Question 4 of 7: Composite steel-concrete floor system (14 + 6 marks)

Question 4: Composite steel-concrete floor system (14 + 6 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. A 12 m by 22 m floor with a 160 mm deck slab on steel beams at 2.5 m centres, built unshored, carrying 8 kPa of live load with full interaction assumed between slab and beam.

Find. (a) The steel beam size, checked for the construction stage on the bare steel and for the final stage on the composite section; (b) the number of headed studs per beam.

Approach. Unshored construction splits the design into two stages: the bare steel carries the wet slab, and the composite section carries the total factored load. Size on the composite ultimate moment, then confirm that the construction-stage strength and deflection are acceptable, and finally develop the full flange force with studs.

1. Part (a) — establish the two load stages. Ignoring the steel self weight as instructed, $$w_f=1.25(9.60)+1.5(20.0)=42.0\ \text{kN/m}\ \Rightarrow\ M_f=\frac{42.0(12)^2}{8}=\boxed{756\ \text{kN}\cdot\text{m}},\quad V_f=252\ \text{kN}$$ and during construction the bare beam carries only the wet slab, $M_{f,\text{constr}}=1.25(9.6)(12)^2/8=216$ kN·m.
2. Fix the effective slab width. Clause 17.4.1 limits the flange to the smaller of a quarter of the span and the beam spacing: $$b_{\text{eff}}=\min\left(\frac{12\,000}{4},\,2500\right)=\boxed{2500\ \text{mm}}$$ The slab compressive resistance is then $C_r=0.85\phi_cf'_cb_{\text{eff}}t_s=6630$ kN, far above anything the steel can deliver, so the plastic neutral axis will lie inside the slab.
3. Select the beam on the composite resistance. Trying W460×68 (459 × 154, 15.4 mm flange, 9.1 mm web, $A=8640$ mm²), $$T_r=\phi A F_y=0.9(8640)(350)=2722\ \text{kN}\ <\ C_r,\qquad a=\frac{T_r}{0.85\phi_cf'_cb_{\text{eff}}}=65.7\ \text{mm}<160\ \text{mm}$$ so the whole steel section yields in tension and $$M_r=T_r\!\left(\frac{d}{2}+t_s-\frac{a}{2}\right) =2722\,(229.5+160-32.9)=\boxed{971\ \text{kN}\cdot\text{m}}\ \ge\ 756\ \text{kN}\cdot\text{m}$$
4. Check the construction stage. The bare beam has $Z=1.47\times10^6$ mm³, so $\phi M_p=463$ kN·m, comfortably above the 216 kN·m of wet concrete; the deck and its formwork brace the compression flange, consistent with the question's bracing note. Deflection is the real construction-stage control: $$\Delta_{D}=\frac{5w_DL^4}{384EI_x} =\frac{5(9.6)(12\,000)^4}{384(200\,000)(293\times10^6)}=\boxed{44.2\ \text{mm}} =\frac{L}{271}$$ This is why W460×68 is chosen rather than the lighter W460×60, which satisfies strength ($M_r=848$ kN·m) but deflects 51.6 mm, i.e. $L/232$, outside the customary $L/240$ limit for wet concrete. Camber the beams 40 mm.
5. Check the in-service deflection and shear. With $n=E_s/E_c=8.11$ the transformed section has its neutral axis 122 mm below the slab top and $I_{tr}=1102\times10^6$ mm⁴, so the live-load deflection is $$\Delta_L=\frac{5(20.0)(12\,000)^4}{384(200\,000)(1102\times10^6)}=24.5\ \text{mm} =\frac{L}{490}\ \le\ \frac{L}{360}$$ The web at $h/w=47.1$ is stocky, so $F_s=0.66F_y$ and $V_r=\phi dwF_s=868$ kN against $V_f=252$ kN.
6. Part (b) — size the shear connection. With full interaction the studs between the point of maximum moment and the support must develop the smaller of $C_r$ and $T_r$, i.e. $V_h=2722$ kN. For a 19 mm headed stud in normal-density 30 MPa concrete, $$q_r=0.5\phi_{sc}A_{sc}\sqrt{f'_cE_c} =0.5(0.80)(283.5)\sqrt{30(24\,648)}=97.5\ \text{kN}$$ which is below the cap $\phi_{sc}A_{sc}F_u=102$ kN, so $$n=\frac{2722}{97.5}=27.9\ \rightarrow\ \boxed{28\ \text{studs per half span},\ 56\ \text{per beam}}$$
7. Detail the studs and the floor layout. Place the 56 studs in a single line at 200 mm centres over each 6 m half, which satisfies the 4d minimum longitudinal spacing and the 800 mm maximum. Studs 100 mm long give >4d embedment with 60 mm of cover in the 160 mm slab. Across the 22 m floor width, 2.5 m spacing does not divide evenly, so provide nine bays at 2.44 m (eight interior beams plus two edge beams); the reduced spacing keeps the tributary width below the 2.5 m used above, so the design remains conservative.