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How do I represent load distribution of over hanging object

  1. Dec 3, 2015 #1
    I am trying to make a FBD for equipment which is half out of the beam, I realise I can represent it with a single line of force.

    But how if I want to go for distribute load? which one is the correct one. (purposely not drawing reaction force on floor.)

    https://qph.is.quoracdn.net/main-qimg-75c17d5dcf6179df24c60cfe0eb70165?convert_to_webp=true [Broken]
     
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Dec 3, 2015 #2

    PhanthomJay

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    Assuming uniformly distributed equipment weight with its cg at the center, your first FBD is ok if you are looking at the beam/equipment system (but with floor reactions required). Note however your text and dimensions don't agree with your figure , if the equipment overhangs half the beam then it's cg should be at the end of the beam (plate). Also note that if your looking at a FBD of the beam, then neither diagram is correct, because the beam just sees the precariously balanced equipment load as a Normal force point load at its free end, not as a distributed load.
     
  4. Dec 3, 2015 #3
    I'm intending to draw shear and moment diagram for the beam, but first I need to get the FBD correct first to be used in Shear and moment diagram. assuming I have drawn reaction force from the floor and reaction force of bolt holding the plate at the floor (Left side).
    On the end of beam (Right) should I use point load or distributed load?
    If i'm using distributed load at the end beam, my shear diagram for beam will not be zero at the end because plate is finish but not the distributed force,
    But if I'm using point load, it is acceptable?

    Thank you.
     
  5. Dec 4, 2015 #4

    PhanthomJay

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    With half the equipment overhanging the beam, the beam just sees a 10 N point load at its end. Remember that the equipment weight is the force of gravity between the equipment and the earth, not the force on the beam. The beam sees the normal contact force from the equipment , which in this case is the point load at the free end as if it was a fulcrum on a see-saw. If the equipment was fully seated on the beam, then the normal force would be a uniformly distributed load like the weight force,of 1N/cm, but this is not the case here.
     
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