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  {
   "cells": [
    {
     "cell_type": "heading",
     "level": 1,
     "metadata": {},
     "source": [
      "Chapter 7 : Forces in beams and cable"
     ]
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 7.1  Page No : 335"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math \n",
      "\n",
      "# Given Data\n",
      "P = 2400.;\t\t\t#N, Vertical Force applied at D\n",
      "AB = 2.7;\t\t\t#m, perpendicular dismath.tance between A and B\n",
      "BE = 2.7;\t\t\t#m, perpendicular dismath.tance between E and B\n",
      "BK = 1.5;\t\t\t#m, perpendicular dismath.tance between B and K\n",
      "AJ = 1.2;\t\t\t#m, perpendicular dismath.tance between A and J\n",
      "EF = 4.8;\t\t\t#m, perpendicular dismath.tance between E and F\n",
      "BD = 3.6;\t\t\t#m, perpendicular dismath.tance between D and B\n",
      "#For entire truss\n",
      "#By free body diagram we get the force at A, B , c\n",
      "A = 1800.;\t\t\t#N\n",
      "B = 1200.;\t\t\t#N\n",
      "C = 3600.;\t\t\t#N\n",
      "\n",
      "# Calculations and Results\n",
      "alpha = math.degrees(math.atan(EF/(AB+BE)));\t\t\t#rad\n",
      "#a. Internal forces at j\n",
      "#Applying sum(M_J) = 0\n",
      "M = A*AJ;\t\t\t#N.m,Couple on member ACF at J\n",
      "#Applying sum(Fx) = 0\n",
      "F = A*math.cos(math.radians(alpha));\t\t\t#N, Axial force at J\n",
      "#Applying sum(Fy) = 0\n",
      "V = A*math.sin(math.radians(alpha));\t\t\t#N, shearing force at J\n",
      "print \"Thus, Internal forces at J are equivalent to  Couple M  =  %.0f N.m  \\\n",
      "\\nAxial force F =  %.0f N  \\\n",
      "\\nShearing force V =  %.0f N\"%(M,F,V);\n",
      "\n",
      "#a. Internal forces at K\n",
      "#Applying sum(M_K) = 0\n",
      "M = B*BK;\t\t\t#N.m,Couple on frame\n",
      "#Applying sum(Fx) = 0\n",
      "F = 0;\t\t\t#N, Axial force at J\n",
      "#Applying sum(Fy) = 0\n",
      "V = -B;\t\t\t#N, shearing force at J\n",
      "print \"Thus, Internal forces at K are equivalent to  Couple M  =  %.0f N.m  \\\n",
      "\\nAxial force F =  %.0f N  \\\n",
      "\\nShearing force V =  %.0f N\"%(M,F,V);\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Thus, Internal forces at J are equivalent to  Couple M  =  2160 N.m  \n",
        "Axial force F =  1345 N  \n",
        "Shearing force V =  1196 N\n",
        "Thus, Internal forces at K are equivalent to  Couple M  =  1800 N.m  \n",
        "Axial force F =  0 N  \n",
        "Shearing force V =  -1200 N\n"
       ]
      }
     ],
     "prompt_number": 4
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 7.2  Page No : 344"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "%matplotlib inline\n",
      "import math \n",
      "from matplotlib.pyplot import plot,suptitle,xlabel,ylabel\n",
      "#Drawing of shear and bending moment diagram\n",
      "print \"Given problem is for drawing diagram, this diagram is drawn by step by step manner. \"\n",
      "\n",
      "# Given Data\n",
      "F_A = -20.;\t\t\t#kN, force applied at A\n",
      "F_C = -40.;\t\t\t#kN, force applied at C\n",
      "AB = 2.5;\t\t\t#m, perpendicular dismath.tance between A and B\n",
      "BC = 3.;\t\t\t#m, perpendicular dismath.tance between C and B\n",
      "CD = 2.;\t\t\t#m, perpendicular dismath.tance between C and D\n",
      "#By free body of entire beam\n",
      "#By sum(m_D) = 0\n",
