Solving 5 Unknowns with 6 Equations
Solving 5 Unknowns with 6 Equations
(OP)
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.
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.





RE: Solving 5 Unknowns with 6 Equations
Good on ya,
Goober Dave
RE: Solving 5 Unknowns with 6 Equations
RE: Solving 5 Unknowns with 6 Equations
1: 0 = F2*cos(phi1 + theta6) - F1*cos(theta6) + To*cos(lamda4) - To*cos(lamda5)
2 : 0 = F2*sin(phi1 + theta6) - F1*sin(theta6) + To*sin(lamda4) + To*sin(lamda5)
3 : 0 = - F2*cos(phi1 + theta6) - F3*cos(phi2 + theta7) - Fs*cos(psiL)
4 : 0 = - F2*sin(phi1 + theta6) - F3*sin(phi2 + theta7) - Fs*sin(psiL)
5 : 0 = F3*cos(phi2 + theta7) - F4*cos(theta7) + To*cos((3*pi)/2 + lamda2) + To*cos(pi/2 - lamda3)
6 : 0 = To*sin(pi/2 - lamda3) + F3*sin(phi2 + theta7) - F4*sin(theta7) + To*sin((3*pi)/2 + lamda2)
The variables are F1, F2, F3, F4, Fs and each equation pair is the force summation at the three respective joints. i.e. equation 3 is the sum of the forces in the x-direction and equation 4 is the sum of the forces in the y-direction for Joint B
RE: Solving 5 Unknowns with 6 Equations
Your post is meaningless without a diagram of the problem.
RE: Solving 5 Unknowns with 6 Equations
So you can use eq 1,2,3,5,6 and get your answer.
When you get F2 and F3 from these equations, either eq3 or eq 4 will yield F6 uniquely. Try it.
RE: Solving 5 Unknowns with 6 Equations
use eq 1,2 5,6 to get
F1, F2, F3, F4
Then use either eq 3 or Eq 4 to get F5
RE: Solving 5 Unknowns with 6 Equations
RE: Solving 5 Unknowns with 6 Equations
Let's say we have 3 vectors A, B, and C, represented by complex numbers with magnitudes -F2, -F3 and -Fs respectively and angles in the complex plane of phi1+theta6, phi2+theta7 and psiL, respectively.
A = -F2* {cos(phi1+theta6) + i * sin(phi1+theta6)}
B = - F3*{cos(phi2 + theta7) + i * sin(phi2 + theta7)
C = - Fs*{cos(psiL) + i * sin(psiL)}
We can see that equation 3 tells us the real part of the vector sum (A+B+C) is 0. In contrast equation 4 tells us the imaginary part of the vector sum (A+B+C) is 0. The vector sum has two independent coordinates, there is no reason that one has to be zero if the other one is zero.
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RE: Solving 5 Unknowns with 6 Equations
Right, just solve the 5x5.
RE: Solving 5 Unknowns with 6 Equations
it looks like you're assuming an applied force (To) to be the toatl joint reaction in a direction ?
not so sure i agree that eqns 3 & 4 are redundant (ie equivalent, a multiple of one another ... looks to like sum of forces in orthogonal directions.
3 joints = 6 equations (3*sum forces in two direction).
how many members ? 4? (F1 to F4) and Fs and To ??
RE: Solving 5 Unknowns with 6 Equations
Let's say we have 3 vectors A, B, and C, represented by complex numbers with magnitudes -F2, -F3 and -Fs respectively and angles in the complex plane of phi1+theta6, phi2+theta7 and psiL, respectively.
A = -F2* {cos(phi1+theta6) + i * sin(phi1+theta6)}
B = - F3*{cos(phi2 + theta7) + i * sin(phi2 + theta7)
C = - Fs*{cos(psiL) + i * sin(psiL)}
We can see that equation 3 tells us the real part of the vector sum (A+B+C) is 0. In contrast equation 4 tells us the imaginary part of the vector sum (A+B+C) is 0. The vector sum has two independent coordinates, there is no reason that one has to be zero if the other one is zero."
