Exercise set

Marine Structures and Hull Integrity Exercises

Worked marine structure exercises for hull stress, torsion, shear flow, slamming, deck cargo, docking, fatigue, watertight load and repair release.

Branch: Naval and Marine Engineering
Content: Exercise set
Updated: Jul 03, 2026
Revision: v1.2.6 · reviewed

These exercises practise marine structure and hull integrity calculations as engineering evidence. They cover hull girder stress, residual section modulus, torsional shear flow, shear flow, combined equivalent stress, pressure panels, slamming stiffener loads, deck cargo grillage reactions, docking block pressure, watertight closure load, buckling with corrosion wastage, survey release, welded fatigue, crack-growth inspection intervals, machinery-foundation vibration, and guarded repair decisions.

The goal is not only to compute a stress or margin. The goal is to decide whether the loading condition, structural model, material condition, inspection evidence, and operating restriction support a defensible vessel decision.

Assume simplified screening models unless an exercise states otherwise. Final marine structural decisions require approved loading manuals, class rules, direct strength analysis where required, material allowables, survey records, repair drawings, NDT evidence, and vessel-specific operating limits.

Release Evidence Notes

Use these worked problems as structural screens until vessel-specific evidence proves the same load basis. A credible hull-integrity release package should connect loading condition, draft, trim, ballast state, corrosion readings, material grade, weld detail, survey scope, repair drawing, class basis, operating restriction and validation record.

For marine structural decisions, the strongest evidence is agreement between calculation, inspection and vessel condition:

loading manual, route restriction, weather state, cargo condition, docking plan and local load case match the checked stress boundary;
measured thickness, corrosion trend, coating state, wastage deduction, NDT coverage and repair history support the section properties used in the calculation;
fatigue, crack-growth and fracture checks use the same weld detail, stress range, environment, inspection method and consequence class;
deck cargo, temporary equipment, sea fastenings and grillage supports are checked as load paths rather than only as footprint pressures;
drydock support reactions, block contact, ballast sequence, side-block layout and local keel structure are verified before bearing pressure is accepted;
watertight closures, dogs, hinges, seals, coamings and penetrations are checked against the flooded head and damage-control consequence;
release records state whether the outcome is unrestricted service, restricted service, repair redesign, shorter inspection interval or hold point.

When those evidence paths disagree, do not release from a clean nominal stress alone. Identify the wrong loading condition, missing thickness survey, weak NDT coverage, hidden fatigue detail, inaccurate support reaction, ineffective repair boundary or unapproved operating restriction before accepting the vessel decision.

How to Use These Exercises

For each exercise, define:

the loading condition, draft, trim, displacement, ballast state, cargo state, and operating mode;
whether the check is global hull girder strength, combined stress, local pressure, docking support, buckling, fatigue, fracture, vibration, survey release, or repair release;
whether stresses are nominal, hot-spot, local notch, equivalent, or finite-element stresses;
which thickness, corrosion, weld detail, and inspection assumptions are being used;
what engineering action follows if the margin is weak.

The common failure is not arithmetic. It is using a simple formula without checking whether the load basis, stress definition, corrosion state, fatigue data, and inspection method match the vessel decision.

Engineering Boundary Notes

Marine structural calculations are only defensible when the load boundary and stress definition are explicit. Hull girder bending, torsion, local panel pressure, deck cargo reaction, slamming, docking, fatigue and repair release are different problems even when they reuse familiar stress formulas.

Keep nominal stress, hot-spot stress, local notch stress, equivalent stress and finite-element stress separate. A fatigue detail that is acceptable under nominal stress may be weak under hot-spot stress; a local panel that passes bending may still fail a weld, support reaction or corrosion allowance. The exercise result should therefore state which boundary has been checked and which structural modes remain outside the calculation.

Inspection evidence is part of the boundary. Thickness readings, NDT coverage, coating condition, weld history, repair drawings, temporary supports, cargo grillage details and drydock block contact can change the capacity more than a rounding choice. When the checked margin depends on measured condition, the calculation should not be released without survey traceability.

Common Release Mistakes

using gross scantlings after corrosion wastage is known or suspected;
accepting average thickness while the governing reading is a local minimum at the critical load path;
applying a beam formula to a panel, bracket, weld toe or grillage detail with a different stress definition;
releasing deck cargo from footprint pressure while ignoring stool reactions and grillage load paths;
checking drydock bearing pressure without confirming which blocks are actually active;
using a fatigue or crack-growth result without NDT detectability and inspection access;
treating a repair as restored capacity before material, weld, NDT and class evidence are complete.

Scenario Map

Scenario	Core calculation	Engineering decision
Hull girder section modulus	Design bending moment divided by deck and bottom section modulus	Treat global stress as a screen before shear, buckling, openings, fatigue and class checks.
Hull torsion and closed-box shear flow	Torsional moment divided by twice enclosed area, then checked against wasted plate thickness	Decide whether quartering-sea, asymmetric loading or damaged box-girder conditions require derating or more detailed analysis.
Longitudinal web shear flow	Vertical shear multiplied by $Q/I$ and split between webs	Confirm the structural model and local details before using average web stress.
Combined bending and shear	Von Mises equivalent stress from normal stress and web shear	Check whether individually acceptable stresses become marginal when combined.
Local pressure panel	Hydrostatic head factored by dynamic and load factors	Make dynamic pressure factors visible because local panels and welds may govern.
Plate buckling with wastage	Elastic plate buckling stress compared with compression demand	Track thickness-squared sensitivity and require survey coverage where reserve is weak.
Stiffener buckling	Euler load compared with compressive demand	Include attached plating, tripping, eccentricity, end brackets and model boundary stiffness.
Thickness survey release	Minimum reading minus uncertainty and future corrosion	Choose release interval from guarded thickness, not average measured thickness.
Welded bracket fatigue	Miner damage summed over sea-state blocks	Set inspection interval before accumulated damage exceeds the chosen trigger.
Crack-growth inspection	Critical crack size and factored crack-growth life	Tie interval to NDT detectability, access, stress range and cycle counting.
Foundation vibration	Natural frequency compared with blade-passing frequency	Avoid resonance release without mode-shape, damping and sea-trial evidence.
Guarded repair release	Guarded capacity divided by guarded demand	Prefer restricted loading when unrestricted margin is technically positive but not robust.
Residual section modulus	Deck wastage deducted from hull-girder section properties	Check whether corrosion changes global bending stress enough to affect release.
Pressure-panel stress	Distributed design pressure converted to plate bending stress	Decide whether local plate wastage leaves enough stress reserve.
Slamming stiffener load	Bottom impact pressure converted to partial-span stiffener bending	Decide whether an impact patch controls local renewal or restriction.
Deck cargo support	Factored cargo weight, stool reaction, bearing pressure and grillage reaction	Decide whether deck cargo can be released with the current support layout or needs extra load spread.
Drydock block pressure	Docking weight divided by active keel-block contact area with reaction factor	Decide whether the docking plan still works if some supports are ineffective.
Watertight closure load	Flooded-compartment head converted to door force and dog reaction	Decide whether a closure can preserve subdivision after local flooding.

