Hydrostatic forces on a submerged inclined plane surface
Setup and assumptions
Fluid:
We assume the fluid is static (no motion) and incompressible (its density does not change with depth).
This means the only variation in pressure comes from the weight of the fluid above a point.
The density is denoted by \(\rho\) (kg/m³).
Gravity:
The fluid is under the influence of gravity of magnitude \(g\) (≈ 9.81 m/s² on Earth).
We take the vertical direction as positive downward, so pressure increases with depth.
Surface:
The surface under consideration is a flat plane of total area \(A\).
Its centroid (geometric center) lies at a vertical depth \(h_c\) below the free surface of the fluid.
The plane is inclined at an angle \(\theta\) to the horizontal.
- If \(\theta = 0^\circ\), the surface is horizontal.
- If \(\theta = 90^\circ\), the surface is vertical.
The orientation affects how pressure distributes across the surface.
Pressure variation:
Hydrostatic pressure increases linearly with depth according to the law:
\[
p(h) = p_0 + \rho g h
\]
where:
\(h\) = vertical depth below the free surface,
\(p_0\) = reference pressure at the free surface (often atmospheric pressure),
\(\rho g h\) = pressure due to the fluid column above depth \(h\).
In many engineering problems, we use gauge pressure (measured relative to the free surface), so \(p_0\) is taken as zero.
Horizontal plate at depth h
Horizontal plate: At constant depth h, pressure is uniform. Force magnitude F = ρ g h · A (acts downward normal to the plate).
Vertical plate of height h (top at free surface)
Vertical plate (top at free surface): Pressure increases linearly with depth. Average pressure p̄ = ρ g h / 2.
Force magnitude F = p̄ · A = (ρ g h / 2) · (b · h). Center of pressure is at h/3 from the bottom.
Diagram
Fluid: Static, incompressible; pressure increases linearly with depth.
Inclined surface: Angle θ to horizontal; along-plane coordinate y relates to depth via h = y·sin(θ).
Resultant force F: Magnitude F = ρ g A h_c; acts normal through the center of pressure.
Depth markers: h_c is vertical depth of centroid; CoP is deeper due to nonuniform pressure.
Center of pressure (depth):
\[
h_{\mathrm{cp}} = h_c + \frac{I_G \sin^2 \theta}{A \, h_c}
\]
Larger \(I_G\) or steeper \(\theta\) pushes the CoP deeper.
Resultant hydrostatic force
A submerged surface carries a distributed pressure load that increases with vertical depth.
The hydrostatic resultant is the single force that produces the same net effect (force and moment) as that distribution.
It acts normal to the surface and its magnitude depends on fluid properties, area, and how deep the surface lies.
Step 1: Pressure at depth (hydrostatic law)
At vertical depth \(h\) below the free surface:
\[
p(h) = p_0 + \rho g h
\]
where \(p_0\) is the free-surface (reference) pressure, \(\rho\) is fluid density, and \(g\) is gravitational acceleration.
In gauge form (pressures measured relative to the free surface), set \(p_0 = 0\) so \(p(h) = \rho g h\).
Pressure units: Pa; \([\rho] = \text{kg/m}^3\), \([g] = \text{m/s}^2\), \([h] = \text{m}\).
Step 2: Differential force on an area element
A small element \(dA\) at depth \(h\) experiences a normal force:
\[
dF = p(h)\, dA \quad \Rightarrow \quad dF = \rho g h\, dA \quad (\text{gauge})
\]
The direction of \(dF\) is perpendicular to the surface (pointing from fluid into the solid).
Step 3: Integrate the distribution to get the resultant
Integrate over the wetted area \(A\):
\[
F = \int_A \rho g h\, dA = \rho g \int_A h\, dA
\]
The integral \(\int_A h\, dA\) is the first moment of area with respect to the free surface depth.
Dividing by area gives the average depth of the surface:
\[
\frac{1}{A}\int_A h\, dA = h_c
\]
where \(h_c\) is the vertical depth of the centroid of the area.
