# 1.2: Elastic Deformation

The deformation recorded in the rock record provides a direct measurement of the total strain that the rock has undergone (if we can infer something about its original size). However, it is important to note two things. First, knowing the total strain does not give us information about the strain path. For example, I can stretch and compress a spring many times and then stop. I can then measure the final strain, but that measurement does not tell me how much I stretched and compressed the rock beforehand. In other words, we know the final strain but not the strain (deformation) history.

Second, rocks record permanent deformation. Again thinking about the spring. I can stretch and compress the spring many times, but it always returns to its original shape. There is no permanent deformation so there is no record of the deformation. However, if pull on the spring very hard I can cause it to stay partly stretched even when I let go. This is permanent deformation, and in the case of the spring it occurs when the force applied causes the spring to deform plastically. The same is true for rocks, low stress in a rock causes elastic deformation that is not permanent (instead, it is recoverable). This elastic deformation is not recorded in the rock record. The deformation recorded in the rocks occurs when the rocks break (fail) or flow (a type of crystal-scale failure)

Unlike the geologic record, when rocks are deformed in a laboratory experiment or when computer simulation are used to model the deformation, we can record (or impose) the stress in the rock and we can relate the applied stresses to the observed strain. We will not go into a full description of how this is done, but we will touch on a few key aspects that help a geologist think about the state of stress in the crust or lithosphere. Elastic deformation is the dominant form of deformation at shallow depths in the crust and lithosphere because both the temperature and the pressure is low. However the crust and lithosphere are also brittle and when the stress is big enough, failure occurs. Viscous deformation occurs deeper and at much higher pressures and temperatures than elastic deformation.

## Elastic Deformation

While the spring is an appropriate model for linear strain, in order to consider the three-dimensional deformation of rocks, we need to move to consider a block of material or a volume of rock. For the spring the elastic response is determined by the spring constant (the elasticity of the spring). For the rock, we need (at least) two parameters: the Young's modulus and Poisson's ratio.

Young's modulus, $$E$$, quantifies the relationship between an applied linear stress $$\sigma_n$$ and the resulting linear strain $$\epsilon_n$$ when no other stresses act on the material:

$E=\frac{\sigma_n}{\epsilon_n}$

Since stress has units of Pascals and strain is dimensionless, Young's modulus has the same units as stress (Pa). Essentially, it measures the ability of an object to resist deformation when a force is applied. It determines to what degree a material will deform in response to elastic stress. Young's modulus for rocks is measured in the laboratory by taking small cylinders of rock, compressing them in a vice with a known force and measuring the change in length. The stress and strain are determined (knowing the area of the top and bottom of the cylinder and its length) and then Young's modulus is calculated. Note that when this experiment is done there is no stress acting on the curved sides of the cylinder, only on the top and bottom. This state of stress is referred to a uni-axial. However, note that even though there is no stress on the sides of the cylinder, there is a strain perpendicular to the sides because the compression of the cylinder in one direction causes it to bulge out in the perpendicular directions. The amount that the material bulges out is determined by the second elastic property, the Poisson's ratio.

Poisson's ratio, $$\nu$$, quantifies the amount a material deforms in one direction relative to the amount it deforms in the perpendicular direction:

$\nu=\frac{-\varepsilon_{\perp}}{\varepsilon_{\parallel}}$

where , $$\varepsilon_{\perp}$$ is referred to as lateral strain and $$\varepsilon_{\parallel}$$ is referred to as the longitudinal strain.

Note from the above figure, we can say that:

$e_n=\frac{l-L}{L}=\frac{\Delta L}{L}\; and \; e_n=e_{\parallel}$

## Elastic Deformation in 3D

The elastic response of an applied stress in 2D can be extended to 3D. Here, for simplicity, we will only consider the normal (linear) components of strain and stress. Consider the normal (linear) strain in one direction resulting from normal stresses applied in three orthogonal directions. As noted above, for an elastic solid, a stress applied in one direction will also cause a strain in the perpendicular directions. Therefore, the linear strain in the x direction depends on the strain caused by the stress in the x direction, but it also depends on the strain caused by the stresses acting in the y and z direction also. Therefore, the components of linear (normal) strain in an elastic solid, acting in three orthogonal directions are given by:

$\varepsilon_{xx}=\left(\frac{1}{E}\right)(\sigma_{xx}-\nu(\sigma_{yy}+\sigma_{zz}))$

for the x direction,

$\varepsilon_{yy}=\left(\frac{1}{E}\right)(\sigma_{yy}-\nu(\sigma_{zz}+\sigma_{xx}))$

for the y direction, and

$\varepsilon_{zz}=\left(\frac{1}{E}\right)(\sigma_{zz}-\nu(\sigma_{xx}+\sigma_{yy}))$

for the z direction.

