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\[ \epsilon_{\text{True}} = \ln \left( {L_F \over L_o} \right) \]

for an object undergoing tension and/or compression.

\[ x(t) = x_{\text{(t=0)}} \left[ 1 + \left( { 0.05 \over 0.50}\right) \left( {t \over 2} \right) \right] \]

This works because at \(t=0\), it reduces to \(x(t) = x\), and at \(t=2\), it gives \(x(t) = x_{\text{(t=0)}}(0.55/0.50)\). And yes, the equation is in terms of \(x\), but it's \(x\) at \(t=0\), so that is the same as \(\bf{X}\).

Between 2 and 4 seconds, the wire is stretched to 0.60 m long.

\[ x(t) = x_{\text{(t=2)}} \left[ 1 + \left( { 0.05 \over 0.55}\right) \left( {t - 2 \over 2} \right) \right] \]

And between 4 and 6 seconds, it is stretched to 0.65 m long.

\[ x(t) = x_{\text{(t=4)}} \left[ 1 + \left( { 0.05 \over 0.60}\right) \left( {t - 4 \over 2} \right) \right] \]

In order to calculate a velocity gradient, we first need velocities. So take the time derivative of each equation.

\[ \begin{eqnarray} 0 < t < 2: & \quad & v_x = \left( {x \over 2} \right) \left( { 0.05 \over 0.50}\right) \\ \\ 2 < t < 4: & \quad & v_x = \left( {x \over 2} \right) \left( { 0.05 \over 0.55}\right) \\ \\ 4 < t < 6: & \quad & v_x = \left( {x \over 2} \right) \left( { 0.05 \over 0.60}\right) \\ \end{eqnarray} \]

And the velocity gradient during each 2 second time step is

\[ \begin{eqnarray} 0 < t < 2: & \quad & D_{11} = \left( {1 \over 2} \right) \left( { 0.05 \over 0.50}\right) \\ \\ 2 < t < 4: & \quad & D_{11} = \left( {1 \over 2} \right) \left( { 0.05 \over 0.55}\right) \\ \\ 4 < t < 6: & \quad & D_{11} = \left( {1 \over 2} \right) \left( { 0.05 \over 0.60}\right) \\ \end{eqnarray} \]

So far, so good, but nothing special. But now numerically integrate \(\int D \, dt\).

\[ \int D \, dt = {0.05 \over 0.50} + {0.05 \over 0.55} + {0.05 \over 0.60} = 0.274 \]

The important point is to recognize the connection to true strain. To see this, note that

\[ D \, dt \; = \; \left( {\partial v \over \partial x} \right) dt \; = \; {\partial (v\,dt) \over \partial x} \; = \; {\partial \,(dx) \over \partial x} \; = \; {dl \over l} \]

So

\[ \epsilon_{\text{True}} \; = \; \int D \, dt \; = \; \int {dl \over l} \; = \; \ln \left( {L_F \over L_o} \right) \; = \; \ln \left( {0.65 \over 0.50} \right) \; = \; 0.262 \]

The only reason that the numerical integration gave a slightly different result than \(\epsilon_{\text{True}}\) is because the time steps were relatively large.

This example demonstrates that \(\int D\,dt = \epsilon_{\text{True}}\) and \(D = \dot \epsilon_{\text{True}}\).

True strain does have an attractive property that no other strain definition possesses. That is, its range spans from \(- \infty\) to \(+ \infty\). It is compared to engineering strain in the figure below. Note that the relationship is

\[ \epsilon_{\text{True}} = \ln (1 + \epsilon_{\text{Eng}}) \]

because

\[ 1 + \epsilon_{\text{Eng}} \; \; = \; \; 1 + {\Delta L \over L_o} \; \; = \; \; {L_o \over L_o} + {\Delta L \over L_o} \; \; = \; \; {L_F \over L_o} \]

