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Homework Help: Analysis of structure problem

  1. Apr 17, 2004 #1
    I am overwhelmed by this HW problem I have. Coming back from spring break, our professor blasted through the entire chapter covering analysis of structures, and then assigned this problem for HW (see attached).

    I am completely stuck. I basically don't know if I am on the right track.

    I started with part a) by trying to find the force (tension) pulling on member AB. I did this by calculating the moment about point D.
    Md = 6kN * .3m = 1.8kNm (ccw)

    I took this torque and used it to compute the x component of the tension at AB.
    Fab=Md / distance
    Fab= 1.8kNm * .320m = 5.625kN to the left

    Using this tension, I tried to find the x component of the force at point C by using the moment created at E by tension AB.
    Me = .460m * 5.625kN = 2.59kNm (ccw)

    Fcx = Me / distance
    Fcx = 2.59kNm / .300m = 8.63kN

    Finally, I try to compute the force along CD using the x component I just found (or think I found). I do this using the right triangle created between D, C and the horizontal from D. This gives me .4 along the horz, and .16 along the vertical. That gives a hypotenuse of 80*sqrt(29). Also, the angle created by DC with the horz is 21.8 degrees. By using similar triangles, I find the force along CD to be
    Fcd=(80*sqrt(29)/400) * Fcx = 9.29kN

    This answer is wrong. The book has 10.72kN at 21.8 degrees. Atleast I found the angle correctly. I am close, but not correct.
    Also, I don't feel like I followed any process that we were given for the analysis of structures. Ofcourse, the class examples were painfully simple, and I left class with a great feeling that this was a piece of cake. This problem is changing my mind. Maybe this is simple too, but I am just not seeing it.

    I am in over my head with this problem. Can someone atleast give me push towards shore? Maybe toss me a life preserver even?
    I have not even considered part b yet!

    Thanks in advance.

    Attached Files:

  2. jcsd
  3. Apr 18, 2004 #2
    Can no one even help me with this?

    I have spent more time on this question, and I suspect my problem is in treating the segment BCE. Since it is not straight, does that change calculation of the moment about point E? I don't think so since we are only concerned with the perpendicular distance to B.
  4. Apr 19, 2004 #3
    Alas, it appears that Physics Forums has failed me.

    48 hours has elapsed without a single response to my post.

    Is my problem that difficult? :confused:

    Or perhaps it is too trivial for anyone to consider worth their time. :wink:

    This hurts guys and gals. :frown:
  5. Apr 20, 2004 #4


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    Patience grasshopper. You posted on a Sunday night, arguably the slowest on the week... and the problem is a bit nasty.

    I don't have time to look through it right now. I'll help out in a few hours.
    Last edited: Apr 20, 2004
  6. Apr 20, 2004 #5
    Thanks Enima. So there is hope after all.
    I will be home tonight working on other physics HW (after the kids are in bed), so I will check in from time to time.
    Thanks again.
  7. Apr 20, 2004 #6


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    :biggrin: I'm an enigma, not an enima... :eek:

    Here's where your mistake is:

    Anytime the vectors aren't at 90 degrees to each other, you need to take special care. A moment is Force*Dist only when they are at 90 degrees to each other.

    In this case, you'll have Force x Dist = 2.59e3 Nm, or FDsin(theta) where theta is the angle between them, and Dist is the length from E to C.

    For the part above where you found the moment due to the force in AB, the calculation was OK, but I'm not sure if you caught why. The actual moment arm is the line EB, not .460m. The reason you got the right answer for that part is because EB cos (theta) is the same as making the (shorter) moment arm [E->perpendicular to AB], or .460.

    If you want to shortcut it in a similar way for the DC branch, you'd need to find the perpendicular distance to the line DC, and use that. It'll be the same number either way. I personally prefer to plug in the vectors and run a cross product calculation rather than take the shortcut.

    See if that fixes the problem.
    Last edited: Apr 20, 2004
  8. Apr 21, 2004 #7
    Sorry for the typo, and it was a typo. No harm intended.

    So the fact that BCE has an angle to it changes the torque calculations?
    I suspected that it did, but than I figured the only thing that really mattered was perpendicular distance. I thought I could extend the force vector at B to the left, and use the distance straight up from E. Same thing for force at C.

    Thanks for the help. I will continue to work through it with your tips and let you know how I do.

  9. Apr 21, 2004 #8
    Still not getting it

    I see what you were talking about with the moments. I have have reworked it but I am still not getting the correct answer as per the text.

