*Linear elastic materials deform twice as much when you double the load, and all deformation is recovered when the load is removed. All stable materials are approximately linearly elastic for suitably small deformations.

**Isotropic materials look the same from all directions. Examples? Amorphous ceramics such as window glass have no strongly preferred direction and are thus considered isotropic. Metals are generally polycrystalline, and their grain orientation in bulk form is often sufficiently random that any directional influence is negligible, so they can also often be considered isotropic. What materials aren’t isotropic? The single-crystal silicon used in wafers for integrated circuit manufacturing, for example. Single-crystal turbine blades and metal whiskers. Thin polycrystalline films or strongly deformed bulk metals in which the grains have a preferred direction. Composites, in which certain stress-bearing materials are often oriented in specific directions. You get the idea.

Understanding normal and shear stresses in 3D

Consider two distinct ways to apply a surface force on an object: perpendicular (or normal) to the surface (to produce a normal stress) or parallel with the surface (to produce a shear stress):

Elasticity involves static conditions only; all forces in every direction—and even around every axis—must balance. (Even the simple state of static equilibrium leaves us enough to worry about.) (Created in Python using this code.)

But in both cases, to prevent the object from accelerating away or beginning to spin, we need to add some additional forces to describe the complete stress state. And although we started with the example of a surface, we can think of these stress states appearing anywhere in a body; thus, the 2D square shown above represents a very small region (a differential element) in any 3D orientation and in any location.

The combinations of possible stresses in 3D are distinguished by an index \(i\) for the direction of the force and an index \(j\) for the direction perpendicular to the surface; thus, \(\sigma_{12}\) represents the stress from a force in the 1 direction pulling on a surface oriented in the 2 direction. Static equilibrium conditions let us drop pairs of minus signs (thus, \(\sigma_{(-1)(-1)}=\sigma_{11}\), for example, as in the correctly drawn normal stress example above). They also require \(\sigma_{ij}=\sigma_{ji}\) to prevent rotation. (Try drawing the schematics for \(\sigma_{ij}\) and \(\sigma_{ji}\), including all the surface forces necessary to prevent acceleration. You’ll find that the resulting schematics are identical.)

The resulting set of 3D stresses for \(i, j =1\mbox{–}3\) is a tensor, meaning that it couples algebraic objects (here, corresponding to forces and areas). And because those algebraic objects are each vectors themselves (specifically, a force vector and a surface normal vector), or first-order tensors, the stress tensor that couples them is in turn specifically a second-order tensor, which can be written in matrix form:

$$\boldsymbol{\sigma}=\sigma_{ij}=\left[\begin{array}{c} \sigma_{11} & \sigma_{12} & \sigma_{13}\\ & \sigma_{22} & \sigma_{23}\\ \mbox{sym.}& &\sigma_{33}\\ \end{array}\right]. $$

The numbers of rows and columns equal the dimensionality of the problem, which is 3. (For clarity, vectors and higher-order tensors are written in boldface in this note.)

Understanding normal and shear strains in 3D

First up is the case of normal strain, quantified by the relative elongation of a line pointing in a certain direction. Note especially the movement and direction of the displacement (let’s call it \(\color{green}{\boldsymbol{u}}\)) relative to the orientation and length of the original vector \(\color{blue}{\boldsymbol{x}}\).

A vector \(\color{blue}{\boldsymbol{x}}\) undergoing normal strain. Below, the initial length of the vector, \(\color{blue}{x_1}\), and the displacement \(\color{green}{u_1}\) of its tip relative to its tail.

Normal strains tend to change differential element side lengths. We could handle this case with a simple ratio of vector magnitudes if we wished. (That is, we could divide \(|\color{green}{\boldsymbol{u}}|\) by \(|\color{blue}{\boldsymbol{x}}|\), and this scheme works regardless of whether we fix any edge or point of the differential element.) Things become more ambiguous, however, when we consider shear strains, which instead tend to change element corner angles:

Shear strain: two orthogonal original vectors and the displacements of their tips relative to their tails. The results now differ depending on which edge or point we consider fixed. But any discrepancy can be eliminated if we consider the average relative deformation of both vectors in the 1 and 2 directions. (Another useful definition is the engineering shear stress \(\gamma\), which is simply the reduction in the top-right or bottom-left angle.)

