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Homework Help: Statics - Method of Members

  1. Mar 27, 2008 #1
    What is the easiest way to determine the two force member in a frame question?

    When I look at a frame and see the pinned connections, I automatically assume that there is a reaction in the x and y directions.

    How can I tell at which pin there is only one reaction, and what direction it is in?

    All of the examples in my textbook state which member is the two force member and wether it is a pulling horizontally or vertically on the pin, whereas the problems do not give this information.

  2. jcsd
  3. Mar 27, 2008 #2
    The reactions in the x and y-directions are components of a single force. If two of these unknown forces act on one piece of the frame, that piece is a two-force member. It does not matter if there is an extra load of known force on the member, as long as you have two unknown forces acting on that particular piece of the frame, it is considered a two-force member. Does that make sense?
  4. Mar 27, 2008 #3
    Thanks for the reply. I understand what you are saying but I have trouble applying it.
    Could you maybe explain to me, on the attached truss where the two force member is and how you came to that conclusion? I would really appreciate it.

    Attached Files:

  5. Mar 27, 2008 #4
    I would like to but your attachment is "pending." It might be faster if you obtain a link for your image and post it.
    Last edited: Mar 27, 2008
  6. Mar 27, 2008 #5
  7. Mar 27, 2008 #6
    You dont have to get approval, somebody has to come along and ok it. It's a strange process. Anyways..

    I take it that points B, C, and D have pins, correct?
  8. Mar 27, 2008 #7
    Yes they do.
  9. Mar 28, 2008 #8
    You can break up the frame into two pieces: member ABD, and member BC. We have four different forces acting on the entire frame (neglecting its weight). Call these forces A, B, C, and D, where A is the 6kN load. Can you tell me which member has only two unknown forces acting on it (this will be the two force member)?
  10. Mar 28, 2008 #9
    That cleared things up alot thanks..
    So member BC is the 2 force member because it only has the forces B and C acting on it?

    Member BC is in tension so it can only pull on the pins at B and C? Is this correct?

    Because of this, when I draw the FBD of member BC the reactions are only in the X direction right?
  11. Mar 28, 2008 #10
    From the looks of things, member BC would most likely be in tension, yes. But I wouldn't draw conclusions about whether a member is in tension or compression until you know the magnitudes and directions of the unknown forces on the member.

    It's best to pick an arbitrary direction for each component of any unknown force. If you find the component to be a negative value, this means that the component points opposite to the direction you chose. I wouldn't say that the only reactions on BC are in the x-direction. The load acts vertically downward, and the frame is in equilibrium, where does the opposing force act?

    If the diagram you've posted is for a particular problem you're trying to work out, be sure to state all of the known information about the problem and what its asking for. I may be able to help more if I know more about the problem.

    Edit: And yes, BC is the two-force member.
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