path: root/Mechanics_of_Materials_by_R._C._Hibbeler/Chapter1.ipynb

{
 "metadata": {
  "name": ""
 },
 "nbformat": 3,
 "nbformat_minor": 0,
 "worksheets": [
  {
   "cells": [
    {
     "cell_type": "heading",
     "level": 1,
     "metadata": {},
     "source": [
      "Chapter 1:Equilibrium of a deformable body"
     ]
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 1.1 Page no 11"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Given\n",
      "w_varying = 270.0     #N/m, torque\n",
      "l_crossection = 9.0   #m , length\n",
      "l_cb = 6.0            #m length\n",
      "l_ac = 2.0             #m length\n",
      "\n",
      "#Calculation\n",
      "w_c = (w_varying/l_crossection) * l_cb #By proportion, load at C is found.\n",
      "f_resultant_c = 0.5* w_c *l_cb \n",
      "#Balancing forces in the x direction:\n",
      "n_c = 0\n",
      "#Balncing forces in the y direction:\n",
      "v_c = f_resultant_c\n",
      "# Balncing the moments about C:\n",
      "m_c = - (f_resultant_c*l_ac)\n",
      "\n",
      "#Results\n",
      "print\"The resultant force at C  is  \",f_resultant_c,\"N\"\n",
      "print\"The horizontal force at C is  \",n_c,\"N\"\n",
      "print\"The vertical force at C   is  \",v_c,\"N\"\n",
      "print\"The moment about C        is  \",m_c,\"Nm\"\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The resultant force at C  is   540.0 N\n",
        "The horizontal force at C is   0 N\n",
        "The vertical force at C   is   540.0 N\n",
        "The moment about C        is   -1080.0 Nm\n"
       ]
      }
     ],
     "prompt_number": 4
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 1.2 Page no 12"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Given\n",
      "f_d = 225         #N\n",
      "w_uniform = 800   # N/m\n",
      "l_ac = 0.200      #m\n",
      "l_cb = 0.05+0.1   #m\n",
      "l_bd = 0.100      #m\n",
      "l_bearing = 0.05    #m\n",
      "f_resultant = w_uniform*l_cb #120N\n",
      "l_f_resultant_b = (l_cb/2)+ l_bearing #0.125m\n",
      "l = l_ac + l_cb + l_bearing + l_bd  \n",
      "\n",
      "\n",
      "#Calculation\n",
      "m_b = 0                                                        # Net moment about B is zero for equilibrium \n",
      "a_y = -((f_d*l_bd) - (f_resultant*l_f_resultant_b))/ (l - l_bd) # finding the reaction force at A\n",
      "\n",
      "# Refer to the free body diagram in Fig.1-5c.\n",
      "f_c = 40                        #N\n",
      "#Balancing forces in the x direction\n",
      "n_c = 0\n",
      "#Balncing forces in the y direction\n",
      "v_c = a_y - f_c            #-18.75N - 40N-Vc = 0\n",
      "# Balncing the moments about C\n",
      "m_c =  ((a_y * (l_ac + 0.05)) - f_c*(0.025) ) # Mc+40N(0.025m)+ 18.75N(0.250m) = 0\n",
      "\n",
      "\n",
      "# Result\n",
      "print'The horizontal force at C =',n_c,\"N\"\n",
      "print'The vertical force at C   =',v_c,\"N\"\n",
      "print'The moment about C        =',round(m_c,2),\"NM\"\n",
      "\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The horizontal force at C = 0 N\n",
        "The vertical force at C   = -58.75 N\n",
        "The moment about C        = -5.69 NM\n"
       ]
      }
     ],
     "prompt_number": 3
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 1.3 Page no 13"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "# Given\n",
      "l_ac = 2     #m.,length\n",
      "l_cb = 1     #m.\n",
      "l_ad = 1.5   #m.\n",
      "#r_a = 0.125  #m.\n",
      "#r_d = 0.125  #m.\n",
