--- title: "Stability checks of 50 m trapezoidal dam" subject: "Hydro Power" chapter: 4 key: "semester_6/hydro_power:trapezoidal-dam-50m-checks" source: https://www.bctengineer.com/q/semester_6/hydro_power/trapezoidal-dam-50m-checks site: BCT Engineering --- # Stability checks of 50 m trapezoidal dam Hydro Power · Chapter 4: Dam Engineering - Asked in: 2073 Bhadra, 8 marks - Video: [Design of Gravity dam numerical || stability check of gravity dam numerical](https://www.youtube.com/watch?v=A0W93k5bjvs) — CIVIL ZONE - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/hydro_power/trapezoidal-dam-50m-checks - Whole chapter: https://www.bctengineer.com/md/semester_6/hydro_power/chapter-4 ## Question A concrete dam of trapezoidal section is of height 50 m with upstream vertical face. Top width, base width and freeboard are 4.5 m, 42 m and 5 m respectively. Check the stability of the dam against (i) Overturning (ii) sliding (iii) Tension and (iv) Crushing. The unit weight of water is 10 kN/m³, the unit weight of concrete is 25 kN/m³ and Crushing strength of the material 10 kg/cm². Take angle of internal friction (ϕ = 37°), uplift coefficient 0.8. Consider self-weight, hydraulic pressure and uplift pressure only. Mention your recommendation. ## Short answer FS overturning 2.23, sliding 1.60, eccentricity 0.26 m (no tension), maximum principal stress 829.6 kN/m² below permissible 981 kN/m². Safe; section is conservative. ## Answer #### Data (per metre length) - Dam height 50 m, freeboard 5 m, so water depth $h = 45$ m (no tailwater). - Top width 4.5 m, base 42 m, upstream face vertical; downstream slope $\tan\alpha = 37.5/50 = 0.75$. - $\gamma_w = 10$, $\gamma_c = 25$ kN/m³; $\mu = \tan 37° = 0.754$; uplift coefficient C = 0.8. - Crushing strength $= 10$ kg/cm² $= 10 \times 9.81 \times 10^4$ N/m² $= 981$ kN/m². ```text 4.5 +--+ --- | |\ 5 m FB 45 |W1| \ m | |W2\ 50 m -> | | \ heel+--+----\ toe <-- 42 m --> ``` #### Forces and moments about the toe | Force | Value (kN) | Arm (m) | Moment (kN·m) | |---|---|---|---| | W1 = 4.5×50×25 | 5,625 ↓ | 39.75 | +223,594 | | W2 = ½×37.5×50×25 | 23,437.5 ↓ | 25.00 | +585,938 | | P = ½×10×45² | 10,125 → | 15.00 | −151,875 | | U = ½×0.8×10×45×42 | 7,560 ↑ | 28.00 | −211,680 | $$ \begin{aligned} \sum V &= 5{,}625 + 23{,}437.5 - 7{,}560 = 21{,}502.5\ \text{kN} \\ \sum M_R &= 809{,}531, \quad \sum M_O = 363{,}555\ \text{kN·m} \end{aligned} $$ #### (i) Overturning $$ FS_o = \frac{809{,}531}{363{,}555} = 2.23 > 1.5 \quad \text{(safe)} $$ #### (ii) Sliding $$ FS_s = \frac{0.754 \times 21{,}502.5}{10{,}125} = 1.60 > 1 \quad \text{(safe)} $$ #### (iii) Tension $$ \begin{aligned} \bar{x} &= \frac{809{,}531 - 363{,}555}{21{,}502.5} = 20.74\ \text{m from toe} \\ e &= 21 - 20.74 = 0.26\ \text{m} < B/6 = 7\ \text{m} \end{aligned} $$ No tension (safe). #### (iv) Crushing $$ \begin{aligned} p_{toe} &= \frac{21{,}502.5}{42}\left(1 + \frac{6\times0.26}{42}\right) = 511.97 \times 1.037 = 530.9\ \text{kN/m}^2 \\ p_{heel} &= 511.97 \times 0.963 = 493.0\ \text{kN/m}^2 \\ \sigma_1 &= p_{toe}\sec^2\alpha = 530.9 \times 1.5625 = 829.6\ \text{kN/m}^2 \end{aligned} $$ Both are below 981 kN/m² (safe). | Check | Value | Limit | Result | |---|---|---|---| | Overturning | 2.23 | ≥ 1.5 | Safe | | Sliding | 1.60 | ≥ 1.0 | Safe | | Tension | e = 0.26 m | ≤ 7 m | Safe | | Crushing | 829.6 kN/m² | ≤ 981 kN/m² | Safe | **Recommendation:** The dam is safe in all four checks. The resultant is almost at the centre of the base, so stresses are nearly uniform; the section is on the heavy side and the downstream slope could be steepened to save concrete, after re-checking with earthquake and silt loads. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Height 50 m and freeboard 5 m; phi = 37 deg. What is used? A. h = 45 m and mu = tan 37 = 0.754 B. h = 50 m and mu = sin 37 C. h = 55 m and mu = cos 37 D. h = 45 m and mu = 0.37 Answer: A. Water stands 5 m below the crest, and mu = tan(phi). (Key point: Water depth and friction) 2. How is 10 kg/cm2 converted? A. 10 kN/m2 B. 981 kN/m2 C. 9,810 kN/m2 D. 98.1 kN/m2 Answer: B. 10 x 9.81 x 10^4 N/m2 = 981 kN/m2. (Key point: Crushing strength unit) 3. e = 0.26 m against B/6 = 7 m. What does this indicate? A. Resultant outside the middle third B. Tension at heel C. No tension, stresses nearly uniform D. Overturning Answer: C. e is far below B/6 so no tension and the resultant is near the centre. (Key point: Tension check) 4. sigma1 = p_toe sec^2(alpha) = 829.6 kN/m2. What is the conclusion? A. Unsafe in sliding B. It must be redesigned wider C. It exceeds 981, so unsafe D. It is below 981 kN/m2, so the dam is safe and slightly heavy, so the slope could be steepened Answer: D. All four checks pass, and the near-central resultant suggests economy by steepening the downstream face. 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