Fluid Mechanics & Hydraulics

Hydrostatic Force on a Submerged Surface

Resultant hydrostatic force and center of pressure on a fully submerged plane surface (rectangle, circle, or triangle; vertical or inclined): F = γ·hc·A with ycp = yc + Ic/(yc·A) in slant coordinates, reported both along the plate and as vertical depth. Use for: force on a gate/dam face/window, 'where does the resultant act', center of pressure problems. Not for: curved surfaces, surfaces piercing the free surface, or layered fluids.

Direct solving is free and needs no account. Have a word problem instead? Submit it as text.

Inputs

Units

Load a sample problem:

surfacetype

widthPlate width (horizontal edge) [dimension: length]heightPlate height measured ALONG the plate (slant dimension for inclined plates) [dimension: length]

top edge depthVertical depth of the plate's TOP edge below the free surface (zero = top edge at the surface) [dimension: length]inclination deg (optional)Plate inclination from the HORIZONTAL free surface in degrees; 90 = vertical plate (default)

fluidFluid definition — give at most ONE of specificGravity, unitWeight, or density; omit all three for water (γ = 9.81 kN/m³)

specific gravity (optional)Specific gravity SG relative to water (γw = 9.81 kN/m³): 13.6 mercury, ~0.85 oil, 1 water. Give at most one of specificGravity, unitWeight, density.unit weight (optional)Fluid unit weight γ [dimension: unitweight]density (optional)Fluid mass density ρ (converted internally via γ = ρ·g) [dimension: density]

Worked examples

These are the solver's own reference problems — the answers come from published sources or independent hand computation, never from the solver itself. The solution below is the live solver output for each.

Example 1

SI: vertical 2 m × 3 m rectangular gate, top edge 1 m deep

Given

surface: shape rectanglewidth 2 mheight 3 m
top edge depth: 1 m
inclination deg: 90

Assumptions

No fluid was specified — fresh water (γ = 9.81 kN/m³) was assumed.
The fluid is static with constant unit weight, and the plate is fully submerged (top edge at or below the free surface).
Forces are gage-pressure resultants: atmospheric pressure acts on both sides of the plate and cancels.

Solution steps

Plate geometry
Area and centroidal second moment of area of the rectangular plate.
A = w·h; Ixc = w·h³/12
A = 6 m^2; I_xc = 4.5 m^4
Locate the centroid in slant coordinates
Distances y are measured ALONG the plate from where its plane meets the free surface. The top edge sits at y = h_top/sin θ, and the centroid a further distance down the plate. For a vertical plate sin θ = 1 and slant distances equal vertical depths.
yc = h_top/sin θ + d_c; hc = yc·sin θ
θ = 90 deg; y_top = 1 m; centroid 1.5 m down the plate from the top edge → y_c = 2.5 m, vertical depth h_c = 2.5 m
Resultant hydrostatic force
The resultant equals the pressure at the centroid times the plate area.
FR = γ·hc·A
F_R = 9.81 kN/m^3 × 2.5 m × 6 m^2 = 147.2 kN
= 147.2 kN
Center of pressure (slant coordinate)
Pressure grows with depth, so the resultant acts BELOW the centroid. The transfer formula uses slant distances — applying it to vertical depths is only correct for vertical plates.
ycp = yc + Ixc/(yc·A)
y_cp = 2.5 m + 4.5 m^4 / (2.5 m × 6 m^2) = 2.8 m
= 2.8 m
Center of pressure as a vertical depth
Project the slant location back to a vertical depth below the free surface.
hcp = ycp·sin θ
h_cp = 2.8 m × sin θ = 2.8 m
= 2.8 m

Results

Resultant hydrostatic force FR

147.2kN

Vertical depth of the centroid hc

2.5m

Vertical depth of the center of pressure hcp

2.8m

Slant distance to the centroid yc (along the plate)

2.5m

Slant distance to the center of pressure ycp (along the plate)

2.8m

Plate area

6m^2

Where this answer was checked

source: Plane-surface formulas F = γ·hc·A, ycp = yc + Ic/(yc·A) (NCEES FE Reference Handbook) — vertical rectangle, water
verified by: hand-recomputed
derivation: hc = 1 + 3/2 = 2.5 m; A = 6 m²; F = 9810·2.5·6 = 147 150 N = 147.15 kN. Ic = 2·3³/12 = 4.5 m⁴; ycp = 2.5 + 4.5/(2.5·6) = 2.8 m (vertical plate: slant ≡ vertical).

Example 2

US: vertical circular gate D = 4 ft, top edge 6 ft deep, γ = 62.4 pcf

Given

surface: shape circlediameter 4 ft
top edge depth: 6 ft
inclination deg: 90
fluid: unit weight 62.4 pcf

Assumptions

The fluid is static with constant unit weight, and the plate is fully submerged (top edge at or below the free surface).
Forces are gage-pressure resultants: atmospheric pressure acts on both sides of the plate and cancels.

