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Vertical reactions and UDLs

  1. Sep 4, 2008 #1
    I have to calculate the vertical reactions at point A and C.
    I know that the sum of all vertical forces is equal to zero but i can't solve the equation since there's two variables.
    [​IMG]
    Bv=50
    Cv=55
    Dv=45
    w=80
     
  2. jcsd
  3. Sep 5, 2008 #2

    tiny-tim

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    Hi Ry122! :smile:
    general rule … when you have a unknown force you can't get rid of, take moments about a point (strictly, an axis) through the line of that force.​

    In this case, take moments about A and about C.

    (Alternatively, find the point about which the moment of all the given forces is zero :wink:)
     
  4. Sep 5, 2008 #3
    Is this correct for the vertical reaction at C?
    0=((15/2)*80)-(50x5)-(55x10)+(Cyx10)-(45x14)
     
  5. Sep 6, 2008 #4

    tiny-tim

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    Hi Ry122! :smile:

    Yes … you've taken moments about A …

    and everything's correct (except you need a minus in front of the 15/2*80 :wink:) :smile:

    (I suggest you take moments about C also, just for practice … but of course, as I expect you've noticed, you don't need to, because you already know the total reaction for A and C together :wink:)
     
  6. Sep 8, 2008 #5
    for the UDL wouldnt the moment created about point A be the load (80) multiplied by the distance it covers (15m) then multipied by the half way point (7.5m) giving you a moment of 9000 KN/M about point a?

    rather than what you have with 7.5 x 80 giving you a moment of 600 KN/M about point a?
     
  7. Sep 8, 2008 #6

    tiny-tim

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    Welcome to PF!

    Hi badeany! Welcome to PF! :smile:

    ooh … you're right!

    I didn't notice the small print ("kN/m") on the diagram! :redface:

    Yes, 80 is the force-per-length, so it does have to be multiplied by the length. :wink:

    Thanks, and very well spotted! :biggrin:
     
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