Step 4: Compact result and its direction
Substituting \(\int_A h\, dA = A h_c\) yields the standard magnitude:
\[
F = \rho g\, A\, h_c
\]
The resultant acts normal to the plane (unit normal \(\mathbf{n}\)).
In vector form:
\[
\mathbf{F} = \big(\rho g\, A\, h_c\big)\, \mathbf{n}
\]
For an inclined plane at angle \(\theta\) to the horizontal, if \(y\) measures distance along the plane from its intersection with the free surface, then \(h = y \sin\theta\). This relation is useful when switching between along‑plane and vertical coordinates.
Physical meaning and checks
Shape independence (magnitude): The total force depends on area \(A\) and centroid depth \(h_c\), not on the detailed shape—because hydrostatic pressure varies linearly with depth.
Depth sensitivity: Doubling the centroid depth \(h_c\) doubles the resultant force; increasing \(\rho\) (e.g., oil vs. water) scales force proportionally.
Direction: Always normal to the surface; components along global axes can be found by projecting \(\mathbf{n}\).
Sanity check: For a horizontal surface at uniform depth \(h\), pressure is uniform, so \(F = (\rho g h)\, A\) (consistent with \(h_c = h\)).
Common pitfalls
Using edge depth instead of centroid depth: The formula requires \(h_c\), not the top or bottom edge depth.
Wrong coordinate for depth: On inclined planes, use vertical depth \(h\), not along‑plane distance \(y\), unless converted via \(h = y \sin\theta\).
Ignoring partial submergence: Only integrate over the wetted area; if a plate is partially submerged, recompute \(A\), \(h_c\) for the wetted portion.
Engineering use
The magnitude \(F = \rho g A h_c\) provides the resultant load on gates, hatches, dams, tank walls, and hull plating.
For design, combine this magnitude with the center of pressure (line of action) to compute moments about supports and to size hinges, seals, and reinforcements.
Center of pressure (line of action)
The center of pressure (CoP) is the exact point on a submerged surface where the single
resultant hydrostatic force can be considered to act. While the magnitude of the resultant force
depends only on the centroid depth \(h_c\), the location of this force is shifted below the centroid
because pressure increases linearly with depth. In other words, the pressure distribution is triangular
(or trapezoidal for partial submergence), so the resultant force acts deeper than the centroid.
Derivation idea:
- Each elemental force \(dF = \rho g h\, dA\) acts at its own depth \(h\).
- To find the line of action of the resultant, we equate the moment of the distributed forces about a reference axis
to the moment of the single resultant force.
- This ensures that the resultant not only produces the same net force but also the same turning effect (moment).
Mathematical expression:
For a plane surface inclined at angle \(\theta\) to the horizontal, the vertical depth of the center of pressure is:
\[
h_{\mathrm{cp}} = h_c + \frac{I_G \sin^2 \theta}{A h_c}
\]
where:
\(h_c\) = vertical depth of the centroid,
\(A\) = area of the surface,
\(I_G\) = second moment of area (area moment of inertia) about a horizontal axis through the centroid and parallel to the free surface,
\(\theta\) = inclination angle of the plane surface with respect to the horizontal.
Alternative along‑plane form:
Sometimes it is convenient to measure distances along the plane itself.
If \(y_c\) is the distance of the centroid along the plane from the free surface, then:
\[
y_{\mathrm{cp}} - y_c = \frac{I_G}{A y_c}
\]
The relation between vertical depth and along‑plane distance is \(h = y \sin\theta\).
Physical interpretation:
The CoP is always below the centroid for inclined or vertical surfaces, because deeper portions of the surface experience higher pressure and “pull” the resultant downward.
The shift between centroid and CoP depends on the geometry of the surface (through \(I_G\)) and its orientation (through \(\sin^2\theta\)).
