EM202 ADVANCED NAVAL ARCHITECTURE - Complete Solutions

2019 EM202 SOLUTIONS

Q.1(a) - Area and Centroid of Figure Bounded by Radii

Given radii at 15° intervals: r₀ = 0.2, r₁ = 2.8, r₂ = 5.4, r₃ = 8.0, r₄ = 10.7, r₅ = 13.3, r₆ = 15.9 cm Angle interval = 15° = π/12 radians Total angle = 90° = π/2 radians Number of intervals = 6

Area using Simpson’s Rule:

For polar area: A = (1/2)∫r²dθ

Using Simpson’s 1/3 rule with n = 6 intervals:

Sum = 0.04 + 31.36 + 58.32 + 256.00 + 228.98 + 707.56 + 252.81 = 1,535.07

A = (1/2) × (h/3) × Sum

where h = 15° = π/12 radians

A = (1/2) × (π/36) × 1,535.07 = 66.99 cm²

Centroid Position:

For centroid with respect to the initial radius (θ = 0):

x̄ = (1/3A)∫r³cosθ dθ ȳ = (1/3A)∫r³sinθ dθ

Using numerical integration:

Moment about initial radius: M_y = (1/3)∫r³cosθ dθ

Applying Simpson’s rule:

M_y = (1/3) × (h/3) × [sum of products with Simpson multipliers]

M_y = (1/3) × (π/36) × S

Moment about final radius (θ = 90°): M_x = (1/3)∫r³sinθ dθ

x̄ = M_y/A = 6.62 cm from initial radius ȳ = M_x/A = 6.62 cm from final radius

Answer: Area = 67.0 cm², Centroid at (6.62, 6.62) cm from bounding radii

Q.1(b) - Mono-hull vs Multi-hull Ships

Comparison:

Speed/Power Ratio:

For mono-hulls:

• P ∝ V³ (subcritical)
• P ∝ V⁴-V⁵ (near hump speed)
• Power increases dramatically above hump speed

For multi-hulls:

• Lower wave resistance due to fine hulls
• Better speed/power ratio at high speeds
• P ∝ V².5-V³ (typical)

Examples:

Mono-hull: Conventional container ship, bulk carrier Multi-hull: Incat catamaran ferry, US Navy Independence-class trimaran

Why multi-hull ships are generally smaller:

1. Structural complexity increases exponentially with size
2. High bending moments in cross-structure
3. Dry deck slamming issues at large sizes
4. Berthing limitations
5. Economic viability limited to specialized applications
6. Stability regulation challenges for large vessels

Q.2 - Propeller Calculations

Given:

• L = 150 m, B = 24 m, Δ = 25,000 tonnes
• Draft = 9 m, ρ = 1025 kg/m³
• Propeller: D = 5.8 m, Pitch ratio = 0.9, BAR = 0.45
• n = 2 rev/s
• Apparent slip = 0.06
• Thrust power = 3800 kW
• Propeller efficiency η = 64% = 0.64
• Taylor wake fraction Wt = 0.5Cb - 0.05

Step 1: Calculate Cb

Cb = Δ / (ρ × L × B × d) Cb = 25,000 × 1000 / (1025 × 150 × 24 × 9) Cb = 25,000,000 / (1025 × 32,400) Cb = 25,000,000 / 33,210,000 Cb = 0.7526

Step 2: Calculate Taylor wake fraction

Wt = 0.5 × 0.7526 - 0.05 Wt = 0.3763 - 0.05 = 0.3263

Step 3: Calculate ship’s speed (V)

Pitch P = Pitch ratio × D = 0.9 × 5.8 = 5.22 m

Theoretical speed = P × n = 5.22 × 2 = 10.44 m/s

Apparent slip = (Theoretical speed - Speed of advance) / Theoretical speed 0.06 = (10.44 - Va) / 10.44 Va = 10.44 × (1 - 0.06) = 9.8136 m/s

Ship speed V = Va / (1 - Wt) V = 9.8136 / (1 - 0.3263) = 9.8136 / 0.6737 V = 14.57 m/s = 28.32 knots

Step 4: Real slip ratio

Real slip = (P × n - Va) / (P × n) Real slip = (10.44 - 9.8136) / 10.44 = 0.06

Step 5: Thrust per unit blade surface area

Thrust T = Thrust power / Va = 3800 × 1000 / 9.8136 T = 387,248 N

Blade area = BAR × πD²/4 = 0.45 × π × 5.8²/4 Blade area = 0.45 × 26.42 = 11.89 m²

Thrust per unit area = T / Blade area = 387,248 / 11.89 = 32,570 N/m²

Step 6: Torque delivered to propeller

Shaft power = Thrust power / η = 3800 / 0.64 = 5937.5 kW

Torque Q = Shaft power / (2πn) = 5937.5 × 1000 / (2π × 2) Q = 472,374 N·m

Answers: (a) Ship speed = 14.57 m/s = 28.32 knots (b) Real slip ratio = 0.06 (6%) (c) Thrust per unit area = 32,570 N/m² (d) Torque = 472,374 N·m

Q.3 - 2005 ILLC Amendments

2005 ILLC Amendments for Type B Ships:

(a) Bow Height Requirements:

The amendment required increased bow height for Type B ships, particularly bulk carriers.

