07-Str-A5 · Undated paper

Question 4 of 7: B1: Reinforced concrete section for the beam of Figure 3

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 4 — B1: Reinforced concrete section for the beam of Figure 3 (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 simply supported beam of 4.5 m span (pin at A, roller at B) carrying a uniform load W1 over the whole span and a triangular surcharge W2 that falls from its peak at A to zero 2.5 m along.

Given data — Figure 3
Quantity	Symbol	Value
Span, simply supported	$L$	4.5 m
W1 — dead	$w_D$	4.5 kN/m over the full span
W1 — live/snow	$w_S$	3.0 kN/m over the full span
W1 — wind	$w_W$	3.6 kN/m over the full span
W2 — snow (drift)	$w_{S2}$	6.0 kN/m at A falling linearly to 0 at 2.5 m
Concrete / reinforcement	$f'_c$ / $f_y$	25 MPa / 400 MPa

Find. Beam width, depth, flexural reinforcement and shear reinforcement satisfying CSA A23.3 at ultimate, with deflection controlled by span/depth.

[Figure not reproduced: Figure 3 as printed: the triangular W2 dies out at 2.5 m, so the peak moment sits left of mid-span. See the official exam paper.]

Approach. Trial a section, add its self weight to the dead load, find the governing NBCC combination and the peak moment (which is not at mid-span), then size the tension steel and confirm that the concrete alone carries the shear.

Trial section and self weight. Take $b_w = 250$ mm, $h = 450$ mm. Then $w_{sw} = 0.25(0.45)(24) = 2.70$ kN/m, so the total dead load is 7.20 kN/m. The span/depth ratio $L/h = 10$ clears the A23.3 Table 9.2 limit of $L/16 = 281$ mm, so no explicit deflection calculation is needed.
Select the load combination. Three cases are evaluated. With snow leading (case 3), $1.25D + 1.5S + 0.4W$ gives a uniform intensity $$w_f = 1.25(7.20) + 1.5(3.0) + 0.4(3.6) = 14.94\ \text{kN/m}$$ plus a factored drift peaking at $1.5(6.0) = 9.0$ kN/m. Wind leading, $1.25D + 1.4W + 0.5S$, produces a peak moment of 40.9 kN$\cdot$m and $1.4D$ only 21.4 kN$\cdot$m — neither governs.
Find the peak moment. The drift resultant is $\tfrac12(9.0)(2.5) = 11.25$ kN acting 0.833 m from A, so $$R_A = \frac{14.94(4.5)}{2} + 11.25\left(\frac{4.5 - 0.833}{4.5}\right) = 33.62 + 9.17 = 42.78\ \text{kN}$$ Setting $\mathrm{d}M/\mathrm{d}x = 0$ on $M(x) = 42.78x - 7.47x^2 - 1.5(3x^2 - 0.4x^3)$ locates the peak at $x = 2.127$ m, giving $$\boxed{M_f = 42.6\ \text{kN}\cdot\text{m},\qquad V_f = 42.8\ \text{kN}\ \text{at A}}$$
Size the tension steel. With 40 mm cover, 10M stirrups and 15M bars, $d = 450 - 40 - 10 - 10 = 390$ mm. For $f'_c = 25$ MPa, $\alpha_1 = 0.85 - 0.0015f'_c = 0.8125$ and $\beta_1 = 0.97 - 0.0025f'_c = 0.9075$. Trying 2–15M ($A_s = 400$ mm$^2$), $$a = \frac{\phi_sA_sf_y}{\alpha_1\phi_cf'_cb_w} = \frac{0.85(400)(400)}{0.8125(0.65)(25)(250)} = 41.2\ \text{mm}$$ $$M_r = \phi_sA_sf_y\left(d - \frac{a}{2}\right) = 0.85(400)(400)(390 - 20.6)/10^6 = \boxed{50.2\ \text{kN}\cdot\text{m}} \ \ge\ 42.6\ \text{kN}\cdot\text{m}$$
Check ductility and minimum steel. The neutral axis sits at $c = a/\beta_1 = 45.4$ mm, so $c/d = 0.116$ — far below the 0.5 that would signal a compression-controlled section, and the steel yields well before the concrete crushes. Minimum reinforcement under Cl. 10.5.1.2 is $$A_{s,\min} = \frac{0.2\sqrt{f'_c}}{f_y}b_th = \frac{0.2(5)}{400}(250)(450) = 281\ \text{mm}^2 \ \lt\ 400\ \text{mm}^2$$
Check shear. The critical section is $d_v$ from the support, with $d_v = \max(0.9d,\ 0.72h) = 351$ mm. Marching in 0.39 m, $V_f = 33.7$ kN. Using the A23.3 simplified method with minimum stirrups present ($\beta = 0.18$), $$V_c = \phi_c\lambda\beta\sqrt{f'_c}\,b_wd_v = 0.65(1.0)(0.18)(5)(250)(351)/10^3 = 51.3\ \text{kN} \ \ge\ 33.7\ \text{kN}$$ so the concrete alone suffices. Crushing of the web is nowhere near: $V_{r,\max} = 0.25\phi_cf'_cb_wd_v = 356$ kN.
Detail. Provide 2–15M bottom bars, hooked at both supports for anchorage, plus 2–10M top bars as stirrup carriers, and 10M closed stirrups at 175 mm throughout — nominal, but they are what justifies $\beta = 0.18$ and they hold the cage. Clear spacing of the two 15M bars in a 250 mm web is $250 - 2(40) - 2(10) - 2(16) = 118$ mm, comfortably above the 25 mm and $1.4d_b$ minima.

