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.

 

[rb1957]

math is one thing ... reality another.

the problem as defined implies a statically indeterminate problem which requires a different approach to solve correctly.

i think, but i could be wrong ...

 

[zekeman]

Newreynolds,

Not enough unknowns, I suspect you missed a link somewhere.

Maybe at joint C?

 

[newreynolds]

K guys, sorry I haven't gotten back sooner...been busy with actual work.

So this is just something I saw in a random statics book when I en devoured to UofT's library for a copy of Dudley's "When Splines Need Stress Control". I wish I could read German so I could just use the DIN standard =(

So originally I analyzed it by taking the moments of the individual members and I was able to get all of the member forces, but I could not solve for the spring force so I tried to solve it like a truss since it is kind of like a truss and that is where those equations came from. Looking them over now, I suspect some negative signs may be off. I don't often (read never) analyze trusses so I wasn't sure if I was just doing something wrong.

According to the original problem, each of the "To" forces are tension forces caused by 4 groups of people pulling on the structure with rope. They are each equal to 200N.

The problem asks to determine the spring force that keeps the structure at Theta6 = 100deg and Theta7 = 20deg

The diagram is obviously not to scale and Link 1 is 9m long, Link 2 is 5m long, Link 3 is 7m long, Link 4 is 9m long

Lamda2 = 50deg
lamda3 = 55deg
lamda4 = 50deg
lamda5 = -20 deg

Everything else i had to calculate.

Sorry I didn't post the original problem in the first place but to be honest, I wanted to solve it on my own as much as possible. It seems that many are interested as well though so have at it.

 

[FeX32]

OP, Don't tell me you are a student.....

rb, I think you mean overdeterminente (i know overdeterminante is not a regularly used word) . Indeterminate suggests not enough equations, to which does mathematically imply a parametric solution (infinite amount of solutions).
Overdeterminent can imply no solutions, as is the case with more equations than unknowns. Sorry to be anal here , you probably meant that.
Everyone,
Basically, the OP has something wrong. Otherwise he CANNOT solve the system (given the independence of the equations).
I think the eng forum is over thinking the problem. A quick glance at that figure tells me there is something not right. But I won't go ahead and analyze it as the OP may be a student.... but I will say that 1 of 2 things can be wrong.
Variable 6 is in the figure.... ...or there is only 5 equations.....have fun.





Fe

 

[zekeman]

Your problem is that the external forces in T0 have a net moment so it is not in static equilibrium.

That may be the only problem with the analysis.

 

[electricpete]

I would say PsiL should be treated as a variable.

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(2B)+(2B)' ?

 

[electricpete]

I guess we should look closely at all three angles PsiL, Phi1, Phi2. I'm sure they are not independent variables. Maybe there are two equations that can be brought to defining the relationship of these three new angle variables, which would bring number of equations to 8 and number of variables to 8?

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

 

[electricpete]

I'll go back to my 19 Oct 11 1:01 answer. Phi1 and Phi2 can be determined from (theta6-theta7) knowing the lengths of the four bars. PsiL remains a variable (it cannot be arbitrarily specified)

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

 

[electricpete]

Nope maybe not. PsiL can be solved from the lengths and theta6 and theta7 as well.

The whole assembly should be able to pivot by some angle delta. Theta6 changes to Theta6+delta, Theta7 change to Theta7+delta, PsiL changes to PsiL+delta. Delta would be the 6th variable.

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(2B)+(2B)' ?

 

[electricpete]

Disregard my last 4 comments, which were putting the cart before the horse and still not going in the right direction.

See zekeman's comments 19 Oct 11 0:46.

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(2B)+(2B)' ?

