{ "cells": [ { "cell_type": "markdown", "metadata": {}, "source": [ "# Chapter 3 : Torsion" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Example 3.01, Page number 150" ] }, { "cell_type": "code", "execution_count": 2, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Largest permissible torque that can be applied to the shaft = 4.084070 kN\n", "Minimum shearing stress that can be applied to the shaft = 80.000000 MPa\n" ] } ], "source": [ "import math\n", "\n", "#Variable declaration\n", "l=1.5 # length of the cylindrical shaft\n", "Tmax=120 # Maximum allowable torque\n", "c1=0.02 # Inner radius \n", "c2=0.03 # Outer radius\n", "\n", "\n", "\n", "#Calculation \n", "#Case(a)\n", "J=(1/2.0)*(math.pi)*(c2**4-c1**4) # Polar moment of inertia \n", "c=c2 # Letting c equal to c2\n", "T=((J*Tmax*(pow(10,6)))/(c))/(1000.0) # Largest Permissible Torque\n", "#Case(b)\n", "Tmin=(c1/c2)*(Tmax) # Minimum Shearing Stress\n", "\n", "#Result\n", "print ('Largest permissible torque that can be applied to the shaft = %lf kN' %T)\n", "print ('Minimum shearing stress that can be applied to the shaft = %lf MPa' %Tmin)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## SAMPLE PROBLEM 3.1, Page number 152" ] }, { "cell_type": "code", "execution_count": 3, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Maximum shearing stress in shaft BC = 86.229980 MPa\n", "Minimum shearing stress in shaft BC = 64.650000 MPa\n", "Required diamter of the shafts AB and CD = 78.353203 mm\n" ] } ], "source": [ "import math\n", "from sympy import symbols, solve\n", "\n", "#Variable declaration\n", "Din=90/(1000.0) # Inner diamter of shaft BC(m) \n", "Dout=120/(1000.0) # Outer diamter of shaft BC(m)\n", "Tab=symbols('Tab')\n", "Tbc=symbols('Tbc')\n", "\n", "#Calculation \n", "#Equations of statics\n", "Tab=solve(6-Tab,Tab) # Torque in the shaft AB(kN.m)\n", "Tbc=14+6 # Torque in the shaft BC(kN.m)\n", "\n", "#Case(a) Shaft BC\n", "J=((math.pi)/2.0)*((pow(Dout/2,4)-pow(Din/2,4))) # Polar moment of inertia(m**4)\n", "# Maximum Shearing Stress\n", "Tmax=((Tbc*(Dout/2))/(J))/(1000.0) # Maximum shearing stress(MPa)\n", "#Minimum Shearing Stress\n", "Tmin=(45/60.0)*86.2 # Minimum shearing stress(MPa) \n", "#Case(b) Shafts AB and CD\n", "c=((6*4)/(65*(math.pi)))**1/3 # Radius of the shafts(m)\n", "\n", "#Result\n", "print ('Maximum shearing stress in shaft BC = %lf MPa' %Tmax) \n", "print ('Minimum shearing stress in shaft BC = %lf MPa' %Tmin)\n", "print ('Required diamter of the shafts AB and CD = %lf mm' %(c*2*1000))" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## SAMPLE PROBLEM 3.2, Page number 153" ] }, { "cell_type": "code", "execution_count": 6, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Case(a):Maximum torque that can be transmitted by the shaft as designed = 408.407045 kip.in\n", "Case(b):Maximum torque that can be transmitted by the shaft of equal weight = 210.744442 kip.in\n", "Case(c):Maximum torque that can be transmitted by the hollow shaft of equal weight and 8 in outer diameter = 636.000000 kip.in\n" ] } ], "source": [ "import math\n", "\n", "#Variable declaration\n", "din=4 # Inner diamter of shaft (in)\n", "dout=6 # Outer diamter of shaft (in)\n", "STRESS=12 # Shearing stress(ksi) \n", "\n", "#Calculation \n", "#Hollow Shaft as Designed. \n", "J=((math.pi)/2.0)*(pow((dout/2),4)-pow((din/2),4)) # Polar moment of inertia(in**4)\n", "Th=(J*STRESS)/3.0 # Allowable shearing stress(kip.in) \n", "\n", "#Solid Shaft of Equal Weight\n", "rad=math.sqrt((dout/2)**2-(din/2)**2) # Radius of solid shaft of equal weight(in)\n", "Te=(12*(math.pi)*(pow(rad,3)))/2.0 # Maximum allowable torque(kip.in)\n", "\n", "#Hollow Shaft of 8-in. Diameter.