Question 7 of 7: Glulam timber column ABC for a farm building

Question C1: Glulam timber column ABC for a farm building (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. Same frame and loads as Figure B2 (N=100 kN, M=400 kN·m unfactored, governing over B–C); Douglas Fir-Larch glulam 20f-E (f_b=25.6 MPa, f_c=30.2 MPa, E=12 400 MPa, E₀₅=9500 MPa); permanent load duration (K_D=0.65); wet service (K_Sb=0.80, K_Sc=0.73, K_SE=0.833); untreated (K_T=1.0).

Find. A rectangular glulam cross-section for column ABC satisfying the CSA O86 combined bending and axial compression check.

Approach. Reuse the frame analysis from Question B2 (N_f, M_f are the same physical loads, factored the same way). Compute wood design values F_b, F_c with the specified duration/service/treatment factors, then size a trial section by the CSA O86 Cl. 6.5.12 beam-column interaction, checking slenderness (C_c≤50) along the way.

1. (a) Design forces and material values. As in B2: $$N_f=\boxed{150\text{ kN}},\qquad M_f=\boxed{600\text{ kN}\cdot\text{m}}$$ Factored strengths: $$F_b=f_bK_DK_HK_{Sb}K_T=25.6(0.65)(1.0)(0.80)(1.0)=\boxed{13.31\text{ MPa}}$$ $$F_c=f_cK_DK_HK_{Sc}K_T=30.2(0.65)(1.0)(0.73)(1.0)=\boxed{14.33\text{ MPa}}$$
2. (b) Trial section and slenderness. The frame sways (roller at D), so take K_e=2.0, same as the cantilever pole of A2; over the full L=8 m, $$L_e=16\,000\text{ mm}$$. Try 365×1140 mm (30 laminations at 38 mm): $$C_c=\frac{L_e}{\min(b,d)}=\frac{16\,000}{365}=\boxed{43.8} \le 50\ \checkmark$$ Axial resistance with the slenderness factor K_c: $$K_c=\left[1+\frac{F_cK_{Zcp}C_c^3}{35E_{05}K_{SE}K_T}\right]^{-1}=0.657,\qquad P_r=\phi_cF_cAK_{Zcp}K_c=0.8(14.33)(416\,100)(1)(0.657)=\boxed{890\text{ kN}}$$
3. (c) Bending resistance and interaction. Assuming the weak axis is braced by girts (K_L=1.0) and taking the size factor conservatively as K_Zb=0.9: $$M_r=\phi_bF_bSK_{Zb}K_L=0.9(13.31)\left(\frac{365(1140)^2}{6}\right)(0.9)(1.0)=\boxed{853\text{ kN}\cdot\text{m}}$$ Critical buckling load for the amplification term: $$P_E=\frac{\pi^2E_{05}K_{SE}K_TI}{L_e^2}=\boxed{13\,750\text{ kN}}$$ Interaction (CSA O86 Cl. 6.5.12): $$\frac{N_f}{P_r}+\frac{M_f}{M_r(1-N_f/P_E)}=\frac{150}{890}+\frac{600}{853(1-150/13750)}=\boxed{0.88 \le 1.0\ \checkmark}$$