From d36fc3b8f88cc3108ffff6151e376b619b9abb01 Mon Sep 17 00:00:00 2001 From: kinitrupti Date: Fri, 12 May 2017 18:40:35 +0530 Subject: Revised list of TBCs --- .../chapter07_11.ipynb | 551 --------------------- 1 file changed, 551 deletions(-) delete mode 100755 Engineering_Mechanics_by_Tayal_A.K./chapter07_11.ipynb (limited to 'Engineering_Mechanics_by_Tayal_A.K./chapter07_11.ipynb') diff --git a/Engineering_Mechanics_by_Tayal_A.K./chapter07_11.ipynb b/Engineering_Mechanics_by_Tayal_A.K./chapter07_11.ipynb deleted file mode 100755 index aae60e42..00000000 --- a/Engineering_Mechanics_by_Tayal_A.K./chapter07_11.ipynb +++ /dev/null @@ -1,551 +0,0 @@ -{ - "metadata": { - "name": "chapter7.ipynb" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "heading", - "level": 1, - "metadata": {}, - "source": [ - "Chapter 7: Application Of Friction" - ] - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 7.7-1,Page No:152" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math\n", - "\n", - "# Initilization of variables\n", - "\n", - "d1=24 # cm # diameter of larger pulley\n", - "d2=12 # cm # diameter of smaller pulley\n", - "d=30 #cm # seperation betweem 1st & the 2nd pulley\n", - "pie=3.14\n", - "\n", - "# Calcuations\n", - "\n", - "r1=d1*0.5 #cm # radius of 1st pulley #1/2=0.5\n", - "r2=d2*0.5 #cm # radius of 2nd pulley #1/2=0.5\n", - "theta=(arcsin((r1-r2)/d))*(180/pi) #degrees \n", - "\n", - "# Angle of lap\n", - "beta_1=180+(2*theta) #degree # for larger pulley\n", - "beta_2=180-(2*theta) #degree # for smaller pulley\n", - "L=pie*(r1+r2)+(2*d)+((r1-r2)**(2/d)) #cm # Length of the belt\n", - "\n", - "# Results\n", - "\n", - "print\"The angle of lap for the larger pulley is \",round(beta_1,2),\"degree\"\n", - "print\"he angle of lap for the smaller pulley is \",round(beta_2,2),\"degree\"\n", - "print\"he length of pulley required is \",round(L,2),\"cm\"\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "The angle of lap for the larger pulley is 203.07 degree\n", - "he angle of lap for the smaller pulley is 156.93 degree\n", - "he length of pulley required is 117.52 cm\n" - ] - } - ], - "prompt_number": 22 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 7.7-2,Page No:153" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math\n", - "\n", - "# Initilization of variables\n", - "\n", - "d1=0.6 #m # diameter of larger pulley\n", - "d2=0.3 #m diameter of smaller pulley\n", - "d=3.5 #m #separation between the pulleys\n", - "pie=3.14\n", - "\n", - "# Calculations\n", - "\n", - "r1=d1*0.5 #m # radius of larger pulley #1/2=0.5\n", - "r2=d2*0.5 #m # radius of smaller pulley #1/2=0.5\n", - "theta=arcsin((r1+r2)/d)*(180/pi) #degree\n", - "\n", - "# Angle of lap for both the pulleys is same, i.e\n", - "\n", - "beta=180+(2*theta) # degree\n", - "L=((pie*(r1+r2))+(2*d)+(((r1+r2)**2)/d)) #cm # Length of the belt\n", - "\n", - "# Results\n", - "\n", - "print\"The angle of lap for the pulley is \",round(beta,1),\"degree\"\n", - "print\"The length of pulley required is \",round(L,3),\"m\"\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "The angle of lap for the pulley is 194.8 degree\n", - "The length of pulley required is 8.471 m\n" - ] - } - ], - "prompt_number": 29 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 7.7-4,Page No:161" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math\n", - "\n", - "# Initilization of variables\n", - "\n", - "W1=1000 #N\n", - "mu=0.25 #coefficient of friction\n", - "pie=3.14\n", - "beta=pie\n", - "T1=W1 # Tension in the 1st belt carrying W1\n", - "e=2.718 #constant\n", - "\n", - "# Calculations\n", - "\n", - "T2=T1/(e**(mu*beta)) #N # Tension in the 2nd belt\n", - "W2=T2 #N\n", - "\n", - "# Results\n", - "\n", - "print\"The minimum weight W2 to keep W1 in equilibrium is \",round(W2),\"N\"\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "The minimum weight W2 to keep W1 in equilibrium is 456.0 N\n" - ] - } - ], - "prompt_number": 31 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 7.7-5,Page No:162" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math\n", - "\n", - "# Initilization of variables\n", - "\n", - "mu=0.5 # coefficient of friction between the belt and the wheel\n", - "W=100 #N\n", - "theta=45 #degree\n", - "e=2.718\n", - "Lac=0.75 #m # ength of the lever\n", - "Lab=0.25 #m\n", - "Lbc=0.50 #m\n", - "r=0.25 #m\n", - "pie=3.14 # constant\n", - "\n", - "# Calculations\n", - "\n", - "beta=((180+theta)*pie)/180 # radian # angle of lap\n", - "\n", - "# from eq'n 2\n", - "T1=(W*Lbc)/Lab #N \n", - "T2=T1/(e**(mu*beta)) #N # from eq'n 1\n", - "\n", - "# consider the F.B.D of the pulley and take moment about its center, we get Braking Moment (M)\n", - "M=r*(T1-T2) #N-m\n", - "\n", - "# Results\n", - "\n", - "print\"The braking moment (M) exerted by the vertical weight W is \",round(M,2),\"N-m\"\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "The braking moment (M) exerted by the vertical weight W is 42.97 N-m\n" - ] - } - ], - "prompt_number": 32 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 7.7-6,Page No:163" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math\n", - "\n", - "# Initiization of variables\n", - "\n", - "W= 1000 #N # or 1kN\n", - "mu=0.3 # coefficient of friction between the rope and the cylinder\n", - "e=2.718 # constant\n", - "pie=3.14 # constant\n", - "alpha=90 # degree # since 2*alpha=180 egree\n", - "\n", - "# Calculations\n", - "\n", - "beta=2*pie*3 # radian # for 3 turn of the rope\n", - "\n", - "# Here T1 is the larger tension in that section of the rope which is about to slip\n", - "T1=W #N\n", - "F=W/e**(mu*(1/(sin(alpha*(pi/180))))*(beta)) #N # Here T2=F\n", - "\n", - "# Results\n", - "\n", - "print\"The force required to suport the weight of 1000 N i.e 1kN is \",round(F,1),\"N\"\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "The force required to suport the weight of 1000 N i.e 1kN is 3.5 N\n" - ] - } - ], - "prompt_number": 34 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 7.7-7,Page No:163" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math\n", - "\n", - "# Initilization of variables\n", - "\n", - "Pw=50 #kW\n", - "T_max=1500 #N\n", - "v=10 # m/s # velocity of rope\n", - "w=4 # N/m # weight of rope\n", - "mu=0.2 # coefficient of friction \n", - "g=9.81 # m/s^2 # acceleration due to gravity\n", - "e=2.718 # constant\n", - "pie=3.14 # constant\n", - "alpha=30 # degree # since 2*alpha=60 \n", - "\n", - "# Calcuations\n", - "\n", - "beta=pie # radian # angle of lap\n", - "T_e=(w*v**2)/g # N # where T_e is the centrifugal tension\n", - "T1=(T_max)-(T_e) #N\n", - "T2=T1/(e**(mu*(1/sin(alpha*(pi/180)))*(beta))) #N # From eq'n T1/T2=e^(mu*cosec(alpha)*beta)\n", - "P=(T1-T2)*v*(10**-3) #kW # power transmitted by a single rope\n", - "N=Pw/P # Number of ropes required\n", - "\n", - "# Results\n", - "\n", - "print\"The number of ropes required to transmit 50 kW is \",round(N),\"Nos\"\n", - "# approx no of ropes is 5\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "The number of ropes required to transmit 50 kW is 5.0 Nos\n" - ] - } - ], - "prompt_number": 36 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 7.7-8,Page No:164" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math\n", - "\n", - "# Initilization of variables\n", - "\n", - "d1=0.45 #m # diameter of larger pulley\n", - "d2=0.20 #m # diameter of smaller pulley\n", - "d=1.95 #m # separation between the pulley's\n", - "T_max=1000 #N # or 1kN which is the maximum permissible tension\n", - "mu=0.20 # coefficient of friction\n", - "N=100 # r.p.m # speed of larger pulley\n", - "e=2.718 # constant\n", - "pie=3.14 # constant\n", - "T_e=0 #N # as the data for calculating T_e is not given we assume T_e=0\n", - "\n", - "# Calculations\n", - "\n", - "r1=d1*0.5 #m # radius of larger pulley #1/2=0.5\n", - "r2=d2*0.5 #m # radius of smaller pulley #1/2=0.5\n", - "theta=arcsin((r1+r2)/d)*(180/pi) # degree\n", - "\n", - "# for cross drive the angle of lap for both the pulleys is same\n", - "beta=(180+(2*(theta)))*(pi/180) #radian\n", - "T1=T_max-T_e #N\n", - "T2=T1/(e**(mu*(beta))) #N # from formulae, T1/T2=e^(mu*beta)\n", - "v=((2*pie)*N*r1)/60 # m/s # where v=velocity of belt which is given as, v=wr=2*pie*N*r/60\n", - "P=(T1-T2)*v*(10**-3) #kW # Power\n", - "\n", - "# Results\n", - "\n", - "print\"The power transmitted by the cross belt drive is \",round(P,1),\"kW\"\n", - "# the approx answer is 1.3 kW The answer given in the book (i.e 1.81kW) is wrong.