07-Str-A5 · May 2015

Question 7 of 7: Member AB designed as a beam–column

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

Notes on this paper

Paper format. National Exams, May 2015 — 07-Str-A5 Advanced Structural Design. Three hours, "closed book" (textbooks and design handbooks permitted, no notes). Seven questions; any five constitute a complete paper and all are of equal value, the printed mark split being 20 marks each. All loads shown on Figures 1–4 are unfactored. All seven questions are answered here.

Design data printed on page 1 and used throughout. Concrete $f'_c = 30\ \text{MPa}$; structural steel $F_y = 350\ \text{MPa}$; reinforcing steel $f_y = 400\ \text{MPa}$; prestressed concrete $f'_{ci} = 35\ \text{MPa}$ at transfer and $f'_c = 50\ \text{MPa}$ in service, $n = 6$; $f_{pu} = 1750\ \text{MPa}$, $f_{py} = 1450\ \text{MPa}$, $f_{p,\text{initial}} = 1200\ \text{MPa}$, losses $= 240\ \text{MPa}$, hence $f_{pe} = 960\ \text{MPa}$ and the loss ratio $\eta = 0.80$.

Check: load factors. The paper says only that "all loads shown are unfactored" and gives no load classification. Every printed load is therefore treated as a specified live load and factored at 1.5; member self weight is treated as dead and factored at 1.25 (NBCC 2020 Case 2, $1.25D + 1.5L$). Resistance factors are $\phi = 0.90$ (steel), $\phi_w = 0.67$ and the extra 0.67 of CSA S16 Cl. 13.13.2.2 (welds), $\phi_{sc} = 0.80$ (shear studs), $\phi_c = 0.65$, $\phi_s = 0.85$, $\phi_p = 0.90$ (CSA A23.3). In an examination you would state exactly this and proceed.

Reference texts for 07-Str-A5.

CSA S16:19, Design of Steel Structures, with the CISC Handbook of Steel Construction (11th ed.) — Cl. 13.3 (compression), 13.5–13.6 (flexure and lateral–torsional buckling), 13.7 (bracing at plastic hinges), 13.8 (beam–columns), 13.13 (welds), 14.3–14.6 (plate girders), Cl. 17 (composite beams).
CSA A23.3:19, Design of Concrete Structures, with the CAC Concrete Design Handbook (4th ed.) — Cl. 10 (flexure and columns), Cl. 11 (shear), Cl. 18 (prestressed concrete).
Salmon, Johnson and Malhas, Steel Structures: Design and Behavior, 5th ed. — plastic analysis (Ch. 10), plate girders (Ch. 11), composite construction (Ch. 16).
MacGregor and Bartlett, Reinforced Concrete: Mechanics and Design, Canadian edition — frames, beam–columns, detailing.
Collins and Mitchell, Prestressed Concrete Structures — cable zone, transfer and service stress checks, harped profiles.
National Building Code of Canada 2020 for load combinations; CSA S6 (CHBDC) for the pedestrian-bridge serviceability and vibration criteria.

Question 7: Member AB designed as a beam–column (12 + 4 + 4 = 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. From the analysis in Question 6: factored axial compression $C_f = 1269\ \text{kN}$, moment $2076\ \text{kN}\cdot\text{m}$ at the top $B$ and $-1038\ \text{kN}\cdot\text{m}$ at the fixed base $A$ (double curvature), shear 389.3 kN over an 8 m storey height; $f'_c = 30\ \text{MPa}$, $f_y = 400\ \text{MPa}$, $\phi_c = 0.65$, $\phi_s = 0.85$.

Find. Column dimensions, longitudinal steel, ties, and the slenderness verdict.

Approach. The eccentricity $e = M_f/C_f = 2076/1269 = 1.64\ \text{m}$ is far larger than any sensible column dimension, so the member is a flexural element carrying a modest useful axial load, not a compression member. Check slenderness first (it turns out not to matter), then locate the neutral axis for the given axial load and take moments about the plastic centroid.

