16-Civ-A2 Elementary Structural Design · May 2016

Question 1 of 7: A1 — Maximum factored load on a free-standing tubular post

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

Notes on this paper

Paper format. National Examinations — May 2016, 98-Civ-A2 Elementary Structural Design. Three-hour closed-book examination (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), 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 in total, all of equal value. All seven are solved here, because the set is a study resource rather than an examination script. Page 1 states that all loads shown are unfactored, so the load factors are applied within each solution.

Reference texts.

CSA S16:19, Design of Steel Structures — Clauses 11 (class of section), 13.3 (compressive resistance), 13.5–13.6 (bending), 13.8 (axial compression and bending), 13.13 (welds).
CISC, Handbook of Steel Construction, 12th ed. — section property tables, weld tables.
CSA A23.3:19, Design of Concrete Structures — Clauses 10 (flexure and axial load), 11 (shear), 7.6 (ties).
CSA O86:19, Engineering Design in Wood, with the Canadian Wood Council Wood Design Manual 2020 — Clause 7 (glued-laminated timber).
NBCC 2020 — Table 4.1.3.2 load combinations.

Check — assumptions common to the whole paper. (i) Every load drawn in Figures A1, A2, A3, B3 and quoted in C1 is a specified (unfactored) load; unless a question names the load type, the point loads are treated as live (factor 1.5) and concrete/steel self-weight as dead (factor 1.25), per NBCC Table 4.1.3.2 case 2. (ii) Steel is CSA G40.21 350W, so F_y = 350 MPa, E = 200 000 MPa. (iii) Question B3 states f'_c = 35 MPa and f_y = 400 MPa; because B1 and B2 quote no material strengths, the same pair is adopted for them and the assumption is stated in each solution. (iv) Section properties are recomputed from the nominal plate dimensions of each rolled shape rather than read off a table, so every number below is reproducible; they agree with the CISC Handbook to better than 1 %.

Question 1: A1 — Maximum factored load on a free-standing tubular post (8 + 12)

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
Circular hollow section, outside diameter D	219.1 mm
Wall thickness t	7.95 mm
Steel, G40.21 350W Class H	F_y = 350 MPa, E = 200 000 MPa, n = 2.24
Height of post L	8.0 m (top cable row at 8 m, lower row at 6 m)
Each cable row (from Figure A1)	5 kN at 2.0 m one side, 2 kN at 1.0 m the other side
Support conditions	fixed at base, free at top ⇒ K = 2.0

Find. The largest factored vertical load the post can carry when the cables are hung in the pattern of Figure A1 — that is, the factored axial force P_F at which the S16 Clause 13.8 axial-plus-bending interaction reaches unity, together with the check that the loads actually shown are safely below it.

[Figure not reproduced: Figure A1 (redrawn from the examination paper). Each cable row applies a downward force on both sides of the post; the two forces are unequal and act at unequal eccentricities, so each row delivers both an axial force and a net torque-free bending couple about the post axis. See the official exam paper.]

Approach. Establish the axial force and the bending moment the cable pattern delivers to the base, note that their ratio (the eccentricity) is fixed by the geometry, compute the tubular section’s axial and flexural resistances for K = 2.0, and scale the load pattern until the S16 Clause 13.8.2 interaction equations are satisfied exactly.

