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.

Strain

In an earlier blog, the one with the example of stretched rubber tubing, we included a description of the concept of strain.  Now, we try to make it more plausible, more precise, and to include shearing, as well as normal strain.

We'll start with the one-dimensional case again.  We choose a fixed coordinate system, and consider a surface somewhere inside a material body, at a point x = (x,y,z), where the plane tangent to the surface is the y-z plane.  That is, the normal vector n(x) = (1,0,0). 


Under the action of forces applied to the surface of the body, let the displacement of the point x in the x direction be u(x).  Next, consider a point at a distance a further along the x-axis,  at x+a.  What is the displacement u (x+a)?  We can find out by expanding u(x+a) in a Taylor series (Wikipedia has a good article on the latter).  The expansion is

 

where u' is the first derivative of u with respect to x, u" is the second derivative, and so on.  If we take a to be small enough, then we can drop all the terms with higher powers of a, keeping only the first two.  Then, for a sufficiently small,  we have [u(x+a) - u(x)]/a = du/dx.  This is called the normal strain in the x-direction.  (It's also simply the definition of the derivative of u(x)). Later we will use the observation that [u(x+a)-u(x)]/a  is equal to the fractional change in length of the line segment (originally of length a) joining the points x and x+a.


The material is said to be in tension when the normal strain is positive.  Note that the displacement is then larger at x+a than it is at x.  So as we noted with the rubber tubing, the displacement must increase in the direction of increasing x.  Furthermore, we noted that for materials that follow Hooke's law, the normal displacement is a linear function of x.  


In the previous blog, we found that for a body in mechanical equilibrium, the divergence of the stress tensor is zero, which means that the stress is constant (i.e., doesn't depend on x) for this one-dimensional example.  Therefore each point in the material along x must be in tension.  


For example, when you pull on the rubber tubing at one end (holding the left end fixed in space), the material at the point x moves to the right.  Therefore the material at the point x+a must also move to the right - otherwise the material at x+a must also move to the right, since it is under the same stress.  So the material at x+a must move the same distance to the right as the material at x, plus an additional distance in order to be in tension. For a material following Hooke's law, the fractional  (i.e., percentage) change in position of each point is constant.  


The purpose of all this discussion is to make Hooke's law seem plausible.  That is, if stress and strain are both constant in a one-dimensional linearly elastic material, it makes sense that they should be proportional to each other, with the coefficient of proportionality being a property of the material.  


As an aside, note that we are assuming the material is isotropic, i.e. that its properties are the same in all directions.  For example if the elasticity varied so that, for example, the rubber was "stiffer" at x+a than at x, it would be possible for the displacement at x+a to be less than at x.  


However, things get more complicated in the three-dimensional case, as we'll see in the next blog.  Since stress is a tensor, it seems reasonable that strain is also, and we need to find a linear relationship between these two tensors.

The divergence of the stress tensor

The law of conservation of momentum usually says that the net force on an object is equal to its rate of change of momentum.  But for a region inside the object, there has to be an additional term, to account for the possible flux of momentum through the enclosing surface.  So the correction to the last equation from the previous blog, for conservation of momentum, must be as shown below.  The arrows over P and n are to remind us that the momentum and unit normal are vectors, and similarly for the tilde over the stress tensor.  This equation states that the rate of change of momentum of material within the volume V is equal to the combined effects of 1) a source term, given by the force density f, and 2) a flux term, given by the stress tensor dotted into the unit vector normal to the surface.  

 This equation fits the general form a conservation law in continuum mechanics, with the net rate of change of a quantity  (mass, momentum, energy) being the sum of a source within the region being considered, and the flux of the quantity in through the surface.  We have a minus sign for the flux term, because we take n to be the outer normal vector.  I should add that "continuum mechanics" is the study of materials whose properties, like mass, momentum and energy, are continuous functions of their spatial coordinates, i.e., x, y and z.  One more comment:  The volume V and its enclosing surface S are fixed in space.  It's hard to imagine much of a change in momentum in a solid object being slowly deformed - by stretching, compressing, bending, shearing, etc.  Yet the force required to cause the deformation can be very large, so this equation must represent the small difference between two very large numbers.

In calculating the work done in deforming a body, we assume that the rate of change of momentum, dP/dt, is so small it can be neglected, and therefore we can take the terms on the right hand side of the above equation as being equal.  

We now continue the discussion of work.  To get started, we make use of a formula from vector calculus, known as the divergence theorem (you can read about it in Wikipedia).  It states that the surface integral above can be transformed into a volume integral, by taking the divergence of the stress tensor:

The integrands on both sides are vectors, because the stress is a tensor.  By combining this this result with the previous equation (and with the rate of change of momentum equal to zero), we find:

Since the volume of integration is arbitrary, the integrands must be equal everywhere, and this leads to the second and third equations above (this time omitting the arrow and tilde).  The vector force acting on an infinitesimal volume of material is equal to the gradient of the stress tensor within that same small volume.  For a material in equilibrium, whose properties are the same in all directions (i.e., is isotropic), the divergence of the stress tensor is zero.  

