16-Civ-A2 Elementary Structural Design · May 2015

Question 7 of 7: C1 — Design of the oblique sawn timber purlins

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

Notes on this paper

Paper format. National Examinations, May 2015 — 98-Civ-A2 Elementary Structural Design, 3-hour duration, closed book (textbooks and design handbooks permitted; one approved Casio or Sharp calculator). Seven questions in three parts: Part A (A1–A3, structural steel, CSA S16), Part B (B1–B3, reinforced concrete, CSA A23.3) and Part C (C1, timber, CSA O86). A candidate answers two from Part A, two from Part B and the one question in Part C — five solutions, all of equal value. All seven are solved here so the set works as a complete study resource.

Reference texts. CSA S16, Design of Steel Structures, with the CISC Handbook of Steel Construction; CSA A23.3, Design of Concrete Structures, with the CAC Concrete Design Handbook; CSA O86, Engineering Design in Wood, with the CWC Wood Design Manual; National Building Code of Canada (load combinations); Kulak & Grondin, Limit States Design in Structural Steel; MacGregor & Bartlett, Reinforced Concrete: Mechanics and Design.

Check — load factors. Note 6 on page 1 states “All loads shown are unfactored”, but the paper never splits the figure loads into dead and live components. Every applied figure load in Parts A and B is therefore treated as live load and factored by 1.5; concrete self-weight, where it is not expressly excluded, would be dead load at 1.25 (NBCC 4.1.3.2, combination 1.25D + 1.5L). Part C names its components, so 1.25D + 1.5L is applied directly there. If an examiner intended a different split, the method below is unchanged — only the numerical factor moves.

Check — figure dimensions. All four figures are hand sketches on page 3; every dimension below was read from the printed figures. Figure A2 reads 80 – 300 – 80 across the bottom plate (overall 460 mm wide) with the 300 mm dimension spanning the two web centrelines, and 250 mm overall depth. Figure B1 reads 3.0 m top, 2.5 m bottom, 2.0 m deep, walls “300 constant” and cover “70 typical”. Figure B2 reads 5 m + 5 m span, 4 m + 4 m column height, roller at A and hinge at D.

Question 7: C1 — Design of the oblique sawn timber purlins (10 + 5 + 5 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.

Quantity	Value
Purlin spacing	2.5 m
Purlin span	5.5 m, single span, simply supported
Roof pitch	20.0°
Specified dead load	1.0 kPa (includes the purlin's own weight)
Specified live load	2.0 kPa
Material	D. Fir-L, Select Structural, untreated, dry service

Find. A sawn timber purlin size that satisfies biaxial bending, shear, and deflection under CSA O86.

Oblique purlin on a 20° roof. The vertical gravity line load resolves into a normal component bending the purlin about its strong axis and a tangential component bending it about its weak axis; both must be carried simultaneously.

Approach. Convert the area loads to a line load on one purlin, resolve that gravity load into components normal and tangential to the roof plane (the defining feature of an oblique purlin), then check biaxial bending by linear interaction, shear, and the resultant deflection.

Line load on one purlin. Each purlin carries a tributary width equal to the spacing: $$ w_{\text{spec}} = (1.0 + 2.0)(2.5) = 7.50~\text{kN/m} $$ $$ w_f = \left[1.25(1.0) + 1.5(2.0)\right](2.5) = 4.25(2.5) = 10.625~\text{kN/m} $$
Resolve onto the principal axes. Gravity acts vertically but the purlin's axes are rotated with the roof, so the load splits into a component normal to the roof (strong-axis bending) and one along the slope (weak-axis bending): $$ w_n = w_f\cos 20^\circ = 10.625(0.9397) = 9.984~\text{kN/m} $$ $$ w_t = w_f\sin 20^\circ = 10.625(0.3420) = 3.634~\text{kN/m} $$ This tangential component is the whole point of the question — ignoring it makes the purlin appear to pass when it does not.
Factored moments. For a simple span, $M = wL^2/8$ with $L^2/8 = 5.5^2/8 = 3.781$ m²: $$ M_{fx} = 9.984(3.781) = 37.75~\text{kN}\cdot\text{m}, \qquad M_{fy} = 3.634(3.781) = 13.74~\text{kN}\cdot\text{m} $$
Material properties and modification factors. From CSA O86 Table 6.3.1B for D. Fir-L Select Structural in the Beam and Stringer size class: $f_b = 19.5$ MPa, $f_v = 1.5$ MPa, $E = 12\,000$ MPa. Beam and Stringer sizes are graded green, so the wet-service factors are $K_{Sb} = K_{Sv} = K_{SE} = 1.0$; the load duration is standard so $K_D = 1.0$; the timber is untreated so $K_T = 1.0$; and a single member gives $K_H = 1.0$. The roof sheathing is fixed to the top of every purlin, so lateral stability is assured and $K_L = 1.0$.
Try 191 × 292 mm and reject it. This size gives $S_x = 2.715\times10^6$ and $S_y = 1.775\times10^6$ mm³, hence $M_{rx} = 47.6$ and $M_{ry} = 31.2$ kN·m. The linear interaction is $$ \frac{37.75}{47.6} + \frac{13.74}{31.2} = 0.79 + 0.44 = 1.23 \;>\; 1.0 $$ It fails by 23 percent. Note that a strong-axis-only check would have read 0.79 and passed — the tangential term is what rejects the section.
Adopt 191 × 343 mm. Check it qualifies as a Beam and Stringer: $b = 191 \ge 114$ mm and $d = 343 > b + 51 = 242$ mm. The section moduli are $S_x = 3.745\times10^6$ and $S_y = 2.086\times10^6$ mm³. The size factor is $K_{Zb} = (305/d)^{1/9} = (305/343)^{1/9} = 0.987$ for strong-axis bending, and 1.0 for weak-axis bending (where the bending dimension is 191 mm, so the formula returns a value above unity and is capped): $$ M_{rx} = \phi F_b S_x K_{Zb} K_L = 0.90(19.5)(3.745\times10^6)(0.987)\times10^{-6} = 64.9~\text{kN}\cdot\text{m} $$ $$ M_{ry} = 0.90(19.5)(2.086\times10^6)(1.0)\times10^{-6} = 36.6~\text{kN}\cdot\text{m} $$ $$ \frac{M_{fx}}{M_{rx}} + \frac{M_{fy}}{M_{ry}} = \frac{37.75}{64.9} + \frac{13.74}{36.6} = 0.58 + 0.38 = \boxed{0.96 \le 1.0} $$
Shear (5 marks). The factored end shears are $V_{fx} = w_n L/2 = 27.5$ kN and $V_{fy} = w_t L/2 = 10.0$ kN. With $A = 191(343) = 65\,513~\text{mm}^2$, O86 Clause 7.5.7 gives $$ V_r = \phi F_v \frac{2A}{3} = 0.90(1.5)\frac{2(65\,513)}{3}\times10^{-3} = 59.0~\text{kN} $$ which exceeds both components with a utilisation of 0.47 on the governing normal direction. Shear does not control, as is usual for a span-to-depth ratio of 16.
Deflection (5 marks). Deflections are computed at specified (unfactored) load, with $w_n = 7.048$ and $w_t = 2.565$ kN/m, $I_x = 642.3\times10^6$ and $I_y = 199.2\times10^6$ mm⁴: $$ \Delta = \frac{5wL^4}{384 E I} \;\Rightarrow\; \Delta_n = 10.9~\text{mm}, \quad \Delta_t = 12.8~\text{mm} $$ The in-plane deflection is the larger of the two even though its load is smaller, because I_y is only a third of I_x. Combining them vectorially, $$ \Delta = \sqrt{10.9^2 + 12.8^2} = 16.8~\text{mm} \;<\; \frac{L}{240} = 22.9~\text{mm} $$ and the live-load share alone is 11.2 mm against the $L/360 = 15.3$ mm limit. Both criteria are met.

