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Drive line inertia and torsional effects exam question

  1. May 7, 2016 #1
    upload_2016-5-7_13-44-52.png
    Hi can anyone help me with this question on inertial forces in gears?
    Cheers
    Ali :)
     

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  2. jcsd
  3. May 7, 2016 #2

    jack action

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    For every shaft [itex]i[/itex], you get:
    [tex]\sum T_i = I_i \alpha_i[/tex]
    And for every gear [itex]i[/itex] connected to gear [itex]j[/itex], you get:
    [tex]\frac{T_{i\ out}}{T_{j\ in}} = \frac{N_i}{N_j}[/tex]
    [tex]\frac{\alpha_i}{\alpha_j} = \frac{N_j}{N_i}[/tex]

    This gives you 7 equations with 7 unknowns ([itex]T_{A\ out}[/itex], [itex]T_{B\ in}[/itex], [itex]T_{B\ out}[/itex], [itex]T_{C\ in}[/itex], [itex]\alpha_A[/itex], [itex]\alpha_B[/itex], [itex]\alpha_C[/itex]).
     
  4. May 7, 2016 #3
    i would be tempted to apportion an effective moment of inertia to shafts B & C (this takes into account the relative rotation rates)

    Select an arbitrary rotation rate for shaft A (say 10 rad/sec)
    Calculate the rotational KE of each shaft at this rate
    The effective moment of inertia of shafts B and C you calculate by comparing the KE values to that of shaft A
    For example :
    The KE of shaft A at 10 rad / sec = 75 joules
    The KE of shaft B at 5 rad / sec = 0.625 joules, so its effective moment of inertia = 1.5 * ( 0.625 / 75 ) = 0.0125 kg . m^2
    Repeat for shaft C then add A, B and C for total effective moment of inertia of the system
     
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