\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "The power transmitted by the cross belt drive is 1.2 kW\n" - ] - } - ], - "prompt_number": 48 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 7.7-9,Page No:165" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math\n", - "\n", - "# Initilization of variabes\n", - "\n", - "b=0.1 #m #width of the belt\n", - "t=0.008 #m #thickness of the belt\n", - "v=26.67 # m/s # belt speed\n", - "pie=3.14 # constant\n", - "beta=165 # radian # angle of lap for the smaller belt\n", - "mu=0.3 # coefficient of friction\n", - "sigma_max=2 # MN/m^2 # maximum permissible stress in the belt\n", - "m=0.9 # kg/m # mass of the belt\n", - "g=9.81 # m/s^2\n", - "e=2.718 # constant\n", - "\n", - "# Calculations\n", - "\n", - "A=b*t # m^2 # cross-sectional area of the belt\n", - "T_e=m*v**2 # N # where T_e is the Centrifugal tension\n", - "T_max=(sigma_max)*(A)*(10**6) # N # maximum tension in the belt\n", - "T1=(T_max)-(T_e) # N \n", - "T2=T1/(e**((mu*pie*beta)/180)) #N # from formulae T1/T2=e^(mu*beta)\n", - "P=(T1-T2)*v*(10**-3) #kW # Power transmitted\n", - "T_o=(T1+T2)/2 # N # Initial tension\n", - "\n", - "# Now calculations to transmit maximum power\n", - "\n", - "Te=T_max/3 # N # max tension\n", - "u=(T_max/(3*m))**0.5 # m/s # belt speed for max power\n", - "T_1=T_max-Te # N # T1 for case 2\n", - "T_2=T_1/(e**((mu*pie*beta)/180)) # N \n", - "P_max=(T_1-T_2)*u*(10**-3) # kW # Max power transmitted\n", - "\n", - "# Results\n", - "\n", - "print\"The initial power transmitted is \",round(P,2),\"kW\"\n", - "print\"The initial tension in the belt is \",round(T_o,1),\"N\"\n", - "print\"The maximum power that can be transmitted is \",round(P_max,2),\"kW\" # the answer is approx 15.017kW wheres the answer 14.99kW given in book is wrong \n", - "print\"The maximum power is transmitted at a belt speed of \",round(u,2),\"m/s\"\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "The initial power transmitted is 14.8 kW\n", - "The initial tension in the belt is 682.3 N\n", - "The maximum power that can be transmitted is 15.02 kW\n", - "The maximum power is transmitted at a belt speed of 24.34 m/s\n" - ] - } - ], - "prompt_number": 54 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 7.7-10,Page No:169" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math\n", - "\n", - "# Initilization of variables\n", - "\n", - "p=0.0125 # m # pitch of screw\n", - "d=0.1 #m # diameter of the screw\n", - "r=0.05 #m # radius of the screw\n", - "l=0.5 #m # length of the lever\n", - "W=50 #kN # load on the lever\n", - "mu=0.20 # coefficient of friction \n", - "pie=3.14 #constant\n", - "\n", - "# Calculations\n", - "\n", - "theta=arctan(p/(2*pie*r))*(180/pi) #degree # theta is the Helix angle\n", - "phi=arctan(mu)*(180/pi) # degree # phi is the angle of friction\n", - "\n", - "# Taking the leverage due to handle into account,force F1 required is,\n", - "F1=(W*(tan(theta*(pi/180)+phi*(pi/180))))*(r/l) #kN\n", - "\n", - "# To lower the load\n", - "F2=(W*(tan(theta*(pi/180)-phi*(pi/180))))*(r/l) #kN # -ve sign of F2 indicates force required is in opposite sense\n", - "E=(tan(theta*(pi/180))/tan(theta*(pi/180)+phi*(pi/180)))*100 # % # here E=eata=efficiency in %\n", - "\n", - "# Results\n", - "\n", - "print\"The force required (i.e F1) to raise the weight is \",round(F1,3),\"kN\" #due to decimal variance answer varies by 0.004kN \n", - "print\"The force required (i.e F2) to lower the weight is \",round(F2,3),\"kN\"\n", - "print\"The efficiency of the jack is \",round(E,2),\"percent\"\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "The force required (i.e F1) to raise the weight is 1.209 kN\n", - "The force required (i.e F2) to lower the weight is -0.795 kN\n", - "The efficiency of the jack is 16.47 percent\n" - ] - } - ], - "prompt_number": 58 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 7.7-11,Page No:172" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math\n", - "\n", - "# Initilization of variabes\n", - "\n", - "P=20000 #N #Weight of the shaft\n", - "D=0.30 #m #diameter of the shaft\n", - "R=0.15 #m #radius of the shaft\n", - "mu=0.12 # coefficient of friction\n", - "\n", - "# Calculations\n", - "\n", - "# Friction torque T is given by formulae,\n", - "T=(0.666)*P*R*mu #N-m #2/3=0.666\n", - "M=T #N-m\n", - "\n", - "# Results\n", - "\n", - "print\"The frictional torque is \",round(M),\"N-m\"\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "The frictional torque is 240.0 N-m\n" - ] - } - ], - "prompt_number": 61 - } - ], - "metadata": {} - } - ] -} \ No newline at end of file -- cgit