In these three equations, the subscript $$xx$$ indicates a strain or stress in the x direction acting perpendicular to a plane whose normal directions is also in the x directions. Therefore, a repeated index indicates normal (linear) components of deformation. Likewise, a subscript $$xy$$ would indicated a shear strain or shear stress acting in the x direction, but on a plane whose normal points in the y direction.

Using lithostatic Stress to Calculate Strain

We want to calculate the strain in the z direction resulting from the lithostatic stress. Let's assume the rock has elastic properties with $$E=5$$ GPa, $$\nu=0.25$$, and the lithostatic stress is $$\sigma_{lith}=200$$, MPa.

The strain in the vertical, z, direction depends on the stress applied in all three directions.

$\varepsilon_{zz} =[\frac{1}{E}](\sigma_{zz}-\nu(\sigma_{xx}+\sigma_{yy}))$

In the absence of any applied tectonic stress, the pressure at a depth of z is equal to the lithostatic stress. Therefore, the normal stresses in 3 orthogonal directions will all be equal to the lithostatic stress. That is $$\sigma_{zz} = \sigma_{xx} = \sigma_{yy}$$.

\begin{align*} \varepsilon_{zz} &=\frac{1}{E}(\sigma_{zz}-\nu(\sigma_{xx}+\sigma_{yy})) \\[4pt] &= \sigma_{lith}(\frac{1-2\nu}{E}) \\[4pt] &= (200\times10^6) (\frac{1 - 2 (0.25)}{5 \times 10^9}) \\[4pt] &=20 (\frac{10^6}{10^9}) \\[4pt] &=0.02 \end{align*}

The strain in the vertical direction resulting from the lithostatic stress is 0.02. What does this mean? If we started out with a length of $$L = 100$$ m, multiplying by 0.2 we find that the change in length is 2 m. Relative to 1000 m, 2 m is small, but not insignificant. The rock is actually shortened. This means that processes, such a erosion, which act to decrease the lithostatic stress result in an expansion of the rock beneath as it is unloaded. Likewise a process that adds to the lithostatic load, such as growth of a volcano, with compress the rock and cause it shorten.

## Tectonic Stress and Faults

Usually one can assume that the important stresses acting on a rock are vertical⇒lithostatic stress $$(\sigma_{lith}$$) and horizontal⇒tectonic stress $$(\sigma_{t}$$). Compression is a positive tectonic stress and extension is a negative tectonic stress.

Figure $$\PageIndex{3}$$: Positive and Negative Tectonic Stresses (MIB: Note need to correct figure: arrows for extension should point out)

The forces that create tectonic stresses are density anomalies in the mantle that drive convection and move the plates. $$\sigma_{H} > \sigma_{v}$$ in the case of compression and $$\sigma_{H} < \sigma_{t}$$ in the case of extension. Except in cases of very shallow rock, all the components of stress in the rock are positive due to the increase in the lithostatic stress with depth. Thus, at very shallow depths where the stress is in fact negative (causes opening of cracks), this means that the tectonic stress is either negative or it is positive but less than the magnitude of lithostatic stress at the depth.

In regions of shear deformation, there are two tectonic stresses, instead of just one in a normal stress region (extension or compression). In this case $$\sigma_{lith}$$ becomes an intermediate stress, with a magnitude between the other two stresses. A famous example of shearing is the San Andreas Fault (SAF).

Note in the SAF figure that on the fault there are normal stresses and shear stresses as well.

As drawn in the figure the three sets of stresses $$\sigma_{lith}, \sigma_{H_1}, \sigma_{H_2}$$ are the principal stresses. The principal stress directions are defined as three orthogonal directions for which there are no shear stresses acting on the planes perpendicular to these directions. Therefore, the principal stress are normal stresses: they act perpendicular to the plane. For any other plane in the rock, the magnitude of the normal and shear stresses will depend on the orientation of the plane. The principal stresses are a convenient (and actually complete) way to represent the state of stress in a rock. (Side note: for those who have had linear algebra, $$\sigma_{lith}, \sigma_{H_1}, \sigma_{H_2}$$ are principal stresses, which are defined by eigenvectors (direction) and eigenvalues (magnitude). )

When thinking about faulting, the faults can any orientation, and most of these orientation will result in both shear stress and normal stress acting on the fault plane. (There are 3 and only 3 possible plane orientation that will have a zero shear $$\sigma_{s}=0$$: the planes perpendicular to the principal stress directions). Since motion across a fault is caused by shear stress, the 3 planes with zero stress are not likely to be in the same orientation as active faults. Note, in the example above, that the San Andreas Fault is at an angle to the principal stress directions.