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 --- .../chapter03_3.ipynb | 598 +++++++++++++++++++++ 1 file changed, 598 insertions(+) create mode 100755 Engineering_Mechanics_by_Tayal_A.K./chapter03_3.ipynb (limited to 'Engineering_Mechanics_by_Tayal_A.K./chapter03_3.ipynb') diff --git a/Engineering_Mechanics_by_Tayal_A.K./chapter03_3.ipynb b/Engineering_Mechanics_by_Tayal_A.K./chapter03_3.ipynb new file mode 100755 index 00000000..cdcc5a68 --- /dev/null +++ b/Engineering_Mechanics_by_Tayal_A.K./chapter03_3.ipynb @@ -0,0 +1,598 @@ +{ + "metadata": { + "name": "chapter3.ipynb" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 3 : Parallel Forces In A Plane" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.3-1,Page No:64" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Initilization of variables\n", + "\n", + "W=1000 #N\n", + "Lab=1 #m\n", + "Lac=0.6 #m\n", + "theta=60 #degree #angle made by the beam with the horizontal\n", + "\n", + "#Calculations\n", + "\n", + "Q=(W*Lac*cos(theta*(pi/180)))/(Lab*cos(theta*(pi/180))) #N # from eq'n 2\n", + "P=W-Q #N # from eq'n 1\n", + "\n", + "#Results\n", + "\n", + "print\"The load taken by man P is \",round(P),\"N\"\n", + "print\"The load taken by man Q is \",round(Q),\"N\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The load taken by man P is 400.0 N\n", + "The load taken by man Q is 600.0 N\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.3-2,Page No:64" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Initilization of variables\n", + "\n", + "F=1000 #N\n", + "Lab=1 #m\n", + "Lbc=0.25 #m\n", + "Lac=1.25 #m\n", + "\n", + "#Calculations\n", + "\n", + "Rb=(F*Lac)/Lab #N # from eq'n 2\n", + "Ra=Rb-F #N # fom eq'n 1\n", + "\n", + "#Results\n", + "\n", + "print\"The reaction (downwards)at support A is \",round(Ra),\"N\"\n", + "print\"The reaction (upwards)at support B is \",round(Rb),\"N\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The reaction (downwards)at support A is 250.0 N\n", + "The reaction (upwards)at support B is 1250.0 N\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.3-3,Page No:65" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Inilitization of variables\n", + "\n", + "Lab=12 #m\n", + "Mc=40 #kN-m \n", + "Md=10 #kN-m\n", + "Me=20 #kN-m\n", + "Fe=20 #kN #force acting at point E\n", + "\n", + "#Calculations\n", + "\n", + "Xa=-(Fe) #kN #take sum Fx=0\n", + "a=Me+Md-Mc #N #take moment at A\n", + "Rb=a*(Lab)**-1\n", + "Ya=-Rb #N #take sum Fy=0\n", + "\n", + "#Results\n", + "\n", + "print\"The vertical reaction (upwards) at A is \",round(Ya,3),\"kN\"\n", + "print\"The horizontal reaction (towards A) is \",round(Xa,2),\"kN\"\n", + "print\"The reaction (downwards) at B is \",round(Rb,3),\"kN\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The vertical reaction (upwards) at A is 0.833 kN\n", + "The horizontal reaction (towards A) is -20.0 kN\n", + "The reaction (downwards) at B is -0.833 kN\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.3-5,Page No:66" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "import numpy as np\n", + "\n", + "#Initilization of variables\n", + "\n", + "W=1000 #N\n", + "Lad=7.5 #m\n", + "Lae=1.5 #m\n", + "La1=3.75 #m #distance of 1st 1000N load from pt A\n", + "La2=5 #m #distance of 2nd 1000N load from pt A\n", + "La3=6 #m # distance of 3rd 1000N load from pt A\n", + "\n", + "# Calculations (part1)\n", + "\n", + "#using matrix to solve the given eqn's 1 & 2\n", + "\n", + "A=np.array([[1 ,-2.5],[3.5 ,-5]])\n", + "B=np.array([1000,7250])\n", + "C=np.linalg.solve(A,B)\n", + "\n", + "#Resuts\n", + "\n", + "print\"The reaction at F i.e Rf is \",round(C[0]),\"N\"\n", + "print\"The reaction at D i.e Rd is \",round(C[1]),\"N\"\n", + "\n", + "#Calculations (part 2)\n", + "#Consider combined F.B.D of beams AB,BC &CD. Take moment at A\n", + "\n", + "Re=((W*La1)+(W*La2)+(W*La3)+(C[1]*Lad)-(C[0]*La3))/Lae #N\n", + "Ra=C[1]-Re-C[0]+(3*W) #N #Taking sum of forces in Y direction\n", + "\n", + "#Results\n", + "\n", + "print\"The reaction at pt E i.e Re is \",round(Re),\"N\"\n", + "print\"The reaction at pt A i.e Ra is \",round(Ra),\"N\" #acting