07-Str-A2 · December 2013

Question 3 of 7: Maximum factored load on a free-standing sign standard

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

Notes on this paper

Paper format. National Examinations, December 2013 — 07-Str-A2, Elementary Structural Design. Three hours; a "CLOSED BOOK" examination in which handbooks and textbooks are permitted. Seven questions in three parts — Part A steel (CAN/CSA-S16), Part B reinforced concrete (CAN/CSA-A23.3), Part C timber (CAN/CSA-O86). A candidate answers two questions from Part A, two from Part B and the single question in Part C, five in all, and page 1 records that all questions are of equal value, so each carries 20 marks with the split A1 (6 + 7 + 7), A2 (5 + 10 + 5), A3 (4 + 4 + 12), B1 (6 + 8 + 6), B2 (10 + 10), B3 (4 + 4 + 12), C1 (10 + 10) printed in the marking scheme. All seven questions are solved below.

Reference texts. CSA S16, Design of Steel Structures, used with the CISC Handbook of Steel Construction (section tables, Class H HSS, bolt and weld tables); CSA A23.3, Design of Concrete Structures, used with the CAC Concrete Design Handbook; CSA O86, Engineering Design in Wood, used with the CWC Wood Design Manual; National Building Code of Canada (load combinations); MacGregor and Bartlett, Reinforced Concrete: Mechanics and Design (Canadian edition); Kulak and Grondin, Limit States Design in Structural Steel.

Check — load factors used throughout. Page 1 note 6 states that all loads shown are unfactored but the paper nowhere splits them into dead and live. Every applied load in Figures A2, B1 and B3 is therefore treated as a specified live load and factored by 1.5, and self-weight, where a question asks for it, by 1.25, following NBCC load combination case 2, 1.25D + 1.5L. Question B3 names its 80 kN horizontal load as wind, so that question is additionally checked under case 4, 1.25D + 1.4W + 0.5L. If a grader intends a different dead/live split every factored action scales linearly and no design step below changes.

Check — steel grade in Figure A1. The A1 stem attaches "G40.21 300W" to the plates and is silent on the grade of the hollow section. All three components are taken as 300W, F_y = 300 MPa; if the HSS is in fact 350W the plates still yield first and the plastic moments quoted rise by less than 8 per cent. Corner radii of the hollow section are ignored (square corners), which is the usual hand-calculation idealisation and overstates the HSS area by about 4 per cent relative to the tabulated value.

Question A3: Maximum factored load on a free-standing sign standard (20 marks: 4 + 4 + 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. A 10 m free-standing circular hollow post carrying two downward cantilevered loads on opposite sides of its axis.

Given data (Figure A3 and the question stem)
Quantity	Value
Section	round HSS 273.1 × 12.7, G40.21M 350W Class H
Height above the foundation	10.0 m
Load 2P_f	3.0 m to one side of the post axis
Load P_f	2.0 m to the opposite side
End conditions	fixed at the base, free at the top (K = 2.0)
F_y, E	350 MPa, 200 000 MPa

Find. The largest value of the factored load P_f the post can carry.

Figure A3 — the sign standard. Both cantilevered loads act downwards, so they add in axial compression but partly cancel in bending.

Approach. Express the axial force and the moment in terms of P_f, compute the compressive and flexural resistances of the section, and solve the CSA S16 beam-column interaction equation for the value of P_f that makes it equal to unity.

(a) Section properties and classification (4 marks)

Properties of the circular hollow section with $D = 273.1$ mm, $t = 12.7$ mm and $D_i = 247.7$ mm: $$A = \frac{\pi}{4}\left(D^2 - D_i^2\right) = 10\,389\ \text{mm}^2 \qquad I = \frac{\pi}{64}\left(D^4 - D_i^4\right) = 88.27\times10^{6}\ \text{mm}^4$$ $$r = \sqrt{I/A} = 92.2\ \text{mm} \qquad S = 646\times10^{3}\ \text{mm}^3 \qquad Z = \frac{D^3 - D_i^3}{6} = 862\times10^{3}\ \text{mm}^3$$
Class. For a circular hollow section $D/t = 273.1/12.7 = 21.5$, against a Class 1 limit of $13\,000/F_y = 37.1$ and an axial-compression limit of $23\,000/F_y = 65.7$. The section is Class 1 in bending and is not slender in compression, so the plastic moment is available and no local-buckling reduction applies.

(b) Factored actions (4 marks)

Axial force and moment. Both loads act downwards, so they add axially, while their moments about the post axis subtract because the brackets are diametrically opposite: $$C_f = 2P_f + P_f = 3P_f \qquad M_f = 2P_f(3.0) - P_f(2.0) = 4P_f\ \ (\text{kN}\cdot\text{m})$$ There is no horizontal load, so the shear is zero everywhere and the moment is constant from the head of the post to the foundation — the worst possible moment gradient for stability, since every cross-section is equally stressed.
Effective length. A free-standing post fixed at the base has $K = 2.0$, so $$\frac{KL}{r} = \frac{2.0(10\,000)}{92.2} = 217$$

Check — the section as given exceeds the S16 slenderness limit. Clause 10.4.2.1 states that the slenderness ratio of a compression member shall not exceed 200, and $KL/r = 217$ breaches it. The calculation below is carried through with the section the question specifies, because the question asks what that section can carry; but a real sign standard of this height would be detailed with a larger diameter (a 323.9 mm section brings $KL/r$ below 180) or with the base moment reduced by making the two brackets equal and opposite. The answer should be quoted with that observation attached.

