--- title: "Stability of trapezoidal gravity dam (RL 250 m)" subject: "Hydro Power" chapter: 4 key: "semester_6/hydro_power:trapezoidal-dam-stability-250m" source: https://www.bctengineer.com/q/semester_6/hydro_power/trapezoidal-dam-stability-250m site: BCT Engineering --- # Stability of trapezoidal gravity dam (RL 250 m) Hydro Power · Chapter 4: Dam Engineering - Asked in: 2080 Asoj, 10 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-stability-250m - Whole chapter: https://www.bctengineer.com/md/semester_6/hydro_power/chapter-4 ## Question Check the stability of a trapezoidal concrete gravity dam having maximum reservoir level of 250.0 m, base level of 200.0 m, tail water elevation of 210.0 m, base width of 40 m, top width of 6 m, location of drainage gallery 10 m from u/s face. Take sp. wt and permissible compressive strength of the dam material are 24 kN/m³ and 300 kN/m², respectively. ## Short answer For the dam (H=50 m, base 40 m, top 6 m, tailwater 10 m, drainage gallery 10 m from u/s, γc=24 kN/m3, γw=9.81 kN/m3, μ=0.7 assumed): FS overturning = 1.76 (safe); FS sliding = 1.16 (safe); eccentricity e = 3.77 m within middle-third, no tension; toe compressive stress = 760.6 kN/m2, principal stress = 1066.9 kN/m2 exceed permissible 300 kN/m2, crushing failure (if 3000 kN/m2 intended, safe). ## Answer The dam is checked for full reservoir against overturning, sliding, tension and compression (crushing), per metre length. #### Data and assumptions - Upstream water depth $h_1 = 250 - 200 = 50$ m; tailwater depth $h_2 = 210 - 200 = 10$ m. - Dam height is not given, so the crest is taken at the maximum reservoir level: $H = 50$ m. - Upstream face vertical; top width 6 m; downstream face slopes straight from the crest to the toe, so $\tan\alpha = (40-6)/50 = 0.68$. - $\gamma_w = 9.81$ kN/m³, $\gamma_c = 24$ kN/m³. Coefficient of friction (not given) assumed $\mu = 0.7$. - Uplift with drainage gallery (IS 6512 rule): heel $\gamma_w h_1$, at gallery $\gamma_w[h_2 + \frac{1}{3}(h_1 - h_2)]$, toe $\gamma_w h_2$. Uplift intensities: heel $= 9.81 \times 50 = 490.5$ kN/m²; gallery $= 9.81 \times (10 + 40/3) = 9.81 \times 23.33 = 228.9$ kN/m²; toe $= 98.1$ kN/m². ```text 6 m +--+ El. 250 (crest = MRL) | |\ P1 |W1| \ W2 50 m -> | | \ | | \ ~~ TW El. 210 (10 m) heel+--+----\ toe El. 200 <10>|<- 30 -> gallery B = 40 m ``` #### Forces and moments about the toe | Force | Value (kN) | Arm (m) | Moment (kN·m) | |---|---|---|---| | W1 = 6×50×24 | 7,200 ↓ | 37.000 | +266,400 | | W2 = ½×34×50×24 | 20,400 ↓ | 22.667 | +462,400 | | Tailwater wedge = ½×6.8×10×9.81 | 333.5 ↓ | 2.267 | +756 | | P1 = ½×9.81×50² | 12,262.5 → | 16.667 | −204,375 | | P2 = ½×9.81×10² | 490.5 ← | 3.333 | +1,635 | | U1 = 228.9×10 | 2,289 ↑ | 35.000 | −80,115 | | U2 = ½×261.6×10 | 1,308 ↑ | 36.667 | −47,960 | | U3 = 98.1×30 | 2,943 ↑ | 15.000 | −44,145 | | U4 = ½×130.8×30 | 1,962 ↑ | 20.000 | −39,240 | $$ \begin{aligned} \sum U &= 8{,}502\ \text{kN} \\ \sum V &= 7{,}200 + 20{,}400 + 333.5 - 8{,}502 = 