16-Civ-A2 Elementary Structural Design · December 2019
Question 7 of 7: C1 — Built-up sawn-lumber beam in wet service
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Examinations, December 2019 — 16-Civ-A2 Elementary Structural Design. Three hours, CLOSED BOOK. 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 single Part C question — all of equal value. All seven are solved below, because this set is a study resource rather than an exam script. Note 6 on page 1 states that all loads shown are unfactored unless otherwise stated, so every load pattern is factored here before it meets a resistance; Note 7 fixes G40.21 300W steel and 400W reinforcement unless a question says otherwise.
Reference texts.
CSA S16:19, Design of Steel Structures, and CISC Handbook of Steel Construction, 11th ed. — Clauses 13.3 (compression), 13.13 (welds), 17 (composite beams), and the angle/W-shape section tables.
CSA A23.3:19, Design of Concrete Structures, and the Cement Association of Canada Concrete Design Handbook, 4th ed. — Clauses 10 (flexure and columns) and 11 (shear).
CSA O86:19, Engineering Design in Wood, and the Canadian Wood Council Wood Design Manual — Clause 6 (sawn lumber) and the service-condition and size-factor tables.
Kulak & Grondin, Limit States Design in Structural Steel, 11th ed., Chs. 4, 5, 8, 9.
National Building Code of Canada, Table 4.1.3.2 (load combinations).
Check — assumptions declared under Note 1 of the paper. (i) Figure A1 shows a solid vertical between F and C as well as the load arrow at C; that member is required, because without it the truss counts m + r = 12 + 3 = 15 against 2n = 16 and is a mechanism. With it, m + r = 16 = 2n and the truss is determinate, which is how it is solved here. (ii) In A3 the number of shear studs is reported as a bonus; the question asks only for the ultimate moment. (iii) In B1 the two stacked dimensions read as a sum, so the tee is 100 mm flange plus 400 mm of stem, i.e. 500 mm overall, and the beam spacing of 1250 mm governs the effective flange width because no span is given. (iv) In C1 the 1.5 kPa / 1.0 kPa area loads are converted to line loads on the 0.9 m tributary width, and the deflection limits adopted are L/240 on the back span and 2L/180 on the cantilever.
Question 7: C1 — Built-up sawn-lumber beam in wet service (20 marks)
Given. The design starts from the beam of Figure C1 — a 3.0 m span with a 1.0 m overhang, a 1.5 kPa live / 1.0 kPa dead area load over the whole length and a 2.5 kN live / 3.0 kN dead tip load — at 0.9 m spacing, built up from SPF No.1/No.2 dimension lumber in wet service.
Given data
Item
Value
Geometry
3.0 m span A–B plus a 1.0 m overhang beyond B
Area load W1
1.5 kPa live, 1.0 kPa dead
Tip load P1
2.5 kN live, 3.0 kN dead
Tributary width
0.9 m
Material
SPF No.1/No.2 dimension lumber, built up
Service condition
wet
Deflection limits adopted
L/240 on the span, 2L/180 on the cantilever
Find. A built-up dimension-lumber section that satisfies bending, shear, bearing and deflection under CSA O86.
Figure C1 rebuilt: 3.0 m back span with a 1.0 m overhang carrying the tip load P1; the uniform load runs over the full 4.0 m.
Approach. Convert the area loads to line loads on the 0.9 m tributary width, factor them, solve the overhanging beam by statics, then select a built-up section and apply the O86 modification factors — the wet-service factors are the ones this question is really testing.
Line loads. On a 0.9 m tributary width, $$w_D = 1.0(0.9) = 0.90\ \text{kN/m},\qquad w_L = 1.5(0.9) = 1.35\ \text{kN/m}$$
Factored loads. The governing combination is $1.25D + 1.5L$ (the $1.4D$ case gives 1.26 kN/m and 4.2 kN, both smaller): $$w_f = 1.25(0.90) + 1.5(1.35) = \boxed{3.15\ \text{kN/m}}$$ $$P_f = 1.25(3.0) + 1.5(2.5) = \boxed{7.5\ \text{kN}}$$
Reactions. Taking moments about A over the full 4.0 m of loaded beam, $$R_B = \frac{3.15(4.0)(2.0) + 7.5(4.0)}{3.0} = \frac{25.2 + 30.0}{3.0} = 18.40\ \text{kN}$$ $$R_A = 3.15(4.0) + 7.5 - 18.40 = 1.70\ \text{kN}$$ The reaction at A is very small, because the overhang loads are trying to lever the back span upward.