If A, B, C were arbitrary vectors, you would be right.
However, if you look at it in the context of a joint, these are not arbitrary vectors but sum to zero.
Given that, if you know 2 of the vectors, then the amplitude of the third requires only one equation.
RE: Solving 5 Unknowns with 6 Equations
A+B+C=0
RE: Solving 5 Unknowns with 6 Equations
Two scalar equations are required to make those vectors sum to zero. Just because you know they sum to zero doesn't change the fact that these 2 equations must be satisfied to make them sum to zero.
So, my conclusion remains that Equations 3 and 4 by themselves constitute 2 independent scalar equations in 3 scalar unknowns (F2, F2, Fs), assuming all the angles are known.
HOWEVER, assuming there is a solution, I agree with you there must be redundancy somewhere among the 6 equations. It certainly may be that equation 3 or equation 4 can be dropped. But as for myself, I don't think that can be determined without looking at the other four equations.
I notice Equation 5 and 6 have the same form as equations 3 and 4 (vectors sum to 0). However, equations 1 and 2 are different in that the magnitude T0 appears twice. So it is only a vector sum to 0 equation if there are included two different vectors with exactly the same magnitude T0, but with different directions (described by lambda4 and lambda5). Does that part make sense for the physical problem?
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RE: Solving 5 Unknowns with 6 Equations
I'm just asking. I don't know the answer. I'm not real familiar with the joint method.
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RE: Solving 5 Unknowns with 6 Equations
RE: Solving 5 Unknowns with 6 Equations
From a strictly mathematical point of view you have a system of equations and the number of equations exceeds the number of unknowns so:
1. If you don't have two equations linearly dependent amongst your five, your system is impossible
2. Upon condition you have two equations linearly dependent amongst your five, if your equations are not compatible the system is impossible
3. Upon condition you have two equations linearly dependent amongst your five, and your equations are compatible, the system has, at least, one solution.
4. If your equations are compatible and you have more than two equations linearly dependent amongst your five, the system has infinite solutions.
RE: Solving 5 Unknowns with 6 Equations
If we attempt to analyse in absence of drawing, we notice:
Eq1 and 2 have unknowns: F1, F2
Eq2 and 3 have unknowns F2, F3, Fs
Eq5 and 6 have unknowns F3, F4
ASSUMING the equations are correct as written, and the unknowns are F1, F2, F3, F4, Fs, then if we wanted to solve it manually, we could manually solve two ways (below). The first way suggests we can ignore either equation 1 or 2. The second way suggests we can ignore either equation 5 or 6. I could not find any straightforward way to suggest we can ignore equations 3 or 4:
FIRST WAY:
Solve Eq1 and Eq2 For F2 and F1.
Use results of above (F2) to solve Eq3 and Eq4 for F3 and Fs.
Now we are left with 2 equations (Eq5 and Eq6) with only one unknown (F4). In this case (assuming equations are correct) I think it is safe to ignore 5 or 6 (use the ignored one as a double-check).
SECOND WAY:
Solve Eq5 and Eq6 For F3 and F4.
Use results of above (F3) to solve Eq3 and Eq4 for F2 and Fs.
Now we are left with 2 equations (Eq1 and Eq1) with only one unknown (F1). In this case (assuming equations are correct) I think it is safe to ignore 1 or 2 (use the ignored one as a double-check).
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RE: Solving 5 Unknowns with 6 Equations
THIRD WAY:
Use 1, 2, 5, 6 to solve F1, F2, F3, F4.
Then we have left equations 3 and 4 with only one unknown (Fs).
Ignore 3 or 4.
This suggests we can pick any of the 6 equations and ignore it (assuming the equations are correct and define a unique solution)
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RE: Solving 5 Unknowns with 6 Equations
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RE: Solving 5 Unknowns with 6 Equations
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
RE: Solving 5 Unknowns with 6 Equations
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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RE: Solving 5 Unknowns with 6 Equations
phi1 + theta6 = phi2 + theta7 = psiL = 45° (or 225°)
RE: Solving 5 Unknowns with 6 Equations
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.
