Solving 5 Unknowns with 6 Equations 1

[newreynolds]

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

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

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

 

[newreynolds]

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.

 

[FeX32]

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.




Fe

 

[Cockroach]

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

 

[newreynolds]

Who said anything about summing moments?

 

[zekeman]

"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.

 

[FeX32]

Lets make it longer


Fe

 

[electricpete]

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.

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)' ?

 

[Occupant]

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

 

[electricpete]

Because it's there.

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

 

[FeX32]

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 . But, who am I to say...haha, I had fun reading all the different ideas and analysis's.


Fe

 

[rb1957]

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.

 

[FeX32]

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.


Fe