Validation Package Checklist

Before treating a hull-integrity calculation as release evidence, collect:

vessel condition, loading case, draft, trim, ballast state and route or operation;
checked structural boundary: global hull, local panel, support, weld, fatigue, fracture, docking or closure;
load source, factor basis, corrosion deduction and material allowable;
thickness survey, NDT scope, coating state, weld detail and repair history;
stress definition, model assumption, support condition and interaction effect;
class rule, loading manual, repair drawing or approved analysis reference;
inspection interval, operating restriction or hold point for marginal cases;
final release decision with responsible reviewer and evidence location.

Exercise 1: Hull Girder Section Modulus and Stress

A loading condition for a coastal vessel gives still-water hogging moment:

M_{SW}=42\ \text{MN m}

The wave bending moment used for this screening case is:

M_W=28\ \text{MN m}

Use:

M_d=1.0M_{SW}+1.1M_W

Available section moduli are:

Z_{deck}=0.56\ \text{m}^3,\quad Z_{bottom}=0.62\ \text{m}^3

The allowable stress for this screening basis is:

\sigma_{allow}=155\ \text{MPa}

Calculate deck and bottom bending stress and decide whether both pass the elastic screen.

Solution

Design bending moment:

M_d=1.0(42)+1.1(28)=72.8\ \text{MN m}

Convert:

M_d=72.8\times10^6\ \text{N m}

Deck stress:

\displaystyle \sigma_{deck}=\frac{M_d}{Z_{deck}}=\frac{72.8\times10^6}{0.56}=130\times10^6\ \text{Pa}=130\ \text{MPa}

Bottom stress:

\displaystyle \sigma_{bottom}=\frac{72.8\times10^6}{0.62}=117\ \text{MPa}

Required section modulus at the allowable stress is:

\displaystyle Z_{req}=\frac{72.8\times10^6}{155\times10^6}=0.470\ \text{m}^3

Both available section moduli exceed $0.470\ \text{m}^3$ , so both pass this simplified elastic screen.

Engineering Comment

The deck is more critical because it has the smaller section modulus. This result is only a global strength screen. A release decision still needs shear, buckling, openings, corrosion deductions, fatigue-sensitive details, loading-manual consistency, and class-rule checks for the actual vessel.

Plausibility Check

The combined design moment is 72.8 MN m, producing 130 MPa at the deck and 117 MPa at the bottom. Both are below the 155 MPa screening allowable, and the required section modulus is 0.470 m3 versus available values of 0.56 m3 and 0.62 m3. The deck remains the controlling side because it has the smaller section modulus and higher utilization.

Exercise 2: Shear Flow Split Between Longitudinal Webs

At a hull section, vertical shear is:

V=2.4\ \text{MN}

The section-property term at a checked connection is:

\displaystyle \frac{Q}{I}=0.31\ \text{m}^{-1}

The total shear flow is shared by two webs. The port web carries 60 percent of the flow and has thickness:

t_p=10\ \text{mm}

The starboard web carries 40 percent and has thickness:

t_s=8\ \text{mm}

Calculate average shear stress in both webs.

Solution

Total shear flow:

\displaystyle q=V\frac{Q}{I}=2.4\times10^6(0.31)=744000\ \text{N/m}

Port web shear flow:

q_p=0.60(744000)=446400\ \text{N/m}

Starboard web shear flow:

q_s=0.40(744000)=297600\ \text{N/m}

Port web stress:

\displaystyle \tau_p=\frac{q_p}{t_p}=\frac{446400}{0.010}=44.6\ \text{MPa}

Starboard web stress:

\displaystyle \tau_s=\frac{297600}{0.008}=37.2\ \text{MPa}

Engineering Comment

The thinner web does not control here because it carries less flow. In a real hull, the split should come from the structural model, not convenience. Cutouts, brackets, welds, shear lag, corrosion, and local buckling can make local stresses higher than this average shear-flow screen.

Plausibility Check

The total shear flow is 744,000 N/m. The port web carries 60 percent of that flow, so even with 10 mm thickness it reaches 44.6 MPa, while the 8 mm starboard web reaches 37.2 MPa because it carries only 40 percent. The result is plausible, but the flow split must come from the hull model and local detail, not from a convenient assumption.

Exercise 3: Local Pressure Panel With Dynamic Factor

A tank boundary panel is checked under seawater head:

h=6.5\ \text{m}

Use:

\rho=1025\ \text{kg/m}^3,\quad g=9.81\ \text{m/s}^2

The project basis applies a dynamic pressure factor:

C_d=1.30

and load factor:

\gamma_p=1.15

The tributary area to the stiffener is:

A=0.72\ \text{m}^2

Find static pressure, design pressure, and design force.

Solution

Static gauge pressure:

p_g=\rho gh=1025(9.81)(6.5)=65331\ \text{Pa}=65.3\ \text{kPa}

Design pressure:

p_d=\gamma_p C_d p_g=1.15(1.30)(65.3)=97.6\ \text{kPa}

Design force:

F_d=p_dA=97.6(0.72)=70.3\ \text{kN}

If the dynamic and load factors were ignored, the force would be:

F_{static}=65.3(0.72)=47.0\ \text{kN}

Engineering Comment

The factored design force is about 50 percent higher than the static pressure force. The calculation record should make this visible because local panels, stiffener brackets, welds, and boundaries may be governed by dynamic or rule pressure rather than calm static head.