RE: Solving 5 Unknowns with 6 Equations
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, ...)
RE: Solving 5 Unknowns with 6 Equations
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RE: Solving 5 Unknowns with 6 Equations
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
RE: Solving 5 Unknowns with 6 Equations
Fe
RE: Solving 5 Unknowns with 6 Equations
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.
RE: Solving 5 Unknowns with 6 Equations
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
RE: Solving 5 Unknowns with 6 Equations
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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RE: Solving 5 Unknowns with 6 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 ...
RE: Solving 5 Unknowns with 6 Equations
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.
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RE: Solving 5 Unknowns with 6 Equations
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.)
RE: Solving 5 Unknowns with 6 Equations
RE: Solving 5 Unknowns with 6 Equations
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.
RE: Solving 5 Unknowns with 6 Equations
RE: Solving 5 Unknowns with 6 Equations
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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RE: Solving 5 Unknowns with 6 Equations
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
RE: Solving 5 Unknowns with 6 Equations
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RE: Solving 5 Unknowns with 6 Equations
the problem as defined implies a statically indeterminate problem which requires a different approach to solve correctly.
i think, but i could be wrong ...
RE: Solving 5 Unknowns with 6 Equations
Not enough unknowns, I suspect you missed a link somewhere.
Maybe at joint C?
RE: Solving 5 Unknowns with 6 Equations
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.
RE: Solving 5 Unknowns with 6 Equations
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
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....
Fe
RE: Solving 5 Unknowns with 6 Equations
That may be the only problem with the analysis.
RE: Solving 5 Unknowns with 6 Equations
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RE: Solving 5 Unknowns with 6 Equations
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RE: Solving 5 Unknowns with 6 Equations
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RE: Solving 5 Unknowns with 6 Equations
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)' ?
RE: Solving 5 Unknowns with 6 Equations
See zekeman's comments 19 Oct 11 0:46.
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RE: Solving 5 Unknowns with 6 Equations
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 ?
RE: Solving 5 Unknowns with 6 Equations
Fe
RE: Solving 5 Unknowns with 6 Equations
RE: Solving 5 Unknowns with 6 Equations
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?
RE: Solving 5 Unknowns with 6 Equations
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
RE: Solving 5 Unknowns with 6 Equations
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
Fe
RE: Solving 5 Unknowns with 6 Equations
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.
RE: Solving 5 Unknowns with 6 Equations
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RE: Solving 5 Unknowns with 6 Equations
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.
RE: Solving 5 Unknowns with 6 Equations
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RE: Solving 5 Unknowns with 6 Equations
Then when you try to find the required force of the spring to balance the force applied by bars 2 and 3 at the center joint, you will find you need a vector force. But the directio of that vector will not align with the pre-defined PsiL (it is pre-defined by virtue of specified values for theta6, theta7, L1, L2, L3, L4 are all specified).
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RE: Solving 5 Unknowns with 6 Equations
the line of action of the spring is defined by the geometry. if and that's If) the problem is static then it Should work, the spring force should balance the upper links.
If it doesn't balance (ie one variable will satisfy only one equation) then it isn't in static equilibrium and the angles will change untill it does (as the geometry changes, the length of the spring changes, and so the load.; but note we not working with the spring stiffness).
RE: Solving 5 Unknowns with 6 Equations
Attached computes the moment associated with applied forces assuming T0=1. It is 1.97-54.6 which does not come out to zero.
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RE: Solving 5 Unknowns with 6 Equations
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RE: Solving 5 Unknowns with 6 Equations
...from your solution...then 1.97-5.46=I*alpha, where I is the mass moment of inertia of the entire mechanism about the pivot point.
