Rock MechanicsFREE· Reviewed Jun 2026

Geomechanics & Ground Control

Rock-mass classification (RQD, RMR, Q-system), the Mohr–Coulomb failure envelope, in-situ stress and pillar safety factors — the analytical core of every ground-control question.

Section 1

Recent Trend Analysis (2017–2026)

Geomechanics is the single largest scoring block in GATE MN, reliably worth 6–10 marks and spanning both 1-mark concept checks and the hardest 2-mark NATs on the paper.

The decade-long shift in question style:
- 2017–2019 — definition-heavy recall: state the RQD thresholds, the five RMR parameters, or the six Q-system factors. Pure-memory MCQs dominated.
- 2020–2022 — a move to single-formula computation: compute RQD from a core run, convert **

Q \leftrightarrow R M R

, or solve a one-step tributary-area pillar stress**.
- 2023–2026 — full multi-variable NAT territory: complete pillar safety-factor chains (in-situ stress → tributary pillar stress → empirical pillar strength → SF), Mohr–Coulomb principal-stress failure with the

N_{ϕ}

form, and stress concentrations around circular openings (Kirsch boundary stresses).

Exact recurring themes you must own:
- Pillar Safety Factor by the tributary-area method — the examiners' favourite 2-mark chain, with the

(W + B)^{2} / W^{2}

square-pillar trap.
- Mohr–Coulomb in principal-stress form

σ_{1} = σ_{3} N_{ϕ} + 2 c N_{ϕ}

N_{ϕ} = tan^{2} (4 5^{\circ} + ϕ /2)

.
- RQD from volumetric joint count

J_{v}

, and the **

R M R \approx 9 ln Q + 44

** bridge.
- Kirsch boundary stresses — roof/floor vs sidewall stress concentration in a biaxial field.
- A rising count of MSQs asking which support/classification statements are simultaneously valid — one wrong tick zeroes the mark (no partial credit).

Section 2

Master Formula Matrix & Derivations

Rock Quality Designation (RQD)

RQD is the percentage of a core run made up of **sound pieces

\geq 10 c m

** long, measured along the core axis — a direct index of fracturing.

R Q D = \frac{\sum ( core pieces \geq 10 c m )}{total core run length} \times 100 %

⚡ Exam shortcut ·

Without core, estimate from the volumetric joint count

J_{v}

R Q D = 115 - 3.3 J_{v}

(valid

4.5 \leq J_{v} \leq 30

; take

R Q D = 100

J_{v} < 4.5

and

R Q D = 0

J_{v} > 35

Variable Index (SI units)

Symbol	Meaning	SI unit
$R Q D$	Rock Quality Designation	$%$ (0–100)
$J_{v}$	Volumetric joint count (joints per $m^{3}$ )	$j o in t s / m^{3}$

Bieniawski RMR (1989)

The Rock Mass Rating sums five primary parameter ratings and applies a joint-orientation adjustment to grade the mass (Class I–V).

R M R = R_{U C S} + R_{R Q D} + R_{s p a c in g} + R_{co n d i t i o n} + R_{w a t er} + R_{or i e n t}

⚡ Exam shortcut ·

Classes:

R M R 81 - 100

= I (very good) …

< 21

= V (very poor). The orientation adjustment

R_{or i e n t}

is always

\leq 0

(it only penalises).

Variable Index (SI units)

Symbol	Meaning	SI unit
$R M R$	Total rock mass rating	dimensionless (0–100)
$R_{U C S}$	Rating for intact strength (0–15)	dimensionless
$R_{R Q D}$	Rating for RQD (3–20)	dimensionless
$R_{or i e n t}$	Joint-orientation adjustment ( $\leq 0$ )	dimensionless

Barton Q-system

The NGI tunnelling quality

Q

multiplies three ratios: block size, inter-block shear strength, and active stress.

Q = \frac{R Q D}{J _{n}} \times \frac{J _{r}}{J _{a}} \times \frac{J _{w}}{S R F}

⚡ Exam shortcut ·

Bridge to RMR with

R M R \approx 9 ln Q + 44

Q

spans

0.001

(exceptionally poor) to

1000

(exceptionally good) on a log scale.

Variable Index (SI units)

Symbol	Meaning	SI unit
$R Q D$	Rock Quality Designation	$%$
$J_{n}$	Joint set number	dimensionless
$J_{r}, J_{a}$	Joint roughness / alteration number	dimensionless
$J_{w}$	Joint water reduction factor	dimensionless
$S R F$	Stress Reduction Factor	dimensionless

Mohr–Coulomb Failure Criterion

Shear failure occurs when shear stress on a plane reaches cohesion plus frictional resistance; recast in principal stresses for the failure envelope.

