Basic Hookes law question

I am diving more into the FE capabilities of Karamba and wanted to start with a simple test: Tension on a rectangular steel plate modelled via (triangular) shell elements.

I am using the benefits of symmtry and applied corresponding boundary conditions, with the respective degrees of freedom. (Hope my logic is right here…)

A load of 1000 kN should be applied, but due to the symmetrical nature, the point load on the loaded vertex on the symmetry axis is halved, which equals to a total load of about 974 kN.

With Hookes law (sigma=E*epsilon), one can find the theoretical deformation:

epsilon=(LF)/(EA) = 0.325 mm

L … initial length
F … Force
E … E-Modulus
A … Area

The max. deformation according to Karamba is however 0.309 mm. About 5% wrong. Is that a normal thing when working with these TRIC elements, or is there sth that I am missing?

Thanks a lot, cheers,
Rudi

karambaTest.gh (62.6 KB)

Poisson ratio effects?

From my knowledge when using Karamba3D it takes EVERYTHING into account, so you are probably missing something. Could be a unit rounding as your dealing with small numbers, or even something a trivial as gravity or a starting force in a direction. Another possibility is with the meshing. If you have your point load at 0,0,0, and then mesh subdivides where the center node is .5,0,0, you wont catch it if the length is say 500 in X. It could be something as small as that like snapping to a grid node.

This does not include Poisson ratio effects. I do not know if Poisson ratio effects would be large enough for this amount of tension to have a 5% effect on elongation.

Hello @rudolf.neumerkel,

how did you divide the the piece in sections for the hand calculation?

When dividing the specimen into three sections to approximate the non-uniform cross section I get:
A1 = 100cm², L1 = 43cm, A2 ~ 150cm², L2 = 7cm, A3 = 200cm², L3 = 20cm.
With E=21000kN/cm2 and an axial load of 1000kN the approximated displacement is
u = 1000/21000 x (43/100+7/150+20/200) = 0.027cm
With A = 100cm² and L = 70cm the result is
u = 1000/21000 x 70 / 100 = 0.033cm

Since the stress distribution in the thicker part is not uniform the above calculations are only approximate.

Did you test a uniform geometry as well?

– Clemens

Exactly. I did not, excuse me… Sometimes I forget to look at the problem from a distance.

Looking at a uniform geometry, everything works fine (max deformation nearly identical (+0,1%) to the calculated one) and a homogeneous stress state.

Yes of course. Sorry, my rational thinking to have been shut off yesterday…

Another mistake I made, is, that I divided the individual point loads uniformly. However, if the mesh edges are not uniform, then one has to calculate the correct equivalent point loads, that approximate the line load. Uniform point loads would otherwise lead to unrealistic stress concentrations.

Excuse me for the confusion, and thank you for the help!
Cheers, Rudi

E modulus is 1D property. It is called engineering modulus of elasticity. It is calculated neglecting actual cross section reduction during straining (it is referred to original area of cross section). True modulus of elasticity which is material property is E/(1-ν)

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