Dynamic force analysis was one of my favourite topics in uni. .. and still is
Fe
RE: Solving 5 Unknowns with 6 Equations
RE: Solving 5 Unknowns with 6 Equations
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RE: Solving 5 Unknowns with 6 Equations
RE: Solving 5 Unknowns with 6 Equations
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RE: Solving 5 Unknowns with 6 Equations
i wonder if the problem also told you the unstrained length of the spring and it's stiffness, so you'd determine the spring load from F = k*x ?
RE: Solving 5 Unknowns with 6 Equations
I think given one thing, both of you are correct.
With giving the values for phi1 and phi2 (along with all the other angles you can get 4 equations and 4 unknowns which solve for F1,F2,F3,F4. With these you can solve for Fs.
(this is neglecting the non-zero moment, that whoever came up with the problem didn't really notice/ care about...)
For example, not in vectoral form the 2 equations for the left side can be simply: -To*cos(lam5)+To*cos(lam4)=F1*cos(90-theta6)+F2*cos(-theta6-phi1) and
-To*sin(lam5)+To*sin(lam4)=-F1*sin(90-theta6)-F2*sin(-theta6-phi1)
bascially we get 4 equations like the OP's equations 1,2,5,6, which can be solved for F1,F2,F3,F4. You then use F2 and F4 to easily solve for Fs.
But, if you don't have phi1 and phi2 then you need 2 more equations. agree?
So, if that's the case then we can do the following. Determine the 6 equations similar to the OP's (if not the same as I haven't checked all of them). Assume that phi2=180-phi1 and we have gotten rid of one of the 7 variables (F1 to F4, Fs, phi1, phi2), and we now have 6 equations and 6 unknowns with 1 unique solution.
Done deal.
Fe
RE: Solving 5 Unknowns with 6 Equations
Fe
RE: Solving 5 Unknowns with 6 Equations
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RE: Solving 5 Unknowns with 6 Equations
If we could vary any of the lamda's or theta's, there might be a solution but since it's a linkage there's limited range.
All of the theta's and the phi's are related by the linkage closure equations. I guess I was the one that over-thought the problem.
RE: Solving 5 Unknowns with 6 Equations
RE: Solving 5 Unknowns with 6 Equations
Fe
RE: Solving 5 Unknowns with 6 Equations
The spring is no problem; you make it a solid link and fix lamda 3 as I suggested. Now you have a system in equilibrium and each of its parts are in equilibrium.
RE: Solving 5 Unknowns with 6 Equations
RE: Solving 5 Unknowns with 6 Equations
you have a solution for the link loads, based on force equilibirum at the loaded points. you could solve these "old school" using a force polygon (= triangle) diagram.
you said you couldn't find the spring force; but i think you can from force balance of the upper spring attmt point.
is this the complete problem ? were you told anything else about the spring ?
RE: Solving 5 Unknowns with 6 Equations
RE: Solving 5 Unknowns with 6 Equations
Starting at the bottom, we can traverse L1 and L2 or L4 and L3 to arrive at same point:
L1*exp(i*theta6)+L2*exp(i*<theta6+phi1>) = L4*exp(i*theta7)+L3*exp(i*<theta7+phi2>)
Taking the real and imaginary parts yields two scalar equations in 2 scalar unknowns phi1 and phi2
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RE: Solving 5 Unknowns with 6 Equations
[quote]The problem must be malformed, moment equilibrium of the whole structure can't be satisfied with the spring fixed to the pivot. [./quote]
.... and if you changed the pivot type, you would have bending moments which completely change the approach.
And we don't know for sure if the angle psiL happens to be the single exact one required one to satisfy our force balance (until we do the calc), but it would seem a very artificial situation where we analyse a geometry that happens to have exact right value of psiL (along with the fact that we are ingoring lack of moment balance which also would make it artificial).
I agree, the problem seems malformed to me.
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RE: Solving 5 Unknowns with 6 Equations
The angles given together with the link lengths determine the location of pivot points at intersections of L1- L2 and L3-l4.
