--- title: "Gravity dam stresses, principal stress, FOS (2079)" subject: "Hydropower Engineering" chapter: 4 key: "semester_7/hydropower_engineering:gravity-dam-stresses-2079-bhadra" source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/gravity-dam-stresses-2079-bhadra site: BCT Engineering --- # Gravity dam stresses, principal stress, FOS (2079) Hydropower Engineering · Chapter 4: Headworks of Storage Plants - Asked in: 2079 Bhadra, 10 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-stresses-2079-bhadra - Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-4 ## Question The section of the gravity dam is shown below. i) Calculate the maximum vertical stresses at the heel and toe of the dam. ii) Calculate the major principal stress at the toe of the dam. iii) Calculate the factor of safety against overturning and sliding. Take γc = 24 kN/m³ and σa = 2500 kN/m². [Figure: crest width 8 m; 5 m from the crest to the water level; water depth 85 m; base width 60 m; centre line of drainage gallery 10 m from the upstream face; vertical upstream face; sloping downstream face; tail water depth 10 m.] ## Short answer Using crest width 8 m, base 60 m, and uplift with drainage gallery, the maximum vertical stress at the toe is 1690 kN/m², at the heel 203 kN/m². Major principal stress at the toe is 2222 kN/m². Factor of safety against overturning is 1.76, against sliding 1.14 (with assumed μ=0.70). ## Answer #### Data and assumptions - Crest width 8 m; water level 5 m below the crest; $H_w=85$ m, dam height $=90$ m; tail water $H_t=10$ m; base width $B=60$ m. - U/s face vertical; the d/s face is taken as a straight slope from the d/s edge of the crest to the toe (horizontal run 52 m over 90 m, $\tan\phi_d=0.578$). - $\gamma_c=24$ kN/m³ (given), $\gamma_w=9.81$ kN/m³, $\sigma_a=2500$ kN/m². The friction coefficient is not given, so **$\mu=0.70$ is assumed**. - Uplift: $\gamma_wH_w=833.9$ kN/m² at the heel, $\gamma_wH_t=98.1$ kN/m² at the toe, and at the gallery (10 m from the u/s face) $98.1+\frac13(833.9-98.1)=343.4$ kN/m². ```text crest 8 m +--+ 5 m | freeboard ~~~~~| water 85 m |\ | \ d/s slope G | \ | \ ~~ tail water 10 m +----\ B = 60 m ``` #### Forces and moments about the toe | Force | Magnitude (kN/m) | Arm from toe (m) | Moment about toe (kN·m/m) | |---|---|---|---| | **Vertical (resisting)** | | | | | Rectangle 8 × 90 | 17,280.0 | 56.00 | 967,680 | | D/s triangle (52 × 90) | 56,160.0 | 34.67 | 1,946,880 | | Tail water on d/s face | 283.4 | 1.93 | 546 | | **ΣW (downward)** | **73,723.4** | | **2,915,106** | | **Horizontal / uplift (overturning)** | | | | | Water thrust $P_w=\frac12\gamma_w H_w^2$ | 35,438.6 | 28.33 | 1,004,094 | | Uplift part 1 (upward) | 3,433.5 | 55.00 | 188,842 | | Uplift part 2 (upward) | 2,452.5 | 56.67 | 138,975 | | Uplift part 3 (upward) | 4,905.0 | 25.00 | 122,625 | | Uplift part 4 (upward) | 6,131.2 | 33.33 | 204,375 | | **Uplift U** | **16,922.2** | | | | Tail water thrust (resisting) | 490.5 | 3.33 | 1,635 | | **Total overturning** | | | **ΣM_O = 1,658,912** | #### Results - $\Sigma V = \Sigma W - U = 73,723.4 - 16,922.2 = 56,801.1$ kN/m - $\Sigma H = 35,438.6 - 490.5 = 34,948.1$ kN/m - $\Sigma M_R = 2,915,106+1,635 = 2,916,741$ kN·m/m (weights plus the tail water thrust); $\Sigma M_O = 1,658,912$ kN·m/m **Overturning:** $FS_o = \Sigma M_R/\Sigma M_O = 2,916,741/1,658,912 = 1.76$ — safe (> 1.5). **Sliding (friction):** $FS_s = \mu\Sigma V/\Sigma H = 0.7\times56,801.1/34,948.1 = 1.14$ — safe (> 1). **Resultant:** $\bar x = (\Sigma M_R-\Sigma M_O)/\Sigma V = 1,257,829/56,801.1 = 22.14$ m from the toe; $e = B/2-\bar x = 30-22.14 = 7.86$ m; $B/6 = 10.00$ m, so the resultant is inside the middle third (no tension). $$ \sigma_{toe} = \frac{\Sigma V}{B}\left(1+\frac{6e}{B}\right) = 946.7\,(1+0.786) = 1690.4\ \text{kN/m}^2 $$ $$ \sigma_{heel} = 946.7\,(1-0.786) = 203.0\ \text{kN/m}^2 $$ The toe stress of 1690 kN/m² is less than the allowable 2500 kN/m², so the dam is safe against crushing. The heel stress of 203 kN/m² is compressive, so there is no tension. **Shear and principal stress at the toe** ($\tan\phi_d=0.578$, $p'=98.1$ kN/m²): $$ \tau_{toe} = (\sigma_{toe}-p')\tan\phi_d = (1690.4-98.1)\times0.578 = 920.0\ \text{kN/m}^2 $$ $$ \sigma_{1,toe} = \sigma_{toe}\sec^2\phi_d - p'\tan^2\phi_d = 1690.4\,(1+0.334) - 98.1\times0.334 = 2221.9\ \text{kN/m}^2 $$ **Answer: (i) maximum vertical stresses: toe = 1690 kN/m², heel = 203 kN/m²; (ii) major principal stress at the toe = 2222 kN/m² (< 2500 kN/m²); (iii) $FS$ overturning = 1.76, $FS$ sliding = 1.14.