16-Civ-B2 Advanced Structural Design · Undated paper
Question 5 of 7: Square reinforced concrete column for the Figure 2 frame with a rigid deck
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. 16-Civ-B2 Advanced Structural Design, 3 hours, closed book (handbooks and textbooks permitted). Seven design questions of 20 marks each in three parts — Part A (steel, do two of three), Part B (reinforced concrete, do two of three) and Part C (prestressed concrete, do question C1); five solutions constitute a complete paper. All seven are solved here, because the set is a study resource rather than an examination script.
Design data given on page 1. Solutions to the latest CAN/CSA S16 (steel), CAN/CSA A23.3 (concrete) and CAN/CSA O86 (timber). All loads shown on the figures are unfactored. Structural steel is G40.21 300W unless noted, so $F_y = 300\ \text{MPa}$ and $E = 200\,000\ \text{MPa}$; reinforcement is 400W, so $f_y = 400\ \text{MPa}$. Concrete strengths are stated question by question (25 MPa in B1 and B3, 35 MPa in B2, 40 MPa in C1). Load combinations follow NBCC Table 4.1.3.2.
Reference texts. CISC, Handbook of Steel Construction, 12th ed. (CSA S16-19 with commentary) — Parts 1, 4 and 5; Kulak & Grondin, Limit States Design in Structural Steel, 10th ed.; MacGregor & Bartlett, Reinforced Concrete: Mechanics and Design (Canadian edition); CAC, Concrete Design Handbook, 4th ed. (A23.3 with explanatory notes); Collins & Mitchell, Prestressed Concrete Structures; CPCI, Design Manual, 5th ed.; NBCC Part 4 for loads and load combinations.
Every value quoted here was read from the printed figure of the original page: B1 uses 25 MPa, B2 uses 35 MPa, B3 uses 150 kPa allowable / 225 kPa ultimate with 25 MPa concrete, and C1 refers to Figure 3, not Figure 2.
Question B2: Square reinforced concrete column for the Figure 2 frame with a rigid deck (20 marks)
Find. A square reinforced concrete column section, with its longitudinal steel and ties, adequate for the axial force and the sway moment that the rigid-deck idealisation delivers.
Check — the label “C-D” in the question does not name a column. In Figure 2, C is the mid-point of the deck and D is the head of the right-hand column, so C–D is a beam segment; the only columns are A–B and D–F. The question is answered as the design of the column whose head is at D, that is D–F, since B3 separately deals with column A–B. Both RC columns are taken as the same square section, which is what makes the rigid-deck stiffness split determinate, and the chosen section is checked for A–B as well so that the answer covers either reading of the label.
Approach. With a rigid deck the column tops neither rotate nor move relative to one another, so the storey shear splits in proportion to $3EI/h^{3}$; take the column moments from that split, magnify them for sway with A23.3 Clause 10.16, and check the section on a strain-compatibility interaction diagram.
Establish the sway stiffness of each column. A column pinned at the base and held against rotation at the top by a rigid deck has a sway stiffness $k = 3EI/h^{3}$. With both columns the same square section the $EI$ cancels and only the cube of the height survives:
$$k_{AB} : k_{DF} = \frac{1}{12^{3}} : \frac{1}{8^{3}} = 1 : 3.375$$
$$f_{AB} = 0.2286, \quad f_{DF} = 0.7714$$
Distribute the horizontal load and get the first-order moments.
$$V_{AB} = 0.2286(75) = 17.14\ \text{kN}, \quad V_{DF} = 0.7714(75) = 57.86\ \text{kN}$$
Each column carries zero moment at its pinned base, so the moment at the head is simply $V h$:
$$M_B = 17.14(12) = 205.7\ \text{kN}\!\cdot\!\text{m}, \quad M_D = 57.86(8) = 462.9\ \text{kN}\!\cdot\!\text{m}$$
The shorter column is the stiffer one and takes three-quarters of the shear, which is the opposite of what the flexible steel frame of A2 did — the difference is entirely the rigid-deck assumption, which suppresses the gravity-induced sway.
Get the axial forces from global equilibrium. Taking moments about A for the whole frame, remembering that base F sits 4 m above base A:
$$12F_y = 750(6) + 375(12) - 75(12) + 4(57.86)$$
$$\boxed{F_y = 694.3\ \text{kN\ (column D--F)}, \quad A_y = 1500 - 694.3 = 805.7\ \text{kN}}$$
The horizontal load relieves the right-hand column by about 100 kN and adds the same to the left.
