# 1.1: Stress and Strain (1-D)

When a force is applied to a rock, this produces stress. Stress is a force per unit area applied. In other words, stress is force divided by area, or $$\sigma=\frac{F}{A}$$. An important differentiation must be made between stress and traction. Stress is a tensor, that is it is two parameters being the intensity of the force and the plane on which that force is acting. However, traction is a vector because it has only one parameter, the intensity of the force on the particular plane.

In geology, stress is usually quantified using units of Pascals (Pa) which is the equivalent of one Newton per meter squared $$\frac{N}{m^2}$$ or in SI units $$\frac{kg}{ms^2}$$.

## Components of Stress

There are two types of stress, differentiated based on which direction the force is applied relative to the surface it is acting on. Normal stress acts at a right angle (perpendicular) to a surface, and is can either push a rock together (compression) or pull it apart (tension). In geology, compressional stress is positive and tensional stress is negative. In engineering, the opposite it sign convention is adopted. In geological settings compressional stress is typically found in convergent margins where tectonic plates collide. Compressional stress also results from the weight of the rock itself pushing down on the rock below it. Tensional stress is typically found in divergent margins where tectonic plates are pulling the plates apart. Figure $$\PageIndex{1}$$: Normal Stress (Tension) (CC BY-SA 3.0; Sanpaz, via Wikimedia)

The second type of stress is a shear stress, which is a stress that acts tangential (parallel) to a surface. If you slide your hand across the top of your desk, the resistance you feel is a shear stress between your hand and the desk. Similar if you slide two blocks past each other, the frictional resistance to sliding generates a shear stress. Shear stresses are typically found transform boundaries as tectonic plates slide past each other. The stress that resists slip on a fault is a shear stress. Figure $$\PageIndex{2}$$: Shear Stress

Generally speaking the the stress state at particular location in the crust, lithosphere or mantle, is not simply described by two opposing stresses, but rather a three-dimensional stress-state. To completely describe such a three-dimensional stress requires three sets of normal stress oriented in three orthogonal directions (i.e., x, y, z) and the six sets of shear stress acting the planes that are perpendicular to the normal stresses. There are six shear stresses because on one plane there can be a component of shear stress in the horizontal direction and a component shear stress in the vertical direction. Such a three dimensional stress-state can also be completely represented by special set of three orthogonal normal stress called principal stresses. The orientation of these three stress are special because in this orientation the shear stress components all have a value of zero (no shear stress). We will return these stress components in the next section.

If all three normal stress components are equal, the stress state is referred to as a pressure. (Think about being in the air, air presses in on you with the same stress from all different directions. This is why we talk about the pressure in the air.) In geology, when there is no applied tectonic stress, the stress in the crust is due to the weight of the rock above that point. Because the rock around that point is pushing back with the same stress (assuming there is no active deformation), the magnitude of the stress is the same in all directions (there is a pressure). This pressure is referred to as the lithostatic stress or pressure (litho ⇒ rock and static ⇒ not moving).

Calculating Stress

Let's try another real world example to summarize our knowledge about stress. Looking at the figure below, let's calculate the stress 10 km below the surface. Figure $$\PageIndex{6}$$: Calculating Stress

First, we will assume that gravity is 10 $$\frac{m}{s^2}$$ and $$\rho$$ (density) is 2000 $$\frac{kg}{m^3}$$. We can derive an equation to calculate the stress through the following: the mass of the column of rock is m=v•$$\rho$$. The volume is length•width•height, thus m=LWH$$\rho$$. We also know that F=m•g, thus F=LWH$$\rho$$g. Now that we know the force and the area, we can build our stress equation.

\begin{align*} \sigma &=\frac{LWH\rho g}{LW} \\[4pt] &=H\rho g. \end{align*}

\begin{align*} \sigma &=2000 \frac{kg}{m^3} 10 \frac{m}{s^2}\, 10000 \,m \\[4pt] &=2\times 10^8 \frac{kg}{ms^2} \\[4pt] &=200\, MPa. \end{align*}

At 100 km, 10x the previous depth, the stress would be 2000 MPa or 2GPa.

## Strain

When this stress is applied to a solid material (a rock), the material deforms or squishes. Think of pressing your fingers with a sponge in between. This squishing is called strain and it is how we describe the internal deformation of rocks. Just like stress, there are two types of strain: normal (or linear) strain and shear strain. Figure $$\PageIndex{3}$$: Extensional Strain (CC BY-SA 3.0; Svjo (edited by Taryn Lausch), via Wikimedia)

Normal (linear) strain is likely already a familiar concept as it can be seen when a spring is stretched. The strain of a spring is described by noting the initial and final lengths of the spring. Thus, normal strain is the quantified as the change in length of the spring relative to is original length:

$e=\frac{l-L}{L}=\frac{\Delta L}{L}$

where $$L$$ is the original length of the material and $$l$$ is the final length. The greater the stretching of the spring, the larger the strain. Note that strain is a dimensionless quantity since both the numerator and denominator have the units of length. The strain is also often expressed as a percentage ($$100 \times e$$%). Note that the strain is positive if the spring lengthens and negative if it shortens.

Strain in rocks is used to quantify the magnitude and directions of permanent deformation recorded in rocks from the out-crop scale (centimenters) to the plate boundary scale (100s of kilometers). Measurements of change in the relative distance between two permanent GPS locations can be used to calculate the strain occurring in the rock in between.

The second type of strain is shear strain, which results from a shear stress. The shear strain is given by the displacement in one direction (x) occurring across a region in the other direction (y). In the figure below, and applied shear stress slides the top of the layer of thickness y, side-ways by an amount $$\Delta x$$. The shear strain is then given by,

$\epsilon_{s}=\frac{\Delta x}{y}$ Figure $$\PageIndex{4}$$: Shear Strain Variables

It is also common to see the shear strain define with respect the shear angle $$\gamma$$. The shear angle is defined as

$\tan{\gamma}=\frac{\delta_{x}}{L}$

where $$\gamma$$ is the change in the right angle of a square object, $$L$$ is the initial length of the object, and $$\delta_{x}$$ is the change in length of the object. Note, that different variable are used (L instead of Y, $$\delta_x$$ instead of $$\Delta_x$$, but the same lengths are being measure. It is common to see different variable use to denote these variables throughout the geology literature. Similarly, strain is sometimes denoted by the letter $$e$$ or by the greek symbol epsilon $$\epsilon$$. Figure $$\PageIndex{5}$$: Shear Strain

Note, that $$\delta x$$ is again the displacement perpendicular to and across the layer with thickness y. Therefore, $$\epsilon_s = \tan{\gamma}$$.

Convention states that if the angle between the x and y axes decreases, the shear strain is positive, and if the angle increases, the shear strain in negative. Like normal strain, shear strain is a ratio and is thus dimensionless. This type of strain is commonly seen is shear zones that occur deep within crust below faults where tectonic plates slide past each other.

Calculating Shear Strain

To better understand how to calculate shear strain, let's look at the following right lateral fault. Figure $$\PageIndex{5}$$: Right Lateral Fault

If $$\Delta x$$=2 km and y=1 km, then $$\epsilon_{s}=\frac{2}{1}$$ or $$\epsilon_{s}$$=2. The percent strain is then

$\epsilon_{s} \times 100\%=200\%. \nonumber$

Note: Units

A quick refresher:

kPa x103 kilo

MPa x106 mega

GPa x109 Giga