Question 3 of 7: Composite Steel–Concrete Warehouse Floor

Question 3: Composite Steel–Concrete Warehouse Floor (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.

Find. (a) a steel section acting compositely with the slab that satisfies the factored moment, the construction-stage moment and deflection; (b) the number of headed studs.

Approach. Take the interior beam as the critical one, establish the effective slab width, size the section on the fully plastic composite resistance, then check the bare steel beam under wet concrete and finish with the horizontal shear to be transferred.

1. Part (a) — assemble the loads on an interior beam. Over a 2.5 m tributary width the slab contributes $0.15 \times 24 = 3.6$ kPa and the superimposed dead 1.0 kPa; adding 1.25 kN/m for the beam itself, $w_D = 12.75$ kN/m and $w_L = 16 \times 2.5 = 40$ kN/m. Hence $$w_f = 1.25(12.75) + 1.5(40) = 75.9\ \text{kN/m}, \qquad \boxed{M_f = \frac{w_f L^2}{8} = 1860\ \text{kN}\cdot\text{m}, \quad V_f = 532\ \text{kN}}$$
2. Determine the effective slab width. Clause 17.4.1 takes the least of a quarter of the span and the beam spacing: $$b_{\text{eff}} = \min(14\,000/4,\ 2500) = 2500\ \text{mm}$$ so the full tributary width is effective — the beams are close enough that shear lag does not bite. (The edge beams, at $0.75 + 1.25 = 2.0$ m, are lighter loaded and are covered by the interior design.)
3. Size the composite section. Try W610 × 125 ($A = 15\,793$ mm², $d = 612$ mm, modelled from nominal plate dimensions). The steel can deliver $$T_r = \phi A F_y = 0.9(15\,793)(350) = 4975\ \text{kN}$$ while the slab in compression can deliver $C_r = 0.85\phi_c f'_c b_{\text{eff}} t = 6216$ kN. Since $T_r < C_r$ the plastic neutral axis lies inside the slab, at $$a = \frac{T_r}{0.85\phi_c f'_c b_{\text{eff}}} = 120.1\ \text{mm} \;<\; 150\ \text{mm}$$ Taking moments about the centre of the concrete block, $$M_r = T_r\left(\frac{d}{2} + t - \frac{a}{2}\right) = 4975\times10^3(306 + 150 - 60.0) = \boxed{1970\ \text{kN}\cdot\text{m}}$$ against $M_f = 1860$ kN·m — a utilisation of 0.94.
4. Check the construction stage. Unshored means the bare steel beam alone carries the wet concrete. With $w = 1.25(9.0 + 1.25) + 1.5(0.5 \times 2.5) = 14.7$ kN/m the construction moment is 360 kN·m, against the bare-steel resistance $\phi Z_x F_y = 1145$ kN·m. The section is Class 1 in that condition ($b/t = 5.84 \le 7.75$; $h/w = 48.1 \le 58.8$), so the plastic modulus is available and the check passes with a factor of three. Web shear, $V_r = \phi(0.66F_y)dw = 1514$ kN, dwarfs the 532 kN.
5. Check deflection. With $E_c = 4500\sqrt{30} = 24\,648$ MPa and $n = 8.11$, the transformed slab width is $2500/8.11 = 308$ mm and the composite second moment of area is $I_{tr} = 2770 \times 10^6$ mm⁴. The live-load deflection is $$\Delta_L = \frac{5w_LL^4}{384EI_{tr}} = 36.1\ \text{mm} = \frac{L}{388} \;\le\; \frac{L}{360} = 38.9\ \text{mm}$$ The bare steel deflects 26 mm under the wet slab, so specify that camber.
6. Part (b) — horizontal shear to be transferred. With full interaction the connectors between the support and the point of maximum moment must transfer the smaller of the two forces computed in step 3, $V_h = 4975$ kN.
7. Size and count the studs. For a 19 mm headed stud, $A_{sc} = 283.5$ mm², Clause 17.7.2.1 gives $$q_r = 0.5\phi_{sc}A_{sc}\sqrt{f'_c E_c} = 97.5\ \text{kN} \quad\left(\le \phi_{sc}A_{sc}F_u = 102\ \text{kN}\right)$$ so $$n = \frac{V_h}{q_r} = \frac{4975}{97.5} = 51.0 \;\rightarrow\; \boxed{52\ \text{studs per half span, i.e. 104 per beam}}$$ Place them as 26 pairs at 269 mm centres over each half span. Two studs fit across the 229 mm flange (transverse spacing 4d = 76 mm minimum), the longitudinal spacing exceeds the 6d = 114 mm minimum, and 269 mm is well below the 8t = 1200 mm maximum.

Final Results