--- title: "Gravity dam section: shear stress near toe (2082)" subject: "Hydropower Engineering" chapter: 4 key: "semester_7/hydropower_engineering:gravity-dam-stability-2082-baishakh" source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/gravity-dam-stability-2082-baishakh site: BCT Engineering --- # Gravity dam section: shear stress near toe (2082) Hydropower Engineering · Chapter 4: Headworks of Storage Plants - Asked in: 2082 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-stability-2082-baishakh - Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-4 ## Question The figure below (all dimensions are in m) 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 stresses on a horizontal plane near toe and heel. Assume the unit weight of concrete is 24 kN/m³. The allowable stress and average shear strength in the concrete may be taken as 2500 kN/m² and 2200 kN/m², respectively. Take μ = 0.75. [Figure: crest width 10.0 m; 5.0 m from the top to the water level; upstream face vertical for 40.0 m of water depth below the 5.0 m, then flared for the lower 20.0 m with a horizontal run of 15.0 m to the heel; downstream face vertical for the top 10.0 m, then sloping to the toe with a horizontal run of 35.0 m; tail water depth 8.0 m; a drainage gallery in the body of the dam (5.0 m dimension marked below the gallery).] ## Short answer Using force/moment equilibrium with uplift, FS overturning=2.24, sliding=1.64, SFF=9.1, resultant inside middle third; toe/heel stresses 976 and 314 kN/m2, both below 2500, shear stresses 570.2 (toe) and 214.4 kN/m2 (heel), below 2200, so dam is stable. ## Answer #### Assumptions (from the figure description) - Dam height $=5+40+20=65$ m. Crest width 10 m. U/s face vertical down to 20 m above the base, then flared with a horizontal run of 15 m to the heel. Water level 5 m below the crest, so $H_w=60$ m. D/s face vertical for the top 10 m, then sloping (35 m horizontal run over 55 m height) to the toe. Base width $B=15+10+35=60$ m. - Tail water depth $H_t=8$ m. $\gamma_c=24$ kN/m³, $\gamma_w=10$ kN/m³, $\mu=0.75$. - Uplift: full hydrostatic head at the heel ($\gamma_wH_w=600$ kN/m²) and at the toe ($\gamma_wH_t=80$ kN/m²). The drainage gallery (assumed with its centre 5 m from the heel) reduces the uplift at the gallery to $80+\frac13(600-80)=253.3$ kN/m². Earthquake is neglected. ```text crest 10 m +--------+ ~|~~5 m~~~| water level | | d/s vertical 10 m | 40 m \ | \ slope: 35 m run in 55 m | 20 m \ \__________\_ tail water 8 m 15 10 35 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)** | | | | | Flare triangle (15 × 20) | 3,600.0 | 50.00 | 180,000 | | Rectangle 10 × 65 | 15,600.0 | 40.00 | 624,000 | | D/s triangle (35 × 55) | 23,100.0 | 23.33 | 539,000 | | Water on u/s face | 7,500.0 | 53.00 | 397,500 | | Tail water on d/s face | 203.6 | 1.70 | 346 | | **ΣW (downward)** | **50,003.6** | | **ΣM_R = 1,740,846** | | **Horizontal / uplift (overturning)** | | | | | Water thrust $P_w=\frac12\gamma_w H_w^2$ | 18,000.0 | 20.00 | 360,000 | | Uplift part 1 (upward) | 1,266.7 | 57.50 | 72,833 | | Uplift part 2 (upward) | 866.7 | 58.33 | 50,556 | | Uplift part 3 (upward) | 4,400.0 | 27.50 | 121,000 | | Uplift part 4 (upward) | 4,766.7 | 36.67 | 174,778 | | **Uplift U** | **11,300.0** | | | | Tail water thrust (resisting) | 320.0 | 2.67 | 853 | | **Total overturning** | | | **ΣM_O = 779,167** | (The tail water also acts on