Question 7 of 7: The same column designed as a Douglas-fir glulam member

Question C1: The same column designed as a Douglas-fir glulam member (20 marks: 10 + 10)

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 Question B3 column — 6 m long, fixed at both ends, 300 kN axial at the head and 80 kN of horizontal wind at mid-height — now to be built in Douglas-fir glulam under permanent load duration, dry service and untreated.

Find. A rectangular glulam cross-section that satisfies the CSA O86 combined bending-and-compression interaction for the same loading.

Approach. Modify the specified strengths for load duration, service and treatment; compute the compressive and bending resistances of a trial section including the glulam size factors and the slenderness factor K_C; and check the O86 interaction equation, whose second-order term carries its own amplification.

(a) Material properties, modification factors and section resistances (10 marks)

1. Trial section. Take 265 × 418 mm, that is eleven 38 mm laminations on a standard 265 mm glulam width, oriented with the 418 mm depth in the plane of the wind load. Then $$A = 110\,770\ \text{mm}^2, \quad S = \frac{bd^2}{6} = 7.717\times10^{6}\ \text{mm}^3, \quad I = 1.613\times10^{9}\ \text{mm}^4$$
2. Modified strengths. With $K_D = 0.65$, $K_S = 1.0$ and $K_T = 1.0$, $$F_b = f_bK_DK_SK_T = 25.6(0.65) = 16.64\ \text{MPa}, \qquad F_c = 30.2(0.65) = 19.63\ \text{MPa}$$ The permanent-duration factor alone removes a third of the material strength, which is the single largest influence on the size of this member.
3. Size factors. For glulam the volume-based factors are $K_{Zcg} = 0.68Z^{-0.13} \le 1.0$ with the member volume $Z = 0.665\ \text{m}^3$, and $K_{Zbg} = 1.03(BL)^{-0.18} \le 1.0$ with $B = 0.265$ m and $L = 6.0$ m: $$K_{Zcg} = 0.717, \qquad K_{Zbg} = 0.948$$
4. Compressive resistance. Fixed at both ends gives an effective-length factor of 0.65 for design, so $L_e = 3\,900$ mm and the governing slenderness is about the weak axis, $C_c = L_e/b = 3\,900/265 = 14.72$, safely under the limit of 50. Then $$K_C = \left[1 + \frac{F_cK_{Zcg}C_c^3}{35E_{05}K_SK_T}\right]^{-1} = 0.895$$ $$P_r = \phi_cF_cAK_{Zcg}K_C = 0.80(19.63)(110\,770)(0.717)(0.895) = 1\,116\ \text{kN}$$
5. Bending resistance. $$M_r = \phi_bF_bS\min(K_{Zbg},K_L) = 0.90(16.64)\left(7.717\times10^{6}\right)(0.948) = 109.5\ \text{kN}\cdot\text{m}$$ and the Euler load in the plane of bending is $P_E = \pi^2E_{05}I/L_e^2 = 11\,408\ \text{kN}$.

(b) Interaction check and the final section (10 marks)

1. Factored actions. The applied loads are those of Question B3; only the self-weight changes, glulam at about $5.5\ \text{kN/m}^3$ giving $0.665(5.5) = 3.7$ kN for the whole column against 17.6 kN in concrete. Hence $$\text{case 2: } P_f = 1.5(300) + 1.25(3.7) = 454.6\ \text{kN}, \quad M_f = 24.0\ \text{kN}\cdot\text{m}$$ $$\text{case 4: } P_f = 150\ \text{kN}, \quad M_f = 84.0\ \text{kN}\cdot\text{m}$$ The first-order moments are used directly, because the O86 interaction equation carries its own amplification term.
2. Combined bending and axial compression (CSA O86 Clause 7.5.12): $$\left(\frac{P_f}{P_r}\right)^2 + \frac{M_f}{M_r}\cdot\frac{1}{1 - P_f/P_E} \le 1.0$$ For the governing wind case, $$\left(\frac{150}{1\,116}\right)^2 + \frac{84.0}{109.5}\cdot\frac{1}{1 - 150/11\,408} = 0.018 + 0.777 = 0.795$$ and for the gravity case $$\left(\frac{454.6}{1\,116}\right)^2 + \frac{24.0}{109.5}\cdot\frac{1}{1 - 454.6/11\,408} = 0.166 + 0.228 = 0.394$$ $$\boxed{\text{265} \times \text{418 mm 20f-EX D.Fir-L glulam, governing interaction } 0.795 < 1.0}$$
3. Shear. The end shear is $1.4(80)/2 = 56\ \text{kN}$, against $$V_r = \phi_vF_v\left(\tfrac{2}{3}\right)A = 0.90(2.0 \times 0.65)\left(\tfrac{2}{3}\right)(110\,770) = 86.4\ \text{kN}$$ so shear is satisfied with 35 per cent in hand.
4. Comparing the two materials. The glulam column needs $265 \times 418 = 0.111\ \text{m}^2$ of section where the concrete column needed $0.123\ \text{m}^2$ — almost the same plan area, but at roughly a fifth of the weight. The comparison is only that close because the wind case governs both: bending is where timber competes best, since its strength-to-weight ratio in bending is excellent, while the permanent-duration factor of 0.65 is what stops it from being smaller still. Connections at the two built-in ends, not the section, would be the real difficulty in timber, because developing a 84 kN·m moment through a glulam end connection requires a substantial steel shoe or a glued-in-rod detail.