path: root/Mechanics_of_Materials_by_James_M._Gere/chapter3.ipynb

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  {
   "cells": [
    {
     "cell_type": "heading",
     "level": 1,
     "metadata": {},
     "source": [
      "Chapter 3: Torsion"
     ]
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.1, page no. 196"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "\n",
      "import math \n",
      "\n",
      "#initialisation\n",
      "d = 1.5                         # diameter of bar in inch\n",
      "L = 54.0                        # Length of bar in inch\n",
      "G = 11.5e06                     # modulus of elasticity in psi \n",
      "\n",
      "#calculation\n",
      "\n",
      "# Part (a)\n",
      "T = 250.0                           # torque\n",
      "t_max = (16*T*12)/(math.pi*(d**3))  # maximum shear stress in bar\n",
      "Ip = (math.pi*(d**4))/32            # polar miment of inertia \n",
      "f = (T*12*L)/(G*Ip)                 # twist in radian\n",
      "f_ = (f*180)/math.pi                # twist in degree\n",
      "print \"Maximum shear stress in the bar is \", round(t_max), \" psi\"\n",
      "print \"Angle of twist is\", round(f_,2), \" degree\"\n",
      "\n",
      "#Part (b)\n",
      "t_allow = 6000                          # allowable shear stress\n",
      "T1 = (math.pi*(d**3)*t_allow)/16        #allowable permissible torque in lb-in\n",
      "T1_ = T1*0.0831658                      #allowable permissible torque in lb-ft\n",
      "f_allow = (2.5*math.pi)/180             # allowable twist in radian\n",
      "T2 = (G*Ip*f_allow)/L                   # allowable stress via a another method\n",
      "T2_ = T2*0.0831658                      #allowable permissible torque in lb-ft\n",
      "T_max = min(T1_,T2_)                    # minimum of the two\n",
      "print \"Maximum permissible torque in the bar is\", round(T_max), \" lb-ft\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Maximum shear stress in the bar is  4527.0  psi\n",
        "Angle of twist is 1.62  degree\n",
        "Maximum permissible torque in the bar is 331.0  lb-ft\n"
       ]
      }
     ],
     "prompt_number": 2
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.2, page no. 197"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "import math \n",
      "\n",
      "#initialisation\n",
      "T = 1200.0                                # allowable torque in N-m\n",
      "t = 40e06                               # allowable shear stress in Pa\n",
      "f = (0.75*math.pi)/180.0                # allowable rate of twist in rad/meter\n",
      "G = 78e09                               # modulus of elasticity\n",
      "\n",
      "#calculation\n",
      "\n",
      "# Part (a) : Solid shaft\n",
      "d0 = ((16.0*T)/(math.pi*t))**(1.0/3.0)\n",
      "Ip = T/(G*f)                            # polar moment of inertia\n",
      "d01 = ((32.0*Ip)/(math.pi))**(1.0/4.0)        # from rate of twist definition\n",
      "print \"The required diameter of the solid shaft is \", round(d0,5), \"m\"\n",
      "\n",
      "# Part (b) : hollow shaft\n",
      "d2 = (T/(0.1159*t))**(1.0/3.0)              # Diamater of hollow shaft in meter\n",
      "d2_ = (T/(0.05796*G*f))**(1.0/4.0)          # Another value of d2 by definition of theta(allow), f = T/(G*Ip)\n",
      "d1 = 0.8*d2_                                # because rate of twist governs the design\n",
      "print \"The required diameter of the hollow shaft is \", round(d2,5), \"m\"\n",
      "\n",
      "# Part (c) : Ratio of diameter and weight\n",
      "r1 = d2_/d01                            # diameter ratio\n",
      "r2 = ((d2_**2.0)-(d1**2.0))/(d01**2.0)        # Weight Ratio\n",
      "print \"Ratio of the diameter of the hollow and solid shaft is\", round(r1,2)\n",
      "print \"Ratio of the weight of the hollow and solid shaft is\", round(r2,2)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " The required diameter of the solid shaft is  0.05346 m\n",