Plausibility Check

The static seawater head is 65.3 kPa, and the combined factor $1.15\times1.30=1.495$ raises the design pressure to 97.6 kPa. Over 0.72 m2, that gives 70.3 kN instead of the calm static 47.0 kN. The factor increase is large enough that omitting it would materially understate panel and stiffener demand.

Exercise 4: Plate Buckling With Corrosion Wastage

A steel plate panel under compression has:

E=200\ \text{GPa},\quad \nu=0.30,\quad k=4.0

Unsupported plate breadth:

b_p=0.55\ \text{m}

Measured effective thickness:

t=6.8\ \text{mm}

Compressive demand:

\sigma_{comp}=72\ \text{MPa}

Use:

\displaystyle \sigma_{cr}=\frac{k\pi^2E}{12(1-\nu^2)}\left(\frac{t}{b_p}\right)^2

Calculate buckling reserve factor. Then estimate the reserve if future wastage reduces thickness to 6.4 mm.

Solution

For 6.8 mm:

t=0.0068\ \text{m}

\displaystyle \sigma_{cr}=\frac{4\pi^2(200\times10^9)}{12(1-0.30^2)}\left(\frac{0.0068}{0.55}\right)^2=110.5\ \text{MPa}

Buckling reserve factor:

\displaystyle RF_b=\frac{110.5}{72}=1.53

For 6.4 mm, use thickness-squared sensitivity:

\displaystyle \sigma_{cr,6.4}=110.5\left(\frac{6.4}{6.8}\right)^2=97.9\ \text{MPa}

The future reserve factor is:

\displaystyle RF_{b,6.4}=\frac{97.9}{72}=1.36

Engineering Comment

Both screens pass, but the reserve falls as thickness is lost. This is a structural integrity issue, not only a corrosion maintenance issue. If readings are localized, uncertain, or near welds and cutouts, the engineer should require closer survey coverage or a more detailed panel assessment.

Plausibility Check

The 6.8 mm plate gives a buckling stress of 110.5 MPa and reserve factor 1.53. Reducing thickness to 6.4 mm is only a 5.9 percent thickness loss, but the buckling stress falls to 97.9 MPa because the relation is thickness-squared. The reserve drops to 1.36, so corrosion wastage directly affects structural capacity.

Exercise 5: Stiffener Buckling With Section Loss

A longitudinal stiffener is approximated as a column with:

E=200\ \text{GPa}

I=1.3\times10^{-7}\ \text{m}^4

Unsupported length:

L=1.4\ \text{m}

Effective length factor:

K=0.8

The compressive demand is:

P_{demand}=120\ \text{kN}

Use Euler buckling:

\displaystyle P_{cr}=\frac{\pi^2EI}{(KL)^2}

Calculate reserve factor. Then estimate reserve if corrosion and ineffective plating reduce $I$ by 20 percent.

Solution

Initial buckling load:

\displaystyle P_{cr}=\frac{\pi^2(200\times10^9)(1.3\times10^{-7})}{(0.8\times1.4)^2}=205\ \text{kN}

Initial reserve factor:

\displaystyle RF_b=\frac{205}{120}=1.71

With a 20 percent reduction in effective $I$ :

P_{cr,loss}=0.80(205)=164\ \text{kN}

Reduced reserve:

\displaystyle RF_{b,loss}=\frac{164}{120}=1.37

Engineering Comment

The simplified column still passes, but the reserve is no longer generous. A real stiffener check should also evaluate attached plating, tripping, end brackets, eccentricity, welding distortion, local plate buckling, and whether the finite-element model uses realistic boundary stiffness.

Plausibility Check

The initial Euler load is 205 kN, giving a reserve factor of 1.71 against 120 kN demand. A 20 percent loss of effective inertia reduces the critical load directly to 164 kN and the reserve to 1.37. That margin can be consumed by eccentricity, tripping, poor end restraint, welding distortion or ineffective attached plating.

Exercise 6: Thickness Survey Release Decision

A hull survey measures a plate required to retain at least:

t_{req}=8.3\ \text{mm}

Four ultrasonic readings in the critical region are:

8.9,\quad 9.2,\quad 8.7,\quad 9.0\ \text{mm}

Use the minimum reading for release. Measurement uncertainty is:

u_t=0.12\ \text{mm}

Coverage factor:

k=2

Expected corrosion rate:

r_c=0.06\ \text{mm/year}

Check whether the plate can be released for 4 years and for 2 years using:

t_{guard}=t_{measured}-ku_t-r_c t_{future}

Solution

Minimum measured thickness:

t_{measured}=8.7\ \text{mm}

For 4 years:

t_{guard,4}=8.7-2(0.12)-0.06(4)=8.22\ \text{mm}

Comparison:

8.22<8.3

The guarded 4-year release fails.

For 2 years:

t_{guard,2}=8.7-2(0.12)-0.06(2)=8.34\ \text{mm}

Comparison:

8.34>8.3

The guarded 2-year release passes narrowly.

Engineering Comment

The arithmetic supports a shorter release interval, not unrestricted service. A responsible survey decision would document the critical location, coating condition, corrosion mechanism, nearby readings, repair threshold, and whether operating restrictions are needed before the next inspection.

Plausibility Check

The minimum measured thickness is 8.7 mm, but the guarded calculation first subtracts 0.24 mm for expanded measurement uncertainty. Four years of expected corrosion subtracts another 0.24 mm, leaving 8.22 mm and failing the 8.3 mm requirement by 0.08 mm. The 2-year case leaves 8.34 mm, only 0.04 mm above the requirement, so the release is narrow and should be documented as such.

Exercise 7: Welded Bracket Fatigue Damage

A welded machinery-foundation bracket is screened for one year of service using three stress-range bins:

Load block	Applied cycles $n_i$	S-N life $N_i$
normal wave vibration	$2.2\times10^6$	$8.0\times10^7$
heavy-weather operation	$3.0\times10^5$	$1.2\times10^7$
rare slam events	$2.0\times10^4$	$1.0\times10^6$

Use Miner’s rule:

\displaystyle D=\sum\frac{n_i}{N_i}

If focused inspection is required before accumulated damage exceeds 0.20, estimate the maximum inspection interval.

