Pressure and resulting stress on surface

[Pingen]

Hi,

In essence, I'm simulating a plate which is supported by bars that run along the plate's edges. A pressure is applied on the plate's surface and the system is exposed to a gravitational field.

I run the simulation and then look at the stresses in the plate's normal direction on the plate's surface. Shouldn't these stresses be equal to the applied pressure?

Pingen

 

[GBor]

Do you have a copy of Roark's Formulas for Stress and Strain? This is a problem right out of his book and he provides some simple formulas that will give you what your results should be.

It sounds like you are using an isotropic plate, but have you done a mesh sensitivity analysis? What are your boundary conditions? How is the load applied? And, what software are you using? How thick is your plate?

There are many things that go in to producing an FEA result. Without a few of these questions being answered, it is difficult to determine whether your results are correct.

Garland E. Borowski, PE
Borowski Engineering & Analytical Services, Inc.
Lower Alabama SolidWorks Users Group
Magnitude The Finite Element Analysis Magazine for the Engineering Community

 

[Pingen]

Hi Garland,

I don't have the book, but I just ordered it.

-The plate is a curved rectangle, shaped like a half-pipe
-The pressure is unevenly distributed on the concave surface
-Two surfaces are fixed. If the plate was extended to a full cylinder, then the fixed surfaces would be located by the cylinder's openings. (I hope that explanation can be understood)
-It's a relatively thin plate, yet I'm using solid elements
-It's a fine mesh and the stresses appear in smooth transitions. I have not done a mesh sensitivity analysis though
-I'm using ANSYS
-The model is currently linear

Regards,
Pingen

 

[rb1957]

if the plate is curved then it'll support the normal pressure with in-plane membrane hoop stresses = pR/t. the applied pressure should also produce shear stresses thru the thickness, as tho' the pressure is trying to shear the plate along the perimeter of the area (of the pressure) ... if it was a circular area the applied force is p*(pi*R^2), and the shear stress is p(pi*R^2)/(pi*2R*t) = pR/t (that's the first time i did that calc ... didn't expect that result !)

 

[GBor]

I think that's pR/2t.

Sorry I missed your last post, Pingen, but rb1957 is more than qualified to solve your problem.

Sounds like you have something similar to half a pressure vessel, which is where the pR/t and pR/2t equations would come from.

 

[rb1957]

doh !

GBor's right about the shear stress (funnily enough it equals the longitudinal load in the pressure vessel), but the hoop stress is pR/t

but the structure is more like a doubly cantilevered beam.

i guess we could turn the question around and ask why would you think the stresses in a beam should be equal to the applied pressure ?

 

[pja]

You are right..if you are applying the pressure to a flat plate where the normal is the z direction than sigma_zz on the surface should be the same as the applied pressure..however many shell elements used in commercial codes condense the constituitive equations down to a plane stress assumption hence you may see zeroes for sigma_zz.

 

[rb1957]

pja,

i'd say that's right if you're close to an axis of symmetry ... the further away for this the plate sigma zz has to react the sum of the applied pressure (effectively the integral of the applied pressure). in any case, i'd expect that this stress would be much less than sigma xx due to bending.

 

[pja]

I stand by what I said..the Cauchy equations for the boundary tell you that if a traction is applied in the z direction, which for a FLAT PLATE would be the pressure applied, the normal component of the stress would be equal in magnitude to the pressure:

Z=sig_zz * n + sig_xz * l + sig_yz * m

where Z is the z traction and n,l and m are the direction cosines which for the upper and bottom surfaces of a flat plate have values of +-1, 0 and 0.

Whether or not this guy has a flat plate I have no idea..he didn't say one way or the other.

 

[pja]

Oh he did say whether it was curved..my fault I didn't read his second post. My equation is still correct though..: ) and the others are correct about the fact that the load will be predominantly handled via shear stresses. The above formulas are correct for thin-walled pipes, ie if R/t >10

 

[prost]

Wasn't the original question about the computed normal stress on the surface (some people call these 'tractions')? I understood the question to be nothing about the internal stress state (pR/t, pR/2t for thin shells, thick shells have bigger formulae of course) but only about the external stress state, that is, the tractions. Did I miss something?

If the plates are curved, if they are flat, doesn't matter--if the load is only a pressure (indicating a stress normal to the surface everywhere) and no shear tractions are applied, then the normal stress has to be equal to the pressure; in this case the direction cosine of the surface is the same whether you are computing pressure or normal traction. Now the reasons for the discrepancy in your FE model can only be speculated. Bad mesh comes to mind first, that is not enough elements or poorly placed elements. Are you doing this with plates or shell elements, or are you doing this with 3D brick elements? It will help if you identify the element as precisely as possible--not just "brick" but how many nodes are in the brick.

 

[prost]

I humbly beg your pardon; surface tractions are handled different ways, depending your whether are using plates/shells or 3D elements. I don't do much work with shells, forgot about that little bit. Sorry.

To see why it makes a difference, it might help to think of the shell as the two dimensional version of a beam; certainly the equations that describe equilibrium in shells and beams are similar, with similar approximations in the models. Now think of a beam, simply supported, loaded by a constant pressure 'q' along the beam (say in the 'y' direction, the x direction is aligned with the axis of the straight beam). Certain approximations are made to make this an easy problem to compute. Now try to calculate the complete stress state, sxx, syy, szz, sxy, sxz, syz in this beam. szz is zero of course, so are sxz and syz. sxx is from the classical My/I, shear from VQ/It. What is syy? Zero! Completely counterintuitive, nevertheless the right answer for the model, that is, a product of the formulation (beam equations).