Rock Mechanics and Geomechanics

Pillar stress and extraction ratio

Extraction ratio R is the fraction of the seam area removed in a bord and pillar layout, so the pillar area fraction is (1-R). Under tributary area theory, each pillar carries the in situ vertical stress \gamma H acting over the whole panel, which gives the average pillar stress \sigma_p=\gamma H/(1-R). This relation is convex: pillar stress rises gently at low extraction and grows steeply as R approaches 1, so a straight line or a decreasing curve cannot represent it. The tributary area assumption also implies a nearly uniform vertical stress across the width of each pillar, which is why the flat stress profile is the one matched to this method. Changing the extraction ratio from R_1 to R_2 multiplies the pillar stress by (1-R_1)/(1-R_2), so a rise in extraction from 0.15 to 0.25 increases the pillar stress by about 13.3%. The safety factor is the pillar strength divided by the pillar stress, SF=S_p/\sigma_p. For a rectangular pillar of width w, length l and height h, an empirical strength formula S_p=S_1(0.64+0.54\,w/h-0.18\,w^2/(lh)) is used, and for a square pillar it reduces to S_1(0.64+0.36\,w/h), the Bieniawski form. Splitting a pillar into halves lowers the strength factor relative to the original square pillar, so the safety factor of each half falls even though the load per unit area is redistributed. When a panel is split into several stooks with galleries between them, the reduction in safety factor can exceed half of the original value, so the gallery width and the centre-to-centre spacing must be used consistently when computing the new pillar width. For a target safety factor, the extraction ratio is found by equating p/(1-R) to S_p/SF, where p=\gamma H. For example, with p=5 MPa and S_p=10 MPa, a safety factor of 1.25 requires \sigma_p=8 MPa and therefore R=0.375.

Elastic stresses around circular opening

For a circular opening in an elastic medium, the boundary stresses follow Kirsch's solution. In a hydrostatic field of magnitude p_0, the radial displacement at the wall of radius a under internal support pressure p_i is u_a=(1+\nu)a(p_0-p_i)/E. This relation shows that the support pressure needed to limit wall convergence increases with the in situ stress and the radius and decreases with the modulus, so a stiffer rock needs less support for the same allowable closure. In a biaxial field with vertical stress p and horizontal stress kp, the hoop stress at the wall is \sigma_\theta=(1+k)p-2(1-k)p\cos 2\theta, where \theta is measured from the vertical axis. The ratio of tangential stresses at two boundary points therefore depends only on k, so a measured ratio can be inverted to find the lateral stress ratio. Tangential stress at one boundary point and at another point at a different angular position are linked by the same \cos 2\theta dependence, so a known value at one point allows the value at another to be obtained. In a hydrostatic field the boundary tangential stress is uniform and equal to twice the in situ stress, which is a useful check on any boundary stress calculation.

Unit weight and moisture content

Moisture content is defined as the ratio of water mass to solid mass, w=W_w/W_s, and the total unit weight is related to the dry unit weight by \gamma=\gamma_d(1+w). Hence \gamma_d=\gamma/(1+w). For example, a sample at 10% moisture with total unit weight 16.5 kN/m^3 has a dry unit weight of 15.0 kN/m^3, so moisture always makes the total unit weight larger than the dry value. For an irregular rock lump, Archimedes' principle gives the volume as the volume of displaced liquid, V=m_d/(SG_l\,\rho_w), where m_d is the displaced liquid mass and SG_l its specific gravity. The unit weight then follows as \gamma=(m/m_d)\,SG_l\,\gamma_w, with \gamma_w=9.81 kN/m^3.

Effective stress in saturated rock

Terzaghi's effective stress principle states that \sigma'=\sigma-p, where p is the pore pressure. Full saturation under pore pressure p reduces both principal effective stresses by p, so the Mohr circle of a dry rock translates along the normal stress axis without changing its diameter, because the principal stress difference is unaffected by pore pressure. In undrained triaxial tests the total stress is set by the cell pressure and the axial loading, so any pore pressure developed in the specimen directly reduces its effective stress and strength. A specimen that fails at a lower deviatoric stress than another under the same confining pressure must have developed a higher pore pressure.

Rock failure criterion and safety factor

The safety factor against failure is commonly defined as the major principal stress at failure divided by the actual major principal stress, both taken at the same minor principal stress, FS=\sigma_{1f}/\sigma_1. A stress state lying inside the failure envelope has FS>1, while a state lying outside it has FS<1, so the envelope gives a direct check on stability. For a sliding block on a single joint, the safety factor is the resisting force divided by the driving force. For a dry slope with no tension crack or water pressure, FS=(cL+N\tan\phi)/(W\sin\psi_p), where N=W\cos\psi_p, W is the block weight, \psi_p is the joint dip and L is the joint length.

Mohr-Coulomb failure envelope

The Mohr-Coulomb criterion is \tau=c+\sigma_n\tan\phi. In principal stress form, \sigma_1=\sigma_3\dfrac{1+\sin\phi}{1-\sin\phi}+2c\dfrac{\cos\phi}{1-\sin\phi}. The ratio N=(1+\sin\phi)/(1-\sin\phi) is inverted by \sin\phi=(N-1)/(N+1), so the friction angle follows directly from the slope of the principal stress plot. In a triaxial test the failure plane is inclined at 45^\circ+\phi/2 to the minor principal stress direction, which is the standard Coulomb shear plane geometry. When the envelope is given in \tau-\sigma_n form, the intercept c and the slope \tan\phi can be read directly and used to find the shear stress on the failure plane at a given confining pressure, with \tau=\tfrac{\sigma_1-\sigma_3}{2}\cos\phi at failure. The normal stress on that plane is then checked by substituting back into the envelope equation.

Maximum shear stress from principal stresses

In two dimensions the maximum shear stress is \tau_{max}=(\sigma_1-\sigma_3)/2, acting on planes inclined at 45^\circ to the principal planes. When the minor principal stress is tensile and equal to a fraction of the major one, the maximum shear stress grows with the difference. For \sigma_3=-\sigma_1/2, \tau_{max}=\tfrac{3}{4}\sigma_1. The normal stress on the plane of maximum shear stress is (\sigma_1+\sigma_3)/2. When \sigma_3=-\sigma_1, the maximum shear stress equals \sigma_1 and the normal stress on that same plane is zero.

Lauffer stand-up time chart

The modified Lauffer chart relates the unsupported span of an excavation, the rock mass rating RMR and the stand-up time, which is the period an opening stays stable without support. Stand-up time increases with RMR and decreases with span, so for a given span a higher RMR gives a much longer stand-up time. The increase is steep, so reading the chart at two RMR values for the same span gives a large multiplying factor in stand-up time, and the chart is used as an approximate guide rather than an exact formula.

Borehole extensometer displacement readings

A multi-point borehole extensometer records the displacement of each anchor relative to the collar. Comparing the increments of each anchor between two dates shows how the rock between the collar and each anchor is moving. A discontinuity lies between two anchors where the differential movement, the difference in their incremental displacements, is significantly larger than elsewhere. Increments must be compared rather than absolute readings, because absolute readings include the common movement of the whole borehole column.

Uniaxial compressive strength failure plane stresses

In a uniaxial compressive test the axial stress equals the peak load divided by the cross-sectional area A=\pi D^2/4. If \theta is the angle between the normal to the failure plane and the axial direction, the normal stress on the plane is \sigma_n=\sigma_1\cos^2\theta and the shear stress is \tau=\sigma_1\sin\theta\cos\theta. For a failure plane inclined at 60° to the horizontal, the normal stress is one quarter of the axial stress, and the shear stress is larger than the normal stress in magnitude.