Now consideration is given to changes of the geometrical deforming tensors \(\mathbf {\tilde {F},\tilde {U}}\) and \(\mathbf {\tilde {V}}\) when the body is in its final deformed state and then subjected to a pure rigid body rotation \(\mathbf {\tilde {Q}}\), and to what happens to the various stress tensors derived above under the same \(\mathbf {\tilde {Q}}\).
Polar decomposition of \(\mathbf {\tilde {F}}\) is given by \[ \fbox {$\mathbf {\tilde {F}=\tilde {R}\cdot \tilde {U}}$}\] where \(\mathbf {\tilde {F}}\) is the deformation gradient tensor and \(\mathbf {\tilde {U}}\) is the stretch before rotation \(\mathbf {\tilde {R}}\) tensor, and \(\mathbf {\tilde {R}}\) is the rotation tensor. The polar decomposition of \(\mathbf {\tilde {F}}\) is \[ \boxed { \mathbf {\tilde {F}=\tilde {V}\cdot \tilde {R}} } \] where \(\mathbf {\tilde {V}}\) is the stretch after rotation \(\mathbf {\tilde {R}}\) tensor.
The effect of applying pure rigid body rotation \(\mathbf {\tilde {Q}}\) on \(\mathbf {\tilde {F},\tilde {U}}\) and \(\mathbf {\tilde {V} }\) is now determined.
In each of the following derivations the following setting is assumed to be in place: There is a body originally in the undeformed state \(B\) and loads are applied on the body. The body undergoes deformation governed by the deformation gradient tensor \(\mathbf {\tilde {F}}\) resulting in the body being in the final deformed state state \(b\) with a stress tensor \(\mathbf {\tilde {\tau }}\) at point \(p\). If the body is considered to be first under the effect of \(\mathbf {\tilde {U}}\) (stretch), then the new state will be called \(B^{\ast }\), and after applying the effect of \(\mathbf {\tilde {R}}\) (point to point rotation tensor), then the state will be called \(b\) (which is the final deformation state).
If however \(\mathbf {\tilde {R}}\) (rotation) is applied first, then the new state will also be called \(B^{\ast }\) and then when applying the stretch \(\mathbf {\tilde {V}}\) the state will becomes \(b\) (which is the final deformation state).
From state \(b\), which is the final deformation state, a pure rigid body rotation tensor \(\mathbf {\tilde {Q}}\) is applied to the whole body (with its fixed supports if any). Hence there will be no changes in the body shape, and the new state is called \(q\).
Is also possible to consider the change of state from state \(B\) to state \(q\) to be the result of a new deformation gradient tensor which is called \(\mathbf {\tilde {F}}_{q}\). The polar decomposition of \(\mathbf {\tilde {F}}_{q}\) can also be written as \[ \boxed {\mathbf {\tilde {F}}_{q}\mathbf {=\tilde {R}}_{q}\mathbf {\cdot \tilde {U}}_{q} } \] or as \[ \boxed { \mathbf {\tilde {F}}_{q}\mathbf {=\tilde {V}}_{q}\mathbf {\cdot \tilde {R}}_{q} } \]
\(\mathbf {\tilde {F}}\) is compared to \(\mathbf {\tilde {F}}_{q}\), and \(\mathbf {\tilde {U}}\) is compared to \(\mathbf {\tilde {U}}_{q}\) and \(\mathbf {\tilde {V}}\) is compared to \(\mathbf {\tilde {V}}_{q}\) in order to see the effect of the rigid body rotation on these tensors.
The above diagram show that \[ \boxed { \mathbf {\tilde {F}}_{q}=\mathbf {\tilde {Q}\cdot \tilde {F}} } \]
Given that
Similarly, \[ \mathbf {\tilde {U}}_{q}\mathbf {=}\left ( \mathbf {\tilde {F}}_{q}^{T}\mathbf {\cdot \tilde {F}}_{q}\right ) ^{\frac {1}{2}}\]
Where \(\mathbf {\tilde {F}}_{q}\) \(=\mathbf {\tilde {Q}\cdot \tilde {F}}\), hence the above becomes
Using linear algebra it follows that \(\left ( \mathbf {\tilde {Q}\cdot \tilde {F}}\right ) ^{T}=\mathbf {\tilde {F}}^{T}\cdot \mathbf {\tilde {Q}}^{T}\). Therefore the above becomes
But \(\mathbf {\tilde {Q}}^{T}\mathbf {\cdot \tilde {Q}=\tilde {I}}\) since \(\mathbf {\tilde {Q}}\) is orthogonal. Hence the above becomes
Comparing (1) and (2) shows they are the same. Hence\[ \boxed { \mathbf {\tilde {U}}_{q}\mathbf {=\tilde {U}} } \] Therefore
.
