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Total moment caused by an unbalanced mass on a rotating shaft

  1. Aug 1, 2017 #1
    1. The problem statement, all variables and given/known data

    The shaft connecting A to D is rotating at 1200RPM clockwise and has a concentrated unbalanced load of W2 of 8lb at a radius of 2in. in a plane midway between points A and D as shown in the figure. A constant vertical load W1 of 1500 lb is applied by means of two anti-friction bearings B and C. If the shaft material has an ultimate strength of 60,000 PSI and yield strength 45,000 PSI, determine the required diameter of the shaft at the mid section for a design factor of safety N=2. Only consider the mid point.

    upload_2017-8-1_19-35-40.png
    2. Relevant equations
    W1=1500 lb
    W2=8 lb

    3. The attempt at a solution
    The book gives an answer, and part of the answer is this : "(1)Show that the bending moment due to the rotating load varies from -6540 in-lb to 6540 in-lb. (2) show that the bending moment due to the constant vertical load is 7500 in-lb. (3) The combined bending moment then varies from 960 in-lb to 14,040 in-lb."

    I can't figure out how the book got (1) or (2). Here is my work :
    To get forces at B and C due to load W1:
    $$B+C=-1500\\\\14(1500)-20C=0\\\\B=-450\\\\C=-1050$$
    To get forces at A and D:
    $$A-450-8-1050+D=0\\\\450*10+8*20+1050*30-D*40=0\\\\A=604\\\\D=904$$

    This gives me this moment diagram:
    upload_2017-8-1_19-48-30.png

    7580 total moment max at the 20" mark.

    Okay, so the book says the bending moment due to the rotating load varies from -6540 to 6540 in-lb.
    I don't understand where they got this from. Is this ONLY the rotating load? So, that would mean solve a force diagram where W1 is removed? That makes no sense because then you would have A=4 lb B = 4 lb holding up W1=8 lb. The bending moment would be 8*20 = 160

    So, I try to take the rotation into account.
    It is rotating at w=1200RPM $$w=[1200[RPM]*\frac{2\pi}{60}[\frac{Rads}{RPM * Sec}] = 125.7[\frac{Rads}{Sec}]$$
    This would give the mass a velocity of RwCos(wt) = 2*125.7= 251.4
    Also an acceleration dv/dt = -Rw^2sin(wt), with a min and max of +-Rw^2.
    But... ##R*w^2=2*125.7^2=31601 >> 6540## ... So this has to be wrong. Not even close to 6540.

    Here is another failed attempt :
    If I look at that section of the beam as a 2D section, it would be a circle with a mass 2 inches from the center. Sum of torque = change in angular momentum :
    $$T= dH/dt \\\\
    H=I_{zz}w \hat{k} + [-R\hat{i} x (-8Rw\hat{j})]$$
    mass of the shaft is 0, so I_zz vanishes. ##H=8R^2w\hat{k}##
    dH/dt=0 because w is constant. Sum of Torques = 0. That would mean r cross F = 0, no forces in the tangential direction, nothing that would point down..

    What am I doing wrong!? Please help
     

    Attached Files:

  2. jcsd
  3. Aug 3, 2017 #2

    CWatters

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    Not really my field, and I normally work in Metric but you posted this yesterday and haven't had a reply yet so here goes...

    That would be my understanding.

    I agree ω is 125.7 rads/s

    I'm really struggling to remember how to use imperial units but I think that's where the issue is....

    The centripetal force is..
    Fc = mω²r
    where
    mass m = 8lbs = 0.2484 slugs
    ω = 125.7 rads/s
    r = 2/12 = 0.1667ft

    Fc = 0.2484 * 125.72 * 0.1667
    = 654.3 lbf

    So now we have a beam AD length 40" (3.33ft) with a 654.3lbf force in the middle. Then calculate the bending moment. Note their answer is in "in-lbs" not "ft-lb". Again I'm very rusty but I think in this simple case it's

    20" * 0.5 * 654.3 = 6543 in-lb

    Close enough.
     
  4. Aug 3, 2017 #3

    CWatters

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    Q2 asks..
    but you wrote
    ..which includes W2 the 8lb rotating mass. I think you should ignore that in which case you get..

    A−450−1050+D=0
    450∗10+1050∗30−D∗40=0
    (4500 + 31500)/40 = D
    A=600
    D=900
     
  5. Aug 5, 2017 #4
    Thanks... wow... it was the units... LOL.

    Hey, why did you multiply the torque equation by 0.5? (20 inches * 0.5 * 654.3 lbF) instead of (20 inches * 654.3 lbF) Is it an average because its a rotating?

    PS Thanks so much for helping! I really really appreciate it! Thought this one was not gonna get an answer!
     
  6. Aug 5, 2017 #5

    CWatters

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    No. As I recall when you work out the bending moment you first work out the reaction force at each end of the beam. The 654.3lbf load is in the middle so the reaction force at each end is half of that.
     
  7. Aug 5, 2017 #6
    thanks!
    ?u=http%3A%2F%2Fmemecrunch.com%2Fmeme%2F2TQQO%2Fhero-of-the-day%2Fimage.png
     
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