", + "q_s=40 #N/mm**2 #Max torsional shear stress\n", + "G=84*10**3 #N/mm**2 #Modulus of rigidity\n", + "\n", + "#Calculations\n", + "\n", + "#P=2*pi*n*T*(60)**-1 #Equation of Power transmitted\n", + "T=P*60*(2*pi*n)**-1 #N-mm #torsional moment\n", + "\n", + "#From Consideration of shear stress\n", + "d1=(T*16*(pi*40)**-1)**0.333333 \n", + "\n", + "#From Consideration of angle of twist\n", + "d2=(T*L*32*180*(pi*84*10**3*pi)**-1)**0.25\n", + "\n", + "#result\n", + "print\"Diameter of solid shaft is\",round(d1,2),\"mm\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Diameter of solid shaft is 124.09 mm\n" + ] + } + ], + "prompt_number": 21 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.6.3,Page No.227" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Initilization of Variables\n", + "\n", + "G=80*10**3 #N/mm**2 #Modulus of rigidity\n", + "q_s=80 #N/mm**2 #Max sheare stress\n", + "P=736*10**6 #N-mm/sec #Power transmitted\n", + "n=200\n", + "\n", + "#Calculations\n", + "\n", + "T=P*60*(2*pi*n)**-1 #N-mm #Torsional moment\n", + "\n", + "#Now From consideration of angle of twist\n", + "theta=pi*180**-1\n", + "#L=15*d\n", + "\n", + "d=(T*32*180*15*(pi**2*G)**-1)**0.33333\n", + "\n", + "#Now corresponding stress at the surface is\n", + "q_s2=T*32*d*(pi*2*d**4)**-1\n", + "\n", + "#Result\n", + "print\"Max diameter required is\",round(d,2),\"mm\"\n", + "print\"Corresponding shear stress is\",round(q_s2,2),\"N/mm**2\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Max diameter required is 156.66 mm\n", + "Corresponding shear stress is 46.55 N/mm**2\n" + ] + } + ], + "prompt_number": 27 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.6.4,Page No.228" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Initilization of Variables\n", + "\n", + "d=25 #mm #Diameter of steel bar\n", + "p=50*10**3 #N #Pull\n", + "dell_1=0.095 #mm #Extension of bar\n", + "l=200 #mm #Guage Length\n", + "T=200*10**3 #N-mm #Torsional moment\n", + "theta=0.9*pi*180**-1 #angle of twist\n", + "L=250 #mm Length of steel bar\n", + "\n", + "#Calculations\n", + "\n", + "A=pi*4**-1*d**2 #Area of steel bar #mm**2\n", + "E=p*l*(dell_1*A)**-1 #N/mm**2 #Modulus of elasticity \n", + "\n", + "J=pi*32**-1*d**4 #mm**4 #Polar modulus\n", + "\n", + "G=T*L*(theta*J)**-1 #Modulus of rigidity #N/mm**2\n", + "\n", + "#Now from the relation of Elastic constants\n", + "mu=E*(2*G)**-1-1\n", + "\n", + "#result\n", + "print\"The Poissoin's ratio is\",round(mu,3)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The Poissoin's ratio is 0.292\n" + ] + } + ], + "prompt_number": 30 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.6.5,Page No.229" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Initilization of Variables\n", + "\n", + "L=6000 #mm #Length of circular shaft\n", + "d1=100 #mm #Outer Diameter\n", + "d2=75 #mm #Inner Diameter\n", + "R=100*2**-1 #Radius of shaft\n", + "T=10*10**6 #N-mm #Torsional moment\n", + "G=80*10**3 #N/mm**2 #Modulus of Rigidity\n", + "\n", + "#Calculations\n", + "\n", + "J=pi*32**-1*(d1**4-d2**4) #mm**4 #Polar Modulus \n", + "\n", + "#Max Shear stress produced\n", + "q_s=T*R*J**-1 #N/mm**2\n", + "\n", + "#Angle of twist\n", + "theta=T*L*(G*J)**-1 #Radian\n", + "\n", + "#Result\n", + "print\"MAx shear stress produced is\",round(q_s,2),\"N/mm**2\"\n", + "print\"Angle of Twist is\",round(theta,2),\"Radian\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "MAx shear stress produced is 74.5 N/mm**2\n", + "Angle of Twist is 0.11 Radian\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.6.6,Page No.229" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Initilization of Variables\n", + "\n", + "d1=200 #mm #External Diameter of shaft\n", + "t=25 #mm #Thickness of shaft\n", + "n=200 #rpm\n", + "theta=0.5*pi*180**-1 #Radian #angle of twist\n", + "L=2000 #mm #Length of shaft\n", + "G=84*10**3 #N/mm**2\n", + "d2=d1-2*t #mm #Internal Diameter of shaft\n", + "\n", + "#Calculations\n", + "\n", + "J=pi*32**-1*(d1**4-d2**4) #mm**4 #Polar Modulus \n", + "\n", + "#Torsional moment\n", + "T=G*J*theta*L**-1 #N/mm**2 \n", + "\n", + "#Power Transmitted\n", + "P=2*pi*n*T*60**-1*10**-6 #N-mm\n", + "\n", + "#Max shear stress transmitted\n", + "q_s=G*theta*(d1*2**-1)*L**-1 #N/mm**2 \n", + "\n", + "#Result\n", + "print\"Power Transmitted is\",round(P,2),\"N-mm\"\n", + "print\"Max Shear stress produced is\",round(q_s,2),\"N/mm**2\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Power Transmitted is 824.28 N-mm\n", + "Max Shear stress produced is 36.65 N/mm**2\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.6.7,Page No.230" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Initilization of Variables\n", + "\n", + "P=3750*10**6 #N-mm/sec\n", + "n=240 #Rpm\n", + "q_s=160 #N/mm**2 #Max shear stress\n", + "\n", + "#Calculations\n", + "\n", + "#d2=0.8*d2 #mm #Internal Diameter of shaft\n", + "\n", + "#J=pi*32**-1*(d1**4-d2**4) #mm**4 #Polar modulus\n", + "#After substituting value in above Equation we get\n", + "#J=0.05796*d1**4\n", + "\n", + "T=P*60*(2*pi*n)**-1 #N-mm #Torsional moment\n", + "\n", + "#Now from Torsion Formula\n", + "#T*J**-1=q_s*R**-1 ......................................(1)\n", + "\n", + "#But R=d1*2**-1 \n", + "\n", + "#Now substituting value of R and J in Equation (1) we get\n", + "d1=(T*(0.05796*q_s*2)**-1)**0.33333\n", + "\n", + "d2=d1*0.8\n", + "\n", + "#Result\n", + "print\"The size of the Shaft is:d1\",round(d1,3),\"mm\"\n", + "print\" :d2\",round(d2,3),\"mm\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The size of the Shaft is:d1 200.362 mm\n", + " :d2 160.289 mm\n" + ] + } + ], + "prompt_number": 22 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.6.8,Page No.231" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Initilization of Variables\n", + "\n", + "P=245*10**6 #N-mm/sec #Power transmitted\n", + "n=240 #rpm\n", + "q_s=40 #N/mm**2 #Shear stress\n", + "theta=pi*180**-1 #radian #Angle of twist\n", + "L=1000 #mm #Length of shaft\n", + "G=80*10**3 #N/mm**2\n", + "\n", + "#Tmax=1.5*T\n", + "\n", + "#Calculations\n", + "\n", + "T=P*60*(2*pi*n)**-1 #N-mm #Torsional Moment\n", + "Tmax=1.5*T\n", + "\n", + "#Now For Solid shaft\n", + "#J=pi*32*d**4\n", + "\n", + "#Now from the consideration of shear stress we get\n", + "#T*J**-1=q_s*(d*2**-1)**-1\n", + "#After substituting value in above Equation we get\n", + "#T=pi*16**-1*d**3*q_s\n", + "\n", + "#Designing For max Torque\n", + "d=(Tmax*16*(pi*40)**-1)**0.33333 #mm #Diameter of shaft\n", + "\n", + "#For max Angle of Twist\n", + "#Tmax*J**-1=G*theta*L**-1 \n", + "#After substituting value in above Equation we get\n", + "d2=(Tmax*32*180*L*(pi**2*G)**-1)**0.25\n", + "\n", + "#For Hollow Shaft\n", + "\n", + "#d1_2=Outer Diameter\n", + "#d2_2=Inner Diameter\n", + "\n", + "#d2_2=0.5*d1_2\n", + "\n", + "# Polar modulus\n", + "#J=pi*32**-1*(d1_2**4-d2_2**4)\n", + "#After substituting values we get\n", + "#J=0.092038*d1_2**4\n", + "\n", + "#Now from the consideration of stress\n", + "#Tmax*J**-1=q_s*(d1_2*2**-1)**-1\n", + "#After substituting values and further simplifying we get\n", + "d1_2=(Tmax*(0.092038*2*q_s)**-1)**0.33333\n", + "\n", + "#Now from the consideration of angle of twist\n", + "#Tmax*J**-1=G*theta*L**-1\n", + "#After substituting values and further simplifying we get\n", + "d1_3=(Tmax*180*L*(0.092038*G*pi)**-1)**0.25\n", + "\n", + "d2_2=0.5*d1_2\n", + "\n", + "#result\n", + "print\"Diameter of shaft is:For solid shaft:d\",round(d,2),\"mm\"\n", + "print\" :For Hollow