Solution

Annual damage:

\displaystyle D=\frac{2.2\times10^6}{8.0\times10^7}+\frac{3.0\times10^5}{1.2\times10^7}+\frac{2.0\times10^4}{1.0\times10^6}

D=0.0275+0.0250+0.0200=0.0725

Maximum interval from the damage limit:

\displaystyle t_{inspect}=\frac{0.20}{0.0725}=2.76\ \text{years}

Engineering Comment

A 2-year inspection interval is more defensible than a 3-year interval because the calculation is a screen. The result depends on hot-spot stress definition, weld detail category, corrosion, residual stress, sea-state scatter, slam counting, and whether the selected NDT method can inspect the actual weld toe.

Plausibility Check

The annual damage contributions are 0.0275, 0.0250 and 0.0200, for total annual damage of 0.0725. A 0.20 inspection trigger divided by that damage rate gives 2.76 years. Rounding down to a 2-year inspection interval is reasonable because 3 years would accumulate about 0.2175 damage and exceed the trigger.

Exercise 8: Crack-Growth Inspection Interval

A repaired deck opening is assessed with:

K_c=60\ \text{MPa}\sqrt{\text{m}}

Geometry factor:

Y=1.15

Nominal tensile stress:

\sigma=150\ \text{MPa}

Critical crack size is estimated from:

\displaystyle a_c=\frac{1}{\pi}\left(\frac{K_c}{Y\sigma}\right)^2

A crack-growth analysis from the reliably detectable crack size to $a_c$ gives:

N_{growth}=9.0\times10^5\ \text{cycles}

Use inspection factor:

F=3

and service cycling:

150000\ \text{cycles/year}

Find the inspection interval.

Solution

Critical crack size:

\displaystyle a_c=\frac{1}{\pi}\left(\frac{60}{1.15(150)}\right)^2=0.0385\ \text{m}=38.5\ \text{mm}

Inspection interval in cycles:

\displaystyle N_{inspect}\leq\frac{N_{growth}}{F}=\frac{9.0\times10^5}{3}=3.0\times10^5\ \text{cycles}

Convert to years:

\displaystyle t_{inspect}=\frac{3.0\times10^5}{150000}=2.0\ \text{years}

Engineering Comment

The calculated interval is not just a calendar number. It assumes the NDT method can reliably find the starting flaw size, the stress range is representative, the repair geometry factor is valid, and operating cycles are counted honestly. If access is poor or crack detectability is uncertain, the interval should be shorter or the repair should be redesigned.

Plausibility Check

The critical crack size is 38.5 mm under the stated toughness, geometry factor and nominal stress. The growth-life factor of 3 limits the inspection interval to 300,000 cycles, which is 2.0 years at 150,000 cycles/year. The result is only defensible if the NDT method can reliably detect the starting flaw size and access permits repeatable inspection.

Exercise 9: Propeller Blade-Passing Frequency Near a Foundation Mode

A local foundation has estimated stiffness:

k=115\ \text{MN/m}

and effective participating mass:

m=18000\ \text{kg}

Use:

\displaystyle f_n=\frac{1}{2\pi}\sqrt{\frac{k}{m}}

A five-bladed propeller turns at:

150\ \text{rpm}

Compute natural frequency, blade-passing frequency, and separation. Then estimate natural frequency if stiffness is increased by 45 percent.

Solution

Natural frequency:

\displaystyle f_n=\frac{1}{2\pi}\sqrt{\frac{115\times10^6}{18000}}=12.7\ \text{Hz}

Shaft frequency:

\displaystyle f_{shaft}=\frac{150}{60}=2.5\ \text{Hz}

Blade-passing frequency:

f_{BPF}=5(2.5)=12.5\ \text{Hz}

Initial separation:

\displaystyle S_f=\frac{|12.7-12.5|}{12.5}=0.016=1.6\%

If stiffness is increased by 45 percent:

f_{n,new}=12.7\sqrt{1.45}=15.3\ \text{Hz}

New separation:

\displaystyle S_{f,new}=\frac{|15.3-12.5|}{12.5}=0.224=22.4\%

Engineering Comment

The original frequency is essentially coincident with blade-passing excitation and should not be released without further evidence. Increasing stiffness appears to move the mode away from the propeller order, but the engineer still needs a mode-shape check, damping estimate, sea-trial vibration data, and a review against other engine, gear, shaft, and electrical excitation frequencies.

Plausibility Check

The original natural frequency is 12.7 Hz and blade-passing frequency is 12.5 Hz, leaving only 1.6 percent separation. Increasing stiffness by 45 percent raises natural frequency by the square root of 1.45, giving 15.3 Hz and 22.4 percent separation from blade passing. That improvement is plausible, but it must not create another coincidence with engine, gear, shaft or electrical orders.

Exercise 10: Guarded Repair Release Margin

A repaired longitudinal detail has nominal allowable stress:

C_{nom}=165\ \text{MPa}

Nominal demand under the proposed unrestricted loading condition:

D_{nom}=137\ \text{MPa}

Capacity uncertainty:

u_C=7\ \text{MPa}

Demand uncertainty:

u_D=6\ \text{MPa}

Use coverage factor:

k=2

and:

C_g=C_{nom}-ku_C

D_g=D_{nom}+ku_D

\displaystyle MS_g=\frac{C_g}{D_g}-1

Check the guarded margin. Then check a restricted loading condition that reduces nominal demand by 10 percent.

Solution

Guarded capacity:

C_g=165-2(7)=151\ \text{MPa}

Guarded unrestricted demand:

D_g=137+2(6)=149\ \text{MPa}

Guarded unrestricted margin:

\displaystyle MS_g=\frac{151}{149}-1=0.0134=1.34\%

For the restricted loading condition:

D_{nom,restricted}=0.90(137)=123.3\ \text{MPa}

Guarded restricted demand:

D_{g,restricted}=123.3+2(6)=135.3\ \text{MPa}

Guarded restricted margin:

\displaystyle MS_{g,restricted}=\frac{151}{135.3}-1=0.116=11.6\%

Engineering Comment

The unrestricted release is technically positive but too thin to be convincing for a repaired hull detail. The restricted condition has a more defensible guarded margin, but it still needs repair traceability, inspection access, operating limits, and a planned follow-up survey. A small positive stress margin should not be confused with long-term hull integrity.

Plausibility Check

The guarded unrestricted capacity is 151 MPa and guarded demand is 149 MPa, giving only 1.34 percent margin. Reducing nominal demand by 10 percent lowers the guarded demand to 135.3 MPa and improves the guarded margin to 11.6 percent. That supports a restricted release rather than treating the unrestricted case as robust.