Since \begin{equation} \mathbf {\tilde {F}}_{q}=\mathbf {\tilde {Q}\cdot \tilde {F}} \tag {1}\end{equation}
And \(\mathbf {\tilde {F}=\tilde {R}}\cdot \mathbf {\tilde {U}}\) by polar decomposition on \(\mathbf {\tilde {F}}\) the above can be written as
Applying polar decomposition on \(\mathbf {\tilde {F}}_{q}\) results in \(\mathbf {\tilde {F}}_{q}=\mathbf {\tilde {R}}_{q}\cdot \mathbf {\tilde {U}}_{q}\) and the above becomes
It was found earlier that \(\mathbf {\tilde {U}}_{q}=\mathbf {\tilde {U}}\), therefore the above becomes
Now the second form of polar decomposition on \(\mathbf {\tilde {F}}_{q}\) is utilized giving
Substituting (2) into the above equation results in
Substituting (1) into the above gives
Since \(\mathbf {\tilde {Q}\cdot \tilde {R}}\) is invertible (need to check), the above can be written as
But \(\left ( \mathbf {\tilde {Q}\cdot \tilde {R}}\right ) ^{-1}=\mathbf {\tilde {R}}^{T}\mathbf {\cdot \tilde {Q}}^{T}\) (Since \(\mathbf {\tilde {Q}\cdot \tilde {R}}\) is an orthogonal matrix. (check). Hence the above becomes
From polar decomposition it is known that \(\mathbf {\tilde {F}=\tilde {V}\cdot \tilde {R}}\), hence \(\mathbf {\tilde {V}=\tilde {F}\cdot \tilde {R}}^{-1}\), but \(\mathbf {\tilde {R}}^{-1}=\mathbf {\tilde {R}}^{T}\) since it is an orthogonal matrix, therefore \begin{equation} \mathbf {\tilde {V}=\tilde {F}\cdot \tilde {R}}^{T} \tag {4}\end{equation}
Substituting (4) in (3) gives
This above is how \(\mathbf {\tilde {V}}\) transforms due to rigid rotation \(\mathbf {\tilde {Q}}\).
Now that the transformation of \(\mathbf {\tilde {F},\tilde {U}}\) and \(\mathbf {\tilde {V}}\) was obtained, the next step is to find how each one of the stress tensors derived earlier transforms due to \(\mathbf {\tilde {Q}}\).
The stress \(\mathbf {\tilde {\tau }}_{q}\) (Cauchy stress in state \(q\)) is calculated at the point \(b_{q}\). Since this is a rigid body rotation, the area \(da\) will not change, only the unit normal vector \(\mathbf {n}\) will change to \(\mathbf {n}_{q}\)
The tensor \(\mathbf {\tilde {Q}}\) maps the vector \(\mathbf {df}\) to the vector \(\mathbf {df}_{q}\)
But in state \(b\) (the deformed state), the Cauchy stress tensor is given by\begin{equation} \mathbf {df=}\left ( da\ \mathbf {n}\right ) \cdot \mathbf {\tilde {\tau }} \tag {2}\end{equation} Substituting (2) into (1) gives \[ \mathbf {df}_{q}=\mathbf {\tilde {Q}}\cdot \left ( da\ \mathbf {n}\right ) \cdot \mathbf {\tilde {\tau }}\]
Exchanging the order of \(\mathbf {\tilde {\tau }}\) and \(da\ \mathbf {n}\) and using the transpose of \(\mathbf {\tilde {\tau }}\) gives
Therefore since the tensor \(\mathbf {\tilde {Q}}\) maps the oriented area \(\left ( da\ \mathbf {n}\right ) \) to the oriented area \(\left ( da\ \mathbf {n}_{q}\right ) \) then
Or
Substituting (4) into (3) gives
However, the stress \(\mathbf {\tilde {\tau }}_{q}\) in state \(q\) is given by \(\mathbf {df}_{q}=\left ( da\ \mathbf {n}_{q}\right ) \cdot \mathbf {\tilde {\tau }}_{q}\), hence the above equation becomes
Or
This implies
Comparing the above transformation result with the deformation tensors transformation results in
The transformation of the first Piola-Kirchhoff stress tensor \(\mathbf {\tilde {t}}\) is given below.
