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CELE Hydraulics & Fluid Mechanics — Buoyancy and FlotationDetailed Explanation

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Detailed explanations for CELE Hydraulics & Fluid Mechanics — Buoyancy and Flotation. This page treats you like a serious reviewer: we unpack the concepts thoroughly, show worked examples of how Professional Regulation Commission (PRC) — Board of Civil Engineering frames Buoyancy and Flotation questions, and explain the underlying reasoning that gets you to the right answer every time.

Exam context

Professional Regulation Commission (PRC) — Board of Civil Engineering runs the Civil Engineer Licensure Examination on May and November 2026. Its Hydraulics & Fluid Mechanics section sits under a "Core" weighting, and Buoyancy and Flotation is the 3rd chapter in the 10-chapter CELE Hydraulics & Fluid Mechanics rotation. The CELE passing mark is 70% weighted average, no sub-test below 50%, and the most recent 2026 paper drew about a meaningful share of questions from Hydraulics & Fluid Mechanics.

Buoyancy and Flotation - Detailed Explanation

Buoyancy and Flotation is a fundamental chapter in Hydraulics and Fluid Mechanics that consistently appears in the PRC Civil Engineer Licensure Examination. The principles covered here — Archimedes' Principle, flotation, draft calculation, and metacentric height for stability — directly apply to the design of barges, pontoons, caissons, cofferdams, and marine structures. As a Philippine civil engineer, you may work on port infrastructure, river crossings, or temporary works involving floating platforms. Mastery of this chapter requires both conceptual understanding and proficiency in the standard board-exam formula set: F_B = γV_disp, d = V_disp/A, BM = I/V_disp, and GM = BM − BG. This explanation covers every concept from first principles through board-level worked problems, with emphasis on the common pitfalls that cost examinees points.

Concepts

Archimedes' Principle and Buoyant Force

Archimedes' Principle states that any body partially or fully submerged in a fluid experiences an upward buoyant force (F_B) equal to the weight of the fluid it displaces. This is derived from the pressure distribution on the body's surfaces: the net upward pressure force on the bottom exceeds the net downward pressure force on the top, producing a resultant upward force. Mathematically: F_B = γ_fluid × V_displaced where: γ_fluid = specific weight of the fluid (kN/m³ or N/m³) V_displaced = volume of fluid displaced by the body (m³) For a FULLY SUBMERGED body: V_displaced equals the total volume of the body. For a FLOATING (partially submerged) body: V_displaced equals only the submerged portion of the body's volume. The buoyant force acts upward through the CENTER OF BUOYANCY (point B), which is the centroid of the displaced fluid volume. For fresh water: γ_w = 9.81 kN/m³ = 9810 N/m³ For seawater: γ_sw = 10.05 kN/m³ ≈ 1.025 × 9.81 (specific gravity ≈ 1.025) When F_B > W (body weight): the body rises (body is less dense than the fluid). When F_B = W: the body is in equilibrium (floats). When F_B < W: the body sinks (body is denser than the fluid). APPARENT WEIGHT: When a body is submerged in a fluid, it appears lighter. The apparent weight is: W_apparent = W_actual − F_B = γ_body × V − γ_fluid × V = V(γ_body − γ_fluid) For a solid steel object in water: W_apparent = V × (γ_steel − γ_water) — it still sinks but weighs less when measured under water.

Examples

Since SG_concrete = 2.40 > 1.0 (SG of water), the block sinks and remains fully submerged. The displaced volume equals the total body volume. The apparent weight is the force you would measure if you weighed it on a scale while submerged — it is always less than the actual weight for a body denser than water.

Scenario

A solid concrete block (SG = 2.40) measures 0.5 m × 0.5 m × 0.4 m. It is fully submerged in fresh water. Find: (a) actual weight, (b) buoyant force, and (c) apparent weight.

Solution

Volume of block: V = 0.5 × 0.5 × 0.4 = 0.10 m³ (a) Actual weight: γ_concrete = 2.40 × 9.81 = 23.544 kN/m³ W = 23.544 × 0.10 = 2.354 kN (b) Buoyant force (fully submerged, V_disp = V_body): F_B = γ_w × V = 9.81 × 0.10 = 0.981 kN (c) Apparent weight: W_app = W − F_B = 2.354 − 0.981 = 1.373 kN

The cable must support the apparent weight of the sphere in seawater. Note that in seawater, the buoyant force is slightly larger than in fresh water due to the higher specific weight of seawater. Always use the correct fluid unit weight in F_B calculations.

Scenario

A steel sphere of diameter 0.6 m and SG = 7.85 is suspended by a cable from a crane while fully submerged in seawater (SG = 1.025). Find the tension in the cable.

