path: root/Theory_Of_Machines_by__B._K._Sarkar/Chapter3.ipynb

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  {
   "cells": [
    {
     "cell_type": "heading",
     "level": 1,
     "metadata": {},
     "source": [
      "Chapter3-FRICTION"
     ]
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Ex1-pg102"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "##CHAPTER 3 ILLUSRTATION 1 PAGE NO 102\n",
      "##TITLE:FRICTION\n",
      "##FIRURE 3.16(a),3.16(b)\n",
      "import math\n",
      "##===========================================================================================\n",
      "##INPUT DATA\n",
      "P1=180.##                        PULL APPLIED TO THE BODY IN NEWTONS\n",
      "theta=30.##                      ANGLE AT WHICH P IS ACTING IN DEGREES\n",
      "P2=220.##                        PUSH APPLIED TO THE BODY IN NEWTONS\n",
      "##Rn=                           NORMAL REACTION\n",
      "##F=                            FORCE OF FRICTION IN NEWTONS\n",
      "##U=                            COEFFICIENT OF FRICTION\n",
      "##W=                            WEIGHT OF THE BODY IN NEWTON\n",
      "##==========================================================================================\n",
      "##CALCULATION\n",
      "F1=P1*math.cos(theta/57.3)##              RESOLVING FORCES HORIZONTALLY FROM 3.16(a)\n",
      "F2=P2*math.cos(theta/57.3)##              RESOLVING FORCES HORIZONTALLY FROM 3.16(b)\n",
      "##                               RESOLVING FORCES VERTICALLY  Rn1=W-P1*sind(theta) from 3.16(a)\n",
      "##                               RESOLVING FORCES VERTICALLY  Rn2=W+P1*sind(theta) from 3.16(b)\n",
      "##                               USING THE RELATION F1=U*Rn1    &     F2=U*Rn2  AND SOLVING FOR W BY DIVIDING THESE TWO EQUATIONS\n",
      "X=F1/F2##                        THIS IS THE VALUE OF   Rn1/Rn2\n",
      "Y1=P1*math.sin(theta/57.3)\n",
      "Y2=P2*math.sin(theta/57.3)\n",
      "W=(Y2*X+Y1)/(1-X)##                BY SOLVING ABOVE 3 EQUATIONS\n",
      "U=F1/(W-P1*math.sin(theta/57.3))##          COEFFICIENT OF FRICTION\n",
      "##=============================================================================================\n",
      "##OUTPUT\n",
      "print'%s %.1f %s %.1f %s '%('WEIGHT OF THE BODY =',W,'N''THE COEFFICIENT OF FRICTION =',U,'')\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "WEIGHT OF THE BODY = 989.9 NTHE COEFFICIENT OF FRICTION = 0.2  \n"
       ]
      }
     ],
     "prompt_number": 1
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Ex2-pg103"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "##CHAPTER 3 ILLUSRTATION 2 PAGE NO 103\n",
      "##TITLE:FRICTION\n",
      "##FIRURE 3.17\n",
      "import math\n",
      "##===========================================================================================\n",
      "##INPUT DATA\n",
      "THETA=45##                ANGLE OF INCLINATION IN DEGREES\n",
      "g=9.81##                   ACCELERATION DUE TO GRAVITY IN N/mm**2\n",
      "U=.1##                     COEFFICIENT FRICTION\n",
      "##Rn=NORMAL REACTION\n",
      "##M=MASS IN NEWTONS\n",
      "##f=ACCELERATION OF THE BODY\n",
      "u=0.##                      INITIAL VELOCITY\n",
      "V=10.##                     FINAL VELOCITY IN m/s**2\n",
      "##===========================================================================================\n",
      "##CALCULATION\n",
      "##CONSIDER THE EQUILIBRIUM OF FORCES PERPENDICULAR TO THE PLANE\n",
      "##Rn=Mgcos(THETA)\n",
      "##CONSIDER THE EQUILIBRIUM OF FORCES ALONG THE PLANE\n",
      "##Mgsin(THETA)-U*Rn=M*f.............BY SOLVING THESE 2 EQUATIONS \n",
      "f=g*math.sin(THETA/57.3)-U*g*math.cos(THETA/57.3)\n",
