From d36fc3b8f88cc3108ffff6151e376b619b9abb01 Mon Sep 17 00:00:00 2001 From: kinitrupti Date: Fri, 12 May 2017 18:40:35 +0530 Subject: Revised list of TBCs --- .../RohitPhadtare_version_backup/chapter.6.ipynb | 1494 ++++++++++++++++++++ sample_notebooks/RohitPhadtare/chapter_1.ipynb | 438 ++++++ sample_notebooks/RohitPhadtare/chapter_1_som.ipynb | 438 ------ sample_notebooks/RohitPhadtare/chapter_no.6.ipynb | 1494 -------------------- 4 files changed, 1932 insertions(+), 1932 deletions(-) create mode 100755 sample_notebooks/RohitPhadtare/RohitPhadtare_version_backup/chapter.6.ipynb create mode 100755 sample_notebooks/RohitPhadtare/chapter_1.ipynb delete mode 100755 sample_notebooks/RohitPhadtare/chapter_1_som.ipynb delete mode 100755 sample_notebooks/RohitPhadtare/chapter_no.6.ipynb (limited to 'sample_notebooks/RohitPhadtare') diff --git a/sample_notebooks/RohitPhadtare/RohitPhadtare_version_backup/chapter.6.ipynb b/sample_notebooks/RohitPhadtare/RohitPhadtare_version_backup/chapter.6.ipynb new file mode 100755 index 00000000..4657aa0b --- /dev/null +++ b/sample_notebooks/RohitPhadtare/RohitPhadtare_version_backup/chapter.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 diff --git a/sample_notebooks/RohitPhadtare/chapter_1.ipynb b/sample_notebooks/RohitPhadtare/chapter_1.ipynb new file mode 100755 index 00000000..ff3fcb22 --- /dev/null +++ b/sample_notebooks/RohitPhadtare/chapter_1.ipynb @@ -0,0 +1,438 @@ +{ + "metadata": { + "name": "chapter 1 som.ipynb" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 1:Centre Of Gravity\n" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem 1.1,Page No.8" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Initilization of Variables\n", + "\n", + "#Rectangle-1\n", + "a_1=37.5 #cm**2 \n", + "y_1=26.25 #cm \n", + "\n", + "#Rectangle-2\n", + "a_2=50 #cm**2 \n", + "y_2=15 #cm \n", + "\n", + "#Rectangle-3\n", + "a_3=150 #cm**2 \n", + "y_3=2.5 #cm \n", + "\n", + "\n", + "#Calculation\n", + "\n", + "\n", + "Y_bar=(a_1*y_1+a_2*y_2+a_3*y_3)*(a_1+a_2+a_3)**-1 #cm \n", + "\n", + "#Result\n", + "print\"The centroid of the section is\",round(Y_bar,2),\"cm\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The centroid of the section is 8.88 cm\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem 1.2,Page No.9" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Initilization of variables\n", + "\n", + "#Area-1\n", + "a_1=6 #cm**2 \n", + "x_1=3 #cm\n", + "y_1=0.5 #cm\n", + "\n", + "#Area-2\n", + "a_2=6 #cm**2\n", + "x_2=2.671 #cm\n", + "y_2=3 #cm\n", + "\n", + "#Area-3\n", + "a_3=16 #cm**2\n", + "x_3=1 #cm\n", + "y_3=5 #cm\n", + "\n", + "\n", + "#Calculation\n", + "\n", + "\n", + "X_bar=(a_1*x_1+a_2*x_2+a_3*x_3)*(a_1+a_2+a_3)**-1 #cm\n", + "Y_bar=(a_1*y_1+a_2*y_2+a_3*y_3)*(a_1+a_2+a_3)**-1 #cm\n", + "\n", + "\n", + "#Result\n", + "print\"The centre of gravity of section is\",round(X_bar,2),\"cm\"\n", + "print\"The centre of gravity of section is\",round(Y_bar,2),\"cm\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The centre of gravity of section is 1.79 cm\n", + "The centre of gravity of