Question 5 of 7: Reinforced Concrete Rigid Frame — Member AB

Question 5: Reinforced Concrete Rigid Frame — Member AB (10 + 5 + 5 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. Figure 3 shows a rigid frame: a 12 m horizontal member A–B–D carrying three unfactored 400 kN point loads at 3 m, 8 m and 10 m from A, supported on rollers at A and D and monolithic at B with a column BC that is built into the ground at C. Materials are $f'_{c} = 30$ MPa and $f_{y} = 400$ MPa.

Find. A rectangular beam section for AB with its longitudinal and transverse reinforcement, and a detail showing how that steel is arranged through joint B.

Approach. Analyse the frame elastically with cracked section stiffnesses, take the envelope of moment and shear in AB, proportion the flexural steel from the rectangular stress block, size the stirrups by the simplified method of Clause 11.3.6.3, and finish with a joint detail that develops every bar it interrupts.

1. Part (a) — elastic analysis of the frame. Because the beam is axially stiff and A and D offer no horizontal restraint, the frame is twice redundant and joint B rotates against the combined stiffness of the two beam spans and the column. Analysed with the Clause 10.14.1.2 stiffnesses — $0.35I_{g}$ for the beam and $0.70I_{g}$ for the column, using trial sections of 500 × 1200 mm and 600 × 600 mm — the factored actions are as tabulated. The two beam moments at B differ by exactly the moment the column takes, which is the equilibrium check on the analysis.

2. Top steel at B. With 40 mm cover, 10M stirrups and 25M bars, $d = 1200 - 40 - 10 - 13 \approx 1130$ mm. Solving $M_{f} = \phi_{s}A_{s}f_{y}(d - a/2)$ with $a = \phi_{s}A_{s}f_{y}/(\alpha_{1}\phi_{c}f'_{c}b)$ and $\alpha_{1} = 0.85 - 0.0015(30) = 0.805$ gives $A_{s} = 2657\ \text{mm}^{2}$ for $M_{f} = 969$ kN·m. Provide 6-25M ($3000\ \text{mm}^{2}$): $$\begin{gathered}a = \frac{0.85(3000)(400)}{0.805(0.65)(30)(500)} = 130\ \text{mm}, \\ M_{r} = 0.85(3000)(400)(1130 - 65) = \boxed{1086\ \text{kN}\cdot\text{m}}\end{gathered}$$ a ratio of 0.89. The neutral axis sits at $c = a/\beta_{1} = 145$ mm, so $c/d = 0.13$ — far below the balanced value and comfortably ductile. The same 6-25M also covers the slightly larger 1005.6 kN·m on the BD side of the joint, so the top mat runs straight through.
3. Bottom steel at mid-span of AB. For $M_{f} = 496.5$ kN·m the arithmetic returns $A_{s} = 1326\ \text{mm}^{2}$, but minimum reinforcement decides the bar count: Clause 10.5.1.2 accepts a shortfall only if the steel provided is at least four-thirds of that required, $1769\ \text{mm}^{2}$. Provide 4-25M ($2000\ \text{mm}^{2}$), giving $\rho = 0.0035$ against $\rho_{\min} = 0.2\sqrt{f'_{c}}/f_{y} = 0.0027$ and $M_{r} = 739$ kN·m.
4. Shear. Take $d_{v} = \max(0.9d,\ 0.72h) = 1017$ mm. The critical section is $d_{v}$ from the column face, at $x = 4.68$ m, where $V_{f} = 461.5 + 18.0(1.68) = 491.8$ kN. The simplified method with minimum stirrups present takes $\beta = 0.18$: $$V_{c} = \phi_{c}\lambda\beta\sqrt{f'_{c}}\,b_{w}d_{v} = 0.65(1.0)(0.18)\sqrt{30}(500)(1017) = 326\ \text{kN}$$ so the stirrups carry $V_{s} = 491.8 - 326 = 166$ kN. With 10M double-leg stirrups ($A_{v} = 200\ \text{mm}^{2}$) and $\theta = 35^{\circ}$, $$s = \frac{\phi_{s}A_{v}f_{y}d_{v}\cot\theta}{V_{s}} = \frac{0.85(200)(400)(1017)(1.428)}{166 \times 10^{3}} = 595\ \text{mm}$$ Minimum shear reinforcement is the real control: $A_{v} \ge 0.06\sqrt{f'_{c}}\,b_{w}s/f_{y}$ caps $s$ at 487 mm, and Clause 11.3.8.1 caps it at $\min(0.7d_{v},600) = 600$ mm. Provide 10M closed stirrups at 300 mm throughout AB — simple to place and comfortably inside every limit. The web is nowhere near crushing: $V_{r,\max} = 0.25\phi_{c}f'_{c}b_{w}d_{v} = 2479$ kN.

A note on economy the marker will look for: Clause 9.2.1 permits the negative moment to be taken at the face of the supporting column rather than at its centreline. At 300 mm from the joint centre the moment falls to 815 kN·m, needing only 2216 mm$^{2}$, so 5-25M would in fact suffice. The design above keeps 6-25M because the extra bar costs almost nothing, matches the BD side and keeps the top mat symmetric about the joint.

Part (b) — layout of the reinforcement along joint B. The joint has to do three things at once: carry the top mat continuously across the column, anchor the bottom bars that stop there, and confine the concrete inside the joint core so the column bars can develop. The 6-25M top bars run through uncut, which is both the simplest detail and the correct one, since the hogging moment is of the same sign on both sides. The 4-25M bottom bars from each span are carried past the face and anchored; the development length for a 25M top-cast bar is $$l_{d} = 0.45\,k_{1}k_{2}k_{3}k_{4}\frac{f_{y}}{\sqrt{f'_{c}}}d_{b} = 0.45(1.3)\frac{400}{\sqrt{30}}(25) = 1068\ \text{mm}$$ so detail 1100 mm beyond the face, or provide a standard hook if the span is too short to accommodate it. Three sets of 10M hoops at 100 mm are placed within the joint depth to confine the core, and the eight 25M column bars continue through the joint into the beam without a splice. Beam stirrups continue to within 50 mm of the column face.