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Forces on a beam, replace with force and couple

  1. Feb 9, 2015 #1
    1. The problem statement, all variables and given/known data
    http://oi58.tinypic.com/2a5d4w3.jpg


    2. Relevant equations

    So I can just sum up all the forces then sum the moments?


    3. The attempt at a solution


    FBx = 0 lb
    FCx = 800 cos30 = 692.82 lb
    FEx = 0 lb

    FBy = -2500 lb
    FCy = 800sin30 = 400 lb
    FEy = -1200 lb

    Resultant force:
    FAx = 0 + 692.82 + 0 = 692.82 lb
    FAy = -2500 + 400 - 1200 = -3300 lb

    Moments about point C:
    MOB = (2500)(2) = 5000 lb*ft counterclockwise
    MOC = 0 lb*ft
    MOE = having trouble with this one


    Does everything seem ok here so far?
     
  2. jcsd
  3. Feb 9, 2015 #2

    SteamKing

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    It's not clear why you are calculating moments about point C. The problem statement clearly states that the three forces are to be replaced by a force-couple combination acting at point A.
     
  4. Feb 9, 2015 #3
    I thought I could calculate moments anywhere?
     
  5. Feb 9, 2015 #4

    SteamKing

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    Yes, you can calculate moments anywhere, but when you are asked to find them at a particular point, it saves time to use that point as the reference for your moment calculations.

    BTW, your moment calculations about point C were incorrect, anyway. Remember, the force must act perpendicular to the moment arm in order to calculate the magnitude of the moment.
     
  6. Feb 9, 2015 #5
    So moment about A:

    B: 2500*4 = 10000 ft*lb

    That's correct, right?

    The reason why I wanted to take them about C is because of the angling on C and to simplify it more.
     
  7. Feb 9, 2015 #6

    SteamKing

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    The magnitude of this moment is correct, but you need to establish a sign convention for your moment directions. Moments can be either clockwise or counterclockwise, and failing to distinguish between these two directions will lead to erroneous results.

    The angle of the arm BCD w.r.t. the horizontal only affects the moment due to the force at C.
     
  8. Feb 9, 2015 #7
    Oh B should be counterclockwise, right?

    so then C should be Moc = (800)(6)? If I get what you're saying?
     
  9. Feb 9, 2015 #8

    SteamKing

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    Right.

    Point C is not 6' from point A. Remember the definition of what a moment is.

    moment.PNG
     
  10. Feb 9, 2015 #9
    So because it's on an angle we want only the horizontal component of where C is?
     
  11. Feb 9, 2015 #10

    SteamKing

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    Yes.

    But it's not that simple. Because the force at C is also directed at an angle, it must be broken up into its components in order to calculate the moment acting about point A.
     
  12. Feb 9, 2015 #11
    so we want the cos of the angle? 2*cos(30) = 1.732 ft
     
  13. Feb 9, 2015 #12

    SteamKing

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    The distance from A to C would be 4' + 2' * cos (30°), but the force at C is 800 lbs. acting at 60° to the horizontal. You've got to compute the components of that force. Each force component will create a moment acting about point A.
     
  14. Feb 9, 2015 #13
    4+ 2 cos(30) = 5.73

    Moc = (5.73) * 800cos60 = (5.73)*400 = 2292 clockwise, right?

    If I'm understanding you, that is.
     
  15. Feb 10, 2015 #14

    SteamKing

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    Nope. You should draw the components of the force at C on your picture and see how they produce their moments.
     
  16. Feb 10, 2015 #15
    I made a mistake:

    4+ 2 cos(30) = 5.73

    Moc = (5.73) * 800sin60 = (5.73)*692.82 = 3969.86 clockwise, right?

    Does that seem better?
     
  17. Feb 10, 2015 #16

    SteamKing

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    The magnitude looks OK. Are you sure the direction of this moment is clockwise w.r.t. point A?

    What about the moment due to the horizontal component of the force at C?
     
  18. Feb 10, 2015 #17
    No it should be counterclockwise I believe. Is there a better way to tell despite intuition?

    In the case of C:
    10.46 * 1200 = 12552 lb*ft
     
  19. Feb 10, 2015 #18

    SteamKing

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    This is actually the moment due to the force applied at point E.

    What about the moment due to the horizontal component of the force applied at point C?

    Examination of the direction of the force in relation to the reference point is fine. There are different methods of calculating moments which take care of having to visualize which direction the moment will take, but you will learn those later.
     
  20. Feb 10, 2015 #19
    Sorry, I mistyped and meant E :p

    Wouldn't the horizontal component of C cause a counterclockwise rotation also? Since it would push against the angle and push it upwards.
     
  21. Feb 10, 2015 #20

    SteamKing

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    Yes, it would.
     
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