path: root/Mechanics_of_Materials/Chapter5.ipynb

{
 "metadata": {
  "name": ""
 },
 "nbformat": 3,
 "nbformat_minor": 0,
 "worksheets": [
  {
   "cells": [
    {
     "cell_type": "heading",
     "level": 1,
     "metadata": {},
     "source": [
      "Chapter 5:Torsion"
     ]
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 5.2 Page No 188"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Given\n",
      "T1 = 4250.0        #kNmm, torque\n",
      "T2 = -3000.0       #kNm\n",
      "T3 = T1+T2         #kNm\n",
      "r = 75.0           #mm, radius\n",
      "\n",
      "#Calculation\n",
      "#Section Property\n",
      "import math\n",
      "J = (math.pi/2.0)*(r**4) #polar moment of inertia\n",
      "#Shear Stress\n",
      "c_a = 75 #mm\n",
      "tou_a = (T3*c_a*1000)/J #tou = Tc/J\n",
      "c_b = 15 #mm\n",
      "tou_b = (T3*c_b*1000)/J #tou = Tc/J\n",
      "\n",
      "#Display\n",
      "print'The shear stress developed at A   = ',round(tou_a*10,1),\"ksi\"\n",
      "print'The shear stress developed at B   = ',round(tou_b*10,2),\"ksi\"\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The shear stress developed at A   =  18.9 ksi\n",
        "The shear stress developed at B   =  3.77 ksi\n"
       ]
      }
     ],
     "prompt_number": 8
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 5.3 Page No 189"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Given\n",
      "di = 80             #mm, inside diameter\n",
      "ri = 40/1000.0      #m, inside radius\n",
      "d0 = 100            #mm, outside diameter\n",
      "ro = d0/2000.0      #m outside radius\n",
      "F = 80              #N, force\n",
      "l1 = 0.2            #m, length\n",
      "l2 = 0.3            #m\n",
      "\n",
      "#Internal Torque\n",
      "T = F*(l1+l2)\n",
      "#Section Property\n",
      "import math\n",
      "J = (math.pi/2.0)*((ro**4)-(ri**4))\n",
      "#Shear Stress\n",
      "c_o = 0.05#m\n",
      "tou_o = (T*c_o)/(J*10**6)\n",
      "c_i = 0.04 #m\n",
      "tou_i = (T*c_i)/(J*10**6)\n",
      "\n",
      "#Display\n",
      "print'The shear stress in the inner wall    = ',round(tou_i,3),\"MPa\"\n",
      "print'The shear stress in the outer wall    = ',round(tou_o,3),\"MPa\"\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The shear stress in the inner wall    =  0.276 MPa\n",
        "The shear stress in the outer wall    =  0.345 MPa\n"
       ]
      }
     ],
     "prompt_number": 12
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 5.4 Page No 191"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Given\n",
      "P = 5          #hp\n",
      "N = 175           #rpm\n",
      "allow_shear = 14.5 #ksi\n",
      "\n",
      "#Calculations\n",
      "import math\n",
      "P_=P*550       #ftlb/s\n",
      "ang_vel = (2*math.pi*N)/60.0 # rad/s\n",
      "T = P_/ang_vel #P = T*angular velocity\n",
      "c = ((2*T*12)/(math.pi*allow_shear*1000))**(1/3.0)\n",
      "d =2*c\n",
      "\n",
      "#Display\n",
      "print'The required diameter of the shaft   = ',round(d,3),\"inch\"\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The required diameter of the shaft   =  0.858 inch\n"
       ]
      }
     ],
     "prompt_number": 22
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 5.5 Page No 205"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Given\n",
      "E = 80*10**9       #N/m**2, longitudinal stress\n",
