Question 5 of 7: B2: Square reinforced concrete column for the frame of Figure 2
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.
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 5 — B2: Square reinforced concrete column for the frame of Figure 2 (20 marks)
Check — which column the question means. Figure 2 as printed has only two columns, A–B and D–F; "C–D" is a length of the horizontal deck, not a column. The label is a leftover from an earlier version of the figure in which the beam ran B–C and the right column was C–D; adding the mid-span point C pushed the labels along to D, E and F without the question text being updated. Since question B3 explicitly designs the footing for column A–B, "column C–D" must be the other one, and the right-hand column D–F is designed here. This assumption should be written on the answer paper, as NOTE 1 of the exam invites.
Given. The frame and loads of Figure 2 (see Question 2), but with the deck taken as rigid in flexure so that the column tops cannot rotate. Both columns are square reinforced concrete of the same size; $f'_c = 35$ MPa and $f_y = 400$ MPa.
Find. The square section and longitudinal reinforcement of the 8 m column D–F, including slenderness effects.
Approach. Re-analyse the frame with a rigid deck — which redistributes the storey shear towards the shorter, stiffer column — then magnify for sway under A23.3 Cl. 10.16 and check the section on its factored P–M interaction diagram.
Re-analyse with a rigid deck. Making the deck flexurally rigid removes the joint rotations, so each column behaves as a member pinned at the base and restrained against rotation at the top. Sway stiffness then varies as $3EI/h^3$, and the 8 m column is $(12/8)^3 = 3.4$ times stiffer than the 12 m one. Taking a trial 600 mm square with the cracked stiffness $0.7I_g$ that Cl. 10.14.1.2 requires, and $E_c = 4\,500\sqrt{35} = 26\,622$ MPa, the analysis gives
$$V_{DF} = 56.4\ \text{kN},\qquad V_{AB} = 18.6\ \text{kN},\qquad V_{DF} + V_{AB} = 75.0\ \text{kN}\ \checkmark$$
Extract the column actions. With no moment at the pinned base, the whole moment appears at the deck:
$$C_f = 693.8\ \text{kN},\qquad M_{f,1} = V_{DF}\,h = 56.4(8.0) = 451.3\ \text{kN}\cdot\text{m}$$
Note how different this is from Question 2: the rigid deck triples the right column's share of the storey shear and hands it the larger moment, whereas in the flexible steel frame the tall column took nearly all of it.
Test for sway sensitivity. The factored deck sway is 50.6 mm, so the stability index of Cl. 10.13.4 is
$$Q = \frac{\sum P_f\,\Delta_o}{V_f\,l_c} = \frac{693.8(50.6)}{56.4(8\,000)} = 0.0777$$
Since $Q \gt 0.05$ the storey is a sway storey and second-order effects must be included; since $Q \lt 1/3$ the magnifier method is still permitted.
Check that no separate member magnification applies. For the braced (non-sway) case, $\beta_d = 1.25(359.8)/693.8 = 0.648$, so
$$EI = \frac{0.4E_cI_g}{1 + \beta_d} = 6.98\times10^{13}\ \text{N}\cdot\text{mm}^2,\qquad P_c = \frac{\pi^2EI}{(kl_u)^2} = 16\,816\ \text{kN}$$
with $k = 0.8$. Then $\delta_b = C_m/(1 - P_f/0.75P_c) = 0.6/0.945 = 0.635 \lt 1.0$, so $\delta_b = 1.0$ and no further magnification is needed.
Choose the section and reinforcement. The design eccentricity is $e = 489.4/693.8 = 705$ mm — larger than the section itself, so this is a flexure-dominated member and the reinforcement must be placed on the two faces perpendicular to the bending axis. Take a 600 mm square with 8–30M ($A_{st} = 5\,600$ mm$^2$, $\rho = 1.56\%$), arranged three bars per face with 40 mm cover and 10M ties.
Build the interaction diagram. Strain compatibility with $\varepsilon_{cu} = 0.0035$, $\alpha_1 = 0.7975$, $\beta_1 = 0.8825$, $\phi_c = 0.65$ and $\phi_s = 0.85$, deducting displaced concrete for bars inside the stress block, gives at $e = 705$ mm
$$\boxed{P_r = 943\ \text{kN},\qquad M_r = 664\ \text{kN}\cdot\text{m}}$$
against $C_f = 694$ kN and $M_f = 489$ kN$\cdot$m — a utilisation of 0.74 on the radial line through the design point.
Check the code limits. The steel ratio 1.56% lies inside the 1–8% band of Cl. 10.9.1. The maximum axial resistance under Cl. 10.10.4 is
$$P_{r,\max} = 0.80\left[0.85\phi_cf'_c(A_g - A_{st}) + \phi_sf_yA_{st}\right] = 7\,006\ \text{kN}$$
far above the applied 694 kN, confirming that this column is governed entirely by moment.
Detail the ties. 10M ties at the least of 16 longitudinal bar diameters (478 mm), 48 tie diameters (542 mm) and the least column dimension (600 mm) — so 10M ties at 450 mm, closed to 135° hooks, with every alternate bar tie-supported. Tie spacing tightens to 150 mm over the top 600 mm where the moment is greatest.
