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\[ \epsilon_{Hyd} = {\epsilon_{11} + \epsilon_{22} + \epsilon_{33} \over 3} \]

And there are many alternative ways to write this.

\[ \epsilon_{Hyd} \; = \; {1 \over 3} \text{tr}(\boldsymbol{\epsilon}) \; = \; {1 \over 3} I_1 \; = \; {1 \over 3} \epsilon_{kk} \]

It is a scalar quantity, although it is regularly used in tensor form as

\[ \boldsymbol{\epsilon_{Hyd}} = \left[ \matrix{\epsilon_{Hyd} & 0 & 0 \\ 0 & \epsilon_{Hyd} & 0 \\ 0 & 0 & \epsilon_{Hyd} } \right] \]

\[ {\bf E} = \left[ \matrix{ 0.50 & \;\;\; 0.30 & \;\;\;0.20 \\ 0.30 & -0.20 & -0.10 \\ 0.20 & -0.10 & \;\;\;0.10 } \right] \]

The hydrostatic strain is

\[ E_{Hyd} \, = \, {0.50 + (-0.20) + 0.10 \over 3} \, = \, 0.133 \]

which can be written as

\[ {\bf E}_{Hyd} = \left[ \matrix{ 0.133 & 0 & 0 \\ 0 & 0.133 & 0 \\ 0 & 0 & 0.133 } \right] \]

And that's all there is to it.

\[ \boldsymbol{\epsilon}_{Hyd} = \left[ \matrix{\epsilon_{Hyd} & 0 & 0 \\ 0 & \epsilon_{Hyd} & 0 \\ 0 & 0 & \epsilon_{Hyd} } \right] \]

contains equal amounts of strain in all three directions. In fact, it contains equal strain in all directions. And since the tensor does not change under any transformation, this means that no shear strains ever arise, so every direction is a principal direction with \(\epsilon_{Hyd}\) strain.

Consider the object in the sketch here. We will look at its volume change. Its initial volume is \(V_o = W_o D_o H_o\) and its final volume is \(V_F = W_F D_F H_F\).

It can be thought of as being aligned with the principal strain tensor and this is why there are no shears. Let the principal strains be defined as

\[ \epsilon_i = {\Delta L \over L_o} \]

where \(L\) can be \(W\), \(D\), or \(H\). So

\[ \begin{eqnarray} W_F & = & W_o ( 1 + \epsilon_1) \\ \\ D_F & = & D_o ( 1 + \epsilon_2) \\ \\ H_F & = & H_o ( 1 + \epsilon_3) \\ \end{eqnarray} \]

Now look at the ratio of \(V_F\) to \(V_o\).

\[ {V_F \over V_o} = {W_F D_F H_F \over W_o D_o H_o } \]

Substituting the strain equations gives

\[ {V_F \over V_o} = ( 1 + \epsilon_1) ( 1 + \epsilon_2) ( 1 + \epsilon_3) \]

and expanding this out gives

\[ {V_F \over V_o} = 1 + \epsilon_1 + \epsilon_2 + \epsilon_3 + \epsilon_1 \epsilon_2 + \epsilon_2 \epsilon_3 + \epsilon_3 \epsilon_1 + \epsilon_1 \epsilon_2 \epsilon_3 \]

Now, if the strains are small, then the products will be negligible and can be ignored. This leaves

\[ {V_F \over V_o} \approx 1 + \epsilon_1 + \epsilon_2 + \epsilon_3 \qquad \text{(small strains)} \]

which can be manipulated to give

\[ {V_F \over V_o} \approx 1 + 3 \left( {\epsilon_1 + \epsilon_2 + \epsilon_3 \over 3} \right) = 1 + 3 \epsilon_{Hyd} \]

So \(\epsilon_{Hyd}\) is directly related to volume change when the strains are small. Also,

\[ \epsilon_{Vol} = {\Delta V \over V_o} \]

and this leads to

\[ \epsilon_{Vol} = \epsilon_1 + \epsilon_2 + \epsilon_3 \qquad \text{(small strains)} \]

Note that the sum of principal strains here is an invariant, so the sum can in fact be made at anytime, not just when the normal strains are principals.

