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Alternatively, you can change the definition of shear strain by a factor of two, and use mathematics that doesn't need any "special" definitions, just standard vector calculus. The second way has obvious advantages if you want to combine continuum mechanics with other phenomena such as fluid dynamics, or with special or general relativity.
As we see a small triangular wedge with the base $\ W= shear\ strain$ is subtracted from the left side but the same volume is added to the right side. So under small angle (linear) shear strain, the volume remains constant while the coordinates of x or Y change.
1) Shear Strain due to Pure Shear. 2) Shear Strain due to Shear Deformation. Note, shear deformation becomes a concern when L/d (Span/beam depth) $\leq$ 10. Otherwise, it is usually ignored. Edit: For typical beams that follow the beam theory (Euler-Bernoulli or Timoshenko), as stated above, the shear deformation is very small thus usually ignored.
If one looks up the meaning of shear strain, most sources talk about this situation: That is, a force acts along a surface of an element at it is elongated, turning a rectangle into a rhombus. Apparently this is called "simple shear". But then I found out there is something called "pure shear", which is illustrated in this picture:
While the shear loads for a square cross-section can be though of as uniform, then normal loads (which are more significant and generate cracks) are greater away from the neutral axis. So you have a combination of stresses, and (in the majority of the cases) the normal stresses are more severe.
Thus the true strain is $\frac{1}{0.997}$ times larger than engineering strain at the elastic limit, or about $1.003$ times, or about $0.3\%$ larger. Keep in mind this is at the elastic limit of an exceptionally strong linearly elastic material, and so is a reasonably conservative estimate of the difference between true strain and engineering ...
The maximum shear stress is always positive. (or zero for hydrostatic case) Note this is an invarient, you can report maximum shear without specifying a coordinate system. absolute shear stress is probably the absolute value of the shear stress on a specified orientation which may or may not be the special orientation that gives the maximum.
One way to derive the strain tensor is from geometry. The diagonal (normal) components $\epsilon_{rr}$, $\epsilon_{θθ}$, and $\epsilon_{zz}$ represent the change of length of an infinitesimal element. The non-diagonal (shear) components describe the change of angles. normal strains normal strain in radial direction $\epsilon_{rr}$
Now, the stresses pair forms a couple, which in turn must be countered by a pair of shear stresses on the two vertical edges/faces to maintain rotational equilibrium. Note that this proof holds true for the other two perpendicular planes too, which, in combination, validates the relationship of the stresses stated above.
Transverse shear is caused by the fact that since the moment felt by the beam is higher when we move to the left, the normal forces inside the beam when considering a small box inside the beam (as in the above picture) are at imbalance and thus a transverse shear has to be present to maintain equilibrium.