Solution

Volume of sphere: V = (π/6) × d³ = (π/6) × (0.6)³ = 0.1131 m³ γ_steel = 7.85 × 9.81 = 77.00 kN/m³ Actual weight: W = 77.00 × 0.1131 = 8.709 kN γ_seawater = 1.025 × 9.81 = 10.055 kN/m³ Buoyant force: F_B = 10.055 × 0.1131 = 1.137 kN Cable tension (equilibrium: T + F_B = W): T = W − F_B = 8.709 − 1.137 = 7.572 kN

Applications

  • Determining the crane lifting capacity for submerged concrete precast elements during underwater construction.
  • Calculating the net downward force on buried pipelines subjected to high groundwater levels (buoyancy uplift check).
  • Designing concrete anchor blocks for underwater pipelines and cables.
  • Verifying the stability of caissons during sinking operations in port construction.
  • Checking whether temporary cofferdams will float during construction dewatering.

Misconceptions

  • WRONG: Using the body's specific weight in F_B = γ × V. The γ in the formula is ALWAYS the fluid's specific weight.
  • WRONG: Using the total body volume for a floating (partially submerged) body. Use only the submerged volume.
  • WRONG: Thinking buoyancy depends on depth. Buoyancy depends only on the volume of fluid displaced, not how deep the body is submerged.
  • WRONG: Ignoring the difference between fresh water and seawater — a ship displaces less volume in seawater than in fresh water for the same weight.

Related Concepts

  • Hydrostatic pressure distribution
  • Specific gravity and specific weight
  • Flotation and draft
  • Stability of floating bodies
  • Pressure forces on submerged surfaces

Common Exam Questions

Example

A wooden log (SG = 0.65) with volume 0.8 m³ is held fully submerged in fresh water. Find F_B and the force needed to hold it down. F_B = 9.81 × 0.8 = 7.848 kN; W = 0.65 × 9.81 × 0.8 = 5.101 kN; Hold-down force = F_B − W = 2.747 kN upward.

Approach

Identify whether the body is fully or partially submerged. For fully submerged: F_B = γ_fluid × V_body. Compute V from geometry. Use γ of the fluid (not the body). Apparent weight = W_actual − F_B.

Question Type

Find the buoyant force or apparent weight

Example

A body floats at the interface of oil (SG=0.85) and water. If 30% is in oil and 70% in water, with V=0.2 m³, F_B = 0.85×9.81×0.06 + 9.81×0.14 = 0.500 + 1.373 = 1.873 kN.

Approach

Split V_displaced into portions in each fluid layer. F_B_total = γ_1 × V_1 + γ_2 × V_2. Set equal to weight for flotation problems.

Question Type

Multi-fluid or layered fluid buoyancy

Key Points To Remember

  • F_B = γ_fluid × V_displaced — use the FLUID's specific weight, not the body's.
  • For floating bodies, V_displaced is ONLY the submerged volume, not the total body volume.
  • The buoyant force acts upward through the center of buoyancy B (centroid of displaced volume).
  • Apparent weight = Actual weight − Buoyant force = V(γ_body − γ_fluid).
  • A body floats if its average specific gravity is less than the fluid's specific gravity.
  • For seawater (SG ≈ 1.025), the buoyant force is about 2.5% greater than in fresh water — relevant for ship draft calculations.
  • The buoyant force is independent of the depth of submersion (only depends on volume displaced).

Flotation and Draft

Flotation is the equilibrium condition of a body floating on a fluid surface, where the upward buoyant force exactly equals the body's weight: F_B = W → γ_fluid × V_disp = W The DRAFT (d) is the depth to which the body sinks below the fluid surface. For a prismatic (constant cross-section) floating body with waterplane area A: V_disp = A × d → d = V_disp / A = W / (γ_fluid × A) For a HOMOGENEOUS body of specific gravity s floating in water (SG_water = 1.0), the fraction of height submerged equals s: d/H = s → d = s × H where H is the total height of the body. This is an extremely important and frequently tested result. A wooden block of SG = 0.7 floats with 70% of its height submerged, regardless of its cross-sectional shape. For floating in a fluid of specific gravity s_f (not necessarily water): d/H = SG_body / SG_fluid = s_body / s_fluid FREEBOARD is the portion of the body above the waterline: Freeboard = H − d. CHANGE IN DRAFT: When a load is added to a floating body, the additional draft is: Δd = W_added / (γ_fluid × A) where A is the waterplane area. This is directly applicable to loading barges and ships.

Examples

The maximum buoyant force at the limiting draft equals the total weight that can be supported. The cargo weight is simply the difference. Note the draft limit (1.5 m) is less than the total depth (2.0 m), providing 0.5 m freeboard — a safety margin against waves overtopping the sides.

Scenario

A rectangular wooden barge is 8 m long, 3 m wide, and 2 m deep. The barge (empty) weighs 120 kN. How much additional cargo (kN) can it carry if the maximum allowable draft is 1.5 m? (fresh water)

Solution

Maximum displaced volume at draft 1.5 m: V_disp_max = 8 × 3 × 1.5 = 36 m³ Maximum buoyant force: F_B_max = γ_w × V_disp_max = 9.81 × 36 = 353.16 kN By flotation: F_B_max = W_barge + W_cargo W_cargo = 353.16 − 120 = 233.16 kN

When a log floats with its axis horizontal, you must work with the cross-sectional geometry. For exam purposes, if the cross-section is rectangular (e.g., a timber plank), simply use d = (SG_body / SG_fluid) × height. The circular cross-section problem is less common but shows how the general principle applies.

