--- title: "Gravity dam stability (SG 2.4, RL 0-70)" subject: "Hydropower Engineering" chapter: 4 key: "semester_7/hydropower_engineering:gravity-dam-stability-2082-bhadra" source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/gravity-dam-stability-2082-bhadra site: BCT Engineering --- # Gravity dam stability (SG 2.4, RL 0-70) Hydropower Engineering · Chapter 4: Headworks of Storage Plants - Asked in: 2082 Bhadra, 12 marks - 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-2082-bhadra - Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-4 ## Question A concrete gravity dam with a specific gravity of 2.4 has the dimensions as shown in the figure below. Neglect all other forces except self-weight, hydrostatic force, and uplift pressure. The friction coefficient between the base and foundation of the dam is 0.70, the coefficient of uplift pressure is 0.5, and the allowable crushing strength and tensile strength of dam materials are 30 kgf/cm² and 5 kgf/cm², respectively. Check the stability of the dam. [Figure: dam section with base at RL 0 m and crest at RL 70 m, crest width 5 m; water level at 65 m; upstream face vertical from the crest down to RL 30 m, then flaring outward with a 1.0 : 1.0 slope to the base at RL 0 m; downstream face vertical for the top 10 m, then sloping at 0.6 : 1.0 to the base.] ## Short answer For the 71 m base, 65 m head dam, overturning FS = 2.35 (>1.5), sliding FS = 1.61 (>1.0). Eccentricity e = 7.61 m < B/6 = 11.83 m, so no tension. Max compressive stress at toe = 1103 kN/m² (< 2943 kN/m² allowable); heel stress = 240 kN/m² (compression). The dam is stable. ## Answer #### Assumptions - Base at RL 0 m. U/s face: vertical from the crest (RL 70) down to RL 30, then a 1:1 flare out to the heel (horizontal projection 30 m). D/s face: vertical for 10 m (to RL 60), then 0.6 (H) : 1 (V) to the toe (horizontal projection $0.6\times60=36$ m). - Base width $B=30+5+36=71$ m. Water depth $H_w=65$ m, no tail water. Per metre length of dam. - $\gamma_c = 2.4\times9.81=23.54$ kN/m³, $\gamma_w=9.81$ kN/m³. - Uplift: triangular, intensity at the heel $=0.5\,\gamma_wH_w = 0.5\times9.81\times65 = 318.8$ kN/m², zero at the toe (uplift coefficient 0.5). - Allowable crushing $30$ kgf/cm² $=2943$ kN/m²; allowable tension $5$ kgf/cm² $=490.5$ kN/m². ```text RL70 +-----+ | | RL65 ~|~~~~~| water level | | d/s vertical to RL60 | \ RL30 | \ 0.6 : 1 /| \ RL0 /_|__________\ heel toe B = 71 m ``` #### Forces and moments about the toe | Force | Magnitude (kN/m) | Arm from toe (m) | Moment about toe (kN·m/m) | |---|---|---|---| | **Vertical (resisting)** | | | | | Flare triangle (30 × 30) | 10,594.8 | 51.00 | 540,335 | | Rectangle 5 × 70 | 8,240.4 | 38.50 | 317,255 | | D/s triangle (36 × 60) | 25,427.5 | 24.00 | 610,260 | | Water on u/s face | 14,715.0 | 57.50 | 846,112 | | **ΣW (downward)** | **58,977.7** | | **ΣM_R = 2,313,963** | | **Horizontal / uplift (overturning)** | | | | | Water thrust $P_w=\frac12\gamma_w H_w^2$ | 20,723.6 | 21.67 | 449,012 | | Uplift part 1 (upward) | 11,318.3 | 47.33 | 535,732 | | **Uplift U** | **11,318.3** | | | | **Total