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diff --git a/Machine_Design_by_U.C._Jindal/Ch18_1.ipynb b/Machine_Design_by_U.C._Jindal/Ch18_1.ipynb new file mode 100644 index 00000000..edb3ac36 --- /dev/null +++ b/Machine_Design_by_U.C._Jindal/Ch18_1.ipynb @@ -0,0 +1,460 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Ch:18 Rolling bearings" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## exa 18-1 - Page 507" + ] + }, + { + "cell_type": "code", + "execution_count": 21, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Pf is 53.08 W \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import sqrt, pi\n", + "Pr=16*10**3#\n", + "u=0.0011#\n", + "F=u*Pr#\n", + "r=20*10**-3#\n", + "#Let frictional moment be M\n", + "M=F*r#\n", + "N=1440#\n", + "w=2*pi*N/60#\n", + "Pf=M*w#\n", + "print \"Pf is %0.2f W \"%(Pf)#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## exa 18-2 - Page 508" + ] + }, + { + "cell_type": "code", + "execution_count": 22, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "L1 is 57.3 million revolutions \n", + "\n", + "L2 is 43.46 million revoltions \n" + ] + } + ], + "source": [ + "C=5590#\n", + "Ca=2500#\n", + "Pa=625#\n", + "Pr=1250#\n", + "V=1#\n", + "X=0.56#\n", + "Y=1.2#\n", + "P1=(X*V*Pr)+(Y*Pa)#\n", + "L1=(C/P1)**3#\n", + "V=1.2#\n", + "P2=(X*V*Pr)+(Y*Pa)#\n", + "L2=(C/P2)**3#\n", + "print \"L1 is %0.1f million revolutions \"%(L1)#\n", + "print \"\\nL2 is %0.2f million revoltions \"%(L2)#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## exa 18-4 - Page 509" + ] + }, + { + "cell_type": "code", + "execution_count": 23, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Lh is 199.424 hrs \n" + ] + } + ], + "source": [ + "P=20*10**3#\n", + "Co=22400#\n", + "C=41000#\n", + "Ln=(C/P)**3#\n", + "Lh=Ln*10**6/(720*60)#\n", + "print \"Lh is %0.3f hrs \"%(Lh)#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## exa 18-5 - Page 510" + ] + }, + { + "cell_type": "code", + "execution_count": 24, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Designation of Bearing B1 is 61805 \n", + "\n", + "d1 is 25 mm \n", + "\n", + "D1 is 37 mm \n", + "\n", + "B1 is 7 mm \n", + "\n", + "C1 is 3120 N \n", + "\n", + "Designation of Bearing B2 is 16005 \n", + "\n", + "d2 is 25 mm \n", + "\n", + "D2 is 47 mm \n", + "\n", + "B2 is 8 mm \n", + "\n", + "C2 is 7620 N \n", + "Bearing 61805 at B1 and 16005 at B2 can be installed.\n" + ] + } + ], + "source": [ + "from math import sqrt, pi\n", + "R1x=120#\n", + "R1y=250#\n", + "R2x=300#\n", + "R2y=400#\n", + "Lh=8000#\n", + "N=720#\n", + "Ln=Lh*60*N*10**-6#\n", + "R1=sqrt(R1x**2+R1y**2)#\n", + "R2=sqrt(R2x**2+R2y**2)#\n", + "#Let load factor be Ks\n", + "Ks=1.5#\n", + "P1=R1*Ks#\n", + "P2=R2*Ks#\n", + "C1=P1*(Ln**(1/3))#\n", + "C2=P2*(Ln**(1/3))#\n", + "#let designation,d,D,B,C at bearing B1 be De1,d1,D1,B1,C1\n", + "d1=25#\n", + "D1=37#\n", + "B1=7#\n", + "C1=3120#\n", + "De1=61805#\n", + "#let designation,d,D,B,C at bearing B2 be De2,d2,D2,B2,C2\n", + "d2=25#\n", + "D2=47#\n", + "B2=8#\n", + "C2=7620#\n", + "De2=16005#\n", + "print \"Designation of Bearing B1 is %0.0f \"%(De1)#\n", + "print \"\\nd1 is %0.0f mm \"%(d1)#\n", + "print \"\\nD1 is %0.0f mm \"%(D1)#\n", + "print \"\\nB1 is %0.0f mm \"%(B1)#\n", + "print \"\\nC1 is %0.0f N \"%(C1)#\n", + "print \"\\nDesignation of Bearing B2 is %0.0f \"%(De2)#\n", + "print \"\\nd2 is %0.0f mm \"%(d2)#\n", + "print \"\\nD2 is %0.0f mm \"%(D2)#\n", + "print \"\\nB2 is %0.0f mm \"%(B2)#\n", + "print \"\\nC2 is %0.0f N \"%(C2)#\n", + "print 'Bearing 61805 at B1 and 16005 at B2 can be installed.'