Question 5 of 7: Reinforced concrete frame — limit states design of member BC

Question 5: Reinforced concrete frame — limit states design of member BC (14 + 6 = 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 2 built in reinforced concrete, all members of equal $EI$; pinned at A and D; $h_{AB}=6$ m, $h_{CD}=8$ m; beam B–O–C of $8+8=16$ m; unfactored loads 80 kN horizontal at B and 500/400/500 kN vertical at B/O/C. Materials $f'_c=30$ MPa ($\phi_c=0.65$), $f_y=400$ MPa ($\phi_s=0.85$). A trial beam of 600 × 1500 mm weighs $0.6(1.5)(24)=21.6$ kN/m, i.e. 27.0 kN/m factored.

Find. A rectangular cross-section for the beam B–O–C together with the flexural and shear reinforcement, using an elastic frame analysis and A23.3 limit states design.

Approach. Analyse the one-degree-redundant frame elastically by the stiffness method with equal $EI$ throughout and the factored loads, take the peak hogging and sagging moments in the beam, size the section by the rectangular stress block, then check shear against the A23.3 simplified method.

1. Factored loading. $H_f=120$ kN at B; $V_B=V_C=750$ kN; $V_O=600$ kN; plus the beam's own weight $w_f=1.25(21.6)=27.0$ kN/m along BOC. The 750 kN loads sit directly over the columns, so they enter the beam only through the joints.
2. Elastic analysis. Solving the frame (four members, equal $EI$, pins at A and D) gives the factored beam moments $$M_B=-710\ \text{kN}\cdot\text{m},\qquad M_O=+1956\ \text{kN}\cdot\text{m},\qquad M_C=-1907\ \text{kN}\cdot\text{m}$$ (sagging positive). The distribution is strongly unsymmetric because the two columns have different heights: the 8 m column CD is the more flexible of the two, but it also attracts the larger end moment once the 120 kN sway load is superposed on the gravity case.
3. Sanity check on the analysis. A simply supported 16 m beam under the same loads would carry $\frac{P L}{4}+\frac{w_fL^{2}}{8}=\frac{600(16)}{4}+\frac{27(16)^{2}}{8}=3264$ kN·m at mid-span. The frame lifts the ends to $-710$ and $-1907$, whose average is 1308, and $$3264-1308=1956\ \text{kN}\cdot\text{m}$$ which reproduces $M_O$ exactly — the analysis is internally consistent.
4. Beam shears. The same solution gives end shears of 441 kN at B, 591 kN at C, with 225 kN and 375 kN each side of O; the design shear is therefore $V_f=591$ kN at the C end.

1. Choose the section. A span/depth ratio of about $L/11$ suits a heavily loaded frame beam, so keep the trial $b=600$ mm, $h=1500$ mm. With 40 mm cover, 10M stirrups and 25M bars in two layers, the effective depth is $$d=1500-40-11.3-\frac{25.2}{2}-\left(25.2+25\right)=1386\ \text{mm}$$
2. Flexural steel at O (sagging, 1956 kN·m). With $\alpha_1=0.85-0.0015f'_c=0.805$ and $\beta_1=0.97-0.0025f'_c=0.895$, iterate on $$M_r=\phi_sA_sf_y\left(d-\frac{a}{2}\right),\qquad a=\frac{\phi_sA_sf_y}{\alpha_1\phi_cf'_cb}$$ Convergence gives $A_s=4403$ mm$^2$. Provide 9 – 25M ($A_s=4500$ mm$^2$) in two layers, 5 + 4, then $$a=\frac{0.85(4500)(400)}{0.805(0.65)(30)(600)}=162.4\ \text{mm},\qquad M_r=0.85(4500)(400)\left(1386-81.2\right)=\boxed{1996\ \text{kN}\cdot\text{m}}$$ which covers $M_O=1956$ kN·m.
3. Flexural steel at B and C (hogging). $M_C=1907$ kN·m is the larger of the two, and the same 9 – 25M arrangement in the top of the section gives the same $M_r=1996$ kN·m. Run all nine top bars continuously from B through O to C and lap them at mid-span where the top-face moment is smallest; run the nine bottom bars full length as well, so the same section serves the whole member and the joint reinforcement of Question 6 is continuous.
4. Ductility check. $c=a/\beta_1=162.4/0.895=181.5$ mm, so $$\frac{c}{d}=\frac{181.5}{1386}=0.131\ \ll\ \frac{700}{700+f_y}=0.636$$ The section is comfortably tension-controlled: the steel yields long before the concrete crushes, which is what limit states design requires of a frame beam that must also rotate at its joints. Bar spacing also works: the clear distance between the five bars in the bottom layer is $\left[600-2(40)-2(11.3)-5(25.2)\right]/4=93$ mm, well over the $1.4d_b=35$ mm minimum.
5. Shear — concrete contribution. Using the A23.3 simplified method with $\beta=0.18$ and $\theta=35^{\circ}$, and $d_v=\max(0.9d,\ 0.72h)=1247$ mm, $$V_c=\phi_c\lambda\beta\sqrt{f'_c}\,b_wd_v=0.65(1)(0.18)\sqrt{30}(600)(1247)\times10^{-3}=480\ \text{kN}$$ so stirrups must carry $V_s=591-480=111$ kN.
6. Shear — stirrups. For 10M double-leg stirrups ($A_v=200$ mm$^2$), $$s=\frac{\phi_sA_vf_yd_v\cot\theta}{V_s}=\frac{0.85(200)(400)(1247)(1.428)}{111\times10^{3}}=1091\ \text{mm}$$ Strength does not govern. The minimum-steel rule does: $$\frac{A_v}{s}\ge0.06\sqrt{f'_c}\frac{b_w}{f_y}=0.493\ \text{mm}^{2}/\text{mm}\ \Rightarrow\ s\le406\ \text{mm}$$ together with $s\le\min(0.7d_v,600)=600$ mm. Provide 10M double-leg stirrups at 400 mm throughout, closed up to 200 mm over a length $2h=3.0$ m each side of B and C where the beam must be able to rotate. Crushing is not an issue: $V_{r,max}=0.25\phi_cf'_cb_wd_v=3648$ kN.