16-Civ-A2 Elementary Structural Design · May 2013

Question 1 of 7: A1 — Moments of Resistance of a Built-Up Channel-and-Plate Section

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

Notes on this paper

Paper format. National Examinations, May 2013 — 98-Civ-A2 Elementary Structural Design, 3-hour duration, CLOSED BOOK (handbooks and textbooks permitted, no notes). Seven questions are printed: Part A (A1–A3, steel), Part B (B1–B3, reinforced concrete) and Part C (C1, timber). The candidate answers two from Part A, two from Part B and the one question in Part C — five solutions in total, all of equal value. All seven are solved here, because the set is a study resource. All loads shown on the figures are unfactored; NBCC load combination 1.25D + 1.5L is applied throughout. Solutions are to CSA-S16 (steel), CAN/CSA-A23.3 (concrete) and CSA-O86 (timber), as the paper directs.

Reference texts.

CSA S16, Design of Steel Structures — Clauses 11 (classification), 13.3 (compression), 13.5 (bending), 13.8 (beam-columns), 13.13 (welds).
CISC, Handbook of Steel Construction — section-property tables for C-shapes, W-shapes and hollow structural sections.
CSA A23.3, Design of Concrete Structures — Clauses 10 (flexure and axial load) and 11 (shear, general and simplified methods).
Brzev & Pao, Reinforced Concrete Design: A Practical Approach — Chapters 3–6 and 9.
CSA O86, Engineering Design in Wood — Clause 6 (sawn lumber), Tables 6.3.1B and 6.4.5.
Canadian Wood Council, Wood Design Manual — beam selection and modification factors.

Check — assumptions carried through this paper. (1) The 60 kN, 50 kN, 40 kN, 200 kN, 250 kN and 300 kN forces on Figures A1–B3 are specified live loads, so a load factor of 1.5 is used; concrete self-weight is factored at 1.25. (2) Steel members are assumed continuously laterally supported unless an unbraced length is printed — no unbraced lengths appear on this paper. (3) In Figure B2 the "65 typical" dimension is taken as the distance from the tension face to the centroid of the bottom bars, and the 15M stirrups are assumed at 300 mm on centre in each stem (the figure gives no spacing). (4) CSA O86 Table 6.3.1B Beam-and-Stringer specified strengths for D.Fir-L No.1 are used in C1.

Question 1: A1 — Moments of Resistance of a Built-Up Channel-and-Plate Section (10 + 10 marks)

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. Two C310×45 channels placed back to back (webs in contact on the vertical centreline, flanges projecting outward), welded to a 20 mm × 260 mm plate on the underside. Steel G40.21 350W, $F_y = 350$ MPa, $E = 200\,000$ MPa, $\phi = 0.90$.

Property (per C310×45, CISC)	Value
Area $A$	5 690 mm²
Depth $d$	305 mm
Flange width $b_f$ / thickness $t_f$	80.5 mm / 12.7 mm
Web thickness $w$	13.0 mm
$I_x$ (own horizontal axis)	67.4 × 10⁶ mm⁴
$I_y$ (own vertical axis)	2.13 × 10⁶ mm⁴
$\bar{x}$ from back of web	17.1 mm

Find. The factored moments of resistance $M_{rx}$ and $M_{ry}$ of the fabricated section about its two centroidal axes.

Figure A1 — fabricated section: two C310×45 channels back to back (overall 305 mm deep × 161 mm wide) welded to a 20 × 260 mm cover plate. Overall depth 325 mm.

Approach. Classify the plate elements to CSA S16 Clause 11, locate the elastic centroid and compute $I_x$, $I_y$ by the parallel-axis theorem, then — because every element proves Class 1 — compute the plastic moduli $Z_x$, $Z_y$ and take $M_r = \phi Z F_y$ about each axis.

