A
1
5
m high rock slope with an overall slope angle of
6
8
°
is intersected by a fully developed planar discontinuity extending from the toe to the crest. The discontinuity has a dip of
4
5
°

Question

A 
1
5
 ﻿m high rock slope with an overall slope angle of 
6
8
°
 ﻿is intersected by a fully developed planar discontinuity extending from the toe to the crest. The discontinuity has a dip of 
4
5
°
 
(
strike parallel to the slope face
)
,
 ﻿a friction angle of 
3
2
°
,
 ﻿and a cohesion of 
1
5
 ﻿kN
/
m
^
2
.
 ﻿The rock unit weight is 
2
5
 ﻿kN
/
m
^
3
,
 ﻿and the slope section can be assumed to have a unit thickness of 
1
 ﻿m
.
 
1
.
 ﻿Using a limit equilibrium approach, calculate the Factor of Safety 
(
FoS
)
 ﻿for the potential planar failure mode. Clearly justify any assumptions you make in your derivation and calculation. 
2
.
 ﻿Critically evaluate the calculated FoS with respect to acceptable design criteria. If the FoS is less than 
1
.
3
,
 ﻿propose and justify an appropriate reinforcement scheme 
(
such as rock bolts, shotcrete, or anchors
)
.
 ﻿Your design must include assumptions on reinforcement spacing, length, and capacity, and a discussion of how the reinforcement interacts with the slope geometry and discontinuity. 
3
.
 ﻿Using RocPlane 
(
or equivalent slope stability software
)
,
 ﻿verify your analytical calculation of the FoS before reinforcement. Provide screen

VIPSolutions · Accepted Answer

# Planar Failure Analysis of a Rock Slope

## Problem Statement

- **Slope height** ($ h $): 15 m  
- **Slope angle** ($ \beta $): 68°  
- **Discontinuity dip** ($ \theta $): 45° (strike parallel to face)  
- **Friction angle ($ \phi $)**: 32°  
- **Cohesion ($ c $)**: 15 kN/m²  
- **Unit weight ($ \gamma $)**: 25 kN/m³  
- **Section thickness**: 1 m

---

## 1. Analytical Calculation of Factor of Safety (FoS)

### Assumptions

- **Planar failure** (block sliding along a discontinuity).
- **Dry slope** (no water pressure).
- **No external surcharge loads**.
- **Discontinuity daylighting at slope face** (fully developed).
- **Unit thickness (1 m)**: All calculations per meter out of the section.

### Planar Failure Limit Equilibrium Equation

$$
\text{FoS} = \frac{\text{Resisting Forces}}{\text{Driving Forces}}
$$

#### Geometry

- **Length of failure plane** ($ L $):

$$
L = \frac{h}{\sin\theta} = \frac{15}{\sin 45^\circ} = \frac{15}{.7071} \approx 21.21\ \text{m}
$$

- **Area of sliding surface** ($ A $):  
$$
A = L \times \text{thickness} = 21.21 \times 1 = 21.21\ \text{m}^2
$$

- **Weight of block ($ W $)**:

$$
W = \gamma \times \text{area (base)} \times \text{thickness} = \gamma \times (h/\tan\beta) \times h/2 \times 1
$$
However, for a prismatic block sliding on a plane from crest to toe, the weight is:

$$
W = \gamma \cdot h \cdot (L \cdot \cos\theta) \cdot 1
$$

But since the width is 1 m, and the block is defined by the triangle between the crest and toe, the length along the slope face is:

$$
\text{Horizontal projection of slope} = h / \tan\beta
$$
But for the planar failure, the block is the prism above the discontinuity, so:

$$
\text{Area (plan view)} = 1 \text{ m (thickness)} \times L \cdot \cos\theta
$$

