Question 7 of 7: Reinforced-concrete tee beam — member AB (14 + 6 marks)

Question 7: Reinforced-concrete tee beam — member AB (14 + 6 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 2 beam, now in reinforced concrete: span AB of 12 m, continuous over support B, carrying two 400 kN loads at 4 m and 8 m, cast monolithically with a floor slab so that the section acts as a tee.

Find. The flexural steel at mid-span and over the interior support, the stirrup arrangement, and the curtailment layout along AB.

Approach. The flange is in compression at mid-span and in tension over the support, so the two regions behave quite differently: a very wide, lightly stressed tee at mid-span and a plain 500 mm rectangle at the support. Analyse elastically for the envelope, design each region on its own behaviour, then set the curtailment from the point of inflection.

1. Analyse the continuous beam including self weight. The point loads give $M_B=-2400$ kN·m as in Question 2, and the factored self weight adds $w_fL^2/8$: $$M_B=-2400-\frac{21.75(12)^2}{8}=\boxed{-2791.5\ \text{kN}\cdot\text{m}}$$ $$R_A=\left(\frac{600(8)+600(4)}{12}+\frac{21.75(12)}{2}\right)+\frac{M_B}{12} =497.9\ \text{kN}$$ so the largest sagging moment sits under the first load, $$M(4\ \text{m})=497.9(4)-\frac{21.75(4)^2}{2}=\boxed{1817.5\ \text{kN}\cdot\text{m}}$$ and the shear just inside B is $V_f=|497.9-1200-261|=\boxed{963\ \text{kN}}$.
2. Fix the effective flange width. Clause 10.3.3 limits the overhang each side to the least of a tenth of the span, twelve slab thicknesses and half the clear distance to the next web: $$\min(1200,\ 1800,\ 1250)=1200\ \text{mm}\ \Rightarrow\ b_{\text{eff}}=500+2(1200)=2900\ \text{mm}$$ That width matters only in the sagging region, where the flange is in compression.
3. Design the sagging steel as a tee. With 40 mm cover, 10M stirrups and one layer of 30M bars, $d=1534$ mm. Assuming rectangular behaviour over the full flange width, $$A_s=3515\ \text{mm}^2\ \Rightarrow\ a=\frac{\phi_sA_sf_y}{\alpha_1\phi_cf'_cb_{\text{eff}}}=26\ \text{mm}\ <\ h_f=150\ \text{mm}$$ so the compression block lies entirely inside the flange and the assumption holds. Provide 6–30M (4200 mm²), which fits in a single layer in the 500 mm web with 41.9 mm clear spacing, giving $$a=31.4\ \text{mm},\qquad M_r=\boxed{2168\ \text{kN}\cdot\text{m}}\ \ge\ 1818\ \text{kN}\cdot\text{m}$$ with $c/d=0.023$: the tee is enormously under-reinforced in sagging, which is typical when a wide slab acts as the compression flange.
4. Design the hogging steel as a rectangle. Over the support the flange is in tension and only the 500 mm web resists compression: $$A_s=5834\ \text{mm}^2\ \Rightarrow\ \text{provide}\ \textbf{9--30M}\ (6300\ \text{mm}^2)$$ $$a=\frac{0.85(6300)(400)}{0.805(0.65)(30)(500)}=272.9\ \text{mm},\qquad M_r=\boxed{2993\ \text{kN}\cdot\text{m}}\ \ge\ 2792\ \text{kN}\cdot\text{m}$$ with $c/d=0.199$, comfortably ductile. Clause 10.5.3.1 requires this tension steel to be spread across the smaller of the effective flange width and a tenth of the span, so distribute the nine bars over 1200 mm of slab at 150 mm centres — which is also what keeps them in one layer and preserves $d=1534$ mm. The minimum, $A_{s,\min}=2191$ mm², is not close to governing at either location.
5. Design the shear reinforcement. With $d_v=\max(0.9d,0.72h)=1380$ mm and the simplified method, $$V_c=0.65(0.18)\sqrt{30}(500)(1380)/10^3=442\ \text{kN},\qquad V_s=963-442=521\ \text{kN}$$ $$s=\frac{\phi_sA_vf_yd_v\cot\theta}{V_s} =\frac{0.85(400)(400)(1380)(1.428)}{521\times10^3}=515\ \text{mm}$$ Provide 15M closed stirrups at 400 mm within 4 m of support B. Away from that zone the shear falls below the 889 kN that 15M at the 600 mm maximum spacing carries, so 15M at 600 mm suffices over the rest of the span, comfortably above $A_{v,\min}=246$ mm². The web crushing limit $0.25\phi_cf'_cb_wd_v=3363$ kN is nowhere approached.
6. Set the curtailment layout. The moment diagram crosses zero at x = 9.0 m from A. Clause 12.12.3 requires the negative steel to run at least $d$, $12d_b$ or $L_n/16$ — here 1534 mm — past that point, so carry the 9–30M top bars from B back to x = 7.4 m, and continue at least a third of them (3–30M) to $x=6.5$ m as nominal top steel and hanger bars for the stirrups. Of the 6–30M bottom bars, run four the full length into both supports for the Clause 12.11.1 positive-moment anchorage, and stop the other two where the four-bar resistance plus a development length ($l_d\approx1000$ mm for a 30M bottom bar in 30 MPa concrete) is still exceeded, about 1.2 m from each support. Provide 10M skin reinforcement at 300 mm on each face over the lower half of the 1600 mm web, as Clause 10.6.2 requires for deep members.