# Displacement and Strain

While the relation between the displacement field u(r) and the local strain tensor eij is rather elementary, it does not hurt to recall the decisive points.
Let's take the simple example from the backbone and consider a rod that is uniformly elongated; i.e. u(r) = ux(x) = a · x; a is some constant.
In other words, the vector u only has a component in x-direction, which only depends on x as variable. The geometry than looks like this:
At any point in the rod a little cube will be deformed into a cuboid - the side in x-direction is somewhat longer than the others.
What kind of strain do we have to put on a cube positioned a x, to produce the cuboid?
Well, since there is only strain in x-direction, we simply write down the elementary formula for strain
exx  = ex =  l  –  l0
l0
=   ux(x + dx) – ux(x)
dx
=   dux
dx
If we deform in all three directions, we get corresponding expressions for eyy and ezz.
Since we also might have displacement components in x-direction that depend on y or z, e.g. ux(x, y, z) = a · y, we may, in general, also form mixed (partial) derivatives; e.g. ux(x, y, z)/y. What do those derivatives signify?
Shear stresses, of course. A little less easy to see, perhaps, but there can be no doubt about it.
You may want to try to show that for yourself with the simple displacement field given above and the equations in the backbone as a guideline for what you are looking for.

5.2.1 Elasticity Theory, Energy and Forces

5.2.2 Stress Field of a Straight Dislocation

© H. Föll (Defects - Script)