Nonlinear finite elements/Kinematics - time derivatives and rates

Time derivatives and rate quantities

Material time derivatives

Material time derivatives are needed for many updated Lagrangian formulations of finite element analysis.

Recall that the motion can be expressed as

$𝐱=𝝋(𝐗,t)\text{or}𝐗={𝝋}^{-1}(𝐱,t)$

If we keep $𝐗$ fixed, then the velocity is given by

$𝐕(𝐗,t)=\frac{\partial 𝝋}{\partial t}(𝐗,t)$

This is the material time derivative expressed in terms of $𝐗$.

The spatial version of the velocity is

$𝐯(𝐱,t)=𝐕({𝝋}^{-1}(𝐱,t),t)$

We will use the symbol $𝐯$ for velocity from now on by slightly abusing the notation.

We usually think of quantities such as velocity and acceleration as spatial quantities which are functions of $𝐱$ (rather than material quantities which are functions of $𝐗$).

Given the spatial velocity $𝐯(𝐱,t)$, if we want to find the acceleration we will have to consider the fact that $𝐱\equiv 𝐱(𝐗,t)$, i.e., the position also changes with time. We do this by using the chain rule. Thus

$\frac{D𝐯(𝐱,t)}{Dt}=𝐚(𝐱,t)=\frac{\partial 𝐯(𝐱,t)}{\partial t}+\frac{\partial 𝐯(𝐱,t)}{\partial 𝐱}\cdot \frac{\partial 𝝋(𝐗,t)}{\partial t}=\frac{\partial 𝐯}{\partial t}+𝜵𝐯\cdot 𝐕=\frac{\partial 𝐯}{\partial t}+𝜵𝐯\cdot 𝐯$

Such a derivative is called the material time derivative expressed in terms of $𝐱$. The second term in the expression is called the convective derivative..

Velocity gradient

Let the velocity be expressed in spatial form, i.e., $𝐯(𝐱,t)$. The spatial velocity gradient tensor is given by

$𝒍:=\frac{\partial 𝐯(𝐱,t)}{\partial 𝐱}=𝜵𝐯$

The velocity gradient $𝒍$ is a second order tensor which can expressed as

$𝒍=l_{ij}{𝐞}_i\otimes {𝐞}_j=\frac{\partial v_i}{\partial x_j}{𝐞}_i\otimes {𝐞}_j$

The velocity gradient is a measure of the relative velocity of two points in the current configuration.

Time derivative of the deformation gradient

Recall that the deformation gradient is given by

$𝑭=\frac{\partial 𝝋}{\partial 𝐗}$

The time derivative of $𝑭$ (keeping $𝐗$ fixed) is

$\dot{𝑭}=\frac{\partial }{\partial t}\left(\frac{\partial 𝝋}{\partial 𝐗}\right)=\frac{\partial }{\partial 𝐗}\left(\frac{\partial 𝝋}{\partial t}\right)=\frac{\partial 𝐯}{\partial 𝐗}={𝜵}_{\circ }𝐯$

Using the chain rule

$\dot{𝑭}=\frac{\partial 𝐯}{\partial 𝐱}\cdot \frac{\partial 𝐱}{\partial 𝐗}=\frac{\partial 𝐯}{\partial 𝐱}\cdot \frac{\partial 𝝋}{\partial 𝐗}=𝒍\cdot 𝑭$

Form this we get the important relation

$𝒍=\dot{𝑭}\cdot {𝑭}^{-1}.$

Time derivative of strain

Let $d{𝐗}_1$ and $d{𝐗}_2$ be two infinitesimal material line segments in a body. Then

$d{𝐱}_1=𝑭\cdot d{𝐗}_1;d{𝐱}_2=𝑭\cdot d{𝐗}_2$

Hence,

$d{𝐱}_1\cdot d{𝐱}_2=(𝑭\cdot d{𝐗}_1)\cdot (𝑭\cdot d{𝐗}_2)=d{𝐗}_1\cdot ({𝑭}^T\cdot 𝑭)\cdot d{𝐗}_2=d{𝐗}_1\cdot 𝑪\cdot d{𝐗}_2=d{𝐗}_1\cdot (2𝑬+1)\cdot d{𝐗}_2$

