A ship's hull is one very long, very thin beam, and almost every scantling decision — plate thickness, stiffener spacing, bracket radius — traces back to how that beam bends, shears and buckles under load. This chapter builds the full stress picture, from the hull girder bending moment down to the weld toe where a fatigue crack actually starts.
Start by forgetting the hull is a ship and treat it as a beam. At any point along the length, weight per metre and buoyancy per metre are both known curves — weight from the loading condition, buoyancy from the immersed sectional area at the waterline. Subtract one from the other and the result is a net load curve running the length of the vessel: positive where weight exceeds buoyancy locally, negative where buoyancy wins.
Integrate the load curve once and you get the shear force diagram; integrate again and you get the bending moment diagram. Nothing exotic is happening here — it is the same beam theory as a simply supported joist, just applied to a structure that is itself the support. In still water the load curve is set by how cargo, ballast and machinery weight are distributed against the smooth buoyancy curve, which is why an uneven loading plan (heavy ore in alternate holds, say) can drive bending moment even in flat calm.
Superimposed on the still-water curve is the wave-induced component: a wave crest sitting amidships concentrates buoyant support under the middle of the ship relative to the ends, bending the hull so the middle arches up — hogging. A wave trough amidships does the opposite — the ends are better supported than the middle, and the hull sags. Because shear force is the first integral of load and bending moment the second, shear force tends to peak away from midships, typically near the quarter-length points, while bending moment peaks near amidships where the accumulated area under the shear diagram is greatest.
Shear and bending moment do not peak at the same station, and still-water and wave-induced components must each be evaluated along the full length and combined at every station — the worst combination is not necessarily at midships once local openings, tank boundaries or bulkheads are considered.
Once the design bending moment at a station is known, the question becomes: how much stress does it put into the steel? The answer follows ordinary beam theory — stress is proportional to distance from the neutral axis, and the constant of proportionality is the moment of inertia of the cross-section.
The neutral axis itself is found from a first-moment balance — sum each structural element's area times its distance from an arbitrary baseline (usually the keel), divide by the total area, and that gives the height of the axis about which the section bends with zero stress. Only continuous longitudinal material is allowed into this calculation: plating, decks, longitudinal girders and stiffeners that run essentially the full length of the ship. A short deckhouse or an engine casing that stops after two or three frame spaces is not structurally continuous and must be left out, even though it visibly adds steel to the section — including it flatters the section modulus and understates the true stress.
Because deck and keel sit at different distances from the neutral axis, they do not share a single section modulus even though they share the same moment of inertia. The one with the smaller Z is the one that sees the higher stress for a given moment, and it is that fibre — not necessarily the extreme fibre you would guess by eye — that governs the check. On most conventional hull forms the deck modulus tends to be the more critical of the two, but that is a consequence of geometry, not a rule to apply blindly.
Compute deck and keel stress separately from the same M and I; the smaller section modulus governs, and only continuous longitudinal material may be counted toward I.
Bending moment tells you the direct stress in the plating running fore and aft; shear force tells you something different — the tendency of horizontal layers of the section to slide past one another, resisted by shear stress. In a thin-walled section such as a ship's midship body, that shear stress is usually expressed as shear flow.
Q is zero at the extreme fibres — there is no area left outboard of the deck or keel to generate a first moment — and grows toward the neutral axis, where Q is largest, so shear stress is greatest at the neutral axis and essentially zero at deck and keel. This is the reverse of the bending stress distribution, which is exactly why the two checks are done separately rather than added at the same point on the section. In practice, the side shell plating — sitting near the neutral axis and continuous over the full depth — carries the bulk of the vertical shear, while deck and bottom plating do comparatively little shear work but carry most of the bending.
This split has a practical consequence: because shear force peaks near the quarter-length and bending moment peaks near amidships, the side shell is most heavily worked in shear away from midships, while deck and bottom are most heavily worked in bending at midships. A structural check at a single station should use the shear force and bending moment that actually apply there, not the peak of each diagram regardless of location.
