16-Civ-A2 Elementary Structural Design · May 2018

Question 7 of 7: C1 — Glulam column under dead, live and wind loads

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

Notes on this paper

Paper format. National Examinations, May 2018 — 16-Civ-A2 Elementary Structural Design. Three hours, CLOSED BOOK (handbooks and textbooks permitted). Seven questions in three parts: Part A (A1–A3, steel), Part B (B1–B3, reinforced concrete), Part C (C1, timber). 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 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, Table 7 minimum fillet sizes).
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.
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 2015, Table 4.1.3.2 (load combinations).

Check — one dimension is missing from the source. Question A2 describes a stub cantilever welded to a column but the paper contains no Figure A2 and never states the cantilever projection. The projection is therefore taken as L = 1.5 m from the column face throughout question A2; every result below is also given in the general form so any other projection can be substituted directly. All other data are read from the printed text and from Figure A3, B1, B2 and B3 on page 3.

Question 7: C1 — Glulam column under dead, live and wind loads (8 + 6 + 6 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
Section	130 × 228 mm, 20f-EX Spruce-Pine glulam
Specified strengths (O86 Table 7.3)	f_b = 25.6, f_c = 25.2, f_v = 1.75 MPa; E = 10 300, E₀₅ = 8600 MPa
Height, L	5.0 m, pin-ended, restrained in the weak direction
Axial loads (specified)	10 kN dead + 35 kN live
Wind	10 kN concentrated at mid-height
Service conditions	dry, untreated: K_S = K_T = K_H = 1.0

Find. P_f, P_r, M_f and M_r for the dead-plus-live-plus-wind case, and the resulting interaction check.

Figure 7.1 — Glulam column: axial load at the head, wind as a point load at mid-height bending the member about its strong (228 mm) axis.

Approach. Form the NBCC load combinations that contain all three loads, take the governing one, compute the compressive and bending resistances with the appropriate duration factor, then apply the O86 combined-loading interaction.

Load combinations (NBCC 2015 Table 4.1.3.2). Two combinations contain dead, live and wind together:
Case 3 — 1.25D + 1.5L + 0.4W: P_f = 1.25(10) + 1.5(35) = 65.0 kN, M_f = 0.4(10)(5)/4 = 5.0 kN·m;

Case 4 — 1.25D + 1.4W + 0.5L: P_f = 1.25(10) + 0.5(35) = 30.0 kN, M_f = 1.4(10)(5)/4 = 17.5 kN·m.
Both are evaluated below; Case 4, with wind as the principal load, proves to govern by a wide margin. The moment comes from a central point load on a pin-ended member, M = WL/4.
Section properties and duration factor. A = 130(228) = 29 640 mm², S = bd²/6 = 1.126 × 10⁶ mm³, I = bd³/12 = 128.4 × 10⁶ mm⁴. With wind as the principal load the duration factor is K_D = 1.15, so the factored strengths are $$F_{c}=f_{c}K_{D}=25.2(1.15)=28.98\ \text{MPa},\qquad F_{b}=f_{b}K_{D}=25.6(1.15)=29.44\ \text{MPa}$$
Compressive resistance (O86 Cl. 7.5.8). The column is restrained in the weak direction, so buckling can only occur about the strong axis and the slenderness ratio uses the 228 mm dimension: $$C_{c}=\frac{L_{e}}{d}=\frac{5000}{228}=21.9\ \ (\le 50)$$ The size factor for glulam in compression uses the member volume Z = 0.130(0.228)(5.0) = 0.148 m³: $$K_{Zcg}=0.68Z^{-0.13}=0.68(0.148)^{-0.13}=0.872$$ $$K_{c}=\left[1+\frac{F_{c}K_{Zcg}C_{c}^{3}}{35E_{05}}\right]^{-1}=\left[1+\frac{28.98(0.872)(10\,545)}{35(8600)}\right]^{-1}=0.531$$ $$P_{r}=\phi F_{c}AK_{Zcg}K_{c}=0.80(28.98)(29\,640)(0.872)(0.531)=\boxed{318\ \text{kN}}$$
Bending resistance (O86 Cl. 7.5.6). Because the member is restrained in the weak direction, lateral-torsional buckling is prevented and K_L = 1.0. The bending size factor $$K_{Zbg}=\left(\frac{130}{b}\right)^{0.1}\left(\frac{610}{d}\right)^{0.1}\left(\frac{9100}{L}\right)^{0.1}=1.17\ \Rightarrow\ \text{taken as }1.0$$ so the resistance is governed by strength: $$M_{r}=\phi F_{b}SK_{Zbg}K_{L}=0.90(29.44)(1.126\times10^{6})=\boxed{29.8\ \text{kN}\cdot\text{m}}$$
Combined loading (O86 Cl. 7.5.12). The Euler load about the bending axis is $$P_{E}=\frac{\pi^{2}E_{05}I}{L_{e}^{2}}=\frac{\pi^{2}(8600)(128.4\times10^{6})}{5000^{2}}=436\ \text{kN}$$ and the interaction equation for a glulam member in combined bending and compression is $$\left(\frac{P_{f}}{P_{r}}\right)^{2}+\frac{M_{f}}{M_{r}\left(1-\dfrac{P_{f}}{P_{E}}\right)}\le 1.0$$ $$\left(\frac{30.0}{318}\right)^{2}+\frac{17.5}{29.8\left(1-\dfrac{30.0}{436}\right)}=0.009+0.630=\boxed{0.64\ \le 1.0}$$
Check the companion combination. For Case 3 the duration factor drops to K_D = 1.0, giving P_r = 294 kN and M_r = 26.0 kN·m against P_f = 65.0 kN and M_f = 5.0 kN·m, so the interaction returns 0.049 + 0.226 = 0.28. Case 4 therefore governs, and the higher K_D of the wind case does not offset the far larger moment.