Scenario

A homogeneous cylindrical log (SG = 0.75, diameter = 0.4 m, length = 3 m) floats in seawater (SG = 1.025) with its axis horizontal. Find the draft.

Solution

Since the axis is horizontal, 'draft' is actually how deep the bottom of the circular cross-section is submerged. This becomes a circular segment problem. Fraction of volume submerged = SG_log / SG_seawater = 0.75 / 1.025 = 0.7317 = 73.17% For a circular cross-section, we need the depth d such that the submerged circular segment area / total circle area = 0.7317. Let r = 0.2 m. The submerged area fraction f = 0.7317. Using the circular segment formula: f = (1/2π)[θ − sin θ] where θ is the central angle. By trial/interpolation: θ ≈ 4.20 rad → d = r(1 − cos(θ/2)) = 0.2(1 − cos 2.10) = 0.2(1 − (−0.5048)) = 0.2(1.5048) = 0.3010 m Note: In board exams, this is often simplified — for prismatic bodies with rectangular cross-section, the fraction approach is straightforward.

Hollow structures (caissons, ships, pontoons) float despite using heavy materials because the enclosed air space reduces the average specific weight of the entire structure below that of water. Always compute the actual weight (material volume × γ_material) and use it in the flotation equation.

Scenario

A concrete caisson (hollow box) is 6 m × 4 m × 3 m (L×W×H). The walls and bottom are 0.2 m thick. The caisson floats in fresh water. Find the draft. (γ_concrete = 23.5 kN/m³)

Solution

External volume: 6 × 4 × 3 = 72 m³ Internal void: (6−0.4)(4−0.4)(3−0.2) = 5.6 × 3.6 × 2.8 = 56.448 m³ Concrete volume: V_conc = 72 − 56.448 = 15.552 m³ Weight of caisson: W = 23.5 × 15.552 = 365.47 kN Displaced volume at flotation: V_disp = W / γ_w = 365.47 / 9.81 = 37.26 m³ Waterplane area: A = 6 × 4 = 24 m² Draft: d = V_disp / A = 37.26 / 24 = 1.552 m Freeboard = 3 − 1.552 = 1.448 m (caisson floats with ample freeboard, suitable for towing to site)

Applications

  • Determining the safe loading capacity of river barges used in Philippine waterway transportation.
  • Calculating the draft of caissons used as bridge pier foundations (common in Philippine river bridge projects).
  • Designing floating docks and pontoon bridges for temporary river crossings.
  • Checking if a floating concrete pontoon for a Ro-Ro ramp has adequate freeboard.
  • Determining how much ballast to add to a floating structure to achieve the desired draft.

Misconceptions

  • WRONG: Using the full body volume in V_disp for a floating body. Only the submerged portion is displaced.
  • WRONG: Forgetting to check freeboard — draft must be less than the body height or the caisson/barge will sink.
  • WRONG: Applying d = s × H only to rectangular cross-sections when the body has a different shape in cross-section.
  • WRONG: Using γ = 9.81 kN/m³ when the problem specifies seawater or a different fluid.

Related Concepts

  • Archimedes' Principle
  • Stability and metacentric height
  • Hydrostatic pressure
  • Specific gravity relationships
  • Load analysis for marine structures

Common Exam Questions

Example

A 0.3m × 0.3m × 0.3m wood block, SG = 0.6, in fresh water: d = 0.6 × 0.3 = 0.18 m. This is the standard board exam shortcut for homogeneous blocks.

Approach

Use d = W/(γ_fluid × A). For a homogeneous body, W = SG_body × γ_w × V_total, which simplifies to d = SG_body × H. Always verify d < H (body must not sink completely).

Question Type

Find the draft of a floating prismatic body

Example

10m × 5m barge, max draft 1.8 m, barge weight 200 kN in seawater (γ=10.05 kN/m³): F_B_max = 10.05 × 90 = 904.5 kN; W_cargo = 904.5 − 200 = 704.5 kN.

Approach

W_total_max = γ_fluid × V_disp_at_max_draft = γ_fluid × A × d_max. Then W_load = W_total_max − W_barge.

Question Type

Find the load a barge can carry

Example

A 5m × 3m barge in fresh water: additional 50-kN load causes Δd = 50 / (9.81 × 15) = 0.340 m additional draft.

Approach

Δd = W_added / (γ_fluid × A). Waterplane area A is the plan area of the body at the waterline.

Question Type

Change in draft after adding load

Key Points To Remember

  • Flotation condition: F_B = W, meaning γ_fluid × V_disp = W.
  • Draft formula: d = W / (γ_fluid × A) for prismatic bodies.
  • For homogeneous body of SG = s floating in water: d = s × H (fraction submerged = SG of body).
  • For floating in fluid of SG = s_f: fraction submerged = SG_body / SG_fluid.
  • Freeboard = Total height − Draft.
  • Additional load causes additional draft: Δd = W_load / (γ_fluid × A).
  • A body can only float if its average SG < SG of fluid; hollow steel ships float because of their average SG < 1.