      "\n",
      "# Calculations and Results\n",
      "R_B = -(CD*F_C+(AB+BC+CD)*F_A)/(BC+CD);\t\t\t#kN, Reaction atB\n",
      "#By sum(m_A) = 0\n",
      "R_D = -(BC*F_C-(AB)*F_A)/(BC+CD);\t\t\t#kN, Reaction atB\n",
      "#For section 1\n",
      "#Applying sum(Fy) = 0\n",
      "V1 = F_A;\t\t\t#kN\n",
      "#Applying sum(M1) = 0\n",
      "M1 = V1*0;\t\t\t#kN.m\n",
      "\n",
      "#For section 2\n",
      "#Applying sum(Fy) = 0\n",
      "V2 = F_A;\t\t\t#kN\n",
      "#Applying sum(M1) = 0\n",
      "M2 = F_A*AB;\t\t\t#kN.m\n",
      "\n",
      "#For section 3\n",
      "#Applying sum(Fy) = 0\n",
      "V3 = R_B+F_A;\t\t\t#kN\n",
      "#Applying sum(M1) = 0\n",
      "M3 = F_A*AB;\t\t\t#kN.m\n",
      "\n",
      "#For section 4\n",
      "#Applying sum(Fy) = 0\n",
      "V4 = R_B+F_A;\t\t\t#kN\n",
      "#Applying sum(M1) = 0\n",
      "M4 = F_A*(AB+BC)+R_B*BC \t\t\t#kN.m\n",
      "\n",
      "#For section 5\n",
      "#Applying sum(Fy) = 0\n",
      "V5 = R_B+F_A+F_C;\t\t\t#kN\n",
      "#Applying sum(M1) = 0\n",
      "M5 = F_A*(AB+BC)+R_B*BC\t\t\t#kN.m\n",
      "\n",
      "#For section 6\n",
      "#Applying sum(Fy) = 0\n",
      "V6 = R_B+F_A+F_C;\t\t\t#kN\n",
      "#Applying sum(M1) = 0\n",
      "M6 = V6*0\t\t\t#kN.m\n",
      "X = [0,2.5,2.5,5.5,5.5,7.5]\n",
      "\n",
      "V = [V1,V2,V3,V4,V5,V6];\t\t\t#Shear matrix\n",
      "M = [M1,M2,M3,M4,M5,M6];\t\t\t#Bending moment matrix\n",
      "plot(X,V);\t\t\t#Shear diagram\n",
      "plot(X,M,'r');\t\t\t#Bending moment diagram\n",
      "suptitle( 'Shear and bending moment diagram')\n",
      "xlabel('X axis')\n",
      "ylabel( 'Y axis') ;\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Given problem is for drawing diagram, this diagram is drawn by step by step manner. \n"
       ]
      },
      {
       "metadata": {},
       "output_type": "display_data",
       "png": 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Ko4+CRPMXimP8eBg0CObOdR2JSNEpKQSN6guF1dgIEyfaKqi9e7uORqToVFMI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\n",
       "text": [
        "<matplotlib.figure.Figure at 0x10ef97650>"
       ]
      }
     ],
     "prompt_number": 7
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 7.3  Page No : 345"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "%matplotlib inline\n",
      "import math \n",
      "from matplotlib.pyplot import plot,suptitle,xlabel,ylabel\n",
      "\n",
      "#Drawing of shear and bending moment diagram\n",
      "#Values taken in N and m instead of lb and in\n",
      "print \"Given problem is for drawing diagram%( this diagram is drawn by step by step manner. \"\n",
      "\n",
      "# Given Data\n",
      "F_AC = 40.;\t\t\t#lb/in, distributed load applied at A to C\n",
      "F_E = 400.;\t\t\t#lb, force applied at E\n",
      "AC = 12.;\t\t\t#in, perpendicular dismath.tance between A and B\n",
      "CD = 6.;\t\t\t#in, perpendicular dismath.tance between C and D\n",
      "DE = 4.;\t\t\t#in, perpendicular dismath.tance between E and D\n",
      "EB = 10.;\t\t\t#in, perpendicular dismath.tance between E and B\n",
      "AB = 32.;\t\t\t#in, length of beam AB\n",
      "\n",
      "# Calculations and Results\n",
      "F = F_AC*AC;\t\t\t#N, Force due to districuted load at AC/2\n",
      "#By free body of entire beam\n",
      "#By sum(m_A) = 0\n",
      "By = (F*(AC/2)+F_E*(AC+CD+DE))/AB;\t\t\t#N,Y componet of Reaction at B\n",
