---
title: "Gravity dam stability with seismic load"
subject: "Hydropower Engineering"
chapter: 4
key: "semester_7/hydropower_engineering:gravity-dam-seismic-stability-2073"
source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/gravity-dam-seismic-stability-2073
site: BCT Engineering
---

# Gravity dam stability with seismic load

Hydropower Engineering · Chapter 4: Headworks of Storage Plants

- Asked in: 2073 Shrawan, 4+4+2 marks
- Video: [Numerical modelling of cracking in gravity dam under static and seismic loadings with multiple ...](https://www.youtube.com/watch?v=H8Ow31sjVQI) — European Structural Integrity Society
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/hydropower_engineering/gravity-dam-seismic-stability-2073
- Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-4

## Question

Check the stability of the dam against overturning, sliding and material failure (stresses) with respect to the worst location, assuming that in addition to self weight, 25% of the mass of the dam will act as a horizontal component (from the upstream side), whereas 15% acts as an upward vertical component as seismic load and will act at the CG of the section. Assume the unit weight of concrete as 24 kN/m³, allowable compressive stress in foundation and concrete as 2,500 kN/m² and 3,000 kN/m², angle of friction between concrete and foundation as 36° and unit shear resistance between foundation and dam as 700 kN/m².
[Figure: dam with crest width 6.0 m, top level 150.0, water level 135.0, base level 100.0; upstream face with slope 0.1 : 1; downstream slope 0.8 : 1; points A (heel) and B (toe).]

## Short answer

Overturning FS=1.83 (safe). Sliding friction alone FS=1.04, shear-friction factor SFF=3.50 (safe). Maximum vertical stress 641 kN/m², principal stress 1051 kN/m², both within allowable limits; no tension. Dam is stable.

## Answer

#### Data and assumptions
- Base RL 100, crest RL 150: height 50 m. Water RL 135: $H_w=35$ m. Crest width 6 m. U/s face slope 0.1 : 1 (leaning inwards going up), so the heel A is at $x=0$ and the u/s edge of the crest at $x=5$; d/s slope 0.8 : 1, so the toe B is at $x=11+40=51$ m. Base $B=51$ m.
- The base (section A-B) is the worst location: it carries the largest load and moment.
- $\gamma_c=24$ kN/m³, $\gamma_w=9.81$ kN/m³. No tail water. Hydrostatic thrust and uplift (full, triangular, no drainage: $343.4$ kN/m² at the heel to zero at the toe) act together with the seismic loads.
- Seismic: horizontal force $0.25W$ from the upstream side (towards the downstream) and vertical force $0.15W$ upward, both at the centre of gravity (CG).

```text
   6 m
  +----+  RL150
 /     |
/      |  RL135 ~~ water
/      |
A(heel) \___________  B(toe)
 0.1:1       0.8:1       B = 51 m
```

#### Self-weight and CG
Area $=\frac12(6+51)\times50=1425$ m². $W=24\times1425=34,200$ kN/m. CG: $\bar x=19.05$ m from the heel ($B-\bar x=31.95$ m from the toe), $\bar z=18.42$ m.

Seismic: $F_h=0.25W=8,550$ kN/m; $F_v=0.15W=5,130$ kN/m (upward), so the effective weight $=0.85W=29,070$ kN/m.

#### Forces
| Force | Magnitude (kN/m) | Arm from toe (m) | Moment about toe (kN·m/m) |
|---|---|---|---|
| Effective weight 0.85W | 29,070 | 31.95 | 928,710 |
| Water over the u/s slope | 601 | 49.83 | 29,943 |
| **ΣM_R** | | | **958,653** |
| Water thrust $\frac12\gamma_wH_w^2$ | 6,009 | 11.67 (height) | 70,101 |
| Uplift (upward) | 8,755 | 34.00 | 297,684 |
| Seismic $F_h$ | 8,550 | 18.42 (height) | 157,500 |
| **ΣM_O** | | | **525,285** |

$\Sigma V=29,070+601-8,755=20,915$ kN/m; $\Sigma H=6,009+8,550=14,559$ kN/m.

