Question 4 of 7: Composite steel–concrete bridge cross-section and shear connectors

Question 4: Composite steel–concrete bridge cross-section and shear connectors (15 + 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. From Figure 3: a 12 m wide deck with a 275 mm reinforced concrete slab on two built-up steel box girders at $4.5 + 4.5 = 9$ m centres with 1.5 m overhangs each side. The notes on the figure give a design span of 18 m and a design live load of 16 kPa. Materials are $f'_c = 30$ MPa and $F_y = 350$ MPa.

Find. (a) plate sizes for one built-up box girder such that the composite section resists the factored sagging moment; (b) the number and spacing of 19 mm headed studs in each shear span.

[Figure not reproduced: Figure 3 as printed: 12 m deck, 275 mm slab, two built-up steel box girders at 9 m centres with 1.5 m overhangs. Design span 18 m, design live load 16 kPa. See the official exam paper.]

Approach. Take one girder with its 6 m tributary width, build the factored load, fix the effective slab width from Cl 17.4, size the box on the plastic composite resistance with the neutral axis in the slab, check web shear, then convert the flange force into a stud count with the Cl 17.7 resistance.

1. Part (a) — loads on one girder. The slab weighs $0.275(24) = 6.6$ kPa, so over 6 m of deck it delivers 39.6 kN/m. Allowing 1.5 kN/m for the railing and, after the section below is settled, 3.13 kN/m for the girder itself (its area is $37\,000$ mm$^2$, plus 10 per cent for diaphragms), $$w_D = 39.6 + 3.13 + 1.5 = 44.2\ \text{kN/m},\qquad w_L = 16(6.0) = 96.0\ \text{kN/m}.$$
2. Factored moment and shear. With $1.25D + 1.5L$, $$w_f = 1.25(44.2) + 1.5(96.0) = 199.3\ \text{kN/m},$$ $$M_f = \frac{w_f L^2}{8} = \frac{199.3(18)^2}{8} = 8071\ \text{kN}\cdot \text{m},\qquad V_f = \frac{w_f L}{2} = 1794\ \text{kN}.$$ The live load dominates: it contributes 72 per cent of $w_f$.
3. Effective slab width. Cl 17.4.1 limits the width acting with each girder to $L/4$, and separately to the girder spacing. Here $L/4 = 4500$ mm governs against the 9000 mm spacing, so $$b_e = 4500\ \text{mm}.$$ Note the deck is 12 m wide but only $2 \times 4.5 = 9$ m of it is mobilised as compression flange — the rest is load, not resistance.
4. Trial box girder. Take a bottom flange $600 \times 20$, two webs $900 \times 10$ and two top flanges $250 \times 14$: $$A_s = 600(20) + 2(900)(10) + 2(250)(14) = 37\,000\ \text{mm}^2,$$ overall steel depth $934$ mm, and taking moments of area about the soffit the steel centroid sits $\bar y = 407.3$ mm up. The mass is 290 kg/m, confirming the 3.13 kN/m assumed in step 1.
5. Locate the plastic neutral axis. The steel can deliver $$T = \phi A_s F_y = 0.90(37\,000)(350)/10^3 = 11\,655\ \text{kN},$$ while the slab can supply $C_{max} = 0.85\phi_c f'_c b_e t_s = 0.85(0.65)(30)(4500)(275)/10^3 = 20\,512$ kN. Since $T \lt C_{max}$ the whole steel section yields in tension and the neutral axis lies inside the slab, with $$a = \frac{T}{0.85\phi_c f'_c b_e} = \frac{11\,655\times10^3}{74\,588} = 156.3\ \text{mm} \;\lt\; 275\ \text{mm}\quad\checkmark$$
6. Moment resistance. The lever arm runs from the centroid of the compression block to the steel centroid: $$d_1 = \left(934 + 275 - \frac{156.3}{2}\right) - 407.3 = 723.6\ \text{mm},$$ $$M_r = T d_1 = 11\,655(0.7236) = \boxed{8434\ \text{kN}\cdot\text{m}} \;\gt\; M_f = 8071\ \text{kN}\cdot\text{m}\quad\checkmark$$ a utilisation of 0.96 — a tight, economical section.
7. Web shear. The two webs give $A_w = 2(900)(10) = 18\,000$ mm$^2$ with $h/w = 90$. Unstiffened ($k_v = 5.34$), $F_{cri} = 180\,000(5.34)/90^2 = 118.7$ MPa, so $$V_r = 0.90(18\,000)(118.7)/10^3 = 1922\ \text{kN} \;\gt\; 1794\ \text{kN} \quad\checkmark$$ at 0.93 — adequate without intermediate stiffeners, though bearing stiffeners and internal diaphragms are still required at the supports.
8. Part (b) — horizontal shear to be transferred. For full interaction Cl 17.9.4 requires the connectors between the point of maximum moment and the support to carry the smaller of the two flange capacities: $$V_h = \min(11\,655,\ 20\,512) = 11\,655\ \text{kN}$$ per half span, i.e. the whole steel yield force.
9. Resistance of one stud. For a 19 mm headed stud, $A_{sc} = 283.5$ mm$^2$ and $E_c = 4500\sqrt{30} = 24\,648$ MPa, so Cl 17.7.2 gives $$q_r = 0.5\phi_{sc}A_{sc}\sqrt{f'_c E_c} = 0.5(0.80)(283.5)\sqrt{30(24\,648)}/10^3 = 97.5\ \text{kN},$$ below the ceiling $\phi_{sc}A_{sc}F_u = 102$ kN, so $q_r = 97.5$ kN.
10. Number and spacing. $$n = \frac{V_h}{q_r} = \frac{11\,655}{97.5} = \boxed{120\ \text{studs per half span}}\ (240\ \text{per girder}).$$ With four studs per row — two on each of the two top flanges — that is 30 rows over 9 m, i.e. a uniform pitch of 300 mm. Check the detailing limits: the pitch is under the $8t_s = 2200$ mm and 600 mm maxima, the transverse spacing of 125 mm exceeds $4d = 76$ mm, and 100 mm studs give $h/d = 5.3 \ge 4$.