      "#By sum(m_B) = 0\n",
      "#print (By)\n",
      "A = (F*(AB-AC/2)+F_E*EB)/AB;\t\t\t#N, Reaction at A\n",
      "#by sum(Fx) = 0\n",
      "#print (A)\n",
      "Bx = 0;\t\t\t#N, xcomponent of rection at B\n",
      "#Diagrams\n",
      "#For section A to C\n",
      "#Applying sum(Fy) = 0\n",
      "\n",
      "i = 0;\n",
      "X = []\n",
      "V = []\n",
      "M = []\n",
      "for x in range(0,13,2):\n",
      "    i = i+1;\n",
      "    X.append(x);\n",
      "    V.append(A-F*x);\t\t\t#N\n",
      "    #Applying sum(M1) = 0\n",
      "    M.append(A*x-F/2*x**2);\t\t\t#N.m\n",
      "\n",
      "#For section Cto D\n",
      "#Applying sum(Fy) = 0\n",
      "for x in range(12,19,2):\n",
      "    i = i+1;\n",
      "    X.append(x);\n",
      "    V.append(A-F);\t\t\t#N\n",
      "    #Applying sum(M1) = 0\n",
      "    M.append( A*x-F*(x-0.15));\t\t\t#N.m\n",
      "\n",
      "for x in range(18,33,2):\n",
      "    i = i+1;\n",
      "    X.append(x);\n",
      "    #Applying sum(Fy) = 0\n",
      "    V.append(A-F-F_E);\t\t\t#N\n",
      "    #Applying sum(M1) = 0\n",
      "    M.append(A*x-F*(x-0.15)+F_E*DE-F_E*(x-0.045));\t\t\t#N.m\n",
      "\n",
      "plot(X,V,'r');\t\t\t#Shear diagram\n",
      "plot(X,M,'-');\t\t\t#Bending moment diagram\n",
      "suptitle( 'Shear and bending moment diagram')\n",
      "xlabel('X axis')\n",
      "ylabel('Y axis') ;\n",
      "\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Given problem is for drawing diagram%( this diagram is drawn by step by step manner. \n"
       ]
      },
      {
       "metadata": {},
       "output_type": "display_data",
       "png": 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ALKdyk382xq6q5rnHY4FebrsH8KzbngCc57YvBKZhCWYbMB1LUpJChxwCvXvD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\nAElFTkSuQmCC\n",
       "text": [
        "<matplotlib.figure.Figure at 0x1090b1690>"
       ]
      }
     ],
     "prompt_number": 9
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 7.4  Page No : 346"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "%matplotlib inline\n",
      "import math \n",
      "from matplotlib.pyplot import plot\n",
      "\n",
      "#Drawing of shear and bending moment diagram\n",
      "# Given Data\n",
      "print \"Given problem is for drawing diagram%( this diagram is drawn by step by step manner. \"\n",
      "F_B = 500.;\t\t\t#N, force applied at B\n",
      "F_C = 500.;\t\t\t#N, force applied at C.\n",
      "F_DE = 2400.;\t\t\t#N/m, distributed load applied at D to E\n",
      "AB = 0.4;\t\t\t#m, perpendicular dismath.tance between A and B\n",
      "BC = 0.4;\t\t\t#m, perpendicular dismath.tance between C and B\n",
      "CD = 0.4;\t\t\t#m, perpendicular dismath.tance between C and D\n",
      "DE = 0.3;\t\t\t#m, perpendicular dismath.tance between E and D\n",
      "F_E = F_DE*DE;\t\t\t#N, force exerted at DE/2 from E\n",
      "\n",
      "#By free body of entire beam\n",
      "#By sum(m_D) = 0\n",
      "A = (CD*F_C+(BC+CD)*F_B-F_E*DE/2)/(AB+BC+CD);\t\t\t#N, Reaction at A\n",
      "#By sum(Fy) = 0\n",
      "Dy = F_C+F_B+F_E-A;\t\t\t#N,Y component of  Reaction at D\n",
      "#By sum(Fx) = 0\n",
      "Dx = 0;\t\t\t#N,Y component of  Reaction at D\n",
      "#For section 1\n",
      "#Applying sum(Fy) = 0\n",
      "V1 = A;\t\t\t#N, shear force from A to B\n",
      "\n",
      "#For section 2\n",
