# Geometric derivation of the infinitesimal strain tensor Consider a two-dimensional deformation of an infinitesimal rectangular material element with dimensions $d x$! by $d y$ (Figure 1), which after deformation, takes the form of a rhombus. From the geometry of Figure 1 we have

\begin{align} \overline {ab} &= \sqrt{\left(dx+\frac{\partial u_x}{\partial x}dx \right)^2 + \left( \frac{\partial u_y}{\partial x}dx \right)^2} \\ &= dx\sqrt{1+2\frac{\partial u_x}{\partial x}+\left(\frac{\partial u_x}{\partial x}\right)^2 + \left(\frac{\partial u_y}{\partial x}\right)^2} \end{align}\,

For very small displacement gradients, i.e., $\|\nabla \mathbf u\| \ll 1 \,$, we have

$\overline {ab} \approx dx +\frac{\partial u_x}{\partial x}dx\,$

How is the y-axis change equal to $\frac{\partial u_x}{\partial y}dx$?

Can someone derive it? It is taken from:

• Wikipedia contributors. "Infinitesimal strain theory." Wikipedia, The Free Encyclopedia. Wikipedia, The Free Encyclopedia, 29 Dec. 2016. Web. 13 Feb. 2017.

$$\overline {ab} = dx\sqrt{1+2\frac{\partial u_x}{\partial x}+\left(\frac{\partial u_x}{\partial x}\right)^2 + \left(\frac{\partial u_y}{\partial x}\right)^2}$$

And we can consider that $\partial u_x/\partial x \gg \partial u_x/\partial x^2$, since those are second order terms, thus

$$\overline {ab} \approx dx\sqrt{1+2\frac{\partial u_x}{\partial x}}$$

Then you can expand in the result in a Taylor series

$$\overline {ab} \approx dx\left[1 + \frac{\partial u_x}{\partial x} - \frac{1}{2}\left(\frac{\partial u_x}{\partial x}\right)^2 + \frac{1}{2}\left(\frac{\partial u_x}{\partial x}\right)^3 + \cdots\right]$$

and just taking the first two terms, you obtain

$$\overline {ab} \approx dx\left[1 + \frac{\partial u_x}{\partial x}\right]$$

That is the result that they show.

• I think he is asking why $\dfrac{\partial u_y}{\partial x}$ appears in the last term in the radical of your first line. The reason being that $b=B+u(B)=A+(dx,0) + u(A+(dx,0))=A+(dx,0) + u(A) + (\dfrac{\partial u_x}{\partial x}dx,\dfrac{\partial u_y}{\partial x}dx)$, and since $a=A+u(A)$, $|b-a|=|(dx + \dfrac{\partial u_x}{\partial x}dx,\dfrac{\partial u_y}{\partial x}dx)|$ – Brian Moths Feb 13 '17 at 2:59