Hi there.
I meshed a cubic model in the Hypermesh, and I imported it in Abaqus. I want to apply a shear load (in step 2) and a tensile displacement (in step 1) to it. I have 2 questions, and I will be grateful to answer my questions?
1- in order to apply a shear load, I chose all nodes on the top and bottom of the model, and I constrained all nods except U2 and defined a positive and a negative displacement in the U1 direction in the top and bottom, respectively. and then, I chose all nodes on the right and left of the model, and I constrained all nods except U1 and defined a positive and a negative displacement in the U2 direction in the right and left, respectively. Is this method correct to apply a shear load? If it is not, what is the correct method?
I did not constrain U2 on the top and bottom because it interferes with e.g. the edge connecting right face and bottom face.
2- I don't want that the result in the first step (the tensile load) affects the result of the second step (the shear load) because they are independent. therefore, I deactivate the first step in the second step. However, in the visualization module, the contour of stress in step 1-increment 1 is the same as the contour of that in step 2-increment 0. I think It means the boundary condition affects the second step because the contour of stress in step 2-increment 0 should be zero. What is the solution?
Hi Amirhosein Naghizadeh,
1. To create shear deformation, select all nodes on the boundary of the cube, set U1=1,U2=U3=0, and define Distribution as any linear function of Y using Create Analytical Field (see f(x) on the right hand of the Distribution in your 1.jpg). Replace Uniform by (A) AnalyticalField-1.
2. To delete all deformations obtained in the first step (provided that material behaviour is elastic), create Step-2 with all displacement BCs equal to zero. Then create Step-3 with shear deformation.
Go through this tutorial : https://www.youtube.com/watch?v=AHezCUtatZs it will help you.
thank you Mr Victor J. Аdlucky. I have a question about your answer. I will be grateful to be answered again.
1-Am I supposed to repeat this method for other faces? I mean I should set U1=-1, U2=U3=0, and define Distribution for the bottom and set U1=U3=0, U2=-1 for the right and so on.
2-What is the difference between the method you mentioned and my method? because the Y of the top face is constant, creating a linear function of Y doesn't make a difference with setting a constant U1 and U2 with help of the blank box
provided.
@ Victor J. Аdlucky
Dear Amirhosein Naghizadeh,
I have suggested to apply one BC for all 6 faces simultaneously. Of course you can apply your own BCs for each face separately. For example, 1) top face: U1=u, U2 - unconstrained, U3=0; 2) bottom face: U1=-u, U2 - unconstrained, U3=0; 3) right face: U1 - unconstrained, U2=v, U3=0; 4) left face: U1 - unconstrained, U2=-v, U3=0; 5) front face: U1,U2 – unconstrained, U3=0; 6) back face: U1,U2 – unconstrained, U3=0. u and v are arbitrary values. This way you’ll get pure shear stress-strain state with nonzero strains E12=E21=2*(u+v)/L, where L is the edge length. My case is v=0, your case is u=v. The exceptional case u=-v corresponds to the rigid body rotation without shear deformation.
Hi try this tutorial for applying shear load
Link: https://www.youtube.com/watch?v=YOm1BYFVqN4
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