Any scheme of simply taking the ratio of two vectors now runs into a problem: which initial vector should we pick—the one pointing in the 1 direction or the one pointing in the 2 direction? This problem is resolved by defining the tensorial shear strain as XXX

$$\varepsilon_{ij}=\frac{1}{2}\left(\frac{\partial \color{green}{u_i}}{\partial \color{blue}{x_j}}+\frac{\partial \color{green}{u_j}}{\partial \color{blue}{x_i}}\right),$$

which works because the two scenarios are averaged together. The connection with angular changes comes about because when \(\color{green}{\boldsymbol{u}}\) is perpendicular to \(\color{blue}{\boldsymbol{x}}\), \(|\color{green}{\boldsymbol{u}}|/|\color{blue}{\boldsymbol{x}}|\) equals the tangent of the included angle, which for small \(\color{green}{\boldsymbol{u}}\) is approximately equal to the angle itself (in radians).

Even better, this definition works for normal strains too, where \(i=j\), because

$$\varepsilon_{11}=\frac{1}{2}\left(\partial \color{green}{u_1}/\partial\color{blue}{x_1}+\partial \color{green}{u_1}/\partial\color{blue}{x_1}\right)=\partial \color{green}{u_1}/\partial\color{blue}{x_1},$$

for direction 1, for example, so we have a definition we can apply to both scenarios. Note that the additive tensorial strain definition also provides symmetry: \(\varepsilon_{ij}=\varepsilon_{ji}\). So both the stress and strain tensors are symmetric; the latter is

$$\boldsymbol{\varepsilon}=\varepsilon_{ij}=\left[\begin{array}{c} \varepsilon_{11} & \varepsilon_{12} & \varepsilon_{11}\\ & \varepsilon_{22} & \varepsilon_{23}\\ \mbox{sym.}& &\varepsilon_{31}\\ \end{array}\right]. $$

(The engineering shear strain \(\gamma\) is also widely used; it’s defined simply as the reduction in a right angle of a differential element as shown above. The engineering shear strain is twice the tensorial shear strain.)

Defining and assembling the constitutive equations

In fact, generalized Hooke’s Law ultimately describes six (!) separate deformation situations with one equation. The first three situations couple normal stresses with normal strains. For a general 3D chunk of isotropic material, in any direction \(i\), we can expect longitudinal deformation from a normal stress oriented in the \(i\) direction and lateral (Poisson) deformation from normal stresses oriented in the other two directions—the \(j\) and \(k\) directions. According to the definitions of Young’s modulus and Poisson’s ratio above, we can express the normal strains as

$$\varepsilon_{11}=\frac{1}{E}\sigma_{11}-\frac{\nu }{E}\sigma_{22}-\frac{\nu }{E}\sigma_{33};$$

$$\varepsilon_{22}=\frac{1}{E}\sigma_{22}-\frac{\nu }{E}\sigma_{11}-\frac{\nu }{E}\sigma_{33};$$

$$\varepsilon_{33}=\frac{1}{E}\sigma_{33}-\frac{\nu }{E}\sigma_{11}-\frac{\nu }{E}\sigma_{22}.$$

That is, in the 3D framework of generalized Hooke’s Law, any positive/negative normal stress in any direction causes elongational/contractile strain in that direction (as mediated by Young’s modulus) and lateral strain in the other directions (as mediated by Young’s modulus and Poisson’s ratio).

The second three situations couple shear stresses with shear strains and require some geometrical craftiness to apply our two primary assumptions. Consider the deformation state shown below on the left for pure shear. We’ll want to transform this stress state into one involving a normal stress so that we can apply our initial assumptions involving Young’s modulus and Poisson’s ratio. An equivalent stress state is shown on the right that involves only normal stresses in an orientation (\(1^\prime\)-\(2^\prime\)) that’s tilted by 45° from the original one:

Caption ***FIX ARROWS AND MAKE SIZES CONSISTENT WITH THE OTHER FIGURES; SHIFT COORDINATE SYSTEMS TO THEIR RESPECTIVE SIDES***

From symmetry, we could hypothesize that the magnitude of \(\sigma_{1^\prime 1^\prime}\) should equal that of \(\sigma_{2^\prime 2^\prime}\). Although it’s outside the scope of this note, a Mohr’s circle analysis would confirm this hypothesis and also show that \(\sigma_{12}\), \(\sigma_{1^\prime 1^\prime}\), and \(\sigma_{2^\prime 2^\prime}\) all have the same magnitude.

