Why is Guyan reduced Stiffness matrix zero?

[Su5hi]

When a Guyan reduction is applied to a finite element model and reduced to a statically determinate number of degrees of freedom (1 node/6dof), why does the reduced stiffness matrix become zero? I've been trying to wrap my head around this all day, and I understand the intuition behind it. The definition of a statically determinate system is that the system's equilibrium equations are fully sufficient to describe its behavior; therefore, the stiffness matrix is simply redundant and not necessary to define the system's behavior. Additionally, if you displace that one node, everything should move rigidly and forces should be zero. Henece, the K matrix must be zero. That all makes sense. But I don't understand how the math works that it comes out to be zero.

K_reduced =

So, in this case, the left (boundary/preserved) and right (internal/omitted) stiffness terms are exact. Thus, K_reduced = 0.

Working out the math to get the result isn't difficult, but why does K_mm = k_ms*K_ss^-1*K_sm only when the size of K_mm is statically deteremiant is puzzling me. I would appreicate any insights you guys could provide. Thanks

 

[rb1957]

your explanation of this analysis approach doesn't jive with what looks to be the source (presumably) ...

https://arc.aiaa.org/doi/abs/10.2514/3.2874

(written by Guyan) where the matrix is reduced by removing nodes that have zero force. How you find these in the first place is a bit of a mystery, but the idea makes sense (a node with zero force won't contribute to the strain energy of the structure).

Not sure where the determinate structure comes in ?

"Hoffen wir mal, dass alles gut geht !"
General Paulus, Nov 1942, outside Stalingrad after the launch of Operation Uranus.

 

[Su5hi]

Thank you for the response! Yes, that is the source! The equation in that paper is the same as what I showed above. You don't necessarily have to remove all the nodes with zero force. In fact, you shouldn't. You must carefully pick the nodes to reasonably capture the mass distribution. The reduction process preserves the stiffness, not the mass.

Sorry if my explanation wasn't clear. When indicating statically determinate or not, I am not referring to the full system itself, but to the number of DOF the structure is reduced down to.

If you take an FEM where all nodes have 6 DOF, and you use Guyan reduction to reduce it down to just one node (6 DOF), the K matrix will be zero. Now, since the model boundary/interface only has 1 node/6 DOF, I refer to this as statically determinate boundary. If you reduce it down to 2 nodes/12 DOF, the reduced K matrix won't be zero.

The reasoning behind why the K matrix should be zero when reduced down to one node makes sense. Why the math works out that way to give you zero is what I'm having a hard time working out.

 

[rb1957]

sorry but (as I'm used to saying) "words have meaning", particularly technical words.

does the K matrix redistribute forces in answer to the model's redundancy ?
if so, that should answer your question.

another part of the answer may be that a model with only 6 dof would still require 6 dof constraint (for rigid body motion) ... which is back again to be statically determinate.

"Hoffen wir mal, dass alles gut geht !"
General Paulus, Nov 1942, outside Stalingrad after the launch of Operation Uranus.

 

[dmapguru]

Static condensation starts by moving the points in the M set each, one at a time, by a value of 1.0 while all other points in the M set are fixed. The result is M displacements of the sm DOF according to these unit displacements. This computes the result of this action on the S set so that once you know how the S set moves relative to these unit displacements, then you can work out how the S set moves relative to any scaled combination of these unit displacements. Along with the stiffness on the S set, the unit motion of the M set defines a static transformation from M to S, i.e. the transformation is the term Kss^-1.Ksm.

When you move the M set with the unit displacements, they push on the S set DOF with a force equal to the terms in the Ksm partition, that is, the Ksm partition contains the magnitudes of the loads applied on the S set due to unit motion of the M set.

The static transformation Kss^-1.Ksm contains value in displacement units, it is a force applied to an inverted stiffness. To compute the reaction forces on the M set it needs to be multiplied by Kms in the same way as restraint forces are computed once the displacement on the unrestrained points are available, so the reaction forces on the M set are Kms.Kss^-1.Ksm. But these reaction forces come from a set of unit displacements on the M set, so if F=KU, and U=1, F=K. The reaction forces are the sign reversed values of the stiffness that needs to be added to the M set to represent the contribution to the M set stiffness from the reduced S set stiffness. For a statically determinate M set, the values Kmm and Kms.Kss^-1.Ksm are equal so result in total zero stiffness.

DG

 

[Su5hi]

dmapguru,

I understand that Kms.Kss^-1.Ksm represents the reaction forces on the m-set from the reduced s-set. This makes sense. But why do we then subtract it from kmm, which represents the existing forces on the m-set (again, assuming unit displacement on the m-set so F = K)? Why isn't it kmm + Kms.Kss^-1.Ksm? That way, you take the existing forces on the m-set and add the missing contribution from the reduced s-set. Why subtract it?

Also, if we follow the same logic, for a statically indeterminate m-set, why is kmm not equal to Kms.Kss^-1.Ksm? This is where my understanding hits a dead end.

Thank you!

Also, I like your username. lol. I started to get into DMAP to understand existing alters but paused because I was so discouraged by how arduous it was. And, I'm someone who enjoys programming. I could really use a DMAP guru in my work life. haha

 

[dmapguru]

The term Kms.Kss^-1.Ksm is subtracted because it computes reaction forces. Just as an applied load is equal and opposite to a reaction force in linear statics, so the total behaviour must turn the reaction force into internal loads by reversing the sign.

For a determinate system, when any one of the DOF in the boundary is moved while the others remain fixed, no forces are generated at the fixed DOF - this is a requirement of a statically determinate system. For an indeterminate system, this is not true. The motion of each boundary DOF creates reaction forces at the other boundary DOF; the sum of all the forces, including the (indeterminate) reaction forces, must be in equilibrium.

Try it out on a very small problem with no condensation defined, so it is easy to see what is happening. Define the DOF on the boundary as fixed DOF (SPC), then define as many load cases as you have DOF in the boundary. In each load case, define enforced motion of 1.0 at each boundary DOF in turn. Look at the reaction forces at the other fixed DOF. For a determinate system, these should be zero and non-zero for an indeterminate system.

Yes DMAP is quirky, but if you use MSC Nastran a lot, it is very useful. There is a book about it available from (let me look...) Lulu press by Mark Robinson - google dmap robinson lulu (don't waste your time trying Amazon - their bot scrapers make it look like it is available to pull you into their web site, but when you get to their pages it says currently unavailable and it's been that way for years).

DG

 

[Su5hi]

I did end up playing around with a small model to better understand why the math works the way it does. After that, going through your explanations again, it all makes sense —great explanation! Thank you!

I also have a copy of that book and have some training materials from MSC, so I’m planning to revisit those and dive back into DMAP. Wish me luck! haha