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Truss Analysis and Method of Sections

  1. Jun 5, 2008 #1
    Hello all, first post.

    I have a mechanics test coming up and a the lecturer gave a very strong hint that a question like this with a scissor truss will be on the exam so how would you go about solving a question like this (finding force in all members).

    I worked out the reaction forces as:
    [tex]\sum[/tex]F(y) = 0 = A(y) + F(y) - 500 - 500 ; 1000 = A(y) + F(y) [eq1] .
    [tex]\sum[/tex]F(x) = 0 = A(x)
    [tex]\sum[/tex]M(A) = 0 = 15[F(y)] - (5x500) - (10x500) ; 15[F(y)] = 7500 ; F(y) = 500
    Subbing F(y) = 500 into eq1 gives ; A(y) + 500 = 1000 ; A(y) = 500

    So A(y) = F(y) = 500 and A(x) = 0 , with all forces being in kilo newtons.

    I know that it has to be done using the method of slices because theres no joint with only two unknowns.

    Now my question is where do I take the slice? Because the only two places I can think of where there are only three unknown forces are long the dotted lines in the picture but that gives the problems of not being able to do a sum of moments equation. The reason for this is that all three unknowns would be concurrent from the same joint so no moment equation can be done around that point (please refer to the FBD I have so eloquently drawn).

    Please help

    Attached Files:

  2. jcsd
  3. Jun 6, 2008 #2
    Do you go to QUT? i missed out on the exam heads up. can you tell me what else he said would be on the exam?
    i dont wana hijack your thread so send me a PM. Thanks
    Last edited: Jun 6, 2008
  4. Jun 6, 2008 #3
    yeah i didnt actually go to that either. basically 80% of the marks are gonna be centroids, truss questions and shear force & bending moment type stuff with a few tidbits of 3d concurrent forces thrown in for good measure
  5. Jun 6, 2008 #4


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  6. Jun 6, 2008 #5
    No I wasn't aware that you could do that. Would that sort-of be equivalent to using the section on the other side of the cut and then summing the moments about D? As in cutting along that same line however instead using the section that contains A (i.e. the left side of the cut) and using the section that contains D (i.e. the other side of the cut)?
  7. Jun 6, 2008 #6
    that wouldn't be any different because you would still have the same forces exposed.

    Is the length of AB the same as the length of AD? If so, after summing moments about point D to find force A(y) you could then determine the height of point B from point A and then sum moments about B to determine a(x).
  8. Jun 7, 2008 #7
    I was able to find the forces in the members by setting up a series of simultaneous equations for four of the joints. Is there a quicker way to do it though?
  9. Jun 7, 2008 #8
    yeah i think i've got it now, I just worked it out using the other section. I just used a series of simultaneous equations and plugged them into the graphics calculator and thats how we were taught. I guess a quicker way would be to plug them into a truss modeler or something anyway thanks for the help guys
  10. Jun 7, 2008 #9
    Hey, I'm at QUT doing this aswell.. But so is the truss the same height? Because otherwise we'll have to do quite a bit of trig to get all the lengths out. And do you guys know which forces we'll have to find? is it all of them or just those leading up to the supports?
  11. Jun 7, 2008 #10
    Hrm actually sorry I got the length for AB quite simply... Since BAF is 60d and BFA is 30, ABF must be 90d. So then 15 sin 30 = 7.5 = AB.. But the rest of my questions stand :D
  12. Jun 7, 2008 #11
    I'm assuming that itll be the reaction forces and also the force in every member. One last question, would it be incorrect to take the sum of the moments around A or F seeing as how A is a pinjoint and F is a roller joint ans these supports cant have reaction moments?
  13. Jun 7, 2008 #12
    my understanding is that the sum of moments about any point is 0 regardless of whether or not it has a moment reaction. for an example of this look at page 320 of our engineering text book.
    Even though there is no moment reaction along the bending axis of a hinged joint, the moment equilibrium equation is still applied to point b about the x axis as can be seen on page 321 in the 2nd eqation down.
  14. Apr 13, 2009 #13
    the solution is simple - first off, analysis by method of joints is much faster to solve for all member forces (axial forces) and not method of sections which focuses on specific members.

    If you look at BE (or CD) you can see you can solve for this member by using the force at B (summing vertical forces = 0) by doing that, you can use one of the following methods:

    (1) Use many simultaneous equations and solve through matrix calculation

    resolving member forces at each remaining join into 2 simultaneous equations (F(h) = 0 F(v) = 0).

    proceed to complete for all joins and members until you reach a point where you see (which should take only 2 or 3 more joins) that number of equations = number of unknowns

    (2) Use the method of sections accordingly.

    DC, DE, AC and AB where DE is known, continue to solve.

    Hope this helps

    Do the same for member CD (or BE)
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