Exercise 11: Residual Section Modulus After Deck Wastage

A hull-girder screen originally used deck section modulus:

Z_{deck,0}=0.560\ \text{m}^3

The deck extreme-fiber distance from the neutral axis is:

y_{deck}=3.20\ \text{m}

A thickness survey finds broad deck wastage over an effective deck breadth:

b=8.0\ \text{m}

with average thickness loss:

\Delta t=1.2\ \text{mm}

The affected deck material is approximately:

y=3.0\ \text{m}

from the neutral axis. Estimate the lost second moment of area using:

\Delta I=A_{loss}y^2

where:

A_{loss}=b\Delta t

Then estimate the reduced deck section modulus and deck stress under:

M_d=72.8\ \text{MN m}

Compare with allowable stress:

\sigma_{allow}=155\ \text{MPa}

Solution

Lost deck area:

A_{loss}=8.0(0.0012)=0.0096\ \text{m}^2

Lost second moment of area:

\Delta I=0.0096(3.0)^2=0.0864\ \text{m}^4

Equivalent section-modulus loss at the deck:

\displaystyle \Delta Z=\frac{\Delta I}{y_{deck}}=\frac{0.0864}{3.20}=0.0270\ \text{m}^3

Reduced deck section modulus:

Z_{deck}=0.560-0.0270=0.533\ \text{m}^3

Deck bending stress:

\displaystyle \sigma_{deck}=\frac{72.8\times10^6}{0.533}=136.6\ \text{MPa}

Stress utilization:

\displaystyle U=\frac{136.6}{155}=0.881

The residual section still passes this simplified elastic stress screen.

Engineering Comment

The result is acceptable only as a broad corrosion deduction screen. Real residual strength assessment needs the surveyed thickness grid, effective breadth rules, local buckling, openings, wasted longitudinals, neutral-axis shift, class corrosion criteria and whether the loading condition is still valid for the vessel’s operating restriction.

Plausibility Check

The wastage removes about $4.8\%$ of the original deck section modulus:

\displaystyle \frac{0.0270}{0.560}=4.8\%

The deck stress rises from about $130\ \text{MPa}$ in the original screen to $136.6\ \text{MPa}$ . That increase is significant but not enough to exceed the $155\ \text{MPa}$ allowable in this simplified calculation.

Exercise 12: Side-Shell Pressure Panel Stress With Wastage

A side-shell plate panel is checked as a simply supported strip between longitudinal stiffeners. The design pressure is:

p=38\ \text{kPa}

Stiffener spacing is:

s=0.45\ \text{m}

Current plate thickness is:

t=7.0\ \text{mm}

Use a one-way strip screen:

\displaystyle M'=\frac{ps^2}{8}

where $M'$ is moment per meter width, and:

\displaystyle \sigma=\frac{6M'}{t^2}

Compare the stress with allowable stress:

\sigma_{allow}=155\ \text{MPa}

Then estimate stress if future wastage reduces thickness to:

t=6.4\ \text{mm}

Solution

Moment per meter width:

\displaystyle M'=\frac{38000(0.45)^2}{8}=962\ \text{N m/m}

Stress at $7.0\ \text{mm}$ :

\displaystyle \sigma_{7.0}=\frac{6(962)}{0.007^2}=117.8\ \text{MPa}

Utilization:

\displaystyle U_{7.0}=\frac{117.8}{155}=0.760

Stress at $6.4\ \text{mm}$ :

\displaystyle \sigma_{6.4}=\frac{6(962)}{0.0064^2}=140.9\ \text{MPa}

Future utilization:

\displaystyle U_{6.4}=\frac{140.9}{155}=0.909

Both cases pass the simplified strip stress screen, but the future thickness leaves much less local reserve.

Engineering Comment

This pressure-panel screen does not replace a class plate rule or finite-element check. It is useful because it shows the squared sensitivity to thickness. A survey release should connect the panel stress to actual boundary restraint, corrosion pattern, coating condition, local dents, weld toes, adjacent stiffener condition and future wastage allowance.

Plausibility Check

The pressure and stiffener spacing produce about $962\ \text{N m/m}$ of strip moment. Reducing thickness from $7.0$ to $6.4\ \text{mm}$ increases stress by:

\displaystyle \left(\frac{7.0}{6.4}\right)^2=1.20

so the stress rises from $117.8$ to about $140.9\ \text{MPa}$ . That matches the expected thickness-squared behavior.

Exercise 13: Bottom Slamming Patch on a Longitudinal Stiffener

A bottom longitudinal stiffener is checked for a simplified slamming pressure patch. The design slamming pressure is:

p_s=110\ \text{kPa}

The stiffener spacing is:

s=0.50\ \text{m}

The stiffener span between transverse floors is:

L=1.40\ \text{m}

The high-pressure patch is centered on the span and has length:

a=0.90\ \text{m}

Treat the stiffener as simply supported and use line load:

w=p_ss

For a centered partial-span uniform load, use:

\displaystyle M_{max}=\frac{wa(2L-a)}{8}

The current stiffener section modulus is:

Z=82\ \text{cm}^3

Allowable bending stress for this screen is:

\sigma_{allow}=155\ \text{MPa}

Calculate load, reaction, bending stress and utilization. Then estimate utilization if corrosion and ineffective attached plating reduce section modulus by $15\%$ .

Solution

Convert pressure:

p_s=110000\ \text{Pa}

Line load on the stiffener:

w=p_ss=110000(0.50)=55000\ \text{N/m}

Total patch load:

F=wa=55000(0.90)=49500\ \text{N}=49.5\ \text{kN}

Symmetric support reaction:

\displaystyle R=\frac{F}{2}=24.75\ \text{kN}

Maximum bending moment:

\displaystyle M_{max}=\frac{55000(0.90)(2(1.40)-0.90)}{8}

M_{max}=11756\ \text{N m}=11.76\ \text{kN m}

Convert section modulus:

Z=82\ \text{cm}^3=82\times10^{-6}\ \text{m}^3

Bending stress:

\displaystyle \sigma=\frac{M_{max}}{Z}

\displaystyle \sigma=\frac{11756}{82\times10^{-6}}=143.4\ \text{MPa}

Utilization:

\displaystyle U=\frac{143.4}{155}=0.925

The current stiffener passes this simplified screen, but the margin is not large.