Design moment. The overhang hogs over B: $$M_B = -\left[\frac{3.15(1.0)^{2}}{2} + 7.5(1.0)\right] = \boxed{-9.08\ \text{kN}\cdot\text{m}}$$ In the back span the shear vanishes at $x = R_A/w_f = 1.70/3.15 = 0.540$ m, where the sagging moment is only $1.70(0.540) - 3.15(0.540)^{2}/2 = 0.46$ kN·m. The cantilever therefore governs by a factor of twenty — a 1 m overhang with a 7.5 kN tip load is doing all the work.
Design shear. Just left of B, $V = 1.70 - 3.15(3.0) = -7.75$ kN; just right of B it jumps to $-7.75 + 18.40 = 10.65$ kN. So $$V_f = 10.65\ \text{kN}$$
Trial section and modification factors. Take three 38 × 235 members nailed together, so $b = 114$ mm and $d = 235$ mm: $$S = \frac{114(235)^{2}}{6} = 1.049\times10^{6}\ \text{mm}^{3},\quad A = 26\,790\ \text{mm}^{2},\quad I = 1.233\times10^{8}\ \text{mm}^{4}$$ SPF No.1/No.2 gives $f_b = 11.8$ MPa, $f_v = 1.5$ MPa, $E = 9500$ MPa and $f_{cp} = 5.3$ MPa. The wet-service factors for dimension lumber are $K_{Sb} = 0.84$, $K_{Sv} = 0.96$ and $K_{SE} = 0.94$; the load-duration factor is $K_D = 1.0$ (standard term); and the built-up system factor for three members is $K_H = 1.10$ in bending.
Bending resistance. With the size factor $K_{Zb} = 1.1$ for a 235 mm deep member and $K_L = 1.0$ (the deck laterally supports the compression edge), $$F_b = f_b K_D K_H K_{Sb} = 11.8(1.0)(1.10)(0.84) = 10.90\ \text{MPa}$$ $$M_r = \phi F_b S K_{Zb} K_L = 0.9(10.90)(1.049\times10^{6})(1.1)/10^{6} = \boxed{11.33\ \text{kN}\cdot\text{m}}$$ against 9.08 kN·m required, a utilisation of 0.80.
Shear resistance. Taking $K_H = 1.0$ conservatively for shear, $$F_v = f_v K_D K_{Sv} = 1.5(1.0)(0.96) = 1.44\ \text{MPa}$$ $$V_r = \phi F_v \frac{2A_g}{3} = 0.9(1.44)\frac{2(26\,790)}{3}\cdot\frac{1}{10^{3}} = 23.1\ \text{kN}$$ against 10.65 kN, a utilisation of 0.46. Unlike a glulam under permanent load, shear is comfortable here because the standard-term duration factor keeps bending at full value.
Deflection at specified loads. With $E_s = 9500(0.94) = 8930$ MPa, $EI = 1.10\times10^{12}$ N·mm², $w_s = 2.25$ N/mm and $P_s = 5500$ N, the tip deflection is the sum of three effects: $$\Delta_{\text{tip}} = \frac{P a^{2}(L+a)}{3EI} + \frac{w a^{3}(4a + 3L)}{24EI} - \frac{w L^{3} a}{24EI} = 6.66 + 1.11 - 2.30 = \boxed{5.47\ \text{mm}}$$ against the 2L/180 = 11.1 mm limit.
Back-span deflection, and its sign. The hogging moment carried in from the overhang, $M_{\text{end}} = -6.63$ kN·m at specified load, lifts the span: $$\Delta_{\text{mid}} = \frac{5wL^{4}}{384EI} + \frac{M_{\text{end}}L^{2}}{16EI} = 2.16 - 3.39 = -1.23\ \text{mm}$$ i.e. 1.2 mm upward, well inside the L/240 = 12.5 mm limit. The sign is worth reporting: the back span cambers up, so the floor finish detail should not assume it sags.
Bearing at B. $$A_{b,\text{req}} = \frac{R_B}{\phi F_{cp} b} = \frac{18.40\times10^{3}}{0.9(5.3)(0.96)(114)} = 35\ \text{mm of length}$$ so a standard 89 mm wide bearing plate or a doubled stud pack is more than sufficient.
