---
title: "Design elementary profile of masonry dam"
subject: "Hydropower Engineering"
chapter: 4
key: "semester_7/hydropower_engineering:elementary-profile-stone-masonry-design"
source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/elementary-profile-stone-masonry-design
site: BCT Engineering
---

# Design elementary profile of masonry dam

Hydropower Engineering · Chapter 4: Headworks of Storage Plants

- Asked in: 2071 Chaitra, 8 marks
- Video: [Analysis And Design of Masonary Dam](https://www.youtube.com/watch?v=lDGH2J4Y52s) — Career Research With Ranjeet
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/hydropower_engineering/elementary-profile-stone-masonry-design
- Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-4

## Question

Design an elementary profile of a gravity dam made of stone masonry using the following data: RL of base of dam = 198 m; HFL = 228 m; specific gravity of masonry = 2.4; safe compressive stress in masonry = 1200 kN/m²; tan φ = 0.70; seepage coefficient = 1.

## Answer

#### Data
$H=228-198=30$ m, $S_c=2.4$, $K=1$ (full uplift), $\tan\phi=\mu=0.70$, permissible stress $1200$ kN/m², $\gamma_w=9.81$ kN/m³. Reservoir full (HFL at the apex).

#### Base width from stability conditions
**No tension:**
$$
B_1=\frac{H}{\sqrt{S_c-K}}=\frac{30}{\sqrt{2.4-1}}=25.35\ \text{m}
$$

**Sliding ($FS=1$):**
$$
B_2=\frac{H}{\mu(S_c-K)}=\frac{30}{0.70\times(2.4-1)}=30.61\ \text{m}
$$

The larger value governs: **adopt $B=31$ m**, with the vertical u/s face and the d/s slope $B/H=1.03:1$.

#### Check at $B=31$ m (per metre)
- $W=\frac12\times31\times30\times2.4\times9.81=10,948$ kN; $U=\frac12\times1\times9.81\times30\times31=4,562$ kN; $P=\frac12\times9.81\times30^2=4,414$ kN
- $\Sigma V=W-U=6,386$ kN; $\Sigma M_R=W\cdot\frac23B=226,258$; $\Sigma M_O=P\frac{H}{3}+U\frac23B=138,419$ kN·m
- $\bar x=\dfrac{87,839}{6,386}=13.75$ m from the toe, $e=B/2-\bar x=1.75$ m $<B/6=5.17$ m: no tension.
- Sliding: $\mu\Sigma V/P=0.7\times6,386/4,414=1.01\ge1$ — safe.
- Overturning: $\Sigma M_R/\Sigma M_O=1.63$.

$$
\sigma_{toe}=206.0\left(1+\frac{6\times1.75}{31}\right)=275.6\ \text{kN/m}^2,\qquad \sigma_{heel}=136.4\ \text{kN/m}^2
$$

Principal stress at the toe $\sigma_1=\sigma_{toe}(1+\tan^2\phi_d)=275.6\,(1+1.068)=569.9$ kN/m² $<1200$ kN/m² — safe.

```text
  HFL 228 ^   apex
          |\
          | \
     30 m |  \   d/s slope
          |   \
  198 ----+----\
         B = 31 m
```

**Result:** elementary profile with base width $B=31$ m, height 30 m, vertical u/s face. For the practical section add a crest width (about 3-4 m) and a freeboard above HFL (2-3 m).

## Check yourself

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

1. For an elementary profile with uplift coefficient K, which expression gives the base width that avoids tension?
   A. B = H/(mu(Sc-K))
   B. B = H/sqrt(Sc-K)
   C. B = H/sqrt(Sc+K)
   D. B = H(Sc-K)
   Answer: B. No tension needs the resultant at the lower third point, which gives B = H/sqrt(Sc-K) = 25.35 m here. (Key point: No-tension base width)

2. The no-tension width is 25.35 m and the sliding width is 30.61 m. Which width is adopted and why?
   A. 25.35 m, because the smaller value is economical
   B. The average of the two widths
   C. 31 m (about 30.61 m), because the larger value satisfies both conditions
   D. 25.35 m, because sliding is never critical
   Answer: C. The larger value governs so that both no-tension and sliding conditions are satisfied; B is rounded up to 31 m. (Key point: Governing width)

3. In this design, which formula gives the base width required for sliding with FS = 1?
   A. B = H/(mu(Sc-K))
   B. B = mu H/(Sc-K)
   C. B = H/sqrt(mu(Sc-K))
   D. B = H(Sc-K)/mu
   Answer: A. Equating mu*(W-U) to the water thrust gives B = H/(mu(Sc-K)) = 30/(0.7x1.4) = 30.61 m. (Key point: Sliding width formula)

4. After finding B = 31 m, how is the toe stress verified against the 1200 kN/m^2 limit?
   A. Using sigma = (SigmaV/B)(1+6e/B), then principal stress sigma1 = sigma(1+tan^2 phi_d) is compared with 1200
   B. Only checking that the weight W is less than 1200 kN
   C. Comparing the water thrust P with 1200 kN/m^2
   D. Checking only that e is greater than B/6
   Answer: A. The vertical toe stress is computed from the eccentricity, and the principal stress at the toe (569.9 kN/m^2) is below 1200. (Key point: Toe stress check)

## Related in Hydropower Engineering

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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
- [Minimum safe width of 18 m elementary profile](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Aelementary-profile-min-width-18m) — Hydropower Engineering, chapter 4
- [Gravity dam equilibrium and overtopping discharge coefficient](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Agravity-dam-reservoir-capacity-curve-hrwl) — Hydropower Engineering, chapter 4
- [Gravity dam stability and shear stress at toe (2081)](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Agravity-dam-stability-2081-bhadra) — Hydropower Engineering, chapter 4
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