Stability of Floating Bodies and Metacentric Height

A floating body is STABLE if, when tilted slightly, it tends to return to its original upright position. Stability analysis is critical for designing ships, pontoons, and floating platforms. KEY REFERENCE POINTS (measured from the keel — the very bottom of the body): K = Keel (reference point, bottom of body) B = Center of Buoyancy = centroid of the displaced volume (at d/2 for a rectangular body) G = Center of Gravity of the body M = Metacenter = point where the buoyancy line of action intersects the original vertical axis after a small tilt HEIGHTS ABOVE KEEL: KB = distance from keel to B = d/2 (for rectangular cross-section) KG = distance from keel to G (given or computed) BG = |KG − KB| = distance between B and G METACENTRIC RADIUS (BM): BM = I / V_disp where: I = second moment of area (moment of inertia) of the WATERPLANE AREA about the axis of tilting V_disp = displaced volume For a rectangular waterplane (L × B): Rolling (about longitudinal axis): I = L × B³ / 12 Pitching (about transverse axis): I = B × L³ / 12 Note: Rolling uses the BEAM (B) cubed — this is the critical formula to get right. METACENTRIC HEIGHT (GM): GM = BM − BG = BM − (KG − KB) Equivalently: GM = KB + BM − KG STABILITY CONDITIONS: GM > 0 (M above G): STABLE — body returns to upright after tilt GM = 0 (M coincides with G): NEUTRAL STABILITY GM < 0 (M below G): UNSTABLE — body will capsize RIGHTING MOMENT for small heel angle θ: RM = W × GM × sin θ For small angles, sin θ ≈ θ (in radians). SPECIAL CASE — G below B: This always gives GM > 0 regardless. This is the case for submarines and submerged bodies, where G is physically below B. Any floating body with G below B is unconditionally stable. EFFECT OF ADDING FREE SURFACE (liquid cargo): A free liquid surface in a tank reduces the effective GM by a correction term I_tank / V_disp. This is the free surface effect — important for tanker ships.

Examples

The positive GM confirms the pontoon is stable for rolling. The metacenter M is located at KB + BM = 0.60 + 1.111 = 1.711 m above the keel, which is above G at 1.0 m. The righting moment for a 5° heel: RM = 470.88 × 0.711 × sin 5° = 470.88 × 0.711 × 0.0872 = 29.17 kN·m — this corrective moment would push the pontoon back to upright.

Scenario

A rectangular pontoon is 10 m long, 4 m wide, and 2 m deep. Its total weight is 470.88 kN and its center of gravity is 1.0 m above the keel. It floats in fresh water. Find GM for rolling (tilt about the long axis) and determine if it is stable.

Solution

Step 1 — Find draft: V_disp = W / γ_w = 470.88 / 9.81 = 48.0 m³ A_waterplane = 10 × 4 = 40 m² d = V_disp / A = 48 / 40 = 1.20 m Step 2 — Find KB: KB = d / 2 = 1.20 / 2 = 0.60 m (for rectangular cross-section) Step 3 — Find BM (rolling about longitudinal axis, so use B = 4 m): I = L × B³ / 12 = 10 × 4³ / 12 = 10 × 64 / 12 = 53.33 m⁴ BM = I / V_disp = 53.33 / 48.0 = 1.111 m Step 4 — Find BG: KG = 1.0 m (given) BG = KG − KB = 1.0 − 0.60 = 0.40 m (G is above B) Step 5 — Find GM: GM = BM − BG = 1.111 − 0.40 = 0.711 m GM = +0.711 m > 0 → STABLE

This is a classic board exam type — a tall, narrow floating body tends to be unstable. The BM is very small (small I, large V_disp) while BG is relatively large (high G, low B). In practice, ballast is added at the bottom to lower G, increasing GM. For the cylinder to be stable upright, it needs GM > 0, requiring BM > BG. This is not possible with this geometry without modification.

Scenario

A solid cylindrical buoy (SG = 0.8, diameter D = 1.0 m, height H = 1.5 m) floats upright in fresh water. Check stability for tilting about a diametric axis.

Solution

Step 1 — Find draft: d = SG × H = 0.8 × 1.5 = 1.20 m Step 2 — Find KB: KB = d / 2 = 0.60 m Step 3 — Find BM: I of circular waterplane = π D⁴ / 64 = π (1.0)⁴ / 64 = 0.04909 m⁴ V_disp = π D²/4 × d = π(1.0)²/4 × 1.20 = 0.9425 m³ BM = I / V_disp = 0.04909 / 0.9425 = 0.05209 m Step 4 — Find KG (for a homogeneous solid cylinder): KG = H / 2 = 0.75 m Step 5 — Find BG: BG = KG − KB = 0.75 − 0.60 = 0.15 m Step 6 — Find GM: GM = BM − BG = 0.05209 − 0.15 = −0.0979 m GM = −0.098 m < 0 → UNSTABLE The cylinder will tip over and float on its side.