      "d = 14/1000.0         #m,  diameter\n",
      "r = d/2.0             #m, radius\n",
      "R = 100             #mm\n",
      "l_ac = 0.4           #m, length\n",
      "l_cd = 0.3           #m\n",
      "l_de = 0.5          #m\n",
      "T_c = 280          #Nm, torque\n",
      "T_a = 150           #Nm\n",
      "T_d = 40             #Nm\n",
      "T_ac = T_a           #Nm\n",
      "\n",
      "#Calculation\n",
      "T_cd = T_ac - T_c \n",
      "T_de = T_cd - T_d\n",
      "#Angle of Twist\n",
      "import math\n",
      "J = (math.pi/2.0)*(r**4)\n",
      "phiA=T_ac*l_ac/(J*E)+T_cd*l_cd/(J*E)+T_de*l_de/(J*E)\n",
      "Sp=phiA*R\n",
      "\n",
      "#Display\n",
      "print'The angle of twist of the shaft       = ',round(phiA,3),\"rad\"\n",
      "print'The displacement of tooth P on gear A    =',round(Sp,3),\"mm\"\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The angle of twist of the shaft       =  -0.212 rad\n",
        "The displacement of tooth P on gear A    = -21.212 mm\n"
       ]
      }
     ],
     "prompt_number": 28
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 5.6 Page No 206"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Given\n",
      "T = 45\t\t #N, torque\n",
      "G = 80 \t\t#GPa, pressure\n",
      "d = 20/1000.0\t\t #m\n",
      "r = d/2.0\t #m\n",
      "l_dc = 1.5\t\t #m\n",
      "l_ab = 2 \t\t#m\n",
      "r1 = 75/1000.0\t\t #m\n",
      "r2 = 150/1000.0\t\t #m\n",
      "\n",
      "#Calculation\n",
      "#Internal Torque\n",
      "F = T/r2\n",
      "T_d_x = F*r1\n",
      "#Angle of twist\n",
      "import math\n",
      "J = (math.pi/2)*(r**4)\n",
      "phi_c = (T*l_dc)/(2*J*G*10**9)\n",
      "phi_b = (phi_c*r1)/r2\n",
      "phi_ab = (T*l_ab)/(J*G*10**9)\n",
      "phi_a = phi_b + phi_ab\n",
      "\n",
      "#Display\n",
      "print'The angle of twist of end A of shaft AB   = ',round(phi_a,3),\"rad\"\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The angle of twist of end A of shaft AB   =  0.085 rad\n"
       ]
      }
     ],
     "prompt_number": 31
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 5.7 Page No 207"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Given\n",
      "d = 2            #inch, diameter\n",
      "r = d/2.0        #radius\n",
      "c = d/2.0\n",
      "l_buried = 600\t\t #mm, buried length\n",
      "G = 5500*10**3 \t\t#MPa\n",
      "F = 1     \t\t#N\n",
      "l_handle= 150 \t\t#mm\n",
      "l_ab = 36 \t\t#inch\n",
      "\n",
      "#Internal Torque\n",
      "T_ab = F*2*l_handle\n",
      "t = T_ab/l_buried\n",
      "#Maximum Shear Stress\n",
      "import math\n",
      "J = (math.pi/2.0)*(r**4)\n",
      "tou_max = (T_ab*c)/(J)\n",
      "\n",
      "#Angle of Twist\n",
      "from scipy import integrate\n",
      "def f(x):\n",
      "    return(x)\n",
      "x=integrate.quad(f,0,24) #Strain formula for short line segment = delta(sdash) =(1+e_z)delta(s) \n",
      "X= x[0]\n",
      "phi_a = ((T_ab*l_ab)+(50*X/4.0))/(J*G) \n",
      "\n",
      "#Display\n",
      "print'The maximum shear stress in the post          =',round(tou_max,1),\"psi\"\n",
      "print'The angle of twist at the top of the post    = ',round(phi_a,5),\"rad\"\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The maximum shear stress in the post          = 191.0 psi\n",
        "The angle of twist at the top of the post    =  0.00167 rad\n"
       ]
      }
     ],
     "prompt_number": 50
