3 Stress Measures

3.1 Introduction

Before outlining the different stress measures, the different entities involved are described and illustrated.

Given the undeformed state \(B\), let \(P\) a point in \(B\) where its location in the deformed state becomes \(b\) (Lagrangian description). Let \(dA\) be a differential area at point \(P\) on the surface of \(B\) where \(\mathbf {dN}\) is a unit vector normal to this area in \(B\). After deformation, this differential area will is deformed to a new differential area \(da\) in the deformed state \(b\). Let \(\mathbf {dn}\) be the unit vector normal to \(da\) in \(b\).

Let \(\mathbf {df}\) be the differential force vector which represents the resultant of the total internal forces acting on \(da\) in the deformed state \(b\).

The following diagram illustrates the above.

3.2 Cauchy stress measure

The Cauchy stress measure \(\mathbf {\tilde {\tau }}\) is a measure of

force per unit area in the deformed state

It is called the true measure of stress. The followng follows from the above definition \[ \boxed { \mathbf {df}=\left ( da\ \mathbf {n}\right ) \cdot \mathbf {\tilde {\tau }} } \]

Cauchy stress tensor is in general (in absence of body couples) a symmetric tensor.

3.3 First Piola-Kirchhoff or Piola-Lagrange stress measure

The above diagram shows that this stress \(\mathbf {\tilde {t}}\) can be regarded as

The force in the deformed body per unit undeformed area

.

The following shows the derivation of this stress tensor. Starting by moving the vector \(\mathbf {df}\) (the result of internal forces in the deformed state) which acts on the deformed area \(da\) in a parallel transport to the image of \(da\) in the undeformed state, which will be the differential area \(dA\)

Hence in the undeformed state the following results \begin{equation} \mathbf {df}=\left ( dA\ \mathbf {N}\right ) \cdot \mathbf {\tilde {t}} \tag {1}\end{equation}

Given that \[ \mathbf {N}\ dA=\frac {1}{J}\left ( da\ \mathbf {n}\right ) \cdot \mathbf {\tilde {F}}\]

Which is a relationship derived from geometrical consideration [2], then from the above equation the following results \[ da\ \mathbf {n}=J\ \left ( \mathbf {N}\ dA\right ) \cdot \mathbf {\tilde {F}}^{-1}\]

Since \(\mathbf {df}=\left ( da\ \mathbf {n}\right ) \cdot \tilde {\tau }\), then using the above equation gives

\begin{align} \mathbf {df} & =\left ( J\ \left ( \mathbf {N}\ dA\right ) \cdot \mathbf {\tilde {F}}^{-1}\right ) \cdot \mathbf {\tilde {\tau }}\nonumber \\ & =\mathbf {N}\ dA\cdot \left ( J\mathbf {\tilde {F}}^{-1}\cdot \mathbf {\tilde {\tau }}\right ) \tag {2}\end{align}

Comparing (1) to (2) gives \[ \left ( dA\ \mathbf {N}\right ) \cdot \mathbf {\tilde {t}}\ =\left ( \mathbf {N}\ dA\right ) \cdot J\mathbf {\tilde {F}}^{-1}\cdot \mathbf {\tilde {\tau }}\] Hence \[ \fbox {$\mathbf {\tilde {t}}=J\ \mathbf {\tilde {F}}^{-1}\cdot \mathbf {\tilde {\tau }}$}\] First Piola-Kirchhoff stress tensor in general is unsymmetrical.

3.4 Second Piola-Kirchhoff stress tensor

From the above diagram stress \(\tilde {s}_{1}\) can be regarded as

Modified version of forces in the deformed body per unit undeformed area

.

The stress measure \(\tilde {s}_{1}\) is similar to the first Piola-Kirchhoff stress measure, except that instead of parallel transporting the force \(\mathbf {df}\) from the deformed state to the undeformed state, a force vector \(\mathbf {d\hat {f}}\) is first created which is derived from \(\mathbf {df}\) and then parallel transport this new vector is made.

Everything else remains the same. The purpose of this is that the second Piola-Kirchhoff stress tensor will now be a symmetric tensor while the first Piola-Kirchhoff stress tensor was nonsymmetric.

\begin{equation} \mathbf {d\hat {f}}=\mathbf {\tilde {F}}^{-1}\cdot \mathbf {df} \tag {1}\end{equation}

Hence in the undeformed state (after parallel transporting \(\mathbf {d\hat {f}}\) to \(dA\)) the following relationship results

\begin{equation} \mathbf {d\hat {f}}=\left ( dA\ \mathbf {N}\right ) \cdot \mathbf {\tilde {s}}_{1} \tag {2}\end{equation}

In the deformed state the following relation applies

\begin{equation} \mathbf {df}=\left ( da\ \mathbf {n}\right ) \cdot \mathbf {\tilde {\tau }} \tag {3}\end{equation}

As before, an expression for \(\mathbf {\tilde {s}}_{1}\) in terms of the Cauchy stress tensor \(\tilde {\tau }\) is now found.

