Question 7 of 7: C1 — The frame ABC designed in glued-laminated timber

Question 7: C1 — The frame ABC designed in glued-laminated timber (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. The Figure B2 frame (pin at A, 7 m column AB, 8 m beam BC on a roller at C; 100 kN horizontal at B and 400 kN vertical at midspan, unfactored), now to be built in Douglas Fir-Larch glulam. Grade 24f-EX is adopted: f_b = 30.6 MPa, f_v = 2.0 MPa, f_c = 30.2 MPa, E = 12 800 MPa, E₀₅ = 10 900 MPa, density 5.5 kN/m³. Permanent load duration gives K_D = 0.65; dry service gives K_S = 1.0; preservative treatment in dry service gives K_T = 1.0.

Find. A single Douglas-fir glulam rectangular section that satisfies bending, shear and combined axial-plus-bending for the whole frame.

Approach. Re-run the Question 5 statics with the much lighter glulam self-weight, then size one section for the governing beam moment under a K_D of 0.65, check shear (which is often the real governor in a deep glulam) and finally check the column under combined compression and bending.

1. Trial section and self-weight. Take 425 mm wide × 1520 mm deep — forty 38 mm laminations, a standard glulam width and depth: $$w_{sw}=0.425(1.520)(5.5)=3.55\ \text{kN/m}\ \Rightarrow\ 1.25(3.55)=4.44\ \text{kN/m}$$ which is barely a fifth of the concrete beam's self-weight and noticeably reduces the reactions.
2. Statics. With the same factored loads (H_f = 150 kN, P_f = 600 kN), $$\begin{aligned} &C_y=\frac{600(4)+150(7)+4.44(8)^2/2}{8}=449.0\ \text{kN} \\ &A_y=186.5\ \text{kN} \end{aligned}$$ $$\begin{aligned} &M_B=150(7)=1050\ \text{kN}\cdot\text{m} \\ &M_{max}=\boxed{1760\ \text{kN}\cdot\text{m}}\ \text{at midspan} \end{aligned}$$
3. Bending resistance. \(S=bd^2/6=425(1520)^2/6=163.7\times10^{6}\) mm³. The glulam size factor for bending is $$K_{Zbg}=\left(\frac{130}{b}\right)^{0.1}\left(\frac{610}{d}\right)^{0.1}\left(\frac{9100}{L}\right)^{0.1}=0.9(0.913)(1.013)=0.821$$ The specified strength is modified by \(K_D K_H K_{Sb} K_T=0.65\), so \(F_b=30.6(0.65)=19.89\) MPa and $$M_r=\phi F_bSK_{Zbg}=0.9(19.89)(163.7\times10^{6})(0.821)=\boxed{2406\ \text{kN}\cdot\text{m}}$$ $$\frac{M_f}{M_r}=\frac{1760}{2406}=0.73$$ The compression edge is assumed laterally supported by the roof decking, so \(K_L=1.0\); if it were not, the lateral stability factor would have to be computed and would likely govern on a member this deep.
4. Shear resistance. Taking the shear at a distance d from the support, \(V_f=449.0-4.44(1.52)=442.3\) kN. For a glulam member O86 Clause 7.5.7 gives $$V_r=\phi F_v\left(\frac{2A_g}{3}\right)=0.9\left[2.0(0.65)\right]\frac{2(425)(1520)}{3}=\boxed{504\ \text{kN}}$$ $$\frac{V_f}{V_r}=\frac{442.3}{504}=0.88$$

Shear, not bending, is what actually sizes this member: the bending check has 27 % in hand while shear has only 12 %. That is the usual pattern for a heavily loaded glulam under permanent duration, because K_D = 0.65 penalises both strengths equally but the shear demand is concentrated at the support where the depth buys nothing. Narrowing the section to the next standard width, 365 mm, would drop V_r to 433 kN and fail.

5. Column: compressive resistance. The frame sways, so the pinned-base column takes \(L_e=2(7000)=14\,000\) mm in plane; out of plane the wall framing is assumed to brace it at mid-height. The larger slenderness ratio is $$C_c=\frac{L_e}{d}=\frac{14\,000}{1520}=9.21\ \ (<50)$$ With member volume \(Z=0.425(1.520)(7)=4.52\ \text{m}^3\), \(K_{Zcg}=0.68Z^{-0.13}=0.559\), and $$K_C=\left[1+\frac{F_cK_{Zcg}C_c^{3}}{35E_{05}K_{SE}K_T}\right]^{-1}=0.978$$ $$P_r=\phi F_cAK_{Zcg}K_C=0.9(19.63)(646\,000)(0.559)(0.978)=6238\ \text{kN}$$
6. Column: combined loading. \(N_f=186.5+4.44(7)=217.6\) kN and \(M_f=1050\) kN·m. The Euler load is $$P_E=\frac{\pi^2E_{05}I}{L_e^2}=\frac{\pi^2(10\,900)(1.244\times10^{11})}{14\,000^2}=68\,267\ \text{kN}$$ and the size factor over the 7 m column length gives \(M_r=2438\) kN·m. O86 Clause 7.5.12 then requires $$\left(\frac{P_f}{P_r}\right)^{2}+\frac{M_f}{M_r}\cdot\frac{1}{1-P_f/P_E}=(0.0349)^{2}+\frac{1050}{2438}(1.003)=\boxed{0.433}\ \le\ 1.0$$

One section therefore serves the whole frame: 425 × 1520 mm D.Fir-L 24f-EX glulam, governed by shear in the beam at 0.88 and only 43 % utilised in the column. The column is so lightly stressed axially that the squared axial term is negligible and the check is essentially a bending check — exactly as in the concrete solution, where the same 1050 kN·m against 385 kN of axial load gave an eccentricity of 2.7 m.