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authorTrupti Kini2016-11-23 23:30:49 +0600
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+{
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "# Chapter9-Beams"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "# example 9.1 page number 286"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 1,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "RB= 18.8684 KN\n",
+ "RA= 29.989 KN\n",
+ "alpha= 25.21 °\n"
+ ]
+ }
+ ],
+ "source": [
+ "from math import pi,atan,sqrt,cos,sin\n",
+ "\n",
+ "#variable declaration\n",
+ "\n",
+ "#summation of all horizontal forces is zero & vertical forces is zero.\n",
+ "P1=float(10) #Vertical down Load at 4m from A,KN\n",
+ "P2=float(15) #Inclined down Load at angle 30° at 6m from A,KN\n",
+ "P3=float(20) #Inclined down Load at angle 45° at 10m from A,KN\n",
+ "theta2=30\n",
+ "theta3=45\n",
+ "#horizontal,vertical component at A is Ha,Va respectively.\n",
+ "\n",
+ "Ha=P2*cos(theta2*pi/180)+P3*cos(theta3*pi/180)\n",
+ "Rb=(P1*4+P2*6*sin(theta2*pi/180)+P3*10*sin(theta3*pi/180))/12 #reaction at B point,KN\n",
+ "\n",
+ "print \"RB=\",round(Rb,4),\"KN\"\n",
+ "\n",
+ "#now vertical component\n",
+ "Va=P2*sin(theta2*pi/180)+P3*sin(theta3*pi/180)+P1-Rb\n",
+ "\n",
+ "Ra=sqrt(pow(Ha,2)+pow(Va,2))\n",
+ "\n",
+ "print \"RA=\",round(Ra,4),\"KN\"\n",
+ "\n",
+ "alpha=(atan(Va/Ha))*180/pi\n",
+ "\n",
+ "print \"alpha=\",round(alpha,2),\"°\""
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "# example 9.2 page number 287"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 2,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "RB= 100.4475 KN\n",
+ "RA= 87.0172 KN\n",
+ "alpha= 79.45 °\n"
+ ]
+ }
+ ],
+ "source": [
+ "from math import pi,atan,sqrt,cos,sin\n",
+ "\n",
+ "#variable declaration\n",
+ "\n",
+ "#summation of all horizontal forces is zero & vertical forces is zero.\n",
+ "P1=float(60) #inclined down to right Load at angle 60 at 1m from A,KN\n",
+ "P2=float(80) #Inclined down to left Load at angle 75° at 3m from A,KN\n",
+ "P3=float(50) #Inclined down to left Load at angle 60° at 5.5m from A,KN\n",
+ "theta1=60 \n",
+ "theta2=75\n",
+ "theta3=60\n",
+ "thetaRb=60\n",
+ "#horizontal,vertical component at A is Ha,Va respectively.\n",
+ "\n",
+ "Rb=(P1*1*sin(theta1*pi/180)+P2*3*sin(theta2*pi/180)+P3*5.5*sin(theta3*pi/180))/(6*sin(thetaRb*pi/180)) #reaction at B point,KN\n",
+ "Ha=-P1*cos(theta1*pi/180)+P2*cos(theta2*pi/180)-P3*cos(theta3*pi/180)+Rb*cos(thetaRb*pi/180)\n",
+ "print \"RB=\",round(Rb,4),\"KN\"\n",
+ "\n",
+ "#now vertical component\n",
+ "Va=P1*sin(theta1*pi/180)+P2*sin(theta2*pi/180)+P3*sin(theta3*pi/180)-Rb*sin(thetaRb*pi/180)\n",
+ "\n",
+ "Ra=sqrt(pow(Ha,2)+pow(Va,2))\n",
+ "\n",
+ "print \"RA=\",round(Ra,4),\"KN\"\n",
+ "\n",
+ "alpha=(atan(Va/Ha))*180/pi\n",
+ "\n",
+ "print \"alpha=\",round(alpha,2),\"°\"\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "# example 9.3 page number 288\n"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 3,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "RA= 91.6503 KN\n",
+ "HB= 42.4264 KN\n",
+ "VB= 90.7761 KN\n",
+ "RB= 100.2013 KN\n",
+ "alpha= 64.95 °\n"
+ ]
+ }
+ ],
+ "source": [
+ "from math import pi,atan,sqrt,cos,sin\n",
+ "\n",
+ "#variable declaration\n",
+ "\n",
+ "#summation of all horizontal forces is zero & vertical forces is zero.\n",
+ "P1=float(20) #vertical down Load at 2m from A,KN\n",
