16-Civ-A2 Elementary Structural Design · December 2019

Question 5 of 7: B2 — Square reinforced concrete column under wind

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

Notes on this paper

Paper format. National Examinations, December 2019 — 16-Civ-A2 Elementary Structural Design. Three hours, CLOSED BOOK. 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 single Part C question — all of equal value. All seven are solved below, because this set is a study resource rather than an exam script. Note 6 on page 1 states that all loads shown are unfactored unless otherwise stated, so every load pattern is factored here before it meets a resistance; Note 7 fixes G40.21 300W steel and 400W reinforcement unless a question says otherwise.

Reference texts.

CSA S16:19, Design of Steel Structures, and CISC Handbook of Steel Construction, 11th ed. — Clauses 13.3 (compression), 13.13 (welds), 17 (composite beams), and the angle/W-shape section tables.
CSA A23.3:19, Design of Concrete Structures, and the Cement Association of Canada Concrete Design Handbook, 4th ed. — Clauses 10 (flexure and columns) and 11 (shear).
CSA O86:19, Engineering Design in Wood, and the Canadian Wood Council Wood Design Manual — Clause 6 (sawn lumber) and the service-condition and size-factor tables.
Kulak & Grondin, Limit States Design in Structural Steel, 11th ed., Chs. 4, 5, 8, 9.
MacGregor & Bartlett, Reinforced Concrete: Mechanics and Design (Canadian edition), Chs. 4, 6, 11.
National Building Code of Canada, Table 4.1.3.2 (load combinations).

Check — assumptions declared under Note 1 of the paper. (i) Figure A1 shows a solid vertical between F and C as well as the load arrow at C; that member is required, because without it the truss counts m + r = 12 + 3 = 15 against 2n = 16 and is a mechanism. With it, m + r = 16 = 2n and the truss is determinate, which is how it is solved here. (ii) In A3 the number of shear studs is reported as a bonus; the question asks only for the ultimate moment. (iii) In B1 the two stacked dimensions read as a sum, so the tee is 100 mm flange plus 400 mm of stem, i.e. 500 mm overall, and the beam spacing of 1250 mm governs the effective flange width because no span is given. (iv) In C1 the 1.5 kPa / 1.0 kPa area loads are converted to line loads on the 0.9 m tributary width, and the deflection limits adopted are L/240 on the back span and 2L/180 on the cantilever.

Question 5: B2 — Square reinforced concrete column under wind (20 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. The design starts from a 6.0 m square column pinned at both ends, carrying 300 kN dead and 150 kN live axially together with a 5 kN/m uniform wind load, in 25 MPa concrete with interior exposure.

Given data
Item	Value
Height / end conditions	6.0 m, pinned both ends (k = 1.0)
Lateral load	5 kN/m uniformly distributed (wind)
Axial dead / live	300 kN / 150 kN
Concrete	f′_c = 25 MPa, interior exposure
Reinforcement	400W, f_y = 400 MPa
Cover	40 mm clear, interior exposure

Find. A square tied column that satisfies every NBCC Table 4.1.3.2 combination, including the slenderness effects that the 6 m pinned height forces on it.

Question B2: pin-ended 6 m column with a uniform lateral wind load; the first-order moment peaks at mid-height as wL2/8.

Approach. Build the four relevant load combinations, magnify the first-order wind moment for slenderness, and test each factored (P, M) pair against the interaction diagram of a trial section by strain compatibility. Start from the smallest section that satisfies the 1 % minimum-steel rule and work up.

Load combinations (NBCC Table 4.1.3.2). With no snow and a lateral wind load, four cases matter. Because both ends are pinned and the lateral load is uniform, the first-order moment is $M_0 = wL^{2}/8$ at mid-height and the end moments are zero.
First-order actions
Case P_f (kN) w (kN/m) M₀ (kN·m)
1.4D 420 0 0
1.25D + 1.5L + 0.4W 600 2.0 9.00
1.25D + 1.4W + 0.5L 450 7.0 31.50
0.9D + 1.4W 270 7.0 31.50
Trial section. Take a 350 mm square with 4-25M. $A_g = 122\,500$ mm², $A_{st} = 2000$ mm², so $\rho = 1.63\;\%$ — above the A23.3 Clause 10.9.1 minimum of 1 % and below the 8 % maximum. The pure-axial cap is $$P_{r,\max} = 0.80\left[\alpha_1\phi_c f′_c (A_g - A_{st}) + \phi_s f_y A_{st}\right] = 1817\ \text{kN}$$
Slenderness. With $k = 1.0$ and $r = 0.3h = 105$ mm, $$\frac{k l_u}{r} = \frac{6000}{105} = 57.1$$ For a braced member with transverse loading the end moments are zero, so $M_1/M_2 = 0$ and the A23.3 Clause 10.15.2 threshold is $34 - 12(0) = 34$. Since $57.1 > 34$, the column is slender and the moment magnifier is mandatory — it cannot be waved away.
Moment magnifier, A23.3 Clause 10.15.3. With $E_c = 4500\sqrt{25} = 22\,500$ MPa and $I_g = 350^{4}/12 = 1.25\times10^{9}$ mm⁴, use $EI = 0.4E_c I_g/(1 + \beta_d)$ where $\beta_d$ is the ratio of factored sustained axial load to total factored axial load. For the governing wind case $\beta_d = 375/450 = 0.833$: $$EI = \frac{0.4(22\,500)(1.25\times10^{9})}{1.833} = 6.14\times10^{12}\ \text{N}\cdot\text{mm}^{2}$$ $$P_c = \frac{\pi^{2} EI}{(k l_u)^{2}} = \frac{\pi^{2}(6.14\times10^{12})}{6000^{2}} = 1683\ \text{kN}$$ $$\delta_b = \frac{C_m}{1 - P_f/(\phi_m P_c)} = \frac{1.0}{1 - 450/(0.75\times1683)} = 1.554$$ with $C_m = 1.0$ because the member carries transverse load between its supports.