But this is just

\[ \epsilon_{Vol} = 3 \, \epsilon_{Hyd} \]

\[ {V_F \over V_o} = ( 1 + \epsilon_1) ( 1 + \epsilon_2) ( 1 + \epsilon_3) \]

This equation is exact for all strains as long as the principal strains are defined as \(\Delta L / L_o\). The equation is in fact the determinant of \({\bf I} + \boldsymbol{\epsilon}_P\). But \({\bf I} + \boldsymbol{\epsilon}_P\) is just \({\bf U}\) in the principal orientation, so it is also the determinant of \({\bf U}\). And this means that \(\text{det}({\bf U})\) is \(V_F / V_o\).

Recall that computing \({\bf U}\) is pretty difficult, so one might initially think that there is no easy way to find \(V_F / V_o\). But one would be wrong! It turns out that \(\text{det}({\bf U})\) is exactly equal to \(\text{det}({\bf F})\) because \({\bf R}\) does not alter the determinant. So

\[ {V_F \over V_o} = \text{det}({\bf F}) \qquad \text{(all strains, always)} \]

This is such an important result that \(\text{det}({\bf F})\) is given a special symbol, \(J\), and a special name, the

At large strains, hydrostatic strain loses its link to volume because it is no longer an approximation of its change. It is reduced to a mere mathematical property of a strain tensor. One can get a feel for this using the two examples below.

\[ {V_F \over V_o} = ( 1 + \epsilon_1) ( 1 + \epsilon_2) ( 1 + \epsilon_3) = 1 \]

Select "1" as the tension/compression direction and call that strain \(\epsilon_{Eng}\). The other two strains must be equal. Call both of them, \(\epsilon_{other}\). This gives

\[ ( 1 + \epsilon_{Eng}) ( 1 + \epsilon_{other})^2 = 1 \]

Solving for \(\epsilon_{other}\) gives

\[ \epsilon_{other} = \sqrt{1 \over 1 + \epsilon_{Eng}} - 1 \]

and combining two of these with one \(\epsilon_{Eng}\) to get \(\epsilon_{Hyd}\) gives

\[ \epsilon_{Hyd} = {1 \over 3} \epsilon_{Eng} + {2 \over 3} \left[ \sqrt{1 \over 1 + \epsilon_{Eng}} - 1 \right] \]

The function is graphed below. You can see that \(\epsilon_{Hyd}\) is very close to zero at small strains, correctly reflecting that the volume does not change. But at large strains, \(\epsilon_{Hyd}\) drifts into positive territory, even though the volume is not changing.

\[ \boldsymbol{\epsilon}_P = \left[ \matrix{ \gamma/2 & 0 & 0 \\ 0 & -\gamma/2 & 0 \\ 0 & 0 & \epsilon_3 } \right] \] And the ratio of initial to final volume is

\[ {V_F \over V_o} = ( 1 + \gamma / 2) ( 1 - \gamma / 2) ( 1 + \epsilon_3) = 1 \]

Solve for \(\epsilon_3\) to get

\[ \epsilon_3 = {1 \over 1 - \left( { \gamma \over 2} \right)^2 } - 1 \]

And the hydrostatic strain is

\[ \epsilon_{Hyd} = {1 \over 3} \left[ {1 \over 1 - \left( { \gamma \over 2} \right)^2 } - 1 \right] \]

The deviation from zero is much less severe in this case than for tension and compression. Nevertheless, it is still present because hydrostatic strain is only an approximation of volume change, not an exact measure.