+ "P2=float(30) #uniform distributed load from 2m to 6m from A,KN/m(in 4m of span)\n",
+ "P3=float(60) #Inclined down to right Load at angle 45° at 7m from A,KN\n",
+ "\n",
+ "theta3=45\n",
+ "#horizontal,vertical component at B is Hb,Vb respectively.\n",
+ "\n",
+ "Ra=(P1*7+P2*4*5+P3*2*sin(theta3*pi/180))/(9) #reaction at B point,KN\n",
+ "\n",
+ "print \"RA=\",round(Ra,4),\"KN\"\n",
+ "\n",
+ "Hb=P3*cos(theta3*pi/180)\n",
+ "print \"HB=\",round(Hb,4),\"KN\"\n",
+ "#now vertical component\n",
+ "Vb=P1+P2*4+P3*sin(theta3*pi/180)-Ra\n",
+ "print \"VB=\",round(Vb,4),\"KN\"\n",
+ "\n",
+ "Rb=sqrt(pow(Hb,2)+pow(Vb,2))\n",
+ "\n",
+ "print \"RB=\",round(Rb,4),\"KN\"\n",
+ "\n",
+ "alpha=(atan(Vb/Hb))*180/pi\n",
+ "\n",
+ "print \"alpha=\",round(alpha,2),\"°\""
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "# example 9.4 page number 288"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 4,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "no horizontal force HA=0\n",
+ "VA= 74.0 KN\n",
+ "MA= 148.0 KN-m\n"
+ ]
+ }
+ ],
+ "source": [
+ "#variable declaration\n",
+ "#Let the reactions at A be Ha, Va and Ma\n",
+ "#summation of all horizontal forces is zero & vertical forces is zero.\n",
+ "\n",
+ "P1=float(20) #vertical down Load at 2m from A,KN\n",
+ "P2=float(12) #vertical down Load at 3m from A,KN \n",
+ "P3=float(10) #vertical down Load at 4m from A,KN\n",
+ "Pu=float(16) #uniform distributed load from A to 2m from A,KN/m(in 2m of span)\n",
+ "##horizontal,vertical component at A is Ha,Va respectively.\n",
+ "print\"no horizontal force \",\"HA=0\"\n",
+ "Va=Pu*2+P1+P2+P3\n",
+ "print \"VA=\", round(Va,2),\"KN\"\n",
+ "Ma=Pu*2*1+P1*2+P2*3+P3*4\n",
+ "print \"MA=\", round(Ma,2),\"KN-m\"\n",
+ "\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "# example 9.5 page number 288\n"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 5,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "no horizontal force HA=0\n",
+ "VA= 65.0 KN\n",
+ "MA= 165.0 KN-m\n"
+ ]
+ }
+ ],
+ "source": [
+ "#variable declaration\n",
+ "#Let the reactions at A be Va and Ma\n",
+ "#summation of all horizontal forces is zero & vertical forces is zero.\n",
+ "\n",
+ "P1=float(15) #vertical down Load at 3m from A,KN\n",
+ "P2=float(10) #vertical down Load at 5m from A,KN \n",
+ "M=float(30) #CW moment at 4m distance from A, KN-m\n",
+ "Pu=float(20) #uniform distributed load from A to 2m from A,KN/m(in 2m of span)\n",
+ "##horizontal,vertical component at A is Ha,Va respectively.\n",
+ "print\"no horizontal force \",\"HA=0\"\n",
+ "Va=Pu*2+P1+P2\n",
+ "print \"VA=\", round(Va,2),\"KN\"\n",
+ "Ma=Pu*2*1+P1*3+P2*5+M\n",
+ "print \"MA=\", round(Ma,2),\"KN-m\"\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "# example 9.6 page number 289"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 6,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "RB= 100.0 KN\n",
+ "RA= 30.0 KN\n"
+ ]
+ }
+ ],
+ "source": [
+ "#variable declaration\n",
+ "\n",
+ "#As supports A and B are simple supports and loading is only in vertical direction, the reactions RA and RB are in vertical directions only.\n",
+ "\n",
+ "#summation of all horizontal forces is zero & vertical forces is zero.\n",
+ "\n",
+ "P1=float(30) #vertical down Load at 1m from A,KN\n",
+ "P2=float(40) #vertical down Load at 6.5m from A,KN \n",
+ "Pu=float(20) #uniform distributed load from 2m to 5m from A,KN/m(in 3m of span).\n",
+ "\n",
+ "Rb=(Pu*3*3.5+P1*1+P2*6.5)/5\n",
+ "print \"RB=\", round(Rb,2),\"KN\"\n",