      "s=(V**2-u**2)/(2*f)##                  DISTANCE ALONG THE PLANE IN metres\n",
      "##==============================================================================================\n",
      "##OUTPUT\n",
      "print'%s %.1f %s'%('DISTANCE ALONG THE INCLINED PLANE=',s,' m')\n",
      "\n",
      "\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "DISTANCE ALONG THE INCLINED PLANE= 8.0  m\n"
       ]
      }
     ],
     "prompt_number": 2
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Ex3-pg104"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "##CHAPTER 3 ILLUSRTATION 3 PAGE NO 104\n",
      "##TITLE:FRICTION\n",
      "##FIRURE 3.18\n",
      "import math\n",
      "##===========================================================================================\n",
      "##INPUT DATA\n",
      "W=500.##                  WEGHT IN NEWTONS\n",
      "THETA=30.##               ANGLE OF INCLINATION IN DEGRESS\n",
      "U=0.2##                  COEFFICIENT FRICTION\n",
      "S=15.##                   DISTANCE IN metres\n",
      "##============================================================================================\n",
      "Rn=W*math.cos(THETA/57.3)##       NORMAL REACTION IN NEWTONS\n",
      "P=W*math.sin(THETA/57.3)+U*Rn##   PUSHING FORCE ALONG THE DIRECTION OF MOTION\n",
      "w=P*S\n",
      "##============================================================================================\n",
      "##OUTPUT\n",
      "print'%s %.1f %s'%('WORK DONE BY THE FORCE=',w,' N-m')\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "WORK DONE BY THE FORCE= 5048.8  N-m\n"
       ]
      }
     ],
     "prompt_number": 3
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Ex4-pg104"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "##CHAPTER 3 ILLUSRTATION 4 PAGE NO 104\n",
      "##TITLE:FRICTION\n",
      "##FIRURE 3.19(a)  &  3.19(b)\n",
      "import math\n",
      "##===========================================================================================\n",
      "##INPUT DATA\n",
      "P1=2000.##           FORCE ACTING UPWARDS WHEN ANGLE=15 degrees IN NEWTONS\n",
      "P2=2300.##           FORCE ACTING UPWARDS WHEN ANGLE=20 degrees IN NEWTONS\n",
      "THETA1=15.##         ANGLE OF INCLINATION IN 3.19(a)\n",
      "THETA2=20.##         ANGLE OF INCLINATION IN 3.19(b)\n",
      "##F1=               FORCE OF FRICTION IN 3.19(a)\n",
      "##Rn1=              NORMAL REACTION IN 3.19(a)\n",
      "##F2=               FORCE OF FRICTION IN 3.19(b)\n",
      "##Rn2=              NORMAL REACTION IN 3.19(b)\n",
      "##U=                 COEFFICIENT OF FRICTION\n",
      "##===========================================================================================\n",
      "##CALCULATION\n",
      "##P1=F1+Rn1             RESOLVING THE FORCES ALONG THE PLANE\n",
      "##Rn1=W*cosd(THETA1)....NORMAL REACTION IN 3.19(a)\n",
      "##F1=U*Rn1\n",
      "##BY SOLVING ABOVE EQUATIONS P1=W(U*cosd(THETA1)+sind(THETA1))---------------------1\n",
      "##P2=F2+Rn2             RESOLVING THE FORCES PERPENDICULAR TO THE PLANE\n",
      "##Rn2=W*cosd(THETA2)....NORMAL REACTION IN 3.19(b)\n",
      "##F2=U*Rn2\n",
      "##BY SOLVING ABOVE EQUATIONS P2=W(U*cosd(THETA2)+sind(THETA2))----------------------2\n",
      "##BY SOLVING EQUATIONS 1 AND 2\n",
      "X=P2/P1\n",
      "U=(math.sin(THETA2/57.3)-(X*math.sin(THETA1/57.3)))/((X*math.cos(THETA1/57.3)-math.cos(THETA2/57.3)))##        COEFFICIENT OF FRICTION\n",
      "W=P1/(U*math.cos(THETA1/57.3)+math.sin(THETA1/57.3))\n",
      "##=============================================================================================\n",
      "##OUTPUT\n",
      "##print'%s %.1f %s'%('%f',X)\n",