Solution steps

Plate geometry
Area and centroidal second moment of area of the circular plate.
A = πD²/4; Ixc = πD⁴/64
A = 12.6 ft^2; I_xc = 12.6 ft^4
Locate the centroid in slant coordinates
Distances y are measured ALONG the plate from where its plane meets the free surface. The top edge sits at y = h_top/sin θ, and the centroid a further distance down the plate. For a vertical plate sin θ = 1 and slant distances equal vertical depths.
yc = h_top/sin θ + d_c; hc = yc·sin θ
θ = 90 deg; y_top = 6 ft; centroid 2 ft down the plate from the top edge → y_c = 8 ft, vertical depth h_c = 8 ft
Resultant hydrostatic force
The resultant equals the pressure at the centroid times the plate area.
FR = γ·hc·A
F_R = 62.4 pcf × 8 ft × 12.6 ft^2 = 6273 lbf
= 6273 lbf
Center of pressure (slant coordinate)
Pressure grows with depth, so the resultant acts BELOW the centroid. The transfer formula uses slant distances — applying it to vertical depths is only correct for vertical plates.
ycp = yc + Ixc/(yc·A)
y_cp = 8 ft + 12.6 ft^4 / (8 ft × 12.6 ft^2) = 8.125 ft
= 8.125 ft
Center of pressure as a vertical depth
Project the slant location back to a vertical depth below the free surface.
hcp = ycp·sin θ
h_cp = 8.125 ft × sin θ = 8.125 ft
= 8.125 ft

Results

Resultant hydrostatic force FR

6.273kip

Vertical depth of the centroid hc

8ft

Vertical depth of the center of pressure hcp

8.125ft

Slant distance to the centroid yc (along the plate)

8ft

Slant distance to the center of pressure ycp (along the plate)

8.125ft

Plate area

12.6ft^2

Where this answer was checked

source: Plane-surface formulas in US customary units — vertical circle, water at 62.4 lbf/ft³
verified by: hand-recomputed
derivation: hc = 6 + 2 = 8 ft; A = π·4²/4 = 12.5664 ft²; F = 62.4·8·12.5664 = 6273.1 lbf. Ic = πD⁴/64 = 4π = 12.5664 ft⁴; Ic/(yc·A) = 4π/(8·4π) = 1/8 exactly → ycp = 8.125 ft.

Example 3

SI: rectangle 1.2 m × 2 m on a 60° incline, top edge 1 m deep

Given

surface: shape rectanglewidth 1.2 mheight 2 m
top edge depth: 1 m
inclination deg: 60

Assumptions

No fluid was specified — fresh water (γ = 9.81 kN/m³) was assumed.
The fluid is static with constant unit weight, and the plate is fully submerged (top edge at or below the free surface).
Forces are gage-pressure resultants: atmospheric pressure acts on both sides of the plate and cancels.

Solution steps

Plate geometry
Area and centroidal second moment of area of the rectangular plate.
A = w·h; Ixc = w·h³/12
A = 2.4 m^2; I_xc = 0.8 m^4
Locate the centroid in slant coordinates
Distances y are measured ALONG the plate from where its plane meets the free surface. The top edge sits at y = h_top/sin θ, and the centroid a further distance down the plate. For a vertical plate sin θ = 1 and slant distances equal vertical depths.
yc = h_top/sin θ + d_c; hc = yc·sin θ
θ = 60 deg; y_top = 1.15 m; centroid 1 m down the plate from the top edge → y_c = 2.15 m, vertical depth h_c = 1.87 m
Resultant hydrostatic force
The resultant equals the pressure at the centroid times the plate area.
FR = γ·hc·A
F_R = 9.81 kN/m^3 × 1.87 m × 2.4 m^2 = 43.93 kN
= 43.93 kN
Center of pressure (slant coordinate)
Pressure grows with depth, so the resultant acts BELOW the centroid. The transfer formula uses slant distances — applying it to vertical depths is only correct for vertical plates.
ycp = yc + Ixc/(yc·A)
y_cp = 2.15 m + 0.8 m^4 / (2.15 m × 2.4 m^2) = 2.309 m
= 2.309 m
Center of pressure as a vertical depth
Project the slant location back to a vertical depth below the free surface.
hcp = ycp·sin θ
h_cp = 2.309 m × sin θ = 2 m
= 2 m

Results

Resultant hydrostatic force FR

43.93kN

Vertical depth of the centroid hc

1.87m

Vertical depth of the center of pressure hcp

Slant distance to the centroid yc (along the plate)

2.15m

Slant distance to the center of pressure ycp (along the plate)

2.309m

Plate area

2.4m^2

Where this answer was checked

source: Inclined plane-surface analysis in slant coordinates (any fluids text) — rectangle on a 60° slope; slant vs vertical center of pressure
verified by: hand-recomputed
derivation: sin 60° = 0.866025. y_top = 1/0.866025 = 1.154701 m; yc = 1.154701 + 1 = 2.154701 m; hc = yc·sin 60° = 1.866025 m; A = 2.4 m²; F = 9810·1.866025·2.4 = 43 933.7 N = 43.934 kN. Ic = 1.2·2³/12 = 0.8 m⁴; ycp = 2.154701 + 0.8/(2.154701·2.4) = 2.309401 m (= 4/√3); hcp = ycp·sin 60° = 2.000 m exactly.