The deviatoric strain will be represented by \(\boldsymbol{\epsilon}'\), or \({\bf E}'\), or \({\bf e}'\) depending on what the starting strain tensor is. For example

\[ \boldsymbol{\epsilon}' = \boldsymbol{\epsilon} - \boldsymbol{\epsilon}_{Hyd} \]

In tensor notation, it is written as

\[ \epsilon'\!_{ij} = \epsilon_{ij} - {1 \over 3} \delta_{ij} \epsilon_{kk} \]

\[ {\bf E} = \left[ \matrix{ 0.50 & \;\;\; 0.30 & \;\;\;0.20 \\ 0.30 & -0.20 & -0.10 \\ 0.20 & -0.10 & \;\;\;0.10 } \right] \]

The hydrostatic strain is

\[ E_{Hyd} \, = \, {0.50 + (-0.20) + 0.10 \over 3} \, = \, 0.133 \]

which can be written as

\[ {\bf E}_{Hyd} = \left[ \matrix{ 0.133 & 0 & 0 \\ 0 & 0.133 & 0 \\ 0 & 0 & 0.133 } \right] \]

Subtracting the hydrostatic strain tensor from the total strain gives

\[ {\bf E}' = \left[ \matrix{ 0.50 & \;\;\; 0.30 & \;\;\;0.20 \\ 0.30 & -0.20 & -0.10 \\ 0.20 & -0.10 & \;\;\;0.10 } \right] - \left[ \matrix{ 0.133 & 0 & 0 \\ 0 & 0.133 & 0 \\ 0 & 0 & 0.133 } \right] = \left[ \matrix{ 0.367 & \;\;\; 0.300 & \;\;\;0.200 \\ 0.300 & -0.333 & -0.100 \\ 0.200 & -0.100 & -0.033 } \right] \]

Note that the result is

Once again, note that just because all the normal strains are zero, this does not mean that the strain tensor represents a constant volume deformation. A simple example demonstrating this is the case of pure shear. Starting from

\[ \boldsymbol{\epsilon} = \left[ \matrix{ 0 & \gamma / 2 & 0 \\ \gamma / 2 & 0 & 0 \\ 0 & 0 & 0 } \right] \]

The principal strain tensor is

\[ \boldsymbol{\epsilon}_P = \left[ \matrix{ \gamma / 2 & \;0 & 0 \\ 0 & -\gamma / 2 & 0 \\ 0 & \;0 & 0 } \right] \]

And so the determinant of \({\bf I} + \boldsymbol{\epsilon}_P\) equals

\[ {V_F \over V_o} = ( 1 + \gamma / 2) ( 1 - \gamma / 2) = 1 - (\gamma / 2)^2 \]

which decreases with increasing shear strain.

\[ \gamma = {D \over T} \]

For this case, the deformation gradient is

\[ {\bf F} = \left[ \matrix{ 1 & 0 & 0 \\ \gamma & 1 & 0 \\ 0 & 0 & 1 } \right] \]

and its determinant is indeed 1 regardless of the value of \(\gamma\), so it truly is a constant volume deformation. (Recall that the determinant of \({\bf F}\)

The Green strain tensor for this case is

\[ {\bf E} = \left[ \matrix{ \gamma^2 / 2 & \gamma / 2 & 0 \\ \gamma / 2 & 0 & 0 \\ 0 & 0 & 0 } \right] \]

So the hyrdrostatic strain tensor is

\[ {\bf E}_{Hyd} = \left[ \matrix{ \gamma^2 / 6 & 0 & 0 \\ 0 & \gamma^2 / 6 & 0 \\ 0 & 0 & \gamma^2 / 6 } \right] \]

And the deviatoric strain tensor is

\[ {\bf E}' = \left[ \matrix{ \gamma^2 / 3 & \gamma / 2 & 0 \\ \gamma / 2 & -\gamma^2 / 6 & 0 \\ 0 & 0 & -\gamma^2 / 6 } \right] \]

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Bob McGinty

bmcginty@gmail.com

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