Uniaxial Strain and the Resting Earth
Learning objectives
- Explain why laterally confined burial generates horizontal stress at all
- Derive the K0 ratio: effective horizontal over effective vertical stress equals nu over one minus nu
- Compute the canon example: nu of one quarter gives K0 of exactly one third, 12.5 MPa from 37.4
- Say honestly where the elastic K0 story fails, and why measured stresses usually sit above it
Why Is There Horizontal Stress?
Gravity pushes down. Why should a rock at depth be squeezed from the side? The answer is Poisson's ratio plus confinement. A rod loaded from above with free sides bulges sideways as it shortens. A rock element in a laterally extensive basin has no such freedom: its neighbors are trying to bulge into it exactly as hard as it bulges into them, so the net lateral strain of every element is zero. The element still wants to expand sideways; forbidding the strain requires a stress. That geometry, vertical load with zero lateral strain, is called uniaxial strain, and it is the natural resting state of a quiet sedimentary basin.
Run Hooke's law with the lateral strains pinned to zero and the required side stress falls out: , the ratio known as . It applies to effective stresses, the ones the frame feels, which is why the pore pressure of Part 1.5 rode along silently. For the canon , exactly: a third of the effective overburden comes back as horizontal squeeze.
Load the piston in the figure and watch the side gauges climb. At the canon depth the effective overburden is 37.4 MPa (hydrostatic pressure), so the elastic resting earth predicts an effective horizontal stress of about 12.5 MPa. Slide toward 0.5, the fluid limit, and runs to one: a material with no shear stiffness pushes sideways as hard as it is pushed down, which is just the hydrostatic state of a fluid seen from a new angle. Slide it toward zero and the side stress dies: cork, famously near zero , is easy to push into a bottle for exactly this reason.
Where the Story Fails, Honestly
Now the disclaimer that separates a teaching model from a field method. The elastic derivation assumes the basin is a linear elastic body that has never done anything but subside quietly. Real rocks creep over geologic time, relaxing toward stresses higher than the elastic ratio; tectonics adds sideways push the model has no slot for; faulting caps what the crust can carry, as Part 5 will show; and unloading by erosion leaves horizontal stresses locked in. The upshot, borne out by measurement after measurement: real minimum horizontal stresses usually sit well above the elastic K0 line. Our canon of 46 MPa total, 10.7 effective, is essentially at the elastic prediction, marking it as a relatively relaxed basin, but the number the course trusts is the one the LOT ledger of Part 8 measures. Use K0 to understand why horizontal stress exists; never to predict its value unmeasured.
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.