16-Civ-A2 Elementary Structural Design · December 2015

Question 7 of 7: C1 — Timber joist — maximum concentrated load

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

Notes on this paper

Paper format. National Examinations, December 2015 — 98-Civ-A2 Elementary Structural Design, 3-hour closed-book paper (textbooks and design handbooks permitted). Seven questions in three parts: Part A (A1–A3, steel), Part B (B1–B3, reinforced concrete), Part C (C1, timber). The 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, because the set is a study resource rather than an exam script.

Reference texts.

CSA S16, Design of Steel Structures — Clauses 11 (classification), 13.3 (compression), 13.5–13.6 (flexure), 13.8 (axial + bending), 13.11–13.12 (bolts, block shear).
CISC, Handbook of Steel Construction — section property tables for W-shapes and hollow structural sections.
CSA A23.3, Design of Concrete Structures — Clauses 10 (flexure and axial), 11 (shear), 10.15 (slender columns).
CSA O86, Engineering Design in Wood — Clause 6 (modification factors), Clause 7 (sawn lumber, bending and shear).
Canadian Wood Council, Wood Design Manual — specified strengths for Beam and Stringer grades.

Check — load factoring. Note 6 on page 1 states “all loads shown are unfactored”, so every load taken from a figure is a specified load and is factored here as 1.25D + 1.5L (NBCC Table 4.1.3.2, Case 2), treating the drawn point loads as live and member self-weight as dead. Where a question states a load directly in its text without a dead/live split (B3), that load is taken as already factored; this is stated again at that question.

Question 7: C1 — Timber joist — maximum concentrated load (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.

Quantity	Value
Member 130 × 304 mm, D.Fir-L No. 2, Beam and Stringer grade	$f_b = 9.0$, $f_v = 1.5$ MPa, $E = 9500$ MPa
Span, simply supported, restrained at bearings	5.0 m
Duration of load / service condition	permanent (dead) / wet, unseasoned
Treatment	fire-retardant
Deflection limit	8 mm

Find. The largest specified central concentrated load $P$ satisfying bending, shear and deflection.

The joist and its section. Each of the three conditions — bending, shear and deflection — yields a candidate load; the smallest governs.

Approach. Assemble the O86 modification factors for a green-graded timber in wet service with a fire-retardant treatment under permanent load, compute $M_r$ and $V_r$, then solve each limit state for the allowable $P$ after deducting self-weight.

Grade, size and section properties. At 130 mm thick and 304 mm deep the member is a Beam and Stringer, so O86 Table 6.3.1B applies: $f_b = 9.0$ MPa, $f_v = 1.5$ MPa, $E = 9500$ MPa for D.Fir-L No. 2. Then $$S=\frac{bd^{2}}{6}=2.002\times10^{6}\ \text{mm}^{3},\qquad I=\frac{bd^{3}}{12}=3.044\times10^{8}\ \text{mm}^{4},\qquad A=39\,520\ \text{mm}^{2}$$
Modification factors. Duration: the load is permanent, so $K_D = 0.65$. System: the joist is isolated, so $K_H = 1.0$. Service condition: because Beam and Stringer sizes are graded green, timbers thicker than 89 mm take $K_S = 1.0$ even in wet service — the 0.84 wet-service factor belongs to dimension lumber and does not apply here. Treatment: a fire-retardant treatment in wet service gives $K_T = 0.75$ on strength and 0.80 on modulus. Size: $K_{Zb}=(305/304)^{1/9}=1.00$. Lateral stability: with $d/b = 2.34 < 4$ and the ends held against rotation, $K_L = 1.0$.
Factored bending and shear strengths. $$F_b=f_b\left(K_DK_HK_{Sb}K_T\right)=9.0(0.65)(1.0)(1.0)(0.75)=4.388\ \text{MPa}$$ $$F_v=f_v\left(K_DK_HK_{Sv}K_T\right)=1.5(0.65)(0.75)=0.731\ \text{MPa}$$ The permanent duration and the fire-retardant treatment together strip away half the published bending strength — this is the dominant effect in the whole question.
Member resistances. With $\phi = 0.9$, $$M_r=\phi F_bSK_{Zb}K_L=\boxed{7.91\ \text{kN}\cdot\text{m}},\qquad V_r=\phi F_v\left(\frac{2A}{3}\right)=\boxed{17.34\ \text{kN}}$$
Self-weight. Taking a green density of 5.5 kN/m³, $$w=0.130(0.304)(5.5)=0.217\ \text{kN/m},\qquad w_f=1.25(0.217)=0.272\ \text{kN/m}$$
Bending limit. For a central point load plus a uniform self-weight, $$M_f=\frac{1.25P(5)}{4}+\frac{0.272(5)^{2}}{8}=1.5625P+0.849\le 7.91\;\Longrightarrow\;P=4.52\ \text{kN}$$
Shear limit. $$V_f=\frac{1.25P}{2}+\frac{0.272(5)}{2}=0.625P+0.679\le 17.34\;\Longrightarrow\;P=26.7\ \text{kN}$$ Shear is nowhere near critical, as is typical for a slender solid-sawn bending member.
Deflection limit. Deflection is checked at specified (unfactored) load with $E'=E K_{SE}K_T=9500(1.0)(0.80)=7600$ MPa. The self-weight alone contributes $$\Delta_w=\frac{5wL^{4}}{384E'I}=0.765\ \text{mm}$$ leaving 7.235 mm for the point load: $$\Delta_P=\frac{PL^{3}}{48E'I}\le 7.235\ \text{mm}\;\Longrightarrow\;P=6.43\ \text{kN}$$
Governing answer. Comparing 4.52, 26.7 and 6.43 kN, bending governs: $$\boxed{P_{\max}=4.52\ \text{kN}}$$ At that load the actual deflection is 5.85 mm, comfortably inside the 8 mm limit.

