Elementary profile and 40 m gravity dam stability

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.

        |\
  P --> | \   40 m
  13.33 |W \
   m    |   \
   heel |____\ toe
        <-28 m->

Forces and moments

$$\begin{aligned} \sum V &= 13{,}440 - 3{,}360 = 10{,}080\ \text{kN} \\ \sum H &= 8{,}000\ \text{kN} \\ \sum M_R &= 250{,}880\ \text{kN·m} \\ \sum M_O &= 106{,}667 + 62{,}720 = 169{,}387\ \text{kN·m} \end{aligned}$$

(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.