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Math Help - Rotating shaft

  1. #1
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    Rotating shaft

    A shaft 2 m long rotates at 1500 rpm between bearings. The bearings experience forces of 5 kN and 3 kN acting in the same plane.
    The shaft is to be balanced using two masses (m1 and m2)placed 0.5 m and 1.5 m from end A and 180 from the direction of the bearing reactions, each on radius arms 100 mm long. Calculate the sizes of m1 and m2.

    Now through a process of elimination I know what the masses m1 and m2 should be and also what these forces f1 and f2 are.

    My problem is taking the 2 m shaft and how I got to these values without just plugging in different numbers till I got there. I think it must be something about the summing of the moments from the right hand or left hand end. Hope someone can point me in the correct direction with this.....
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  2. #2
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    Re: Rotating shaft

    Think it maybe something like F1+F2 = 2 ( Ra-Rb ) But not sure and if this was correct how I narrow it down to solve for one of the variables it being either F1 or F2. Could really be doing with a shove in the right direction with this as I have been at it for days!!
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  3. #3
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    Re: Rotating shaft

    Surely someone must have a clue for this???
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  4. #4
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    Re: Rotating shaft

    How about this as a solution and see if anyone agrees......
    Ra=5kN Rb=3kN
    Ra + (Rb) = F1+F2
    Ra = F1+F2-Rb
    Leaves
    5kN=F1+F2-3kN
    F1-F2=5kN-3kN
    F1=F2+2kN
    Now
    F1+F2 must equal 8kN
    So F1=6kn
    F2=2kN
    Would this be an acceptable route??
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  5. #5
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    Re: Rotating shaft

    I looked at your problem and didn't get it. Could you upload a sketch of what's going on? These arms the balancing weights are mounted on stick out of the shaft 100 mm?
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  6. #6
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    Re: Rotating shaft

    The image shows the dimensions please not though f1 and f2 are downwards though and the reactions upwards.
    The shaft is 2 metres long.
    Bearing reaction A is place at 0 metres and is 5kN upwards
    Bearing reaction B is placed at 2meres and is 3kN upwards
    There are point masses downwards at 0.5 metres from the left hand end and a point mass at 1.5 metres also downwards
    Hope this helps....
    Attached Thumbnails Attached Thumbnails Rotating shaft-attachment.jpg  
    Last edited by Jock; February 27th 2014 at 07:03 AM.
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  7. #7
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    Re: Rotating shaft

    The image shows the dimensions please not though f1 and f2 are downwards though and the reactions upwards.
    huh?
    What about the "radius arms" that are 100mm each. They won't usually give dimensions to something in a problem unless it's important.

    The balancing weights are placed 180 deg from the bearing reactions? Don't these weights rotate with the shaft? How is that 180 deg maintained?

    Still not getting it.
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  8. #8
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    Re: Rotating shaft

    I used the radius arms to find the masses once I worked out what the forces of F1 and F2 were. My problem was justifying how I got the forces F1 and F2 rather than just plugging in various different values so the radius arms are irrelevant for what I am trying achieve. This below is what I am trying to solve then from there I solve for the masses using the radius arms....
    Ra=5kN Rb=3kN
    Ra + (Rb) = F1+F2
    Ra = F1+F2-Rb
    Leaves
    5kN=F1+F2-3kN
    F1-F2=5kN-3kN
    F1=F2+2kN
    Now
    F1+F2 must equal 8kN
    So F1=6kn
    F2=2kN
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  9. #9
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    Re: Rotating shaft

    Quote Originally Posted by Jock View Post
    F1-F2=5kN-3kN
    where does this come from?
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  10. #10
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    Re: Rotating shaft

    This is where I need the help Iam not sure how to get the F2 = 2kn and F1 = 6kn as required. Any ideas????
    I thought by moving the F1+F2 from the right hand side of the equation to left I could change them from addition to subtraction
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  11. #11
    MHF Contributor ebaines's Avatar
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    Re: Rotating shaft

    It appears that you have determined forces F1 and F2 that cause reactions Ra and Rb for a weightless horizontal stationary beam. You've assumed the shaft is not rotating at all - it's a beam with the two masses F1 and F2 and two simple supports at Ra and Rb. So by setting sum of vertical forces = 0 and sum of moment about either end = 0 you can determine that F1= 6N and F2=2 N.

    However, for a rotating shaft you want to balance the centripetal forces of the two masses with the bearing reaction forces, which gives:

     \sum F = 0 = R_a+R_b - m_1 \omega^2 R - m_2 \omega^2 R

     \sum T_a = 0 = 0.5 (m_1 \omega^2 R) + 1.5(m_2 \omega^2 R) - 2(R_b)

    Now solve for m_1 and m_2.
    Last edited by ebaines; February 27th 2014 at 08:26 AM.
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  12. #12
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    Re: Rotating shaft

    Surely you need to know the force if you don't know the mass though??
    I follow the rest of the answer though
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  13. #13
    MHF Contributor ebaines's Avatar
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    Re: Rotating shaft

    Quote Originally Posted by Jock View Post
    Surely you need to know the force if you don't know the mass though??
    I have no idea what you mean. From your previous work you know that F_1= 6N; this force equals m_1\omega^2R, and F_2= 3N = m_2 \omega^2 R
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  14. #14
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    Re: Rotating shaft

    I only got the F1 and F2 values through crunching numbers for a day then in turn I got the masses I needed . I need to find a better route to getting to that point rather than telling someone it was through trial and error
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  15. #15
    MHF Contributor ebaines's Avatar
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    Re: Rotating shaft

    As I indicated back in post #11: there are two equations in two unknowns:

    Sum of vertical forces = 0: \sum F = 0 = R_a + R_b -(F1 + F_2)

    Sum of moments about one end or the other = 0. Let's take moments about point A:  \sum T_a = 0 = 0.5F1 + 1.5F_2- 2R_b

    Now you can solve for F1 and F2, and you should find F1=6N and F2=2N.

    BTW, what class are you doing this for? Using this method is introductory mechanics....
    Thanks from Jock
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