CELE Hydraulics & Fluid Mechanics — Hydrostatic Pressure and Forces on SurfacesStudy Notes

In a static fluid (at rest), pressure increases linearly with depth below the free surface. This is expressed by the fundamental hydrostatic equation: **p = γh** where: - p = gauge pressure (kPa) - γ = specific weight of fluid (kN/m³) - h = vertical depth below free surface (m) For water at 15°C, γ = 9.81 kN/m³ (often rounded to 9.8 or 10 for quick calculations). **Key Principles:** 1. **Pressure Head:** The equivalent height of fluid column producing the pressure is h = p/γ, measured in metres of fluid. 2. **Pascal's Law:** Pressure acts equally in all directions and perpendicular to any surface. 3. **Absolute vs. Gauge Pressure:** - Gauge pressure: p_g = pressure above atmospheric (the value used in hydrostatic force problems) - Absolute pressure: p_abs = p_g + p_atm - For most submerged-surface problems, use gauge pressure (atmospheric cancels out). 4. **Horizontal Plane Property:** All points on a horizontal plane in a connected static fluid experience the same pressure. **Example Calculation:** At a depth of 5 m below a water surface: p = 9.81 × 5 = 49.05 kPa Pressure head = 49.05 / 9.81 = 5 m (verification) This linear relationship is the foundation for all subsequent hydrostatic force calculations. When designing spillways, reservoir gates, or tank foundations, engineers begin by understanding that deeper points experience greater pressure—and thus greater force.

A manometer measures pressure by balancing fluid columns in a tube or U-tube. Two types are common in engineering: **A. Simple (Open) Manometer** One end connects to the pressure source; the other is open to atmosphere. The pressure difference equals the manometer fluid weight: **p_gauge = γ_manometer × h_deflection** **B. Differential Manometer** Both ends are closed or connected to different pressure sources. Solution requires walking along the tube systematically: - **Add** γh when moving downward in the tube - **Subtract** γh when moving upward in the tube - Equate pressure at the same horizontal level in the same fluid **Practical Method for U-Tube:** If a heavier (denser) gage fluid is used (e.g., mercury, s = 13.6), the manometer deflection is much smaller than with water, making it suitable for measuring small pressure differences. **Important Note:** When using a manometer gage fluid with specific gravity s: **p = (s × γ_water) × h_deflection = 9.81s × h** For example, mercury (s = 13.6): p = 9.81 × 13.6 × h = 133.4h (in kPa if h in metres) This is why mercury manometers are compact—a small mercury column represents a large pressure. **Board-Exam Pitfall:** Students often confuse the gage fluid with the fluid being measured. Always identify which fluid fills which section of the manometer.

When a plane surface (flat gate, dam section, or wall) is submerged, the distributed pressure loading creates a total force. Because pressure increases with depth, the force is not uniformly distributed—it is concentrated lower than the centroid of the area. **3.1 Total Force on a Plane Surface** The magnitude of the total hydrostatic force is: **F = γ × h̄ × A** where: - γ = specific weight of fluid (kN/m³) - h̄ = vertical depth of the **centroid** of the area below the free surface (m) - A = area of the submerged surface (m²) **Critical Insight:** The force depends on the **depth of the centroid**, not the area's position along the plane. For an inclined surface, h̄ = ȳ sin θ, where ȳ is the distance along the plane to the centroid and θ is the angle of inclination from horizontal. **3.2 Location of the Force: Center of Pressure** The total force does NOT act at the centroid. Because deeper areas experience higher pressure, the resultant force acts **lower**. Its location is the **center of pressure (y_p)**, measured along the plane from the free surface: **y_p = ȳ + (I_g / (ȳ × A))** where: - ȳ = distance along the plane from the surface to the centroid (m) - I_g = second moment of area (centroidal moment of inertia) about the horizontal axis through the centroid (m⁴) - A = area (m²) The term I_g / (ȳ × A) is always positive, so **y_p > ȳ always**—the center of pressure is always below (deeper than) the centroid. **3.3 Second Moments of Common Shapes** For calculations, recall: - **Rectangle** (width b, height h): I_g = bh³/12 - **Triangle** (base b, height h): I_g = bh³/36 - **Circle** (radius r): I_g = πr⁴/4 **3.4 Vertical vs. Inclined Surfaces** **Vertical Surface (θ = 90°):** - h̄ = ȳ (the vertical depth equals the distance along the plane) - Force acts perpendicular to the surface (horizontal) - Example: a vertical dam face or reservoir gate **Inclined Surface (θ < 90°):** - h̄ = ȳ sin θ (vertical depth is less than the slant distance) - Measured distance along the plane: ȳ = h̄ / sin θ - Force acts perpendicular to the inclined surface - Example: a sloped spillway or embankment **3.5 Surface Partially Submerged** For a surface with its top edge at or above the water surface: - Consider only the submerged portion in the area A and centroid depth h̄ - The freeboard portion contributes no force **3.6 Board-Exam Pitfalls** 1. **Confusing ȳ with h̄:** For a vertical surface they are equal; for inclined surfaces, h̄ = ȳ sin θ. 2. **Using area centroid instead of pressure centroid:** The force magnitude uses the geometric centroid, but the force location uses the pressure distribution (center of pressure). 3. **Placing force at centroid:** Always place the resultant at y_p, not ȳ. 4. **Forgetting the I_g term:** A common error is to assume y_p = ȳ. The difference I_g / (ȳ A) is often small for large depths but critical for accuracy.

