3:2 Fluid Forces on Structures - Point of Action, C of P = y_R, Failure Mechanisms

Overview

Derek Elsworth develops a practical method for finding both the magnitude and point of action of hydrostatic forces on submerged gates and other structures. He derives the resultant-force relation and center-of-pressure formula, then applies them to a circular gate and a compound gate, using moments about hinges to calculate the forces needed to prevent rotation; historical dam failures and aircraft fatigue illustrate why structural loading matters.

Key takeaways

For any plane submerged surface, the hydrostatic resultant magnitude is F_R = γh_cA, where h_c is the centroid’s vertical depth—not its distance along an inclined plate.
The center of pressure is below the centroid when pressure varies with depth; its location is y_R = y_c + I_xc/(y_cA).
A rectangular plate’s centroidal second moment is ba³/12, and a circle’s is πr⁴/4; these shape properties determine the offset between centroid and center of pressure.
A gate’s required retaining force cannot be found from the fluid resultant alone: moments must be balanced about the hinge or other axis associated with the possible failure mechanism.
For the 5 m-radius circular gate centered 10 m below the water surface, the resultant is about 7,700 kN and acts roughly 0.625 m below the centroid.
In compound-gate problems, calculate each pressure resultant and its line of action separately, then sum their moments about the hinge before solving for the applied force.

Chapters

0:00 Hydrostatic Loading and the Malpasset Dam Failure

The lesson connects increasing water pressure with depth to the task of calculating total forces on structures.
The 1959 Malpasset Dam failure in southern France followed movement of its rock abutments; the released reservoir water killed people downstream.
The example frames accurate fluid-force analysis as an engineering responsibility for public safety.

4:00 Vajont Dam, Comet Fatigue, and Structural Risk

At Italy’s Vajont Dam, a landslide displaced reservoir water over the dam in 1963, killing about 2,000 people even though the dam itself remained intact.
Early de Havilland Comet airliners developed fatigue cracks around square windows, where corners concentrated stress under repeated cabin pressurization cycles.
A water-tank test recreated pressure differences and cycling to investigate fatigue failure.

8:00 Dam Busters and the Need to Locate Fluid Forces

The dam-busting example explains that a submerged explosion could transfer more energy into a dam because surrounding water confined the blast.
Elsworth recaps that pressure increases with depth and that points at the same elevation in a connected fluid have equal pressure.
The analysis must find both the resultant force on a structure and where that force acts.

12:00 Resultant Force from the Pressure at the Plate Centroid

For a plane surface, the resultant hydrostatic force is the pressure at the area centroid multiplied by the plate area.
For water, pressure at the centroid is the unit weight times its vertical depth, giving the form F_R = γ h_c A.
A linearly increasing pressure distribution can be represented as a uniform rectangular component plus a triangular component.

17:00 Finding the Center of Pressure by Taking Moments

The rectangle’s resultant acts at its midpoint, while a triangular pressure distribution acts two-thirds of the way from its zero-pressure tip.
Taking moments of the rectangular and triangular components about the water surface gives the location y_R of the combined force.
Because pressure is greater at the deeper edge, the center of pressure generally lies below the plate centroid.

23:00 Second Moment of Area Formula for Center of Pressure

The standard center-of-pressure relation is y_R = y_c + I_xc/(y_c A), where I_xc is the area’s second moment about its centroidal axis.
For a rectangular plate of width b and depth a, I_xc = ba³/12 and A = ba.
For an inclined plate, y_c is the distance along the plate to its centroid, while h_c remains the centroid’s vertical depth used to calculate force.

29:00 Inclined Plates and How Pressure Shifts the Resultant

The center-of-pressure formula applies to vertical and inclined surfaces when the geometry uses the correct slope distance and vertical depth.
On a horizontal plate at constant depth, pressure is uniform and the center of pressure coincides with the centroid.
As a plate becomes more vertical, its pressure variation with depth increases and shifts the resultant farther below the centroid.

35:00 Circular Gate: Resultant Force and Center-of-Pressure Depth

For a circular gate with radius 5 m and centroid 10 m below the surface, the area is πr² and the centroidal second moment is πr⁴/4.
The center-of-pressure shift is I/(y_c A) = 25/40 = 0.625 m, placing the resultant about 10.625 m below the surface.
The water force is approximately 7,700 kN, calculated as γh_cA for water with γ ≈ 9.81 kN/m³.

41:00 Hinge Moments and the Force Needed to Keep a Gate Shut

For the circular gate, the hinge is at the gate’s center and the resultant acts about 0.625 m below the hinge.
Taking moments about the hinge balances the water-force moment against the applied force at the gate’s top, 5 m from the hinge.
The lecture gives an approximate closing force of 954 kN and emphasizes that the free-body diagram must match the gate’s rotation or failure mechanism.

47:00 Compound Gate: Combining Two Pressure Resultants

The final gate example separates the loading into two fluid-pressure resultants, each calculated from γh_cA.
For the 4 m by 3 m vertical gate section, the centroid is 5 m below the surface; its rectangular second moment is used to locate the horizontal force’s center of pressure.
The second section has uniform pressure and a 1 m lever arm; moments about the hinge combine both fluid forces with the applied force 4 m from the hinge.

Summary, takeaways, and chapters were generated by AI from the video's transcript and may contain errors. The video belongs to its creator, Derek Elsworth.