16-Civ-A2 Elementary Structural Design · May 2015

Question 2 of 7: A2 — Check of the W460×106 column AB

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

Notes on this paper

Paper format. National Examinations, May 2015 — 98-Civ-A2 Elementary Structural Design, 3-hour duration, closed book (textbooks and design handbooks permitted; one approved Casio or Sharp calculator). Seven questions in three parts: Part A (A1–A3, structural steel, CSA S16), Part B (B1–B3, reinforced concrete, CSA A23.3) and Part C (C1, timber, CSA O86). A candidate answers two from Part A, two from Part B and the one question in Part C — five solutions, all of equal value. All seven are solved here so the set works as a complete study resource.

Reference texts. CSA S16, Design of Steel Structures, with the CISC Handbook of Steel Construction; CSA A23.3, Design of Concrete Structures, with the CAC Concrete Design Handbook; CSA O86, Engineering Design in Wood, with the CWC Wood Design Manual; National Building Code of Canada (load combinations); Kulak & Grondin, Limit States Design in Structural Steel; MacGregor & Bartlett, Reinforced Concrete: Mechanics and Design.

Check — load factors. Note 6 on page 1 states “All loads shown are unfactored”, but the paper never splits the figure loads into dead and live components. Every applied figure load in Parts A and B is therefore treated as live load and factored by 1.5; concrete self-weight, where it is not expressly excluded, would be dead load at 1.25 (NBCC 4.1.3.2, combination 1.25D + 1.5L). Part C names its components, so 1.25D + 1.5L is applied directly there. If an examiner intended a different split, the method below is unchanged — only the numerical factor moves.

Check — figure dimensions. All four figures are hand sketches on page 3; every dimension below was read from the printed figures. Figure A2 reads 80 – 300 – 80 across the bottom plate (overall 460 mm wide) with the 300 mm dimension spanning the two web centrelines, and 250 mm overall depth. Figure B1 reads 3.0 m top, 2.5 m bottom, 2.0 m deep, walls “300 constant” and cover “70 typical”. Figure B2 reads 5 m + 5 m span, 4 m + 4 m column height, roller at A and hinge at D.

Question 2: A2 — Check of the W460×106 column AB (8 + 12 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.

Quantity	Value
Section	W460×106, G40.21 350W (F_y = 350 MPa)
Properties	A = 13 500 mm², Z_x = 2390×10³ mm³, r_x = 190 mm, r_y = 43.1 mm
	I_x = 488×10⁶, I_y = 25.1×10⁶ mm⁴, J = 1450×10³ mm⁴, C_w = 1250×10⁹ mm⁶
Height	8.0 m, fixed base at A, free to sway at the top
Axial loads (unfactored)	450 kN on the column head + 150 kN beam reaction at B
Lateral load (unfactored)	80 kN horizontal, 4.0 m above the base

Find. Whether the W460×106 satisfies the S16 Clause 13.8.2 beam-column interaction checks under the factored loads, and if not, what change makes it satisfactory.

Approach. Factor the loads, build the axial-force and bending-moment diagrams for the cantilever column, classify the section, compute the compressive and flexural resistances for the governing effective lengths, and apply all three Clause 13.8.2 interaction equations.

