Give step-by-step solution with explanation and final answer:
1
A
1
5
m high rock slope with an overall slope angle of
6
8
°
is intersected by a fully developed planar discontinuity extending

Question

Give step-by-step solution with explanation and final answer:
1
 ﻿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 anal

VIPSolutions · Accepted Answer

# 1. Planar Failure Mode: Analytical Factor of Safety Calculation

## **Given Data**

- **Slope height, $ H $**:  m  
- **Slope angle $ \beta $**: 68°  
- **Discontinuity dip, $ \theta $**: 45° (strike parallel to slope)  
 **Friction, $ \phi $**: 32°  
- **Cohesion, $ c $**: 15 kN/m²  
- **Unit weight, $ \gamma $**: 25 kN/m³  
- **Slope thickness (width out of plane), $ b $**: 1 m  
- **Assume dry conditions, no water or surcharge.**

---

## **Assumptions**

- **Planar failure**: Sliding occurs along the discontinuity (no toppling, etc.).
- **Block geometry**: The slide mass is a parallelogram with height $ H $ and thickness $ b $.
- **Limit equilibrium**: Forces are resolved parallel and perpendicular to the slip plane.
- **Unit thickness**: All calculations per meter into the slope.

---

## **Step 1: Geometry of the Sliding Mass**

### **Length of Discontinuity (L)**

The length along the discontinuity from crest to toe is:
$$
L = \frac{H}{\sin\theta}
$$
$$
L = \frac{15}{\sin 45^\circ} = \frac{15}{.7071} = 21.21 \text{ m}
$$

---

## **Step 2: Forces on the Sliding Mass**

### **Weight ($ W))**
\[
W = \gamma \times H \times L_{\text{horizontal}} \times b
\]
But for a planar failure, we use the area of the sliding mass projected onto the slope. For a unit thickness:
\[
\text{Area of slip surface} = L \times b
\]
\[
W = \gamma \times \text{Area} \times \text{horizontal width (projected)}
\]

But since the width is not given, for simplicity and as is common, we analyze per 1 m width:
\[
\text{Block volume} = H \times 1 \times 1 = 15 \text{ m}^3
\]
\[
W = 25 \times 15 = 375 \text{ kN}
\]

---

## **Step 3: Factor of Safety (FoS) Equation**

The factor of safety for planar slides:
\[
\text{FoS} = \frac{\text{Resisting Forces}}{\text{Driving Forces}}
\]

- **Resisting Forces:** Cohesion + Friction
    - \( c $ acts along the slip surface area
    - Friction acts on the normal component of weight on the slip surface

- **Driving Force:** Component of weight down the slip surface

### **Slip Surface Area (A)**
$$
A = L \cdot b = 21.21 \times 1 = 21.21 \text{ m}^2
$$

### **Normal ($ N $) and Shear ($ S $) Components**

- Weight ($ W $) acts vertically.
- Inclined slip plane at $ \theta = 45^\circ $.

**Normal to slip plane:**
$$
N = W \cos \theta
$$

**Down dip (parallel to slip plane):**
$$
S = W \sin \theta
$$

---

### **Resisting Forces**

$$
R = c \cdot A + N \cdot \tan \phi
$$

### **Driving Forces**

$$
D = S
$$

---

### **Plug in the numbers**

- $ W = 375 $ kN  
- $ \theta = 45^\circ $, $ \sin 45^\circ = .7071 $, $ \cos 45^\circ = .7071 $
- $ c = 15 $ kN/m²  
- $ A = 21.21 $ m²  
- $ \phi = 32^\circ $, $ \tan 32^\circ = .6249 $

**Normal force:**
$$
N = W \cos \theta = 375 \times .7071 = 265.16 \text{ kN}
$$

**Shear force:**
$$
S = W \sin \theta = 375 \times .7071 = 265.16 \text{ kN}
$$

**Cohesion:**
$$
c \cdot A = 15 \times 21.21 = 318.15 \text{ kN}
$$

**Frictional resistance:**
$$
N \times \tan \phi = 265.16 \times .6249 = 165.67 \text{ kN}
$$

**Total Resisting Force:**
$$
R = 318.15 + 165.67 = 483.82 \text{ kN}
$$

**Driving Force:**
$$
D = 265.16 \text{ kN}
$$

---

### **Factor of Safety**

$$
\text{FoS} = \frac{R}{D} = \frac{483.82}{265.16} = 1.83
$$

---

## **Final Analytical Answer**

$$
\boxed{\text{FoS} = 1.83}
$$

---

# 2. Evaluation of Calculated FoS and Reinforcement Proposal

## **Evaluation vs. Design Criteria**

- **Typical minimum design FoS:**  
    - Permanent slopes: $ \geq 1.3 $–1.5  
    - Temporary slopes: $ \geq 1.1 $–1.2  
- **Calculated FoS:** 1.83 (> 1.3)

### **Conclusion**

- The slope **meets acceptable design criteria**.
- **No reinforcement is strictly required**.

## **If FoS < 1.3: Reinforcement Scheme (Hypothetical)**

If the FoS were less than 1.3, consider the following reinforcement:

### **Rock Bolts**

- **Purpose:** Increase normal force and frictional resistance along the discontinuity.

#### **Design Assumptions**
- **Bolt Length:** $ 2/3 $ of slope height $ = 10 $ m (to anchor into stable rock).
- **Bolt Spacing:** 2 m (horizontal and vertical grid).
  - Number of bolts along slip surface $ = \frac{L}{2} = 10.6 $ → round up to 11 bolts.
- **Bolt Capacity:** 100 kN per bolt (conservative design).
- **Total Capacity:** $ 11 \times 100 = 110 $ kN

#### **Interaction with Slope**

- Bolts are installed **perpendicular to the discontinuity**, increasing effective normal stress, and thus frictional resistance.
- **Anchorage beyond the slip surface** is essential to ensure load transfer.

---

# 3. Numerical Verification with RocPlane

## **Input Parameters (for RocPlane or Equivalent Software)**

| Parameter           | Value         |
|:--------------------|:-------------|
| Slope height        | 15 m         |
| Slope angle         | 68°          |
| Discontinuity dip   | 45°          |
| Unit weight         | 25 kN/m³     |
| Cohesion            | 15 kN/m²     |
| Friction angle      | 32°

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