Question 6 of 7: Composite steel-concrete pedestrian bridge

Question 6: Composite steel-concrete pedestrian bridge

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. Simply supported span 16 m, deck 12 m wide, slab 240 mm deep on five beams at 3 m centres (four spaces make up the 12 m). Unshored construction; the beams are adequately braced. $f'_c = 30$ MPa, $F_y = 350$ MPa.

Find. (a) the composite cross-section; (b) the number of shear connectors.

Approach. Take an interior beam with a 3 m tributary width, design the bare steel section for the wet-concrete stage, check the composite section for strength and live-load deflection, then size the stud connection from the horizontal shear that has to cross the interface.

1. Establish the loads on one interior beam. The slab weighs $0.240 \times 24 = 5.76$ kPa, i.e. 17.28 kN/m over a 3 m tributary width. Add 1.49 kN/m of steel beam, 3.00 kN/m of superimposed dead load for railings and wearing surface, and the pedestrian live load of 5.0 kPa (CSA S6:19 Cl 3.8.9), i.e. 15.00 kN/m.
Deck cross-section. The interior beams each take a 3 m tributary width; the edge beams take half of that.
2. Design for the construction stage. Unshored construction means the bare steel beam carries the wet concrete alone. With a construction live load of 0.5 kPa, $w_f = 1.25(17.28 + 1.49) + 1.5 \times 1.5 = 25.71$ kN/m and $$M_f = \frac{w_f L^2}{8} = 823 \text{ kN}\cdot\text{m}$$ Trial W690x152: $Z_x = 4940 \times 10^3$ mm$^3$ so $\phi M_p = 1556$ kN·m, a utilisation of 0.529. The beam is adequately braced, so lateral torsional buckling does not reduce this.
3. Check the wet-concrete deflection, which actually chooses the beam. Under the unfactored slab and steel weight the bare beam deflects $$\Delta = \frac{5 w L^4}{384 E I_x} = 53.9 \text{ mm} = L/297$$ That is the real reason for the section size: on strength alone a much lighter beam would do, but it would settle into the wet slab and change the deck thickness. Specify a camber of 55 mm so that the deck finishes level.
4. Find the effective slab width. Cl 17.4.1 limits the effective width to the smaller of the beam spacing and a quarter of the span: $b_e = \min(3000,\ 16000/4) = 3000$ mm.
5. Check the composite section for strength. The slab in compression can deliver $C_c = 0.85 f'_c b_e t = 18360$ kN whereas the steel in tension can deliver only $T = A F_y = 6713$ kN, so the plastic neutral axis lies inside the slab and the depth of the stress block is $a = T/(0.85 f'_c b_e) = 87.7$ mm. The couple acts over $d/2 + t - a/2 = 540.1 \text{ mm}$, giving $$\boxed{M_r = \phi T \left(\frac{d}{2} + t - \frac{a}{2}\right) = 3263 \text{ kN}\cdot\text{m}}$$ against the composite-stage demand $w_f = 49.71$ kN/m, $M_f = 1591$ kN·m — a utilisation of 0.488. The end shear 398 kN is trivial against $V_r = 1874$ kN.
Plastic composite section. Because the neutral axis lies in the slab, the whole steel section is at yield in tension.
6. Check the live-load deflection. Transforming the slab by $n = E_s/E_c = 8.11$ gives a composite second moment of $5.3074 \times 10^{9}$ mm$^4$, nearly 3.6 times the bare beam. Under the pedestrian load alone $$\Delta_{LL} = \frac{5 w_{LL} L^4}{384 E I_{tr}} = 12.1 \text{ mm} = L/1327$$ comfortably inside the $L/800$ that CSA S6 applies to a bridge carrying pedestrians.
7. Part (b) — size one stud. Take 19 mm diameter headed studs, 100 mm long, $A_{sc} = 283.5$ mm$^2$, $F_u = 450$ MPa. Cl 17.7.2 gives $$q_r = \min\!\left(0.5 \phi_{sc} A_{sc} \sqrt{f'_c E_c},\ \phi_{sc} A_{sc} F_u\right) = 97.5 \text{ kN per stud}$$
8. Count the studs. With full shear connection the horizontal force crossing the interface between the point of zero moment and the point of maximum moment is the smaller of $C_c$ and $T$, i.e. $V_h = 6713$ kN. Hence $$n = \frac{V_h}{q_r} = 69 \text{ studs per half span} \;\Rightarrow\; \boxed{138 \text{ studs per beam}}$$ Arranged in pairs that is 35 transverse rows over each 8 m half, a pitch of 229 mm, which sits between the Cl 17.7.3 minimum of $4d = 76$ mm and the maximum of $8t = 1920$ mm. Space them uniformly: the plastic method assumes the connectors redistribute, so uniform spacing is both permitted and simpler to build.