CELE Hydraulics & Fluid Mechanics — Hydrostatic Pressure and Forces on SurfacesRevision Notes
Revision notes for CELE Hydraulics & Fluid Mechanics — Hydrostatic Pressure and Forces on Surfaces. Short, focused, and designed for the week before exam day. Use these when you are already familiar with the chapter and need a quick refresh on the high-yield items Professional Regulation Commission (PRC) — Board of Civil Engineering tests.
Exam context
The Civil Engineer Licensure Examination is conducted by Professional Regulation Commission (PRC) — Board of Civil Engineering and is scheduled for May and November 2026. The Hydraulics & Fluid Mechanics subtest is marked as "Core" in the official pattern, and Hydrostatic Pressure and Forces on Surfaces appears in position 2nd of 10 in the CELE Hydraulics & Fluid Mechanics review rotation. Passing mark: 70% weighted average, no sub-test below 50%. Recent CELE 2026 papers have drawn roughly a meaningful share of questions from this subject.
Hydrostatic Pressure and Forces on Surfaces - Revision Notes
Hydrostatics — the study of fluids at rest — is one of the most heavily tested topics in the PRC Civil Engineer Licensure Examination (Hydraulics & Fluid Mechanics portion). Every dam gate, retaining wall, tank panel, and submerged structure must be analyzed for the magnitude and location of hydrostatic forces before it can be safely designed. This chapter consolidates the three pillars of hydrostatics: (1) pressure variation with depth, (2) force on plane surfaces and its point of application, and (3) force on curved surfaces resolved into horizontal and vertical components. Mastery of these concepts, combined with careful formula application, is the key to perfect marks on this examination topic.
Sections
Formulas
Example
At the base of a 12 m water tank: p = 9.81 × 12 = 117.72 kPa. Pressure head = 12 m of water.
Formula
p = γh
Variables
p = gauge pressure (kPa); γ = unit weight of fluid (kN/m³), γ_water = 9.81 kN/m³; h = depth below free surface (m)
Application
Compute the gauge pressure at any point submerged in a static liquid. Used for tank bottom pressures, foundation loads under reservoirs, and manometer calculations.
Example
p_abs = 49.05 + 101.325 = 150.375 kPa at 5 m depth in water.
Formula
p_abs = p_gauge + p_atm
Variables
p_abs = absolute pressure (kPa); p_gauge = gauge pressure (kPa); p_atm ≈ 101.325 kPa
Application
Required when working with gas laws, cavitation checks, or any problem where vacuum pressures appear.
Example
A pressure of 50 kPa in water: h = 50/9.81 = 5.097 m of water.
Formula
h = p / γ
Variables
h = pressure head (m); p = pressure (kPa); γ = unit weight (kN/m³)
Application
Convert pressure to an equivalent fluid column height — used extensively in manometry and hydraulic grade line (HGL) computations.
Exam Tips
- Memorize γ_water = 9.81 kN/m³; γ_seawater ≈ 10.05 kN/m³; γ_mercury = 13.6 × 9.81 = 133.4 kN/m³.
- When the problem says 'gauge pressure,' the free surface is at p = 0 — measure depth h from that surface directly.
- For problems with multiple fluid layers, compute the pressure step-by-step at each interface: p₂ = p₁ + γ₂h₂.
- In board exams, g = 9.81 m/s² is standard; some problems use g = 9.80 or 10 m/s² — read the problem data carefully.
Key Points
- Gauge pressure increases linearly with depth: p = γh, where γ = unit weight of fluid (9.81 kN/m³ for water at standard conditions).
- Absolute pressure = gauge pressure + atmospheric pressure (p_atm ≈ 101.325 kPa at sea level).
- Pascal's Law: pressure at a point in a static fluid acts equally in all directions and is transmitted undiminished throughout the fluid.
- Pressure is constant on any horizontal plane within a connected body of the same fluid — the fundamental basis for manometry.