\n", "c5=math.sqrt(4**2 + 2**2 -3**2) # Inner radius of hallow shaft(in)\n", "J8=((math.pi)*(4**4-3.317**4))/2.0 # Polar moment of inertia(in**4) \n", "Tor=((212)*(12))/4.0 \n", "\n", "# Result\n", "print ('Case(a):Maximum torque that can be transmitted by the shaft as designed = %lf kip.in' %Th) \n", "print ('Case(b):Maximum torque that can be transmitted by the shaft of equal weight = %lf kip.in' %Te)\n", "print ('Case(c):Maximum torque that can be transmitted by the hollow shaft of equal weight and 8 in outer diameter = %lf kip.in' %Tor)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Example 3.02, Page number 160" ] }, { "cell_type": "code", "execution_count": 8, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Maximum torque that can be transmitted by the shaft as designed = 1.829501 kN.m\n" ] } ], "source": [ "import math\n", "\n", "#Variable declaration\n", "G=77*(pow(10,9)) # Modulus of rigidity(GPa) \n", "L=1.5 # length of the shaft(m)\n", "TWIST=2 # Allowable twist\n", "\n", "#Calculation \n", "#Case(a)\n", "phy=(2)*((2*(math.pi))/(360)) # Angle of twist(rad)\n", "#Case(b)\n", "J=1.021*(pow(10,-6)) # Polar moment of inertia(m**4)\n", "T=(((J*G)/(L))*(phy))/(1000) # Torque to be applied to the end of shaft(kN.m)\n", "\n", "# Result\n", "print ('Maximum torque that can be transmitted by the shaft as designed = %lf kN.m' %T) " ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Example 3.03, Page number 160" ] }, { "cell_type": "code", "execution_count": 4, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Maximum torque that can be transmitted by the shaft as designed = 3.906530 degree\n" ] } ], "source": [ "import math\n", "\n", "#Variable declaration\n", "tmin=70*(pow(10,6)) # Shearing stress(Pa) \n", "G=77*(pow(10,9))*(1.0) # Modulus of rigidity(Pa)\n", "L=1500 # ength of arc AA'(mm)\n", "c1=20 # inner radius(mm)\n", "\n", "#Calculation \n", "#Case(a)\n", "Ymin=tmin/G # shearing strain on the inner surface of the shaft\n", "#Case(b)\n", "phy=((L*Ymin)/(c1))*(360/(2*(math.pi))) # Angle of twist(degrees)\n", "\n", "# Result\n", "print ('Maximum torque that can be transmitted by the shaft as designed = %lf degree' %phy) " ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Example 3.04, Page number 163" ] }, { "cell_type": "code", "execution_count": 11, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Angle of rotation of end E\n", "5*L*T/(G*J)\n" ] } ], "source": [ "import math\n", "from sympy import symbols\n", "\n", "#Variable declaration\n", "rA=symbols('rA') # Variable declaration \n", "rB=symbols('rB') # Variable declaration \n", "Tad=symbols('Tad') # Variable declaration\n", "T=symbols('T') # Variable declaration\n", "L=symbols('L') # Variable declaration \n", "J=symbols('J') # Variable declaration\n", "G=symbols('G') # Variable declaration\n", "\n", "#Calculation \n", "rA=2*rB # Given relation \n", "Tad=2*T # Given relation \n", "phya=(Tad*L)/(J*G) # Angle