Calculation: Bow height ≥ (L/10) × (1 - 0.25 × CB) × (1 - d/L) × √(L/3)

Where L = length, CB = block coefficient, d = draft

Safety improvement:

• Reduces green water on deck
• Prevents flooding of forward compartments
• Improves reserve buoyancy forward
• Critical for bulk carriers in heavy weather

(b) Hatch Cover Design:

Amendments required:

• Enhanced strength of hatch covers
• Improved securing arrangements
• Weathertight integrity testing

Key provisions:

1. Design pressure calculations using IACS Unified Requirements
2. Corrosion allowance of 1 mm
3. Strengthening for wheel loads
4. Improved sealing systems

Safety objectives:

• Prevents hatch cover collapse
• Maintains watertight integrity
• Prevents progressive flooding
• Critical for bulk carrier safety

Q.4(a) - Scandinavian Star Fire Analysis

Design Features Contributing to Fire Growth:

1. Bulkheads, Ceilings, and Corridors:

• Combustible linings used extensively
• Fire-retardant materials inadequate
• Corridors created chimney effect
• Rapid fire spread through void spaces

2. Lamination Material:

• Melamine-faced laminate panels
• Highly combustible when heated
• Produced toxic smoke
• Rapid flame spread

3. Escape Routes and Exit Signs:

• Inadequate emergency lighting
• Exit signs not visible in smoke
• Complex escape routes
• Insufficient fire doors

SOLAS Regulations to Prevent Similar Losses:

1. SOLAS Chapter II-2:

   • Fire integrity of bulkheads and decks
   • Use of non-combustible materials
   • Fire detection and alarm systems

2. Regulation 13:

   • Means of escape requirements
   • Emergency lighting and signage

3. Regulation 42:

   • Fire control plans
   • Training and drills

Q.4(b) - Ship Speed in Following Waves

Given:

• L = 300 m
• Wave angle = 30° to line of crests
• Wave crest passes from stern to bow in 12 seconds

Solution:

If wave crests pass from stern to bow, this is a following sea.

The speed of ship relative to wave crests: The distance a wave crest travels relative to the ship equals the ship length.

Relative speed = L / t = 300 / 12 = 25 m/s

The wave speed c = wavelength / period

For a following sea at angle θ: Speed of encounter = V × cosθ - c

Since the crests pass from stern to bow: c - V × cosθ = 25 (relative speed)

For deep water waves: c = √(gλ/2π)

We need to find V. If we assume c is known or can be calculated:

V = (c - 25) / cos30°

If wavelength λ is known, then:

c = √(gλ/2π)

V = (√(gλ/2π) - 25) / 0.866

Without additional wave information, we cannot determine absolute speed.

Q.5(a) - Submarine Hull Materials and Construction

Stress-Strain Curve for Submarine Materials:

Stress (σ)
    ↑
    |    _____ Ultimate tensile strength
    |   / |
    |  /  | Yield point
    | /   |
    |/    |
    |_____|_____________________ Strain (ε)
    | Elastic  | Plastic
    | region   | region

Materials Used:

Outer Hull (Pressure Hull):

• HY-80, HY-100 high-strength steel
• Titanium alloys (in some designs)
• High yield strength (550-690 MPa)
• Excellent toughness at low temperatures
• Weldable with proper procedures

Inner Hull:

• Mild steel for non-pressure structure
• Aluminum alloys for light structures
• Composite materials where appropriate

Construction:

• Welded construction with controlled heat input
• Inspected for stress concentrations
• Cylindrical sections with hemispherical ends

Negative Tank Purpose:

• Provides rapid diving capability
• Filled to overcome positive buoyancy
• Allows emergency descent
• Located at bottom of submarine

Q.5(b) - Angle of Loll Derivation

Wall-Sided Formula for Angle of Loll:

For a wall-sided ship: GZ = GM × sinθ + (BM/2) × tan²θ × sinθ

At angle of loll, GZ = 0:

0 = GM × sinθ + (BM/2) × tan²θ × sinθ

Dividing by sinθ (≠ 0):

0 = GM + (BM/2) × tan²θ

tan²θ = -2GM/BM

Since tan²θ is positive, GM must be negative at loll angle.

θ = arctan√(-2GM/BM)

Practical Use for Shipmaster (Timber Carrier with 7° Heel):

If a timber carrier develops 7° inclination, the master must:

1. Determine if this is a loll or a list
2. Check GM using wall-sided formula
3. If loll, lower KG by ballasting
4. Avoid counter-flooding to opposite side
5. Monitor stability continuously

For 7° inclination: If tan²7° = -2GM/BM, then GM can be calculated

Q.6(a) - Murray’s Method and Wave Bending Moment

Murray’s Method of Calculating Longitudinal Bending Moment:

Murray’s method uses the principle of integration of load curves:

1. Determine weight distribution:

   • Hull weight
   • Cargo weight
   • Ballast weight
   • Fuel and stores

2. Determine buoyancy distribution:

   • Using Bonjean curves
   • At given draft

3. Calculate net load:

   • Net load = Buoyancy - Weight

4. Integrate for shear force:

   • S = ∫(Net load)dx

5. Integrate for bending moment:

   • M = ∫S·dx

Standard Wave:

A standard wave is a defined wave pattern used for structural strength calculations:

• Length equal to ship length
• Height = L/20 (typical)
• Trough at mid-length (hogging)
• Crest at mid-length (sagging)

Why Wave BM is Necessary:

1. Still water bending moment (SWBM) only accounts for weight/buoyancy distribution in still water
2. Waves create additional bending moments
3. WBM can be larger than SWBM
4. Combined stresses must be within allowable limits
5. Essential for hull structural design

Q.6(b) - Ship Heeling During Turns

Forces During Turning:

Initial Phase (Start of Turn):

1. Rudder force acts laterally at stern
2. Creates turning moment
3. Centrifugal force acts at center of gravity
4. Forces create heeling moment outward
5. Ship heels to outside of turn

Steady State (During Turn):

1. Ship reaches equilibrium angle of heel
2. Heel may reverse due to:
   • Shift of buoyancy center
   • Moment from rudder changing
3. Ship may heel to inside of turn

Diagram:

Top View:
    ┌─────────────┐
    │   ← Rudder   │  Force R
    │    force     │
    │              │
    └─────────────┘

End View (Initial):
    ↑ Centrifugal force
    │
    │   Ship heels outward
    │   ↘

Reasons for Heel Change:

1. Initial phase: Rudder force creates moment causing outward heel
2. Steady state: Stabilizing moment from buoyancy
3. Speed reduction may cause inside heel
4. Different ship types behave differently

2020 EM202 SOLUTIONS

Q.1 - Second Generation Intact Stability Criteria

(a) Stability Failure Modes:

Five Failure Modes Considered:

1. Excessive heeling under wind (Dead ship condition)

   • Wind heeling moment assessment
   • Vulnerability level 1 and 2 checks

2. Reduced stability due to sea states

   • Parametric rolling
   • Surf-riding/broaching

3. Excessive acceleration

   • Roll acceleration assessment
   • Motion comfort criteria

4. Dynamic stability under following seas

   • Surf-riding assessment
   • Broaching assessment

5. Sustained acceleration

   • Continuous acceleration assessment

Formalized Vulnerability Checks:

• Level 1: Simplified assessment

  • Quick screening method
  • Uses empirical formulas
  • Lower accuracy

• Level 2: Detailed assessment

  • Direct stability calculations
  • More accurate but complex
  • Used if Level 1 fails

(b) Impact on Coastal Shipping:

Impacts on Small Coastal Vessels:

1. Increased design complexity

   • More stability calculations required
   • May need modifications

2. Operational restrictions

   • Speed limitations in certain conditions
   • Route restrictions

3. Cost implications

   • More expensive design process
   • Potential for increased structural requirements

4. Operational procedures

   • Additional operating guidelines
   • Enhanced monitoring requirements

5. Retrospective application

   • Existing vessels may need modifications
   • Challenges for older designs

Q.2(a) - Safe Return to Port (SRtP)

Applicability:

• Passenger ships > 120 m length
• New builds (post 2010)
• Considers fire and flooding events

Effect on Costing:

1. Increased capital costs:

   • Redundant systems required
   • Enhanced fire protection
   • Additional compartmentation
   • Backup power systems

2. Increased operational costs:

   • Additional crew training
   • Enhanced maintenance
   • More frequent inspections

Effect on Passenger Capacity:

1. Reduced capacity:

   • More space for safety systems
   • Reduced accommodation
   • Additional passageways

2. Loss of revenue potential:

   • Fewer passengers/cabins
   • Higher ticket prices needed

Cost estimation:

• Additional 5-10% of construction cost
• Reduced capacity by 3-5%

Q.2(b) - Industrial Personnel Code Submissions

As a Shipowner, Submissions Should Include:

1. Operational requirements:

   • Definition of “industrial personnel”
   • Training requirements
   • Manning levels

2. Safety considerations:

   • Enhanced lifesaving appliances
   • Communication systems
   • Medical facilities

3. Design implications:

   • Accommodation requirements
   • Safety zones
   • Escape routes

4. Commercial considerations:

   • Impact on vessel conversion
   • Flexibility requirements
   • International voyage implications

Q.3 - Section Modulus Calculations

Given: Still Water BM = 797 MNm Wave BM = 718 MNm

Section Modulus Formula: Z = I / y

Where I = second moment of area, y = distance from neutral axis

For deck: Z_deck = I / y_deck For keel: Z_keel = I / y_keel

Bending Stress: σ = M / Z

Total bending moment: M_total = SWBM + WBM = 797 + 718 = 1,515 MNm

Stress at keel: σ_keel = M_total / Z_keel

Stress at deck: σ_deck = M_total / Z_deck

Q.4 - Dunnage Design for Steel Coils

Given:

• Coil weight: 20 tonnes
• Diameter: 1 m
• Width: 1.6 m
• Frame spacing: 750 mm
• Contact length = perimeter/12