Now it is easy to see that the point L2-L3 is determined by swinging two arcs (with radii L2 and L3) from the pivot points noted above; so you get TWO intersections or two different L2-L3 points; you can pick one to complete the geometry and determine ph1 and ph2. And treat the spring as a solid link, since its stretched length is determined and the problem leaves you with no other choice.
If you want to speculate further on this, it could be that the author was mistaken about theta6 and theta 7 and all of the lambdas are correct.So instead of fixing lambda3, fix these thetas for rotational equilibrium.But why ..?
RE: Solving 5 Unknowns with 6 Equations
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Fe
RE: Solving 5 Unknowns with 6 Equations
I did not look at the equations but generally when you have an overdetermined linear equation system you can use a least-squares method. Just write it in matrix form and solve it in matlab, or look up the matlab c=A\y and see what it does for overdetermined systems.
Without further thought what I would do with an non-linear system say Fn(x) = 0 where Fn represents a set of functions F1-FN and x is a vector of variables is to solve
min Y, where Y = F1(x)^2 + F2(x)^2 + .. + FN(x)^2
You can put in weighting factors above for specific use...
br
Drex
RE: Solving 5 Unknowns with 6 Equations
however, the question in my mind is can you apply static equilibrium to points on the structure when it is not in moment equilibrium ? you can apply static equilibrium to the whole structure (i think) and determine the ground reactions; but i doubt the results of applying static equilibrium to say the RH load point (can you determine the loads in the two links given the applied loads ?)
RE: Solving 5 Unknowns with 6 Equations
NewReynolds, you need to be factual and accurate in the forum and not post interpretations of problems as you have done in the start of this thread. Like some sort of exam problem, write the problem out accurately and give us the input information. You don't start by saying something like 5 equations in 6 unknowns, this presumes a ton of work which may be in error.
Okay, you got said linkage, four "To" means ropes applying tensile load to the system at 200 N each, lamba two thru five exclusively are 50, 55, 50 and -20 degrees. The linkage one thru four are 9, 5, 7 and 9 meters long. The question is to compute spring force required keeping theta 6 at 100 and theta 7 at 20 degrees.
Is this correct? This is the original problem?
Kenneth J Hueston, PEng
Principal
Sturni-Hueston Engineering Inc
Edmonton, Alberta Canada
RE: Solving 5 Unknowns with 6 Equations
I'd be happy to solve the model as given (in a dynamic fashion). But, I think this could expand this thread to oblivion.
Fe
RE: Solving 5 Unknowns with 6 Equations
RE: Solving 5 Unknowns with 6 Equations
Fe
RE: Solving 5 Unknowns with 6 Equations
I'd be happy to solve the model as given (in a dynamic fashion). But, I think this could expand this thread to oblivion"
Fex32,
Oh, really. If you could do that accurately I would eat all these posts and post a picture of it.
I don't think that even Matlab could help you.
--------------------------------------
Drexl,
Get real, how do you solve a 5X6 linear set with Matlab?
RE: Solving 5 Unknowns with 6 Equations
Use the first 4 equations previously presented.
Instead of using eq 5 and 6, consider the following
Point L3-L4 (intersection of L3 and l4) is in equilibrium, so instead of using the T0 inputs at that point, use -F3 and -F4 as the external vector forces that put that point in equilibrium and now write the moment equation in vector form
eq5 T01xL1=-F3xL4-F3x4=-F3xL3
x is the cross product symbol (see Pete's explanation)
which is now eq 5
where T01 is the vector force of the rope at point L1-L2 and note that F3xL3=0Now you have the 5x5linear set which has a solution.
And if you have a valid problem , -F3 -F4 must be equal to the rope vector force at the point L3-L4 .
RE: Solving 5 Unknowns with 6 Equations
I specialize in dynamic analysis. This would not be a problem. Even not assuming the spring is a rod it is numerically possible. It also may be analytically possible using a powerful Udwadia analytic method....