- Pressure head h = p/γ expresses pressure as an equivalent height of fluid.
- For a fluid of specific gravity s, γ = s × 9.81 kN/m³.
- The hydrostatic pressure equation p = γh is valid for incompressible fluids (liquids) — applicable to nearly all civil engineering problems.
- Pressure is always perpendicular (normal) to any surface in contact with a static fluid.
Definitions
Term
Gauge Pressure
Definition
Pressure measured above atmospheric pressure. p_gauge = p_abs − p_atm. A negative gauge pressure indicates partial vacuum (sub-atmospheric).
Importance
Almost all hydrostatic force calculations use gauge pressure — structural loads are due to pressure above atmospheric, since atmospheric acts on both sides of most structures.
Term
Unit Weight (Specific Weight) γ
Definition
Weight of fluid per unit volume. For water: γ = 9.81 kN/m³ (or 9810 N/m³). For a fluid of specific gravity s: γ = s × 9.81 kN/m³.
Importance
Every hydrostatic force formula involves γ — always confirm the fluid type (water, seawater γ = 10.05 kN/m³, mercury γ = 133.4 kN/m³).
Term
Pascal's Law
Definition
An externally applied pressure is transmitted equally and undiminished to every point in a confined static fluid.
Importance
Foundation for hydraulic jacks, hydraulic brakes, and manometer analysis. Also justifies treating pressure as the same on any horizontal level within connected fluid.
Term
Pressure Head
Definition
Equivalent height h = p/γ of fluid that would produce the given pressure at its base.
Importance
Used in manometer problems to equate pressure at a datum level — mixing pressure units is a top exam error source.
Section Title
Pressure Variation with Depth and Pascal's Law
Common Mistakes
- Using absolute pressure instead of gauge pressure in hydrostatic force formulas — remember: atmospheric pressure cancels out for submerged surfaces open to atmosphere on both sides.
- Forgetting to multiply by the correct specific gravity when the fluid is not water (e.g., seawater, oil, mercury).
- Mixing units: kPa vs Pa vs N/m² — always carry units explicitly and convert early.
- Assuming pressure only acts downward — pressure in a static fluid acts equally in ALL directions (Pascal's Law).
- Using depth to a corner or edge instead of depth to the CENTROID when computing total force.
Formulas
Example
U-tube with mercury (s=13.6), deflection x = 250 mm = 0.25 m, water above: p_A = 13.6×9.81×0.25 = 33.35 kPa (gauge)
Formula
p_A + γ₁h₁ − γ_m·x − γ₂h₂ = p_B
Variables
p_A, p_B = pressures at points A and B (kPa); γ₁, γ₂ = unit weights of process fluids; γ_m = unit weight of manometric fluid; h₁, h₂ = heights of respective fluid columns (m); x = manometer deflection (m)
Application
General equation for a differential U-tube manometer connecting points A and B. Set p_B = 0 for a simple manometer open to atmosphere.
Exam Tips
- Always start at the known pressure side (often p_A known or = 0 for open-ended tube) and walk systematically to the unknown side.
- Sketch the manometer and label every fluid column with its height and fluid type before writing the equation.
- For mercury U-tube with water: p_A (kPa) = 13.6 × 9.81 × x(m) − 9.81 × h_water. Remember the factor of 13.6.
- Board exam shortcut: for mercury-water manometer, p_A ≈ 133.4x − 9.81h (all in consistent units).
Key Points
- Manometers measure pressure by equating fluid column heights at a common datum (usually the lower meniscus in a U-tube).
- Walking rule: moving DOWN the tube ADD γh; moving UP the tube SUBTRACT γh. The total sum from one end to the other equals zero (open end = atmospheric = 0 gauge).
- Simple (piezometer) manometer: open tube filled with the process fluid — reads gauge pressure directly as h.
- U-tube manometer: uses a gage fluid (usually mercury, s = 13.6) to measure higher pressures. The deflection x of gage fluid gives pressure difference.
