5. Superimposed Rotation

In preparation for the constitutive formulations in later chapters, this short chapter will derive how different tensors behave under superimposed rigid body motion. This will enable us, for example, to show that \boldsymbol{\sigma} is a function of \mathbf{B} in the beginning of the next chapter. The idea of “work conjugate” pairs will be introduced in this chapter. We will see that the Cauchy stress tensor, \boldsymbol{\sigma}, for example, changes with rigid body rotation, while \mathbf{E} does not. Thus, \boldsymbol{\sigma} and \mathbf{E} are not work-conjugate, and therefore it would not be appropriate to use them as a pair when forming a constitutive relationship. This was first recognized by Hill [Hill].

Let’s define the vector \mathbf{dx^*} as follows:

\mathbf{dx^* = Q \cdot dx}, where \mathbf{Q} is a superimposed rigid body motion, sometimes called a Euclidean transformation (see Fig).


We know that \mathbf{dx=F \cdot dX}

What about “\mathbf{F^*}”?

Consider the following square body (see Fig), paying careful attention to the heavily bolded edge in order to track the motion of the body.


Clearly, if \mathbf{dX=dX^*} (which is what we’re going to want), and, say, \mathbf{F^*=F},
then \mathbf{dx^* \ne F^* \cdot dX^*} (this should be clear from the above Fig).

Since we want to find \mathbf{F^*} such that \mathbf{dx^* = F^* \cdot dX^*}, we can conclude that \mathbf{F^* \ne F}.

What we do know is the following:

\mathbf{dX^*=dX} ; \mathbf{dx^*=Q \cdot dx} ; \mathbf{dx^*=F^* \cdot dX^*}

So, \mathbf{Q \cdot dx = F^* \cdot dX}, which gives:

(1)   \begin{equation*} \mathbf{F^*=Q \cdot F} \end{equation*}

Eq. 1 is what we wanted to find, and it should be expected. Recall that \mathbf{R}\cdot\mathbf{U} is physically understood to be a deformation (axial strains and shear), \mathbf{U}, followed by a rotation, \mathbf{R}. Thus, \mathbf{Q \cdot F} is the total deformation + rotation, \mathbf{F}, followed by an explicit rigid body rotation, \mathbf{Q}, – i.e. a rotation that is superimposed on \mathbf{dx}.

Now, we can see how different measures of strain and stress behave under superimposed rigid body motion, by substituting the above expression for \mathbf{F^*}.

(2)   \begin{equation*} \mathbf{C=F}^{T} \cdot \mathbf{F} \longrightarrow \mathbf{C^*=F}^{*T} \cdot \mathbf{F^*}=\mathbf{F}^{T} \cdot \underbrace{\mathbf{Q}^{T} \cdot \mathbf{Q}}_{\mathbf{I}} \cdot \mathbf{F}=\mathbf{F}^{T} \cdot \mathbf{F=C} \end{equation*}

Thus, \mathbf{C} is invariant to rigid body motion.

note: \mathbf{A}^{-1} \cdot \mathbf{A} = \mathbf{I} always ; \mathbf{Q}^{T} \cdot \mathbf{Q} = \mathbf{I} since \mathbf{Q} is an orthogonal tensor

note: \mathbf{F^{*T}}=(\mathbf{Q \cdot F})^{T}=\mathbf{F}^{T} \cdot \mathbf{Q}^{T}
informal proof: \mathbf{B}=B_{ik}\mathbf{e_ie_k} ; \mathbf{C}=C_{nj}\mathbf{e_ne_j}
\mathbf{A=B \cdot C} = B_{ik}\mathbf{e_ie_k} \cdot C_{nj}\mathbf{e_ne_j}=B_{ik}C_{kj}\mathbf{e_ie_j}=A_{ij}\mathbf{e_ie_j}

\mathbf{A}^{T} NOT = \mathbf{B}^{T} \cdot \mathbf{C}^{T}=B_{ki}\mathbf{e_ie_k} \cdot C_{jn}\mathbf{e_ne_j}=B_{ki}C_{jk}\mathbf{e_ie_j} which does not match either of the above expressions for \mathbf{A}^{T}
(or \mathbf{B}^{T} \cdot \mathbf{C}^{T}=B_{ik}\mathbf{e_ke_i} \cdot C_{nj}\mathbf{e_je_n}=B_{ik}C_{ni}\mathbf{e_ke_n}=B_{ki}C_{jk}\mathbf{e_ie_j} which is similarly \ne \mathbf{A}^{T})

