Question 6 of 7: Reinforced-concrete continuous girder, limit states design

Question 6: Reinforced-concrete continuous girder, limit states design (10 + 5 + 5 = 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 girder of Figure 1 again — fixed at A and C, simply supported at B, two 8 m spans, a 500 kN specified load at the centre of each — now to be built in reinforced concrete with \(f'_c = 30\ \text{MPa}\) and \(f_y = 400\ \text{MPa}\).

Find. The flexural reinforcement at the supports and at mid-span, the shear reinforcement, and a drawn layout.

Approach. Analyse elastically (limit states design of reinforced concrete uses an elastic distribution of actions, unlike Question 4), design the tension steel from the rectangular stress block at the two critical sections, then design stirrups by the CSA A23.3 simplified method at \(d\) from the support face.

1. Actions. A \(450 \times 1000\) section weighs 10.8 kN/m. Symmetry again makes each span fixed-ended, so with \(P_f = 750\ \text{kN}\) and \(w_f = 13.5\ \text{kN/m}\) $$M_f^{\text{sup}} = \frac{P_fL}{8} + \frac{w_fL^2}{12} = 750 + 72 = \boxed{822\ \text{kN}\cdot\text{m}}$$ $$M_f^{\text{mid}} = 750 + 36 = 786\ \text{kN}\cdot\text{m}, \qquad V_f = 375 + 54 = 429\ \text{kN}$$ The same hogging moment applies at A, at C and over the interior support B.
2. Flexural steel at the supports. Solving \(M_r = \phi_sA_sf_y\left(d - a/2\right)\) with \(a = \phi_sA_sf_y/(\alpha_1\phi_cf'_cb)\) for \(M_r = 822\ \text{kN}\cdot\text{m}\) gives \(A_s = 2803\ \text{mm}^2\). Provide \(\boxed{5-30\text{M}}\) (3500 mm²) in the top face, which develops $$a = \frac{0.85(3500)(400)}{0.805(0.65)(30)(450)} = 168.5\ \text{mm}, \qquad M_r = 1190\times10^3(930 - 84.2) = 1006\ \text{kN}\cdot\text{m}$$ Four bars would give only 821 kN·m, marginally short of the 822 kN·m demand, so the fifth bar is needed. Ductility is ample: \(c/d = 0.202\), far below the \(700/(700+f_y) = 0.636\) balanced limit.
3. Flexural steel at mid-span. For \(M_f = 786\ \text{kN}\cdot\text{m}\), \(A_s = 2670\ \text{mm}^2\); provide \(\boxed{4-30\text{M}}\) (2800 mm²) in the bottom face, \(M_r = 821\ \text{kN}\cdot\text{m}\). Both exceed the minimum \(A_{s,\min} = 0.2\sqrt{f'_c}\,b_th/f_y = 1232\ \text{mm}^2\).
4. Shear demand and concrete contribution. Take \(d_v = \max(0.9d,\ 0.72h) = 837\ \text{mm}\). At \(d\) from the support the demand falls only by the distributed load, since the point load sits at mid-span: $$V_f = 429 - 13.5(0.93) = \boxed{416\ \text{kN}}$$ With at least minimum stirrups, the simplified method takes \(\beta = 0.18\) and \(\theta = 35^\circ\), so $$V_c = \phi_c\lambda\beta\sqrt{f'_c}\,b_wd_v = 0.65(0.18)(5.477)(450)(837) \times 10^{-3} = 241.4\ \text{kN}$$
5. Stirrups. The steel must supply the remaining 175 kN. Using 10M closed double-leg stirrups (\(A_v = 200\ \text{mm}^2\)) at \(s = 250\ \text{mm}\), $$V_s = \frac{\phi_sA_vf_yd_v\cot\theta}{s} = \frac{0.85(200)(400)(837)(1.428)}{250} \times 10^{-3} = 325\ \text{kN}$$ $$V_r = V_c + V_s = \boxed{566\ \text{kN}} \;>\; 416\ \text{kN}$$ The spacing satisfies \(s \le \min(0.7d_v, 600) = 586\ \text{mm}\), and the section is far below the crushing limit \(0.25\phi_cf'_cb_wd_v = 1836\ \text{kN}\). Because the shear beside the central point load is almost as large (375 kN), carry the same 250 mm spacing right through the spans rather than opening it out in the middle.
6. Layout. Run the 4–30M bottom bars continuously from A to C and anchor them into the end walls. Place 5–30M top bars over each of A, B and C, extending them into the spans past the point of contraflexure (about 2.6 m each side of a support here) plus the development length; two of the five continue the full length as stirrup hangers. All stirrups are 10M closed at 250 mm, with the first at 125 mm from each support face.