{ "cells": [ { "cell_type": "markdown", "metadata": {}, "source": [ "# Chapter 3 Parallel forces in a plane" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "# Example 3.1 Resultant of Forces in a Plane" ] }, { "cell_type": "code", "execution_count": 4, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "The load taken by man P is 400 N\n", "The load taken by man Q is 600 N\n" ] } ], "source": [ "import math\n", "#Initilization of variables\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", "#Calculations\n", "Q=(W*Lac*math.cos(theta*180/math.pi))/(Lab*math.cos(theta*180/math.pi)) #N # from eq'n 2\n", "P=W-Q #N # from eq'n 1\n", "#Results\n", "print('The load taken by man P is %d N'%P)\n", "print('The load taken by man Q is %d N'%Q)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "# Example 3.2 Resultant of forces in a Plane" ] }, { "cell_type": "code", "execution_count": 6, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "The reaction (downwards)at support A is 250 N\n", "The reaction (upwards)at support B is 1250 N\n" ] } ], "source": [ "#Initilization of variables\n", "F=1000 #N\n", "Lab=1 #m\n", "Lbc=0.25 #m\n", "Lac=1.25 #m\n", "#Calculations\n", "Rb=(F*Lac)/Lab #N # from eq'n 2\n", "Ra=Rb-F #N # fom eq'n 1\n", "#Results\n", "print('The reaction (downwards)at support A is %d N'%Ra)\n", "print('The reaction (upwards)at support B is %d N'%Rb)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "# Example 3.3 Resultant of Forces in a Plane" ] }, { "cell_type": "code", "execution_count": 7, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "The vertical reaction (upwards) at A is 0.833333 kN\n", "The horizontal reaction (towards A) is -20.000000 kN\n", "The reaction (downwards) at B is -0.833333 kN\n" ] } ], "source": [ "#Inilitization of variables\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", "#Calculations\n", "Xa=-(Fe) #kN #take sum Fx=0\n", "Rb=(Md+Me-Mc)/Lab #N #take moment at A\n", "Ya=-Rb #N #take sum Fy=0\n", "#Results\n", "print('The vertical reaction (upwards) at A is %f kN'%Ya)\n", "print('The horizontal reaction (towards A) is %d kN'%Xa)\n", "print('The reaction (downwards) at B is %f kN'%Rb)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "# Example 3.5 Resultant of Forces in a Plane" ] }, { "cell_type": "code", "execution_count": 11, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "The reaction at F i.e Rf is 3500.000000 N\n", "The reaction at D i.e Rd is 1000.000000 N\n", "The reaction at pt E i.e Re is 833.333333 N\n", "The reaction at pt A i.e Ra is -333.333333 N\n" ] } ], "source": [ "import numpy\n", "#Initilization of variables\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", "# Calculations (part1)\n", "#using matrix to solve the given eqn's 1 & 2\n", "A=numpy.matrix('1 -2.5;3.5 -5')\n", "B=numpy.matrix('1000;7250')\n", "C=numpy.linalg.inv(A)*B\n", "#Calculations (part 2)\n", "#Consider combined F.B.D of beams AB,BC &CD. Take moment at A\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", "#Results\n", "print('The reaction at F i.e Rf is %f N'%C[0])\n", "print('The reaction at D i.e Rd is %f N'%C[1])\n", "print('The reaction at pt E i.e Re is %f N'%Re)\n", "print('The reaction at pt A i.e Ra is %f N'%Ra) " ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "# Example 3.6 Resultant of forces in a plane" ] }, { "cell_type": "code", "execution_count": 12, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "The angle theta is 15.945396 degrees\n" ] } ], "source": [ "# Initilization of variables\n", "W=100 # N #force acting at D\n", "AB=50 # N # weight of bar ab\n", "CD=50 # N # weight of bar cd\n", "# Calculations\n", "# From the derived expression the value of the angle is given as,\n", "theta=math.degrees(math.atan(5/17.5)) #degrees\n", "# Results\n", "print('The angle theta is %f degrees'%theta)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "# Example 3.7 Resultant of forces in a plane" ] }, { "cell_type": "code", "execution_count": 13, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "The reaction at wheel A is 0.950000 kN\n", "The reaction at wheel B is 0.950000 kN\n", "The reaction at wheel C is 0.600000 kN\n" ] } ], "source": [ "#Initilization of variables\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", "#Calculations\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", "#Results\n", "print('The reaction at wheel A is %f kN'%Ra)\n", "print('The reaction at wheel B is %f kN'%Rb)\n", "print('The reaction at wheel C is %f kN'%Rc)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "# Example 3.8 Resultant of Forces in a Plane" ] }, { "cell_type": "code", "execution_count": 15, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "The reaction for single force is -60 N\n", "The distance of Ry from A is 0.783333 m\n", "The moment at A is -47 N-m\n", "The moment at B is 25.020000 N-m\n" ] } ], "source": [ "#Initilization of variables\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", "#Calculations\n", "#(a) A single force\n", "Ry=W1-W2+W3-W4 #N #take sum Fy=0\n", "x=((-W2*Lac)+(W3*Lad)-(W4*Lab))/(Ry) #m\n", "# (b) Single force moment at A\n", "Ma=(Ry*x) #N-m\n", "# Single force moment at B\n", "Mb=W2*Leb #N-m\n", "#Results\n", "print('The reaction for single force is %d N'%Ry)\n", "print('The distance of Ry from A is %f m'%x)\n", "print('The