7.2 Hydrostatic Pressure on Plane and Curved Surfaces

7.2 Hydrostatic Pressure on Plane and Curved Surfaces

A critical competency tested in the CELE Hydraulics and Geotechnical Engineering examination is calculating the magnitude, direction, and precise location of the resultant hydrostatic thrust acting upon hydraulic gates, retainers, and gravity dam cross-sections.

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1. Hydrostatic Force on Submerged Plane Surfaces

Consider a plane surface of arbitrary geometry and total area $A$ submerged in a static liquid of specific weight $\gamma$, inclined at an angle $\theta$ relative to the horizontal liquid surface. Let $y$ be the inclined coordinate along the plane measured from the line of intersection with the free surface, such that vertical depth is $h = y \sin \theta$.

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Resultant Force Magnitude

The elemental hydrostatic thrust on area $dA$ is $dF = P dA = (\gamma h) dA = (\gamma y \sin \theta) dA$. Integrating over the entire area: $F = \int_A \gamma y \sin \theta dA = \gamma \sin \theta \int_A y dA$ Recalling the first moment of area $\int_A y dA = \bar{y} A$, where $\bar{y}$ is the inclined distance to the area centroid: $F = \gamma (\bar{y} \sin \theta) A = \gamma \bar{h} A$

• Rule: The total hydrostatic force on any submerged plane surface equals the pressure at its centroid multiplied by the total submerged area, acting strictly perpendicular to the surface.

Location of the Center of Pressure ($CP$)

Because pressure increases with depth, the resultant force does not act at the centroid ($CG$), but at a deeper point designated the Center of Pressure ($CP$) $(x_p, y_p)$. Taking moments of elemental forces about the surface intersection axis ($x$-axis) and applying Varignon's Theorem: $y_p F = \int_A y dF = \int_A y (\gamma y \sin \theta dA) = \gamma \sin \theta \int_A y^2 dA = \gamma \sin \theta I_x$ Substituting $F = \gamma \bar{y} \sin \theta A$ and using the Parallel Axis Theorem ($I_x = I_{cg} + A \bar{y}^2$): $y_p = \frac{I_x}{\bar{y} A} = \frac{I_{cg} + A \bar{y}^2}{\bar{y} A} = \bar{y} + \frac{I_{cg}}{\bar{y} A}$

• The distance $e = y_p - \bar{y} = \frac{I_{cg}}{\bar{y} A}$ is the centroidal eccentricity of the center of pressure.
• Lateral coordinate $x_p$ is determined via product of inertia: $x_p = \bar{x} + \frac{I_{xy,cg}}{\bar{y} A}$ If the plane has a vertical axis of symmetry passing through $CG$, $I_{xy,cg} = 0$, so $x_p = \bar{x}$.

Geometric Properties of Common Cross-Sections

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2. The Pressure Prism Method

For planar rectangular surfaces of uniform width $b$, the linear hydrostatic pressure distribution forms a three-dimensional volume known as the pressure prism.

• Volume of Pressure Prism: The total resultant force equals the volume of this prism: $F = \text{Volume} = \left(\frac{P_{\text{top}} + P_{\text{bottom}}}{2}\right) A$
• Line of Action: The resultant force passes directly through the centroid of the pressure prism. For a vertical rectangular gate extending from the liquid surface ($P_{\text{top}} = 0$) to depth $h$, the prism is triangular, so the resultant acts at $\frac{2}{3}h$ from the surface (or $h/3$ from the base).

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3. Hydrostatic Forces on Curved Surfaces

Because the orientation of the surface normal varies across a curved boundary, direct area integration of pressure vectors is mathematically tedious. The standard engineering approach resolves the resultant force into orthogonal horizontal and vertical components.

Horizontal Component ($F_h$)

The horizontal force on any curved surface equals the hydrostatic force exerted on the projection of the curved surface onto a vertical plane: $F_h = \gamma \bar{h}_{proj} A_{proj}$ where:

• $A_{proj}$ is the projected area of the curved surface onto a vertical plane perpendicular to the direction of interest.
• $\bar{h}_{proj}$ is the vertical depth to the centroid of that projected area.
• $F_h$ acts horizontally through the center of pressure of the projected vertical area ($y_{p,proj} = \bar{y}_{proj} + \frac{I_{cg,proj}}{\bar{y}_{proj} A_{proj}}$).

Vertical Component ($F_v$)

The vertical hydrostatic force equals the weight of the column of liquid (real or virtual) positioned directly above the curved surface, bounded by the surface and the elevation of the free liquid surface: $F_v = \gamma V$ where:

• $V$ is the volume of the vertical fluid prism extending from the curved boundary up to the free surface datum.
• Direction of $F_v$:
  • If real liquid rests above the curved surface, $F_v$ acts downward.
  • If liquid lies below the curved surface, $F_v$ acts upward (uplift force equal to the weight of the imaginary/virtual liquid volume displaced above the surface up to the extended free surface level).
• Line of Action: $F_v$ acts vertically through the centroid of the liquid volume $V$.

Resultant Force and Line of Action

$R = \sqrt{F_h^2 + F_v^2}, \quad \theta = \arctan\left(\frac{F_v}{F_h}\right)$

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4. Dam Stability Analysis

Concrete gravity dams resist external water and silt thrust primarily through their own self-weight. In CELE board problems, dam stability is evaluated per unit length ($1.0\text{ m}$ strip along the crest).