This barge is stable but with a relatively small GM. In practice, a GM below about 0.3–0.5 m for a small barge is considered marginal. Adding top-heavy cargo would increase KG, reducing GM. The righting moment of 4.05 kN·m at 8° heel is the restoring couple that would return the barge to upright.

Scenario

A 3 m × 6 m × 1.5 m (W×L×D) barge floats at 0.9 m draft. The center of gravity is 1.1 m above the keel. Find: (a) GM for rolling, (b) righting moment at 8° heel.

Solution

(a) GM for rolling: V_disp = 3 × 6 × 0.9 = 16.2 m³ KB = 0.9 / 2 = 0.45 m I_rolling = L × B³ / 12 = 6 × 3³ / 12 = 6 × 27 / 12 = 13.5 m⁴ BM = 13.5 / 16.2 = 0.833 m BG = KG − KB = 1.1 − 0.45 = 0.65 m GM = 0.833 − 0.65 = 0.183 m > 0 → STABLE (marginally) (b) Total weight: W = γ_w × V_disp = 9.81 × 16.2 = 158.92 kN Righting moment at 8°: RM = W × GM × sin 8° = 158.92 × 0.183 × 0.1392 = 4.052 kN·m

Applications

  • Design of ferry boat hulls and river barges for inter-island and river transport in the Philippines.
  • Stability analysis of pontoon bridges used as temporary crossings during emergency operations.
  • Design of floating dry docks in Philippine shipyards.
  • Checking stability of construction barges carrying heavy equipment (cranes, concrete batchers) during piling operations.
  • Marine engineering for offshore platform pontoons in Philippine waters.

Misconceptions

  • WRONG: Using I = B×L³/12 (pitching) instead of I = L×B³/12 (rolling) when rolling stability is asked. B = beam (width), L = length. Rolling uses the smaller dimension cubed when B < L.
  • WRONG: Taking BG = KB − KG when G is above B. Always BG = KG − KB when KG > KB (the usual case for surface vessels).
  • WRONG: Forgetting that GM = BM − BG (not BM + BG). If G is below B (unusual for surface ships), BG is subtracted from BM with appropriate sign.
  • WRONG: Using the body's full volume instead of V_disp (the displaced volume) in BM = I/V_disp.
  • WRONG: For a fully submerged body, applying the BM formula with a waterplane area — there is no waterplane, so BM = 0, and stability requires G below B.

Related Concepts

  • Second moment of area (moment of inertia) of plane figures
  • Center of gravity determination
  • Buoyant force and center of buoyancy
  • Free surface effect on stability
  • Ship design and naval architecture fundamentals

Common Exam Questions

Example

Rectangular pontoon 6m × 2m × 1.5m, W=176.58 kN, KG=0.9m in fresh water. d=3 m³/(12m²)=1.5m, KB=0.75, I=6×8/12=4 m⁴, V=12m³, BM=0.333, BG=0.15, GM=0.183 m — stable.

Approach

Step 1: Find V_disp = W/γ or = A×d. Step 2: KB = d/2 (rectangular). Step 3: I = L×B³/12 for rolling. Step 4: BM = I/V_disp. Step 5: BG = KG − KB. Step 6: GM = BM − BG. Check sign.

Question Type

Find GM and determine stability

Example

For a pontoon with KB=0.6m and BM=1.0m: KG_max = 0.6 + 1.0 = 1.6 m. If KG > 1.6 m, the pontoon is unstable.

Approach

Set GM = 0 as the limiting condition: BM = BG = KG − KB. Solve: KG_max = KB + BM.

Question Type

Find the maximum KG for stability

Example

W = 500 kN, GM = 0.8 m, θ = 10°: RM = 500 × 0.8 × sin 10° = 500 × 0.8 × 0.1736 = 69.44 kN·m.

Approach

First find GM, then RM = W × GM × sin θ. W is the total weight (= γ × V_disp).

Question Type

Righting moment calculation

Key Points To Remember

  • Three key points: B (center of buoyancy at d/2 for rectangle), G (center of gravity of body), M (metacenter above B).
  • BM = I / V_disp, where I is the waterplane's second moment of area about the TILT AXIS.
  • For rolling of a rectangular pontoon: I = L × B³ / 12 (B = beam/width).
  • GM = BM − BG = KB + BM − KG. Stable if GM > 0 (M above G).
  • Righting moment = W × GM × sin θ.
  • A wide, shallow draft body has large BM → more stable (wide beam is key to pontoon stability).
  • A tall, narrow body has small BM and high KG → tends to be unstable.
  • For a fully submerged body, BM = 0 (no waterplane area). Stability requires G below B.