        "The required diameter of the hollow shaft is  0.06373 m\n",
        "Ratio of the diameter of the hollow and solid shaft is 1.14\n",
        "Ratio of the weight of the hollow and solid shaft is 0.47\n"
       ]
      }
     ],
     "prompt_number": 5
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.3, Page number 200"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "\n",
      "import math\n",
      "\n",
      "#Variable declaration\n",
      "R1 = 0.6*1    #assumption for simplicity in calculations(for fig. a)\n",
      "R2 = 1        #assumption for simplicity in calculations(for fig. b)\n",
      "\n",
      "#Calculations\n",
      "Ip1 = (math.pi/2)*(1-(R1**4))\n",
      "Ip2 = (math.pi*R2**4)/2\n",
      "\n",
      "B1 = Ip2/Ip1\n",
      "B2 = Ip2/Ip1\n",
      "\n",
      "Wh = (math.pi*R2**2)*(1-R1**2)\n",
      "Ws = math.pi*R2**2\n",
      "B3 = Wh/Ws\n",
      "\n",
      "print \"Maximum shear stress in the hollow shaft to that in the solid shaft is\",round(B1,2)\n",
      "print \"The ratio of the angles of twists is\",round(B2,2)\n",
      "print \"The ratio of the weight of the hollow shaft to the weight of the solid shaft is\",B3\n",
      "\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Maximum shear stress in the hollow shaft to that in the solid shaft is 1.15\n",
        "The ratio of the angles of twists is 1.15\n",
        "The ratio of the weight of the hollow shaft to the weight of the solid shaft is 0.64\n"
       ]
      }
     ],
     "prompt_number": 11
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.4, page no. 205"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "import math \n",
      "\n",
      "#initialisation\n",
      "d = 0.03                                # diameter of the shaft in meter\n",
      "T2 = 450.0                                # Torque in N-m\n",
      "T1 = 275.0  \n",
      "T3 = 175.0  \n",
      "Lbc = 0.5                               # Length of shaft in meter\n",
      "Lcd = 0.4                               # Length of shaft in meter\n",
      "G = 80e09                               # Modulus of elasticity\n",
      "\n",
      "#calculation\n",
      "Tcd = T2-T1                             # torque in segment CD\n",
      "Tbc = -T1                               # torque in segment BC\n",
      "tcd = (16.0*Tcd)/(math.pi*(d**3))         # shear stress in cd segment\n",
      "\n",
      "print \"Shear stress in segment cd is\", round(tcd/1000000,1), \" MPa\"\n",
      "tbc = (16.0*Tbc)/(math.pi*(d**3))         # shear stress in bc segment\n",
      "\n",
      "#answer given in the textbook for tbc is wrong\n",
      "print \"Shear stress in segment bc is\", round(tbc/1000000,1), \" MPa\"\n",
      "Ip = (math.pi/32)*(d**4)                # Polar monent of inertia\n",
      "fbc = (Tbc*Lbc)/(G*Ip)                  # angle of twist in radian\n",
      "fcd = (Tcd*Lcd)/(G*Ip)                  # angle of twist in radian\n",
      "fbd = fbc + fcd                         # angle of twist in radian\n",
      "print \"Angles of twist in section BD\", round(fbd,3), \" radian\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Shear stress in segment cd is 33.0  MPa\n",
        "Shear stress in segment bc is -51.9  MPa\n",
        "Angles of twist in section BD -0.011  radian\n"
       ]
      }
     ],
     "prompt_number": 6
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.6, page no. 214"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "\n",
      "import math\n",
      "\n",
      "#initialisation \n",
      "d1 = 0.06                             # Inner diameter in meter\n",
      "d2 = 0.08                             # Outer diameter in meter\n",
      "r = d2/2.0                            # Outer radius\n",
      "G = 27e09                             # Modulus of elasticity\n",
      "T = 4000.0                            # Torque in N-m\n",
      "\n",
      "#calculation\n",