We know that an elastic material, according to Cauchy's definition (see the blog about stretching a piece of rubber tubing), is one in which the stress depends only on the strain.  So this means we should be able to find an expression for the strain, set equal to zero.  

Since the strain depends on the gradient of the deformation (see the blog just referred to), we expect to find a set of three second order differential equations (for i = 1,2,3) for the deformation vector u of an isotropic material.  These equations, plus boundary conditions, should have a unique solution.

Unfortunately, nearly all boundary value problems in elasticity are very difficult to solve analytically, so there aren't very many examples that can be used just to clarify the concepts.  

Next time, we'll talk about Hooke's Law, the one we used for the rubber tubing example, generalized to three dimensions.

The strain tensor, and the stress-strain relationship

In order to proceed from the simple example of tension in a rod to a more general formulation of stress and strain, we need to consider the work done in deforming an elastic material.

We start with the definition: work is equal to force times distance - for example, think of rolling a ball uphill.  You multiply the required force by the distance traveled to obtain the work done.  Note that the work is recoverable,  since you can let the ball roll back downhill.  In pushing the ball uphill, you are storing potential energy, which can be recovered as kinetic energy by letting it roll back down.  The stored energy will be completely recovered only if none of it is needed to overcome friction.  More generally, the process is reversible if none of the stored energy is converted into heat.

Similarly, the work required to deform an elastic body is the force acting at each point in the body times the displacement of that point from its position when no force was applied.  The deformation is reversible if the body returns to its original state when the force is removed.  Note that this force is the limit of the force per unit volume, as the volume shrinks to a point, and not the force per unit area.

This force was introduced in an earlier blog, "The stress principle of Euler and Cauchy," with the first equation shown below.  The stress vector was given in terms of the stress tensor in the blog "Cauchy's Stress Theorem,"  as shown in the second equation below.  This latter formula can be written in tensor notation as given in the third equation, with the convention that repeated subscripts are summed.  So when you see a repeated subscript, imagine a summation sign, with the sum going from k=1 to k=3.  In this case, the equation means that the ith component of the stress vector depends on all three components of the surface normal vector, and also on three components of the stress tensor.

Finally, substituting the right hand side of the third equation in place of the stress vector into the first equation will gives us the relationship between the force density vector and the stress tensor shown in the fourth equation.  Recall that the area S is the entire area enclosing the volume V, and the equation holds for any volume chosen within the material body.  Note that the integrand on the right is given by the ith component of a vector, which has (at most) three terms, given by the summation over the subscript k.  

The meaning of this last equation is the same as the first equation:  It states that the forces acting on the interior of a part of the material body may be entirely determined from the forces acting on the enclosing surface.

But wait a minute!  In the earlier blogs on the balance of forces, and on the stretched piece of rubber, we learned that the total force acting on any portion of a body in equilibrium is zero.  Since this is true for any volume, it follows that f =0 at each point in the material.  You can imagine a point in the body being pulled in various directions, and if the net force were not zero, the body would deform in the neighborhood of that point.  

The two sides of the last equation above are non-zero only if the material is in the process of being deformed. Just as with the ball on the hill, you're only working while you're rolling it, at least as far as the ball is concerned.  (The energy your body requires to hold it there is another story).  

So there is a term missing from the equation.  To find it, we need to consider the conservation of momentum, which we'll do in the next blog.

A simple example of strain - a stretched piece of rubber tubing

The practical goal of elasticity theory is often to calculate the deformation of an elastic body under an applied stress. We want to know how much force can be applied to the wing of an airplane before it breaks, or how much force it would take to crumple a car body in a collision, or how much weight a bridge can carry.

The situations I just described represent large deformations and the eventual failure, or rupture, of the object. But elasticity theory usually applies only to small deformations (say, 3% or less). Often, however, it only takes a small deformation to cause trouble. For calculations involving the strength of materials, one wants to know how much stress can be applied before the material yields, i.e. is no longer elastic.

An elastic material is one which has a "natural" state, that it tries to return to when an external force is applied, and will return to when that force is released. This statement applies to each point in the body. (A material that does not return to its initial state, but remains deformed, is said to be "plastic.")

Consider a point in an elastic body which has no applied stress - no force is being exerted on its surface. We think of this point as being attached to the material, so that if the body is deformed in some way by an applied stress, this "material point" may move, relative to a coordinate system that is fixed in space.