Given that \[ da\ \mathbf {n}=J\ \left ( \mathbf {N}\ dA\right ) \cdot \mathbf {\tilde {F}}^{-1}\]

Substituting the above in (3) gives

\[ \mathbf {df}=\left ( J\ \left ( \mathbf {N}\ dA\right ) \cdot \mathbf {\tilde {F}}^{-1}\right ) \cdot \mathbf {\tilde {\tau }}\]

From (1) \(\mathbf {df}=\mathbf {\tilde {F}}^{T}\cdot \mathbf {d\hat {f}}\) , hence the above equation becomes

\[ \mathbf {\tilde {F}}^{T}\cdot \mathbf {d\hat {f}}=\left ( J\ \left ( \mathbf {N}\ dA\right ) \cdot \mathbf {\tilde {F}}^{-1}\right ) \cdot \mathbf {\tilde {\tau }}\]

Therefore

\begin{align} \mathbf {d\hat {f}} & \mathbf {=}\left ( J\ \left ( \mathbf {N}\ dA\right ) \cdot \mathbf {\tilde {F}}^{-1}\right ) \cdot \mathbf {\tilde {\tau }}\cdot \mathbf {\tilde {F}}^{-T}\nonumber \\ & =\left ( \mathbf {N}\ dA\right ) \cdot \left ( J\mathbf {\tilde {F}}^{-1}\cdot \mathbf {\tilde {\tau }}\cdot \mathbf {\tilde {F}}^{-T}\right ) \tag {4}\end{align}

Comparing (4) with (2) gives

\begin{align*} \mathbf {d\hat {f}} & =\left ( \mathbf {N}\ dA\right ) \cdot \overset {\tilde {s}_{1}\ \ 2^{nd}\ PK\ stress\ tensor}{\overbrace {\left ( J\mathbf {\tilde {F}}^{-1}\cdot \mathbf {\tilde {\tau }}\cdot \mathbf {\tilde {F}}^{-T}\right ) }\ \ }\ \ \ \ \ \ \ \ \ \ \ \ \\ & =\left ( dA\ \mathbf {N}\right ) \cdot \mathbf {\tilde {s}}_{1}\end{align*}

Therefore the second Piola-Kirchhoff stress tensor is

\[ \boxed {\mathbf {\tilde {s}}_{1}=J\ \mathbf {\tilde {F}}^{-1}\cdot \mathbf {\tilde {\tau }}\cdot \mathbf {\tilde {F}}^{-T} } \]

The second Piola-Kirchhoff stress tensor is in general symmetric.

3.5 Kirchhoff stress tensor

Kirchhoff stress tensor \(\mathbf {\tilde {\sigma }}\) is a scalar multiple of the true stress tensor \(\mathbf {\tilde {\tau }}\). The scale factor is the determinant of \(\mathbf {\tilde {F}}\), the deformation gradient tensor.

Hence \[ \boxed { \mathbf {\tilde {\sigma }}=J\ \mathbf {\tilde {\tau }} } \]

\(\mathbf {\tilde {\sigma }}\) is symmetric when \(\mathbf {\tilde {\tau }}\) is symmetric which is in general the case.

3.6 \(\tilde {\Gamma }\) stress tensor

The \(\boldsymbol {\tilde {\Gamma }}\) stress tensor is a result of internal forces generated due to the application of the stretch tensor only. Hence this stress acts on the area deformed due to stretch only. Therefore this stress represents

forces due to stretch only in the stretched body per unit stretched area

.

