From Force to Traction to Stress

Part 0: The Bridge

Learning objectives

Distinguish a force from a traction: the same load spread over different areas does different things
Resolve the traction on a plane into its normal and shear components
Show that in a uniaxially loaded body the shear traction peaks on the 45 degree plane at exactly half the load
State what the stress tensor is: the machine that takes a plane orientation and returns the traction acting on it

A Force Needs an Area

Push on a rock with a thumbtack and with a thumb, same force, and the outcomes differ completely, because what rock feels is not force but force per area. That quantity, a traction when it acts on a surface, is the currency of this whole course, and its unit is the megapascal: one newton per square millimeter, about 145 psi. The numbers worth calibrating your intuition on: a car tire holds about 0.2 MPa, a firm handshake is a few tenths, and the rock at 3000 m in the previous section carries 67.7 MPa of overburden. Depth turns modest densities into enormous loads, and everything downhole lives at tens of megapascals.

Traction Depends on the Plane

Here is the subtlety that makes stress richer than pressure. In a fluid at rest, the push on a surface is the same from every direction. In a loaded solid it is not. Take a block squeezed along one axis by a stress $\sigma$ and slice it with an imaginary plane. If the plane faces the load squarely, it feels the full $\sigma$ , all of it normal, pressing the faces together. Tilt the plane and the traction on it both shrinks and rotates: part still presses across the plane, the normal component $\sigma_n = \sigma \cos^2\beta$ n=sigmacos2beta, and part now drags along it, the shear component $\tau = \sigma \sin\beta \cos\beta$ , where $\beta$ is the angle between the load axis and the plane's normal.

Rotate the plane in the figure and watch the split. The shear component starts at zero, rises, and peaks on the plane whose normal sits at $45^\circ$ to the load, at exactly half the applied stress: $\tau_{max} = \sigma/2$ max=sigma/2. That one identity is the seed of everything in Part 3, because rocks do not usually fail by being squeezed; they fail by sliding, and sliding is driven by shear. A rock that fails in compression almost always breaks on a plane oblique to the load, near the angle where shear is winning, tilted from 45 degrees only by friction's preference.

Stress Is the Machine

So the traction is not one number: every plane through the same point feels a different one. The object that holds all of them at once is the stress tensor. Feed it a plane orientation and it returns the traction vector on that plane; that is its whole job. In three dimensions it takes nine components, three tractions on three reference faces, and balance of moments makes it symmetric, leaving six independent numbers. Geomechanics adds one convention worth stating now: compression is positive, opposite to the elasticity texts, because in the crust essentially everything is being squeezed. Part 1 opens the machine properly, finds the three special planes that feel no shear at all, and draws the picture, the Mohr circle, that turns this algebra into something you can see.

References

Jaeger, J. C., Cook, N. G. W., & Zimmerman, R. W. (2007). Fundamentals of Rock Mechanics (4th ed.). Blackwell.
Fjaer, E., Holt, R. M., Horsrud, P., Raaen, A. M., & Risnes, R. (2008). Petroleum Related Rock Mechanics (2nd ed.). Elsevier.
Zoback, M. D. (2007). Reservoir Geomechanics. Cambridge University Press.