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## exa 18-6 - Page 511" + ] + }, + { + "cell_type": "code", + "execution_count": 25, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Lh1 is 14585 hrs \n", + "\n", + "Lh2 is 24216 hrs \n" + ] + } + ], + "source": [ + "from math import tan, pi,sqrt\n", + "P=7500#\n", + "N=1440#\n", + "w=2*pi*N/60#\n", + "T=P/w#\n", + "r=0.2#\n", + "#Let T1-T2=t\n", + "t=T/r#\n", + "T2=t/2.5#\n", + "T1=3.5*T2#\n", + "R=0.125#\n", + "Ft=T/R#\n", + "Fr=Ft*tan(20*pi/180)#\n", + "# RD & RA are reaction forces calculated in vertical and horizontal directions from FBD by force equilibrium\n", + "RDv=186.5#\n", + "RAv=236.2#\n", + "RDh=36.2#\n", + "RAh=108.56#\n", + "RA=sqrt(RAv**2+RAh**2)#\n", + "RD=sqrt(RDv**2+RDh**2)#\n", + "Ks=1.4#\n", + "P1=RA*Ks#\n", + "P2=RD*Ks#\n", + "#let designation,d,D,B,C at bearing B1 be De1,d1,C1\n", + "d1=25#\n", + "C1=3120#\n", + "De1=61805#\n", + "#let designation,d,D,B,C at bearing B2 be De2,d2,C2\n", + "d2=25#\n", + "\n", + "C2=2700#\n", + "De2=61804#\n", + "L1=(C1/P1)**3#\n", + "Lh1=L1*10**6/(720*60)#\n", + "L2=(C2/P2)**3#\n", + "Lh2=L2*10**6/(720*60)#\n", + "print \"Lh1 is %0.0f hrs \"%(Lh1)#\n", + "print \"\\nLh2 is %0.0f hrs \"%(Lh2)#\n", + "#Incorrect value of P2 is taken in the book while calculating L2." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## exa 18-7 - Page 511" + ] + }, + { + "cell_type": "code", + "execution_count": 26, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "C is 44589 N \n" + ] + } + ], + "source": [ + "from math import log\n", + "P=3500#\n", + "Lh=6000#\n", + "N=1400#\n", + "R98=0.98#\n", + "R90=0.9#\n", + "L98=Lh*60*N/10**6#\n", + "x=(log(1/R98)/log(1/R90))**(1/1.17)#\n", + "L90=L98/x#\n", + "C=P*L90**(1/3)#\n", + "print \"C is %0.0f N \"%(C)#\n", + "#The difference in the value of C is due to rounding-off of value of L." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## exa 18-8 - Page 512" + ] + }, + { + "cell_type": "code", + "execution_count": 27, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "C is 6.337 kN \n" + ] + } + ], + "source": [ + "from math import log\n", + "n=3#\n", + "P=3#\n", + "#Let Reliability of system be R\n", + "R=0.83#\n", + "L94=6#\n", + "R94=(R)**(1/n)#\n", + "x=(log(1/R94)/log(1/0.90))**(1/1.17)#\n", + "L90=L94/x#\n", + "C=P*L90**(1/3)#\n", + "print \"C is %0.3f kN \"%(C)#\n", + "#The difference in the value of C is due to rounding-off of value of L." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## exa 18-9 - Page 512" + ] + }, + { + "cell_type": "code", + "execution_count": 28, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "C is 34219 N \n" + ] + } + ], + "source": [ + "P1=3000#\n", + "P2=4000#\n", + "P3=5000#\n", + "N1=1440#\n", + "N2=1080#\n", + "N3=720#\n", + "t1=1/4#\n", + "t2=1/2#\n", + "t3=1/4#\n", + "n1=N1*t1#\n", + "n2=N2*t2#\n", + "n3=N3*t3#\n", + "N=(n1+n2+n3)#\n", + "Pe=(((n1*P1**3)+(n2*P2**3)+(n3*P3**3))/N)**(1/3)#\n", + "Lh=10*10**3#\n", + "L=Lh*60*N/10**6#\n", + "C=Pe*L**(1/3)#\n", + "print \"C is %0.0f N \"%(C)#\n", + "#The difference in the value of C is due to rounding-off of value of Pe" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## exa 18-10 - Page 513" + ] + }, + { + "cell_type": "code", + "execution_count": 29, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Lh is 227.66 hrs \n" + ] + } + ], + "source": [ + "Co=695#\n", + "C=1430#\n", + "Pa1=200#\n", + "Pr1=600#\n", + "x=Pa1/Co#\n", + "y=Pa1/Pr1#\n", + "e=0.37+((0.44-0.37)*0.038/0.28)#\n", + "X=1#\n", + "Y=0#\n", + "P1=600#\n", + "Pa2=120#\n", + "Pr2=300#\n", + "X=0.56#\n", + "Y=1.2-(0.2*0.042/0.12)#\n", + "P2=(X*Pr2)+(Y*Pa2)#\n", + "N1=1440#\n", + "N2=720#\n", + "t1=2/3#\n", + "t2=1/3#\n", + "n1=N1*t1#\n", + "n2=N2*t2#\n", + "N=(n1+n2)#\n", + "Pe=(((n1*P1**3)+(n2*P2**3))/N)**(1/3)#\n", + "L=(C/Pe)**3#\n", + "Lh=L*10**6/(N*60)#\n", + "print \"Lh is %0.2f hrs \"%(Lh)#\n", + "#The difference in the value of Lh is due to rounding-off of value of Pe" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} |