- Differential manometer: connected between two points to measure pressure difference directly — no need to know absolute pressures at either point.
- For connected vessels at the same elevation with the same fluid, pressures must balance at the connecting level.
- The gage fluid must be immiscible with the process fluid and have a higher density (for most applications).
Definitions
Term
Manometric Fluid (Gage Fluid)
Definition
The heavier liquid inside the U-tube (commonly mercury, s = 13.6, or carbon tetrachloride, s = 1.59) used to amplify small pressure differences into readable column heights.
Importance
Must correctly identify the gage fluid and use its specific γ — using γ_water for mercury is a fatal error in manometry problems.
Term
Differential Manometer
Definition
A U-tube manometer connected between two pressure taps to measure the pressure DIFFERENCE (p_A − p_B) directly, without needing the absolute value at either point.
Importance
Commonly used in pipe flow problems (Venturi, orifice meter) appearing in later chapters — the manometry skill is foundational.
Section Title
Manometry
Common Mistakes
- Forgetting the direction rule — ADDING when going down, SUBTRACTING when going up. A wrong sign flips the answer.
- Omitting the fluid column ABOVE the gage fluid on either side of the manometer.
- Using x (mercury deflection) as h (water column) — they represent different fluids at different γ values.
- Not converting mm to m before computing γh products.
Formulas
Example
3 m × 4 m vertical gate, top edge 2 m below surface: ȟ = 2 + 4/2 = 4 m; F = 9.81 × 4 × 12 = 470.88 kN
Formula
F = γ·ȟ·A
Variables
F = total hydrostatic force (kN); γ = unit weight of fluid (kN/m³); ȟ = depth to centroid of area below free surface (m); A = total area of the surface (m²)
Application
Compute the resultant force on any plane surface (vertical, inclined, or horizontal) submerged in a static liquid.
Example
For the gate above: ȳ = ȟ = 4 m (vertical); I_g = 3(4)³/12 = 16 m⁴; y_p = 4 + 16/(4×12) = 4 + 0.333 = 4.333 m below surface
Formula
y_p = ȳ + I_g/(ȳ·A)
Variables
y_p = distance from free surface to center of pressure along the inclined plane (m); ȳ = distance from free surface to centroid along the plane (m); I_g = centroidal moment of inertia (m⁴); A = area (m²)
Application
Locate the center of pressure (point of application of the resultant force) on any plane surface.
Example
From above: e = 16/(4×12) = 0.333 m below centroid (at 4 m), so center of pressure is at 4.333 m.
Formula
e = I_g / (ȳ·A)
Variables
e = eccentricity = distance from centroid to center of pressure (m); always positive (center of pressure is below centroid for downward-increasing pressure)
Application
Quick check for how far below the centroid the force acts. Important for moment calculations about hinge points.
Example
60° inclined gate, 2 m × 4 m, top at surface: ȳ = 2 m, ȟ = 2 sin60° = 1.732 m; F = 9.81 × 1.732 × 8 = 135.9 kN
Formula
F = γ·(ȳ sinθ)·A [inclined surface]
Variables
θ = angle of inclined surface from horizontal; ȳ sinθ = ȟ = depth to centroid
Application
For inclined gates or chute walls — always compute ȟ (vertical depth) from ȳ (along-plane distance) using ȟ = ȳ sinθ.
Exam Tips
- Memorize the 'two-thirds rule': for a rectangular gate with its TOP edge at the free surface, center of pressure = 2/3 of the gate height from the top. This is a direct board exam shortcut.
- For deeply submerged surfaces (large ȳ), the eccentricity I_g/(ȳA) → 0 and y_p ≈ ȳ — the center of pressure approaches the centroid.
- Always draw a clear diagram: mark the free surface, the surface dimensions, depth to top edge, centroid location, and center of pressure. This prevents nearly all setup errors.
- For gates hinged at one edge with a stop at the other, take moments about the hinge to find the stop reaction. Use y_p to locate F, then compute moments correctly.
