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diff --git a/Engineering_Mechanics_by_Tayal_A.K./chapter05_4.ipynb b/Engineering_Mechanics_by_Tayal_A.K./chapter05_4.ipynb new file mode 100755 index 00000000..efb681de --- /dev/null +++ b/Engineering_Mechanics_by_Tayal_A.K./chapter05_4.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": {}
+ }
+ ]
+}
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