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Solving 5 Unknowns with 6 Equations

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.

RE: Solving 5 Unknowns with 6 Equations

Tell us your equations.

Good on ya,

Goober Dave

RE: Solving 5 Unknowns with 6 Equations

post up the actual problem

 

RE: Solving 5 Unknowns with 6 Equations

(OP)
Thanks guys, the equations are as follows:

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

hi

Your post is meaningless without a diagram of the problem.

 

RE: Solving 5 Unknowns with 6 Equations

Eq 3 and 4 are redundant.
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

Correction

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

Plenty of matrix solvers out there waiting for input.

RE: Solving 5 Unknowns with 6 Equations

Looking at the math without knowing the underlying problem, I don't see Equations 3 and 4 as being dependent. They seem independent to me.

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.
 

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

RE: Solving 5 Unknowns with 6 Equations

"Plenty of matrix solvers out there waiting for input."

Right, just solve the 5x5.

RE: Solving 5 Unknowns with 6 Equations

give us a pic ...

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

"Looking at the math without knowing the underlying problem, I don't see Equations 3 and 4 as being dependent. They seem independent to me.

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

Another way of looking at this is the OP eqs 3 and 4 are precisely the statement  that
A+B+C=0

 

RE: Solving 5 Unknowns with 6 Equations

Quote:

Another way of looking at this is the OP eqs 3 and 4 are precisely the statement  thatA+B+C=0
Yes. That is a 2-D vector equation, which is equivalent to 2 scalar equations.

Quote:

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.
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?  

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

RE: Solving 5 Unknowns with 6 Equations

Quote:

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?
Is it possible there should be another unknown magnitude substituted for one of those T0's? (That would give 6 equations in 6 unknowns)

I'm just asking. I don't know the answer.  I'm not real familiar with the joint method.

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

RE: Solving 5 Unknowns with 6 Equations

I agree with those who say we need the actual problem rather than a set of equations.

RE: Solving 5 Unknowns with 6 Equations

Are you trying to solve a truss? Post a drawing and you'll get an answer.

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

I agree that having a drawing describing the problem is better than trying to guess from the equations.  By guessing I made a mistake assuming "To" was an unknown.  Therefore disregard my message 17 Oct 11 6:00.

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).
 

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

RE: Solving 5 Unknowns with 6 Equations

Add the following way:
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)
 

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

RE: Solving 5 Unknowns with 6 Equations

My apologies zekeman for disagreeing with you on that point.

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

RE: Solving 5 Unknowns with 6 Equations

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

RE: Solving 5 Unknowns with 6 Equations

Quote:

...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.   

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

RE: Solving 5 Unknowns with 6 Equations

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

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

RE: Solving 5 Unknowns with 6 Equations

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.

  

RE: Solving 5 Unknowns with 6 Equations

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, ...)

RE: Solving 5 Unknowns with 6 Equations

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) ?  

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

RE: Solving 5 Unknowns with 6 Equations

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.

cheers

peace
Fe

RE: Solving 5 Unknowns with 6 Equations

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.  

peace
Fe

RE: Solving 5 Unknowns with 6 Equations

(OP)
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.

RE: Solving 5 Unknowns with 6 Equations

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

RE: Solving 5 Unknowns with 6 Equations

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.  

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

RE: Solving 5 Unknowns with 6 Equations

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 ...

RE: Solving 5 Unknowns with 6 Equations

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

Quote (electricpete):

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)'  ?

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

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.

RE: Solving 5 Unknowns with 6 Equations

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.

RE: Solving 5 Unknowns with 6 Equations

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.
 

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

RE: Solving 5 Unknowns with 6 Equations

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

RE: Solving 5 Unknowns with 6 Equations

Quote:

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

Quote:

... 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?
 

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

RE: Solving 5 Unknowns with 6 Equations

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 ...

RE: Solving 5 Unknowns with 6 Equations

Newreynolds,

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

Maybe at joint C?
 

RE: Solving 5 Unknowns with 6 Equations

(OP)
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.

RE: Solving 5 Unknowns with 6 Equations

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 smile, 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....smile ...or there is only 5 equations.....have fun.

cheers
 

peace
Fe

RE: Solving 5 Unknowns with 6 Equations

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.

RE: Solving 5 Unknowns with 6 Equations

I would say PsiL should be treated as a variable.  

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

RE: Solving 5 Unknowns with 6 Equations

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)'  ?

RE: Solving 5 Unknowns with 6 Equations

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)'  ?

RE: Solving 5 Unknowns with 6 Equations

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.

