Question 4 of 7: Composite floor beam for a warehouse (15 + 5 marks)
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams, May 2013, 98-Civ-B2 Advanced Structural Design; 3 hours, closed book (handbooks and textbooks permitted); seven design questions, any five of which constitute a complete paper, all of equal value (20 marks each). All seven are worked below, because the set is a study resource rather than a timed sitting. Design data given on page 1 and used throughout: concrete $f^{\prime}_{c}=30\ \text{MPa}$, structural steel $F_{y}=350\ \text{MPa}$, reinforcing bar $f_{y}=400\ \text{MPa}$; for the prestressed girder $f^{\prime}_{ci}=35\ \text{MPa}$, $f^{\prime}_{c}=50\ \text{MPa}$, $n=6$, $f_{pu}=1750\ \text{MPa}$, $f_{py}=1450\ \text{MPa}$, $f_{pi}=1200\ \text{MPa}$ and losses of 240 MPa. All loads shown on the figures are unfactored.
Reference texts. CSA S16, Design of Steel Structures; CISC, Handbook of Steel Construction; Kulak & Grondin, Limit States Design in Structural Steel; CSA A23.3, Design of Concrete Structures; MacGregor & Bartlett, Reinforced Concrete: Mechanics and Design (Canadian edition); Collins & Mitchell, Prestressed Concrete Structures; Beedle, Plastic Design of Steel Frames; National Building Code of Canada.
Check — load factor. The paper states that the loads shown are unfactored but does not separate dead from live. Every design below therefore applies a single load factor of 1.5 to the given loads, which is the NBCC live-load factor and is the conservative reading when the split is unknown. Self weight, where it is generated by a member being sized here (the reinforced concrete frame of Questions 5 and 6, and the prestressed girder of Question 7), is carried separately at 1.25. If the examiner intended a different split, only the magnitudes change — every mechanism, section classification and interaction equation is unaffected.
Question 4: Composite floor beam for a warehouse (15 + 5 marks)
Given. A heavily loaded warehouse floor in composite construction: steel beams at 2.5 m centres spanning 16 m, a design live load of 18 kPa, unshored construction, full interaction assumed between the steel and the slab, and adequate bracing to the steel beams.
Given data — composite floor beam
Quantity
Symbol
Value
Beam spacing
$s$
2.5 m
Design span
$L$
16 m
Design live load
$w_{L}$
18 kPa
Slab thickness assumed
$t_{s}$
150 mm normal density
Concrete strength
$f^{\prime}_{c}$
30 MPa
Steel yield strength
$F_{y}$
350 MPa
Construction
—
unshored, full interaction
Find. (a) the composite cross-section, and (b) the number of shear connectors required.
Composite section: 150 mm slab acting with a W760x147 through 19 mm headed studs; effective width 2 500 mm.
Approach. Build the factored load from the tributary width, then locate the plastic neutral axis by comparing the slab compressive resistance with the steel tensile resistance, take moments about the tension resultant for the composite moment resistance, and finish with the two checks unshored construction forces on you: the bare beam under wet concrete, and live-load deflection of the composite section.
Part (a) — loads on one beam. A 150 mm slab weighs $0.15(24)=3.6\ \text{kPa}$, so over a 2.5 m tributary width the dead load is $9.0\ \text{kN/m}$ plus about $1.44\ \text{kN/m}$ of steel, and the live load is $18(2.5)=45\ \text{kN/m}$. Then $w_{f}=1.25(10.44)+1.5(45)=80.6\ \text{kN/m}$, so $M_{f}=w_{f}L^{2}/8=2578\ \text{kN}\cdot\text{m}$ and $V_{f}=644\ \text{kN}$. The live load carries almost 84 percent of the total, which is why the 18 kPa figure dominates every subsequent decision.
Part (a) — effective width and slab resistance. $b_{e}=\min(L/4,\ s)=\min(4000,\ 2500)=2500\ \text{mm}$. The slab can deliver $$C_{r}=0.85\phi_{c}f^{\prime}_{c}b_{e}t_{s}=0.85(0.65)(30)(2500)(150)=6216\ \text{kN}$$
Part (a) — trial section and the neutral axis. Try a W760x147 ($A=18\,501\ \text{mm}^{2}$, $d=753\ \text{mm}$). Its tensile resistance $T_{r}=\phi AF_{y}=5828\ \text{kN}$ is less than $C_{r}$, so the plastic neutral axis lies inside the slab and the whole steel section yields in tension. The depth of the compression block is $a=T_{r}/(0.85\phi_{c}f^{\prime}_{c}b_{e})=140.6\ \text{mm}\le150\ \text{mm}$, confirming it.
Part (a) — composite moment resistance. Taking moments about the steel centroid, the lever arm between the two resultants is $e=d/2+t_{s}-a/2=376.5+150-70.3=456.2\ \text{mm}$, so $$M_{rc}=T_{r}e=5828(0.4562)=\boxed{2659\ \text{kN}\cdot\text{m}\ \ge\ M_{f}=2578\ \text{kN}\cdot\text{m}}$$ Web shear is never in question: $h/w=54.5$ gives $F_{s}=230\ \text{MPa}$ and $V_{r}=2059\ \text{kN}$ against $V_{f}=644\ \text{kN}$.
Part (a) — the unshored checks. Before the slab cures the bare steel carries everything: $w_{f}=1.25(10.44)+1.5(2.5)=16.8\ \text{kN/m}$ gives $M_{f}=538\ \text{kN}\cdot\text{m}$ against $\phi Z_{x}F_{y}=1582\ \text{kN}\cdot\text{m}$, with the formwork restraining the top flange. Dead-load deflection at that stage is 27 mm, or $L/585$, so camber is optional. For the composite section with $n=E/E_{c}=8.11$ the transformed inertia is $4409\times10^{6}\ \text{mm}^{4}$ and the live-load deflection is $43.5\ \text{mm}=L/367$, just inside the customary $L/360$.
Part (b) — shear connector resistance. Full interaction requires the smaller of $C_{r}$ and $T_{r}$ to cross the interface between the support and mid-span, so $V_{h}=5828\ \text{kN}$ per half span. For a 19 mm headed stud with $A_{sc}=283.5\ \text{mm}^{2}$, $E_{c}=4500\sqrt{30}=24\,648\ \text{MPa}$ and $F_{u}=450\ \text{MPa}$, $$q_{r}=0.5\phi_{sc}A_{sc}\sqrt{f^{\prime}_{c}E_{c}}=97.5\ \text{kN}\ \le\ \phi_{sc}A_{sc}F_{u}=102\ \text{kN}$$ so $q_{r}=97.5\ \text{kN}$ governs.
Part (b) — number and layout. $n=V_{h}/q_{r}=5828/97.5=59.8$, so use $\boxed{60\ \text{studs per half span},\ 120\ \text{in total}}$. Placed in pairs the 30 rows give a 267 mm pitch, comfortably between the 6 stud diameters (114 mm) minimum and the 800 mm maximum, with the pair at 100 mm across the 265 mm flange. Because the section is compact and the load uniform, the studs may be spaced uniformly rather than following the shear diagram.
