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Truss,zero force member

  1. Sep 29, 2007 #1
    1. The problem statement, all variables and given/known data
    using joint method,indicate all the zero force members in the truss
    (Note that G is a roller, and A is a pin)

    2. Relevant equations
    [tex]\sum[/tex]Fx, Fy=0

    3. The attempt at a solution

    my solution:
    consider D, by summation Fx,Fy=0, F(cd),F(de)=0
    consider E since[tex]\sum[/tex]Fx=0,=> F(ef)=0
    consider C ,take x-axis along BCD, since bar CD is neglected, => F(CB)=0
    i am not sure if it is correct ,since the whole structure will be so strange if BC,CD,ED,FE dissappear.
    My final answer is HB,BC,CD,DE,FE are zero force members
    Can anyone tell me if my assumption is correct .i am so confused.

    Attached Files:

  2. jcsd
  3. Sep 29, 2007 #2
    I'm not sure how you got BC,CD,FE,DE as zero force members.

    Here are couple more you may be able to find-

    Sum up the forces in x and y directions in joint E. What does that tell you about CE?

    Then sum up the forces at joint C. What does that tell you about CF?
  4. Sep 30, 2007 #3
    You need to sum all the forces(at each pt), and if distances are given, take moments.

    I'm nearly positive, that DE, FE(and also most likely not BC, CD) are not zero force.

    Generally, if there is more than three forces at a point, 1 is a zero force....and it is usually the one which comes in at a 90 degree angle.

    You must prove then other ones....show some work, and I'll try to help ya along.
  5. Sep 30, 2007 #4
    let me explain my solution in detail in this way:

    consider FBD of joint D,take x-axis along bar CD, as there is no other force to balance
    F(cd) so F(cd)=0.And, F(de)sin[tex]\theta[/tex]=0 ,=> F(de)=0
    Thus , CD DE become zero force members which can be neglected in the structure

    Then, consider the FBD of joint E, as load P is applied at E, there cannot be any zero force member at joint E.

    consider joint H, take x-axis along Bar AHF, => F(ha)=F(hf)=0
    take y-axis along Bar HB, F(hb)=0, so HB is a zero force member.

    Therefore, HB, CD, ED are zero force members.
    Can anyone tell me is my solution correct or not.
  6. Sep 30, 2007 #5


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    Those three are correct. There is one more.
  7. Sep 30, 2007 #6
    One more? i cannot come up with any ideas pls give me some hints!
  8. Sep 30, 2007 #7
    I didn't check/read all of your reply.....but I did check the CD part(summing forces about pt D).

    How do you figure CD=0? If the x-axis is along cd...then, wouldn't there be a force from DE....seens how they aren't at 90 degree angles(like the x and y axis would be).....?? There should be a force from DE....on CD.
  9. Sep 30, 2007 #8
    No, there are two more.
  10. Sep 30, 2007 #9


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    Take a look at joint G.
    Brad: If there were a force in CD, then there'd be a vertical component of it at joint D, but no way for it to be balanced, since DE cannot take a vertical force. So both are zero, as if they were not there.
  11. Sep 30, 2007 #10
    Oh....yeah, now I see it Jay.....thanks! I didn't actually draw and work the problem out...guess I made a stupid mistake. Been about a year since we've done these.....
  12. Oct 1, 2007 #11
    i guess F(gf) should be equal to zero as there is no other force to balance it in x- direction.Am i right? However ,Joint G is applied with a reaction by the roller, i am not sure whether zero member can exist at the joint which is applied with reaction.

    Some textbooks state that zero force member may not exist at the joint which is applied with a load, what about with a reaction ? That's the point i am struggling now!
  13. Oct 1, 2007 #12
    Zero force member CAN exist even if there is reaction. This is an example of it.

    If you sum the forces on G, you know there's a force acting on the X direction by the roller, but not the Y direction. therefore,

    Sum of force X = F_GF + F_Roller = 0

    And if you try to sum the forces in the Y direction, there is only one force acting in that way, which is F_AG, and it has to equal 0 to maintain static equilibrium.

    As shuh suggested earlier, make sure you take a look at CE and CF more closely, because those are easy to forget.
  14. Oct 1, 2007 #13
    why is there a x-component of F_roller?The roller should be placed vertically as my tutor finally adjusted the question.(sorry! i missed to state in the question!)Doesn't the roller only act a y-direction force on G?i.e. Sum of Fy=F_roller+F_AG=0 and Sum of Fx=F_GF=0

    i have tried to take a look at CE and CF,it seems that Fcf can be solved into either x or y direction, so how can i sum up all components ?Fce is balanced by load P,isn't it?

    till now, my answer is :HB,CD,DE,GF are zero members, im not quite sure about the last one.Can anyone tell me the final answer?
    Last edited: Oct 1, 2007
  15. Oct 1, 2007 #14
    OOPS, I didn't see the load there. forget about CE and CF.

    If the roller is placed below the joint G, then in the Y direction, you know the summation of force turns out to be

    F_AG + F_Roller = 0

    So we can't say AG is a zero force member.

    But, in the X direction, there's only one force you can add up, and it is F_GF = 0, so you are right.
  16. Oct 1, 2007 #15


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    HB,CD, and DE are correct. GF is incorrect. When your tutor noted that the rolller is placed vertically, he or she meant that it is free to slide vertically, such that it cannot take a vertical force, but that it can take a horizontal (or x) component. This must be so, because you should be able to see that when you sum the moment of the applied P force about point A, then there must be a horizontal reaction at G to balance the moment. So thus it is AG that is the zero force member, in addition to HB, CD, and DE. These are the only 4 zero force members; CE is not a zero force member because it supports the load P, as you noted; and CF is not a zero force member because it supports part of the load from CE.
  17. Oct 1, 2007 #16
    nope the tutor showed that the wheels of the roller are placed in x-direction(horizontally) , under this condition can the moment still be balanced?If contradiction occurs , i think i' ve made a mistake!
    Last edited: Oct 1, 2007
  18. Oct 1, 2007 #17


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    No, under this condition, the truss would be unstable. There must be a horizontal force at G to counteract the moment about A caused by P, for equilibrium. There is a misunderstanding somewhere.
  19. Oct 1, 2007 #18
    THANKS A LOT PHANTHOMJAY!Oh! i must have made a stupid mistake!
    i think the roller should be placed in such a way that it can slide vertically!
  20. Oct 2, 2007 #19


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    Don't forget EF, it is also a zero force member as you said earlier wkh023002.
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