Reinforced Concrete Design of the Rigid Frame: 16-Civ-B2 Advanced Structural Design, December 2015, Q7

Question 7: Reinforced Concrete Design of the Rigid Frame (6 + 5 + 5 + 4 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 Figure 4 frame in reinforced concrete: column AB 10 m built in at A, beam BC 9 m built in at C, factored loads 900 kN at B, 675 kN at 3 m and at 6 m, and 300 kN horizontally at mid-height. Concrete $f'_c = 30$ MPa ($E_c = 4500\sqrt{30} = 24\,648$ MPa), rebar $f_y = 400$ MPa.

Find. (a) the size and reinforcement of the beam-column AB, including a slenderness assessment; (b) the horizontal deflection at mid-height under sustained load.

Approach. Trial member sizes fix the stiffness ratio, so run an elastic frame analysis with the Clause 10.14.1.2 effective stiffnesses, design the column on the resulting axial–moment pair using a strain-compatibility interaction, then compute the mid-height sway and multiply the sustained part by a creep factor.

Part (a) — adopt trial members and analyse the frame. Take the column 600 × 800 (800 mm deep in the plane of the frame) and the beam 500 × 1200. Following Clause 10.14.1.2 the analysis uses $0.70I_g$ for the column and $0.35I_g$ for the beam, with the members' own weight included at 1.25. A stiffness analysis of the two-member frame, fixed at A and at C, gives $$\boxed{N_f = 1681\ \text{kN}, \quad M_B = 776\ \text{kN}\cdot\text{m}, \quad M_{5\text{m}} = 274\ \text{kN}\cdot\text{m}, \quad M_A = 176\ \text{kN}\cdot\text{m}}$$ and, for reference, a beam moment of 1851 kN·m at C. Repeating with gross sections shifts $M_B$ to 595 kN·m, so the cracked-stiffness values are the more onerous and are used for design.
Assess slenderness. The clear height is $l_u = 10\,000 - 600 = 9400$ mm; with a fixed base and a top restrained against translation, $k = 0.8$ and $r = 0.3h = 240$ mm, so $kl_u/r = 31.3$. The Clause 10.15.2 threshold for a braced column is $$\frac{25 - 10(M_1/M_2)}{\sqrt{P_f/(f'_cA_g)}} = \frac{25 + 10(0.227)}{\sqrt{1681\times10^3/(30 \times 480\,000)}} = 79.9$$ Since $31.3 < 79.9$ the column is non-slender and no moment magnification is required; the design moment stays at 776 kN·m. (Evaluating the magnifier anyway confirms it: $P_c = 25\,200$ kN gives $\delta_b = 1.00$.)
Design the reinforcement. Place bars in two layers at the faces perpendicular to the bending axis, 70 mm from each face. Building the $P$–$M$ interaction from strain compatibility ($\varepsilon_{cu} = 0.0035$, $\alpha_1 = 0.805$, $\beta_1 = 0.895$, concrete displaced by the compression steel deducted), 8 – 30M (4 per face, 5600 mm²) gives at $N_f = 1681$ kN a resistance $$M_r = \boxed{1146\ \text{kN}\cdot\text{m}} \;\ge\; M_f = 776\ \text{kN}\cdot\text{m}$$ a utilisation of 0.68. The reinforcement ratio is $\rho = 5600/480\,000 = 1.17\ \%$, above the 1 % minimum of Clause 10.9.1 and far below the 8 % maximum. The pure-axial ceiling, $P_{r,\max} = 0.80[\alpha_1\phi_cf'_c(A_g - A_{st}) + \phi_sf_yA_{st}] = 7481$ kN, is not approached.
Detail the column. Ties must be at least 10M for 30M longitudinals, spaced at the least of $16d_b = 478$ mm, $48d_{\text{tie}} = 542$ mm and the least column dimension 600 mm: use 10M ties at 400 mm, closed, with every alternate bar restrained by a corner or a cross-tie. Because the base is built in and the joint at B transfers the full beam moment, the ties are tightened to 150 mm over a distance h = 800 mm from the base and from the underside of the beam, and the column bars are fully anchored into the footing and hooked into the beam.
Part (b) — the immediate deflection has two competing parts. At service load the 200 kN horizontal load pushes the mid-height of AB towards the beam, while the gravity load on BC hogs the joint at B and rotates the column top the other way. Analysing each separately with the same effective stiffnesses: $$\Delta_{\text{lateral}} = +3.07\ \text{mm}, \qquad \Delta_{\text{gravity}} = -2.62\ \text{mm}, \qquad \Delta_{\text{net}} = +0.45\ \text{mm}$$ The near cancellation is real but fragile, so the answer is best given as both components and their envelope rather than as the net figure alone.
Allow for creep. Sustained load causes the concrete to creep, which is handled either by the effective modulus $E_{\text{eff}} = E_c/(1 + \phi_{cr})$ or, equivalently, by multiplying the sustained deflection by $(1 + \phi_{cr})$. Taking $\phi_{cr} = 2.0$ for a 30 MPa member of this size at normal humidity:

Case Immediate Long term (× 3.0 on the sustained part)

All loads sustained +0.45 mm +1.4 mm

Gravity sustained, 200 kN transient +0.45 mm −4.8 mm

Gravity alone (lateral load removed) −2.62 mm −7.9 mm

so the mid-height of AB moves within a band of about 8 mm either way over the life of the frame: $$\boxed{|\Delta_{\text{long term}}| \le 7.9\ \text{mm} \approx \frac{h}{1270}}$$ comfortably inside the usual $h/500 = 20$ mm serviceability limit for a 10 m storey.

Case	Immediate	Long term (× 3.0 on the sustained part)
All loads sustained	+0.45 mm	+1.4 mm
Gravity sustained, 200 kN transient	+0.45 mm	−4.8 mm
Gravity alone (lateral load removed)	−2.62 mm	−7.9 mm

Check — sustained-load split. The paper gives no dead/live breakdown, so part (b) is answered by bracketing: the creep multiplier $(1 + \phi_{cr}) = 3.0$ is applied first to everything, then to the gravity actions alone with the 200 kN treated as transient. Whichever split is assumed, the long-term movement stays below 8 mm, so the conclusion does not depend on the assumption. If the 200 kN is a wind or crane action it should also be checked at its own load level against the drift limit, which it passes with an immediate 3.1 mm.

Reinforced-concrete beam-column AB.

Final Results

Item	Value
Trial members	column 600 × 800; beam 500 × 1200
Design actions on AB	N_f = 1681 kN, M_f = 776 kN·m (at B)
Slenderness	kl_u/r = 31.3 < 79.9 — non-slender
Reinforcement	8 – 30M (ρ = 1.17 %), 10M ties at 400 mm
Resistance	M_r = 1146 kN·m at N_f — utilisation 0.68
Immediate deflection at mid-height	+3.07 mm (lateral), −2.62 mm (gravity), +0.45 mm net
Long-term deflection (φ_cr = 2.0)	within ±8 mm, i.e. h/1270

Topic. Reinforced-concrete beam-column design and long-term lateral deflection of a braced frame.

Key relations. Clause 10.14.1.2 analysis stiffnesses $0.70I_g$ (columns) and $0.35I_g$ (beams); Clause 10.15.2 non-slender limit $(25 - 10M_1/M_2)/\sqrt{P_f/(f'_cA_g)}$; strain-compatibility $P$–$M$ interaction with $\varepsilon_{cu} = 0.0035$; long-term deflection $\Delta_{lt} = (1 + \phi_{cr})\Delta_{\text{sustained}}$.

Why this works. A concrete frame's internal forces depend on the relative stiffness of its members, and concrete stiffness depends on how cracked each member is — which is why the code prescribes different reductions for beams and columns rather than letting the designer use gross sections everywhere. The effect is not cosmetic: halving the beam stiffness relative to the column here moved the design moment at B from 595 to 776 kN·m, a 30 % change. On the deflection side, the two-component result is the physically interesting part of this question. Frames are usually pictured swaying under lateral load, but a gravity-loaded beam framing into a column also rotates the joint, and in an L-frame with only one column that rotation drives the column the other way. Creep amplifies only the sustained part, so the sign of the long-term deflection can be opposite to the sign of the immediate one.

Common pitfall. Applying a moment magnifier reflexively. Checking the Clause 10.15.2 limit first costs one line and here shows that the column is short, saving a great deal of work — and, more importantly, avoiding a magnified moment that would have driven an unnecessarily large section. The other pitfall is reporting the net immediate deflection as “the” answer: with two components that nearly cancel, a small change in the assumed dead/live split flips the sign, so the defensible answer is the envelope.

Source. CSA A23.3:19 Cl 10.9, 10.10, 10.13–10.16, and Cl 9.8 (deflections); MacGregor & Bartlett, Reinforced Concrete, Ch. 11 (columns) and Ch. 12 (slender columns); Collins & Mitchell, Ch. 9.

Question 7 of 7: Reinforced Concrete Design of the Rigid Frame

Question 7: Reinforced Concrete Design of the Rigid Frame (6 + 5 + 5 + 4 marks)