of rotation phyA of gear A\n", "phyb=(rA/rB)*(phya) # Angle of rotation phyB of gear B\n", "phyEB=(T*L)/(J*G) # Angle of rotation phyEB through which end E rotates with respect to end B\n", "phyE=phyb+phyEB # Angle of rotation of end E\n", "\n", "# Result\n", "print ('Angle of rotation of end E') \n", "print (phyE) " ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Example 3.05, Page number 164" ] }, { "cell_type": "code", "execution_count": 12, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Torque exerted on the shaft by support A. = 51.736527 lb.ft\n", "Torque exerted on the shaft by support B. = 38.285030 lb.ft\n" ] } ], "source": [ "import math\n", "from sympy import symbols, solve\n", "\n", "#Variable declaration\n", "L=10 # Outer Length(in)\n", "r=(7/16.0) # Outer radius(in)\n", "Li=5 # Internal Length(in)\n", "ri=(5/16.0) # Inner radius(in)\n", "L1=5 # Equivalent to internal length(in)\n", "L2=5 # Equivalent to internal length(in)\n", "Ta=symbols('Ta') # Torque at a\n", "Tb=symbols('Tb') # Torque at b \n", "L1=symbols('L1') # Length\n", "L2=symbols('L2') # Length \n", "J1=symbols('J1') # polar moment of inertia\n", "J2=symbols('J2') # polar moment of inertia\n", "G=symbols('G') # modulus of rigidity of steel.\n", "\n", "#Calculation \n", "Tb=solve(((Ta*L1)/(J1*G))-((Tb*L2)/(J2*G)),Tb) # Internal torque\n", "Tb[0]=Tb[0].subs([ (J1, 57.6*(pow(10,-3))), (J2, 42.6*(pow(10,-3))),(L1, 5), (L2, 5)]) \n", "Ta=solve(90-Tb[0]-Ta,Ta) # Internal torque(lb.ft)\n", "Tb=0.740*Ta[0] # Internal torque(lb.ft) \n", "\n", "# Result\n", "print ('Torque exerted on the shaft by support A. = %lf lb.ft' %Ta[0])\n", "print ('Torque exerted on the shaft by support B. = %lf lb.ft' %Tb)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## SAMPLE PROBLEM 3.3, Page number 165" ] }, { "cell_type": "code", "execution_count": 5, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Angle of twist at end A = 2.310432 degree\n" ] } ], "source": [ "import math\n", "from sympy import symbols, solve\n", "\n", "#Variable declaration\n", "Tab=symbols('Tab') # Torque on shaft AB\n", "Tbc=symbols('Tbc') # Torque on shaft BC \n", "Tcd=symbols('Tcd') # Torque on shaft CD \n", "\n", "#Calculation\n", "#Statics\n", "Tab=solve(250-Tab,Tab) # Using the free body shown(N.m)\n", "Tbc=2000+250 # Torque on BC(N.m)\n", "Tcd=Tbc # Torque on CD(N.m)\n", "\n", "#Polar Moments of Inertia\n", "Jab=((math.pi)*((0.015)**4))/2.0 # Polar moment of inertia AB(m**4) \n", "Jbc=((math.pi)*((0.030)**4))/2.0 # Polar moment of inertia BC(m**4)\n", "Jcd=((math.pi)*((0.030)**4-(0.022)**4))/2.0 # Polar moment of inertia CD(m**4) \n", "\n", "#Angle of Twist\n", "AngleA=((1/77.0)*(((250*0.4)/(0.0795))+(((2250)*(0.2))/(1.272))+(((2250)*(0.6))/(0.904))))/((pow(10,-6))*(pow(10,9))*(1.0)) # Angle of twist(rad)\n", "AngleA=AngleA*(360/(2*math.pi)) # convertiong radian into degree\n", "\n", "# Result\n", "print ('Angle of twist at end A = %lf degree' %AngleA)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## SAMPLE PROBLEM 3.4, Page number 166" ] }, { "cell_type": "code", "execution_count": 14, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Case(a): Maximum permissible torque = 561.0 lb.in\n", "Case(b): The corresponding angle through which end A of shaft AB rotates = 10.5 degrees\n" ] } ], "source": [ "import math\n", "from sympy import symbols \n", "\n", "#Variable declaration\n", "T0=symbols('T0') # Variable declaration\n", "Tab=symbols('Tab') # Variable declaration \n", "F=symbols('F') # Force\n", "phyB=symbols('phyB') # Variable declaration\n", "phyC=symbols('phyC') # Variable declaration\n", "Smax=8000 # Maximum permissible shearing stress(psi) \n", "cAB=0.375 # Radius of shaft AB(in)\n", "cCD=0.5 # Radius of shaft CD(in)\n", "rC=2.45 # Radius at C(in)\n", "rB=0.875 # Radius at B(in) \n", "Tcd=symbols('Tcd') # Variable declaration\n", "\n", "# Calculation\n", "# Statics\n", "T0=F *0.875 # Equation 1 by Submission of Mb\n", "Tcd=F*2.45 # Equation 2 by Submission of Mc\n", "phyB=phyC*(rC/rB) # By noting that the peripheral motions of the gears are equal\n", "#Shaft AB\n", "Tab=T0 # Variable declaration\n", "T0=round((8000)*(1/2.0)*(math.pi)*(0.375)**3) # Largest torque that must be applied to end A of shaft AB(lb.in) \n", "#Shaft CD\n", "T0=symbols('T0') # Variable declaration\n", "Tcd=2.8*T0 # Equation \n", "T0=round((8000*(1/2.0)*(math.pi)*(0.5)**3)/2.8) # Largest torque that must be applied to end C of shaft CD(lb.in)\n", "#Angle of Rotation at End A\n", "#Shaft AB\n", "phyAB=(((561)*(24))/((1/2.0)*(math.pi)*((0.375)**4)*(11.2*pow(10,6))))*(360/(2*math.pi)) \n", "#Shaft CD\n", "phyCD=(((561)*(36)*(2.8))/((1/2.0)*(math.pi)*((0.5)**4)*(11.2*pow(10,6))))*(360/(2*math.pi)) \n", "phyC=phyCD\n", "phyB=2.8*(phyC)\n", "phyA=phyB+phyAB\n", "\n", "#Result\n", "print('Case(a): Maximum permissible torque = %.1f lb.in' %T0)\n", "print('Case(b): The corresponding angle through which end A of shaft AB rotates = %.1f degrees' %phyA)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## SAMPLE PROBLEM 3.5, Page number 167" ] }, { "cell_type": "code", "execution_count": 16, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Case(a): Maximum permissible torque = 6.3 lb.in\n" ] } ], "source": [ "import math\n", "from sympy import symbols \n", "\n", "#Variable declaration\n", "T0=symbols('T0') # Variable declaration\n", "T1=symbols('T1') # Variable declaration \n", "T2=symbols('T2') # Variable declaration \n", "phy1=symbols('phy1') # Variable declaration\n", "phy2=symbols('phy2') # Variable declaration\n", "Ts=120 # Maximum allowable stress in stell(MPa)\n", "Ta=70 # Maximum allowable stress in aluminum(MPa)\n", "Gs=77 # Modulus of regidity for steel(GPa) \n", "Ga=27 # Modulus of regidity for aluminium(GPa) \n", "T0=T1+T2 # Equation \n", "phy1=phy2 # Equation\n", "T2=T1*(77/27)*(0.614/2.003) # Since both the tube and the shaft are connected to the rigid disk\n", "# Shearing Stresses\n", "T1=round(((70*pow(10,6))*(2.003*pow(10,-6)))/(0.038)) # Torque(N.m)\n", "T2=(0.874)*(3690) # Torque(N.m)\n", "tsteel=((3225)*(0.025)*(pow(10,-6)))/(0.614*pow(10,-6)) # Maximum shearing stress in steel shaft(MPa) \n", "T2=round(((120)*pow(10,6)*(0.614*pow(10,-6)))/(0.025),-1) # Reobtaining torque as our earlier assumption was wrong(N.m)\n", "T1=round(2950/0.874) # Torque(N.m) \n", "T0=(T1+T2)/(1000) # maximum torque(N.m) \n", "\n", "#Result\n", "print('Case(a): Maximum permissible torque = %.1f lb.in' %T0)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Example 3.06, Page number 178" ] }, { "cell_type": "code", "execution_count": 6, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Case(a): Size