25-Nav-A1 Fundamentals of Naval Architecture · May-98-Nav-A1 2016
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
National Examinations, May 2016 — 98-Nav-A1 Fundamentals of Naval Architecture, 3 hours, closed book (5 questions constitute a complete exam paper, first five as they appear in the answer book are marked, each of equal value; all 7 answered below for full study coverage).
Reference texts: Tupper, Introduction to Naval Architecture; Lewis (ed.), Principles of Naval Architecture (PNA); IMO, International Code on Intact Stability (IS Code).
Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.
When a tank carrying liquid is not completely full (it is "slack"), the liquid surface remains horizontal as the ship heels, so the liquid shifts toward the low side. This shifted liquid moves the ship's centre of gravity horizontally toward the low side as well, which reduces the righting arm at every angle of heel — equivalent to a virtual rise of $G$ by the free-surface correction $FSC=(\rho_{liq}/\rho_{sw})(i/\nabla)$, where $i$ is the tank's own free-surface second moment of area. The effect depends only on the free-surface plan area of the slack tank, not on the quantity of liquid in it (a nearly-full or nearly-empty tank has almost the same free surface as a half-full one, and hence almost the same loss of stability) — which is why ships subdivide large tanks with centreline washplates/bulkheads to shrink $i$, and why a full ("pressed-up") tank has zero free-surface effect.
Given. A double-bottom tank is loaded completely full of oil (a pressed-up tank — no free surface, so this is a pure added-weight bodily-sinkage-and-trim problem), off-centred longitudinally from the centre of flotation.
| Quantity | Symbol | Value |
|---|---|---|
| Length | $L$ | 150.00 m |
| Draft forward (before) | $d_F$ | 8.20 m |
| Draft aft (before) | $d_A$ | 8.90 m |
| Tonnes per cm immersion | $TPC$ | 28 t/cm |
| Moment to change trim 1 cm | $MCT1cm$ | 260 tonnef·m |
| Centre of flotation | $LCF$ | 1.5 m aft of midship |
| Tank (full, oil SG 0.8) | $l\times b\times h$ | 20.00 × 10.00 × 1.2 m |
| Tank centre | — | 50.00 m aft of midship, on centreline |
Find. New forward draft $d_F'$ and new aft draft $d_A'$.
Approach. Find the weight added and its lever from $F$, compute parallel sinkage and total change of trim, then apportion the trim between the two ends using each end's distance from $F$.
| Quantity | Value |
|---|---|
| Weight of oil added | $192.0$ tonnef |
| Parallel sinkage | $\approx 6.86$ cm |
| Change of trim | $\approx 35.82$ cm (by the stern) |
| New forward draft | $\approx 8.086$ m |
| New aft draft | $\approx 9.144$ m |
