Solving 5 Unknowns with 6 Equations 1

[newreynolds]

Probably a stupid question but I am doing a static force analysis using a Method of Joints. There are 3 joints of interest, so 6 equations but I only need to solve for 5 unknowns.

The last unknown is in the last 2 equations so I solve one and then do a check with the final equation but it's never zero. My question is then, should that last equation be satisfied since I only really needed 5 equations?

I believe it should but after looking over my algebra and not finding an error, I'm questioning things.

 

[zekeman]

Electricpete,

Almost everything you suggested is problematical because as we have established
equations 3 and 4 are NOT INDEPENDENT.

Moreover anytime you have a linear set with more equations than unknowns you know there can be at most 5 independent equations.

So you solve the 5x5 using either sets of

1,2,3,5,6
or
1,2,4,5,6

You don't need 3 ways to solve this simple system.Your 3rd way is the only correct way.

take your choice

 

[electricpete]

...as we have established
equations 3 and 4 are NOT INDEPENDENT.

I disagree. If we consider equation 3 and 4 in absence of other four equations (and we disallow the possibility that the argument of cos is +/-90 degrees or the argument of sin is zero... which would turn some coefficients to 0 and change the form of the problem), then equations 3 and 4 form two independent linear scalar equations in three scalar unknowns (F2, F3, Fs). This can be shown two ways:

We can view that in simple 2-D graphical space: The sum of the three 2-D vectors A B C disussed above is itself a 2-D vector. Being a 2-D vector, it has 2 independent coordinates.

We can view this in simple linear algebra space: If equation 3 and 4 were not independent, then you should be able to provide a way to express equation 4 as a scalar multiplier (could be a trigonetric function of the known angles) of equation 3. I am pretty sure no-one here can come up with a scalar multiplier to turn equation 3 into equation 4.

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(2B)+(2B)' ?

 

[ione]

Equations no. 3 and 4 are linearly dependent only upon condition:

phi1 + theta6 = phi2 + theta7 = psiL = 45° (or 225°)

 

[zekeman]

Well, looking more closely at your remarks ,it looks like you are not convinced so I will make one last effort.

Look at eq 3 and 4

Let F2, F3 and F5 be vectors with amplitudes f2,f3,f5.

eq 3 says that the projection of the sum along the x axis is zero

eq 4 says that the projection of the sum along the y axis is zero

which says that the vector loop of F1, F2 and F5 is closed and physically represents a joint with no net force.

Therefore the equations 1,2 5,6 can get you f1,f2,f3,f4 for sure and they are independent since they contain T0 .
Since you have f2 and f3, therefore you have the 2 vectors F2 and F3 and loop closure of F2,F3 and F6 now only requires that the scalor , f6 be obtained ; either eq 3 or eq 4 will be used , not both; therefore they are not independent.
Pete, I welcome your comments.

 

[rb1957]

a different tack ...

if your equations are correct and you have 6 equations defining 5 unknowns, then you can use matrix math to determine the minimum error (least squares) solution. google "least squares solution for over defined equations".

of course this is a mathematical answer. the engineering answer is that the structure is redundant and you need to use energy methods to augment equations of equilibrium for a solution.

have you checked you structure for redunancy ? (number of members, number of joints, ...)

 

[electricpete]

Thanks zekeman. It’s a good discussion. I agree with everything you said, up until the punchline that equation 3 and 4 are not independent. You worked through your sequence that I had called the “third way”, and came to the conclusion that equations 3 and 4 are not independent. I say it proves only that the set of 6 equations are not independent..... because you brought information from the other equations 1, 2, 5, 6 into your analysis of equations 3 and 4 (you did not analyse equation 3 and 4 in a vacuum). What would be wrong with using the “first way” or “second way” that I suggested (which ends up discarding one of the other equations 1, 2, 5, 6) ?

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(2B)+(2B)' ?

 

[FeX32]

Static problems like this are "statically overdeterminate"....
The OP should only have 5 eqns and 5 unknowns.
For the OP, try disregarding eq 4 to solve, then try disregarding eq 3 and solve (or try another eqn as pete suggests). The answers should be the same. If not, you have a serious problem.




Fe

 

[FeX32]

Of course, if you insist that the equations are correct, you could formulate an optimization problem and actually solve for the 5 unknowns with all 6 equations. Depending on convergence and constraints this would also tell you if it is actually possible or not to attain good values for a solution.


Fe

 

[newreynolds]

Wow thanks guys. My question is a particularly mathematical question.

I had used electricpete's first method and chose to ignore equation 5 and use equation 6 as a check and it did not match up. My question has to do with whether or not I can, in fact, ignore the equation even if said equation is not satisfied.

To find the linearly dependant equations, I have a modified rref() matlab function that tells me which rows 0 out when reduced to 'reduced row echelon form'. Haven't gotten around to putting it into matrix form yet.

 

[zekeman]

Pete,

In looking this thing over, I see your point and from what I can make of it , given the funny T0 inputs it looks like it is a linkage problem ( not a truss)and the angles cannot be possibly known, with a tensile string woven thru 2 of the end joints.

I reconstructed a linkage with the joints indicated and come to te conclusion that there is one more unknown and question the whole analysis.

