Question 7 of 7: C1 — Glulam column ABC for the same frame

Question 7: C1 — Glulam column ABC for the same frame (8 + 6 + 6 = 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.

Find. A rectangular glulam section (a whole number of 38 mm laminations deep) that satisfies the O86 combined bending-and-compression interaction, plus a shear check.

Approach. The frame is unchanged, so reuse the factored actions from Question 5; size the section on bending (which will dominate at $e$ = 4 m), then compute $P_r$ with the slenderness factor and close with the O86 6.5.12 interaction and a shear check.

1. Carry the actions across. The geometry, supports and loads are those of Figure B2, so the column still carries $P_f = 150$ kN with $M_f = 600$ kN·m constant from B to C, and a shear of 150 kN below B. Timber handles second-order effects inside the interaction equation rather than through a separate magnifier, so the first-order moment is the right input here.
2. Size on bending. The factored bending strength is$$F_b = f_b(K_DK_HK_{Sb}K_T) = 30.6(0.65)(1.0)(0.80)(1.0) = 15.91\ \text{MPa}$$ so before size effects the section modulus must exceed $600\times10^{6}/(0.9\times15.91) = 41.9\times10^{6}$ mm³. Try 365 × 912 mm (a standard 365 mm glulam width, 24 laminations of 38 mm), giving $S = bd^2/6 = 50.6\times10^{6}$ mm³ and $A = 332.9\times10^{3}$ mm².
3. Size factor and moment resistance. The member volume is $0.365(0.912)(8.0) = 2.66$ m³, so the glulam size factor is$$K_{Zbg} = 1.03(bdL)^{-0.18} = 1.03(2.66)^{-0.18} = 0.864$$ Taking the lesser of $K_{Zbg}$ and the lateral-stability factor $K_L$ (which is 1.0 with the column restrained by the frame and cladding in the weak direction):$$M_r = \phi F_bSK_{Zbg} = 0.90(15.91)(50.6\times10^{6})(0.864)/10^6= \boxed{626\ \text{kN}\cdot\text{m}}\;>\;600\ \text{kN}\cdot\text{m}$$
4. Compressive resistance. With $F_c = f_cK_DK_{Sc}K_T = 30.2(0.65)(0.69) = 13.55$ MPa and $K_{Zcg} = 0.68Z^{-0.13} = 0.599$, the slenderness ratios are $L_e/d = 2(8000)/912 = 17.5$ in the plane of the frame (sway, $k = 2$) and $L_e/b = 8000/365 = 21.9$ out of plane (pin-ended). The larger governs:$$K_C = \left[1.0+\frac{F_cK_{Zcg}C_c^3}{35E_{05}K_{SE}K_T}\right]^{-1}= \left[1.0+\frac{13.55(0.599)(21.9)^3}{35(10\,900)(0.90)}\right]^{-1}=0.801$$$$P_r = \phi F_cAK_CK_{Zcg} = 0.80(13.55)(332.9\times10^{3})(0.801)(0.599)/10^3= 1729\ \text{kN}$$
5. Combined bending and compression, O86 6.5.12. The Euler load in the plane of bending, with $I = bd^3/12 = 23.07\times10^{9}$ mm⁴, is$$P_E = \frac{\pi^2E_{05}K_{SE}I}{(2L)^2} = 8726\ \text{kN}$$ and the interaction, in which the bracketed term is the P-delta amplifier, gives$$\left(\frac{P_f}{P_r}\right)^2+\frac{M_f}{M_r}\cdot\frac{1}{1-P_f/P_E}=\left(\frac{150}{1729}\right)^2+\frac{600}{626}(1.017)=0.008+0.976=\boxed{0.98\le 1.0\;\checkmark}$$ The section is 98 % utilised, and essentially all of that is bending — the axial term contributes under 1 %.
6. Shear check. Glulam shear uses two-thirds of the gross area:$$F_v = f_vK_DK_{Sv}K_T = 2.0(0.65)(0.87) = 1.131\ \text{MPa}$$$$V_r = \phi F_v\left(\frac{2A_g}{3}\right)= 0.90(1.131)\left(\frac{2(332.9\times10^{3})}{3}\right)/10^3= 226\ \text{kN}\;>\;150\ \text{kN}\;\checkmark$$ Adopt a 365 × 912 mm D.Fir-L 24f-EX glulam column, with a pinned base shoe at A detailed to transfer 150 kN of horizontal thrust and a moment-resisting steel connection at the corner C.