Recall that the ratio of initial to final volume is

\[ {V_F \over V_o} \, = \, \left( {W_F \over W_o} \right) \left( {D_F \over D_o} \right) \left( {H_F \over H_o} \right) \, = \, 1 \]

Now take the natural log of this equation to get

\[ \ln \left( {W_F \over W_o} \right) + \ln \left( {D_F \over D_o} \right) + \ln \left( {H_F \over H_o} \right) = 0 \]

But each log term is just the true strain. In fact, the sum is the trace of the true strain tensor.

\[ \epsilon^\text{True}_1 + \epsilon^\text{True}_2 + \epsilon^\text{True}_3 = 0 \qquad \text{(incompressible materials)} \]

Unlike small strains and Green strains, the above relationship applies to true strains even when the strains are finite.

Also, since the sum is zero, the rate of change of the sum will also always be zero for incompressible materials. So

\[ \dot \epsilon^\text{True}_1 + \dot \epsilon^\text{True}_2 + \dot \epsilon^\text{True}_3 = 0 \qquad \text{(incompressible materials)} \]

But since \(D = \dot \epsilon_{True}\), then the above equation can be written equally well as

\[ D_{11} + D_{22} + D_{33} = 0 \qquad \text{(incompressible materials)} \]

And furthermore, since each of these equations is the first invariant of its associated strain or rate of deformation tensor, the above identities are not limited to principal orientations and principal strains. They will apply even if shears are present.

\[ \ln \left( {V_F \over V_o} \right) \; = \; \ln \left( {W_F \over W_o} \right) + \ln \left( {D_F \over D_o} \right) + \ln \left( {H_F \over H_o} \right) \; = \; \epsilon^\text{True}_{\text{Vol}} \]

So

\[ \epsilon^\text{True}_1 + \epsilon^\text{True}_2 + \epsilon^\text{True}_3 = \epsilon^\text{True}_{\text{Vol}} \]

except this time, this applies for all strains, not just small ones.

Taking the time derivative again gives

\[ \dot \epsilon^\text{True}_1 + \dot \epsilon^\text{True}_2 + \dot \epsilon^\text{True}_3 = \dot \epsilon^\text{True}_{\text{Vol}} \]

and in terms of the rate of deformation tensor components...

\[ D_{11} + D_{22} + D_{33} = \dot \epsilon^\text{True}_{\text{Vol}} \]

Once again, this applies for finite strains, not just infinitesimal ones, and not just in principal orientations.

At \(t = 0\), the object is being stretched along the x-axis, and shrinking along the y-axis due to Poisson's effect. The rate of deformation tensor could be

\[ {\bf D} = \left[ \matrix{ 3.0 & \;\;\;0.0 \\ 0.0 & -2.0 } \right] \]

And if this takes place for 0.1 sec, then the integration of \(\int {\bf D}\,dt\) gives

\[ \boldsymbol{\epsilon}_{\text{True}} = \left[ \matrix{ 0.3 & \;\;\;0.0 \\ 0.0 & -0.2 } \right] \]

At some instant later, the object has rotated 45°, and continues to stretch. At this point, because it is at 45°, this deformation shows up as shear. So \({\bf D}\) could be

\[ {\bf D} = \left[ \matrix{ 0.0 & 1.5 \\ 1.5 & 0.0 } \right] \]

And if this occurs for another 0.1 sec, then the

\[ \Delta \boldsymbol{\epsilon}_{\text{True}} = \left[ \matrix{ 0.00 & 0.15\\ 0.15 & 0.00 } \right] \]

and the total strain is the sum of the two.

\[ \boldsymbol{\epsilon}_{\text{True}} = \left[ \matrix{ 0.30 & \;\;\;0.15 \\ 0.15 & -0.20 } \right] \]

Then at a later time, the object has rotated 90° and is still being pulled in tension such that

\[ {\bf D} = \left[ \matrix{ -1.0 & 0.0 \\ \;\;\;0.0 & 1.5 } \right] \]

So the increment for this step is

\[ \Delta \boldsymbol{\epsilon}_{\text{True}} = \left[ \matrix{ -0.10 & 0.00\\ \;\;\;0.00 & 0.15 } \right] \]

Remember that it has been stretched to a longer length now, so \(dl/l\) is decreasing because the denominator is increasing.