(c) Member resistances and the interaction check (12 marks)

Compressive resistance (Clause 13.3.1, with $n = 2.24$ for a hot-formed Class H hollow section): $$\lambda = \frac{KL}{r}\sqrt{\frac{F_y}{\pi^2 E}} = 217\sqrt{\frac{350}{\pi^2(200\,000)}} = 2.889$$ $$C_r = \phi A F_y\left(1 + \lambda^{2n}\right)^{-1/n} = 0.90(10\,389)(350)\left(1 + 2.889^{4.48}\right)^{-1/2.24} = 391\ \text{kN}$$ The squash load $\phi A F_y = 3\,273$ kN is reduced to an eighth of itself by buckling, which is what a slenderness of 217 costs.
Flexural resistance (Clause 13.5(a); a circular section has no weak axis, so lateral-torsional buckling cannot occur): $$M_r = \phi Z F_y = 0.90\left(862\times10^{3}\right)(350) = 271.5\ \text{kN}\cdot\text{m}$$
Second-order amplification. The Euler load in the plane of bending, using the same effective length, is $$C_e = \frac{\pi^2 EI}{(KL)^2} = \frac{\pi^2(200\,000)(88.27\times10^{6})}{(20\,000)^2} = 436\ \text{kN}$$ With a constant moment along the member $\omega_1 = 1.0$, so the amplification factor is $U_{1} = 1/(1 - C_f/C_e)$. The post is only nominally stiffer than its own Euler load, so this factor will be substantial — the physical reason a tall sign standard is governed by P-delta effects rather than by strength.
Solve the interaction equation. Clause 13.8.2 requires, for the overall member strength of a Class 1 section, $$\frac{C_f}{C_r} + \frac{U_{1}M_f}{M_r} \le 1.0 \qquad\Longrightarrow\qquad \frac{3P_f}{391} + \frac{1}{1 - 3P_f/436}\cdot\frac{4P_f}{271.5} = 1.0$$ Solving this single non-linear equation gives $$\boxed{P_f = 36.5\ \text{kN}}$$ at which $C_f = 109.6$ kN, $M_f = 146.1\ \text{kN}\cdot\text{m}$ and $U_1 = 1.336$. The applied loads are therefore 73.1 kN and 36.5 kN.
Cross-sectional strength check. Repeating Clause 13.8.2 with the squash load and $K = 1.0$ in the amplifier gives $109.6/3\,273 + 1.067(146.1)/271.5 = 0.61 < 1.0$, so the member buckling case governs, as expected for so slender a post. Had the amplification been ignored altogether the answer would have come out at 44.6 kN, a 22 per cent overestimate — the whole difference between a safe pole and one that leans progressively further under load.

Final results — Question A3
Quantity	Value
A / r / Z	10 389 mm² / 92.2 mm / 862 × 10³ mm³
D/t = 21.5	Class 1
KL/r (K = 2.0), λ	217 (exceeds the 200 limit), 2.889
C_r	391 kN
M_r	271.5 kN·m
C_e, U₁	436 kN, 1.336
Maximum factored load P_f	36.5 kN (loads of 73.1 and 36.5 kN)
Governing action at that load	C_f = 109.6 kN, M_f = 146.1 kN·m

Topic. Beam-columns: combined axial compression and bending in a slender free-standing member.

Key relations. $\lambda = (KL/r)\sqrt{F_y/\pi^2E}$; $C_r = \phi AF_y(1+\lambda^{2n})^{-1/n}$ with $n = 2.24$ for Class H and 1.34 for Class C; $M_r = \phi ZF_y$; $C_e = \pi^2EI/(KL)^2$; $U_1 = \omega_1/(1 - C_f/C_e)$ with $\omega_1 = 0.6 - 0.4\kappa \ge 0.4$; and the interaction $C_f/C_r + U_1M_f/M_r \le 1.0$.

Why this works. A column that is also bent deflects sideways, and the axial force then acts through that deflection to produce extra moment, which produces more deflection. The factor $U_1$ is the closed-form sum of that geometric series, and it blows up as $C_f$ approaches the Euler load $C_e$. Here $C_e$ is only 436 kN, so even a modest axial force costs a third of the moment capacity. The interaction equation adds the two demands as fractions of their separate resistances, which is a linear approximation to a curved failure surface but is conservative for the range of slenderness that governs practical design.

Common pitfall. The two most damaging errors here are geometric, not numerical: reading the loads as acting on the same side of the post, which gives $M_f = 8P_f$ instead of $4P_f$ and halves the answer; and taking $K = 1.0$ because the base is "fixed", when a free top means the member is a cantilever and $K = 2.0$. The third is to check strength only and never form $C_e$, which drops the P-delta amplification entirely.

Source. CSA S16, Clauses 10.4 (slenderness), 13.3 (compression), 13.5 (bending) and 13.8 (members in combined bending and compression); CISC Handbook of Steel Construction, Part 4; Kulak and Grondin, Limit States Design in Structural Steel, Chapter 8.

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