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Statics - Pin jointed structure - Equations of equilibrium and reactive forces

  1. Aug 4, 2012 #1
    1. The problem statement, all variables and given/known data

    Using the pin-jointed structure in the image:
    a) Set up ALL equations of equilibrium for the structure.
    b) Determine ALL reactive forces within the structure, including the directions and locations where they act.

    I have attempted the problem however i can't seem to find a probable method to go about solving the questions. Any help on finding a method to solve the above would be fantastic!

    2qbdls5.jpg
     
  2. jcsd
  3. Aug 4, 2012 #2

    PhanthomJay

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    I guess joints B, C, and D are pinned connections, but A must be a fixed support or else this structure will tumble over. It is important to recognize so called '2-force' members which are members subject to forces at each end and nothing in between. BD and CD are 2-force members, and as such, take axial forces only directed along the member. Support A is fixed and can supply horizontal and vertical reactions and a moment couple. If you take moments about A = 0 and sum forces in x and y directions = 0, you can easily get these reactions. The forces in the horiz and vert members you can find as you would for a truss.
     
  4. Aug 5, 2012 #3
    I think the question is a bit unfair. If B were a pin joint then triangle BCD would rotate freely around B. What is more likely intended is that ABC is a continuous member and the joint B at the end of BD is pinned to the side of ABC. So this is not like the normal pin-jointed truss carrying axial loads only. There is bending and shear in some of the members.
     
  5. Aug 8, 2012 #4
    Hi, can someone post a Free body diagram of this pin jointed structure. Struggling with this whole question and any help would be appreciated.
    Thanks
     
  6. Aug 8, 2012 #5

    PhanthomJay

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    It would not be within the spirit of PF rules to post Free Body Diagrams without some effort being shown. I do, however, wish to make some clarification to this problem along the lines suggested by pongo38. The frame is fixed at the base (Joint A). There can be vertical and horizontal forces, and a moment couple, at that support.
    BD and CD are joined together with a pin at D. BC and CD are joined together with a pin at C. BD is joined to to member ABC with a pin at B. ABC is one continuous member. Because BD and CD are so called '2-force' members which are members subject to forces at each end and nothing in between, they can take axial loads only, directed along the member length. No shears. Member ABC, however, is not a 2-force member, because it takes a moment at the end as well as a force at B in between, so it can take both axial forces and shear forces, and bending moments. Once you first get the support reactions at A, you might want to look at a FBD of ABC. There can be only a horiz force at C, since CD supports axial loads only, and at B, there is a diagonal force with components Bx and By. If you sum forces in y direction, you can solve for By, and continue.....
     
  7. Aug 8, 2012 #6
    Thank you very much
     
  8. Aug 8, 2012 #7
    Hi guys,

    From reading the above, would this be faeseable as a FBD for the structure?

    Thanks, appreciate the help.
     

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  9. Aug 8, 2012 #8

    PhanthomJay

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    Not quite. Your first order of business us to break up the applied force F into its x and y components. Then sum moments of those force components about A= 0 to solve for M_A at the base. Put the value down or you'll get lost as to what is known and unknown . Then please sum forces in x and y direction to solve numerically for the support force reactions A_x and A_y.
    Now you can draw a FBD of
    ABC.......which is one continuous member, don't break it at B.....you have the known reactions at A, the unknown forces B_x and B_y acting at B, and the unknown force C_x acting at C...there is no C_y remember because CD is 2-force member.... So you can solve now for B_y right? And then get B_x from trig, then solve C_x ...and the puzzle starts to unwind...watch directions of forces and don't forget Newton 3...
    Edit: Looks like the value of F is not given, so you will have to determine reactions and internal pin forces as a function of F, which can lead to confusion unfortunately. You can if you want assume F = 1, then when you get the reactions and forces, multiply each by F for the results.
     
    Last edited: Aug 8, 2012
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