07-Str-A2 · May 2016

Question 7 of 7: C1 — Glued-laminated timber column

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

Notes on this paper

Paper format. National Examinations — May 2016, 07-Str-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 7: C1 — Glued-laminated timber column (8 + 6 + 6)

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
Axial dead load	12 kN
Axial snow load	36 kN
Axial wind load	8 kN
Lateral load at mid-height	12 kN (wind)
Length / end conditions	6.0 m, hinged top and bottom
Restraint	at mid-height, about the weak axis
Assumed grade	20f-E D.Fir-L glulam: f_b = 25.6, f_c = 30.2, f_v = 2.0 MPa, E₀₅ = 10 900 MPa
Service conditions (assumed)	dry service, untreated: K_S = K_T = 1.0

Find. A glulam section — width and a whole number of 38 mm laminations — satisfying the CSA O86 combined axial-and-bending interaction for the governing load combination.

The mid-height restraint acts perpendicular to the lateral load: it halves the effective length for weak-axis buckling to 3.0 m while leaving the strong-axis bending span the full 6.0 m.

Approach. Test both NBCC combinations that can govern — snow principal and wind principal — compute the O86 axial and flexural resistances of a trial section with their respective duration factors, and check the parabolic-plus-amplified-bending interaction.

Interpret the restraint. A member restrained at mid-height and loaded there must be restrained perpendicular to the load; otherwise the restraint would simply absorb the 12 kN and there would be no bending at all. Accordingly the strong-axis bending span is the full 6.0 m and the weak-axis buckling length is 3.0 m.
Assemble the load combinations. NBCC 2020 Table 4.1.3.2 gives two candidates: $$\text{Case 3 (snow principal):}\quad P_{f} = 1.25(12)+1.5(36)+0.4(8) = 72.2\ \text{kN}, \quad W_{f} = 0.4(12) = 4.8\ \text{kN}$$ $$\text{Case 4 (wind principal):}\quad P_{f} = 1.25(12)+1.4(8)+0.5(36) = 44.2\ \text{kN}, \quad W_{f} = 1.4(12) = 16.8\ \text{kN}$$ For a point load at mid-span of a pin-ended member, M_f = W_fL/4: $$\begin{aligned} M_{f,\text{snow}} &= \frac{4.8(6.0)}{4} = 7.2\ \text{kN}\cdot\text{m} \\ M_{f,\text{wind}} &= \frac{16.8(6.0)}{4} = \boxed{25.2\ \text{kN}\cdot\text{m}} \end{aligned}$$
Choose the duration factor. O86 Clause 5.3.2 gives K_D = 1.15 for short-term (wind) loading and K_D = 1.0 for standard-term (snow), so the two cases must be checked with different material strengths — the wind case gets stronger wood as well as more moment.
Try a 130 × 228 mm section (six 38 mm laminations). $$\begin{aligned} A &= 29\,640\ \text{mm}^{2} \\ S &= \frac{130(228)^{2}}{6} = 1.126\times10^{6}\ \text{mm}^{3} \\ I_{x} &= 128.4\times10^{6}\ \text{mm}^{4} \end{aligned}$$ The governing slenderness is the larger of the two: $$C_{c} = \max\left(\frac{6000}{228},\ \frac{3000}{130}\right) = \max(26.3,\ 23.1) = 26.3 \;<\; 50\ \checkmark$$
Compressive resistance for the wind case. With F_c = 30.2(1.15) = 34.7 MPa, the glulam size factor $$K_{Zcg} = 0.68Z^{-0.13} = 0.68(0.178)^{-0.13} = 0.851 \quad (Z = 0.130 \times 0.228 \times 6.0\ \text{m}^{3})$$ and the slenderness factor $$K_{c} = \left[1+\frac{F_{c}K_{Zcg}C_{c}^{3}}{35E_{05}}\right]^{-1} = \left[1+\frac{34.7(0.851)(26.3)^{3}}{35(10\,900)}\right]^{-1} = 0.415$$ $$P_{r} = \phi F_{c}AK_{Zcg}K_{c} = 0.8(34.7)(29\,640)(0.851)(0.415) = \boxed{291\ \text{kN}}$$
Flexural resistance. The lateral-stability factor is unity because the restrained length gives a slender-beam ratio well inside the short-beam range: $$C_{B} = \sqrt{\frac{L_{e}d}{b^{2}}} = \sqrt{\frac{3000(228)}{130^{2}}} = 6.4 \;<\; 10 \quad\Longrightarrow\quad K_{L} = 1.0$$ With K_Zbg = 1.0 and F_b = 25.6(1.15) = 29.4 MPa, $$M_{r} = \phi F_{b}SK_{Zbg}K_{L} = 0.9(29.4)\left(1.126\times10^{6}\right) = \boxed{29.8\ \text{kN}\cdot\text{m}}$$
Combined axial load and bending — wind case. O86 Clause 7.5.12 uses a squared axial term and a P-delta amplified bending term, with $$P_{E} = \frac{\pi^{2}E_{05}I_{x}}{L^{2}} = \frac{\pi^{2}(10\,900)\left(128.4\times10^{6}\right)}{6000^{2}} = 384\ \text{kN}$$ $$\left(\frac{P_{f}}{P_{r}}\right)^{2} + \frac{M_{f}}{M_{r}\left(1-P_{f}/P_{E}\right)} = \left(\frac{44.2}{291}\right)^{2} + \frac{25.2}{29.8\left(1-44.2/384\right)} = 0.023 + 0.955 = \boxed{0.98} \le 1.0\ \checkmark$$
Combined axial load and bending — snow case. Repeating with K_D = 1.0 gives P_r = 274 kN and M_r = 25.9 kN·m, so $$\left(\frac{72.2}{274}\right)^{2} + \frac{7.2}{25.9\left(1-72.2/384\right)} = 0.069 + 0.342 = 0.41 \le 1.0\ \checkmark$$ The wind combination governs decisively, even though it carries only 61 % of the axial load: bending, not compression, is what sizes this member.
Shear. The maximum shear is half the lateral load: $$V_{r} = \phi F_{v}\left(\frac{2A}{3}\right) = 0.9(2.0)(1.15)\left(\frac{2(29\,640)}{3}\right) = 40.9\ \text{kN} \;>\; V_{f} = 8.4\ \text{kN}\ \checkmark$$ Adopt a 130 × 228 mm 20f-E D.Fir-L glulam column (six laminations), with the mid-height restraint detailed to act about the weak axis.