shaft:d1_2\",round(d1_2,3),\"mm\"\n", + "print\" : :d2_2\",round(d2_2,3),\"mm\" " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Diameter of shaft is:For solid shaft:d 123.01 mm\n", + " :For Hollow shaft:d1_2 125.69 mm\n", + " : :d2_2 62.845 mm\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.6.11,Page No.235" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Initilization of Variables\n", + "\n", + "P=250*10**6 #N-mm/sec #Power transmitted\n", + "n=100 #rpm\n", + "q_s=75 #N/mm**2 #Shear stress\n", + "\n", + "#Calculations\n", + "\n", + "#From Equation of Power we have\n", + "T=P*60*(2*pi*n)**-1 #N-mm #Torsional moment\n", + "\n", + "#Now from torsional moment equation we have\n", + "#T=j*q_s*(d/2**-1)**-1\n", + "#After substituting values in above equation and further simplifying we get\n", + "#T=pi*16**-1**d**3*q_s\n", + "d=(T*16*(pi*q_s)**-1)**0.3333 #mm #Diameter of solid shaft\n", + "\n", + "#PArt-2\n", + "\n", + "#Let d1 and d2 be the outer and inner diameter of hollow shaft\n", + "#d2=0.6*d1\n", + "\n", + "#Again from torsional moment equation we have\n", + "#T=pi*32**-1*(d1**4-d2**4)*q_s*(d1/2)**-1\n", + "d1=(T*16*(pi*(1-0.6**4)*q_s)**-1)**0.33333\n", + "d2=0.6*d1\n", + "\n", + "#Cross sectional area of solid shaft\n", + "A1=pi*4**-1*d**2 #mm**2\n", + "\n", + "#cross sectional area of hollow shaft\n", + "A2=pi*4**-1*(d1**2-d2**2)\n", + "\n", + "#Now percentage saving in weight\n", + "#Let W be the percentage saving in weight\n", + "W=(A1-A2)*100*A1**-1\n", + "\n", + "#Result\n", + "print\"Percentage saving in Weight is\",round(W,3),\"%\"\n", + "print\"Size of shaft is:solid shaft:d\",round(d,3),\"mm\"\n", + "print\" :Hollow shaft:d1\",round(d1,3),\"mm\"\n", + "print\" : :d2\",round(d2,3),\"mm\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Percentage saving in Weight is 29.735 %\n", + "Size of shaft is:solid shaft:d 117.418 mm\n", + " :Hollow shaft:d1 123.031 mm\n", + " : :d2 73.818 mm\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.6.12,Page No.237" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Initilization of Variables\n", + "d=100 #mm #Diameter of solid shaft\n", + "d1=100 #mm #Outer Diameter of hollow shaft\n", + "d2=50 #mm #Inner Diameter of hollow shaft\n", + "\n", + "#Calculations\n", + "\n", + "#Torsional moment of solid shaft\n", + "#T_s=J*q_s*(d*2**-1)**-1 \n", + "#After substituting values in above equation and further simplifying we get\n", + "#T_s=pi*16*d**3*q_s ...............(1)\n", + "\n", + "#torsional moment for hollow shaft is\n", + "#T_h=J*q_s*(d1**4-d2**4)**-1*(d1*2**-1)\n", + "#After substituting values in above equation and further simplifying we get\n", + "#T_h=pi*32**-1*2*d1**-1*(d1**4-d2**4)*q_s ...........(2)\n", + "\n", + "#Dividing Equation 2 by 1 we get\n", + "#Let the ratio of T_h*T_s**-1 Be X\n", + "X=1-0.5**4\n", + "\n", + "#Loss in strength \n", + "#Let s be the loss in strength\n", + "#s=T_s*T_h*100*T_s**-1\n", + "#After substituting values in above equation and further simplifying we get\n", + "s=(1-0.9375)*100\n", + "\n", + "#Weight Ratio \n", + "#Let w be the Weight ratio\n", + "#w=W_h*W_s**-1\n", + "\n", + "A_h=pi*32**-1*(d1**2-d2**2) #mm**2 #Area of Hollow shaft\n", + "A_s=pi*32**-1*d**2 #mm**2 #Area of solid shaft\n", + "\n", + "w=A_h*A_s**-1 \n", + "\n", + "#Result\n", + "print\"Loss in strength is\",round(s,2)\n", + "print\"Weight ratio is\",round(w,2)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Loss in strength is 6.25\n", + "Weight ratio is 0.75\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.6.13,Page No.239" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Initilization of Variables\n", + "T=8 #KN-m #Torque \n", + "d=100 #mm #Diameter of portion AB\n", + "d1=100 #mm #External Diameter of Portion