From c47cbe85dfcce41214aeabdf4c639354cc32c813 Mon Sep 17 00:00:00 2001 From: Jean-Paul Pelteret Date: Fri, 12 Aug 2022 00:16:31 +0200 Subject: [PATCH] Fix some theory in step-44 introduction The deviator tensor involves the 4th-order symmetric identity tensor, not the "normal" 4th-order identity tensor. --- examples/step-44/doc/intro.dox | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/examples/step-44/doc/intro.dox b/examples/step-44/doc/intro.dox index 30660874aa..edd216beb7 100644 --- a/examples/step-44/doc/intro.dox +++ b/examples/step-44/doc/intro.dox @@ -260,7 +260,7 @@ Similarly, the Kirchhoff stress can be decomposed into volumetric and isochoric \boldsymbol{\tau}_{\text{iso}} &= 2 \mathbf{b} \dfrac{\partial \Psi_{\textrm{iso}} (\overline{\mathbf{b}})}{\partial \mathbf{b}} \\ - &= \underbrace{( \mathcal{I} - \dfrac{1}{3} \mathbf{I} \otimes \mathbf{I})}_{\mathbb{P}} : \overline{\boldsymbol{\tau}} \, , + &= \underbrace{( \mathcal{S} - \dfrac{1}{3} \mathbf{I} \otimes \mathbf{I})}_{\mathbb{P}} : \overline{\boldsymbol{\tau}} \, , @f} where $p \dealcoloneq \dfrac{\partial \Psi_{\text{vol}}(J)}{\partial J}$ is the pressure response. -- 2.39.5