16-Civ-A2 Elementary Structural Design · December 2018

Question 7 of 7: C1 — Glulam design of the beam in Figure B2

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

Notes on this paper

Paper format. National Examinations, December 2018 — 16-Civ-A2 Elementary Structural Design. Three hours, CLOSED BOOK (handbooks and textbooks permitted). Seven questions in three parts: Part A (A1–A3, steel to CSA S16), Part B (B1–B3, reinforced concrete to CSA A23.3), Part C (C1, timber to CSA O86). A candidate submits five solutions — two from Part A, two from Part B and the one question in Part C — all of equal value. All seven questions are solved below, because this set is a study resource rather than an exam script. Page 1 states that all loads shown are unfactored, so every load case is factored here (1.5 on the specified live/imposed loads, 1.25 on self weight) before any resistance is compared against it.

Reference texts.

CSA S16:19, Design of Steel Structures — and CISC, Handbook of Steel Construction, 11th ed. (section tables, Table 2 class limits, Clause 13.8 interaction, Clause 13.5 effective widths).
CSA A23.3:19, Design of Concrete Structures — and Cement Association of Canada, Concrete Design Handbook, 4th ed.
CSA O86:19, Engineering Design in Wood — and Canadian Wood Council, Wood Design Manual (glulam selection tables).
Kulak & Grondin, Limit States Design in Structural Steel, 11th ed., Chs. 4, 5, 8, 9.
MacGregor & Bartlett, Reinforced Concrete: Mechanics and Design (Canadian edition), Chs. 4, 6, 11.
National Building Code of Canada, Table 4.1.3.2 (load combinations).

Check — assumptions declared under Note 1 of the paper. (i) In A2 the single dimension h is the wall thickness of the whole fabricated panel (top plate and both stems), which is how the figure dimensions it. (ii) In A3 the bolted joint at the column is a simple (shear-only) connection and the steel tie is a two-force member, which is what makes the stub beam determinate. (iii) In B1/B3 the left column A–B is taken as 6 m high, level with the roller at D, as Figure B1 draws it; the beam self weight is included, the column self weight is not (it does not change any design action at C). (iv) In B2 and C1 the beam is drawn with a tapered soffit, but part (a) asks for a uniform cross-section, so a prismatic member is designed.

Question 7: C1 — Glulam design of the beam in Figure B2 (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
Geometry and loads	as Figure B2: 5.0 m span A to B, 3.0 m overhang to C, 100 kN at 2.0 m and 3.0 m, 20 kN at C
Material	Douglas-fir glulam, 24f-EX D.Fir-L: $f_b$ = 30.6 MPa, $f_v$ = 2.0 MPa, $E$ = 12 800 MPa
Service condition (a)	dry — $K_{Sb} = K_{Sv} = K_{SE} = 1.0$
Load duration (b)	permanent — $K_D = 0.65$
Glulam density	5.5 kN/m³; laminations 38 mm

Find. A rectangular glulam section for ABC satisfying bending, shear and deflection to CSA O86.

Approach. Reanalyse the Figure B2 beam with the (much lighter) timber self weight, apply the O86 modification factors for permanent duration and dry service, and size the section — expecting shear rather than bending to govern, because the permanent-duration factor cuts both strengths equally while shear resistance depends only on gross area.

