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FRAME work ( statics)

  1. May 8, 2012 #1
    1. every member is 0.45m and all the angle is 45degree....
    C is fixed support, D is roller support









    3. Ʃfx = 0 , Cx = 0
    Ʃfy = 0 , Cy + Dy = 20N

    ƩMc = 0 , 20(0.45) + Dy(0.45) = 0
    Dy = -20N


    OR
    Ʃfx = 0 , Cx - Dx = 0
    Cx = Dx
    Ʃfy = 0, Cy -20N = 0
    Cy = 20N
    ƩMc = 0 , 20(0.45) + Dx(0.45) = 0
    Dx= -20N

    may i know which one is correct??
    help me....please....
     

    Attached Files:

  2. jcsd
  3. May 8, 2012 #2

    PhanthomJay

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    The second one is correct.

    I think you meant to say that C is pinned (it can take forces in the x and y directions), and that D is a roller support (it can only take a load in one direction at a right angle to the rollers (in the negative x direction in this case)).
     
  4. May 8, 2012 #3
    i dont understand about roller support there....
    how to know the force is in which direction?
    i thought the roller support 's force is always oppose to the force applied(20N)......
    thank you
     
  5. May 8, 2012 #4

    PhanthomJay

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    No, not always, and this problem is an example: the roller must rest on the side wall, not the floor; otherwise, the frame would be unstable, since there would be no way to balance the horizontal force at the other support. When you assumed this in your first incorrect solution, you made an error when summing moments about C: You said
    , but a Dy force cannot produce a moment about C (there is no moment arm) , so your equation should have said
    , which is nonsensical, and therefore incorrect.
     
  6. May 9, 2012 #5
    ok.....i got it...thank you

    now these are my answer......could you help me check and are they correct?

    C joint
    (vertical)
    Cy - Fce sin45 = 0
    20 - Fce sin 45= 0
    Fce = 28.28N

    (horizontal)
    Cx + Fbc - Fce cos 45 = 0
    -20 + Fbc - 28.28cos 45 = 0
    Fbc= 40N

    joint D(horizontal)
    Dx = Fde
    Fde = -20N

    joint E(horizontal)
    Fde + Fce cos45 -Fae = 0
    -20 + 28.28cos45 - Fae = 0
    Fae = 0

    (vertical)
    Fbe + Fce cos45 -20 = 0
    Fbe = 0

    joint A
    Fab =0

    i know they are wrong....but i dont know which one is wrong and how come will be wrong.........help me....i'm lost in statics!!! thanks
     
  7. May 9, 2012 #6

    PhanthomJay

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    Yes, this value is correct, good work, but you must also indicate whether it is tension force or compression force. Member tension forces always pull away from the joints on which they act, and member compression forces always push toward the joint on which they act.
    No, this is not correct. You have to be very careful of force and force reaction directions and the use of the plus and minus sign. Which way does Cx reaction force on the joint point, to the left or right, and is it thus plus or minus? You can sum moments about D to get its direction.
    Yes, tension or compression?
    excellent, now you are cookin'
    Yeahhh, buddy!:smile:
    Yes, again!
    Actually, you did yourself proud...the only errors are whether the forces are T or C, and the force in BC was wrong. Try that one again.
     
  8. May 9, 2012 #7
    thank you....

    Fbc = 20N??

    Fbc = Fce cos 45
    Fbc = 20N

    but why just now what i did is wrong? i feel like is correct because
    Fx = 0
    Fx contain Cx, Fbc, Fce cos 45, that's why i formed a equation.


    Fab = tension
    Fae= tension
    Fbe = compression
    Fde= compression
    Fce= tension
    Fbc = compression
     
  9. May 9, 2012 #8

    PhanthomJay

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    Yes, but you formed it wrong. You wrote:
    When you should have wrote, per my earlier hint on signage
    [itex] C_x + F_{bc} - F_{ce} cos 45 = 0 [/itex]
    [itex] +20 + F_{bc} - F_{ce} cos 45 = 0 [/itex]
    Now solve for F_bc. The plus and minus sign will bite you every time if you let it. Do you see your error?
    but these you have identified as all zero force members, neither in tension nor compression
    Yes!
    Yes!
    NO!

    You only got 2 out of 6 correct :frown::smile:
    (I need to work on my LateX)
     
  10. May 9, 2012 #9
    Fbc should be tension.....i see it.......

    my understanding about zero force members......

    zero force members are a member which cannot carry load. hence there is no force in these members. these members do not have compression force and tension force.
    the existence of zero force members are to stabilise other members.

    my understanding is correct or no?
     
  11. May 9, 2012 #10

    PhanthomJay

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    NO, that it is not right. Did you work out the equation?
    +20 + F_{bc} - F_{ce} cos 45 = 0
    + 20 + F_{bc} - 20 = 0
    F_{bc} = ____???___
    Not quite. The zero force members in this problem are not stabilizing anything. You can take them out of the frame as if they didn't exist, and it doesn't change the solution or stability in any way.
     
  12. May 9, 2012 #11
    Fbc = 0N??
    is not tension and compression...

    then why this phenomenon happenned? i mean zero force members.....
     
  13. May 10, 2012 #12

    PhanthomJay

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    They only have no force in them because of the way the truss is loaded with only a force appied at E...If a load was applied downward at A, all the members would have forces in them. In other words, for the given problem and loading, the members are not needed, they are just extras in case some day you wished to apply at load at A...right now, they just go along for a ride .....
     
  14. May 10, 2012 #13
    but when i doing this in lab, my result from lab is quite big difference to the theoretical result. the aim of the lab is to determine the forces in members and the model is same as the diagram. the force applied is also same. What is the possible source of errors?
    Do we need to consider the weight of the members?
     
  15. May 10, 2012 #14

    PhanthomJay

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    My assumption was that the weight of the members could be neglected. However, if the member weights are significant (and they might be since you are only applying a 20N load), then yes, you would have to consider their weight. For the puposes of the calculation, you may assume each member's weight is distributed 1/2 to each of its end joints, and applied as a load at those joints.
     
  16. May 10, 2012 #15
    thank you very much......you help me a lot......appreciate your help.......(bow)
     
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