--- title: "Gravity dam forces and stability checks (2081 Baishakh)" subject: "Hydropower Engineering" chapter: 4 key: "semester_7/hydropower_engineering:gravity-dam-forces-stresses-2081-baishakh" source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/gravity-dam-forces-stresses-2081-baishakh site: BCT Engineering --- # Gravity dam forces and stability checks (2081 Baishakh) Hydropower Engineering · Chapter 4: Headworks of Storage Plants - Asked in: 2081 Baishakh, 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-forces-stresses-2081-baishakh - Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-4 ## Question Calculate the forces and the principal stress and the shear stress at the toe and heel of the gravity dam section shown below. Check the dam against sliding, crushing, overturning and tension. Do not consider forces other than self-weight, hydrostatic pressure and uplift pressure. Assume the allowable compressive stress for the material of the foundation is 50 kg/cm², the allowable crushing stress for the material of the dam body is 10.5 kg/cm², the friction coefficient is 0.70 and the uplift coefficient is 0.45. [Figure: dam height 50 m with water against the upstream face; crest width 5 m with 2 m and 3 m dimensions at the top; upstream face flared with slope 1 : 1 near the heel (horizontal 15 m); downstream slope about 2V : 1H; total base width 57 m; tail water depth 5 m; a drainage gallery near the heel.] ## Short answer Using self-weight, hydrostatic and 0.45 uplift, FS overturning=2.60 and sliding=1.80; resultant inside middle third. Toe/heel stresses 644.1 and 364.5 kN/m2, no tension; toe shear 440.4, heel shear 106.4, principal stresses 970.0 and 258.0 kN/m2, all below allowables. ## Answer #### Assumptions (figure not available in full) - Dam height 50 m; crest width 5 m; water level 2 m below the crest, $H_w=48$ m; tail water $H_t=5$ m. - U/s face vertical from the crest to 15 m above the base, then flared at 1 : 1 to the heel (15 m horizontal run). D/s face runs straight from the d/s edge of the crest to the toe; the total base width $B=57$ m gives a d/s run of $57-15-5=37$ m over 50 m ($\tan\phi_d=0.74$). - $\gamma_c=24$, $\gamma_w=9.81$ kN/m³, $\mu=0.70$, uplift coefficient $K=0.45$: heel ordinate $=\gamma_wH_t+K\gamma_w(H_w-H_t)=238.9$ kN/m², toe ordinate $=\gamma_wH_t=49.1$ kN/m² (the gallery near the heel is the reason for this reduced uplift). - Allowable stresses: foundation 50 kgf/cm² $=4905$ kN/m²; dam body 10.5 kgf/cm² $=1030$ kN/m². ```text 5 m +--------+ ~~~|~~~~~~~~| water (48 m) | |\ | | \ d/s slope | | \ /| | \ ~~ tail water 5 m /_|________|____\ 15 m flare B = 57 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 (15 × 15) | 2,700.0 | 47.00 | 126,900 | | Rectangle 5 × 50 | 6,000.0 | 39.50 | 237,000 | | D/s triangle (37 × 50) | 22,200.0 | 24.67 | 547,600 | | Water on u/s face | 5,959.6 | 49.96 | 297,758 | | Tail water on d/s face | 90.7 | 1.23 | 112 | | **ΣW (downward)** | **36,950.3** | | **ΣM_R = 1,209,370** | | **Horizontal / uplift (overturning)** | | | | | Water thrust $P_w=\frac12\gamma_w H_w^2$ | 11,301.1 | 16.00 | 180,818 | | Uplift part 1 (upward) | 2,795.9 | 28.50 | 79,682 | | Uplift part 2 (upward) | 5,410.0 | 38.00 | 205,579 | | **Uplift U** | **8,205.8** | | | | Tail water thrust (resisting) | 122.6 | 1.67 | 204 | | **Total overturning** | | | **ΣM_O = 466,078** | #### Checks - $\Sigma V = \Sigma W - U = 36,950.3 - 8,205.8 = 28,744.5$ kN/m - $\Sigma H = 11,301.1 - 122.6 = 11,178.5$ kN/m - $\Sigma M_R = 1,209,574$ kN·m/m (tail water thrust moment included); $\Sigma M_O = 466,078$ kN·m/m **Overturning:** $FS_o = \Sigma M_R/\Sigma M_O = 