      "#Applying sum(Fy) = 0\n",
      "V2 = A-F_B;\t\t\t#N, shear force from B to C\n",
      "\n",
      "#For section 3\n",
      "#Applying sum(Fy) = 0\n",
      "V3 = A-F_B-F_C;\t\t\t#N, shear force  from C to D\n",
      "\n",
      "#For section 4\n",
      "#Applying sum(Fy) = 0\n",
      "V4 = A-F_B-F_C+Dy;\t\t\t#N, shear force At D\n",
      "\n",
      "#For section 5\n",
      "#Applying sum(Fy) = 0\n",
      "V5 = 0;\t\t\t#N, shear force at A\n",
      "#Area under bending curve is change in bending moment of that 2 points\n",
      "MA = 0;\t\t\t#N.m\n",
      "MB = MA+V1*AB;\t\t\t#N.m\n",
      "MC = MB+V2*BC;\t\t\t#N.m\n",
      "MD = MC+V3*CD;\t\t\t#N.m\n",
      "ME = MD+1./2*V4*AB;\t\t\t#N.m\n",
      "\n",
      "\n",
      "X = [0,0.4,0.4,0.8,0.8,1.2,1.2,1.5];\n",
      "V = [V1,V1,V2,V2,V3,V3,V4,V5];\t\t\t#Shear matrix,\n",
      "\n",
      "plot(X,V);\t\t\t#Shear diagram\n",
      "X = [0,AB,AB+BC,AB+BC+CD,AB+BC+CD+DE];\n",
      "M = [MA,MB,MC,MD,ME];\t\t\t#Bending moment matrix\n",
      "plot(X,M,'r');\t\t\t#Bending moment diagram\n",
      "suptitle( 'Shear and bending moment diagram')\n",
      "xlabel('X axis')\n",
      "ylabel('Y axis') ;\n",
      "\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Given problem is for drawing diagram%( this diagram is drawn by step by step manner. \n"
       ]
      },
      {
       "metadata": {},
       "output_type": "display_data",
       "png": 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       "text": [
        "<matplotlib.figure.Figure at 0x10efd1a50>"
       ]
      }
     ],
     "prompt_number": 12
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 7.5  Page No : 347"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "%matplotlib inline\n",
      "import math \n",
      "from matplotlib.pyplot import plot\n",
      "#Drawing of shear and bending moment diagram\n",
      "\n",
      "# Given Data\n",
      "w = 20.;\t\t\t#kN/m, distributed load applied at D to E\n",
      "AB = 6.;\t\t\t#m, perpendicular dismath.tance between A and B\n",
      "BC = 3.;\t\t\t#m, perpendicular dismath.tance between C and B\n",
      "\n",
      "# Calculations and Results\n",
      "F_B = w*AB;\t\t\t#kN, force exerted at AB/2 from A\n",
      "\n",
      "#By free body of entire beam\n",
      "#By sum(m_C) = 0\n",
      "RA = (F_B*(AB/2+BC))/(AB+BC);\t\t\t#kN, Reaction at A\n",
      "\n",
      "#By sum(m_A) = 0\n",
      "RC = (F_B*(AB/2)/(AB+BC));\t\t\t#kN, Reaction at C\n",
      "\n",
      "#For section 1\n",
      "#Applying sum(Fy) = 0\n",
      "VA = RA;\t\t\t#N, shear force just to right to A\n",
      "\n",
      "#For section 2\n",
      "#Applying sum(Fy) = 0\n",
      "VB = VA-F_B;\t\t\t#kN, shear force just left to B\n",
      "\n",
      "#For section 3\n",
      "#Applying sum(Fy) = 0\n",
      "VC = VB;\t\t\t#kN, shear force  from B to C\n",
      "\n",
      "\n",
      "#Bending moment at each end is zero\n",
      "# Maximum bending moment is at D where V = 0\n",
      "VD = 0;\t\t\t#kN\n",
      "\n",
      "x = -(VD-VA)/w;\t\t\t#m, location of maximum bending moment\n",
      "print \"Maximum bending moment is at D x =  %.0f m from A\"%(x);\n",
      "MA = 0;\t\t\t#kN.m\n",
      "MD = MA+1/2*VA*x;\t\t\t#kN.m, maximum bending moment is at D\n",
      "MB = MD+1/2*VB*(AB-x);\t\t\t#N.m\n",
      "MC = MB+VB*BC;\t\t\t#N.m\n",
      "\n",