      "W = 500      # N, force\n",
      "g=9.81\n",
      "\n",
      "#Calculation\n",
      "#Balancing forces in the x direction\n",
      "n_c = (W*g*(l_ac+l_cb))/(3*2/5.0) # N\n",
      "#Balncing forces in the y direction\n",
      "v_c = n_c*(4/5.0)                 #N\n",
      "# Balncing the moments about C\n",
      "m_c = n_c*(3/5.0)-(W*g)\n",
      "\n",
      "\n",
      "# Result\n",
      "print'The horizontal force at C   =  ',-v_c/1000,\"KN\"\n",
      "print'The vertical force at C     =  ',-m_c/1000,\"KN\"\n",
      "print'The moment about C          =  ',-m_c/1000,\"KNm\"\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The horizontal force at C   =   -9.81 KN\n",
        "The vertical force at C     =   -2.4525 KN\n",
        "The moment about C          =   -2.4525 KNm\n"
       ]
      }
     ],
     "prompt_number": 21
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 1.4 Page no 14"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "# Given\n",
      "l_ag = 2           #Length of AG is 1m.\n",
      "l_gd = 1         #Length of GD is 1m.\n",
      "l_de = 3         #Length of DE is 1m.\n",
      "f_a = 1500       #Force at A is 1500N.\n",
      "l_ec = 1.5       #Length of EC is 1m.\n",
      "l = l_ag +l_gd +l_de\n",
      "w_uniform_varying = 600 #Nm.\n",
      "f_ba = 7750      #N\n",
      "f_bc = 6200      #N\n",
      "f_bd = 4650      #N\n",
      "\n",
      "#Calculation\n",
      "w_resultant = 0.5*l_de*w_uniform_varying\n",
      "# calling point of action of resultant as P\n",
      "l_ep = (2/3.0)*l_de \n",
      "l_ap = l - l_ep        \n",
      "\n",
      "#Free Body Diagram Using the result for Fba, the left section AG of the beam is shown in Fig 1-7d.\n",
      "# Equations of equilibrium\n",
      "#Balancing forces in the x direction\n",
      "n_g = -f_ba * (4/5.0)    # N\n",
      "#Balncing forces in the y direction\n",
      "v_g = -f_a + f_ba*(3/5.0) #N\n",
      "\n",
      "# Balncing the moments about C\n",
      "m_g = (f_ba * (3/5.0)*l_ag) - (f_a * l_ag) #Nm\n",
      "\n",
      "# Result\n",
      "print'The horizontal force at G =  ',n_g,\"lb\"\n",
      "print'The vertical force at G   =  ',v_g,\"lb\"\n",
      "print'The moment about G        =  ',m_g,\"lb-ft\"\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The horizontal force at G =   -6200.0 lb\n",
        "The vertical force at G   =   3150.0 lb\n",
        "The moment about G        =   6300.0 lb-ft\n"
       ]
      }
     ],
     "prompt_number": 31
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 1.5 Page no 15"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Given\n",
      "f_a = 50       #N, force\n",
      "m_a = 70       # Moment at A in Nm\n",
      "l_ad = 1.25    #Length of AD in m.\n",
      "l_bd = 0.5     #Length of BD in m.\n",
      "l_cb = 0.75    #Length of BC in m.\n",
      "w_l  = 2       #Kg/m\n",
      "g = 9.81       #N/kg- acceleration due to gravity\n",
      "\n",
      "#Free Body Diagram \n",
      "import math\n",
      "w_bd = w_l*l_bd*g #in N. \n",
      "w_ad = w_l*l_ad*g\n",
      "# Equations of Equilibrium\n",
      "#Balancing forces in the x, y and z direction\n",
      "f_b_x = 0     #  N\n",
      "f_b_y = 0     #N\n",
      "f_b_z = g + w_ad + f_a      #N\n",
      "# Balancing Moments in the x,y and z direction\n",
      "m_b_x = - m_a + (f_a*l_bd) + (w_ad*l_bd) + (l_bd/2.0)*g #Nm\n",
      "m_b_y = - (w_ad*(l_ad/2.0)) - (f_a*l_ad) #Nm\n",
      "m_b_z = 0                       #Nm\n",
