Principal Stresses and Rotation

Part 1, Part 1: Stress, the Tensor and the Circle

Learning objectives

  • Rotate a 2D stress state and read how each component transforms with the double angle
  • Find the principal directions, where the shear vanishes, and the principal stresses they carry
  • Compute the principal pair from center plus-or-minus radius and locate the angle from the double-angle arctangent
  • Say why the subsurface hands us a principal frame for free: one principal stress is very nearly vertical

Rotation and the Double Angle

Take the 2D state (sigmaxx,sigmayy,tauxy)(\sigma_{xx}, \sigma_{yy}, \tau_{xy})xx,sigmayy,tauxy) and rotate your axes by theta\theta. The components transform as sigmaxx′=tfracsigmaxx+sigmayy2+tfracsigmaxx−sigmayy2cos2theta+tauxysin2theta\sigma_{xx}' = \tfrac{\sigma_{xx}+\sigma_{yy}}{2} + \tfrac{\sigma_{xx}-\sigma_{yy}}{2}\cos 2\theta + \tau_{xy}\sin 2\thetaxx+sigmayy2+tfracsigmaxx−sigmayy2cos2theta+tauxysin2theta and tauxy′=−tfracsigmaxx−sigmayy2sin2theta+tauxycos2theta\tau_{xy}' = -\tfrac{\sigma_{xx}-\sigma_{yy}}{2}\sin 2\theta + \tau_{xy}\cos 2\thetaxx−sigmayy2sin2theta+tauxycos2theta. Notice the machinery runs on 2theta2\theta, not theta\theta: rotate the axes a quarter turn and the components complete a half cycle, swapping the two normal stresses and flipping the shear's sign. That doubled angle is not a curiosity; it is the reason the picture in the next section will be a circle.

The Shear-Free Frame

Because tauxy′(theta)\tau_{xy}'(\theta)xy′(theta) is a sinusoid, it must cross zero, and where it does the axes line up with the principal directions: the orientations whose planes feel pure push and no drag. Setting the shear to zero gives tan2thetap=dfrac2tauxysigmaxx−sigmayy\tan 2\theta_p = \dfrac{2\tau_{xy}}{\sigma_{xx}-\sigma_{yy}}xysigmaxx−sigmayy, and the normal stresses there are the extremes the state can offer, the principal stresses sigma1gesigma3\sigma_1 \ge \sigma_3, computed as center plus-or-minus radius: tfracsigmaxx+sigmayy2pmsqrtleft(tfracsigmaxx−sigmayy2right)2+tauxy2\tfrac{\sigma_{xx}+\sigma_{yy}}{2} \pm \sqrt{\left(\tfrac{\sigma_{xx}-\sigma_{yy}}{2}\right)^2 + \tau_{xy}^2}xx+sigmayy2pmsqrtleft(tfracsigmaxx−sigmayy2right)2+tauxy2.

Principal StressesInteractive figure, enable JavaScript to interact.

Run the rotation in the figure with the teaching state (30,10,10)(30, 10, 10). The shear curve crosses zero at \theta_p = 22.5^\circ, the normal components peak and trough at sigma_1=34.14\sigma_1 = 34.14 and sigma3=5.86\sigma_3 = 5.86 MPa, and the two curves' sum rides the flat line at 40 MPa the whole way around: the invariant again. Ninety degrees past the first principal direction sits the second, always; principal directions come as a perpendicular set.

Why the Earth Cooperates

Finding principal axes in general takes an eigenvalue computation. The subsurface, mercifully, hands us most of the answer: the ground surface can carry no shear traction, so near it, and to a good approximation well below it, one principal stress stands vertical and the other two lie horizontal. That single observation, formalized by Anderson in Part 5, is why this course can speak of SvS_vv, SHmaxS_{Hmax}Hmax, and ShminS_{hmin}hmin as the three subsurface stresses, and why measuring "the stress state" reduces to finding two horizontal magnitudes and one azimuth. The vertical one, as section 1.6 will show, you can compute from a density log before lunch.

References

  • Jaeger, J. C., Cook, N. G. W., & Zimmerman, R. W. (2007). Fundamentals of Rock Mechanics (4th ed.). Blackwell.
  • Anderson, E. M. (1951). The Dynamics of Faulting (2nd ed.). Oliver & Boyd.
  • Zoback, M. D. (2007). Reservoir Geomechanics. Cambridge University Press.

This page is prerendered for SEO and accessibility. The interactive widgets above hydrate on JavaScript load.