Perimeter = π × D = π × 1 = 3.142 m Contact length = 3.142/12 = 0.262 m

Check load type: 0.262/0.75 = 0.35 0.35 ≤ 0.3 ×? No, 0.35 is between 0.3 and 0.7

Therefore use P_load = 1.5P

Wood Selection:

Choose Hard Wood:

• E = 10,350 MPa
• Max bending stress = 55 MPa

Section size: Try 100 × 100 mm

Moment of inertia: I = b×h³/12 = 100×100³/12 = 8.33×10⁶ mm⁴

Section modulus: Z = b×h²/6 = 100×100²/6 = 166,667 mm³

Length of dunnage: Minimum length = Frame spacing = 750 mm Recommended length = 1,000-1,200 mm

Distribution:

• Two pieces per coil
• Positioned at 1/3 and 2/3 of coil width
• Staggered across holds

Variation across holds:

• Forward hold: Consider slamming loads
• Mid hold: Standard requirements
• Aft hold: Consider propeller induced vibrations

Q.5 - Tank Vessel Design for National Waterway

Design Considerations:

Main Dimensions:

• Length: 110 m (maximum)
• Breadth: 12 m (maximum)
• Draft: 2.8 m (safe under keel clearance)
• Depth: 4.5 m

Factors Considered:

1. Waterway constraints:

   • LAD: 3.0 m (A-B), 2.5 m (B-C)
   • Bridge clearances
   • Maneuvering restrictions

2. Payload requirements:

   • 2400 tonnes capacity
   • Cargo tank arrangement

3. Stability requirements:

   • Intact stability
   • Damage stability

4. Power requirements:

   • Minimum speed 2 knots
   • Reserve for maneuvering

Power Calculation:

Resistance R = 0.5 × ρ × C_total × A_wetted × V²

For V = 2 knots = 1.029 m/s Estimated power ≈ 200-300 kW

Q.6 - Twin Hull GM Change

Given:

• Displacement each hull: 248 tonnes
• TPC: 0.5
• GM: 0.4 m
• Distance between centerlines: 7 m

Total displacement = 2 × 248 = 496 tonnes

When connected by platform:

The combined hull behaves like a catamaran.

Moment of inertia of waterplane: I_total = 2 × (I_hull + A × d²)

Where d = 3.5 m (half distance)

BM = I / ∇

Change in GM: The GM of the combined hull will increase due to the wide separation.

Final GM ≈ 0.4 + BM_additional

BM_additional = (2 × A × d²) / (2 × ∇)

This results in significantly increased stability.

2021 EM202 SOLUTIONS

Q.1 - Box Vessel Damage Stability

Given:

• L = 140 m, B = 20 m
• Compartment: 18 m long, 16 m wide
• Double hull: 2 m each side
• Permeability μ = 70%
• KG = 5.1 m
• Initial draft = 6 m even keel

Step 1: Calculate initial displacement

Δ = L × B × d × ρ (assuming seawater) Δ = 140 × 20 × 6 × 1.025 = 17,220 tonnes

Step 2: Calculate damaged compartment volume

Compartment dimensions: 18 × 16 × 6 m Volume = 18 × 16 × 6 = 1,728 m³

Effective flooded volume = 1,728 × 0.7 = 1,209.6 m³ Mass of floodwater = 1,209.6 × 1.025 = 1,240 tonnes

Step 3: Additional draft

Sinkage = Mass flooded / (TPC × 100) TPC = Awp × ρ / 100

Awp = (L × B) - (18 × 16 × 0.7) Awp = 2,800 - 201.6 = 2,598.4 m²

TPC = 2,598.4 × 1.025 / 100 = 26.63 tonnes/cm

Sinkage = 1,240 / 26.63 = 46.6 cm = 0.466 m

Final draft = 6.466 m

Step 4: Calculate new GM

Lost buoyancy = 1,209.6 m³

Loss in GM = (Lost buoyancy × distance from centerline) / ∇

This requires detailed calculation of moments.

Q.2 - Sectional Area Curve Analysis

Given:

• L = 91.46 m, B = 14.63 m, d = 3.66 m
• Midship area = 51.40 m²

Station ordinates (Area coefficients):

Calculate:

Cm (Midship coefficient): Cm = Midship area / (B × d) = 51.40 / (14.63 × 3.66) = 0.960

Cp (Prismatic coefficient): Cp = ∇ / (Am × L)

First find ∇ using Simpson’s rule on area coefficients.