section is 3.61 cm\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem 1.3,Page no.10" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Initilization of variables\n", + "\n", + "#Area-1\n", + "a_1=93.75 #cm**2 \n", + "y_1=6.25 #cm\n", + "\n", + "#Area-2\n", + "a_2=93.75 #cm**2 \n", + "y_2=6.25 #cm\n", + "\n", + "#Area-3\n", + "a_3=375 #cm**2 \n", + "y_3=9.375 #cm\n", + "\n", + "#Area-4\n", + "a_4=353.43 #cm**2\n", + "y_4=6.366 #cm\n", + "\n", + "\n", + "#Calculation\n", + "\n", + "Y_bar=(a_1*y_1+a_2*y_2+a_3*y_3-a_4*y_4)*(a_1+a_2+a_3-a_4)**-1 #cm\n", + "\n", + "\n", + "#Result\n", + "print\"The centre of gravity lies at a distance of \",round(Y_bar,2),\"cm\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The centre of gravity lies at a distance of 11.66 cm\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem 1.4,Page no.10\n", + "\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Initilization of variables\n", + "\n", + "\n", + "a_1=36*pi #cm**2 #Area of Quadrant of a circle\n", + "x_1=16/pi #cm \n", + "y_1=16*pi**-1 #cm\n", + "\n", + "\n", + "a_2=18*pi #cm**2 #Area of the semicircle\n", + "x_2=6 #cm\n", + "y_2=8*pi**-1 #cm\n", + "\n", + "\n", + "#Calculation-1\n", + "\n", + "X_bar=(a_1*x_1-a_2*x_2)*(a_1-a_2)**-1 #cm\n", + "\n", + "#Calculation-2\n", + "#To calculate Y_bar,taking AB as the Reference line\n", + "\n", + "Y_bar=(a_1*y_1-a_2*y_2)*(a_1-a_2)**-1 #cm\n", + "\n", + "#Result\n", + "\n", + "print\"The centre of gravity is \",round(X_bar,2),\"cm\"\n", + "print\"The centre of gravity is\",round(Y_bar,2),\"cm\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The centre of gravity is 4.19 cm\n", + "The centre of gravity is 7.64 cm\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem 1.5,Page no.11" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Initilization of variables\n", + "\n", + "#Circle-1 \n", + "a_1=100*pi #cm**2\n", + "x_1=10 #cm\n", + " \n", + "#Square-2 \n", + "a_2=50 #cm**2\n", + "x_2=15 #cm\n", + " \n", + "#Calculation\n", + "\n", + "X_bar=(a_1*x_1-a_2*x_2)*(a_1-a_2)**-1 #cm\n", + "\n", + "\n", + "#Result\n", + "print\"The centre of gravity is\",round(X_bar,2),\"cm\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The centre of gravity is 9.05 cm\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem 1.6,Page no.12\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#intilization of variables \n", + "\n", + "#Rectangle-1\n", + "a_1=51200 #mm**2 \n", + "x_1=160 #mm\n", + "y_1=80 #mm\n", + "\n", + "#Triangle-2\n", + "a_2=6400 #mm**2\n", + "x_2=80*3**-1 #mm\n", + "y_2=320*3**-1 #mm\n", + "\n", + "#Semicircle-3\n", + "a_3=1250*pi #mm**2\n", + "x_3=210 #mm\n", + "y_3=(160-(4*50-(3*pi)**-1)) #mm\n", + "\n", + "\n", + "#Calculation\n", + "\n", + "X_bar=(a_1*x_1-a_2*x_2-a_3*x_3)*(a_1-a_2-a_3)**-1 #mm\n", + "Y_bar=(a_1*y_1-a_2*y_2-a_3*y_3)*(a_1-a_2-a_3)**-1 #mm\n", + "\n", + "#Result\n", + "print\"The centroid of the given area is\",round(X_bar,2),\"mm\"\n", + "print\"The centroid of the given area is\",round(Y_bar,2),\"mm\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The centroid of the given