I think it would behoove the OP to post the problem because the formulation is seriously flawed and he is wasting a lot of his and our time in pursuing it; there is no way he can get a solution as presented.Just try taking the determinant of 5 of the six equations which means you have 6 sets. If you have a real problem, then one of these should yield a determinant of 0.

I come to this conclusion by starting with the first 2 eq describing the 1st joint. This yields the vectors F1 and F2. The second and 3rd equations ( joint 2) show the closed vector sum F2, F3 and F5. Since all af the angles are known and F2 is known then it is now easy to see that F5 and F3 are determined . Finally to the 3rd joint we have F3 and the angles for F4 and T0 which is overspecified, meaning that it would yield another T0, which not possible.








NewReynolds

 

[electricpete]

Thanks zekeman. I'll take your word for it. I have a hard time visualizing what the underlying physical problem is.

fwiw, attached I converted the equations to matrix form using mupad (included in Matlab). Results of ignoring one equation at a time give different result for every single equation ignored (no two solutions match)

3 possibilities come to mind.
1 - I made an error in programming. Always a possibility.
2 - there is a specific required relationship among the angle variables which was not captured in my arbitrary choice of angles. But if you (op) got the same result using correct angles, that makes this unlikely.
3 - zekeman is right. there is a flaw in the equation set.

With so many people asking, I cast my vote for posting original problem as well. The knowledgeable M.E.'s here might be able to help better if you give them a bigger view of the problem.

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(2B)+(2B)' ?

 

[rb1957]

i don't think you can ignore one equation and hope that the solution of the remaining equations will solve the shunned one. unless you can prove it is a redundnat equation (a multiple of existing equations).

you could obtain 6 different solutions, by using different sets of 5 equations, then the average of these solutions should be a pretty good estimate.

but like others above, give a picture of the problem ...

 

[electricpete]

rb1957 – Just to explain where I was coming from. My stated assumption was:

This suggests we can pick any of the 6 equations and ignore it (assuming the equations are correct and define a unique solution)

If the assumption was valid that there was a single unique solution (consisting of a value for each of 5 forces), then the equations must be redundant and by the logic previously shown we could pick any one to ignore (and should get the same result).

I agree the validity of that assumption of unique solution is in question and in fact seems disproven by my previous attachment. That was part of the purpose of the exercise.... to explore the characteristics of the equation set.

I’m not sure I follow your logic of seeking a “best-fit” for this type of problem.

newreynolds – if you want to post your angles, I can easily plug them in to see if I get the same values on equations 5 and 6 as you did. That would just a double-check that neither of us made an error in our programming.

=====================================
(2B)+(2B)' ?

 

[zekeman]

From an earlier post I remarked

"..... there is no way he can get a solution as presented.Just try taking the determinant of 5 of the six equations which means you have 6 sets. If you have a real problem, then one of these should yield a determinant of 0."



what I meant to say was:

there is no way he can get a solution as presented.Just try taking the determinant of 5 of the six equations which means you have 6 sets. If you have a real problem here, then two of these should yield a determinant different from 0, noting that a 0 determinant means redundancy in the set. Either of these 2 sets would yield the same answer.

On another note, I erroneously commented that eq3 and eq 4 were not independent, when in fact they are (thank you Pete for correctly sticking to your guns on this). They are both needed for loop closure at the 2nd joint (equilibrium)for this static problem.)

 

[Occupant]

I got a solution - compare it to yours. Obviously, as was said before, I had to remove one of the equations.

 

[zekeman]

Occupant
If you were paying attention, you just can't take ANY 5 equations of a flawed set of 6 equations and have a VALID solution.

 

[Occupant]

No, that's not true. You have five unknowns, you need five equations. It doesn't matter which ones you pick as long as they are independent. Of course, a valid solution reqires a valid set of equations. Garbage in - garbage out.

 

[electricpete]

Occupant’s solution matches the last scenario in my attachment where equation 6 was ignored. It’s a useful datapoint for me, in that it helps convince me there are no silly math errors hiding in my attachment. ...which tends to reinforce the conclusion that zekeman reached – there is no a single solution that satisfies all six equations... i.e. the set of 6 equations is not self-consistent.

Side note – since there were no errors generated during computation of the inverses of the 5x5 A matrices of my attachment, we know that each group of 5 equations has a single unique solution.... but again for this particular 6-equation set, the solutions are different depending on which group of 5 we pick.


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(2B)+(2B)' ?

 

[Cockroach]

I would concur with Zekeman, he is correct, equations 3 and 4 are dependent. Taking the derivative of 3, I get 4. So actually you have five equations in five unknowns.

If you are getting other errors, you have a modelling issue, check those equations again.

Just my two cents worth.

Kenneth J Hueston, PEng
Principal
Sturni-Hueston Engineering Inc
Edmonton, Alberta Canada

 

[electricpete]

3 : 0 = - F2*cos(phi1 + theta6) - F3*cos(phi2 + theta7) - Fs*cos(psiL)

... equations 3 and 4 are dependent. Taking the derivative of 3, I get 4.

I’m not sure what role derivatives have in discussion of linear dependence/independence of equations. But I have to ask: derivative with respect to what variable?


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(2B)+(2B)' ?