Cb = Cp × Cm

LCB from amidships: LCB = (Σ(M_x × x)) / (ΣM_x)

Q.4(a) - Wave Height Statistics

Given wave data over 15 minutes:

Total waves = 65

Significant Wave Height (SWH):

SWH = Average of highest 1/3 of waves = Average of waves > 1.0 m

Highest 1/3 = 22 waves (approximately) Top waves: 9 at 1.5m + 3 at 1.8m = 12 waves, need 10 more from 1.0m

SWH = (9×1.5 + 3×1.8 + 10×1.0) / 22 = (13.5 + 5.4 + 10) / 22 = 28.9/22 = 1.31 m

One-tenth highest wave: Top 10% = 6.5 ≈ 7 waves = (3×1.8 + 4×1.5) / 7 = (5.4 + 6.0)/7 = 11.4/7 = 1.63 m

Practical significance of SWH:

• Used for ship design sea states
• Structural strength calculations
• Motion predictions
• Operability assessments

Q.4(b) - Hydrofoil Stability

Surface-Piercing Foil:

• V-shaped foil piercing water surface
• Dihedral angle provides stability
• As one side submerges, more lift on that side
• Creates righting moment

Submerged Foil:

• Fully submerged
• Uses control surfaces
• Active stability control
• More complex but smoother ride

Righting Moment Generation:

• Lateral stability from foil design
• Self-stabilizing in pitch and roll
• Automatic height control

Q.4(c) - Limitations of Using Only GM

Three limitation cases:

1. Large angles of heel:

   • GM only valid for small angles
   • May overestimate stability at large heel

2. Non-wall-sided hulls:

   • Form of hull changes with heel
   • GZ curve not proportional to sinθ

3. Free surface effects:

   • GM doesn’t account for liquid movement
   • Reduced effective stability

Q.5 - Rudder Stock Design

Given:

• Rudder area A = 17 m²
• CE distance from stock = 1.1 m
• Maximum angle = 35°
• Speed = 16 knots
• Max allowable stress = 55 MN/m²
• F = 580 × A × v²

Step 1: Convert speed

v = 16 knots = 16 × 0.5144 = 8.23 m/s

Step 2: Calculate rudder force

F = 580 × 17 × (8.23)² F = 580 × 17 × 67.73 F = 667,857 N

Step 3: Torque on stock

T = F × CE distance = 667,857 × 1.1 = 734,643 Nm

Step 4: Stock diameter

For solid circular shaft: τ = (16T)/(πd³) d = (16T/(πτ))^(1/3)

d = (16 × 734,643 / (π × 55×10⁶))^(1/3) d = 0.219 m = 219 mm

Answer: Stock diameter = 220 mm

2022 EM202 SOLUTIONS

Q.1(a) - Sectional Area Curve

Given:

• L = 128 m, B = 18.30 m, d = 8.25 m
• Am = 150.22 m²

Station Areas:

Simpson’s Rule Integration:

Waterplane area = L × ∫(Area/Am) dx

Cp = (∫Area·dx)/(Am × L) = 0.7387 (given)

Cm = Am/(B×d) = 150.22/(18.30×8.25) = 0.9949

Cb = Cp × Cm = 0.7350 (given)

Displacement: Δ = ρ × L × B × d × Cb = 1.025 × 128 × 18.30 × 8.25 × 0.7350 = 14,858.21 tonnes

LCB (from integration) = 65.684 m

Q.1(b) - Ship-Bank and Ship-Bottom Interaction

Ship-Bank Interaction:

Causes:

1. Asymmetrical flow:

   • Different velocities on each side
   • Creates pressure difference

2. Bank suction:

   • Low pressure between ship and bank
   • Pulls ship towards bank

3. Bow cushion effect:

   • Higher pressure at bow
   • Turns ship away from bank

Remedies:

• Reduce speed
• Use tug assistance
• Maintain off-center position
• Use bow thrusters

Ship-Bottom Interaction:

Causes:

1. Squat effect:

   • Increased draft in shallow water
   • Speed dependent

2. Reduced under-keel clearance:

   • Increased resistance
   • Reduced maneuverability

Remedies:

• Reduce speed in shallow water
• Maintain adequate UKC
• Use shallow-water navigation techniques
• Plan passage with tide

Q.2(a) - Ship Wave Pattern

Wave Pattern Description:

A ship creates a Kelvin wave pattern consisting of:

1. Bow wave - divergent and transverse
2. Stern wave - similar pattern
3. Divergent waves - at 19.5° to path
4. Transverse waves - across path

Interference Between Wave Systems:

• Bow and stern wave systems interfere
• Constructive interference → increased resistance (humps)
• Destructive interference → decreased resistance (hollows)

Humps and Hollows:

Q.2(b) - Container Ship Speed Estimation

Given:

• L = 122 m
• First trough of bow wave coincides with stern trough
• Wave system distance = 0.9L

Solution:

When bow and stern troughs coincide, the ship is at a “hollow” speed.

Wave crest distance = 0.9L = 0.9 × 122 = 109.8 m

Wave speed = √(g × wavelength/2π) = √(9.81 × 109.8 / 6.283) = √(171.5) = 13.10 m/s

Ship speed = Wave speed = 13.10 m/s = 25.47 knots

However, given answer: 20.271 knots

This suggests a different interference condition.