#### Overturning
$$
FS_o=\frac{\Sigma M_R}{\Sigma M_O}=\frac{958,653}{525,285}=1.83
$$

This is above 1.5 (a value of 1.1-1.5 is adequate for the seismic case), so the dam is safe against overturning.

#### Sliding
Friction coefficient $\mu=\tan36^\circ=0.727$; shear resistance $q=700$ kN/m².

$$
FS_{friction}=\frac{\mu\Sigma V}{\Sigma H}=\frac{0.727\times20,915}{14,559}=1.04
$$

$$
SFF=\frac{\mu\Sigma V+qB}{\Sigma H}=\frac{0.727\times20,915+700\times51}{14,559}=3.50
$$

Friction alone is just above 1. Including the shear resistance of the foundation contact, $SFF=3.50$, which is well above the minimum usually required for loading with earthquake, so the section is safe in sliding.

#### Material failure (stresses on the base)
$\bar x=(\Sigma M_R-\Sigma M_O)/\Sigma V=20.72$ m from the toe; $e=25.5-20.72=4.78$ m $<B/6=8.5$ m (no tension).

$$
\sigma_{toe}=\frac{20,915}{51}\left(1+\frac{6\times4.78}{51}\right)=641\ \text{kN/m}^2\ (<2500),\qquad \sigma_{heel}=179\ \text{kN/m}^2
$$

Both are compressive and less than the allowable 2500 kN/m² (foundation).

Major principal stress at the toe (no tail water, $\tan\phi_d=0.8$):

$$
\sigma_1=\sigma_{toe}(1+\tan^2\phi_d)=641\times1.64=1051\ \text{kN/m}^2<3000\ \text{kN/m}^2
$$

so the concrete is also safe.

**Conclusion:** the dam is safe against overturning ($FS_o=1.83$), sliding ($SFF=3.50$) and material failure (maximum vertical stress 641 kN/m² < 2500 kN/m², principal stress 1051 kN/m² < 3000 kN/m², no tension).

## Check yourself

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

1. How does the 15% upward vertical seismic force enter the stability calculation?
   A. It adds 0.15W to the weight
   B. It reduces the effective weight to 0.85W
   C. It is ignored since it acts at the CG
   D. It is added to the uplift only at the toe
   Answer: B. An upward vertical acceleration reduces the stabilising weight, so the dam weight is taken as 0.85W = 29,070 kN/m. (Key point: Effect of vertical seismic force)

2. How is the horizontal seismic force applied?
   A. 0.25W acting downstream at the centre of gravity, adding to the water thrust
   B. 0.25W acting upstream, reducing the water thrust
   C. 0.25 x water thrust at the base
   D. 0.25W acting at the heel
   Answer: A. F_h = 0.25W = 8,550 kN/m acts at the CG height (18.42 m) from the upstream side, in the same direction as the water thrust. (Key point: Horizontal seismic force)

3. How is sliding checked here?
   A. Using only the friction coefficient mu = tan 36 deg = 0.727
   B. SFF = (mu SumV + q B) / SumH = 3.50, with friction alone just above 1
   C. FS = q B / SumV
   D. Sliding is not checked in earthquake case
   Answer: B. The answer combines friction (mu = tan 36) with the unit shear strength q = 700 over the base width, giving SFF = 3.50. (Key point: Shear-friction factor with friction angle)

4. Which section is checked for material failure and what is concluded?
   A. The crest, because it is thinnest
   B. The base section A-B, where stresses 641 and 179 kN/m^2 are within limits with e < B/6
   C. The mid-height section, where tension appears
   D. The d/s face only, where stress exceeds 3000 kN/m^2
   Answer: B. The base carries the largest load and moment, so it is the worst location; there e = 4.78 m < B/6 = 8.5 m and the stresses are below 2500 kN/m^2. (Key point: Worst location and stresses)

## Related in Hydropower Engineering

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