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Rigid plate suspended by 4 wires

  1. Feb 4, 2014 #1
    Hi everyone! This is my first post here, i hope to not make a fool of myself...

    1. The problem statement, all variables and given/known data

    I was asked to find the tension forces in the 4 wires by which a cabinet is suspended from the ceiling. The cabinet's centre of gravity does not coincide with its centroid, but is located off-centre. The 4 wires are joined in the same point where they are attached to the ceiling, and the attachment point is located right above the centre of gravity.

    This is in my opinion a statically indeterminate problem, as there are 4 wires where 3 would be enough for equilibrium. I would like to try to solve it using the flexibility method with which i am familiar.

    However, i am having some problems with this model and i am getting confused. For my first attempt, i have modeled the problem as in the attached pdf: a rigid plate with the 4 wire tensions applied at its corners (decomposed along x-, y- and z-direction), the weight of the cabinet applied at the centre of gravity (for clearity omitted from figure).

    This is where the confusion starts: drawing the free body diagram, only the 4 wire tensions are present. So this is not a statically indeterminate problem at all?

    2. Relevant equations
    3. The attempt at a solution

    Thanks up front for your thoughts,
    Mark
     

    Attached Files:

  2. jcsd
  3. Feb 4, 2014 #2

    SteamKing

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    Yes it is statically indeterminate: You have 4 unknown tensions but only two equations of static equilibrium. Therefore, additional equations involving the strains in the wires are required to solve for the unknown tensions.
    If the c.g. of the cabinet lay on an axis of symmetry w.r.t. the wire arrangement, that might provide sufficient simplification, but in your figure, it appears this is not the case.
     
  4. Feb 4, 2014 #3
    Thanks for your reply SteamKing.

    Indeed i expected to need compatibility equations, but i do not understand why there are only 2 equilibrium equations. which 2 are those? i thought i could actually just state all 6... do 4 of them become identically zero?
     
  5. Feb 4, 2014 #4

    SteamKing

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    As always, there are only 2 equations of static equilibrium: sum of the forces and sum of the moments.

    For example, with a simply supported beam, you have two unknown reactions, so the eqns. of static equilibrium are sufficient to determine these reactions. However, if you put a third support between the first two, the beam becomes indeterminate, since you have three unknown reactions but still only two equilibrium equations.
     
  6. Feb 5, 2014 #5
    Ah ok, i am used to think of the equilibrium equations in terms of their x-, y- and z-components so i always think there are 6 of them. So, viewed in that light would you agree that there are 4 tensions in 12 unknown components and 6 component equations of equilibrium?
    That would mean i need 6 strain related compatibility equations:
    1. delta(A)=delta(B)
    2. delta(A)=delta(C)
    3. delta(A)=delta(D)
    4. delta(B)=delta(C)
    5. delta(B)=delta(D)
    6. delta(D)=delta(C).
    And using delta=FL/EA as constitutive equation to relate the displacements to the tensions.

    Btw, you are correct about the c.g. not being positioned on any symmetry axis.
     
  7. Feb 5, 2014 #6
    Ok, so what I did is to state the 6 compatibility equations in terms of the tensions and corner displacements. Again, these 6 equations have 3 cartesian components each. So there are 18 component compatibility eqs of which only 6 are needed.

    As i need 6 component equations to complete a 12x12 matrix (the 6 equilibrium component equations form the first 6 eqs) i could choose:
    -either use for example delta(A)=delta(B) and delta(A)=delta(C) with 3 components each, or
    -use only 1 component of all 6 compatibility eqs.

    Now, both methods should yield a set of 6 independent equations imho (if one or more of them would be dependent, the resulting matrix would turn out singular, fail to have an inverse and so no solution would be found). So i used the latter method.
    In excel i take the inverse of the matrix and left-multiply with the (12x1) force component vector.

