Nastran linear analysis problem 1

[kristinj]

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.

 

[pjhype]

For a linear system, a change in the material modulus will not change the stress in the system. Recall that the fundamental definition of stress is the intensity of the forces distributed over a given section (sigma = force divided by area). There is no modulus in this equation. Modulus enters into the picture when you want to calculate strain or displacement (i.e. epsilon = sigma divided by modulus). So the results that you are getting out of Nastran are valid. The calculated change in displacement will be inversely proportional to the change in modulus.

pj

 

[brad]

pjhype is mostly correct, but missed an important point:
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

 

[brad]

One addition--
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

 

[bjpil]

I guess I do not understand either. Do not most codes first solve for displacement/strain, create a the stiffness matrix, using two of the three(the E, G, and Poisson's ratio), and then solve for stress using Hooke's Law? If this is so then changing the modules should change the stress. Please advise.

Thanks

 

[brad]

bjpil,

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

 

[pjhype]

bjpil,

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

 

[bjpil]

I got it. Thanks.

 

[kristinj]

thanks for the replies. i did not consider my system statically determinite so i was expecting the stress to change. but taking another look at it in that light, the results do make sense. thank you.

 

[brad]

Kristinj,
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