\mathbf{A}^{T}=\mathbf{C}^{T} \cdot \mathbf{B}^{T}=C_{jn}\mathbf{e_ne_j} \cdot B_{ki}\mathbf{e_ie_k}=C_{in}B_{ki}\mathbf{e_ne_k}=C_{kj}B_{ik}\mathbf{e_je_i} \longrightarrow matches 1^{st} expression for \mathbf{A}^{T}
\mathbf{A}^{T}=\mathbf{C}^{T} \cdot \mathbf{B}^{T}=C_{nj}\mathbf{e_je_n} \cdot B_{ik}\mathbf{e_ke_i}=C_{kj}B_{ik}\mathbf{e_je_i}=C_{ki}B_{jk}\mathbf{e_ie_j} \longrightarrow matches 2^{nd} expression for \mathbf{A}^{T}

formal proof: see Appendix A.1

(3)   \begin{equation*} \mathbf{B=F \cdot F}^{T} \longrightarrow \mathbf{B^*=F^* \cdot F}^{*T}=\mathbf{Q} \cdot \underbrace{\mathbf{F \cdot F}^{T}}_{\mathbf{B}} \cdot \mathbf{Q}^{T}=\mathbf{Q \cdot B \cdot Q}^{T} \end{equation*}

What about \mathbf{U^*}, \mathbf{V^*}, and \mathbf{R^*}?

\mathbf{F=V \cdot R = R \cdot U} \longrightarrow \mathbf{F^*=V^* \cdot R^*=R^* \cdot U^*}

Recall, \mathbf{F^*=Q \cdot F = Q\cdot V \cdot R}

Insert \mathbf{Q}^{T} \cdot \mathbf{Q} = \mathbf{I} into the above expression \longrightarrow \mathbf{F^*}=\underbrace{\mathbf{Q \cdot V} \cdot \mathbf{Q}^{T}}_{\mathbf{V^*}}\cdot \underbrace{\mathbf{Q} \cdot \mathbf{R}}_{\mathbf{R^*}}

So, \mathbf{V^*=Q \cdot V \cdot Q}^{T} ; \mathbf{R^*=Q \cdot R} (\mathbf{V^*} is symmetric ; \mathbf{R^*} is orthogonal)

Also, \mathbf{U^*=U} (skipped work)

How about rates?

\mathbf{F}^*=\mathbf{Q \cdot F} \quad ; \quad \dot{\mathbf{F}}^*=\dot{\mathbf{Q}} \cdot \mathbf{F}+\mathbf{Q} \cdot \dot{\mathbf{F}}

Recall also, \mathbf{L}=\dot{\mathbf{F}} \cdot \mathbf{F}^{-1} \quad ; \quad \mathbf{L}^*=\dot{\mathbf{F}}^* \cdot \mathbf{F}^{*-1} \quad ; \quad \mathbf{F}^{*-1}=\mathbf{F}^{-1} \cdot \mathbf{Q}^{-1}

So, \mathbf{L}^*=(\dot{\mathbf{Q}} \cdot \mathbf{F}+\mathbf{Q} \cdot \dot{\mathbf{F}}) \cdot \mathbf{F}^{-1} \cdot \mathbf{Q}^{-1}=\dot{\mathbf{Q}} \cdot \mathbf{Q}^{-1}+\mathbf{Q} \cdot (\underbrace{\dot{\mathbf{F}} \cdot \mathbf{F}^{-1}}_{\mathbf{L}}) \cdot \mathbf{Q}^{T}

note: (\mathbf{Q}^{-1}=\mathbf{Q}^{T} since \mathbf{Q} is orthogonal)

(4)   \begin{equation*} \mathbf{L}^*=\dot{\mathbf{Q}} \cdot \mathbf{Q}^{-1}+\mathbf{Q} \cdot (\mathbf{D}+\mathbf{W}) \cdot \mathbf{Q}^{T}=\underbrace{\mathbf{Q \cdot D \cdot Q}^{T}}_{\mathbf{D}^*}+\underbrace{\dot{\mathbf{Q}} \cdot \mathbf{Q}^{-1}+\mathbf{Q \cdot W \cdot Q}^{T}}_{\mathbf{W}^*} \end{equation*}

note: \mathbf{W}^* is antisymmetric, and indeed, the two terms that make up \mathbf{W}^* are each antisymmetric

How about stress?