With a $15\%$ section-modulus reduction:

Z_{loss}=0.85(82)=69.7\ \text{cm}^3

Z_{loss}=69.7\times10^{-6}\ \text{m}^3

Reduced-section stress:

\displaystyle \sigma_{loss}=\frac{11756}{69.7\times10^{-6}}=169\ \text{MPa}

Reduced-section utilization:

\displaystyle U_{loss}=\frac{169}{155}=1.09

The corrosion-reduced stiffener fails this simplified slamming screen.

Engineering Comment

Slamming checks can be controlled by local patch size, not only by average hydrostatic pressure. A stiffener that passes in calm pressure service may become marginal under impact pressure after wastage, ineffective attached plating or poor end brackets. A real release needs the rule pressure basis, load patch definition, boundary condition, attached plating effectiveness, bracket details, local dents, survey readings and operating restrictions for heavy-weather or high-speed service.

Plausibility Check

The line load is $55\ \text{kN/m}$ and the impact patch is shorter than the span, so the total load is about $50\ \text{kN}$ and each support reaction is about $25\ \text{kN}$ . Current bending stress is near the allowable at $143\ \text{MPa}$ , so a $15\%$ loss of section modulus should raise stress by about $1/0.85=1.18$ times to roughly $169\ \text{MPa}$ , which matches the calculation.

Exercise 14: Combined Bending and Shear Equivalent Stress

A hull web connection is screened at a section where the global bending calculation gives normal stress:

\sigma_x=132\ \text{MPa}

The shear-flow calculation gives average web shear stress:

\tau=44.6\ \text{MPa}

Use the plane-stress von Mises screen:

\sigma_{vm}=\sqrt{\sigma_x^2+3\tau^2}

The equivalent-stress allowable for this simplified release check is:

\sigma_{allow}=155\ \text{MPa}

Calculate equivalent stress and utilization. Then repeat the calculation if local load-path redistribution increases shear stress by $10\%$ .

Solution

Initial equivalent stress:

\sigma_{vm}=\sqrt{132^2+3(44.6)^2}=152.9\ \text{MPa}

Initial utilization:

\displaystyle U=\frac{152.9}{155}=0.987

The initial combined-stress screen passes, but only narrowly.

With $10\%$ higher shear:

\tau_{10}=1.10(44.6)=49.1\ \text{MPa}

Updated equivalent stress:

\sigma_{vm,10}=\sqrt{132^2+3(49.1)^2}=157.0\ \text{MPa}

Updated utilization:

\displaystyle U_{10}=\frac{157.0}{155}=1.01

The redistributed load case fails the equivalent-stress screen.

Engineering Comment

This exercise shows why bending and shear should not always be reviewed as isolated pass/fail numbers. A web detail can look acceptable under global bending and average shear separately, while the combined equivalent stress has almost no reserve. A real release would also check local web openings, weld toes, bracket geometry, shear lag, finite-element stress extraction, corrosion deductions and whether von Mises stress is the correct acceptance basis for the detail.

Plausibility Check

The normal stress is already close to the allowable, so adding a shear term should push the equivalent stress near the limit. The initial equivalent stress of $152.9\ \text{MPa}$ is just below $155\ \text{MPa}$ , while a modest $10\%$ shear increase raises it to $157.0\ \text{MPa}$ . That sensitivity is plausible for a nearly fully utilized hull web connection.

Exercise 15: Drydock Keel-Block Bearing Pressure

A vessel entering drydock has supported mass:

m=3200\ \text{t}

The docking plan uses:

N=28

active keel blocks. Use:

g=9.81\ \text{m/s}^2

and a reaction concentration factor:

C_R=1.25

to account for uneven block contact. Each block has contact footprint:

b=0.30\ \text{m},\quad l=1.20\ \text{m}

The simplified allowable bearing pressure for the support interface is:

p_{allow}=4.0\ \text{MPa}

Calculate average block reaction, design block reaction, bearing pressure and utilization. Then repeat the pressure check if two blocks are ineffective and only $26$ blocks carry load.

Solution

Convert supported mass to weight:

W=3200(1000)(9.81)=31.392\times10^6\ \text{N}=31.4\ \text{MN}

Average reaction per block:

\displaystyle R_{avg}=\frac{W}{N}=\frac{31.392\times10^6}{28}=1.12\ \text{MN}

Design reaction per block:

R_d=C_RR_{avg}=1.25(1.12)=1.40\ \text{MN}

Contact area per block:

A_b=bl=0.30(1.20)=0.360\ \text{m}^2

Bearing pressure:

\displaystyle p_b=\frac{R_d}{A_b}=\frac{1.40\times10^6}{0.360}=3.89\ \text{MPa}

Utilization:

\displaystyle U_p=\frac{3.89}{4.0}=0.973

The nominal docking plan passes, but it is close to the simplified pressure limit.

If only $26$ blocks are effective:

\displaystyle R_{avg,26}=\frac{31.392\times10^6}{26}=1.21\ \text{MN}

R_{d,26}=1.25(1.21)=1.51\ \text{MN}

\displaystyle p_{b,26}=\frac{1.51\times10^6}{0.360}=4.19\ \text{MPa}

Updated utilization:

\displaystyle U_{p,26}=\frac{4.19}{4.0}=1.05

The ineffective-block case fails the simplified bearing-pressure screen.

Engineering Comment

A drydock plan is a structural load case, not just a logistics drawing. The keel blocks, side blocks, ballast state, hull girder bending, local keel structure, shell thickness, previous repairs and block-height survey must be consistent. A support plan that passes only when every block is effective should be treated as fragile unless the yard can prove block contact, load distribution and vessel condition.

Plausibility Check

The total supported weight is about $31\ \text{MN}$ , so a 28-block plan gives roughly $1.1\ \text{MN}$ per block before concentration factors. Applying a $1.25$ factor raises the checked pressure to $3.89\ \text{MPa}$ , just below a $4.0\ \text{MPa}$ limit. Losing two effective blocks increases pressure to $4.19\ \text{MPa}$ , so failure under the degraded support case is plausible.

