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diff --git a/The_Theory_of_Machines_by_T._Bevan/ch6.ipynb b/The_Theory_of_Machines_by_T._Bevan/ch6.ipynb new file mode 100644 index 00000000..3639fce9 --- /dev/null +++ b/The_Theory_of_Machines_by_T._Bevan/ch6.ipynb @@ -0,0 +1,325 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 6: Friction" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1, Page 185" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable declaration\n", + "theta=60.#degrees\n", + "u1=0.15#between surfaces A annd B\n", + "u2=0.10#for the guides\n", + "\n", + "#Calculations\n", + "phi=math.degrees(math.atan(u1))\n", + "phi1=math.degrees(math.atan(u2))\n", + "alpha=(theta+phi+phi1)/2#from 6.22, maximum efficiency is obtained at alpha\n", + "#from 6.23, maximum efficiency is given by nmax=(cos(theta+phi+phi1)+1)/(cos(theta-phi-phi1)+1)\n", + "nmax=(math.cos((theta+phi+phi1)*math.pi/180)+1)/(math.cos((theta-phi-phi1)*math.pi/180)+1)\n", + "\n", + "#Results\n", + "print \"Maximum efficiency = %.2f %% and it is obtained when alpha = %.2f degrees\"%((nmax*100),alpha)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Maximum efficiency = 74.90 % and it is obtained when alpha = 37.12 degrees\n" + ] + } + ], + "prompt_number": 23 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3, Page 193" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable declaration\n", + "#from equation 6.36 we know, M=(2/3)*u*W*(ri^3-r2^3)/(r1^2-r2^2)\n", + "u=0.04\n", + "W=16#tons\n", + "w=W*2240#lbs\n", + "r1=8#in\n", + "r2=6#in\n", + "N=120\n", + "P=50#lb/in^2\n", + "\n", + "#Calculations\n", + "M=(2./3)*u*w*(r1**3-r2**3)/(r1**2-r2**2)\n", + "hp=M*2*math.pi*N/(12*33000)#horse power absorbed\n", + "#from fig 137,effective bearing surface per pad is calsulate from the dimensions to be 58.5 in^2\n", + "A=58.5#in^2\n", + "n=w/(A*P)\n", + "x=math.floor(n)\n", + "\n", + "#Results\n", + "print \"Horsepower absorbed = %.2f\\nNumber of collars required = %.f\"%(hp,x)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Horsepower absorbed = 19.24\n", + "Number of collars required = 12\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4, Page 195" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable declaration\n", + "ratio=1.25\n", + "u=.675\n", + "P=12#hp\n", + "\n", + "#Calculations\n", + "#W=P*%pi*(r1^2-r2^2); Total axal thrust.\n", + "#M=u*W*(r1+r2); Total friction moemnt \n", + "#reducing the two equations and using ratio=1.25(r1=1.25*r2) we get, M=u*21.2*r2^3\n", + "ReqM=65#lb ft \n", + "RM=ReqM*12#lb in\n", + "r2=(RM/(u*P*math.pi*(1.25**2-1)))**(1./3)\n", + "r1=1.25*r2\n", + "d1=r1*2\n", + "d2=r2*2\n", + "\n", + "#Results\n", + "print \"The dimensions of the friction surfaces are:\\nOuter Diameter= %.1f in\\nInner Diameter= %.1f in\"%(d1,d2)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The dimensions of the friction surfaces are:\n", + "Outer Diameter= 9.5 in\n", + "Inner Diameter= 7.6 in\n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5, Page 196" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable declaration\n", + "P=20#lb/in^2\n", + "u=0.07#friction coefficient\n", + "N=3600#rpm\n", + "H=100#hp\n", + "r1=5#in\n", + "\n", + "#Calculations\n", + "r2=0.8*r1#given\n", + "A=math.pi*(r1**2-r2**2)#the area of each friction surface\n", + "W=A*P#total axial thrust on plates\n", + "M=(1./2)*u*W*(r1+r2)#friction moment for each pair of contacts\n", + "T=H*33000*12/(2*math.pi*N)#total torque to be transmitted\n", + "x=(T/M)#effective friction surfaces required\n", + "\n", + "#Results\n", + "print \"\\nNumber of effective friction surfaces required= %.f\"%x\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "Number of effective friction surfaces required= 10\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7, Page 198" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable declaration\n", + "P=6 #tons\n", + "u=0.05\n", + "theta=60#degrees\n", + "CP=80\n", + "Stroke=16#in\n", + "OC=Stroke/2\n", + "r1=7#in\n", + "r2=15#in\n", + "r3=4.4#in\n", + "\n", + "#Calculations\n", + "#Radius of friction circle\n", + "ro=u*r1\n", + "rc=u*r2\n", + "rp=u*r3\n", + "phi=math.degrees(math.asin(OC*math.sin(theta*math.pi/180)/CP))\n", + "alpha=math.degrees(math.asin((rc+rp)/CP))\n", + "#a) without friction\n", + "Qa=P/math.cos((phi)*math.pi/180)\n", + "Xa=OC*math.cos((30-phi)*math.pi/180)#tensile force transmitted along the eccentric rod when friction is NOT taken into account\n", + "Ma=Qa*Xa/12\n", + "#b) with friction\n", + "Qb=P/math.cos((phi-alpha)*math.pi/180)#tensile force transmitted along the eccentric rod when friction is taken into account\n", + "Xb=OC*math.cos((30-(phi-alpha))*math.pi/180)-(rc+ro)\n", + "Mb=Qb*Xb/12\n", + "n=Mb/Ma\n", + "\n", + "#Results\n", + "print \"Turning moment applied to OC:\\na)Without friction= %.2f ton.ft\\nb)With friction(u=0.05)= %.2f ton.ft\"%(Ma,Mb)\n", + "print \"\\nThe efficiency of the mechanism is %.2f %%\"%(n*100)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Turning moment applied to OC:\n", + "a)Without friction= 3.64 ton.ft\n", + "b)With friction(u=0.05)= 3.06 ton.ft\n", + "\n", + "The efficiency of the mechanism is 84.17 %\n" + ] + } + ], + "prompt_number": 21 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8, Page 201" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "import numpy as np\n", + "\n", + "#Variable declaration\n", + "stroke=4#in\n", + "d=11.5#in\n", + "ds=4#in\n", + "dp=14#in\n", + "theta=math.pi\n", + "u1=.25\n", + "T1=350#lb\n", + "u2=0.1\n", + "\n", + "#Calculations\n", + "k=math.e**(u1*theta)\n", + "T2=T1/k\n", + "Tor=(T1-T2)*(dp/2)#total resisting torque\n", + "#total resisting torque is also given by P*(r+2*(cos%pi/6))+u2*R*(ds/2)\n", + "#equating and putting values we get the following quadratic equation\n", + "p = [1,-1163,334200]\n", + "a=np.roots(p)\n", + "\n", + "#Results\n", + "print \"P=\",a\n", + "print \"The larger of two values is inadmissible. \\n It corresponds to a negative sign in front of the second\" \\\n", + " \"term on the \\n right hand side of equation (1)\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "P= [ 644.28733949 518.71266051]\n", + "The larger of two values is inadmissible. \n", + " It corresponds to a negative sign in front of the secondterm on the \n", + " right hand side of equation (1)\n" + ] + } + ], + "prompt_number": 22 + } + ], + "metadata": {} + } + ] +}
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