--- title: "Stresses in gravity dam with drain holes" subject: "Hydro Power" chapter: 4 key: "semester_6/hydro_power:gravity-dam-stresses-figure" source: https://www.bctengineer.com/q/semester_6/hydro_power/gravity-dam-stresses-figure site: BCT Engineering --- # Stresses in gravity dam with drain holes Hydro Power · Chapter 4: Dam Engineering - Asked in: 2081 Chaitra, 12 marks - Video: [Analysis of Gravity Dams Sample Problem](https://www.youtube.com/watch?v=xyxUo_aGfNQ) — Jose Antonio Rodavia - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/hydro_power/gravity-dam-stresses-figure - Whole chapter: https://www.bctengineer.com/md/semester_6/hydro_power/chapter-4 ## Question For the concrete gravity dam shown in the figure below, calculate the maximum vertical stresses at the heel and toe, the major principal stresses at the toe, and the shear stress intensity on a horizontal plane near the toe. Assume the unit weight of concrete is 24 kN/m³ and the unit weight of water is 10 kN/m³. [Figure: gravity dam with vertical upstream face and top width 12 m; the downstream face is vertical for the top 15 m and then slopes down to the toe; upstream water depth is 65 m with the water surface 5 m below the crest; drain holes are 10 m from the heel and the toe is a further 40 m beyond them; tailwater depth is 5 m.] ## Short answer For the dam (height 70 m, base 50 m, top width 12 m, tailwater 5 m, uplift with drains, γc=24 kN/m3, γw=10 kN/m3): maximum vertical stress at toe = 1076.1 kN/m2, at heel = 316.9 kN/m2 (compressive); major principal stress at toe = 1565.9 kN/m2; shear stress near toe = 708.9 kN/m2. ## Answer #### Geometry and assumptions (per metre length) - Water depth $h_1$ = 65 m, water surface 5 m below crest, so dam height = 70 m. - Top width 12 m; downstream face vertical for the top 15 m, then sloping from 55 m height down to the toe. - Base width $B$ = 10 + 40 = 50 m (drain holes 10 m from heel). Tailwater depth $h_2$ = 5 m. - Downstream slope: horizontal 50 − 12 = 38 m over vertical 55 m, so $\tan\alpha = 38/55 = 0.6909$ (α measured from vertical). - Uplift with drains (usual IS/textbook rule): heel $= \gamma_w h_1$; at drains $= \gamma_w[h_2 + \frac{1}{3}(h_1 - h_2)]$; toe $= \gamma_w h_2$. - Silt, wave and earthquake forces are ignored. Moments are taken about the toe. ```text 12 m +----+ crest (70 m) ~~~|~~~~| 5 m | | 15 m vertical d/s 65m | + (55 m) wat | W1 \ er | W2 \ | \ ~~~ tailwater 5 m heel +-----------\ toe <10>|<---40---> drains B = 50 m ``` #### Uplift pressures - Heel: $10 \times 65 = 650$ kN/m² - Drains: $10 \times [5 + (65 - 5)/3] = 10 \times 25 = 250$ kN/m² - Toe: $10 \times 5 = 50$ kN/m² #### Forces and moments about the toe | Force | Value (kN) | Arm (m) | Moment (kN·m) | |---|---|---|---| | W1 = 12×70×24 | 20,160 ↓ | 44.000 | +887,040 | | W2 = ½×38×55×24 | 25,080 ↓ | 25.333 | +635,360 | | Tailwater on slope = ½×3.455×5×10 | 86.4 ↓ | 1.152 | +99 | | P1 = ½×10×65² | 21,125 → | 21.667 | −457,708 | | P2 = ½×10×5² | 125 ← | 1.667 | +208 | | U1 = 50×40 | 2,000 ↑ | 20.000 | −40,000 | | U2 = ½×200×40 | 4,000 ↑ | 26.667 | −106,667 | | U3 = 250×10 | 2,500 ↑ | 45.000 | −112,500 | | U4 = ½×400×10 | 2,000 ↑ | 46.667 | −93,333 | (Tailwater wedge width on the slope $= 5 \times 0.6909 = 3.455$ m.) $$ \begin{aligned} \sum U &= 10{,}500\ \text{kN} \\ \sum V &= 20{,}160 + 25{,}080 + 86.4 - 10{,}500 = 34{,}826.4\ \text{kN} \\ \sum H &= 21{,}125 - 125 = 21{,}000\ \text{kN} \\ \sum M_R &= 1{,}522{,}708\ \text{kN·m} \\ \sum M_O &= 457{,}708 + 352{,}500 = 810{,}208\ \text{kN·m} \end{aligned} $$ #### Position of resultant $$ \begin{aligned} \bar{x} &= \frac{1{,}522{,}708 - 810{,}208}{34{,}826.4} = 20.459\ \text{m from toe} \\ e &= 25 - 20.459 = 4.541\ \text{m} < B/6 = 8.333\ \text{m (no tension)} \end{aligned} $$ #### (a) Maximum vertical stresses $$ \begin{aligned} \frac{\sum V}{B} &= \frac{34{,}826.4}{50} = 696.53\ \text{kN/m}^2, \quad \frac{6e}{B} = 0.545 \\ p_{v,toe} &= 696.53(1 + 0.545) = 1076.1\ \text{kN/m}^2 \\ p_{v,heel} &= 696.53(1 - 0.545) = 316.9\ \text{kN/m}^2 \end{aligned} $$ #### (b) Major principal stress at the toe With tailwater pressure at the toe $p' = 10 \times 5 = 50$ kN/m², $\tan^2\alpha = 0.4774$, $\sec^2\alpha = 1.4774$: $$ \begin{aligned} \sigma_1 &= p_v \sec^2\alpha - p'\tan^2\alpha \\ &= 1076.1 \times 1.4774 - 50 \times 0.4774 \\ &= 1565.9\ \text{kN/m}^2 \end{aligned} $$ #### (c) Shear stress on a horizontal plane near the toe $$ \begin{aligned} \tau &= (p_v - p')\tan\alpha \\ &= (1076.1 - 50) \times 0.6909 = 708.9\ \text{kN/m}^2 \end{aligned} $$ **Answer: Vertical stress at toe = 1076.1 kN/m², at heel = 316.9 kN/m² (both compressive); major principal stress at toe = 1565.9 kN/m²; shear stress near toe = 708.9 kN/m².** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is the base width taken as B = 50 m in this problem? A. Drain holes are 10 m from the heel and the toe is 40 m beyond them, so 10 + 40 B. Tailwater depth multiplied by 10 C. Water depth 65 m minus 15 m D. Top width 12 m plus dam height 70 m minus water depth Answer: A. The geometry gives 10 m from heel to drains plus 40 m to the toe, so B = 50 m. (Key point: Base width) 2. With drains, what is the uplift pressure at the drain line? A. 400 kN/m2 B. 250 kN/m2, from gamma_w[h2 + (h1 - h2)/3] C. 650 kN/m2 (full head) D. 50 kN/m2 (tailwater head) Answer: B. Head at drains = 5 + 60/3 = 25 m, giving 250 kN/m2. (Key point: Uplift at drains) 3. Which expression gives the major principal stress at the toe? A. p_v x tan(alpha) B. p_v / sec^2(alpha) C. p_v sec^2(alpha) - p' tan^2(alpha) D. (p_v - p') tan(alpha) Answer: C. sigma1 = p_v sec^2(alpha) - p' tan^2(alpha), where p' is the tailwater pressure at the toe. (Key point: Principal stress at toe) 4. What are the shear stress formula near the toe and the tension result? A. tau = p_v x tan(alpha); tension at heel since e > B/6 B. tau = p' x tan(alpha); tension at toe C. tau = (p_v - p') / tan(alpha); no tension since e > B/6 D. tau = (p_v - p') tan(alpha); no tension since e = 4.54 m < B/6 = 8.33 m Answer: D. Shear on a horizontal plane is (p_v - p') tan(alpha) = 708.9 kN/m2, and e is inside B/6 so both stresses are compressive. (Key point: Shear stress and tension) ## Related in Hydro Power - [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 - [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 - [Stability of gravity dam (crest 1530 m)](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Agravity-dam-stability-1530) — 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 - [Specific gravity, weight and Newtonian fluids](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Anewtonian-non-newtonian-fluids) — Hydro Power, chapter 1 - [Stability of trapezoidal gravity dam (RL 250 m)](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Atrapezoidal-dam-stability-250m) — Hydro Power, chapter 4 ## Related in other subjects - [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 - [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 - [Gravity dam stresses, principal stress, FOS (2079)](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Agravity-dam-stresses-2079-bhadra) — 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