What we may want from the theory is a way to calculate the displacement, the distance and direction that any material point in the body moves due to an applied force. Alternatively, we may want to know the distribution of forces within the body, given a certain deformation.

To get started, imagine that you are holding an exercise band - or any length of rubber tubing with a handle at each end. Put one handle on the floor and hold it there with your foot, and use one hand to pull the other handle up away from the floor, as you might to exercise your bicep. We refer to the end of the band attached to the handle on the floor as "the foot end," and the other end as "the hand end."

Imagine a cross sectional area through the tubing, and the total force acting over that area on the part of tubing that is above that cross section - between it and the hand end. The total force, the integral of the stress vector over that area, must equal the force you applying with your hand, since you are hold the band very still.

Since this is true for any cross section through the band, it must be that the force of the material acting on itself (the integral of the stress tensor over the surface) is constant along the band. Because the rubber is uniform, it seems reasonable that the stress tensor is uniform over the cross section, and is therefore itself constant along the length of the band.

Suppose the band has the length l when no force is applied. Consider a one dimensional coordinate system, with the foot end of the band at x=0, and the hand end at x=l. Now when the band is stretched to some length l+dl, every point material point fixed in the band moves away from the foot end. The hand end moves the distance dl, but the foot end doesn't move at all.

Similarly, a material point in the band that was at the position x in the natural state will move to a new position x.' The displacement of the band due to the stretching is u=x'-x. We see that u(0)=0, and u(l) =dl.* Since we are considering a uniform, or homogeneous piece of rubber, whose elastic properties are the same throughout, it seems reasonable to expect the displacement to be a linear function of position. That is, u is a constant times x, and from the boundary conditions just given, u=(dl/l)x.

You can try this, by marking the band with a piece of chalk 1" from one end. Hold the band with both hands on a flat surface, alongside a tape measure, with the chalk mark near the left end. Suppose the band is 20 inches long. Hold the left end fixed at the zero end of the tape measure. Now move the right end 5" to the right, so that it has moved 25% of the initial ("natural") distance from the left end. You will see that the chalk mark has moved 1/4," which also represents a 25% stretch from its initial position.

It seems plausible that the fractional movement of each point along the band under an applied force will be the same, as long as the elastic properties of the band are uniform. On the other hand, if there were a weak region in the material, it would stretch more there.

We already know how to calculate the stress as a function of the applied force. In fact, in this simple case, the stress is constant, and equal and opposite to the applied force per unit area, from the discussion above. But how can we determine the displacement from the applied force, i.e., from the stress?

We see that the stress cannot be an explicit function of the displacement, since the displacement is not constant, but increases linearly from the foot end, along the band. However, the derivative of the displacement is constant, since du/dx=dl/l. This derivative is called the strain.

So we will postulate that for this piece of rubber, the stress is equal to the strain times a proportionality constant, which is a property of the material.  (This postulate is known as Hooke's Law, originally developed for a spring).  We write the equation below, where sigma is the normal stress on a surface cut through the band normal to x (the y-z plane), dl/l is the strain (which is dimensionless), and E is the Young's Modulus of elasticity. Note that E has the same units as the stress, i.e. force per unit area

Since the stress is constant and equal to the applied force per unit area, say P, this completes the solution to this simple problem. The strain is constant, and equal to the applied force per unit area, divided by the Young's Modulus. The strain is a constant fraction of the total length of the band. For rubber this fraction can be fairly large without causing permanent damage - 20 % or more. For other materials, like metals, the strain for elastic behavior must remain quite small.

From the equation above, the applied force must behave correspondingly - i.e., not exceed around 20% of the Young's Modulus for the rubber material used to make the exercise tubing. (Actually, you can get them with different wall thicknesses, and or elastic properties). In my case, I didn't have to pull very hard to stretch the band 5", probably with no more than 25 pounds of force. The tubing is 3/8" diameter, with a wall thickness of 1/16", so the cross-sectional area is a little more than 0.07 sq in. The force per unit area, which is the stress in the material, is therefore about 350 psi. So from the equation above, the elastic modulus for this particular piece of rubber is about 1750 psi.

According to the literature (Table 6-1, Barrett, Nix and Tetelman, The Principles of Engineering Materials, Prentice-Hall, 1973, P. 197), the elastic modulus for natural rubber varies from 1,000 to 10,000 psi.

Other materials have much larger moduli - Aluminum's, for example is 10 million psi. If I applied the same force to a piece of aluminum tubing with the same length and cross-section, its length would increase by 1750 x 10E-7, which is less than 0.2 mil, or 0.0002".

