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Pin Connection with one support reaction?

  1. Apr 14, 2007 #1
    Please click on the image to make it bigger.


    At pin connnection G shouldn't there be two support reactions just like the pin connection at A?

    If the solution in the diagram is correct why is it correct?
    Last edited: Apr 14, 2007
  2. jcsd
  3. Apr 14, 2007 #2


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    The link is not working. I can't access your diagram.
  4. Apr 14, 2007 #3
    It is working for me. Do you have a pop-up blocker on? The link pops up in a new tab for me( runnning Firefox).

    If that wont work, right click on the image and select copy image location and paste that in to your browser.
  5. Apr 14, 2007 #4


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    Sorry, there's no response from the site whatever method I use.

    I'm outside the USA, maybe that's it.
  6. Apr 14, 2007 #5
    um, I posted the same question to another board. You can check it out athttp://www.sosmath.com/CBB/viewtopic.php?p=135924#135924

    I'd like to know if you have the same problem there as well.
  7. Apr 14, 2007 #6


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    OK, I got the pic. There's only one member connected to G, so only one reaction, but two on A. Is that what's troubling you ?
  8. Apr 14, 2007 #7
    Yes, so if there is only one member connectecd then only one reaction is needed? I thought that pin connection always had two reactions regardless of the members.

    BTW thank you.
    Last edited: Apr 14, 2007
  9. Apr 14, 2007 #8


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    In a statics problem, one resolves the forces at each node with the constraint that the total is zero because nothing is moving.

    At A there are two forces to be resolved into the horizontal reaction, at G there's only one.

    I don't see what is causing you a problem. Am I missing something.
  10. Apr 14, 2007 #9
    No, I was missing something. You cleared it up for me :O
  11. Aug 3, 2009 #10
    i m not still clear plz help me out. Both A and G are pin connected then why there is one support reaction at G and two at A
  12. Aug 3, 2009 #11


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    Members subject to forces at their hinges only (a force at each end, with no forces in between the ends) are known, appropriately, as "two-force" members, and in order for the member to be in equilibrium, the forces acting on the member at each end must be equal and opposite and co-linear, acting along the longitudinal 'axial' axis of the member. Since the sole member at G is horizontal, the force acting on it at the pin, G, must be horizontal, with no vertical component.
  13. Aug 5, 2009 #12
    Thanks alot i got it :)
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