First-order actions
Case	P_f (kN)	w (kN/m)	M₀ (kN·m)
1.4D	420	0	0
1.25D + 1.5L + 0.4W	600	2.0	9.00
1.25D + 1.4W + 0.5L	450	7.0	31.50
0.9D + 1.4W	270	7.0	31.50

Magnified moments. Applying the same procedure case by case:

Second-order actions and utilisations
Case	P_f (kN)	δ_b	M_f (kN·m)	e = M/P (mm)	Utilisation
1.4D	420	1.000	0	0	0.23
1.25D + 1.5L + 0.4W	600	1.728	15.55	26	0.33
1.25D + 1.4W + 0.5L	450	1.554	48.95	109	0.40
0.9D + 1.4W	270	1.386	43.67	162	0.32

Capacity check by strain compatibility. For each case the capacity is taken as the point on the interaction diagram lying on the ray of constant eccentricity $e = M_f/P_f$, found by bisecting on the neutral-axis depth until the ratio $M/P$ matches. Every case falls inside the envelope, and the wind combination $1.25D + 1.4W + 0.5L$ governs at $$\boxed{\text{utilisation} = 0.40}$$
Why not a smaller column? A 300 mm square with 4-20M was tried and fails: its slenderness rises to $kl_u/r = 66.7$, the magnifier for the wind case reaches 2.95, and the demand exceeds the capacity by 33 %. The 350 mm square is therefore the smallest practical answer, and its apparently low utilisation is the familiar consequence of the 1 % minimum-steel rule and slenderness, not of a conservative load estimate.
Detailing. Ties are 10M at the smaller of $16d_b = 403$ mm, $48d_{tie} = 542$ mm and the least column dimension 350 mm, so use 10M ties at 350 mm, with the first tie 175 mm from each end. Clear cover 40 mm suits interior exposure per A23.3 Table 17.

Final results
Quantity	Value
Section selected	350 mm square, 4-25M (ρ = 1.63 %), 10M ties @ 350 mm
P_r,max (pure axial)	1817 kN
Slenderness kl_u/r	57.1 (threshold 34 — slender)
Governing combination	1.25D + 1.4W + 0.5L
P_f / M₀ / δ_b / M_f	450 kN / 31.50 kN·m / 1.554 / 48.95 kN·m
Governing utilisation	0.40
Gravity case 1.25D + 1.5L + 0.4W	P = 600 kN, M_f = 15.55 kN·m, utilisation 0.33
300 mm square trial	fails — utilisation 1.33

Topic. Slender tied column under combined axial load and transverse wind.

Key relations. $M_0 = wL^{2}/8$ for a pinned-pinned member with a uniform lateral load; A23.3 10.15.2 threshold $34 - 12(M_1/M_2)$; $EI = 0.4E_c I_g/(1 + \beta_d)$; $P_c = \pi^{2}EI/(kl_u)^{2}$; $\delta_b = C_m/[1 - P_f/(\phi_m P_c)]$ with $\phi_m = 0.75$; $P_{r,\max} = 0.80[\alpha_1\phi_c f′_c(A_g - A_{st}) + \phi_s f_y A_{st}]$.

Why this works. A slender column deflects under its transverse load, and the axial force then acts through that deflection to produce extra moment, which produces more deflection. The magnifier sums that geometric series, and it blows up as $P_f$ approaches the critical load — which is exactly why halving the section from 350 to 300 mm is catastrophic rather than merely uneconomic: $P_c$ scales with $h^{4}$, so a 15 % reduction in depth removes 45 % of the buckling load. Because the axial and moment demands come from different combinations, every case must be plotted against the interaction diagram; the one with the largest $P$ is not necessarily the critical one.

Common pitfall. Checking only the gravity case is the most common failure. Here $1.25D + 1.5L + 0.4W$ has the largest axial load (600 kN) but the wind case at 450 kN governs, because its eccentricity is four times larger. A second trap is using the theoretical $k = 0.5$ or assuming "short column" without computing $kl_u/r$; the question states pinned ends, so $k = 1.0$ and the member is unambiguously slender.

Source. CSA A23.3:19 Clauses 10.9.1, 10.10, 10.15.2 and 10.15.3; NBCC Table 4.1.3.2; MacGregor & Bartlett, Chs. 11 and 12.

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