07-Str-A2 · December 2015

Question 1 of 7: A1 — Light standard post — maximum factored load

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

Notes on this paper

Paper format. National Examinations, December 2015 — 07-Str-A2 Elementary Structural Design, 3-hour closed-book paper (textbooks and design handbooks permitted). Seven questions in three parts: Part A (A1–A3, steel), Part B (B1–B3, reinforced concrete), Part C (C1, timber). The candidate answers two from Part A, two from Part B and the one question in Part C — five solutions, all of equal value. All seven are solved here, because the set is a study resource rather than an exam script.

Reference texts.

CSA S16, Design of Steel Structures — Clauses 11 (classification), 13.3 (compression), 13.5–13.6 (flexure), 13.8 (axial + bending), 13.11–13.12 (bolts, block shear).
CISC, Handbook of Steel Construction — section property tables for W-shapes and hollow structural sections.
CSA A23.3, Design of Concrete Structures — Clauses 10 (flexure and axial), 11 (shear), 10.15 (slender columns).
CSA O86, Engineering Design in Wood — Clause 6 (modification factors), Clause 7 (sawn lumber, bending and shear).
Canadian Wood Council, Wood Design Manual — specified strengths for Beam and Stringer grades.

Check — load factoring. Note 6 on page 1 states “all loads shown are unfactored”, so every load taken from a figure is a specified load and is factored here as 1.25D + 1.5L (NBCC Table 4.1.3.2, Case 2), treating the drawn point loads as live and member self-weight as dead. Where a question states a load directly in its text without a dead/live split (B3), that load is taken as already factored; this is stated again at that question.

Question 1: A1 — Light standard post — maximum factored load (10 + 10 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
Round hollow section (Class H), outside diameter $D$	323.9 mm
Wall thickness $t$	12.7 mm
Steel G40.21 350W: $F_y$, $E$	350 MPa, 200 000 MPa
Post height $L$ (fixed base, free top)	10 m
Cantilever arm loads (from Figure A1)	2P at 3 m one side, 2.5P at 4 m the opposite side
Horizontal load at the head	0.25P

Find. The largest factored load multiplier $P_F$ for which the post satisfies every S16 Clause 13.8 beam-column check.

[Figure not reproduced: Figure A1 (redrawn from the examination paper): free-standing 10 m post carrying two diametrically opposed cantilever loads and a horizontal head load. See the official exam paper.]

Approach. Compute the hollow-section properties and its S16 Clause 11 class, resolve the head loads into a factored axial force and a factored base moment, then apply the three Clause 13.8.2 beam-column checks and solve each for the load multiplier.

