{ "metadata": { "name": "chapter_3.ipynb" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "Chapter 3:Coplanar Parallel forces" ] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.1,Page No.48" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#Declaration Of Variables\n", "\n", "#Lengths\n", "L_AB=L_BC=L_CD=L_DA=2 #m\n", "\n", "#Loads\n", "F_B=10 #N\n", "F_C=20 #N\n", "F_D=30 #N\n", "F_A=40 #N\n", "\n", "#Calculations\n", "\n", "#Taking Moment at point A\n", "#As the Forces F_A & F_B pass through point A,these Forces will be zero\n", "#Resultant Moment of all Forces\n", "M_A=-(-F_D*L_DA-F_C*L_CD) #N.m\n", "\n", "#Result\n", "print\"Resultant Moment about point A is\",round(M_A,2),\"N.m (Anticlockwise)\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Resultant Moment about point A is 100.0 N.m (Anticlockwise)\n" ] } ], "prompt_number": 1 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.2,Page No.50" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "from math import sin, cos, tan, radians, pi\n", "\n", "#Declaration Of Variables\n", "F_A=100 #N\n", "OCB=60 #Degrees\n", "L_OC=3 #m\n", "\n", "#Calculations\n", "\n", "#Triangle OBC is a right Angled triangle\n", "L_OB=L_OC*sin(60*pi*180**-1)\n", "\n", "#Moment of force 100 #N about o\n", "M_O=F_A*L_OB #Nm\n", "\n", "#Result\n", "print\"Moment of Force about O is\",round(M_O,2),\"Nm\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Moment of Force about O is 259.81 Nm\n" ] } ], "prompt_number": 2 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.3,Page No.54" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#Declaration Of Variables\n", "\n", "#Lengths\n", "L_AB=30 #cm\n", "L_BC=40 #cm\n", "\n", "#Loads\n", "F_A=100 #N\n", "F_B=200 #N\n", "F_C=300 #N\n", "\n", "#Calculations\n", "\n", "#resultant of all forces\n", "R=F_A+F_B+F_C #N\n", "\n", "#Let resultant be acting at distance x from point A\n", "\n", "#Now taking moments at point A\n", "M_A=-(-F_C*(L_AB+L_BC)-F_B*L_AB) #N.m\n", "\n", "#Moment of resultant R about A\n", "#M_R=R*x\n", "\n", "#But algebraic sum of moments of all forces about A = Moment of resultant about A\n", "x=M_A*R**-1\n", "\n", "#Result\n", "print\"Resultant is\",round(R,2),\"N\"\n", "print\"Distance of resultant from point A\",round(x,2),\"cm\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Resultant is 600.0 N\n", "Distance of resultant from point A 45.0 cm\n" ] } ], "prompt_number": 3 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.4,Page No.54" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#Declaration Of Variables\n", "\n", "#Lengths\n", "L_AC=4 #m\n", "L_CD=3 #m\n", "L_AD=7 #m\n", "\n", "#Forces\n", "F_A=50 #N\n", "F_D=100 #N\n", "F_B=200 #N\n", "R=250 #N #Resultant \n", "\n", "#Calculations\n", "\n", "#Part-1\n", "#Magnitude of Force F_B\n", "F=R-F_A-F_D #N\n", "\n", "#Part-2\n", "#Distance from pt A\n", "#Taking Moment of all forces at pt A\n", "#M_A=0\n", "#Now moments of all forces = Moment of Resultant\n", "#After simplifying further we get\n", "x=(R*L_AC-F_D*L_AD)*F_B**-1 #m\n", "\n", "#Result\n", "print\"Magnitude of Force F is\",round(F,2),\"N\"\n", "print\"Distance of force f from A is\",round(x,2),\"m\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Magnitude of Force F is 100.0 N\n", "Distance of force f from A is 1.5 m\n" ] } ], "prompt_number": 1 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.5,Page No.55" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#Declaration Of Variables\n", "\n", "#Lengths\n", "L_AB=0.9 #m\n", "L_BC=1.2 #m\n", "L_CD=0.75 #m\n", "L=L_AB+L_BC+L_CD #m\n", "\n", "#Forces\n", "F_A=100 #N\n", "F_B=150 #N\n", "F_C=25 #N\n", "F_D=200 #N\n", "\n", "#Calculations\n", "\n", "#Part-1\n", "#Magnitude of Resultant\n", "R=F_A-F_B-F_C+F_D #N\n", "\n", "#Part-2\n", "#Let x be the distance of Resultant from A\n", "x=-(F_B*L_AB+F_C*(L_AB+L_BC)-F_D*L)*R**-1 #m\n", "\n", "#Result\n", "print\"Magnitude of Resultant\",round(R,2),\"N\"\n", "print\"Distance of Resultant from x is\",round(x,2),\"m\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Magnitude of Resultant 125.0 N\n", "Distance of Resultant from x is 3.06 m\n" ] } ], "prompt_number": 5 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.6,Page No.57" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#Declaration Of Variables\n", "\n", "#Lengths\n", "L_AC=L_CD=1 #m\n", "L_DB=1.5 #m\n", "L=3.5 #m\n", "\n", "#Forces\n", "F_A=32.5 #N\n", "F_C=150 #N\n", "F_D=67.5 #N\n", "F_B=10 #N\n", "\n", "#Calculations\n", "\n", "#Part-1\n", "\n", "#Single Force