area is 176.07 mm\n", + "The centroid of the given area is 87.34 mm\n" + ] + } + ], + "prompt_number": 27 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem 1.8,Page no.12" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Initilization of variables\n", + "\n", + "\n", + "alpha=pi/2 #degree #In case of semicircle\n", + "\n", + "#Semicircle-1\n", + "r_1=20 #cm #radius of semicircle \n", + "y_1=4*r_1*(3*pi)**-1 #cm #distance from the base\n", + "a_1=(pi*r_1**2)*2**-1 #cm**2 #area of semicircle\n", + "\n", + "#Semicircle-2\n", + "r_2=16 #cm #radius of semicircle\n", + "y_2=4*r_2*(3*pi)**-1 #cm #distance from the base\n", + "a_2=(pi*r_2**2)*2**-1 #cm**2 #area of semicircle\n", + "\n", + "#Calculations\n", + "\n", + "\n", + "Y_bar=(a_1*y_1-a_2*y_2)*(a_1-a_2)**-1 #cm #centroid\n", + "\n", + "\n", + "#Result\n", + "print\"The centroid of the area is \",round(Y_bar,2),\"cm\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The centroid of the area is 11.51 cm\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem no1.12,Page no.16" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Initilization of variables\n", + "\n", + "#Right Circular Cyclinder\n", + "#m_1=(16*pi*h*rho_1) #gm \n", + "#y_1=4+h*2**-1 #cm\n", + "\n", + "#Hemisphere\n", + "#m_2=256*pi*rho_1 #gm \n", + "y_2=2.5 #cm \n", + "\n", + "Y_bar=4 #cm\n", + "r=4 #cm\n", + "\n", + "#Calculation\n", + "\n", + "#Y_bar=(m_1*y_1+m_2*y_2)*(m_1+m_2)**-1 #cm #Centroid\n", + "h=(402.114*25.132**-1)**0.5\n", + "\n", + "#Result\n", + "print\"The value of h is\",round(h,2),\"cm\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of h is 4.0 cm\n" + ] + } + ], + "prompt_number": 2 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/sample_notebooks/RohitPhadtare/chapter_1_som.ipynb b/sample_notebooks/RohitPhadtare/chapter_1_som.ipynb deleted file mode 100755 index ff3fcb22..00000000 --- a/sample_notebooks/RohitPhadtare/chapter_1_som.ipynb +++ /dev/null @@ -1,438 +0,0 @@ -{ - "metadata": { - "name": "chapter 1 som.ipynb" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "heading", - "level": 1, - "metadata": {}, - "source": [ - "Chapter 1:Centre Of Gravity\n" - ] - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Problem 1.1,Page No.8" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math\n", - "\n", - "#Initilization of Variables\n", - "\n", - "#Rectangle-1\n", - "a_1=37.5 #cm**2 \n", - "y_1=26.25 #cm \n", - "\n", - "#Rectangle-2\n", - "a_2=50 #cm**2 \n", - "y_2=15 #cm \n", - "\n", - "#Rectangle-3\n", - "a_3=150 #cm**2 \n", - "y_3=2.5 #cm \n", - "\n", - "\n", - "#Calculation\n", - "\n", - "\n", - "Y_bar=(a_1*y_1+a_2*y_2+a_3*y_3)*(a_1+a_2+a_3)**-1 #cm \n", - "\n", - "#Result\n", - "print\"The centroid of the section is\",round(Y_bar,2),\"cm\"" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "The centroid of the section is 8.88 cm\n" - ] - } - ], - "prompt_number": 5 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Problem 1.2,Page No.9" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math\n", - "\n", - "#Initilization of variables\n", - "\n", - "#Area-1\n", - "a_1=6 #cm**2 \n", - "x_1=3 #cm\n", - "y_1=0.5 #cm\n", - "\n", - "#Area-2\n", - "a_2=6 #cm**2\n", - "x_2=2.671 #cm\n", - "y_2=3 #cm\n", - "\n", - "#Area-3\n", - "a_3=16 #cm**2\n", - "x_3=1 #cm\n", - "y_3=5 #cm\n", - "\n", - "\n", - "#Calculation\n", - "\n", - "\n", - "X_bar=(a_1*x_1+a_2*x_2+a_3*x_3)*(a_1+a_2+a_3)**-1 #cm\n", - "Y_bar=(a_1*y_1+a_2*y_2+a_3*y_3)*(a_1+a_2+a_3)**-1 #cm\n", - "\n", - "\n", - "#Result\n", - "print\"The centre of gravity of section is\",round(X_bar,2),\"cm\"\n", - "print\"The centre of gravity of section is\",round(Y_bar,2),\"cm\"\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "The centre of gravity of section is 1.79 cm\n", - "The centre of gravity of section is 3.61 cm\n" - ] - } - ], - "prompt_number": 7 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Problem 1.3,Page no.10" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math\n", - "\n", - "#Initilization of variables\n", - "\n", - "#Area-1\n", - "a_1=93.75 #cm**2 \n", - "y_1=6.25 #cm\n", - "\n", - "#Area-2\n", - "a_2=93.75 #cm**2 \n", - "y_2=6.25 #cm\n", - "\n", - "#Area-3\n", - "a_3=375 #cm**2 \n", - "y_3=9.375 #cm\n", - "\n", - "#Area-4\n", - "a_4=353.43 #cm**2\n", - "y_4=6.366 #cm\n", - "\n", - "\n", - "#Calculation\n", - "\n", - "Y_bar=(a_1*y_1+a_2*y_2+a_3*y_3-a_4*y_4)*(a_1+a_2+a_3-a_4)**-1 #cm\n", - "\n", - "\n", - "#Result\n", - "print\"The centre of gravity lies at a distance of \",round(Y_bar,2),\"cm\"\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "The centre of gravity lies at a distance of 11.66 cm\n" - ] - } - ], - "prompt_number": 8 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Problem 1.4,Page no.10\n", - "\n" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math\n", - "\n", - "#Initilization of variables\n", - "\n", - "\n", - "a_1=36*pi #cm**2 #Area of Quadrant of a circle\n", - "x_1=16/pi #cm \n", - "y_1=16*pi**-1 #cm\n", - "\n", - "\n", - "a_2=18*pi #cm**2 #Area of the semicircle\n", - "x_2=6 #cm\n", - "y_2=8*pi**-1 #cm\n", - "\n", - "\n", - "#Calculation-1\n", - "\n", - "X_bar=(a_1*x_1-a_2*x_2)*(a_1-a_2)**-1 #cm\n", - "\n", - "#Calculation-2\n", - "#To calculate Y_bar,taking AB as the Reference line\n", - "\n", - "Y_bar=(a_1*y_1-a_2*y_2)*(a_1-a_2)**-1 #cm\n", - "\n", - "#Result\n", - "\n", - "print\"The centre of gravity is \",round(X_bar,2),\"cm\"\n", - "print\"The centre of gravity is\",round(Y_bar,2),\"cm\"\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "The centre of gravity is 4.19 cm\n", - "The centre of gravity is 7.64 cm\n" - ] - } - ], - "prompt_number": 9 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Problem 1.5,Page no.11" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math\n", - "\n", - "#Initilization of variables\n", - "\n", - "#Circle-1 \n", - "a_1=100*pi #cm**2\n", - "x_1=10 #cm\n", - " \n", - "#Square-2 \n", - "a_2=50 #cm**2\n", - "x_2=15 #cm\n", - " \n", - "#Calculation\n", - "\n", - "X_bar=(a_1*x_1-a_2*x_2)*(a_1-a_2)**-1 #cm\n", - "\n", - "\n", - "#Result\n", - "print\"The centre of gravity is\",round(X_bar,2),\"cm\"\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "The centre of