Question 7 of 7: Design of oblique sawn timber purlins
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Examinations,
May 2014 — 07-Str-A2 Elementary Structural Design, 3 hours, closed book
(handbooks and textbooks permitted). Seven questions in three parts: Part A steel
(three questions), Part B reinforced concrete (three questions), Part C timber
(one question). Candidates answer two from A, two from B and the one in C — five
solutions in all, every question of equal value (20 marks). All seven are worked
below.
Reference texts for 07-Str-A2.
CSA S16:19, Design of Steel Structures — named on the paper.
CSA A23.3:19, Design of Concrete Structures — named on the paper.
CSA O86:19, Engineering Design in Wood — named on the paper.
CISC, Handbook of Steel Construction (section tables and beam-diagram
formulae; the paper directs the candidate to it in question A1).
CWC, Wood Design Manual (O86 specified strengths and modification factors).
MacGregor & Bartlett, Reinforced Concrete: Mechanics and Design
(Canadian edition).
Kulak & Grondin, Limit States Design in Structural Steel (CISC).
NBCC 2020 Part 4 for the load combinations.
Check — load factors, stated once for the
whole paper. Page 1 note 6 says “All loads shown are
unfactored”, and the paper nowhere splits dead load from live. Every applied
load below is therefore taken as specified live load with
$\alpha_L = 1.5$ (NBCC 2020 Table 4.1.3.2, principal-load case 2); member
self-weight is small against the applied actions and is neglected except where noted.
A candidate who instead assumed a dead/live split would obtain proportionally different
factored actions but identical method and identical section proportions.
Find. A sawn timber section that satisfies biaxial bending,
shear and deflection for the stated roof.
“Oblique” purlins: laid normal to the roof surface, their principal axes are tilted $20^{\circ}$ from the vertical, so the gravity load resolves into components about both axes and the member is in biaxial bending.
Check — how the area loads are
applied. Both the 1.00 kPa and the 2.6 kPa are taken to act on the
sloping roof area, so the line load on a purlin is the pressure times the 2.2 m
spacing measured up the slope. That is unambiguous for the dead load and conservative for
the live load: if the 2.6 kPa were instead a snow load specified on the horizontal
projection, the live line load would fall by $\cos 20^{\circ}$ to 5.38 kN/m, the
factored load to 10.82 kN/m, and the interaction from 0.84 to 0.81 — the same
section.
Approach. Convert the area loads to a line load on one purlin,
factor them, resolve into components normal to and along the roof plane because the purlin
sits oblique, get the two bending moments, then select a section that satisfies the O86
biaxial interaction together with shear and deflection.
Line loads on one purlin.
$$w_D = 1.00 \times 2.2 = 2.20\ \text{kN/m},\qquad
w_L = 2.6 \times 2.2 = 5.72\ \text{kN/m}$$
$$w_f = 1.25 w_D + 1.5 w_L = 1.25(2.20) + 1.5(5.72) = 2.75 + 8.58 = 11.33\ \text{kN/m}$$
(NBCC 2020 load combination 2; the dead load is a companion here, so both factors
apply.)
Resolve onto the purlin’s own axes. An oblique purlin is laid
with its faces parallel and perpendicular to the roof plane, so its principal axes are
rotated $20^{\circ}$ from the vertical and the vertical load splits:
$$w_{fx} = w_f\cos 20^{\circ} = 11.33(0.9397) = 10.65\ \text{kN/m}\ \text{(normal to the roof)}$$
$$w_{fy} = w_f\sin 20^{\circ} = 11.33(0.3420) = 3.88\ \text{kN/m}\ \text{(down the slope)}$$
The second component is what makes this question different from an ordinary purlin: it
bends the member about its weak axis.
Modification factors, and the wet-service trap. For sawn
timbers (any dimension greater than 89 mm) O86 Table 6.4.2 gives service-condition
factors of 1.00 even in wet service, because timbers are supplied
unseasoned and the specified strengths already reflect that — unlike dimension
lumber, where $K_{Sb} = 0.84$ would apply. Standard-term loading gives $K_D = 1.0$; the
purlins are spaced at 2.2 m so there is no case-1 system factor, $K_H = 1.0$; the
treatment is preservative without incising, so $K_T = 1.0$. Because $d/b \leq 4$ for the
section chosen below, O86 Cl.7.5.6.4 gives $K_L = 1.0$ without needing lateral support
calculations. Only the size factor is not unity:
$$K_{Zb} = \left(\frac{305}{d}\right)^{1/9} \leq 1.0$$
Biaxial interaction.
$$\frac{M_{fx}}{M_{rx}} + \frac{M_{fy}}{M_{ry}}
= \frac{33.27}{64.88} + \frac{12.11}{36.60} = 0.513 + 0.331$$
$$\boxed{0.844 \leq 1.0 \quad \text{— 191} \times \text{343 mm is adequate}}$$
The next section down, 191 × 292, returns 1.087 and fails, and 140 × 394
returns 1.075 and also fails, so 191 × 343 is the economical choice among the
standard sizes.
Shear. The resultant end reaction is
$\sqrt{10.65^2 + 3.88^2}(5.0)/2 = 28.33$ kN, and for sawn timber
$$V_r = \phi F_v \frac{2A}{3} = 0.9(1.5)\frac{2(191)(343)}{3} \times 10^{-3}
= 58.96\ \text{kN} \;>\; 28.33\ \text{kN}$$
comfortable at 48 per cent.
Deflection. With $I_x = 642.3 \times 10^6$ and
$I_y = 199.2 \times 10^6$ mm$^4$ and $E = 12\,000$ MPa (again with no wet-service
reduction for a timber), the specified live load gives
$$\Delta_x = \frac{5 w_{Lx} L^4}{384 EI_x} = 5.68\ \text{mm},\qquad
\Delta_y = 6.66\ \text{mm},\qquad \Delta = \sqrt{5.68^2 + 6.66^2} = 8.75\ \text{mm}$$
against $L/360 = 13.9$ mm. Under total specified load the resultant is 12.1 mm against
$L/240 = 20.8$ mm. Both satisfied — note that the weak-axis component is the larger
of the two, another consequence of the oblique arrangement.
Design note worth making. The weak-axis term consumes a third of the
capacity. A single row of sag rods at mid-span would quarter $M_{fy}$ to
3.03 kN·m, drop the interaction to $0.513 + 0.083 = 0.60$, and allow the smaller
191 × 292 section. The question specifies single-span purlins with no such
restraint, so 191 × 343 is the answer given — but the observation is what a
practising designer would put on the drawing.
