16-Civ-A2 Elementary Structural Design · May 2016

Question 5 of 7: B2 — Reinforced concrete column CD of the determinate frame

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

Notes on this paper

Paper format. National Examinations — May 2016, 98-Civ-A2 Elementary Structural Design. Three-hour closed-book examination (handbooks and textbooks permitted). Seven questions in three parts: Part A (A1–A3, structural steel to CSA S16), Part B (B1–B3, reinforced concrete to CSA A23.3), Part C (C1, timber to CSA O86). A candidate answers two from Part A, two from Part B and the one question in Part C — five solutions in total, all of equal value. All seven are solved here, because the set is a study resource rather than an examination script. Page 1 states that all loads shown are unfactored, so the load factors are applied within each solution.

Reference texts.

CSA S16:19, Design of Steel Structures — Clauses 11 (class of section), 13.3 (compressive resistance), 13.5–13.6 (bending), 13.8 (axial compression and bending), 13.13 (welds).
CISC, Handbook of Steel Construction, 12th ed. — section property tables, weld tables.
CSA A23.3:19, Design of Concrete Structures — Clauses 10 (flexure and axial load), 11 (shear), 7.6 (ties).
CSA O86:19, Engineering Design in Wood, with the Canadian Wood Council Wood Design Manual 2020 — Clause 7 (glued-laminated timber).
NBCC 2020 — Table 4.1.3.2 load combinations.

Check — assumptions common to the whole paper. (i) Every load drawn in Figures A1, A2, A3, B3 and quoted in C1 is a specified (unfactored) load; unless a question names the load type, the point loads are treated as live (factor 1.5) and concrete/steel self-weight as dead (factor 1.25), per NBCC Table 4.1.3.2 case 2. (ii) Steel is CSA G40.21 350W, so F_y = 350 MPa, E = 200 000 MPa. (iii) Question B3 states f'_c = 35 MPa and f_y = 400 MPa; because B1 and B2 quote no material strengths, the same pair is adopted for them and the assumption is stated in each solution. (iv) Section properties are recomputed from the nominal plate dimensions of each rolled shape rather than read off a table, so every number below is reproducible; they agree with the CISC Handbook to better than 1 %.

Question 5: B2 — Reinforced concrete column CD of the determinate frame (10 + 6 + 4)

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. The frame geometry and loads of Figure A3 (see Question 3): 250 kN at 2 m, 6 m and 8 m along the beam; 60 kN horizontal 4 m below C; roller at A; hinge at B; fixed base at D; column CD 8.0 m long. Materials f'_c = 35 MPa and f_y = 400 MPa are carried over from B3.

Find. A cross-section, longitudinal reinforcement and tie arrangement for column CD, checked for combined axial load and bending including slenderness effects.

Approach. Repeat the Question 3 statics with a concrete beam’s self-weight, add the column’s own weight, carry the beam’s hogging moment into the column head and add the 60 kN load’s contribution down to the base, check slenderness, then verify the section by strain compatibility at the actual axial load.

Re-run the frame statics for a concrete beam. Assuming a 400 × 900 mm reinforced concrete beam, w = 0.4(0.9)(24) = 8.64 kN/m, factored to 10.8 kN/m. Repeating Steps 3–5 of Question 3: $$A_{y} = 209.1\ \text{kN}, \qquad M_{C} = \boxed{1673\ \text{kN}\cdot\text{m}}, \qquad V_{C} = 1002\ \text{kN}$$
Carry the moment down the column. Since the roller at A transmits no horizontal force, the column head carries zero shear; the moment therefore stays constant at M_C from C down to the 60 kN load, and grows below it: $$M_{f,\text{base}} = M_{C} + H_{f}(4.0) = 1673 + 90(4.0) = \boxed{2033\ \text{kN}\cdot\text{m}}$$ Adding the column’s own factored weight for a trial 600 × 1400 section, 1.25(0.6)(1.4)(24)(8) = 202 kN, $$P_{f} = 1002 + 202 = \boxed{1204\ \text{kN}}$$
Recognise what kind of member this is. The load eccentricity is $$e = \frac{M_{f}}{P_{f}} = \frac{2033}{1204} = 1.69\ \text{m}$$ more than the depth of any plausible section. The member is overwhelmingly flexural, so it must be designed by strain compatibility at the actual P_f; the pure-axial expression P_r,max = 0.80[α₁φ_cf'_c(A_g − A_st) + φ_sf_yA_st] is meaningless here.
Check slenderness. The roller at A leaves the frame with no lateral restraint at C, so the column is an unbraced cantilever with k = 2.0. For a 1400 mm depth, r ≈ 0.3h = 420 mm: $$\frac{k\ell_{u}}{r} = \frac{2.0(8000)}{420} = 38.1 \;>\; 22$$ so A23.3 Clause 10.15.2 requires slenderness to be considered. With E_c = 4500√35 = 26 620 MPa and EI = 0.4E_cI_g, $$P_{c} = \frac{\pi^{2}EI}{(k\ell_{u})^{2}} = 56\,300\ \text{kN}, \qquad \delta = \frac{1}{1-P_{f}/(0.75P_{c})} = 1.029$$ $$M_{f,\text{design}} = 1.029(2033) = \boxed{2092\ \text{kN}\cdot\text{m}}$$ The magnification is only 3 % — the section is so deep that it is stocky despite the 8 m height.
Verify the trial section. Take 600 mm wide × 1400 mm deep, reinforced with 12–30M: six bars in the face nearest the tension side and six in the opposite face, at 65 mm to bar centres. $$\rho = \frac{12(700)}{600(1400)} = 1.00\ \%$$ which meets the A23.3 Clause 10.9.1 minimum of 1 % exactly and is far below the 8 % maximum. Solving force equilibrium at P_f = 1204 kN gives a neutral axis at c = 141 mm; the tension steel is far past yield and the compression steel reaches 368 MPa. Summing moments about mid-depth: $$M_{r} = \boxed{2582\ \text{kN}\cdot\text{m}} \;>\; 2092\ \text{kN}\cdot\text{m} \quad\text{(utilisation 0.81)}\ \checkmark$$
Detail the ties. A23.3 Clause 7.6.5 limits tie spacing to the least of 16 longitudinal bar diameters, 48 tie diameters and the least column dimension: $$s \le \min\left[16(29.9),\ 48(16),\ 600\right] = \min\left[478,\ 768,\ 600\right] = 478\ \text{mm}$$ Specify 15M closed ties at 400 mm centres, reduced to 200 mm over the top and bottom 1400 mm where the moment gradient and the joint congestion are greatest. With six bars per face, every alternate bar needs a cross-tie or the corner of an overlapping tie, so use one perimeter tie plus two cross-ties per set.
Sketch of the reinforcement. The section and elevation below record the design.

