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

Question

Give step-by-step solution with explanation and final answer:
﻿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 analyti

VIPSolutions · Accepted Answer

# Step-by-Step Solution: Planar Failure Analysis of a Rock Slope

## Problem Recap

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

---

## 1. Limit Equilibrium Factor of Safety (FoS) Calculation

### **Assumptions**
- Planar failure along continuous joint.
- Discontinuity daylighting at slope face (no toe restraint).
- Water pressure is neglected (dry conditions).
- Slope height = length of joint trace on slope face.

### **Step 1: Geometry of Failure Plane**

**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 failure surface (per meter thickness):**
$$
A = L \times 1 = 21.21\, \text{m}^2
$$

### **Step 2: Weight of Sliding Block**

**Horizontal width ($ w $):**
$$
w = \frac{H}{\tan \beta} = \frac{15}{\tan 68^\circ} = \frac{15}{2.475} \approx 6.06\, \text{m}
$$

**Block volume:**
$$
V = \frac{1}{2} \times H \times w \times 1 = .5 \times 15 \times 6.06 \times 1 = 45.45\, \text{m}^3
$$

**Weight ($ W $):**
$$
W = V \times \gamma = 45.45 \times 25 = 1,136.36\, \text{kN}
$$

### **Step 3: Forces on the Failure Plane**

**Normal to the plane:**
- Weight is resolved into components **perpendicular** and **parallel** to the plane.

$$
\text{Normal component:} \quad W_n = W \cos(\theta - \beta)
$$
$$
\text{Shear component:} \quad W_s = W \sin(\theta - \beta)
$$

But the geometry needs to be handled carefully as the block is on the plane dipping at 45°, but the slope face is 68°.

**Alternatively, the driving force is the downslope component of the weight parallel to the discontinuity:**

$$
\text{Driving force (downslope):} \quad W \sin(\theta)
$$

$$
\text{Normal force:} \quad W \cos(\theta)
$$

### **Step 4: Shear Strength (Resisting Force)**

$$
\text{Shear strength} = cA + (W \cos \theta) \tan \phi
$$

### **Step 5: Factor of Safety (FoS) Expression**

$$
\boxed{
\text{FoS} = \frac{cA + (W \cos \theta) \tan \phi}{W \sin \theta}
}
$$

### **Step 6: Plug in Values**

- $ c = 15 $ kN/m²  
- $ A = 21.21 $ m²  
- $ W = 1,136.36 $ kN  
- $ \theta = 45^\circ $  
- $ \phi = 32^\circ $

Calculate:

- $ \sin 45^\circ = .7071 $
- $ \cos 45^\circ = .7071 $
- $ \tan 32^\circ = .6249 $

Now,

$$
\text{Numerator} = cA + (W \cos \theta) \tan \phi = 15 \times 21.21 + (1,136.36 \times .7071) \times .6249
$$
$$
= 318.15 + (803.48 \times .6249) = 318.15 + 502.29 = 820.44~\text{kN}
$$

$$
\text{Denominator} = W \sin \theta = 1,136.36 \times .7071 = 803.48~\text{kN}
$$

$$
\text{FoS} = \frac{820.44}{803.48} = 1.02
$$

---

### **Final Answer (Analytical FoS)**

$$
\boxed{ \text{FoS} = 1.02 }
$$

---

## 2. Critical Evaluation and Reinforcement Proposal

### **Design Criteria**
- **Acceptable FoS:** Typically ≥ 1.3 for permanent slopes in rock engineering.

**Calculated FoS (1.02) < Design minimum (1.3):**  
**Slope does NOT meet stability criteria.**

---

### **Reinforcement Scheme Proposal**

#### **Reinforcement Type: Rock Bolts**
- **Purpose:** Increase normal force across the discontinuity and enhance shear resistance.

#### **Assumptions for Design**
- Bolts act perpendicular to failure plane.
- Uniform spacing along the discontinuity.
- Each bolt: 25 mm diameter, fully grouted, capacity 200 kN.
- Bolt length: 2/3 × slope height (to anchor in stable rock): $ \approx 10~\text{m} $
- Spacing: 2. m horizontally and vertically along the failure plane.

#### **Number of Bolts**
- Failure plane area: $ 21.21~\text{m}^2 $
- Area per bolt: $ 2. \times 2. = 4.~\text{m}^2 $
- Number of bolts: $ 21.21 / 4. \approx 6 $

#### **Total Bolt Capacity**
- $ 6 $ bolts × $ 200~\text{kN} $ = $ 1,200~\text{kN} $
- Assume only the component acting perpendicular to the failure plane contributes to resisting sliding (conservative).

#### **Revised FoS Calculation (with bolts)**

Shear resistance is now:

$$
\text{Resisting force} = cA + (W \cos \theta + P_{bolts}) \tan \phi
$$
Where $ P_{bolts} $ is the **sum of bolt forces normal to the plane**.

$$
P_{bolts} = 1,200~\text{kN}
$$

$$
\text{Numerator} = 318.15 + ((803.48 + 1,200) \times .6249)
$$
$$
= 318.15 + (2,003.48 \times .6249) = 318.15 + 1,252.45 = 1,570.60~\text{kN}
$$

$$
\text{Denominator} = 803.48~\text{kN}
$$

$$
\text{FoS (reinforced)} = \frac{1,570.60}{803.48} = 1.95
$$

### **Discussion**

- **FoS after reinforcement:** 1.95 > 1.3 (meets criterion)
- **Justification:** Bolts are spaced to provide even support. Their length ensures anchorage in stable rock.
- **Economics:** 6 bolts is a modest number for a 15 m high slope (per meter length along the slope), making this a practical scheme.

---

## 3. Verification Using RocPlane (or Equivalent)

### **Input Parameters Screenshot**

_Example i

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