Special Cases: Submerged Bodies and Free Surface Effect

FULLY SUBMERGED BODIES: For a completely submerged body (like a submarine), the waterplane area is zero. Therefore: BM = I / V_disp = 0 (no waterplane) Stability of a submerged body depends entirely on the relative positions of B and G: - If G is BELOW B: Stable (righting moment acts to restore upright position) - If G is ABOVE B: Unstable (overturning moment) - If G coincides with B: Neutral Submarines are made stable by placing heavy equipment (batteries, ballast) low and buoyant chambers high. FREE SURFACE EFFECT: When a floating body contains a tank that is PARTIALLY FILLED with a liquid (free surface), the liquid shifts during tilting, raising the effective center of gravity. The reduction in metacentric height due to free surface effect is: GM_effective = GM_solid − (i × γ_liquid) / (V_disp × γ_body) For a simpler form used in most board problems: GM_corrected = GM_calculated − i / V_disp (when same fluid) where i = second moment of area of the FREE SURFACE of the liquid in the tank about its own tilting axis. This effect ALWAYS reduces stability — partially filled tanks are dangerous. In exam problems, the free surface correction is often given directly. BODY FLOATING BETWEEN TWO FLUIDS: For a body floating at the interface between two fluids of specific gravities s_1 (lower) and s_2 (upper, s_2 < s_1): Let x = fraction of body in the lower fluid, (1−x) = fraction in upper fluid Equilibrium: s_body = s_2(1−x) + s_1(x) Solve for x: x = (s_body − s_2) / (s_1 − s_2)

Examples

When G is below B (the unusual but ideal case for a submerged body), the buoyant force (through B) and weight (through G) create a righting couple when tilted. The formula still works: GM = BM − BG = 0 − (2.1 − 2.5) = +0.4 m (positive because BG is negative in the formula when G is below B).

Scenario

A submarine (fully submerged) has B at 2.5 m above keel and G at 2.1 m above keel. Is it stable?

Solution

For a fully submerged body, BM = 0. GM = BM − BG = 0 − (KG − KB) = 0 − (2.1 − 2.5) = 0 − (−0.4) = +0.4 m Since G is BELOW B (KG < KB), GM = +(KB − KG) = +0.4 m > 0 → STABLE

This result is physically correct — a block of SG = 0.9 floats in fresh water and would not penetrate into denser mercury. The block simply floats on the water surface with 90% submerged in water. To float at a water-mercury interface, the body's SG must be between SG_water and SG_mercury.

Scenario

A block of wood (SG = 0.9, volume = 0.1 m³) floats at the interface of mercury (SG = 13.6) and fresh water (SG = 1.0). Find the fraction of the block in mercury.

Solution

Let x = fraction in mercury, (1−x) = fraction in water. Flotation: weight = buoyant force from both fluids SG_block × γ_w × V = SG_mercury × γ_w × xV + SG_water × γ_w × (1−x)V Divide by γ_w × V: 0.9 = 13.6x + 1.0(1−x) 0.9 = 13.6x + 1.0 − x 0.9 − 1.0 = 12.6x −0.1 = 12.6x x = −0.1/12.6 = −0.00794 Negative x means the block does NOT dip into mercury at all — it floats entirely on water, resting on the mercury surface. Re-check: SG_block (0.9) < SG_water (1.0), so the block floats on water, not at the water-mercury interface.

Applications

  • Submarine and underwater vehicle design.
  • Stability analysis of ships carrying liquid cargo (tankers) with partially filled tanks.
  • Design of oil storage vessels that float on water.
  • Analysis of partially submerged buoys and navigation markers.

Misconceptions

  • WRONG: Applying BM = I/V_disp to a fully submerged body — there is no waterplane, so I = 0 and BM = 0.
  • WRONG: Thinking free surface effect can ever increase stability — it always DECREASES GM.
  • WRONG: Not checking if the computed fraction x is between 0 and 1 in two-fluid problems.

Related Concepts

  • Archimedes' Principle
  • Metacentric height and stability
  • Hydrostatic pressure in stratified fluids
  • Submarine ballast tank design

Common Exam Questions

Example

SG_block=0.5 floating between oil(SG=0.8) and water(SG=1.0): 0.5 = 0.8(1−x) + 1.0(x) → 0.5 = 0.8 + 0.2x → x = −1.5 (impossible) → block floats entirely in oil layer (SG_block < SG_oil).

Approach

Write flotation equilibrium including buoyant forces from both fluids. Express volumes in terms of fraction x. Solve for x. Check if 0 ≤ x ≤ 1.

Question Type

Body floating between two liquids

Key Points To Remember

  • Fully submerged body: BM = 0, stable only if G is below B.
  • Surface vessels: BM = I/V_disp > 0, stable if G is below M (GM > 0).
  • Free surface effect always reduces the effective GM — partially filled tanks reduce stability.
  • Correction for free surface: subtract i/V_disp from GM (where i = waterplane MOI of the liquid surface in the tank).
  • Body at interface of two fluids: use weighted average specific gravity concept to find how much is in each layer.

Practice Problems

The cube sinks because SG = 7.85 >> 1.0. The cable supports the apparent weight, which equals actual weight minus buoyant force. This is the fundamental apparent weight problem — common in board exams. Always identify whether the body is fully or partially submerged.

