Structural Mechanics: 25-Nav-A4 Ship Structure and Strength of Ships, May 2013, Q3

Question 3: Structural Mechanics (18 marks)

(a) Centroid, moment of inertia and section modulus of a frame section [8]

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. Built-up steel section, all dimensions mm: bottom (frame) flange 200 wide × 25 thick; web 200 deep × 25 thick, centred on the flange; top plate (attached hull plating) 300 wide × 20 thick, on top of the web. Overall depth $=25+200+20=245$ mm.

Element	$b$ (mm)	$t$ (mm)	Area $A$ (mm²)	$y$ to own centroid, from bottom (mm)
Bottom flange	200	25	5000	12.5
Web	25 (thk.)	200 (deep)	5000	125.0
Top plate	300	20	6000	235.0

Find. Centroid $\bar y$ from the bottom, moment of inertia $I$ about the centroidal (neutral) axis, and section modulus at top and bottom fibres.

Figure 3.1 — Frame + hull-plating section; the neutral axis (red) sits below mid-depth because the heavier bottom flange + web area pulls the centroid down against the wide but thin top (hull) plate.

Approach. Split the section into three rectangles, locate the composite centroid by the first-moment-of-area rule, then sum each rectangle's own inertia plus its parallel-axis transfer term.

Composite centroid. Summing $A_i y_i$ over the three rectangles and dividing by total area: $$\bar y=\frac{\sum A_iy_i}{\sum A_i}=\frac{5000(12.5)+5000(125.0)+6000(235.0)}{5000+5000+6000}=\frac{2{,}097{,}500}{16{,}000}=\boxed{131.1\ \text{mm from the bottom}}.$$ That is part (i) — note it sits below mid-depth (122.5 mm) because the heavier, more-concentrated bottom flange+web pulls the axis down against the wide but thin top (hull) plate.
Moment of inertia about the N.A. Each rectangle contributes its own inertia $bt^3/12$ plus $A(\bar y-y_i)^2$: $$I=\sum\left[\frac{bt^3}{12}+A(\bar y-y_i)^2\right] = 152.4\times10^{6}\ \text{mm}^4\ \boxed{(I_{NA}=1.524\times10^{8}\ \text{mm}^4)}.$$ (Bottom-flange term $\approx 70.5\times10^6$, web term $\approx 16.9\times10^6$, top-plate term $\approx 65.0\times10^6\ \text{mm}^4$, each including its small own-inertia contribution.)
Section modulus, top and bottom fibres. With $c_{top}=245-131.1=113.9$ mm and $c_{bot}=131.1$ mm: $$Z_{top}=\frac{I}{c_{top}}=\frac{152.4\times10^6}{113.9}=\boxed{1.338\times10^{6}\ \text{mm}^3},\qquad Z_{bot}=\frac{I}{c_{bot}}=\frac{152.4\times10^6}{131.1}=\boxed{1.163\times10^{6}\ \text{mm}^3}.$$ That is part (iii); the smaller $Z_{bot}$ shows the bottom flange fibre is the more highly stressed extreme fibre for a given applied moment.

Quantity	Result
Centroid $\bar y$ (from bottom)	131.1 mm
$I_{NA}$	$152.4\times10^6\ \text{mm}^4$
$Z_{top}$	$1.338\times10^6\ \text{mm}^3$
$Z_{bot}$	$1.163\times10^6\ \text{mm}^3$

(b) Beam bending — propped cantilever with partial UDL [10]

Given. Propped cantilever, span $L=4$ m: fixed support at A ($x=0$), roller at C ($x=4$ m); UDL $w=20$ kN/m acts only over B–C ($x=2$ to $4$ m), nothing on A–B; $EI=1.0\ \text{kN}\cdot\text{m}^2$ (a deliberately small value chosen for arithmetic convenience, not a realistic steel stiffness — see the Verify note).

Find. (i) The qualitative shape of the $V$, $M$, slope and deflection diagrams; (ii) the numerical deflection at B (the beam's midpoint, $x=2$ m).

Approach. This is a one-degree statically-indeterminate propped cantilever. Use the force (compatibility) method: remove the roller at C to get a primary cantilever fixed at A, compute the tip deflection under the applied UDL using the cantilever deflection tables/direct integration, then find the redundant prop reaction $R_C$ that cancels the tip deflection using the tabulated unit-tip-load cantilever formula $\delta=PL^3/(3EI)$. Everything else (reactions, the deflection at B) follows by superposition of the same two cases.

