Hooke's law in component form

 We start with Hooke's law for the strain tensor, and then substitute the E and nu material properties for K and mu.

The result is

We see that there is still a compression term, now on the right, but not a pure shear term, and both terms contain both material constants.  

We mention here that  nu is between 0.2 and 0.3 for many materials.  Applying this range of values to the expressions for the bulk and shear moduli above, we see that K, the bulk modulus value is about 60 to 80 percent of the Young's modulus, and mu, the shear modulus is about is about 40 % of the Young's modulus.  The bulk modulus (of compression) is therefore about 1.5 to 2 times larger than the shear modulus.  

Since K and mu are in the denominators of the compression and shear terms of the strain tensor, respectively, and in the numerators of the corresponding terms of the stress tensor, we see that most materials are more resistant to compressive forces than to shear.  This effect is magnified further since the denominator for compression (expansion) is 9K, whereas the denominator for pure shear is only 2 mu.

As I may have mentioned earlier, this seems reasonable from a microscopic point of view, that it would be harder to push atoms or molecules closer together, or farther apart, than it would be to move them around without changing the volume of the body.

We can now write Hooke's law in a convenient component form:

The shear terms at the bottom are analogous to the shear terms with the K, G (or K, mu) formulation, with (1 + nu)/E replaced by 1/(2 mu) or 1/(2G).  Recall that the shear modulus is designated by either G or mu.  The numerical results should be the same in either formulation, of course.

You could say that with the appropriate use of Poisson's ratio (using the formulas given in the last blog), we can transform Young's modulus into either the bulk or shear modulus.  One nice thing about the Young's modulus/Poisson's ratio formulation of Hooke's law is that it makes the role of the material properties so clear:  Young's modulus for compression/expansion, and Poisson's ratio for transverse compression/longitudinal extension, or vice versa.

You can write these component equations with the bulk and shear moduli, but the coefficients of each term are combinations of the two constants, so you don't see their role so well.  Of course, if you keep the pure compression and shear terms separate, then the moduli are also separated, so that's another way to look at it, I guess.  Also, you can always write the 1/E and sigma/E coefficients in terms of K and mu.  Wikipedia has a  conversion table for elastic constants at http://en.wikipedia.org/wiki/Elastic_modulus.

The equations for the normal strains in the above set display clearly that there can be strain components perpendicular to the  stress components, even in the absence of  of the corresponding stress.  I think this is another indication of why you need tensors to describe elastic behavior.

It would be hard to imagine a vector formulation that could describe this behavior, because I think in the end one would have to match up x-components of a vector strain with x-components of vector stress, etc., and there would be no way to relate, say, x-components of one with y-components of the other.

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.

small angle approximation

The magnitude of an angle (in radians) is given by the fraction of the circle's circumference that it subtends. Thus a 360 degree angle subtends an arc whose length is 2 pi times the radius of the circle (this is the definition of pi - circumference divided by diameter). Dividing this by the radius gives the magnitude of the angle, 2 pi radians.  Similarly, a 180 degree angle is equivalent to pi radians, 90 degrees to pi/2 radians, and so on.

In general, the magnitude of the angle is given by the arc length subtended by the angle of a circle centered at its vertex, divided by the circle's  radius, i.e 2 pi l/2 pi r, or l/r,  where l is the arc length subtended by the angle, and r is the radius of the circle centered at the angle's vertex.  Thus, it is a dimensionless quantity.

Now consider the drawing above, with a circle whose radius is approximately equal to a, centered at V, and an angle whose sine is x/a.   For small angles, the length of the arc subtended by this angle can be approximated by x.  Hence x/a also approximates the magnitude of the angle, as defined above.

