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Forces in framework

  1. Dec 2, 2006 #1
    i've to design a 3 or 5 bar statically determinate frame, out of Al, so that it'll support a certain load. now I thought it would be a veritable piece of cake to figure out the forces in the individual members, but so far i've tried everything under the sun, and my answers from resolving at joints (see diagram) for B1 will just not agree.

    i'm trying to minimise the bar forces by using excel to calculate the best design from a number of cases (specifying the angles and lengths using a coordinate system), and frankly i've no idea if this is a genuinely hard problem or if i'm just being an idiot.

    i end up with:

    B1 = R2/(sin(a)+tan(c)cos(a))

    by resolving at joint C, and:

    B1 = R1/(tan(b)cos(a)-sin(a))

    by resolving at joint A. (ie resolving vertically and horizontally, and substituting in, just like every other simultaneous equation.) i've checked my sign convention, checked issues with positive and negative angle values, and still nothing!

    these give diff. answers in excel, and if you boil them down to pure trig they also are not equal. so what gives? this is a basic issue ffs.
     

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    Last edited: Dec 2, 2006
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  3. Dec 2, 2006 #2

    radou

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    Too bad the attachment doesn't work.
     
  4. Dec 2, 2006 #3
    erm...anyone else see it? I can see it fine, but might be different as im the OP...
     
  5. Dec 2, 2006 #4
    I see it just fine but I don't know enough to help sorry
     
  6. Dec 2, 2006 #5
    S is the load, thought i'd clarify
     
  7. Dec 2, 2006 #6

    Danger

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    I can see the angles fine, but can't read any of the text. That's irrelevant, though, because I'm as useless as tits on a nun when it comes to math. My approach is just to always build it twice or thrice as strong as you think that is should be. (Of course, I only build things for myself without bugetary or temporal restrictions.)
     
  8. Dec 2, 2006 #7

    radou

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    Btw, since it's aluminium, you may want to avoid compression if/wherever possible.
     
  9. Dec 3, 2006 #8
    lol brilliant phrase

    can't rebuild it at all.....it has to be built from scratch mm-perfect, then it has to slot into the rig on testing day. no fit-testing beforehand.

    and obviously compression is bad due to buckling problems, but why especially with aluminium (i lied kinda, its an alloy if that makes a diff)
     
  10. Dec 3, 2006 #9

    Danger

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    Ehhh... it's a Detroit thing. :biggrin:

    One thing that you didn't specify (unless it's one of the things that I can't read in the picture) is what form of aluminum it is. Square tubing will not be the same as round tubing, which won't be the same as angle or I-beam.
    This is not to say that I can help you even if I know that, but it might make it easier for others.
    Your forces and such will still be the same, but what you need to deal with them should be dependent upon the shape of the members.
     
    Last edited: Dec 3, 2006
  11. Dec 3, 2006 #10
    well we've a choice of angle section, box section, plain rectangle section, and can join two angles to make a T, in 3 different alloys, and varying cross sectional areas. but its really this basic, stupid, annoying bar force problem which we need to overcome. its like A-level trig for christs sake, why is it not working?

    is my writing bad or something? (concerning you not being able to read the text)
     
  12. Dec 3, 2006 #11

    Danger

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    No problem with your writing as far as I can tell; I just can't see it. Combination of bad eyes and a not-great monitor. From what I can make out, though, I wouldn't be able to understand it anyhow. It would appear to be math, of which I know nothing. :redface:
     
  13. Dec 4, 2006 #12
    aaaanyone else then? (the more mathematically inclined maybe :))
     
  14. Dec 4, 2006 #13

    PhanthomJay

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    You have so many unknowns that the letters get very confusing, and you quickly get lost in the alphabet and with the trig. This is a statically determinate problem, so you don't need the delta stuff. You ought to be able to get the correct result for the members in the forces by first determining the reactions R1 and R2, then using the method of joints. I would throw in some simple values for alpha, beta, gamma, L1, L2, L3 ,S, then prove it to yourself that the method works. Then clear up the results using the letters.
     
  15. Dec 4, 2006 #14

    Mech_Engineer

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    I have a question about points (a) and (c) in terms of their nature and mounting. From what I see, it would seem these are the two hard-mounting points in your trussed frame, so are they allowed to slide left and right, but not up and down (which would explain the necessity of truss [a-c])? If this is the case, you'll want to know that truss [a-c] will always be in compression given the geometry presented in your drawing. Are points (a) (b) and (c) fixed in space, or are you able to move any of them around in any directions?

    Another question- are you trying to minimize the force in each member, or the stress in each member?
     
  16. Dec 6, 2006 #15
    -the delta are for strains and deflections which we need to calculate later on

    -so far all i've been doing is the joint equilibrium method using the reactions, nout else.

    -a is a attachment point which can supply vertical and horizontal components, but c cannot provide a horizontal component as it is free sliding, hence for total equi., a cannot provide a horizontal component.

    - i realise that a-c will be in compression

    - i can move point a (ie. there are multiple mounting holes), and i can move point b within a certain vertical range

    - im trying to minimise force at this point....we have a choice of bar thickness and cross sections we can use, so worrying about stress will take place after I've found this )"&R% truss forces.
     
  17. Dec 6, 2006 #16

    PhanthomJay

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    And point c is not movable? Just exactly how many unknowns are you dealing with?? I mean like if angle a and b are 45 degrees, (and the top chord horizontal), then the load S is shared equally by the diagonals (.707S tension in each), and there is 0.5S compressive force in the top chord. But better still, let angles a and b each be equal to 89.9999 degrees, then the diagonal force is 0.5S each, and nothing in the top chord, you can't get much cheaper than that, as long as you're not looking at any lateral loads. But is this layout contrary to what you're trying to do??? I don't get it. And in any case, the truss is solvable regardless of lengths and angles, so you just have to use method of joints and painfully, carefully, grind out the results. Check your trig, there's a lot of variables to deal with.
     
    Last edited: Dec 6, 2006
  18. Dec 29, 2006 #17
    i got it to agree in the end....i had to resolve parallel and perp. to the bars instead of the vertical and horizontal. never figured out why it didn't work the normal way, neither did my lab partner, a final year mech eng....silly really.

    just have to build it now.
     
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