      "print \"Maximum bending moment is at MD =  %.0fkN. m from A\"%(MD);\n",
      "X = [0,x,AB,AB+BC];\t\t\t#m, \n",
      "V = [VA,VD,VB,VC];\t\t\t#kN,Shear matrix,\n",
      "\n",
      "plot(X,V);\t\t\t#Shear diagram\n",
      "X = [0,x,AB,AB+BC];\t\t\t#m\n",
      "M = [MA,MD,MB,MC];\t\t\t#kN.m,Bending moment matrix\n",
      "plot(X,M,'r');\t\t\t#Bending moment diagram\n",
      "suptitle( 'Shear and bending moment diagram')\n",
      "xlabel('X axis')\n",
      "ylabel('Y axis') ;\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Maximum bending moment is at D x =  4 m from A\n",
        "Maximum bending moment is at MD =  0kN. m from A\n"
       ]
      },
      {
       "metadata": {},
       "output_type": "display_data",
       "png": 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h2SdJqpJuFXzvO4Cd2tj/VeA37bzmb8AuwHJibOMWYO+KRCdJqil3s+Ggd3uP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\nYII=\n",
       "text": [
        "<matplotlib.figure.Figure at 0x10a6f7550>"
       ]
      }
     ],
     "prompt_number": 13
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 7.8  Page No : 348"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math \n",
      "from numpy.linalg import solve\n",
      "from numpy import array\n",
      "\n",
      "# given data\n",
      "F_B = 30.;\t\t\t#kN, Vertical Force applied at B\n",
      "F_C = 60.;\t\t\t#kN, Vertical Force applied at C\n",
      "F_D = 20.;\t\t\t#kN, Vertical Force applied at D\n",
      "AB = 6.;\t\t\t#m, perpendicular dismath.tance between A and B\n",
      "BC = 3.;\t\t\t#m, perpendicular dismath.tance between C and B\n",
      "CD = 4.5;\t\t\t#m, perpendicular dismath.tance between c and D\n",
      "DE = 4.5;\t\t\t#m, perpendicular dismath.tance between D and E\n",
      "AE = 6.;\t\t\t#m, vertical perpendicular dismath.tance between A and E\n",
      "AC = 1.5;\t\t\t#m, vertical perpendicular dismath.tance between A and C\n",
      "#For entire cable\n",
      "#Sum(M_E) = 0, AB*Ax-Ay*(AB+BC+CD+DE)+F_B*(BC+CD+DE)+F_C*(CD+DE)+F_D*(DE) = 0\n",
      "\n",
      "# Calculations and Results\n",
      "#Free body ABC \n",
      "#Sum(M_c) = 0 gives -Ax*AC-Ay*(AB+BC)+F_B*BC = 0\n",
      "#we get 2 equations in Ax and Ay\n",
      "A = [[AB,-(AB+BC+CD+DE)],[-AC,-(AB+BC)]];\t\t\t#Matrix of coeficients\n",
      "B = [[-(F_B*(BC+CD+DE)+F_C*(CD+DE)+F_D*(DE))],[-F_B*BC]]; \n",
      "X = solve(A,-array(B));\t\t\t#kN, Solution matrix\n",
      "Ax = X[0];\t\t\t#kN, X component of reaction at A\n",
      "Ay = X[1];\t\t\t#kN, Y component of reaction at A\n",
      "\n",
      "#a. Elevation of points B and D\n",
      "#Free body AB\n",
      "#sum(M_B) = 0\n",
      "yB = -Ay*AB/Ax;\t\t\t#m, below A\n",
      "print \"Elevation of point B is %.2f m below A\"%(yB);\n",
      "#free body ABCD\n",
      "#sum(M_D) = 0\n",
      "yD = (Ay*(AB+BC+CD)-F_B*(BC+CD)-F_C*CD)/Ax;\t\t\t#m, above A\n",
      "print \"Elevation of point D is %.2f m above A\"%(yD);\n",
      "\n",
      "#Maximum slope and maximum tension\n",
      "theta = math.atan((AE-yD)/DE);\t\t\t#rad \n",
      "Tmax = -Ax/math.cos(theta);\t\t\t#kN, maximum tension\n",
      "theta = theta/math.pi*180;\t\t\t#degree\n",
      "\n",
      "print \"Maximum slope is theta =  %.1f degree and maximum tension in the cable is Tmax =  %.1f kN \"%(theta,Tmax);"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Elevation of point B is 1.67 m below A\n",
