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Too Many Unknowns

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takekjer

Mechanical
Sep 20, 2012
1
Hello Everyone and thank you for allowing me to show my ignorance. I am new to the forum and am unaware of proper rules/etiquet - therefore I hope this breaks none. Enough with formality, know that I greatly appreciate your input however.

I constantly run into the issue of having too many unknowns for the number of equations I have available (see the attached). What is the proper way of going about solving for these unknowns? I've summed the Forces in X, and Y, and a Moment about the point 'o'. As you can see, I have 4 unknowns with only three equations in which to describe them. I seem to remember being able to make multiple moment equations about different points on the same FBD, but I know this doesn't work in all cases. Any ideas? I'd certainly appreciate it. Thanks in advance.
 
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1) ...
if you have Ax = b, where A is not square then ...
Least squares solution to over-defined set of linear equations
In practice, Equation ( 48 ) is solved in a least squares sense: a number of angles, R, are chosen at which the Yc and Ys terms are evaluated from Equations ( 50 ) and ( 51 ). The number of multipoles, M, is chosen such that M < R, and the f2m terms are evaluated, according to Equation ( 53 ), for each of the multipoles at each of the angles. Thus there are more linear equations than unknowns. It may be shown that the least squares solution to this system of equations may be expressed, in matrix form, as in Equation ( 50 ). This system may then easily be solved by Gauss elimination or any other matrix solving method, such as Gauss Seidel or SOR (successive over-relaxation).

ATAx = AT b
where AT is the transpose of A, so that ATA is a square matrix.

2) in your example, i think you can solve it making assumptions (plane sections remain plane, the structure connecting the loas to the reactions is rigid, ...
if, for example, you're saying you've got three shear reactions, assume each is 1/3rd of the applied shear. you might create your free body with this assumption, find your critical fastener, then in the fastener analysis, add a factor for "loading uncertainities"

 
Unless I am missing something, its a statically determinate problem. You should know the bloom weight from the customers specification (or your specification), and from there you should have an idea of the weight of the rollers needed to support the bloom (I am assuming that that is what Wconv represents). At that point, you simply take the sum of the moments equal to zero around one of the unknowns, say F1, which would give you the moment around F2. Divide the moment you calculated by the distance to get the resultant force, and use sum(Fx) and sum (Fy) equal to zero. If I misread the diagram please feel free to correct me.
 
About telling us what this drawing is actually representing. The details shows a dynamic system as you are indicating an acceleration component but yet OP and responses indicate a static problem.
 
If you have too few equations for a system, you either have not found the remaining equation(s) or the problem is indeterminate.
For the indeterminate case you make assumptions usually strain related to get a reasonable relationship(s)
No other method will give rational answers,no matter how esoteric. You just can't make something out of nothing.




 
Now having seen your "sketch", are we to guess what it is?
Would you kindly furnish some background to get a reasonable answer.
You can't get serious answers to a problem that may not even be formulated correctly.
I say this because, for openers, you show acceleration with no mass shown, so the dynamics are already questionable.

 
"Those involved in banging their heads on this brick wall, please take a look at and send any comments or revisions to me. Thank you

Cheers"

Good job, but I would suggest, as a practical matter, using an orthogonal coordinate system instead of "non parallel" which is correct but less expeditious.
 
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