Classify the plate elements (S16 Cl. 11, Table 2). The channel flanges are outstanding elements supported along one edge: $$\frac{b}{t}=\frac{80.5}{12.7}=6.34 \quad\text{versus}\quad \frac{145}{\sqrt{F_y}}=\frac{145}{\sqrt{350}}=7.75$$ The two webs in contact act as a single 26 mm plate over the clear depth $305-2(12.7)=279.6$ mm, giving $h/w = 279.6/26 = 10.8$ against a Class 1 limit of $1100/\sqrt{350}=58.8$. The cover plate projects $(260-161)/2 = 49.5$ mm past the channel flanges, so $b/t = 49.5/20 = 2.48$. Every ratio is far inside its Class 1 limit, so $$\boxed{\text{the section is Class 1 about both axes}}$$
Locate the elastic centroid (measured up from the underside of the plate). The plate contributes $A_p = 260(20) = 5\,200$ mm² at $y = 10$ mm; the two channels contribute $2(5\,690) = 11\,380$ mm² at $y = 20 + 305/2 = 172.5$ mm. $$\bar{y}=\frac{\sum A_i y_i}{\sum A_i}=\frac{5\,200(10)+11\,380(172.5)}{16\,580}=121.5\ \text{mm}$$ The overall depth is $20 + 305 = 325$ mm, so the extreme fibres lie 121.5 mm below and 203.5 mm above the neutral axis.
Second moment of area about x–x. Applying the parallel-axis theorem to each part, $$I_x=\left[\frac{260(20)^3}{12}+5\,200(111.5)^2\right]+\left[2(67.4\times10^6)+11\,380(51.0)^2\right]$$ $$I_x=64.9\times10^6+164.4\times10^6=229\times10^6\ \text{mm}^4$$ The elastic section moduli follow as $S_{x,\text{bot}} = 229\times10^6/121.5 = 1\,886\times10^3$ mm³ and $S_{x,\text{top}} = 229\times10^6/203.5 = 1\,127\times10^3$ mm³, the smaller of which controls an elastic check.
Plastic neutral axis and $Z_x$. For a Class 1 section the resistance is set by equal-area division, not by the centroid. With a total idealised area of 16 559 mm², each half is 8 280 mm². The plate alone supplies 5 200 mm², the two channel bottom flanges $2(80.5)(12.7) = 2\,045$ mm², leaving 1 035 mm² to be taken from the 26 mm-thick webs, a height of $1\,035/26 = 39.8$ mm. Hence the plastic neutral axis sits $$y_{PNA}=20+12.7+39.8=72.5\ \text{mm}$$ above the underside of the plate. Summing the first moments of all areas about that axis, $$Z_x=\sum A_i\,|y_i-y_{PNA}|=1\,691\times10^3\ \text{mm}^3$$
Moment of resistance about x–x. With full lateral support the Class 1 resistance is the plastic moment reduced by $\phi$: $$M_{rx}=\phi Z_x F_y=0.90\,(1\,691\times10^3)(350)$$ $$\boxed{M_{rx}=533\ \text{kN}\cdot\text{m}}$$ For comparison, an elastic (Class 3) evaluation on the smaller modulus would give $0.90(1\,127\times10^3)(350) = 355$ kN·m — the shape factor of this monosymmetric section is a substantial 1.50.
Properties about y–y. The section is symmetric about the vertical axis, so no centroid search is needed. The plate bends about its own strong direction and each channel is displaced 17.1 mm from the centreline: $$I_y=\frac{20(260)^3}{12}+2\left[2.13\times10^6+5\,690(17.1)^2\right]=29.3\times10^6+7.59\times10^6=36.9\times10^6\ \text{mm}^4$$ with $S_y = 36.9\times10^6/130 = 284\times10^3$ mm³. The plastic modulus takes each half-plate at its own centroid and each whole channel at $\bar{x}$: $$Z_y=2\left[(130)(20)(65)\right]+2\left[5\,690(17.1)\right]=338\times10^3+195\times10^3=533\times10^3\ \text{mm}^3$$
Moment of resistance about y–y. Weak-axis bending of a Class 1 section cannot buckle laterally, so the plastic moment governs directly: $$M_{ry}=\phi Z_y F_y=0.90\,(533\times10^3)(350)$$ $$\boxed{M_{ry}=168\ \text{kN}\cdot\text{m}}$$ The corresponding elastic value is $0.90(284\times10^3)(350) = 89.4$ kN·m. The cover plate is what makes the weak axis viable: without it $I_y$ would be only 7.6 × 10⁶ mm⁴.

The two answers differ by a factor of about three, which is the practical message of the question: this fabricated girder is efficient in the plane of the web and comparatively weak transversely, so bracing that prevents twisting is essential if the strong-axis capacity is to be realised.

Quantity	Value
Section classification	Class 1, both axes
Total area / overall depth	16 580 mm² / 325 mm
Centroid $\bar{y}$ above plate soffit	121.5 mm
$I_x$ / $Z_x$	229 × 10⁶ mm⁴ / 1 691 × 10³ mm³
$I_y$ / $Z_y$	36.9 × 10⁶ mm⁴ / 533 × 10³ mm³
$M_{rx}$ (about x–x)	533 kN·m
$M_{ry}$ (about y–y)	168 kN·m

Topic. Flexural resistance of a fabricated monosymmetric steel section.

Key relations. Element classification $b/t \le 145/\sqrt{F_y}$ (Class 1 outstand); centroid $\bar{y} = \sum A_iy_i/\sum A_i$; parallel-axis theorem $I = \sum(I_i + A_id_i^2)$; equal-area plastic neutral axis; $M_r = \phi Z F_y$ for Class 1 and 2, $M_r = \phi S F_y$ for Class 3.

Why this works. Classification decides whether the cross-section can reach and hold the fully plastic stress distribution before any plate buckles locally. Because every outstand and web here is stocky, the whole area can yield, and the resistance is the plastic moment: the first moment of area about the equal-area axis multiplied by $F_y$. The equal-area axis is not the elastic centroid on a monosymmetric section — the heavy bottom plate pulls the elastic neutral axis down to 121.5 mm but the plastic axis down to only 72.5 mm, because the plastic distribution needs equal areas, not equal first moments.

Common pitfall. Using $\bar{x} = 17.1$ mm as though the channels were toe-to-toe. Back to back means the flat backs of the webs touch, so each channel centroid is 17.1 mm from the centreline and the flanges point outward — the opposite arrangement would place the centroids at $80.5 - 17.1 = 63.4$ mm and nearly triple $I_y$. The second common slip is reporting $\phi S_x F_y$ after correctly proving the section is Class 1, which throws away a third of the true capacity.

Source. CSA S16 Clauses 11.2 and 13.5; CISC Handbook of Steel Construction, channel property tables; Kulak & Grondin, Limit States Design in Structural Steel, Ch. 5.

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