Slenderness. For a 700 × 1000 section bent about its strong axis, $r = 0.3(1000) = 300\ \text{mm}$, and with $k = 1.2$ for the unbraced frame $kL_u/r = 32 > 22$, so slenderness must be examined. With $\beta_d = 0.113$ and $EI = 0.4E_cI_g/(1+\beta_d) = 5.17\times10^{14}\ \text{N}\cdot\text{mm}^2$,
$$P_c = \frac{\pi^2EI}{(kL_u)^2} = 55\,313\ \text{kN},\qquad C_m = 0.6 + 0.4\frac{M_1}{M_2} = 0.6 + 0.4(-0.5) = 0.40$$ $$\delta_{ns} = \frac{C_m}{1 - C_f/(0.75P_c)} = \frac{0.40}{1 - 0.031} = 0.41 \ \rightarrow\ \text{taken as } 1.0$$
The axial load is only 2 % of $P_c$, so second-order effects are negligible and the member is designed for the first-order moment. The minimum moment $C_f(15 + 0.03h) = 57\ \text{kN}\cdot\text{m}$ is irrelevant here.
Choose a trial section and steel. Take 700 mm × 1000 mm with the 1000 mm dimension in the plane of bending, and symmetric reinforcement of 6–35M in each of the two faces perpendicular to bending ($A_s = A'_s = 6000\ \text{mm}^2$, total $12\,000\ \text{mm}^2$, $\rho = 1.71\ \%$, between the 1 % and 4 % limits of A23.3 Cl. 10.9). With 40 mm cover and 10M ties, $d = 930.9\ \text{mm}$ and $d' = 69.2\ \text{mm}$.
Locate the neutral axis for $C_f = 1269\ \text{kN}$. Axial equilibrium requires $C_c + C_s - T = C_f$, where the compression steel may not have yielded:
$$C_c = \alpha_1\phi_cf'_cb(\beta_1c),\quad C_s = \phi_sA'_s(f'_s - \alpha_1\phi_cf'_c),\quad f'_s = \min\!\left(f_y,\ \frac{0.0035(c-d')}{c}E_s\right),\quad T = \phi_sA_sf_y$$
Solving gives $c = 149.5\ \text{mm}$, $a = 133.8\ \text{mm}$ and $f'_s = 376.2\ \text{MPa}$ — the compression steel is just short of yielding, which is the usual outcome when the neutral axis is this shallow. The three forces are $C_c = 1470$, $C_s = 1839$ and $T = 2040\ \text{kN}$, and they sum to 1269 kN as required. The tension steel strain is 0.0183, over nine times yield, so the section is firmly tension-controlled.
Moment resistance about the plastic centroid. With symmetric steel the plastic centroid is the geometric centre, 500 mm from either face:
$$M_r = C_c\left(\frac{h}{2} - \frac{a}{2}\right) + C_s\left(\frac{h}{2} - d'\right) + T\left(d - \frac{h}{2}\right)$$ $$M_r = 1470(0.433) + 1839(0.431) + 2040(0.431) = \boxed{2308\ \text{kN}\cdot\text{m}\ \ge\ M_f = 2076\ \text{kN}\cdot\text{m}}$$
Utilisation 0.900. The axial load is helping: the pure-flexure resistance of the same section would be smaller, and the ultimate axial capacity $P_{r,\max} = 0.80[\alpha_1\phi_cf'_c(A_g - A_{st}) + \phi_sf_yA_{st}] = 11\,904\ \text{kN}$ is nine times the applied load, confirming the member sits deep on the tension-controlled branch of the interaction diagram.
Shear and ties. $V_f = 389.3\ \text{kN}$; with $d_v = 837.8\ \text{mm}$, $V_c = 375.8\ \text{kN}$, so nominal ties finish the job: 10M double-leg ties at 400 mm give $V_r = 579\ \text{kN}$. The tie-spacing limit of Cl. 7.6.5 is $\min(16d_b, 48d_{\text{tie}}, \text{least dimension}) = 542\ \text{mm}$, so 400 mm governs, tightened to 150 mm over the top and bottom 1.0 m where the moments peak and the bars must be confined.
Reinforcing details. The 35M bars run full height and are spliced only at mid-height, away from the maximum-moment regions at $A$ and $B$. At the base they extend into the footing as dowels with 90° hooks developed for $f_y$ in tension; at the top they continue through the joint and lap with the column above or, at a roof joint, hook into the beam. Because the moment reverses 2.67 m above the base, both faces need the full 6–35M — symmetric reinforcement is not a convenience here but a requirement.

