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Help with stress tensor interpretation on very easy beam example

Posted 28 févr. 2021, 04:48 UTC−5 Structural Mechanics 3 Replies

Peter Zellinger
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Hi,

I wanted to look into shear stress in an extremely easy example, a beam with x=1mm, y=20mm and z=50mm, which is pulled in z-Direction via boundary load of 10kN (I have attached some pictures for easier explanation of the model) and has a fixed constraint at x=0.

The material was modified to have an extremely high Young modulus of 1e15 and a poisson ratio of 0 to only have a vertical deformation. The resulting shear stress in the yz-plane at x=0 should therefore be tau = F/A, which would be 10E7 N/m^2 (With an area of 20x50mm). When i evaluate surface averages at the fixed boundary x=0, y=20, z=50, for different stress tensors solid.syz, solid.sz,... , only the solid.sxz tensor results in 1E07 N/m^2 (I have attached results as pictures as well). Souldn't the resulting shear stress of 1E07 be in the yz plane, therefore solid.syz? Why is it that the solid.sxz tensor component gives the real result and all other values are completely different? I have tried it in different orientations and with real parameter values as well, but i cant seem to find the right tensor to display the shear stress. The bending momentum should be nearly 0 in this case as the youngs modulus is so high and poisson ratio is 0, so i expected stresses in all other directions than yz to be near 0 as well.

Thanks in Advance for any help! Peter

3 Replies Last Post 4 mars 2021, 17:02 UTC−5
Henrik Sönnerlind COMSOL Employee

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Posted: 4 years ago 1 mars 2021, 04:03 UTC−5

Your force acts in the Z-direction. The normal to the boundary on which you are measuring the stress is in the X-direction. Thus, only the xz component of the shear stress is significanct and matches F/A.

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Henrik Sönnerlind
COMSOL
Your force acts in the Z-direction. The normal to the boundary on which you are measuring the stress is in the X-direction. Thus, only the xz component of the shear stress is significanct and matches F/A.
Peter Zellinger
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Posted: 4 years ago 4 mars 2021, 09:45 UTC−5
Updated: 4 years ago 4 mars 2021, 09:49 UTC−5

Thanks for the clarification! What if the geometry would be in horizontal position, so the plate is lying on the ground, fixed constraint is on the ground and force is applied on the top layer in x-direction, the stress tensor in COMSOL would be solid.szx, but in COMSOL, i can't evaluate that value. The stress tensor is symmetric, so can one assume that solid.sxz is equal to solid.szx if i want to evaluate the shear stress in a vertical structure, even if it is a complicated stress impact in all directions like thermal expansion? Thank you!

Thanks for the clarification! What if the geometry would be in horizontal position, so the plate is lying on the ground, fixed constraint is on the ground and force is applied on the top layer in x-direction, the stress tensor in COMSOL would be solid.szx, but in COMSOL, i can't evaluate that value. The stress tensor is symmetric, so can one assume that solid.sxz is equal to solid.szx if i want to evaluate the shear stress in a vertical structure, even if it is a complicated stress impact in all directions like thermal expansion? Thank you!
Henrik Sönnerlind COMSOL Employee

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Posted: 4 years ago 4 mars 2021, 17:02 UTC−5

Yes, we assume that the stress tensor is symmetric, so only the sxy, sxz, and szx shear components are available. For a complicated stress state, you will thus have six values, including the direct stresses sx, sy, and sz.

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Henrik Sönnerlind
COMSOL
Yes, we assume that the stress tensor is symmetric, so only the sxy, sxz, and szx shear components are available. For a complicated stress state, you will thus have six values, including the direct stresses sx, sy, and sz.

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