1,209,574/466,078 = 2.60$ — safe (> 1.5). **Sliding (friction):** $FS_s = \mu\Sigma V/\Sigma H = 0.7\times28,744.5/11,178.5 = 1.80$ — safe (> 1). **Resultant:** $\bar x = (\Sigma M_R-\Sigma M_O)/\Sigma V = 743,496/28,744.5 = 25.87$ m from the toe; $e = B/2-\bar x = 28.5-25.87 = 2.63$ m; $B/6 = 9.50$ m — inside the middle third (no tension). $$ \sigma_{toe} = \frac{\Sigma V}{B}\left(1+\frac{6e}{B}\right) = 504.3\,(1+0.277) = 644.1\ \text{kN/m}^2 $$ $$ \sigma_{heel} = 504.3\,(1-0.277) = 364.5\ \text{kN/m}^2 $$ The toe stress of 644 kN/m² is below the dam-body allowable of 1030 kN/m². The heel stress of 364 kN/m² is compressive, so there is no tension. **Shear and principal stress at the toe** ($\tan\phi_d=0.740$, $p'=49.1$ kN/m²): $$ \tau_{toe} = (\sigma_{toe}-p')\tan\phi_d = (644.1-49.1)\times0.740 = 440.4\ \text{kN/m}^2 $$ $$ \sigma_{1,toe} = \sigma_{toe}\sec^2\phi_d - p'\tan^2\phi_d = 644.1\,(1+0.548) - 49.1\times0.548 = 970.0\ \text{kN/m}^2 $$ **Shear and principal stress at the heel** ($\tan\phi_u=1.000$, $p=470.9$ kN/m²): $$ \tau_{heel} = (p-\sigma_{heel})\tan\phi_u = (470.9-364.5)\times1.000 = 106.4\ \text{kN/m}^2 $$ $$ \sigma_{1,heel} = \sigma_{heel}\sec^2\phi_u - p\tan^2\phi_u = 258.0\ \text{kN/m}^2 $$ - The maximum principal stress of 970 kN/m² (at the toe) is below the dam-body allowable of 1030 kN/m², and the toe stress is far below the foundation allowable of 4905 kN/m². **Conclusion:** sliding ($FS_s=1.80$), overturning ($FS_o=2.60$), tension (none) and crushing (below 1030 and 4905 kN/m²) are all satisfied, so the dam section is safe. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the uplift ordinate at the heel found when a drainage gallery reduces the uplift (K = 0.45)? A. Full gamma_w x H_w at the heel with no reduction B. gamma_w x H_t + K x gamma_w x (H_w - H_t) C. K x gamma_w x H_w only D. gamma_w x H_t only Answer: B. The heel ordinate is the tail water pressure plus K times the extra head (H_w - H_t), giving 238.9 kN/m^2; the toe ordinate is gamma_w x H_t. (Key point: Uplift ordinates) 2. How is the factor of safety against overturning obtained, and what result does the answer get? A. Sum of horizontal forces / sum of vertical forces = 0.30 B. mu x SumV / SumH = 1.80, safe C. Sum of resisting moments / sum of overturning moments about the toe = 2.60, safe (> 1.5) D. Sum of overturning moments / sum of resisting moments = 0.38, unsafe Answer: C. Moments about the toe are compared: resisting 1,209,574 over overturning 466,078 gives 2.60, above the required 1.5. (Key point: Overturning check) 3. Why is there no tension at the heel? A. Because e = 2.63 m is less than B/6 = 9.50 m, so the resultant lies in the middle third B. Because the uplift is zero at the heel C. Because the toe stress is below the allowable stress D. Because the friction factor of safety exceeds 1 Answer: A. When the eccentricity e is below B/6 the base stress is compressive everywhere, so the heel stress (364.5 kN/m^2) is positive. (Key point: Middle-third rule (tension)) 4. Which expressions give the shear stress and major principal stress at the toe (tail water pressure p', d/s face angle phi_d)? A. tau = sigma_toe x tan(phi_d); sigma_1 = sigma_toe x tan^2(phi_d) B. tau = (sigma_toe - p') x tan(phi_d); sigma_1 = sigma_toe x sec^2(phi_d) - p' x tan^2(phi_d) C. tau = (sigma_toe + p') x tan(phi_d); sigma_1 = sigma_toe - p' D. tau = (sigma_toe - p') / tan(phi_d); sigma_1 = sigma_toe x cos^2(phi_d) Answer: B. The d/s face is a plane carrying the water pressure p', so the shear is (sigma - p') tan(phi) and sigma_1 = sigma sec^2(phi) - p' tan^2(phi). 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