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-{
- "metadata": {
- "name": "",
- "signature": "sha256:75a895f6b6e2b52911a88eac6dae9689200eafdf77449669257c302ec104ee3c"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Ch:23 Brakes"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "exa 23-1 - Page 618"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "from math import asin, pi\n",
- "W=20e3#\n",
- "m=W/9.81#\n",
- "#diameter of brake drum\n",
- "Db=0.6#\n",
- "p=1#\n",
- "Vi=1#\n",
- "Vf=0#\n",
- "D=1#\n",
- "R=0.5#\n",
- "wi=Vi/R#\n",
- "wf=0#\n",
- "w=1#\n",
- "Vav=0.5#\n",
- "S=2#\n",
- "t=S/Vav#\n",
- "#angle turned by by hoist drum=theta\n",
- "theta=0.5*wi*t#\n",
- "K_E=0.5*m*Vi**2#\n",
- "P_E=2*W#\n",
- "T_E=K_E+P_E#\n",
- "T=T_E/theta#\n",
- "P=wi*T*10**-3#\n",
- "Rb=Db/2#\n",
- "Ft=0.5*T*p/Rb#\n",
- "u=0.35#\n",
- "N=Ft/u#\n",
- "#contact area of brake lining=A\n",
- "A=N/p#\n",
- "b=0.3*Db#\n",
- "L=A*10**-6/(b)#\n",
- "#angle subtended at brake drum centre=theta2\n",
- "theta2=2*(asin(L/Db))#\n",
- "theta2=theta2*180/pi# # converting radian to degree\n",
- "print \"T is %0.1f Nm \"%(T)#\n",
- "print \"\\nP is %0.4f kW \"%(P)#\n",
- "print \"\\nb is %0.2f m \"%(b)#\n",
- "print \"\\nL is %0.3f m \"%(L)#\n",
- "print \"\\ntheta2 is %0.0f deg \"%(theta2)#\n",
- " "
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "T is 10254.8 Nm \n",
- "\n",
- "P is 20.5097 kW \n",
- "\n",
- "b is 0.18 m \n",
- "\n",
- "L is 0.271 m \n",
- "\n",
- "theta2 is 54 deg \n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "exa 23-2 - Page 618"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import exp\n",
- "b=80#\n",
- "t=2#\n",
- "theta=225*pi/180#\n",
- "u=0.22#\n",
- "#F1/F2=e**(u*theta)\n",
- "#let F1/F2=x#\n",
- "x=exp(u*theta)#\n",
- "#maximum tensile stress in steel tape is siga\n",
- "siga=60#\n",
- "A=b*t#\n",
- "F1=siga*A#\n",
- "F2=F1/x#\n",
- "r=0.2#\n",
- "T=(F1-F2)*r#\n",
- "OA=30#\n",
- "OB=100#\n",
- "OC=350#\n",
- "P=((F2*OB)+(F1*OA))/OC#\n",
- "OA=F2*OB/F1#\n",
- "print \"F1 is %0.0f N \"%(F1)#\n",
- "print \"\\nF2 is %0.1f N \"%(F2)#\n",
- "print \"\\nT is %0.2f Nm \"%(T)#\n",
- "print \"\\nOA is %0.2f mm \"%(OA)#"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "F1 is 9600 N \n",
- "\n",
- "F2 is 4046.4 N \n",
- "\n",
- "T is 1110.72 Nm \n",
- "\n",
- "OA is 42.15 mm \n"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "exa 23-3 - Page 619"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import sin\n",
- "theta=pi/3#\n",
- "r=160#\n",
- "u=0.3#\n",
- "pmax=0.9#\n",
- "b=40#\n",
- "R=(4*r*sin(theta))/((2*theta)+sin(2*theta))#\n",
- "#frictional torque is T\n",
- "T=2*u*pmax*b*(r**2)*sin(theta)#\n",
- "T=2*T*10**-3#\n",
- "Rx=0.5*pmax*b*r*((2*theta)+(sin(2*theta)))*10**-3#\n",
- "Ry=u*Rx#\n",
- "print \"T is %0.2f Nmm \"%(T)#\n",
- "print \"\\nR is %0.3f mm \"%(R)#\n",
- "print \"\\nRx is %0.3f kN \"%(Rx)#\n",
- "print \"\\nRy is %0.2f kN \"%(Ry)#"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "T is 957.75 Nmm \n",
- "\n",
- "R is 187.222 mm \n",
- "\n",
- "Rx is 8.526 kN \n",
- "\n",
- "Ry is 2.56 kN \n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "exa 23-4 - Page 620"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import sin, cos, pi, sqrt\n",
- "d=320#\n",
- "r=d/2#\n",
- "b=50#\n",
- "u=0.3#\n",
- "pmax=1#\n",
- "c=115*2#\n",
- "# From to fig. 23-9, distance OA=R is calculated.\n",
- "R=sqrt(115**2+66.4**2)#\n",
- "C=115*2#\n",
- "theta1=0#\n",
- "theta2=120*pi/180#\n",
- "theta0=120*pi/180#\n",
- "thetamax=pi/2#\n",
- "Tr=u*pmax*b*r**2*(cos(theta1)-cos(theta2))/sin(thetamax)*10**-3#\n",
- "#the notation 'r' is used for moments of right hand shoe, similarly 'l' for the left shoe.\n",