But for simplicity, assume unit width (1 m), so:

$$
W = \gamma \cdot h \cdot 1 = 25 \cdot 15 \cdot 1 = 375\ \text{kN}
$$

#### Resisting Forces

$$
\text{Shear Strength} = c \cdot A + N \cdot \tan\phi
$$

- **Normal force ($ N $)**:

$$
N = W \cdot \cos(\theta - \beta)
$$

- **Driving force (downslope component of weight)**:

$$
S = W \cdot \sin(\theta - \beta)
$$

#### Calculate angles:

$$
\theta = 45^\circ,\ \beta = 68^\circ \rightarrow \theta - \beta = 45^\circ - 68^\circ = -23^\circ
$$

$$
\cos(-23^\circ) = .9205
$$
$$
\sin(-23^\circ) = -.3907
$$

#### Calculate forces:

$$
N = 375 \cdot .9205 = 345.2\ \text{kN}
$$
$$
S = 375 \cdot (-.3907) = -146.5\ \text{kN}
$$

(Negative sign indicates driving force is downslope.)

#### Calculate resisting shear strength:

$$
\text{Shear Strength} = c \cdot A + N \cdot \tan\phi
$$
$$
c = 15\ \text{kN/m}^2,\ A = 21.21\ \text{m}^2,\ \phi = 32^\circ\ (\tan 32^\circ = .6249)
$$
$$
c \cdot A = 15 \cdot 21.21 = 318.15\ \text{kN}
$$
$$
N \cdot \tan\phi = 345.2 \cdot .6249 = 215.6\ \text{kN}
$$
$$
\text{Total Shear Strength} = 318.15 + 215.6 = 533.75\ \text{kN}
$$

#### Calculate FoS

$$
\text{FoS} = \frac{\text{Resisting Shear Strength}}{\text{Driving Force}}
$$
$$
|\text{Driving Force}| = |S| = 146.5\ \text{kN}
$$

$$
\text{FoS} = \frac{533.75}{146.5} = 3.64
$$

---

## 2. Evaluation of Calculated FoS

### Acceptable Design Criteria

- **Typical minimum FoS for permanent slopes:** 1.3–1.5 (varies by code and risk).

### Assessment

- **Calculated FoS = 3.64** (significantly above 1.3).
- **No reinforcement required** under dry, static conditions for this geometry and rock mass properties.

#### If FoS < 1.3

If the FoS were less than 1.3, a reinforcement scheme such as rock bolts could be proposed:

#### Example Reinforcement Scheme (if needed):

- **Rock bolt type:** Fully grouted rebar bolts
- **Length:** At least 2/3 the slope height, e.g., 10 m (to anchor beyond the failure surface)
- **Capacity per bolt:** 100 kN (typical for standard bolts)
- **Spacing:** 2 m horizontal and vertical (grid pattern)
- **Number of bolts:** For 15 m x 1 m section, 8 bolts per row, 2–3 rows = 16–24 bolts
- **Interaction:** Bolts cross the failure plane, increase normal force, mobilize greater friction and cohesion, and resist sliding.

---

## 3. RocPlane Verification (Before Reinforcement)

### Input Parameters

- **Slope height:** 15 m
- **Slope angle:** 68°
- **Discontinuity dip:** 45°
- **Friction angle:** 32°
- **Cohesion:** 15 kN/m²
- **Unit weight:** 25 kN/m³
- **No water, no reinforcement**

### Expected RocPlane Output

- **FoS should be close to analytical value (3.6–3.7).**

### [Insert Screenshot Here]
*Alt text: RocPlane input window showing slope geometry, shear strength parameters, and computed FoS.*

#### Possible Discrepancies

- **Minor differences** due to:
  - More precise geometry modeling
  - Automated calculation of block weight and area
  - Slightly different failure surface identification

---

## 4. RocPlane Analysis (With Reinforcement)

### Proposed Reinforcement

- **Rock bolts:** 10 m long, 2 m spacing (grid), 100 kN capacity per bolt
- **Assume bolts act perpendicular to the sliding plane**

### RocPlane Input Adjustment

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