07-Str-A2 · May 2017

Question 7 of 7: C1 — Oblique sawn-timber roof purlins

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Question 7: C1 — Oblique sawn-timber roof purlins (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.

Find. A sawn timber size that satisfies biaxial bending, shear and deflection for the oblique purlin.

Approach. Convert the areal loads to a line load on one purlin, resolve gravity into components normal and tangential to the roof plane, design for biaxial bending using the linear interaction of O86, then check shear and the resultant deflection at specified load.

1. Line loads. Each purlin carries a 2.5 m tributary width of roof: $$w_D = 1.5(2.5) = 3.75\ \text{kN/m}, \qquad w_L = 3.0(2.5) = 7.5\ \text{kN/m}$$ $$w_f = 1.25(3.75) + 1.5(7.5) = \boxed{15.94\ \text{kN/m}}, \qquad w_{spec} = 11.25\ \text{kN/m}$$
2. Resolve onto the roof axes. Gravity acts vertically but the purlin's principal axes are tilted with the roof, so the load splits into a component normal to the roof (strong-axis bending) and one down the slope (weak-axis bending): $$w_{fx} = w_f \cos 18^\circ = 15.94(0.9511) = 15.16\ \text{kN/m}$$ $$w_{fy} = w_f \sin 18^\circ = 15.94(0.3090) = 4.92\ \text{kN/m}$$ The tangential component is the whole point of the word "oblique" — a purlin laid normal to a pitched roof bends about both axes at once.
3. Design moments. Simply supported over 5 m: $$M_{fx} = \frac{w_{fx}L^{2}}{8} = \frac{15.16(25)}{8} = \boxed{47.37\ \text{kN}\cdot\text{m}}, \qquad M_{fy} = \frac{4.92(25)}{8} = \boxed{15.39\ \text{kN}\cdot\text{m}}$$
4. Modification factors. Standard duration gives K_D = 1.0; dry service K_S = 1.0; the purlins at 2.5 m centres are not a load-sharing system so K_H = 1.0; treated but not incised, so K_T = 1.0; and the roof sheathing holds the compression edge, so K_L = 1.0. The size factor is the only one that bites: $$K_{Zb} = \left(\frac{305}{d}\right)^{1/9} \le 1.0$$
5. Try a 191 × 394 mm Beam-and-Stringer section. The section moduli are $$S_x = \frac{bd^{2}}{6} = \frac{191(394)^{2}}{6} = 4.942\times10^{6}\ \text{mm}^{3}, \qquad S_y = \frac{db^{2}}{6} = 2.396\times10^{6}\ \text{mm}^{3}$$ with K_Zb = (305/394)^1/9 = 0.972 for strong-axis bending and 1.0 for weak-axis bending (because b = 191 mm < 305 mm caps the factor at unity). Hence $$M_{rx} = \phi f_b S_x K_{Zb} = 0.9(19.5)(4.942\times10^{6})(0.972) = 84.29\ \text{kN}\cdot\text{m}$$ $$M_{ry} = 0.9(19.5)(2.396\times10^{6})(1.0) = 42.04\ \text{kN}\cdot\text{m}$$
6. Biaxial interaction. With no axial load, O86 uses the linear form: $$\frac{M_{fx}}{M_{rx}} + \frac{M_{fy}}{M_{ry}} = \frac{47.37}{84.29} + \frac{15.39}{42.04} = 0.562 + 0.366 = \boxed{0.93 \le 1.0}\ \checkmark$$ The tangential term contributes 39 % of the total. Checking strong-axis bending alone would return 0.56 and pass a section that is really at 93 % of capacity — and a lighter size would fail outright.
7. Shear. Using the O86 sawn-timber form with the two-thirds gross area, $$V_f = \frac{w_{fx}L}{2} = \frac{15.16(5)}{2} = 37.9\ \text{kN}$$ $$V_r = \phi f_v \left(\frac{2A_g}{3}\right) = 0.9(1.5)\left(\frac{2(191)(394)}{3}\right) = 67.7\ \text{kN} \;>\; 37.9\ \text{kN}\ \checkmark$$ a utilisation of 0.56, so bending governs on this member.
8. Deflection at specified load. Deflection is checked unfactored. With I_x = 973.5×10⁶ and I_y = 228.8×10⁶ mm⁴, $$\Delta_x = \frac{5w_xL^{4}}{384EI_x} = 7.45\ \text{mm}, \qquad \Delta_y = \frac{5w_yL^{4}}{384EI_y} = 10.30\ \text{mm}$$ Note that the in-plane deflection is the larger of the two despite carrying less than a third of the load, because I_y is 4.3 times smaller. Combining them vectorially, $$\Delta = \sqrt{7.45^{2} + 10.30^{2}} = \boxed{12.7\ \text{mm}} \;<\; \frac{L}{240} = 20.8\ \text{mm}\ \checkmark$$

Design. Use 191 × 394 mm D.Fir-L Select Structural sawn purlins at 2.5 m centres, single span 5 m, laid with the 394 mm dimension normal to the roof plane, the compression edge restrained by the roof sheathing, and bearing seats at both ends detailed for the 37.9 kN reaction.