Fea on single plate element-image attached

[chandran]

I just modeled a single square plate of size 50x50 and thick 1 and applied a force on two nodes(total force is hence 100). The software i used is nastran. The force vector is shown as arrow in the image.

My doubt is all the points in the plate should have the stress=100/area=100/(50*1)=2. But in the four nodes, two have a stress of 1.90 and two others have 2.03. Any body can explain this?

 

[PerennialII]

Sounds like a somewhat typical difference arising from stress interpolation, from your point loading contrary to smooth boundary traction and overall accuracy of the FEM solution is typically worth a percent or two ... you got a bilinear element and the other end is fixed right? Check stress values at integration points, they are most accurate at those locations and should be closer to your reference.

 

[chandran]

yes, perennial the other end is fixed. But then why this? how to find the integration point?

 

[PerennialII]

I don't use Nastran myself but in FE codes you can typically specify from which location you want your output extracted. Usually the options are nodes, element centroids and integration points. So requestion, via some output option, a listed (numerical) output such that the location of the output is integration points should give you the specific results. I'll be interesting to see how much difference there is.

In bilinear isoparametric plates if you're using reduced 1*1 integration you've your single point at point (0,0) (the center of your element), alternatively you may have (and probably have) used a 2*2 grid integration scheme, where the points are located symmetrically at $1/\sqrt{3}$ (symmetrically with respect to both coordinates axes, 4 points in total).

 

[chandran]

perennial,
I have modeled a single plate element with 50x50 dimensions and thickness of 1. I fix the two nodes at one side and apply 50 force at each node at the other side. After running the static test
the result of von mises stress for that element is 288.


Manually if i consider the plate as a beam and apply the classical mechanics formula of stress=moment*y/moment of inertia(where y is 0.5(half the thk) in this case) i get the result for the stress as 600. Where I have gone wrong


Note:
moment=50(distance)*(50+50)=5000
y=0.5
moment of inertia=4.16


I have attached the image

 

[PerennialII]

First make sure you've comparable stresses, the beam theory estimate you've calculated compares to the uniaxial stress component of the plate rather than the von Mises stress. Then we can compare it to analytical plate theory reference using both biaxial plane stress components.