"I don't think that even Matlab could help you."....
This honestly makes me laugh.
Also,
"Get real, how do you solve a 5X6 linear set with Matlab?"
This is possible within numerical tolerances....
Fe
RE: Solving 5 Unknowns with 6 Equations
Fex32,
"specialize in dynamic analysis. This would not be a problem. Even not assuming the spring is a rod it is numerically possible. It also may be analytically possible using a powerful Udwadia analytic method...."
Powerful methods only depend on accurately modeling the components. That was my point; the dangling spring would be a huge problem and also where would you start the problem at t=0 in this case? he only gave you what he thought was the equilibrium position.
RE: Solving 5 Unknowns with 6 Equations
Fe
RE: Solving 5 Unknowns with 6 Equations
the engineering solution to th eproblem is quite different to the mathematical approach, it is easy to solve the equations given; possibly there isn't a unique solution, but i'm sure there is a solution (mathematically). but from the eningeering viewpoint, i think we've seen that the basic formulation of the equations is in error.
on a different tack, what was the answer (ie the approach to derive Ps)
RE: Solving 5 Unknowns with 6 Equations
As a purely academic/programming excercize, attached I have now verified that the solution of PsiL will not be correct for static equilibrium of the middle joint..... i.e. the required line of action of spring force is not in the direction along the spring (all of this based on static analysis which ignores the lack of moment balance).
The result is shown in graphical form on page 6 of 6.
The bars are in blue.
Tension vectors are in green.
Applied force vectors are in red.
Spring force vector is in purple.
From the graphical solution, you can re-create the calculation for yourself:
Tension forces at left and right pin have to balance the force vectors.
Then spring force vector in the center pin has to balance the tension vectors.
The direction of the purple spring force vector is nowhere close to being toward the origin.
I apologize for prolonging a thread which some of you seeem impatient with (will it hit 100 posts?). As for me, it doesn't bother me how long it goes. Everyone has their own slant, and that's what makes it interesting.
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RE: Solving 5 Unknowns with 6 Equations
(that brings us one lcoser to 100)
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RE: Solving 5 Unknowns with 6 Equations
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RE: Solving 5 Unknowns with 6 Equations
http:
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RE: Solving 5 Unknowns with 6 Equations
4 eqns 4 unknows, then the mid joint utilized to solve for Fs. Exactly how we (me, you, zeke, rb (some more directly than others)) mentioned many posts above.
About the direction of the force, could we not just assume the spring pivot (bottom side) could change to suit the direction
Seems we have completely destroyed it. haha.
Quote of the discussion goes to pete
Cheers to that.
Fe
RE: Solving 5 Unknowns with 6 Equations
Fe
RE: Solving 5 Unknowns with 6 Equations
Cheers
Greg Locock
New here? Try reading these, they might help FAQ731-376: Eng-Tips.com Forum Policies http://eng-tips.com/market.cfm?
RE: Solving 5 Unknowns with 6 Equations
Four ropes hold a four bar linkage in equilibrium. A spring is to maintain orientation of linkage 4 & 1 at a particular angle to the right hand horizontal datum. Given the rope tensile load and orientation, we are asked to compute that spring force.
So what is with "dynamics"? The system is in equilibrium, there is not motion, not velocity, no acceleration.
This problem has been a nightmare since improperly posted 104 entries ago!
Kenneth J Hueston, PEng
Principal
Sturni-Hueston Engineering Inc
Edmonton, Alberta Canada
RE: Solving 5 Unknowns with 6 Equations
I couldn't agree with you more.
The first post to the thread should have read
'Mr Newreynolds
6 equations in 5 unknowns could be entirely correct.
But at least one of them is a linear combination of the 5 others. If you formulated it correctly, you should have found at least a few sets of 5 equations and obtained the same answers. You did not, so the problem is either you did not formulate it properly OR the problem you received was not in static equilibrium.
Please post the original problem and you can get help here. If you choose not to post the problem then we cannot help you.'
That's hindsight!