Column AB: 700 × 1000 mm with 12–35M, six in each face perpendicular to the plane of bending.

Result	Value
Design actions	$C_f = 1269$ kN, $M_f = 2076$ kN·m (top), −1038 kN·m (base), $V_f = 389$ kN
Eccentricity	1636 mm — flexure-dominated
Slenderness	$kL/r = 32$; $P_c = 55\,313$ kN, $\delta_{ns} = 1.0$ (no magnification)
Section	700 mm × 1000 mm (1000 mm in the plane of bending)
Longitudinal steel	12–35M, 6 per face, $\rho = 1.71$ %
Neutral axis / compression steel stress	$c = 149.5$ mm, $f'_s = 376$ MPa (not yielded)
$M_r$ at $C_f = 1269$ kN	2308 kN·m (utilisation 0.900)
$P_{r,\max}$	11 904 kN
Ties	10M at 400 mm; 150 mm within 1.0 m of $A$ and $B$

Topic. Reinforced-concrete beam–column design on the tension-controlled branch of the interaction diagram.

Key relations. Strain compatibility with $\varepsilon_{cu} = 0.0035$; $f'_s = 0.0035(c-d')E_s/c \le f_y$; axial equilibrium $C_c + C_s - T = P_f$; moment about the plastic centroid; $P_{r,\max} = 0.80[\alpha_1\phi_cf'_c(A_g-A_{st}) + \phi_sf_yA_{st}]$ (Cl. 10.10.4); moment magnification $\delta_{ns} = C_m/(1 - P_f/0.75P_c) \ge 1.0$ with $P_c = \pi^2EI/(kL_u)^2$ and $EI = 0.4E_cI_g/(1+\beta_d)$ (Cl. 10.15).

Why this works. Below the balance point, adding axial compression increases the moment resistance, because the concrete compression block that the axial load requires also supplies part of the flexural couple. At $C_f = 1269$ kN the neutral axis is only 150 mm deep in a 1000 mm section, so the section behaves almost exactly like a doubly reinforced beam, and the design reduces to finding the one value of $c$ that satisfies axial equilibrium and then taking moments. Checking slenderness before anything else is worth the two minutes it costs: here $P_f/P_c = 0.02$ kills the magnifier outright, and a candidate who starts by amplifying moments wastes time on a correction of no consequence.

Common pitfall. Assuming the compression steel has yielded. With $c = 149.5$ and $d' = 69.2\ \text{mm}$ its strain is 0.00188, just under the 0.0020 yield strain; taking $f'_s = f_y$ overstates $C_s$ by 6 % and, worse, makes the axial equilibrium iteration oscillate instead of converge. The second pitfall is taking moments about the tension steel or the neutral axis rather than the plastic centroid, which is the only reference for which $P$ and $M$ form a valid interaction pair.

Source. CSA A23.3:19 Cl. 10.9–10.15 and Cl. 7.6.5; MacGregor and Bartlett, Reinforced Concrete: Mechanics and Design, Ch. 11 (columns) and Ch. 12 (slender columns); CAC Concrete Design Handbook, column interaction charts.

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Result	Value
Design actions	\(C_f = 1269\) kN, \(M_f = 2076\) kN·m (top), −1038 kN·m (base), \(V_f = 389\) kN
Eccentricity	1636 mm — flexure-dominated
Slenderness	\(kL/r = 32\); \(P_c = 55\,313\) kN, \(\delta_{ns} = 1.0\) (no magnification)
Section	700 mm × 1000 mm (1000 mm in the plane of bending)
Longitudinal steel	12–35M, 6 per face, \(\rho = 1.71\) %
Neutral axis / compression steel stress	\(c = 149.5\) mm, \(f'_s = 376\) MPa (not yielded)
\(M_r\) at \(C_f = 1269\) kN	2308 kN·m (utilisation 0.900)
\(P_{r,\max}\)	11 904 kN
Ties	10M at 400 mm; 150 mm within 1.0 m of \(A\) and \(B\)