The column is adequate, working at 64 % of capacity, and the result is overwhelmingly a bending problem: the axial term contributes barely 1 % of the unity check because it enters squared and the applied load is under a tenth of P_r. Reducing the depth would be the first economy to explore, but 228 mm is already close to the minimum that keeps the interaction below unity.

Quantity	Case 4 (governs)	Case 3
Combination	1.25D + 1.4W + 0.5L	1.25D + 1.5L + 0.4W
K_D	1.15	1.00
P_f	30.0 kN	65.0 kN
P_r	318 kN	294 kN
M_f	17.5 kN·m	5.0 kN·m
M_r	29.8 kN·m	26.0 kN·m
Interaction	0.64 — adequate	0.28

Topic. Glulam column under combined axial compression and transverse bending to CSA O86.

Key relations. P_r = φF_cAK_ZcgK_c with φ = 0.8, K_Zcg = 0.68Z^−0.13 and K_c = [1 + F_cK_ZcgC_c³/(35E₀₅)]⁻¹; M_r = φF_bSK_ZbgK_L with φ = 0.9; P_E = π²E₀₅I/L_e²; interaction (P_f/P_r)² + M_f/[M_r(1 − P_f/P_E)] ≤ 1.0.

Why this works. Timber design multiplies a specified strength by a chain of modification factors, each capturing one physical effect: load duration (K_D), moisture (K_S), treatment (K_T), member size (K_Z) and slenderness (K_c or K_L). Wood is stronger under short-duration load because creep-rupture has no time to develop, which is why K_D = 1.15 for a wind-governed case and only 0.65 for permanent load. The interaction equation squares the axial term because a slender wood column's failure envelope is convex, and the amplification factor 1/(1 − P_f/P_E) handles the P-delta effect inside the bending term rather than through a separate magnifier.

Common pitfall. Using the same K_D for both combinations. The two cases have different principal loads, so Case 3 must be evaluated at K_D = 1.0; borrowing the wind value of 1.15 overstates its resistance by 15 %. The second pitfall is computing C_c from the 130 mm dimension — the phrase “restrained in the weak direction” removes that buckling mode entirely, so the slenderness that matters is 5000/228.

Source. CSA O86:19 Cl. 7.5.6, 7.5.8, 7.5.12 and Table 7.3; Canadian Wood Council, Wood Design Manual, glulam column and beam-column sections; NBCC 2015 Table 4.1.3.2.

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