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author | kinitrupti | 2017-05-12 18:40:35 +0530 |
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committer | kinitrupti | 2017-05-12 18:40:35 +0530 |
commit | d36fc3b8f88cc3108ffff6151e376b619b9abb01 (patch) | |
tree | 9806b0d68a708d2cfc4efc8ae3751423c56b7721 /sample_notebooks/AdityaR/Chapter_5-Sample_Notebook.ipynb | |
parent | 1b1bb67e9ea912be5c8591523c8b328766e3680f (diff) | |
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diff --git a/sample_notebooks/AdityaR/Chapter_5-Sample_Notebook.ipynb b/sample_notebooks/AdityaR/Chapter_5-Sample_Notebook.ipynb deleted file mode 100755 index a77ec491..00000000 --- a/sample_notebooks/AdityaR/Chapter_5-Sample_Notebook.ipynb +++ /dev/null @@ -1,279 +0,0 @@ -{ - "cells": [ - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "# Chapter 5 General Case of Forces in a plane" - ] - }, - { - "cell_type": "markdown", - "metadata": { - "collapsed": true - }, - "source": [ - "# Example 5.2 Equations of equilibrium" - ] - }, - { - "cell_type": "code", - "execution_count": 7, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "('The reaction at P is', 5656.85424949238, 'N')\n", - "('The reaction at Q is ', 4000.0, 'N')\n" - ] - } - ], - "source": [ - "import math\n", - "\n", - "#Initialization of Variables\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", - "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=math.sqrt(Xp**2+Yp**2) #N\n", - "\n", - "#Resuts\n", - "print('The reaction at P is' ,Rp ,'N')\n", - "print('The reaction at Q is ',Rq ,'N')" - ] - }, - { - "cell_type": "markdown", - "metadata": { - "collapsed": true - }, - "source": [ - "# Example 5.3 Equations of equilibrium" - ] - }, - { - "cell_type": "code", - "execution_count": 6, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "('The reaction at A i.e Ra is ', matrix([[ 120.27406887]]), 'N')\n", - "('The reaction at B i.e Rb is ', matrix([[ 35.13703443]]), 'N')\n", - "('The required tension in the string is ', matrix([[ 40.57275258]]), 'N')\n" - ] - } - ], - "source": [ - "import math,numpy\n", - "#Initilization of vaiables\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", - "#Using matrix to solve the simultaneous eqn's 3 & 4\n", - "A=numpy.matrix('2 -4; 1 -0.577')\n", - "B=numpy.matrix('100;100')\n", - "C=numpy.linalg.inv(A)*B\n", - "\n", - "#Results\n", - "print('The reaction at A i.e Ra is ',C[0] ,'N')\n", - "print('The reaction at B i.e Rb is ',C[1] ,'N')\n", - "\n", - "#Calculations\n", - "T=C[1]/math.cos(math.radians(alpha)) #N # from (eqn 1)\n", - "\n", - "#Results\n", - "print('The required tension in the string is ',T, 'N')" - ] - }, - { - "cell_type": "markdown", - "metadata": { - "collapsed": true - }, - "source": [ - "# Example 5.4 Equations of Equilibrium" - ] - }, - { - "cell_type": "code", - "execution_count": 4, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "('The reaction at B i.e Rb is ', 25.0, 'N')\n", - "('The horizontal reaction at A i.e Xa is ', 21.650635094610966, 'N')\n", - "('The vertical reaction at A i.e Ya is ', 112.5, 'N')\n" - ] - } - ], - "source": [ - "import math\n", - "#Initilization of variables\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", - "Lab=Lob*(1/math.cos(math.radians(alpha))) #m length of the ladder\n", - "Lad=Lcd*math.tan(math.radians(alpha)) #m\n", - "Rb=(W*Lad)/Lab #N take moment at A\n", - "Xa=Rb*math.sin(math.radians(theta)) #N From eq'n 1\n", - "Ya=W+Rb*math.cos(math.radians(theta)) #N From eq'n 2\n", - "\n", - "#Results\n", - "print('The reaction at B i.e Rb is ',Rb, 'N')\n", - "print('The horizontal reaction at A i.e Xa is ',Xa, 'N')\n", - "print('The vertical reaction at A i.e Ya is ',Ya,'N')\n" - ] - }, - { - "cell_type": "markdown", - "metadata": { - "collapsed": true - }, - "source": [ - "# Example 5.5 Equations of Equilibrium" - ] - }, - { - "cell_type": "code", - "execution_count": 3, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "('The horizontal reaction at A i.e Xa is ', 28.867513459481287, 'N')\n", - "('The vertical reaction at A i.e Ya is ', 100, 'N')\n", - "('The reaction at B i.e Rb is ', 28.867513459481287, 'N')\n" - ] - } - ], - "source": [ - "import math\n", - "#Initilization of variables\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", - "# using the equiblirium equations\n", - "Ya=W #N From eq'n 2\n", - "Lad=Lcd*math.tan(math.radians(alpha)) #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", - "print('The horizontal reaction at A i.e Xa is ',Xa, 'N')\n", - "print('The vertical reaction at A i.e Ya is ',Ya,'N' )\n", - "print('The reaction at B i.e Rb is ',Rb ,'N')\n" - ] - }, - { - "cell_type": "markdown", - "metadata": { - "collapsed": true - }, - "source": [ - "# Example 5.6 Equations of Equilibrium" - ] - }, - { - "cell_type": "code", - "execution_count": 2, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "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 ', 8.089499366462674, 'N')\n", - "('The reaction at D i.e Rd is ', 4.8, 'N')\n" - ] - } - ], - "source": [ - "import math\n", - "#Initilization of variables\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", - "theta=math.degrees(math.atan(h/d)) #angle which the bar makes with the horizontal\n", - "Lad=math.sqrt(d**2+h**2) #m Lad is the length of the bar from point A to point B\n", - "Rd=(W*h*(math.cos(theta*math.pi/180)))/Lad #N Taking moment at A\n", - "Xa=Rd*(math.sin(theta*math.pi/180)) #N sum Fx=0.... From eq'n 1\n", - "Ya=W-(Rd*(math.cos(theta*math.pi/180))) #N sum Fy=0..... From eq'n 2\n", - "Ra=math.sqrt(Xa**2+Ya**2) #resultant of Xa & Ya\n", - "\n", - "#Results\n", - "print('The horizontal reaction at A i.e Xa is ',Xa, 'N')\n", - "print('The vertical reaction at A i.e Ya is ',Ya, 'N')\n", - "print('Therefore the reaction at A i.e Ra is ',Ra,'N')\n", - "print('The reaction at D i.e Rd is ',Rd,'N')" - ] - } - ], - "metadata": { - "kernelspec": { - "display_name": "Python [Root]", - "language": "python", - "name": "Python [Root]" - }, - "language_info": { - "codemirror_mode": { - "name": "ipython", - "version": 2 - }, - "file_extension": ".py", - "mimetype": "text/x-python", - "name": "python", - "nbconvert_exporter": "python", - "pygments_lexer": "ipython2", - "version": "2.7.12" - }, - "widgets": { - "state": {}, - "version": "1.1.2" - } - }, - "nbformat": 4, - "nbformat_minor": 0 -} |