16-Civ-A2 Elementary Structural Design · May 2017

Question 3 of 7: A3 — Maximum factored load on the built-up section used as a column

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

Notes on this paper

Paper format. National Examinations, May 2017 — 16-Civ-A2 Elementary Structural Design. Three hours, closed book (handbooks and textbooks permitted). Seven questions in three parts: Part A (A1–A3, structural steel to CSA S16), Part B (B1–B3, reinforced concrete to CSA A23.3) and Part C (C1, timber to CSA O86). A candidate answers two from Part A, two from Part B and the one question in Part C — five solutions of equal value. All seven are solved here, because the set is a study resource rather than a sitting.

Reference texts.

CSA S16:19, Design of Steel Structures — Clauses 11 (section class), 13.3 (compression), 13.5 (bending), 13.8 (axial force and bending), 13.13 (welds), 14.3 (concentrated forces).
CISC, Handbook of Steel Construction — rolled-section dimensions and properties, weld tables.
CSA A23.3:19, Design of Concrete Structures — Clauses 10 (flexure and axial load), 11 (shear).
CSA O86:19, Engineering Design in Wood — Clause 6 (sawn lumber), Tables 6.3.1B and 6.4.5.
Canadian Wood Council, Wood Design Manual; National Building Code of Canada 2020 (load combinations).

Load factors. NOTE 6 on page 1 states that all loads shown are unfactored. The point loads drawn on Figures A1, B1 and B2 are occupancy loads, so they are factored by 1.5; the only dead load anywhere in this paper is the concrete self-weight of the Question B2 beam, which takes 1.25. This follows NBCC combination 2, 1.25D + 1.5L.

Check — figure page. Two figure readings are flagged where they arise: in Figure B1 the 350 kN load acts at midspan of beam BA (3.5 m from B), not at joint B; and the 80 kN horizontal load acts 4 m below B, i.e. 6 m above the pin at C.

Question 3: A3 — Maximum factored load on the built-up section used as a column (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 (from A2)	A = 37 500 mm², r_x = 118.5 mm, r_y = 192.1 mm, Z_x = 4.094×10⁶ mm³, Class 1
Height and end conditions	L = 9000 mm, fixed top and bottom → K = 0.65 (S16 recommended design value)
Load position	at O, on the y–y axis, e = 60 mm from the centroid C
Material	F_y = 350 MPa, E = 200 000 MPa, φ = 0.90

Find. The largest factored axial load P_f the member can carry, given that it acts at 60 mm eccentricity and therefore delivers a moment as well as a thrust.

Figure A3 — 9 m fixed-ended column, and the cross-section showing the load point O on the y–y axis.

Approach. Recognise that an offset along y–y bends the column about x–x, compute C_r about the governing axis and M_rx (unreduced, because x–x is the weak axis), then write S16 Clause 13.8.2 with M_fx = P_fe and solve the resulting single-unknown interaction.