Choose a trial size from the slenderness limit, not from strength. A23.3 Clause 10.13.2 requires a full second-order analysis once $k\ell_u/r$ exceeds 100. For a column pinned at one end, rotationally fixed at the other and free to sway, $k = 2.0$; for a square section $r = 0.3h$:
$$\text{column A--B: } \frac{2(12\,000)}{0.3h} \le 100 \ \Rightarrow\ h \ge 800\ \text{mm}$$
Adopt $h = 800\ \text{mm}$ square, which gives $k\ell_u/r = 100.0$ for A–B and 66.7 for D–F.
Compute the sway magnifier. With the reduced stiffness of Clause 10.15.3, $EI = 0.4E_cI_g/(1 + \beta_d)$ and $\beta_d = 0$ because the lateral load is entirely live:
$$EI = 0.4(26\,622)\frac{800^{4}}{12} = 3.635\times10^{5}\ \text{kN}\!\cdot\!\text{m}^{2}$$
$$k_{AB} = \frac{3EI}{12^{3}} = 631\ \text{kN/m}, \quad k_{DF} = \frac{3EI}{8^{3}} = 2130\ \text{kN/m}$$
$$\Delta_o = \frac{75}{631 + 2130} = 27.2\ \text{mm}$$
Including the column self weight ($0.8^{2}(24) = 15.4\ \text{kN/m}$), the stability index and magnifier are
$$Q = \frac{\sum P_f/h}{\sum k} = \frac{1036/12 + 848/8}{2761} = 0.0697, \quad \delta_s = \frac{1}{1 - Q} = 1.075$$
$$\boxed{M_{f,D} = 1.075(462.9) = 497.5\ \text{kN}\!\cdot\!\text{m}, \quad M_{f,B} = 221.1\ \text{kN}\!\cdot\!\text{m}}$$
Select the longitudinal steel. Clause 10.9.1 sets a minimum of one per cent of the gross area:
$$A_{st,min} = 0.01(800)^{2} = 6400\ \text{mm}^2$$
Provide 12-30M ($A_{st} = 8400\ \text{mm}^2$, $\rho = 1.31$ per cent), arranged four to a face with the corner bars shared, which is the natural layout for a square column in a frame that can sway either way.
Build the interaction diagram and read off the resistance. The section was analysed by strain compatibility with $\varepsilon_{cu} = 0.0035$, the rectangular block $\alpha_1 = 0.7975$, $\beta_1 = 0.8825$, each bar layer taking $\phi_s f_s$ less the displaced concrete where it lies inside the block. At the applied axial load:
$$\boxed{M_r = 1193.8\ \text{kN}\!\cdot\!\text{m at } N_f = 694.3\ \text{kN} \ \Rightarrow\ \frac{497.5}{1193.8} = 0.417}$$
For column A–B, at $N_f = 805.7\ \text{kN}$ the resistance is 1226.3 kN·m, so its utilisation is 221.1/1226.3 = 0.180.
Check the axial ceiling. Clause 10.10.4 caps the resistance of a tied column at
$$P_{r,max} = 0.80\left[\alpha_1\phi_c f_c'\left(A_g - A_{st}\right) + \phi_s f_y A_{st}\right] = 11\,452\ \text{kN}$$
The applied 694 kN is six per cent of that, confirming what the interaction curve already shows: this column is in the tension-controlled region and behaves essentially as a beam with a modest axial assist.
Detail the ties and state what sets the size. Clause 7.6.5.2 limits tie spacing to the least of 16 longitudinal bar diameters ($16 \times 29.9 = 478\ \text{mm}$), 48 tie diameters ($48 \times 11.3 = 542\ \text{mm}$) and the least column dimension (800 mm), so 10M ties at 450 mm govern, with every alternate bar tied by a corner or a cross-tie. The column is at 42 per cent utilisation because slenderness, not strength, fixed its size — trimming it to suit the moment would push $k\ell_u/r$ of the companion 12 m column past 100 and force a full second-order analysis, and would raise $Q$ towards the 0.20 value at which Clause 10.16.3 stops being applicable.
Column section for D–F (and equally for A–B): 800 mm square with 12-30M.