RE: Solving 5 Unknowns with 6 Equations
I do agree with zekeman's response, as that would have been perfectly appropriate.
Fe
RE: Solving 5 Unknowns with 6 Equations
Attached is a scaled layout of the problem. The 4 Bar Linkage can be seen as labelled in black, length and angles preserved to the datum (magenta, horizontal right). Same with the ropes (yellow), 200 N each, four in total, orientated as shown. I have arbitarily set the intersection of links 1 & 4 as the origin, (0,0).
The red circles are the position of links 2 & 3. There are two intersection points, the spring stretches from the lower to upper point. That spring is drawn in at the lower position in green.
So the problem then is to compute the spring force required to maintain equilibrium, i.e. motion impending.
Kenneth J Hueston, PEng
Principal
Sturni-Hueston Engineering Inc
Edmonton, Alberta Canada
RE: Solving 5 Unknowns with 6 Equations
Kenneth J Hueston, PEng
Principal
Sturni-Hueston Engineering Inc
Edmonton, Alberta Canada
RE: Solving 5 Unknowns with 6 Equations
This is regardless of the linkage moving or not. It will move whether we know the inertias ect. or not. So, if it were a real problem, and the linkage started to move but you didn't know the inertias you would consider it a static problem still? I hope not.
I don't want to be controversial. It's just that many of us think that with the given applied loads by the OP, the linkage would rotate about that pin joint. Of course we know it is given as a static problem.
We can still solve it as electricpete did above. But it shows that the spring must exhibit a force and moment for it to be in equilibrium.
Fe
RE: Solving 5 Unknowns with 6 Equations
Fe
RE: Solving 5 Unknowns with 6 Equations
RE: Solving 5 Unknowns with 6 Equations
It's typical of engineers to go nuts on details. Some of us combine that with stubbornness and you know what happens next..
Fe
RE: Solving 5 Unknowns with 6 Equations
The previous attachment covered the solution: (phi1 = 3.70, phi2 = 2.494 radians)
Attached is revised attachment which covers the other solution: (phi1 = -1.95, phi2 = 2.04 radians)
The conclusion is the same.
I have no further comments
ht
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RE: Solving 5 Unknowns with 6 Equations
Fe
RE: Solving 5 Unknowns with 6 Equations
Having trouble posting attachments. It may be my system, but I am trying.
Kenneth J Hueston, PEng
Principal
Sturni-Hueston Engineering Inc
Edmonton, Alberta Canada
RE: Solving 5 Unknowns with 6 Equations
I forgot to mention, compressive 276.2 N @ 71 degrees. The spring is in a state of compression. All other members are in tensile loading.
This solution is for Pts A, the spring being 5.42692 m long. The second solution is for Pts B were the spring is 8.52523 m long.
There you have it. No dynamic motion, forget dynamic model. If I were use an extremely heavy spring, stiff as a structural steel member, would you still argue dynamic model? There is no motion.
But agreed, a very interesting problem.
Kenneth J Hueston, PEng
Principal
Sturni-Hueston Engineering Inc
Edmonton, Alberta Canada
RE: Solving 5 Unknowns with 6 Equations
Thanks for the input. It has now become more interesting. You have approached the problem totally different that I would have. It is also a much different approach to what electricpete posted. I do agree with electricpete's calculations. Yours I also find no fault from 5 minutes of following it through. However, you state in your calculations that; "since no dynamic information is given, the solution is static with motion impending." Impending? I agree here. This solution is the instantaneous static solution right before it moves
If you don't agree, then how come you don't have a moment balance, which would still show an unbalanced moment about that pin joint. Regardless, good discussion.
Fe
RE: Solving 5 Unknowns with 6 Equations
Fe
RE: Solving 5 Unknowns with 6 Equations
All we need is perspective. Personally, I would have gown with equilibrium.
Fe
RE: Solving 5 Unknowns with 6 Equations
Also, I believe this is a 5 bar linkage since the ground link counts as a link.