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

RE: Solving 5 Unknowns with 6 Equations

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.

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

RE: Solving 5 Unknowns with 6 Equations

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 ?

RE: Solving 5 Unknowns with 6 Equations

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.

cheers

peace
Fe

RE: Solving 5 Unknowns with 6 Equations

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".  

RE: Solving 5 Unknowns with 6 Equations

(OP)
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?

RE: Solving 5 Unknowns with 6 Equations

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? smile
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.

cheers

peace
Fe

RE: Solving 5 Unknowns with 6 Equations

Quote:

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 pipe. If I have time later I will do it.  

peace
Fe

RE: Solving 5 Unknowns with 6 Equations

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.

RE: Solving 5 Unknowns with 6 Equations

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)'  ?

RE: Solving 5 Unknowns with 6 Equations

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.

RE: Solving 5 Unknowns with 6 Equations

Quote:

they should all give the same answer
I doubt it. Give it a try.

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(2B)+(2B)'  ?

RE: Solving 5 Unknowns with 6 Equations

I agree you can solve the left and right joints to find a tension in bars 1, 2, 3, 4.
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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(2B)+(2B)'  ?

RE: Solving 5 Unknowns with 6 Equations

not all the links will be in tension; i think link4 is in compression.

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

Quote:

if and that's If) the problem is static then it Should work,
That's the point. The applied moments do not sum to zero. There is no restraining moment. Net moment is not zero. Static equilibrium is not satisfied.

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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(2B)+(2B)'  ?

RE: Solving 5 Unknowns with 6 Equations

[i]1.97-54.6[i] Should'be been 1.97-5.46

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(2B)+(2B)'  ?

RE: Solving 5 Unknowns with 6 Equations

pete beat me to it wink

...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 smile

peace
Fe

RE: Solving 5 Unknowns with 6 Equations

agreed, but the force in the spring is calculatable given the geometry, and i think the result of plotting the two load vectors and the ground reaction vector will show that the three forces don't intersect and so it isn't in balance ... it is in force equilibrium ('cause we've used sum Forces to determine the results) but fails moment equilibrium (as you've shown).

RE: Solving 5 Unknowns with 6 Equations

You're right: force equilibrium and moment equilibrium are two different topics that perhaps I mixed up. Neither will be satisfied in the problem as specified.   

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(2B)+(2B)'  ?

RE: Solving 5 Unknowns with 6 Equations

no, i believe you can satisfy force equilirbium (in that the ground is going to prevent the structure from translating), but as you've pointed out there is no moment equilibrium (so it'll want to rotate).  still the problem can be answered, the load in the spring can be determined from the geometry.

RE: Solving 5 Unknowns with 6 Equations

The problem with the force equilibrium results from specifying theta6, theta7, L1, L2, L3, L4, which in effect specifies psiL.  The angle of psiL does not correspond to the angle of spring force required for static force equilibrium.

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

RE: Solving 5 Unknowns with 6 Equations

that's a problem !  maybe that's a symptom of the moment imbalance, in that this joint (and the others neither ?) isn't in force balance ... 'cause if the links are rotating, the loading points will be translating.

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

Quote:

The problem asks to determine the spring force that keeps the structure at Theta6 = 100deg and Theta7 = 20deg
Jeeze. I neglected to read that. Haha.

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.

 

peace
Fe

RE: Solving 5 Unknowns with 6 Equations

In real life I would do a dynamic analysis however.... smile

peace
Fe

RE: Solving 5 Unknowns with 6 Equations

A correction to my own comments - I stated the value of PsiL would not be correct for static equilibrium of the middle joint (balancing the forces F2 and F3 computed from left and right joints), but I never verified that  (I was assuming based on my previous calcs using ficticious angles... never tried the real PsiL).  

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(2B)+(2B)'  ?

RE: Solving 5 Unknowns with 6 Equations

(OP)
The problem must be malformed, moment equilibrium of the whole structure can't be satisfied with the spring fixed to the pivot.

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

I think somebody ought to take the bold step of MAKING   lambda3=29.665 deg ( correct me if I'm wrong)  forcing the contraption into equilibrium so one of us can  mercifully put this thing to bed with a single unique solution.




 

RE: Solving 5 Unknowns with 6 Equations

I concur sleeping2

peace
Fe

RE: Solving 5 Unknowns with 6 Equations

"The problem must be malformed, moment equilibrium of the whole structure can't be satisfied with the spring fixed to the pivot.'

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

(OP)
Well, I assumed that the loads wouldn't change and that it was the linkage that needed to be adjusted to go into equilibrium.

