---
title: "Sliding and overturning check of gravity dam"
subject: "Hydro Power"
chapter: 4
key: "semester_6/hydro_power:gravity-dam-sliding-overturning-figure"
source: https://www.bctengineer.com/q/semester_6/hydro_power/gravity-dam-sliding-overturning-figure
site: BCT Engineering
---

# Sliding and overturning check of gravity dam

Hydro Power · Chapter 4: Dam Engineering

- Asked in: 2071 Magh, 9 marks
- Video: [Safety Factor Analysis of Dam: Sliding & Overturning Conditions](https://www.youtube.com/watch?v=F8pD1SPdblU) — Civil learning online
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/hydro_power/gravity-dam-sliding-overturning-figure
- Whole chapter: https://www.bctengineer.com/md/semester_6/hydro_power/chapter-4

## Question

Check safety against sliding and overturning for the gravity dam shown in figure. Assume, unit weight of water = 9.81 kN/m³, unit weight of concrete = 24 kN/m³, angle of internal friction ϕ = 37°, uplift coefficient = 80%. Consider self-weight, hydrostatic pressure and uplift pressure only. Mention your recommendations. [Figure: gravity dam with vertical upstream face and top width 2.0 m; the water surface is 1.0 m below the crest and the water depth is 30 m; the downstream face is vertical for the top 1.5 m and then slopes at 0.75 (H) : 1.0 (V) down to the toe.]

## Short answer

Sliding FS 1.11, overturning FS 1.67, no tension. Safe, though sliding margin is narrow; recommend drainage gallery and foundation key to increase shear resistance.

## Answer

#### Geometry and data (per metre length)
- Water depth 30 m, water surface 1 m below crest, so dam height $= 31$ m.
- Top width 2 m; downstream face vertical for 1.5 m, then slopes 0.75H : 1V for the remaining $31 - 1.5 = 29.5$ m, giving horizontal projection $0.75 \times 29.5 = 22.125$ m.
- Base width $B = 2 + 22.125 = 24.125$ m. No tailwater.
- $\gamma_w = 9.81$, $\gamma_c = 24$ kN/m³; $\mu = \tan37° = 0.754$; uplift C = 0.8 (triangular, $0.8\gamma_w h$ at heel to zero at toe).

```text
     2 m
    +--+    1 m above WL
    |  +    1.5 m vertical
 30 |   \
 m  |W1  \  0.75H:1V
 -> |  W2 \       29.5 m
heel+------\ toe
    <-24.125->
```

#### Forces and moments about the toe

| Force | Value (kN) | Arm (m) | Moment (kN·m) |
|---|---|---|---|
| W1 = 2×31×24 | 1,488 ↓ | 23.125 | +34,410 |
| W2 = ½×22.125×29.5×24 | 7,832.3 ↓ | 14.750 | +115,526 |
| P = ½×9.81×30² | 4,414.5 → | 10.000 | −44,145 |
| U = ½×0.8×9.81×30×24.125 | 2,840.0 ↑ | 16.083 | −45,677 |

$$
\begin{aligned}
\sum V &= 1{,}488 + 7{,}832.3 - 2{,}840.0 = 6{,}480.3\ \text{kN} \\
\sum H &= 4{,}414.5\ \text{kN} \\
\sum M_R &= 149{,}936, \quad \sum M_O = 89{,}822\ \text{kN·m}
\end{aligned}
$$

#### Safety against sliding

$$
FS_s = \frac{\mu\sum V}{\sum H} = \frac{0.754 \times 6{,}480.3}{4{,}414.5} = 1.11 > 1
$$

#### Safety against overturning

$$
FS_o = \frac{149{,}936}{89{,}822} = 1.67 > 1.5
$$

Check of resultant: $\bar{x} = (149{,}936 - 89{,}822)/6{,}480.3 = 9.28$ m from toe, $e = 12.06 - 9.28 = 2.79$ m $< B/6 = 4.02$ m, so there is no tension at the heel (toe stress 454.7 kN/m², heel stress 82.5 kN/m²).

**Answer: FS against sliding = 1.11 and FS against overturning = 1.67; the dam is safe.**

**Recommendations:** the sliding margin is small (just above 1), so:
- provide foundation drainage holes/gallery to reduce uplift,
- key the base into sound rock or give it a slight upstream slope to increase shear resistance,
- check the shear friction factor using the shear strength of the rock–concrete joint.

## Check yourself

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

1. Water depth 30 m, water surface 1 m below the crest. What are the dam height and base width?
   A. 31 m and 24.125 m
   B. 31 m and 22.125 m
   C. 29.5 m and 24 m
   D. 30 m and 24.125 m
   Answer: A. Height = 30 + 1 = 31 m; base = 2 + 0.75 x 29.5 = 24.125 m. (Key point: Geometry)

2. What is mu and the FS against sliding?
   A. mu = 1; FS_s = 2
   B. mu = tan 37 = 0.754; FS_s = 1.11
   C. mu = 0.37; FS_s = 0.5
   D. mu = 0.754; FS_s = 1.67
   Answer: B. FS_s = 0.754 x 6,480.3 / 4,414.5 = 1.11. (Key point: Friction)

3. FS overturning = sum M_R / sum M_O. What is it?
   A. 2.4
   B. 0.9
   C. 1.67, above 1.5
   D. 1.11
   Answer: C. 149,936 / 89,822 = 1.67. (Key point: Overturning)

4. The sliding margin is small. Which is a suggested remedy?
   A. Increasing the uplift coefficient
   B. Removing the foundation
   C. Raising the reservoir level
   D. Drainage holes and keying the base into sound rock
   Answer: D. Drains reduce uplift and keying raises shear resistance. (Key point: Recommendation)

## Related in Hydro Power

- [Overturning and sliding check with shear strength](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Agravity-dam-overturning-sliding) — Hydro Power, chapter 4
- [Elementary profile and 40 m gravity dam stability](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Agravity-dam-elementary-40m) — Hydro Power, chapter 4
- [Stability checks of 50 m trapezoidal dam](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Atrapezoidal-dam-50m-checks) — Hydro Power, chapter 4
- [Triangular gravity dam 35 m high](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Atriangular-dam-35m) — Hydro Power, chapter 4
- [Stability of gravity dam (crest 1530 m)](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Agravity-dam-stability-1530) — Hydro Power, chapter 4
- [Stresses in gravity dam with drain holes](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Agravity-dam-stresses-figure) — Hydro Power, chapter 4

## Related in other subjects

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- [Gravity dam forces and stability checks (2081 Baishakh)](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Agravity-dam-forces-stresses-2081-baishakh) — Hydropower Engineering, chapter 4