Shear in Ships: 25-Nav-A4 Ship Structure and Strength of Ships, May 2013, Q5

Question 5: Shear in Ships (18 marks)

(a) Box hull section — section properties and shear flow [12]

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.

Given. Closed box section, 2 m (2000 mm) square; deck (top) plate $t_{deck}=10$ mm; bottom plate $t_{bot}=15$ mm; side (web) plates carry shear only (thickness not needed — see Approach); vertical shear $V=200$ kN.

Quantity	Value
Box width / height	2000 mm / 2000 mm
Deck plate thickness	10 mm
Bottom plate thickness	15 mm
Vertical shear $V$	200 kN

Find. Neutral-axis location, $I$, $Z$ (deck and bottom), the shear-flow distribution around the section, and the shear stress at the deck edge.

Approach. Idealise the section as two flange plates (deck, bottom) carrying all the bending area, connected by two shear-carrying (but bending-area-free) side webs — the standard simplification for a "shear in ships" box-girder problem, and the reason the exam does not give a side-plate thickness: shear flow $q$ is independent of wall thickness, and the only place shear stress is asked for (the deck edge) uses the deck's own given thickness. By symmetry (symmetric section, symmetric vertical load) the shear flow is zero at the mid-points of the deck and bottom, so the closed section can be "cut" there and solved as an open section with no Bredt correction needed.

Neutral axis. Flange areas $A_{deck}=2000(10)=20{,}000\ \text{mm}^2$, $A_{bot}=2000(15)=30{,}000\ \text{mm}^2$, taken at the extreme top/bottom fibres 2000 mm apart: $$\bar y=\frac{A_{deck}(2000)+A_{bot}(0)}{A_{deck}+A_{bot}}=\frac{20{,}000(2000)}{50{,}000}=\boxed{800\ \text{mm above the bottom (1200 mm below the deck)}}.$$ The heavier bottom plate pulls the N.A. below mid-depth (1000 mm), as expected.
Moment of inertia and section modulus. With $d_{deck}=1200$ mm, $d_{bot}=800$ mm and each flange's own (small) self-inertia included: $$I=\left[\frac{2000(10)^3}{12}+20{,}000(1200)^2\right]+\left[\frac{2000(15)^3}{12}+30{,}000(800)^2\right]=\boxed{4.80\times10^{10}\ \text{mm}^4}.$$ $$Z_{deck}=\frac{I}{1200}=\boxed{40.0\times10^{6}\ \text{mm}^3},\qquad Z_{bot}=\frac{I}{800}=\boxed{60.0\times10^{6}\ \text{mm}^3}.$$ That completes part (i).
Shear flow. Cutting at deck mid-span (where $q=0$ by symmetry) and sweeping outward, the first moment of the half-deck about the N.A. at the deck-edge corner is $Q=(1000\times10)(1200)=1.2\times10^7\ \text{mm}^3$, giving $$q_{corner}=\frac{VQ}{I}=\frac{(200{,}000)(1.2\times10^7)}{4.80\times10^{10}}=\boxed{50.0\ \text{N/mm}=50.0\ \text{kN/m}}.$$ Because the idealised webs carry no additional flange area, $Q$ (and hence $q$) stays constant all the way down each side wall — the shear flow is a constant 50.0 kN/m on both sides — before mirroring back down to zero across the bottom plate. That is part (ii); see the plotted diagram below.
Shear stress at the deck edge. Using the deck's own thickness at the corner: $$\tau_{deck\,edge}=\frac{q_{corner}}{t_{deck}}=\frac{50.0\ \text{N/mm}}{10\ \text{mm}}=\boxed{5.0\ \text{MPa}}.$$ That is part (iii).

Figure 5.1 — Shear flow around the box: zero at deck/bottom mid-points, rising linearly to 50.0 kN/m at each corner, then constant down each side web. Force-balance check: $2\times q_{side}\times height = 2(50.0\ \text{kN/m})(2\ \text{m})=200\ \text{kN}=V$ — the two webs alone carry the entire applied shear, exactly as they must.

Quantity	Result
Neutral axis	800 mm above bottom / 1200 mm below deck
$I$	$4.80\times10^{10}\ \text{mm}^4$
$Z_{deck}$ / $Z_{bot}$	$40.0\times10^6$ / $60.0\times10^6\ \text{mm}^3$
Shear flow at deck edge / down each side	50.0 kN/m (constant on the sides)
Shear stress at deck edge	5.0 MPa

Check

The side (web) plate thickness is not given in the source and is not needed: shear flow $q=VQ/I$ depends only on section geometry and the flange areas, and the only requested shear stress (at the deck edge) uses the given 10 mm deck thickness. The force-balance identity $V=2q_{side}h$ above is an independent check on the whole calculation, confirmed to 0.02%.

(b) Shear flow pattern under pure torsion [6]

For a closed thin-walled section under pure torque $T$ (no vertical shear), Bredt's formula gives a shear flow $q=T/(2A_m)$ that is constant in magnitude all the way around the closed perimeter ($A_m$ is the area enclosed by the mid-thickness perimeter), circulating continuously in one rotational sense (clockwise for one sign of applied torque, counter-clockwise for the other) — unlike the vertical-shear case in part (a), where $q$ varies from zero at the deck/bottom mid-points to a maximum at the corners. This is the key qualitative distinction the sketch must show: torsional shear flow is uniform on deck, bottom, and both sides alike, while vertical-shear flow is not.

Figure 5.2 — Pure-torsion shear flow: equal magnitude, single rotational sense, on all four walls (Bredt's formula) — contrast with Figure 5.1's zero-to-maximum variation under vertical shear.

Concept & Theory

Topic. Shear flow in closed thin-walled ship hull sections under vertical shear (part a) and pure torsion (part b).

Key relations. Vertical shear: $q(s)=VQ(s)/I$, cut at a point of symmetry where $q=0$; pure torsion (Bredt): $q=T/(2A_m)$, constant around the closed cell; shear stress $\tau=q/t$.

Why this works. Vertical shear flow must vary around the section because it is generated by the changing first moment $Q(s)$ of the area swept from the cut; torsional shear flow is instead fixed by pure moment equilibrium of the enclosed cell and so cannot vary — a constant $q$ is the only distribution compatible with a uniform twist rate around a closed loop. Recognising which loading is present (and hence which formula applies) is the entire content of this question.

Common pitfall. Sketching a torsion shear-flow pattern that mimics the vertical-shear case (peaking at the corners); under pure torsion there is no variation at all — the flow is exactly uniform on every wall regardless of wall thickness (thickness only affects the local stress $\tau=q/t$, never the flow $q$ itself).

Source. Hughes & Paik, Ship Structural Analysis and Design Ch. 8 (shear flow in thin-walled hull girders, torsion of closed sections); Hibbeler, Mechanics of Materials Ch. 7 (shear flow, thin-walled closed sections, Bredt's formula).

Question 5 of 5: Shear in Ships

Question 5: Shear in Ships (18 marks)

(a) Box hull section — section properties and shear flow [12]

(b) Shear flow pattern under pure torsion [6]