**Problem 911**

A concrete dam has the profile shown in Figure P-911. If the density of concrete is 2400 kg/m^{3} and that of water is 1000 kg/m^{3}, determine the maximum compressive stress at section *m-n* if the depth of the water behind the dam is *h* = 15 m.

**Solution 911**

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$W_1 = 180,000 ~ \text{kg}$

$W_2 = 2400 \times 3(25)(1)$

$W_2 = 180,000 ~ \text{kg}$

$F_w = 1000(7.5) \times 15(1)$

$F_w = 112,500 ~ \text{kg}$

Moment About *m*

*RM*

$RM = 4W_1 + 7.5W_2 = 4(180,000) + 7.5(180,000)$

$RM = 2,070,000 ~ \text{kg}\cdot\text{m}$

Overturning Moment, *OM*

$OM = 5F_w = 5(112,500)$

$OM = 562,500 ~ \text{kg}\cdot\text{m}$

Reactions at the Base

$R_x = F_w = 112,500 ~ \text{kg}$

Location of *R _{y}*

$\bar{x}(360,000) = 2,070,000 - 562,500$

$\bar{x} = 4.1875 ~ \text{m}$

Eccentricity

$e = 0.3125 ~ \text{m}$

$M = R_y e = 360,000(0.3125)$

$M = 112,500 ~ \text{kg}\cdot\text{m}$

$\sigma_a = \dfrac{P}{A} = \dfrac{360,000}{1(9)}$

$\sigma_a = 40,000 ~ \text{kg/m}^2$

$\sigma_f = \dfrac{6M}{bd^2} = \dfrac{6(112,500)}{1(9^2)}$

$\sigma_f = 8,333.33 ~ \text{kg/m}^2$

$q_{max} = \sigma_a + \sigma_f = 40,000 + 8,333.33$

$q_{max} = 48,333.33 ~ \text{kg/m}^2$ *answer*