07-Str-A2 · May 2016

Question 6 of 7: B3 — Reinforced concrete girder with two overhangs

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Question 6: B3 — Reinforced concrete girder with two overhangs (8 + 12)

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.

Find. A concrete section, longitudinal reinforcement top and bottom, and stirrup arrangement satisfying flexure, minimum steel and shear.

[Figure not reproduced: Figure B3 (redrawn). The loading and geometry are symmetric about mid-span, so both reactions are equal and the hogging moments at the two supports are identical. See the official exam paper.]

Approach. Exploit symmetry to get the reactions, build the bending-moment envelope for the factored point loads plus factored self-weight, size the section from the larger of the two peaks, then check minimum steel, ductility and shear.

1. Factor the loads. Trial section 400 × 750 mm gives w = 0.4(0.75)(24) = 7.2 kN/m: $$P_{f,\text{tip}} = 1.5(50) = 75\ \text{kN}, \quad P_{f,\text{int}} = 1.5(100) = 150\ \text{kN}, \quad w_{f} = 1.25(7.2) = 9.0\ \text{kN/m}$$
2. Reactions by symmetry. The structure and its loading are symmetric about the 7.5 m point, so each support takes half of everything: $$R = \frac{2(75)+2(150)+9.0(15.0)}{2} = \boxed{292.5\ \text{kN}}$$
3. Hogging moment at the supports. Take the left overhang as a free body — only the tip load and the self-weight act on it: $$M^{-} = -\left[75(3.0)+\frac{9.0(3.0)^{2}}{2}\right] = \boxed{-265.5\ \text{kN}\cdot\text{m}}$$
4. Sagging moment in the span. Between the two interior point loads the shear from the point loads alone vanishes, so the sagging peak sits at mid-span where the self-weight shear also changes sign: $$M^{+} = -75(7.5) + 292.5(4.5) - 150(1.5) - \frac{9.0(7.5)^{2}}{2} = \boxed{+275.6\ \text{kN}\cdot\text{m}}$$ The two peaks are nearly equal, which is what makes the symmetric top-and-bottom reinforcement below so efficient.
5. Effective depth. With 40 mm cover, 10M stirrups and 25M main bars in one layer: $$d = 750 - 40 - 10 - \frac{25}{2} = \boxed{687.5\ \text{mm}}$$
6. Flexural reinforcement. Try 3–25M (A_s = 1500 mm²). Force equilibrium of a rectangular section gives $$a = \frac{\phi_{s}f_{y}A_{s}}{\alpha_{1}\phi_{c}f'_{c}b} = \frac{0.85(400)(1500)}{0.7975(0.65)(35)(400)} = 70.3\ \text{mm}$$ $$M_{r} = \phi_{s}f_{y}A_{s}\left(d-\frac{a}{2}\right) = 0.85(400)(1500)\left(687.5-35.1\right) = \boxed{333\ \text{kN}\cdot\text{m}}$$ This exceeds both peaks (utilisation 0.83 on the sagging case), so provide 3–25M bottom through the span and 3–25M top through the supports and overhangs, each set lapped well past its point of contraflexure.
7. Ductility and minimum steel. The neutral axis sits at c = a/β₁ = 79.6 mm, so $$\varepsilon_{t} = 0.0035\left(\frac{687.5-79.6}{79.6}\right) = 0.027 \;>\; 0.005$$ — strongly tension-controlled. A23.3 Clause 10.5.1.2 requires $$A_{s,\min} = \frac{0.2\sqrt{35}}{400}(400)(750) = 887\ \text{mm}^{2} \;<\; 1500\ \text{mm}^{2}\ \checkmark$$
8. Shear. The critical shear is just inside each support: $$\begin{aligned} V_{f} &= 190.5\ \text{kN} \\ d_{v} &= \max\left(0.9d,\ 0.72h\right) = 618.8\ \text{mm} \end{aligned}$$ With minimum stirrups present the simplified method takes β = 0.18 and θ = 45°: $$V_{c} = \phi_{c}\lambda\beta\sqrt{f'_{c}}\,b_{w}d_{v} = 0.65(1.0)(0.18)\sqrt{35}(400)(618.8) = 171\ \text{kN}$$ $$V_{s} = \frac{\phi_{s}A_{v}f_{y}d_{v}}{s} = \frac{0.85(200)(400)(618.8)}{300} = 140\ \text{kN}$$ $$V_{r} = 171+140 = \boxed{312\ \text{kN}} \;>\; 190.5\ \text{kN}\ \checkmark$$ The web-crushing cap 0.25φ_cf'_cb_wd_v = 1408 kN is nowhere near critical. Provide 10M closed stirrups at 300 mm throughout, tightened to 150 mm over 1.5 m each side of both supports.