Room-and-Pillar Design for a Uranium Orebody: 24-MMP-B1 Applied Rock Mechanics, May 2014, Q4

Question 4: Room-and-Pillar Design for a Uranium Orebody (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.

4.1 — Factor of safety of the planned layout

Given. Depth H = 1000 m; overburden unit weight γ = 25 kN/m³; room span W_o = 6.0 m; square pillar width w_p = 5.0 m; pillar (mining) height h = 3.0 m (full orebody thickness); $S=133.2h^{-0.75}w_p^{0.5}$ MPa.

Find. The tributary-area pillar stress, the pillar strength, and the resulting factor of safety.

Plan view of the room-and-pillar layout: 5×5 m square pillars on a 6 m room span (centre-to-centre pillar spacing 11 m).

Approach. Tributary-area theory assigns each pillar the full overburden load carried by its share of the plan area (pillar plus half the surrounding rooms); compare the resulting pillar stress to the empirical pillar strength.

Vertical (overburden) stress. $\sigma_v=\gamma H=25\times1000=25{,}000\ \text{kPa}=\boxed{25.0\ \text{MPa}}$.
Tributary-area pillar stress. For square pillars on a square room grid, $\sigma_p=\sigma_v\left(\dfrac{w_p+W_o}{w_p}\right)^2=25\left(\dfrac{5+6}{5}\right)^2=25(2.2)^2=\boxed{121.0\ \text{MPa}}$.
Pillar strength. $S=133.2\,h^{-0.75}w_p^{0.5}=133.2(3)^{-0.75}(5)^{0.5}=133.2(0.4387)(2.236)=\boxed{130.7\ \text{MPa}}$.
Factor of safety. $\mathrm{FS}=S/\sigma_p=130.7/121.0=\boxed{1.08}$ — below any acceptable design value (typically 1.5–2.0 for a permanent pillar layout), so the planned 5 m pillars are inadequate and must be redesigned.

4.2 — Redesign for FS = 1.6 at maximum extraction

Given. Target factor of safety 1.6; same depth, unit weight, pillar height and strength formula as above.

Check: the redesign holds the 6.0 m room span fixed (set by equipment/ventilation access, not by the strength calculation) and solves for the smallest pillar width that meets FS = 1.6 — the smallest such pillar is exactly the layout that maximises the extraction ratio, since a larger pillar than necessary only sacrifices ore for no safety benefit and a smaller one violates the target FS.

Find. The redesigned pillar width and the resulting extraction ratio.

Approach. Both the tributary stress and the pillar strength are functions of pillar width w_p (room span W_o=6 m fixed); solve $S(w_p)/\sigma_p(w_p)=1.6$ for the minimum w_p, then compute the extraction ratio.

Set up the governing equation. $\dfrac{133.2(3)^{-0.75}w_p^{0.5}}{25\left(\dfrac{w_p+6}{w_p}\right)^2}=1.6$. Solving numerically (bisection) gives $\boxed{w_p=6.47\ \text{m}}$ (verified: at this width $S=133.2(0.4387)\sqrt{6.47}=148.5$ MPa, $\sigma_p=25\left(\frac{12.47}{6.47}\right)^2=92.8$ MPa, ratio $=1.60$).
Extraction ratio, before and after. Extraction ratio $e=1-\left(\dfrac{w_p}{w_p+W_o}\right)^2$. Original (w_p=5 m): $e=1-(5/11)^2=\boxed{79.3\%}$ at FS=1.08 (unsafe). Redesigned (w_p=6.47 m): $e=1-(6.47/12.47)^2=\boxed{73.1\%}$ at FS=1.60 (safe).
Justification of assumptions. Increasing the pillar width (rather than narrowing the rooms) is the only lever that increases both S (via $w_p^{0.5}$) and reduces σ_p simultaneously, and it preserves the 6 m room span needed for equipment access; this necessarily sacrifices extraction ratio (from 79.3% to 73.1%) to buy the required safety margin — the two objectives (maximum extraction, FS≥1.6) are in direct tension, and 73.1% is the highest extraction ratio achievable subject to the FS constraint and the fixed room span.

Question 4 — summary
Quantity	Original layout (w_p=5 m)	Redesigned layout (w_p=6.47 m)
Pillar stress, σ_p	121.0 MPa	92.8 MPa
Pillar strength, S	130.7 MPa	148.5 MPa
Factor of safety	1.08 (inadequate)	1.60 (target met)
Extraction ratio	79.3%	73.1%

Topic. Tributary-area pillar design and the extraction-ratio/safety-factor trade-off in room-and-pillar mining.

Key relations. Tributary stress $\sigma_p=\sigma_v\left(\frac{w_p+W_o}{w_p}\right)^2$; empirical pillar strength $S=133.2h^{-0.75}w_p^{0.5}$; extraction ratio $e=1-(w_p/(w_p+W_o))^2$.

Why this works. Tributary-area theory assumes each pillar carries the full weight of the overburden column above its own plan area plus half of each adjoining room — a reasonable approximation for a regular, extensive room-and-pillar layout away from abutments. Because pillar strength scales only with the square root of width while tributary stress scales with the square of the (w_p+W_o)/w_p ratio, a fairly large increase in pillar width is needed to close a modest FS shortfall, and extraction ratio always falls as FS is increased this way.

Common pitfall. Using the ORIGINAL orebody (in-situ) thickness for pillar height when part of the seam is left unmined, and confusing depth-based σ_v (a vertical stress) with the compressive pillar strength S (measured on a specimen/pillar of given height-to-width ratio) — the empirical strength formula already embeds the shape (h,w_p) effect, so h must be the actual mined pillar height, not a nominal specimen size.

Source. Brady & Brown, Rock Mechanics for Underground Mining, 3rd ed., ch. 10 (pillar design, tributary-area theory, empirical pillar-strength formulae).

Question 4 of 6: Room-and-Pillar Design for a Uranium Orebody

Question 4: Room-and-Pillar Design for a Uranium Orebody (20 marks)

4.1 — Factor of safety of the planned layout

4.2 — Redesign for FS = 1.6 at maximum extraction