07-Str-A5 · December 2014

Question 7 of 7: Reinforced concrete column CE

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

Notes on this paper

Paper format. 07-Str-A5 Advanced Structural Design, National Exams December 2014. Three hours, open-book (handbooks and textbooks permitted, no notes). Seven questions of equal value; any five constitute a complete paper, and only the first five presented are marked. All solutions below answer all seven, because the set is a study resource rather than an exam script. All loads shown on the figures are unfactored.

Design data supplied on the paper (SI). Concrete $f'_c = 30\ \text{MPa}$; structural steel $F_y = 350\ \text{MPa}$; rebar $f_y = 400\ \text{MPa}$. Prestressed concrete: $f_{ci} = 35\ \text{MPa}$ at transfer, $f'_c = 50\ \text{MPa}$, $n = 6$, $f_{ult} = 1750\ \text{MPa}$, $f_{y} = 1450\ \text{MPa}$, $f_{initial} = 1200\ \text{MPa}$, losses $= 240\ \text{MPa}$, hence $f_{pe} = 960\ \text{MPa}$.

Reference texts.

CSA S16:19, Design of Steel Structures — Cl. 13 (member resistances), Cl. 14 (plate girders), Cl. 17 (composite beams), Cl. 21 (welds).
CSA A23.3:19, Design of Concrete Structures — Cl. 10 (flexure, axial, slenderness), Cl. 11 (shear), Cl. 15 (footings), Cl. 18 (prestressed concrete).
National Building Code of Canada 2020, Part 4 — load combinations $1.25D + 1.5L$.
Salmon, Johnson & Malhas, Steel Structures: Design and Behavior, 5th ed. — Ch. 11 (plate girders), Ch. 12 (composite), Ch. 10 (plastic design).
MacGregor & Bartlett, Reinforced Concrete: Mechanics and Design (Canadian ed.) — Ch. 4–6, Ch. 11 (slender columns), Ch. 15 (footings), Ch. 19 (prestressed).
Handbook of Steel Construction (CISC), 11th ed. — Class limits, Cl. 13.8 interaction tables.

Check: load classification. The paper prints the loads as unfactored but does not split them between dead and live. Throughout this solution every printed load is treated as a specified live load and factored at 1.5, while member self-weight is treated as dead and factored at 1.25 (NBCC 2020 combination 2, $1.25D + 1.5L$). If a different split is stated on exam day, re-run the same arithmetic with the stated factors — the method is unchanged.

Check: Figure 4 geometry. Read from the drawing on page 4, the beam $AC$ is $4 + 8 + 4 = 16\ \text{m}$ long with the rigid joint $C$ at its right-hand end, directly over the column; the 400 kN acts at $C$, and the 200 kN loads act at 4 m and 12 m from $A$. Support $A$ is drawn with the same circle-on-hatching symbol used for the rollers in Figures 1 and 2, so it is taken as a roller (vertical reaction only); base $E$ is fixed. The frame is therefore indeterminate to the first degree.

Question 7: Reinforced concrete column CE (14 + 6 = 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.

Quantity	Value
Column $CE$	10 m clear, fixed at $E$, rigidly framed to the beam at $C$
Frame type	Sway-permitted (roller at $A$ gives no lateral restraint)
Concrete / reinforcement	$f'_c = 30\ \text{MPa}$; $f_y = 400\ \text{MPa}$; $E_c = 24\,648\ \text{MPa}$
Beam from Question 6	500 × 1100 mm, $I = 5.546\times10^{10}$ mm⁴, far end on a roller
Trial column	500 × 800 mm

Find. A rectangular column section with its longitudinal reinforcement, checked for strength (axial–moment interaction) and stability (slenderness and second-order sway amplification).

Approach. Extract the factored axial force and the moment envelope from the same elastic frame analysis used in Question 6, split the base moment into its non-sway and sway parts, magnify only the sway part by the CSA A23.3 Cl. 10.16 sway factor, and check the magnified pair against the column's $P\!-\!M$ interaction diagram.

