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.

In the last blog, we saw that two material constants are required to describe elastic behavior in solids - the shear modulus G and the modulus of compression K, which is also called the bulk modulus.  (Timoshenko and Goodier use the letter G for the shear modulus, while Landau and Lifshitz use the Greek letter mu).

Apparently, materials may have a different response to compressive forces  than they do to shear forces.  This seems plausible from a microscopic viewpoint, that squeezing atoms or molecules closer together would require more force than moving them around without changing the volume.  However, we will see later that for many materials, K and G have about the same value.

Continuing from the previous blog, we find an expression for the strain tensor in terms of stress, following Landau and Lifshitz.  First we find the  trace of the stress tensor in terms of strain from the formula for Hooke's law obtained last time:

Setting i = j in this formula, we find the parenthetical expression vanishes (the trace of the Kronecker delta is equal to 3), and the trace of the stress tensor is 3K times the trace of the strain tensor.  Inverting this and substituting in Hooke's law,

Solving for the strain tensor gives the converse form of Hooke's law:

We observe that the trace term contains only the bulk modulus, and the shear term contains only the shear modulus, in both versions of Hooke's law, as they should.  There is another way to look at Hooke's law, though, and to find it we'll use this last formulation to revisit the stretched rubber tube problem, discussed earlier (in the March 13 blog), which can also be found in Chapter 1, Section 5 in Landau and Lifshitz, Theory of Elasticity.

Hooke's Law

In order to combine the shear and compression terms into one expression for Hooke's law, Landau and Lifshitz make use of the Kronecker delta symbol.  For the compression term, this is straightforward:

If i = j, the delta symbol is equal to one, and we have the expression given for compression in the previous blog. For the shear term, we need to subtract the trace components from the strain tensor, leaving only the off-diagonal shear components:

Note the reason for the factor 1/3:  When i = j, the terms with repeated indices are summed, so that the delta symbol is equal to 3.  As a result, the trace of the stress tensor vanishes in the shear term.

Finally, the formula for the stress tensor in terms of the strain, for Hooke's law (in Cartesian coordinates) is written

This covers all the possibilities.  However, before we write this out in component form, we will obtain the converse formula, for the strain tensor in terms of the stress.  The latter expression is more easily understood, and allows us to define a different pair of material constants which are often used in practice.

Relative volume change and the trace of the strain tensor

Suppose we have a cube subject only to normal stresses on its six faces, as shown below.  We expect the cube to expand, to form a slightly larger body with parallel sides (although not necessarily a cube).  

Let the cube have sides of dimension lx, ly and lz in the x, y, and z directions, and the displacements of the x, y and z planes (those with normals in the x, y, and z directions, respectively) be ux, uy, and uz.  Then the relative change in volume due to the normal stresses will be

Since these are the diagonal terms of the strain tensor, the right hand side is called the trace of the strain tensor.

From the drawing at the top, it might appear that Hooke's law would have a term in which the trace of the stress tensor was a linear function (i.e., a constant times) the trace of the strain tensor (and vice versa).  In fact, this is the case, and

In the next blog, we'll look at how to combine the shear and compression terms into a single expression for Hooke's law that covers all cases.

More on shear, and Hooke's Law

Hooke's Law requires only one material property to describe shear, i.e. one constant, which is G, called the modulus of rigidity, or modulus of elasticity in shear:

One example of this would be the strain due to the force in the x direction on the y plane, that is

In the previous blog we asked why the force on one plane would cause displacement in two planes - in this case the force on the x plane causes displacement in the x and y planes.  In an earlier blog (the one about shearing strain and the change in angle), we used a 2-dimensional drawing which I've modified as shown below:

This drawing shows an example of what is called "pure shear," where there are no normal forces.  The change in angle (from 90 degrees) between the two planes which intersect at (x,y), which are the x-z and y-z planes, is shown by the two blue-shaded triangular areas.

Hooke's Law (in the first equation above) says that the total change in angle for these two planes is determined by the x-y component of the stress tensor, i.e. the force applied to the upper and lower horizontal y-z planes (whose normal vectors are in the + y and - y directions).

But we know from symmetry that the x-y and y-x components of the stress and strain tensors are equal.  For the strain tensor, this is an identity, but for the stress, symmetry is a consequence of conservation of angular momentum.  So in Hooke's Law, it doesn't matter which stress component you use - the strain shown by the total angle in the drawing may be thought of as caused by either the x-y (horizontal) or y-x (vertical) stresses.

If you imagine the action of these forces on the block of material, it makes sense that Hooke's law would only apply to homogeneous, isotropic materials - those in which the material properties are constant throughout.