      "v_b_shear = sqrt(f_b_z **2 + 0) #Shear Force in N\n",
      "t_b = - m_b_y                   #Torsional Moment in Nm\n",
      "m_b = math.sqrt(m_b_x **2+ 0)   # Bending moment in Nm\n",
      "\n",
      "# Result\n",
      "print' The weight of segment BD                =',w_bd,\"N\"\n",
      "print' The weight of segment AD                =',w_ad,\"N\"\n",
      "print' The force at B in the Z direction       =',f_b_z,\"N\"\n",
      "print' The moment about B in the X direction   =',m_b_x,\"Nm\"\n",
      "print' The moment about G in the Y direction   =',m_b_y,\"Nm\"\n",
      "print' The Shear Force at B                    =',v_b_shear,\"N\"\n",
      "print' The Torsional Moment at B               =',t_b,\"Nm\"\n",
      "print' The Bending Moment at B                 =',m_b,\"Nm\"\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " The weight of segment BD                = 9.81 N\n",
        " The weight of segment AD                = 24.525 N\n",
        " The force at B in the Z direction       = 84.335 N\n",
        " The moment about B in the X direction   = -30.285 Nm\n",
        " The moment about G in the Y direction   = -77.828125 Nm\n",
        " The Shear Force at B                    = 84.335 N\n",
        " The Torsional Moment at B               = 77.828125 Nm\n",
        " The Bending Moment at B                 = 30.285 Nm\n"
       ]
      }
     ],
     "prompt_number": 34
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 1.6 Page no 28"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Given\n",
      "netf_b = 18*(10 **3)      #N Net force at B.\n",
      "netf_c = 8*(10**3)        #N Net force at C.\n",
      "f_a = 12 *(10**3)         #N Force at A.\n",
      "f_d = 22* (10**3)         #N Force at D.\n",
      "w = 35.0                    #mm Width.\n",
      "t = 10.0                    #mm Thickness.\n",
      "\n",
      "#calculations\n",
      "p_bc = netf_b + f_a          #N Net force in region BC.\n",
      "a = w*t                      #m**2 The area of the cross section.\n",
      "avg_normal_stress = p_bc/a   #Average Normal Stress.\n",
      "\n",
      "# Result\n",
      "print'The Average Normal Stress in the bar when subjected to load = ',round(avg_normal_stress,1),\"MPa\"\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The Average Normal Stress in the bar when subjected to load =  85.7 MPa\n"
       ]
      }
     ],
     "prompt_number": 37
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 1.7 Page no: 29"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Given :\n",
      "m_lamp = 80   #Mass of lamp in Kg.\n",
      "d_ab = 10     # Diameter of AB in mm.\n",
      "d_bc = 8      # Diameter of BC in mm.\n",
      "ab_h = 60 *(math.pi/180.0) # In degrees - Angle made by AB with the horizontal.\n",
      "w = m_lamp*9.81 #N\n",
      "a_bc = (math.pi/4.0)*(d_bc**2) #m**2 Area of cross section of rod BC\n",
      "a_ab = (math.pi/4.0)*(d_ab**2) #m**2 Area of cross section of rod AB\n",
      "\n",
      "\n",
      "#calculation\n",
      "import math   # Equations of equilibrium: Solving equilibrium equations simultaneously ,\n",
      "f_bc=395\n",
      "f_ab=f_bc*(4/5.0)/(math.cos(60*3.14/180.0))\n",
      "avg_normal_stress_a = f_ab / a_ab #Mpa Average Normal Stress in AB\n",
      "avg_normal_stress_c = f_bc/ a_bc# Mpa   Average Normal Stress in BC\n",
      "\n",
      "# Displaying results:\n",
      "print\"The Average Normal Stress in AB when subjected to load = MPa\",round(avg_normal_stress_a,2),\"MPa\"\n",
