3:1 Fluid Forces on Structures - Resultant F_R

Overview

Derek Elsworth develops the hydrostatic resultant force on a submerged surface, connecting pressure distributions to structural safety examples including the Malpasset Dam, the de Havilland Comet, and Aloha Airlines Flight 243. The central results are that the force magnitude is F_R = γ h_c A, acting normal to the surface, and that its location is the center of pressure, found using the area’s centroid and second moment of area; for a vertical rectangle with its top at the waterline, the force acts 2H/3 below the surface.

Key takeaways

The hydrostatic resultant on a plane surface has magnitude F_R = γh_cA, where γ is fluid unit weight, h_c is centroid depth measured vertically, and A is surface area.
Hydrostatic pressure acts normal to a surface, and its resultant generally acts below the centroid because pressure increases with depth.
For a vertical rectangle of height H whose top coincides with the waterline, the resultant acts at depth 2H/3, not at the centroid depth H/2.
The center of pressure for a plane area is found from y_R = y_c + I_G/(y_cA), using distances along the plate and the centroidal second moment of area.
Structural checks must consider both force magnitude and line of action: the combined water force and dam weight can be assessed for overturning relative to the dam toe.
Repeated pressure cycles can cause fatigue damage: the de Havilland Comet’s square window corners concentrated stress, while frequent short flights raised fatigue concerns for Aloha Airlines Flight 243.

Chapters

0:00 Dam Failures Motivate Calculating Fluid Loads

The Malpasset Dam in Provence was a thin arch dam whose canyon abutments were critical to resisting reservoir pressure.
Fault movement in the abutments displaced the dam and released the reservoir; the disaster killed about 423 people.
The example distinguishes calculating static water forces on structures from predicting the downstream flood using flow dynamics.

4:40 Comet Window Fatigue and Aloha Airlines Flight 243

The de Havilland Comet’s square windows concentrated stress at their corners, where cracks could initiate and grow under repeated pressurization.
At about 30,000 feet, cabin pressure cycles substantially below the roughly 100 kPa sea-level atmosphere, loading the aircraft skin on each flight.
Aloha Airlines Flight 243 illustrates fatigue and corrosion concerns in an aircraft making frequent short flights between Hawaiian islands.

9:00 The Dam Busters and Pressure at the Dam Base

Operation Chastise used bouncing bombs designed to pass over defenses and reach the base of hydroelectric dams.
The example highlights how water confinement and the dam’s strength affect where an impact must apply force to breach a structure.
Elsworth frames fluid-force calculations as useful both for public safety and for understanding historic engineering operations.

11:00 Review: Hydrostatic Pressure Changes with Depth

Pressure changes vertically according to dp/dz = −γ when the vertical coordinate is positive upward.
For an incompressible fluid, moving downward by depth H adds γH to pressure, while moving upward subtracts it.
At the same elevation in a connected static fluid, pressure is equal even when a wall separates the points.

15:00 Pressure Acts in Every Direction on a Surface

At a point in a fluid, pressure is isotropic, so the pressure components in the x, y, and z directions are equal.
The force on a surface element is pressure multiplied by area, and the component of interest acts normal to the surface.
The ideal gas relation p = ρRT links pressure, density, and temperature, explaining how heating air or using helium can make a balloon buoyant.

20:20 Resultant Force and Structural Stability

The resultant hydrostatic force can be combined with a structure’s weight to assess the location of the total force.
For a dam, the resultant’s position relative to the toe indicates whether the combined loading could cause overturning.
The lesson’s goals are to calculate both the magnitude of the fluid force and the point where it acts.

24:00 Centroid and Pressure Prism on an Inclined Plate

The centroid is the balance point of the plate’s area, while the pressure varies with vertical depth below the free surface.
For a plane surface, the resultant magnitude is F_R = γh_cA, where h_c is the centroid’s vertical depth and A is the plate area.
The plate may be rectangular, triangular, or circular; the force formula stays the same when the correct centroid depth and area are used.

30:00 Integrating the Pressure Distribution on a Vertical Plate

For a vertical rectangular plate of width B, pressure at depth y is p = γy and a thin strip has area B dy.
Integrating dF = pB dy from the surface to depth H gives F_R = γBH²/2.
This equals γh_cA because the centroid is at H/2 and the plate area is BH.

37:00 Center of Pressure and Second Moment of Area

The resultant generally acts below the centroid because deeper portions of the plate experience greater pressure.
Its distance along the plate is y_R = y_c + I_G/(y_cA), where I_G is the second moment of area about the centroidal x-axis.
For a rectangle of width B and length A, the centroidal second moment is I_G = BA³/12.

45:00 Vertical Rectangle Result: Force at Two-Thirds Depth

For a vertical rectangle extending from the water surface to depth H, the centroid depth is H/2 and its area is BH.
Substituting these values into the center-of-pressure equation gives y_R = 2H/3.
To solve other geometries, measure h_c vertically from the free surface for force magnitude, then use the plate-plane geometry and I_G to locate the center of pressure.

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.