      "Ip = (math.pi/32)*((d2**4)-(d1**4))         # Polar moment of inertia\n",
      "t_max = (T*r)/Ip                            # maximum shear stress\n",
      "print \"Maximum shear stress in tube is \", t_max, \" Pa\"\n",
      "s_t = t_max                                 # Maximum tensile stress\n",
      "print \"Maximum tensile stress in tube is \", s_t, \" Pa\"\n",
      "s_c = -(t_max)                              # Maximum compressive stress\n",
      "print \"Maximum compressive stress in tube is \", s_c, \" Pa\"\n",
      "g_max = t_max / G                           # Maximum shear strain in radian\n",
      "print \"Maximum shear strain in tube is \", round(g_max,4), \" radian\"\n",
      "e_t = g_max/2.0                               # Maximum tensile strain in radian\n",
      "print \"radian\",e_t,\"Maximum tensile strain in tube is \", round(e_t,4), \" radian\"\n",
      "e_c = -g_max/2.0                              # Maximum compressive strain in radian\n",
      "print \"Maximum compressive strain in tube is \", round(e_c,4), \" radian\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Maximum shear stress in tube is  58205236.3308  Pa\n",
        "Maximum tensile stress in tube is  58205236.3308  Pa\n",
        "Maximum compressive stress in tube is  -58205236.3308  Pa\n",
        "Maximum shear strain in tube is  0.0022  radian\n",
        "radian 0.00107787474687 Maximum tensile strain in tube is  0.0011  radian\n",
        "Maximum compressive strain in tube is  -0.0011  radian\n"
       ]
      }
     ],
     "prompt_number": 7
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.7, page no. 219"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "\n",
      "import math \n",
      "\n",
      "#initialisation\n",
      "H = 40.0                          # Power in hp\n",
      "s = 6000.0                        # allowable shear stress in steel in psi\n",
      "\n",
      "#calculation\n",
      "# Part (a)\n",
      "n = 500.0                                     # rpm\n",
      "T = ((33000.0*H)/(2*math.pi*n))*(5042.0/420.0)    # Torque in lb-in\n",
      "d = ((16.0*T)/(math.pi*s))**(1.0/3.0)             # diameter in inch\n",
      "print \"Diameter of the shaft at 500 rpm\", round(d,2), \" inch\"\n",
      "\n",
      "# Part (b)\n",
      "n1 = 3000.0                                    # rpm\n",
      "T1 = ((33000.0*H)/(2*math.pi*n1))*(5042.0/420.0)   # Torque in lb-in\n",
      "d1 = ((16*T1)/(math.pi*s))**(1.0/3.0)            # diameter in inch\n",
      "print \"Diameter of the shaft at 3000 rpm\", round(d1,2), \" inch\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Diameter of the shaft at 500 rpm 1.62  inch\n",
        "Diameter of the shaft at 3000 rpm 0.89  inch\n"
       ]
      }
     ],
     "prompt_number": 8
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.8, page no. 221"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "\n",
      "import math \n",
      "\n",
      "#initialisation\n",
      "\n",
      "d = 0.05                        # diameter of the shaft\n",
      "Lab = 1.0                         # Length of shaft ab in meter\n",
      "Lbc = 1.2                       # Length of shaft bc in meter\n",
      "Pa = 50000.0                      # Power in Watt at A\n",
      "Pb = 35000.0                      # Power in Watt at B\n",
      "Ip = (math.pi/32)*(d**4)        # Polar moment of inertia\n",
      "Pc = 15000.0                      # Power in Watt at C\n",
      "G = 80e09                       # Modulus of elasticity\n",
      "f = 10.0                          # frequency in Hz \n",
      "\n",
      "#Calculations\n",
      "Ta = Pa/(2*math.pi*f)           # Torque in N-m at A\n",
      "Tb = Pb/(2*math.pi*f)           # Torque in N-m at B\n",
      "Tc = Pc/(2*math.pi*f)           # Torque in N-m at B\n",
      "Tab = Ta                        # Torque in N-m in shaft ab\n",
      "Tbc = Tc                        # Torque in N-m in shaft bc\n",