You can check this out with a table of trigonometric functions.  Up to about 5 degrees, the angle in radians and its sine are equal to within 3 or 4 decimal places.  Even at 10 degrees, the angle is 0.1745 radians, and the sine is 0.1736.  At 20 degrees, the angle is 0.3491 and the sine is 0.3420.

shearing strain and the change in angle caused by deformation

In this post we are looking at the change in angle of one of the vertices after a deformation, for the rectangle discussed in the previous blog.  (Click on the drawing to enlarge it).  The original angle for the vertex at (x,y) is, of course, 90 degrees, or pi/2 radians.  After deformation, all four vertices may be displaced, as shown below.  As before, we use the vector u to represent the displacement.  

The y-component of the displacement of the point (x,y) is in the positive y direction, as shown on the right side of the figure.  Similarly the y-component of the point (x+a,y) is in the negative y direction.  Thus the difference in displacement between the two points is

Now consider the angle of the deformed figure for the vertex which was initially at (x,y).  It is larger than 90 degrees, due to the combined displacements of three of the vertices of the deformed figure:  the ones at (x,y), (x+a,y), and (x,y+b).  

The change in the original 90 degree angle due to the above expression is the one formed by the dashed horizontal line and the lower side of the deformed figure.  The sine of that angle is given by

provided the displacements are small enough (see footnote below).  As we've seen in earlier blogs, this expression is simply the derivative of the y-component of the with respect to x, i.e., one part of the x-y (or y-x) shear component of the strain tensor.  

To get this result, we have to keep the signs straight, bearing in mind that the displacement at x+a is in the negative y-direction.  Strictly speaking, the change in angle is also negative, but we are looking for magnitudes here, so the minus sign that would otherwise be in front of the expression above was dropped.

Finally, we note that for a small angle, the sine of the angle is nearly equal to the angle itself, measured in radians.  I'll discuss this useful approximation in the next blog.  But for now, we need to examine the other portion of the change in the angle at (x,y), i.e. the x-displacement of the points at (x,y) and (x,y+b).  

These displacements are also in opposite directions, so we have to remember that the displacement of the upper vertex is in the negative x-direction.  By the same process as used above, we find the sine of this angle to be

which is the other piece of the x-y component of the strain tensor.  As before, this is also the sine of the angle being considered, and equal to the angle itself, provided the angle is small enough.

The total change in angle is the sum of these two terms, i.e., twice the x-y component of the strain tensor.  This is the second reason for defining the strain tensor the way it is.  (The factor 1/2 in the strain tensor will hopefully become clear later).

The above discussion is an elaboration of a similar one in the first chapter of Theory of Elasticity, by Timoshenko and Goodier.

After a small detour to talk about the approximation that the sine of a small angle is approximately equal to the  magnitude of the angle, in radians, we'll tackle the question of relating stress and strain.

Footnote:  The use of a in the denominator of the expression for the sine of the angle above is an approximation which can be justified as follows:  The actual length for the long side of the angle is given in the denominator of the expression below, for the sine of the angle (which replaces the previous equation for the sine):

Using the definition of a derivative as before, the sine becomes

After cancelling a in the fraction, the denominator has the form 1/(1+x), which can be expanded in a Taylor series (or see Pierce and Foster, A Short Table of Integrals):

Applying this to the previous expression gives 

where we have neglected higher order terms, as usual.  Multiplying out, we see that the second term, a product of two strain terms, will be small compared to the first term, with the result that the sign of the angle is one part of the x-y strain tensor, as stated in the main part of this blog.

Shearing Strain

The next step in coming up with a strain tensor is to look at the deformation caused by a shear force.  We start with a 2-dimensional description.  Imagine a square or rectangular region within a material body, with a coordinate system such that the x-axis is horizontal, and the y-axis is vertical.  Now add stress vectors acting on the rectangle, parallel to the top and sides.  These are the shear forces.  It helps to imagine this region as embedded in a beam, with one end attached to a wall, and the other end with a load pulling it down.  (We'll see later that tensile, compressive and shear forces are all operating within the beam).

The above drawing shows the rectangle before it deforms under stress.

