16-Civ-A2 Elementary Structural Design · Undated paper
Question 7 of 7: C1 — Design of a built-up dimension-lumber beam
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. 16-Civ-A2 Elementary Structural Design, May 2019 sitting, 3 hours, closed book with handbooks and textbooks permitted. Seven questions of 20 marks each: Part A (A1–A3, structural steel), Part B (B1–B3, reinforced concrete), Part C (C1, timber). The candidate answers two from Part A, two from Part B and one from Part C — all seven are solved here.
Reference texts. CSA S16 Design of Steel Structures and the CISC Handbook of Steel Construction (section properties, Tables 4-4 and 7); CSA A23.3 Design of Concrete Structures with Brzev & Pao, Reinforced Concrete Design: A Practical Approach; CSA O86 Engineering Design in Wood with the Canadian Wood Council Wood Design Manual; NBCC Part 4 for load combinations.
Page 1 of the paper specifies steel grade G40.21 300W unless noted, reinforcement 400W, and all loads shown unfactored unless otherwise stated.
Question 7: C1 — Design of a built-up dimension-lumber beam (20 marks)
Given. A 3.5 m simply supported timber beam at 1.2 m centres, in wet service, carrying a uniform area load W1 over the whole span plus a triangular snow surcharge W2 that peaks at A and falls to zero at 2.5 m.
Given data — C1
Quantity
Symbol
Value
Span / spacing
$L$ / $s$
3.5 m / 1.2 m
W1 dead
$D$
1.5 kPa
W1 live/snow
$L$
1.0 kPa
W1 wind
$W$
1.2 kPa
W2 snow, triangular over 2.5 m
$S$
2.0 kPa at A, zero at 2.5 m
Service condition
—
Wet
Material
—
D.Fir-L No.1/No.2 dimension lumber
Find. A built-up dimension-lumber section adequate in bending, shear and deflection under the governing NBCC load combination.
Figure 7 — C1 beam: uniform W1 over the full 3.5 m span plus the triangular W2 snow surcharge peaking at support A.
Check: load classification. The W1 legend reads “LIVE/SNOW 1.0 kPa”. It has been treated as a live load and the separate 2.0 kPa triangular W2 as the snow load, which is the reading that produces distinct $L$ and $S$ terms for the NBCC combinations. Treating W1's 1.0 kPa as additional snow instead moves the governing case to $1.25D+1.5S+0.4W$ and changes $M_f$ by under 4 %, so the section chosen is unaffected. Full lateral support from the decking is assumed, so $K_L=1.0$.
Approach. Convert the area loads to line loads on the 1.2 m tributary width, evaluate every NBCC combination with its own duration factor $K_D$, identify the governing case on a $M_f/K_D$ basis, then check a trial built-up section for bending, shear and deflection using the wet-service modification factors.
Convert to line loads. Multiplying each area load by the 1.2 m tributary width,
$$w_D=1.8,\quad w_L=1.2,\quad w_W=1.44\ \text{kN/m (uniform)},\qquad w_{S,0}=2.4\ \text{kN/m at A}$$
with the snow falling linearly to zero at 2.5 m from A.
Set up the combinations with their duration factors. The duration factor rewards short-term loading, so a combination must be judged on the ratio $M_f/K_D$ rather than on $M_f$ alone:
NBCC load combinations — C1
Combination
Uniform (kN/m)
Triangular peak (kN/m)
$K_D$
$M_f$ (kN·m)
$M_f/K_D$
$1.4D$
2.52
—
0.65
3.86
5.94
$1.25D+1.5L+0.5S$
4.05
1.20
1.00
6.80
6.80
$1.25D+1.5L+0.4W$
4.63
—
1.00
7.08
7.08
$1.25D+1.5S+0.5L$
2.85
3.60
1.00
6.19
6.19
$1.25D+1.4W+0.5S$
4.27
1.20
1.15
7.13
6.20
$1.25D+1.4W+0.5L$
4.87
—
1.15
7.45
6.48
Identify the governing case. The wind combinations produce the largest raw moments, but their $K_D=1.15$ discounts them. Ranking on $M_f/K_D$, the governing case is
$$1.25D+1.5L+0.4W:\qquad M_f=\boxed{7.08\ \text{kN}\cdot\text{m}},\qquad V_f=8.10\ \text{kN},\qquad K_D=1.0$$
This is the standard-term case, and it is worth noting that the triangular snow surcharge — the visually dominant feature of the figure — never governs, because the combinations that amplify it must discount the live load.
Establish the modification factors. For dimension lumber in wet service the three moisture factors are different and must not be collapsed into one:
$$K_{Sb}=0.84,\qquad K_{Sv}=0.96,\qquad K_{SE}=0.94$$
The built-up system factor applies to bending only: $K_{Hb}=1.10$ for three plies, $K_{Hv}=1.0$. Take $K_L=1.0$ (decking provides continuous lateral support) and $K_T=1.0$ (untreated).
Adopt a trial section and compute the bending resistance. Try three 38 × 235 mm plies, giving $b=114$ mm, $d=235$ mm:
$$S=\frac{bd^{2}}{6}=\frac{114(235)^{2}}{6}=1.049\times10^{6}\ \text{mm}^3$$
With $f_b=10.0$ MPa for D.Fir-L No.1/No.2 and $K_{Zb}=1.1$ at $d=235$ mm,
$$F_b=f_b\left(K_DK_{Hb}K_{Sb}K_T\right)=10.0(1.0)(1.10)(0.84)=9.24\ \text{MPa}$$
$$M_r=\phi F_bSK_{Zb}K_L=0.9(9.24)(1.049\times10^{6})(1.1)=\boxed{9.60\ \text{kN}\cdot\text{m}\ >\ 7.08}$$
Utilisation 0.74.
Check shear. With $f_v=1.9$ MPa,
$$F_v=1.9(1.0)(1.0)(0.96)=1.824\ \text{MPa},\qquad A_g=114(235)=26790\ \text{mm}^2$$
$$V_r=\phi F_v\left(\frac{2A_g}{3}\right)=0.9(1.824)\left(\frac{2(26790)}{3}\right)=29.3\ \text{kN}\ \gg\ 8.10\ \text{kN}\quad\checkmark$$
Shear is not remotely critical here, in contrast to permanent-duration glulam problems, because $K_D=1.0$ restores the full bending strength while the shear demand stays small.
Check deflection at specified load. With $E_s=E(K_{SE})=12500(0.94)=11750$ MPa and $I=bd^{3}/12=1.233\times10^{8}$ mm⁴, superposing the uniform dead-plus-live load with the partial triangular snow load gives
$$\Delta=5.10\ \text{mm}\ <\ \frac{L}{240}=\frac{3500}{240}=14.6\ \text{mm}\quad\checkmark$$
and the variable-load component alone is well inside $L/360=9.7$ mm.
Confirm three plies are necessary. A two-ply 38 × 235 section gives $M_r=6.40$ kN·m, 10 % short of the 7.08 kN·m demand, and a three-ply 38 × 184 gives 6.42 kN·m. Three 38 × 235 plies is therefore the smallest standard built-up member that works.
Design summary: a three-ply built-up 38 × 235 mm D.Fir-L No.1/No.2 beam, nailed or bolted together in accordance with the built-up member provisions of O86, at 1.2 m centres. The design is bending-governed at a utilisation of 0.74 with shear at 0.28 and deflection at 0.35 — a well-proportioned result, and the wet-service penalty on bending (16 %) is what pushes the answer from two plies to three.