Problem

PROBLEM 1 (Board-Type): A steel cube of side 0.20 m (SG = 7.85) is suspended in fresh water by a cable. Find: (a) the actual weight, (b) the buoyant force, and (c) the tension in the cable.

Solution

Volume: V = (0.20)³ = 0.008 m³ (a) Actual weight: W = SG × γ_w × V = 7.85 × 9.81 × 0.008 = 0.6160 kN = 616.0 N (b) Buoyant force (fully submerged): F_B = γ_w × V = 9.81 × 0.008 = 0.07848 kN = 78.48 N (c) Cable tension (equilibrium: T + F_B = W): T = W − F_B = 616.0 − 78.48 = 537.5 N

Ship displacement is expressed in mass (tonnes) in naval architecture. Convert to weight force first: W = m × g. Then V_disp = W / γ_fluid. Note that in seawater, the displaced volume is smaller than in fresh water for the same ship weight — ships ride higher in seawater, which is why draft marks (Plimsoll lines) indicate different drafts for different water densities.

Problem

PROBLEM 2 (Board-Type): A ship displaces 5,000 tonnes (metric) in seawater (SG = 1.03). Find the displaced volume.

Solution

Weight of ship: W = 5,000 tonnes × 9.81 kN/tonne = 49,050 kN Note: 1 metric tonne = 1,000 kg; W = 5,000,000 kg × 9.81 m/s² = 49,050,000 N = 49,050 kN Specific weight of seawater: γ_sw = 1.03 × 9.81 = 10.1043 kN/m³ Displaced volume: V_disp = W / γ_sw = 49,050 / 10.1043 = 4,854.3 m³

This problem covers all three stability calculations — GM, stability check, and righting moment. The barge is stable but with marginal GM (0.183 m). For a small barge, this is acceptable but adding cargo high up would reduce GM further. Note: for rolling, use I = L × B³/12 where B is the beam (3 m). Many examinees mistakenly use I = B × L³/12 — this would give an incorrect (much larger) BM for the pitching direction.

Problem

PROBLEM 3 (Board-Type): A 3 m × 6 m rectangular barge floats at a draft of 0.9 m in fresh water. The center of gravity G is 1.1 m above the keel. (a) Find GM for rolling. (b) Is the barge stable? (c) What is the righting moment at a 6° heel?

Solution

(a) GM for rolling (tilt about the 6-m long axis): V_disp = 3 × 6 × 0.9 = 16.2 m³ KB = d/2 = 0.9/2 = 0.45 m I_rolling = L × B³ / 12 = 6 × 3³ / 12 = 6 × 27 / 12 = 13.5 m⁴ BM = I / V_disp = 13.5 / 16.2 = 0.8333 m BG = KG − KB = 1.1 − 0.45 = 0.65 m GM = BM − BG = 0.8333 − 0.65 = 0.1833 m (b) GM = +0.183 m > 0 → STABLE (M is above G) (c) Total weight: W = γ_w × V_disp = 9.81 × 16.2 = 158.9 kN Righting moment at 6°: RM = W × GM × sin 6° = 158.9 × 0.1833 × 0.10453 = 3.046 kN·m

The tall, narrow cylinder is unstable when floating upright — it will tip over and float on its side. Part (d) shows the stability criterion: a cylinder floating upright is stable only if SG is very small (nearly hollow, floating high) or very large (dense, floating with little freeboard). For SG = 0.8, the cylinder is in the unstable range. This result is counterintuitive but important for design of floating cylindrical buoys.

Problem

PROBLEM 4 (Board-Type): A solid cylinder with SG = 0.8, diameter = 1.0 m, and height = 1.5 m floats upright in fresh water. (a) Find the draft. (b) Find GM. (c) Is it stable? (d) What is the minimum SG for it to be stable when floating upright?

Solution

(a) Draft: d = SG × H = 0.8 × 1.5 = 1.20 m (b) GM: V_disp = π(1.0)²/4 × 1.20 = 0.9425 m³ KB = d/2 = 0.60 m I = πD⁴/64 = π(1.0)⁴/64 = 0.04909 m⁴ BM = 0.04909 / 0.9425 = 0.05208 m KG = H/2 = 0.75 m (homogeneous solid, G at midheight) BG = KG − KB = 0.75 − 0.60 = 0.15 m GM = 0.05208 − 0.15 = −0.0979 m (c) GM < 0 → UNSTABLE. The cylinder will tip over. (d) For minimum SG to be stable (set GM = 0 → BM = BG): BM = I/V_disp = (πD⁴/64) / (πD²d/4) = D²/(16d) BG = H/2 − d/2 = (H−d)/2 d = SG × H BM = D²/(16 × SG × H) BG = H(1 − SG)/2 Set BM = BG: D²/(16 × SG × H) = H(1 − SG)/2 D²/(8H²) = SG(1 − SG) For this cylinder: D=1.0m, H=1.5m: 1/(8×2.25) = SG(1−SG) 0.05556 = SG − SG² SG² − SG + 0.05556 = 0 SG = [1 ± √(1 − 4×0.05556)]/2 = [1 ± √0.7778]/2 = [1 ± 0.8819]/2 SG₁ = 0.9409 or SG₂ = 0.0591 The cylinder is stable for SG < 0.0591 or SG > 0.9409 (floating upright). At SG = 0.8: unstable (confirmed by part c).

