Hello all,
I'm currently working on setting-up a 2D bone remodeling problem of a femur from literature. I take a 2D femur geometry and give it an out-of plane plane stress/plane thickness in ABAQUS. Now, there are three loads which are provided in literature as concentrated loads. Since I give thickness, if I choose to apply this concentrated load as distributed load over 3-4 elements, I get the area as length*thickness, where length is the in-plane edge length over the 3-4 elements.
Once I get the area, I get the distributed load as F/A and use surface traction (normal) to apply it as distributed load. I choose surface traction instead of pressure because I need to apply each load at an angle with the vertical axis (i.e., not necessarily perpendicular to the surface at the area of application). Is this the correct way to convert point loads to distributed loads in 2D problem with plane stress/plane strain thickness? I really appreciate any help that I can get.
Thank you,
Balavignesh
I think that value of thickness in the case of ordinary plane strain (i. e. U3=0) has no influence on results, and distributed load should be equal to F/(length*1).
Thank you Dr. Adlucky, how does this change if I choose to apply force F as concentrated loads distributed on multiple nodes? I believe that in that case, I will have to scale it as F*t/n, where 't' is the thickness and 'n' is the number of nodes. Am I correct?
Dear Bala Vignesh,
Could you show a little sketch of your problem.
Dear Dr. Victor J. Аdlucky ,
I have summarized my questions in the attached picture. It'd be great if you can take a look and help.
Bala
Dear Bala Vignesh,
If you want to distribute concentrated force Q between n nodes, then each statically equivalent nodal force will be equal to Q/n. If you want to use tractions, then it will be equal to Q/L. All this is true if we assume that Q is a concentrated force uniformly distributed in out-of plane direction, i. e. line force. In this case resultant 3-D concentrated force is equal to Q*t.
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