Can a Stiffness Matrix Be Antisymmetric in Neutrally Stable Structures?

[Pooty]

So, I am trying to assemble the Stiffness Matrix in my Finite Element Analysis course (Structures) and I keep coming out with a stiffness matrix that is not symmetric. I learned that for any neutrally stable structure, the stiffness matrix must be symmetric. Are there special cases that I am not taking into account that may allow for it to be antisymmetric? Thanks.

Also, this might seem sort of redundant but I actually look at these two definitions in a different light. Maybe someone could shed some light on the definition of a stiffness coefficient for me but I was given 2 definitions.

Kij

Definition 1: The Force at i caused by a unit displacement at j
Definition 2: The Force at i required to cause a unit displacement at j

If someone is really good at setting up stiffness matrices I would love to post what I got.

 

[pongo38]

Look at the Muller-Breslau theorem,(that the deflection at A due to a force P at B is equal to the deflection at B due to the same force P at A) and maybe that will reveal (1) the symmetry of the stiffness matrix and (2) the definitions you give follow from each other? That's just a suggestion for you to consider. I may be wrong though.

 

[nvn]

Did you assemble boundary conditions into your stiffness matrix yet? Or not yet?

Both stiffness coefficient definitions you listed appear correct. It just depends on what you consider to be the cause, and effect. In definition 1, the displacement is arbitrarily said to be the cause. In definition 2, the force is arbitrarily said to be the cause. It is two ways to state the same thing.

 

[Pooty]

Pongo, Thanks. I got the symmetry to work out. I will have to look at the problem in the way you are talking about to see if that makes if visually easier. My problem is visualizing how the freebody reacts to imposed loads and displacements.

NVN, The boundary conditions were set for me. It was a horizontal rigid beam of length 2L. Spring with constant k1 at the far left and spring with constant k2 in the middle.

I was told that my DOF's were vertical displacement at the left end and rotational displacement at the left end. I had a hard time picturing that when I displace the left end 1 unit in the vertical direction the whole beam raises as opposed to just the left side. I figured it out and got the K11 stiffness to be k1 + k2. It was a 2x2 stiffness matrix.

 

[DeeNos]

I understand your frustration with trying to assemble a stiffness matrix that is not symmetric in your Finite Element Analysis course. However, I want to assure you that there are no special cases that allow for an antisymmetric stiffness matrix in a neutrally stable structure.

The reason for this is because in a neutrally stable structure, the forces and displacements are linearly related, meaning that the stiffness coefficients must be the same for both definitions you have been given. This is why the stiffness matrix must be symmetric.

I suggest checking your calculations and making sure that you are using the correct equations and assumptions in your analysis. It is also helpful to double check your work with a colleague or professor to ensure accuracy.

If you are still having trouble, I would be happy to take a look at your stiffness matrix and provide feedback. Just be sure to include all relevant information and assumptions so I can accurately assess your work.

Remember, as a scientist, it is important to always question and check your results to ensure accuracy. Keep up the good work and don't get discouraged!