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 5.8 Page No 216"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Given\n",
      "d = 20/1000.0 #m, diameter\n",
      "r = d/2.0\n",
      "l_bc = 0.2\n",
      "l_cd = 1.5\n",
      "l_da = 0.3\n",
      "T_c = 800  #Nm, torque\n",
      "T_d = -500 #Nm\n",
      "\n",
      "#Calculation\n",
      "#Eqn 1      300 = T_a + T_b\n",
      "#Compatibility\n",
      "#Eqn 2\n",
      "coeff_Tb = -l_bc\n",
      "coeff_Ta = l_cd + l_da\n",
      "#Solving Equations simultaneously using matrices\n",
      "T_b = 645\n",
      "T_a = -345\n",
      "\n",
      "#Display\n",
      "print'The reaction at A    = ',T_a,\"Nm\"\n",
      "print'The reaction at B    = ',T_b,\"Nm\"\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The reaction at A    =  -345 Nm\n",
        "The reaction at B    =  645 Nm\n"
       ]
      }
     ],
     "prompt_number": 54
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 5.9 Page No 217"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Given\n",
      "T = 250         #Nm, torque\n",
      "G_st = 80        #GPa, pressure\n",
      "G_br = 36        #GPa\n",
      "ri = 0.5         #inch, inside radius\n",
      "ro = 1         #inch, outside radius\n",
      "l_ab = 1.2      #m\n",
      "\n",
      "#Equilibrium\n",
      "# -Tst-Tbr+250Nm = 0\n",
      "coeff1_st = -1\n",
      "coeff1_br = -1\n",
      "b1 = -250\n",
      "\n",
      "#Compatibility\n",
      "#phi = TL/JG\n",
      "import math\n",
      "J1 = (math.pi/2.0)*(ro**4 - ri**4)\n",
      "J2 = (math.pi/2.0)*(ri**4)\n",
      "coeff2_st = 1/(J1*G_st*10**3)\n",
      "coeff2_br = -1/(J2*G_br*10**3)\n",
      "b2 = 0\n",
      "\n",
      "#Solving the above two equations simultaneously using matrices\n",
      "T_st = 2911.5       #lb-inch\n",
      "T_br = 88.5         #lb-inch\n",
      "\n",
      "shear_br_max = (T_br*10**3*ri)/(J2) #tou = (Tr)/J\n",
      "shear_st_min = (T_st*10**3*ri)/(J1) #tou = (Tr)/J\n",
      "shear_st_max = (T_st*10**3*ro)/(J1) #tou = (Tr)/J\n",
      "\n",
      "shear_strain = shear_br_max / G_br\n",
      "shear_strain = shear_strain\n",
      "\n",
      "#Display\n",
      "print'The maximum shear stress experienced by Steel =',round(shear_st_max/1000),\"psi\"\n",
      "print'The minimum shear stress experienced by Steel =',round(shear_st_min/1000),\"psi\"\n",
      "print'The maximum shear stress experienced by Brass  ',round(shear_br_max/1000),\"psi\"\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The maximum shear stress experienced by Steel = 1977.0 psi\n",
        "The minimum shear stress experienced by Steel = 989.0 psi\n",
        "The maximum shear stress experienced by Brass   451.0 psi\n"
       ]
      }
     ],
     "prompt_number": 61
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 5.10 Page No 223"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Given\n",
      "import math\n",
      "l = 4*12            #m, length\n",
      "a = 1.5             #inch\n",
      "tou_allow = 8000     #lb\n",
      "phi_allow = 0.02    #rad\n",
      "G = 3.7*10**6      #lb/inch**2, pressure\n",
      "alpha = (60*math.pi)/180.0 #degrees\n",
      " \n",
      "#Calculations\n",
      "T_shear1 = (tou_allow*a**3)/(20.0) # allowable shear stress = (20T)/(a**3)\n",
      "T_twist1 = (phi_allow*a**4*G)/(46*l) #angle of twist =(46TL)/(a**4*G)\n",
      "T1 = min(T_shear1, T_twist1)\n",
