16-Civ-A2 Elementary Structural Design · Undated paper

Question 1 of 7: A1 — Axial capacity of a back-to-back channel column

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

Notes on this paper

Paper format. 16-Civ-A2 Elementary Structural Design, May 2019 sitting, 3 hours, closed book with handbooks and textbooks permitted. Seven questions of 20 marks each: Part A (A1–A3, structural steel), Part B (B1–B3, reinforced concrete), Part C (C1, timber). The candidate answers two from Part A, two from Part B and one from Part C — all seven are solved here.

Reference texts. CSA S16 Design of Steel Structures and the CISC Handbook of Steel Construction (section properties, Tables 4-4 and 7); CSA A23.3 Design of Concrete Structures with Brzev & Pao, Reinforced Concrete Design: A Practical Approach; CSA O86 Engineering Design in Wood with the Canadian Wood Council Wood Design Manual; NBCC Part 4 for load combinations.

Page 1 of the paper specifies steel grade G40.21 300W unless noted, reinforcement 400W, and all loads shown unfactored unless otherwise stated.

Question 1: A1 — Axial capacity of a back-to-back channel column (20 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. The strut is two rolled channels placed back to back about a common axis of symmetry, connected at intervals by welded spacer plates, and pinned at both ends about both axes.

Given data — A1
Quantity	Symbol	Value
Channel (each of two)	—	C310x37
Area, one channel	$A_1$	4740 mm²
Second moment, strong axis	$I_{x1}$	$59.9\times10^{6}$ mm⁴
Second moment, weak axis	$I_{y1}$	$1.86\times10^{6}$ mm⁴
Weak-axis radius of gyration	$r_{y1}$	19.8 mm
Centroid from web back	$\bar{x}$	17.1 mm
Clear gap between webs	$g$	25 mm
Spacer spacing	$a$	500 mm
Column height, pinned both ends	$KL$	6000 mm ($K=1.0$)
Steel grade (page 1, Note 7)	$F_y$	300 MPa

Find. The maximum factored axial compressive load the built-up column can carry, i.e. its factored compressive resistance $C_r$ to CSA S16.

Figure 1 — A1 built-up cross-section: two C310x37 channels back to back with a 25 mm gap bridged by welded spacer plates. The flanges point outward and the webs face each other across the gap.

Approach. Assemble the built-up section properties about both principal axes, decide which axis governs, inflate the governing slenderness for the discrete spacer connection using S16 Cl. 19.1.4, confirm no element is Class 4, and evaluate $C_r$ from Cl. 13.3.1.