Question 4 — results
Quantity	Value
Governing combination	$1.25D + 1.5S + 0.4W$
Factored uniform load / drift peak	14.94 kN/m / 9.0 kN/m
Reaction and end shear at A	42.8 kN
Peak moment (at 2.13 m from A)	42.6 kN$\cdot$m
Section	250 mm × 450 mm, $f'_c = 25$ MPa
Flexural reinforcement	2–15M bottom ($A_s = 400$ mm$^2$), $d = 390$ mm
$M_r$ / $c/d$ / $A_{s,\min}$	50.2 kN$\cdot$m / 0.116 / 281 mm$^2$
Shear at $d_v$ / $V_c$ / $V_{r,\max}$	33.7 / 51.3 / 356 kN
Stirrups	10M closed at 175 mm

Topic. Singly reinforced rectangular beam under a combined uniform and triangular (drift) load.

Key relations. $a = \phi_sA_sf_y/(\alpha_1\phi_cf'_cb)$; $M_r = \phi_sA_sf_y(d - a/2)$; $A_{s,\min} = 0.2\sqrt{f'_c}\,b_th/f_y$; simplified shear $V_c = \phi_c\lambda\beta\sqrt{f'_c}\,b_wd_v$ with $d_v = \max(0.9d,\,0.72h)$.

Why this works. The triangular W2 has the shape of an NBCC snow drift against an obstruction, and treating it as snow puts it in the same load group as the 3 kN/m uniform, which is why case 3 with a 1.5 factor on both is the governing combination. Its asymmetry moves the peak moment 0.12 m to the left of mid-span — a small shift, but locating it by differentiation rather than assuming mid-span is the habit the question is testing. Once the moment is known the section is routine: the beam is very lightly loaded, so the answer is set by minimum reinforcement and constructability, not by strength.

Common pitfall. Forgetting to add self weight, which for a 250 × 450 section is 2.7 kN/m — 60% of the given dead load, and enough to move $M_f$ from 34.1 to 42.6 kN$\cdot$m. The second is integrating the drift as if it ran the whole 4.5 m rather than dying out at 2.5 m, which overstates the reaction at A by more than 40%.

Source. CSA A23.3:19 Cl. 10.1, 10.5.1.2, 11.3.6.3 and Table 9.2; NBCC 2020 Cl. 4.1.6.2 (drift); MacGregor & Bartlett, Reinforced Concrete, Canadian ed., Ch. 4–6.

← Question 3 Next: Question 5 →