Identify the bending axis. Point O lies on the y–y axis, 60 mm from C, so the offset is in the y-direction of the cross-section and the couple it produces acts about the x–x axis: $$M_{fx} = P_f \times 0.060\ \text{m}$$ Since I_x < I_y, x–x is the weak axis. A member bent about its weak axis cannot buckle laterally, so no lateral-torsional reduction applies and M_rx = 1290 kN·m carries straight over from Question A2.
Slenderness. With KL = 0.65 × 9000 = 5850 mm, $$\frac{KL}{r_x} = \frac{5850}{118.5} = 49.4, \qquad \frac{KL}{r_y} = \frac{5850}{192.1} = 30.5$$ so buckling is governed by the x–x axis — conveniently the same axis that carries the moment. Every element also clears the Class 3 uniform-compression limits (200/√F_y = 10.7 for an outstand, 670/√F_y = 35.8 for a supported plate), so the section is not Class 4 and no effective-area reduction is needed.
Compressive resistance. The non-dimensional slenderness is $$\lambda = \frac{KL}{r_x}\sqrt{\frac{F_y}{\pi^{2}E}} = 49.4\sqrt{\frac{350}{\pi^{2}(200\,000)}} = 0.657$$ and with n = 1.34, $$C_r = \phi A F_y \bigl(1 + \lambda^{2n}\bigr)^{-1/n} = 0.90(37\,500)(350)(1.3254)^{-0.746} = \boxed{9575\ \text{kN}}$$ The cross-sectional value, used in Clause 13.8.2(a), is C_r0 = φAF_y = 11 813 kN.
Euler load for the amplifier. The moment amplifier uses the actual unbraced length with K = 1: $$C_{ex} = \frac{\pi^{2}EA}{(L/r_x)^{2}} = \frac{\pi^{2}(200\,000)(37\,500)}{(75.96)^{2}} = 12\,829\ \text{kN}$$
Write the interaction. This is a plate assembly, not an I-shaped member, so the 0.85 coefficient of Clause 13.8.2(a) does not apply — β = 1.0 in every term. With equal end eccentricities in single curvature, ω₁ = 1.0 and $$U_{1x} = \frac{\omega_1}{1 - P_f/C_{ex}} \;\ge\; 1.0$$ The two checks are $$\text{(a)}\quad \frac{P_f}{C_{r0}} + \frac{U_{1x}(0.060 P_f)}{M_{rx}} \le 1.0, \qquad \text{(b)}\quad \frac{P_f}{C_{r}} + \frac{U_{1x}(0.060 P_f)}{M_{rx}} \le 1.0$$ Check (c), lateral-torsional buckling, does not apply for weak-axis bending.
Solve each check. Both are nonlinear in P_f because U_1x grows with the load; iterating (or bisecting) to unity gives $$\text{(a)}: P_f = 5868\ \text{kN}, \qquad \text{(b)}: P_f = \boxed{5409\ \text{kN}}$$ The overall member check governs, as it must whenever slenderness is appreciable.
Confirm the governing case. At P_f = 5409 kN the amplifier is U_1x = 1/(1 − 5409/12 829) = 1.729 and the applied moment is M_fx = 5409 × 0.060 = 324.5 kN·m, so $$\frac{5409}{9575} + \frac{1.729 \times 324.5}{1290} = 0.565 + 0.434 = 0.999 \approx 1.0 \checkmark$$ The axial term carries 57 % of the interaction and the amplified moment 43 %. The 60 mm eccentricity, small though it looks against a 300 mm-deep section, costs the column 44 % of the 9575 kN it could carry concentrically.

Quantity	Value
Effective length, KL	5850 mm (K = 0.65)
KL/r_x / KL/r_y	49.4 / 30.5 — x–x governs
Non-dimensional slenderness, λ	0.657
C_r (member) / C_r0 (section)	9575 / 11 813 kN
M_rx (no LTB — weak-axis bending)	1290 kN·m
C_ex	12 829 kN
P_f from check 13.8.2(a)	5868 kN
P_f from check 13.8.2(b) — governs	5409 kN
At that load: U_1x, M_fx	1.729, 324.5 kN·m

Topic. Beam-columns: combined axial compression and uniaxial bending under CSA S16 Clause 13.8.

Key relations. C_r = φAF_y(1 + λ²ⁿ)^−1/n with n = 1.34; λ = (KL/r)√(F_y/π²E); U₁ = ω₁/(1 − C_f/C_e); interaction C_f/C_r + βU₁M_f/M_r ≤ 1.0.

Why this works. An eccentric axial load is statically identical to a concentric load plus a couple Pe, but the two effects are not independent: as the column shortens and bows, the thrust acts on the lateral deflection and magnifies the moment. U₁ is the closed-form amplification of the first-order moment, and it runs away as P_f approaches the Euler load C_e — which is why the answer must be found by iteration rather than by rearranging the interaction once. Clause 13.8.2 then applies the same equation three times with different resistances: (a) the cross-section at full yield, (b) the member with its buckling resistance, (c) the lateral-torsional mode. Whichever gives the lowest capacity governs, and for a member of moderate slenderness that is almost always (b).

Common pitfall. Two on this question. First, mistaking which axis the eccentricity bends the section about — an offset measured along y–y produces a moment about x–x. Second, importing the 0.85 and 0.6 coefficients of β that apply to Class 1 I-shaped members; a welded plate assembly is not an I-shape, so β = 1.0 and the moment term bites harder. A third, subtler error is solving the interaction once with U₁ = 1.0, which here returns 6624 kN — 22 % unconservative.

Source. CSA S16:19 Clauses 13.3.1, 13.8.2 and 13.8.4; CISC Handbook of Steel Construction, Part 4.

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