A pillar or a stiffener loaded in compression does not necessarily fail by the steel crushing — if it is slender enough, it fails by buckling sideways at a stress well below yield. Euler's formula gives that critical load for an ideal, pin-ended elastic column, and effective length adapts it to other end conditions.
L_e is not the physical length of the member — it is the length of an equivalent pin-ended column that would buckle at the same load, and it takes one of a handful of standard values: L for pinned-pinned, roughly 0.7L for pinned-fixed, 0.5L for fixed-fixed, and 2L for fixed-free (a cantilevered pillar, fixed at one end and completely unrestrained at the other). Because L_e is squared in the formula, an error in choosing the right end condition is squared in its effect on the answer — this is one of the easiest ways to lose marks on an otherwise correct calculation.
Slenderness must be checked before Euler is trusted. For a short, stocky member, Euler's formula predicts a critical stress above the material's yield stress — which is meaningless, because the member will simply squash in compression before it ever reaches that elastic buckling stress. Comparing σ_cr against σ_y tells you which failure mode actually governs: if σ_cr is below yield, the member is slender enough for elastic buckling to govern and Euler applies directly; if σ_cr comes out above yield, the member is stocky and a yield-based check governs instead.
A plate panel between stiffeners buckles by a related but distinct mechanism. Unlike a column pinned only at its ends, a panel is restrained along all four edges by the surrounding stiffeners and plating, so it buckles into a wave pattern across the panel rather than a single sideways bow.
The critical point is that plate buckling strength scales with the square of thickness-to-spacing ratio, not with slenderness. Halving the stiffener spacing has the same effect as doubling the plate thickness, and a small loss of thickness to corrosion erodes buckling strength faster than a linear reading of the formula would suggest — which is exactly why corrosion margins matter more for buckling than for simple yield checks.
Column buckling load falls off with the square of effective length; plate buckling stress falls off with the square of the spacing-to-thickness ratio. Both punish a small geometric error heavily.
Deflection checks are usually the least demanding of the structural checks, but they still come up — for a panel or a stiffener idealised as a simply supported beam under a uniformly distributed lateral load, the standard result applies.
Fatigue is a different kind of check altogether, because it is not driven by the nominal stress calculated from M/Z or by a single overload — it is driven by repeated cycling of stress at a geometric discontinuity: a hatch corner, a bracket toe, a weld termination, a scallop. At each of these, the local stress is higher than the nominal stress in the surrounding plating by a stress concentration factor, K_t, and it is that amplified local stress range that a fatigue crack actually responds to.
Each weld detail has a characteristic S–N curve relating stress range to cycles-to-failure, with a slope typically taken around m ≈ 3 for welded steel details. Because a ship sees many different stress ranges over its life — one per sea state, essentially — fatigue damage from each range is assessed separately and then summed using Miner's rule: the ratio of cycles actually experienced to cycles that would cause failure at that range alone, added across every range, with failure predicted once the sum reaches 1.0.
A detail that is fatigue-critical is not fixed by specifying a higher-strength steel: the S–N curve for a welded joint is governed by the weld geometry and the residual stress left by welding, not by the parent plate's yield strength. The only effective fixes are geometric — increase a toe radius, grind or peen a weld toe, fit a soft-toe bracket — or reduce the stress range reaching the detail in the first place.
Everything above can be done with hand calculation because it idealises the hull as a beam and each panel as a simple plate. Finite element analysis does not replace that reasoning — it replaces the hand-estimated inputs (a stress concentration factor read from a chart, an assumed load distribution) with a direct numerical answer for a specific geometry.
A coarse global model of the whole hull, built from shell and beam elements representing the major structure, is mainly used as a sanity check on the beam theory result — does the overall stress distribution from the model line up with σ = M/Z at midships? A fine local model, meshed only around one detail such as a bracket toe or hatch corner and often loaded using boundary displacements taken from the coarse global model, gives the actual hot-spot stress at that detail directly, which can then be fed straight into the fatigue check in place of an assumed K_t.
