From 74d1be80a3de58f2f0c560c435f3a825c968735e Mon Sep 17 00:00:00 2001 From: Wolfgang Bangerth Date: Tue, 3 Mar 2020 18:23:58 -0700 Subject: [PATCH] Use \mathbf C to denote the stress-strain tensor. --- examples/step-8/doc/intro.dox | 14 +++++++------- 1 file changed, 7 insertions(+), 7 deletions(-) diff --git a/examples/step-8/doc/intro.dox b/examples/step-8/doc/intro.dox index 6d1f929394..f6c9085163 100644 --- a/examples/step-8/doc/intro.dox +++ b/examples/step-8/doc/intro.dox @@ -26,16 +26,16 @@ as @f[ - \text{div}\, - (C \nabla \mathbf{u}) + ({\mathbf C} \nabla \mathbf{u}) = \mathbf f, @f] where $\mathbf u$ is the vector-valued displacement at each point, -$\mathbf f$ the force, and $C$ is a rank-4 tensor (i.e., it has four +$\mathbf f$ the force, and ${\mathbf C}$ is a rank-4 tensor (i.e., it has four indices) that encodes the stress-strain relationship -- in essence, it represents the "spring constant" in -Hookes law that relates the displacement to the forces. $C$ will, in many +Hookes law that relates the displacement to the forces. ${\mathbf C}$ will, in many cases, depend on $\mathbf x$ if the body whose deformation we want to simulate is composed of different materials. @@ -57,15 +57,15 @@ then read as @f[ - \text{div}\, - (C \varepsilon(\mathbf u)) + ({\mathbf C} \varepsilon(\mathbf u)) = \mathbf f, @f] which you can think of as the more natural generalization of the Laplace equation to vector-valued problems. (The form shown first is equivalent to -this form because the tensor $C$ has certain symmetries, namely that -$C_{ijkl}=C_{ijlk}$, and consequently $C \varepsilon(\mathbf u)_{kl} -= C \nabla\mathbf u$.) +this form because the tensor ${\mathbf C}$ has certain symmetries, namely that +$C_{ijkl}=C_{ijlk}$, and consequently ${\mathbf C} \varepsilon(\mathbf u)_{kl} += {\mathbf C} \nabla\mathbf u$.) One can of course alternatively write these equations in component form: @f[ -- 2.39.5