---
title: "Triangular gravity dam 35 m high"
subject: "Hydro Power"
chapter: 4
key: "semester_6/hydro_power:triangular-dam-35m"
source: https://www.bctengineer.com/q/semester_6/hydro_power/triangular-dam-35m
site: BCT Engineering
---

# Triangular gravity dam 35 m high

Hydro Power · Chapter 4: Dam Engineering

- Asked in: 2080 Chaitra, 8 marks
- Video: [DHS Module 4 [Part 06] - Stability Analysis of Gravity Dams - Problem](https://www.youtube.com/watch?v=NV6EbVvtpv8) — AM CIVIL ACADEMY
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/hydro_power/triangular-dam-35m
- Whole chapter: https://www.bctengineer.com/md/semester_6/hydro_power/chapter-4

## Question

Water stand on the u/s side of a gravity dam of triangular section up to the full height of 35 m. The base width of dam is 26 m. If the uplift pressure coefficient is 0.5, calculate a) Eccentricity of the resultant from the toe. b) Factor of safety against sliding. c) The crushing strength developed in the dam. d) Factor of safety against overturning. Take coefficient of friction between base and foundation is 0.75 and the unit weight of dam material is 2400 kg(f)/m³.

## Short answer

Resultant is 9.07 m from toe (eccentricity 3.93 m from centre), FS sliding 1.06, max crushing stress 622.2 kN/m² (965.5 kN/m² principal), FS overturning 1.71.

## Answer

#### Data (per metre length)
H = 35 m, B = 26 m, uplift coefficient C = 0.5, μ = 0.75. Unit weight of concrete $= 2400$ kgf/m³ $= 2400 \times 9.81/1000 = 23.544$ kN/m³; water $= 9.81$ kN/m³. Upstream face vertical, water up to the top. Uplift varies from $C\gamma_w H$ at the heel to zero at the toe. Moments are about the toe.

```text
        |\
  P --> | \    35 m
        | W\
   heel |___\ toe
        <26 m>
```

#### Forces

| Force | Value (kN) | Arm from toe (m) | Moment (kN·m) |
|---|---|---|---|
| W = ½×26×35×23.544 | 10,712.5 ↓ | 2×26/3 = 17.333 | +185,683.7 |
| P = ½×9.81×35² | 6,008.6 → | 35/3 = 11.667 | −70,100.6 |
| U = ½×0.5×9.81×35×26 | 2,231.8 ↑ | 17.333 | −38,684.1 |

$$
\begin{aligned}
\sum V &= 10{,}712.5 - 2{,}231.8 = 8{,}480.7\ \text{kN} \\
\sum M_R &= 185{,}683.7, \quad \sum M_O = 108{,}784.7\ \text{kN·m}
\end{aligned}
$$

#### a) Eccentricity

$$
\begin{aligned}
\bar{x} &= \frac{185{,}683.7 - 108{,}784.7}{8{,}480.7} = 9.067\ \text{m from toe} \\
e &= \frac{26}{2} - 9.067 = 3.933\ \text{m}
\end{aligned}
$$

$e < B/6 = 4.333$ m, so the resultant lies in the middle third (no tension).

#### b) Factor of safety against sliding

$$
FS_s = \frac{\mu\sum V}{\sum H} = \frac{0.75 \times 8{,}480.7}{6{,}008.6} = 1.06
$$

#### c) Maximum compressive (crushing) stress

$$
\begin{aligned}
p_{toe} &= \frac{\sum V}{B}\left(1 + \frac{6e}{B}\right) = \frac{8{,}480.7}{26}\left(1 + \frac{6 \times 3.933}{26}\right) \\
&= 326.18 \times 1.9075 = 622.2\ \text{kN/m}^2 \ (\approx 63.4\ \text{t/m}^2)
\end{aligned}
$$

Heel stress $= 326.18(1 - 0.9075) = 30.2$ kN/m² (compression). Major principal stress at toe, with $\tan\alpha = 26/35$: $\sigma_1 = 622.2 \times (1 + 0.5518) = 965.5$ kN/m².

#### d) Factor of safety against overturning

$$
FS_o = \frac{185{,}683.7}{108{,}784.7} = 1.71
$$

**Answer: Resultant at 9.067 m from the toe (e = 3.933 m from centre); FS sliding = 1.06; maximum vertical stress at toe = 622.2 kN/m² (principal stress 965.5 kN/m²); FS overturning = 1.71.** The dam is safe in overturning and tension; FS against sliding is just above 1, so the margin is small.

## Check yourself

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

1. How is the unit weight of dam material 2400 kgf/m3 used?
   A. Converted to 2400 x 9.81 / 1000 = 23.544 kN/m3
   B. Converted to 2400 / 9.81 kN/m3
   C. It is used directly as 2400 kN/m3
   D. Converted to 2.4 kN/m3
   Answer: A. kgf is converted to kN by multiplying by 9.81 and dividing by 1000. (Key point: Unit weight conversion)

2. Which expression gives the uplift force for this triangular dam with coefficient C = 0.5?
   A. U = C gamma_w H B
   B. U = 1/2 C gamma_w H B, acting at 2B/3 from the toe
   C. U = gamma_w H^2 / 2
   D. U = C gamma_c H B
   Answer: B. Uplift is triangular from C gamma_w H at heel to zero at toe, so U = 1/2 x 0.5 x 9.81 x 35 x 26 = 2,231.8 kN. (Key point: Uplift force)

3. With x_bar = 9.067 m from the toe and B = 26 m, what is e and what does it show?
   A. e = 9.067 m, tension at the heel
   B. e = 13 m, the dam overturns
   C. e = 3.933 m < B/6 = 4.333 m, resultant in middle third, no tension
   D. e = 3.933 m > B/6, tension at the heel
   Answer: C. e = B/2 - x_bar = 13 - 9.067 = 3.933 m, inside B/6, so no tension. (Key point: Eccentricity and tension)

4. FS against sliding is 0.75 x 8,480.7 / 6,008.6. What is the conclusion?
   A. 1.71, same as overturning
   B. 2.5, very safe
   C. 0.60, unsafe
   D. 1.06, only just above 1 so margin is small
   Answer: D. FS_s = 1.06, barely above 1. (Key point: Sliding margin)

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