Check — assumed data. The question invites assumptions, and four have been made: (i) the 12 kN lateral load is wind, so it pairs with the 8 kN axial wind load in the same combination; (ii) grade 20f-E D.Fir-L, the common Canadian glulam stress grade; (iii) dry service and untreated, giving K_S = K_T = 1.0; (iv) the mid-height restraint acts about the weak axis only. The section is exactly at 98 % utilisation, so any softening of these assumptions — wet service, a lower grade, or a restraint that does not act — requires the next lamination depth, 130 × 266 mm, which brings the interaction down to about 0.74.

Result	Value
Snow-principal combination P_f / M_f	72.2 kN / 7.2 kN·m
Wind-principal combination P_f / M_f	44.2 kN / 25.2 kN·m (governs)
Trial section	130 × 228 mm (six 38 mm laminations)
Slenderness ratio C_c	26.3 (< 50)
K_Zcg / K_c / K_L	0.851 / 0.415 / 1.0
P_r / M_r / P_E (wind case)	291 kN / 29.8 kN·m / 384 kN
Interaction, wind case	0.98 — adequate
Interaction, snow case	0.41
V_f / V_r	8.4 kN / 40.9 kN
Adopted section	130 × 228 mm 20f-E D.Fir-L glulam

Topic. Combined axial compression and bending in a glued-laminated timber member, with load-duration and size effects.

Key relations. O86 Clause 7.5.8 P_r = φF_cAK_ZcgK_c with K_c = [1 + F_cK_ZcgC_c³/(35E₀₅)]⁻¹; Clause 7.5.6 M_r = φF_bSK_ZbgK_L; Clause 7.5.12 interaction (P_f/P_r)² + M_f/[M_r(1 − P_f/P_E)] ≤ 1.0; K_D = 1.15 short term, 1.0 standard term; glulam size factors K_Zcg = 0.68Z^−0.13 and K_Zbg = 1.03(BL)^−0.18, both capped at 1.0.

Why this works. Timber design differs from steel and concrete in that the material strength itself depends on how long the load acts, so a combination with less load can govern if it also brings a shorter duration — except here the effect runs the other way: the wind case gets the 15 % strength bonus and still governs, because it triples the moment. Squaring the axial term in the interaction, rather than using it linearly as steel does, reflects the fact that timber columns are not much weakened by a modest axial load when bending dominates. The 1/(1 − P_f/P_E) amplifier is the standard second-order magnification, and with P_f/P_E = 0.115 it adds about 13 % to the bending demand — small but not negligible.

Common pitfall. Checking only the combination with the largest axial load. Here that combination gives an interaction of 0.41 while the true governing case reaches 0.98, so a candidate who checks only the snow case will size the member at roughly half of what is needed. Two further traps: using the same K_D for both combinations, and taking the depth as a free variable rather than a whole number of 38 mm laminations.

Source. CSA O86:19, Engineering Design in Wood, Clauses 5.3.2 (K_D), 7.5.6, 7.5.8 and 7.5.12; Canadian Wood Council, Wood Design Manual 2020, glulam selection tables; NBCC 2020 Table 4.1.3.2.

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