- Check: Is the answer for y_p greater than ȳ? If yes, you're likely correct. If y_p < ȳ, you have a sign or formula error.
- Board exam shortcut for rectangle with top at depth d, height H: y_p = d + H/2 + H²/[12(d + H/2)] — derive this once and remember the form.
Key Points
- The total hydrostatic force on any plane surface equals the product of unit weight, depth to centroid, and area: F = γ·ȟ·A.
- This force acts not at the centroid but at the CENTER OF PRESSURE, which is always BELOW the centroid (for a surface with positive depth).
- The center of pressure depth (along the inclined plane): y_p = ȳ + I_g/(ȳ·A), where ȳ is measured along the inclined plane from the free-surface line.
- For a VERTICAL surface: ȳ = ȟ (depth to centroid equals distance along plane to centroid).
- For an INCLINED surface at angle θ from horizontal: ȟ = ȳ·sin θ — the depth and the along-plane distance differ.
- I_g = centroidal moment of inertia of the plane area about its horizontal centroidal axis.
- The eccentricity (distance below centroid to center of pressure) = I_g/(ȳ·A). As depth increases, eccentricity decreases — for very deep surfaces, force approaches the centroid.
- For a rectangle (width b, height h): I_g = bh³/12. For a circle (diameter d): I_g = πd⁴/64. For a triangle (base b, height h): I_g = bh³/36.
- The resultant force on any plane surface always passes through the center of pressure.
- For a surface touching the free surface (top edge at waterline), center of pressure is at 2/3 depth for a rectangle.
- This topic applies to: sluice gates, dam faces, tank panels, manhole covers, bulkheads.
Definitions
Term
Center of Pressure
Definition
The point on a submerged plane surface through which the resultant hydrostatic force acts. Located at y_p = ȳ + I_g/(ȳA), always below the centroid of the area.
Importance
The most tested output in hydrostatic force problems. Gate hinge reactions, overturning moments, and stability analyses all depend on the correct location of the center of pressure.
Term
Centroidal Moment of Inertia (I_g)
Definition
The second moment of the plane area about its horizontal centroidal axis. Rectangle: bh³/12; Circle: πd⁴/64; Triangle: bh³/36. Units: m⁴.
Importance
The only geometric property that differs between shapes in the center-of-pressure formula. Memorize I_g for rectangle, circle, and triangle — these three cover virtually all board exam problems.
Term
Centroid Depth (ȟ)
Definition
The vertical depth from the free surface to the geometric centroid of the submerged area. For a vertical rectangle: ȟ = (depth to top) + height/2.
Importance
Governs the MAGNITUDE of the resultant force. A common error is using the top or bottom edge depth instead of the centroid depth.
Section Title
Hydrostatic Force on Plane Surfaces
Common Mistakes
- Using the depth to the top edge or bottom edge instead of the depth to the CENTROID for computing F = γȟA.
- Confusing ȳ (along-plane distance) with ȟ (vertical depth) for inclined surfaces — they are equal only for vertical surfaces.
- Forgetting to add eccentricity e to ȳ to get y_p — some examinees report the centroid location as the center of pressure.
- Using the wrong I_g formula — rectangle is bh³/12 where h is the dimension parallel to the pressure gradient (vertical for vertical surfaces).
- For submerged (not surface-piercing) surfaces: ȳ ≠ h/2 — you must add the submergence depth to h/2.
- Taking moments about the wrong point when finding hinge reactions — always use y_p for moment arm, not ȟ.
Formulas
Example
Quarter-circle gate, radius R = 1.5 m, width w = 1 m, water above: A_v = R×w = 1.5 m², ȟ_v = R/2 = 0.75 m; F_H = 9.81×0.75×1.5 = 11.04 kN
Formula
F_H = γ·ȟ_v·A_v
Variables
F_H = horizontal component (kN); ȟ_v = depth to centroid of vertical projection (m); A_v = area of vertical projection of curved surface (m²)
Application
Compute horizontal hydrostatic force on any curved surface — use the standard plane-surface method on the vertical projection.