Along line bc, based on the equations linking normal stress and normal strain, we expect an elongational strain from the tensile stress \(\sigma_{1^\prime 1^\prime}\) and another one (a Poisson-mediated one) from the compressive strain \(\sigma_{2^\prime 2^\prime}\) of the same magnitude:

$$\varepsilon_{1^\prime 1^\prime}=\frac{1+\nu}{E}\sigma_{1^\prime 1^\prime}.$$

For line ad, we expect an contractile strain from \(\sigma_{2^\prime 2^\prime}\) and another one (a Poisson-mediated one) from \(\sigma_{1^\prime 1^\prime}\):

$$\varepsilon_{2^\prime 2^\prime}=-\frac{1+\nu}{E}\sigma_{1^\prime 1^\prime}=-\varepsilon_{1^\prime 1^\prime}.$$

Applying trigonometry to the diagram above, we find that the angle \(\theta\), which was initially 45° or \(\pi/4\) (satisfying \(\arctan\left(\frac{|ad|/2}{|bc|/2}\right)=\arctan(1)\) because lines bc and ad were of equal length) now satisfies

$$\theta=\arctan\left(\frac{(1+\varepsilon_{2^\prime 2^\prime})/2}{(1+\varepsilon_{1^\prime 1^\prime})/2}\right)=\arctan\left(\frac{1-\varepsilon_{1^\prime 1^\prime}}{1+\varepsilon_{1^\prime 1^\prime}}\right).$$

Multiplication by \(\frac{1-\varepsilon_{1^\prime 1^\prime}}{1-\varepsilon_{1^\prime 1^\prime}}\) within the arctangent argument and dropping any \(\varepsilon^2\) terms as neglibly small gives

$$\theta=\arctan(1-2\varepsilon_{1^\prime 1^\prime}). $$

This looks like the arctangent of 1 minus a small value—a great candidate for Taylor series expansion. (I try to never miss a chance to apply a Taylor series expansion.) Specifically, a Taylor series restates any function \(f\) evaluated at \(x_0+\delta x\) as

$$f(x_0+\delta x)=f(x_0)+f^\prime (x_0)\delta x+\frac{1}{2!}f^{\prime\prime} (x_0)(\delta x)^2+\frac{1}{3!}f^{\prime\prime\prime} (x_0)(\delta x)^3\cdots.$$

where the prime notation indicates a derivative. This expansion is useful when we know \(f\) and its derivatives at \(x_0\) but not \(f(x_0+\delta x\).

(Expressed in words, this series of terms means simply that near \(x_0\), any smooth function takes a value similar to that at \(x_0\) but with a slope-based linear correction for the distance between \(x_0\) and \(x_0+\delta x\), and we might add a slight correction for the quadratic curvature of \(f\), and maybe the tiniest correction for any additional cubic character, and so on.)

The smaller \(\delta x\) is, the fewer terms we need for a good approximation. In this case, let’s evaluate \(\arctan(1-2\varepsilon_{1^\prime 1^\prime})\) with just two terms:

$$\arctan(1-2\varepsilon_{1^\prime 1^\prime})\approx\arctan(1)+\frac{1}{1+1^2}(-2\varepsilon_{1^\prime 1^\prime})=\frac{\pi}{4}-\varepsilon_{1^\prime 1^\prime}$$

because the derivative of \(\arctan(x)\) is \(1/(1+x^2)\). Thus, the decrease in the 45° angle, corresponding to the tensorial shear stress and half the engineering shear stress, is \(\varepsilon_{1^\prime 1^\prime}=(1+\nu)\sigma_{1^\prime 1^\prime}/E=(1+\nu)\sigma_{12}/E\).

Since this relation can be applied in the 1-2, 1-3, and 2-3 directions, we have three new equations:

$$\varepsilon_{12}=\frac{1+\nu}{E}\sigma_{12}; $$

$$\varepsilon_{13}=\frac{1+\nu}{E}\sigma_{13}; $$

$$\varepsilon_{23}=\frac{1+\nu}{E}\sigma_{23}. $$

Generalized Hooke’s Law, then, needs to encompass all of this information (six equations coupling normal and shear stresses with normal and shear strains in 3D) in a single equation, and it does so with the help of two notational tools:

1. The repetition of a dummy index within a certain term implies summation over the range of indices; i.e., \(\sigma_{kk}=\sigma_{11}+\sigma_{22}+\sigma_{33}\), for example. (This is called Einstein summation.)

2. The Kronecker delta \(\delta_{ij}\) is 1 if \(i=j\) and 0 otherwise.

Thus, we can write

$$\varepsilon_{ij}=\frac{1+\nu}{E}\sigma_{ij}-\frac{\nu}{E}\sigma_{kk}\delta_{ij}.$$

Take a second to recognize how this equation simplifies in the normal strain case (when \(\delta_{ij}=1\)) and in the shear strain case (when \(\delta_{ij}=0\)) to obtain each of the six individual equations above.

