From d36fc3b8f88cc3108ffff6151e376b619b9abb01 Mon Sep 17 00:00:00 2001 From: kinitrupti Date: Fri, 12 May 2017 18:40:35 +0530 Subject: Revised list of TBCs --- .../chapter03.ipynb | 598 --------------------- 1 file changed, 598 deletions(-) delete mode 100755 Engineering_Mechanics_by_Tayal_A.K./chapter03.ipynb (limited to 'Engineering_Mechanics_by_Tayal_A.K./chapter03.ipynb') diff --git a/Engineering_Mechanics_by_Tayal_A.K./chapter03.ipynb b/Engineering_Mechanics_by_Tayal_A.K./chapter03.ipynb deleted file mode 100755 index cdcc5a68..00000000 --- a/Engineering_Mechanics_by_Tayal_A.K./chapter03.ipynb +++ /dev/null @@ -1,598 +0,0 @@ -{ - "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