Example
Quarter-circle gate (concave), R = 1.5 m, w = 1 m: V_above = (π/4)R²·w = (π/4)(1.5²)(1) = 1.767 m³; F_V = 9.81×1.767 = 17.33 kN (downward)
Formula
F_V = γ·V_above
Variables
F_V = vertical component (kN); γ = unit weight of fluid (kN/m³); V_above = volume of fluid (real or imaginary) directly above the curved surface up to the free surface (m³)
Application
Compute vertical hydrostatic force. For a concave-up surface with fluid above, V_above is the real fluid volume. For convex-up with fluid below, use the imaginary fluid volume.
Example
F = √(11.04² + 17.33²) = √(121.9 + 300.3) = √422.2 = 20.55 kN at α = arctan(17.33/11.04) = 57.5° from horizontal
Formula
F = √(F_H² + F_V²)
Variables
F = resultant hydrostatic force (kN); F_H = horizontal component; F_V = vertical component
Application
Combine orthogonal force components to get the resultant for any curved surface.
Example
From above: α = arctan(17.33/11.04) = 57.5°. The resultant of 20.55 kN acts at 57.5° below horizontal toward the center of curvature.
Formula
α = arctan(F_V / F_H)
Variables
α = angle of resultant from horizontal (degrees); F_V = vertical component; F_H = horizontal component
Application
Determine the direction of the resultant force on a curved surface. For circular-arc surfaces, verify that the resultant passes through the center of curvature.
Exam Tips
- Always draw the free body diagram, label the free surface, shade the fluid volume above the curved surface, and identify the vertical projection rectangle.
- For a quarter-circle gate of radius R retaining water to the full height: V_above = (πR²/4)·w (quarter-circle area × width). F_H uses A_v = R·w with ȟ_v = R/2.
- The direction of F_V (up or down) depends on which side the fluid is on — ask: 'Is there fluid (real or imaginary) above this surface?' If yes, F_V is downward (or upward for imaginary).
- Circular-arc gate board shortcut: if asked for the moment about the center of curvature, it is ZERO — F passes through that point. This simplifies hinge reaction problems enormously.
- Resultant direction check: arctan(F_V/F_H). Verify the angle is physically reasonable (between 0° and 90° for quarter-circle).
Key Points
- For curved surfaces, the pressure direction varies point-to-point, so direct integration is impractical. Instead, RESOLVE into horizontal and vertical components.
- Horizontal component F_H: equals the force on the VERTICAL PROJECTION of the curved surface. Use the plane-surface method: F_H = γ·ȟ_v·A_v, where A_v is the vertically projected area and ȟ_v is its centroid depth.
- Vertical component F_V: equals the weight of the real or imaginary fluid column above the curved surface up to the free surface. F_V = γ·V_above.
- If fluid is ABOVE the surface: F_V acts DOWNWARD (weight of real fluid above pushes down).
- If fluid is BELOW the surface (e.g., bottom of a curved gate convex upward): F_V acts UPWARD (buoyancy concept — weight of imaginary fluid above).
- Resultant: F = √(F_H² + F_V²), inclined at angle α = arctan(F_V/F_H) from horizontal.
- For a CIRCULAR-ARC (cylindrical) curved surface: all pressure force vectors are radial (perpendicular to the surface and pointing toward the center of curvature). Therefore the RESULTANT PASSES THROUGH THE CENTER OF CURVATURE.
- This 'passes through center' property is a major board exam shortcut for circular-arc gates (Tainter gates, drum gates).
- The location of F_H uses the same y_p formula as plane surfaces; the location of F_V is at the centroid of the volume above.
- Three-dimensional (non-cylindrical) curved surfaces: same component method applies; V_above may require integration or geometric decomposition.