      "print'%s %.1f %s  %.1f %s '%('COEFFICIENT OF FRICTION=',U,'' 'WEIGHT OF THE BODY=',W,' N')\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "COEFFICIENT OF FRICTION= 0.3 WEIGHT OF THE BODY=  3927.0  N \n"
       ]
      }
     ],
     "prompt_number": 4
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Ex5-pg105"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "##CHAPTER 3 ILLUSRTATION 5 PAGE NO 105\n",
      "##TITLE:FRICTION\n",
      "import math\n",
      "##===========================================================================================\n",
      "##INPUT DATA\n",
      "d=5.##                     DIAMETER OF SCREW JACK IN cm\n",
      "p=1.25##                  PITCH IN cm\n",
      "l=50.##                    LENGTH IN cm\n",
      "U=.1##                    COEFFICIENT OF FRICTION\n",
      "W=20000.##                 LOAD IN NEWTONS\n",
      "PI=3.147\n",
      "##=============================================================================================\n",
      "##CALCULATION\n",
      "ALPHA=math.atan((p/(PI*d)/57.3))\n",
      "PY=math.atan(U/57.3)\n",
      "P=W*math.tan((ALPHA+PY)*57.)\n",
      "P1=P*d/(2.*l)\n",
      "##=============================================================================================\n",
      "##OUTPUT\n",
      "print'%s %.1f %s '%('THE AMOUNT OF EFFORT NEED TO APPLY =',P1,' N')\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "THE AMOUNT OF EFFORT NEED TO APPLY = 180.4  N \n"
       ]
      }
     ],
     "prompt_number": 5
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Ex6-pg106"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "##CHAPTER 3 ILLUSRTATION 6 PAGE NO 106\n",
      "##TITLE:FRICTION\n",
      "import math\n",
      "##===========================================================================================\n",
      "##INPUT DATA\n",
      "d=50.##                 DIAMETER OF SCREW IN mm\n",
      "p=12.5##               PITCH IN mm\n",
      "U=0.13##               COEFFICIENT OF FRICTION\n",
      "W=25000.##              LOAD IN mm\n",
      "PI=3.147\n",
      "##===========================================================================================\n",
      "##CALCULATION\n",
      "ALPHA=math.atan((p/(PI*d))/57.3)\n",
      "PY=math.atan(U/57.3)\n",
      "P=W*math.tan((ALPHA+PY)/57.3)##         FORCE REQUIRED TO RAISE THE LOAD IN N\n",
      "T1=P*d/2.##                   TORQUE REQUIRED IN Nm\n",
      "P1=W*math.tan((PY-ALPHA)/57.3)##        FORCE REQUIRED TO LOWER THE SCREW IN N\n",
      "T2=P1*d/2.##                  TORQUE IN N\n",
      "X=T1/T2##                     RATIOS REQUIRED\n",
      "n=math.tan((ALPHA/(ALPHA+PY))/57.3)##    EFFICIENCY\n",
      "##============================================================================================\n",
      "print'%s %.1f %s'%('RATIO OF THE TORQUE REQUIRED TO RAISE THE LOAD,TO THE TORQUE REQUIRED TO LOWER THE LOAD =',X,'')\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "RATIO OF THE TORQUE REQUIRED TO RAISE THE LOAD,TO THE TORQUE REQUIRED TO LOWER THE LOAD = 4.1 \n"
       ]
      }
     ],
     "prompt_number": 6
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Ex7-pg107"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "##CHAPTER 3 ILLUSRTATION 7 PAGE NO 107\n",
      "##TITLE:FRICTION\n",
      "import math\n",
      "##===========================================================================================\n",
      "##INPUT DATA\n",
      "d=39.##                 DIAMETER OF THREAD IN mm\n",
      "p=13.##                 PITCH IN mm\n",
      "U=0.1##                COEFFICIENT OF FRICTION\n",
      "W=2500.##               LOAD IN mm\n",
      "PI=3.147\n",
      "##===========================================================================================\n",
