---
title: "Elementary profile base width formula proof"
subject: "Hydro Power"
chapter: 4
key: "semester_6/hydro_power:elementary-profile-base-width-proof"
source: https://www.bctengineer.com/q/semester_6/hydro_power/elementary-profile-base-width-proof
site: BCT Engineering
---

# Elementary profile base width formula proof

Hydro Power · Chapter 4: Dam Engineering

- Asked in: 2077 Chaitra, 7 marks
- Video: [[HYDROPOWER #2] Base width of Elementary Profile of Dam | Why it is Theoretical concept?](https://www.youtube.com/watch?v=wrwxs_0pQBM) — Just Learn Civil
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/hydro_power/elementary-profile-base-width-proof
- Whole chapter: https://www.bctengineer.com/md/semester_6/hydro_power/chapter-4

## Question

Explain elementary profile of dam and prove that B = H/√(G − K) where B is base width, H is height of water at the upstream side of the dam, G is specific gravity of dam material and K is uplift pressure intensity coefficient.

## Short answer

The elementary profile is a right triangle with vertical upstream face and water to apex. For no tension, resultant passes through outer third point; moment equilibrium of weight, uplift and water pressure gives B = H/√(G−K).

## Answer

#### Elementary profile
The **elementary profile** of a gravity dam is a right-angled triangle with a vertical upstream face, zero top width and base width B, with water up to its apex (height H). Water pressure is zero at the top and grows linearly with depth, so a triangle is the ideal shape: it uses the least material and, if the base is wide enough, has no tension under full or empty reservoir. Practical profiles are made by adding top width and freeboard to it.

```text
        |\
        | \
  P --> |  \      H
  (H/3) | W \
        |    \
   heel |__.__\ toe
        ^  ^   ^
        U (KγwH at heel -> 0 at toe)
        <--B-->
```

#### Forces per metre length
With $G$ = specific gravity of dam material, $K$ = uplift intensity coefficient:

$$
\begin{aligned}
W &= \frac{1}{2}BH\,G\gamma_w \quad (\text{at } B/3 \text{ from heel}) \\
P &= \frac{1}{2}\gamma_w H^2 \quad (\text{at } H/3 \text{ above base}) \\
U &= \frac{1}{2}K\gamma_w H B \quad (\text{at } B/3 \text{ from heel})
\end{aligned}
$$

#### Proof of $B = H/\sqrt{G-K}$ (no-tension condition)
For no tension at the heel with full reservoir, the resultant must pass through the outer middle-third point, i.e. at $2B/3$ from the heel ($B/3$ from the toe). Taking moments about this point:
- W and U act at $B/3$ from the heel, so their lever arm is $2B/3 - B/3 = B/3$.
- P has lever arm $H/3$.

For the resultant to pass through this point, the net moment must be zero:

$$
\begin{aligned}
(W - U)\frac{B}{3} &= P\frac{H}{3} \\
\left(\frac{1}{2}BHG\gamma_w - \frac{1}{2}K\gamma_w HB\right)B &= \frac{1}{2}\gamma_w H^2 \cdot H \\
\frac{1}{2}\gamma_w B^2 H (G - K) &= \frac{1}{2}\gamma_w H^3 \\
B^2 (G - K) &= H^2 \\
B &= \frac{H}{\sqrt{G - K}}
\end{aligned}
$$

Hence proved. Without uplift ($K = 0$), $B = H/\sqrt{G}$; for concrete ($G = 2.4$) this gives $B = 0.645H$, and with full uplift ($K = 1$) $B = 0.845H$.

#### No-sliding condition (for comparison)
$\mu(W - U) = P$ gives $B = \dfrac{H}{\mu(G - K)}$. The larger of the two widths is adopted.

## Check yourself

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

1. Why is the triangle the ideal shape?
   A. Water pressure grows linearly with depth, zero at top, so it uses least material
   B. Triangles are cheaper to form
   C. Uplift is zero
   D. The top width is large
   Answer: A. The profile matches the pressure diagram. (Key point: Why triangle)

2. Which condition is used to prove B = H/sqrt(G - K)?
   A. Stress zero at toe
   B. Resultant passes through the outer middle-third point, so no tension at heel
   C. FS sliding = 1
   D. Resultant at the centre
   Answer: B. No tension means e = B/6, resultant at B/3 from the toe. (Key point: Condition used)

3. About that point, what are the lever arms of W-U and P?
   A. 2B/3 and H/3
   B. B/3 and H
   C. B/3 and H/3
   D. B/2 and H/2
   Answer: C. Point at 2B/3 from heel and W, U act at B/3 from heel, so the arm is B/3; P acts at H/3. (Key point: Moment arms)

4. For concrete G = 2.4, what base width results with K = 0?
   A. B = 0.845 H
   B. B = 1.55 H
   C. B = H
   D. B = 0.645 H
   Answer: D. B = H/sqrt(2.4) = 0.645H; with K = 1, B = H/sqrt(1.4) = 0.845H. (Key point: Resulting widths)

## Related in Hydro Power

- [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
- [Elementary profile width by no tension and no sliding](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Aelementary-profile-no-tension-sliding) — Hydro Power, chapter 4, asked 2×
- [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
- [Sliding and overturning check of gravity dam](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Agravity-dam-sliding-overturning-figure) — 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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- [Elementary profile of gravity dam](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Aelementary-profile-definition-proportioning) — Hydropower Engineering, chapter 4
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