    So far, all is good. Now the part that discombobulates me: the outcome is for a cabinet weight of 98100 Newtons (1000 kgf)
    Fa,z=47.36 kN
    Fb,z=50.74 kN
    Fc,z=-50.74 kN
    Fd,z= 50.74 kN.
    The forces nicely add up to a total of 98100 N, but still this cannot be correct, can it? All forces are directed in positive z-direction, so the wires have become vertical in stead of configured as a pyramid. Also, it would mean that wire C supports a compressive force...

    I think this has something to do with the choice of displacement compatibility equations. Clearly, to keep the plate level this set of wire forces is needed. Apparently, in reality the cabinet would rotate about the line BD with point A dropping and point C rising.

    My question remains: how to repair my model??
    Can it be done by using a different set of compatibility equations? The only alternative I can think of is to restrain the c.g. to remain directly below the sling hub, but that only gives me 3 component equations, and i need 6...

    Should the compatibility equations express the directions of the wires?

    Should I abandon the rigid plate with c.g. in the plane and model the cabinet as a plate with the c.g. located BELOW it?

    So obviously something is wrong with my model... I just can't figure it out...

    I have thoroughly underestimated this problem, it looked so simple...
     
    Last edited: Feb 5, 2014
  8. Feb 6, 2014 #7
    anyone any ideas? would be much appreciated.
     
  9. Feb 6, 2014 #8
    First of all, you're right. Only three supports are required.
    You have four unknowns, I'll call them Fa, Fb, Fc, and Fd.
    What I would do is:
    1) Set up the three linear equations:
    Faz+Fbz+Fcz+Fdz = sin(α)Fa + sin(ß)Fb + sin(y)Fc + sin(δ)Fd = Rez
    Fax+Fbx+Fcx+Fdx = cos(α)cos(αxy)Fa + cos(ß)sin(ßxy)Fb - cos(y)cos(y)Fc - cos(δ)sin(δ)Fd = 0
    Fay+Fby+Fcy+Fdy = cos(α)sin(αxy)Fa - cos ... = 0

    2) Next, I would create four solutions, one each for Fa=0, Fb=0, Fc=0, and Fd=0. Then I would discard any solutions where any of the tension forces were less than 0.

    3) The remaining solutions will give you the range of value that each of the tensions can take.

    In order to pick a particular solution, you would need some additional criteria.

    I realize this may or may not line up with whatever you're learning in class, but I hope this helps.
     
    Last edited: Feb 6, 2014
  10. Feb 6, 2014 #9
    Other than Rez=98.1KN, Rey=0, Rex=0, which parameters are known?

    For example, have you been given specific values for H, W, L, Xcg, and Ycg?
     
    Last edited: Feb 6, 2014
  11. Feb 9, 2014 #10
    Hi .Scott,

    thanks for helping out! That is an interseting approach.
    -If I understand correctly you set up the three equilibrium equations of force, which i also started with. They look exactly the same as my own work. But why don't you use the 3 equations of moment too?
    -Then you would have 6 equations and when you set one tension equal to zero, 3 tensions in 3 components = 9 unknowns. So 3 additional equations are needed.
    -I choose disp(B)=disp(C), disp(B)=disp(D) and disp(C)=disp(D), this works out to:
    -Fbz=((sin(beta)^2)/(sin(gamma)^2))*(LCF/LBF)*Fcz
    -Fbz=((sin(beta)^2)/(sin(delta)^2))*(LDF/LBF)*Fdz
    -Fcz=((sin(gamma)^2)/(sin(delta)^2))*(LDF/LCF)*Fdz

    but when i plug the 6 equilibrium equations and the displacement compatibility equations into a matrix system, again i end up with a singular matrix...

    seems there is still the problem of incorrect compatibility equations.

    also i find it counterintuitive that when setting one of the tensions to zero (and expecting that you should get a statically determinate system) you still have only 6 equations for 9 unknowns and still having a statically undeterminate system... so i still needed to set up 3 additional displacement equations.

    am i overlooking something or is this a common misconception due to the fact that all that is in the regular books is 2D???
     
  12. Feb 9, 2014 #11

    AlephZero

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    You don't have 9 independent unknowns. The tension in a wire can only act along the direction of the wire. So there are really only 3 unknowns, the tensions in the 3 wires.