      "tab = (16*Tab)/(math.pi*(d**3))         # shear stress in ab segment\n",
      "fab = (Tab*Lab)/(G*Ip)                  # angle of twist in radian\n",
      "tbc = (16*Tbc)/(math.pi*(d**3))         # shear stress in ab segment\n",
      "fbc = (Tbc*Lbc)/(G*Ip)                  # angle of twist in radian\n",
      "fac = (fab+fbc)*(180.0/math.pi)           # angle of twist in degree in segment ac\n",
      "tmax = Tab                              # Maximum shear stress\n",
      "\n",
      "#Result\n",
      "print \"The maximum shear stress tmax in the shaft\", round(tmax), \" Nm\"\n",
      "print \"Angle of twist in segment AC\", round(fac,2), \" degree\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The maximum shear stress tmax in the shaft 796.0  Nm\n",
        "Angle of twist in segment AC 1.26  degree\n"
       ]
      }
     ],
     "prompt_number": 9
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.10, page no. 230"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "\n",
      "import math \n",
      "\n",
      "#initialisation\n",
      "Ta = 100.0                          # Torque in N-m at A\n",
      "Tb = 150.0                          # Torque in N-m at B\n",
      "L = 1.6                             # Length of shaft in meter\n",
      "G = 80e09                           # Modulus of elasticity\n",
      "Ip = 79.52e-09                      # polar moment of inertia in m4\n",
      "\n",
      "#calculation\n",
      "\n",
      "Ua = ((Ta**2)*L)/(2*G*Ip)           # Strain energy at A\n",
      "print \"Torque acting at free end\", round(Ua,2), \" joule\"\n",
      "Ub = ((Tb**2)*L)/(4*G*Ip)           # Strain energy at B\n",
      "print \"Torque acting at mid point\", round(Ub,2), \" joule\"\n",
      "a = (Ta*Tb*L)/(2*G*Ip)              # dummy variabble\n",
      "Uc = Ua+a+Ub                        # Strain energy at C\n",
      "print \"Total torque\", round(Uc,2), \"joule\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Torque acting at free end 1.26  joule\n",
        "Torque acting at mid point 1.41  joule\n",
        "Total torque 4.56 joule\n"
       ]
      }
     ],
     "prompt_number": 7
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.11, page no. 231"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "\n",
      "import math \n",
      "t = 480.0                             # Torque of consmath.tant intensity\n",
      "L = 144.0                             # Length of bar\n",
      "G = 11.5e06                           # Modulus of elasticity in Psi\n",
      "Ip = 17.18                            # Polar moment of inertia\n",
      "\n",
      "#Calculation\n",
      "U = ((t**2)*(L**3))/(G*Ip*6)          # strain energy in in-lb\n",
      "\n",
      "#Result\n",
      "print \"The strain energy for the hollow shaft is\", round(U), \"in-lb\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The strain energy for the hollow shaft is 580.0 in-lb\n"
       ]
      }
     ],
     "prompt_number": 11
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "example 3.14, page no. 242"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "import math\n",
      "\n",
      "#Variable declaration\n",
      "r = 1     #assuming r = 1 for simplicity in calculations\n",
      "\n",
      "#Calculations\n",
      "Am2 = (math.pi*r)**2/4\n",
      "Am1 = math.pi*r**2\n",
      "\n",
      "T1_T2 = Am2/Am1\n",
      "\n",
      "t = 1   #assuming t = 1 for simplicity in calculations\n",
      "J2 = ((math.pi*r)**3*t)/8\n",
      "J1 = 2*math.pi*r**3*t\n",
      "phi1_phi2 = J2/J1\n",
      "\n",
      "#Results\n",
      "print \"Ratio of shear stress is \", round(T1_T2,2)\n",
      "print \"Ratio of angle of twist is \", round(phi1_phi2, 2)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " Ratio of shear stress is  0.79\n",
        "Ratio of angle of twist is  0.62\n"
       ]
      }
     ],
     "prompt_number": 13
    }
   ],
   "metadata": {}
  }
 ]
}