I try to simulate a sphere inside a cube in Abaqus. The cube is allowed to expand in Z direction only, so I used symmetry on all faces except one XY face (which located on Z=70). The load is temperature increment from 300 K to 700 K. I used “Tie constraint“ in one simulation and “Contact Properties” in an Other one and I get the same behavior.
I can’t find any reason for:
1. The stress changes very rapidly at interface of the two material.
2. The Region of cube after sphere in Z direction has low stress.
Best regards.
Sphere properties:
Elastic modulus: 200 GPa
Poisson Ratio: 0.316
Thermal Expansion Coefficient: 9E-6 (K^-1)
Linear Plastic behavior:
Stress-Strain
400 MPa, 0.0 Plastic strain
700 MPa, 0.1 Plastic strain
cube properties:
Elastic modulus: 80 GPa
Poisson Ratio: 0.36
Thermal Expansion Coefficient: 5.6E-6 (K^-1)
Linear Plastic behavior:
Stress-Strain
130 MPa, 0.0 Plastic strain
200 MPa, 0.1 Plastic strain
Dear Saeed Mirzaee
I have reviewed your statement. Based on the differences between the two Young modulus it makes sense to have rapid changes in the stress graph. Remember that, because of the continuity law in a solid volume it is the strain that has to be smooth. Also, if you had a finer size of elements you would have smoother changes of stress.
Hi Saeed Mirzaee
I agree with Ali Golbaf
When you use tie interaction, the nodal displacement at the interface will be the same for both materials. Correspondingly, the strain at the interface is also the same. But the stress is computed from mechanical properties and is different for each part. So you see a high variation in stress.
For your second question, I think it is because of the boundary conditions that you have applied that the stress value is low.
Moreover, I personally prefer to use "tie" interaction in this problem (Contact interaction makes the problem very computationally expensive to solve and I do not see any physical sense to use contact interaction in this problem).
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