Mesh density matters more than it might seem: an under-refined mesh smooths over exactly the stress peak the analysis was run to find, so the result looks acceptable while the real detail is not. The usual discipline is to refine the mesh around a hot spot until the peak stress stops changing significantly between refinements — if it is still rising each time the mesh is tightened, the answer is not yet trustworthy. For exam purposes, the concept to hold onto is what FEA is being used for at each stage — global sanity check or local hot-spot stress — since that is what determines whether the beam-theory checks above are still the ones that decide the outcome.
The three examples below move from a global hull girder check, through a local column buckling check, to a fatigue life estimate at a structural detail — the same progression, in miniature, that a real structural assessment follows from the whole ship down to a single weld.
A 190 m general cargo/bulk vessel is being checked amidships in the full-load departure condition. The classification society's rule still-water sagging bending moment is 150 MN·m and the rule wave-induced sagging bending moment is 210 MN·m. The midship section has a net (continuous-material) moment of inertia about the neutral axis of 48.0 m⁴, with the neutral axis 6.0 m above the keel and the strength deck 8.0 m above the neutral axis. The permissible bending stress for this steel grade is 175 MPa. Check whether the section is adequate, and identify which fibre governs.
M_sw = 150 MN·m (sagging) M_wave = 210 MN·m (sagging) I_NA = 48.0 m⁴ y_deck = 8.0 m, y_keel = 6.0 m σ_perm = 175 MPa
Check whether the section is adequate, and identify which fibre governs
Sagging still-water and wave bending moments act the same way on the girder.
Both hog the ends down relative to the middle — so for this check the design bending moment is their direct sum.
Deck and keel have different section moduli even though they share the same I.
Because they sit at different distances from the neutral axis. Divide I by each distance separately.
Bending stress at each fibre follows σ = M/Z.
| Fibre | Z (m³) | σ = M/Z (MPa) | σ_perm (MPa) | Margin |
|---|---|---|---|---|
| Deck | 6.0 | 60.0 | 175 | 66% |
| Keel | 8.0 | 45.0 | 175 | 74% |
Apply the same design moment to both section moduli and compare each against the permissible stress.
The deck has the smaller section modulus.
So for an identical bending moment it always carries the higher stress. It is the deck that governs this check, not the keel, even though both are within the permissible stress here.
AnswerDeck governs: σ_deck = 60.0 MPa against 175 MPa permissible — the section is adequate, with about 66% margin at the deck and 74% at the keel.
The trap: quoting a single 'section modulus' for the whole midship section — deck and keel sit at different distances from the neutral axis, so they must be checked separately, and it is the smaller Z (here the deck) that governs, not the larger.
A machinery-space pillar is welded rigidly to the tank top at its base and supports a deck girder at its unbraced head, with no lateral restraint at the top — effectively fixed at the bottom and free at the top. The pillar is 2.5 m long, with cross-sectional area 20 cm² and second moment of area 500 cm⁴, in mild steel (E = 200 GPa, σ_y = 235 MPa). The design axial load carried by the pillar is 140 kN, and a working factor of safety of 2.5 is applied to the elastic critical load. Check whether the pillar is adequate.
L = 2.5 m, fixed base / free head (K = 2.0) A = 20 cm², I = 500 cm⁴ E = 200 GPa, σ_y = 235 MPa Design load P = 140 kN, required FoS = 2.5
Check whether the pillar is adequate
Use the effective length, not the actual length.
A fixed-free column buckles as if it were the top half of a pinned-pinned column of twice the length, so its end-condition factor is K = 2.0 — one of the four standard cases (K = 1, 0.5, 0.7 or 2 depending on end fixity).
Find the radius of gyration from the given section properties.
Then the slenderness ratio — this decides whether Euler's formula can be trusted at all.
Compute the Euler critical stress at this slenderness.
Compare it with the yield stress. Had the Euler value come out above σ_y, the pillar would squash before it could buckle elastically, and Euler's formula would not apply.