gravity is 9.05 cm\n" - ] - } - ], - "prompt_number": 10 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Problem 1.6,Page no.12\n" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math\n", - "\n", - "#intilization of variables \n", - "\n", - "#Rectangle-1\n", - "a_1=51200 #mm**2 \n", - "x_1=160 #mm\n", - "y_1=80 #mm\n", - "\n", - "#Triangle-2\n", - "a_2=6400 #mm**2\n", - "x_2=80*3**-1 #mm\n", - "y_2=320*3**-1 #mm\n", - "\n", - "#Semicircle-3\n", - "a_3=1250*pi #mm**2\n", - "x_3=210 #mm\n", - "y_3=(160-(4*50-(3*pi)**-1)) #mm\n", - "\n", - "\n", - "#Calculation\n", - "\n", - "X_bar=(a_1*x_1-a_2*x_2-a_3*x_3)*(a_1-a_2-a_3)**-1 #mm\n", - "Y_bar=(a_1*y_1-a_2*y_2-a_3*y_3)*(a_1-a_2-a_3)**-1 #mm\n", - "\n", - "#Result\n", - "print\"The centroid of the given area is\",round(X_bar,2),\"mm\"\n", - "print\"The centroid of the given area is\",round(Y_bar,2),\"mm\"" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "The centroid of the given area is 176.07 mm\n", - "The centroid of the given area is 87.34 mm\n" - ] - } - ], - "prompt_number": 27 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Problem 1.8,Page no.12" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math\n", - "\n", - "#Initilization of variables\n", - "\n", - "\n", - "alpha=pi/2 #degree #In case of semicircle\n", - "\n", - "#Semicircle-1\n", - "r_1=20 #cm #radius of semicircle \n", - "y_1=4*r_1*(3*pi)**-1 #cm #distance from the base\n", - "a_1=(pi*r_1**2)*2**-1 #cm**2 #area of semicircle\n", - "\n", - "#Semicircle-2\n", - "r_2=16 #cm #radius of semicircle\n", - "y_2=4*r_2*(3*pi)**-1 #cm #distance from the base\n", - "a_2=(pi*r_2**2)*2**-1 #cm**2 #area of semicircle\n", - "\n", - "#Calculations\n", - "\n", - "\n", - "Y_bar=(a_1*y_1-a_2*y_2)*(a_1-a_2)**-1 #cm #centroid\n", - "\n", - "\n", - "#Result\n", - "print\"The centroid of the area is \",round(Y_bar,2),\"cm\"\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "The centroid of the area is 11.51 cm\n" - ] - } - ], - "prompt_number": 11 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Problem no1.12,Page no.16" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math\n", - "\n", - "#Initilization of variables\n", - "\n", - "#Right Circular Cyclinder\n", - "#m_1=(16*pi*h*rho_1) #gm \n", - "#y_1=4+h*2**-1 #cm\n", - "\n", - "#Hemisphere\n", - "#m_2=256*pi*rho_1 #gm \n", - "y_2=2.5 #cm \n", - "\n", - "Y_bar=4 #cm\n", - "r=4 #cm\n", - "\n", - "#Calculation\n", - "\n", - "#Y_bar=(m_1*y_1+m_2*y_2)*(m_1+m_2)**-1 #cm #Centroid\n", - "h=(402.114*25.132**-1)**0.5\n", - "\n", - "#Result\n", - "print\"The value of h is\",round(h,2),\"cm\"\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "The value of h is 4.0 cm\n" - ] - } - ], - "prompt_number": 2 - } - ], - "metadata": {} - } - ] -} \ No newline at end of file diff --git a/sample_notebooks/RohitPhadtare/chapter_no.6.ipynb b/sample_notebooks/RohitPhadtare/chapter_no.6.ipynb deleted file mode 100755 index 4657aa0b..00000000 --- a/sample_notebooks/RohitPhadtare/chapter_no.6.ipynb +++ /dev/null @@ -1,1494 +0,0 @@ -{ - "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