    I recalculated the moment about E due the force at B using the cross product. I confirmed, as you said, that I had the right number, 2.59kNm, albeit for the wrong reasons.

    I find the angle at point C made by ECCx to be 71.6 degrees.
    Theta = arctan(.3/.1)

    I then proceeded to find the force of Cx:
    Me = F D sin(theta)
    Me = Cx D sin(theta)
    Cx = Me / D*sin(theta) = 2.59k/.3*sin(71.6) = 9.1kN to the left

    I think this is wrong, it needs to be higher.
    Anyway, using that number, I calculate:
    Fcd = Cx / cos(theta) where cos(theta) is 21.8 degrees

    Now I get 9.8kN at 21.8 degrees. The books answer is 10.72kN.
    What am I doing wrong?
  10. Apr 21, 2004 #9


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    the theta is the angle between the Dist and F, not the angle between Dist and horizontal

    Element EC has an angle of 71.5 above the horizontal, and DC has an angle of 28.1 above the horizontal. So your theta is 71.5-28.1 clockwise.

    Also, the distance isn't .3, it is sqrt(.3^2+.1^2)

    edit: hrm... dunno. my answer doesn't match the book either. Hopefully, I haven't been leading you astray.
    Last edited: Apr 21, 2004
  11. Apr 21, 2004 #10
    I Got It!!!

    :biggrin: Great and thank you. I finally pieced it together and got the correct answer.
    That is, for part (a) anyway.

    Thanks for your help, Enigma. It seems like I completely threw out everything I learned about torques and moment arms. I guess I let the overall size of the problem knock me out of the basics.

    Now it is on to part (b), so stay tuned!!! :eek:
  12. Apr 21, 2004 #11
    Part b!!! Ahhhhh!

    I am starting to seriously hate this problem.

    I think that I should start part b by finding the moment about G.
    Is this the correct approach? If so, I am having trouble finding it. How do I treat the weight? As 6kN straight down applied where? At point D? This seems too simple.
    Any guidance on this portion of the question?
  13. Apr 21, 2004 #12


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    You need to find the moment about G, yes.

    For this one, since you don't know the Y position of the force, you need to define the moment arm as the line from point G perpendicular to the force (.3m to the left of D).

    How did you get the right answer for part A? Did I make a mistake somewhere as well?
  14. Apr 21, 2004 #13
    Well, if you made an error, I think it is in the angle EC makes with the horizontal. You mentioned it is 71.5, but I found it to be 71.6. This small difference makes the final answer work out to 10.74kN instead of 10.72kN. Is that what you were talking about.
    Other than that, I basically followed along the lines you lead me, and it all makes sense now. I could scan and email you the sheet that I worked it all out on. I tried to scan and post it, but I can't seem to get it into a format and size the forums will accept, and keep it legible.

    For part b, let me just say thanks for hanging through this with me. I was afraid you might be getting tired of this problem by now.

    I will work on part b a little more tonight, then hit it again in the morning. I am pretty beat right now.
  15. Apr 21, 2004 #14
    Moment at G

    I am getting 3kNm as the moment at point G. I think this is too small.

    I have defined the moment arm (r) to be -1.62i+.5j and performed a cross product with force of 6kN down, -6kNi.
    Mg = r X F = 3kNm

    I pressed on from there anyway:
    Mg = Ffh D sin(theta) where Ffh is the force on member FH, D is the moment arm along FG which is sqrt(.120^2+.480^2) and theta is 30.97 degrees (45-14.03)

    Solving for Ffh, I get 11.78kN. Intuitively this is way too low. The book answer is 35.4kN at 45 degrees. I think this method is correct (that doesn't mean much) but I think that my moment about G is too low. That is supposed to be the easy part, right?
  16. Apr 22, 2004 #15
    Here is a scan of my work so far...

    Attached Files:

  17. Apr 22, 2004 #16


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    The force is -6kN j, not i
  18. Apr 22, 2004 #17
    :eek: DOH!!!

    And to think I posted it here for all to see.

    I will continue my efforts later and let you know.
  19. Apr 22, 2004 #18
    OK, I got it all. Thanks Enigma for the help on this problem. You really helped me to re-explore what I have learned. This was a cool problem because it helped me to apply a lot of concepts that we have learned all semester.
    Thanks again for the help, and for sticking through it with me.
  20. Apr 23, 2004 #19


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    No problem
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