Exercise 16: Hull Torsional Shear Flow With Wastage

A ship hull is screened as a thin-walled closed box under torsional moment from asymmetric loading and quartering seas. The checked closed-cell area is:

A_m=12.0\ \text{m}^2

The nominal torsional moment is:

T=13.0\ \text{MN m}

The original effective plating thickness along the torsional shear path is:

t_0=8.0\ \text{mm}

A survey and future-wastage allowance reduce the guarded thickness to:

t_g=7.0\ \text{mm}

The release screen applies torsional load factor:

\gamma_T=1.15

and allowable shear stress:

\tau_{allow}=85\ \text{MPa}

Use the simplified closed-section shear-flow relation:

\displaystyle q=\frac{T}{2A_m}

Calculate nominal shear flow, nominal shear stress at $t_0$ , guarded torsional moment, guarded shear stress at $t_g$ , shear-stress margin, maximum nominal torsional moment allowed by the guarded screen, and the required torsional-moment reduction.

Solution

Nominal shear flow:

\displaystyle q=\frac{13.0\times10^6}{2(12.0)}=541{,}667\ \text{N/m}

q=542\ \text{kN/m}

Nominal shear stress at the original thickness:

\displaystyle \tau_0=\frac{q}{t_0}=\frac{541{,}667}{0.0080}=67.7\ \text{MPa}

Guarded torsional moment:

T_g=\gamma_TT=1.15(13.0)=14.95\ \text{MN m}

Guarded shear flow:

\displaystyle q_g=\frac{14.95\times10^6}{2(12.0)}=622{,}917\ \text{N/m}

Guarded shear stress at the wasted thickness:

\displaystyle \tau_g=\frac{622{,}917}{0.0070}=89.0\ \text{MPa}

Shear-stress margin:

M_\tau=85.0-89.0=-4.0\ \text{MPa}

The guarded case fails the simplified torsional shear screen.

Maximum nominal torsional moment allowed by the guarded screen:

\displaystyle T_{max}=\frac{2A_mt_g\tau_{allow}}{\gamma_T}

\displaystyle T_{max}=\frac{2(12.0)(0.0070)(85\times10^6)}{1.15}=12.4\ \text{MN m}

Required torsional-moment reduction:

\Delta T=13.0-12.4=0.6\ \text{MN m}

The vessel should not be released for the checked asymmetric loading case without derating the torsional load, confirming greater effective thickness, adding structural reinforcement, or replacing this screen with an approved direct-strength assessment.

Engineering Comment

Closed-box torsion is a load-path check. A low nominal shear stress can become marginal when load factor and measured wastage are applied together. Real release evidence should verify the effective closed cell, openings, deck and side-shell continuity, corrosion map, hatch-corner details, shear-lag effects, local buckling, warping restraint, loading manual condition and whether asymmetric ballast, cargo or wave loading matches the checked torsional moment.

Plausibility Check

The nominal $8.0\ \text{mm}$ path gives $67.7\ \text{MPa}$ , safely below the $85\ \text{MPa}$ screen. Applying the load factor and reducing thickness to $7.0\ \text{mm}$ increases stress to $89.0\ \text{MPa}$ , just above the limit. A required reduction of about $0.6\ \text{MN m}$ is therefore consistent with a narrow guarded failure rather than a completely unsuitable section.

Exercise 17: Watertight Closure Load After Local Flooding

A watertight door in an internal bulkhead is checked for a local flooding case. The estimated water head at the door centroid is:

h=1.40\ \text{m}

Use seawater density:

\rho=1025\ \text{kg/m}^3

and:

g=9.81\ \text{m/s}^2

The door clear opening is:

b=0.90\ \text{m},\quad H=1.80\ \text{m}

The closure has:

n=4\ \text{dogs}

The flooded-condition load factor is:

\gamma_f=1.20

and the dog-load concentration factor is:

C_d=1.35

Each dog has allowable design load:

R_{dog,allow}=8.5\ \text{kN}

Compute hydrostatic pressure at the centroid, total door force, factored door force, design force per dog and release decision. Then repeat the dog-load check if the closure is redesigned with six effective dogs.

Solution

Hydrostatic pressure at the door centroid:

p=\rho gh

p=(1025)(9.81)(1.40)=14{,}077\ \text{Pa}

p=14.1\ \text{kPa}

Door area:

A=bH=0.90(1.80)=1.62\ \text{m}^2

Total unfactored hydrostatic force:

F=pA

F=(14{,}077)(1.62)=22{,}805\ \text{N}

F=22.8\ \text{kN}

Factored door force:

F_d=\gamma_fF

F_d=1.20(22.8)=27.4\ \text{kN}

Design force per dog, including concentration:

\displaystyle R_{dog}=\frac{C_dF_d}{n}

\displaystyle R_{dog}=\frac{1.35(27.4)}{4}=9.24\ \text{kN}

The four-dog closure fails the simplified dog-capacity screen:

9.24\ \text{kN}>8.5\ \text{kN}

With six effective dogs:

\displaystyle R_{dog,6}=\frac{1.35(27.4)}{6}=6.16\ \text{kN}

The six-dog case passes the dog-capacity screen:

6.16\ \text{kN}<8.5\ \text{kN}

The closure should not be released in the four-dog configuration for this flooding case without stronger dogs, more effective dogs, a reduced accepted head, or approved detailed closure evidence.

Engineering Comment

Watertight integrity is structural integrity. A door that looks serviceable in normal operation can fail a flooding head if dogs, hinges, seals, frames, coamings, corrosion, misalignment or operating procedure are weak. The release record should state the flooded-compartment scenario, water head, closure type, dog engagement evidence, hinge/frame condition, gasket compression, closing time, inspection status and damage-control consequence.

Plausibility Check

A $1.4\ \text{m}$ seawater head gives about $14\ \text{kPa}$ , which is modest pressure but acts over a $1.62\ \text{m}^2$ door. The resulting force is tens of kilonewtons. Splitting that force over four dogs with a concentration factor gives a per-dog load just above the allowable; adding two effective dogs lowers the load by one third, so the pass/fail change is plausible.

