# 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.

With frame

5.2.1 Elasticity Theory, Energy and Forces

5.2.2 Stress Field of a Straight Dislocation

© H. Föll (Defects - Script)