vertically downwards" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The reaction at F i.e Rf is 3500.0 N\n", + "The reaction at D i.e Rd is 1000.0 N\n", + "The reaction at pt E i.e Re is 833.0 N\n", + "The reaction at pt A i.e Ra is -333.0 N\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.3-7,Page No:69" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Initilization of variables\n", + "\n", + "Ws=2 #kN #weight of scooter\n", + "Wd=0.5 #kN #weight of driver\n", + "Lab=1 #m\n", + "Led=0.8 #m\n", + "Leg=0.1 #m\n", + "\n", + "#Calculations\n", + "\n", + "Rc=((2*Leg)+(Wd*Led))/Lab #kN #take moment at E\n", + "Ra=(2+Wd-Rc)/2 #kN # as Ra=Rb,(Ra+Rb=2*Ra)\n", + "Rb=Ra # kN\n", + "\n", + "#Results\n", + "\n", + "print\"The reaction at wheel A is \",round(Ra,2),\"kN\"\n", + "print\"The reaction at wheel B is \",round(Rb,2),\"kN\"\n", + "print\"The reaction at wheel C is \",round(Rc,2),\"kN\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The reaction at wheel A is 0.95 kN\n", + "The reaction at wheel B is 0.95 kN\n", + "The reaction at wheel C is 0.6 kN\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.3-8,Page No:69" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Initilization of variables\n", + "\n", + "W1=15 #N #up\n", + "W2=60 #N #down\n", + "W3=10 #N #up\n", + "W4=25 #N #down\n", + "Lab=1.2 #m\n", + "Lac=0.4 #m\n", + "Lcd=0.3 #m\n", + "Ldb=0.5 #m\n", + "Lad=0.7 #m\n", + "Leb=0.417 #m #Leb=Lab-x\n", + "\n", + "#Calculations\n", + "\n", + "#(a) A single force\n", + "\n", + "Ry=W1-W2+W3-W4 #N #take sum Fy=0\n", + "x=((-W2*Lac)+(W3*Lad)-(W4*Lab))/(Ry) #m\n", + "\n", + "#(b) Single force moment at A\n", + "\n", + "Ma=(Ry*x) #N-m\n", + "\n", + "# Single force moment at B\n", + "\n", + "Mb=W2*Leb #N-m\n", + "\n", + "#Results\n", + "\n", + "print\"The reaction for single force (a) is \",round(Ry,2),\"N\"\n", + "print\"The distance of Ry from A is \",round(x,3),\"m\"\n", + "print\"The moment at A is \",round(Ma,2),\"N-m\"\n", + "print\"The moment at B is \",round(Mb,2),\"N-m\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The reaction for single force (a) is -60.0 N\n", + "The distance of Ry from A is 0.783 m\n", + "The moment at A is -47.0 N-m\n", + "The moment at B is 25.02 N-m\n" + ] + } + ], + "prompt_number": 25 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.3-9,Page No:71" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "import numpy as np\n", + "\n", + "#Initilization of variables\n", + "\n", + "Ra=5000 #N\n", + "Ma=10000 #Nm\n", + "alpha=60 #degree #angle made by T1 with the pole\n", + "beta=45 #degree #angle made by T2 with the pole\n", + "theta=30 #degree #angle made by T3 with the pole\n", + "Lab=6 #m\n", + "Lac=1.5 #m\n", + "Lcb=4.5 #m\n", + "\n", + "#Calculations\n", + "\n", + "T3=Ma/(4.5*sin(theta*(pi/180))) #N #take moment at B\n", + "\n", + "# Now we use matrix to solve eqn's 1 & 2 simultaneously,\n", + "\n", + "A=np.array([[-0.707, 0.866],[0.707, 0.5]])\n", + "B=np.array([2222.2,8848.8])\n", + "C=np.linalg.solve(A,B)\n", + "\n", + "#Results\n", + "\n", + "print\"Tension in wire 1 i.e T1 is \",round(C[1],1),\"N\" #answer may vary due to decimal variance\n", + "print\"Tension in wire 2 i.e T2 is \",round(C[0],1),\"N\"\n", + "print\"Tension in wire 3 i.e T3 is \",round(T3,1),\"N\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Tension in wire 1 i.e T1 is 8104.7 N\n", + "Tension in wire 2 i.e T2 is 6784.2 N\n", + "Tension in wire 3 i.e T3 is 4444.4 N\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.3-10,Page No:76" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Initilization of variables\n", + "\n", + "w=2000 #N/m\n", + "Lab=3 #m\n", + "\n", + "#Calculations\n", + "\n", + "W=w*Lab/2 #N# Area under the curve\n", + "Lac=(0.6666)*Lab #m#centroid of the triangular load system\n", + "Rb=(W*Lac)/Lab #N #sum of moment at A\n", + "Ra=W-Rb #N\n", + "\n", + "#Results\n", + "\n", + "print\"The resultant of the distibuted load lies at \",round(Lac),\"m\"\n", + "print\"The reaction at support A is \",round(Ra),\"N\"\n", + "print\"The reaction at support B is \",round(Rb),\"N\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The