19{,}431.5\ \text{kN} \\ \sum H &= 12{,}262.5 - 490.5 = 11{,}772\ \text{kN} \\ \sum M_R &= 731{,}191\ \text{kN·m}, \quad \sum M_O = 415{,}835\ \text{kN·m} \end{aligned} $$ #### (i) Overturning $$ FS_o = \frac{731{,}191}{415{,}835} = 1.76 > 1.5 \quad \text{(safe)} $$ #### (ii) Sliding $$ FS_s = \frac{\mu\sum V}{\sum H} = \frac{0.7 \times 19{,}431.5}{11{,}772} = 1.16 > 1 \quad \text{(safe)} $$ #### (iii) Tension $$ \begin{aligned} \bar{x} &= \frac{731{,}191 - 415{,}835}{19{,}431.5} = 16.23\ \text{m from toe} \\ e &= 20 - 16.23 = 3.77\ \text{m} < B/6 = 6.67\ \text{m} \end{aligned} $$ The resultant is within the middle third, so **no tension**. #### (iv) Compression $$ \begin{aligned} \frac{\sum V}{B} &= 485.8\ \text{kN/m}^2, \quad \frac{6e}{B} = 0.566 \\ p_{toe} &= 485.8(1 + 0.566) = 760.6\ \text{kN/m}^2 \\ p_{heel} &= 485.8(1 - 0.566) = 211.0\ \text{kN/m}^2 \end{aligned} $$ Major principal stress at the toe ($p' = 98.1$ kN/m²): $$ \sigma_1 = p_v\sec^2\alpha - p'\tan^2\alpha = 760.6(1.4624) - 98.1(0.4624) = 1066.9\ \text{kN/m}^2 $$ These exceed the permissible 300 kN/m², so the dam **fails in crushing** as the data stand. **Answer: FS overturning = 1.76 (safe); FS sliding = 1.16 (safe, μ = 0.7 assumed); e = 3.77 m < B/6 (no tension); toe stress 760.6 kN/m² and principal stress 1066.9 kN/m² > 300 kN/m² (unsafe in compression).** A permissible strength of 300 kN/m² is very low for concrete; if 3000 kN/m² was intended, the dam is safe in all four checks. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The dam height is not given. What is assumed for the analysis? A. The crest is at maximum reservoir level, so H = 50 m B. The crest is at tail water level C. The dam is 100 m high D. The height equals the base width of 40 m Answer: A. Crest is taken at the MRL, so H = 250 - 200 = 50 m and the downstream slope is tan(alpha) = 34/50 = 0.68. (Key point: Assumed height) 2. What head is used for uplift at the drainage gallery? A. h2 = 10 m B. h2 + (h1 - h2)/3 = 23.33 m C. h1/2 = 25 m D. h1 = 50 m Answer: B. The IS 6512 rule gives the tailwater head plus one-third of the difference, 10 + 40/3 = 23.33 m. (Key point: Uplift with gallery) 3. Which statement matches the results for overturning, sliding and tension? A. e = 3.77 m > B/6 so tension at the heel B. All four checks fail C. FS overturning 1.76 safe, FS sliding 1.16 safe, e = 3.77 m < B/6 so no tension D. FS overturning 0.8 unsafe, sliding safe, tension present Answer: C. FS_o = 1.76 > 1.5, FS_s = 1.16 > 1 and e = 3.77 < 6.67 m. (Key point: Stability results) 4. Toe stress is 760.6 kN/m2 against permissible 300 kN/m2. What is concluded? A. Safe, since 760.6 > 300 shows reserve strength B. Safe in crushing because e is small C. Crushing is not checked for gravity dams D. The dam fails in crushing as the data stand; 300 kN/m2 is unrealistically low for concrete Answer: D. Stresses exceed the permissible value so it fails; the answer notes 3000 kN/m2 would make it safe. 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