CELE Hydraulics & Fluid Mechanics — Buoyancy and FlotationRevision Notes

Formulas

Exam Tips

• Memorize γ_water = 9.81 kN/m³ = 9810 N/m³ = 1000 kgf/m³ and γ_seawater ≈ 10.1 kN/m³.
• Quick check: if SG of body < SG of fluid → floats; fraction submerged = SG_body / SG_fluid.
• For 'find the specific gravity using apparent weight' problems: SG = W_actual / (W_actual − W_app).
• Board exams frequently give the weight of a floating object and ask for displaced volume: V_disp = W / γ_fluid.

Key Points

• The buoyant force FB acts vertically upward through the center of buoyancy B, which is the centroid of the displaced fluid volume.
• FB equals the weight of the fluid displaced — NOT the weight of the fluid above the body.
• For a FULLY SUBMERGED body: V_disp = total volume of the body.
• For a FLOATING body: V_disp = volume of the body below the waterline (submerged portion only).
• A body sinks if its average unit weight exceeds the fluid's unit weight; it floats if less; neutral buoyancy if equal.
• Apparent weight W_app = W_actual − FB (weight measured while submerged in fluid).
• The principle applies to any fluid: water (γ = 9.81 kN/m³), seawater (γ ≈ 10.1 kN/m³, SG ≈ 1.03), oil, air, etc.

Definitions

Section Title

1. Archimedes' Principle and Buoyant Force

Common Mistakes

• Using the TOTAL body volume for a floating body instead of only the submerged volume.
• Confusing γ_fluid with γ_body — always identify which specific weight belongs to which material.
• Forgetting that F_B acts at the center of buoyancy (centroid of displaced volume), not at the centroid of the whole body.
• Using γ_water = 9.81 kN/m³ when the problem states seawater (use γ_sw ≈ 10.1 kN/m³ or SG × 9.81).
• Not converting units — mixing kN and N, or m³ and liters, leads to errors by a factor of 1000.

Formulas

Exam Tips

• Shortcut: for a homogeneous block, fraction submerged = SG_body / SG_fluid (works for ANY fluid, not just water).
• When asked 'how much additional load can the pontoon carry before sinking,' compute the remaining freeboard volume × γ_fluid.
• Ship displacement problems: 5000 tonnes in seawater (SG = 1.03): V_disp = 5,000,000 / (1.03 × 1000) = 4854.4 m³.
• Always draw a quick sketch showing the waterline, draft, freeboard, and center of buoyancy — it prevents volume errors.

Key Points

• A body floats when the buoyant force equals the body's weight: F_B = W.
• The DRAFT (d) is the vertical depth of the body below the waterline — i.e., the depth of the submerged portion.
• For a prismatic (constant cross-section) body of plan area A: d = V_disp / A.
• A homogeneous rectangular block of specific gravity s floating in water: d = s × total height h.
• If a floating body changes fluid (e.g., freshwater to seawater), its draft changes: d₂ = d₁ × (γ₁ / γ₂).
• Reserve buoyancy = freeboard volume above waterline × γ_fluid — the additional load capacity before sinking.
• For a ship displacing mass m_ship in seawater: V_disp = m_ship / ρ_seawater (use ρ_sw ≈ 1025 kg/m³).

Definitions

Section Title

2. Flotation Condition and Draft

Common Mistakes

• Applying d = s × h only for homogeneous rectangular blocks — this formula does NOT apply to cylinders or irregular shapes without adjustment.
• For a hollow vessel, the 'body volume' is the external volume, not the material volume — use the full external dimensions.
• Failing to convert ship displacement from tonnes to kN: 1 metric tonne = 9.81 kN (weight), or divide mass in kg by 1000 for tonnes.
• Neglecting load on a floating body — total weight W includes the vessel weight PLUS all cargo and live loads.

