07-Str-A5 · December 2015

Question 7 of 7: Reinforced-concrete rigid frame — beam–column AB and its deflection

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

Notes on this paper

Paper format. National Exams, December 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 and figure data. The paper states only that "all loads shown are unfactored" and gives no load classification, so every printed load is treated as a specified live load and factored at 1.5, while member and element 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), $\phi_w = 0.67$ with the additional 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). Figure 2 has a free tip at A, a roller at B and a pin at C; Figure 4 has a hatched fixed base at A and a hatched fixed end at C, with the 200 kN arrow pointing towards the frame at mid-height of the column.

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), 13.4 (shear), 13.5–13.6 (flexure and lateral–torsional buckling), 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. 9 (deflections), Cl. 10 (flexure, columns, slenderness), Cl. 11 (shear), Cl. 13 (two-way 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, slender 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 and deflection limits.

Question 7: Reinforced-concrete rigid frame — beam–column AB and its deflection (6 + 5 + 5 + 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. The frame of Figure 4 built in reinforced concrete: $f'_c = 30\ \text{MPa}$, $f_y = 400\ \text{MPa}$, $E_c = 4500\sqrt{30} = 24\,648\ \text{MPa}$. Take a beam BC of $400 \times 900$ mm and a column AB of $500 \times 700$ mm, the 700 mm dimension lying in the plane of the frame.

Given data
Quantity	Value
Column AB / beam BC	500 × 700 mm, 10 m / 400 × 900 mm, 9 m
Loads (specified)	200 kN horizontal at 5 m; 600 kN at B; 450 kN at 3 m and 6 m
$I_c$ / $I_b$ (gross)	$1.429 \times 10^{10}$ / $2.430 \times 10^{10}$ mm&sup4;
Cover / bars	40 mm; 30M longitudinal, 10M ties; $d = 632.5$, $d' = 67.5$ mm

Find. (a) the longitudinal reinforcement and ties for AB, including slenderness; (b) the long-term horizontal deflection at mid-height of AB.

Approach. The frame has only one degree of kinematic freedom — the rotation of joint B — because C is fixed and the members are axially rigid, so a single slope-deflection equation gives all four end moments; design the column from strain compatibility with the moment magnified for slenderness, then integrate the column curvature to get its mid-height deflection and multiply for cracking and creep.

Fixed-end moments (factored). With $H_f = 300\ \text{kN}$ and the two beam loads at 675 kN, $$\text{FEM}_{AB} = -\frac{H_fL}{8} = -375, \quad \text{FEM}_{BA} = +375\ \text{kN}\cdot\text{m}$$ $$\text{FEM}_{BC} = -\sum\frac{Pab^2}{L^2} = -(900 + 450) = -1350, \quad \text{FEM}_{CB} = +1350\ \text{kN}\cdot\text{m}$$
Solve the single joint rotation. Moment equilibrium at B, $M_{BA} + M_{BC} = 0$, gives $$\theta_B\left(\frac{4E_cI_c}{10} + \frac{4E_cI_b}{9}\right) = 1350 - 375 = 975\ \text{kN}\cdot\text{m} \;\Rightarrow\; \theta_B = 2.395 \times 10^{-6}\ \text{rad}$$ Back-substituting, $$M_B = \boxed{712.5\ \text{kN}\cdot\text{m}}, \quad M_A = -206.3, \quad M_{\text{mid}} = +290.6, \quad M_C = 1668.8\ \text{kN}\cdot\text{m}$$ Note that the column bends in double curvature, its moment changing sign at about 2 m above the base.
Axial load on the column. The beam shear at B is $675 - (M_C - M_B)/9 = 568.7\ \text{kN}$, so with the 900 kN at the joint and the column's own factored weight (105 kN), $$C_f = 900 + 568.7 + 105 = \boxed{1574\ \text{kN}}$$
(a) Slenderness. Taking $k = 0.7$ for a braced column fixed at the base and restrained by a stiff beam at the top, $kl_u/r = 0.7(10\,000)/(0.3 \times 700) = 33.3$, against the limit $34 - 12(M_1/M_2) = 30.5$. Slenderness must therefore be considered. With $EI = 0.4E_cI_g/(1 + \beta_d)$ and $\beta_d = 0.07$, $P_c = \pi^2EI/(kl_u)^2 = 26\,524\ \text{kN}$, and with $C_m = 1.0$ because a transverse load acts between the ends, $$\delta_b = \frac{1}{1 - C_f/(\phi_mP_c)} = 1.086 \;\Rightarrow\; M_c = 1.086(712.5) = \boxed{774\ \text{kN}\cdot\text{m}}$$
(a) Reinforcement from strain compatibility. Try 4–30M in each of the two 500 mm faces ($A_s = A_s' = 2800\ \text{mm}^2$). Axial equilibrium $C_c + C_s - T = C_f$ with both bar layers yielding gives $$C_c = 1618\ \text{kN} \;\Rightarrow\; a = \frac{C_c}{\alpha_1\phi_cf'_cb} = 206.1\ \text{mm}, \qquad c = \frac{a}{\beta_1} = 230.3\ \text{mm}$$ Both strains confirm yielding ($\varepsilon_s' = 0.00247$, $\varepsilon_t = 0.00611$, against $\varepsilon_y = 0.0020$). Moments about the section centroid then give $$M_r = C_c\left(\frac{h}{2} - \frac{a}{2}\right) + C_s\left(\frac{h}{2} - d'\right) + T\left(d - \frac{h}{2}\right) = \boxed{925\ \text{kN}\cdot\text{m}} \;>\; 774\ \text{kN}\cdot\text{m}$$ Add two mid-side bars for detailing, giving 10–30M in all ($\rho = 0.020$, above the 0.01 minimum), with 10M ties at 400 mm ($\le 16d_b = 478$ mm).
(b) Immediate horizontal deflection. Repeat the analysis at specified loads: $\theta_B = 1.597 \times 10^{-6}$, $M_A = -137.5$, $M_B = 475.0\ \text{kN}\cdot\text{m}$, base shear $V_A = 66.25\ \text{kN}$. The column base is fixed, so integrating the curvature from A upwards, $$\Delta_5 = \int_0^{5}\frac{(5-x)\,M(x)}{E_cI_g}\,dx = \frac{1}{E_cI_g}\left[-137.5(12.5) + 66.25(20.83)\right] = \boxed{0.96\ \text{mm}}$$ This is far less than the 2.96 mm a fixed-ended column would give under the same 200 kN, because the beam's large hogging moment at B bends the column back against the sway.
(b) Cracking and creep. The column cracks at both ends, so take $I_e \approx 0.7I_g$ (A23.3 Cl. 10.14.1.2), raising the immediate value to 1.37 mm. For the sustained part of the load, the long-term multiplier is $$s = \frac{\zeta}{1 + 50\rho'} = \frac{2.0}{1 + 50(0.00886)} = 1.386$$ so, treating the whole loading as sustained, $$\Delta_{\text{long-term}} = 1.37(1 + 1.386) = \boxed{3.3\ \text{mm}}$$ which is only $h/3000$, comfortably inside the usual $h/500 = 20\ \text{mm}$ drift limit. A column this deep in a frame restrained at both ends is governed by strength, not by deflection.