Definitions
Term
Vertical Projection
Definition
The shadow of the curved surface cast on a vertical plane — a flat rectangle (or other shape) representing where the curved surface would appear if viewed from the side. Area A_v = width × vertical height of curved surface.
Importance
This projection transforms a complex curved-surface problem into a simple plane-surface problem for the horizontal component.
Term
Imaginary Fluid Volume
Definition
For curved surfaces where fluid is on the convex side (below the surface), there is no real fluid above it. The vertical force is computed as if a column of fluid occupied the space from the surface up to the free surface — this imaginary volume gives the upward F_V.
Importance
The concept of imaginary fluid is frequently tested; students who confuse 'fluid above' with 'fluid below' will flip the direction of F_V.
Term
Center of Curvature
Definition
The geometric center of the circle (or sphere) of which the curved surface is an arc. For cylindrical curved gates (Tainter gates, drum gates), the resultant hydrostatic force passes through this point.
Importance
Board exam shortcut: for circular-arc surfaces, no moment analysis is needed to find the line of action — it always passes through the center of curvature.
Section Title
Hydrostatic Force on Curved Surfaces
Common Mistakes
- Treating F_V as the weight of fluid DIRECTLY above the FLAT horizontal projection — V_above must be the actual volume above the CURVED surface, which may include triangular or circular-sector portions.
- Forgetting to check the direction of F_V — downward if fluid is above the surface, upward if fluid is below.
- Using the curved-surface area (arc length × width) instead of the VERTICAL PROJECTION area for F_H.
- Not accounting for the volume of a triangular or sector region when computing V_above for a quarter-circle gate.
- Forgetting that the resultant on a circular arc always points toward the center — this is a useful check and shortcut.
- Ignoring atmospheric pressure on both sides (it cancels for open systems — only apply if one side is pressurized above atmospheric).
Formulas
Example
2 m wide, 3 m tall gate: I_g = 2(3)³/12 = 54/12 = 4.5 m⁴
Formula
I_g (rectangle) = bh³/12
Variables
b = width (perpendicular to pressure gradient, m); h = height (parallel to pressure gradient / depth direction, m)
Application
Most common shape in board exam gate problems. The h in I_g must be the dimension in the VERTICAL direction.
Example
Circular gate d = 1 m: I_g = π(1)⁴/64 = 0.04909 m⁴; A = π/4 = 0.7854 m²
Formula
I_g (circle) = πd⁴/64 = πr⁴/4
Variables
d = diameter (m); r = radius (m)
Application
Circular hatches, portholes, pipe ends. The circle's centroid is at its geometric center.
Example
Triangular gate base 2 m, height 3 m: I_g = 2(3)³/36 = 54/36 = 1.5 m⁴
Formula
I_g (triangle) = bh³/36
Variables
b = base (m); h = height (m). Centroid is at h/3 from base.
Application
Triangular weir notches, inclined triangular panels. Note centroid at 1/3 from base, not mid-height.
Exam Tips
- Write the four I_g values on your scratch paper at the start of every board exam session: rectangle bh³/12, circle πd⁴/64, triangle bh³/36. These three cover >95% of problems.
- Double-check: I_g/(ȳA) must have units of m⁴/(m·m²) = m. If your units don't work out to meters, recheck the formula.
- For a surface-piercing rectangle (top at free surface): y_p = ȳ + I_g/(ȳA) = H/2 + (bH³/12)/[(H/2)(bH)] = H/2 + H/6 = 2H/3. Memorize this classic result.
Key Points
- Rectangle (b wide, h tall — h is the dimension in the pressure direction): I_g = bh³/12; centroid at h/2 from either edge; A = bh.
- Circle (diameter d): I_g = πd⁴/64; centroid at center; A = πd²/4.
- Triangle (base b, height h — vertex at top in typical gate problems): I_g = bh³/36; centroid at h/3 from base (2h/3 from vertex); A = bh/2.
