Question 3 of 6: Plane Failure Analysis of a Jointed Rock Slope

Question 3: Plane Failure Analysis of a Jointed Rock Slope (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.

3.1 — Factor of safety, dry slope

Given. Slope height H = 25 m; face angle ψ_f = 55°; failure-plane dip ψ_p = 30°; tension crack depth z = 6.8 m located b = 10 m behind the crest (upper slope surface assumed horizontal, since no upper-surface slope angle is given); c′ = 2.05 kPa, φ′ = 35°; γ_rock = 27 kN/m³, γ_water = 9.81 kN/m³.

Find. The factor of safety against sliding on the 30° failure plane, dry.

Approach. Set up slope coordinates with the toe at the origin; locate the crest, the tension crack top/bottom, and the point where a 30° plane through the crack base daylights on the 55° face; take the sliding wedge as the resulting quadrilateral, compute its weight, then apply the plane-failure factor-of-safety equation.

1. Locate the crest and tension crack. Horizontal distance toe–crest $=H/\tan\psi_f=25/\tan55^\circ=17.51\ \text{m}$, so crest = (17.51, 25). The crack sits b=10 m further back on the horizontal upper surface: crack top = (27.51, 25); crack bottom = (27.51, 25−6.8) = (27.51, 18.2).
2. Find where the failure plane daylights on the face. The 30° plane through the crack base meets the 55° face line where $18.2-\tan30^\circ(27.51-x)=x\tan55^\circ$, giving $\boxed{x=2.73\ \text{m},\ y=3.89\ \text{m}}$ — i.e. the sliding surface actually daylights 3.89 m up the face, not at the toe.
   Check: the given z and b do not place the crack base exactly on a 30° plane through the slope toe (that plane would need z≈9.1 m for b=10 m); rather than silently forcing the plane through the toe, the daylight point is solved for directly from the given geometry, which is the more defensible reading of the data as given.
3. Wedge geometry. Failure-plane length (toe daylight point to crack base) $A=\sqrt{(27.51-2.73)^2+(18.2-3.89)^2}=\boxed{28.61\ \text{m}}$. The sliding wedge is the quadrilateral (2.73, 3.89) – (27.51, 18.2) – (27.51, 25) – (17.51, 25); by the shoelace formula its cross-sectional area is $189.8\ \text{m}^2$, so the weight per metre of slope length is $W=\gamma_{rock}\times\text{area}=27\times189.8=\boxed{5124\ \text{kN/m}}$.
4. Factor of safety (dry). For a dry plane failure, $\mathrm{FS}=\dfrac{c'A+W\cos\psi_p\tan\phi'}{W\sin\psi_p}=\dfrac{2.05(28.61)+5124\cos30^\circ\tan35^\circ}{5124\sin30^\circ}=\dfrac{58.7+3107.2}{2562}=\boxed{\mathrm{FS}=1.24}$.

3.2 — Factor of safety with a water-filled tension crack

Given. As above, but the tension crack is completely filled with water (depth of water z_w = z = 6.8 m), creating an uplift force U along the failure plane and a horizontal thrust V in the crack.

Find. The revised factor of safety with the water-filled crack.

Approach. Water pressure is triangular, peaking at $\gamma_w z_w$ at the crack base and dissipating to zero at the point the plane daylights; integrate this distribution over the crack (giving V, horizontal) and over the full plane length A (giving U, normal to the plane), then use the general plane-failure FS equation with U and V included.

1. Water forces. $U=\tfrac12\gamma_w z_w A=\tfrac12(9.81)(6.8)(28.61)=\boxed{954\ \text{kN/m}}$; $V=\tfrac12\gamma_w z_w^2=\tfrac12(9.81)(6.8)^2=\boxed{227\ \text{kN/m}}$.
2. Factor of safety. $\mathrm{FS}=\dfrac{c'A+(W\cos\psi_p-U-V\sin\psi_p)\tan\phi'}{W\sin\psi_p+V\cos\psi_p}=\dfrac{58.7+(4437-954-113)\tan35^\circ}{2562+196}=\dfrac{58.7+2359.5}{2758.4}=\boxed{\mathrm{FS}=0.88}$.
3. Interpretation. Filling the tension crack with run-off water drops the factor of safety from 1.24 to below 1.0 — the slope is predicted to fail once the crack fills, which is exactly why surface-water control (crest drainage, crack sealing/backfilling) is a primary, low-cost stabilisation measure for this class of failure.

3.3 — Factor of safety with tensioned rock bolt reinforcement

Given. Reinforcing the dry slope of 3.1 with rock bolts installed perpendicular to the sliding plane, total anchor load T = 350 kN per linear metre of slope.

Find. The factor of safety with bolt reinforcement.

1. Add the bolt force to the resisting term. Because T acts at right angles to the sliding plane, it adds entirely to the normal force and none to the driving (shear) force: $\mathrm{FS}=\dfrac{c'A+(W\cos\psi_p+T)\tan\phi'}{W\sin\psi_p}=\dfrac{58.7+(4437+350)\tan35^\circ}{2562}=\dfrac{58.7+3351}{2562}=\boxed{\mathrm{FS}=1.33}$.
2. Interpretation. 350 kN/m of tensioned, perpendicular anchor load raises FS from 1.24 to 1.33, a modest but useful gain typical of reinforcement installed normal (rather than at some flatter, more favourable angle) to the sliding plane.

3.4 — Bolt layout for a target load of 400 kN/m

Given. Working load per bolt = 200 kN; target total anchor load = 400 kN per metre of slope length.

Find. The number of bolts (and spacing) per metre of slope required.

1. Bolts per metre. $n=\dfrac{400}{200}=\boxed{2\ \text{bolts per linear metre of slope}}$, each stressed to its 200 kN working load. A single row at 0.5 m centres along the strike would satisfy this arithmetically but is impractically tight for drilling and plate bearing.
2. Practical layout. Spread the same 2 bolts/m over several rows up the face. With N rows and a horizontal (along-strike) spacing s, the delivered load per metre is $200N/s$; setting this to 400 kN/m requires $s=N/2$ m. A convenient pattern is $\boxed{N=4\ \text{rows at}\ s=2.0\ \text{m}\ \text{horizontal spacing}}$ (check: $200\times4/2.0=400$ kN/m), i.e. four horizontal rows of 200 kN bolts at 2.0 m centres along the strike, all drilled normal to the 30° bedding plane and anchored well beyond it. Note that the load per metre depends only on N and s, not on the vertical row spacing, so the rows are placed where they do most good — distributed up the lower part of the face, above the level at which the sliding plane daylights ($y=3.89$ m) and below the crest.