197.4 MPa is below the 235 MPa yield stress.
So the pillar is slender enough that elastic (Euler) buckling governs, not yielding — the Euler load can be used directly.
Convert the critical stress to a critical load.
Then apply the required factor of safety to get the allowable working load.
AnswerP_allow ≈ 157.9 kN against a design load of 140 kN — the pillar is adequate, with roughly a 13% margin.
The trap: using the actual 2.5 m length instead of the effective length — for a fixed-free pillar L_e = 2L = 5.0 m, and because P_cr is inversely proportional to L_e², using the true length here would have overstated the critical load by a factor of four.
A bracket toe on a longitudinal stiffener is exposed to two dominant sea states over the vessel's life. In moderate seas the nominal stress range at the bracket is 25 MPa, occurring about 400,000 cycles per year; in heavier seas the nominal stress range rises to 50 MPa, occurring about 50,000 cycles per year. The bracket toe has a stress concentration factor of 2.0, and the detail's S–N curve is N = C·S⁻³ with C = 1.0×10¹² (S in MPa). The vessel's required fatigue design life is 25 years. Estimate the fatigue life of this detail and say whether it meets the design requirement.
Bin 1: ΔS_nom = 25 MPa, n₁ = 400,000 cycles/yr Bin 2: ΔS_nom = 50 MPa, n₂ = 50,000 cycles/yr K_t = 2.0 S–N curve: N = C·S⁻³, C = 1.0×10¹² Required design life = 25 years
Estimate the fatigue life of this detail and say whether it meets the design requirement
Fatigue is driven by the local.
Hot-spot stress, not the nominal stress in the plating. Multiply each nominal stress range by the stress concentration factor to get the range the weld toe actually experiences.
Use the S–N curve to find the cycles to failure at each of these local stress ranges.
Miner's rule sums a damage fraction for each stress range.
| Bin | n (cycles/yr) | N (cycles to failure) | n/N (damage/yr) |
|---|---|---|---|
| 1 — moderate seas | 400,000 | 8,000,000 | 0.05 |
| 2 — heavy seas | 50,000 | 1,000,000 | 0.05 |
| Total | 0.10 /yr |
The cycles actually applied against the cycles that would cause failure at that range alone.
Fatigue failure is predicted once the cumulative damage sum reaches 1.0.
So invert the annual damage rate to get the fatigue life.
Ten years is less than half the 25-year design life the detail must meet.
So it fails as designed. Because the weld toe, not the parent plate, controls fatigue strength, specifying a higher steel grade would not help here — the fix has to reduce K_t geometrically, for example a larger toe radius or a soft-toe bracket, or reduce the stress range itself.
AnswerPredicted fatigue life ≈ 10 years against a 25-year requirement — the detail fails; improve it by softening the toe geometry (lower K_t), not by upgrading the steel.
The trap: assuming a higher-strength steel fixes a fatigue-critical detail — S–N curves for welded joints are set by weld geometry and residual stress, not by parent-plate yield strength, so a stronger steel grade gives no extra fatigue life here.
σ = M·y/I = M/ZCheck deck and keel separately — the smaller Z gives the higher stressI = Σ(I₀ + A·d²)Parallel axis theorem, summed about the neutral axisNA from ΣA·y / ΣANeutral axis height — continuous longitudinal material onlyτ = VQ/(It)Shear flow, greatest at the neutral axis; side shell carries most of the shearλ = L_e/rSlenderness ratio — check before trusting EulerP_cr = π²EI/L_e²L_e = L, 0.5L, 0.7L or 2L depending on end fixityσ_cr = kπ²E/(12(1−ν²))·(t/b)²Plate panel buckling, k ≈ 4 for simply supported edgesδ = 5wL⁴/384EIUDL, simply supported — deflection benchmarkN = C·S^(−m), m ≈ 3S–N curve for a welded detailΣ(n_i/N_i) = 1.0Miner's rule — cumulative fatigue damage at failure