# Why is tensorial shear strain half of engineering shear strain?

Small deformation shear strain $$\gamma$$ in engineering can be expressed as

$$\gamma_{xy}=\alpha +\beta ={\frac {\partial u_{y}}{\partial x}}+{\frac {\partial u_{x}}{\partial y}}$$

where $$u$$ represents the displacement of the edges of the infinetesimal element. But in the strain tensor we have defined

$$\epsilon_{xy} = 1/2 \gamma_{xy}$$

The shear strain represents the change in the angle between the lines of the infinitesimal element. So why do we define the tensorial component as half of this angle? Why not just use the angle itself?

• Search for engineering shear strain on here... – Solar Mike Jun 7 '19 at 19:08
• @SolarMike I did that before asking this question. – S. Rotos Jun 7 '19 at 19:14
• So you must have found at least one then... – Solar Mike Jun 7 '19 at 19:15
• Have a look at engineering.stackexchange.com/q/6020/10902 – Solar Mike Jun 7 '19 at 19:17
• @SolarMike I don't see how that's relevant to my question. I'm asking why there is a factor of one half in the definition of shear strain, not why we use engineering strain. – S. Rotos Jun 7 '19 at 19:24