--- title: "Gravity dam stability: flotation, overturning, sliding" subject: "Hydropower Engineering" chapter: 4 key: "semester_7/hydropower_engineering:gravity-dam-stability-flotation-shear-resistance" source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/gravity-dam-stability-flotation-shear-resistance site: BCT Engineering --- # Gravity dam stability: flotation, overturning, sliding Hydropower Engineering · Chapter 4: Headworks of Storage Plants - Asked in: 2075 Ashwin, 8 marks - Video: [SOLVED CE BOARD GRAVITY DAM PROBLEM | FLUID MECHANICS | DE LA CRUZ TUTORIALS](https://www.youtube.com/watch?v=qoNnwPnzvYA) — Engr. Jobert S. De La Cruz - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/hydropower_engineering/gravity-dam-stability-flotation-shear-resistance - Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-4 ## Question A concrete gravity dam of the given profile is proposed by a designer for implementation. The unit shear resistance and angle of resistance are 500 kN/m² and 35° respectively. $\gamma_{con} = 24$ kN/m³. Check the stability of the dam against flotation, overturning and sliding. [Figure: dam section with crest width 6 m, 5 m dimension at the top, 15 m dimension on the downstream face, dam height 60 m, tail water depth 40 m, base width 26 m.] ## Short answer The dam is safe against flotation (FS 1.89) but unsafe against overturning (FS 1.07 < 1.5). Sliding safety factor using shear-friction is 2.94, marginal; tension develops at the heel. The base width is insufficient for stability. ## Answer #### Assumptions - Height 60 m, crest width 6 m, base width $B=26$ m. Water level 5 m below the crest, so $H_w=55$ m. Tail water $H_t=40$ m. U/s face vertical; the d/s face is vertical for the top 15 m, then slopes to the toe (20 m run over 45 m). - $\gamma_c=24$, $\gamma_w=9.81$ kN/m³. Uplift (no drainage): $\gamma_wH_w=539.6$ kN/m² at the heel and $\gamma_wH_t=392.4$ kN/m² at the toe (linear). - Unit shear resistance $q=500$ kN/m², angle of shear resistance $\phi=35^\circ$. ```text 6 m +----+ | | 5 m freeboard ~|~~~~| water 55 m | | | \ 15 m vertical d/s | \ | \ slope (20 m run) | \ ~~ tail water 40 m +--------\ B = 26 m ``` #### Forces and moments about the toe | Force | Magnitude (kN/m) | Arm from toe (m) | Moment about toe (kN·m/m) | |---|---|---|---| | **Vertical (resisting)** | | | | | Rectangle 6 × 60 | 8,640.0 | 23.00 | 198,720 | | D/s triangle (20 × 45) | 10,800.0 | 13.33 | 144,000 | | Tail water on d/s face | 3,488.0 | 5.93 | 20,670 | | **ΣW (downward)** | **22,928.0** | | **363,390** | | **Horizontal / uplift (overturning)** | | | | | Water thrust $P_w=\frac12\gamma_w H_w^2$ | 14,837.6 | 18.33 | 272,023 | | Uplift part 1 (upward) | 10,202.4 | 13.00 | 132,631 | | Uplift part 2 (upward) | 1,913.0 | 17.33 | 33,158 | | **Uplift U** | **12,115.4** | | | | Tail water thrust (resisting) | 7,848.0 | 13.33 | 104,640 | | **Total overturning** | | | **ΣM_O = 437,812** | Total resisting moment: $\Sigma M_R=363,390+104,640=468,030$ kN·m/m (weights plus the tail water thrust). #### Checks **1. Flotation (uplift):** $$ FS_f=\frac{\Sigma W}{U}=\frac{22,928.0}{12,115.4}=1.89>1 $$ The dam does not float (its weight is 1.89 times the uplift). **2. Overturning:** $$ FS_o=\frac{\Sigma M_R}{\Sigma M_O}=\frac{468,030}{437,812}=1.07 $$ This is far below the required 1.5, so the dam is **not safe against overturning**. **3. Sliding (shear-friction):** $\Sigma V=22,928.0-12,115.4=10,812.6$ kN/m and $\Sigma H=14,837.6-7,848.0=6,989.6$ kN/m. $$ SFF=\frac{qB+\Sigma V\tan\phi}{\Sigma H}=\frac{500\times26+10,812.6\times\tan35^\circ}{6,989.6}=\frac{13000+7,571.1}{6,989.6}=2.94 $$ Friction alone gives $\tan35^\circ\,\Sigma V/\Sigma H=1.08$. With the shear resistance, $SFF=2.94$, which is below the usual 3-5, so sliding is also marginal. **4. Position of the resultant:** $\bar x=30,218/10,812.6=2.79$ m from the toe, $e=10.21$ m $>B/6=4.33$ m, so tension develops at the heel ($\sigma_{heel}=-564$ kN/m², $\sigma_{toe}=1395$ kN/m²). **Conclusion:** the profile is safe against flotation but **unsafe against overturning**, and it is marginal in sliding and has tension at the heel. The base must be widened or the uplift reduced by drainage. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the safety against flotation checked? A. Total weight of dam / total uplift force, which must exceed 1 B. Uplift / weight, which must exceed 1 C. Water thrust / weight D. Resisting moment / overturning moment Answer: A. The dam floats if uplift exceeds the weight; FS_f = SumW / U = 1.89 > 1, so it does not float. (Key point: Flotation check) 2. What is the conclusion on overturning and why? A. Safe, because FS_o = 1.07 is more than 1 B. Not safe, because FS_o = 1.07 is below the required 1.5 C. Safe, because tail water adds resisting moment D. Cannot be checked without cohesion Answer: B. Overturning needs SumM_R / SumM_O of at least 1.5; the section gives only 1.07. (Key point: Overturning result) 3. Which expression is used for sliding with unit shear resistance q and friction angle phi? A. SFF = (SumV tan(phi)) / q B B. SFF = (q B + SumV tan(phi)) / SumH C. SFF = (q + SumV) / SumH D. SFF = SumH / (q B + SumV tan(phi)) Answer: B. The shear-friction factor adds the shear strength over the base (qB) to the friction (SumV tan phi), divided by the horizontal thrust. (Key point: Shear-friction factor) 4. Why does tension develop at the heel? A. The tail water is too deep B. The eccentricity e = 10.21 m exceeds B/6 = 4.33 m, so the resultant is outside the middle third C. The toe stress is above the allowable stress D. Friction coefficient is too low Answer: B. The resultant is only 2.79 m from the toe, outside the middle third, giving a negative (tensile) heel stress. 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