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Archived Power transfer through shaft

  1. Mar 15, 2013 #1
    A shaft runs at 300 rev/min and transmits power from a pulley A at one end to two
    pulleys, B and C, which each drive a machine in a workshop. The distance between
    pulleys A and B is 3 m and that between B and C is 2.4 m (i.e. the total length of the
    shaft is 5.4 m). The shaft has a diameter of 50 mm between A and B and a diameter
    of 40 mm between B and C. If the maximum permissible shear stress in the shaft is 80
    MPa, calculate the maximum power which may be supplied from each of the pulleys B
    an C, assuming that both machines would be in operation at the same time. Also
    calculate the total angle of twist of one en
    d of the shaft relative to the other when
    running on full load. G = 80 (GPa)

    Answers:
    (30.2 kW; 31.4 kW; 13.7°)


    Attached is my working, i can arrive at the answer of 31.4KW for the pulley at C at the smaller diameter step, however when i try the same calculate with the second moment of area (J) for the larger diameter i get an answer in excess of 60000KW. How would i calculate the torque for the larger diameter section?

    as you can see from my working ive used the standard formula rearranged for torque, then used the standard power equation.

    Thanks
     

    Attached Files:

  2. jcsd
  3. Feb 24, 2016 #2

    CWatters

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    Science Advisor
    Homework Helper

    Here is a diagram of the set up..
    Pulleys.jpg

    Relevant Equations:
    Power = T * ω .........................(1)
    where
    T = Torque in Nm
    ω = Angular velocity in Rads/S

    The Polar Second Moment of Area (J) for a solid round shaft is
    J = πd4/32........................(2)
    where d is the diameter in meters.

    Max Stress τmax = Td/2J.................(3)

    Substitute (2) into (3) and rearrange to give an equation for the Max Torque in terms of the Max Stress..

    T = (τmax * π * d3)/16...............(4)

    Shaft from B to C

    Now we can calculate the max torque and power allowed in the section of shaft from B to C...

    Using (4) and...

    τmax = 80 * 106Pa
    d = 0.04m

    T = (80*106 * π *0.043)/16
    = 1005Nm

    Then using (1) we can calculate the max power

    PowerC = 1005 * 300 * 2π/60
    = 31.6kW
    This is slightly higher than the book answer of 31.4kW. Looks like they rounded the Torque down to 1000Nm.

    Shaft from A to B

    Now we can calculate the max torque in the section of shaft from A to B and the power available at B...

    Using (4) and...

    τmax = 80 * 106Pa
    d = 0.05m

    T = (80*106 * π *0.053)/16
    = 1963Nm

    However this time to calculate the power available at B we must subtract the torque required at C ..

    T = 1963 - 1005 = 958NM

    Then using (1) we can calculate the max power that can be delivered to machine B..

    PowerB = 958 * 300 * 2π/60
    = 30.1kW
    This is fractionally lower than the book answer of 30.2kW. I haven't tried to account for the difference.
     
  4. Feb 24, 2016 #3

    CWatters

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    Angle of Twist

    Relevant Equations

    The angle of twist in each section is given by
    θ = TL/GJ............................(5)

    where
    T = Torque
    L = Length
    J = Polar Second Moment of Area of that section

    From my post above..
    J = πd4/32........................(2)

    Substituting for J in (5)...

    θ = 32TL/Gπd4

    For the section AB

    T = 1963Nm
    L = 3m
    d = 0.05m
    G = 80 * 109Pa

    so

    θAB = (32 * 1963 * 3) / (80 * 109 π * 0.054)

    = 0.12 Rads
    = 6.8 degrees

    For the section BC

    T = 1000Nm
    L = 2.4m
    d = 0.04m
    G = 80 * 109Pa

    so

    θAB = (32 * 1000 * 2.4) / (80 * 109 π * 0.044)

    = 76800 / 643398
    = 0.12 rADS
    = 6.8 degrees same as section AB

    Total Twist = 6.8 + 6.8 = 13.6 degrees
     
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