07-Str-A5 · Undated paper

Question 1 of 7: A1: Wide-flange section for the two-span beam of Figure 1

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

Notes on this paper

Paper format. National Exams — 07-Str-A5 Advanced Structural Design, printed footer 07-Str-A5/May 2019. Three hours, "closed book" with handbooks and textbooks permitted. Seven questions of equal value (20 marks each) in three parts: A1–A3 (do two), B1–B3 (do two), C1 (do one) — five solutions make a complete paper. All seven are solved here, because the set is a study resource rather than an exam attempt.

Design data given on page 1. Solutions to CAN/CSA S16 (steel), CAN/CSA A23.3 (concrete) and CAN/CSA O86 (timber). All loads shown on the figures are unfactored. All structural steel is G40.21 300W — $F_y = 300$ MPa, $F_u = 450$ MPa, $E = 200\,000$ MPa, $G = 77\,000$ MPa. All reinforcement is 400W — $f_y = 400$ MPa. Concrete strength is stated per question. Steel sections are W-shapes unless noted otherwise.

Reference texts.

CSA S16:19, Design of Steel Structures — Cl. 11 (section classification), 13.3 (compression), 13.4 (shear), 13.5–13.6 (bending and lateral–torsional buckling), 13.8 (beam-columns), 13.13 (welds), 14.3.2 (web crippling), 8.4.3 (stability effects).
CSA A23.3:19, Design of Concrete Structures — Cl. 8.6 (material properties), 10 (flexure and axial load, slenderness), 11 (shear), 15 (footings), 18 (prestressed concrete).
National Building Code of Canada 2020, Part 4 — Table 4.1.3.2 load combinations, 4.1.6 snow and drift loads.
CISC, Handbook of Steel Construction, 11th ed. — W-shape dimensions and properties.
Salmon, Johnson & Malhas, Steel Structures: Design and Behavior, 5th ed. — Ch. 9 (LTB), 12 (beam-columns), 13 (connections).
MacGregor & Bartlett, Reinforced Concrete: Mechanics and Design, Canadian ed. — Ch. 4–6 (flexure), 11 (columns), 12 (slender columns), 15 (footings).
Collins & Mitchell, Prestressed Concrete Structures — Ch. 4 (service stresses), 6 (flexural strength); PCI Design Handbook, 8th ed. — multiplier method for long-term camber.

Check — load combinations. The figures split the loads into DEAD, LIVE, SNOW and WIND, so NBCC 2020 Table 4.1.3.2 is applied literally. Throughout Part A the governing case is $1.25D + 1.5L$ (case 2 with no companion snow or wind on the figure). For Figures 3 the governing case is $1.25D + 1.5S + 0.4W$ (case 3), with $1.25D + 1.4W + 0.5S$ and $1.4D$ checked and shown not to govern. The "LIVE/SNOW" label on W1 of Figure 3 is read as a single 3 kN/m specified roof load that may be either occupancy live or balanced snow, and the triangular W2 is read as the additional snow-drift surcharge that its shape implies.

Question 1 — A1: Wide-flange section for the two-span beam of Figure 1 (20 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. A two-span continuous beam A–B–C, pinned at A and on rollers at B and C, with a uniformly distributed load over span A–B only and a single point load at the middle of span B–C; lateral restraint exists only at A, B and C.

Given data — Figure 1
Quantity	Symbol	Value
Span A–B and span B–C	$L_1, L_2$	5.0 m each
Supports	—	A pin, B roller, C roller
UDL on A–B (dead / live)	$w_D, w_L$	15.0 / 10.0 kN/m
Point load, 2.5 m from C (dead / live)	$P_D, P_L$	160 / 100 kN
Unbraced length of each segment	$L_u$	5 000 mm
Steel	$F_y$	300 MPa (G40.21 300W)

Find. The lightest W-shape whose factored moment, lateral–torsional buckling, shear and deflection resistances all satisfy the two-span beam.

[Figure not reproduced: Figure 1 as printed: two equal 5 m spans, UDL confined to A–B, point load at the centre of B–C. See the official exam paper.]