Trial section and self weight. Try 265 mm x 874 mm, i.e. 23 laminations of 38 mm. Its self weight is $0.265(0.874)(5.5) = 1.27$ kN/m, so $w_f = 1.25(1.27) = 1.59$ kN/m — an order of magnitude lighter than the concrete alternative in B2.
Factored actions. Repeating the Question B2 analysis with this self weight, $$M_f=\boxed{265.9\ \text{kN}\cdot\text{m}}\ \text{(sagging, at 2.0 m)},\qquad V_f=173.4\ \text{kN}\ \text{(just left of B)}$$
Modification factors. Dry service and permanent duration give $$F_b=f_b\left(K_DK_HK_{Sb}K_T\right)=30.6(0.65)(1.0)(1.0)(1.0)=19.89\ \text{MPa}$$ $$F_v=f_v\left(K_DK_HK_{Sv}K_T\right)=2.0(0.65)(1.0)(1.0)(1.0)=1.30\ \text{MPa}$$ The permanent-duration factor of 0.65 is severe — it removes 35 % of both strengths — and it is what makes this a large section for a 5 m span.
Section properties and the size factor. $A_g = 265(874) = 231.6\times10^{3}$ mm² and $S = bd^{2}/6 = 33.74\times10^{6}$ mm³. The glulam size factor is $$K_{Zbg}=1.03\left(b\,d\,L\right)^{-0.18}=1.03(0.265\times0.874\times5.0)^{-0.18}=1.003\ \Rightarrow\ \text{use }1.0$$ with $b$, $d$ and $L$ in metres. Lateral stability: the beam is assumed restrained against lateral displacement by the storage deck it supports, so $K_L = 1.0$.
Bending resistance. $$M_r=\phi F_b S K_{Zbg}=0.90(19.89)\left(33.74\times10^{6}\right)(1.0)=\boxed{604\ \text{kN}\cdot\text{m}}$$ against $M_f = 265.9$ kN·m — a utilisation of only 0.44.
Shear resistance — the governing check. For glulam, $$V_r=\phi F_v\left(\frac{2A_g}{3}\right)=0.90(1.30)\left(\frac{2\times231.6\times10^{3}}{3}\right)=\boxed{181\ \text{kN}}$$ against $V_f = 173.4$ kN, a utilisation of 0.96. Shear is the binding limit state, and it is what sets the size: the area required is $$A_{\text{req}}=\frac{3V_f}{2\phi F_v}=\frac{3(173.4\times10^{3})}{2(0.90)(1.30)}=222\,300\ \text{mm}^{2}$$ The next lamination down, 265 x 836 (22 laminations), gives only $V_r = 172.8$ kN against a shear of 173.2 kN, so it fails — 23 laminations is genuinely the minimum.
Deflection. Under the specified (unfactored) loads with $E_{05}$ taken as $0.9E$, the second moment of area is $I = 265(874)^{3}/12 = 1.474\times10^{10}$ mm⁴ and the peak deflection sums to about 2.8 mm. Permanent loading requires the deflection to be doubled for creep (O86 Clause 7.4.4), giving 5.7 mm against the $L/240 = 20.8$ mm limit. Deflection is not close to governing.

This is the characteristic shape of a permanent-duration glulam design: bending is easy and shear is desperate. Bending resistance grows with $bd^{2}$ while shear resistance grows only with $bd$, so making a beam deeper barely helps its shear capacity. The correct response when a glulam beam fails in shear is to add width or accept more laminations for area, never to chase depth.

Result	Value
Factored moment $M_f$ / shear $V_f$	265.9 kN·m / 173.4 kN
Factored strengths $F_b$ / $F_v$ ($K_D$ = 0.65, dry)	19.89 MPa / 1.30 MPa
Section adopted	265 mm x 874 mm 24f-EX D.Fir-L glulam (23 laminations)
Bending resistance $M_r$ / utilisation	604 kN·m / 0.44
Shear resistance $V_r$ / utilisation — governs	181 kN / 0.96
Gross area required by shear	222 300 mm²
Creep-adjusted deflection / limit $L/240$	5.7 mm / 20.8 mm

Topic. Glulam beam design to CSA O86: modification factors, and why shear governs under permanent load.

Key relations. $F_b = f_b K_DK_HK_{Sb}K_T$ and $F_v = f_v K_DK_HK_{Sv}K_T$; $M_r = \phi F_b S K_{Zbg}K_L$ with $K_{Zbg}=1.03(bdL)^{-0.18} \le 1.0$; $V_r = \phi F_v\left(2A_g/3\right)$; $\phi = 0.90$ for glulam; depths are whole multiples of the 38 mm lamination.

Why this works. The duration-of-load factor is the distinctive feature of timber design: wood loses strength under sustained stress, so a load that sits on a member for years is resisted at 65 % of the short-term value. Because $K_D$ multiplies bending and shear strengths alike, it does not change which limit state governs — the geometry does. $M_r \propto bd^{2}$ but $V_r \propto bd$, so as spans get short and loads get heavy the shear check overtakes bending, and the section ends up sized by area.

Common pitfall. Sizing on bending and then merely "checking" shear. In a heavily loaded short-span glulam that ordering wastes a design cycle; size on the required gross area from shear first, round up to a whole number of laminations, then confirm bending and deflection.

Source. CSA O86:19, Clauses 7.5.6 (glulam bending, size factor), 7.5.7 (shear) and Table 7.3 (24f-EX D.Fir-L strengths); Canadian Wood Council, Wood Design Manual, glulam beam selection tables.

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