A concrete dam has the cross-section shown in fig. Determine the maximum and minimum stresses at the base of the dam (section m-n). The concrete density=24 kN/m^3 and unit weight of water γw = 9.81 kN/m^3. Consider the length of the dam as 1 m. Draw the Stress Distribution Diagram at the Dam Base.

Length of Dam = 1 m

$\gamma_w = 9.81 \ KN/m^3$

$\gamma_c = 24 \ KN/m^3$

1] Total water pressure load (P).

2] Self weight of Dam = $\gamma_c \times Volume \ of \ concrete$

$W_2 = 24 \times (\frac{1}{2} \times 1 \times 25) \times 1 = 300 \ KN$

$W_1 = 24 \times (3 \times 25) \times 1 = 1800 \ KN$

$w_2$ will act at C.G of [1] i.e. $\frac{3}{2} = 1.5 \ m$

$w_1$ will act at C.G of [2] i.e. $\frac{1}{3} = 0.33 \ m$

3] Resultant of P, $w_1 \ and \ w_2$ =

$\sum fx = -73.57 \ KN (\rightarrow) = 73.57 \ KN (\leftarrow)$

$\sum fy = -300 – 1800 = -2100 \ KN (\downarrow)$

$R = \sqrt{(-73.57)^2 + (-2100)^2}$

R = 2101.28 KN

$\alpha = tan^{-1} (\frac{2100}{73.57}) = 87.99°$

Using Varigon’s Theorem:

$\sum M_A = \sum fy \ x \ \bar{x}$

$300 \times (0.33 + 3) + 1800 \times (1.5) + 73.57 \times 5$

$= 2100 \times \bar{x}$

$\therefore \ X = 1.93 \ m$

Now, $e = \bar{x} - \frac{b}{2}$

$e = 1.93 - \frac{4}{2} = -0.062 \ m$

$p_d = \frac{\sum fy}{A} = \frac{2100}{b \times 1} = \frac{2100}{4 \times 1} = 525 \ KN/m^2 $ [c]

$P_b = \pm \frac{M}{I} \times y = \pm \frac{2100 \times 0.062}{1 \times \frac{4^3}{12}} \times \frac{4}{2} = \pm 48.825 \ KN/m^2$ [c]

Where $I = \frac{1 \times b^3}{12}$

$y = \frac{b}{2}$

$P_r]_{max} = 525 + 48.825 = 573.32 \ KN/m^2$ [c]

$P_r]_{min} = 525 – 48.825 = 476.17 \ KN/m^2$ [c]