True strain, engineering strain, strain gauges I've been somewhat confused over the concepts of true and engineering strain, and I just want to see if I am understanding this correctly. 
Let us denote true and engineering strain as $\epsilon_t$ and $\epsilon_e$. 
I know that $\epsilon_t = \ln(1+\epsilon_e)$.
For small deformations, we have the following relationship between strain and displacement:
$\epsilon = 0.5[\nabla u + (\nabla u)^T]$
This relationship is for the engineering strain, and not the true strain right? 
And I also understand that strain gauges measure the engineering strain.
If all of this is true, then what is the point of true strain?
 A: As described here, some advantages of true strain are:
It’s the exact value, not an approximation predicated on small values, which makes it useful when characterizing materials that deform by large amounts (considerable fractions of their length up to many times their length). Examples include hydrogels, elastomers, biological tissue, and fabric. 
Sequential strains can be added, which is not the case with engineering strain. 
It’s geometrically symmetric: that is, if the strain associated with being stretched to n times the original length is e, then the strain associated with being compressed to 1/n the original length is −e.
A: The second formula approximates the true strain, but is not equal to it. On the one hand the true strain is:
$$\varepsilon_t = \frac{\text{d}L_f}{\text{d}L_0} - 1$$
On the other hand the second formula proceeds from defining the right Cauchy–Green deformation tensor as:
$$\mathbf{C} = (\mathbf{1} + \nabla\boldsymbol{u})^T(\mathbf{1} + \nabla\boldsymbol{u})$$
This tensor is related to the true strain by:
$$(1+\varepsilon_t)^2 = \left(\frac{\text{d}L_f}{\text{d}L_0}\right)^2 = \boldsymbol{n}\cdot (\mathbf{C}\boldsymbol{n})$$
where $\boldsymbol{n}$ is a unit vector in the direction in which you are calculating the strain. The above formula implies that:
$$1+\varepsilon_t = \sqrt{\boldsymbol{n}\cdot \mathbf{C}\cdot \boldsymbol{n}} = \sqrt{ [(\mathbf{1} + \nabla\boldsymbol{u})\boldsymbol{n}]^T[(\mathbf{1} + \nabla\boldsymbol{u})\boldsymbol{n}]}$$
Using a Taylor series development to linearize the above expression, we obtain:
$$1+\varepsilon_t \approx 1 + \boldsymbol{n}\cdot \left(\frac{(\nabla\boldsymbol{u})^T + \nabla\boldsymbol{u}}{2}\right) \cdot \boldsymbol{n}$$
Therefore, the infinitesimal tensor $[(\nabla\boldsymbol{u})^T + \nabla\boldsymbol{u}]/2$ does not coincide with either the true strain or the engineering strain. If we had not done this last linerization we would have obtained the true strain, but computationally the formula obtained is disadvantageous.
A: The second formula you show is for the true strain, this is the physical strain in the material. It is is different in every location of the material, and is practically impossible to measure. In order to have something more practical to work with, the engineering strain has been invented. This is nothing more than some measure for the average strain in a standard tensile test specimen. It has very little physical meaning, but because everybody uses it the same way, it is very useful for comparing material properties.
Strain gauges cant really measure strain, they can only measure their own average elongation. This gives information about the true strain of the material it is attached to, in one direction only.  
This link gives a good explanation:
http://www.engineeringarchives.com/les_mom_truestresstruestrainengstressengstrain.html