      "print\"The Average Normal Stress in BC when subjected to load = MPa\",round(avg_normal_stress_c,2),\"MPa\"\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The Average Normal Stress in AB when subjected to load = MPa 8.04 MPa\n",
        "The Average Normal Stress in BC when subjected to load = MPa 7.86 MPa\n"
       ]
      }
     ],
     "prompt_number": 15
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 1.8 Page no 30"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "#Given\n",
      "h_above_ab = 2.75        #ft , height\n",
      "h_below_ab = 0.2 \n",
      "d_a = 0.75               #diameter ft \n",
      "d_b = 0.1 \n",
      "sp_w = 490 \n",
      "\n",
      "# Equation of Equilibrium\n",
      "import math\n",
      "a = math.pi* (d_a**2)  # Area of cross section in m**2\n",
      "p = sp_w * h_above_ab * a\n",
      "avg_comp_stress = p/a # The average compressive stress in kN/m**2\n",
      "\n",
      "#Display\n",
      "print'The average compressive stress = ',avg_comp_stress,\"lb/ft**2\"\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The average compressive stress =  1347.5 lb/ft**2\n"
       ]
      }
     ],
     "prompt_number": 42
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 1.9 Page no 31"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Determine Distance of force from AB\n",
      "\n",
      "#Given \n",
      "f = 3000 #N Force acting at distance x from AB.\n",
      "l_ac = 200 #Length of AC in mm.\n",
      "a_ab = 400 #Cross sectional area of AB in mm**2.\n",
      "a_c = 650 # area of C in mm**2.\n",
      "\n",
      "#Calculation\n",
      "#Equations are\n",
      "#Fab+Fc=3000\n",
      "#-3000*x+Fc*200=0\n",
      "Fc=1857\n",
      "Fab=3000-Fc\n",
      "x=Fc*200/3000\n",
      "\n",
      "#Display\n",
      "print'Distance of force from AB =  ',x,\"mm\"\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Distance of force from AB =   123 mm\n"
       ]
      }
     ],
     "prompt_number": 44
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 1.10 Page no 35"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Given\n",
      "af =30\t\t #kN Axial force along centroidal axis\n",
      "t1 = 2 \t\t#m thickness of square cross section\n",
      "t2=4\n",
      "#Internal loading\n",
      "import math\n",
      "Fb=af*t1/((t1+t2)*(4/5.0))\n",
      "Ax=Fb*(3/5.0)\n",
      "Ay=af-(Fb*(4/5.0))\n",
      "Fa=math.sqrt(Ax**2+Ay**2)\n",
      "Va=Fa/t1\n",
      "Vb=Fb\n",
      "Aa=(math.pi/4.0)*((t1/100.0)**2)\n",
      "Ab=(math.pi/4.0)*(((t1+t2)/200.0)**2)\n",
      "Ta=Va/Aa\n",
      "Tb=Vb/Ab\n",
      "\n",
      "#Display\n",
      "print'The Average Normal Stress for section b-b   =  ',round(Ta/1000,0),\"Mpa\"\n",
      "print'The Average Shear Stress for section b-b    =  ',round(Tb/1000,1),\"Mpa\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The Average Normal Stress for section b-b   =   34.0 Mpa\n",
        "The Average Shear Stress for section b-b    =   17.7 Mpa\n"
       ]
      }
     ],
     "prompt_number": 68
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 1.11 Page no 36"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Given \n",
      "f = 6          #kN\n",
      "d_rod = 0.1        #Diameter of steel rod in mm.\n",
      "l_bc = 0.125        #Length of side bc in mm.\n",
      "l_ac=0.15\n",
      "#Calculation\n",
      "F=f/2.0\n",
      "#Appling Force balance\n",
      "Va=F\n",
      "Vb=F\n",
      "a_rod=d_rod*l_ac\n",
      "a_strut=l_bc*l_ac\n",