For a horizontal surface (\(\theta = 0^\circ\)), \(\sin^2\theta = 0\), so \(h_{\mathrm{cp}} = h_c\). The resultant passes through the centroid because pressure is uniform.
For a vertical surface (\(\theta = 90^\circ\)), the shift is maximum:
\[
h_{\mathrm{cp}} = h_c + \frac{I_G}{A h_c}.
\]
Geometric influence:
Deeper centroid: As \(h_c\) increases, the denominator \(A h_c\) grows, reducing the shift. Thus, for very deep surfaces, the CoP approaches the centroid.
Slender or wide shapes: A larger second moment of area \(I_G\) increases the shift, moving the CoP further below the centroid.
Orientation: The \(\sin^2\theta\) factor shows that tilting the surface closer to vertical increases the shift, while making it more horizontal reduces it.
Engineering significance:
Knowing the CoP is essential for structural design.
- On gates, hatches, and dams, the line of action determines the moment about hinges or supports.
- If the CoP is not accounted for, designers may underestimate the overturning or bending moments.
- For floating bodies or ship hulls, the CoP helps in analyzing stability and rolling resistance.
- In laboratory experiments, locating the CoP validates hydrostatic theory by balancing moments on submerged plates.
Summary:
- The resultant hydrostatic force magnitude is simple: \(F = \rho g A h_c\).
- The center of pressure corrects for the non‑uniform distribution of pressure.
- Its depth is always at or below the centroid, never above.
- The exact shift depends on geometry (\(I_G\)), area, centroid depth, and inclination angle.
Common second moments of area
The second moment of area (also called the area moment of inertia) is a geometric property
that measures how an area is distributed about an axis. It is not the same as mass moment of inertia,
but it plays a similar role in determining how forces and pressures are distributed.
In hydrostatics, the second moment of area appears in the formula for the center of pressure:
\[
h_{\mathrm{cp}} = h_c + \frac{I_G \sin^2 \theta}{A h_c}
\]
where \(I_G\) is taken about a horizontal axis through the centroid and parallel to the free surface.
A larger \(I_G\) means the area is spread further from the axis, which increases the shift between centroid and center of pressure.
Below are the most common shapes encountered in engineering, with their centroidal second moments of area:
Rectangle (width \(b\), height \(a\)) about centroidal axis parallel to width:
Imagine a vertical rectangle submerged with its base horizontal. The axis of interest runs horizontally through the centroid, parallel to the base.
The second moment of area is:
\[
I_G = \frac{b\, a^3}{12}
\]
Explanation: The cubic dependence on height \(a^3\) shows that tall rectangles resist bending or pressure shifts much more strongly than wide but shallow ones.
Circle (radius \(R\)) about any centroidal diameter:
For a circular plate, symmetry means all centroidal diameters are equivalent. The second moment of area is:
\[
I_G = \frac{\pi R^4}{4}
\]
Explanation: The \(R^4\) dependence means that even small increases in radius greatly increase the second moment, so large circular hatches or windows have significantly deeper centers of pressure than small ones.
Triangle (base \(b\), height \(a\)) about centroidal axis parallel to base:
For a triangular plate with base horizontal and height vertical, the centroidal axis is parallel to the base and passes through the centroid (at one‑third the height from the base).
The second moment of area is:
\[
I_G = \frac{b\, a^3}{36}
\]
Explanation: Compared to a rectangle of the same base and height, the triangular area is less “spread out” about the centroidal axis, so its second moment is smaller (note the denominator 36 instead of 12).
Key observations:
All formulas involve the cube or fourth power of a dimension, showing that geometry strongly influences the second moment.
For rectangles and triangles, the vertical dimension (height) dominates because pressure varies with depth.
For circles, the radius to the fourth power makes size a critical factor in hydrostatic design.
Engineering significance:
- These values are used to calculate the center of pressure for gates, hatches, and submerged surfaces.
- They also appear in structural analysis (beam bending, deflection) and fluid mechanics (hydrostatic stability).