ststem\n", "R=-(F_A-F_C+F_D-F_B) #N\n", "\n", "#Let x be the distance of Resultant from A\n", "x=-(F_C*L_AC-F_D*(L_AC+L_CD)+F_B*L)*R**-1 #m\n", "\n", "#Part-2\n", "\n", "#Single Force is given By R\n", "\n", "#Now moment of couple at pt A\n", "M_A=-R*round(x,2) #N.m\n", "\n", "#Part-3\n", "\n", "#Now couple at B\n", "L_BE=L+x\n", "M_B=R*round(L_BE,3) #N.m\n", "\n", "#Result\n", "print\"Single Force is\",round(R,2),\"N\"\n", "print\"Couple at A\",round(M_A,2),\"N.m\"\n", "print\"Couple at B\",round(M_B,2),\"N.m\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Single Force is 60.0 N\n", "Couple at A 49.8 N.m\n", "Couple at B 160.02 N.m\n" ] } ], "prompt_number": 6 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.7,Page No.59" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#Declaration Of Variables\n", "\n", "#Lengths\n", "L_AB=L_DE=0.6 #m\n", "L_BC=0.9 #m\n", "L_CD=1.2 #m\n", "\n", "#Forces\n", "F_A=4 #N\n", "F_B=8 #N\n", "F_C=8 #N\n", "F_D=16 #N\n", "F_E=12 #N\n", "\n", "#Calculations\n", "\n", "#Resultant Of All Forces\n", "R=-F_A+F_B-F_C+F_D-F_E #N\n", "\n", "#As the Resultatn Force is zero,tere will be two possibilities.The system will have a resultant coup\n", "#Algebraic sum of moments of all forces about A\n", "M_A=-F_B*L_AB+F_C*(L_AB+L_BC)-F_D*(L_AB+L_BC+L_CD)+F_E*(L_AB+L_BC+L_CD+L_DE) #N.m\n", "\n", "#As the algebraic sum of moments of all forces is not zero,the ststem will have couple of magnitude 3.6 #N.m\n", "\n", "#Result\n", "print\"Resultant of Parallel Forces is\",round(R,2),\"N\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Resultant of Parallel Forces is 0.0 N\n" ] } ], "prompt_number": 7 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.8,Page No.59" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#Declaration Of Variables\n", "\n", "L_AB=L_DE=2 #m\n", "L_BC=0.5 #m\n", "L_CD=0.5 #m\n", "\n", "#Forces\n", "F_A=20 #N\n", "F_B=20 #N\n", "F_C=40 #N\n", "F_D=30 #N\n", "F_E=10 #N\n", "\n", "\n", "#Calculations\n", "\n", "#Resultant Of All Forces\n", "R=-F_A+F_B+F_C-F_D-F_E #N\n", "\n", "#As the Resultant is zero and also the resultant force on the body is zero,the body will be in equilibrium\n", "\n", "#Result\n", "print\"Resultant of Parallel Forces is\",round(R,2),\"N\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Resultant of Parallel Forces is 0.0 N\n" ] } ], "prompt_number": 8 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.9,Page No.60" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "from math import sin, cos, tan, radians, pi\n", "\n", "#Declaration Of Variables\n", "\n", "#Distances\n", "L_OA=200 #mm\n", "L_OB=100 #mm\n", "L_BC=200 #mm\n", "\n", "COA=90 #Degrees\n", "\n", "#Forces\n", "F_A=2000 #N\n", "F_B=1500 #N\n", "F_C=1000 #N\n", "\n", "#Calculations\n", "\n", "#Resolving FOrce A in two components\n", "F_A1=F_A*cos(30*pi*180**-1) #Component along x-axis\n", "F_A2=F_A*sin(30*pi*180**-1) #Component along y-axis\n", "\n", "#Resolving all forces along X-axis\n", "F_x=F_A1-F_B-F_C\n", "F_y=F_A2\n", "\n", "#Resultant \n", "R=(F_x**2+F_y**2)**0.5 #N\n", "\n", "#Taking Moments of all forces about pt O\n", "M_o=-(F_y*L_OA-F_B*L_OB-F_C*(L_BC+L_OB)) #N.mm\n", "\n", "#Result\n", "print\"Equivalent system through point O is:Resultant\",round(R,2),\"N\"\n", "print\" :Moment\",round(M_o,2),\"N.mm\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Equivalent system through point O is:Resultant 1260.85 N\n", " :Moment 250000.0 N.mm\n" ] } ], "prompt_number": 9 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.10,Page No.61" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#Declaration Of Variables\n", "\n", "#Lengths\n", "L_AC=1 #m\n", "L_CB=1.5 #m\n", "L_CD=0.8 #m\n", "L_DB=0.7 #m\n", "L=2.5 #m\n", "\n", "#Forces\n", "F_C=4000 #N\n", "F_B=2500 #N\n", "M_D=2000 #N*m\n", "\n", "#Calculations\n", "\n", "#Resultant of all forces\n", "R=-F_C+F_B #N\n", "\n", "#As the Force is acting in downward direction,so negative sign\n", "R2=-R\n", "\n", "#Sum of Moments of all Forces\n", "M=(F_C*L_AC+M_D-F_B*L)\n", "M2=-M #Anticlockwise\n", "\n", "#Now distance of resultant from x is\n", "x=(F_C*L_AC+M_D-F_B*L)*R2**-1\n", "\n", "#Negative sign indicates that it acts left of A\n", "x2=-x\n", "\n", "#Result\n", "print\"Resultant of the system\",round(R2,2),\"N\"\n", "print\"Equivalent system through A\",round(M2,2),\"N.m (Anticlockwise)\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Resultant of the system 1500.0 N\n", "Equivalent system through A 250.0 N.m (Anticlockwise)\n" ] } ], "prompt_number": 10 } ], "metadata": {} } ] }