25-Nav-B3 Finite Element Analysis for Ship Structures · May 2016

Question 4 of 7: Strength of Ship Structure

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Exams, May 2016 — 98-Nav-B3, 3 hours, closed book, non-communicating calculator permitted (any five of the seven questions constitute a complete paper, all equal value; all seven answered below for full study coverage).

Reference texts: Tupper, Introduction to Naval Architecture, 5th ed.; Lewis (ed.), Principles of Naval Architecture (PNA), 3 vols.; International Code on Intact Stability (IMO IS Code), 2008; Canada Shipping Act / Transport Canada Marine Safety.

Check: this paper, although listed under Finite Element Analysis for Ship Structures, is headed “98-Nav-B3, Small Commercial Ships”; it is a broad small-craft naval-architecture survey paper — propeller open-water performance, ship stability and the inclining experiment, hull/propulsion selection trade-offs, longitudinal shear/bending of a floating body, structural loads and hull materials, fishing-vessel stability regulation, and main-engine selection for a tug — with no finite-element-analysis content whatsoever. It is solved as the exam it actually is.

Question 4: Strength of Ship Structure (20 marks: a–10, b–10)

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.

Quantity	Symbol	Value
Block length / breadth / depth	$L,\,B,\,d_0$	1.0 m, 0.30 m, 0.15 m
Block mass (uniform, over $L$)	$m_b$	20 kg
Added weight (uniform over central 0.5 m)	$m_a$	10 kg
Fresh water density	$\rho_w$	$1000\ \text{kg}\,\text{m}^{-3}$

Find. The shear force $V(x)$ and bending moment $M(x)$ distributions along the block's length, with all key values labelled.

Approach. This is the classic ship "still-water longitudinal strength" problem in miniature: find the (uniform) draft from overall buoyancy/weight balance, form the net load per unit length by subtracting the local weight distribution from the uniform buoyancy support, then integrate the net load once for shear and again for bending moment, using the free (unsupported) ends as the integration constants.

Equilibrium draft. Total weight $=(m_b+m_a)g=(30)(9.81)=294.3\ \text{N}$. Since the block floats level (the added load is symmetric about mid-length) at uniform draft $d$: $\rho_w g\,B L\,d=294.3\ \Rightarrow\ d=\dfrac{294.3}{1000\times9.81\times0.30\times1.0}=\boxed{0.100\ \text{m}}$ — well within the 0.15 m depth, confirming the block does not swamp.
Load intensities. Block self-weight, uniform over the full length: $q_b=m_bg/L=20(9.81)/1=196.2\ \text{N/m}$. Added weight, uniform over its central 0.5 m: $q_a=m_ag/0.5=10(9.81)/0.5=196.2\ \text{N/m}$. Buoyant support, uniform over the full length at the computed draft: $q_{buoy}=\rho_w g B d=1000(9.81)(0.30)(0.10)=294.3\ \text{N/m}$.
Net load per unit length (upward positive). Over $0\le x<0.25$ and $0.75
(a) Shear force (integrate $q_{net}$, $V(0)=0$ at the free end). $V(x)=\displaystyle\int_0^x q_{net}(\xi)\,d\xi$, piecewise linear: $V(0.25)=98.1(0.25)=\boxed{+24.5\ \text{N}}$; falling linearly through zero at mid-length, $V(0.5)=0$; continuing to $V(0.75)=24.5-98.1(0.5)=\boxed{-24.5\ \text{N}}$; rising back to $V(1.0)=-24.5+98.1(0.25)=0$, confirming the free-end boundary condition.
(b) Bending moment (integrate $V$, $M(0)=0$). $M(x)=\displaystyle\int_0^x V(\xi)\,d\xi$, piecewise parabolic: $M(0.25)=\tfrac12(98.1)(0.25)^2=\boxed{3.07\ \text{N}\!\cdot\!\text{m}}$; peaking at mid-length where $V=0$: $M(0.5)=3.07+\big[24.5(0.25)-\tfrac12(98.1)(0.25)^2\big]=\boxed{6.13\ \text{N}\!\cdot\!\text{m}}$; symmetric back down to $M(0.75)=3.07\ \text{N}\!\cdot\!\text{m}$ and $M(1.0)=0$, again confirming the free ends.

Load, shear-force and bending-moment diagrams along the block's length. The net load reverses sign at $x=0.25$ and $0.75$ m, giving a piecewise-linear shear force (peaking at ±24.5 N) and a piecewise-parabolic, symmetric sagging bending moment (peaking at 6.13 N·m amidships) — a miniature analogue of a ship hull sagging under a concentrated deck cargo.

Location	Shear $V$	Moment $M$
$x=0$ (end)	0	0
$x=0.25$ m	+24.5 N	3.07 N·m
$x=0.50$ m (mid-length)	0	6.13 N·m (maximum)
$x=0.75$ m	−24.5 N	3.07 N·m
$x=1.0$ m (end)	0	0

Topic: Longitudinal still-water shear and bending moment from a non-uniform weight/buoyancy distribution.

Key relations: $dV/dx=q_{net}(x)$; $dM/dx=V(x)$; free (unsupported) ends give $V=M=0$ as natural boundary conditions, which is also a built-in arithmetic check on the whole calculation.

Why this works: A floating body's local buoyancy support does not automatically match its local weight; the difference is exactly the "load" that produces internal shear and bending, the same mechanism (at a much larger scale) that produces hogging/sagging bending moments in an actual ship hull loaded unevenly along its length.

Common pitfall: Assuming the buoyant support is concentrated under the added weight rather than spread uniformly along the whole waterplane at the (single) equilibrium draft — a floating body of uniform cross-section supports load uniformly per unit length, regardless of where the weight sits.

Source: PNA Vol. I, Ch. 6 (Hull girder longitudinal strength); Tupper, Introduction to Naval Architecture, Ch. 3 (Static equilibrium, trim, and simple hull-girder loading).

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