Question 5 of 7: Reinforced-concrete rigid frame — beam BC (10 + 5 + 5 marks)

Question 5: Reinforced-concrete rigid frame — beam BC (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. A reinforced-concrete rigid frame with fixed bases, a 12 m beam BC between an 8 m column at A and a 10 m column at D, loaded by 100 kN horizontally at B together with 600 kN at each joint and 500 kN at mid-span.

Find. (a) A rectangular beam section for BC with its flexural and shear reinforcement; (b) the long-term mid-span deflection; (c) the bar layout along the member.

Approach. Analyse the frame by the stiffness method using the Clause 10.14.1.2 cracked-section stiffnesses (0.35I_g for the beam, 0.70I_g for the columns), design the beam for the resulting envelope, then recompute the service deflection on Branson's effective moment of inertia and amplify it for creep.

1. Part (a) — set the trial section and member stiffnesses. Try a beam 600 mm wide by 1400 mm deep and columns 600 × 800 mm. With $E_c=4500\sqrt{30}=24\,648$ MPa and Clause 10.14.1.2 stiffnesses, $$I_b=0.35\frac{0.6(1.4)^3}{12}=0.0480\ \text{m}^4,\qquad I_c=0.70\frac{0.6(0.8)^3}{12}=0.0179\ \text{m}^4$$ The beam self weight is $0.6(1.4)(24)=20.16$ kN/m, factored to 25.2 kN/m.
2. Analyse the frame. Solving the three-member frame with fixed bases at two different levels gives the beam end actions $$M_B=-328.8,\qquad M_{\text{mid}}=+2040.3,\qquad M_C=-997.8\ \text{kN}\cdot\text{m}$$ $$V_B=470.5\ \text{kN},\qquad V_C=-581.9\ \text{kN}$$ Horizontal equilibrium confirms the analysis: the base shears of 20.6 kN at A and 170.6 kN at D sum to the applied 150 kN. Note how unequal the two ends are — the taller, more flexible column at D attracts nearly three times the beam end moment of the stiffer 8 m column at A, because the sway caused by the horizontal load relieves joint B and aggravates joint C.
3. Design the mid-span sagging steel. With 40 mm cover, 10M stirrups and 30M bars in two layers, $d=1304$ mm and, with $\alpha_1=0.805$, $$A_s=\frac{M_f}{\phi_sf_y(d-a/2)}\ \Rightarrow\ A_s=4941\ \text{mm}^2$$ Provide 8–30M (5600 mm²), for which $$a=\frac{\phi_sA_sf_y}{\alpha_1\phi_cf'_cb}=202\ \text{mm},\qquad M_r=\phi_sA_sf_y\!\left(d-\frac{a}{2}\right)=\boxed{2290\ \text{kN}\cdot\text{m}} \ \ge\ 2040\ \text{kN}\cdot\text{m}$$ with $c/d=0.195$, so the section is firmly tension-controlled.
4. Design the support hogging steel. The larger support moment, 997.8 kN·m at C, needs only $A_s=2326$ mm², and at B the demand falls to 750 mm². Both are close to or below $$A_{s,\min}=\frac{0.2\sqrt{f'_c}\,b_th}{f_y} =\frac{0.2\sqrt{30}(600)(1400)}{400}=2300\ \text{mm}^2$$ so provide 4–30M top at both supports (2800 mm²), giving $M_r=1193$ kN·m. At joint B the minimum-steel rule, not strength, sets the top mat.
5. Design the shear reinforcement. With $d_v=\max(0.9d,\,0.72h)=1173$ mm and the simplified method ($\beta=0.18$, $\theta=35^\circ$), $$V_c=\phi_c\lambda\beta\sqrt{f'_c}\,b_wd_v =0.65(0.18)\sqrt{30}(600)(1173)/10^3=451\ \text{kN}$$ so at C the stirrups carry only $V_s=581.9-451.2=131$ kN, which 10M double-leg stirrups would satisfy at 871 mm centres. Detailing therefore governs: the maximum spacing is $\min(0.7d_v,600)=600$ mm and $A_{v,\min}=0.06\sqrt{f'_c}b_ws/f_y$ caps 10M stirrups at 400 mm. Provide 10M closed stirrups at 400 mm, tightened to 250 mm within 1.5 m of each support. The upper limit $0.25\phi_cf'_cb_wd_v=3432$ kN is not approached.
6. Part (b) — find the service moment and the effective inertia. Re-running the frame with unfactored loads gives a mid-span service moment of 1400.8 kN·m, well above the cracking moment $$M_{cr}=\frac{f_rI_g}{y_t}=\frac{0.6\sqrt{30}\,(137.2\times10^9)}{700} =644\ \text{kN}\cdot\text{m}$$ With $n=8.11$ and 8–30M, the cracked neutral axis lies 394 mm below the top and $I_{cr}=49.7\times10^9$ mm⁴, so Branson's relation gives $$I_e=\left(\frac{M_{cr}}{M_a}\right)^3I_g +\left[1-\left(\frac{M_{cr}}{M_a}\right)^3\right]I_{cr} =\boxed{58.2\times10^9\ \text{mm}^4}$$
7. Compute the immediate and long-term deflections. Integrating the curvature along BC and adding the vertical movement of joints B and C gives an immediate mid-span deflection of $$\Delta_i=\boxed{11.9\ \text{mm}}$$ Treating the whole load as sustained — the safe reading, since page 1 gives no dead/live split — and taking $s=2.0$ for five years or more with $\rho'=2800/(600\times1304)=0.0036$, $$\lambda_\Delta=\frac{s}{1+50\rho'}=\frac{2.0}{1.179}=1.70,\qquad \Delta_{LT}=\Delta_i(1+\lambda_\Delta)=\boxed{32.1\ \text{mm}}=\frac{L}{374}$$ That is inside the $L/240$ limit of Table 9.3 for members not supporting deflection-sensitive construction; if the 500 kN mid-span load were transient the long-term value would fall to about 20 mm, so 32 mm is the upper bound of the envelope.
8. Part (c) — lay the reinforcement out along BC. Run the 4–30M top mat continuously from face to face — it is the minimum steel the member must carry anyway, it covers both support moments, and it doubles as the compression steel that halves the creep multiplier at mid-span. Carry six of the eight bottom 30M bars into both supports for the positive-moment anchorage required by Clause 12.11.1, and stop the remaining two 1.5 m either side of mid-span, where the moment has fallen below the six-bar resistance plus a development length. Because the beam is 1400 mm deep, Clause 10.6.2 also calls for skin reinforcement: provide 10M side bars at 300 mm over the lower half of each face.