From 4a1f703f1c1808d390ebf80e80659fe161f69fab Mon Sep 17 00:00:00 2001 From: Thomas Stephen Lee Date: Fri, 28 Aug 2015 16:53:23 +0530 Subject: add books --- .../chapter05_1.ipynb | 285 +++++++++++++++++++++ 1 file changed, 285 insertions(+) create mode 100755 Engineering_Mechanics_by_Tayal_A.K./chapter05_1.ipynb (limited to 'Engineering_Mechanics_by_Tayal_A.K./chapter05_1.ipynb') diff --git a/Engineering_Mechanics_by_Tayal_A.K./chapter05_1.ipynb b/Engineering_Mechanics_by_Tayal_A.K./chapter05_1.ipynb new file mode 100755 index 00000000..efb681de --- /dev/null +++ b/Engineering_Mechanics_by_Tayal_A.K./chapter05_1.ipynb @@ -0,0 +1,285 @@ +{ + "metadata": { + "name": "chapter5.ipynb" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 5: General Case Of Forces In Plane" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.5-2,Page No:111" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Inilization of variables\n", + "\n", + "W=2000 #N\n", + "Lab=2 #m #length of the member from the vertical to the 1st load of 2000 N\n", + "Lac=5 #m #length of the member from the vertical to the 2nd load of 2000 N\n", + "Lpq=3.5 #m\n", + "\n", + "#Calculations\n", + "\n", + "Rq=((W*Lab)+(W*Lac))/Lpq #N #take moment abt. pt P\n", + "Xp=Rq #N #sum Fx=0\n", + "Yp=2*W #N #sum Fy=0\n", + "Rp=(Xp**2+Yp**2)**0.5 #N\n", + "\n", + "#Resuts\n", + "\n", + "print\"The reaction at P is \",round(Rp,1),\"N\"\n", + "print\"The reaction at Q is \",round(Rq),\"N\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The reaction at P is 5656.9 N\n", + "The reaction at Q is 4000.0 N\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.5-3,Page No:112" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "import numpy as np\n", + "#Initilization of vaiables\n", + "\n", + "W=25 #N # self weight of the ladder\n", + "M=75 #N # weight of the man standing o the ladder\n", + "theta=63.43 #degree # angle which the ladder makes with the horizontal\n", + "alpha=30 #degree # angle made by the string with the horizontal\n", + "Loa=2 #m # spacing between the wall and the ladder\n", + "Lob=4 #m #length from the horizontal to the top of the ladder touching the wall(vertical)\n", + "\n", + "#Calculations\n", + "\n", + "#Using matrix to solve the simultaneous eqn's 3 & 4\n", + "\n", + "A=np.array([[2 ,-4],[1, -0.577]])\n", + "B=np.array([100,100])\n", + "C=np.linalg.solve(A,B)\n", + "\n", + "#Results\n", + "\n", + "print\"The reaction at A i.e Ra is \",round(C[0],2),\"N\"\n", + "print\"The reaction at B i.e Rb is \",round(C[1],2),\"N\"\n", + "\n", + "#Calculations\n", + "\n", + "T=C[1]/cos(alpha*(pi/180)) #N # from (eqn 1)\n", + "\n", + "#Results\n", + "\n", + "print\"The required tension in the string is \",round(T,2),\"N\"\n", + "\n", + "#answer may vary due to decimal variance" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The reaction at A i.e Ra is 120.27 N\n", + "The reaction at B i.e Rb is 35.14 N\n", + "The required tension in the string is 40.57 N\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.5-4,Page No:113" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Initilization of variables\n", + "\n", + "W=100 #N\n", + "theta=60 #degree #angle made by the ladder with the horizontal\n", + "alpha=30 #degree #angle made by the ladder with the vertical wall\n", + "Lob=4 #m # length from the horizontal to the top of the ladder touching the wall(vertical)\n", + "Lcd=2 #m # length from the horizontal to the centre of the ladder where the man stands\n", + "\n", + "#Calculations\n", + "\n", + "Lab=Lob*(1/cos(alpha*(pi/180))) #m #length of the ladder\n", + "Lad=Lcd*tan(alpha*(pi/180)) #m\n", + "Rb=(W*Lad)/Lab #N #take moment at A\n", + "Xa=Rb*sin(theta*(pi/180)) #N # From eq'n 1\n", + "Ya=W+Rb*cos(theta*(pi/180)) #N #From eq'n 2\n", + "\n", + "#Results\n", + "\n", + "print\"The reaction at B i.e Rb is \",round(Rb),\"N\"\n", + "print\"The horizontal reaction at A i.e Xa is \",round(Xa,2),\"N\"\n", + "print\"The vertical reaction at A i.e Ya is \",round(Ya,1),\"N\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The reaction at B i.e Rb is 25.0 N\n", + "The horizontal reaction at A i.e Xa is 21.65 N\n", + "The vertical reaction at A i.e Ya is 112.5 N\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.5-5,Page No:114" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Initilization of variables\n", + "\n", + "W=100 #N #self weight of the man\n", + "alpha=30 #degree # angle made by the ladder with the wall\n", + "Lob=4 #m # length from the horizontal to the top of the ladder touching the wall(vertical)\n", + "Lcd=2 #m\n", + "\n", + "#Calculations\n", + "\n", + "# using the equiblirium equations\n", + "\n", + "Ya=W #N # From eq'n 2\n", + "Lad=Lcd*tan(alpha*(pi/180)) #m #Lad is the distance fom pt A to the point where the line from the cg intersects the horizontal\n", + "Rb=(W*Lad)/Lob #N # Taking sum of moment abt A\n", + "Xa=Rb #N # From eq'n 1\n", + "\n", + "#Results\n", + "\n", + "print\"The horizontal reaction at A i.e Xa is \",round(Xa,2),\"N\"\n", + "print\"The vertical reaction at A i.e Ya is \",round(Ya),\"N\"\n", + "print\"The reaction at B i.e Rb is \",round(Rb,2),\"N\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The horizontal reaction at A i.e Xa is 28.87 N\n", + "The vertical reaction at A i.e Ya is 100.0 N\n", + "The reaction at B i.e Rb is 28.87 N\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.5-6,Page No:115" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Initilization of variables\n", + "\n", + "d=0.09 #m #diametre of the right circular cylinder\n", + "h=0.12 #m #height of the cyinder\n", + "W=10 #N # self weight of the bar\n", + "l=0.24 #m #length of the bar\n", + "\n", + "#Calculations\n", + "\n", + "theta=arctan(h/d)*(180/pi) # angle which the bar makes with the horizontal\n", + "Lad=(d**2+h**2)**0.5 #m # Lad is the length of the bar from point A to point B\n", + "Rd=(W*h*cos(theta*(pi/180)))/Lad #N # Taking moment at A\n", + "Xa=Rd*sin(theta*(pi/180)) #N # sum Fx=0.... From eq'n 1\n", + "Ya=W-(Rd*cos(theta*(pi/180))) #N # sum Fy=0..... From eq'n 2\n", + "Ra=(Xa**2+Ya**2)**0.5 #resultant of Xa & Ya\n", + "\n", + "#Results\n", + "\n", + "print\"The horizontal reaction at A i.e Xa is \",round(Xa,2),\"N\"\n", + "print\"The vertical reaction at A i.e Ya is \",round(Ya,2),\"N\"\n", + "print\"Therefore the reaction at A i.e Ra is \",round(Ra,2),\"N\"\n", + "print\"The reaction at D i.e Rd is \",round(Rd,2),\"N\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The horizontal reaction at A i.e Xa is 3.84 N\n", + "The vertical reaction at A i.e Ya is 7.12 N\n", + "Therefore the reaction at A i.e Ra is 8.09 N\n", + "The reaction at D i.e Rd is 4.8 N\n" + ] + } + ], + "prompt_number": 10 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file -- cgit