+ "Ra=Pu*3+P1+P2-Rb\n",
+ "print \"RA=\", round(Ra,2),\"KN\""
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "# example 9.7 page number 289\n"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 7,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "no horizontal force HA=0\n",
+ "VB= 50.0 KN\n",
+ "VA= 70.0 KN\n"
+ ]
+ }
+ ],
+ "source": [
+ "#variable declaration\n",
+ "#Let the reactions at A be Va and Ma.\n",
+ "#summation of all horizontal forces is zero & vertical forces is zero.\n",
+ "\n",
+ "P1=float(60) #vertical down Load at 4m from A to right,KN\n",
+ "P2=float(20) #vertical down Load at 11m from A to right,KN \n",
+ "M=float(30) #CW moment at 7m distance from A to right, KN-m\n",
+ "Pu=float(20) #uniform distributed load from A to 2m from A to left ,KN/m(in 2m of span)\n",
+ "##horizontal,vertical component at A is Ha,Va respectively.\n",
+ "print\"no horizontal force \",\"HA=0\"\n",
+ "Vb=(-Pu*2*1+P1*4+P2*11+M)/9\n",
+ "print \"VB=\", round(Vb,2),\"KN\"\n",
+ "Va=Pu*2+P1+P2-Vb\n",
+ "print \"VA=\", round(Va,2),\"KN\"\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "# example 9.8 page number 290\n"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 8,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "RB= 71.011 KN\n",
+ "(Negative sign show that the assumed direction of VA is wrong. In other words, VA is acting vertically downwards). \n",
+ "RA= 23.3666 KN\n",
+ "alpha= 24.79 °\n"
+ ]
+ }
+ ],
+ "source": [
+ "from math import pi,atan,sqrt,cos,sin\n",
+ "\n",
+ "#variable declaration\n",
+ "\n",
+ "#summation of all horizontal forces is zero & vertical forces is zero.\n",
+ " \n",
+ "P1=float(30) #Inclined down Load at angle 45° to left at 5m from A,KN\n",
+ "Pu=float(20) #uniformly distributed load from 6m to 8m from A ,KN,(2m of span)\n",
+ "theta1=45\n",
+ "M=40 #ACW moment at 3m from A, KN-m\n",
+ "#horizontal,vertical component at A is Ha,Va respectively.\n",
+ "\n",
+ "Rb=(M+P1*5*sin(theta1*pi/180)+Pu*2*7)/6 #reaction at B point,KN\n",
+ "\n",
+ "print \"RB=\",round(Rb,4),\"KN\"\n",
+ "\n",
+ "Ha=P1*cos(theta1*pi/180)\n",
+ "\n",
+ "#now vertical component\n",
+ "Va=P1*sin(theta1*pi/180)-Rb+Pu*2\n",
+ "\n",
+ "Ra=sqrt(pow(Ha,2)+pow(Va,2))\n",
+ "\n",
+ "print \"(Negative sign show that the assumed direction of VA is wrong. In other words, VA is acting vertically downwards). \"\n",
+ "\n",
+ "Va1=-1*Va\n",
+ "print \"RA=\",round(Ra,4),\"KN\"\n",
+ "\n",
+ "alpha=(atan(Va1/Ha))*180/pi\n",
+ "\n",
+ "print \"alpha=\",round(alpha,2),\"°\""
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "# example 9.9 page number 290\n"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 9,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "X= 5.0 m\n"
+ ]
+ }
+ ],
+ "source": [
+ "#variable declaration\n",
+ "\n",
+ "#summation of all horizontal forces is zero & vertical forces is zero.\n",
+ " \n",
+ "#Let the left support C be at a distance x metres from A. \n",
+ "\n",
+ "P1=float(30) #vertical down load at A,KN\n",
+ "Pu=float(6) #uniform distributed load over whole span,KN/m,(20m of span)\n",
+ "P2=float(50) #vertical down load at B, KN\n",
+ "\n",
+ "#Rc=Rd(given) reaction at C & D is equal.\n",
+ "\n",
+ "Rc=(P1+P2+Pu*20)/2\n",
+ "Rd=Rc\n",
+ "\n",
+ "#taking moment at A \n",
+ "x=(((Pu*20*10+P2*20)/100)-12)/2\n",
+ "\n",
+ "print \"X=\", round(x,2),\"m\"\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"
+ }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}