RE: Solving 5 Unknowns with 6 Equations

@ the OP, since you wanted to solve this problem,

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

Oh, somehow I missed that ph1 and ph2 were not given so you must assume some values to complete the geometry and use a solid link for the spring.

RE: Solving 5 Unknowns with 6 Equations

Quote:

somehow I missed that ph1 and ph2 were not given so you must assume some values to complete the geometry
In effect, they are specified.  We have theta1 and thet2 specified as well as L1, L2, L3, L4. This means phi1 and phi2 can be solved.

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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(2B)+(2B)'  ?

RE: Solving 5 Unknowns with 6 Equations

Quote:

If we could vary any of the lamda's or theta's, there might be a solution but..
.... but it would seem a contradition to the stated problem which was supposed to achieve target values of theta6 and theta7  (and by extension the associated phi1, phi2, psiL).

[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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(2B)+(2B)'  ?

RE: Solving 5 Unknowns with 6 Equations

You are right,but there are 2 solution sets for ph1 and ph2. Using geometry:

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

Yes, I did realize there were 2 solutions.  It didn't seem relevant to the point I was driving to which was that the angle psiL is not a continuous variables that can be freely chosen to suit force balance, rather it is pre-defined by the problem statement. But good to clarify.   

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(2B)+(2B)'  ?

RE: Solving 5 Unknowns with 6 Equations

Good point pete. I didn't realize at first you could solve for phi1 and phi2. This means what I said about the 2nd method is not the way. But, the first one is valid.

cheers

peace
Fe

RE: Solving 5 Unknowns with 6 Equations

Hi,

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

in this particular problem i think equations 5 and 6 are inter-dependent ... i think they both solve for Ps given the loads in the other links.

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

Oh man, this is all wrong.

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

This is great. My last input is all "dynamic" smile.
I'd be happy to solve the model as given (in a dynamic fashion). But, I think this could expand this thread to oblivion.

cheers

peace
Fe

RE: Solving 5 Unknowns with 6 Equations

Since there's a moment about the fixed pin, no spring force applied to the fixed pin will be able to keep Theta6 and Theta7 to 100 and 20 degrees.  If we were mathematicians instead of engineers we might have realized that when we reduce our equations down to something like Fs != Fs, then we have proven that the original premise (Theta6=100 and Theta7=20 in a static condition) is impossible.  Thanks, newreynolds, for taking me back to school.  Sometimes old dogs can relearn old tricks.

RE: Solving 5 Unknowns with 6 Equations

Hence dynamic smile

peace
Fe

RE: Solving 5 Unknowns with 6 Equations

"This is great. My last input is all "dynamic" smile.
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

I have a proposal to handle the dilemma posed by the 5X6 linear set

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

Relax Zeke.
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....

peace
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

Fair enough Zekeman. There would be some assumptions in the analysis. (IC's ect.)

cheers

peace
Fe

RE: Solving 5 Unknowns with 6 Equations

maybe now the OP understands why we need to see the problem, rather than be given a peice of it to solve.

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

Dynamic solution would require knowledge of mass per length of each bar and knowledge of the spring constant of the spring (and by the way, do we assume the bars are rigid = infinite spring constant)  And I'm pretty sure it would never come to equilibrium without damping.  If anyone can do it, I'm sure FeX32 can. I'm not sure exactly what this author had in mind, but I think we'd all agree it wasn't dynamic analysis.

Quote (electricpete):

A correction to my own comments - I stated the value of PsiL would not be correct for static equilibrium of the middle joint (balancing the forces F2 and F3 computed from left and right joints), but I never verified that
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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(2B)+(2B)'  ?

RE: Solving 5 Unknowns with 6 Equations

Perfect pete.
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 smile? So it would move way left to a position calculable from the force direction. This also proves that the author of the problem didn't really think about the problem at all. The problem is totally flawed from the beginning.
Seems we have completely destroyed it. haha.

Quote of the discussion goes to pete

Quote:

Everyone has their own slant, and that's what makes it interesting.

Cheers to that.  

peace
Fe

RE: Solving 5 Unknowns with 6 Equations

The site is a bit wacky this evening (glitchy). (admin: feel free to kill this post later if you like pipe)

peace
Fe

RE: Solving 5 Unknowns with 6 Equations

I still shake my head.  This problem is an engineering STATICS issue, what's with the dynamic model?

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

Cockroach,
 
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 thought we agreed that as stated the problem is not in static equilibrium... the external forces simply create a moment about that fixed pivot that cannot be reacted upon by the mechanism. Thus, if those loads were applied the system would move...hence dynamic.