Assemble the section about the strong axis. Both channels bend about their own $x$–$x$ axis, which is also the built-up $x$–$x$ axis, so the second moments simply add and there is no transfer term: $$A=2A_1=2(4740)=9480\ \text{mm}^2,\qquad I_x=2I_{x1}=119.8\times10^{6}\ \text{mm}^4$$ $$r_x=\sqrt{\frac{119.8\times10^{6}}{9480}}=112.4\ \text{mm}$$
Assemble the section about the weak axis, including the transfer term. Each channel centroid sits half the gap plus $\bar{x}$ away from the axis of symmetry: $$e=\frac{g}{2}+\bar{x}=\frac{25}{2}+17.1=29.6\ \text{mm}$$ Applying the parallel-axis theorem to each channel, $$I_y=2\left(I_{y1}+A_1e^{2}\right)=2\left(1.86\times10^{6}+4740(29.6)^{2}\right)=12.03\times10^{6}\ \text{mm}^4$$ $$r_y=\sqrt{\frac{12.03\times10^{6}}{9480}}=35.6\ \text{mm}$$ The transfer term supplies more than two thirds of $I_y$: the whole point of separating the channels is to lift the weak-axis radius of gyration from 19.8 mm to 35.6 mm.
Identify the governing axis. With $K=1.0$ about both axes, $$\left(\frac{KL}{r}\right)_x=\frac{6000}{112.4}=53.4,\qquad \left(\frac{KL}{r}\right)_y=\frac{6000}{35.6}=168.5$$ The weak axis governs by a wide margin, so the column is a slender strut and the answer will be far below the squash load.
Inflate the weak-axis slenderness for the discrete connectors. A back-to-back pair joined only at spacers is not fully composite: between spacers each channel buckles alone. CSA S16 Cl. 19.1.4 gives the modified slenderness $$\left(\frac{KL}{r}\right)_m=\sqrt{\left(\frac{KL}{r}\right)_o^{2}+\left(\frac{a}{r_{i}}\right)^{2}}$$ where $a=500$ mm is the spacer spacing and $r_i=r_{y1}=19.8$ mm is the least radius of gyration of one channel. First confirm the code limit on the local panel, $a/r_i \le 0.75(KL/r)_o$: $$\frac{a}{r_i}=\frac{500}{19.8}=25.3\ \le\ 0.75(168.5)=126.4\quad\checkmark$$ $$\left(\frac{KL}{r}\right)_m=\sqrt{168.5^{2}+25.3^{2}}=\boxed{170.3}$$ The penalty is only 1.1 %, because the spacers are close enough that the individual channel is nowhere near its own buckling load.
Check the element slenderness (Class 4 screening). A Class 4 section would need an effective-area reduction: $$\frac{b_f}{t_f}=\frac{77.4}{12.7}=6.09\ <\ \frac{200}{\sqrt{300}}=11.55\quad\checkmark$$ $$\frac{h}{w}=\frac{305-2(12.7)}{9.83}=28.4\ <\ \frac{670}{\sqrt{300}}=38.7\quad\checkmark$$ Every element is at worst Class 3, so the gross area may be used.
Evaluate the compressive resistance. S16 Cl. 13.3.1 with $n=1.34$: $$\lambda=\frac{(KL/r)_m}{\pi}\sqrt{\frac{F_y}{E}}=\frac{170.3}{\pi}\sqrt{\frac{300}{200000}}=2.099$$ $$C_r=\phi AF_y\left(1+\lambda^{2n}\right)^{-1/n}=0.90(9480)(300)\left(1+2.099^{2.68}\right)^{-1/1.34}$$ $$C_r=2559.6\times0.206=\boxed{527\ \text{kN}}$$
Confirm the strong axis is not critical. Repeating step 6 with $(KL/r)_x = 53.4$ gives $\lambda=0.658$ and $C_r=2074$ kN — almost four times the weak-axis value. The weak axis governs, as anticipated.

The answer is therefore 527 kN. It is worth noticing how little of the material is actually working: the squash load $\phi AF_y$ is 2560 kN, so at $KL/r = 170$ the column realises only about a fifth of its cross-sectional strength. If the 25 mm gap were opened to roughly 90 mm the two axes would become equally critical and the same weight of steel would carry close to 2000 kN, which is why real back-to-back columns are detailed with a much wider separation whenever the connection geometry allows it.

Final results — A1
Quantity	Value
Built-up area $A$	9480 mm²
$r_x$ / $r_y$	112.4 mm / 35.6 mm
Governing slenderness $(KL/r)_m$	170.3
$C_r$ about $y$–$y$ (governs)	527 kN
$C_r$ about $x$–$x$	2074 kN

Topic. Compressive resistance of a laced or battened built-up column, CSA S16 Clauses 13.3.1 and 19.1.4.

Key relations. Parallel-axis theorem $I=\sum(I_i+A_id_i^{2})$; modified slenderness $(KL/r)_m=\sqrt{(KL/r)_o^{2}+(a/r_i)^{2}}$; column curve $C_r=\phi AF_y(1+\lambda^{2n})^{-1/n}$ with $\lambda=(KL/r)\sqrt{F_y/E}/\pi$ and $n=1.34$ for hot-rolled shapes.

Why this works. Two channels separated by a gap behave as one member only to the extent that the connectors force them to deform together. Between spacers each channel is free to bow about its own weak axis, so the built-up strut carries a second, shorter buckling mode superimposed on the overall one. The square-root-of-sum-of-squares rule in Cl. 19.1.4 is an interaction formula between those two modes: when the spacers are close the local term vanishes and the member behaves as a single unit; when they are far apart the local term dominates and the assembly degenerates into two independent channels. The rest of the calculation is ordinary column design, and the class check matters because a Class 4 element would buckle locally before the member curve is ever reached.

Common pitfall. Two errors recur. The first is measuring the transfer distance $e$ from the web face rather than from the channel centroid — forgetting $\bar{x}$ understates $I_y$ by roughly a factor of three here. The second is skipping Cl. 19.1.4 entirely and reporting 530 kN as if the pair were monolithic; that happens to be nearly right at 500 mm spacing, but at 1500 mm spacing the same shortcut overstates capacity by more than 15 %.

Source. CSA S16, Cl. 13.3.1 and Cl. 19.1.4; CISC Handbook of Steel Construction, channel property tables and Part 4 column tables.

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