Question 7 of 7: Glulam Column Equivalent to the Concrete Column

Question C1: Glulam Column Equivalent to the Concrete Column (10 + 10 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. The same member as B3: 6.0 m long, effectively fixed at both ends, factored axial load $P_f = 375$ kN at the top and factored moment $M_f = 84.0$ kN·m from the mid-height wind load. Material 20f-E Douglas-fir–Larch glulam, with the assumed specified strengths $f_b = 25.6$ MPa, $f_c = 30.2$ MPa and $E_{05} = 10\,900$ MPa. Service condition factors: permanent load duration $K_D = 0.65$, dry service $K_S = 1.0$, untreated $K_T = 1.0$, $K_H = 1.0$.

Find. A rectangular glulam section that satisfies the CSA O86 combined axial-and-bending interaction.

Approach. Modify the specified strengths for load duration and service conditions, compute the compressive resistance including the column-stability factor about the weaker direction, compute the bending resistance, then test the O86 interaction expression, which carries its own second-order amplification term.

1. Design strengths. The question dictates permanent load duration, so $K_D = 0.65$ and, with dry service and untreated material, $$F_b = f_b K_D K_S K_T = 25.6(0.65) = 16.64 \text{ MPa}, \qquad F_c = 30.2(0.65) = 19.63 \text{ MPa}$$ Applying the permanent-duration factor to the wind case as well is conservative; a short-term combination would use $K_D = 1.15$.
2. Effective length. CSA O86 recommends a design value $K_e = 0.65$ for a member effectively fixed at both ends, rather than the theoretical 0.5, so $$L_e = 0.65(6000) = 3900 \text{ mm}$$ and the slenderness ratio about the weak (215 mm) direction is $C_c = 3900/215 = 18.1$, within the O86 limit of 50.
3. Trial section and size factor. Try 215 × 456 mm (twelve 38 mm laminations), $A = 98\,040$ mm², $S = 7.451 \times 10^6$ mm³. The size factor for compression is $$K_{Zc} = 6.3\,(d L)^{-0.13} = 6.3\,(215 \times 6000)^{-0.13} = 1.012 \; (\le 1.3)$$
4. Column stability factor and $P_r$. The slenderness factor is $$K_C = \left[1 + \frac{F_c K_{Zc} C_c^3}{35 E_{05} K_{SE} K_T}\right]^{-1} = \left[1 + \frac{19.63(1.012)(18.14)^3}{35(10\,900)}\right]^{-1} = 0.763$$ $$P_r = \phi F_c A K_{Zc} K_C = 0.8(19.63)(98\,040)(1.012)(0.763)/10^3 = \boxed{1188 \text{ kN}}$$ Buckling has removed about a quarter of the crushing capacity — far less severe than in the steel post of A3 because the fixed ends give a short effective length.
5. Bending resistance. With the glulam size factor $$K_{Zbg} = \left(\tfrac{130}{b}\right)^{0.1}\left(\tfrac{610}{d}\right)^{0.1}\left(\tfrac{9100}{L}\right)^{0.1} = 1.021$$ and lateral support assumed at the ends (no reduction for lateral stability on a member of this aspect ratio), $$M_r = \phi F_b S K_{Zbg} = 0.9(16.64)(7.451\times10^6)(1.021)/10^6 = 113.9 \text{ kN}\cdot\text{m}$$
6. Euler load and interaction. About the axis of bending, $$P_E = \frac{\pi^2 E_{05} I}{L_e^2} = \frac{\pi^2(10\,900)(1.699\times10^9)}{3900^2} = 12\,016 \text{ kN}$$ CSA O86 Cl. 6.5.12 combines the two actions as $$\left(\frac{P_f}{P_r}\right)^2 + \frac{M_f}{M_r}\left(\frac{1}{1 - P_f/P_E}\right) \le 1.0$$ $$\left(\frac{375}{1188}\right)^2 + \frac{84.0}{113.9}\left(\frac{1}{1 - 375/12\,016}\right) = 0.100 + 0.761 = \boxed{0.861 \le 1.0 \quad \text{OK}}$$
7. Final design. A 215 × 456 mm 20f-E Douglas-fir–Larch glulam column, bending about the 456 mm axis, satisfies the combined loading at 86 % utilisation. The axial term contributes only a tenth of the total — as in B3, the wind moment governs — so the depth rather than the width is the sensitive dimension. Note that the timber section is roughly 60 % larger in area than the 350 mm concrete square, but weighs about a fifth as much, which is the usual trade when a concrete column is converted to glulam.