      "##CALCULATION\n",
      "ALPHA=math.atan((p/(PI*d))/57.3)\n",
      "PY=math.atan(U/57.3)\n",
      "P=W*math.tan((ALPHA+PY)*57.3)##         FORCE IN N\n",
      "T1=P*d/2.##                   TORQUE REQUIRED IN Nm\n",
      "T=2.*T1##                      TORQUE REQUIRED ON THE COUPLING ROD IN Nm\n",
      "K=2.*p##                      DISTANCE TRAVELLED FOR ONE REVOLUTION\n",
      "N=20.8/K##                   NO OF REVOLUTIONS REQUIRED\n",
      "w=2.*PI*N*T/100.##                 WORKDONE BY TORQUE\n",
      "w1=w*(7500.-2500.)/2500.##      WORKDONE TO INCREASE THE LOAD FROM 2500N TO 7500N\n",
      "n=math.tan(ALPHA/57.3)/math.tan((ALPHA+PY)/57.3)##    EFFICIENCY\n",
      "##============================================================================================\n",
      "##OUTPUT\n",
      "print'%s %.1f %s %.1f %s %.1f %s '%('workdone against a steady load of 2500N=',w,' N' 'workdone if the load is increased from 2500N to 7500N=',w1,' N' 'efficiency=',n,'')\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "workdone against a steady load of 2500N= 1025.5  Nworkdone if the load is increased from 2500N to 7500N= 2050.9  Nefficiency= 0.5  \n"
       ]
      }
     ],
     "prompt_number": 7
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Ex8-pg107"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "##CHAPTER 3 ILLUSRTATION 8 PAGE NO 107\n",
      "##TITLE:FRICTION\n",
      "import math\n",
      "##===========================================================================================\n",
      "##INPUT DATA\n",
      "W=50000.##                        WEIGHT OF THE SLUICE GATE IN NEWTON\n",
      "P=40000.##                        POWER IN WATTS\n",
      "N=580.##                          MAX MOTOR RUNNING SPEEED IN rpm\n",
      "d=12.5##                         DIAMETER OF THE SCREW IN cm\n",
      "p=2.5##                          PITCH IN cm\n",
      "PI=3.147\n",
      "U1=.08##                          COEFFICIENT OF FRICTION for SCREW\n",
      "U2=.1##                           C.O.F BETWEEN GATES AND SCREW\n",
      "Np=2000000.##                     NORMAL PRESSURE IN NEWTON\n",
      "Fl=.15##                       FRICTION LOSS\n",
      "n=1.-Fl##                       EFFICIENCY\n",
      "ng=80.##                        NO OF TEETH ON GEAR\n",
      "##===========================================================================================\n",
      "##CALCULATION\n",
      "TV=W+U2*Np##                     TOTAL VERTICAL HEAD IN NEWTON\n",
      "ALPHA=math.atan((p/(PI*d))/57.3)##         \n",
      "PY=math.atan(U1/57.3)##                 \n",
      "P1=TV*math.tan((ALPHA+PY)*57.3)##            FORCE IN N\n",
      "T=P1*d/2./100.##                    TORQUE IN N-m\n",
      "Ng=60000.*n*P*10**-3./(2.*PI*T)##              SPEED OF GEAR IN rpm\n",
      "np=Ng*ng/N##                       NO OF TEETH ON PINION\n",
      "##=========================================================================================\n",
      "##OUTPUT\n",
      "print'%s %.1f %s %.1f %s '%('NO OF TEETH ON PINION =',np,' say ',np+1,'')\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "NO OF TEETH ON PINION = 19.8  say  20.8  \n"
       ]
      }
     ],
     "prompt_number": 8
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Ex9-pg108"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "##CHAPTER 3 ILLUSRTATION 9 PAGE NO 108\n",
      "##TITLE:FRICTION\n",
      "import math\n",
      "##===========================================================================================\n",
      "##INPUT DATA\n",
      "d=5.##                         MEAN DIAMETER OF SCREW IN cm\n",
      "p=1.25##                      PITCH IN cm\n",