the sloping d/s face; its vertical component is included as a resisting load, and its horizontal thrust acts at $H_t/3$ above the base.) #### Stability - $\Sigma V=50,003.6-11,300.0=38,703.6$ kN/m; $\Sigma H=18,000.0-320.0=17,680.0$ kN/m - $\Sigma M_R=1,741,699$ kN·m/m (including tail water); $\Sigma M_O=779,167$ kN·m/m **Overturning:** $FS_o=1,741,699/779,167=2.24$ — safe (> 1.5). **Sliding (friction only):** $FS_s=\mu\Sigma V/\Sigma H = 0.75\times38,703.6/17,680.0=1.64$ — safe (> 1). **Shear-friction factor:** $SFF=\dfrac{\mu\Sigma V+\tau B}{\Sigma H}=\dfrac{0.75\times38,703.6+2200\times60}{17,680.0}=9.1$ — very safe (> 3 to 5). **Resultant and stresses:** $\bar x=962,532/38,703.6=24.87$ m from the toe; $e=30-24.87=5.13$ m; $B/6=10$ m (inside the middle third). $$ \sigma_{toe}=645.1\,(1+0.513)=976.0\ \text{kN/m}^2,\quad \sigma_{heel}=645.1\,(1-0.513)=314.1\ \text{kN/m}^2 $$ Both are compressive and below the allowable 2500 kN/m². #### Shear stress on a horizontal plane near the toe and heel Face slopes: d/s $\tan\phi_d=35/55=0.636$; u/s (flare) $\tan\phi_u=15/20=0.75$. Water pressures at the faces: $p'=\gamma_wH_t=80$ kN/m² at the toe and $p=\gamma_wH_w=600$ kN/m² at the heel. $$ \tau_{toe}=(\sigma_{toe}-p')\tan\phi_d=(976.0-80)\times0.636=570.2\ \text{kN/m}^2 $$ $$ \tau_{heel}=(p-\sigma_{heel})\tan\phi_u=(600-314.1)\times0.75=214.4\ \text{kN/m}^2 $$ Both are much smaller than the average shear strength of 2200 kN/m². Principal stresses: $\sigma_{1,toe}=\sigma_{toe}\sec^2\phi_d-p'\tan^2\phi_d=1338.9$ kN/m², $\sigma_{1,heel}=\sigma_{heel}\sec^2\phi_u-p\tan^2\phi_u=153.3$ kN/m². **Conclusion:** the dam is safe against overturning (2.24), sliding (1.64; shear-friction factor 9.1), tension (none: the minimum stress of 314 kN/m² is compressive) and crushing (maximum principal stress 1339 kN/m² < 2500 kN/m²). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How does the drainage gallery change the uplift pressure? A. It raises uplift to the heel value B. It reduces the pressure at the gallery to tail-water head + 1/3 of the difference (253.3 kN/m²) C. It removes uplift altogether D. It has no effect on uplift Answer: B. Drainage cuts pressure at the gallery to 80 + (600 - 80)/3 = 253.3 kN/m², lowering the uplift force. (Key point: Gallery effect on uplift) 2. What does the shear-friction factor (SFF) include beyond simple sliding factor? A. Only the friction μΣV B. Friction plus shear strength over the base: (μΣV + τB)/ΣH C. Only the shear strength τB D. Overturning moments divided by ΣH Answer: B. SFF = (μΣV + τB)/ΣH accounts for both friction and shear strength; here it equals 9.1. (Key point: Shear-friction factor) 3. How is the shear stress on a horizontal plane near the toe calculated? A. τ = (σ_toe - p') tan φ_d, with p' the tail water pressure B. τ = σ_toe x μ C. τ = σ_heel tan φ_u D. τ = (p - σ_heel) tan φ_d Answer: A. At the toe, τ = (σ_toe - p') tan φ_d = (976 - 80) x 0.636 = 570 kN/m². (Key point: Shear stress at toe) 4. What is the outcome of the stability check for this dam? A. Unsafe against overturning B. Tension appears at the heel C. Crushing governs because toe stress exceeds 2500 D. Safe: FSo 2.24, FSs 1.64, no tension (e = 5.13 m < 10 m), max stress 976 < 2500 kN/m² Answer: D. All checks pass, and the shear stresses (570 and 214 kN/m²) are well below the 2200 kN/m² shear strength. 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