- Semicircle (diameter d, flat side up): I_g = 0.1098r⁴ ≈ 0.11r⁴; centroid at 4r/(3π) from flat side; A = πr²/2.
- Parallel Axis Theorem (not needed for center-of-pressure formula — I_g is already the CENTROIDAL value): I = I_g + Ad², used only when shifting from centroid to another axis.
- For compound shapes: decompose into rectangles/triangles, find I_g for each, use parallel axis theorem to combine about a common axis if needed.
- The term I_g/(ȳA) is the eccentricity e — always has units of length (m).
Definitions
Term
Eccentricity (e = I_g / ȳA)
Definition
The vertical distance by which the center of pressure lies BELOW the centroid of the area. e = y_p − ȳ. Always positive for surfaces with downward-increasing pressure.
Importance
Quick way to find y_p without re-deriving from scratch: y_p = ȳ + e. Also shows that deeper surfaces (larger ȳ) have smaller eccentricity.
Section Title
Moment of Inertia Reference and Area Properties
Common Mistakes
- Using I about the base edge (bh³/3) instead of the centroidal I_g (bh³/12) — the formula y_p = ȳ + I_g/(ȳA) uses CENTROIDAL I only.
- For a triangle, confusing the centroid at h/3 from BASE vs h/3 from vertex — the centroid is at h/3 from the base (closer to base).
- Forgetting that b and h in I_g formulas refer to specific orientations — b is perpendicular to depth, h is in the depth direction.
Connections
- Buoyancy and Archimedes' Principle (next chapter): F_V on a curved surface is the hydrostatic vertical force — buoyancy is the net upward F_V when the surface completely encloses a volume submerged in fluid.
- Fluid Statics and Pressure Diagrams: The trapezoidal or triangular pressure distribution on dam faces integrates to F = γȟA — connecting the distributed load to the resultant force concept used in structural analysis (NSCP 2015 Section 206 for lateral fluid pressures on walls).
- Hydraulic Machinery: Pascal's Law (p = γh transmitted everywhere) is the operating principle for hydraulic jacks, presses, and brakes — connects to fluid power topics.
- Open Channel Flow (later chapter): The hydrostatic force framework is applied to gates, weirs, and culvert headwalls — the gate force calculation you learn here directly applies to sluice gate flow analysis.
- Structural Engineering: Lateral hydrostatic pressure on retaining walls, basement walls, and tanks is a load combination in NSCP 2015 Section 206 — 'Flood and Hydrostatic Pressure (H)' load category.
- Soil Mechanics: Pore water pressure in saturated soils follows p = γ_w·h — direct application of the hydrostatic pressure equation to geotechnical effective stress analysis.
- Surveying and Leveling: Pressure head h = p/γ is the basis for hydraulic leveling and water-surface datum establishment — used in engineering surveys.
- Manometry connects to Bernoulli's Equation: Manometers measure pressure differences that appear in the Bernoulli equation for pipe flow and Venturi meters in the Flow in Closed Conduits chapter.
Exam Strategy
For the PRC CE Board Exam Hydraulics portion, allocate roughly 15–20 minutes on hydrostatics problems. Approach every problem in this sequence: (1) DRAW — sketch the surface, mark free surface, depth to top edge, dimensions, and fluid type; (2) IDENTIFY — plane or curved surface? Vertical or inclined? (3) COMPUTE — for plane surfaces: ȟ → F = γȟA → y_p = ȳ + I_g/(ȳA); for curved surfaces: decompose → F_H = γȟ_vA_v → F_V = γV_above → F = √(F_H²+F_V²); (4) CHECK — Is y_p > ȟ? Is F_V direction correct? Does the resultant angle make sense? Memorize the five essential I_g formulas (rectangle, circle, triangle) and the three classic shortcuts: (a) surface-piercing rectangle → y_p = 2H/3; (b) circular-arc gate → resultant passes through center; (c) deeply submerged gate → e ≈ 0. In multi-part problems, budget 60% of time on the setup (diagram and identification) — computational errors usually stem from wrong setup. If time is short, confirm the force magnitude (F = γȟA) first — partial credit is better than a blank. For manometry problems, use the walking rule every time without exception; never try to shortcut by intuition alone. Finally, always report units (kN, kPa, m) — unit errors cost points even with a correct numerical answer.
