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From Wikipedia article on deformation, the shear strain is defined as the angle of the deformation. I had always thought of it as the limiting ratio of the difference in perpendicular displacement of the beginning and end of a line element with the length of that line element.

$$\frac{\partial{u_y}}{\partial{x}}$$

Where $u_y$ is the displacement of a point in the $y$ direction. In this sense, the definition would follow similarly to that of normal strain. That is, a ratio of change in length to that of original length. Why is it that it is defined as the angle instead?

I understand that in very small lengths (differential lengths) the two are the same. For pure shear stress,

$$\frac{\partial{u_y}}{\partial{x}}=\tan{(\alpha)}\approx\alpha\;\;;\alpha\approx0$$

But why is it defined as the angle and not the ratio?

enter image description here

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    $\begingroup$ Because it was all defined in the linear, small strain limit from the get-go. $\endgroup$ – Jon Custer Jul 12 '18 at 22:58
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Let (x,y) be the coordinates of an arbitrary material point in the undeformed configuration of the material, and let u(x,y) and v(x,y) be the displacements of this material point in the x and y directions, respectively. Then the coordinates of the material point in the deformed configuration of the material are (x+u,y+v). The differential position vector between two closely neighboring points in the deformed configuration of the material will be: $$\mathbf{ds}=\left(dx+\frac{\partial u}{\partial x}dx+\frac{\partial u}{\partial y}dy\right)\mathbf{i}+\left(dy+\frac{\partial v}{\partial x}dx+\frac{\partial v}{\partial y}dy\right)\mathbf{j}$$ The square of this differential position vector (in the deformed configuration of the body) is given, to linear terms in the displacements, by:$$(ds)^2=(dx)^2+(dy)^2+2\frac{\partial u}{\partial x}(dx)^2+2\frac{\partial u}{\partial y}dxdy+2\frac{\partial v}{\partial y}(dy)^2+2\frac{\partial v}{\partial x}dxdy$$ If we subtract the square of the length of the differential position vector in the undeformed configuration of the material, we obtain:$$(ds)^2-(ds)_0^2=2\frac{\partial u}{\partial x}(dx)^2+2\left(\frac{\partial u}{\partial y}+\frac{\partial v}{\partial x}\right)dxdy+2\frac{\partial v}{\partial y}(dy)^2$$where $$(ds)^2_0=(dx)^2+(dy)^2$$Next, if we divide by $(ds)^2_0$, we obtain:$$\frac{(ds)^2-(ds)_0^2}{(ds)^2_0}=2\left[\frac{\partial u}{\partial x}\cos^2{\alpha}+\left(\frac{\partial u}{\partial y}+\frac{\partial v}{\partial x}\right)\cos{\alpha}\sin{\alpha}+\frac{\partial v}{\partial y}\sin^2{\alpha}\right]$$ The term in parenthesis is the strain in the differential position vector between the undeformed and deformed configurations of the material:

$$\epsilon=\epsilon_{xx}\cos^2{\alpha}+2\epsilon_{xy}\cos{\alpha}\sin{\alpha}+\epsilon_{yy}\sin^2{\alpha}$$ This illustrates how the partial derivatives of the displacements (including the shear components) are related to the changes in length of the material elements.

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  • $\begingroup$ cool, that actually makes a lot of conceptual sense when you see it all derived out from scratch like that. Thank you! $\endgroup$ – Colin Hicks Jul 12 '18 at 23:46

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