BC\n", + "d2=75 #mm #Internal Diameter of Portion BC\n", + "G=80 #KN/mm**2 #Modulus of Rigidity\n", + "L1=1500 #mm #Radial Distance of Portion AB\n", + "L2=2500 #mm #Radial Distance ofPortion BC\n", + "\n", + "#Calculations\n", + "\n", + "R=d*2**-1 #mm #Radius of shaft\n", + "\n", + "#For Portion AB,Polar Modulus\n", + "J1=pi*32**-1*d**4 #mm**4 \n", + "\n", + "#For Portion BC,Polar modulus \n", + "J2=pi*32**-1*(d1**4-d2**4) #mm**4\n", + "\n", + "#Now Max stress occurs in portion BC since max radial Distance is sme in both cases\n", + "q_max=T*J2**-1*R*10**6 #N/mm**2 \n", + "\n", + "#Let theta1 be the rotation in Portion AB and theta2 be the rotation in portion BC\n", + "theta1=T*L1*(G*J1)**-1 #Radians\n", + "theta2=T*L2*(G*J2)**-1 #Radians\n", + "\n", + "#Total Rotational at end C\n", + "theta=(theta1+theta2)*10**3 #Radians\n", + "\n", + "#Result\n", + "print\"Max stress induced is\",round(q_max,2),\"N/mm**2\"\n", + "print\"Angle of Twist is\",round(theta,3),\"radians\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Max stress induced is 59.6 N/mm**2\n", + "Angle of Twist is 0.053 radians\n" + ] + } + ], + "prompt_number": 23 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.6.14,Page No.240" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Initilization of Variables\n", + "\n", + "q_b=80 #N/mm**2 #Shear stress in Brass\n", + "q_s=100 #N/mm**2 #Shear stress in Steel\n", + "G_b=40*10**3 #N/mm**2 \n", + "G_s=80*10**3 \n", + "L_b=1000 #mm #Length of brass shaft\n", + "L_s=1200 #mm #Length of steel shaft\n", + "d1=80 #mm #Diameter of brass shaft\n", + "d2=60 #mm #Diameter of steel shaft\n", + "\n", + "#Calculations\n", + "\n", + "#Polar modulus of brass rod\n", + "J_b=pi*32**-1*d1**4 #mm**4 \n", + "\n", + "#Polar modulus of steel rod\n", + "J_s=pi*32**-1*d2**4 #mm**4\n", + "\n", + "#Considering bras Rod:AB\n", + "T1=J_b*q_b*(d1*2**-1)**-1 #N-mm \n", + "\n", + "#Considering Steel Rod:BC\n", + "T2=J_s*q_s*(d2*2**-1)**-1 #N-mm\n", + "\n", + "#Max Torque that can be applied\n", + "T2\n", + "\n", + "#Let theta_b and theta_s be the rotations in Brass and steel respectively\n", + "theta_b=T2*L_b*(G_b*J_b)**-1 #Radians\n", + "theta_s=T2*L_s*(G_s*J_s)**-1 #Radians\n", + "\n", + "theta=theta_b+theta_s #Radians #Rotation of free end\n", + "\n", + "#Result\n", + "print\"Total of free end is\",round(theta,3),\"Radians\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Total of free end is 0.076 Radians\n" + ] + } + ], + "prompt_number": 36 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.6.15,Page No.241" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Initilization of Variables\n", + "\n", + "G=80*10**3 #N/mm**2 #Modulus of Rigidity\n", + "d1=100 #mm #Outer diameter of hollow shft\n", + "d2=80 #mm #Inner diameter of hollow shaft\n", + "d=80 #mm #diameter of Solid shaft\n", + "d3=60 #mm #diameter of Solid shaft having L=0.5m\n", + "L1=300 #mm #Length of Hollow shaft\n", + "L2=400 #mm #Length of solid shaft\n", + "L3=500 #mm #LEngth of solid shaft of diameter 60mm\n", + "T1=2*10**6 #N-mm #Torsion in Shaft AB\n", + "T2=1*10**6 #N-mm #Torsion in shaft BC\n", + "T3=1*10**6 #N-mm #Torsion in shaft CD\n", + "\n", + "#Calculations\n", + "\n", + "#Now Polar modulus of section AB\n", + "J1=pi*32**-1*(d1**4-d2**4) #mm**4 \n", + "\n", + "#Polar modulus of section BC\n", + "J2=pi*32**-1*d**4 #mm**4\n", + "\n", + "#Polar modulus of section CD\n", + "J3=pi*32**-1*d3**4 #mm**4\n", + "\n", + "#Now angle of twist of AB\n", + "theta1=T1*L1*(G*J1)**-1 #radians\n", + "\n", + "#Angle of twist of BC\n", + "theta2=T2*L2*(G*J2)**-1 #radians\n", + "\n", + "#Angle of twist of CD\n", + "theta3=T3*L3*(G*J3)**-1 #radians\n", + "\n", + "#Angle of twist\n", + "theta=theta1-theta2+theta3 #Radians\n", + "\n", + "#Shear