Factored actions. The column head carries the 450 kN directly and the bolted joint at B delivers the beam reaction of 150 kN (Question A1), so the axial force is constant over the full height: $$ C_f = 1.5\,(450 + 150) = 900~\text{kN} $$ The roller at C cannot restrain the frame horizontally, so column AB acts as a vertical cantilever. The moment is zero from the base to the 4 m level going down from the top, and reaches its maximum at the fixed base: $$ M_f = 1.5\,(80)(4.0) = \boxed{480~\text{kN}\cdot\text{m}} $$
Classify the section (S16 Table 2). Flange: $b/2t = 97/20.6 = 4.71 < 145/\sqrt{350} = 7.75$. Web in combined compression and bending, with $C_y = AF_y = 4725$ kN: $$ \frac{h}{w} = \frac{428}{12.6} = 33.9 \;<\; \frac{1100}{\sqrt{F_y}}\left(1 - 0.39\frac{C_f}{\phi C_y}\right) = 53.9 $$ The section is Class 1, so its full plastic moment may be used and $M_{rx} = \phi Z_x F_y = 0.90(2390\times10^3)(350) = 753~\text{kN}\cdot\text{m}$.
Effective lengths. In the plane of the frame the column is fixed at the base and free to translate at the top, so $K_x = 2.0$: $$ \frac{K L}{r_x} = \frac{2.0(8000)}{190} = 84.2 $$ Out of plane the beam holds the column laterally at B but nothing restrains it in between, so $K_y = 1.0$ over the full 8.0 m: $$ \frac{K L}{r_y} = \frac{8000}{43.1} = 185.6 $$ This is just inside the S16 Clause 10.4.2.1 limit of 200, but it is an extremely slender minor-axis condition.
Compressive resistance as drawn. With $\lambda = (KL/r)\sqrt{F_y/\pi^2 E} = 185.6\sqrt{350/(\pi^2 \cdot 200\,000)} = 2.47$ and $n = 1.34$: $$ C_r = \phi A F_y \left(1 + \lambda^{2n}\right)^{-1/n} = 0.90(13\,500)(350)(1 + 2.47^{2.68})^{-1/1.34} $$ $$ C_r = \boxed{653~\text{kN}} \;<\; C_f = 900~\text{kN} $$ The column fails on minor-axis flexural buckling under axial load alone, before any moment is applied — the axial utilisation is already 1.38. As drawn, the W460×106 is inadequate.
Identify the cure. The deficiency is entirely a minor-axis slenderness problem: the strong-axis resistance is ample and the flexural resistance is barely used. Adding a single line of lateral bracing at mid-height (4.0 m) halves the minor-axis unbraced length: $$ \frac{KL}{r_y} = \frac{4000}{43.1} = 92.8 \quad\Rightarrow\quad \lambda = 1.236,\; C_r = 1991~\text{kN} $$ and about the strong axis $\lambda = 1.121$ gives $C_r = 2241$ kN, so the braced column is governed by the minor axis at $C_r = 1991$ kN.
Lateral-torsional buckling of the braced segment. The critical segment runs from the fixed base to the new brace, over which the moment falls linearly from 480 to zero, giving $\kappa = 0$ and $\omega_2 = 1.75$: $$ M_u = \frac{\omega_2 \pi}{L}\sqrt{E I_y G J + \left(\frac{\pi E}{L}\right)^2 I_y C_w} = 1588~\text{kN}\cdot\text{m} $$ Since $M_u > 0.67 M_p = 561$ kN·m the inelastic branch applies: $$ M_r = 1.15\,\phi M_p\left(1 - \frac{0.28 M_p}{M_u}\right) = 738~\text{kN}\cdot\text{m} \;\le\; \phi M_p = 753~\text{kN}\cdot\text{m} $$
Apply the three interaction checks (S16 13.8.2). For the cross-sectional check $U_{1x} = \omega_1/(1 - C_f/C_{ex})$ with $\omega_1 = 1.0$ and $C_{ex} = \pi^2 E I_x/L^2 = 15\,051$ kN, so $U_{1x} = 1.064$. For checks (b) and (c) the member is part of a sway system, so Clause 13.8.4 permits $U_{1x} = 1.0$: $$ \text{(a)}\quad \frac{900}{4252} + 0.85(1.064)\frac{480}{753} = 0.212 + 0.576 = 0.79 $$ $$ \text{(b)}\quad \frac{900}{1991} + 0.85\frac{480}{753} = 0.452 + 0.542 = 0.99 $$ $$ \text{(c)}\quad \frac{900}{1991} + 0.85\frac{480}{738} = 0.452 + 0.553 = 1.005 $$ With the mid-height brace in place the member sits essentially exactly at unity: the lateral-torsional buckling check (c) exceeds 1.0 by half a percent, which is within the precision of the assumed section properties but should be resolved by moving the brace slightly below mid-height or by adopting the next heavier section.

The verdict is therefore conditional rather than a simple pass. As detailed on the figure the column is not acceptable; with one line of minor-axis bracing at mid-height it is at the limit of acceptability and the design can be signed off only if the brace is shown on the drawings.

Check	As drawn (no minor-axis brace)	With a brace at mid-height
C_f / M_f	900 kN / 480 kN·m
KL/r_y	185.6	92.8
C_r (governing)	653 kN	1991 kN
M_rx, laterally supported	753 kN·m
M_rx including LTB	–	738 kN·m
13.8.2(a) cross-section	0.79 ✓
13.8.2(b) overall member	1.38 ✗	0.99 ✓
13.8.2(c) lateral-torsional	✗	1.005 (at the limit)
Verdict	Inadequate	Acceptable, no reserve

Topic. Beam-columns in limit states steel design: axial-flexural interaction, effective length and lateral-torsional buckling.

Key relations. $C_r = \phi A F_y (1+\lambda^{2n})^{-1/n}$ with $\lambda = (KL/r)\sqrt{F_y/\pi^2E}$; $M_r = \phi Z F_y$ for a Class 1 laterally supported section; $M_u = (\omega_2\pi/L)\sqrt{EI_yGJ + (\pi E/L)^2 I_y C_w}$; the three Clause 13.8.2 interaction equations.

Why this works. A beam-column has two ways to fail that a pure column and a pure beam do not share, and S16 asks three separate questions to catch them all: can the cross-section itself carry the combination, can the member survive overall buckling in the plane of bending, and can it survive lateral-torsional buckling out of plane. The three checks use different compressive resistances and different flexural resistances precisely because they interrogate different modes. Here the first check is comfortable and the last two are not, which is the diagnostic signature of a slenderness problem rather than a strength problem — and slenderness problems are fixed by adding restraint, not by adding steel. Recognising that distinction is what turns a bare “fails” into a useful engineering answer.

Common pitfall. Checking only the strong axis because the moment acts about the strong axis. The moment is about x–x, but the axial load buckles the member about y–y, and with r_y about a quarter of r_x on a W460 the minor axis controls the compressive resistance by a factor of three. A second trap is taking K = 1.0 in the plane of the frame; the roller at C provides no horizontal restraint, so the column is a true sway cantilever and K = 2.0.

Source. CSA S16, Design of Steel Structures, Clauses 10.4, 13.3, 13.6 and 13.8; CISC Handbook of Steel Construction, Part 4 (compression) and Part 5 (beam-columns); Kulak & Grondin, Limit States Design in Structural Steel, Chapters 4 and 8.

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