      "W=10000.##                     LOAD AVAILABLE IN NEWTONS\n",
      "dc=6.##                        MEAN DIAMETER OF COLLAR IN cm\n",
      "U=.15##                       COEFFICIENT OF FRICTION OF SCREW\n",
      "Uc=.18##                      COEFFICIENT OF FRICTION OF COLLAR\n",
      "P1=100.##                      TANGENTIAL FORCE APPLIED IN NEWTON\n",
      "PI=3.147\n",
      "##============================================================================================\n",
      "##CALCULATION\n",
      "ALPHA=math.atan((p/(PI*d))/57.3)##         \n",
      "PY=math.atan(U/57.3)##                 \n",
      "T1=W*d/2*math.tan((ALPHA+PY)/100)*57.3##         TORQUE ON SCREW IN NEWTON\n",
      "Tc=Uc*W*dc/2./100.##                      TORQUE ON COLLAR IN NEWTON\n",
      "T=T1+Tc##                          TOTAL TORQUE\n",
      "D=2.*T/P1/2.*100.##                       DIAMETER OF HAND WHEEL IN cm\n",
      "##============================================================================================\n",
      "##OUTPUT\n",
      "print'%s %.1f %s'%('SUITABLE DIAMETER OF HAND WHEEL =',D,' cm')\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "SUITABLE DIAMETER OF HAND WHEEL = 111.4  cm\n"
       ]
      }
     ],
     "prompt_number": 9
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Ex10-pg108"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "##CHAPTER 3 ILLUSRTATION 10 PAGE NO 108\n",
      "##TITLE:FRICTION\n",
      "import math\n",
      "##===========================================================================================\n",
      "##INPUT DATA\n",
      "PI=3.147\n",
      "d=2.5##                    MEAN DIA OF BOLT IN cm\n",
      "p=.6##                     PITCH IN cm\n",
      "beeta=55/2.##               VEE ANGLE\n",
      "dc=4.##                     DIA OF COLLAR IN cm\n",
      "U=.1##                       COEFFICIENT OF FRICTION OF BOLT\n",
      "Uc=.18##                      COEFFICIENT OF FRICTION OF COLLAR\n",
      "W=6500.##                     LOAD ON BOLT IN NEWTONS\n",
      "L=38.##                       LENGTH OF SPANNER\n",
      "##=============================================================================================\n",
      "##CALCULATION\n",
      "##LET X=tan(py)/tan(beeta)\n",
      "##y=tan(ALPHA)*X\n",
      "PY=math.atan(U)*57.3\n",
      "ALPHA=math.atan((p/(PI*d)))*57.3\n",
      "X=math.tan(PY/57.3)/math.cos(beeta/57.3)\n",
      "Y=math.tan(ALPHA/57.3)\n",
      "T1=W*d/2.*10**-2*(X+Y)/(1.-(X*Y))##             TORQUE IN SCREW IN N-m\n",
      "Tc=Uc*W*dc/2.*10**-2##                         TORQUE ON BEARING SERVICES IN N-m\n",
      "T=T1+Tc##                                     TOTAL TORQUE \n",
      "P1=T/L*100.##                                      FORCE REQUIRED BY @ THE END OF SPANNER\n",
      "##=============================================================================================\n",
      "##OUTPUT\n",
      "print'%s %.1f %s'%('FORCE REQUIRED @ THE END OF SPANNER=',P1,' N')\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "FORCE REQUIRED @ THE END OF SPANNER= 102.3  N\n"
       ]
      }
     ],
     "prompt_number": 10
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Ex11-pg109"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "##CHAPTER 3 ILLUSRTATION 11 PAGE NO 109\n",
      "##TITLE:FRICTION\n",
      "import math\n",
      "##===========================================================================================\n",
      "##INPUT DATA\n",
      "d1=15.##                                 DIAMETER OF VERTICAL SHAFT IN cm\n",
      "N=100.##                                 SPEED OF THE MOTOR rpm\n",
      "W=20000.##                                LOAD AVILABLE IN N\n",