Taking the derivative with respect to $t$ gives us

$\frac{\partial }{\partial t}(d{𝐱}_1\cdot d{𝐱}_2)=d{𝐗}_1\cdot \frac{\partial 𝑪}{\partial t}\cdot d{𝐗}_2=2d{𝐗}_1\cdot \frac{\partial 𝑬}{\partial t}\cdot d{𝐗}_2$

The material strain rate tensor is defined as

$\dot{𝑬}=\frac{\partial 𝑬}{\partial t}=\frac{1}{2}\frac{\partial 𝑪}{\partial t}=\frac{1}{2}\dot{𝑪}$

Clearly,

$\dot{𝑬}=\frac{1}{2}\frac{\partial }{\partial t}({𝑭}^T\cdot 𝑭)=\frac{1}{2}({\dot{𝑭}}^T\cdot 𝑭+{𝑭}^T\cdot \dot{𝑭}).$

Also,

$\frac{1}{2}\frac{\partial }{\partial t}(d{𝐱}_1\cdot d{𝐱}_2)=d{𝐗}_1\cdot \dot{𝑬}\cdot d{𝐗}_2=({𝑭}^{-1}\cdot d{𝐱}_1)\cdot \dot{𝑬}\cdot ({𝑭}^{-1}\cdot d{𝐱}_2)=d{𝐱}_1\cdot ({𝑭}^{-T}\cdot \dot{𝑬}\cdot {𝑭}^{-1})\cdot d{𝐱}_2$

The spatial rate of deformation tensor or stretching tensor is defined as

$𝒅={𝑭}^{-T}\cdot \dot{𝑬}\cdot {𝑭}^{-1}=\frac{1}{2}{𝑭}^{-T}\cdot \dot{𝑪}\cdot {𝑭}^{-1}$

In fact, we can show that $𝒅$ is the symmetric part of the velocity gradient, i.e.,

$𝒅=\frac{1}{2}(𝒍+{𝒍}^T)$

For rigid body motions we get $𝒅=0$.

Lie derivatives

Most of the operations above can be interpreted as push-forward and pull-back operations. Also, time derivatives of these tensors can be interpreted as Lie derivatives.

Recall that the push-forward of the strain tensor from the material configuration to the spatial configuration is given by

$𝒆={\varphi }_{\ast }[𝑬]={𝑭}^{-T}\cdot 𝑬\cdot {𝑭}^{-1}$

The pull-back of the spatial strain tensor to the material configuration is given by

$𝑬={\varphi }^{\ast }[𝒆]={𝑭}^T\cdot 𝒆\cdot 𝑭$

Therefore, the rate of deformation tensor is a push-forward of the material strain rate tensor, i.e.,

$𝒅={𝑭}^{-T}\cdot \dot{𝑬}\cdot {𝑭}^{-1}={\varphi }_{\ast }[\dot{𝑬}]$

Similarly, the material strain rate tensor is a pull-back of the rate of deformation tensor to the material configuration, i.e.,

$\dot{𝑬}={𝑭}^T\cdot 𝒅\cdot 𝑭={\varphi }^{\ast }[𝒅]$

Now,

$𝑬={\varphi }^{\ast }[𝒆]⟹\dot{𝑬}=\frac{\partial }{\partial t}({\varphi }^{\ast }[𝒆])$

Also,

$𝒅={\varphi }_{\ast }[\dot{𝑬}]={\varphi }_{\ast }\left[\frac{\partial }{\partial t}({\varphi }^{\ast }[𝒆])\right]$

Therefore the rate of deformation tensor can be obtained by first pulling back $𝒆$ to the reference configuration, taking a material time derivative in that configuration, and then pushing forward the result to the current configuration.

Such an operation is called a Lie derivative. In general, the Lie derivative of a spatial tensor $𝐠$ is defined as

${ℒ}_{\varphi }[𝒈]:={\varphi }_{\ast }\left[\frac{\partial }{\partial t}({\varphi }^{\ast }[𝒈])\right].$

Spin tensor

The velocity gradient tensor can be additively decomposed into a symmetric part and a skew part:

$𝒍=\frac{1}{2}(𝒍+{𝒍}^T)+\frac{1}{2}(𝒍-{𝒍}^T)=𝒅+𝒘$

We have seen that $𝒅$ is the rate of deformation tensor. The quantity $𝒘$ is called the spin tensor.