stress in AB From Torsion Equation\n", + "q_s1=T1*(d1*2**-1)*J1**-1 #N/mm**2 \n", + "\n", + "#Shear stress in BC\n", + "q_s2=T2*(d*2**-1)*J2**-1 #N/mm**2 \n", + "\n", + "#Shear stress in CD\n", + "q_s3=T3*(d3*2**-1)*J3**-1 #N-mm**2\n", + "\n", + "#As max shear stress occurs in portion CD,so consider CD\n", + "\n", + "#Result\n", + "print\"Angle of twist at free end is\",round(theta,5),\"Radian\"\n", + "print\"Max Shear stress\",round(q_s3,2),\"N/mm**2\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Angle of twist at free end is 0.00496 Radian\n", + "Max Shear stress 23.58 N/mm**2\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.6.16,Page No.242" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Initilization of Variables\n", + "\n", + "L=1000 #mm #Length of bar\n", + "L1=600 #mm #Length of Bar AB\n", + "L2=400 #mm #Length of Bar BC\n", + "d1=60 #mm #Outer Diameter of bar BC\n", + "d2=30 #mm #Inner Diameter of bar BC\n", + "d=60 #mm #Diameter of bar AB\n", + "T=2*10**6 #N-mm #Total Torque\n", + "\n", + "#Calculations\n", + "\n", + "#Polar Modulus of Portion AB\n", + "J1=pi*32**-1*d**4 #mm*4\n", + "\n", + "#Polar Modulus of Portion BC\n", + "J2=pi*32**-1*(d1**4-d2**4) #mm**4\n", + "\n", + "#Let T1 be the torque resisted by bar AB and T2 be torque resisted by Bar BC\n", + "#Let theta1 and theta2 be the rotation of shaft in portion AB & BC\n", + "\n", + "#theta1=T1*L1*(G*J1)**-1 #radians\n", + "#After substituting values and further simplifying we get \n", + "#theta1=32*600*T1*(pi*60**4*G)**-1\n", + "\n", + "#theta2=T2*L*(J2*G)**-1 #Radians\n", + "#After substituting values and further simplifying we get \n", + "#theta2=32*400*T2*(pi*60**4*(1-0.5**4)*G)**-1 \n", + "\n", + "#Now For consistency of Deformation,theta1=theta2\n", + "#After substituting values and further simplifying we get \n", + "#T1=0.7111*T2 ..................................................(1)\n", + "\n", + "#But T1+T2=T=2*10**6 ...........................................(2)\n", + "#Substituting value of T1 in above equation\n", + "\n", + "T2=T*(0.7111+1)**-1\n", + "T1=0.71111*T2\n", + "\n", + "#Max stress in Portion AB\n", + "q_s1=T1*(d*2**-1)*(J1)**-1 #N/mm**2\n", + "\n", + "#Max stress in Portion BC\n", + "q_s2=T2*(d1*2**-1)*J2**-1 #N/mm**2 \n", + "\n", + "#Result\n", + "print\"Stresses Developed in Portion:AB\",round(q_s1,2),\"N/mm**2\"\n", + "print\" :BC\",round(q_s2,2),\"N/mm**2\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Stresses Developed in Portion:AB 19.6 N/mm**2\n", + " :BC 29.4 N/mm**2\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.6.17,Page No.243" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Initilization of Variables\n", + "\n", + "d1=80 #mm #External Diameter of Brass tube\n", + "d2=50 #mm #Internal Diameter of Brass tube\n", + "d=50 #mm #Diameter of steel Tube\n", + "G_b=40*10**3 #N/mm**2 #Modulus of Rigidity of brass tube\n", + "G_s=80*10**3 #N/mm**2 #Modulus of rigidity of steel tube\n", + "T=6*10**6 #N-mm #Torque\n", + "L=2000 #mm #Length of Tube\n", + "\n", + "#Calculations\n", + "\n", + "#Polar Modulus of brass tube\n", + "J1=pi*32**-1*(d1**4-d2**4) #mm**4 \n", + "\n", + "#Polar modulus of steel Tube\n", + "J2=pi*32**-1*d**4 #mm**4\n", + "\n", + "#Let T_s & T_b be the torque resisted by steel and brass respectively\n", + "#Then, T_b+T_s=T ............................................(1)\n", + "\n", + "#Since the angle of twist will be the same\n", + "#Theta1=Theta2\n", + "#After substituting values and further simplifying we get \n", + "#Ts=0.360*Tb ...........................................