--- title: "Elementary profile and 40 m gravity dam stability" subject: "Hydro Power" chapter: 4 key: "semester_6/hydro_power:gravity-dam-elementary-40m" source: https://www.bctengineer.com/q/semester_6/hydro_power/gravity-dam-elementary-40m site: BCT Engineering --- # Elementary profile and 40 m gravity dam stability Hydro Power · Chapter 4: Dam Engineering - Asked in: 2082 Chaitra, 2+8 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_6/hydro_power/gravity-dam-elementary-40m - Whole chapter: https://www.bctengineer.com/md/semester_6/hydro_power/chapter-4 ## Question What do you mean by elementary profile of gravity dam? A concrete gravity dam with a triangular section has a height of 40 m and a base width of 28 m. The water on the upstream side is up to the crest. Considering only self-weight, hydrostatic pressure, and uplift pressure (with an uplift intensity factor of 0.6), calculate the eccentricity of the resultant force from the toe, the factor of safety against sliding (take coefficient of friction = 0.7), the maximum vertical stress at the toe, and the factor of safety against overturning. Take unit weight of concrete = 24 kN/m³ and unit weight of water = 10 kN/m³. ## Short answer The elementary profile is a right-angled triangle with vertical upstream face and zero top width, water up to the apex. With H=40 m, B=28 m, uplift factor 0.6, μ=0.7, concrete γ=24 kN/m3, water γ=10 kN/m3: eccentricity e=5.915 m (outside middle-third); sliding FS=0.88 (unsafe); max toe vertical stress=816.3 kN/m2; overturning FS=1.48. ## Answer #### Elementary profile The **elementary profile** of a gravity dam is a right-angled triangle with a vertical upstream face, zero top width, and water up to the apex. It is the ideal shape to resist water pressure (which grows linearly with depth) with minimum material and no tension in the base. #### Data (per metre length) H = 40 m, B = 28 m, $\gamma_c$ = 24 kN/m³, $\gamma_w$ = 10 kN/m³, C = 0.6, μ = 0.7. Upstream face vertical; water up to the crest. Moments are taken about the **toe**. ```text |\ P --> | \ 40 m 13.33 |W \ m | \ heel |____\ toe <-28 m-> ``` #### Forces and moments | Force | Value (kN) | Lever arm from toe (m) | Moment (kN·m) | |---|---|---|---| | W = ½×28×40×24 | 13,440 (↓) | 28 − 28/3 = 18.667 | +250,880 | | P = ½×10×40² | 8,000 (→) | 40/3 = 13.333 | −106,667 | | U = ½×0.6×10×40×28 | 3,360 (↑) | 18.667 | −62,720 | $$ \begin{aligned} \sum V &= 13{,}440 - 3{,}360 = 10{,}080\ \text{kN} \\ \sum H &= 8{,}000\ \text{kN} \\ \sum M_R &= 250{,}880\ \text{kN·m} \\ \sum M_O &= 106{,}667 + 62{,}720 = 169{,}387\ \text{kN·m} \end{aligned} $$ #### (a) Position of resultant and eccentricity $$ \begin{aligned} \bar{x} &= \frac{\sum M_R - \sum M_O}{\sum V} = \frac{250{,}880 - 169{,}387}{10{,}080} = 8.085\ \text{m from toe} \\ e &= \frac{B}{2} - \bar{x} = 14 - 8.085 = 5.915\ \text{m} \end{aligned} $$ Since $e = 5.915 > B/6 = 4.667$ m, the resultant lies outside the middle third (towards the toe), so tension develops at the heel. #### (b) Factor of safety against sliding $$ FS_s = \frac{\mu \sum V}{\sum H} = \frac{0.7 \times 10{,}080}{8{,}000} = 0.882 $$ $FS_s < 1$: the section is **unsafe in sliding** (by friction alone). #### (c) Maximum vertical stress at the toe $$ \begin{aligned} p_{toe} &= \frac{\sum V}{B}\left(1 + \frac{6e}{B}\right) = \frac{10{,}080}{28}\left(1 + \frac{6\times5.915}{28}\right) \\ &= 360 \times 2.2676 = 816.3\ \text{kN/m}^2 \end{aligned} $$ At the heel: $p_{heel} = 360(1 - 1.2676) = -96.3$ kN/m² (tension). #### (d) Factor of safety against overturning $$ FS_o = \frac{\sum M_R}{\sum M_O} = \frac{250{,}880}{169{,}387} = 1.48 $$ **Answer: Resultant at 8.085 m from the toe (eccentricity e = 5.915 m from the base centre); FS against sliding = 0.88 (unsafe); maximum toe stress = 816.3 kN/m² (heel tension 96.3 kN/m²); FS against overturning = 1.48 (just below the usual 1.5).** The 28 m base is less than the no-tension width $H/\sqrt{S_c - C} = 40/\sqrt{2.4-0.6} = 29.8$ m, which explains the heel tension. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the elementary profile of a gravity dam? A. A right-angled triangle with vertical upstream face, zero top width and water up to the apex B. A rectangle whose width equals the dam height C. A trapezoid with 6 m top width and freeboard D. An isosceles triangle with both faces sloping Answer: A. The elementary profile is the right-angled triangle with vertical u/s face and zero top width, the ideal shape for linearly increasing water pressure. (Key point: Elementary profile) 2. For H = 40 m, B = 28 m, with uplift factor 0.6, what is the net vertical force used in the stability checks? A. 13,440 kN (weight only) B. 10,080 kN (weight 13,440 minus uplift 3,360) C. 8,000 kN D. 16,800 kN (weight plus uplift) Answer: B. Uplift acts upward, so sum V = W - U = 13,440 - 3,360 = 10,080 kN. (Key point: Net vertical force) 3. Here e = 5.915 m while B/6 = 4.667 m. What does this mean? A. The dam is safe against sliding B. The resultant lies in the middle third, so no tension C. The resultant lies outside the middle third, so tension develops at the heel D. Compression is uniform across the base Answer: C. When e > B/6 the resultant is outside the middle third and the heel goes into tension. (Key point: Tension at heel) 4. Which pair of results is correct for this dam? A. FS sliding = 2.5 and FS overturning = 3.0 B. FS sliding = 1.48 and FS overturning = 0.88 C. Both factors exceed 2 D. FS sliding = 0.88 (unsafe) and FS overturning = 1.48 Answer: D. FS sliding = 0.7 x 10,080 / 8,000 = 0.88 < 1 (unsafe), and FS overturning = 250,880 / 169,387 = 1.48, just under 1.5. 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