Volume and Area Change

For completeness, we’ll derive the volume change and area change:


(1)   \begin{equation*} dV=dV_0det\mathbf{F} \end{equation*}

Derivation starts from the triple scalar product. Cross products and determinants (written in tensor form) is a math topic, so we’ll skip the derivation of volumetric deformation. A quick derivation of volumetric deformation without the use of indices can be found in [Bonet].

note: det\mathbf{F}=1 \longrightarrow dV=dV_0 \longrightarrow “incompressible”

Nanson’s Equation:
Now we know that dV=(det\mathbf{F})dV_0, but what about area?

If dS_0 is the initial area and dS is the final area:
For small volumes, we can say that dV_0=dS_0 \mathbf{dX} and dV=dS \mathbf{dx}.

It was previously shown that dV=det\mathbf{F}dV_0
\therefore dV_0=dS_0 \mathbf{dX} and dV=det\mathbf{F}dV_0=dS \mathbf{dx} \longrightarrow dV_0=\frac{dS \mathbf{dx}}{det\mathbf{F}}

Substituting, we get dS \mathbf{dx}=dS_0 \mathbf{dX}det\mathbf{F}
We also know that \mathbf{dx=F \cdot dX}

\longrightarrow \underbrace{dS \mathbf{F}}_{\mathbf{F}^{T}dS} \cdot \mathbf{dX}=dS_0 \mathbf{dX}det\mathbf{F} \longrightarrow (det\mathbf{F}dS_0-\mathbf{F}^{T}dS) \cdot \mathbf{dX}=0

(2)   \begin{equation*} \longrightarrow dS=dS\mathbf{n}=det\mathbf{F} * \mathbf{F}^{-T}dS_0=det\mathbf{F} * \mathbf{F}^{-T} \cdot \mathbf{N}dS_0 \end{equation*}

where \mathbf{N} and \mathbf{n} are the normal vectors to the surface in the respective initial and final configurations.

Eq. 2 is called Nanson’s Equation and will be useful later when we look at “true” stress versus “nominal” stress, for example.