        "Elevation of point D is -9.25 m above A\n",
        "Maximum slope is theta =  73.6 degree and maximum tension in the cable is Tmax =  -318.0 kN \n"
       ]
      }
     ],
     "prompt_number": 15
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 7.9  Page No : 349"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math \n",
      "\n",
      "# Given Data\n",
      "yB = 0.5;\t\t\t#m, sag of the cable\n",
      "m = 0.75;\t\t\t#kg/m, mass per unit length\n",
      "g = 9.81;\t\t\t#m/s**2, acceleration due to gravity\n",
      "AB = 40.;\t\t\t#m, dismath.tance AB\n",
      "\n",
      "# Calculations and Results\n",
      "#a. Load P\n",
      "w = m*g;\t\t\t#N/m , Load per unit length\n",
      "xB = AB/2;\t\t\t#m, dismath.tance CB\n",
      "W = w*xB;\t\t\t#N, applied at halfway of CB\n",
      "\n",
      "#Summing moments about B\n",
      "#sum(M_B) = 0\n",
      "To = W*xB/2/yB;\t\t\t#N\n",
      "#from force triangle\n",
      "TB = math.sqrt(To**2+W**2);\t\t\t#N,  = P, as tension on each side is same\n",
      "print \"Magnitude of load P =  %.0f N \"%(TB);\n",
      "#slope of cable at B\n",
      "theta = math.atan(W/To);\t\t\t#rad\n",
      "theta = theta*180/math.pi;\t\t\t#degree, conversion to degree\n",
      "print \"Slope of cable at B is theta =  %.1f degree\"%(theta);\n",
      "#length of cable\n",
      "#applying eq. 7.10\n",
      "sB = xB*(1+2./3*(yB/xB)**2);\t\t\t#m\n",
      "\n",
      "print \"Total length of cable from A to B is Length =  %.4f m\"%(2*sB);\n",
      "\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Magnitude of load P =  2947 N \n",
        "Slope of cable at B is theta =  2.9 degree\n",
        "Total length of cable from A to B is Length =  40.0167 m\n"
       ]
      }
     ],
     "prompt_number": 16
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 7.10  Page No : 350"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math \n",
      "\n",
      "# Given Data\n",
      "AB = 150.;\t\t\t#m, dismath.tance AB\n",
      "s = 30.;\t\t\t#m, sag of cable\n",
      "w = 45.;\t\t\t#N/m Uniform weigth per unit length of cable\n",
      "\n",
      "#Equation of cable, by 7.16\n",
      "#Coordinates of B\n",
      "\n",
      "xB = AB/2;\t\t\t#m\n",
      "C = [99,105,98.4,90];\t\t\t#trial values\n",
      "\n",
      "# Calculations and Results\n",
      "for i in range(4):\n",
      "    if ((30/C[i]+1)-math.cosh(xB/C[i]))<0.0001:\n",
      "        c = C[i];\n",
      "        break;\n",
      "\n",
      "yB = s+c;\t\t\t#m\n",
      "\n",
      "#Maximum and minimum values of tension\n",
      "Tmin = w*c;\t\t\t#N, To\n",
      "Tmax = w*yB;\t\t\t#N TB\n",
      "print \"Minimum value of tension in cable is Tmin =  %.0f N\"%(Tmin);\n",
      "print \"Maximum value of tension in cable is Tmax =  %.0f N\"%(Tmax);\n",
      "#Length of cable\n",
      "\n",
      "S_CB = math.sqrt(yB**2-c**2);\t\t\t#m, one halph length by 7.17\n",
      "S_AB = 2*S_CB;\t\t\t#m, full length of cable\n",
      "\n",
      "print \"Fulllength of cable is s_AB =  %.0f m\"%(S_AB);\n",
      "\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Minimum value of tension in cable is Tmin =  4455 N\n",
        "Maximum value of tension in cable is Tmax =  5805 N\n",
        "Fulllength of cable is s_AB =  165 m\n"
       ]
      }
     ],
     "prompt_number": 17
    }
   ],
   "metadata": {}
  }
 ]
}