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-{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "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",
- "calculating maximum shear stress & torque\n",
- "\"\"\"\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",
- "Calculation of required diameter, outer diameter & ratio of diameteres\n",
- "\"\"\"\n",
- "\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 3.4, page no. 205"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "\"\"\"\n",
- "determining maximum shear stress in various parts of the shaft\n",
- "\"\"\"\n",
- "\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",
- "Calculate maximum shear, tensile & compressive stress in the tube.\n",
- "Also, calculate coressponding strains in the tube\n",
- "\"\"\"\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",
- "Required diameters of the shaft at various rpm\n",
- "\"\"\"\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",
- "Find the maximum shear stress & angle of twist in the shaft\n",
- "\"\"\"\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",
- "for various loading cases, obtain formulae and calculate strain energy\n",
- "\"\"\"\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",
- "calculate the strain energy for hollow shaft\n",
- "\"\"\"\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",
- "Ratio of Shear stress and angle of twist for circular & square tube\n",
- "\"\"\"\n",
- "import math\n",
- "\n",
- "r_s = (math.pi/4) #Ratio of shear stress\n",
- "print \"Ratio of shear stress is \", round(r_s,2)\n",
- "r_t = (math.pi**2/16) #Ratio of angle of twit\n",
- "print \"Ratio of angle of twist is \", round(r_t, 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": 4
- }
- ],
- "metadata": {}
- }
- ]
-} \ No newline at end of file
generated by cgit (git 2.34.1) at 2024-12-20 12:02:33 +0000