Question 4 of 5: Tributary-Area Pillar Stress and Elastic Deformation in a Potash Mine

Question 4: Tributary-Area Pillar Stress and Elastic Deformation in a Potash Mine (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Depth $H=335$ m, extraction ratio $r=0.67$, hanging-wall unit weight $\gamma=25.89\ \text{kN/m}^3$, potash $E=13.8$ GPa, $\mu=0.40$, seam (pillar) height $h=5.0$ m, square pillar side $W=7.5$ m; pillars unconfined at all times (no lateral confinement, before or after mining).

Find. (a) Pre- and post-mining vertical stress, and pre-mining horizontal stress (full $K_0$ confinement). (b) Vertical shortening of each pillar from the roof contact to mid-height under the stress increase. (c) Average total transverse (lateral) expansion of the pillar from the same stress increase.

Approach. Pre-mining vertical stress is the overburden weight; pre-mining horizontal stress follows the elastic, laterally-confined ($K_0$) condition $\sigma_h=\sigma_v\,\mu/(1-\mu)$. Post-mining pillar stress follows the tributary-area method, $\sigma_p=\sigma_v/(1-r)$, and because the pillars are stated to be unconfined at all times, the resulting axial and lateral strains from the stress INCREASE are computed as simple uniaxial elastic response ($\varepsilon_z=\Delta\sigma_v/E$, $\varepsilon_x=\mu\varepsilon_z$), not the confined ($K_0$) relation used only for the pre-mining state.

1. Part (a) — pre-mining stresses. $$\sigma_{v,\text{pre}}=\gamma H = (25.89\ \text{kN/m}^3)(335\ \text{m}) = 8673\ \text{kPa} = \boxed{8.673\ \text{MPa}}$$ $$\sigma_{h,\text{pre}}=\sigma_{v,\text{pre}}\frac{\mu}{1-\mu}=8.673\times\frac{0.40}{0.60}=\boxed{5.782\ \text{MPa}}$$
2. Part (a) — post-mining pillar stress (tributary area). $$\sigma_{v,\text{post}}=\frac{\sigma_{v,\text{pre}}}{1-r}=\frac{8.673}{1-0.67}=\frac{8.673}{0.33}=\boxed{26.28\ \text{MPa}}$$ With the pillars stated to be unconfined, the post-mining horizontal (lateral) stress on the pillar sides is $\boxed{\sigma_{h,\text{post}}=0}$ (atmospheric/free lateral boundary).
3. Part (b) — vertical shortening over the upper half-height. The relevant load is the INCREASE in vertical stress, applied to an unconfined (uniaxial) pillar, so $\varepsilon_z=\Delta\sigma_v/E$ with no lateral-confinement correction: $$\Delta\sigma_v=\sigma_{v,\text{post}}-\sigma_{v,\text{pre}}=26.28-8.673=17.61\ \text{MPa}$$ $$\varepsilon_z=\frac{\Delta\sigma_v}{E}=\frac{17.61}{13\,800}=1.276\times10^{-3}$$ Over the upper half-height (roof contact to pillar mid-height, $h/2=2.5$ m): $$\delta_v=\varepsilon_z\times(h/2)=1.276\times10^{-3}\times2.5\ \text{m}=\boxed{3.19\ \text{mm}}$$
4. Part (c) — transverse expansion. For an unconfined element under axial compression, the lateral strain follows Poisson's effect, $\varepsilon_x=\mu\,\varepsilon_z$: $$\varepsilon_x=0.40\times1.276\times10^{-3}=5.104\times10^{-4}$$ $$\Delta W=\varepsilon_x\times W=5.104\times10^{-4}\times7.5\ \text{m}=\boxed{3.83\ \text{mm}}$$