(2)\n", + "\n", + "#After substituting value of Ts in eqn 1 and further simplifying we get \n", + "T_b=T*(0.36+1)**-1 #N-mm\n", + "T_s=0.360*T_b\n", + "\n", + "#Let q_s and q_b be the max stress in steel and brass respectively\n", + "q_b=T_b*(d1*2**-1)*J1**-1 #N/mm**2\n", + "q_s=T_s*(d2*2**-1)*J2**-1 #N/mm**2\n", + "\n", + "#Since angle of twist in brass=angle of twist in steel\n", + "theta_s=T_s*L*(J2*G_s)**-1\n", + "\n", + "#Result\n", + "print\"Stresses Developed in Materials are:Brass\",round(q_b,2),\"N/mm**2\"\n", + "print\" :Steel\",round(q_s,2),\"N/mm**2\"\n", + "print\"Angle of Twist in 2m Length\",round(theta_s,3),\"Radians\" " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Stresses Developed in Materials are:Brass 51.79 N/mm**2\n", + " :Steel 64.71 N/mm**2\n", + "Angle of Twist in 2m Length 0.065 Radians\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.6.18,Page No.245" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Initilization of Variables\n", + "\n", + "d1=60 #mm #External Diameter of aluminium Tube\n", + "d2=40 #mm #Internal Diameter of aluminium Tube\n", + "d=40 #mm #Diameter of steel tube\n", + "q_a=60 #N/mm**2 #Permissible stress in aluminium\n", + "q_s=100 #N/mm**2 #Permissible stress in steel tube\n", + "G_a=27*10**3 #N/mm**2 \n", + "G_s=80*10**3 #N/mm**2 \n", + "\n", + "#Calculations\n", + "\n", + "#Polar modulus of aluminium Tube\n", + "J_a=pi*32**-1*(d1**4-d2**4) #mm**4\n", + "\n", + "#Polar Modulus of steel Tube\n", + "J_s=pi*32**-1*d**4 #mm**4\n", + "\n", + "#Now the angle of twist of steel tube = angle of twist of aluminium tube\n", + "#T_s*L_s*(J_s*theta_s)**-1=T_a*L_a*(J_a*theta_a)**-1\n", + "#After substituting values in above Equation and Further simplifyin we get\n", + "#T_s=0.7293*T_a .....................(1)\n", + "\n", + "#If steel Governs the resisting capacity\n", + "T_s1=q_s*J_s*(d*2**-1)**-1 #N-mm\n", + "T_a1=T_s1*0.7293**-1 #N-mm\n", + "T1=(T_s1+T_a1)*10**-6 #KN-m #Total Torque in steel Tube\n", + "\n", + "#If aluminium Governs the resisting capacity \n", + "T_a2=q_a*J_a*(d1*2**-1) #N-mm\n", + "T_s2=T_a2*0.7293 #N-mm\n", + "T2=(T_s2+T_a2)*10**-6 #KN-m #Total Torque in aluminium tube\n", + "\n", + "#Result\n", + "print\"Steel Governs the torque carrying capacity\",round(T1,2),\"KN-m\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Steel Governs the torque carrying capacity 2.98 KN-m\n" + ] + } + ], + "prompt_number": 19 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.6.19,Page No.247" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Initilization of Variables\n", + "\n", + "P=225*10**6 #N-mm/sec #Power Trasmitted\n", + "q_b=80 #N/mm**2 #Shear stress\n", + "n=200 #Rpm\n", + "q_k=100 #N/mm**2 #PErmissible stress in Keys\n", + "D=300 #mm #Diameter of bolt circle\n", + "L=150 #mm #Length of shear key\n", + "d=16 #mm #Diameterr of bolt\n", + "\n", + "#Calculations\n", + "T=60*P*(2*pi*n)**-1 #N-mm #Torque\n", + "\n", + "#Now From Torsion Formula\n", + "#T*J**-1=q_s*R**-1\n", + "#After substituting values we get\n", + "#T=pi*16*d**3*n\n", + "#After further simplifying we get\n", + "d1=(T*16*(pi*q_s)**-1)**0.33333\n", + "\n", + "#Let b be the width of shear Key\n", + "#T=q_k*L*b*R\n", + "#After simplifying further we get\n", + "b=T*(q_k*L*(d1*2**-1))**-1 #mm\n", + "\n", + "#Let n2 be the no. of bolts required at bolt circle of radius\n", + "R_b=D*2**-1 #mm \n", + "\n", + "n2=T*4*(q_b*pi*d**2*R_b)**-1\n", + "\n", + "#result\n", + "print\"Minimum no. of Bolts Required are\",round(n2,2)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Minimum no. of Bolts Required are 4.45\n" + ] + } + ], + "prompt_number": 26 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.6.20,Page No.250" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Initilization of Variables\n", + "\n", + "T=2*10**6 #N-mm #Torque transmitted\n", + "G=80*10**3 #N/mm**2 #Modulus of rigidity\n", + "d1=40 #mm \n", + "d2=80 #mm\n", + "r1=20 #mm\n", + "r2=40 #mm\n", + "L=2000 #mm #Length of shaft\n", + "\n", + "#Calculations\n", + "\n", + "#Angle of twist \n", + "theta=2*T*L*(r1**2+r1*r2+r2**2)*(3*pi*G*r2**3*r1**3)**-1 #radians\n", + "\n", + "#If the shaft is treated as shaft of average Diameter\n", + "d_avg=(d1+d2)*2**-1 #mm\n", + "\n", + "theta1=T*L*(G*pi*32**-1*d_avg**4)**-1 #Radians\n", + "\n", + "#Percentage Error\n", + "#Let Percentage Error be E\n", + "X=theta-theta1\n", + "E=(X*theta**-1)*100 \n", + "\n", + "#Result\n", + "print\"Percentage Error is\",round(E,2),\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Percentage Error is 32.28 %\n" + ] + } + ], + "prompt_number": 42 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.6.21,Page No.252" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Initilization of Variables\n", + "\n", + "G=80*10**3 #N/mm**2 \n", + "P=1*10**9 #N-mm/sec #Power\n", + "n=300 \n", + "d1=150 #mm #Outer Diameter\n", + "d2=120 #mm #Inner Diameter\n", + "L=2000 #mm #Length of circular shaft\n", + "\n", + "#Calculations\n", + "\n", + "T=P*60*(2*pi*n)**-1 #N-mm\n", + "\n", + "#Polar Modulus \n", + "J=pi*32**-1*(d1**4-d2**4) #mm**4\n", + "\n", + "q_s=T*J**-1*(d1*2**-1) #N/mm**2 \n", + "\n", + "\n", + "#Strain ENergy\n", + "U=q_s**2*(4*G)**-1*pi*4**-1*(d1**2-d2**2)*L\n", + "\n", + "#Result\n", + "print\"Max shear stress is\",round(q_s,2),\"N/mm**2\"\n", + "print\"Strain Energy stored in the shaft is\",round(U,2),\"N-mm\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Max shear stress is 81.36 N/mm**2\n", + "Strain Energy stored in the shaft is 263181.37 N-mm\n" + ] + } + ], + "prompt_number": 51 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.6.22,Page No.254" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Initilization of Variables\n", + "\n", + "d=12 #mm #Diameter of helical spring\n", + "D=150 #mm #Mean Diameter\n", + "R=D*2**-1 #mm #Radius of helical spring\n", + "n=10 #no.of turns\n", + "G=80*10**3 #N/mm**2 \n", + "W=450 #N #Load\n", + "\n", + "#Calculations\n", + "\n", + "#Max shear stress \n", + "q_s=16*W*R*(pi*d**3)**-1 #N/mm**2\n", + "\n", + "#Strain Energy stored\n", + "U=32*W**2*R**3*n*(G*d**4)**-1 #N-mm\n", + "\n", + "#Deflection Produced\n", + "dell=64*W*R**3*n*(G*d**4)**-1 #mm\n", + "\n", + "#Stiffness Spring\n", + "k=W*dell**-1 #N/mm\n", + "\n", + "#Result\n", + "print\"Max shear stress is\",round(q_s,2),\"N/mm**2\"\n", + "print\"Strain Energy stored is\",round(U,2),\"N-mm\"\n", + "print\"Deflection Produced is\",round(dell,2),\"mm\"\n", + "print\"Stiffness spring is\",round(k,2),\"N/mm\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Max shear stress is 99.47 N/mm**2\n", + "Strain Energy stored is 16479.49 N-mm\n", + "Deflection Produced is 73.24 mm\n", + "Stiffness spring is 6.14 N/mm\n" + ] + } + ], + "prompt_number": 53 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.6.23,Page No.255" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Initilization of Variables\n", + "\n", + "K=5 #N/mm #Stiffness\n", + "L=100 #mm #Solid Length\n", + "q_s=60 #N/mm**2 #Max shear stress\n", + "W=200 #N #Max Load\n", + "G=80*10**3 #N/mm**2\n", + "\n", + "#Calculations\n", + "\n", + "#K=W*dell**-1\n", + "#After substituting values and further simplifying we get\n", + "#d=0.004*R**3*n ........(1) #mm #Diameter of wire\n", + "#n=L*d**-1 ........(2)\n", + "\n", + "#From Shearing stress\n", + "#q_s=16*W*R*(pi*d**3)**-1 \n", + "#After substituting values and further simplifying we get\n", + "#d**4=0.004*R**3*n .................