    The neat way to solve the 3-wire problem is take moments about a line through two of the suspension points. That eliminates the tensions in two of the wires.
     
  13. Feb 10, 2014 #12
    You only have 4 independent variables - the tension in each of the four lines.
    The only other limitation is that the value for each tension must be non-negative.

    Since all four tensions are pulling through the same point (F), there is no possibility of inducing any sort of spin on the cabinet. So we don't need to worry about moments.

    Instead, all we need to do is get the forces along X, Y, and Z correct.
    That gives us three equations and four unknowns - plus the requirement that all the tensions must be non-negative.

    One way of expressing the final solution would be to introduces a parameter that can range from 0 to 1 with Fa=FuncA(p), Fb=FuncA(p), Fc=FuncA(p), and Fd=FuncA(p). For p=0 or p=1, at least one of those tensions will be zero. For p<0 or p>1, at least one of those parameters will be negative and therefor not a valid solution. And for p>0 and p<1, all tensions will be positive.
     
    Last edited: Feb 10, 2014
  14. Feb 10, 2014 #13
    So i have a couple of different approaches to use now.

    Scott, when having 4 unknowns, 3 force equations and 2 moment equations (around x- and y-axes, the third would be identically zero, because no spin can occur), why would you use the extra requirement of no compressive force in a wire?
    why not either of the two moment equations?

    is this the reason i keep ending up with singular matrices?
     
  15. Feb 10, 2014 #14
    The cabinet's CG is hanging directly below the single support point "F", so there are no moments at all.
    You have 4 unknowns, 3 force equations, and the requirement that tension values be non-negative, end-of-list.

    Once you have a solution for the tensions that balance out the X and Y forces to zero and support the Z force of the cabinet weight, there will be no possibility of rolling, pitching, or yawing.
     
  16. Feb 10, 2014 #15
    I haven't been trying to reproduce you're math, but with the tension wires set up as they are, I would expect any other linear equations describing this system (such as those based on moments) to be dependent on the 3 force equations - and so they would contribute nothing to the solution. And so your system would remain under-constrained as is commonly indicated by a singular matrix.

    My suggestion is to start with the Fx, Fy, Fz equations I listed below and then solve for each of the four systems where one of the tensions are set to zero: (Fx, Fy, Fz, Fa=0), (Fx, Fy, Fz, Fb=0), (Fx, Fy, Fz, Fc=0), (Fx, Fy, Fz, Fd=0). Then toss out the solutions where some tensions are negative.
     
  17. Feb 10, 2014 #16
    Tonight i'll have time to work on the problem again, .Scott. I will work out your approach for sure!

    One of the things that is (was?) confusing me is that I was under the impression that the cabinet could rotate around the c.g. because of the elongations of the wires...
     
  18. Feb 10, 2014 #17
    Hi .Scott,
    i just worked out your suggestion, but all four solutions feature one or more negative tensions...

    my guess is that this is due to the fact that the cabinet IS actually rotating about its c.g. due to wire elongations.
     
  19. Feb 10, 2014 #18
    You should only get the negative values when the CG falls outside the triangle formed by the "other three lines", the ones not set to zero. Since the CG has to fall on or within at least two of those triangles, the calculations are wrong somewhere along the line.
    If you want to give me the specific numbers, I'll tell you which lines can be safely set to zero. ... After lunch.
     
  20. Feb 11, 2014 #19
    Alright, after some correcting of my equations and debugging of the excel file, everything lines up! Wires A and D may be cut, and all results nicely add up to the applied weight. These results give me an upper bound of the resulting sling forces in the 4 wire statically undetermined problem.

    Thanks SteamKing, AlephZero and .Scott!

    PS: when i have some minor sea of spare time somewhere in the future I might try to find a relation between the displacement of the c.g. and the elongations of the wires in the 4 wire problem.
     

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  21. Feb 15, 2014 #20

    nvn

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    Why? What is the reason causing you to expect this? (I do not doubt your statement, at all. But could you elaborate on the reason(s)?)
     
    Last edited: Feb 16, 2014
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