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Distributed loads

  1. Aug 19, 2011 #1
    1. The problem statement, all variables and given/known data
    what is the difference between a distributed load with arrows pointing up like the once i have attached and just the standard distributed load where the arrow is pointing down. Does it mean the force is acting up the beam instead of down?the value given is -30kN/m. does that mean its actually a load acting downwards

    2. Relevant equations

    3. The attempt at a solution

    Attached Files:

    Last edited: Aug 20, 2011
  2. jcsd
  3. Aug 19, 2011 #2


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    Homework Helper

    That sort of just means that the load is on the underside of the beam.
  4. Aug 19, 2011 #3
    but then would the negative 30kN/m change the direction? meaning its acting downwards on the beam?
  5. Aug 19, 2011 #4


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    Staff: Mentor

    I think in copying the diagram you have lost some of its clarity. Are you sure it's a beam? Maybe it's two beams or bodies, with the lower one supporting the upper one? (Like a man lying on a bed of nails.)

    As you can't have a force without an opposing one, then if the lone arrow represents pressure (and it would be better to have multiple arrows for this), then the multitude of up arrows should indicate the beam's reaction. (A more vivid interpretation would be that this represents the cross-section of a wall of a pressure or vacuum vessel.)

    The fact that you have pressure written as kN/m doesn't give confidence that the diagram is exactly correct. Maybe it isn't pressure; maybe it's a single force. And why a curved arrow?

    You need to supply the context in which the diagram arises.
    Last edited: Aug 19, 2011
  6. Aug 20, 2011 #5
    sorry about that, its just part of a larger question which i have in another thread. I have attached the complete diagram.
  7. Aug 20, 2011 #6


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    Staff: Mentor

    What quantity is measured in kN/m ?

    Do you mean this to be pressure, or moment, or .... ?
  8. Aug 20, 2011 #7
    its a uniformly distributed force/load which can then be simplified to a point load (single force as you mentioned earlier)
  9. Aug 20, 2011 #8


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    Ah, now I understand. Two separate sketches would have been better.

    So has your question been answered? In fact, what was your question?

    Probably this would arise as: what single force applied to the end of the beam would exert the same torque about the beam's point of attachment as does a uniformly distributed load along the length of the bean of x N/m.
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