Assuming these are called \(\mathbf {df}^{\ast }\), then applying this definition results in

\begin{equation} \mathbf {df}^{\ast }=\left ( da\ \mathbf {n}^{\ast }\right ) \cdot \boldsymbol {\tilde {\Gamma }} \tag {1}\end{equation}

When in the final deformed state the following relation applies before

\begin{equation} \mathbf {df}=\left ( da\ \mathbf {n}\right ) \cdot \mathbf {\tilde {\tau }} \tag {2}\end{equation}

The above means that the stretched state can be considered as a partial deformed state, and the final deformed state as the result of applying the rotation tensor on the stretched state. In the final deformed state the result of the internal forces is \(\mathbf {df}\) while in the stretched state, in which all the variables in that state are designated with a star *, the internal forces are called \(\mathbf {df}^{\ast }\)

Therefore \begin{equation} \mathbf {df}=\mathbf {\tilde {R}}\cdot \mathbf {df}^{\ast } \tag {3}\end{equation}

Equation (3) can be written as \(\mathbf {df}^{\ast }=\mathbf {df\cdot \tilde {R}}\). Substituting this into (1) gives

\begin{equation} \overset {\left ( da\ \mathbf {n}\right ) \cdot \tilde {\tau }}{\overbrace {\mathbf {df}}}\mathbf {\cdot \tilde {R}}=\left ( da\ \mathbf {n}^{\ast }\right ) \cdot \boldsymbol {\tilde {\Gamma }} \tag {4}\end{equation}

Substituting for \(\mathbf {df}\) in the above equation the expression for \(\mathbf {df}\) in (2) results in

\begin{equation} \left ( da\ \mathbf {n}\right ) \cdot \mathbf {\tilde {\tau }\cdot \tilde {R}}=\left ( da\ \mathbf {n}^{\ast }\right ) \cdot \boldsymbol {\tilde {\Gamma }} \tag {5}\end{equation}

But \(da\ \mathbf {n=\tilde {R}}\cdot \left ( da\ \mathbf {n}^{\ast }\right )\) hence the above equation becomes

\begin{align*} \mathbf {\tilde {R}}\cdot \left ( da\ \mathbf {n}^{\ast }\right ) \cdot \mathbf {\tilde {\tau }\cdot \tilde {R}} & =\left ( da\ \mathbf {n}^{\ast }\right ) \cdot \boldsymbol {\tilde {\Gamma }}\\ \left ( da\ \mathbf {n}^{\ast }\right ) \cdot \mathbf {\tilde {R}}^{T}\mathbf {\cdot \tilde {\tau }\cdot \tilde {R}} & =\left ( da\ \mathbf {n}^{\ast }\right ) \cdot \boldsymbol {\tilde {\Gamma }}\\ \left ( da\ \mathbf {n}^{\ast }\right ) \cdot \left ( \mathbf {\tilde {R}}^{T}\cdot \mathbf {\tilde {\tau }\cdot \tilde {R}}\right ) & \mathbf {=}\left ( da\ \mathbf {n}^{\ast }\right ) \cdot \boldsymbol {\tilde {\Gamma }}\end{align*}

Therefore

\[ \fbox {$\boldsymbol {\tilde {\Gamma }=\tilde {R}}^{T}\cdot \mathbf {\tilde {\tau }\cdot \tilde {R}}$}\]

3.7 Biot-Lure \(\tilde {r}^{\ast }\) stress tensor

This stress measure exists in the undeformed state as a result of parallel translation of the \(\mathbf {df}^{\ast }\) forces generated in the stretched state back to the undeformed state and applying this force into the image of the stretched area in the undeformed state. Therefore this stress can be considered as

forces due to stretch only applied in the undeformed body per unit undeformed area

In a sense, it is one step more involved than the \(\boldsymbol {\tilde {\Gamma }}\) stress tensor described earlier. The following diagram illustrates the above.

From the above diagram an expression for the Biot-Lure stress tensor is now given

\[ \mathbf {df}^{\ast }=\left ( dA\ \mathbf {N}\right ) \cdot \mathbf {\tilde {r}}^{\ast }\]

Now an expression for \(\mathbf {\tilde {r}}^{\ast }\) is found. Since \(\mathbf {df}^{\ast }=\mathbf {df}\cdot \mathbf {\tilde {R}}\), the above equation becomes

\[ \mathbf {df}\cdot \mathbf {\tilde {R}}=\left ( dA\ \mathbf {N}\right ) \cdot \mathbf {\tilde {r}}^{\ast }\]

Given that \(\mathbf {df}=da\ \mathbf {n\cdot \tilde {\tau }}\), the above equation becomes

\[ \left ( da\ \mathbf {n\cdot \tilde {\tau }}\right ) \cdot \mathbf {\tilde {R}}=\left ( dA\ \mathbf {N}\right ) \cdot \mathbf {\tilde {r}}^{\ast }\]