- Knowing standard formulas saves time and ensures accurate design without re‑deriving integrals each time.
Worked example: rectangular gate
A rectangular gate of height \(a = 2\,\text{m}\) and width \(b = 1.5\,\text{m}\) is fully submerged and inclined at \(\theta = 60^\circ\) to the horizontal.
The top edge is at vertical depth \(h_t = 1\,\text{m}\). Water density \(\rho = 1000\,\text{kg/m}^3\).
Centroid depth: The centroid lies halfway down the plate along its plane.
Along-plane centroid distance from top: \(y_c = a/2 = 1\,\text{m}\). Vertical depth:
\[
h_c = h_t + y_c \sin\theta = 1 + (1)\sin 60^\circ \approx 1 + 0.866 = 1.866\,\text{m}.
\]
Resultant force:
\[
F = \rho g A h_c = (1000)(9.81)(a b)(h_c) = 1000 \cdot 9.81 \cdot (2 \cdot 1.5) \cdot 1.866
\]
\[
F \approx 1000 \cdot 9.81 \cdot 3 \cdot 1.866 \approx 54{,}900\,\text{N} \quad (\approx 54.9\,\text{kN}).
\]
Center of pressure depth (continued):
Using \(A = a b = 2 \times 1.5 = 3.0\,\text{m}^2\) and \(\sin 60^\circ = 0.866\),
\[
h_{\mathrm{cp}} = h_c + \frac{I_G \sin^2 60^\circ}{A h_c}
= 1.866 + \frac{1.0 \times (0.866)^2}{3.0 \times 1.866}
\approx 1.866 + \frac{0.75}{5.598}
\approx 1.866 + 0.134
\approx 2.000\,\text{m}.
\]
So, the line of action of the hydrostatic resultant is at \(2.00\,\text{m}\) below the free surface.
Common second moments of area with derivations
The second moment of area about a given axis measures how the area is distributed relative to that axis.
For hydrostatics, we need the second moment about a horizontal axis through the centroid and parallel to the free surface, denoted \(I_G\).
By definition, for a horizontal centroidal axis:
\[
I_G = \int_{\mathcal{A}} z^2 \, dA,
\]
where \(z\) is the perpendicular distance from the centroidal horizontal axis to the area element \(dA\).
Below are the standard results with derivations (using direct integration). Coordinates are chosen so that the centroidal axis is at \(z=0\).
Place the rectangle so its centroid is at the origin. Let the vertical coordinate \(z\) range from \(-a/2\) to \(a/2\).
Along the width, \(x\) ranges from \(-b/2\) to \(b/2\). The area element is \(dA = dx\,dz\).
The second moment about the horizontal centroidal axis is:
\[
I_G = \int_{-b/2}^{b/2} \int_{-a/2}^{a/2} z^2 \, dz\, dx
= \left( \int_{-b/2}^{b/2} dx \right)\left( \int_{-a/2}^{a/2} z^2 \, dz \right)
= b \cdot \left[ \frac{z^3}{3} \right]_{-a/2}^{a/2}
= b \cdot \frac{2}{3}\left(\frac{a}{2}\right)^3
= \frac{b\,a^3}{12}.
\]
Result:
\[
I_G = \frac{b\, a^3}{12}.
\]
Circle (radius \(R\)) — about any centroidal diameter (horizontal axis through centroid)
Use polar coordinates \((r,\phi)\) with centroid at the origin. For a horizontal axis,
the distance to the axis is \(z = r \sin\phi\). The area element is \(dA = r\,dr\,d\phi\).
Integrate over the disk: \(0 \le r \le R\), \(0 \le \phi \le 2\pi\).
Triangle (base \(b\), height \(a\)) — axis parallel to base (horizontal centroidal axis)
Consider a triangle with base along the \(x\)-axis and height along \(z\), with its centroid at \(z=0\).
For convenience, first compute the second moment about the base, then use the parallel axis theorem to reach the centroidal axis.