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Primary Forces Acting on a Gravity Dam (per linear meter)

1. Self-Weight of Concrete ($W$): Evaluated by dividing the dam cross-section into standard geometric components (triangles and rectangles). Unit weight of plain concrete is typically $\gamma_c = 23.5\text{–}24.0\text{ kN/m}^3$.
2. Upstream Hydrostatic Thrust ($F_h$): $F_h = \frac{1}{2} \gamma_w H^2$, acting horizontally at $H/3$ from the base.
3. Hydrostatic Uplift Pressure ($U$): Underseepage creates an upward pore pressure distribution along the dam-foundation contact interface.
   • Without drainage gallery: Triangular or trapezoidal uplift from $P_{\text{heel}} = \gamma_w H$ to $P_{\text{toe}} = \gamma_w h_{\text{tailwater}}$.
   • Uplift force: $U = \frac{1}{2}(\gamma_w H + \gamma_w h_{tw}) B$, acting at the centroid of the trapezoid.

Stability Evaluation Criteria

1. Factor of Safety Against Overturning ($FS_{ot}$): Taking moments about the downstream toe: $FS_{ot} = \frac{\sum M_{\text{resisting}}}{\sum M_{\text{overturning}}} = \frac{\sum (W_i x_i)}{\sum (F_{h,i} y_i + U x_u)} \ge 1.50$

2. Factor of Safety Against Sliding ($FS_{sl}$): $FS_{sl} = \frac{\mu \sum R_y}{\sum R_x} = \frac{\mu (\sum W - U)}{F_h - F_{\text{tailwater}}} \ge 1.50$ where $\mu$ is the coefficient of friction between the concrete base and bedrock (typically $0.50\text{–}0.70$).

3. Foundation Bearing Pressure and the Middle-Third Rule: The net vertical foundation reaction is $R_y = \sum W - U$. The location of the resultant force from the toe is: $\bar{x} = \frac{\sum M_{\text{toe}}}{\sum R_y} = \frac{\sum M_R - \sum M_{OT}}{\sum R_y}$ The eccentricity from the center of the base ($B/2$) is: $e = \left|\frac{B}{2} - \bar{x}\right|$

   • Case 1: Resultant within Middle-Third ($e \le B/6$): Compression is maintained across the entire base width. Bearing pressure is given by the flexure formula: $q = \frac{R_y}{B}\left(1 \pm \frac{6e}{B}\right)$ $q_{\text{toe}} = \frac{R_y}{B}\left(1 + \frac{6e}{B}\right), \quad q_{\text{heel}} = \frac{R_y}{B}\left(1 - \frac{6e}{B}\right)$
   • Case 2: Resultant outside Middle-Third ($e > B/6$): The heel experiences theoretical tension. Because soil and rock cannot sustain tension, tensile separation (cracking) occurs. The bearing stress redistributes into a purely compressive triangle over an effective contact length of $3\bar{x}$: $q_{\text{max}} = \frac{2 R_y}{3 \bar{x}}, \quad q_{\text{heel}} = 0$

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5. Worked Example: Submerged Inclined Sluice Gate

Problem Statement: A rectangular sluice gate $2.0\text{ m}$ wide ($b = 2.0\text{ m}$) and $3.0\text{ m}$ long ($L = 3.0\text{ m}$) is installed in a reservoir wall inclined at $\theta = 60^\circ$ to the horizontal. The gate is hinged at its top edge, which is located at a vertical depth of $2.50\text{ m}$ below the water surface. Freshwater specific weight is $\gamma_w = 9.81\text{ kN/m}^3$.

1. Determine the total hydrostatic force $F$ acting on the gate.
2. Determine the location of the center of pressure from the hinge.
3. Determine the minimum normal force $P$ applied at the bottom edge required to open the gate.

Step-by-Step Solution:

1. Determine centroidal depth and area:

   • Area of the gate: $A = b \times L = 2.0\text{ m} \times 3.0\text{ m} = 6.00\text{ m}^2$.
   • Centroid is at distance $L/2 = 1.50\text{ m}$ from the hinge along the incline.
   • Vertical depth to the centroid: $\bar{h} = h_{\text{top}} + \left(\frac{L}{2}\right) \sin 60^\circ = 2.50\text{ m} + (1.50\text{ m})(0.8660) = 2.50 + 1.299 = 3.799\text{ m}$
   • Total hydrostatic force: $F = \gamma_w \bar{h} A = (9.81\text{ kN/m}^3)(3.799\text{ m})(6.00\text{ m}^2) = 223.61\text{ kN}$

2. Determine the Center of Pressure ($CP$):

   • Inclined distance from liquid surface to centroid: $\bar{y} = \frac{\bar{h}}{\sin 60^\circ} = \frac{3.799\text{ m}}{0.866025} = 4.387\text{ m}$
   • Centroidal eccentricity $e$: $e = \frac{I_{cg}}{\bar{y} A} = \frac{b L^3 / 12}{\bar{y} (b L)} = \frac{L^2}{12 \bar{y}} = \frac{(3.0\text{ m})^2}{12 (4.387\text{ m})} = \frac{9.0}{52.644} = 0.171\text{ m}$
   • Distance from the hinge to $CP$: $y_{p,\text{hinge}} = \frac{L}{2} + e = 1.50\text{ m} + 0.171\text{ m} = 1.671\text{ m}$

3. Determine required normal opening force $P$:

   • Taking moments about the hinge: $\sum M_{\text{hinge}} = 0$ $P \times L = F \times y_{p,\text{hinge}}$ $P \times (3.00\text{ m}) = (223.61\text{ kN})(1.671\text{ m}) = 373.65\text{ kN}\cdot\text{m}$ $P = \frac{373.65}{3.00} = 124.55\text{ kN}$

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6. CELE Exam Traps & Common Computational Errors