---
title: "Vertical force on stepped dam"
subject: "Fluid Mechanics"
chapter: 3
key: "semester_3/fluid_mechanics:stepped-dam-vertical-force"
source: https://www.bctengineer.com/q/semester_3/fluid_mechanics/stepped-dam-vertical-force
site: BCT Engineering
---

# Vertical force on stepped dam

Fluid Mechanics · Chapter 3: Hydrostatics

- Asked in: 2075 Asoj, 6 marks
- Video: [Analysis of Gravity Dams Sample Problem](https://www.youtube.com/watch?v=xyxUo_aGfNQ) — Jose Antonio Rodavia
- Page: https://www.bctengineer.com/q/semester_3/fluid_mechanics/stepped-dam-vertical-force
- Whole chapter: https://www.bctengineer.com/md/semester_3/fluid_mechanics/chapter-3

## Question

For the geometry shown, what is the vertical force on the dam? The steps are 0.3 m high, 0.3 m deep and 3 m wide. [Figure: stepped dam face with water on the left, full water depth over the steps]

## Short answer

Vertical force equals weight of water above stepped face. For the assumed 3 m water depth with ten 0.3 m treads, the force is 146 kN downward.

## Answer

The number of steps is not legible, so this reading is used. **Assumed:** water depth equals the dam height $H=3$ m, with 10 steps, each 0.3 m high and 0.3 m deep, and the dam is $w=3$ m wide. The water stands against the upstream stepped face.

```text
 water ____
        |_|__          each step 0.3 x 0.3 m
        |   |_|__
        |       |_|__
```

The horizontal force acts on the vertical projection and does not matter here. The **vertical force equals the weight of water above the stepped surface** up to the free surface. Only the horizontal treads carry vertical pressure.

Tread $k$ ($k=1,\ldots,10$) lies at depth $0.3k$ m below the surface, has horizontal area $0.3\times3=0.9\ \text{m}^2$, and carries pressure $\gamma(0.3k)$:

$$
F_V=\gamma\,w\,(0.3)\sum_{k=1}^{10}0.3k=9810(3)(0.3)(0.3)(55)
$$

$$
F_V=9810\times3\times4.95=145.7\ \text{kN}
$$

(Equivalently, the volume of water above the steps is $3\times4.95=14.85\ \text{m}^3$, so $F_V=9.81\times14.85=145.7$ kN.)

**Answer: vertical force on the dam $\approx146$ kN downward** (for 3 m depth and 10 steps). For $n$ steps, $F_V=\gamma w(0.3)^{2}\dfrac{n(n+1)}{2}$.

## Check yourself

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

1. What is the vertical force on the stepped face?
   A. Weight of water vertically above the steps up to the free surface
   B. Force on the vertical projection
   C. Weight of the dam
   D. Pressure at the base x full area
   Answer: A. Vertical force equals the weight of liquid above the surface; only the horizontal treads carry vertical pressure. (Key point: Vertical force principle)

2. Is the horizontal force needed for the vertical force?
   A. Yes, they are added directly
   B. No, it acts on the vertical projection and is a separate component
   C. Yes, it is multiplied by tan θ
   D. No, because it is zero
   Answer: B. The two components are independent. (Key point: Horizontal force)

3. Tread k is at depth 0.3k m, with horizontal area 0.9 m^2. How are loads added?
   A. Use only the lowest tread
   B. Use depth 3 m for each tread
   C. Sum of γ (0.3k)(0.9) over k = 1 to 10 gives water volume 14.85 m^3
   D. Subtract the loads
   Answer: C. Sum of k = 55 gives 3 x 0.3 x 0.3 x 55 = 4.95 x 3 = 14.85 m^3. (Key point: Tread loads)

4. What is the vertical force?
   A. About 146 kN downward
   B. About 1457 kN
   C. About 29 kN
   D. About 88 kN
   Answer: A. F_V = 9.81 x 14.85 = 145.7 kN; in general γ w (0.3)^2 n(n+1)/2. (Key point: Result)

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