Triangular gravity dam 35 m high

Data (per metre length)

H = 35 m, B = 26 m, uplift coefficient C = 0.5, μ = 0.75. Unit weight of concrete $= 2400$ kgf/m³ $= 2400 \times 9.81/1000 = 23.544$ kN/m³; water $= 9.81$ kN/m³. Upstream face vertical, water up to the top. Uplift varies from $C\gamma_w H$ at the heel to zero at the toe. Moments are about the toe.

        |\
  P --> | \    35 m
        | W\
   heel |___\ toe
        <26 m>

Forces

$$\begin{aligned} \sum V &= 10{,}712.5 - 2{,}231.8 = 8{,}480.7\ \text{kN} \\ \sum M_R &= 185{,}683.7, \quad \sum M_O = 108{,}784.7\ \text{kN·m} \end{aligned}$$

a) Eccentricity

$$\begin{aligned} \bar{x} &= \frac{185{,}683.7 - 108{,}784.7}{8{,}480.7} = 9.067\ \text{m from toe} \\ e &= \frac{26}{2} - 9.067 = 3.933\ \text{m} \end{aligned}$$

$e < B/6 = 4.333$ m, so the resultant lies in the middle third (no tension).

b) Factor of safety against sliding

$$FS_s = \frac{\mu\sum V}{\sum H} = \frac{0.75 \times 8{,}480.7}{6{,}008.6} = 1.06$$

c) Maximum compressive (crushing) stress

$$\begin{aligned} p_{toe} &= \frac{\sum V}{B}\left(1 + \frac{6e}{B}\right) = \frac{8{,}480.7}{26}\left(1 + \frac{6 \times 3.933}{26}\right) \\ &= 326.18 \times 1.9075 = 622.2\ \text{kN/m}^2 \ (\approx 63.4\ \text{t/m}^2) \end{aligned}$$

Heel stress $= 326.18(1 - 0.9075) = 30.2$ kN/m² (compression). Major principal stress at toe, with $\tan\alpha = 26/35$: $\sigma_1 = 622.2 \times (1 + 0.5518) = 965.5$ kN/m².

d) Factor of safety against overturning

$$FS_o = \frac{185{,}683.7}{108{,}784.7} = 1.71$$

Answer: Resultant at 9.067 m from the toe (e = 3.933 m from centre); FS sliding = 1.06; maximum vertical stress at toe = 622.2 kN/m² (principal stress 965.5 kN/m²); FS overturning = 1.71. The dam is safe in overturning and tension; FS against sliding is just above 1, so the margin is small.