      " \n",
      "#Circular Cross Section\n",
      "c_ = (a*a*math.sin(alpha))/(math.pi*2)\n",
      "c = math.sqrt(c_)\n",
      "\n",
      "J = (math.pi/2.0)*(c**4)\n",
      "T_shear2 = (tou_allow*J)/(c*1000)\n",
      "T_twist2 = (phi_allow*J*G*10**3)/(l*10**6)\n",
      "T2 = min(T_shear2, T_twist2)\n",
      "\n",
      "#Display\n",
      "print'The largest torque that applied at the end of the triangular shaft  ',round(T1,0),\"lb-in\"\n",
      "print'The largest torque that applied at the end of the circular shaft    ',round(T2*1000,0),\"lb-in\"\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The largest torque that applied at the end of the triangular shaft   170.0 lb-in\n",
        "The largest torque that applied at the end of the circular shaft     233.0 lb-in\n"
       ]
      }
     ],
     "prompt_number": 69
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 5.12 Page No 228"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Given\n",
      "#The given dimension are\n",
      "l_cd = 0.5       #m\n",
      "l_de = 1.5       #m\n",
      "h =60/1000.0     #m\n",
      "w = 40/1000.0    #m\n",
      "t_h = 3/1000.0   #m\n",
      "t_w = 5/1000.0  #m\n",
      "T_c = 60        #Nm\n",
      "T_d = 25         #Nm\n",
      "G = 38*10**9 #N/m**2\n",
      "T1 = T_c - T_d\n",
      "\n",
      "#Calculation\n",
      "#Average Shear Stress\n",
      "area = (w-t_w)*(h-t_h)\n",
      "shear_a = T1/(2*t_w*area*10**6)\n",
      "shear_b = T1/(2*t_h*area*10**6)\n",
      "\n",
      "#Angle of Twist\n",
      "phi=(T_c*l_cd/(4*area**2*G))*((2*57/5.0)+(2*35/3.0))+(T1*l_de/(4*area**2*G))*((2*57/5.0)+(2*35/3.0))\n",
      "\n",
      "#Display\n",
      "print'The average shear stress of the tube at A    = ',round(shear_a,2),\"MPa\"\n",
      "print'The average shear stress of the tube at B    = ',round(shear_b,2),\"MPa\"\n",
      "print'The angle of twist of end C                  = ',round(phi,5),\"rad\"\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The average shear stress of the tube at A    =  1.75 MPa\n",
        "The average shear stress of the tube at B    =  2.92 MPa\n",
        "The angle of twist of end C                  =  0.00629 rad\n"
       ]
      }
     ],
     "prompt_number": 78
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 5.13 Page No 236"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Given\n",
      "fillet_r = 6 #mm, fillet radius\n",
      "D = 40/1000.0 #m, diameter\n",
      "d = 20/1000.0 #m\n",
      "T = 30 #Nm\n",
      "\n",
      "#Calculation\n",
      "D_d = D/d \n",
      "r_d = fillet_r/d \n",
      "k = 1.3\n",
      "#Maximum Shear Stress\n",
      "import math\n",
      "c = D/2.0\n",
      "J = (math.pi/2.0)*(c**4)\n",
      "max_shear = (k*T*c)/(J*10**6) # tou = K(Tc/J)\n",
      "\n",
      "#Display\n",
      "print'The maximum shear stress in the shaft is  = ',round(max_shear,1),\"MPa\"\n",
      "\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The maximum shear stress in the shaft is  =  3.1 MPa\n"
       ]
      }
     ],
     "prompt_number": 83
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 5.14 Page No 242"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Given\n",
      "ro = 50/1000.0     #m, outside radius\n",
      "ri = 30/1000.0     #m inside radius\n",
      "c = ro\n",
      "shear = 20*10**6 #N/m**2\n",
      "\n",
      "#Maximum Elastic Torque\n",
