Find the hydrostatic force and center of pressure on a fully submerged rectangular, circular, or triangular plane, vertical or inclined.
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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
2m
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.