A curved surface (such as a cylindrical dam face, spillway profile, or circular gate) experiences pressure that varies with depth. Because the surface normal changes direction along the curve, the resultant force cannot be found by a single multiplication like F = γh̄A. Instead, the problem is solved by resolving the force into **horizontal** and **vertical** components. **4.1 Horizontal Component** The horizontal component of force on a curved surface equals the force on the **vertical projection** of the surface (as if the surface were "collapsed" onto a vertical plane): **F_H = γ × h̄_vert × A_vert** where: - A_vert = the vertical projection of the curved surface (m²) - h̄_vert = vertical depth to the centroid of this projection (m) This uses the plane-surface formula applied to the vertical projection. The center of pressure of F_H is found using y_p for the projected area. **Key Point:** The horizontal force does not change based on the curvature; it depends only on what you "see" looking at the surface from the side. **4.2 Vertical Component** The vertical component equals the weight of the fluid directly **above** the curved surface (real or virtual): **F_V = γ × V_above** where: - V_above = volume of fluid above the surface (m³) If the surface curves upward (convex, like a normal dam), and the fluid is on top, then V_above is positive and F_V points upward. If the surface curves downward and the fluid is above, V_above is the volume of the region bounded by the curve and the free surface. For a surface entirely underwater, imagine the fluid column directly above it (bounded by vertical lines at the edges). **Critical Insight:** If the curved surface is part of a closed structure (e.g., a gate or conduit), the vertical force is the weight of all fluid directly above it, including the weight of any structures or water bodies on top. **4.3 Resultant Force and Its Direction** The magnitude of the resultant is: **F = √(F_H² + F_V²)** The angle from horizontal: **tan α = F_V / F_H** **4.4 Special Case: Circular-Arc Surface** For a **circular-arc** or **cylindrical** surface, there is an elegant property: **all pressure forces act radially toward the center of curvature**. Therefore, the resultant force passes through the center of the circle (or axis of the cylinder). This greatly simplifies statics problems (equilibrium of gates, for example). **4.5 Application: Dam with Curved Face** For a spillway or dam with a curved downstream face (often circular for structural efficiency): 1. Find F_H using the vertical projection (usually a rectangle or trapezoid). 2. Find F_V by calculating the volume of water above the curve (often using integration or geometric decomposition). 3. Combine using √(F_H² + F_V²). 4. If circular, the resultant passes through the center of curvature; use this for moment equilibrium about the center.

