Compression (or extension) of a rod

Consider a rod with a rectangular cross section, so that Cartesian coordinates can be used in 3 dimensions.  Let the longer dimension of the rod be along the z-axis, with a compressive force per unit area f applied to both ends.  Imagine that the rod is thick enough that it won't bend under the applied force.

We assume that the force is applied evenly to the surface on the ends of the rod, so that there are no changes in the angles of any intersecting planes that might be drawn through the rod.  Therefore, from the definition of shear strain, all the non-diagonal terms in the strain tensor are zero.

From the discussion in the March 10 blog, "the balance of forces within a material under stress," we know that we can slice the rod anywhere to form new plane surfaces perpendicular to the z-axis, and because the forces on each of the two pieces must balance, the force on any such surface is equal to the applied force f.  By the same argument, the force on any plane perpendicular to the x- or y- axes must be zero.


Therefore the magnitude of the stress vector on any z-plane through the rod is equal to f,  and the stress vector on any x- or y- plane must be zero.  From another previous blog, (March 4, "Cauchy's stress theorem"), we know that

Applying this equation to the surface of the rod (or, from the above discussion, any portion of the rod bounded by x-, y- and z-planes) and substituting  f  for the z-component of the stress vector, and since no force is applied to the sides of the rod (which have unit normal vectors in the x and y directions), we have

Since the x component of the stress vector and the shear stress components are all equal to zero, the xx- component of the stress tensor also vanishes.  In a similar manner, the yy- component of the stress tensor is also zero, but the zz-component is equal to the applied force -f .  The minus sign indicates that the force is applied along the inward normal vector.

Now we can use the Hooke's law formula for the strain tensor given in the previous blog:

Substituting the components of the stress tensor, we find

We see that although the non-axial stresses are zero, the non-axial strains are not.  Presumably when we compress the rod from the ends, it gets "thicker," but not by as much as the the ends are compressed (we'll verify this shortly).  This is the part we left out of the earlier discussion of a piece of stretched rubber tubing. Finally, the zz- strain component is clearly negative, indicating compression.

Note that although these are purely normal strains, with no shear, above expressions which will be used to calculate their numerical values contain both the bulk modulus and the shear modulus - not what one might expect from the separation of these material properties in Hooke's law.

Unlike the pure compression model used to write the first term in Hooke's law, the rod is both contracting axially and expanding in the transverse directions.  Perhaps this is why both material properties are required to describe each of the strain components.

In the rubber tubing example, we introduced Young's modulus, which may be written here as

So the coefficient of compression (or extension) - the proportionality constant relating axial stress and axial strain - or Young's modulus, is a combination of bulk and shear moduli.  Following Landau and Lifshitz, we can introduce another material constant, called Poisson's ratio, using the Greek letter nu,  which again makes it possible to write Hooke's law with only two constants.

Poisson's ratio is a dimensionless number defined as the ratio of the transverse expansion (compression) to the longitudinal compression (extension):

For most materials, nu is between 0.2 and 0.3, which tells you the transverse normal strain components are always less than the strain in the direction of the applied forces, as one would expect.

Substituting for the strain components  using the results from the set of six equations listed above, we find Poisson's ratio in terms of K and mu.:

Before we can rewrite Hooke's law, we have one more chore - to find K and mu in terms of E  and nu, which can be done by inverting the two expressions just obtained for the latter pair of terms.

Substituting this last equation in either of the first two in this set eventually gives us the bulk modulus in terms
 of Young's modulus and Poisson's ratio:

In the next blog, we will need the strain tensor in terms of the stress tensor with E and nu,  and then we can write out what is probably the most useful and comprehensible version of Hooke's law in component form.