Earlier it was shown that \[ \mathbf {\tilde {t}}=J\mathbf {\tilde {F}}^{-1}\mathbf {\cdot \tilde {\tau }}\]
Which implies
However, it was found earlier that \(\mathbf {\tilde {\tau }}_{q}=\mathbf {\tilde {Q}\cdot \tilde {\tau }}\cdot \mathbf {\tilde {Q}}^{T}\) therefore the above becomes
Now an expression for \(\mathbf {\tilde {F}}_{q}^{-1}\) is found.
Since \(\mathbf {\tilde {F}}_{q}=\mathbf {\tilde {Q}\cdot \tilde {F}}\), then \(\mathbf {\tilde {F}}_{q}^{-1}=\left ( \mathbf {\tilde {Q}\cdot \tilde {F}}\right ) ^{-1}\), hence \(\mathbf {\tilde {F}}_{q}^{-1}=\mathbf {\tilde {F}}^{-1}\cdot \mathbf {\tilde {Q}}^{-1}\). But since \(\mathbf {\tilde {Q}}\) is orthogonal, then \(\mathbf {\tilde {Q}}^{-1}=\mathbf {\tilde {Q}}^{T}\), therefore \[ \mathbf {\tilde {F}}_{q}^{-1}=\mathbf {\tilde {F}}^{-1}\cdot \mathbf {\tilde {Q}}^{T}\] Substituting the above in (1) gives \begin{align*} \mathbf {\tilde {t}}_{q} & =J\ \mathbf {\tilde {F}}^{-1}\cdot \mathbf {\tilde {Q}}^{T}\mathbf {\cdot \tilde {Q}\cdot \tilde {\tau }}\cdot \mathbf {\tilde {Q}}^{T}\\ & =J\ \mathbf {\tilde {F}}^{-1}\cdot \mathbf {\tilde {\tau }}\cdot \mathbf {\tilde {Q}}^{T}\end{align*}
However, since \(\mathbf {\tilde {t}}=J\mathbf {\tilde {F}}^{-1}\mathbf {\cdot \tilde {\tau }}\) the above simplifies to
Hence \[ \mathbf {\tilde {t}}_{q}=\mathbf {\tilde {Q}}\cdot \mathbf {\tilde {t}}\] By examining how the geometrical tensors transform, results from before showed that \(\mathbf {\tilde {F}}_{q}\) \(=\mathbf {\tilde {Q}\cdot \tilde {F}}\) therefore
\(\displaystyle \mathbf {\tilde {t}} \text {transforms similarly to} \mathbf {\tilde {F}} \)
The transformation of the second Piola-Kirchhoff stress tensor \(\mathbf {\tilde {s}}_{1}\) is given below.
From earlier \[ \mathbf {\tilde {s}}_{1}=J\ \mathbf {\tilde {F}}^{-1}\cdot \mathbf {\tilde {\tau }}\cdot \mathbf {\tilde {F}}^{-T}\]
Hence
An expression for \(\mathbf {\tilde {F}}_{q}^{-1}\) is now found. Since \(\mathbf {\tilde {F}}_{q}=\mathbf {\tilde {Q}\cdot \tilde {F}}\), hence \(\mathbf {\tilde {F}}_{q}^{-1}=\left ( \mathbf {\tilde {Q}\cdot \tilde {F}}\right ) ^{-1}\), hence \(\mathbf {\tilde {F}}_{q}^{-1}=\mathbf {\tilde {F}}^{-1}\cdot \mathbf {\tilde {Q}}^{-1}\). But since \(\mathbf {\tilde {Q}}\) is orthogonal, then \(\mathbf {\tilde {Q}}^{-1}=\mathbf {\tilde {Q}}^{T}\), hence
The above equation becomes
\(\mathbf {\tilde {F}}_{q}^{-T}=\left ( \mathbf {\tilde {F}}_{q}^{-1}\right ) ^{T}\) hence \(\mathbf {\tilde {F}}_{q}^{-T}=\left ( \mathbf {\tilde {F}}^{-1}\cdot \mathbf {\tilde {Q}}^{T}\right ) ^{T}\), therefore \[ \boxed { \mathbf {\tilde {F}}_{q}^{-T}=\mathbf {\tilde {Q}\cdot \tilde {F}}^{-T} } \]
The above equation becomes
From earlier it was found that \(\mathbf {\tilde {\tau }}_{q}=\mathbf {\tilde {Q}\cdot \tilde {\tau }\cdot \tilde {Q}}^{T}\) Therefore the above equation becomes
Therefore
This is the same as \(\mathbf {\tilde {s}}_{1}\), hence \[ \fbox {$\mathbf {\tilde {s}}_{1_{q}}=\mathbf {\tilde {s}}_{1}$}\]
Since it was found earlier that \(\mathbf {\tilde {U}}_{q}=\mathbf {\tilde {U}}\) therefore
Since \(\mathbf {\tilde {\sigma }}\) is a scalar multiple of \(\mathbf {\tilde {\tau }}\) and from earlier it was found that \(\mathbf {\tilde {\tau }}\) is a conjugate pair with \(\mathbf {\tilde {V}}\) then it is concluded that
The transformation of the second \(\boldsymbol {\tilde {\Gamma }}\) stress tensor is shown below.