      "U=.05##                                  COEFFICIENT OF FRICTION\n",
      "PI=3.147\n",
      "##==================================================================================\n",
      "T=2./3.*U*W*d1/2.##                         FRICTIONAL TORQUE IN N-m\n",
      "PL=2.*PI*N*T/100./60.##                         POWER LOST IN FRICTION IN WATTS\n",
      "##==================================================================================\n",
      "##OUTPUT\n",
      "print'%s %.1f %s'%('POWER LOST IN FRICTION=',PL,' watts')\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "POWER LOST IN FRICTION= 524.5  watts\n"
       ]
      }
     ],
     "prompt_number": 11
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Ex12-pg109"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "##CHAPTER 3 ILLUSRTATION 12 PAGE NO 109\n",
      "##TITLE:FRICTION\n",
      "import math\n",
      "##===================================================================================\n",
      "##INPUT DATA\n",
      "PI=3.147\n",
      "d2=.30##                             DIAMETER OF SHAFT IN m \n",
      "W=200000.##                           LOAD AVAILABLE IN NEWTONS\n",
      "N=75.##                              SPEED IN rpm\n",
      "U=.05##                             COEFFICIENT OF FRICTION\n",
      "p=300000.##                          PRESSURE AVAILABLE IN N/m**2\n",
      "P=16200.##                           POWER LOST DUE TO FRICTION IN WATTS\n",
      "##====================================================================================\n",
      "##CaLCULATION\n",
      "T=P*60./2./PI/N##                      TORQUE INDUCED IN THE SHFT IN N-m\n",
      "##LET X=(r1**3-r2**3)/(r1**2-r2**2)\n",
      "X=(3./2.*T/U/W)\n",
      "r2=.15##                             SINCE d2=.30 m\n",
      "c=r2**2.-(X*r2)\n",
      "b= r2-X\n",
      "a= 1.\n",
      "r1=( -b+ math.sqrt (b**2. -4.*a*c ))/(2.* a);##     VALUE OF r1 IN m\n",
      "d1=2*r1*100.##                               d1 IN cm\n",
      "n=W/(PI*p*(r1**2.-r2**2.))\n",
      "##================================================================================\n",
      "##OUTPUT\n",
      "print'%s %.1f %s %.1f %s %.1f %s'%('EXTERNAL DIAMETER OF SHAFT =',d1,' cm''NO OF COLLARS REQUIRED =',n,'' '0 or ',n+1,'')\n",
      "\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "EXTERNAL DIAMETER OF SHAFT = 50.6  cmNO OF COLLARS REQUIRED = 5.1 0 or  6.1 \n"
       ]
      }
     ],
     "prompt_number": 12
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Ex13-pg111"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "##CHAPTER 3 ILLUSRTATION 13 PAGE NO 111\n",
      "##TITLE:FRICTION\n",
      "import math\n",
      "##===================================================================================\n",
      "##INPUT DATA\n",
      "PI=3.147\n",
      "W=20000.##                                LOAD IN NEWTONS\n",
      "ALPHA=120./2.##                               CONE ANGLE IN DEGREES\n",
      "p=350000.##                                INTENSITY OF PRESSURE\n",
      "U=.06\n",
      "N=120.##                                    SPEED OF THE SHAFT IN rpm\n",
      "##d1=3d2\n",
      "##r1=3r2\n",
      "##===================================================================================\n",
      "##CALCULATION\n",
      "##LET K=d1/d2\n",
      "k=3.\n",
      "Z=W/((k**2.-1.)*PI*p)\n",
      "r2=Z**.5##                                  INTERNAL RADIUS IN m\n",
      "r1=3.*r2\n",
      "T=2.*U*W*(r1**3.-r2**3.)/(3.*math.sin(60/57.3)*(r1**2.-r2**2.))##     total frictional torque in N\n",
      "P=2.*PI*N*T/60000.##                            power absorbed in friction in kW\n",