Approach. Factor the loads to NBCC case 2, solve the one-degree-indeterminate beam with the three-moment equation, then size a W-shape against CSA S16 Cl. 13.6 lateral–torsional buckling using the actual moment gradient in each 5 m unbraced segment.

Factor the loads. With dead and live given separately, NBCC 2020 Table 4.1.3.2 case 2 applies: $$w_f = 1.25(15.0) + 1.5(10.0) = 33.75\ \text{kN/m},\qquad P_f = 1.25(160) + 1.5(100) = 350\ \text{kN}$$
Solve the continuous beam. With simple ends at A and C the only redundant is the hogging moment at B. The three-moment equation for a UDL in span 1 and a central point load in span 2 gives $$2M_B(L_1+L_2) = -\left[\frac{w_f L_1^{3}}{4} + \frac{3P_f L_2^{2}}{8}\right] = -\left[\frac{33.75(5)^3}{4} + \frac{3(350)(5)^2}{8}\right]$$ so that $$\boxed{M_B = -216.8\ \text{kN}\cdot\text{m}\quad(\text{hogging})}$$
Recover the reactions. For each span, the simple-beam reaction is corrected by the end moment, $R = R_{\text{simple}} - |M_B|/L$: $$R_A = \tfrac{33.75(5)}{2} - \tfrac{216.8}{5} = 41.02\ \text{kN},\qquad R_C = \tfrac{350}{2} - \tfrac{216.8}{5} = 131.64\ \text{kN}$$ and vertical equilibrium leaves $R_B = 168.75 + 350 - 41.02 - 131.64 = 346.09$ kN.
Locate the design actions. Shear vanishes in span A–B at $x = R_A/w_f = 1.215$ m, where $M = 24.9$ kN$\cdot$m — negligible. Working in from C, the moment under the point load is $$M_{BC} = R_C(2.5) = 131.64(2.5) = \boxed{329.1\ \text{kN}\cdot\text{m}}$$ and the largest shear is just left of the load, $V_f = 175 + 43.36 = 218.4$ kN.
Establish the moment gradient in each unbraced segment. Because the beam is braced only at A, B and C, each segment is 5.0 m long. Using the four-point form of CSA S16 Cl. 13.6, $$\omega_2 = \frac{4M_{\max}}{\sqrt{M_{\max}^2 + 4M_a^2 + 7M_b^2 + 4M_c^2}} \le 2.5$$ segment B–C returns $\omega_2 = 1.325$ (quarter-point moments 164.6, 329.1 and 56.2 kN$\cdot$m), while segment A–B returns 3.12 and is therefore capped at 2.5. Segment B–C, with both the larger moment and the weaker gradient factor, governs.
Try W530×72. From the CISC tables, $A = 9\,100$ mm$^2$, $d = 524$ mm, $b = 207$ mm, $t = 10.9$ mm, $w = 9.0$ mm, $I_x = 399\times10^6$ mm$^4$, $Z_x = 1\,760\times10^3$ mm$^3$, $I_y = 16.1\times10^6$ mm$^4$. From the plate geometry, $J = 0.301\times10^6$ mm$^4$ and $C_w = 1.060\times10^{12}$ mm$^6$. The elastic critical moment is $$M_u = \frac{\omega_2\pi}{L}\sqrt{EI_yGJ + \left(\frac{\pi E}{L}\right)^2 I_y C_w} = 488.2\ \text{kN}\cdot\text{m}$$
Convert to a factored resistance. With $M_p = Z_xF_y = 528.0$ kN$\cdot$m and $M_u \gt 0.67M_p$, Cl. 13.6(a)(i) applies: $$M_r = 1.15\phi M_p\!\left[1 - \frac{0.28M_p}{M_u}\right] = 1.15(0.9)(528.0)\left[1 - \frac{0.28(528.0)}{488.2}\right] = \boxed{381.0\ \text{kN}\cdot\text{m}}$$ which exceeds $\phi M_p = 475.2$ kN$\cdot$m nowhere, so 381.0 kN$\cdot$m stands. Since $381.0 \ge 329.1$ kN$\cdot$m, the section passes at 86% utilisation. Repeating with $\omega_2 = 2.5$ for segment A–B gives $M_r = 458.8$ kN$\cdot$m against 216.8 kN$\cdot$m — comfortable.
Classify the section and check shear. Flange $b/2t = 9.50$ against the Class 1 limit $145/\sqrt{F_y} = 8.37$ and the Class 2 limit $170/\sqrt{F_y} = 9.81$, and web $h/w = 55.8$ against $1100/\sqrt{F_y} = 63.5$: the shape is Class 2, which still develops $Z_xF_y$ under Cl. 13.5. For shear, $h/w = 55.8 \lt 1014/\sqrt{F_y} = 58.5$, so the web yields before it buckles: $$V_r = \phi\,d\,w\,(0.66F_y) = 0.9(524)(9.0)(198)/10^3 = 840.4\ \text{kN} \ \ge\ 218.4\ \text{kN}$$
Check live-load deflection. Under live load alone $M_B = -62.5$ kN$\cdot$m, and superposing the simple-span and end-moment terms on span B–C, $$\Delta_L = \frac{P_LL^3}{48EI_x} - \frac{|M_B|L^2}{16EI_x} = 3.91 - 1.87 = 2.04\ \text{mm} \ \ll\ \frac{L}{360} = 13.9\ \text{mm}$$
Confirm nothing lighter works. W530×74 ($M_r = 320.0$ kN$\cdot$m) and W460×68 ($M_r = 275.3$ kN$\cdot$m) both fail the 329.1 kN$\cdot$m demand — their narrow flanges give $r_y$ near 33 mm and the 5 m unbraced length punishes them. W460×74 works ($M_r = 365.7$ kN$\cdot$m) but is 2 kg/m heavier.