In contrast, suppose the upper layer of the block was much "stiffer" than the rest of the body.  The response of the vertical sides to the stresses applied to them would be quite different than the horizontal ones.  The two tensors would still be symmetric, but the simple linear relationship shown at the beginning of this blog wouldn't apply.  For one thing, you would need at least two material constants.

In the next blog, we'll look at the normal components of Hooke's law.

Stress-Strain relation (Hooke's Law)

The relationship between stress and strain in Hooke's Law is one of simple linear proportionality, with the constant of proportionality characterizing the elastic properties of the material.  For now, let this elastic coefficient be equal to one.  Then the x-y shear term, in Cartesian coordinates, is

In this form, it seems reasonable to say that the x-y component of the stress tensor produces a gradient in the y direction of the x displacement, and the y-x component produces a gradient in the x direction of the y displacement.  In the earlier blog, "Shearing Strain and the change in angle...," we saw that these gradients are each equal to one part of the total change in angle of two perpendicular planes at the point (x,y).  

The right hand side of the equation above is equal to twice the x-y component of the strain tensor, and since the x-y and y-x stress components are equal, the two's cancel, and we see another reason why the factor 1/2 comes into the strain tensor.

When you look at the drawing in the Shearing Strain blog mentioned above, it makes sense (to me, anyway) that the x-y and y-x stress components would each be responsible for the respective displacement gradients, since they act on different planes.  But can this be true?  The strain tensor is defined to contain both terms.  

We know that the stress tensor has to be symmetric (conservation of angular momentum), but we do not know that the two displacement gradients in the x-y strain tensor are equal.  If they were, it seems like it wouldn't be necessary to define the strain as the sum of the two gradients, i.e. to force the symmetry, so to speak.  So we are left with something of a mystery here.  

Somehow the stress tensor, representing the force on one plane, causes a strain affecting the displacement in that plane and the displacement in a perpendicular plane.  Maybe this will become more clear when we work out an example of shear.

Symmetry of Strain and Stress

The strain tensor is symmetric by definition.  Each off-diagonal term is the sum of two terms, which are reversed when the tensor indices are, and the sum remains the same (a+b=b+a).  The two terms represent changes in length, or angle, for lines in two perpendicular planes, e.g. the change in a line in the x-direction embedded in a y-plane, plus the change in a line in the y-direction embedded in the x-plane.  (The nomenclature here identifies the plane by its normal vector)

In contrast, the off-diagonal terms of the stress tensor act in only one direction, in one plane.  E.g., the x-y component of the stress represents the force in the x-direction on a y-plane, and the y-x component represents the force in the y-direction on the x-plane.

In other words, the x-y component of the stress, acting only on the y-plane, produces strain in both the x and y planes, as does the y-x component of the stress.  This may lead us to a third reason for defining the strain tensor as it is.

Two questions are, 1) why are the x-y and y-x components of stress equal, and 2) why are they both linearly related to the x-y component of the strain tensor?

Mathematically, the second question is partly answered just by the symmetry of both tensors, but we may find a physical reason as well.  Another thing to note here, is that the strain tensor is dimensionless, but the stress tensor has units of force/area.  So stress and strain are different animals, so to speak, and any equation relating them linearly must have coefficients with the same units as the stress (or its inverse).

We start with the symmetry question, again considering a rectangle in the x-y plane.  Recall that the components of the stress tensor are to be interpreted as the forces acting on the corresponding planes (e.g., the i-kth component is the force in the i direction on the k plane).  Then we can label the drawing as shown below:

In equilibrium the rectangle is stationary, and the sum of the moments, or torques about the center of the rectangle must equal zero (conservation of angular momentum).  Since the stress components act on the areas, we let the rectangle represent one surface of a cube, whose length in the z direction is c.  Going counterclockwise from the right edge, we have

The stress components here must be taken as the average stresses over their respective areas, and are also assumed to be continuous functions of x,y and z.  Cancelling the area factor abc/2, and if the dimensions of the cube are small enough, the above equation, with some rearranging, becomes (to any desired degree of accuracy),

which shows the symmetry of the x-y components of the stress tensor.  The same thing can be done for the x-z and y-z components.

So the reason for the symmetry seems to be as follows:  the two x-y components are trying to rotate the cube counterclockwise, and the two y-x components are trying to rotate it clockwise. Then equilibrium is obtained by equating the clockwise and counterclockwise forces. As a and b become smaller, the difference between the two x-y (or y-x) components becomes small also, so that they are essentially equal.  Then the balance of forces produces the symmetry.