- "Mfr=u*pmax*b*r*(4*r*(cos(theta1)-cos(theta2))+(R*(cos(2*theta1)-cos(2*theta2))))/(4*sin(thetamax))*10**-3#\n",
- "Mpr=pmax*b*r*R*((2*theta0)-(sin(2*theta2)-(sin(theta1))))/(4*sin(thetamax))*10**-3#\n",
- "F=(Mpr-Mfr)/c*10**3#\n",
- "#Mpl+Mfl=F*c#\n",
- "x=F*c*10**-3#\n",
- "y=(Mpr/pmax)+(Mfr/pmax)#\n",
- "pmax2=x/y#\n",
- "Tl=pmax2*Tr#\n",
- "Mpl=pmax2*Mpr#\n",
- "Mfl=pmax2*Mfr#\n",
- "T=Tl+Tr#\n",
- "print \"Tr is %0.0f Nm \"%(Tr)#\n",
- "print \"\\nMf is %0.2f Nm \"%(Mfr)#\n",
- "print \"\\nMp is %0.2f Nm \"%(Mpr)#\n",
- "print \"\\nTl is %0.1f Nm \"%(Tl)# \n",
- "print \"\\nMfl is %0.2f Nm \"%(Mfl)#\n",
- "print \"\\nMpl is %0.2f Nm \"%(Mpl)#\n",
- "print \"\\nF is %0.1f N \"%(F)#\n",
- "print \"\\nT is %0.1f Nm \"%(T)#\n",
- " \n",
- " #The difference in the answers are due to rounding-off of values.\n",
- " "
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Tr is 576 Nm \n",
- "\n",
- "Mf is 695.51 Nm \n",
- "\n",
- "Mp is 1342.49 Nm \n",
- "\n",
- "Tl is 182.9 Nm \n",
- "\n",
- "Mfl is 220.79 Nm \n",
- "\n",
- "Mpl is 426.18 Nm \n",
- "\n",
- "F is 2812.9 N \n",
- "\n",
- "T is 758.9 Nm \n"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "exa 23-5 - Page 621"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "m=1100#\n",
- "V=65*5/18#\n",
- "t=4#\n",
- "r=0.22#\n",
- "mb=12#\n",
- "C=460#\n",
- "S=0.5*V*t#\n",
- "#Total kinetic energy TE=K.E(vehicle)+K.E(rotating parts).\n",
- "TE=((0.5*m*(V**2))+(0.1*0.5*m*(V**2)))#\n",
- "E=TE/4#\n",
- "w=V/r#\n",
- "theta=S/r#\n",
- "T=E/theta#\n",
- "delT=E/(mb*C)#\n",
- "print \"S is %0.2f m \"%(S)#\n",
- "print \"\\nE is %0.2f Nm \"%(E)#\n",
- "print \"\\nT is %0.2f Nm \"%(T)#\n",
- "print \"\\ndelT is %0.2f \"%(delT)#\n",
- " \n",
- "#The difference in the answers are due to rounding-off of values."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "S is 36.11 m \n",
- "\n",
- "E is 49307.97 Nm \n",
- "\n",
- "T is 300.40 Nm \n",
- "\n",
- "delT is 8.93 \n"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "exa 23-6 - Page 621"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "T=35000#\n",
- "u=0.4#\n",
- "p=0.7#\n",
- "r=200#\n",
- "N=T/(u*r)\n",
- "b=sqrt(N/p)#\n",
- "l=b#\n",
- "#2theta = theta2\n",
- "theta2=2*asin(l/(2*r))#\n",
- "F=u*N#\n",
- "P=((250*N)-(u*N*80))/550#\n",
- "Ry=N-P#\n",
- "Rx=u*N#\n",
- "R=sqrt(Rx**2+Ry**2)#\n",
- "w=2*pi*100/60#\n",
- "# Rate of heat generated is Q\n",
- "Q=u*N*w*r/1000#\n",
- "print \"N is %0.1f N \"%(N)#\n",
- "print \"\\nb is %0.0f mm \"%(b)#\n",
- "print \"\\nP is %0.1f N \"%(P)#\n",
- "print \"\\nR is %0.2f N \"%(R)#\n",
- "print \"\\nQ is %0.2f J/s \"%(Q)#\n",
- "\n",
- "#The answer to Rate of heat generated 'Q' is calculated incorrectly in the book."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "N is 437.5 N \n",
- "\n",
- "b is 25 mm \n",
- "\n",
- "P is 173.4 N \n",
- "\n",
- "R is 316.81 N \n",
- "\n",
- "Q is 366.52 J/s \n"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "exa 23-7 - Page 622"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "Vi=20*5/18#\n",
- "Vf=0#\n",
- "m=80#\n",
- "pmax=1#\n",
- "u=0.1#\n",
- "S=50#\n",
- "KE=0.5*m*Vi**2#\n",
- "N=KE/(u*S*2)#\n",
- "t=sqrt(N/(pmax*3))#\n",
- "b=3*t#\n",
- "print \"KE is %0.1f Nm \"%(KE)#\n",
- "print \"\\nN is %0.2f N \"%(N)#\n",
- "print \"\\nt is %0.1f mm \"%(t)#\n",
- "print \"\\nb is %0.1f mm \"%(b)#\n",
- "\n",
- "#The difference in the answers are due to rounding-off of values."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "KE is 1234.6 Nm \n",
- "\n",
- "N is 123.46 N \n",
- "\n",
- "t is 6.4 mm \n",
- "\n",
- "b is 19.2 mm \n"
- ]
- }
- ],
- "prompt_number": 15
- }
- ],
- "metadata": {}
- }
- ]
-} \ No newline at end of file
generated by cgit (git 2.34.1) at 2025-04-04 14:09:32 +0000