From earlier it is shown that \[ \boldsymbol {\tilde {\Gamma }=\tilde {R}}^{T}\cdot \mathbf {\tilde {\tau }\cdot \tilde {R}}\] Hence\[ \boldsymbol {\tilde {\Gamma }}_{q}\mathbf {=\tilde {R}}_{q}^{T}\cdot \mathbf {\tilde {\tau }}_{q}\mathbf {\cdot \tilde {R}}_{q}\] Since \(\mathbf {\tilde {\tau }}_{q}=\mathbf {\tilde {Q}\cdot \tilde {\tau }\cdot \tilde {Q}}^{T}\) the above becomes
But \(\mathbf {\tilde {F}}_{q}=\mathbf {\tilde {Q}}\cdot \mathbf {\tilde {F}}\) and using polar decomposition results in \[ \mathbf {\tilde {F}}_{q}=\mathbf {\tilde {R}}_{q}\cdot \mathbf {\tilde {U}}_{q} \] hence\begin{align*} \mathbf {\tilde {Q}}\cdot \mathbf {\tilde {F}} & \mathbf {=\tilde {R}}_{q}\cdot \mathbf {\tilde {U}}_{q}\\ \mathbf {\tilde {Q}} & \mathbf {=\tilde {R}}_{q}\cdot \mathbf {\tilde {U}}_{q}\cdot \mathbf {\tilde {F}}^{-1}\end{align*}
But \(\mathbf {\tilde {F}}=\mathbf {\tilde {R}}\cdot \mathbf {\tilde {U}}\ \) hence \(\mathbf {\tilde {F}}^{-1}=\left (\mathbf {\tilde {R}}\cdot \mathbf {\tilde {U}}\right )^{-1}=\mathbf {\tilde {U}}^{-1}\cdot \mathbf {\tilde {R}}^{-1}\), and the above becomes\[ \mathbf {\tilde {Q}}\mathbf {=\tilde {R}}_{q}\cdot \mathbf {\tilde {U}}_{q}\cdot \left ( \mathbf {\tilde {U}}^{-1}\cdot \mathbf {\tilde {R}}^{-1}\right ) \] Since \(\mathbf {\tilde {U}}_{q}=\mathbf {\tilde {U}}\) the above becomes\begin{align*} \mathbf {\tilde {Q}} & \mathbf {=}\mathbf {\tilde {R}}_{q}\cdot \overbrace {\mathbf {\tilde {U}}\cdot \mathbf {\tilde {U}}^{-1}}\cdot \mathbf {\tilde {R}}^{-1}\\ & =\mathbf {\tilde {R}}_{q}\cdot \mathbf {\tilde {R}}^{-1}\end{align*}
But \(\mathbf {\tilde {R}}^{-1}=\mathbf {\tilde {R}}^{T}\) therefore \begin{equation} \mathbf {\tilde {Q}}=\mathbf {\tilde {R}}_{q}\cdot \mathbf {\tilde {R}}^{T} \tag {2}\end{equation}
Hence \begin{align} \mathbf {\tilde {Q}}^{T} & =\left ( \mathbf {\tilde {R}}_{q}\cdot \mathbf {\tilde {R}}^{T}\right ) ^{T}\nonumber \\ & =\mathbf {\tilde {R}\cdot \tilde {R}}_{q}^{T} \tag {3} \end{align}
Substituting (2) and (3) into (1) gives
Since \(\mathbf {\tilde {R}}\) and \(\mathbf {\tilde {R}}_{q}^{T}\) are orthogonal, the above reduces to
But \(\boldsymbol {\tilde {\Gamma }=\tilde {R}}^{T}\cdot \mathbf {\tilde {\tau }\cdot \tilde {R}}\) therefore \[ \boxed { \boldsymbol {\tilde {\Gamma }}_{q}=\boldsymbol {\tilde {\Gamma }} } \]
The transformation of the Biot-Lure stress tensor \(\mathbf {\tilde {r}}^{\ast }\) is given below.
