Question 4 of 7: Unshored composite floor beam for a heavily loaded warehouse

Question 4: Unshored composite floor beam for a heavily loaded warehouse (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. Simply supported composite beams at 2.5 m centres spanning 16.0 m; live load 18 kPa; unshored construction; 100% (full) interaction; $f'_c=30$ MPa, $F_y=350$ MPa. Assumed: a 120 mm solid normal-density slab (2.88 kPa), 0.5 kPa superimposed dead load for finishes and services, and 19 mm headed studs, $F_u=450$ MPa.

Find. (a) A steel section that, acting compositely with the slab, resists the factored moment while the bare steel alone resists the wet-concrete stage; (b) the number and layout of the shear connectors that develop full interaction.

Approach. Compute the factored moment; find the plastic neutral axis by comparing the concrete compressive resistance with the steel tensile resistance; take moments to obtain $M_{rc}$; check the unshored construction stage and the deflections; then divide the horizontal shear at the interface by the resistance of one stud.

(a) The cross-section

1. Design actions. $$M_f=\frac{w_fL^{2}}{8}=\frac{80.2(16)^{2}}{8}=2567\ \text{kN}\cdot\text{m},\qquad V_f=\frac{w_fL}{2}=642\ \text{kN}$$
2. Concrete compressive resistance. $$C_r=0.85\phi_cf'_cb_{eff}t=0.85(0.65)(30)(2500)(120)\times10^{-3}=4972\ \text{kN}$$
3. Try W840 × 176. From the CISC Handbook: $A=22\,400$ mm$^2$, $d=835$ mm, $b_f=292$ mm, $t_f=18.8$ mm, $w=14.0$ mm, $I_x=2460\times10^{6}$ mm$^4$, $Z_x=6810\times10^{3}$ mm$^3$, mass 176 kg/m (1.73 kN/m). Its tensile resistance is $$T_r=\phi AF_y=0.9(22\,400)(350)\times10^{-3}=7056\ \text{kN}\ >\ C_r=4972\ \text{kN}$$ so the plastic neutral axis (PNA) falls inside the steel section, and part of the steel must act in compression.
4. Locate the PNA. The steel compression force is half the surplus, $$C_s=\frac{T_r-C_r}{2}=\frac{7056-4972}{2}=1042\ \text{kN}$$ The top flange alone can supply $\phi b_ft_fF_y=1729$ kN, so the PNA lies in the top flange at $$y=\frac{C_s}{\phi b_fF_y}=\frac{1042\times10^{3}}{0.9(292)(350)}=11.3\ \text{mm}$$ below the top of the steel.
5. Moment resistance. The three internal forces have zero resultant, so moments may be taken about the top of the steel section (levers downward positive; the slab centroid is 60 mm above it): $$M_{rc}=T_r\frac{d}{2}-2C_s\frac{y}{2}+C_r(60)$$ $$M_{rc}=7056(417.5)-2(1042)(5.67)+4972(60)=3.232\times10^{6}\ \text{kN}\cdot\text{mm}=\boxed{3232\ \text{kN}\cdot\text{m}}$$ $M_f/M_{rc}=2567/3232=0.79$ — adequate, with the margin that a warehouse floor deserves.
6. Web shear. With $h/w=(835-2\times18.8)/14=57.0$, which lies between $439\sqrt{k_v/F_y}=54.2$ and $502\sqrt{k_v/F_y}=62.0$ for an unstiffened web ($k_v=5.34$), $$F_s=\frac{290\sqrt{F_yk_v}}{h/w}=220\ \text{MPa}\;\Rightarrow\;V_r=\phi hwF_s=0.9(797)(14)(220)\times10^{-3}=2212\ \text{kN}\ \gg\ 642\ \text{kN}$$
7. The unshored construction stage. Before the slab cures, the bare steel carries the wet concrete and a nominal 0.5 kPa construction live load: $$w_{f,constr}=1.25\left[2.88(2.5)+1.73\right]+1.5\left[0.5(2.5)\right]=13.0\ \text{kN/m}\;\Rightarrow\;M_f=417\ \text{kN}\cdot\text{m}$$ against the bare-steel resistance $\phi Z_xF_y=0.9(6810\times10^{3})(350)=2145$ kN·m. The construction stage is not remotely critical — a useful confirmation, because unshored construction is exactly the case where it sometimes is.
8. Deflections. Dead load deflects the bare steel: $$\Delta_D=\frac{5w_DL^{4}}{384EI_x}=\frac{5(8.93)(16\,000)^{4}}{384(200\,000)(2460\times10^{6})}=15.5\ \text{mm}$$ which is cambered out at the shop. Live load acts on the composite section; with $n=E/E_c=200\,000/24\,650=8.11$ the transformed slab is $2500/8.11=308$ mm wide, the elastic neutral axis lies 120 mm below the top of the steel, and $I_{tr}=5.685\times10^{9}$ mm$^4$, giving $$\Delta_L=\frac{5(45)(16\,000)^{4}}{384(200\,000)(5.685\times10^{9})}=33.8\ \text{mm}=\frac{L}{474}\ <\ \frac{L}{360}$$ Both limits are met.

(b) Shear connectors

1. Resistance of one stud. For a 19 mm headed stud, $A_{sc}=284$ mm$^2$, $\phi_{sc}=0.80$, $F_u=450$ MPa: $$q_r=\min\left[0.5\phi_{sc}A_{sc}\sqrt{f'_cE_c},\ \phi_{sc}A_{sc}F_u\right]=\min\left[97.5,\ 102.1\right]=97.5\ \text{kN}$$ (the concrete-controlled value governs, as it usually does with normal-density 30 MPa concrete).
2. Horizontal shear to be transferred. For full (100%) interaction the connectors between a support and the point of maximum moment must carry the smaller of the two slab/steel forces: $$V_h=\min\left(C_r,\ T_r\right)=\min(4972,\ 7056)=4972\ \text{kN}$$
3. Number of connectors. $$n=\frac{V_h}{q_r}=\frac{4972}{97.5}=51.0\ \Rightarrow\ \boxed{51\ \text{studs per half span}}$$ Because the beam is symmetric and simply supported, both halves need the same number, so $$n_{total}=2(51)=\boxed{102\ \text{studs per beam}}$$
4. Layout. Place them in pairs: 26 rows of two studs over each 8.0 m half span gives 52 per half (104 per beam) at a pitch of $$p=\frac{8000}{26}=308\ \text{mm}$$ This satisfies the minimum longitudinal spacing $6d=114$ mm, the minimum transverse spacing $4d=76$ mm (the pair is set at 140 mm gauge), and the maximum spacing $8t_{slab}=960$ mm. Stud height $\ge 4d=76$ mm and $\ge t+30$; use 100 mm studs, giving at least 20 mm of cover over the head. Uniform spacing is permissible because the section is compact and ductile studs redistribute the interface shear.