Since we are looking for shearing strain, we consider the deformation in the x direction at two points along the y axis. Again making use of the Taylor series expansion, the x components of the displacements at the two points (x,y) and (x,y+b) are related by

The displacement is described by a vector u(x,y), but the x and y components of u are scalars.  These components are functions of two variables, x and y, but we can expand about one while holding the other fixed.  The prime refers to the partial derivative with respect to y, and for sufficiently small values of b, we have

This gives the variation in the y direction of the x-component of the displacement.  Now think of the z direction as coming out of the plane of the paper (or computer screen).  Then the top and bottom horizontal edges of the rectangle in the drawing are lines in planes which are parallel to the x-z plane.  So these displacements in the x direction take place in two nearby x-z planes.  The normal vector for the x-z plane is in the y direction, i.e. n = (0,1,0), so the displacement is perpendicular to the normal vector of the surface in which it occurs.  That's what makes it a shear force, instead of normal a force.

Now consider the vertical edges of the rectangle.  These lines are embedded in two nearby y-z planes, separated by a short distance a in the x direction.  We consider the difference in the y component of the displacement vector in these two planes, for a fixed value of y.  Proceeding in the same manner as previously, we find

We are now in a position to postulate a strain tensor, generalizing to three dimensions.  We are looking  a linear relationship between stress and strain, so since the stress tensor is symmetric, we want the strain tensor to be also.  It should contain both normal and shear strains.  The result is the following definition:

One reason for the 1/2 is so the diagonal components of the strain (xx, yy, and zz) will be given by their respective derivatives of the displacement.  Another is, so the trace (sum of diagonal terms) of the strain tensor is given by the sum of the normal strains, i.e.

                                                       Trace u = du/dx + dv/dy + dw/dz,

where u, v, and z are the displacements in the x,y and z directions of the point at (x,y,z).  (These are of course, partial derivatives, but I don't have a way of using Greek letters without going off into another program).

Using the results obtained from the Taylor series expansion in an earlier blog, we can show that this sum of normal strains is equal to the fractional change in a sufficiently small volume of the material, because the difference in the displacement of two nearby points, say at x and x+a, is simply the change in length of the line segment joining them.  Then a small volume with sides of lengths lx, ly and lz  in the x, y and z directions before deformation would have a new volume, with sides lx+(du/dx)lx, ly+(dv/dy)ly, and lz+(dw/dz)lz, .  The before and after volumes are the products of the  lengths of their three sides:

                                          New volume = lx ly lz (1+du/dx)(1+dv/dy)(1+dw/dz)

                                                             = Old volume (1+du/dx + dv/dy)(1+dw/dz)

                                                             = Old volume (1+du/dx + dv/dy + dw/dz)

where we have neglected the second-order terms.  From this we see that the sum of the normal strains is equal to the fractional change:  (New volume - Old volume)/Old volume.

We chose a coordinate system where all the strains were normal.  But the trace of a tensor, being a scalar, is invariant to an  arbitrary orthogonal rotation of the coordinates, so the result holds generally.  (Just think of a cube, which undergoes pure compression or expansion:  you can always choose a coordinate system with axes parallel to the three sides of the cube).

The other question to be answered is, why are the shear (off-diagonal) terms added together the way they are in the strain tensor?  It does give us a symmetric tensor, which is good since the stress tensor is symmetric, and we want the the two tensors to be linearly related.  But there are a couple of other reasons.  The first is that the pairs of shear terms in the stress tensor are additive. 

 Referring to the diagram of the rectangle at the beginning of this blog, the stress vectors at the two nearby vertical lines (representing y-z planes) are both driving the rectangle in a clockwise direction.  This is quite different from the normal stress vectors, which are equal and opposite, thereby cancelling each other out.  

Similarly for the stress vectors on the two horizontal x-z planes, which are also working together, to drive the rectangle counter-clockwise.  Of course the four vectors must sum to zero in equilibrium, but the ones on opposite sides need not be equal.

I'll save the second reason for the next blog.  It will use the drawing shown at the beginning of this one.