Check. The deflection check above is elastic and instantaneous. For a permanently applied load on an unseasoned member in wet service, O86 requires the long-term deflection to be increased for creep, typically by a factor of 2; applying that factor would reduce the deflection-governed load to about 2.9 kN and make deflection, not bending, the controlling limit state. The question asks for a deflection “not to exceed 8 mm” without qualifying it as long-term, so the elastic value is reported; the creep sensitivity should be raised with the client before the load is accepted.

Limit state	Resistance	Allowable $P$
Bending — governs	$M_r = 7.91\ \text{kN}\cdot\text{m}$	4.52 kN
Shear	$V_r = 17.34$ kN	26.7 kN
Deflection (8 mm, elastic)	$E' = 7600$ MPa	6.43 kN
Deflection at $P = 4.52$ kN	5.85 mm < 8 mm

Topic. CSA O86 modification factors and the three limit states of a solid-sawn bending member.

Key relations. $M_r=\phi F_bSK_{Zb}K_L$ with $F_b=f_b(K_DK_HK_{Sb}K_T)$; $V_r=\phi F_v(2A/3)$; $K_{Zb}=(305/d)^{1/9}\le 1.0$ for Beam and Stringer sizes; central-load deflection $\Delta=PL^{3}/48E'I$.

Why this works. Timber strength is not a single number: the published specified strength assumes a standard-term load on dry, untreated material, and every departure from that reference is handled by a multiplicative factor. Here three departures compound — permanent duration, a strength-reducing chemical treatment, and wet service — and the first two together cut the bending strength to 49 % of its tabulated value. The one factor that does not bite is service condition, because Beam and Stringer sizes are graded in the green condition and their tabulated strengths already reflect wet material. Shear uses two thirds of the gross area rather than the full area, reflecting the parabolic distribution of shear stress across a rectangular section. Deflection is checked under specified loads because serviceability concerns real behaviour, not a factored design event.

Common pitfall. Applying the dimension-lumber wet-service factor $K_{Sb} = 0.84$ to a timber, which double-counts the moisture effect and understates the answer by 16 %. The other frequent error is factoring the loads in the deflection check; serviceability is always assessed at specified load level.

Source. CSA O86, Engineering Design in Wood, Clause 6.4 (modification factors, including Tables 6.4.2 and 6.4.4) and Clause 7.5 (sawn lumber in bending and shear); Canadian Wood Council, Wood Design Manual, Beam and Stringer tables.

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