Dear all mates:
I recently had a talk to someone, and was told a very surprise opinion that If for a bar, theoretically we want to apply compressive pressure to both ends as axial compressive load(as the situation attached), but if I set the boundary condition in that way, the bar in modelling will not be at rest at all, and if I want to make it work, I have to set one end as compressive pressure, another end as a boundary condition that suppress the axial translational DOF, is that the truth? Since I cannot find any clue in ABAQUS documentation, can anyone tell me a bit more about this?
Thanks very much!
Best Regards,
Mengli
Dear Song,
During a finite element static or dynamic analysis, the equation P=K*Δ has to be solved for loads and displacements. Normally, the vectors and matrix in this equation are subdivided into smaller components as follows:
[Pf,Ps]=[Kff,Kfs Ksf,Kss]*[Δf,Δs]
where Pf are the loads imposed on the structure, Ps are the constraint reactions, Δf the displacements at the free dofs, Δs the displacements at the constrained dofs (usually zero), and Kff, Ksf, Kfs, Kss the appropriate submatrices of the stiffness matrix.
So, in your case, all dofs are considered free, and consequently the above equation takes the form:
Pf=Kff*Δf
In your case the loads Pf are known, but the displacements Δf cannot be calculated by the last equation, because the determinant of the stiffness matrix of any system is zero by definition, and therefore not invertible.
To sum up, if no boundary conditions are specified for a model, no displacements can be calculated, and the model will not be at rest at all. In any model, boundary conditions have to be specified, in order to calculate its displacements.
Best regards and hope this helps.
George
Dear Mengli,
I don't understand why you want to do that. If an axial force is applied to one end of the bar, then an axial reaction force of the same size will occur on the other end. Regardless, it is not possible to apply a force and a displacement (with the same degree of freedom) on the same node.
Best regards,
Henry
you need to apply minimum displacement boundary conditions ( a maximum of six for 3D structures) and see that the body is constrained enough against the three translational degrees of freedom and then three rotational DOF. Otherwise, stiffness matrix to be solved is singular and program will give a message to the effect.
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