What does a distributed load with arrows pointing up mean?

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Homework Statement


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


Homework Equations


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The Attempt at a Solution


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Attachments
  • Free-body diagram of a horizontal beam fixed to a wall on the right, with a uniformly distributed load of -30 kN/m shown as upward-pointing arrows along its length.
    sscsa.png
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  • Free-body diagram of a horizontal beam fixed at the right end with reaction Ax and Ay, showing upward-pointing distributed load of 40kN/m, a -50kN point load at C and -30kN/m at the support, with dimensions 3m, 1m and 4m, and empty V and M axes below.
    scan0001.jpg
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but then would the negative 30kN/m change the direction? meaning its acting downwards on the beam?
 
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.
 
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sorry about that, its just part of a larger question which i have in another thread. I have attached the complete diagram.
 
What quantity is measured in kN/m ?

Do you mean this to be pressure, or moment, or ... ?
 
its a uniformly distributed force/load which can then be simplified to a point load (single force as you mentioned earlier)
 
TyErd said:
its a uniformly distributed force/load which can then be simplified to a point load (single force as you mentioned earlier)

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.