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How to apply normal stress on a solid box cube ?

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The thing is that we have to study deformation in solid cube . How to we study ?

3 Replies Last Post 21 apr 2015, 16:16 GMT-4

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Posted: 9 years ago 21 apr 2015, 11:37 GMT-4
Boundary load with a specified force per unit area, right? Then a displacement constraint on at least one boundary to prevent acceleration.

The cube won't be under uniform stress, but then uniform stress would be trivial (a simple function of the compliance matrix).
Boundary load with a specified force per unit area, right? Then a displacement constraint on at least one boundary to prevent acceleration. The cube won't be under uniform stress, but then uniform stress would be trivial (a simple function of the compliance matrix).

Ivar KJELBERG COMSOL Multiphysics(r) fan, retired, former "Senior Expert" at CSEM SA (CH)

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Posted: 9 years ago 21 apr 2015, 13:52 GMT-4
Hi

what about modelling 1/8 of a cube and apply 3 symmetry conditions on 3 adjacent boundaries, that should give you, by loading on one of the free boundaries, a rather symmetric cube load case, no ?

--
Good luck
Ivar
Hi what about modelling 1/8 of a cube and apply 3 symmetry conditions on 3 adjacent boundaries, that should give you, by loading on one of the free boundaries, a rather symmetric cube load case, no ? -- Good luck Ivar

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Posted: 9 years ago 21 apr 2015, 16:16 GMT-4
In the 1/8 cube case, for example the positive octant, if you for example were to apply a unit force per unit area normal to the +z face then implicitly zero to the +x and +y faces, that would be equivalent to applying an opposing force also on the virtual -z face (zero net force), and the symmetry boundary condition would implicitly immobilize the cube, so that would work.
In the 1/8 cube case, for example the positive octant, if you for example were to apply a unit force per unit area normal to the +z face then implicitly zero to the +x and +y faces, that would be equivalent to applying an opposing force also on the virtual -z face (zero net force), and the symmetry boundary condition would implicitly immobilize the cube, so that would work.

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