      "#Average shear stress\n",
      "avg_shear_rod = F/a_rod #for rod in Mpa\n",
      "avg_shear_strut = (f/2)/a_strut #for strut\n",
      "\n",
      "#Display\n",
      "print'The average shear stress for the rod    =  ',avg_shear_rod,\"Kpa\"\n",
      "print'The average shear stress for the strut  =  ',avg_shear_strut,\"Kpa\"\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The average shear stress for the rod    =   200.0 Kpa\n",
        "The average shear stress for the strut  =   160.0 Kpa\n"
       ]
      }
     ],
     "prompt_number": 74
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 1.12 Page no 37"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "#Given\n",
      "l_bc = 1 \t#Length of BC in inc.\n",
      "l_db = 3 \t# in.\n",
      "l_ed = 2 \t# in.\n",
      "l_ab = 1.5 \t# in.\n",
      "f_diagonal = 600 \t#lb\n",
      "\n",
      "#Calculation\n",
      "a1 = l_bc*l_ab \t\t#Area of face AB in mm**2.\n",
      "a2 = l_ab*l_ed  \t#mm**2.\n",
      "a3 = l_db*l_ab  \t# mm**2.\n",
      "#Balancing forces along the x- direction.\n",
      "f_ab = f_diagonal*(3/5.0) #Force on segment AB in N\n",
      "V = f_ab             \n",
      "\n",
      "#Balancing forces along the Y direction.\n",
      "f_bc = f_diagonal*(4/5.0) #Force on segment BC in N.\n",
      "avg_comp_ab = f_ab/a1 # N/mm**2\n",
      "avg_comp_bc = f_bc/a2 # N/mm**2\n",
      "avg_shear = f_ab/a3 # N/mm**2\n",
      "\n",
      "#Display\n",
      "print a1\n",
      "print'The average compressive stress along AB  =  ',avg_comp_ab,\"psi\"\n",
      "print'The average compressive stress along BC  =  ',avg_comp_bc,\"psi\"\n",
      "print'The average shear stress along EDB       =  ',avg_shear,\"psi\"\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "1.5\n",
        "The average compressive stress along AB  =   240.0 psi\n",
        "The average compressive stress along BC  =   160.0 psi\n",
        "The average shear stress along EDB       =   80.0 psi\n"
       ]
      }
     ],
     "prompt_number": 81
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 1.13 Page no 49"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "#Given\n",
      "shear_allow = 3 \t#ksi, stress\n",
      "tensile_allow = 5 \t#ksi, stress\n",
      "l_C1 = 3\t\t #in\n",
      "l_C2 = 2 \t\t #in\n",
      "l_ac=8            #inch\n",
      "Tauallow=8       #ksi\n",
      "\n",
      "#Calculation\n",
      "import math\n",
      "Fab=(tensile_allow*(3/5.0)*(l_C1+l_C2)+shear_allow*l_C1)/(l_ac)\n",
      "Cx=-shear_allow+tensile_allow*(4/5.0)\n",
      "Cy=tensile_allow*(3/5.0)+l_C1\n",
      "Fc=math.sqrt(Cx**2+Cy**2)\n",
      "A=(Fc/2.0)/Tauallow\n",
      "d=2*math.sqrt(A/math.pi)\n",
      "\n",
      "#Result\n",
      "print\" The diameter of rod is =  \",round(d,3),\"inch\"\n",
      "print\"Wse a pin of diameter =\",3/4.0,\"inch\"\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " The diameter of rod is =   0.696 inch\n",
        "Wse a pin of diameter = 0.75 inch\n"
       ]
      }
     ],
     "prompt_number": 87
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 1.14 Page no 50"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Given\n",
      "P= 20.0\t\t      #kN, force\n",
      "d_hole = 40.0 \t  #mm\n",
      "normal_allow = 60.0 #MPa, pressure\n",
      "shear_allow = 35.0  #MPa, pressure\n",
      "\n",
      "\n",
      "#Diameter of Rod\n",
      "import math\n",
      "area1 = (P*10**3)/(normal_allow*10**6) #Area in m**2\n",
      "d = ((math.sqrt((4*area1)/math.pi))*1000) # Area = (math.pi\\4)d**2\n",
