Reinforced Concrete Design of the Continuous Girder: 16-Civ-B2 Advanced Structural Design, December 2015, Q6

Question 6: Reinforced Concrete Design of the Continuous Girder (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. The Figure 1 girder again: two 8 m spans, built in at A and C, propped at B, with 500 kN unfactored at 4 m and 12 m. Concrete $f'_c = 30$ MPa, rebar $f_y = 400$ MPa, $\phi_c = 0.65$, $\phi_s = 0.85$. Stress-block factors $\alpha_1 = 0.85 - 0.0015f'_c = 0.805$ and $\beta_1 = 0.97 - 0.0025f'_c = 0.895$.

Find. A rectangular section, the flexural steel at every critical section, the shear reinforcement, and a bar layout.

Approach. Re-use the elastic analysis of Question 1 with the concrete self weight added, size the section for the hogging moment, then design the stirrups by the CSA A23.3 simplified method and set out the bars.

Adopt a section and recompute the actions. Try 400 mm × 900 mm ($w_{sw} = 8.64$ kN/m, factored 10.80 kN/m). Symmetry again makes each 8 m span fixed-ended, so the point load contributes $PL/8 = 750$ kN·m at the ends and at mid-span, and the distributed self weight $wL^2/12$ and $wL^2/24$: $$\boxed{M_f = 807.6\ \text{kN}\cdot\text{m (hogging at A, B, C)}, \quad M_f = 778.8\ \text{kN}\cdot\text{m (sagging)}, \quad V_f = 418.2\ \text{kN}}$$ Two layers of 30M bars with 40 mm cover and 10 M stirrups give $d = 820$ mm.
Size the flexural steel. Equating the internal couple to the factored moment, with $C = \alpha_1\phi_cf'_cba$ and $T = \phi_sA_sf_y$, $$M_r = \phi_sA_sf_y\left(d - \frac{a}{2}\right), \qquad a = \frac{\phi_sA_sf_y}{\alpha_1\phi_cf'_cb}$$ Solving the resulting quadratic gives $A_s = 3244$ mm² for the hogging sections and 3113 mm² for the sagging ones. Provide 5 – 30M (3500 mm²) at both.
Verify the chosen bars. With $A_s = 3500$ mm², $a = 189.5$ mm and $c = a/\beta_1 = 212$ mm, so $$\frac{c}{d} = 0.258 \;<\; 0.6 \quad\text{(ductile, tension-controlled)}, \qquad M_r = 863.0\ \text{kN}\cdot\text{m} \;\ge\; 807.6\ \text{kN}\cdot\text{m}$$ The minimum steel requirement, $A_{s,\min} = 0.2\sqrt{f'_c}\,b_th/f_y = 986$ mm², is comfortably exceeded. Five 30M bars in one layer need $5(29.9) + 4(40) = 310$ mm of clear width against 400 − 2(40) − 2(11.3) = 297 mm, so place them in two layers of 3 and 2, which is what the 820 mm effective depth already assumes.
Design the shear reinforcement. Take $d_v = \max(0.9d, 0.72h) = 738$ mm. The section at $d_v$ from each support carries $V_f = 410.2$ kN. With minimum stirrups present the simplified method allows $\beta = 0.18$ and $\theta = 35^\circ$: $$V_c = \phi_c\lambda\beta\sqrt{f'_c}\,b_wd_v = 0.65(0.18)\sqrt{30}(400)(738) = 189.2\ \text{kN}$$ $$V_s = 410.2 - 189.2 = 221.0\ \text{kN} \quad\Longrightarrow\quad s = \frac{\phi_sA_vf_yd_v\cot\theta}{V_s} = 324\ \text{mm}$$
Detail the stirrups. Since $V_f = 410$ kN is below $0.125\lambda\phi_cf'_cb_wd_v = 720$ kN, the spacing ceiling is $\min(0.7d_v, 600) = 517$ mm. Adopt 10M closed stirrups at 300 mm through the shear zones, giving $$V_r = 189.2 + \frac{\phi_sA_vf_yd_v\cot\theta}{300} = \boxed{428\ \text{kN}} \;\ge\; 410.2\ \text{kN}$$ and relax to 450 mm in the low-shear region either side of each load point. Web crushing is remote: $V_{r,\max} = 0.25\phi_cf'_cb_wd_v = 1439$ kN.
Set out the layout. Top steel (5 – 30M) runs continuously over each support and is curtailed where the hogging moment falls below the capacity of the remaining bars — roughly 3.2 m each side of A and C and 3.2 m each side of B — with all bars extended a further $\max(d, 12d_b) = 820$ mm beyond the theoretical cut-off. Bottom steel (5 – 30M) runs beneath each 500 kN load, with at least two bars carried into and anchored at the supports for integrity. Top and bottom bars lap over the support at B, and the built-in ends at A and C require full development of the top steel into the supporting wall (a 90° standard hook with $l_{dh} \approx 550$ mm).

Reinforcement layout and cross-section for the RC girder.

Final Results

Item	Value
Section	400 × 900 mm, d = 820 mm
Design moments	807.6 kN·m hogging; 778.8 kN·m sagging
Flexural steel	5 – 30M top at A, B, C; 5 – 30M bottom under each load
Resistance provided	M_r = 863 kN·m, c/d = 0.258
Design shear	V_f = 410.2 kN at d_v from the support
Shear steel	10M closed stirrups at 300 mm (450 mm in low-shear zones)
Shear resistance	V_r = 428 kN

Topic. Limit-states flexural and shear design of a two-span reinforced-concrete girder with built-in ends.

Key relations. $M_r = \phi_sA_sf_y(d - a/2)$ with $a = \phi_sA_sf_y/(\alpha_1\phi_cf'_cb)$; $c/d \le 0.6$ for ductility; $V_c = \phi_c\lambda\beta\sqrt{f'_c}b_wd_v$ and $V_s = \phi_sA_vf_yd_v\cot\theta/s$ with $\beta = 0.18$, $\theta = 35^\circ$.

Why this works. The same symmetry argument that simplified the steel girder applies unchanged in concrete: zero rotation at the prop turns each span into a fixed-ended beam, so the hogging and sagging moments are equal and one bar arrangement serves both. That equality is convenient but it is not an accident of the numbers — it follows from $PL/8$ appearing at both ends and the centre of a fixed-ended span under a central load. On the shear side, CSA A23.3's simplified method is a packaged version of the modified compression field theory: $\beta$ measures the tension the cracked concrete can still carry across the crack, and $\cot\theta$ counts how many stirrup legs a 35° crack crosses. Because $d_v\cot\theta \approx 1050$ mm here, a single stirrup spacing of 300 mm puts three and a half legs across every crack.

Common pitfall. Detailing five 30M bars in a single layer of a 400 mm web — they do not fit at the required clear spacing, and squeezing them in compromises both cover and the assumed $d$. The second pitfall is curtailing the top steel at the theoretical point of contraflexure without the code's extension; the diagonal crack shifts the tension demand towards mid-span and a bar cut off exactly at the theoretical point is a bar cut off too early.

Source. CSA A23.3:19 Cl 10 (flexure), Cl 11.3 (shear, simplified method), Cl 12 (development and curtailment); MacGregor & Bartlett, Reinforced Concrete, Ch. 4, 6 and 8.

Question 6 of 7: Reinforced Concrete Design of the Continuous Girder

Question 6: Reinforced Concrete Design of the Continuous Girder (10 + 5 + 5 marks)