Q8 (10 Marks) Ship Stability
SC&S • Written Exam

(a) Explain why the amplitude of ship motion should be limited. (6)

(b) A ship of 8100 tonne displacement floats upright in seawater, KG = 7.5m and GM = 0.45m. A tank, whose centre of gravity is 0.5m above the keel and 4m from the centreline, contains 100 tonne of water ballast. Neglecting free surface effect, calculate the angle of heel when the ballast is pumped out. (10)

Appeared In: Oct 2024

Verified Model Answer (Text Solution)

Structured for DG Shipping MEO Class II examination scoring criteria.

Exam Ready
Part (a)

Why the amplitude of ship motion should be limited.

Excessive amplitude of ship motion (especially roll, pitch and heave) is undesirable because:

  • It can lead to loss of stability and capsize, particularly in roll resonance or parametric roll.
  • It causes cargo shift, damage to cargo and lashings, and can endanger the crew.
  • It produces large accelerations that cause crew discomfort, seasickness and reduced efficiency, and can injure personnel.
  • It increases structural loading (slamming, racking) and can cause fatigue damage.
  • It increases resistance and fuel consumption, and reduces the ship's speed and course-keeping.
  • It can cause the propeller to emerge and the bow to slam, and can lead to broaching in following seas.

Hence the amplitude is limited by adequate stability (GM), by damping devices (bilge keels, stabilisers, anti-roll tanks), and by operational measures (speed and course changes to avoid resonance).

Part (b)

Angle of heel when ballast is pumped out.

Ship 8,100 t displacement, floats upright in sea water, KG = 7.5 m, GM = 0.45 m. A tank, whose centre of gravity is 0.5 m above the keel and 4 m from the centreline, contains 100 t of water ballast. Neglecting free-surface effect, calculate the angle of heel when the ballast is pumped out.

KM = KG + GM = 7.5 + 0.45 = 7.95 m.

Pumping out 100 t from (z = 0.5 m, y = 4 m):

Transverse shift of G: GG' = w x y/(Delta - w) = 100 x 4/(8100 - 100) = 400/8000 = 0.05 m (towards the opposite side).

Vertical shift of G: removing a low weight raises KG: GGv = w x (KG - z)/(Delta - w) = 100 x (7.5 - 0.5)/8000 = 100 x 7/8000 = 0.0875 m.

New KG = 7.5 + 0.0875 = 7.5875 m. New GM = KM - KG = 7.95 - 7.5875 = 0.3625 m.

Angle of heel: tan(theta) = GG'/GM = 0.05/0.3625 = 0.1379.

theta = atan(0.1379) = 7.85 deg.

Answer: the angle of heel is about 7.9 deg.

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