      "#Thickness of disc\n",
      "V = P\n",
      "area2 = (V*10**3)/(shear_allow*10**6) #Area in m**2\n",
      "thickness = (area2*10**6)/(d_hole*math.pi)# A = pi*d*t\n",
      " \n",
      "\n",
      "#Result\n",
      "print\"The diameter of rode   = \",round(d,2),\"mm\"\n",
      "print\"The thickness of disc  = \",round(thickness,2),\"mm\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The diameter of rode   =  20.6 mm\n",
        "The thickness of disc  =  4.55 mm\n"
       ]
      }
     ],
     "prompt_number": 93
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 1.15 Page no 51"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Given\n",
      "import math\n",
      "bearing_allow = 75.0\t #MPa, pressure\n",
      "tensile_allow = 55.0 \t#MPa, pressure\n",
      "d_shaft = 60.0\t\t #mm\n",
      "r_shaft = d_shaft/2.0 \t#mm\n",
      "area_shaft = math.pi*(r_shaft**2) #Area = pi*r**2\n",
      "d_collar = 80 #mm\n",
      "thick_collar = 20 #mm\n",
      "\n",
      "#Calculation\n",
      "r_collar = d_collar/2.0 #mm\n",
      "area_collar = math.pi*(r_collar**2) #Area = pi*r**2\n",
      "#Normal Stress\n",
      "P1 = (tensile_allow* area_shaft)/3.0 #Tensile stress = 3P/A.\n",
      "P1_kN = P1/1000.0\n",
      "#Bearing Stress\n",
      "bearing_area = area_collar-area_shaft \n",
      "P2 = (bearing_allow*bearing_area)/3 \n",
      "P2_kN= P2/1000.0\n",
      "\n",
      "#Result\n",
      "print\"The load calculated by Normal Stress  \",round(P1_kN,1),\"kN\"\n",
      "print\"The load calculated by Bearing Stress \",round(P2_kN,0),\"kN\"\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The load calculated by Normal Stress   51.8 kN\n",
        "The load calculated by Bearing Stress  55.0 kN\n"
       ]
      }
     ],
     "prompt_number": 96
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 1.16 Page no 52"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Given\n",
      "d_ac= 20.0            #mm\n",
      "area_al = 1800.0       #mm**2, area\n",
      "d_pins = 18.0          #mm\n",
      "st_fail_stress = 680.0 #MPa, pressure\n",
      "al_fail_stress = 70.0 #MPa, pressure\n",
      "shear_fail_pin = 900.0 #MPa\n",
      "fos = 2.0             #Factor of safety\n",
      "l_ab = 2.0            #m\n",
      "l_ap = 0.75           #m\n",
      "\n",
      "#Calculation\n",
      "import math\n",
      "area_pins = math.pi*(d_pins/2.0)**2\n",
      "area_ac = math.pi*(d_ac/2.0)**2 #Area = (math.pi\\4)d**2\n",
      "st_allow= st_fail_stress /fos #MPa\n",
      "al_allow = al_fail_stress/fos #MPa\n",
      "pin_allow_shear = shear_fail_pin/fos #MPa\n",
      "\n",
      "#Rod AC\n",
      "f_ac = (st_allow*area_ac)/1000.0\n",
      "P1 = ((f_ac*l_ab)/(l_ab-l_ap))\n",
      "\n",
      "#Block B\n",
      "f_b =(al_allow*area_al)/1000.0\n",
      "P2 = ((f_b*l_ab)/l_ap)\n",
      "\n",
      "#Pin A or C\n",
      "V = (pin_allow_shear*area_pins)/1000.0\n",
      "P3 = (V*l_ab)/(l_ab-l_ap)\n",
      "\n",
      "#Result\n",
      "print\"The load allowed on rod AC   \",round(P1,0),\"kN\"\n",
      "print\"The load allowed on block B  \",P2,\"kN\"\n",
      "print\"The load allowed on pins A or C \",round(P3,0),\"kN\"\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The load allowed on rod AC    171.0 kN\n",
        "The load allowed on block B   168.0 kN\n",
        "The load allowed on pins A or C  183.0 kN\n"
       ]
      }
     ],
     "prompt_number": 102
    }
   ],
   "metadata": {}
  }
 ]
}