25-Nav-A1 Fundamentals of Naval Architecture · May-98-Nav-A1 2016
Question 1 of 7: Framing Systems & Barge Transverse Stability
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
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 1: Framing Systems & Barge Transverse Stability
Transverse framing builds the hull's structural skeleton from closely spaced frames (ribs) running athwartships — from keel to deck around each cross-section — tied together by widely spaced longitudinal members (a few strong stringers, the keel, and deck girders). It resists local (transverse) loads such as hydrostatic pressure and sloshing well, is simple to build, and was the traditional system for smaller vessels and warships where transverse strength and compartmentation dominate.
Longitudinal framing instead runs the majority of the structural members — longitudinals on the hull plating, deck and bottom — fore-and-aft along the ship's length, tied together by widely spaced transverse web frames and bulkheads. Because the dominant bending load on a ship's hull girder is longitudinal (sagging/hogging in a seaway), this arrangement puts material where the primary bending stress is highest, giving greater longitudinal strength for the same steel weight. It is the standard system for long, slender vessels such as tankers and bulk carriers, where hull-girder bending governs.
Cross-section schematic: transverse framing (left) puts the many closely-spaced members athwartships; longitudinal framing (right) puts them fore-and-aft.
(b) Displacement, transverse GM, and heel after a weight shift
Given. A box-shaped barge floats on even keel; a bottom tank is half full of oil (free-surface liquid), then a deck weight is shifted transversely.
Given data
Quantity
Symbol
Value
Length
$L$
72.00 m
Breadth
$B$
12.00 m
Draft
$T$
4.00 m
Sea-water density
$\rho$
1.025 t/m³
Vertical CG
$KG$
3.50 m
Tank (half full, oil SG 0.8)
$l\times b$
15.00 m × 10.00 m
Weight shifted P→S
$w$
75.00 tonnef
Transverse shift distance
$d$
6.00 m
Find. Displacement $\Delta$, fluid transverse metacentric height $GM$, and heel angle $\theta$ after the weight shift.
Transverse section: K, B, G, M on the centreline before the shift; the tank's free surface pulls the effective metacentre down to $GM_{fluid}$; the deck weight shift then heels the barge toward starboard.
Approach. Compute the box-shape hydrostatics ($\nabla$, $KB$, $BM$, $KM$), subtract the tank's free-surface correction to get the fluid $GM$, then apply the small-angle heeling-moment formula for the weight shift.
KB, BM, KM (solid GM). For a box barge, $KB=T/2=2.00\text{ m}$. The transverse waterplane inertia is $I_T=\dfrac{LB^3}{12}=\dfrac{72.00(12.00)^3}{12}=10{,}368\text{ m}^4$, so $BM=I_T/\nabla=10{,}368/3456.0=3.00\text{ m}$.
$$KM = KB+BM = 2.00+3.00 = 5.00\text{ m}, \qquad GM_{solid}=KM-KG=5.00-3.50=1.50\text{ m}$$
Free-surface correction (half-full tank). The tank's own free-surface inertia is $i=\dfrac{l\,b^3}{12}=\dfrac{15.00(10.00)^3}{12}=1250.0\text{ m}^4$ (its actual liquid depth does not enter — only the free-surface plane). With the tank's oil at SG 0.8:
$$FSC=\frac{\rho_{oil}}{\rho_{sw}}\cdot\frac{i}{\nabla}=\frac{0.8}{1.025}\cdot\frac{1250.0}{3456.0}\approx 0.282\text{ m}$$
$$\boxed{GM_{fluid}=GM_{solid}-FSC = 1.50-0.282 \approx 1.218\text{ m}}$$
Heel from the deck-weight shift. For a small transverse shift, $\tan\theta=\dfrac{w\,d}{\Delta\,GM_{fluid}}$:
$$\tan\theta=\frac{75.00(6.00)}{3542.4(1.218)}=\frac{450.0}{4313.4}\approx0.1043$$
$$\boxed{\theta \approx 5.96^\circ\text{ to starboard}}$$