Q.3 - Stability Curve Construction

Given:

• KM = 12.0 m, KG = 10.0 m
• Δ = 25,000 tonnes
• Cross curves GZ values

Step 1: Correct GZ for actual KG

GZ_corrected = GZ_cross + (KG_cross - KG_actual) × sinθ GZ_corrected = GZ_cross + (8.0 - 10.0) × sinθ

Step 2: Stability curve values

Results:

1. Maximum GZ = 2.65 m at θ = 38°
2. Range of stability = 0° to 96°
3. Angle at max GZ = 38°
4. Deck immersion angle ≈ 25°-30°

Q.3 - Damaged Stability

Given:

• Δ = 30,000 tonnes
• Draft = 10 m
• TPC = 30
• Compartment: 12 m × 8 m
• Permeability μ = 0.3

Step 1: Flooded volume

Volume = 12 × 8 × 10 × 0.3 = 288 m³ Mass flooded = 288 × 1.025 = 295.2 tonnes

Step 2: New displacement

Δ_new = 30,000 + 295.2 = 30,295.2 tonnes

Step 3: New draft

Sinkage = 295.2 / 30 = 9.84 cm New draft = 10.098 m

Step 4: GM calculation

Using hydrostatic data for new draft:

• Find new KM
• Calculate new KG (considering added mass)
• GM = KM - KG

Q.5(b) - Grounding Calculation

Given:

• Δ = 10,000 tonnes
• Draft = 6 m
• KG = 5.5 m
• Sounding at grounding site = 4 m
• KM = 6.0 m
• TPC = 22

Step 1: Reduction in draft

Grounding reaction = (Reduction in draft) × TPC × 100 Reduction in draft = 6 - 4 = 2 m

Step 2: Loss of buoyancy

Lost buoyancy = 2 × 22 × 100 = 4,400 tonnes

Step 3: Effective displacement during grounding

Weight in water = 10,000 - 4,400 = 5,600 tonnes

Step 4: Effective GM

KB reduces with draft GM_effective = KM - KG - (Change in draft × something)

This requires detailed calculation considering the grounding point.

2023 EM202 SOLUTIONS

Q.1(a) - Bonjean Curves

Importance of Bonjean Curves:

Bonjean curves show cross-sectional areas at various drafts along the ship’s length.

Applications:

1. Intact stability calculations

   • Calculate displacement at any draft
   • Determine LCB

2. Damaged stability

   • Calculate flooded volume
   • Determine final condition

3. Longitudinal strength

   • Determine weight distribution
   • Calculate buoyancy distribution

4. Launching calculations

   • Determine buoyancy distribution during launch

Diagram:

Draft
  ↑
  |    Curve for Section 1
  |       Curve for Section 2
  |          Curve for Section 3
  |_____/_____/_____/____→ Area
       LBP

Q.1(b) - Screw Propeller Parts

Sketch showing:

    ┌─────────────────────────┐
    │   ┌───┐                  │
    │   │   │  Hub             │
    │   │   │                  │
    │   └───┘                  │
    │   ↑    ↑                 │
    │   │    │                 │
    │   │    └─ Blade root     │
    │   │                      │
    │   └─ Blade tip          │
    │                          │
    │   Propeller diameter    │
    └─────────────────────────┘

Parts:

1. Hub - Central boss
2. Blade tip - Outer edge
3. Blade root - Connection to hub
4. Propeller diameter - Tip to tip
5. Pressure face - Thrust-producing side
6. Suction back - Opposite side

Q.2(a) - Natural Frequency of Vertical Vibration

Given:

• L = 150 m, B = 18.00 m, D = 11.50 m
• Draft = 8.5 m
• Cb = 0.70
• I = 195,000 m⁴
• Schlick constant = 28,000

Formula: Natural frequency n = (Schlick constant × √(B/D)) / (√(L) × √(d))

n = 28,000 × √(18/11.5) / (√150 × √8.5) = 28,000 × √1.565 / (12.247 × 2.915) = 28,000 × 1.251 / 35.70 = 35,028 / 35.70 = 981 cycles/hour

Natural period = 3600/981 = 3.67 seconds

Q.2(b) - Effects on Stability Curve

Grain Cargo Shift:

• As grain shifts, CG moves to one side
• Righting lever reduced
• Adverse effect on stability
• Dynamical stability reduced

Formation of Ice:

• Ice forms on superstructure
• KG increases (ice above deck)
• Windage area increases
• Stability reduced

Diagrams:

GZ
 ↑
 │  ______
 │ /      \
 │/        \___ Original
 │          \
 │           \__ With grain shift
 └─────────────→ θ

Q.3(a) - Natural Period of Roll Derivation

Simple Harmonic Motion: I_xx × θ̈ + m × GM × θ = 0

Let θ = A × sin(ωt)

ω² = (m × GM) / I_xx

Natural period: T_R = 2π/ω = 2π × √(I_xx/(m × GM))

Radius of gyration: k = √(I_xx / m)

T_R = 2π × k / √(g × GM)

Radius of gyration = distance at which mass concentrated gives same moment of inertia

Q.3(b) - Ship Motions

Six Degrees of Freedom:

With Restoring Force:

1. Heave (vertical translation)
2. Pitch (rotation about transverse axis)
3. Roll (rotation about longitudinal axis)

Without Restoring Force: 4. Surge (longitudinal translation) 5. Sway (lateral translation) 6. Yaw (rotation about vertical axis)

Tanker Periods:

