path: root/Machine_Design_by_U.C._Jindal/Ch18_2.ipynb

{
 "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"
   ]
  }
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