τ = c + σ_{n} tan ϕ \Rightarrow σ_{1} = σ_{3} N_{ϕ} + 2 c N_{ϕ}, N_{ϕ} = tan^{2} (4 5^{\circ} + \frac{ϕ}{2})

⚡ Exam shortcut ·

Unconfined (

σ_{3} = 0

) strength is

σ_{c} = 2 c N_{ϕ}

. The failure plane makes an angle

(4 5^{\circ} + ϕ /2)

with the minor principal stress direction.

Variable Index (SI units)

Symbol	Meaning	SI unit
$τ$	Shear stress at failure	$P a$ (or $M P a$ )
$c$	Cohesion	$P a$ (or $M P a$ )
$σ_{n}$	Normal stress on the plane	$P a$
$ϕ$	Angle of internal friction	degrees (or $r a d$ )
$σ_{1}, σ_{3}$	Major / minor principal stress	$P a$
$N_{ϕ}$	Flow factor $tan^{2} (4 5^{\circ} + ϕ /2)$	dimensionless

In-situ Stress Field

Vertical stress is the weight of overburden; horizontal stress follows from lateral confinement (gravitational case) or tectonics.

σ_{v} = γ H = ρ g H, σ_{h} = K σ_{v}, K = \frac{ν}{1 - ν}

⚡ Exam shortcut ·

Rule of thumb:

σ_{v} \approx 0.025 H

M P a

for

H

in metres (

γ \approx 25 k N / m^{3}

K

can exceed 1 where tectonic stresses dominate.

Variable Index (SI units)

Symbol	Meaning	SI unit
$σ_{v}$	Vertical (overburden) stress	$P a$ (or $M P a$ )
$γ$	Unit weight of rock ( $= ρ g$ )	$N / m^{3}$ (or $k N / m^{3}$ )
$H$	Depth below surface	$m$
$K$	Horizontal-to-vertical stress ratio	dimensionless
$ν$	Poisson's ratio	dimensionless

Pillar Stress & Strength (Tributary Area)

Each pillar carries the overburden of its tributary area; empirical strength formulas relate that to the pillar's width-to-height shape.

σ_{p} = σ_{v} \frac{( W + B ) ^{2}}{W ^{2}}, S_{p} = S_{1} (0.778 + 0.222 \frac{W}{H _{p}})

⚡ Exam shortcut ·

For long rib pillars use the un-squared form

σ_{p} = σ_{v} \frac{W + B}{W}

. Extraction ratio

e = 1 - \frac{W ^{2}}{( W + B ) ^{2}}

, so

σ_{p} = \frac{σ _{v}}{1 - e}

Variable Index (SI units)

Symbol	Meaning	SI unit
$σ_{p}$	Average pillar stress	$M P a$
$σ_{v}$	Vertical in-situ stress $γ H$	$M P a$
$W$	Pillar width	$m$
$B$	Bord (opening) width	$m$
$H_{p}$	Pillar height	$m$
$S_{p}$	Pillar strength	$M P a$
$S_{1}$	Strength of a unit cube of pillar rock	$M P a$

Factor of Safety & Kirsch Boundary Stress

Stability is the ratio of strength to demand; around a circular opening in a biaxial field the boundary tangential stress concentrates at roof and ribs.

F . S . = \frac{S _{p}}{σ _{p}}; σ_{θ, r oo f} = 3 σ_{h} - σ_{v}, σ_{θ, r ib} = 3 σ_{v} - σ_{h}

⚡ Exam shortcut ·

Target

F . S . \geq 1.5 - 2.0

for long-term pillars. For a hydrostatic field (

σ_{h} = σ_{v} = P_{0}

) the boundary stress is a uniform

2 P_{0}

everywhere.

Variable Index (SI units)

Symbol	Meaning	SI unit
$F . S .$	Factor of safety (strength ÷ stress)	dimensionless
$σ_{θ}$	Tangential (hoop) boundary stress	$M P a$
$σ_{v}, σ_{h}$	Vertical / horizontal field stress	$M P a$

