A concrete dam of rectangular section 15 m high and 6m wide contains water up to height of 13 m. Find the maximum and minimum stress at the base. Assume unit weight of water and concrete 10 kN/m^3 and 25kN/m^3.

Consider the length of dam (L) = 1m.

$\gamma_w = 10 \ KN/m^3$ (Density of water).

$\gamma_c = 25 \ KN/m^3$ (Density of concrete).

NOTE:

Effect of water is in the form of triangular loading as shown.

1] Total water pressure load (p) = $\frac{1}{2} \times (13 \times 1) \times 10 = 65 \ KN$

[Which will act at n/3 i.e. 13/3 = 4.33 from base].

2] Self weight of Dam:

$W = \gamma_c \times Volume$

$W = 25 \times 15 \times 6 \times 1 = 2250 \ KN$ [$Volume = b \times h \times l$]

(Which will act at b/2 = 6/2 = 3m).

3] Resultant of p and w (General Force System)

$\sum f_x = 65 \ KN $ $(\rightarrow)$

$\sum f_y = - 2250 = -225-$ ($\downarrow$)

$\therefore$ Resultant $R = \sqrt{ (\sum fx)^2 + (\sum fy)^2} = \sqrt{65^2 + 2250^2}$

= 2250.93 KN

$\alpha = tan^{-1} (\frac{2250}{65})$

$\alpha = 88.34°$

Using Varigon’s Theorem:

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

$\therefore$ $65 \times 4.33 + 2250 \times 3 = 2250 \times \bar{X}$

$\therefore$ $\bar{X} = 3.125 \ m $ from A.

4] Eccentricity of resultant w.r.t. center of base e:

$e = \bar{x} - \frac{b}{2} = 3.125 - \frac{6}{2}$

e = 0.125 m

now,

$P_d = \frac{\sum \ fy}{A} = \frac{2250}{b \times 1} = \frac{2250}{6 \times 1} = 375 \ KN/m^2 $ ( c )

(Direct Stress)

$p_b = \pm \frac{M}{I} \times y = \frac{(\sum \ fy \times e)}{(1 \times \frac{b^3}{12})} \times (\frac{b}{2}) = \frac{2250 \times 0.125}{(1 \times \frac{6^3}{12})} \times \frac{6}{2}$

(Bending stress)

$P_b = \pm 46.875 \ KN/m^2$

$P_r]_{max} = 375 + 46.875 = 421.87 \ KN/m^2$ ( c )

$P_r]_{min} = 375 – 46.875 = 328.125 \ KN/m^2$

$\therefore$ The maximum and minimum stress at base are 421.87 $KN/m^2$ and 328.125 $KN/m^2$.