NAVAL ARCHITECTURE PROBLEMS (TRIM, BUOYANCY & DENSITY)
Prepared for SAMS DNS Marine Students
Q1. Trim Due to Loading/Unloading
Given Value
Ship Length 170 m
Draught Forward 6.85 m
Draught Aft 7.50 m
MCTI 300 tm
TPC 28
LCF 3.5 m forward of midships
Loading/Unloading Details:
+160 t at 63 m aft → Moment = 160 × (-63) = -10080 tm
+200 t at 27 m forward → Moment = 200 × (+27) = +5400 tm
-120 t at 75 m aft → Moment = (-120) × (-75) = +9000 tm
-70 t at 16 m aft → Moment = (-70) × (-16) = +1120 tm
Steps:
1. Total Moment = -10080 + 5400 + 9000 + 1120 = +1440 tm
2. Change in Trim = 1440300=4.8 cm by stern\frac{1440}{300} = \textbf{4.8 cm by stern}
3. Mean Draught = 6.85+7.502=7.175 m\frac{6.85 + 7.50}{2} = 7.175 \text{ m}
4. Trim per metre = 0.048 m0.048 \text{ m}
5. LCF Correction = 4.8100×3.585=0.002 m\frac{4.8}{100} × \frac{3.5}{85} = 0.002 \text{ m}
6. New Draughts: Adjust mean ± (Trim/2 ± LCF correction)
📓 Rough Work:
Q2. Final LCG after Loading
Item Mass (t) LCG (m) Moment (tm)
Lightship 1050 -4.64 (aft) -4872
Cargo 2150 +4.71 (fwd) -10126.5
Fuel 80 32.55 (aft) 2604
Water 15 32.90 (aft) 493.5
Stores 5 -33.60 (fwd) -168
Steps:
1. Total Mass = 3300 t
2. Total Moment = -12069 tm
3. New LCG = −120693300=-3.66 m (fwd)\frac{-12069}{3300} = \textbf{-3.66 m (fwd)}
📓 Use Hydrostatic tables for trim/draughts if LCB known.
Q3. Draught Change: River to Sea Water
Given:
Displacement = 15000 t
Area = 1950 m²
Densities: ρ1=1.005\rho_1 = 1.005, ρ2=1.022\rho_2 = 1.022
Formula: Δd=(DispArea)(1ρ1−1ρ2)\Delta d = \left( \frac{Disp}{Area} \right) \left( \frac{1}{\rho_1} - \frac{1}{\rho_2} \right)
Computation: Δd=7.692×(0.995−0.978)≈0.131 m or 13.1 cm drop\Delta d = 7.692 \times (0.995 - 0.978) \approx 0.131 \text{ m
or 13.1 cm drop}
📓 Effect: Draught decreases when moving to denser water.
Q4. TPC Calculation from Hydrostatic Data
Given:
Displacement at 6.0 m draught = 7450 t
Displacement at 6.2 m draught = 7700 t
Waterplane area = 1500 m²
Formula: TPC=ΔDisplacementΔDraught=7700−74500.2=1250 t/mTPC = \frac{\Delta Displacement}{\Delta Draught} = \frac{7700 -
7450}{0.2} = 1250 \text{ t/m}
Check with: TPC=γ×A100=1.025×1500100=15.375 t/cmTPC = \frac{\gamma \times A}{100} = \frac{1.025 \times 1500}{100} =
15.375 \text{ t/cm}
📓 Use either direct method or formula depending on data.
Q5. Free Surface Effect (FSE) on Stability
Given:
Beam of tank = 20 m
Length of tank = 10 m
Depth of tank = 4 m
Liquid density = 1.025 t/m³
Moment of inertia (I) = l×b312\frac{l \times b^3}{12}
Formula: FSE=ρ⋅IDisplacement=1.025⋅10⋅2031215000FSE = \frac{\rho \cdot I}{Displacement} = \frac{1.025 \cdot \frac{10 \cdot
20^3}{12}}{15000}
📓 FSE reduces GM → affects ship stability.
Q6. Buoyancy Force and Displacement
Given:
Volume submerged = 18000 m³
Water density = 1.025 t/m³
Formula: Displacement=Volume×Density=18000×1.025=18450 tDisplacement = Volume \times Density = 18000 \times 1.025 =
18450 \text{ t}
📓 Buoyancy equals weight of displaced water.
Q7. KM from Hydrostatic Data
Given:
KB = 3.1 m
BM = IV=850013000=0.654 m\frac{I}{V} = \frac{8500}{13000} = 0.654 \text{ m}
Formula: KM=KB+BM=3.1+0.654=3.754 mKM = KB + BM = 3.1 + 0.654 = 3.754 \text{ m}
📓 KM used in GM calculation for stability.
Q8. GM (Metacentric Height) from Stability Test
Given:
W = 5 t shifted by 10 m
G moves 0.05 m laterally
Displacement = 5000 t
Formula: GM=w⋅dΔ⋅tan(θ)=5⋅105000⋅tan(θ)GM = \frac{w \cdot d}{\Delta \cdot tan(\theta)} = \frac{5 \cdot 10}{5000 \cdot tan(\
theta)}
(Find angle θ\theta from shift of G)
📓 GM gives initial stability measure.
Q9. Final Trim using LCB/LCF Method
Given:
LCB aft of midships = 2 m
LCF fwd of midships = 3 m
MCTI = 200 tm
Total Moment = +600 tm
Steps:
1. Trim = 600200=3 cm by stern\frac{600}{200} = 3 \text{ cm by stern}
2. Apply LCF correction to get actual draughts
📓 Use symmetry or apply LCB/LCF adjustment.
Q10. List due to Transverse Loading
Given:
Mass shifted = 30 t
Distance = 6 m across
Displacement = 8000 t
GM = 0.9 m
Formula: tan(θ)=w⋅dΔ⋅GM=30⋅68000⋅0.9≈0.025⇒θ≈1.43°\tan(\theta) = \frac{w \cdot d}{\Delta \cdot GM} = \frac{30 \cdot 6}
{8000 \cdot 0.9} \approx 0.025 \Rightarrow \theta \approx 1.43°
📓 Angle of list shows transverse instability.
Q11. Waterplane Area from TPC
Given:
TPC = 18 t/cm
Water density = 1.025 t/m³
Formula: A=TPC⋅100ρ=18⋅1001.025≈1756.1 m²A = \frac{TPC \cdot 100}{\rho} = \frac{18 \cdot 100}{1.025} \approx 1756.1 \
text{ m²}
📓 Useful for trim and draught changes.
Q12. Mean Draught and Change in Displacement
Given:
Initial Displacement = 10000 t
Final Displacement = 10500 t
Waterplane Area = 1700 m²
Formula: Δd=ΔDArea=5001700=0.294 m\Delta d = \frac{\Delta D}{Area} = \frac{500}{1700} = 0.294 \text{ m}
📓 Rises with unloading, sinks with loading.
Q13. Buoyancy vs Gravity for Floating Condition
Principle:
For ship to float: Weight = Buoyancy
Archimedes’ Principle: Fb=ρ⋅g⋅VF_b = \rho \cdot g \cdot V
📓 If Buoyant Force < Weight → Sinks