Section 3

The "IIT Trap" Warning System

**RQD counts only sound pieces $\geq 10 c m$ measured along the core centre-line**. Drilling-induced fresh breaks must be mentally re-joined before measuring — counting them as natural fractures deflates RQD and is a seeded distractor.
** $N_{ϕ}$ uses $(4 5^{\circ} + ϕ /2)$ in degrees.** Feeding $ϕ$ straight into $tan^{2} ϕ$ , or mixing radians and degrees, is the commonest Mohr–Coulomb error. Also note the failure plane angle is measured from the minor ( $σ_{3}$ ) stress.
**Square pillar uses $(W + B)^{2} / W^{2}$ , NOT $(W + B) / W$ .** The un-squared (rib-pillar) form is always offered as a wrong option for a square-pillar problem — and vice-versa.
Unit weight vs density. $γ = ρ g$ . With $γ$ in $k N / m^{3}$ and $H$ in $m$ , $σ_{v}$ comes out in $k P a$ — convert to $M P a$ (÷1000) before forming the safety factor. Using $ρ$ (in $k g / m^{3}$ ) as if it were $γ$ understates stress ~10×.
Width-to-height, not height-to-width. The strength formula uses $W / H_{p}$ . Inverting the ratio (using $H_{p} / W$ ) collapses the pillar strength and the answer.
** $K$ ratio direction.** $σ_{h} = K σ_{v}$ with $K = ν / (1 - ν)$ for the gravitational case; $K$ can exceed 1 under tectonic stress, so a horizontal stress larger than vertical is not automatically wrong.
RMR orientation adjustment is negative. $R_{or i e n t}$ only ever subtracts (favourable = 0, unfavourable < 0). Adding it as a positive number over-rates the mass.
Kirsch sign at the boundary. At the roof the tangential stress is $3 σ_{h} - σ_{v}$ — if $σ_{v} > 3 σ_{h}$ this is tensile (negative), the classic roof-failure trigger that an MCQ will probe.
Factor of safety threshold. $F . S . < 1$ means failure; a value just above 1 is not 'safe' for a long-life pillar (need $\geq 1.5 - 2.0$ ). Watch MSQs that call $F . S . = 1.1$ 'adequately stable'.

Section 4

High-Fidelity Core Examples

Example 12-mark complexity

A bord-and-pillar panel lies at a depth of

H = 200 m

in rock of unit weight

γ = 25 k N / m^{3}

. Square pillars are

W = 12 m

wide with bords

B = 6 m

wide; pillar height is

H_{p} = 3 m

. The strength of a unit cube of the pillar rock is

S_{1} = 30 M P a

(Obert–Duvall form). Determine the factor of safety of the pillars.

Given Parameters Matrix (clean SI)

Depth ( $H$ )	$200 m$
Unit weight ( $γ$ )	$25 k N / m^{3}$
Pillar width ( $W$ )	$12 m$
Bord width ( $B$ )	$6 m$
Pillar height ( $H_{p}$ )	$3 m$
Unit-cube strength ( $S_{1}$ )	$30 M P a$

Algebraic Derivation Track

Step 1 — Vertical in-situ stress:

σ_{v} = γ H = 25 \times 200 = 5000 k P a = 5.0 M P a

Step 2 — Tributary-area pillar stress (square pillars):

σ_{p} = σ_{v} \frac{( W + B ) ^{2}}{W ^{2}} = 5.0 \times \frac{( 12 + 6 ) ^{2}}{1 2 ^{2}} = 5.0 \times \frac{324}{144} = 5.0 \times 2.25 = 11.25 M P a

Step 3 — Pillar strength (Obert–Duvall), with

W / H_{p} = 12/3 = 4

S_{p} = S_{1} (0.778 + 0.222 \frac{W}{H _{p}}) = 30 (0.778 + 0.222 \times 4) = 30 \times 1.666 = 49.98 M P a

Step 4 — Factor of safety:

F . S . = \frac{S _{p}}{σ _{p}} = \frac{49.98}{11.25} = 4.44

🎯 Final target & accepted range ·

Factor of safety

F . S . \approx 4.44

(accept

4.3 - 4.5

). Well above the

1.5 - 2.0

long-term threshold, so the pillars are stable.

Example 22-mark complexity

A triaxial test on a rock gives cohesion

c = 8 M P a

and an angle of internal friction

ϕ = 3 0^{\circ}

. Using the Mohr–Coulomb criterion, find the major principal stress

σ_{1}

at failure for a confining pressure

σ_{3} = 5 M P a

Given Parameters Matrix (clean SI)

Cohesion ( $c$ )	$8 M P a$
Friction angle ( $ϕ$ )	$3 0^{\circ}$
Confining stress ( $σ_{3}$ )	$5 M P a$

Algebraic Derivation Track

Step 1 — Flow factor:

N_{ϕ} = tan^{2} (4 5^{\circ} + \frac{ϕ}{2}) = tan^{2} (4 5^{\circ} + 1 5^{\circ}) = tan^{2} 6 0^{\circ} = (3)^{2} = 3

Step 2 — Mohr–Coulomb principal-stress form:

σ_{1} = σ_{3} N_{ϕ} + 2 c N_{ϕ} = 5 \times 3 + 2 \times 8 \times 3

Step 3 — Evaluate (

3 = 1.7320

σ_{1} = 15 + 16 \times 1.7320 = 15 + 27.71 = 42.71 M P a

🎯 Final target & accepted range ·

Major principal stress at failure

σ_{1} \approx 42.7 MPa

(accept

42.5 - 43.0 MPa

). The unconfined strength would be

σ_{c} = 2 c N_{ϕ} = 27.7 M P a