of shaft = 0.374342 lb.in\n" ] } ], "source": [ "import math\n", "\n", "#Variable declaration\n", "P=5 # Power(hp)\n", "f=3600 # frequency(rpm) \n", "Tmax=8500 # Maximum torque(psi)\n", "\n", "#Calculation \n", "P=P*(6600) # Converting power into lb/s\n", "f=(3600)/(60.0) # Converting frequency into cycles per second\n", "T=(P)/(2*(math.pi)*f) # Torque exerted on the shaft\n", "Ratio=T/Tmax # Here we are finding the value of J/c\n", "c=(((10.30)*(pow(10,-3))*(2))/(math.pi))**(1/3.0)\n", "d=2*c # Diameter of the shaft that should be used\n", "\n", "#Result\n", "print('Case(a): Size of shaft = %1f lb.in' %d)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Example 3.07, Page number 178" ] }, { "cell_type": "code", "execution_count": 18, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Case(a): Tube thickness that should be used = 4.414195 lb.in\n" ] } ], "source": [ "import math\n", "from sympy import symbols, solve\n", "\n", "#Variable declaration\n", "P=100 # Power(kW)\n", "f=20 # frequency(Hz) \n", "c2=25 # radius(mm) \n", "Smax=60 # Shearing stress maximum(MPa)\n", "c1=symbols('c1')\n", "\n", "#Calculation \n", "T=P/(2*(math.pi)*f) # torque exerted on the shaft\n", "Ratio=T/(Tmax*(pow(10,6))) # Here we are finding the value of J/c2\n", "c1=solve((0.025)**4-(0.050/math.pi)*(13.26*(pow(10,-6)))-(c1)**4,c1) \n", "Thickness=c2-(c1[1]*1000) # Thickness that should be used if the shearing stress is not to exceed\n", "\n", "#Result\n", "print('Case(a): Tube thickness that should be used = %1f lb.in' %Thickness)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## SAMPLE PROBLEM 3.6, Page number 180" ] }, { "cell_type": "code", "execution_count": 19, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Case(a): Maximum power that can be transmitted = 890.000000 hp\n", "Case(b): Percentage in power = 11.000000 \n" ] } ], "source": [ "import math\n", "\n", "#Variable declaration\n", "D=7.5 # Diameter of the bigger shaft(in)\n", "d=3.75 # Diameter of the smaller shaft(in)\n", "r=0.5625 # Inner radius(in)\n", "k=1.33 # Stress concentration factor\n", "\n", "#Calculation \n", "temp1=(D/d) \n", "temp2=(r/d)\n", "T=round((1/2)*(math.pi)*((1.875)**3)*(8/1.33),1) # Maximum torque(ksi) \n", "#Power\n", "f=(900/60) # Frequency(Hz)\n", "Pa=(2*(math.pi)*15*62.3*pow(10,3)) # Power(lb/s)\n", "Pa=round(Pa/6600) # Power(hp) \n", "#Final Design\n", "r=15/16 # Radius(in) \n", "temp2=(0.9375/3.75) \n", "k=1.20 # Stress concentration factor\n", "T=(10.35*(8/1.20)) # Torque(kip.in)\n", "Pb=(2)*(math.pi)*(15)*(69)*(pow(10,3)) # Power(lb/s) \n", "Pb=round(Pb/6600) # Power(hp)\n", "#Percent Change in Power\n", "PC=round(((Pb-Pa)/Pa)*100)\n", "\n", "#Result\n", "print('Case(a): Maximum power that can be transmitted = %1f hp' %Pa)\n", "print('Case(b): Percentage in power = %1f ' %PC)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Example 3.08, Page number 189" ] }, { "cell_type": "code", "execution_count": 20, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Case(a): Radius of elastic core = 15.749013 mm\n", "Case(b): Angle of twist = 8.494327 degree \n" ] } ], "source": [ "import math\n", "from sympy import symbols, solve\n", "\n", "#Variable declaration\n", "l=1.2 # length(m)\n", "d=50 # diameter(mm) \n", "T=4.60 # torque(kN) \n", "G=77 # modulus of