While the relation between the displacement field u(r) and the local strain tensor e_{ij} 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) = u_{x}(x) = a · x; a is some constant.  
In other words, the vector u only has a component in xdirection, 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 xdirection 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 xdirection, we simply write down the elementary formula for strain  


If we deform in all three directions, we get corresponding expressions for e_{yy} and e_{zz}.  
Since we also might have displacement components in xdirection that depend on y or z, e.g. u_{x}(x, y, z) = a · y, we may, in general, also form mixed (partial) derivatives; e.g. ¶u_{x}(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)