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Finding forces in members of frames

  1. Jul 16, 2017 #1
    1. The problem statement, all variables and given/known data
    I'm trying to solve for this problem below.How would you be able to get the force in member ID and GC? Find the reaction forces in member AFGH

    2. Relevant equations
    ∑M = 0
    ∑[F][/y] = 0
    ∑[F][/x] = 0


    3. The attempt at a solution
    I dismembered all of its parts but no matter where member I analyze into there are always three unknowns which makes it impossible to solve using the three equilibrium equations. I got through the part where member GC is a two force member and the reaction forces at pin B and pin D are just equal to the tensions in the pulley, but after that no luck. Any ideas on how should I solve it? I know after solving the forces in the ABCDE member I would be able to get the reactions at the AFGH member but I seem to be lost in the problem.
     

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  2. jcsd
  3. Jul 16, 2017 #2

    haruspex

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    Please post an attempt, as per forum rules.
     
  4. Jul 16, 2017 #3
    My attempts are the free body diagrams. As you can see, the components have subscripts "x" and "y" to indicate their respective axial components.

    The FBDs are labeled accordingly, pulley B for the pulley at point B , pulley E for the pulley at point E. I got the components of these first, which would just be the tension at the pulley. Which is 490N.

    However,after solving the components at the pulleys I don't know where to go next because every member has atleast 3 unknowns which makes it unsolvable with just the three equations of equilibrium.

    I tried solving the components at "FBD 4". For example, I tried getting the moment at point C but I would end up with the components Ay and Dy unknowns, therefore making it impossible to solve for the Ay and Dy components. I tried solving for the summation of X components but Ax, Dx, and the x component of force Fgc are unknowns and the same happens when solving for the summation of y components.

    At FBD 2, I assumed that the member GC is a twi-force member to be able to simplify it's components but I cannot solve for its components without the values at FBD 4. At FBD 3, I tried solving it but without the value of the reactions at D I cannot solve for the reactions at point I.

    Meanwhile, I created an FBD for the pins. At pin B, I assumed the the member AB exerts a vertical and horizontal component at pin B and member BC to exert a horizontal component as well. At pin E, I also assumed the member DE exerts a horizontal and vertical component at pin E.

    I think after solving for the values at other FBDs. The FBD1, would be the last to be solved because without the value of the reactions at point A and G, the reactions at point H would not be solvable.
     

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  5. Jul 16, 2017 #4

    haruspex

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    I think you are making it unnecessarily complicated by breaking it into too many pieces.
    You should be able to write down immediately the x and y components of the forces the pulleys exert on the points B, E, and the force the rope exerts at point F.
    A question worth asking yourself is what would happen if the angle bracket connecting C and G were removed. Would the system collapse?
     
  6. Jul 16, 2017 #5
    A question worth asking yourself is what would happen if the angle bracket connecting C and G were removed. Would the system collapse?[/QUOTE]

    Would it mean that member CG is a zero force member?
     
  7. Jul 16, 2017 #6

    haruspex

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    Perhaps... what do you think CG is doing in the structure? Why is it there?
     
  8. Jul 16, 2017 #7
    Hmm I have an answer key to this problem but it states that there is a forc on the member GC.
     

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  9. Jul 16, 2017 #8

    haruspex

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    Yes, I didn't say there was not a force. I just asked you to think about how the system would move if you took it away.
    Never mind, let's try something else...

    What can you say about the horizontal forces on beam AE?
     
  10. Jul 16, 2017 #9
    I think the horizontal forces on caused by the pins would just cancel out. Thus the remaining horizontal forces would be on point A,C and D. Is it correct?
     
  11. Jul 16, 2017 #10

    haruspex

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    Right. So you have an unknown force CG exerts at G, and an equal and opposite one it exerts at C. Consider its horizontal component.
    What can you say about its relationship to the force the beam AE exerts on beam AH?
    Take moments on AH about H. What do you find?
     
  12. Jul 16, 2017 #11
    If I would sum the moments at point H, the horizontal component of A and G would be the two unknowns. Would I be able to relate this equation to the sum of horizontal forces at the other member AE is it possible?
     
  13. Jul 16, 2017 #12

    haruspex

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    Considering my post #10, can you get that down to one unknown?
     
  14. Jul 16, 2017 #13
    Is it possible?

    Member A-H
    *Considering that counter-clockwise moments are positive.
    ΣM = 0
    0 = (-Ax)(0.21)+(FGCx)(0.7)+(450)(0.8)
    -360 = -0.21Ax+0.7FGCx

    Member A-E
    ∑Fx = 0
    0 = -Ax+FGCx+Dx
    *I didn't include the horizontal components at B and E since they would just cancel.
    I think it would be possible to make it at a one unknown if the horizontal component at D is not included, what should I do next? I was thinking of solving it in systems of equations but I noticed that the component at D is present so now I wouldn't be able to use systems of equations.
     
  15. Jul 16, 2017 #14
    I tried to solve it with neglecting the D component and I got the wrong answer for the other component reactions. It seems that D has indeed a horizontal component, how would I be able to get the value? I imagine taking moments but I think it would not be correct.
     
  16. Jul 16, 2017 #15

    haruspex

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    A few mistakes there.
    Check the 0.21, the 450 and the 0.8.
    Since DI is freely jointed at I, there can be no horizontal force at D.
     
  17. Jul 17, 2017 #16
    Ah yes, I got it now.

    How is it possible for D to not have a horizontal component? I was thinking that D should have a horizontal one since it's support at point I may have a horizontal component?

    I just tried it and I'm so happy it's correct.I noticed that the way we have approached this problem was a littble bit of short because with the way my professor taught me, we were always asked to first dismember all the parts first even before analyzing the whole structure itself in able to be able to do the structural analysis. May I ask is the method that we have just did applicable to any frames and trusses problem?
     
  18. Jul 17, 2017 #17

    haruspex

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    It is freely jointed at I. If there were a horizontal component at D that would have a moment about I that is not balanced by any other force or torque.
    Also, since it is freely jointed at D, there is no torque exerted by DI on the horizontal beam.
    Yes, that is a good methodical approach, but it can be overkill.
    In the present problem I would resolve it into the main components (the four structural beams) and see what I can write down about them. It is reasonably obvious how the rope, pulleys and pins mediate the forces between them. If that is not obvious, then I would analyse those smaller components.
     
  19. Jul 17, 2017 #18
    So a proof of it not having a component is if I sum moments at point I at member DI,
    ΣM = 0
    0 = Dx(2.1)
    Dx=0
    Is it safe to analyze zero force members like this?

    So for this problem the four FBDs I drew were enough? even without the pulley and pin FBDs?
     
  20. Jul 17, 2017 #19
    Thanks your approach kind off gave me a new way to solve frames. Hoping I would be able to use it for solving
     
  21. Jul 17, 2017 #20

    haruspex

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    Yes.
    As long as you are comfortable with that. If you are not sure what forces those will exert on the joints then by all means do separate diagrams for them.
    For pulleys, I find the clearest approach is to regard all parts of the rope that are in contact with the pulley wheel as part of the pulley.
     
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