The same as always: pressure vector = normal vector • stress tensor. Dynamics (vibrations, waves, sound) make volume loads, not surface (contact) ones. This may be seen, for example, via variational approach with Lagrange multipliers

There will be no change. In case of fluid problem and dead loading of an solid block, there will be no change in condition in traction. Yeas, problem of different nature may be arrived. It will be negative when pointing out of surface is there. Strain will be regularized under new radial boundary condition.

Thank you Vadique Myself for the answer

When you wrote pressure vector = normal vector*stress tensor, my problem here is that the pressure is static and the stres is dynamic (so these two can not be equal).

Besides, the pressure maybe "huge" like 100MPa and the stress induced by the wave is generally small so the two sides of the "equation" may never be equal.

once again thank you

thank you Ashish Thakur

I am not sure if I understood what you meant, by "there will be no change".

Do you mean I would have the same boundary conditions as if I had a dynamic loading instead of a static pressure ?

thank you

No I do not mean that. Answer valid only for static only.

Thank you Ashish Thakur

And in this case do you have an idea how to proceed ?

I don’t see any problem here: p = n • σ(t) for any t when p is active, and it is pretty nice boundary condition. “Static” just means constant in time, it doesn’t inhibit using such constants in dynamics

And despite that vibrations may be small in comparison to the pressure, your continuum reacts to this big pressure anyway. You can assume that you have full stress σ as the sum of mean (time-independent) stress σ* and oscillating σ': σ(t) = σ* + σ'(t)