Column CD: 600 × 1400 mm, 12–30M longitudinal bars (six per face), 15M closed ties with two cross-ties per set at 400 mm centres, closed to 200 mm over the end regions.

Check — effective length. k = 2.0 has been used because the roller at A cannot brace the frame laterally, so the column is a true cantilever in sway. This is the conservative reading. If the frame is in fact braced against sidesway by an adjacent structure, k falls to about 0.8, kℓ_u/r drops to 15 and no magnification is required at all — but the section is governed by the primary moment either way, so the design does not change.

Result	Value
Assumed concrete beam for self-weight	400 × 900 mm, 8.64 kN/m
Factored hogging moment at C	1673 kN·m
Factored axial load at the base P_f	1204 kN
Factored base moment M_f (primary)	2033 kN·m
Eccentricity e = M_f/P_f	1.69 m — flexure-dominated
Slenderness kℓ_u/r, magnifier δ	38.1 (> 22), 1.029
Design moment δM_f	2092 kN·m
Column section	600 × 1400 mm, 12–30M (ρ = 1.00 %)
M_r at P_f (c = 141 mm)	2582 kN·m — utilisation 0.81
Ties	15M closed ties + 2 cross-ties at 400 mm (200 mm at the ends)

Topic. Design of a slender reinforced concrete column under a large end moment delivered by a rigidly-framed beam.

Key relations. Equilibrium of the rigid corner: the beam’s end moment passes undiminished into the column head; M_base = M_C + H_fh; A23.3 Clause 10.15.2 slenderness threshold kℓ_u/r > 22 for unbraced frames; Clause 10.16 magnifier δ = C_m/(1 − P_f/0.75P_c) with P_c = π²EI/(kℓ_u)² and EI = 0.4E_cI_g; Clause 10.9.1 reinforcement ratio limits; Clause 7.6.5 tie spacing.

Why this works. A rigid beam-to-column joint has only one moment: whatever the beam delivers, the column must accept. Because there is no horizontal reaction at the roller, the column carries no shear above the 60 kN load, so its moment diagram is flat from C down to that level and only then begins to climb — a shape that is easy to draw and easy to check, since the base value must equal the moment computed from the whole structure about D. The eccentricity of 1.69 m is the number that decides the design method: at e greater than the section depth the concrete compression zone is small and the member behaves as a beam with a modest axial assist, so an interaction-diagram or strain-compatibility solution at the actual P_f is essential.

Common pitfall. Designing the column with the pure-axial capacity formula because the question calls it a column, which can overstate its usable capacity several-fold at this eccentricity. The second is forgetting to add the beam’s hogging moment: taking only H_f × 4 m gives 360 kN·m instead of 2033 kN·m, less than a fifth of the truth.

Source. CSA A23.3:19 Clauses 10.9, 10.10, 10.15, 10.16 and 7.6.5; MacGregor and Bartlett, Reinforced Concrete: Mechanics and Design, Canadian edition, Chapters 11 and 12.

← Question 4 Next: Question 6 →