Question 7 of 7: C1. Design of oblique sawn timber purlins (10 + 10 marks)
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Examinations, May 2014 — 98-Civ-A2
Elementary Structural Design, 3 hours, closed book (handbooks and textbooks permitted).
Seven questions in three parts: Part A (A1–A3, steel), Part B (B1–B3, reinforced concrete)
and Part C (C1, timber). The candidate answers two from Part A, two from Part B and the single
Part C question — five solutions in all, of equal value. All seven are solved here,
because the set is intended as a study resource. Page 1 states that
all loads shown are unfactored, so the load factors are applied below rather than assumed
to be built in.
CISC, Handbook of Steel Construction — section property tables and the standard
beam load/moment diagrams referenced by Question A1.
CSA A23.3, Design of Concrete Structures — Clauses 10 (flexure and axial load) and
11 (shear, simplified method).
CSA O86, Engineering Design in Wood — Clause 6 (sawn lumber), Tables 6.3.1B
(Beam and Stringer specified strengths) and 6.4.2 (service-condition factors).
National Building Code of Canada — load combinations
1.25D + 1.5L.
Check: the figures on page 3 of the paper are hand-drawn; every dimension, support type and load below was read from the printed figures. The 300 kN, 250 kN, 300 kN and 80 kN forces are treated as
live load (factor 1.5) because the paper labels them only as unfactored applied loads;
concrete self-weight is the sole dead load and carries 1.25.
Question C1 — a purlin laid on a 20° roof slope: the gravity load resolves into a component normal to the roof (strong-axis bending) and one in the plane of the roof (weak-axis bending).
Given.
Quantity
Value
Species and grade
D. Fir-L, Select Structural, treated, sawn
Service condition
Wet; standard duration of load
Purlin spacing / span
2.2 m / 5.0 m, single span
Roof pitch
20.0°
Specified dead load
1.00 kPa (includes the purlin)
Specified live load
2.6 kPa
Specified strengths (O86 Table 6.3.1B, B&S)
$f_b = 19.5$, $f_v = 1.5$ MPa, $E = 12\,000$ MPa
Find. A purlin size that satisfies bending (about both axes, because the member is
oblique), shear and deflection.
Approach. Convert the area loads to a line load on one purlin, resolve it into
components normal to and in the plane of the roof, treat the result as biaxial bending on a simply
supported 5 m span, and select the smallest Beam-and-Stringer size that satisfies the interaction
equation, then verify shear and deflection.
Factored and specified line loads. Each purlin picks up a 2.2 m width of roof:
$$w_f = (1.25D + 1.5L)s = [1.25(1.00) + 1.5(2.6)](2.2) = 11.33\ \text{kN/m}$$
$$w_{spec} = (1.00 + 2.6)(2.2) = 7.92\ \text{kN/m}$$
Resolve onto the purlin axes. The purlin is laid on the slope, so gravity acts
neither along nor across its principal axes but at 20° to them:
$$w_n = w_f\cos20^\circ = 11.33(0.9397) = 10.65\ \text{kN/m}, \qquad
w_t = w_f\sin20^\circ = 11.33(0.3420) = 3.875\ \text{kN/m}$$
The first bends the purlin about its strong axis, the second about its weak axis. This is the whole
point of the word oblique in the question.
Design actions. For a simply supported single span,
$$M_{fx} = \frac{w_nL^2}{8} = \frac{10.65(5.0)^2}{8} = 33.27\ \text{kN}\cdot\text{m}, \qquad
M_{fy} = \frac{w_tL^2}{8} = 12.11\ \text{kN}\cdot\text{m}, \qquad
V_f = \frac{w_nL}{2} = 26.62\ \text{kN}$$
Establish the modification factors. Standard duration gives $K_D = 1.0$; the
member is a single purlin so $K_H = 1.0$; it is preservative treated but not incised, so $K_T = 1.0$.
Critically, Beam-and-Stringer sizes are graded wet, so the wet-service factors for timbers
are $K_{Sb} = K_{Sv} = K_{SE} = 1.0$ — unlike dimension lumber, where wet service would cost 16 %
of the bending strength. Lateral stability is provided by the roof sheathing, so $K_L = 1.0$. The
size factor for bending is $K_{Zb} = (305/d)^{1/9} \le 1.0$.
Try 191 mm × 343 mm. The section moduli are
$S_x = 3.745\times10^{6}$ mm$^3$ and $S_y = 2.086\times10^{6}$ mm$^3$, with
$K_{Zb} = (305/343)^{1/9} = 0.987$ about the strong axis and 1.0 about the weak axis:
$$M_{rx} = \phi f_b S_x K_{Zb} = 0.9(19.5)(3.745\times10^{6})(0.987) = 64.88\ \text{kN}\cdot\text{m}$$
$$M_{ry} = 0.9(19.5)(2.086\times10^{6}) = 36.60\ \text{kN}\cdot\text{m}$$
The biaxial interaction check gives
$$\frac{M_{fx}}{M_{rx}} + \frac{M_{fy}}{M_{ry}} = \frac{33.27}{64.88} + \frac{12.11}{36.60}
= 0.513 + 0.331 = 0.844 \le 1.0$$
$$\boxed{\text{Use 191} \times \text{343 mm D. Fir-L Select Structural purlins}}$$
The next size down, 191 × 292, gives an interaction ratio of 1.087 and fails, so 191 × 343 is
the smallest adequate member of that thickness.
Shear check. For sawn timber the shear resistance uses two thirds of the gross
area:
$$V_r = \phi f_v\!\left(\frac{2A}{3}\right) = 0.9(1.5)\frac{2(191)(343)}{3} = 58.96\ \text{kN}
\;>\; 26.62\ \text{kN}$$
a utilisation of only 0.45, which is typical — bending governs a 5 m purlin comfortably.
Deflection at specified loads. Deflections are computed at unfactored load and
must be resolved on both axes, then combined:
$$\Delta_n = \frac{5w_{spec}\cos20^\circ L^4}{384EI_x} = 7.86\ \text{mm}, \qquad
\Delta_t = \frac{5w_{spec}\sin20^\circ L^4}{384EI_y} = 9.22\ \text{mm}$$
$$\Delta_{total} = \sqrt{7.86^2 + 9.22^2} = 12.1\ \text{mm} < \frac{L}{240} = 20.8\ \text{mm}$$
and for live load alone the resultant is 8.75 mm against the $L/360 = 13.9$ mm limit. Both are
satisfied.
Notice that the in-plane deflection is the larger of the two even though the in-plane load is only
a third of the normal load: the weak-axis second moment is about three times smaller, and deflection
scales inversely with it. That is the recurring lesson of oblique purlins — the small tangential
component matters far more than its size suggests, and the usual remedy on longer spans is a line of
sag rods at mid-span to halve the weak-axis span.