Furthermore, these six equations can also be solved for strain instead of stress, giving

$$ \sigma_{ij}=\frac{E}{1+\nu}\varepsilon_{ij}+\frac{\nu E}{(1+\nu)(1-2\nu)}\varepsilon_{kk}\delta_{ij},$$

which we’ll also find useful.

Why aren’t shear strains dependent on normal stresses (or normal strains on shear stresses)?

In the above derivation, normal strains were linked to normal stresses and shear strains to shear stresses, but no cross coupling was discussed. Here’s why. If a normal stress were to cause a shear strain, for example, then we’d expect something like the state of loading and deformation shown below in (a), where the purple dot is used to mark the orientation of the differential element. We could always flip this picture upside down to obtain (b), of course. But since an isotropic material looks the same for all orientations, we should instead expect that the flipped material gives the same shear response as in (a), which is shown in (c). This paradox—the fact that (b) and (c) don’t match—leaves us to conclude that there cannot be any shear strain in this case.

Similar symmetry arguments, involving reflection and rotation, can be applied for various other circumstances and crystal structures to helpfully identify elastic coupling constants that must be zero. In fact, if we briefly drop the condition of isotropy, as assumed elsewhere in this note, a still broader framework applies: \(\boldsymbol{\sigma}=\boldsymbol{C}\boldsymbol{\varepsilon}\), where \(\boldsymbol{C}\) is a fourth-order (!) tensor, called the stiffness tensor, that couples the two second-order tensors of 3D stress and 3D strain. The dimensionality being 3, the fourth-order stiffness tensor has \(3^4=81\) components. Symmetry arguments can be used to show that most of the elements of \(\boldsymbol{C}\) are zero for isotropic materials. For example, \(C_{1112}=0\) because the normal stress \(\sigma_{11}\) is independent of the shear strain \(\varepsilon_{12}\), as shown above through reflection/mirror arguments. What about \(C_{1111}\), which couples the normal stress \(\sigma_{11}\) to the normal strain \(\sigma_{11}\)? What about \(C_{1122}\)? \(C_{2113}\)?

Relating stiffness to the speed of sound (and other seismic waves) in a material

Let’s work out a scaling relation between the stiffness of a material and the speed of an internal wave.

1. Think of a material as a cubic lattice of atoms, each with mass \(m\), connected by springs, each with spring constant \(k\) and original length \(d\), corresponding to the equilibrium atomic spacing (thus, we can expect a density of \(\rho\sim m/d^3\)):

   (Graphic)

2. Now, a spring constant couples a force \(F\) with a displacement \(x\) as \(F=kx\), so let’s divide this equation by \(d^2\) to obtain \(\sigma\sim(k/d)\varepsilon\) (because a force divided by a cross-sectional area gives an effective stress, and a displacement divided by an original length gives an effective strain). This looks like another constitutive stress–strain equation, of course, with an effective modulus of \(E^\prime\sim k/d\).

3. Recall that the frequency of a spring-mass system is \(\sqrt{k/m}\), which we can turn into a velocity by multiplying by a characteristic length (here, our only choice is \(d\), so \(v\sim d\sqrt{k/m}=\sqrt{E^\prime d/m}=\sqrt{E^\prime\rho}\)). Thus, we obtain a scaling relation between the velocity \(v\) of a certain wave mode and the corresponding effective modulus \(E^\prime\). (We also obtain a way to evaluate elastic moduli using time-of-propagation measurements through various materials.)

Sure, this scaling relationship is hand-wavy (because bonded atoms aren’t masses on springs, and almost no materials have a simple cubic structure, and a wave speed isn’t just the maximum speed of a simple oscillator, among other reasons.) But it’s a start. These are the actual definitions used by seismologists and other practitioners. As for using the idea of using springs in a thought experiment to explain the dynamic behavior of spring-like materials, a Richard Feynman comment to a reporter (in the context of discussing magnetic attaction and repulsion) is perhaps relevant:

I can’t explain that attraction in terms of anything else that’s familiar to you. For example, if we said the magnets attract like as if rubber bands, I would be cheating you. Because they’re not connected by rubber bands. I’d soon be in trouble. And secondly, if you were curious enough, you’d ask me why rubber bands tend to pull back together again, and I would end up explaining that in terms of electrical forces, which are the very things that I’m trying to use the rubber bands to explain. So I have cheated very badly, you see.