From earlier
Hence
But \(\mathbf {\tilde {\tau }}_{q}=\mathbf {\tilde {Q}\cdot \tilde {\tau }\cdot \tilde {Q}}^{T}\) therefore
Since \(\mathbf {\tilde {F}}_{q}=\mathbf {\tilde {Q}\cdot \tilde {F}}\) then \(\mathbf {\tilde {F}}_{q}^{-1}=\left ( \mathbf {\tilde {Q}\cdot \tilde {F}}\right ) ^{-1}=\mathbf {\tilde {F}}^{-1}\cdot \mathbf {\tilde {Q}}^{-1}\) but \(\mathbf {\tilde {Q}}\) is orthogonal, hence \(\mathbf {\tilde {Q}}^{-1}=\mathbf {\tilde {Q}}^{T}\), hence \(\mathbf {\tilde {F}}_{q}^{-1}=\mathbf {\tilde {F}}^{-1}\cdot \mathbf {\tilde {Q}}^{T}\). Therefore (2) can be written as
Now \(\mathbf {\tilde {R}}_{q}\) is resolved.
Since \(\mathbf {\tilde {F}}_{q}=\mathbf {\tilde {Q}\cdot \tilde {F}}\) and by polar decomposition \(\mathbf {\tilde {F}}_{q}=\mathbf {\tilde {R}}_{q}\mathbf {\cdot \tilde {U}}_{q}\) then \begin{align} \mathbf {\tilde {R}}_{q}\mathbf {\cdot \tilde {U}}_{q} & =\mathbf {\tilde {Q}\cdot \tilde {F}}\nonumber \\ \mathbf {\tilde {R}}_{q} & =\mathbf {\tilde {Q}\cdot \tilde {F}\cdot \tilde {U}}_{q}^{-1} \tag {4}\end{align}
Substituting (4) into (3) gives
Since \(\mathbf {\tilde {F}=\tilde {R}\cdot \tilde {U}}\) the above becomes
From earlier \(\mathbf {\tilde {U}=\tilde {U}}_{q}\), therefore the above becomes
From (1) gives \[ \fbox {$\mathbf {\tilde {r}}_{q}^{\ast }=\mathbf {\tilde {r}}^{\ast }$}\] And since \(\mathbf {\tilde {U}}_{q}\mathbf {=\tilde {U}}\) therefore
The transformation of the Juamann stress tensor \(\mathbf {\tilde {r}}\) is shown below.
From earlier
From (1) results \[ \mathbf {\tilde {r}}_{q}\mathbf {=}\frac {\left ( \mathbf {\tilde {r}}_{q}^{\ast }+\mathbf {\tilde {r}}_{q}^{\ast T}\right ) }{2}\] Since it was found that \(\mathbf {\tilde {r}}_{q}^{\ast }=\mathbf {\tilde {r}}^{\ast }\) then the above becomes
Therefore
Since \(\mathbf {\tilde {U}}_{q}\mathbf {=\tilde {U}}\) therefore
From earlier
Hence
But \(\mathbf {\tilde {\tau }}_{q}=\mathbf {\tilde {Q}\cdot \tilde {\tau }\cdot \tilde {Q}}^{T}\) and \(\mathbf {\tilde {V}}_{q}=\mathbf {\tilde {Q}\cdot \mathbf {\tilde {V}}\cdot \tilde {Q}}^{T}\) Therefore
Hence
Therefore
Since \(\mathbf {\tilde {T}=}\frac {\left ( \mathbf {\tilde {T}}^{\ast }+\mathbf {\tilde {T}}^{\ast T}\right ) }{2}\) and \(\mathbf {\tilde {T}}^{\ast }\) is conjugate pair with \(\mathbf {\tilde {U}}\) then