But \(da\ \mathbf {n}=J\ \left ( dA\ \mathbf {N}\right ) \cdot \mathbf {\tilde {F}}^{-1}\) hence the above equation becomes

\begin{align*} \left ( J\ \left ( dA\ \mathbf {N}\right ) \cdot \mathbf {\tilde {F}}^{-1}\mathbf {\cdot \tilde {\tau }}\right ) \cdot \mathbf {\tilde {R}} & =\left ( dA\ \mathbf {N}\right ) \cdot \mathbf {\tilde {r}}^{\ast }\\ \ \left ( dA\ \mathbf {N}\right ) \cdot \left ( J\ \mathbf {\tilde {F}}^{-1}\mathbf {\cdot \tilde {\tau }}\cdot \mathbf {\tilde {R}}\right ) & =\left ( dA\ \mathbf {N}\right ) \cdot \mathbf {\tilde {r}}^{\ast }\end{align*}

By comparison it follows that

\[ \fbox {$\mathbf {\tilde {r}}^{\ast }=J\ \mathbf {\tilde {F}}^{-1}\mathbf {\cdot \tilde {\tau }}\cdot \mathbf {\tilde {R}}$}\]

The stress tensor \(\mathbf {\tilde {r}}^{\ast }\) is un-symmetric when \(\mathbf {\tilde {\tau }}\) is symmetric which is in the general is the case.

3.8 Jaumann stress tensor \(\tilde {r}\)

This stress tensor is introduced to create a symmetric stress tensor from the Biot-Lure stress tensor as follows\[ \fbox {$\mathbf {\tilde {r}=}\frac {\left ( \mathbf {\tilde {r}}^{\ast }+\mathbf {\tilde {r}}^{\ast T}\right ) }{2}$}\] No physical interpretation of this stress tensor can be made similar to the Biot-Lure stress tensor.

3.9 The stress tensor \(T^*\)

This stress tensor is defined in the rotated state without any stretch being applied before. The forces that act on the rotated area were parallel transported from the forces that were generated in the final deformed state. Hence this stress can be considered as

forces due to final deformation applied in the rotated body per unit undeformed area

The following diagram illustrates this. Since rotation have been applied before stretch, then the polar decomposition of \(\mathbf {\tilde {F}}\) becomes \[ \mathbf {\tilde {F}=\tilde {U}\cdot \tilde {V}}\]

Where \(\mathbf {\tilde {V}}\) is the rotation tensor (which was called \(\mathbf {\tilde {R}}\) when it was applied after stretch), and \(\mathbf {\tilde {U}}\) is the stretch tensor.

The above diagram shows that \[ \mathbf {df}=\left ( dA\ \mathbf {N}^{\ast }\right ) \cdot \mathbf {\tilde {T}}^{\ast }\]

Since \(\mathbf {df}=da\ \mathbf {n\cdot \tilde {\tau }}\), the above equation becomes

\[ da\ \mathbf {n\cdot \tilde {\tau }}=\left ( dA\ \mathbf {N}^{\ast }\right ) \cdot \mathbf {\tilde {T}}^{\ast }\]

But \(da\ \mathbf {n}=J\left ( dA\ \mathbf {N}^{\ast }\cdot \mathbf {\tilde {V}}^{-1}\right ) \) hence the above equation becomes

\begin{align*} J\left ( dA\ \mathbf {N}^{\ast }\cdot \mathbf {\tilde {V}}^{-1}\right ) \mathbf {\cdot \ \tilde {\tau }} & =\left ( dA\ \mathbf {N}^{\ast }\right ) \cdot \mathbf {\tilde {T}}^{\ast }\\ \left ( dA\ \mathbf {N}^{\ast }\right ) \cdot J\ \mathbf {\tilde {V}}^{-1}\mathbf {\cdot \ \tilde {\tau }} & =\left ( dA\ \mathbf {N}^{\ast }\right ) \cdot \mathbf {\tilde {T}}^{\ast }\end{align*}

Therefore

\[ \fbox {$\mathbf {\tilde {T}}^{\ast }=J\ \mathbf {\tilde {V}}^{-1}\mathbf {\cdot \ \tilde {\tau }}$}\]

\(\mathbf {\tilde {T}}^{\ast }\) is un-symmetric when \(\mathbf {\tilde {\tau }}\) is symmetric.

3.10 The stress tensor \(T\)

This stress tensor is introduced to create a symmetric stress tensor from the \(\mathbf {\tilde {T}}^{\ast }\) stress tensor as follows

\[ \fbox {$\mathbf {\tilde {T}=}\frac {\left ( \mathbf {\tilde {T}}^{\ast }+\mathbf {\tilde {T}}^{\ast T}\right ) }{2}$}\]

No physical interpretation of this stress tensor can be made similar to the \(\mathbf {\tilde {T}}^{\ast }\) stress tensor.