overturning** | | | **ΣM_O = 984,744** | (The water thrust acts at $H_w/3 = 21.67$ m above the base. Uplift is a triangle of area $\tfrac12\times318.8\times71$ acting at $B/3$ from the heel.) #### Stability checks - $\Sigma V = 58,977.7-11,318.3 = 47,659.4$ kN/m; $\Sigma H = 20,723.6$ kN/m - $\Sigma M_R = 2,313,963$ kN·m/m; $\Sigma M_O = 984,744$ kN·m/m **Overturning:** $FS_o = \dfrac{\Sigma M_R}{\Sigma M_O} = \dfrac{2,313,963}{984,744} = 2.35 > 1.5$ — safe. **Sliding:** $FS_s = \dfrac{\mu\,\Sigma V}{\Sigma H} = \dfrac{0.70\times47,659.4}{20,723.6} = 1.61 > 1.0$ — safe (friction only). **Location of resultant:** $\bar x = \dfrac{\Sigma M_R-\Sigma M_O}{\Sigma V} = \dfrac{1,329,219}{47,659.4} = 27.89$ m from the toe. $e = B/2-\bar x = 35.5-27.89 = 7.61$ m $< B/6 = 11.83$ m — resultant is inside the middle third, so no tension. **Base stresses:** $$ \sigma_{toe} = \frac{\Sigma V}{B}\left(1+\frac{6e}{B}\right) = 671.3\,(1+0.643) = 1102.9\ \text{kN/m}^2 $$ $$ \sigma_{heel} = 671.3\,(1-0.643) = 239.6\ \text{kN/m}^2 $$ - Maximum compressive stress $=1103$ kN/m² $< 2943$ kN/m² (allowable) — safe against crushing. - Minimum stress $=240$ kN/m² is positive (compression) — no tension, and it is greater than $-490.5$ kN/m² in any case. - Principal stress at the toe (no tail water): $\sigma_1=\sigma_{toe}(1+\tan^2\phi_d) = 1102.9\,(1+0.6^2)=1500$ kN/m² $<2943$ kN/m² — safe. **Conclusion:** The dam is safe against overturning (2.35), sliding (1.61), tension and crushing. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the base width B = 71 m obtained? A. 30 m flare run + 5 m crest + 36 m d/s run (0.6 x 60) B. 70 + 1 C. 5 + 36 + 30 + 10 D. 30 + 0.6 x 70 Answer: A. The u/s flare gives 30 m horizontally, the crest 5 m, and the d/s slope 0.6 x 60 = 36 m, totalling 71 m. (Key point: Base width and geometry) 2. Which results and formulae give the factors of safety? A. FSo = ΣH/ΣV = 0.44; FSs = ΣM_O/ΣM_R B. FSo = ΣM_R/ΣM_O = 2.35 and FSs = μΣV/ΣH = 1.61, both safe C. FSo = μΣV/ΣH and FSs = ΣM_R/ΣM_O, both below 1 D. FSo = ΣV/ΣH = 2.3 only Answer: B. Overturning: resisting moment over overturning moment = 2.35 (>1.5). Sliding: μΣV/ΣH = 0.7 x 47,659/20,724 = 1.61 (>1). (Key point: Overturning and sliding checks) 3. How is uplift treated when finding ΣV? A. Added to the weight B. Ignored C. Subtracted from the downward weight (ΣV = 58,977.7 - 11,318.3) D. Treated as a horizontal force Answer: C. Uplift acts upward, so it reduces the net vertical load: ΣV = 47,659.4 kN/m. (Key point: Uplift in the vertical force) 4. Why is there no tension, and how do the base stresses compare with allowables? A. e = 7.61 m < B/6 = 11.83 m, so heel stress is 240 kN/m² (compressive) and toe stress 1103 kN/m² < 2943 kN/m² B. e = 7.61 m > B/6, so tension appears at the heel C. e = 35.5 m, so the toe stress is zero D. Stresses are not compared because uplift is neglected Answer: A. The resultant lies within the middle third, so the minimum stress is positive and the maximum is well below the 2943 kN/m² crushing limit. 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