RE: Solving 5 Unknowns with 6 Equations
He should still be able to solve what he did using point A instead. Although, the result will be different.
Still, the fact of the matter is I find nothing wrong with electricpete's equilibrium solution, to which the result is obvious.
Fe
RE: Solving 5 Unknowns with 6 Equations
Method of Joints NewReynolds, never needed sum of moments since the problem is fully determined at the vertex where the Ropes impart load. Infact, I never summed one moment in this method, simply used Vector Mathematics and some fancy algebra. Two equations in two unknowns, I have the old HP41CX program with Linear Algebra module to cut down on the work.
Also, I used AutoCAD to map out the linkage system. Not only did I have the opportunity to check the geometry of the linkage, but you can clearly see the two solution sets imposed by the binomial equations of the circles through the Links 1 & 2, 3 & 4. As correctly mentioned by many previous engineers, there are two solutions, one for each intersection point of those circles.
I suppose we could now specify a spring and have some sort of mechanical dynamic effect as the spring stretches. Clearly the physical nature of that spring would govern such motion, the stiffer the spring, the less motion. But again, we would be reading much more into the problem than the original intent.
Looks like some sort of question I would expect for an assignment in our elementry Statics course. There was a mention about a German textbook, 120 postings ago.
Kenneth J Hueston, PEng
Principal
Sturni-Hueston Engineering Inc
Edmonton, Alberta Canada
RE: Solving 5 Unknowns with 6 Equations
RE: Solving 5 Unknowns with 6 Equations
The previous attachment covered the solution: (phi1 = 3.70, phi2 = 2.494 radians)
Attached is revised attachment which covers the other solution: (phi1 = -1.95, phi2 = 2.04 radians)
The conclusion is the same.
I have no further comments winky smile"
I do, Pete, how come you changed the problem by effectively anchoring the spring to some point other than the bottom pivot?
Who gave you the license? winky smile. Obviously you needed the correcting moment to make the problem static. Very sneaky, indeed.
And Cockroach, how do you get that the spring is in compression?
Will this thread ever end? Guess I'm guilt of prolonging it.
RE: Solving 5 Unknowns with 6 Equations
Fe
RE: Solving 5 Unknowns with 6 Equations
Referring to newreynolds' post 18 Oct 11 21:12, "lamda5 = -20 deg", which would make the force 200 deg CCW from horizontal.
It is tedious to follow someone else's calucation. I sympathize with those who try to follow mine. As far as I can tell, Cockroach's calculation method was equivalent to mine (which was also discussed by many others). With correction of the angle of the force, and correction of mangitude of applied force (I used 3 instead of 200 for plotting convenience),I would expect the same results.
Since we have an extra equation, we have to ignore something, and then see if the result matches the part we ignored. If it did match, we could say the problem was somewhat* well formed (*Except for the lack of moment balance). As it turns out, it didn't match for either solution of ph1,ph2. This was intended to support the conclusion that the original problem is malformed, badly-behaved, improper, ill-mannered, inapropos, objectionable, inappropriate, disfunctional, distasteful, crass, rude, foul, and not very good.
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RE: Solving 5 Unknowns with 6 Equations
RE: Solving 5 Unknowns with 6 Equations
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RE: Solving 5 Unknowns with 6 Equations
Cockroaches soln does have a missing force however. The analyses are not a steady state(SS) solutions. I referred to pete's as being SS due to the fact that I saw an account for the unbalanced moment. However, I now see that it's not there. So neither are SS. But, neither are wrong, the are statically correct. SS is important in the real world though..
This question is certainly improper. I'm not sure if it's " distasteful, crass, rude, foul" however
Fe
RE: Solving 5 Unknowns with 6 Equations
RE: Solving 5 Unknowns with 6 Equations
If we try to determine their deflection we move from statics to mechanics and need things like inertia, cross sections, material properties ect.
On another note, dynamics considering deflection is one of the most difficult applied sciences known to man.
Fe