Question 3 of 5: RMR and Q classification for two tunnel scenarios

Question 3: RMR and Q classification for two tunnel scenarios (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.

Find. RMR (Bieniawski 1989) and Q (Barton NGI) for each rock mass.

Approach. For (a), where RQD is not directly given, estimate it from the volumetric joint count $RQD=115-3.3J_v$ using the stated joint-set spacing; for (b), RQD is given directly. Rate each of RMR's five parameters against Table 1 (page 9 of the exam), apply the tunnel-orientation adjustment, and read the rock class off Table 1C. For Q, select $J_n$, $J_r$, $J_a$, $J_w$ and SRF from the Barton descriptors matching the stated joint-set count, surface condition and groundwater/stress setting, then combine as $Q=\tfrac{RQD}{J_n}\times\tfrac{J_r}{J_a}\times\tfrac{J_w}{SRF}$.

1. (a) Granite — RQD from volumetric joint count. Three joint sets each spaced 0.24 m give $J_v=3/0.24=12.5$ joints/m, so $$RQD=115-3.3(12.5)=\boxed{73.75\%}$$ which falls in Table 1's 50–75% bracket (rating 13).
2. (a) Granite — RMR. UCS 160 MPa → strength rating 12 (100–250 MPa band); spacing 240 mm → rating 10 (200–600 mm band); condition (“smooth… weathered with occasional stains”, between the slightly-rough/highly-weathered and slickensided brackets) → rating 20 (engineering judgement, flagged below); groundwater “wet, not dripping” → rating 7; orientation unstated → assume Fair ($-5$, per Table 1B). $$RMR=12+13+10+20+7-5=\boxed{57}\ \Rightarrow\ \text{Class III (Fair rock)}$$
3. (a) Granite — Q system. Three joint sets → $J_n=9$; smooth surfaces → $J_r=1.0$; slightly altered/stained, non-softening → $J_a=2.0$; wet not dripping → $J_w=1.0$; no abnormal stress at moderate depth → $SRF=1.0$. $$Q=\frac{73.75}{9}\times\frac{1.0}{2.0}\times\frac{1.0}{1.0}=\boxed{4.10}$$ Per the exam's own Rock Classes table (page 14), $Q=4$–10 is class C (“Fair”) — consistent with the RMR Class III result.
4. (b) Sandstone — RMR. UCS 85 MPa → rating 7 (50–100 MPa band); RQD 70% → rating 13; spacing 110 mm → rating 8 (60–200 mm band); condition (“slightly rough…slightly weathered…separation <1 mm”) matches Table 1's own wording exactly → rating 25; groundwater (tunnel 70 m below the water table, no inflow rate stated) assumed “Wet” → rating 7 (flagged below); orientation unstated → Fair ($-5$). $$RMR=7+13+8+25+7-5=\boxed{55}\ \Rightarrow\ \text{Class III (Fair rock)}$$
5. (b) Sandstone — Q system. Two joint sets plus random → $J_n=4$; slightly rough → $J_r=1.5$; unaltered walls, staining only, generally cleaned → $J_a=1.0$; submerged 70 m below the water table, no inflow-rate given → medium inflow/pressure assumed, $J_w=0.66$; SRF=1.0. $$Q=\frac{70}{4}\times\frac{1.5}{1.0}\times\frac{0.66}{1.0}=\boxed{17.3}$$ This lands in the $Q=10$–40 (“Good”) band — one notch better than the RMR Class III result, which is an expected feature of comparing two independently-calibrated empirical systems on judgement-heavy inputs (condition/water descriptors), not a computational inconsistency.