---
title: "Infinite slope seepage, max angle for FoS 2"
subject: "Soil Mechanics"
chapter: 11
key: "semester_4/soil_mechanics:infinite-slope-seepage-max-angle"
source: https://www.bctengineer.com/q/semester_4/soil_mechanics/infinite-slope-seepage-max-angle
site: BCT Engineering
---

# Infinite slope seepage, max angle for FoS 2

Soil Mechanics · Chapter 11: Stability of Slopes

- Asked in: 2075 Baisakh, 4 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/soil_mechanics/infinite-slope-seepage-max-angle
- Whole chapter: https://www.bctengineer.com/md/semester_4/soil_mechanics/chapter-11

## Question

A slope of very large extent of soil with properties c' = 0, e = 0.7, G = 2.7 and $\phi$ = 35° is likely to be subjected to seepage parallel to the slope with water level at the surface. Determine the maximum angle of slope for a factor of safety of 2.0. What will be the factor of safety if the water level were to come down well below the surface for this angle of slope?

## Short answer

With seepage parallel to the surface (c′=0, φ=35°, γ′/γ_sat=0.5), the maximum slope angle for F=2.0 is β = 9.9°. If the water level drops well below the surface, the factor of safety rises to F = tanφ/tanβ = 4.0.

## Answer

**Data:** $c' = 0$, $e = 0.7$, $G = 2.7$, $\phi = 35^\circ$, required $F = 2.0$, seepage parallel to the slope with water at the surface.

**Step 1: Unit weights**

$$
\gamma_{sat} = \frac{(G+e)\gamma_w}{1+e} = \frac{3.4(9.81)}{1.7} = 19.62\ \text{kN/m}^3, \qquad \gamma' = 19.62 - 9.81 = 9.81\ \text{kN/m}^3
$$

$$
\frac{\gamma'}{\gamma_{sat}} = \frac{G-1}{G+e} = \frac{1.7}{3.4} = 0.5
$$

**Step 2: Maximum slope angle for $F=2$**

$$
F = \frac{\gamma'}{\gamma_{sat}}\frac{\tan\phi}{\tan\beta} \;\Rightarrow\; 2.0 = 0.5\,\frac{\tan35^\circ}{\tan\beta}
$$

$$
\tan\beta = \frac{0.5\times0.7002}{2.0} = 0.1750 \;\Rightarrow\; \beta = 9.93^\circ \approx 9.9^\circ
$$

**Step 3: Factor of safety if the water level falls well below the surface** (dry or moist soil, no seepage force):

$$
F = \frac{\tan\phi}{\tan\beta} = \frac{0.7002}{0.1750} = 4.0
$$

**Answer: maximum slope angle $\approx 9.9^\circ$; factor of safety rises to 4.0 when the water level drops well below the surface.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. For G = 2.7 and e = 0.7, what is γ'/γsat?
   A. 0.35
   B. 0.50
   C. 0.65
   D. 0.70
   Answer: B. γ'/γsat = (G-1)/(G+e) = 1.7/3.4 = 0.5. (Key point: Submerged to saturated unit weight ratio)

2. Which equation is used to find the maximum slope angle with seepage at the surface?
   A. F = tanφ/tanβ
   B. F = (γ'/γsat) tanφ/tanβ
   C. F = (γsat/γ') tanφ/tanβ
   D. F = (γ'/γw) tanβ/tanφ
   Answer: B. With parallel seepage at the surface and c' = 0, F = (γ'/γsat)(tanφ/tanβ). (Key point: Governing equation)

3. For F = 2.0, φ = 35°, what is the maximum slope angle?
   A. About 9.9°
   B. About 19.3°
   C. About 17.5°
   D. About 35°
   Answer: A. tanβ = 0.5 x 0.7002 / 2.0 = 0.175, so β ≈ 9.9°. (Key point: Maximum angle)

4. If the water level falls well below the surface for the same angle, what is F?
   A. 2.0, unchanged
   B. 1.0
   C. 4.0
   D. 0.5
   Answer: C. Without seepage F = tanφ/tanβ = 0.7002/0.175 = 4.0, twice the value with seepage. (Key point: Water level falls)

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