Nastran linear analysis problem
Nastran linear analysis problem
(OP)
I am running a straightforward linear stress anaylsis in MSC Nastran for Windows but for some reason when I change the modulus of my material the displacement result changes as expected but the stress does not change at all. Has anyone ever had this happen or have an explanation for it? The structure is basically a box with forces on all sides. The material is isotropic.





RE: Nastran linear analysis problem
pj
RE: Nastran linear analysis problem
For a STATICALLY DETERMINATE system, the stress will not change. There are plenty of linear systems for which a change to the Young's modulus for a portion of the system can radically change the stress distribution within the system.
Kristinj--please don't take this wrong. It is important that you understand what pjhype and I have both stated. This is some fairly fundamental mechanics (sophomore year of college) that, if you don't appreciate, can get you in a lot of trouble when doing CAE work.
Brad
RE: Nastran linear analysis problem
It is to your credit that you are asking the question. Please continue asking such questions. The most dangerous CAE analyst (and I've known many) is the one who doesn't question "why" when something is not understood.
Brad
RE: Nastran linear analysis problem
Thanks
RE: Nastran linear analysis problem
First the code assembles the stiffness matrix using E, G, nu, and geometric factors.
Second, the code solves for deflection.
Third, this is backsubstituted to solve for stress.
Consider for a simple bar (lacking the FEA details, but my point should fall out from this example):
for u=displacement, F=force, L=length, A=Area, K=F/u:
1)K = Area*E/Length
2) Assemble matrix for elements and global matrix, solve for displacement u = F/K = F*L/(A*E)
3) backsubsitute: strain = u/L; stress = E*strain =>
stress = E*u/L = E* (F*L)/(A*E)/L = F/A
If one increases E, then the displacements obtained in step 2 linearly increase, and the stress remains constant (for a statically determinate system). For a statically determinate system E falls out of the stress equation (as does L).
However, go through the same exercise above, but with two simple bars in parallel. Changing "E" (or A) on one bar will change the displacements obtained. But the other bar will still use the same "E" to calculate step three, yet will be operating on a different u (obtained in step 2). In this case the stress in the bar changes. This is the classic static indeterminacy problem from basic Solid Mechanics. The same concepts hold true for FEA.
I hope this makes sense.
Brad
RE: Nastran linear analysis problem
The finite element code will first calculate a global stiffness matrix (which includes the material modulus). Knowing the applied forces, it is possible to use Hooke's law to calculate the nodal displacements (not element stresses). Knowing the relationship between strain and nodal displacement in each element, it is then possible to calculate the element stresses.
Let's say that you double the modulus in your model (everything else constant). Each term of the stiffness matrix doubles, but the calculated displacement is halved. During stress recovery, since the displacements are halved, the strains are also halved. These (halved) strains are then multiplied by the (doubled) modulus, yielding the same stress that was calculated in the original analysis.
pj
RE: Nastran linear analysis problem
RE: Nastran linear analysis problem
RE: Nastran linear analysis problem
One other thing--a system can be statically indeterminate, but portions of it can still be statically determinate (and vice-versa). Kind of a weird concept, but one can come up with many common examples for which this holds true.
Hope this helps,
Brad