--- title: "Gravity dam stability and shear stress at toe (2081)" subject: "Hydropower Engineering" chapter: 4 key: "semester_7/hydropower_engineering:gravity-dam-stability-2081-bhadra" source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/gravity-dam-stability-2081-bhadra site: BCT Engineering --- # Gravity dam stability and shear stress at toe (2081) Hydropower Engineering · Chapter 4: Headworks of Storage Plants - Asked in: 2081 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-2081-bhadra - Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-4 ## Question The figure below shows the section of a concrete gravity dam. Neglecting the effects of earthquakes, check the stability of the dam. Also, calculate the intensity of the shear stress on a horizontal plane near the toe. Assume the unit weights of water and concrete are 10 kN/m³ and 24 kN/m³, respectively. The allowable stress and average shear strength in the concrete may be taken as 2500 kN/m² and 2000 kN/m², respectively. Take μ = 0.70. [Figure: crest width 5 m, top level 390.00 m (printed as 1390.00 m); maximum water level 385.00 m; level 380.00 m where the downstream slope begins; base level 305.00 m; downstream slope 2 : 3; base width 56 m; drainage gallery 8 m from the upstream face; tail water level 311.00 m.] ## Short answer Forces and moments including uplift give FS overturning=1.57, friction sliding=0.95, SFF=4.47; resultant falls outside middle third, so heel stress is -77.4 kN/m2 tension; toe stress 1613 kN/m2 below 2500, toe shear stress 1056 kN/m2 below 2000, so shear-friction safe but friction alone fails. ## Answer #### Assumptions and data - RL of base $=305$ m; top $=390$ m (the printed 1390 m is read as 390 m); dam height $=85$ m. Maximum water level 385 m, so $H_w=385-305=80$ m. Tail water 311 m, so $H_t=6$ m. - Crest width 5 m; u/s face vertical; d/s face vertical from 390 to 380 m, then sloping to the toe. Base width $B=56$ m, so the d/s slope runs 51 m horizontally over 75 m (about 2 : 3 as drawn), $\tan\phi_d=51/75=0.68$. - $\gamma_w=10$, $\gamma_c=24$ kN/m³, $\mu=0.70$, allowable stress 2500 kN/m², shear strength 2000 kN/m². - Uplift: $\gamma_wH_w=800$ kN/m² at the heel, $\gamma_wH_t=60$ kN/m² at the toe; the gallery (8 m from the u/s face) reduces the pressure there to $60+\frac13(800-60)=306.7$ kN/m². Earthquake is neglected. ```text 390 +----+ | | crest 5 m 385~|~~~~| MWL (H_w = 80 m) | | | | vertical to 380 | \ | \ slope 2 : 3 | \ | G \ 311 | \~~ tail water 6 m 305 +---------\ |<-- B = 56 m -->| ``` #### Forces and moments about the toe | Force | Magnitude (kN/m) | Arm from toe (m) | Moment about toe (kN·m/m) | |---|---|---|---| | **Vertical (resisting)** | | | | | Rectangle 5 × 85 | 10,200.0 | 53.50 | 545,700 | | D/s triangle (51 × 75) | 45,900.0 | 34.00 | 1,560,600 | | Tail water on d/s face | 122.4 | 1.36 | 166 | | **ΣW (downward)** | **56,222.4** | | **ΣM_R = 2,106,466** | | **Horizontal / uplift (overturning)** | | | | | Water thrust $P_w=\frac12\gamma_w H_w^2$ | 32,000.0 | 26.67 | 853,333 | | Uplift part 1 (upward) | 2,453.3 | 52.00 | 127,573 | | Uplift part 2 (upward) | 1,973.3 | 53.33 | 105,244 | | Uplift part 3 (upward) | 2,880.0 | 24.00 | 69,120 | | Uplift part 4 (upward) | 5,920.0 | 32.00 | 189,440 | | **Uplift U** | **13,226.7** | | | | Tail water thrust (resisting) | 180.0 | 2.00 | 360 | | **Total overturning** | | | **ΣM_O = 1,344,711** | #### Stability checks - $\Sigma V = \Sigma W - U = 56,222.4 - 13,226.7 = 42,995.7$ kN/m - $\Sigma H = 32,000.0 - 180.0 = 31,820.0$ kN/m - $\Sigma M_R = 2,106,826$ kN·m/m (tail water thrust moment included); $\Sigma M_O = 1,344,711$ kN·m/m **Overturning:** $FS_o = \Sigma M_R/\Sigma M_O = 2,106,826/1,344,711 = 1.57$ — safe (> 1.5). **Sliding (friction):** $FS_s = \mu\Sigma V/\Sigma H = 0.7\times42,995.7/31,820.0 = 0.95$ — less than 1: not safe by friction alone. **Shear-friction factor:** $SFF = (\mu\Sigma V + \tau B)/\Sigma H = (0.7\times42,995.7 + 2000\times56)/31,820.0 = 4.47$ — safe (above the usual minimum of 3 to 5 for normal loading, at the lower end). **Resultant:** $\bar x = (\Sigma M_R-\Sigma M_O)/\Sigma V = 762,115/42,995.7 = 17.73$ m from the toe; $e = B/2-\bar x = 28-17.73 = 10.27$ m; $B/6 = 9.33$ m — outside the middle third (tension at the heel). $$ \sigma_{toe} = \frac{\Sigma V}{B}\left(1+\frac{6e}{B}\right) = 767.8\,(1+1.101) = 1613.0\ \text{kN/m}^2 $$ $$ \sigma_{heel} = 767.8\,(1-1.101) = -77.4\ \text{kN/m}^2 $$ The toe stress of 1613 kN/m² is below the allowable 2500 kN/m², so the dam is safe against crushing. The heel stress is -77 kN/m² (tensile), so tension occurs at the heel. **Shear and principal stress at the toe** ($\tan\phi_d=0.680$, $p'=60.0$ kN/m²): $$ \tau_{toe} = (\sigma_{toe}-p')\tan\phi_d = (1613.0-60.0)\times0.680 = 1056.0\ \text{kN/m}^2 $$ $$ \sigma_{1,toe} = \sigma_{toe}\sec^2\phi_d - p'\tan^2\phi_d = 1613.0\,(1+0.462) - 60.0\times0.462 = 2331.1\ \text{kN/m}^2 $$ **Shear stress near the toe:** $\tau_{toe}=1056$ kN/m², which is about half of the average shear strength of 2000 kN/m², so it is safe. The principal stress of 2331 kN/m² is also below the allowable 2500 kN/m². **Conclusion:** overturning and the shear-friction factor are satisfactory; the friction-only sliding factor (0.95) is below 1, but this is covered by the shear strength of the concrete-rock contact (SFF above). The resultant lies slightly outside the middle third, so a small tension of about 77 kN/m² appears at the heel; better drainage or a slightly wider base would remove it. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the result of the friction-only sliding check, and what does it mean? A. FSs = 0.95 < 1, so friction alone is insufficient B. FSs = 1.57, safe C. FSs = 4.47, safe D. FSs = 2.35, safe Answer: A. μΣV/ΣH = 0.7 x 42,995.7/31,820 = 0.95, below 1, so friction alone does not prevent sliding. (Key point: Sliding by friction only) 2. Why is the dam still considered acceptable against sliding? A. Because uplift is ignored B. Because overturning is safe C. Because the SFF = (μΣV + τB)/ΣH = 4.47 includes shear strength of the contact D. Because the tail water supports the dam Answer: C. The shear-friction factor adds the shear strength of the concrete-rock contact and reaches 4.47, above the 3 to 5 required. (Key point: Role of shear-friction factor) 3. What does e = 10.27 m versus B/6 = 9.33 m indicate? A. The resultant lies in the middle third and the base is wholly compressed B. The resultant is outside the middle third, causing slight tension (-77 kN/m²) at the heel C. The dam will overturn D. The toe stress exceeds the allowable Answer: B. Since e > B/6, the minimum stress (1 - 6e/B) becomes negative at the heel. (Key point: Resultant position) 4. Which formula and value give the toe shear stress? A. τ = σ_toe tan φ = 1097 kN/m² B. τ = (σ_toe - p') tan φ_d = (1613 - 60) x 0.680 = 1056 kN/m², below the 2000 kN/m² strength C. τ = p' tan φ_d = 41 kN/m² D. τ = μΣV/B = 538 kN/m² Answer: B. The toe shear stress subtracts the tail water pressure from the toe stress and multiplies by tan φ_d, giving 1056 kN/m². 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