1. Water Pressure
Pw = ρ*G*H*Ac
Pw = Pressure of water on dam
G = Gravity = 32.2 ft/s2
ρ = Density of liquid = 1.94 slugs/ft3
H = Head of water / Water Depth
Ac = Area of contact / area of the face of the dam in contact with water
Pw = 1.94 slugs/ft3 * 32.2 ft/s2 *H *Ac
Pw = 62.4 lb/ft2*H*Ac
The pressure from the water acting on the inside (wet side) face of the dam, is dependent on the
depth of the water that the dam is holding back. To prevent the dam from overturning, the
moment from the water pressure acting on the bottom outside edge of the dam (Point of interest,
figure 2), must be less than the moment, caused by the weight of the dam’s structure; acting in
the opposite direction on the same point.
Water Body
Dam
Pond Floor
T
H
Figure 1 Dam and Pond Template
H = Water Head / Water Depth
T = Dam Thickness

2. Weight of the Dam’s Structure
Pc = ρ*G*H*Ac
Pc = Pressure of concrete on dam base
G = Gravity = 32.2 ft/s2
ρ = Density of concrete = 4.13 slugs/ft3
( 80 lb bag makes .6 ft3)
H = Head of water / Water Depth
Ac = Area of contact / area of the face of the dam in contact with water
Pc = 1.94 slugs/ft3 * 32.2 ft/s2 *H *Ac
Pc = 133.3 lb/ft2*H*Ac
Water Body
Dam
Pond Floor
T
H
Figure 2 Dam and Pond Template
H = Water Head / Water Depth
T = Dam Thickness
Point of interest
H/3
Water pressure
T/2
Weight of dam

3. Moments About the Point of Interest
The minimum thickness of the dam required so the dam will not overturn, is found by setting the
moment around the point of interest equal to each other.
ΣMp = 0 = Pc*1/2T - Pw*1/3H
(133.3 lb/ft2)*H*Ac*1/2T – (62.4 lb/ft2)*H*Ac*1/3H
𝑻 = [
𝟔𝟐. 𝟒
𝐥𝐛
𝐟𝐭 𝟐 ∗
𝑯
𝟑
𝟏𝟑𝟑. 𝟑
𝐥𝐛
𝐟𝐭 𝟐
] ∗ 𝟐
Also written as..
𝑻 = [
𝝆 𝒘 ∗
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𝟑
𝝆 𝒄
] ∗ 𝟐