Question 7 — final results
Item	Value
Joint rotation $\theta_B$ (factored)	$2.395 \times 10^{-6}$ rad
$M_A$ / $M_{\text{mid}}$ / $M_B$ / $M_C$	206.3 / 290.6 / 712.5 / 1668.8 kN·m
Column axial $C_f$	1574 kN
Slenderness $kl/r$ / $\delta_b$ / $M_c$	33.3 / 1.086 / 774 kN·m
(a) Column AB	500 × 700 mm, 10–30M, 10M ties at 400 mm
(a) $M_r$ at $C_f = 1574$ kN	925 kN·m (ratio 0.84)
(b) Immediate deflection (gross / cracked)	0.96 / 1.37 mm
(b) Long-term deflection with creep	3.3 mm (< h/500 = 20 mm)

Topic. Reinforced-concrete beam–columns: slope-deflection analysis, slenderness magnification, and long-term lateral deflection.

Key relations. $M_{ij} = \frac{2EI}{L}(2\theta_i + \theta_j - 3\psi) + \text{FEM}_{ij}$; $\delta_b = C_m/[1 - P_f/(\phi_mP_c)]$ with $P_c = \pi^2EI/(kl_u)^2$ and $EI = 0.4E_cI_g/(1+\beta_d)$; slenderness may be neglected while $kl_u/r \le 34 - 12(M_1/M_2)$; long-term multiplier $s = \zeta/(1 + 50\rho')$.

Why this works. Recognising that the frame has a single unknown rotation turns what looks like a three-times-indeterminate problem into one equation. It is fixity at C that does this: it stops joint B from translating, so no sway term $\psi$ ever appears. On the column, the axial load rides on the lateral deflection and adds a second-order moment, which the magnifier $\delta_b$ captures without a full $P$-$\Delta$ analysis. For deflection, gross stiffness is optimistic once the ends crack and again once the concrete creeps, so the estimate is built in three steps — elastic, cracked, then crept — each with a defensible factor rather than one lumped guess.

Common pitfall. Assuming sidesway because a horizontal load is drawn, which introduces a spurious $\psi$ and roughly triples the computed deflection; and quoting the immediate gross-section deflection as the answer when the question explicitly asks for creep to be allowed for. A third is skipping the slenderness check because $kl/r$ "looks small" — here it exceeds the limit by only three units, but the magnifier still adds 9 % to the design moment.

Source. CSA A23.3:19 Cl. 9.8, 10.10, 10.14 and 10.15; MacGregor and Bartlett, Reinforced Concrete: Mechanics and Design, Canadian edition, slender-column and deflection chapters.

← Question 6 Back to the paper →

Item	Value
Joint rotation \(\theta_B\) (factored)	\(2.395 \times 10^{-6}\) rad
\(M_A\) / \(M_{\text{mid}}\) / \(M_B\) / \(M_C\)	206.3 / 290.6 / 712.5 / 1668.8 kN·m
Column axial \(C_f\)	1574 kN
Slenderness \(kl/r\) / \(\delta_b\) / \(M_c\)	33.3 / 1.086 / 774 kN·m
(a) Column AB	500 × 700 mm, 10–30M, 10M ties at 400 mm
(a) \(M_r\) at \(C_f = 1574\) kN	925 kN·m (ratio 0.84)
(b) Immediate deflection (gross / cracked)	0.96 / 1.37 mm
(b) Long-term deflection with creep	3.3 mm (< h/500 = 20 mm)