resultant of the distibuted load lies at 2.0 m\n", + "The reaction at support A is 1000.0 N\n", + "The reaction at support B is 2000.0 N\n" + ] + } + ], + "prompt_number": 32 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.3-12,Page No:78" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# Initilization of variables\n", + "\n", + "w1=1.5 #kN/m # intensity of varying load at the starting point of the beam\n", + "w2=4.5 #kN/m # intensity of varying load at the end of the beam\n", + "l=6 #m # ength of the beam\n", + "\n", + "# Calculations\n", + "\n", + "# The varying load distribution is divided into a rectangle and a right angled triangle\n", + "\n", + "W1=w1*l #kN # where W1 is the area of the load diagram(rectangle ABED)\n", + "x1=l/2 #m # centroid of the rectangular load system\n", + "W2=(w2-w1)*l/2 #kN # where W1 is the area of the load diagram(triangle DCE)\n", + "x2=2*l/3 #m # centroid of the triangular load system\n", + "W=W1+W2 #kN # W is the resultant\n", + "x=((W1*x1)+(W2*x2))/W #m # where x is the distance where the resultant lies\n", + "\n", + "#Results\n", + "\n", + "print\"The resultant of the distributed load system is \",round(W),\"kN\"\n", + "print\"The line of action of the resulting load is \",round(x,1),\"m\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The resultant of the distributed load system is 18.0 kN\n", + "The line of action of the resulting load is 3.5 m\n" + ] + } + ], + "prompt_number": 33 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.3-13,Page No:78" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# Initiization of variables\n", + "\n", + "W1=10 #kN #point load acting at D\n", + "W2=20 #kN # point load acting at C at an angle of 30 degree\n", + "W3=5 #kN/m # intensity of udl acting on span EB of 4m\n", + "W4=10 #kN/m # intensity of varying load acting on span BC of 3m\n", + "M=25 #kN-m # moment acting at E\n", + "theta=30 #degree # angle made by 20 kN load with the beam\n", + "Lad=2 #m\n", + "Leb=4 #m\n", + "Laf=6 #m #distance between the resultant of W3 & point A\n", + "Lac=11 #m\n", + "Lag=9 #m #distance between the resultant of W4 and point A\n", + "Lbc=3 #m\n", + "Lab=8 #m\n", + "\n", + "# Calculations\n", + "\n", + "Xa=20*cos(theta*(pi/180)) #kN # sum Fx=0\n", + "Rb=((W1*Lad)+(-M)+(W3*Leb*Laf)+(W2*sin(theta*(pi/180))*Lac)+((W4*Lbc*Lag)/2))/Lab #kN # taking moment at A\n", + "Ya=W1+(W2*sin(theta*(pi/180)))+(W3*Leb)+(W4*Lbc/2)-Rb #kN # sum Fy=0\n", + "Ra=(Xa**2+Ya**2)**0.5 #kN # resultant at A\n", + "\n", + "#Results\n", + "\n", + "print\"The horizontal reaction at A i.e Xa is \",round(Xa,2),\"kN\"\n", + "print\"The vertical reaction at A i.e Ya is \",round(Ya),\"kN\"\n", + "print\"The reaction at A i.e Ra is \",round(Ra),\"kN\"\n", + "print\"The reaction at B i.e Rb is \",round(Rb),\"kN\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The horizontal reaction at A i.e Xa is 17.32 kN\n", + "The vertical reaction at A i.e Ya is 10.0 kN\n", + "The reaction at A i.e Ra is 20.0 kN\n", + "The reaction at B i.e Rb is 45.0 kN\n" + ] + } + ], + "prompt_number": 35 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.3-14,Page No:79" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# Initilization of variables\n", + "\n", + "h=4 #m #height of the dam wall\n", + "rho_w=1000 # kg/m^3 # density of water\n", + "rho_c=2400 # kg/m^3 # density of concrete\n", + "g=9.81 # m/s^2\n", + "\n", + "# Calculations\n", + "\n", + "P=(rho_w*g*h**2)/2 # The resultant force due to water pressure per unit length of the dam\n", + "x=(0.6666)*h #m # distance at which the resutant of the triangular load acts \n", + "b=((2*P*h)/(3*h*rho_c*g))**0.5 # m # eq'n required to find the minimum width of the dam\n", + "\n", + "# Results\n", + "\n", + "print\"The minimum width which is to be provided to the dam to prevent overturning about point B is \",round(b,3),\"m\" " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The minimum width which is to be provided to the dam to prevent overturning about point B is 1.491 m\n" + ] + } + ], + "prompt_number": 37 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file -- cgit