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Distributed load and stress

  1. Feb 8, 2007 #1
    I've attached a picture of the problem - I'm NOT looking for a solution, just an explanation.

    You can see the dimensions x,y, and z. That box is a detail of the connection at the rod interface. There is a distributed load acting across the lower beam of w lb/ft. Given the allowable shear and normal stresses of the blot and rod respectively, I need to find w.

    Again, I don't want a solution to this problem, I want an explanation of how you apply the concept of the distributed load to that angled connecting rod/bolt system. How do the forces go into the components of it? I know that when you're computing internal reactions, you can't find the "center point" of the distributed load, but this is different I guess...

    My thoughts were that you can do a moment about the wall point to find a reaction at the joint. With that, you can then determine what forces will be acting at the rod/bolt interface and then you can find what maximum w will cause failure.

    Help maybe?


    http://img211.imageshack.us/img211/8696/cablejm2.gif [Broken]
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Feb 8, 2007 #2
    The connections at the wall are drawn poorly on my part - those are NOT moment connections (i.e. they're pins...they cannot hold moments)
  4. Feb 8, 2007 #3
    Because the load is even, it acts at the center of the beam. Now you have a concentrated force and a location.

    Now assume the bolt fails in shear, and work to find the value of w.

    Then assume if fails in normal, and find the value of w.

    The smaller value of w is the answer.

    And where are my angles in the other thread?
  5. Feb 8, 2007 #4
    Haha - I'll give you a all my data for that last thread when I get back from the Union here at Purdue.

    As for this problem that's kind of what I'm doing, but for some reason, I'm getting an answer about 0.150 kips/ft too much. I can only assume you sum moments about the wall point there to get w in terms of that force, right?

    And also, the force you find at that connection (being a two force member) is directed along the slanted rod. Is that force there going to be the one that causes shear failure? I'm assuming yes...
  6. Feb 8, 2007 #5
    Your process seems right. Just look at it more and think.
  7. Feb 8, 2007 #6

    I've been using Arctan (4/3) instead of Arctan (3/4) this WHOLE time.

    That's what my problem was. Thank you.

    SO! My question then, if it's a distributed load that's uniform, you can assume it to be acting at that point and solve it like a frame to find reactions and get the failure loads? Right?
  8. Feb 8, 2007 #7
    it acts at the centroid of the distribtion. In your case, the middle of the beam.
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