Hydrostatics of dams – total hydrostatic force per metre length on a vertical wall For a vertical masonry dam wall retaining a liquid of specific weight w to depth H, what is the total hydrostatic force per metre length of the wall (resultant acting horizontally)?

Introduction / Context:
The design of retaining structures like dams requires evaluating the resultant hydrostatic force. On a vertical wall holding a liquid of depth H, the pressure varies linearly with depth, creating a triangular pressure distribution.

Given Data / Assumptions:

• Liquid is static and homogeneous with specific weight w.
• Wall is vertical and of unit breadth (1 m) into the page.
• Free surface is at atmospheric pressure; gauge pressures are considered.

Concept / Approach:
Gauge pressure at depth y is p = w * y. The resultant horizontal force equals the area under the pressure diagram (a triangle with height wH at the base and width H). For unit width, the equivalent area gives the total force.

Step-by-Step Solution:
1) Pressure at depth y: p(y) = w * y.2) Force on a horizontal strip of thickness dy per metre width: dF = p(y) * 1 * dy = w * y * dy.3) Integrate from y = 0 to H: F = ∫(0→H) w * y dy = w * H^2 / 2.4) Line of action occurs at 2H/3 below the free surface (useful for moment checks).

Verification / Alternative check:
The triangular pressure diagram has area (1/2) * base * height = (1/2) * H * (wH) = w * H^2 / 2, identical to the integration result, confirming consistency.

Why Other Options Are Wrong:

• w * H or w * H / 2: these have incorrect dimensional dependence on H (linear rather than quadratic).
• w * H^2 / 4: underestimates the triangular area by a factor of 2.
• 2 * w * H^2: overestimates by a factor of 4.

Common Pitfalls:
Confusing force per metre length with average pressure; forgetting that pressure increases with depth leads to missing the H^2 dependence.

Final Answer:
w * H^2 / 2