Primary-structure tip deflection under the UDL alone. For the cantilever fixed at A with UDL $w$ over $[a,L]=[2,4]$ m, integrating $M(x)/EI$ twice from the fixed end (or equivalently, superposing the tabulated cantilever UDL and moment cases) gives the tip ($x=L$) deflection: $$v_{0}(L) = -546.7\ (\text{downward-negative sign convention, } EI=1).$$
Redundant prop reaction. A unit UPWARD load at the tip of the same cantilever gives, from the standard table, $\delta_{CC}=L^3/(3EI)=4^3/3=21.33$. Compatibility (zero deflection at the roller) gives $$R_C=\frac{-v_0(L)}{\delta_{CC}}=\frac{546.7}{21.33}=\boxed{25.6\ \text{kN, upward}}.$$
Support reactions (equilibrium check). Total applied load $=w(L-a)=20(2)=40$ kN, so $$R_A = 40-R_C = 40-25.6=\boxed{14.4\ \text{kN upward}},\qquad M_A=\boxed{17.5\ \text{kN}\cdot\text{m (hogging, fixed end)}}.$$ ($R_A+R_C=14.4+25.6=40$ kN, matching the total load exactly.)
Deflection at the midpoint B ($x=2$ m). Superposing the UDL case and the now-known $R_C$ load and integrating $M(x)/EI$ over $[0,4]$ (cross-checked independently by a 2-element beam-stiffness solve): $$v_B=\boxed{15.83\ \text{m downward}}\quad (EI=1.0\ \text{kN}\cdot\text{m}^2).$$

Figure 3.2 — Qualitative $V$, $M$, slope and deflection shapes. Shear is constant (+14.4 kN) over the unloaded span A–B, then ramps down through zero (at $x\approx2.72$ m) to $-25.6$ kN at C. Moment is hogging at the fixed end ($-17.5$ kN·m), crosses zero between A and B, and peaks in sagging (+16.4 kN·m) near the zero-shear point before returning to zero at the pin. Slope is zero at A (fixed) and non-zero at C (free rotation on the roller); deflection is zero at both supports and everywhere downward in between, with the marked value at the midpoint B.

Quantity	Result
$R_A$ (vertical)	14.4 kN upward
$M_A$ (fixed-end moment)	17.5 kN·m, hogging
$R_C$ (prop reaction)	25.6 kN upward
Deflection at B ($x=2$ m), $EI=1.0\ \text{kN}\cdot\text{m}^2$	15.8 m, downward

Check

$EI=1.0\ \text{kN}\cdot\text{m}^2$ is far below any real steel section's stiffness (a real ship frame would have $EI$ on the order of $10^5\ \text{kN}\cdot\text{m}^2$ or more); the exam states it "for the section" purely to make the compatibility arithmetic tractable, so the resulting deflection number (metres, not millimetres) is a deliberate feature of the artificial stiffness, not a realistic structural response. The midpoint deflection was cross-checked by two independent methods — force-method/Macaulay integration and a 2-element direct-stiffness solve — which agree to within 0.1%.

Concept & Theory

Topic. Section properties of a built-up beam (part a) and the force method for a statically-indeterminate propped cantilever (part b).

Key relations. Parallel-axis theorem $I=\sum(I_{own}+Ad^2)$; section modulus $Z=I/c$; force-method compatibility $\delta_{0}+R\,\delta_{CC}=0$ for a redundant reaction, using tabulated cantilever tip-deflection formulas ($PL^3/3EI$ for a point load, and the UDL/partial-UDL cases) exactly as supplied on the exam's formula sheet.

Why this works. The force method converts an indeterminate structure into a determinate "primary structure" (here, a simple cantilever) plus one unknown redundant force, found from a single compatibility equation; because superposition holds for a linear-elastic beam, the final internal forces and deflections are just the sum of the two determinate cases.

Common pitfall. Using the fixed-fixed beam formulae from the "Fixed End Loads" table (which assume BOTH ends fixed) for a fixed-roller (propped) beam; the roller end has zero moment, not zero rotation, so the cantilever tip-deflection tables (not the fixed-fixed FEM table) are the correct starting point here.

Source. Hibbeler, Mechanics of Materials Ch. 6 (section properties) and Ch. 9 (statically indeterminate beams, force method); exam-supplied formula sheet (cantilever/simply-supported deflection & slope tables).

Question 3 of 5: Structural Mechanics

Question 3: Structural Mechanics (18 marks)

(a) Centroid, moment of inertia and section modulus of a frame section [8]

(b) Beam bending — propped cantilever with partial UDL [10]