Treat the ship as a beam first and a collection of plates second. The hull girder gives you bending and shear, the panels give you buckling, and the details give you fatigue.
A ship is a hollow box girder several hundred metres long, loaded along its length by two distributions that never match: its own weight, and the buoyancy supporting it.
Weight is where the steel, machinery and cargo happen to be. Buoyancy is where the underwater volume happens to be. At any station the difference between them is a net load per metre, and that load curve is the starting point for everything.
Integrating once gives shear force, twice gives bending moment. Two consequences follow from the shape of typical distributions, and they are worth knowing without doing the integration:
Weight and buoyancy balance in total but never locally. That local mismatch, integrated twice, is the bending moment the hull girder has to carry.
Two loading patterns matter, and the sign convention follows from picturing the deflected shape.
Hogging — the ends droop relative to the middle. The deck is stretched, so it is in tension; the bottom is in compression. Caused by weight at the ends and buoyancy amidships.
Sagging — the middle droops relative to the ends. The deck is in compression, the bottom in tension. Caused by weight amidships and buoyancy at the ends.
The total bending moment has two parts. The still water part comes from the loading condition and is under the operator's control — it is what a loading instrument computes and limits. The wave part is what the seaway adds.
The classical treatment poses the ship on a standing trochoidal wave of its own length, with a height conventionally taken as L/20. Crest amidships produces maximum hogging; trough amidships produces maximum sagging. Modern classification rules replace this with an empirical formula, but the picture is the same one.
The still water moment is the one that can be got wrong in service — which is why exceeding a loading instrument limit is a structural matter, not a paperwork one.
With the bending moment known, the stress follows from simple beam theory.
Three rules decide what goes into that calculation, and each is examined.
Only continuous longitudinal material counts. Deck plating, bottom shell, side shell, longitudinal bulkheads, and longitudinal stiffeners running through the midship region all carry hull girder bending. Transverse frames, floors and beams do not — they resist local loads, not global bending. A superstructure that stops short does not contribute fully and is usually excluded.
Find the neutral axis by taking moments of area. It passes through the centroid of the effective section, which on a ship sits below mid-depth because the bottom structure is heavier than the deck.
There are two section moduli, not one. The distance from the neutral axis to the deck differs from the distance to the keel, so Z(deck) ≠ Z(keel). The smaller one — normally the deck — governs, because that fibre reaches the allowable stress first.
Do not neglect the side shell's own second moment of area. A tall vertical plate has a large I about its own centroid, and leaving it out can cost 10 % of the section's stiffness.
Bending stress is only half the story. Shear force has to be carried too, and in a thin-plated box it flows around the section rather than distributing evenly.
Q is the first moment of the area beyond the point in question, taken about the neutral axis. The consequence is that shear flow is greatest at the neutral axis and falls to zero at the extreme fibres — the opposite distribution to bending stress.
In a ship's section the side shell sits near the neutral axis and is relatively thin, so it carries most of the shear at the highest shear stress. That is why shear buckling of side shell panels near the quarter lengths is a real design case, and why longitudinal bulkheads in a tanker matter as much for shear as for subdivision.
Torsion is a related concern for ships with large deck openings — container ships in particular. A closed section resists torsion efficiently through the Bredt–Batho shear flow; cut a long slot in the deck and the section becomes effectively open, warps under torsion, and needs heavy torsion boxes along the hatch sides.
Compression members do not have to reach yield to fail. They can go sideways first.
Columns. A slender pillar buckles at the Euler load, which depends on how the ends are held.
The effective length is squared, so a mistake in end fixity changes the answer by a factor of four. And Euler's formula only governs for slender members — for a stocky one it predicts a stress above yield, and the member squashes instead. The dividing line is a slenderness ratio of about π√(E/σy), which for mild steel is around 90 to 95.
Plate panels. The plating between stiffeners buckles in a quite different way, and the governing dimension is the spacing, not the panel's length.
Because critical stress depends on (t/b)², halving the stiffener spacing quadruples the buckling strength at no cost in plate thickness. This is the single most useful structural instinct in the chapter: spacing beats thickness.