When the fluid is not water, use fraction submerged = SG_body / SG_fluid. The block sinks more deeply in a lighter fluid (SG = 0.85) than in water (SG = 1.0). This formula is derived directly from the flotation condition and is a frequently tested board exam shortcut.

Problem

PROBLEM 5 (Board-Type): A rectangular wooden block (0.6 m × 0.4 m × 0.3 m, SG = 0.70) floats in a liquid of SG = 0.85. Find the draft and the freeboard.

Solution

The block floats in a liquid of SG = 0.85 (not water). Fraction submerged = SG_block / SG_fluid = 0.70 / 0.85 = 0.8235 If the block floats with height H = 0.3 m: Draft d = 0.8235 × 0.3 = 0.2471 m Freeboard = H − d = 0.3 − 0.2471 = 0.0529 m ≈ 52.9 mm Verification: W = 0.70 × 9.81 × (0.6 × 0.4 × 0.3) = 0.70 × 9.81 × 0.072 = 0.4943 kN V_disp = 0.6 × 0.4 × 0.2471 = 0.05930 m³ F_B = 0.85 × 9.81 × 0.05930 = 0.4943 kN ✓

Exam Preparation Tips

  • MEMORIZE THE FORMULA SEQUENCE: For stability problems, always follow the sequence: (1) Find draft d = W/(γA), (2) KB = d/2, (3) I = LB³/12 for rolling, (4) BM = I/V_disp, (5) BG = KG−KB, (6) GM = BM−BG. Never skip steps.
  • ROLLING vs. PITCHING: Rolling = tilting about the LONG axis (longitudinal). I_rolling = L×B³/12, where B is the SHORT dimension (beam). This is the smaller I and the CRITICAL direction for most pontoons and barges. Pitching uses I = B×L³/12 with L cubed.
  • FLOATING vs. SUBMERGED: For floating bodies, use the submerged volume only in F_B and V_disp. For fully submerged bodies, use the full body volume. This is the most common error in board exams.
  • THE SG SHORTCUT: For a homogeneous solid block floating in water: draft = SG × height. Learn this cold — it saves significant time in board exams. For other fluids: draft/height = SG_body/SG_fluid.
  • UNIT CONSISTENCY: Always use SI units throughout. γ_water = 9.81 kN/m³ = 9810 N/m³. Lengths in meters, forces in kN or N, volumes in m³. Mixed units cause errors.
  • CHECK GM SIGN: If your GM comes out negative, double-check: (a) Did you use the right I formula? (b) Is BG = KG−KB (not KB−KG)? (c) Is V_disp the displaced volume (not total body volume)?
  • SEAWATER DENSITY: Unless stated otherwise, use SG = 1.025 for seawater, giving γ_sw = 10.05 kN/m³. Some problems specify SG = 1.03 — use the given value.
  • RIGHTING MOMENT: RM = W × GM × sin θ. For small angles (≤ 10°), this is accurate. For larger angles, the problem will provide a GZ (righting arm) curve — but this is beyond typical board exam scope.
  • APPARENT WEIGHT PROBLEMS: Apparent weight = W_actual − F_B = V(γ_body − γ_fluid). This directly gives the cable tension or scale reading when a body is weighed in a fluid.
  • PRACTICE THE COMPLETE SOLUTION: In the board exam, always write out all intermediate steps. Partial credit may be given, and showing KB, BM, and BG separately reduces errors. A complete, organized solution is also faster to check.
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In summary

Buoyancy and Flotation is one of the most consistently examined topics in the Hydraulics and Fluid Mechanics portion of the PRC Civil Engineer Licensure Examination. The core formula F_B = γ_fluid × V_displaced and the stability analysis sequence (KB → BM = I/V_disp → GM = BM − BG) cover the vast majority of exam items. The critical success factors for board exam performance in this chapter are: 1. Correctly identifying whether V_displaced equals the full body volume (fully submerged) or only the submerged portion (floating). 2. Using the waterplane moment of inertia I = L×B³/12 for rolling stability — getting the correct dimension cubed. 3. Applying the SG shortcut (d = SG × H for homogeneous bodies in water) for rapid problem solving. 4. Following the systematic GM calculation sequence without skipping steps. 5. Checking the sign of GM and interpreting it correctly: positive = stable, negative = unstable. As future Philippine civil engineers, you will apply these principles in designing port structures, barges, pontoon bridges, and marine foundations — particularly relevant given the Philippines' extensive coastline, numerous river crossings, and island topography. The ability to rapidly assess whether a floating structure is stable is a fundamental professional competency under RA 544 (Civil Engineering Law of the Philippines). Master these formulas and the worked examples in this chapter, and buoyancy problems in the board exam will become among your most confidently solved items.

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