Note that $𝒅$ is symmetric while $𝒘$ is skew symmetric, i.e.,

$𝒅={𝒅}^T;𝒘=-{𝒘}^T.$

So see why $𝒘$ is called a "spin", recall that

$𝒍=\dot{𝑭}\cdot {𝑭}^{-1}$

Therefore,

$𝒘=\frac{1}{2}(\dot{𝑭}\cdot {𝑭}^{-1}-{𝑭}^{-T}\cdot {\dot{𝑭}}^T)$

Also,

$𝑭=𝑹\cdot 𝑼⟹\dot{𝑭}=\dot{𝑹}\cdot 𝑼+𝑹\cdot \dot{𝑼}$

Therefore,

$\dot{𝑭}\cdot {𝑭}^{-1}=(\dot{𝑹}\cdot 𝑼+𝑹\cdot \dot{𝑼})\cdot ({𝑼}^{-1}\cdot {𝑹}^T)=\dot{𝑹}\cdot {𝑹}^T+𝑹\cdot \dot{𝑼}\cdot {𝑼}^{-1}\cdot {𝑹}^T$

and

${𝑭}^{-T}\cdot {\dot{𝑭}}^T=(𝑹\cdot {𝑼}^{-1})\cdot (𝑼\cdot {\dot{𝑹}}^T+\dot{𝑼}\cdot {𝑹}^T)=𝑹\cdot {\dot{𝑹}}^T+𝑹\cdot {𝑼}^{-1}\cdot \dot{𝑼}\cdot {𝑹}^T$

So we have

$𝒘=\frac{1}{2}(\dot{𝑹}\cdot {𝑹}^T+𝑹\cdot \dot{𝑼}\cdot {𝑼}^{-1}\cdot {𝑹}^T-𝑹\cdot {\dot{𝑹}}^T-𝑹\cdot {𝑼}^{-1}\cdot \dot{𝑼}\cdot {𝑹}^T)$

Now

$𝑹\cdot {𝑹}^T=1⟹\dot{𝑹}\cdot {𝑹}^T+𝑹\cdot {\dot{𝑹}}^T=0$

Therefore

$𝒘=\dot{𝑹}\cdot {𝑹}^T+\frac{1}{2}𝑹\cdot (\dot{𝑼}\cdot {𝑼}^{-1}-{𝑼}^{-1}\cdot \dot{𝑼})\cdot {𝑹}^T$

The second term above is invariant for rigid body motions and zero for an uniaxial stretch. Hence, we are left with just a rotation term. This is why the quantity $𝒘$ is called a spin.

The spin tensor is a skew-symmetric tensor and has an associated axial vector $𝝎$ (also called the angular velocity vector) whose components are given by

$𝝎=\left[\begin{array}{c}{w}_1\\ w_2\\ w_3\end{array}\right]$

where

$𝐰=\left[\begin{array}{ccc}0& -w_3& w_2\\ w_3& 0& -w_1\\ -w_2& w_1& 0\end{array}\right]$

The spin tensor and its associated axial vector appear in a number of modern numerical algorithms.

Rate of change of volume

Recall that

$dv=JdV\text{where}J=det𝑭$

Therefore, taking the material time derivative of $dv$ (keeping $𝐗$ fixed), we have

$\frac{d}{dt}(dv)=\dot{J}dV=\frac{\dot{J}}{J}dv$

At this stage we invoke the following result from tensor calculus:

If $𝑨$ is an invertible tensor which depends on $t$ then

$\frac{d}{dt}(det𝑨)=(det𝑨)\text{tr}(\frac{d𝑨}{dt}\cdot {𝑨}^{-1})$

In the case where $𝑨=𝑭,J=det𝑭$ we have

$\frac{d}{dt}(J)=J\text{tr}(\dot{𝑭}\cdot {𝑭}^{-1})$

or,

$\dot{J}=J\text{tr}(𝒍)=J\text{tr}(𝐝)$

Therefore,

$\frac{d}{dt}(dv)=\text{tr}(𝐝)dv$

Alternatively, we can also write

$\dot{J}=\frac{1}{2}J{𝑪}^{-1}:\dot{𝑪}$

These relations are of immense use in numerical algorithms - particularly those which involved incompressible behavior, i.e., when $\dot{J}=0$.

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