Global bending is not the only thing trying to break a ship. Three local load patterns have their own names and their own structural answers.
Fatigue is a different failure mode altogether: cracks grow under cyclic loading at stresses far below yield, and they start at details, not in the middle of a panel. Hatch corners, bracket toes, weld toes and abrupt changes of section are where the stress concentrates and where cracks are found.
The design answers are geometric rather than material: radius the hatch corners, taper bracket toes, grind weld toes smooth, and avoid landing a hard structural end in the middle of an unsupported panel. Steel of higher yield strength does not help — fatigue crack growth is almost independent of yield strength, which is why high-tensile steel structures need more attention to detail, not less.
A section modulus calculation, a shear-force-and-bending-moment problem, and a buckling comparison — between them, most of what this topic asks.
A simplified midship section has the following continuous longitudinal material. Find the position of the neutral axis, the second moment of area about it, and the section modulus at deck and keel. Then find the deck and keel stresses under a hogging bending moment of 850 MN·m.
Item Area (m²) Height above keel (m) Deck plating 0.30 14.00 Deck longitudinals 0.10 13.50 Side shell (both) 0.50 7.00 Bottom longitudinals 0.15 0.50 Bottom shell & keel 0.45 0.05 Side shell own I = A·d²/12 (d = 14.0 m)
Find the position of the neutral axis, the second moment of area about it, and the section modulus at deck and keel
Total area, and the first moment of area about the keel.
| Item | A (m²) | y (m) | A·y (m³) | A·y² (m⁴) |
|---|---|---|---|---|
| Deck plating | 0.30 | 14.00 | 4.200 | 58.800 |
| Deck longitudinals | 0.10 | 13.50 | 1.350 | 18.225 |
| Side shell (both) | 0.50 | 7.00 | 3.500 | 24.500 |
| Bottom longitudinals | 0.15 | 0.50 | 0.075 | 0.038 |
| Bottom shell & keel | 0.45 | 0.05 | 0.023 | 0.001 |
| Side shell own I | — | — | — | 8.167 |
| Σ | 1.50 | 9.148 | 109.731 |
Only continuous longitudinal material is included — a deckhouse that stops short of the ends carries no hull girder bending and must be left out.
The neutral axis passes through the centroid of the section.
Second moment of area about the keel first.
Because that is where the levers were measured. Add each item's A·y², plus the side shell's own second moment about its own centroid — a tall vertical plate has a large one and must not be neglected.
Transfer to the neutral axis with the parallel-axis theorem.
Section modulus is I divided by the distance to the fibre in question.
And the two distances are different, because the neutral axis is not at half depth.
Stresses under 850 MN·m hogging.
| Fibre | y from NA (m) | Z (m³) | σ = M/Z (MPa) | Sense |
|---|---|---|---|---|
| Deck | 7.90 | 6.83 | 124.5 | tension |
| Keel | 6.10 | 8.84 | 96.2 | compression |
Hogging puts the deck in tension and the bottom in compression.
The deck governs, because it has the smaller section modulus.
That is the general result for a ship whose neutral axis sits below mid-depth — which is nearly all of them, since the bottom structure is heavier than the deck.
Answerȳ = 6.10 m, I = 53.92 m⁴, Z(deck) = 6.83 m³, Z(keel) = 8.84 m³; σ = 124.5 / 96.2 MPa
The trap: quoting one section modulus for the section. There are two, they differ, and the smaller one governs. Answering with Z(keel) here understates the deck stress by 23 %.
A box barge 60 m long and 10 m wide floats at even keel in sea water. Its lightweight of 900 t is uniformly distributed over the full length. Cargo of 300 t is loaded uniformly over the middle 20 m. Find the maximum shear force and the maximum bending moment, and say whether the barge hogs or sags.
L = 60 m, B = 10 m, even keel Lightweight 900 t uniform over 60 m Cargo 300 t uniform over the middle 20 m (x = 20 to 40)
Find the maximum shear force and the maximum bending moment, and say whether the barge hogs or sags
Buoyancy first.