Adding this to the previous total strain tensor gives

\[ \boldsymbol{\epsilon}_{\text{True}} = \left[ \matrix{ 0.20 & \;\;\;0.15 \\ 0.15 & -0.05 } \right] \]

The problem here is that the final true strain tensor is just a jumbled mess. It includes shear values, even though the corners remain at 90°, and ends with a negative \(D_{22}\) value even though the object was indeed stretching in the y-direction at the end of the process. This exemplifies the near uselessness of \(\int {\bf D}\,dt\) when rotations are present.

At \(t = 0\), the object is being stretched along the x-axis, and shrinking along the y-axis due to Poisson's effect. The rate of deformation tensor is the same as before.

\[ {\bf D} = \left[ \matrix{ 3.0 & \;\;\;0.0 \\ 0.0 & -2.0 } \right] \]

And it takes place for 0.1 sec, and \({\bf R} = {\bf 0}\) at this point, so the integration of \(\int {\bf D}\,dt\) gives

\[ \boldsymbol{\epsilon}_{\text{True}} = \left[ \matrix{ 0.3 & \;\;\;0.0 \\ 0.0 & -0.2 } \right] \]

After the object has rotated 45°, \({\bf D}\) is now

\[ {\bf D} = \left[ \matrix{ 0.0 & 1.5 \\ 1.5 & 0.0 } \right] \]

But the rotation matrix is

\[ {\bf R} = \left[ \matrix{ 0.7071 & -0.7071 \\ 0.7071 & \;\;\;0.7071 } \right] \]

So \({\bf R}^T \cdot {\bf D} \cdot {\bf R}\) gives

\[ {\bf R}^T \cdot {\bf D} \cdot {\bf R} \; = \; \left[ \matrix{ 1.5 & \;\;\;0.0 \\ 0.0 & -1.5 } \right] \]

Multiplying this by 0.1 sec gives

\[ {\bf R}^T \cdot {\bf D} \cdot {\bf R}\,dt \; = \; \left[ \matrix{ 0.15 & \;\;\;0.0 \\ 0.0 & -0.15 } \right] \]

And adding this to the first result gives

\[ \boldsymbol{\epsilon}_{\text{True}} = \left[ \matrix{ 0.45 & \;\;\;0.00 \\ 0.00 & -0.35 } \right] \]

Finally, the object has rotated 90°. So \({\bf D}\) and \({\bf R}\) are

\[ {\bf D} = \left[ \matrix{ -1.0 & 0.0 \\ \;\;\;0.0 & 1.5 } \right] \qquad \qquad {\bf R} = \left[ \matrix{ 0 & -1 \\ 1 & \;\;\;0 } \right] \]

Computing \({\bf R}^T \cdot {\bf D} \cdot {\bf R}\) gives

\[ {\bf R}^T \cdot {\bf D} \cdot {\bf R} \; = \; \left[ \matrix{ 1.5 & \;\;\;0.0 \\ 0.0 & -1.0 } \right] \]

And multiplying by 0.1 sec gives

\[ {\bf R}^T \cdot {\bf D} \cdot {\bf R}\,dt \; = \; \left[ \matrix{ 0.15 & \;\;\;0.00 \\ 0.00 & -0.10 } \right] \]

And finally, adding this to the previous running total gives

\[ \boldsymbol{\epsilon}_{\text{True}} = \left[ \matrix{ 0.60 & \;\;\;0.00 \\ 0.00 & -0.45 } \right] \]

This result is much more intuitive. It shows that the object was stretched in its

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