Reduce each cable row to a force and a couple at the post axis. The 5 kN and 2 kN loads act on opposite sides, so their moments about the post subtract while their vertical components add: $$N_{\text{row}} = 5 + 2 = 7\ \text{kN}, \qquad M_{\text{row}} = 5(2.0) - 2(1.0) = 8\ \text{kN}\cdot\text{m}$$ Both rows are arranged identically, so the couples act in the same sense and accumulate down the post. At the base $$N = 2(7) = 14\ \text{kN}, \qquad M = 2(8) = 16\ \text{kN}\cdot\text{m}$$ The two arms carry no horizontal load, so the couple is constant over each segment of the post rather than varying linearly — the base is nevertheless the critical section because it carries both couples plus the full axial force.
Fix the load path by its eccentricity. Because every cable load scales together, the ratio of moment to axial force is a property of the geometry alone: $$e = \frac{M}{N} = \frac{16}{14} = \boxed{1.143\ \text{m}}$$ Any factored load level P_F therefore carries with it a factored moment M_f = 1.143 P_F. This is what makes the question answerable: the post has one degree of freedom, not two.
Compute the section properties of the 219.1 × 7.95 tube. With inside diameter d = 219.1 − 2(7.95) = 203.2 mm, $$A = \frac{\pi}{4}\left(D^{2}-d^{2}\right) = \frac{\pi}{4}\left(219.1^{2}-203.2^{2}\right) = 5274\ \text{mm}^{2}$$ $$I = \frac{\pi}{64}\left(D^{4}-d^{4}\right) = 29.43\times10^{6}\ \text{mm}^{4}, \qquad r = \sqrt{I/A} = 74.7\ \text{mm}$$ $$Z = \frac{D^{3}-d^{3}}{6} = \frac{219.1^{3}-203.2^{3}}{6} = 354.6\times10^{3}\ \text{mm}^{3}$$
Classify the section. For a circular hollow section in bending, S16 Table 2 sets the Class 1 limit at 13 000/F_y: $$\frac{D}{t} = \frac{219.1}{7.95} = 27.6 \;<\; \frac{13\,000}{350} = 37.1$$ so the tube is Class 1 and the full plastic moment is available; the axial limit 23 000/F_y = 65.7 is satisfied with a wide margin, so there is no local-buckling reduction on the compressive resistance either.
Compressive resistance for a flagpole (K = 2.0). A member fixed at the base and completely free at the top has an effective length of twice its height: $$KL = 2.0(8000) = 16\,000\ \text{mm}, \qquad \frac{KL}{r} = \frac{16\,000}{74.7} = 214.2$$ $$\lambda = \frac{KL}{r}\sqrt{\frac{F_{y}}{\pi^{2}E}} = 214.2\sqrt{\frac{350}{\pi^{2}(200\,000)}} = 2.852$$ Class H hollow sections use n = 2.24 in S16 Clause 13.3.1, so $$C_{r} = \phi A F_{y}\left(1+\lambda^{2n}\right)^{-1/n} = 0.9(5274)(350)\left(1+2.852^{4.48}\right)^{-1/2.24} = \boxed{203\ \text{kN}}$$ This is only 12 % of the squash load φAF_y = 1661 kN — the post is very slender, which is the heart of the problem.
Flexural resistance and elastic buckling load. With the Class 1 result from Step 4, $$M_{r} = \phi Z F_{y} = 0.9\left(354.6\times10^{3}\right)(350) = \boxed{111.7\ \text{kN}\cdot\text{m}}$$ and the Euler load that drives the moment-amplification term is $$C_{e} = \frac{\pi^{2}EI}{(KL)^{2}} = \frac{\pi^{2}(200\,000)\left(29.43\times10^{6}\right)}{16\,000^{2}} = 227\ \text{kN}$$ Note how close C_e = 227 kN is to C_r = 203 kN: at this slenderness the member is essentially an elastic strut, and the second-order amplification will be severe.
Overall member strength — S16 Clause 13.8.2(b). The post is an unbraced (sway) member, so Clause 13.8.4 sets U_1x = 1.0 for this check, and a tube is not an I-shape so the 0.85 coefficient does not apply: $$\frac{C_{f}}{C_{r}} + \frac{U_{1x}M_{fx}}{M_{rx}} \le 1.0 \quad\Longrightarrow\quad \frac{P_{F}}{203} + \frac{1.143\,P_{F}}{111.7} = 1.0$$ $$P_{F}\left(0.004926 + 0.010233\right) = 1.0 \quad\Longrightarrow\quad P_{F} = 66.0\ \text{kN}$$
Cross-sectional strength — S16 Clause 13.8.2(a). Here C_r is taken with λ = 0 (1661 kN) but the moment is amplified by U_1x = ω₁/(1 − C_f/C_e) with ω₁ = 1.0 for a member carrying a constant couple. Solving the resulting quadratic gives P_F = 66.4 kN, with $$U_{1x} = \frac{1}{1-66.4/227} = 1.41$$ The two checks land within 0.6 % of each other, which is how these examination sections are calibrated. A circular tube has no lateral-torsional buckling mode, so Clause 13.8.2(c) reduces to the same expression as (b) and does not govern.
State the capacity and compare it with the loads drawn. Taking the lower of the two checks, $$\boxed{P_{F} = 66\ \text{kN}}$$ carried at the eccentricity of Figure A1 — equivalent to about 23.6 kN on each 2.0 m arm and 9.4 kN on each 1.0 m arm, or 4.7 times the loads shown. The factored load actually applied is C_f = 1.5(14) = 21 kN, giving an interaction value of 21/66 = 0.32. The post as drawn is comfortably adequate.