Let the base be at \(z=0\) and the apex at \(z=a\). The width at height \(z\) scales linearly:
\[
w(z) = b\left(1 - \frac{z}{a}\right).
\]
Then
\[
I_{\text{base}} = \int_0^a z^2\, w(z)\, dz
= \int_0^a z^2\, b\left(1 - \frac{z}{a}\right)\, dz
= b\left( \int_0^a z^2\, dz - \frac{1}{a}\int_0^a z^3\, dz \right)
= b\left( \frac{a^3}{3} - \frac{a^4}{4a} \right)
= b\left( \frac{a^3}{3} - \frac{a^3}{4} \right)
= b \cdot \frac{a^3}{12}.
\]
The centroid of a triangle is located at \(a/3\) from the base. Using the parallel axis theorem:
\[
I_G = I_{\text{base}} - A d^2,
\]
where \(A = \frac{1}{2} b a\) is the area, and \(d = a/3\) is the distance from base axis to centroid axis.
Thus
\[
I_G = \frac{b a^3}{12} - \left(\frac{1}{2} b a\right)\left(\frac{a}{3}\right)^2
= \frac{b a^3}{12} - \frac{1}{2} b a \cdot \frac{a^2}{9}
= \frac{b a^3}{12} - \frac{b a^3}{18}
= b a^3\left( \frac{1}{12} - \frac{1}{18} \right)
= b a^3 \cdot \frac{1}{36}.
\]
Result:
\[
I_G = \frac{b\, a^3}{36}.
\]
Connecting second moments to the center of pressure formula
The center of pressure depth follows from moment equilibrium of the distributed hydrostatic forces.
Let the local depth be \(h = y \sin\theta\), where \(y\) measures distance along the plane from its free-surface intersection,
and \(\theta\) is the inclination to the horizontal.
The elemental force is \(dF = \rho g h \, dA = \rho g \sin\theta \, y \, dA\).
The resultant \(F = \rho g A h_c\) acts at some along‑plane location \(y_{\mathrm{cp}}\) such that the first moment of the distributed load equals that of the resultant:
\[
\int_A y \, dF = y_{\mathrm{cp}} \, F.
\]
Substituting \(dF\) and \(F\):
\[
\int_A y \left(\rho g \sin\theta \, y \, dA\right) = y_{\mathrm{cp}} \left( \rho g A h_c \right),
\]
\[
\rho g \sin\theta \int_A y^2 \, dA = y_{\mathrm{cp}} \, \rho g A h_c.
\]
Cancelling \(\rho g\) and rearranging:
\[
y_{\mathrm{cp}} = \frac{\sin\theta}{A h_c} \int_A y^2 \, dA.
\]
Using the centroidal second moment along the plane \(I_G' = \int_A (y - y_c)^2 \, dA\) and the identity
\(\int_A y^2 dA = I_G' + A y_c^2\), we obtain
\[
y_{\mathrm{cp}} = \frac{\sin\theta}{A h_c}\left( I_G' + A y_c^2 \right).
\]
Converting to vertical depths via \(h = y \sin\theta\), \(h_c = y_c \sin\theta\), and noting that
\(I_G = I_G' \sin^2\theta\) (the second moment about the horizontal axis), gives the standard form:
\[
h_{\mathrm{cp}} = h_c + \frac{I_G \sin^2\theta}{A h_c}.
\]
This derivation explains why \(I_G\) and \(\theta\) govern the shift: more area spread from the centroidal axis (larger \(I_G\)) and more vertical orientation (larger \(\sin\theta\)) both push the resultant deeper than the centroid.
Practical notes and checks
Units: \(I_G\) has units of \(\text{m}^4\). The shift term \(\frac{I_G \sin^2\theta}{A h_c}\) has units of meters, consistent with depth.
Limiting cases: For \(\theta = 0^\circ\) (horizontal surface), \(\sin^2\theta = 0\) so \(h_{\mathrm{cp}} = h_c\). For \(\theta = 90^\circ\) (vertical), the shift is maximum.