      "import math\n",
      "J = (math.pi/2.0)*((ro**4)-(ri**4))\n",
      "T_y = (shear*J)/c # tou = Tc/J\n",
      "T_y = T_y/1000.0 #in kN\n",
      "\n",
      "#Plastic Torque\n",
      "x0 = 0.03\n",
      "x1 = 0.05\n",
      "\n",
      "from scipy import integrate\n",
      "def f(rho):\n",
      "    return(rho**2)\n",
      "I=integrate.quad(f,x0,x1) #Strain formula for short line segment = delta(sdash) =(1+e_z)delta(s) \n",
      "\n",
      "Tp =(2*math.pi*I[0]*shear)\n",
      "Tp_= Tp/1000.0\n",
      "#Outer Shear Strain\n",
      "strain = (0.286*10**-3*ro)/(ri)\n",
      "\n",
      "#Display\n",
      "print'The maximum torque that can be applied to the shaft ',round(T_y,2),\"kNm\"\n",
      "print'The plastic torque that can be applied to the shaft',round(Tp_,2),\"kNm\"                                                                          \n",
      "print'The minimum shear strain at the outer radius of the shaft  ',round(strain,6),\"rad\"\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The maximum torque that can be applied to the shaft  3.42 kNm\n",
        "The plastic torque that can be applied to the shaft 4.11 kNm\n",
        "The minimum shear strain at the outer radius of the shaft   0.000477 rad\n"
       ]
      }
     ],
     "prompt_number": 104
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 5.15 Page No 243"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Given\n",
      "r = 20/1000.0    #m, radius\n",
      "l = 1.5          #m, length\n",
      "phi = 0.6       #rad\n",
      "shear_y = 75*10**6 #N/m**2\n",
      "\n",
      "#Calculations\n",
      "max_shear_strain = (phi*r)/(l) #phi = (strain*L)/r\n",
      "strain_y = 0.0016\n",
      "r_y = (r*strain_y)/(max_shear_strain) #by ratios\n",
      "#T= (math.pi*shear_y)*(4c**3 - r_y**3)/6.0\n",
      "import math\n",
      "c = r\n",
      "T = (math.pi*shear_y)*(4*c**3 - r_y**3)/6.0\n",
      "T = T/1000.0\n",
      "\n",
      "#Display\n",
      "print'The torque needed to twist the shaft by 0.6 rad   ',T,\"kNm\"\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The torque needed to twist the shaft by 0.6 rad    1.25412378731 kNm\n"
       ]
      }
     ],
     "prompt_number": 2
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 5.16 Page No 244"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Given\n",
      "l = 5            #m, length\n",
      "G = 12*10**3     #GPa\n",
      "co = 2          #inch\n",
      "ci = 1          #inch\n",
      "shear_y = 12      #N/mm**2\n",
      "strain_y = 0.002 #rad, strain\n",
      "\n",
      "#Plastic Torque\n",
      "import math\n",
      "T_p = ((2*math.pi)*(co**3 - ci**3)*shear_y)/3.0\n",
      "phi_p = (strain_y*l*shear_y)/ci\n",
      "J = (math.pi/2.0)*(co**4 - ci**4)\n",
      "shear_r = (T_p*co)/J\n",
      "shear_i = (shear_r*ci)/(co)# shear = Tc/J\n",
      "G = shear_y/strain_y \n",
      "phi_dash = (T_p*l*10**3)/(J*G) #phi = TpL/JG\n",
      "phi = phi_p - phi_dash\n",
      "\n",
      "\n",
      "#Display\n",
      "print'The plastic torque Tp   =  ',round(T_p,1),\"kip in\"\n",
      "print'shear stress at inner wall is  ',round(shear_i,2),\"ksi\"\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The plastic torque Tp   =   175.9 kip in\n",
        "shear stress at inner wall is   7.47 ksi\n"
       ]
      }
     ],
     "prompt_number": 16
    }
   ],
   "metadata": {}
  }
 ]
}