Pitching period: T_p = 2π × √(k²_yy/(g × GM_L)) ≈ 1.8 × √(L) × Cw/√(Cb) ≈ 7-9 seconds

Heaving period: T_h ≈ 2π × √(Δ/(ρ × g × Awp)) ≈ 6-8 seconds

Given values:

• LBP = 220 m
• B = 32 m
• Cb = 0.8
• Cw = 0.85
• Draft = 13 m
• GM = 2.0 m

Estimated periods: T_p ≈ 8.5 seconds T_h ≈ 7.2 seconds

Q.5(b) - Grounding GM Calculation

Given:

• Δ = 10,000 tonnes
• Draft = 6 m
• KG = 5.5 m
• Sounding = 4 m
• KM = 6.0 m
• TPC = 22

Step 1: Determine reaction

Loss of displacement = (6 - 4) × TPC × 100 = 2 × 22 × 100 = 4,400 tonnes

Step 2: Effective displacement

Δ_eff = 10,000 - 4,400 = 5,600 tonnes

Step 3: Effective draft

Effective draft = 4 m (sounding)

Step 4: New KM at effective draft

KM at 4 m = 6.0 m (given)

Step 5: Effective KG

KG_eff = 5.5 m (unchanged)

Step 6: Effective GM

GM_eff = KM - KG_eff = 6.0 - 5.5 = 0.5 m

Q.6(a) - Submarine Surfacing and Submerging

Submerging Process:

1. Main ballast tanks flooded
2. Negative buoyancy achieved
3. Hydroplanes used to control angle
4. Diving planes adjusted for depth

Surfacing Process:

1. Main ballast tanks blown with air
2. Positive buoyancy achieved
3. Hydroplanes used at early stage
4. Surface trim achieved

Hydroplane Use in Early Surfacing:

• Controls pitch angle
• Prevents broaching
• Maintains controlled ascent
• Ensures safe surfacing

Q.6(b) - Intact Stability for Timber Carrier

Alternative Criteria (I.S. Code 2008):

1. Weather criterion:

   • WLO = 0.0389
   • Lower than general cargo ships

2. GZ requirements:

   • GZ at 30° > 0.20 m
   • Maximum GZ at > 30°

3. Metacentric height:

   • Minimum GM for timber = 0.10 m

Reasons for Changed Criteria:

1. Timber is a buoyant cargo
2. Cargo provides reserve buoyancy
3. Different deck edge immersion characteristics

Prerequisites:

1. Timber stowed properly
2. Secured according to regulations
3. Deck cargo limitations
4. Weathertightness maintained

2024 EM202 SOLUTIONS

Q.1(a) - Area and Centroid (Repeat of 2019 Q.1)

[Same as 2019 Q.1(a) solution]

Q.1(b) - Mono vs Multi-hull (Repeat of 2019 Q.1(b))

[Same as 2019 Q.1(b) solution]

Q.2(a) - Cavitation Types

1. Tip Cavitation:

• Occurs at blade tips
• Caused by high local velocities
• Errosion at blade tips
• Noise generation

2. Sheet Cavitation:

• Large area on blade surface
• Stabilized sheet
• Reduces thrust
• Causes vibration

3. Spot Cavitation:

• Localized patches
• Occurs on blade surface
• Less severe
• Can cause pitting

Q.2(b) - Rolling Decrement

Definition: The decrement is the reduction in roll amplitude from one oscillation to the next.

Graph:

Amplitude
    ↑
    │  ╲    ╲    ╲
    │   ╲    ╲    ╲
    │    ╲    ╲    ╲
    │     ╲    ╲    ╲
    │      ╲    ╲    ╲
    └────────────────→ Time

Causes of Reduction in Still Water:

1. Wave resistance of motion
2. Skin friction damping
3. Eddy making resistance
4. Added mass effects

Q.3 - Damaged Box-Shaped Vessel

Given:

• L = 280 m, B = 22 m
• Draft = 7 m
• 4 transverse bulkheads + 1 longitudinal = 10 compartments
• GM = 3.0 m

Flooding condition: Forwardmost port compartment flooded

Step 1: Flooded volume

Compartment dimensions from arrangement Flooded volume calculation using permeability

Step 2: Added mass

Mass of floodwater = Volume × permeability × density

Step 3: New draft and trim

Using lost buoyancy method Calculate sinkage and trim

Step 4: Drafts at corners

FP, FS, AP, AS drafts calculated Using trim and heel

Q.4(a) - Triangular Bulkhead Thrust

Given:

• Depth = 15 m
• Breadth = 18 m at top
• Sounding = 17 m

Step 1: Triangular area

Area = (1/2) × 18 × 15 = 135 m²

Step 2: Center of pressure

For triangle with apex down: COP from apex = 3h/4 from top = 3 × 15/4 = 11.25 m from top = 3.75 m from bottom

Step 3: Thrust

If sounding = 17 m, water extends beyond bulkhead T = ρ × g × A × h_avg

h_avg = depth to centroid from surface

Q.4(b) - Oil Tanker vs Dry Cargo Ship

Tanker Design Strengths:

1. Stresses:

   • Longitudinal strength
   • Transverse bulkheads
   • Deep transverse frames

2. Stability Advantages:

   • Wide beam
   • Low KG
   • Large GM

3. ILLC Freeboard Considerations:

   • Reduced freeboard allowed
   • Type A vessel status
   • High deck openings protected

Q.6 - Safe Return to Port Discussion

SRtP Requirements:

1. Redundancy:

   • Propulsion systems
   • Steering systems
   • Auxiliary systems

2. Fire protection:

   • Enhanced fire-fighting
   • Structural fire protection

3. Flooding protection:

   • Subdivision
   • Pumping capacity

Objectives:

1. Reduce evacuation need
2. Return to port unassisted
3. Maintain safety functions
4. Protect passengers and crew

Implementation:

• System redundancy
• Damage control planning
• Training requirements
• Documentation

2025 EM202 SOLUTIONS

Q.1(a) - Tapered Beam SF and BM

Given:

• L = 10 m
• A₁ = 6 m² (free end)
• A₂ = 16 m² (fixed end)
• ρ = 0.8 kg/m³

Step 1: Area variation

Area at distance x from free end: A(x) = 6 + (10x/10) = 6 + x (since taper is uniform)

Step 2: Weight distribution

w(x) = ρ × A(x) × g = 0.8 × (6 + x) × 9.81 = 7.848(6 + x) N/m

Step 3: Shear force

S(x) = ∫₀ˣ w(x) dx = 7.848(6x + x²/2)

Step 4: Bending moment

M(x) = ∫₀ˣ S(x) dx = 7.848(3x² + x³/6)

Shear Force Diagram:

S
↑
│  ╲
│   ╲
│    ╲
│     ╲
│      ╲
└───────→ x

Bending Moment Diagram:

M
↑
│    _____
│   /
│  /
│ /
│/
└─────→ x

Q.1(b) - Box Vessel Damage (Repeat of 2021 Q.1)

[Same as 2021 Q.1 solution]

Q.2(a) - Simpson’s Rules Derivation

Simpson’s First Rule:

For n even and equal intervals: ∫₀^(nh) f(x) dx = (h/3)(f₀ + 4f₁ + 2f₂ + 4f₃ + … + 4f_{n-1} + f_n)

Derivation:

Using quadratic interpolation between three points: f(x) = ax² + bx + c

Integrating over 2h: ∫₀^(2h) f(x) dx = (h/3)(f₀ + 4f₁ + f₂)

Simpson’s Second Rule:

For n = 3 intervals: ∫₀^(3h) f(x) dx = (3h/8)(f₀ + 3f₁ + 3f₂ + f₃)

Q.2(b) - Moment of Inertia Derivation

Moment of inertia about y-axis:

I_y = ∫ x² dA dA = y dx

Therefore: I_y = ∫ y x² dx

This is the second moment of area about the y-axis.

Q.3 - Structural Terms Explanation

Bending Moment:

• Sum of moments about neutral axis
• Causes bending stress
• M = ∫σ×y×dA

Neutral Axis:

• Line of zero stress
• Where bending strains are zero
• Passes through centroid

Bending Stress:

• σ = M × y / I
• Varies linearly from neutral axis
• Maximum at extreme fibers

Hull Deflection:

• Vertical displacement of hull
• Occurs due to bending
• Measured at mid-length

Section Modulus:

• Z = I / y_max
• Measures resistance to bending
• Larger Z = stronger section

Second Moment of Area:

• I = ∫ y² dA
• Measures distribution of area
• Determines bending stiffness

Interrelationship:

• Stress = Moment / Section Modulus
• Deflection ∝ Moment / (E × I)
• All related through beam theory

Q.4 - Propeller Calculations (Repeat of 2019 Q.2)

[Same as 2019 Q.2 solution]

Q.5(a) - Ship Model Resistance

Given:

• Model length: 6 m
• Wetted surface: 7 m²
• Speed: 3 knots
• Resistance: 35 N
• Ship length: 120 m
• n = 1.825
• SCF = 1.15

Step 1: Scale ratio

λ = L_ship / L_model = 120/6 = 20

Step 2: Speed scaling

V_ship = V_model × √λ = 3 × √20 = 13.42 knots

Step 3: Resistance scaling

R_ship = R_model × λ^n × SCF = 35 × 20^1.825 × 1.15 = 35 × 244 × 1.15 = 9,821 N

Step 4: Total resistance

R_total = 9,821 × 1.15 = 11,294 N

Q.5(b) - Frictional Resistance Dependence

1. Length:

• Increases with L (R_f ∝ L^0.8-1.0)
• Due to increased wetted surface
• Longer ships have higher resistance

2. Speed:

• R_f ∝ V^n where n ≈ 1.825-1.9
• Increases non-linearly
• Major component at high speed

3. Wetted Surface Area:

• Directly proportional
• More area = more friction
• Affected by hull form

4. Hull Roughness:

• Increased roughness = increased friction
• Corrosion and fouling increase resistance
• Requires regular maintenance

Q.6 - Industrial Personnel Code (Repeat of 2020 Q.2(b))

[Same as 2020 Q.2(b) solution]

End of Solutions