Question 3 of 7: Simply supported composite box-girder bridge

Question 3: Simply supported composite box-girder bridge (15 + 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.

Find. (a) Plate sizes for a built-up rectangular box girder that carries the wet slab alone during construction and the full factored moment compositely; (b) the number of shear studs required for full interaction.

Approach. Take the exterior girder — its tributary width of 3.95 m exceeds the interior girder's 3.8 m, so it governs. Check the bare steel section against the factored construction (wet-slab) moment, then check the composite section as a plastic couple between the concrete slab in compression and the whole steel section in tension. The horizontal shear that this couple implies, divided by the factored stud resistance, gives the connector count.

1. Deck geometry and tributary width. The dimension chain closes on the stated overall width, \(1.4 + 1.3 + 2.5 + 1.3 + 2.5 + 1.3 + 1.4 = 11.7\ \text{m}\), confirming three boxes rather than two. Girder centrelines are at 2.05, 5.85 and 9.65 m, a uniform 3.8 m spacing symmetric about the deck centreline. The exterior girder carries from the deck edge to mid-way to its neighbour:
   $$b_{trib} = \frac{2.05 + 5.85}{2} = 3.95\ \text{m} \;>\; 3.80\ \text{m}\ \text{(interior)}$$
   The effective slab width for composite action is \(b_{eff} = \min(b_{trib},\, L/4) = \min(3950,\, 4500) = 3950\ \text{mm}\).
2. Trial box girder. Try an outside width of 1300 mm and depth 700 mm: top plate \(1300 \times 10\), bottom plate \(1300 \times 16\), two webs \(10\ \text{mm} \times 674\). Then
   $$A_s = 47\,280\ \text{mm}^2 \;(371.1\ \text{kg/m}), \quad y_b = 295.3\ \text{mm}, \quad I = 4.349\times10^{9}\ \text{mm}^4$$
   The web slenderness \(h/w = 674/10 = 67.4 \le 1700/\sqrt{350} = 90.9\) keeps the webs Class 2.
3. Loads on the governing girder.
   $$w_{slab} = 0.275(24)(3.95) = 26.07\ \text{kN/m}, \quad w_{girder} = 3.64\ \text{kN/m} \;\Rightarrow\; w_D = 29.71\ \text{kN/m}$$
   $$w_L = 20(3.95) = 79.0\ \text{kN/m}$$
   $$M_{Df} = \frac{1.25(29.71)(18)^2}{8} = 1504.1, \qquad M_{Lf} = \frac{1.5(79.0)(18)^2}{8} = 4799.3\ \text{kN}\cdot\text{m}$$
   $$M_f = 1504.1 + 4799.3 = \boxed{6303.4\ \text{kN}\cdot\text{m}}$$
4. Construction stage — bare steel. Before the slab hardens the steel section alone resists the wet concrete and its own weight. Elastically, \(S_t = I/(h - y_b) = 1.075\times10^{7}\ \text{mm}^3\), so
   $$M_{r,steel} = \phi S_t F_y = 0.90(1.075\times10^{7})(350) = 3385\ \text{kN}\cdot\text{m} \;\ge\; M_{Df} = 1504\ \text{kN}\cdot\text{m}$$
   Comfortable, but the top flange of a box is a plate supported along both edges with \(b/t = 1276/10 = 128\), far outside the Class 3 limit \(670/\sqrt{F_y} = 35.8\). Because that flange is in compression only during casting (afterwards it is embedded against the slab), provide two longitudinal stiffeners on the top plate, giving sub-panels of \(1276/3 = 425\ \text{mm}\) and \(b/t = 42.5\), together with internal cross-diaphragms at 4.5 m centres to control box distortion.
5. Composite flexural resistance. Compare the slab's compressive capacity with the steel section's tensile capacity:
   $$C_r = 0.85\phi_c f'_c b_{eff} t_s = 0.85(0.65)(30)(3950)(275) = 18\,005\ \text{kN}$$
   $$T_r = \phi A_s F_y = 0.90(47\,280)(350) = 14\,893\ \text{kN} \;<\; C_r$$
   The steel yields before the slab crushes, so the plastic neutral axis lies inside the slab. The stress-block depth is
   $$a = \frac{T_r}{0.85\phi_c f'_c b_{eff}} = \frac{14\,893\times10^{3}}{65\,471} = 227.5\ \text{mm} \;<\; 275\ \text{mm}$$
6. Moment resistance. The couple acts between the steel centroid and the centre of the stress block:
   $$\bar{z} = (h - y_b) + t_s - \frac{a}{2} = (700 - 295.3) + 275 - 113.7 = 566.0\ \text{mm}$$
   $$M_{rc} = T_r\,\bar{z} = 14\,893(0.566) = \boxed{8430\ \text{kN}\cdot\text{m}} \;\ge\; M_f = 6303\ \text{kN}\cdot\text{m}$$
   Utilisation 0.75 — the section is governed by the construction-stage depth and web slenderness rather than by composite strength, which is normal for a shallow box.
7. Shear. The end reaction is \(V_f = (1.25 w_D + 1.5 w_L)(L/2) = 1400.8\ \text{kN}\) shared by two webs. With \(h/w = 67.4\), unstiffened (\(k_v = 5.34\)) and \(502\sqrt{k_v/F_y} = 62.0 < 67.4 \le 621\sqrt{k_v/F_y} = 76.7\), the inelastic-buckling branch applies with \(k_a = 0\):
   $$F_s = \frac{290\sqrt{F_y k_v}}{h/w} = 186.0\ \text{MPa}, \qquad V_r = \phi(2 t_w h) F_s = 2257\ \text{kN} \;\ge\; V_f$$
8. (b) Shear connectors. For full interaction the total horizontal shear transferred between the point of zero moment and the point of maximum moment is \(V_h = \min(C_r, T_r) = 14\,893\ \text{kN}\). Using 22 mm headed studs (\(A_{sc} = 380\ \text{mm}^2\), \(F_u = 450\ \text{MPa}\)) with \(E_c = 4500\sqrt{30} = 24\,648\ \text{MPa}\), CSA S16 Cl. 17.7.2.1 gives
   $$q_r = 0.5\phi_{sc}A_{sc}\sqrt{f'_c E_c} = 130.7\ \text{kN} \;\le\; \phi_{sc}A_{sc}F_u = 136.8\ \text{kN}$$
   $$n = \frac{V_h}{q_r} = \frac{14\,893}{130.7} = 113.9 \;\Rightarrow\; \boxed{114\ \text{studs per half span},\ 228\ \text{per girder}}$$
   Placed three to a row across the 1300 mm top flange, 38 rows per half span at \(9000/38 = 237\ \text{mm}\) centres — inside the limits \(4d = 88\ \text{mm}\) minimum and 600 mm maximum.