Problem modeling a simple shear stress 1

[eispiata]

Hi everybody,

I am encountering a very unexpected problem modeling a shear stress in a plane stress membrane.

Thus i got back to a very simple model which accounts for a shear stress. I used only two quad linear elements as you can see on the attached file along with both reduced and standard integration. And the results i got where completely unexpected: the stress was not uniform and the point in the middle of the bottom face didn't remain at his original position after deformation (see attached file).

It looks like Abaqus is unable to model a simple shear stress in an elastic and isotropic rubber material.

I hope you can help me to figure out what is wrong with the computation.

Thanks,

Malik

 

[xerf]

Have you tried higher order hybrid elements ?

 

[eispiata]

Hi,

Whatever i try, it doesn't work....

Actually i talked about that to the Abaqus support. The guy told me that whatever i do i won't be able to get a zero x-displacement of the node 2. According to him, the reason is that there are springs along the diagonal of the quad elements and that since node 2 is free to move along the x-axis, the equilibrium of the force at Node 2 will result in a displacement of this node 2 to the right. I attached a file to illustrate his explanation.

Honestly i don't buy his explanation.

Moreover according to him there is no way you can get a uniform shear stress if you don't constrain all the nodes lying on the bottom horizontal line in the x-direction which is not the case here since Node 2 is free to move along the x-axis.

What do you think about that explanation?

Thanks for your help!

Best regards,

Malik

 

[MrMyers]

what happens of you do not constrain your top line of nodes in the Y direction?
Also, Node 2, should X be free?

Just a guess.

 

[corus]

If, for example, you hadn't restrained Node 6 and applied your shear then I'd expect the whole of the block to compress about the bottom corner node so that nodes 2 and 6 deformed. Your shear could be considered as a direct (compressive) load, plus a bending moment about the bottom face. You've restrained Node 6 so only Node 2 can deform, albeit to a lesser extent than if Node 6 was free.

corus

 

[eispiata]

My question is: due to the fact that both Node 6 and Node 3 are constrained in the model so that they remain fixed all along the deformation process, why is Node 2 which is in the middle slightly moved after deformation.

I mean Node 2 is in the middle, and to me, it should stay in the middle even though the model has been deformed. The material is elastic and isotropic, and thus its middle point on the bottom face should remain fixed? instead of being moved along the x-axis.

Indeed if we just focus on the bottom line, its behavior should be like a beam fixed at his both extremities and thus every point along the beam should remain fixed...

Malik

 

[johnhors]

Malik,

The explanation from Abaqus support is perfectly correct.

You cannot consider the bottom line of nodes to be like a beam, because it isn't a beam ! These nodes are not isolated from the top line of nodes. The springs that Abaqus support talk of, are actually coupling terms in the element stiffness matrices. It then follows that since node 2 has not been restrained in the X direction, the enforced displacement of the top line of nodes in the X direction will drag node 2 along with it.

 

[xerf]

Have you tried to replace the 2 elements with one bi-quadratic element, with the mid-node of bottom edge at node 2 ?

 

[eispiata]

Hi Johnhors,

I truly buy your explanation. I am not very familiar with these things so if you have any document or any idea where i could find a few explanations about that in the Abaqus Documentation, please let me know.

So, as far as i understand, this means that it is definitely not possible to model an homogeneous shear stress by applying the same x-displacement to the top line of nodes, fixing the the extremity points lying on the bottom line and let the other nodes along that bottom horizontal line free to move along the x-axis...?

I mean you definitely have to constrain all the nodes lying within the bottom line of nodes in the x-degree of freedom to get a pure uniform shear stress.

I would like to read up on the spring elements used in Abaqus so let me know if you have any kind of online document about that.

Thanks,

Malik

 

[xerf]

Actually, by analogy with fluid mechanics (simple shear flows), I think you have to contraint all the nodes on the bottom edge to have zero displacement in x-direction. Otherwise some material points will be displaced.

 

[xerf]

Malik,

pure shear is different than simple shear.

If you want pure shear, you have to prescribe the displacements along all 4 boundaries according to:
u1=k*coords(2)
u2=k*coords(1)

where k is a constant.

You can use DISP subroutine. This should give you constant S12 in the entire domain.

 

[eispiata]

Hi xerf,

Johnhors's reply makes me think so. And you just confirmed what he said so i am definitely sure that Abaqus support was right.

I would like to read up now on these springs and on the coupling terms that Johnhors talked about to familiarize myself with these things and don't do the same mistake twice.

Thanks,

Malik