{ "metadata": { "name": "", "signature": "sha256:55c40faafb847932f0fdcda855b3af16f1a2e4ef45941baaf0d7ee692a22c20c" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "Ch:30 Chain drive" ] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "exa 30-1 - Page 778" ] }, { "cell_type": "code", "collapsed": false, "input": [ "from __future__ import division\n", "from math import sqrt, pi\n", "n1=17#\n", "n2=51#\n", "C=300#\n", "p=9.52#\n", "Ln=(2*C/p)+((n1+n2)/2)+((((n2-n1)/(2*pi))**2)*(p/C))#\n", "x=(Ln-((n2+n1)/(2)))**2#\n", "y=8*(((n2-n1)/(2*pi))**2)#\n", "z=Ln-((n1+n2)/2)#\n", "C=(p/4)*(z+(sqrt(x-y)))\n", "\n", "\n", " # printing data in scilab o/p window\n", "print \"C is %0.2f mm \"%(C)#\n", " " ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "C is 300.00 mm \n" ] } ], "prompt_number": 1 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "exa 30-2 - Page 778" ] }, { "cell_type": "code", "collapsed": false, "input": [ "from math import tan\n", "G=4#\n", "n1=17#\n", "n2=n1*G#\n", "N1=2300#\n", "Kc=1.2# #from table 30-2\n", "p=12.7# #fom table 30-1\n", "D1=p*n1#\n", "D2=p*n2#\n", "phi=2*10.6#\n", "x=tan(phi/2)# #phi/2 = 10.6deg, from table 30-3\n", "Da1=(p/x)+(0.6*p)#\n", "Da2=(p/x*4)+(0.6*p)#\n", "Cmin=Kc*((Da1+Da2)/2)#\n", "Ln1=(2*Cmin/p)+((n1+n2)/2)+((((n2-n1)/(2*pi))**2)*(p/Cmin))#\n", "Ln1=80#\n", "print \"Ln is %0.0f \"%(Ln1)#" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Ln is 80 \n" ] } ], "prompt_number": 2 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "exa 30-3 - Page 779" ] }, { "cell_type": "code", "collapsed": false, "input": [ "N1=1000#\n", "N2=500#\n", "P=2.03*10**3# #from table 30-8\n", "K1=1.26#\n", "Ks=1#\n", "#let Pc be the power transmitting capacity of the chain\n", "Pc=P*K1/Ks#\n", "p=9.52#\n", "n1=21#\n", "n2=42#\n", "V=n1*p*N1/(60*10**3)#\n", "#Let the chain tension be T\n", "T=Pc/V#\n", "#Let the breaking load be BL\n", "BL=10700#\n", "FOS=BL/T#\n", "C=50*p#\n", "Ln=(2*C/p)+((n1+n2)/2)+((((n2-n1)/(2*pi))**2)*(p/C))#\n", "L=Ln*p#\n", "Pc=Pc*10**-3#\n", "print \" Pc is %0.2f KW \"%(Pc)#\n", "print \"\\n V is %0.3f m/s \"%(V)#\n", "print \"\\n T is %0.1f N \"%(T)#\n", "print \"\\n FOS is %0.2f \"%(FOS)#\n", "print \"\\n L is %0.2f mm \"%(L)#\n", "\n", "#The difference in the value of L and T is due to rounding-off the values." ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " Pc is 2.56 KW \n", "\n", " V is 3.332 m/s \n", "\n", " T is 767.6 N \n", "\n", " FOS is 13.94 \n", "\n", " L is 1254.01 mm \n" ] } ], "prompt_number": 3 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "exa 30-5 - Page 780" ] }, { "cell_type": "code", "collapsed": false, "input": [ "G=2#\n", "P=5000#\n", "Ks=1.7#\n", "Pd=P*Ks#\n", "K2=1.7#\n", "p=15.88#\n", "n1=17#\n", "n2=n1*G#\n", "D1=n1*p#\n", "D2=n2*p#\n", "C=40*p#\n", "Ln=(2*C/p)+((n1+n2)/2)+((((n2-n1)/(2*pi))**2)*(p/C))#\n", "L=Ln*p#\n", "print \"L is %0.2f mm \"%(L)#\n", "#The difference in the value of L is due to rounding-off the values." ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "L is 1678.25 mm \n" ] } ], "prompt_number": 4 } ], "metadata": {} } ] }