GLY 326 Structural Geology Lecture Stressed out

The Greek alphabet

As a vector, we could resolve the traction in two components σn (normal) and σs (shear) Fn Fs F θ

i.e. it means (if the body is not in accelerated motion), that for every free surface all the forces must sum zero. The same is true for the tractions (obvious innit?)

We now define the surface stress, or each of its components, as a pair of equal (in magnitude) but opposite (in direction) of tractions.

If the normal tractions point towards each other, they define compressive stress. If they point away from each other, then we are talking of tensile stress. We (geologists) consider compressive stresses as positive and tensile stresses as negative (conventional).

Regarding the traction shear components, they define shear stress that can be clockwise or counterclockwise. We consider clockwise shear couples positive and counterclockwise shear couples as negatives. Again conventional.

Traction = Force / Area Action-Reaction normal-shear A Fn Fs So per each coordinate…

In 2-D, we have this situation:

The 3-D situation:

Aij is some transformation matrix, expressing a rotation of the axes In 3D: σ1 > σ2 > σ3 𝛔1, maximum compressive stress 𝛔2, intermediate stress 𝛔3, maximum tensile stress Given the forces; there is a set of axes, the principal axes, in which only normal stresses need to be used to resolve them; one per each axis. Aij is some transformation matrix, expressing a rotation of the axes So… what’s the practicality here?

σ1 σ3 Given a traction… σn σs and a plane… Θ

Given a general stress in blue, red are principal stresses on the control volume, 𝓥; yellow, stresses resolved on an arbitrary surface, 𝓢, such as a fault 𝓥 𝓢

The reverse or inverse fault A crustal block will break according to the relative magnitude of the principal stresses. The break would tend to help it shorten in the sigma-1 direction for a reverse fault σ1 σ3

The normal fault σ1 σ3 Extend in the sigma-3 direction for a normal fault

The strike slip fault σ3 σ1 σ2 … And shorten in the sigma-1 direction or extend in the sigma-3 direction, depending….

Where is σ2?

σ1 σn σ3 σ3 σs Θ Θ =90: σ3=σn; σ1=σs Θ =0: σ1=σn; σ3=σs σ1

Fundamental Stress Equations Normal Stress Shear Stress σn = (σ1 + σ3) +(σ1 - σ3)cos 2Θ 2 2 σs = (σ1 - σ3)sin 2Θ 2

The Mohr-(Coulomb) circle

Let’s begin by defining the “Stress Space”: 2Θ σn

Physical and Stress Space The Mohr’s circle: σs σs σ1 σn 2Θ σ3 σ3 σn σ3 σ1 Θ σ1 Physical Space Stress Space

Differential Stress - diameter of circle 2Θ σn σ3 σ1 (σ1 - σ3)

Deviatoric Stress - radius of circle 2Θ σn σ3 σ1 (σ1 - σ3) 2

Mean Stress - center of circle 2Θ σn σ3 σ1 (σ1 + σ3) 2

The Mohr’s circle: σs σn(p), σs(p) σn(p) σs(p) 2Θ σn σ3 σ1 σ3 σ1 σn(p) = (σ1 + σ3) + (σ1 - σ3)cos 2Θ 2 σs(p) = (σ1 - σ3)sin 2Θ

σs σn(p), σs(p) σn(p) (σ1 - σ3)sin 2Θ 2 2Θ σn σ3 σ1 σn(p) = (σ1 + σ3) + (σ1 - σ3)cos 2Θ 2 σs(p) = (σ1 - σ3)sin 2Θ

σn(p) = (σ1 + σ3) +(σ1 - σ3)cos 2Θ σs(p) = (σ1 - σ3)sin 2Θ σs σn(p), σs(p) 2Θ σn(p) σn σ3 σ1 (σ1 - σ3)cos 2Θ 2 Difference between mean stress and normal stress on plane