moment at A is %d N-m'%Ma)\n", "print('The moment at B is %f N-m'%Mb)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "# Example 3.9 Resultant of Forces in a Plane" ] }, { "cell_type": "code", "execution_count": 17, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Tension in wire 1 i.e T1 is 8101.126884 N \n", "\n", "Tension in wire 2 i.e T2 is 6786.756093 N \n", "\n", "Tension in wire 3 i.e T3 is 4444.444444 N \n", "\n" ] } ], "source": [ "#Initilization of variables\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", "#Calculations\n", "T3=Ma/(4.5*math.sin(theta*math.pi/180)) #N #take moment at B\n", "# Now we use matrix to solve eqn's 1 & 2 simultaneously,\n", "A=numpy.matrix('-0.707 0.8666;0.707 0.5')\n", "B=numpy.matrix('2222.2;8848.8')\n", "C=numpy.linalg.inv(A)*B\n", "#Results\n", "print('Tension in wire 1 i.e T1 is %f N \\n'%C[1])\n", "print('Tension in wire 2 i.e T2 is %f N \\n'%C[0])\n", "print('Tension in wire 3 i.e T3 is %f N \\n'%T3)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "# Example 3.10 Distributed Force in a Plane" ] }, { "cell_type": "code", "execution_count": 19, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "The resultant of the distibuted load lies at 2 m\n", "The reaction at support A is 1000 N\n", "The reaction at support B is 2000 N\n" ] } ], "source": [ "#Initilization of variables\n", "w=2000 #N/m\n", "Lab=3 #m\n", "#Calculations\n", "W=w*Lab/2 #N# Area under the curve\n", "Lac=(2/3)*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", "#Results\n", "print('The resultant of the distibuted load lies at %d m'%Lac)\n", "print('The reaction at support A is %d N'%Ra)\n", "print('The reaction at support B is %d N'%Rb)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "# Example 3.11 Distributed force in a plane" ] }, { "cell_type": "code", "execution_count": 22, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "The resultant is 2000 N and the line of action of the force is 3 m\n" ] } ], "source": [ "#Initiization of variables\n", "w=1500 #N/m\n", "x=4 #m\n", "L=4 #m\n", "#Calculations\n", "k=x**2/w #m**3/N\n", "#Solving the intergral we get\n", "W=L**3/(3*k) #N\n", "x_bar=L**4/(4*k*W) #m\n", "#Result\n", "print(\"The resultant is %d N and the line of action of the force is %d m\"%(W,x_bar))" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "# Example 3.12 Distributed force in a plane" ] }, { "cell_type": "code", "execution_count": 25, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "The resultant of the distributed load system is 18 kN\n", "The line of action of the resulting load is 3.500000 m\n" ] } ], "source": [ "# Initilization of variables\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", "# Calculations\n", "# The varying load distribution is divided into a rectangle and a right angled triangle\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", "#Results\n", "print('The resultant of the distributed load system is %d kN'%W)\n", "print('The line of action of the resulting load is %f m'%x)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "# Example 3.13 Distributed forces in a plane" ] }, { "cell_type": "code", "execution_count": 28, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "The horizontal reaction at A i.e Xa is 17.320508 kN\n", "The vertical reaction at A i.e Ya is 10 kN\n", "The reaction at A i.e Ra is 20 kN\n", "The reaction at B i.e Rb is 45 kN\n" ] } ], "source": [ "# Initiization of variables\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", "# Calculations\n", "Xa=20*math.cos(theta*math.pi/180) #kN # sum Fx=0\n", "Rb=((W1*Lad)+(-M)+(W3*Leb*Laf)+(W2*math.sin(theta*math.pi/180)*Lac)+((W4*Lbc*Lag)/2))/Lab #kN # taking moment at A\n", "Ya=W1+(W2*math.sin(theta*math.pi/180))+(W3*Leb)+(W4*Lbc/2)-Rb #kN # sum Fy=0\n", "Ra=math.sqrt(Xa**2+Ya**2) #kN # resultant at A\n", "#Results\n", "print('The horizontal reaction at A i.e Xa is %f kN'%Xa)\n", "print('The vertical reaction at A i.e Ya is %d kN'%Ya)\n", "print('The reaction at A i.e Ra is %d kN'%Ra)\n", "print('The reaction at B i.e Rb is %d kN'%Rb)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "# Example 3.14 Distributed forces in a plane" ] }, { "cell_type": "code", "execution_count": 29, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "The minimum width which is to be provided to the dam to prevent overturning about point B is 1.490712 m\n" ] } ], "source": [ "# Initilization of variables\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", "# Calculations\n", "P=(rho_w*g*h**2)/2 # The resultant force due to water pressure per unit length of the dam\n", "x=(2/3)*h #m # distance at which the resutant of the triangular load acts \n", "b=math.sqrt((2*P*h)/(3*h*rho_c*g)) # m # eq'n required to find the minimum width of the dam\n", "# Results\n", "print('The minimum width which is to be provided to the dam to prevent overturning about point B is %f m'%b)" ] } ], "metadata": { "kernelspec": { "display_name": "Python 3", "language": "python", "name": "python3" }, "language_info": { "codemirror_mode": { "name": "ipython", "version": 3 }, "file_extension": ".py", "mimetype": "text/x-python", "name": "python", "nbconvert_exporter": "python", "pygments_lexer": "ipython3", "version": "3.5.1" }, "widgets": { "state": {}, "version": "1.1.2" } }, "nbformat": 4, "nbformat_minor": 0 }