Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Examinations, December 2018 — 07-Str-A2 Elementary Structural Design. Three hours, CLOSED BOOK (handbooks and textbooks permitted). Seven questions in three parts: Part A (A1–A3, steel to CSA S16), Part B (B1–B3, reinforced concrete to CSA A23.3), Part C (C1, timber to CSA O86). 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 (1.5 on the specified live/imposed loads, 1.25 on self weight) 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, Clause 13.8 interaction, Clause 13.5 effective widths).
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 (glulam selection tables).
Kulak & Grondin, Limit States Design in Structural Steel, 11th ed., Chs. 4, 5, 8, 9.
National Building Code of Canada, Table 4.1.3.2 (load combinations).
Check — assumptions declared under Note 1 of the paper. (i) In A2 the single dimension h is the wall thickness of the whole fabricated panel (top plate and both stems), which is how the figure dimensions it. (ii) In A3 the bolted joint at the column is a simple (shear-only) connection and the steel tie is a two-force member, which is what makes the stub beam determinate. (iii) In B1/B3 the left column A–B is taken as 6 m high, level with the roller at D, as Figure B1 draws it; the beam self weight is included, the column self weight is not (it does not change any design action at C). (iv) In B2 and C1 the beam is drawn with a tapered soffit, but part (a) asks for a uniform cross-section, so a prismatic member is designed.
Find. The largest factored bracket load $P_f$ that satisfies every applicable CSA S16 Clause 13.8 axial-plus-bending check.
Approach. Compute the section properties and classify the tube, obtain the compressive resistance $C_r$ (Clause 13.3, with $n = 2.24$ because the section is Class H) and the moment resistance $M_r$, then invert the Clause 13.8.2 interaction equations for the single unknown $P_f$, recognising that $M_f = P_f e$ so both terms scale together.
Section properties of the tube. With inside diameter $d = D - 2t = 355.6 - 2(7.95) = 339.7$ mm,
$$A=\frac{\pi}{4}\left(D^{2}-d^{2}\right)=8683\ \text{mm}^{2},\qquad I=\frac{\pi}{64}\left(D^{4}-d^{4}\right)=131.2\times10^{6}\ \text{mm}^{4}$$
so that $r=\sqrt{I/A}=122.9$ mm, the elastic modulus is $S = 2I/D = 738.2\times10^{3}$ mm3 and the plastic modulus is $Z=\left(D^{3}-d^{3}\right)/6=961.0\times10^{3}$ mm3.
Classify the section. For a circular hollow section CSA S16 Table 2 works on the diameter-to-thickness ratio,
$$\frac{D}{t}=\frac{355.6}{7.95}=44.73,\qquad \frac{13\,000}{F_y}=37.1,\qquad \frac{18\,000}{F_y}=51.4$$
The tube exceeds the Class 1 limit but is inside the Class 2 limit, so it is Class 2 in flexure. It is also well inside the axial-compression limit $23\,000/F_y = 65.7$, so no local-buckling reduction applies to the axial term. A Class 2 section develops its full plastic moment, and a circular tube has no lateral-torsional buckling mode.
Moment resistance. Because the section is Class 2 and no LTB reduction applies,
$$M_r=\phi Z F_y=0.90\left(961.0\times10^{3}\right)(350)=\boxed{302.7\ \text{kN}\cdot\text{m}}$$
Effective length and compressive resistance. The column is fixed at the base and hinged (laterally held) at the top, which is the braced pinned–fixed case: the theoretical $K = 0.7$ is raised to the design value $K = 0.8$. Hence $KL = 0.8(10\,000) = 8000$ mm and $KL/r = 8000/122.9 = 65.07$. The slenderness parameter is
$$\lambda=\frac{KL}{r}\sqrt{\frac{F_y}{\pi^{2}E}}=65.07\sqrt{\frac{350}{\pi^{2}(200\,000)}}=0.8665$$
A stress-relieved Class H hollow section uses $n = 2.24$ in Clause 13.3.1 (hot-formed Class C sections would use 1.34), so
$$C_r=\phi A F_y\left(1+\lambda^{2n}\right)^{-1/n}=0.90(8683)(350)\left(1+0.8665^{4.48}\right)^{-1/2.24}=\boxed{2265\ \text{kN}}$$
End-moment distribution and the moment-gradient factor. The bracket applies $M = P_f e$ at the head. For a member fixed at one end and held at the other, an end moment carries over to the far end at half value with reversed sign, so the base moment is $M/2$ and the column bends in double curvature. Therefore $\kappa = +0.5$ and
$$\omega_1=0.6-0.4\kappa=0.6-0.4(0.5)=0.40$$
Clause 13.8.4 places a floor of $U_1 \ge 1.0$ on braced members for the member-strength check, so $U_{1} = 1.0$ there; the elastic buckling load used in $U_1$ is $C_{ex}=\pi^{2}EI/L^{2}=2591$ kN.
Cross-sectional strength, Clause 13.8.2(a). Here $C_r$ is taken with $\lambda = 0$, i.e. $\phi A F_y = 2735$ kN, and $U_1 = \omega_1/(1-P_f/C_{ex})$ may be less than unity. Solving the resulting quadratic gives a capacity of 691.1 kN — far above the member check. Clause 13.8.2(c) is not applicable because a circular tube cannot buckle laterally-torsionally. Check (b) governs.
At the governing load the axial term contributes $412.6/2265 = 0.182$ and the moment term $0.818$: this is a bending-dominated member, which is why the 0.6 m bracket arm costs so much capacity. Applied concentrically the same column would carry 2265 kN, so the bracket eccentricity removes about 82 % of the available strength.
