Re: [deal.II] fourth-order referential deviatoric tensor Dev_P: why is 'S' used in the definition?

Hi Simon, If you compare Holzapfel's definition with that of Wriggers, for instance, one notices the difference in how they define this 4th-order identity tensor. This particular term comes from the derivative d*C*/d*C* (i.e. differentiating a [symmetric] tensor w.r.t itself), and if you do

Re: [deal.II] fourth-order referential deviatoric tensor Dev_P: why is 'S' used in the definition?

On 8/8/22 15:11, Simon Wiesheier wrote: grafik.png grafik.png The only place I use Dev_P is to compute the above double contractions ('P' in the above is what dealii returns as Dev_P) As you can see, the fourth order tangent C_bar consists of outer products of second order symmetric tensors:

Re: [deal.II] fourth-order referential deviatoric tensor Dev_P: why is 'S' used in the definition?

[image: grafik.png] [image: grafik.png] The only place I use Dev_P is to compute the above double contractions ('P' in the above is what dealii returns as Dev_P) As you can see, the fourth order tangent C_bar consists of outer products of second order symmetric tensors: boldsymbol 'I' is the

Re: [deal.II] fourth-order referential deviatoric tensor Dev_P: why is 'S' used in the definition?

On 8/7/22 06:00, Simon wrote: the fourth-order referential deviatoric tensor as returned by Physics::Elasticity::StandardTensors

[deal.II] fourth-order referential deviatoric tensor Dev_P: why is 'S' used in the definition?

Dear all, the fourth-order referential deviatoric tensor as returned by Physics::Elasticity::StandardTensors < dim >::Dev_P includes the fourth-order referential/spatial unit *symmetric*