Rate of Deformation and Spin Tensors

The most obvious first step is to take the time derivative of the deformation gradient, \mathbf{F}, as follows:


For reasons that won’t become clear until the chapter on rate-form constitutive expressions, we would like the “velocity gradient” to be expressed with respect to \mathbf{x} rather than \mathbf{X}. Since \mathbf{x} is a function of \mathbf{X}, we can use the chain rule to arrive at the following important expression:

(1)   \begin{equation*} \dot{\mathbf{F}}=\frac{\partial}{\partial\mathbf{X}}\mathbf{v}=\underbrace{\frac{\partial\mathbf{v}}{\partial\mathbf{x}}}_{\mathbf{L}}\underbrace{\frac{\partial\mathbf{x}}{\partial\mathbf{X}}}_{\mathbf{F}} \end{equation*}

We can see from eq. 1 that \mathbf{\dot{F}}=\mathbf{L \cdot F}
\mathbf{L}=\mathbf{\dot{F}} \cdot \mathbf{F}^{-1}, where \mathbf{L} is the “velocity gradient.”

For our purposes, consider that if we were to have analytical functions of time, t, for each term in \mathbf{F} that describes how \mathbf{R} and \mathbf{U} are changing with time, for example, then we could find \mathbf{L} by noting that \dot{\mathbf{F}}=\frac{d}{dt}\left(\mathbf{R\cdot U}\right)=\dot{\mathbf{R}}\cdot\mathbf{U}+\mathbf{U}\cdot\dot{\mathbf{R}} and \mathbf{L}=\mathbf{\dot{F}} \cdot \mathbf{F}^{-1}. This is quite academic (and potentially quite difficult) compared to the way that \mathbf{L} would be found in FEA, so such an example will not be given. However, \mathbf{L} is a very important quantity as we will see in the chapter on rate-form constitutive relationships.

The velocity gradient, \mathbf{L}, is a very important quantity and can be decomposed as follows:

(2)   \begin{equation*} \mathbf{L}=\underbrace{\frac{1}{2}(\mathbf{L+L}^{T})}_{\mathbf{D}}+\underbrace{\frac{1}{2}(\mathbf{L-L}^{T})}_{\mathbf{W}} \end{equation*}

note: Similar to eq. 2, in linear infinitesimal elasticity, we can split the displacement gradient into a strain tensor and rotation tensor that are symmetric and anti-symmetric, respectively.

note: Any tensor can be similarly split into its symmetric and anti-symmetric parts.

“Rate of Deformation Tensor” = symmetric \longrightarrow \mathbf{D}^{T}=\mathbf{D}
“Spin Tensor” = anti-symmetric or “skew” \longrightarrow \mathbf{W}^{T}=-\mathbf{W}

\mathbf{W} is a pure rigid body rotation. A complete proof of this can be found in [Asaro]. It can also be easily shown that \mathbf{W} is, in fact, skew. This proof can be found in Appendix A.2.

Consider the time rate of change of length of a particular element:
\frac{d}{dt}(ds)^2=\frac{d}{dt}(\mathbf{dx \cdot dx})=\frac{d}{dt}(\mathbf{dX \cdot F}^{T} \cdot \mathbf{F \cdot dX})=\mathbf{dX \cdot}\frac{d}{dt}(\mathbf{F}^{T} \cdot \mathbf{F})\cdot \mathbf{dX}
=\mathbf{dX \cdot \dot{C} \cdot dX}

\frac{d\mathbf{C}}{dt}=\mathbf{\dot{C}}=(\dot{\mathbf{F}^{T} \cdot \mathbf{F}})=\dot{\mathbf{F}}^{T} \cdot \mathbf{F} + \mathbf{F}^{T} \cdot \dot{\mathbf{F}}=\mathbf{F}^{T} \cdot (\dot{\mathbf{F}} \cdot \mathbf{F}^{-1}+\mathbf{F}^{-T} \cdot \dot{\mathbf{F}}^{T}) \cdot \mathbf{F}
=\mathbf{F}^{T} \cdot (\mathbf{L+L}^{T})\cdot \mathbf{F}=2 \cdot \mathbf{F}^{T} \cdot \mathbf{D \cdot F}

So, \frac{d}{dt}(ds)^2=2\mathbf{dX \cdot F}^{T} \cdot \mathbf{D \cdot F \cdot dX}=2\mathbf{dx \cdot D \cdot dx}

\mathbf{dx}=ds\mathbf{n} \longrightarrow \frac{d}{dt}(ds)^2=\cancel{2(ds)}\frac{d}{dt}(ds)=\cancel{2}(ds)^{\cancel{2}}\mathbf{n \cdot D \cdot n}

(3)   \begin{equation*} \longrightarrow \frac{d}{dt}(ds)=D_{nn}ds \end{equation*}

While [Holzapfel] shows that “\mathbf{D} is not pure rate of strain and \mathbf{W} is not a pure rate of rotation,” we can see that the rate of change of length only depends on \mathbf{D}, not \mathbf{W}. Additionally, if \mathbf{D}=0, then we must have rigid body motions only.

note: We could have split up \mathbf{F} into a sum, but we wouldn’t have arrived at anything useful. The way that eq. 3 was derived only works due to the product rule of derivatives. Proving that \mathbf{W} represents “spin” is more difficult and won’t be shown here, although the derivation can be found in many other texts (e.x. [Asaro][Bonet][Holzapfel]).

Similar to the shear strain, E_{mn}, we can again consider how the angle between two vectors, \mathbf{m}, and \mathbf{n}, changes under deformation (see Fig below).


(4)   \begin{equation*} \frac{d\phi}{dt}=2D_{mn}\text{ (skipped work)} \end{equation*}

note: \mathbf{\dot{E}}=\mathbf{F}^{T} \cdot \mathbf{D \cdot F} (Recall also: \mathbf{E}=\mathbf{F}^{T} \cdot \mathbf{e \cdot F})
\mathbf{\dot{E}} reduces to \mathbf{D} for infinitesimal deformation (ignoring rigid body rotations), since \mathbf{F \longrightarrow I}.
In general, though, \mathbf{\dot{E}} \ne \mathbf{\dot{e}} \ne \mathbf{D}. Similarly, in general, \mathbf{\dot{R}} \ne \mathbf{W}. We’ll also look in detail at the physical meaning (or lack thereof) of \boldsymbol{\dot{\sigma}} later on when we get to the chapter on “Rate-Form Constitutive Expressions”