Deeply submerged surfaces: As \(h_c \to \infty\), the shift term tends to zero; the CoP approaches the centroid.
Partial submergence: Recompute \(A\), \(h_c\), and the relevant second moments for the wetted portion only.
Special cases and consistency checks
Vertical plane surface (\(\theta = 90^\circ\)):
\[
h_{\mathrm{cp}} = h_c + \frac{I_G}{A h_c},
\]
which places the center of pressure below the centroid. As \(h_c\) increases, \(h_{\mathrm{cp}} \to h_c\).
Horizontal plane surface (\(\theta = 0^\circ\)):
Uniform hydrostatic pressure over the surface gives
\[
h_{\mathrm{cp}} = h_c,
\]
i.e., the resultant acts at the centroid.
Shallow submergence (\(h_c \to 0\)):
The formula predicts a large shift of the CoP away from the centroid; use the exact integral over the wetted shape rather than centroid-based approximations for partially submerged surfaces.
Alternative derivation via moments
Let the local pressure at depth \(h(y) = y \sin\theta\) be \(p = \rho g y \sin\theta\). The elemental force \(dF = p\,dA\) acts normal to the surface.
Take moments about a reference line (e.g., the top edge along the plane). The center of pressure location \(y_{\mathrm{cp}}\) along the plane satisfies
\[
y_{\mathrm{cp}} = \frac{\int_A y\,p\,dA}{\int_A p\,dA}
= \frac{\rho g \sin\theta \int_A y^2\,dA}{\rho g \sin\theta \int_A y\,dA}
= \frac{\int_A y^2\,dA}{A y_c}
= y_c + \frac{I_G'}{A y_c},
\]
where \(I_G' = \int_A (y - y_c)^2\,dA\) is the second moment about the centroidal axis along the plane. Converting to vertical depth gives
\[
h_{\mathrm{cp}} = y_{\mathrm{cp}} \sin\theta = h_c + \frac{I_G \sin^2\theta}{A h_c},
\]
with \(I_G\) defined about a horizontal axis through the centroid and parallel to the free surface.
Another worked example: circular hatch
A circular hatch of radius \(R = 0.75\,\text{m}\) is centered at depth \(h_c = 2.5\,\text{m}\) in water and inclined at \(\theta = 45^\circ\).
Compute \(F\) and \(h_{\mathrm{cp}}\).
Define geometry: Wetted area \(A\), centroid depth \(h_c\), orientation \(\theta\), second moment \(I_G\) about a horizontal axis through the centroid.
Structural checks: Resolve the resultant into components normal/tangential to the surface, check moments about supports, size hinges/seals/anchors accordingly.
Edge cases: For partial submergence or irregular shapes, perform direct integration over the wetted area or use numerical quadrature.
Quick reference
Hydrostatic pressure:
\[
p(h) = p_0 + \rho g h.
\]
Resultant force:
\[
F = \rho g A h_c.
\]
Center of pressure (depth):
\[
h_{\mathrm{cp}} = h_c + \frac{I_G \sin^2\theta}{A h_c}.
\]
Rectangle \(I_G\):
\[
I_G = \frac{b a^3}{12}.
\]
Circle \(I_G\):
\[
I_G = \frac{\pi R^4}{4}.
\]
Triangle \(I_G\):
\[
I_G = \frac{b a^3}{36}.
\]
Practice problems
Inclined rectangular plate: \(a=3\,\text{m}, b=2\,\text{m}, \theta=30^\circ\). Top edge depth \(h_t=0.5\,\text{m}\). Find \(F\) and \(h_{\mathrm{cp}}\).
Vertical gate: \(a=1.8\,\text{m}, b=1.2\,\text{m}\). Top edge at \(h_t=0.3\,\text{m}\). Compute \(F\) and \(h_{\mathrm{cp}}\) for \(\theta=90^\circ\).