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| author | Thomas Stephen Lee | 2015-08-28 16:53:23 +0530 |
|---|---|---|
| committer | Thomas Stephen Lee | 2015-08-28 16:53:23 +0530 |
| commit | db0855dbeb41ecb8a51dde8587d43e5d7e83620f (patch) | |
| tree | b95975d958cba9af36cb1680e3f77205354f6512 /Engineering_Mechanics_by_Tayal_A.K./chapter03_3.ipynb | |
| parent | 5a86a20b9de487553d4ef88719fb0fd76a5dd6a7 (diff) | |
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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": {}
+ }
+ ]
+}
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