Since \mathbf{df_n^*=Q \cdot df_n}, we can say that \mathbf{t_n^*=Q \cdot t_n} (1)

We also know the following three expressions to be true:
\mathbf{t_n}=\boldsymbol{\sigma} \cdot \mathbf{n} (2)
\mathbf{t_n^*}=\boldsymbol{\sigma^*} \cdot \mathbf{n^*} (3)
\mathbf{n^*=Q} \cdot \mathbf{n} (4)

(4) \longrightarrow (1) \longrightarrow (3) \longrightarrow \mathbf{Q \cdot t_n}=\boldsymbol{\sigma^*} \cdot \mathbf{Q} \cdot \mathbf{n} (5)

(2) \longrightarrow (5) \longrightarrow (\mathbf{Q} \cdot \boldsymbol{\sigma})\cdot \mathbf{n}=(\boldsymbol{\sigma^*} \cdot \mathbf{Q}) \cdot \mathbf{n}

(5)   \begin{equation*} \mathbf{Q} \cdot \boldsymbol{\sigma}=\boldsymbol{\sigma^*} \cdot \mathbf{Q} \longrightarrow \boldsymbol{\sigma^*}=\mathbf{Q} \cdot \boldsymbol{\sigma} \cdot \mathbf{Q}^{T} \end{equation*}

note: Since \boldsymbol{\sigma^*}=\mathbf{Q} \cdot \boldsymbol{\sigma} \cdot \mathbf{Q}^{T} and \mathbf{B^*}=\mathbf{Q \cdot B \cdot Q}^{T}, we can say that \boldsymbol{\sigma} and \mathbf{B} are a “work-conjugate pair.” A more rigorous proof, starting with the “balance of mechanical energy” can be found in [Holzapfel].

How about the nominal stress and the Second Piola-Kirchhoff Stress?

\boldsymbol{\sigma}^0=(det\mathbf{F}) \cdot \mathbf{F}^{-1} \cdot \boldsymbol{\sigma}

Again, we will need to use \mathbf{F}^*=\mathbf{Q \cdot F}

Since \boldsymbol{\sigma}^0 depends on det\mathbf{F}, we will need det\mathbf{F^*}. It turns out, though, that det\mathbf{F^*}=(\underbrace{det\mathbf{Q}}_{\text{``1''}})(det\mathbf{F})=det\mathbf{F}

note: the determinant of an orthogonal tensor is “1”

So, \boldsymbol{\sigma^{0*}}=(det\mathbf{F^*})\mathbf{F}^{*-1} \cdot \boldsymbol{\sigma^*}=det\mathbf{F} \cdot (\mathbf{Q \cdot F})^{-1} \cdot \mathbf{Q} \cdot \boldsymbol{\sigma} \cdot \mathbf{Q}^{T}
=(det\mathbf{F})\mathbf{F}^{-1} \cdot \mathbf{Q}^{-1} \cdot \mathbf{Q} \cdot \boldsymbol{\sigma} \cdot \mathbf{Q}^{T} =\underbrace{(det\mathbf{F})\mathbf{F}^{-1} \cdot \boldsymbol{\sigma}}_{\boldsymbol{\sigma}^0} \cdot \mathbf{Q}^{T}=\boldsymbol{\sigma}^0 \cdot \mathbf{Q}^{T}

How about \hat{\boldsymbol{\sigma}}?

\boldsymbol{\hat{\sigma}}=\boldsymbol{\sigma}^0 \cdot \mathbf{F}^{-T}
\boldsymbol{\hat{\sigma}}^* = \boldsymbol{\sigma}^{0*} \cdot (\mathbf{F^*})^{-T}=\boldsymbol{\sigma}^0 \cdot \mathbf{Q}^{T} \cdot (\mathbf{Q \cdot F})^{-T}=\boldsymbol{\sigma}^0 \cdot \mathbf{Q}^{T} \cdot \mathbf{Q} \cdot \mathbf{F}^{-T}=\boldsymbol{\sigma}^0 \cdot \mathbf{F}^{-T}=\boldsymbol{\hat{\sigma}}

(6)   \begin{equation*} \boldsymbol{\hat{\sigma}}^* = \boldsymbol{\hat{\sigma}} \end{equation*}

\boldsymbol{\hat{\sigma}} is a good measure of stress when there are small (infinitesimal) strains. It will also be used as a “work-conjugate pair” with \mathbf{E} in the following chapters on constitutive relationships.