The barge floats at even keel with a constant cross-section, so buoyancy is uniform along the length and totals the whole displacement.
Weight distribution.
Lightweight is spread evenly; cargo adds to the middle third only.
The load curve is weight minus buoyancy.
Where it is negative, buoyancy exceeds weight.
Shear force is the running integral of the load curve.
Starting from zero at the free end.
Bending moment is the integral of shear force.
Over the first 20 m the load is constant, so shear is linear and the moment is parabolic.
Convert to engineering units and interpret the sign.
Weight concentrated amidships and buoyancy concentrated at the ends means the middle is pushed down relative to the ends — the barge sags.
Note where each maximum falls.
Shear peaks near the quarter lengths, bending moment peaks amidships. That is why the midship section is the one that gets a section modulus calculation and the quarter-length side shell is where shear buckling is checked.
AnswerMax SF = 100 t at the quarter points; max BM = 14.7 MN·m amidships, sagging
The trap: assuming heavy cargo amidships makes a ship hog. It sags — the ends are being held up by buoyancy while the middle is pushed down. Hogging is the opposite loading: weight at the ends, buoyancy amidships.
(a) A tubular pillar 3.0 m long, outside diameter 150 mm and wall thickness 8 mm, is pinned at both ends. Find the Euler buckling load and comment. (b) A deck plate panel 12 mm thick spans 800 mm between longitudinals. Find its critical buckling stress, and the effect of reducing the spacing to 600 mm. Take E = 206 GPa, ν = 0.3, yield = 235 MPa, k = 4.
(a) L = 3.0 m, D = 150 mm, t = 8 mm, pinned–pinned (b) t = 12 mm, b = 800 mm then 600 mm, k = 4 E = 206 GPa, ν = 0.3, σ_yield = 235 MPa
Find the Euler buckling load and comment
Find its critical buckling stress, and the effect of reducing the spacing to 600 mm
Section properties of the tube.
Inside diameter is 150 − 2×8 = 134 mm.
Euler load for pinned ends.
Where the effective length equals the actual length.
Now the comment, which is where the marks are.
Convert that to a stress and compare with yield.
The Euler stress is far above yield, so the pillar will squash before it buckles.
Euler's formula does not govern. The test is slenderness: below the transition value, failure is by yielding.
Part (b): plate panels buckle by a different mechanism.
And the governing dimension is the spacing between supports.
Reduce the spacing to 600 mm and recompute.
Critical stress goes as (t/b)², so it rises with the square of the reduction.
That is the practical lesson.
Closer stiffener spacing raised the panel's buckling strength by 78 % without adding a millimetre of plate thickness — which is why longitudinal spacing, not plate thickness, is the first thing a designer adjusts when a panel fails a buckling check.
Answer(a) P_cr = 2040 kN, but yielding governs (L/r = 60 < 93). (b) 167.6 MPa at 800 mm, 297.9 MPa at 600 mm
The trap: applying Euler's formula without checking slenderness. For a stocky member it returns a buckling load the material can never reach, and quoting it as the failure load is wrong by a factor of more than two here.
σ = M·y/I = M/ZZ differs at deck and keel; the smaller governsȳ = ΣA·y/ΣANeutral axis through the centroid of continuous longitudinal materialI(NA) = Σ(I_own + A·y²) − ΣA·ȳ²Parallel-axis theorem; keep the side shell's own Iq = V·Q/I, τ = q/(I·t)Shear flow greatest at the neutral axisSF = ∫(w − b)dx, BM = ∫SF dxShear peaks at the quarters, BM amidshipsh_wave ≈ L/20Classical standard wave; crest amidships = hoggingP_cr = π²EI/L_e²L_e = L, 0.5L, 0.7L or 2L by end fixityTransition L/r ≈ π√(E/σ_y)≈ 93 for mild steel; below it, yielding governsσ_cr = k·π²E/(12(1−ν²))·(t/b)²Plate panel; k ≈ 4 simply supportedσ_max = K_t·σ_nomFatigue starts at details, not in mid-panel