Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Examinations, May 2018 — 16-Civ-A2 Elementary Structural Design. Three hours, CLOSED BOOK (handbooks and textbooks permitted). Seven questions in three parts: Part A (A1–A3, steel), Part B (B1–B3, reinforced concrete), Part C (C1, timber). A candidate submits five solutions — two from Part A, two from Part B and the one question in Part C — all of equal value. All seven questions are solved below, because this set is a study resource rather than an exam script. Page 1 states that all loads shown are unfactored, so every load case is factored here before any resistance is compared against it.
Reference texts.
CSA S16:19, Design of Steel Structures — and CISC, Handbook of Steel Construction, 11th ed. (section tables, Table 2 class limits, Table 7 minimum fillet sizes).
CSA A23.3:19, Design of Concrete Structures — and Cement Association of Canada, Concrete Design Handbook, 4th ed.
CSA O86:19, Engineering Design in Wood — and Canadian Wood Council, Wood Design Manual.
Kulak & Grondin, Limit States Design in Structural Steel, 11th ed., Chs. 4, 5, 8, 9.
National Building Code of Canada 2015, Table 4.1.3.2 (load combinations).
Check — one dimension is missing from the source. Question A2 describes a stub cantilever welded to a column but the paper contains no Figure A2 and never states the cantilever projection. The projection is therefore taken as L = 1.5 m from the column face throughout question A2; every result below is also given in the general form so any other projection can be substituted directly. All other data are read from the printed text and from Figure A3, B1, B2 and B3 on page 3.
Find. The greatest factored bracket load Pf for which the member satisfies the CSA S16 beam-column interaction equations.
Figure 1.1 — Column elevation. The bracket delivers an axial force Pf together with a moment Pfe at the head; the pinned top and fixed base give a linear moment diagram with a carry-over of one half at the base.
Approach. Compute the section properties and its S16 Table 2 class, obtain the axial resistance Cr and the moment resistance Mr, then substitute Cf = Pf and Mf = 0.6Pf into the S16 Cl. 13.8 interaction checks and solve for Pf.
Section properties of the tube. With outside diameter D and inside diameter d = D − 2t = 355.6 − 2(7.95) = 339.7 mm,
$$A=\frac{\pi}{4}\left(D^{2}-d^{2}\right)=8682\ \text{mm}^{2},\qquad I=\frac{\pi}{64}\left(D^{4}-d^{4}\right)=131.2\times10^{6}\ \text{mm}^{4}$$
so that r = √(I/A) = 122.9 mm, and the plastic modulus of a circular tube is
$$Z=\frac{D^{3}-d^{3}}{6}=\frac{355.6^{3}-339.7^{3}}{6}=960.7\times10^{3}\ \text{mm}^{3}.$$
Classify the wall (S16 Table 2, circular hollow sections). The wall slenderness is D/t = 355.6/7.95 = 44.7. In flexure the limits are 13000/Fy = 37.1 (Class 1) and 18000/Fy = 51.4 (Class 2), so the section is Class 2; in axial compression the Class 3 limit is 23000/Fy = 65.7 > 44.7, so no local-buckling reduction applies to the axial term. A Class 2 section develops its plastic moment:
$$M_{r}=\phi Z F_{y}=0.90\,(960.7\times10^{3})(350)=\boxed{302.6\ \text{kN}\cdot\text{m}}$$
Axial compressive resistance. The top is held laterally and the base is fixed, so this is a braced pinned–fixed column; S16 recommends the design value K = 0.80 (theoretical 0.70). Hence
$$\lambda=\frac{KL}{r}\sqrt{\frac{F_{y}}{\pi^{2}E}}=\frac{0.80(10000)}{122.9}\sqrt{\frac{350}{\pi^{2}(200000)}}=65.1\,(0.01332)=0.867$$
and with n = 2.24 for a Class H (stress-relieved) hollow section,
$$C_{r}=\phi A F_{y}\left(1+\lambda^{2n}\right)^{-1/n}=0.90(8682)(350)(1.527)^{-1/2.24}=\boxed{2264\ \text{kN}}$$
Actions delivered by the bracket. The bracket sits at the head, so Cf = Pf and the applied moment there is Mf = Pfe = 0.6Pf (kN·m with Pf in kN). Slope-deflection for a prismatic member with an applied moment at a pinned end and a fixed far end gives a carry-over of one half, so the base moment is 0.3Pf of opposite sign: the member is in double curvature with κ = +0.5 and
$$\omega_{1}=0.6-0.4\kappa=0.6-0.4(0.5)=0.40\ \ (\ge 0.40).$$
Amplification factor. Using the in-plane Euler load with K = 1 for the cross-sectional check,
$$C_{ex}=\frac{\pi^{2}EI}{L^{2}}=\frac{\pi^{2}(200000)(131.2\times10^{6})}{10000^{2}}=2591\ \text{kN}.$$
For the overall member check in a braced frame S16 Cl. 13.8.4 requires U1x ≥ 1.0, and since ω1/(1 − Cf/Cex) works out below unity at the load found next, U1x = 1.0 governs.
Solve the interaction equation. A circular hollow section cannot buckle laterally — every axis is a principal axis with equal stiffness — so check (c) of Cl. 13.8.2 reduces to check (b), and the overall member strength check governs:
$$\frac{C_{f}}{C_{r}}+\frac{U_{1x}M_{fx}}{M_{rx}}\le 1.0\ \Longrightarrow\ P_{f}\left(\frac{1}{2264}+\frac{0.6}{302.6}\right)\le 1.0$$
Substituting the bracketed terms, 4.418×10−4 + 1.983×10−3 = 2.425×10−3 kN−1, so
$$\boxed{P_{f,\max}=412\ \text{kN}}$$
Confirm the cross-sectional check does not govern. With Cr replaced by φAFy = 2735 kN and U1x = 0.40/(1 − Pf/2591) taken at its calculated value, iteration returns Pf ≈ 690 kN — comfortably above 412 kN. The overall member strength, not the cross-section, limits this column, which is what the 10 m unbraced length and KL/r = 65 would suggest.
At the governing load the axial term contributes 412/2264 = 0.18 and the moment term 0.82 of the unity check, so this is very much a bending-dominated bracket column: doubling the eccentricity would almost halve the permissible load, whereas a 10 % change in Fy would move it only a few percent.
