Croatia Chief Engineer Officer study materials
Croatia Chief Engineer Officer study resources

Croatia Chief Engineer Officer: Ship Construction, Stability and Damage Control

Croatia Chief Engineer Officer · Work through your exam, one topic at a time.

All guide sections

Ship Construction, Stability and Damage Control

What this domain tests

The panel wants proof you can read a ship's structure and its stability book: hull girder strength, watertight subdivision, free surface, GM, GZ curves, damage stability, and the flooding actions a chief engineer influences (pumping, closing, ballasting). Croatian exams often mix a construction question ("what is a cofferdam for?") with a stability calculation.

Core rules and concepts

  • Structural members. Keel, floors, frames, longitudinals, stringers, bulkheads, decks, and the functions of double bottom, cofferdams (separation of dissimilar fluids and a buffer against leakage), collision bulkhead position, and the strength deck.
  • Load line and freeboard. Freeboard deck, assigned summer freeboard, deck line, and the six marks. Loading beyond the mark is a statutory offence.
  • Stability fundamentals. KB, BM = I/V, KM = KB + BM, GM = KM − KG. Positive GM means initial stability. Free surface effect reduces effective GM and must be corrected. GZ curve: area to 30°, to 40°, and to 30–40° must meet the criteria; maximum GZ and its angle; initial GM minimum.
  • Free surface and slack tanks. Slack tanks penalise GM - number and fill level matter. Partially filled tanks are the classic exam trap.
  • Damage control. Watertight subdivision, cross-flooding arrangements, bilge and ballast capability, and the sequence of actions: sound the alarm, muster, close watertight doors, identify the compartment, assess stability, pump or counter-flood only as instructed by the master.
  • Angle of loll versus list. Loll is caused by negative GM and requires correction by filling low tanks to lower KG; list is caused by off-centre weight.

Worked example

Displacement 12,000 t, KG 7.60 m, KM 7.90 m, so GM = 0.30 m. A double-bottom tank of 400 t is filled to 50% with a free surface moment (FSM) of 1,200 t·m. Effective GM = GM − FSM/Δ = 0.30 − 1,200/12,000 = 0.30 − 0.10 = 0.20 m. Reasoning: the tank's weight change is already in the new displacement and KG, so only the free surface correction is applied to GM. Answer with the formula and the units; state that 0.20 m still satisfies initial GM but the GZ curve must be rechecked.

Traps that cost marks

  • Double-counting the free surface: correcting both KG and GM for the same tank.
  • Confusing GM with GZ - GM is initial metacentric height, GZ is the righting lever at an angle.
  • Saying loll is corrected by ballasting high tanks.
  • Forgetting to mention the collision bulkhead is forward and the aft peak is separately bounded.
  • Giving damage control actions without master's authority and without stability recalculation.