The design is dominated by slenderness rather than by strength: at KL/r = 214 the axial resistance has fallen to one-eighth of the squash load, and the bending term consumes roughly two-thirds of the interaction. Halving the effective length — by guying the post at the lower cable row, for instance — would raise P_F several-fold without changing the section.

Result	Value
Axial force / bending moment at the base (unfactored)	14 kN / 16 kN·m
Load eccentricity fixed by the geometry, e	1.143 m
Section properties A / I / r / Z	5274 mm² / 29.43×10⁶ mm⁴ / 74.7 mm / 354.6×10³ mm³
Class of section	Class 1 (D/t = 27.6 < 37.1)
Effective slenderness KL/r, λ	214.2, 2.852
C_r (K = 2.0) / C_r (λ = 0) / M_r / C_e	203 kN / 1661 kN / 111.7 kN·m / 227 kN
Clause 13.8.2(b) member check	P_F = 66.0 kN (governs)
Clause 13.8.2(a) cross-section check	P_F = 66.4 kN
Maximum factored load P_F	66 kN
Applied factored load / utilisation	21 kN / 0.32 — adequate

Topic. Beam-column design of a slender circular hollow section under combined axial compression and a constant couple.

Key relations. S16 Clause 13.3.1 compressive resistance C_r = φAF_y(1 + λ²ⁿ)^−1/n with n = 2.24 for Class H hollow sections; Clause 13.5 M_r = φZF_y for a Class 1 section; Clause 13.8.2 interaction with U_1x = ω₁/(1 − C_f/C_e) for the cross-sectional check and U_1x = 1.0 for a sway member; Euler load C_e = π²EI/(KL)².

Why this works. When every applied load scales by a single multiplier, the axial force and the moment stay in a fixed ratio, so the two-variable interaction equation collapses to one unknown and can be inverted directly for the load level at which the section is exactly fully used. The couple delivered by a pair of unequal cantilevered cable loads is constant along the post because there is no transverse force between the arms; that constancy is what makes ω₁ = 1.0 the right choice and makes the base — where both couples and the whole axial force are present — the single critical section. The factor K = 2.0 follows from the buckled shape of a member restrained in position and direction at one end only: it deflects into half of the pinned-pinned sine wave, so its effective length is twice its physical height.

Common pitfall. Adding the two arm moments instead of subtracting them — the cables hang on opposite sides, so 5(2) − 2(1) = 8, not 12 — which overstates the load capacity by roughly one third. The second is using K = 1.0 or even 0.5 because the base is fixed; the free top means K = 2.0, and since the interaction is essentially controlled by the moment term, that error nearly triples the answer.

Source. CSA S16:19, Design of Steel Structures, Clauses 11.2 (Table 2), 13.3.1, 13.5 and 13.8.2–13.8.4; CISC Handbook of Steel Construction, 12th ed., Part 4 (compressive resistances of hollow sections).

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