Hydrostatic force calculations are essential in the design and analysis of multiple hydraulic structures commonly encountered in Philippine civil engineering practice: **5.1 Spillway Gates and Radial Gates** Radial (or Tainter) gates are curved gates that pivot about a horizontal axis. Because the curved surface is circular, the hydrostatic force (and its resultant) passes through the center of the circular arc. This property simplifies equilibrium analysis: - The hydrostatic force creates no moment about the pivot axis (since the line of action passes through it) - The hoist mechanism (cables, servos) must overcome only the weight of the gate and friction - Design is more efficient than for flat gates **Application:** Many spillways in Philippine dams (e.g., Angat, Magat, Pantabangan) use radial gates to control outflow during high water levels. **5.2 Dam Faces and Concrete Gravity Dams** Dam faces are often curved downstream (convex) for structural efficiency. The hydrostatic force (usually dominant relative to other loads) must be resisted by the dam's weight and by friction on the base. The engineer must: 1. Compute F_H and F_V separately 2. Locate the resultant (or its components) 3. Check overturning stability: the resultant must pass within the base (or within the middle third for no tension) 4. Check sliding stability: friction and shear strength must exceed the horizontal component **5.3 Reservoir Walls and Tank Design** For vertical concrete walls and tank sides: - Use F = γh̄A with h̄ at the centroid of the submerged area - Locate the force at the center of pressure y_p - For tank design, this force determines the bending moments and reinforcement required - Philippine design standards (NSCP 2015) require that the reinforcement resists this moment and the resulting stress distribution **5.4 Submersed Structures (Sluice Gates, Conduits)** For gates and conduits fully submerged or operating under pressure: - The vertical component (buoyancy or weight of water above) is often substantial - For a submerged gate controlling flow, the net vertical force is the weight of water directly above - This force must be considered when designing hoist mechanisms and base support **5.5 Manometers in Field Testing** When testing hydraulic structures (pressure at various depths, verifying design assumptions): - Simple manometers measure gauge pressure at a point - Differential manometers compare pressures between two locations - Mercury manometers provide high sensitivity for small pressure differences (important in laboratory hydraulics) - Multi-fluid manometers must account for each fluid's density when "walking" the tube **5.6 RA 544 (Building Code) and NSCP 2015 Relevance** While RA 544 and NSCP 2015 are primarily structural codes, they contain provisions for: - **Loads from contained fluids:** Hydrostatic pressure is a specified load case for tanks and pools (Section 4.7 of NSCP 2015 covers environmental loads including liquids) - **Pressure distribution:** Design must account for the non-uniform pressure distribution (triangular for fully filled tanks) - **Freeboard and surcharge:** Consideration of surcharge height and safety factors - **Durability:** Materials must withstand sustained fluid pressure and chemical attack **5.7 Common Design Scenarios in Philippine Practice** 1. **Agricultural Irrigation Dams:** Small to medium dams with spillways must be sized for monsoon inflows. Radial gates or sluice gates control discharge. The spillway dam section (often concrete gravity) is designed against the hydrostatic load. 2. **Water Supply Reservoirs:** Large structures like Angat or Benguet dams require precise hydrostatic analysis. The gates are radial or fixed-wheel types that must handle the high pressures from deep water. 3. **Underground Tanks:** In urban areas, basement or underground water storage tanks experience hydrostatic pressure on all sides. Walls must be designed for the full depth, with special attention to the bottom and corners. 4. **Flood Control Structures:** Levees, walls, and gates designed to hold back floodwaters during typhoons must resist the triangular pressure distribution at maximum water level. **5.8 Numerical Verification and Checks** When solving a hydrostatic force problem on the board or in the exam: 1. **Draw a clear diagram:** Show the surface, water surface, depth reference, and any angles. 2. **Identify the surface:** Is it plane or curved? Vertical, horizontal, or inclined? Fully or partially submerged? 3. **For plane surfaces:** - Calculate A (area in m²) and h̄ (centroid depth in m) - Compute F = γh̄A - Find I_g and calculate y_p - Note: y_p > ȳ always 4. **For curved surfaces:** - Find F_H from the vertical projection - Calculate V_above and F_V - Combine: F = √(F_H² + F_V²) 5. **Check units:** All quantities should be in SI units (m, m², m³, kN, kPa) 6. **Reasonableness check:** For a 5 m depth and 10 m² area, F ≈ 500 kN; does your answer fit this scale? 7. **Sign convention:** Typically, horizontal forces are positive toward the structure, vertical forces positive upward; be consistent.