Exercise 18: Deck Cargo Grillage Reaction Gate

A workboat is asked to carry a temporary deck equipment module. The module mass is:

m=24\ \text{t}

The sea-fastening review uses a vertical dynamic factor:

\gamma_v=1.35

The module is supported by:

n=4

grillage stools. Because the support geometry is not perfectly even, apply a load-sharing concentration factor:

C_s=1.25

Each stool bears on a rectangular footprint:

a=0.20\ \text{m},\quad b=0.24\ \text{m}

The simplified local deck bearing limit is:

p_{allow}=2.6\ \text{MPa}

The supporting deck beam below each stool has an allowable concentrated reaction:

R_{beam,allow}=92\ \text{kN}

Calculate factored cargo load, design reaction per stool, footprint bearing pressure, beam reaction margin and release decision. Then repeat the check if the support is changed to six effective stools with the same footprint.

Solution

Factored vertical cargo load:

F_d=\gamma_vmg

F_d=1.35(24000)(9.81)=317844\ \text{N}=317.8\ \text{kN}

Nominal reaction per stool:

\displaystyle R_{nom}=\frac{317.8}{4}=79.5\ \text{kN}

Design reaction per stool with load-sharing concentration:

R_s=C_sR_{nom}=1.25(79.5)=99.3\ \text{kN}

Stool footprint area:

A_s=ab=0.20(0.24)=0.048\ \text{m}^2

Bearing pressure:

\displaystyle p_s=\frac{R_s}{A_s}=\frac{99300}{0.048}=2.07\ \text{MPa}

The local bearing-pressure screen passes because:

2.07\ \text{MPa}<2.6\ \text{MPa}

But the deck-beam reaction margin is:

M_R=R_{beam,allow}-R_s=92.0-99.3=-7.3\ \text{kN}

The four-stool layout should not be released because the supporting beam reaction fails, even though the footprint bearing pressure passes.

With six effective stools:

\displaystyle R_{nom,6}=\frac{317.8}{6}=53.0\ \text{kN}

R_{s,6}=1.25(53.0)=66.2\ \text{kN}

The revised bearing pressure is:

\displaystyle p_{s,6}=\frac{66200}{0.048}=1.38\ \text{MPa}

and the revised beam reaction margin is:

M_{R,6}=92.0-66.2=25.8\ \text{kN}

The six-stool layout passes both simplified screens, provided the stools actually engage the deck, the grillage load path is continuous and the sea-fastening design is approved for horizontal and overturning loads.

Engineering Comment

Deck cargo checks should follow the load path from cargo mass to sea fastening, stool, deck plating, stiffener, beam, girder, bulkhead and hull girder. A footprint pressure pass does not prove the supporting grillage has enough reaction capacity. The release record should state cargo mass, center of gravity, acceleration basis, lashings or welds, load sharing, stool fit-up, deck thickness, supporting member capacity, weld/NDT evidence, corrosion state, route weather limit and inspection after the voyage.

Temporary cargo is especially risky because the structure may be treated as a flat deck surface rather than as a directional grillage. If stools are not shimmed, surveyed or preloaded consistently, one support can carry more load than the drawing assumes.

Plausibility Check

A $24\ \text{t}$ module has static weight near $235\ \text{kN}$ . Applying a $1.35$ vertical factor raises the design load to about $318\ \text{kN}$ , so four supports carry roughly $80\ \text{kN}$ each before load-sharing concentration. The $1.25$ concentration pushes each checked reaction to about $99\ \text{kN}$ , just above a $92\ \text{kN}$ beam limit. Adding two effective supports reduces the checked reaction to $66.2\ \text{kN}$ , so the pass/fail change is consistent with the load path.

Review Checklist

Before accepting a marine structural exercise result, check:

load basis: still-water, wave, slamming impact, tank, docking, accidental, fatigue, or repair;
structural boundary: hull girder, web connection, local panel, stiffener, bracket, foundation, weld, keel block, watertight closure, or repaired opening;
stress definition: nominal, hot-spot, notch, equivalent, or measured;
corrosion basis: measured thickness, uncertainty, trend, coating condition, and future exposure;
pressure or support basis: hydrostatic head, dynamic factor, slamming patch size, flooding head, docking reaction, contact area, load duration, and boundary condition;
torsion basis: effective enclosed area, continuous shear path, openings, wastage, asymmetric loading, and direct-strength evidence;
fatigue basis: stress range, cycle count, S-N curve, weld detail, environment, and inspection method;
deck cargo basis: cargo mass, acceleration factor, stool fit-up, load sharing, footprint pressure, grillage reaction, lashings and route restriction;
validation evidence: class basis, loading record, docking plan, finite-element model, survey data, NDT, strain gauge, vibration measurement, or service feedback;
watertight integrity: closure dogs, hinges, seals, frame condition, gasket compression, closing procedure and damage-control consequence;
decision: unrestricted release, restricted release, shorter inspection interval, repair redesign, further analysis, or hold point.

Common Mistakes

Treating a hull girder, local panel, stiffener, bracket, weld, foundation and docking block as if the same stress definition and acceptance basis applied to all.
Using nominal section properties while measured thickness, corrosion wastage, ineffective plating, coating condition or repair history has changed the real section.
Checking global bending stress and local shear separately when combined equivalent stress or local redistribution controls the detail.
Treating hull torsion as negligible because nominal shear is low, while openings, wastage, asymmetric loading or load factors consume the margin.
Applying still-water or hydrostatic loads to slamming, docking, grounding, tank, fatigue or accidental cases without changing the load basis.
Accepting deck cargo from footprint pressure alone while support reaction, load sharing, lashings, welds and route accelerations are unverified.
Accepting a plate or stiffener buckling screen without imperfection, boundary condition, corrosion, attached plating and class-rule evidence.
Using a fatigue S-N curve without matching weld class, hot-spot stress, environment, cycle count, inspection method and repair consequence.
Setting a crack inspection interval from average growth while ignoring detection threshold, uncertainty, fracture toughness and access constraints.
Treating drydock blocks as uniformly effective when block height, contact, ballast sequence, hull deflection and local keel structure can concentrate load.
Treating a watertight door or hatch as available because it closes, while dog load, seal compression, hinge condition and flooded head are unverified.
Releasing a repair from calculated margin alone when repair drawing, NDT, fit-up, weld procedure, coating and operating restrictions are still open.
Reporting “passes” without stating whether the action is unrestricted operation, restricted operation, retest, repair redesign, shorter inspection or hold point.

The strongest answer is not the one with the most digits. It is the answer that states what can be trusted, what remains uncertain, and what engineering action follows.

REF