 

[rb1957]

it seems odd to me that you have four forces (To) acting in four different directions and not orthogonal.

the spring makes things interesting ... what is it's unstretched length ?

what happens along the dashed line ? (at angle phiL)

the datums for theta1 and theta2 look odd ?

i think that the resultant at the two upper points (where your To loads? are applied) needs to be directed between the two links otherwise there's bending in the links. if the To forces line up with one link, the link reacts the load and there's zero load in the other link.

something to note is that this is a three force body, the three forces (the pairs of To forces at the upper points, and the reaction at the base) will intersect at a point.

it looks like the geometry of the structure is defined for you. if you know the orientation of the links, then you can determine the loads in the links. at a joint, the applied load is the resultant of the pair of To forces and this is balanced by reactions in the two links ... if you know their directions, there is a unique balance (from static equilibrium). the two upper links then join with the spring ... the spring load is the resultant reaction of the two upper link loads, and so finally to the base reaction point (the base reaction is the resultant of the two lower links and the spring force). then check the overall balance of the structure.

is this a student problem ?

 

[FeX32]

zekeman has it IMO. There is the missing variable=> alpha, the angular acceleration of the whole mechanism. Otherwise you can make some assumptions about symmetric forces (so that alpha=0) to try to solve. This should make many of the variables equal to their symmetric counterparts.




Fe

 

[rb1957]

don't the angles provided define the solution ? ie, isn't it a static problem ? i picture it as the static result of the initially dynamic "problem".

 

[newreynolds]

First of all, no I am not a student. I already said that I got the problem from an old statics book so yes, the problem is intended for students.

I don't know where you guys did your undergrad at but I doubt you ever had to design a spline without the professor giving load capacity formulas, hence why I went to the library to get dudleys formulas.

The problem looked interesting so I decided to try an solve it in my free time. I apologize if I'm actually enjoy doing this type of thing.

If it were a school problem, why wouldn't I go ask the prof for help instead of asking randoms?

 

[FeX32]

rb,
I see it as static only if the moment about that fixed pivot point by the external loads is zero (the variable alpha is zero). For this to happen we must make some assumptions about the forces T0.

newreynolds,
The eng-tips community are not randoms. And we also enjoy it or else we would be doing something else with our spare time no?
We just have to check you are not a student, that's all. And about asking prof's questions. Most of them are busy and would not sit down to help a student solve a statics problem from a book; at least that's my general understanding of it...
It's an interesting problem, nonetheless.




Fe

 

[FeX32]

i picture it as the static result of the initially dynamic "problem"

That's an interesting comment. I presume you could simplify it knowing that it's a static problem. So an instantaneous static solution.
Maybe I should just do the problem eh . If I have time later I will do it.


Fe

 

[rb1957]

particularly given "The problem asks to determine the spring force that keeps the structure at Theta6 = 100deg and Theta7 = 20deg."

given theta6 and theta7 and the lengths of the four links, the geometry is fixed, ie the orientation of the upper links can be determined. given this the upper two loading points can be balanced, which'll determine the loads in the links. the spring load is the resultant of the upper two links at the upper spring attmt pt. finally, determine the ground reaction (the lower two links and the spring) and check against the reaction to the applied loads to verify.

 

[electricpete]

If all the angles are fixed as stated/implied, then there is a net moment associated with the applied forces. The moment cannot be resisted by the pinned support. If we imagine instead a fixed support, then we need to consider not just tension/compression in the bars, but also bending.

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(2B)+(2B)' ?

 

[rb1957]

so there are two equations for the RH load pt to give you the loads in links 3 and 4, and two equations at the LH load pt to give you the loads in links 1 and 2. then there are two equations at the upper spring attach pt, but there is only one variable, Ps.

the conclusion i draw is that it doesn't matter which equation you use to derive the spring force (either sumFx or sumFy) 'cause these two are not independent of one another.

you could determine the spring force from the lower attachment, since the ground reaction is determined by the applied loads, so Ps is the only unknown here too.

so really there are 4 equations available to deterine the spring force ... and they should all give the same answer.

 

[electricpete]

they should all give the same answer

I doubt it. Give it a try.

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(2B)+(2B)' ?