From the above discussion, we now have a basis for understanding Cauchy's definition of an elastic material: The stress is determined by the strain only, and that strain is not taken with respect to an arbitrary state but with respect to one particular state, the preferred or 'natural state of the material.'" (Truesdell, The Principles of Continuum Mechanics).

Since stress is a tensor, it seems likely that strain must be a tensor as well, so we'll leave that discussion for the next blog.

The balance of forces within a material under stress

The drawing on the left is intended to represent a 3-dimensional solid body, acted upon by force PR on the right, and PL on the left. Within the body is a surface S.

Imagine that we have a fixed coordinate system (x,y,z) such that the positive x axis is normal to the surface S. We use the label SR for S when it represents a part of the surface of the volume VR, and SL for the part of the surface of the volume VL. (VR and VL together give the total volume).

It seems clear that each component of PR and PL in the x, y and z directions must be equal in magnitude, but with opposite sign. Otherwise, since the net force on a body is equal to the rate of change of its momentum with time, the body would accelerate in the direction of the greater force.

In elasticity theory, it is usually assumed that the material body is somehow held fixed in space, i.e. with balanced applied forces.

Now let's consider the surface S within the body: Here S, SL and SR all occupy the same region in space, but SR has normal vectors pointing into the region VL, and SL's normal vectors point into VR. At each point on S, the normal vectors for SR and SL have components in x, y and z which are equal in magnitude and opposite in direction.

Let's start with SL. We choose a point (x,y,z) on SL, and take the normal vector there to point in the positive x direction (into VR). Then the normal vector has only one non-zero component, as shown in the Cauchy formula on the left.

As an aside, note that we have used the symmetry of the stress matrix, which allows us to equate the off-diagonal components as shown.

Using matrix multiplication, we find that each component of the stress vector contains only one component of the stress tensor, as shown below.

The stress vector is determined by only three components of the stress tensor - the normal stress in the 11 or xx direction, and two shear stresses in the yx and zx directions.

Since the normal vector is of unit length, i.e. n = (1,0,0), that is n1 = 1, the equations on the left show that the x component of the stress vector (that is, of the force acting on the surface SL) is given by the xx component of the stress tensor.

Similarly the y component of the force t is given by the yx component of the stress tensor, and the z component of force is given by the zx component of stress.

Put another way, the yx component of the stress tensor is the y component of the force acting on the x plane. By"x plane," we mean the plane tangent to the surface SL at the point (x,y,z) whose normal vector points in the x direction. It's actually the yz plane, but calling it the x plane makes it easier to identify with the yx component of the stress tensor.

Similarly, the zx component of the stress tensor is the z component of the force acting on the x plane.

We now have a simple example of a particular stress vector acting on a surface within a body. The next step is to consider the stress vector acting on the same plane tangent to the same surface, but now imagining that surface to to belong to the remaining part of the body.

Referring again to the drawing at the top, we use the same coordinate system we had earlier, and consider the surface SR, which is part of the surface of the volume VR. We again choose the same point (x,y,z) that we chose before, with the same plane tangent to SL (and therefore to SR), only this time the normal vector has components (-1,0,0) in our coordinate system.

All we have to do to get stress vector components acting at that point is to replace the x component of n with -1. We get the equations shown below, and find that the components have the same magnitude as before, only they are in the opposite (-x) direction. Since this is true for any point (x,y,z) on the surface drawn anywhere in the body, it must be that the the stress vectors are balanced. Note that the stress components depend only on the location of (x,y,z) and not on the orientation of the surface through this point.

Therefore the stress vectors are balanced. If we choose a different surface through the same point, the components of n will change (the coordinate system is the same), but the stress vectors will still balance.

Truesdell pointed out that this result is actually a consequence of the conservation of momentum: The sum of all the external forces is zero (that PL and PR are equal and opposite). Otherwise, if there is a net force, the body will accelerate, and will no longer be in equilibrium.

Instead, the body will deform as a function of time. The result will be that the surface S will change over time, as will the normal vector: n=n(t).

Furthermore, if the applied forces PL and PR are unequal, then the integrals of t over the surfaces SL and SR will be unequal, since each of those integrals must balance the applied forces on their respective regions VL and VR. From this it is clear that the stress vectors cannot balance, even at a given instant of time. From this we conclude that when the body accelerates, there must be a discontinuity in the stress tensor across the surface S. That is, the value of the stress components will be different if we approach S from the right (VR) or the left (VL) side. This is the subject of the theory of shock waves.

We won't pause now (maybe later) to study shock waves, but suffice it to say that we would expect the acceleration necessary to cause a jump condition within the body of a material would have to be quite large - say of the magnitude that occurs when a massive object from space accelerates towards the earth.

In the next blog, we will return to our main subject - reversible, elastic deformations of homogeneous materials - and introduce some of the concepts needed to actually solve problems.