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. With the water level 5 m below the crest and a dam 90 m high, what is H_w? A. 90 m B. 95 m C. 85 m D. 80 m Answer: C. H_w is the dam height minus the 5 m freeboard, so 90 - 5 = 85 m. (Key point: Water depth vs dam height) 2. How is the uplift at the drainage gallery computed? A. Tail water pressure + one-third of (heel pressure - toe pressure) = 343.4 kN/m^2 B. Half of the heel pressure C. Equal to the toe pressure 98.1 kN/m^2 D. Zero because the gallery drains all water Answer: A. The gallery pressure is the toe (tail water) pressure plus 1/3 of the difference between heel and toe pressures: 98.1 + (833.9 - 98.1)/3. (Key point: Uplift at the gallery) 3. Which formula gives the major principal stress at the toe, and what is the result? A. sigma_1 = sigma_toe / sec^2(phi_d) = 1267 kN/m^2 B. sigma_1 = sigma_toe x sec^2(phi_d) - p' x tan^2(phi_d) = 2222 kN/m^2 C. sigma_1 = tau_toe = 920 kN/m^2 D. sigma_1 = sigma_heel + sigma_toe = 1893 kN/m^2 Answer: B. The toe is on the d/s face, so sigma_1 = sigma sec^2(phi) - p' tan^2(phi), about 2222 kN/m^2, which is below sigma_a = 2500. (Key point: Major principal stress at the toe) 4. The friction coefficient was not given. How does the answer handle this and what is the sliding result? A. Assumes mu = 0.70 and finds FS_s = 1.14 > 1, safe B. Skips the sliding check C. Assumes mu = 1.0 and finds FS_s = 1.6 D. Uses the crushing stress to find mu Answer: A. The answer states mu = 0.7 is assumed and uses FS_s = mu SumV / SumH = 0.7 x 56,801.1 / 34,948.1 = 1.14. (Key point: Assumption of mu and sliding) ## Related in Hydropower Engineering - [Gravity dam section: shear stress near toe (2082)](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Agravity-dam-stability-2082-baishakh) — Hydropower Engineering, chapter 4 - [Principal stresses at toe and heel with silt and earthquake](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Agravity-dam-principal-stress-earthquake-silt) — Hydropower Engineering, chapter 4 - [Gravity dam stresses with drainage gallery uplift](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Agravity-dam-stresses-2080-baishakh) — Hydropower Engineering, chapter 4 - [Derive principal and shear stress equations at toe/heel](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Agravity-dam-principal-stress-derivation-tailwater-eq) — Hydropower Engineering, chapter 4 - [Gravity dam stability: flotation, overturning, sliding](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Agravity-dam-stability-flotation-shear-resistance) — Hydropower Engineering, chapter 4 - [Gravity dam stability and shear stress at toe (2081)](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Agravity-dam-stability-2081-bhadra) — Hydropower Engineering, chapter 4 ## Related in other subjects - [Sliding and overturning check of gravity dam](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Agravity-dam-sliding-overturning-figure) — Hydro Power, chapter 4 - [Stresses in gravity dam with drain holes](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Agravity-dam-stresses-figure) — Hydro Power, chapter 4 - [Elementary profile and 40 m gravity dam stability](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Agravity-dam-elementary-40m) — Hydro Power, chapter 4 - [Triangular gravity dam 35 m high](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Atriangular-dam-35m) — Hydro Power, chapter 4