Question 1 — results
Quantity	Value
Factored UDL / point load	33.75 kN/m; 350 kN
Hogging moment at B	216.8 kN$\cdot$m
Reactions $R_A$ / $R_B$ / $R_C$	41.0 / 346.1 / 131.6 kN
Maximum sagging moment (span B–C)	329.1 kN$\cdot$m
Maximum factored shear	218.4 kN
Moment gradient factor, segment B–C	$\omega_2 = 1.325$
Section selected	W530×72 (Class 2)
$M_r$ (LTB) / $V_r$	381.0 kN$\cdot$m / 840.4 kN
Live-load deflection	2.0 mm (limit 13.9 mm)

Topic. Lateral–torsional buckling of a continuous beam with widely spaced bracing.

Key relations. Three-moment equation $M_AL_1 + 2M_B(L_1+L_2) + M_CL_2 = -\left[w L_1^3/4 + 3PL_2^2/8\right]$; elastic critical moment $M_u = (\omega_2\pi/L)\sqrt{EI_yGJ + (\pi E/L)^2 I_yC_w}$; CSA S16 Cl. 13.6(a) inelastic transition $M_r = 1.15\phi M_p(1 - 0.28M_p/M_u) \le \phi M_p$.

Why this works. A beam braced only at its supports buckles laterally over the whole 5 m span, so the section is chosen by $I_y$, $J$ and $C_w$ rather than by $Z_x$ alone. The gradient factor $\omega_2$ recognises that a segment whose moment varies along its length stores less destabilising energy than one under uniform moment; the four-point formula is used rather than the older $1.75 + 1.05\kappa + 0.3\kappa^2$ because segment B–C carries a point load inside it, so its moment diagram is not linear. Because $M_u$ lands above $0.67M_p$, the section buckles inelastically and the code's straight-line interpolation between $\phi M_p$ and $\phi M_u$ applies.

Common pitfall. Selecting on $Z_x$ alone: W530×74 has a larger plastic modulus than W530×72 yet fails, because its 166 mm flange gives $r_y = 33.1$ mm against 42.1 mm. The second trap is applying $\omega_2 = 1.75$ to segment B–C by treating it as a linear-moment segment, which overstates $M_u$ by a third.

Source. CSA S16:19 Cl. 13.5–13.6 and 13.4.1.1; Salmon, Johnson & Malhas, Steel Structures, 5th ed., Ch. 9.

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