Quick Review Questions
A rectangular gate 1.5 m wide and 2 m tall is vertical with its top edge at the water surface. Compute the total hydrostatic force and the depth of the center of pressure.
ȟ = ȳ = 2/2 = 1.0 m; A = 1.5 × 2 = 3.0 m²; F = 9.81 × 1.0 × 3.0 = 29.43 kN. I_g = 1.5(2)³/12 = 1.0 m⁴; y_p = 1.0 + 1.0/(1.0 × 3.0) = 1.0 + 0.333 = 1.333 m. Alternatively, y_p = 2H/3 = 2(2)/3 = 1.333 m (classic surface-piercing rectangle result).
A 2 m × 2 m vertical square gate has its top edge 4 m below the water surface. Find F and y_p.
ȟ = ȳ = 4 + 1 = 5 m; A = 4 m²; F = 9.81 × 5 × 4 = 196.2 kN. I_g = 2(2)³/12 = 1.333 m⁴; e = 1.333/(5 × 4) = 0.0667 m; y_p = 5 + 0.0667 = 5.067 m. Note: deeply submerged gate → small eccentricity → center of pressure very close to centroid.
Find the gauge pressure (kPa) at the bottom of a tank containing 8 m of oil (specific gravity 0.85).
γ_oil = 0.85 × 9.81 = 8.339 kN/m³; p = γh = 8.339 × 8 = 66.71 kPa gauge.
A U-tube manometer contains mercury (s = 13.6). The mercury deflection is 300 mm with water on one side. What is the gauge pressure at the point of connection?
γ_mercury = 13.6 × 9.81 = 133.4 kN/m³; x = 0.30 m. Assuming the water column above the mercury on the connection side is negligible (or accounted for): p_A = γ_m × x = 133.4 × 0.30 = 40.0 kPa. For the full derivation, apply the walking rule from the open end (p = 0) to point A.
A quarter-circle gate of radius 2 m and width 1 m retains water with the curved surface concave upward. Compute F_H, F_V, and the resultant.
A_v = 2 × 1 = 2 m²; ȟ_v = 1 m (centroid of vertical projection at R/2 from surface); F_H = 9.81 × 1 × 2 = 19.62 kN. V_above = (πR²/4) × w = (π×4/4) × 1 = π m³ = 3.1416 m³; F_V = 9.81 × 3.1416 = 30.82 kN (downward). F = √(19.62² + 30.82²) = √(384.9 + 949.9) = √1334.8 = 36.53 kN ≈ 36.57 kN.
What is the pressure head in meters of water corresponding to a gauge pressure of 75 kPa?
h = p/γ = 75/9.81 = 7.645 m. This means a 7.645 m tall column of water would exert 75 kPa at its base.
For a vertical circular gate of diameter 1.2 m with its center at 3 m depth, find F and y_p.
ȟ = ȳ = 3.0 m; A = π(1.2)²/4 = 1.1310 m²; F = 9.81 × 3.0 × 1.1310 = 33.28 kN ≈ 33.22 kN. I_g = πd⁴/64 = π(1.2)⁴/64 = π(2.0736)/64 = 0.10179 m⁴; e = 0.10179/(3.0 × 1.1310) = 0.10179/3.393 = 0.030 m; y_p = 3.0 + 0.030 = 3.030 m.
State the key difference between computing force on a plane surface and on a curved surface.
On a plane surface, all pressure forces are parallel (perpendicular to the same flat plane), so they combine algebraically. On a curved surface, pressure forces point in different directions at each point, so vector decomposition into horizontal and vertical components is necessary before combining.
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