I do agree with zekeman's response, as that would have been perfectly appropriate.  

peace
Fe

RE: Solving 5 Unknowns with 6 Equations

I don't agree, not a dynamic problem at all since motion is impending.  We know nothing of the linkage mass, inertia, etc, how are you going to pull dynamic information?

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

There is no attachment.

Quote:

We know nothing of the linkage mass, inertia, etc, how are you going to pull dynamic information

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.  

peace
Fe

RE: Solving 5 Unknowns with 6 Equations

IMO, even though this thread has blown up it is still interesting smile

peace
Fe

RE: Solving 5 Unknowns with 6 Equations

(OP)
The original problem was a statics problem.  Thus it was clearly a bad problem.  I think we should move on with our lives, no?

RE: Solving 5 Unknowns with 6 Equations

Haha. Yea, why not pipe.
It's typical of engineers to go nuts on details. Some of us combine that with stubbornness and you know what happens next.. smile

peace
Fe

RE: Solving 5 Unknowns with 6 Equations

As was mentioned, there are two solutions for the pair of angles (psi1,psi2).  

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

http://files.engineering.com/getfile.aspx?folder=c7de07cd-7ae6-4e80-a1a2-8be77affdcbe&file=FILE1c_OtherOrientaiton.pdf

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(2B)+(2B)'  ?

RE: Solving 5 Unknowns with 6 Equations

thumbsup2

peace
Fe

RE: Solving 5 Unknowns with 6 Equations

Spring load is 276.2 N at 71 degrees.  I have used Method of Joints to get the solution, the geometrical interpretation of the 4 Bar Linkage is fully described in my computation.

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

Seems to be uploading today, maybe operator error.  My apologies for the lack of response yesterday.

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

Cockroach,

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 smile. You agree?
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.  

cheers

peace
Fe

RE: Solving 5 Unknowns with 6 Equations

Also, in a real life situation, I believe the dynamic forces in the members would superseded those of the instantaneous static ones. So, from a design perspective (assuming motion is wanted) the dynamic solution still needs to be conducted. No? Treating it as a textbook problem, it's a nonsensical conclusion, and all of us are correct to a certain point.   

peace
Fe

RE: Solving 5 Unknowns with 6 Equations

hmm. Instantaneous vs. equilibrium solutions.
All we need is perspective. Personally, I would have gown with equilibrium.   

peace
Fe

RE: Solving 5 Unknowns with 6 Equations

(OP)
Looked over Cockroach's solution real quick and I don't think I saw any reaction forces when you summed the forces about vertex 0?

Also, I believe this is a 5 bar linkage since the ground link counts as a link.

RE: Solving 5 Unknowns with 6 Equations

Good point OP (newreynolds). At first I thought that point O was point A. But looking again, there is a missing reaction force in Cockroach's solution.
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.

cheers  

peace
Fe

RE: Solving 5 Unknowns with 6 Equations

Correct FeX32, static solution since the sum of forces equals zero.  That would imply motion impending, a split second later, depending on Spring stiffness, which is beyond the scope of the question.  I would presume the point of the problem is to determine the load on that Spring as a result of Rope input tension.

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

(OP)
Who said anything about summing moments?

RE: Solving 5 Unknowns with 6 Equations

"As was mentioned, there are two solutions for the pair of angles (psi1,psi2).  

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

Lets make it longer pipe

peace
Fe

RE: Solving 5 Unknowns with 6 Equations

The location of the left-most force in Cockroach's writeup is 160 degrees CCW from horizontal.  

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.

Quote:

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.
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.
 

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

RE: Solving 5 Unknowns with 6 Equations

Your last comment is obviously correct. One question remains, however, why did you bother?

RE: Solving 5 Unknowns with 6 Equations

Because it's there.

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

RE: Solving 5 Unknowns with 6 Equations

Many comments are correct. 4 eqn's 4 unknowns, then use the middle joint... done deal.
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 pipe. But, who am I to say...haha, I had fun reading all the different ideas and analysis's.     

peace
Fe

RE: Solving 5 Unknowns with 6 Equations

its years (well, decades actually) since i've done a mechanism, but can we apply force equilibrium to the loaded points wihtout including an inertia term ?  the ground reactions can be determined from the overall applied forces.

RE: Solving 5 Unknowns with 6 Equations

Yes we can. We assume the bars are completely rigid so we don't have to deal with their deflection. This is usually a fair assumption in simple structural mechanisms that will deflect very little under load.
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.  

peace
Fe

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