rigidity(GPa)\n", "ty=150 # torque(MPa)\n", "c=25 # radius(mm)\n", "py=symbols('py')\n", "\n", "#Calculation \n", "# Case(a)\n", "J=(1/2)*(math.pi)*(25*(pow(10,-3)))**4 # Polar moment of inertia(m**4) \n", "Ty=(J*ty*(pow(10,6)))/(c*(pow(10,-3))) # Torque(kN.m)\n", "py=solve((py/c)**3-4+((3*4.60)/(3.68)),py) # Radius of elastic core(mm)\n", "# Case(b)\n", "phyY=(Ty*l*(pow(10,3)))/(J*G*(pow(10,9))) # Angle of twist at the onset of yield(rad) \n", "phy=((93.4*(pow(10,-3)))/(0.630))*(360/(2*math.pi)) # Angle of twist(degree)\n", "\n", "# Result\n", "print('Case(a): Radius of elastic core = %1f mm' %py[0])\n", "print('Case(b): Angle of twist = %1f degree ' %phy)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Example 3.09, Page number 190" ] }, { "cell_type": "code", "execution_count": 21, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Case(a): Permanent twist = 1.810362 degree\n", "Case(b): Residual stress = 187.296417 MPa \n" ] } ], "source": [ "import math\n", "\n", "#Variable declaration\n", "T=4.60*(10**3) # Torque(N.m)\n", "L=1.2 # length(m) \n", "G=77*(10**9) # modulus of rigidity(Pa)\n", "J=614*(10**-9) # Polar moment of inertia(m**4)\n", "phy=8.50 \n", "c=25*(10**-3) # radius(m)\n", "\n", "#Calculation \n", "# Case(a)\n", "phyl=((T*L)/(J*G))*(360/(2*(math.pi))) # Lateral twist(degree)\n", "phyp=phy-phyl # Permanent twist(degree)\n", "# Case(b)\n", "Tlmax=((T*c)/(J))/(pow(10,6)) # Residual stresses(MPa) \n", "\n", "# Result\n", "print('Case(a): Permanent twist = %1f degree' %phyp)\n", "print('Case(b): Residual stress = %1f MPa ' %Tlmax)" ] }, { "cell_type": "markdown", "metadata": { "collapsed": true }, "source": [ "## SAMPLE PROBLEM 3.7, Page number 192" ] }, { "cell_type": "code", "execution_count": 23, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Case(a): Magnitude of T when yield first occurs = 5.730000 degree\n", "Case(b): Angle of twist when the deformation has become fully plastic.= 8.590000 MPa \n" ] } ], "source": [ "import math\n", "from sympy import integrate, symbols\n", "\n", "#Variable declaration\n", "p=symbols('p')\n", "G=(11.2*pow(10,6)) # Modulus of rigidity(GPa)\n", "ty=21 # Stress(ksi) \n", " \n", "#Calculation\n", "#Geometric properties\n", "c1=(1/2)*(1.5) # Radius(in)\n", "c2=(1/2)*(2.25) # Radius(in) \n", "J=round((1/2)*(math.pi)*(c2**4-c1**4),2) # Polar moment of inertia(in**4)\n", "#Case(a) Onset of yield\n", "tmax=ty\n", "Ty=round(((21)*(2.02))/1.125,1) # Torque at y(kip.in)\n", "phyY=((21*(10**3))*(60))/(1.125*11.2*pow(10,6)) # Onset of yield(rad)\n", "phyY=round(phyY*(360/(2*math.pi)),2)\n", "#Case(b) Fully Plastic Deformation\n", "Tp=round(2*math.pi*21*integrate(p**2, (p, c1, c2)),1) # Torque(kip.in) \n", "phyF=round((21*pow(10,3)*60)/(0.75*11.2*pow(10,6)),4) # Fully plastic deformation(rad) \n", "phyF=round(phyF*(360/(2*math.pi)),2)\n", "\n", "# Result\n", "print('Case(a): Magnitude of T when yield first occurs = %1f degree' %phyY)\n", "print('Case(b): Angle of twist when the deformation has become fully plastic.