      "##================================================================================\n",
      "print'%s %.1f %s %.1f %s %.1f %s'%('THE INTERNAL DIAMETER OF SHAFT =',r2*100,' cm' 'THE EXTERNAL DIAMETER OF SHAFT =',r1*100,' cm' 'POWER ABSORBED IN FRICTION =',P,' kW')\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "THE INTERNAL DIAMETER OF SHAFT = 4.8  cmTHE EXTERNAL DIAMETER OF SHAFT = 14.3  cmPOWER ABSORBED IN FRICTION = 1.8  kW\n"
       ]
      }
     ],
     "prompt_number": 13
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Ex14-pg111"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "##CHAPTER 3 ILLUSRTATION 14 PAGE NO 111\n",
      "##TITLE:FRICTION\n",
      "import math\n",
      "##===========================================================================================\n",
      "##INPUT DATA\n",
      "PI=3.147\n",
      "P=10000.##                               POWER TRRANSMITTED BY CLUTCH IN WATTS\n",
      "N=3000.##                                SPEED IN rpm\n",
      "p=.09##                                 AXIAL PRESSURE IN N/mm**2\n",
      "##d1=1.4d2                              RELATION BETWEEN DIAMETERS \n",
      "K=1.4##                                 D1/D2\n",
      "n=2.\n",
      "U=.3##                                  COEFFICIENT OF FRICTION\n",
      "##==========================================================================================\n",
      "T=P*60000./1000./(2.*PI*N)##                     ASSUMING UNIFORM WEAR            TORQUE IN N-m\n",
      "r2=(T*2./(n*U*2.*PI*p*10**6.*(K-1.)*(K+1.)))**(1./3.)##            INTERNAL RADIUS\n",
      "\n",
      "##===========================================================================================\n",
      "print'%s %.1f %s %.1f %s '%('THE INTERNAL RADIUS =',r2*100,' cm' 'THE EXTERNAL RADIUS =',K*r2*100,' cm')\n",
      " \n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "THE INTERNAL RADIUS = 5.8  cmTHE EXTERNAL RADIUS = 8.1  cm \n"
       ]
      }
     ],
     "prompt_number": 14
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Ex15-pg111"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "##CHAPTER 3 ILLUSRTATION 14 PAGE NO 111\n",
      "##TITLE:FRICTION\n",
      "\n",
      "\n",
      "\n",
      "##===========================================================================================\n",
      "##INPUT DATA\n",
      "PI=3.147\n",
      "n1=3.##                          NO OF DICS ON DRIVING SHAFTS\n",
      "n2=2.##                          NO OF DICS ON DRIVEN SHAFTS\n",
      "d1=30.##                         DIAMETER OF DRIVING SHAFT IN cm\n",
      "d2=15.##                         DIAMETER OF DRIVEN SHAFT IN cm\n",
      "r1=d1/2.\n",
      "r2=d2/2.\n",
      "U=.3##                          COEFFICIENT FRICTION\n",
      "P=30000.##                       TANSMITTING POWER IN WATTS\n",
      "N=1800.##                        SPEED IN rpm\n",
      "##===========================================================================================\n",
      "##CALCULATION\n",
      "n=n1+n2-1.##                     NO OF PAIRS OF CONTACT SURFACES\n",
      "T=P*60000./(2.*PI*N)##            TORQUE IN N-m\n",
      "W=2.*T/(n*U*(r1+r2)*10.)##           LOAD IN N\n",
      "k=W/(2.*PI*(r1-r2))\n",
      "p=k/r2/100.##                        MAX AXIAL INTENSITY OF PRESSURE IN N/mm**2\n",
      "##===========================================================================================\n",
      "## OUTPUT\n",
      "print'%s %.3f %s'%('MAX AXIAL INTENSITY OF PRESSURE =',p,' N/mm^2')\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "MAX AXIAL INTENSITY OF PRESSURE = 0.033  N/mm**2\n"
       ]
      }
     ],
     "prompt_number": 15
    }
   ],
   "metadata": {}
  }
 ]
}