diff options
Diffstat (limited to 'Theory_Of_Machines_by__B._K._Sarkar/Chapter3.ipynb')
| -rwxr-xr-x | Theory_Of_Machines_by__B._K._Sarkar/Chapter3.ipynb | 782 |
1 files changed, 782 insertions, 0 deletions
diff --git a/Theory_Of_Machines_by__B._K._Sarkar/Chapter3.ipynb b/Theory_Of_Machines_by__B._K._Sarkar/Chapter3.ipynb new file mode 100755 index 00000000..74818ab4 --- /dev/null +++ b/Theory_Of_Machines_by__B._K._Sarkar/Chapter3.ipynb @@ -0,0 +1,782 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:9f3aa65b257e3f2aa586660a443f5f27bf555a236ce21a3b4fb7b3ab1cf26f12"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "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": {}
+ }
+ ]
+}
\ No newline at end of file |