--- title: "Stability of gravity dam (crest 1530 m)" subject: "Hydro Power" chapter: 4 key: "semester_6/hydro_power:gravity-dam-stability-1530" source: https://www.bctengineer.com/q/semester_6/hydro_power/gravity-dam-stability-1530 site: BCT Engineering --- # Stability of gravity dam (crest 1530 m) Hydro Power · Chapter 4: Dam Engineering - Asked in: 2079 Chaitra, 10 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/hydro_power/gravity-dam-stability-1530 - Whole chapter: https://www.bctengineer.com/md/semester_6/hydro_power/chapter-4 ## Question Check the stability of a gravity dam having trapezoidal section with following data: Crest level of dam = 1530 m, Maximum water level at upstream = 1525 m, Top width = 6 m, Base width = 60 m, Tail water depth = 6 m, Bed level = 1470 m. Drainage gallery provided at 10 m from the heel. Consider full uplift condition and permissible compressive strength of concrete 3000 kN/m². Take specific weight of concrete 24 kN/m³. ## Short answer FS overturning 2.74, sliding 1.76, resultant 2.55 m from centre towards heel (no tension), maximum compressive stress 772.7 kN/m² at heel < 3000 kN/m². Dam is safe. ## Answer #### Data and assumptions (per metre length) - Dam height $= 1530 - 1470 = 60$ m; upstream water depth $h_1 = 1525 - 1470 = 55$ m; tailwater depth $h_2 = 6$ m. - Top width 6 m, base 60 m, upstream face vertical, downstream face straight from crest to toe: $\tan\alpha = 54/60 = 0.9$. - $\gamma_w = 9.81$, $\gamma_c = 24$ kN/m³; μ (not given) assumed 0.7. - "Full uplift": full hydrostatic uplift at the heel; with the gallery, IS 6512 rule: head at gallery $= h_2 + \frac{1}{3}(h_1 - h_2) = 6 + 49/3 = 22.33$ m. Uplift intensities: heel 539.6, gallery 219.1, toe 58.9 kN/m². ```text 6 +--+ 1530 | |\ 55 | | \ 60 m m |W1|W2\ | | \ ~~ TW 6 m +--+----\ 1470 <10>|<-50-> gallery B = 60 m ``` #### Forces and moments about the toe | Force | Value (kN) | Arm (m) | Moment (kN·m) | |---|---|---|---| | W1 = 6×60×24 | 8,640 ↓ | 57.0 | +492,480 | | W2 = ½×54×60×24 | 38,880 ↓ | 36.0 | +1,399,680 | | Tailwater wedge = ½×5.4×6×9.81 | 158.9 ↓ | 1.8 | +286 | | P1 = ½×9.81×55² | 14,837.6 → | 18.333 | −272,023 | | P2 = ½×9.81×6² | 176.6 ← | 2.0 | +353 | | U1 = 219.1×10 | 2,190.9 ↑ | 55.0 | −120,500 | | U2 = ½×320.5×10 | 1,602.3 ↑ | 56.667 | −90,797 | | U3 = 58.9×50 | 2,943.0 ↑ | 25.0 | −73,575 | | U4 = ½×160.2×50 | 4,005.8 ↑ | 33.333 | −133,525 | $$ \begin{aligned} \sum U &= 10{,}742\ \text{kN} \\ \sum V &= 8{,}640 + 38{,}880 + 158.9 - 10{,}742 = 36{,}937\ \text{kN} \\ \sum H &= 14{,}837.6 - 176.6 = 14{,}661\ \text{kN} \\ \sum M_R &= 1{,}892{,}799, \quad \sum M_O = 690{,}420\ \text{kN·m} \end{aligned} $$ #### Overturning $$ FS_o = \frac{1{,}892{,}799}{690{,}420} = 2.74 > 1.5 \quad \text{(safe)} $$ #### Sliding $$ FS_s = \frac{0.7 \times 36{,}937}{14{,}661} = 1.76 > 1 \quad \text{(safe)} $$ #### Tension $$ \begin{aligned} \bar{x} &= \frac{1{,}892{,}799 - 690{,}420}{36{,}937} = 32.55\ \text{m from toe} \\ e &= 30 - 32.55 = -2.55\ \text{m} \end{aligned} $$ The resultant is 2.55 m from the centre on the heel side; $|e| < B/6 = 10$ m, so **no tension**. #### Compression $$ \begin{aligned} \frac{\sum V}{B} &= 615.6\ \text{kN/m}^2, \quad \frac{6|e|}{B} = 0.255 \\ p_{heel} &= 615.6(1 + 0.255) = 772.7\ \text{kN/m}^2 \\ p_{toe} &= 615.6(1 - 0.255) = 458.5\ \text{kN/m}^2 \end{aligned} $$ Principal stress at toe $= 458.5(1.81) - 58.9(0.81) = 782.2$ kN/m². All stresses are well below 3000 kN/m² (safe). **Answer: FS overturning = 2.74, FS sliding = 1.76, e = 2.55 m (within middle third, no tension), maximum stress = 772.7 kN/m² at heel < 3000 kN/m². The dam is safe.** (If drains are ignored and full uplift acts over the whole base, FS overturning = 1.98 and FS sliding = 1.42; still safe.) ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Crest 1530, MWL 1525, bed 1470, tailwater depth 6 m. Which is correct? A. Height 60 m, upstream water depth 55 m B. Height 55 m, water depth 55 m C. Height 55 m, water depth 60 m D. Height 60 m, water depth 60 m Answer: A. Height = 1530 - 1470 = 60 m and h1 = 1525 - 1470 = 55 m. (Key point: Depths from levels) 2. What head acts at the gallery? A. 6 m B. 6 + 49/3 = 22.33 m C. 55 m D. 30.5 m Answer: B. Tailwater head plus one-third of (55 - 6) = 22.33 m. (Key point: Uplift head at gallery) 3. e = -2.55 m (x_bar = 32.55 m from the toe, B/2 = 30). What does the negative sign mean and where is maximum stress? A. The dam overturns B. Tension at heel C. Resultant is on the heel side of centre; maximum stress is at the heel (772.7 kN/m2) D. Resultant is on the toe side; maximum stress at the toe Answer: C. x_bar = 32.55 m from the toe exceeds B/2 = 30 m, so the resultant lies on the heel side of centre and the heel carries the larger stress. (Key point: Eccentricity side) 4. Which result summary is correct? A. FS_o = 1.2, unsafe in overturning B. Stress exceeds 3000 kN/m2 C. Tension develops at the heel D. FS_o = 2.74, FS_s = 1.76, no tension, max stress 772.7 < 3000 kN/m2, safe Answer: D. All four checks pass. 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