(4)\n", + "\n", + "#From Equation 1,2,3\n", + "#d**4=0.004*(0.0785*d**3)**3*100*d**-1\n", + "#after further simplifying we get\n", + "d=5168.101**0.25\n", + "n=100*d**-1\n", + "R=(d**4*(0.004*n)**-1)**0.3333\n", + "\n", + "#Result\n", + "print\"Diameter of Wire is\",round(d,2),\"mm\"\n", + "print\"No.of turns is\",round(n,2)\n", + "print\"Mean Radius of spring is\",round(R,2),\"mm\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Diameter of Wire is 8.48 mm\n", + "No.fo turns is 11.79\n", + "Mean Radius of spring is 47.83 mm\n" + ] + } + ], + "prompt_number": 54 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.6.24,Page No.255" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Initilization of Variables\n", + "\n", + "m=5*10**5 #Wagon Weighing\n", + "v=18*1000*36000**-1 \n", + "d=300 #mm #Diameter of Beffer springs\n", + "n=18 #no.of turns\n", + "G=80*10**3 #N/mm**2\n", + "dell=225\n", + "R=100 #mm #Mean Radius\n", + "\n", + "#Calculations\n", + "\n", + "#Energy of Wagon\n", + "E=m*v**2*(9.81*2)**-1 #N-mm\n", + "\n", + "#Load applied\n", + "W=dell*G*d**4*(64*R**3*n)**-1 #N \n", + "\n", + "#Energy each spring can absorb is\n", + "E2=W*dell*2**-1 #N-mm\n", + "\n", + "#No.of springs required to absorb energy of Wagon\n", + "n2=E*E2**-1 *10**7\n", + "\n", + "#Result\n", + "print\"No.of springs Required for Buffer is\",round(n2,2)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "No.of springs Required for Buffer is 4.47\n" + ] + } + ], + "prompt_number": 66 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.6.25,Page No.259" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Initilization of Variables\n", + "\n", + "b=180 #mm #width of flange\n", + "d=10 #mm #Depth of flange\n", + "t=10 #mm #Thickness of flange\n", + "D=400 #mm #Overall Depth \n", + "\n", + "#Calculations\n", + "\n", + "I_xx=1*12**-1*(b*D**3-(b-t)*(D-2*d)**3)\n", + "I_yy=1*12**-1*((D-2*d)*t**3+2*t*b**3)\n", + "\n", + "#If warping is neglected\n", + "J=I_xx+I_yy #mm**4\n", + "\n", + "#Since b/d>1.6,we get\n", + "J2=1*3**-1*d**3*b*(1-0.63*d*b**-1)*2+1*3**-1*t**3*(D-2*d)*(1-0.63*t*b**-1)\n", + "\n", + "#Over Estimation of torsional Rigidity would have been \n", + "T=J*J2**-1\n", + "\n", + "#Result\n", + "print\"Error in assessing torsional Rigidity if the warping is neglected is\",round(T,2)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Error in assessing torsional Rigidity if the warping is neglected is 808.28\n" + ] + } + ], + "prompt_number": 68 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.6.26,Page No.261" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Initilization of Variables\n", + "\n", + "d1=100 #mm #Outer Diameter\n", + "d2=95 #mm #Inner Diameter\n", + "T=2*10**6 #N-mm #Torque\n", + "\n", + "#Calculations\n", + "\n", + "J=pi*32**-1*(d1**4-d2**4) #mm**4 #Polar Modulus\n", + "\n", + "#Shear stress\n", + "q_max=T*J**-1*d1*2**-1 #N/mm**2 \n", + "\n", + "#Now theta*L**-1=T*(G*J)**-1\n", + "#After substituting values and further simplifying we get\n", + "#Let theta*L**-1=X\n", + "X=T*J**-1\n", + "\n", + "#Now Treating it as very thin walled tube\n", + "d=(d1+d2)*2**-1 #mm\n", + "\n", + "r=d*2**-1 \n", + "t=(d1-d2)*2**-1\n", + "q_max2=T*(2*pi*r**2*t)**-1 #N/mm**2\n", + "\n", + "X2=T*(2*pi*r**3*t)**-1 \n", + "\n", + "#Result\n", + "print\"When it is treated as hollow shaft:Max shear stress\",round(q_max,2),\"N/mm**2\"\n", + "print\" :Angle of Twist per unit Length\",round(X,3)\n", + "print\"When it is very thin Walled Tube :Max shear stress\",round(q_max2,2),\"N/mm**2\"\n", + "print\" :Angle of twist per Unit Length\",round(X2,3)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "When it is treated as hollow shaft:Max shear stress 54.91 N/mm**2\n", + " :Angle of Twist per unit Length 1.098\n", + "When it is very thin Walled Tube :Max shear stress 53.57 N/mm**2\n", + " :Angle of twist per Unit Length 1.099\n" + ] + } + ], + "prompt_number": 72 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file -- cgit