--- title: "Gravity dam stresses with drainage gallery uplift" subject: "Hydropower Engineering" chapter: 4 key: "semester_7/hydropower_engineering:gravity-dam-stresses-2080-baishakh" source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/gravity-dam-stresses-2080-baishakh site: BCT Engineering --- # Gravity dam stresses with drainage gallery uplift Hydropower Engineering · Chapter 4: Headworks of Storage Plants - Asked in: 2080 Baishakh, 10 marks - Video: [003_DHS- Elementary Profile of a Gravity Dam](https://www.youtube.com/watch?v=D484z7gorPs) — Athul Mangalath Madhu - 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-2080-baishakh - Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-4 ## Question A concrete gravity dam has the following data: maximum water level = 550.00; bed level = 470.00; RL of top of dam = 554.00; the d/s slope is 0.67 : 1 and starts at RL 545.00; the u/s face is vertical; the centre line of the drainage gallery is 8.0 m from the u/s face. Consider only weight, water pressure and uplift. Calculate the maximum vertical stresses at the toe and heel of the dam. Also calculate the factor of safety against sliding and overturning. Assume 100% uplift pressure at the heel, 50% at the drainage gallery and zero at the toe. Take μ = 0.75. ## Short answer With crest width 8 m, base 58.25 m, the resisting moment is 2,389,982 kN·m/m and overturning moment 1,424,943 kN·m/m. Maximum vertical stress at the toe is 1506 kN/m², at the heel 100 kN/m². Factor of safety against overturning is 1.68, against sliding 1.12 (μ=0.75). ## Answer #### Data and assumptions - Bed RL 470 m, top RL 554 m, so height $=84$ m; MWL 550 m, so $H_w=80$ m; no tail water. - U/s face vertical. D/s face vertical from 554 to 545 m, then 0.67 : 1 to the toe, giving a horizontal run of $0.67\times75=50.25$ m. - Crest width is not given; **8 m is assumed**, so $B=8+50.25=58.25$ m. - $\gamma_c=24$ kN/m³ (assumed), $\gamma_w=9.81$ kN/m³, $\mu=0.75$. - Uplift: 100% of $\gamma_wH_w=784.8$ kN/m² at the heel, 50% ($392.4$ kN/m²) at the gallery (8 m from the u/s face), zero at the toe. ```text 554 +---+ crest 550 |~~~| MWL (H_w = 80 m) | | 545 | \ d/s slope begins | G \ 0.67 : 1 | \ 470 +------\ heel toe ``` #### Forces and moments about the toe | Force | Magnitude (kN/m) | Arm from toe (m) | Moment about toe (kN·m/m) | |---|---|---|---| | **Vertical (resisting)** | | | | | Rectangle 8 × 84 | 16,128.0 | 54.25 | 874,944 | | D/s triangle (50.25 × 75) | 45,225.0 | 33.50 | 1,515,038 | | **ΣW (downward)** | **61,353.0** | | **ΣM_R = 2,389,982** | | **Horizontal / uplift (overturning)** | | | | | Water thrust $P_w=\frac12\gamma_w H_w^2$ | 31,392.0 | 26.67 | 837,120 | | Uplift part 1 (upward) | 3,139.2 | 54.25 | 170,302 | | Uplift part 2 (upward) | 1,569.6 | 55.58 | 87,244 | | Uplift part 3 (upward) | 9,859.1 | 33.50 | 330,278 | | **Uplift U** | **14,567.9** | | | | **Total overturning** | | | **ΣM_O = 1,424,943** | #### Results - $\Sigma V = \Sigma W - U = 61,353.0 - 14,567.9 = 46,785.1$ kN/m - $\Sigma H = P_w = 31,392.0$ kN/m - $\Sigma M_R = 2,389,982$ kN·m/m; $\Sigma M_O = 1,424,943$ kN·m/m **Overturning:** $FS_o = \Sigma M_R/\Sigma M_O = 2,389,982/1,424,943 = 1.68$ — safe (> 1.5). **Sliding (friction):** $FS_s = \mu\Sigma V/\Sigma H = 0.75\times46,785.1/31,392.0 = 1.12$ — safe (> 1). **Resultant:** $\bar x = (\Sigma M_R-\Sigma M_O)/\Sigma V = 965,038/46,785.1 = 20.63$ m from the toe; $e = B/2-\bar x = 29.125-20.63 = 8.50$ m; $B/6 = 9.71$ m — inside the middle third (no tension). $$ \sigma_{toe} = \frac{\Sigma V}{B}\left(1+\frac{6e}{B}\right) = 803.2\,(1+0.875) = 1506.2\ \text{kN/m}^2 $$ $$ \sigma_{heel} = 803.2\,(1-0.875) = 100.1\ \text{kN/m}^2 $$ The toe stress of 1506 kN/m² is well within the usual allowable compressive stress of concrete (about 2500 to 3000 kN/m²; no value is given in the question). The heel stress of 100 kN/m² is compressive, so there is no tension. **Shear and principal stress at the toe** ($\tan\phi_d=0.670$, $p'=0.0$ kN/m²): $$ \tau_{toe} = (\sigma_{toe}-p')\tan\phi_d = (1506.2-0.0)\times0.670 = 1009.2\ \text{kN/m}^2 $$ $$ \sigma_{1,toe} = \sigma_{toe}\sec^2\phi_d - p'\tan^2\phi_d = 1506.2\,(1+0.449) - 0.0\times0.449 = 2182.4\ \text{kN/m}^2 $$ **Answer: maximum vertical stress at the toe = 1506 kN/m²; at the heel = 100 kN/m²; factor of safety against overturning = 1.68; against sliding = 1.12.** (If the actual crest width differs from 8 m, replace the crest rectangle and recompute with the same table.) ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How are the dam height and water head obtained from the levels given (bed 470, MWL 550, top 554)? A. Height = 554 - 470 = 84 m and H_w = 550 - 470 = 80 m B. Height = 550 - 470 = 80 m and H_w = 554 - 470 = 84 m C. Height = 545 - 470 = 75 m and H_w = 80 m D. Height = 84 m and H_w = 84 m Answer: A. Dam height uses the top of dam RL; the water thrust uses the maximum water level, so H_w = 80 m. (Key point: Height and water depth from RLs) 2. How is the base width found? A. B = 0.67 x 84 = 56.3 m B. B = 8 + 0.67 x 75 = 58.25 m, since the slope starts at RL 545 (75 m above the base) C. B = 0.67 x 80 = 53.6 m D. B = 8 + 0.67 x 9 = 14 m Answer: B. The sloping part is 75 m high (545 - 470), giving a run of 50.25 m, which is added to the assumed 8 m crest width. (Key point: Base width from d/s slope) 3. What uplift ordinates are used at the heel, gallery and toe? A. Full gamma_w H_w at heel, zero at gallery, 50% at toe B. 100% at heel, 50% at gallery, zero at toe C. 50% at heel, 100% at gallery, zero at toe D. Uniform 50% over the base Answer: B. As stated in the question: 784.8 kN/m^2 at the heel, 392.4 at the gallery and zero at the toe, so the diagram is made of three parts. 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