Section properties of the round hollow section. With the inside diameter $d = D - 2t = 323.9 - 25.4 = 298.5$ mm, $$A=\frac{\pi}{4}\left(D^{2}-d^{2}\right)=12\,416\ \text{mm}^{2},\qquad I=\frac{\pi}{64}\left(D^{4}-d^{4}\right)=150.6\times10^{6}\ \text{mm}^{4}$$ so that $r=\sqrt{I/A}=110.1$ mm and the plastic modulus is $Z=(D^{3}-d^{3})/6 = 1.231\times10^{6}$ mm³.
Classify the section. For a circular hollow section in flexure, S16 Table 2 sets the Class 1 limit at $13\,000/F_y$: $$\frac{D}{t}=\frac{323.9}{12.7}=25.5\ <\ \frac{13\,000}{350}=37.1$$ The section is therefore Class 1 and the full plastic moment may be developed. A circular tube has no lateral-torsional buckling mode, so no $\omega_2$ or LTB reduction applies.
Factored resistances. Taking $\phi = 0.90$, $$M_r=\phi Z F_y=0.90(1.231\times10^{6})(350)=\boxed{387.6\ \text{kN}\cdot\text{m}}$$ For the cross-sectional check the compressive resistance at $\lambda = 0$ is $C_{r0}=\phi A F_y = 3911$ kN. For the member check, the post is free at the top and fixed at the base, so $K = 2.0$ and $$\frac{KL}{r}=\frac{2.0(10\,000)}{110.1}=181.6,\qquad \lambda=\frac{KL}{r}\sqrt{\frac{F_y}{\pi^{2}E}}=2.418$$ With $n = 2.24$ for a Class H hollow section, $$C_r=\phi A F_y\left(1+\lambda^{2n}\right)^{-1/n}=\boxed{663\ \text{kN}}$$ The slenderness is severe — a 10 m free-standing cantilever loses more than five sixths of its squash load.
Factored load effects at the base. The two arm loads act on diametrically opposite sides, so their moments about the post axis subtract while their axial forces add: $$M_{\text{arms}}=2.5P(4.0)-2P(3.0)=10P-6P=4P\ \text{kN}\cdot\text{m}$$ The horizontal head load bends the post the same way as the heavier arm, and its base moment is $0.25P(10) = 2.5P$. Hence $$C_f=(2+2.5)P=4.5P,\qquad M_f=4P+2.5P=\boxed{6.5P\ \text{kN}\cdot\text{m}}$$ Adding the arm moments instead of subtracting them would nearly double $M_f$ and roughly halve the answer — this is the pivot of the question.
Moment amplification. The Euler load used in the $U_1$ factor is taken on the actual length (S16 Clause 13.8.5), $$C_e=\frac{\pi^{2}EI}{L^{2}}=\frac{\pi^{2}(200\,000)(150.6\times10^{6})}{10\,000^{2}}=2972\ \text{kN}$$ A transverse load acts on the member, so $\omega_1 = 1.0$ and $U_1=1/(1-C_f/C_e)$.
Cross-sectional strength, Clause 13.8.2(a). A plate/tube assembly is not an I-shape, so $\beta = 1.0$ in every interaction term: $$\frac{4.5P}{3911}+U_1\frac{6.5P}{387.6}\le 1.0\;\Longrightarrow\;P=51.7\ \text{kN}$$
Overall member strength, Clause 13.8.2(b). Replacing $C_{r0}$ by the buckling resistance $C_r = 663$ kN, $$\frac{4.5P}{663}+U_1\frac{6.5P}{387.6}\le 1.0\;\Longrightarrow\;\boxed{P_F=40.6\ \text{kN}}$$ The member check governs by about 25 %, exactly as one expects when $KL/r$ approaches 180. Check (c), lateral-torsional buckling, does not apply to a circular tube.
Confirm the governing state. At $P_F = 40.6$ kN the load effects are $C_f = 182.5$ kN and $M_f = 263.7\ \text{kN}\cdot\text{m}$, giving $U_1 = 1.065$; the interaction values are 0.77 for check (a) and 1.00 for check (b). The moment term supplies roughly seven tenths of the total — the post is bending-dominated, not squash-limited.

Result	Value
Section class (flexure)	Class 1, $D/t = 25.5 < 37.1$
Moment resistance $M_r$	387.6 kN·m
Compressive resistance $C_r$ ($K = 2.0$)	663 kN
Load effects	$C_f = 4.5P$, $M_f = 6.5P$
Clause 13.8.2(a) cross-section	$P = 51.7$ kN
Clause 13.8.2(b) member — governs	$P_F = 40.6$ kN

Topic. Beam-columns: combined axial compression and bending in a slender hollow section.

Key relations. $C_r=\phi AF_y(1+\lambda^{2n})^{-1/n}$ with $\lambda = (KL/r)\sqrt{F_y/\pi^{2}E}$; $M_r = \phi Z F_y$ for Class 1 and 2; the Clause 13.8.2 interaction $C_f/C_r + \beta U_1 M_f/M_r \le 1.0$; and $U_1 = \omega_1/(1 - C_f/C_e)$ with $C_e = \pi^{2}EI/L^{2}$.

Why this works. A compression member that also bends fails by an interaction of the two effects rather than by either alone, because axial force acting through the lateral deflection produces additional moment. S16 handles this in two separate places. The $U_1$ factor amplifies the first-order moment to account for the P-delta effect within the member length, and the interaction equation then combines the amplified moment with the axial demand. Two checks are required because the two possible failure modes have different resistances in the axial term: yielding of the most stressed fibre uses the unreduced squash load, while overall buckling of the member uses the reduced $C_r$. On a stocky member the first governs; on a slender one like this 10 m cantilever the second does.

Common pitfall. Adding the two cantilever moments instead of subtracting them — the question says diametrically opposite, so the arms oppose each other in bending even though their axial contributions add. A second frequent error is applying a lateral-torsional-buckling reduction to a circular hollow section; a round tube is torsionally closed and axisymmetric, so it has no LTB mode at all.

Source. CSA S16, Design of Steel Structures, Clauses 11 (Table 2), 13.3.1, 13.5 and 13.8; section properties per the CISC Handbook of Steel Construction, Part 6.

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