= %1f MPa ' %phyF)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## SAMPLE PROBLEM 3.8, Page number 193" ] }, { "cell_type": "code", "execution_count": 26, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Case(a): Residual stress = 0.000000 kip.in\n", "Case(b): Permanent angle of twist= 1.890000 degree \n" ] } ], "source": [ "import math\n", "\n", "#Variable declaration\n", "Tp=44.1\n", "phyF=8.59\n", "\n", "# Calculation\n", "# Elastic Unloading\n", "Tmax=((44.1)*(1.125))/2.02\n", "Tmin=(Tmax)*(0.75/1.125)\n", "phyl=round(((44.1*pow(10,3)*60)*(360/(2*math.pi)))/((2.02)*(11.2*pow(10,6))),2)\n", "phy=phyF-phyl\n", "\n", "# Result\n", "print('Case(a): Residual stress = %1f kip.in' %(0))\n", "print('Case(b): Permanent angle of twist= %1f degree ' %(phy))" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Example 3.10, Page number 202" ] }, { "cell_type": "code", "execution_count": 28, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Case(a): Shearing stress in each wall = 8.346688 ksi\n", "Case(b): Shearing stress in wall AB and AC= 11.128917 ksi \n", "Case(b): Shearing stress in wall BD and CD= 6.677350 ksi \n" ] } ], "source": [ "import math\n", "\n", "#Variable declaration\n", "t=0.160 # thickness(in)\n", "T=24 # Torque(kip.in) \n", "\n", "\n", "#Calculation \n", "# Case(a)\n", "Area=3.84*2.34 # Area bounded by centre line(in**2) \n", "t=(T)/(2*t*Area) # shearing stress in wall(ksi) \n", "# Case(b)\n", "tABAC=0.120\n", "tBDCD=0.200\n", "tAB=(T)/(2*tABAC*Area) # shearing stress in wall(ksi)\n", "tAC=tAB\n", "tBD=(T)/(2*tBDCD*Area) # shearing stress in wall(ksi)\n", "tCD=tBD\n", "\n", "# Result\n", "print('Case(a): Shearing stress in each wall = %1f ksi' %t)\n", "print('Case(b): Shearing stress in wall AB and AC= %1f ksi ' %tAB)\n", "print('Case(b): Shearing stress in wall BD and CD= %1f ksi ' %tCD)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## SAMPLE PROBLEM 3.9, Page number 203" ] }, { "cell_type": "code", "execution_count": 30, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Largest torque on bar with square cross section = 532.480000 N.m\n", "Largest torque on bar with rectangular cross section = 414.400000 N.m\n", "Largest torque on square tube = 555.000000 N.m\n" ] } ], "source": [ "import math\n", "\n", "#Bar with Square Cross Section\n", "#Variable declaration\n", "tALL=40 # Stress(MPa)\n", "\n", "\n", "#Calculation\n", "# Bar with square cross section\n", "a=0.040 # Length(m) \n", "b=0.040 # Length(m)\n", "temp=(a/b) \n", "c1=0.208 # Coefficient\n", "tmax=tALL # Maximum stress(MPa)\n", "T1=(40)*(pow(10,6))*(0.208)*(pow(0.040,3)) # Torque(N.m)\n", "\n", "# Bar with Rectangular Cross Section.\n", "a=0.064 # Length(m)\n", "b=0.025 # Length(m)\n", "temp2=(a/b)\n", "T2=(40)*(pow(10,6))*(0.259)*(0.064)*(pow(0.025,2)) # Torque(N.m)\n", "\n", "#Square Tube\n", "A=(0.034)*(0.034) # Area bounded by the center line of the cross section(m**2)\n", "T3=round((40)*(pow(10,6))*(2)*(0.006)*(1.156)*(pow(10,-3)),0) # Torque(N.m)\n", "\n", "# Result\n", "print('Largest torque on bar with square cross section = %1f N.m' %T1)\n", "print('Largest torque on bar with rectangular cross section = %1f N.m' %T2)\n", "print('Largest torque on square tube = %1f N.m' %T3)" ] } ], "metadata": { "kernelspec": { "display_name": "Python 3", "language": "python", "name": "python3" }, "language_info": { "codemirror_mode": { "name": "ipython", "version": 3 }, "file_extension": ".py", "mimetype": "text/x-python", "name": "python", "nbconvert_exporter": "python", "pygments_lexer": "ipython3", "version": "3.5.1" } }, "nbformat": 4, "nbformat_minor": 0 }