Factored actions. From the frame analysis with the concrete stiffnesses,
$$N_f = 1164.2\ \text{kN}, \qquad M_{f,E} = 795.2\ \text{kN}\cdot\text{m}, \qquad M_{f,C} = 195.2\ \text{kN}\cdot\text{m}$$
The base is by far the critical section. The horizontal shear in the column below $D$ is $1.5(80) = 120\ \text{kN}$ and above $D$ is zero, which is why the column moment is constant from $C$ to $D$ and then grows linearly to $E$.
Slenderness classification. The effective-length factor comes from the sway alignment chart. With the beam's far end on a roller its stiffness is halved for a sway frame:
$$\Psi_C = \frac{I_c/l_c}{0.5 I_b/l_b} = \frac{2.133\times10^{10}/10\,000}{0.5(5.546\times10^{10})/16\,000} = 1.231, \qquad \Psi_E = 1.0\ \text{(fixed, practical)}$$
which gives $k \approx 1.32$. With $r = 0.3h = 240\ \text{mm}$,
$$\frac{k l_u}{r} = \frac{1.32(10\,000)}{240} = \boxed{55.0} \;>\; 22 \quad\text{(A23.3 Cl. 10.15.2 limit for sway frames)}$$
The column is slender and second-order effects must be included.
Critical buckling load. Using Cl. 10.15.3 with $\beta_d$ the ratio of sustained (dead) to total factored axial load, $\beta_d = 257.6/1164.2 = 0.221$:
$$EI = \frac{0.4E_c I_g}{1 + \beta_d} = \frac{0.4(24\,648)(2.133\times10^{10})}{1.221} = 1.722\times10^{14}\ \text{N}\cdot\text{mm}^2$$

$$P_c = \frac{\pi^2 EI}{(kl_u)^2} = \frac{\pi^2(1.722\times10^{14})}{(13\,200)^2} = 9755\ \text{kN}$$
Sway magnification. Separating the base moment into the part from gravity alone and the part caused by the 80 kN lateral load,
$$M_{ns} = 319.7\ \text{kN}\cdot\text{m}, \qquad M_s = 475.5\ \text{kN}\cdot\text{m}, \qquad M_{ns} + M_s = 795.2\ \text{kN}\cdot\text{m}$$

$$\delta_s = \frac{1}{1 - \sum P_f/\sum P_c} = \frac{1}{1 - 1164.2/9755} = 1.136$$

$$M_f = M_{ns} + \delta_s M_s = 319.7 + 1.136(475.5) = \boxed{859.7\ \text{kN}\cdot\text{m}}$$
Second-order effects add 8 % to the design moment — modest, because the axial load is only 12 % of the critical load.
Reinforcement from the interaction diagram. The design eccentricity is $e = 859.7/1164.2 = 738\ \text{mm}$, close to the member depth, so the column behaves as a lightly compressed beam and symmetric reinforcement is appropriate. Try 8 – 30M, four bars in each 500 mm face, $A_{st} = 5600\ \text{mm}^2$, $d = 725\ \text{mm}$, $d' = 75\ \text{mm}$. Solving strain compatibility for the neutral-axis depth that gives $P_r = N_f$:
$$c = 173.1\ \text{mm}, \quad f'_s = 396\ \text{MPa}\ \text{(compression steel just yielding)}, \quad f_s = f_y\ \text{(tension steel yielded)}$$

$$M_r = 994.2\ \text{kN}\cdot\text{m} \;\ge\; M_f = 859.7\ \text{kN}\cdot\text{m} \quad\text{(utilisation } 0.865)$$
Because $c = 173\ \text{mm}$ is far below the balanced depth, the section is on the tension-controlled branch of the interaction diagram — failure would be ductile.
Steel ratio and axial capacity checks.
$$\rho = \frac{5600}{500(800)} = 0.0140 \quad\text{— within the A23.3 Cl. 10.9.1 range } 0.01 \le \rho \le 0.08$$

$$P_{r,max} = 0.80\left[\alpha_1\phi_c f'_c(A_g - A_{st}) + \phi_s f_y A_{st}\right] = 6476\ \text{kN} \;\gg\; N_f = 1164.2\ \text{kN}$$
Axial capacity is nowhere near critical; the column is moment-governed.
Ties and detailing. Tie spacing is the least of 16 longitudinal-bar diameters, 48 tie diameters and the least column dimension:
$$s \le \min\left[16(30),\ 48(11.3),\ 500\right] = \min(480,\ 542,\ 500) = 480\ \text{mm}$$
Provide 10M ties at 400 mm, closed with 135° hooks, tightened to 150 mm over a distance of 1600 mm (twice the member depth) above the base and below the joint, where the plastic demand and the bar splices are concentrated. Longitudinal bars are lap-spliced above the base region, and the beam's bottom bars are anchored into the joint core with standard hooks.
Stability of the frame as a whole. With $\delta_s = 1.136$, the stability index $Q = \sum P_f/\sum P_c = 0.119$ is below 0.20, so the frame is not classed as excessively sway-sensitive and a single first-order analysis with the magnifier is sufficient — a full second-order ($P\!-\!\Delta$) analysis is not required. Had $Q$ exceeded 0.20, the column section would have needed to be increased to stiffen the frame rather than merely to add reinforcement.