Check — assumed data. The question invites assumptions. Three are made: the area loads act on the sloping roof surface (not the horizontal projection); the roof deck is fastened to every purlin, giving continuous lateral support so K_L = 1.0; and the load duration is standard, so K_D = 1.0. If the live load were specified per horizontal projection instead, w_f would fall by about 6 percent and the same section would still govern.

Quantity	Value
Factored line load, w_f	10.625 kN/m
Components	w_n = 9.98 kN/m, w_t = 3.63 kN/m
Factored moments	M_fx = 37.75, M_fy = 13.74 kN·m
Purlin size	191 × 343 mm D. Fir-L Select Structural (Beam and Stringer)
Moment resistances	M_rx = 64.9, M_ry = 36.6 kN·m
Biaxial interaction	0.96 ✓
Shear	V_f = 27.5 kN vs V_r = 59.0 kN ✓
Total-load deflection	16.8 mm vs L/240 = 22.9 mm ✓
Live-load deflection	11.2 mm vs L/360 = 15.3 mm ✓

Topic. Biaxial bending of sloping (oblique) roof purlins in CSA O86 limit states timber design.

Key relations. $w_n = w\cos\theta$, $w_t = w\sin\theta$; $M_r = \phi F_b S K_{Zb} K_L$ with $F_b = f_b K_D K_H K_{Sb} K_T$; size factor $K_{Zb} = (305/d)^{1/9}$; linear interaction $M_{fx}/M_{rx} + M_{fy}/M_{ry} \le 1.0$; $V_r = \phi F_v (2A/3)$; $\Delta = 5wL^4/384EI$.

Why this works. Gravity does not know that the roof is sloped, but the purlin does: its principal axes are tilted with the roof plane, so the vertical load has a component along the weak axis. That component is only about a third of the normal one, yet it consumes almost as much capacity, because the weak-axis section modulus of a 191 × 343 member is little more than half the strong-axis value. Timber uses a linear rather than a quadratic interaction because wood is brittle in bending and does not redistribute stress the way a plastic steel section does, so the two utilisations simply add. Deflection is checked at specified load rather than factored load because serviceability is about how the roof behaves in ordinary use, not at the ultimate limit state, and the two directional deflections combine as a vector because they are simultaneous and orthogonal.

Common pitfall. Checking only the strong axis. A 191 × 292 purlin reads 0.79 on that check alone and looks acceptable, but it is 23 percent overstressed once the tangential term is included. A second trap is applying the dimension-lumber wet-service factor of 0.84 to a Beam and Stringer size — B&S timbers are graded green, so their published values already account for it and K_S = 1.0. A third is combining the two deflections by adding them arithmetically, which overstates the resultant by about 40 percent here.

Source. CSA O86, Engineering Design in Wood, Clauses 6.3, 6.5.6, 7.5.7 and Table 6.3.1B; CWC Wood Design Manual, Section 2 (bending members); National Building Code of Canada 4.1.3.2 (load combinations) and Table 9.4.3.1 (deflection limits).

← Question 6 Back to the paper →