Formulas

Exam Tips

• Systematic approach (KB→BM→BG→GM): (1) Find d from W = γ·V_disp; (2) KB = d/2 for rectangular; (3) I = LB³/12 for rolling; (4) BM = I/V_disp; (5) BG = KG − KB; (6) GM = BM − BG.
• If the problem gives G below B (e.g., heavy keel), then BG is negative in the formula and GM is even larger — this is the safest configuration.
• PRC board exams frequently ask: 'Is the pontoon stable? Determine GM.' — always state the conclusion (stable/unstable) explicitly.
• For a cylinder floating upright (diameter D, draft d): I = π D⁴/64; A = π D²/4; BM = (π D⁴/64) / (π D²/4 × d) = D²/(16d).
• A wide, flat barge has large I (due to B³ term) and is generally stable; a tall, narrow vessel has small I and may be unstable — use this as a quick sanity check.

Key Points

• A floating body is STABLE if, when tilted slightly, a restoring moment returns it upright.
• Three key points: G (center of gravity of body), B (center of buoyancy = centroid of displaced volume), M (metacenter).
• When the body tilts, B shifts but G stays fixed (for a rigid body). M is the intersection of the new buoyancy line with the original vertical axis.
• GM = BM − BG: if GM > 0 (M above G) → STABLE; if GM < 0 (M below G) → UNSTABLE; if GM = 0 → NEUTRAL.
• BM = I / V_disp, where I is the SECOND MOMENT OF AREA (moment of inertia) of the WATERPLANE about the tilting axis.
• For rolling (tilting about longitudinal axis): I = L × B³ / 12 (B = beam/width).
• For pitching (tilting about transverse axis): I = B × L³ / 12 (L = length).
• Righting moment = W × GM × sin θ ≈ W × GM × θ (for small angles θ in radians).
• B is located at d/2 above the keel for a rectangular barge (centroid of rectangular submerged block).
• BG = distance from keel to G minus distance from keel to B = KG − KB, where K = keel.

Definitions

Section Title

3. Stability of Floating Bodies — Metacentric Height

Common Mistakes

• Using I = LB³/12 for PITCHING instead of rolling — the axis of tilt determines which dimension is cubed.
• Computing BG as G − keel instead of |KG − KB|; always take BG = KG − KB and note the sign for GM = BM − BG.
• Using the body's total volume instead of displaced volume V_disp in BM = I/V_disp.
• Forgetting that for an upright floating body, B is at the centroid of the submerged portion — for a rectangular section this is d/2, NOT h/2.
• Assuming a body is stable simply because it floats — flotation requires F_B = W, but stability requires GM > 0 (a separate condition).
• Misidentifying the tilting axis for a rectangular barge: ROLLING tilts about the LONG axis (I uses B³), PITCHING tilts about the SHORT axis (I uses L³).

Formulas

Exam Tips

• For two-fluid problems: set up two equations — (1) force equilibrium F_B1 + F_B2 = W, and (2) geometric constraint V₁ + V₂ = total submerged volume or a given depth relation.
• Maximum additional load before sinking = γ_fluid × A_WP × freeboard (for prismatic vessels).
• PRC exams occasionally use 'tonnes force' — clarify whether they mean kN (1 t = 9.81 kN) or just mass in kg.

Key Points

• LAYERED FLUIDS: A body floating at the interface of two immiscible fluids must satisfy F_B1 + F_B2 = W, where each layer contributes its own displaced volume and specific weight.
• SUBMERGED BODY WITH ATTACHED FLOAT: Analyze combined system — sum of F_B values must equal total system weight.
• HOLLOW BODIES (ships, caissons): Use the full external envelope volume for buoyancy calculations; steel shell weight is separate from buoyancy.
• CHANGE IN LOADING: Adding mass to a floating body increases draft by Δd = ΔW / (γ_fluid × A_WP).
• ROLLING PERIOD: T = 2π × k / √(g × GM), where k = radius of gyration — smaller GM gives longer, gentler rolling (comfort) but less stability.
• WALL-SIDED FORMULA: For large heel angles, GZ = (GM + ½BM·tan²θ)sinθ — beyond scope of most PRC exams but occasionally appears.
• CAISSONS and COFFERDAM STABILITY: Same BM = I/V_disp principle applies to open-top caissons during float-in operations in Philippine marine construction.

Definitions

Section Title

4. Special Cases and Advanced Applications

Common Mistakes

• In layered fluid problems, assuming the body is fully submerged in one fluid — always check which layers the body straddles.
• For hollow structures, including the air volume inside the hull in the material weight calculation instead of treating it as void.
• Using total height instead of draft when calculating KB for a partially-filled hollow barge.