Column CE cross-section: 8 – 30M symmetric, four bars per face, bending about the 800 mm dimension.

Result	Value
Factored actions at $E$	$N_f = 1164.2$ kN; $M_f = 795.2$ kN·m (first order)
Column section	500 × 800 mm (bending about the 800 mm face)
Effective length factor / slenderness	$k = 1.32$; $kl_u/r = 55.0 > 22$ — slender
Critical load / magnifier	$P_c = 9755$ kN; $\delta_s = 1.136$
Magnified design moment	859.7 kN·m
Longitudinal steel	8 – 30M (5600 mm², $\rho = 1.40\%$), four per face
Interaction check	$M_r = 994.2$ kN·m at $N_f$ — utilisation 0.865
Neutral axis / failure mode	$c = 173$ mm — tension-controlled, ductile
$P_{r,max}$	6476 kN ≫ 1164 kN
Ties	10M at 400 mm (limit 480 mm); 150 mm over 1600 mm at base and joint
Stability index $Q$	0.119 < 0.20 — magnifier method acceptable

Topic. Limit-states design of a slender sway column: effective length, the moment magnifier, and the axial–moment interaction diagram.

Key relations. Sway slenderness limit $kl_u/r > 22$; $EI = 0.4E_c I_g/(1 + \beta_d)$; $P_c = \pi^2 EI/(kl_u)^2$; $\delta_s = 1/(1 - \sum P_f/\sum P_c)$; $M_f = M_{ns} + \delta_s M_s$; strain compatibility with $\varepsilon_{cu} = 0.0035$, $\alpha_1\phi_c f'_c b a$ and $\phi_s A_s f_s$; $P_{r,max} = 0.80[\alpha_1\phi_c f'_c(A_g - A_{st}) + \phi_s f_y A_{st}]$.

Why this works. Only the sway component of the moment is magnified, because $P\!-\!\Delta$ effects arise from lateral drift, not from the curvature a gravity load produces in a braced member. Splitting the analysis into a gravity-only run and a lateral-only run is therefore not an accounting convenience — it is what the code's method requires. The eccentricity of 738 mm, nearly the member depth, tells you before any arithmetic that the column will land on the tension-controlled branch, so symmetric steel and a ductile failure are the right target; a compression-controlled design here would be both uneconomical and brittle.

Common pitfall. Using the braced (non-sway) alignment chart, which would give $k \approx 0.75$, a slenderness of 31 and a much smaller magnifier. The roller at $A$ means nothing restrains the frame laterally except the fixed base, so it is unambiguously a sway frame. The other frequent error is magnifying the whole base moment rather than just $M_s$, which over-designs by roughly 5 % here and much more in stiffer frames.

Source. CSA A23.3:19 Cl. 10.9, 10.10, 10.15, 10.16; MacGregor & Bartlett, Reinforced Concrete, Ch. 11 and 12; Handbook of Concrete Design (CAC), Ch. 4.

← Question 6 Back to the paper →

Quantity	Value
Column \(CE\)	10 m clear, fixed at \(E\), rigidly framed to the beam at \(C\)
Frame type	Sway-permitted (roller at \(A\) gives no lateral restraint)
Concrete / reinforcement	\(f'_c = 30\ \text{MPa}\); \(f_y = 400\ \text{MPa}\); \(E_c = 24\,648\ \text{MPa}\)
Beam from Question 6	500 × 1100 mm, \(I = 5.546\times10^{10}\) mm⁴, far end on a roller
Trial column	500 × 800 mm

Result	Value
Factored actions at \(E\)	\(N_f = 1164.2\) kN; \(M_f = 795.2\) kN·m (first order)
Column section	500 × 800 mm (bending about the 800 mm face)
Effective length factor / slenderness	\(k = 1.32\); \(kl_u/r = 55.0 > 22\) — slender
Critical load / magnifier	\(P_c = 9755\) kN; \(\delta_s = 1.136\)
Magnified design moment	859.7 kN·m
Longitudinal steel	8 – 30M (5600 mm², \(\rho = 1.40\%\)), four per face
Interaction check	\(M_r = 994.2\) kN·m at \(N_f\) — utilisation 0.865
Neutral axis / failure mode	\(c = 173\) mm — tension-controlled, ductile
\(P_{r,max}\)	6476 kN ≫ 1164 kN
Ties	10M at 400 mm (limit 480 mm); 150 mm over 1600 mm at base and joint
Stability index \(Q\)	0.119 < 0.20 — magnifier method acceptable