07-Str-A5 · December 2013

Question 7 of 7: Composite steel–concrete floor — section and shear connectors

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. 07-Str-A5 Advanced Structural Design, National Exams, December 2013 — 3 hours, closed book (handbooks and textbooks permitted, no notes). Seven questions of equal value; any five constitute a complete paper. All seven are solved here, because the set is intended as a study resource. All loads printed on the figures are unfactored.

Design data given on the paper (SI). Concrete $f'_c = 30$ MPa; structural steel $F_y = 350$ MPa; reinforcing steel $f_y = 400$ MPa. Prestressed concrete: $f'_{ci} = 35$ MPa at transfer, $f'_c = 50$ MPa, $n = 6$, $f_{pu} = 1750$ MPa, $f_{py} = 1450$ MPa, $f_{initial} = 1200$ MPa, losses $= 240$ MPa (hence $f_{pe} = 960$ MPa).

Reference texts. CSA S16 Design of Steel Structures with the CISC Handbook of Steel Construction — plate girders (Cl. 14), lateral–torsional buckling (Cl. 13.6), plastic design (Cl. 8.6, 13.7), beam-columns (Cl. 13.8), composite beams (Cl. 17), connections (Cl. 21); CSA A23.3 Design of Concrete Structures — flexure and shear (Cl. 10, 11), slenderness (Cl. 10.13–10.16), footings (Cl. 15), prestressed concrete (Cl. 18); NBCC for load combinations; C. G. Salmon, J. E. Johnson & F. A. Malhas, Steel Structures: Design and Behavior; J. G. MacGregor & J. K. Wight, Reinforced Concrete: Mechanics and Design; T. Y. Lin & N. H. Burns, Design of Prestressed Concrete Structures; L. S. Beedle, Plastic Design of Steel Frames.

Check: load factors. The paper states only that the printed loads are unfactored, and gives no dead/live split. Every question below therefore treats each printed concentrated load as specified live load and factors it by $\alpha_L = 1.5$ (NBCC principal case $1.25D + 1.5L$); member self-weight, wherever it matters, is estimated from the trial section and factored at $1.25$. A candidate assuming a different split would obtain proportionally different sizes; the method is what is examined.

Check: section properties. All steel sections selected below are quoted by their plate dimensions (flange $b \times t$, web $h \times w$) and every property — $A$, $I_x$, $I_y$, $Z$, $J$, $C_w$, $r_x$, $r_y$ — is computed from those dimensions with root fillets neglected (slightly conservative). This makes every line checkable without a handbook; a rolled shape of equal or greater properties may be substituted directly.

Check: reading of Figures 3 and 4. On the examination drawing the dimension chain under Figure 3 reads 3 m + 4 m + 7 m + 4 m, and the overall dimension reads 15 m. Reading the drawing directly resolves this: the 15 m dimension line spans B to C ($4 + 7 + 4 = 15$ m) and the 3 m is a cantilever overhang projecting beyond support B to the free end A, which carries the 80 kN load. Similarly in Figure 4 the 10.5 m is the overall width of the floor ($3 \times 3.5$ m beam spacing), not a span; the design span is the 15 m stated in the note. Both readings are self-consistent and are used throughout.

Check: support conditions in Figure 2. The frame drawing shows identical triangle-and-roller symbols at A and D, and the frame must resist the 80 kN horizontal load, so both bases are taken as pinned. The frame is then statically indeterminate to the first degree, which is the classical form of this problem for both the elastic (Questions 2, 3) and the plastic (Questions 5, 6) treatments.

The three welded I-sections selected in this paper, drawn to a common scale.

Question 7: Composite steel–concrete floor — section and shear connectors (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. The design starts from an interior composite beam of 15 m simple span carrying a 3.5 m width of 220 mm slab and a 14 kPa live load.

Given data
Quantity	Symbol	Value
Design span	$L$	15 000 mm
Beam spacing	$s$	3 500 mm
Slab thickness	$t$	220 mm
Design live load	$q_L$	14 kPa
Concrete / steel strength	$f'_c$ / $F_y$	30 MPa / 350 MPa
Shear connectors	—	19 mm diameter headed studs, $F_u = 450$ MPa

Find. (a) the steel section for an interior composite beam, checked for flexure, shear, the construction stage and deflection; (b) the number and spacing of headed shear studs required for full interaction.

Figure 4 — the floor cross-section: 220 mm slab on four beams at 3.5 m centres, 10.5 m overall width.

Approach. Assemble the factored uniform load on one interior beam, find the effective slab width, locate the plastic neutral axis by equating the steel tensile resistance to the concrete compressive resistance, take moments for $M_r$, then check web shear, the unshored construction stage and live-load deflection, and finally size the studs from the horizontal shear that must cross the interface.

Assemble the loads on one interior beam. The slab weighs $0.220(24) = 5.28$ kPa; allow 1.0 kPa for finishes and services. Over a 3.5 m tributary width, with a trial beam self-weight of 1.48 kN/m, $$w_f = 1.25(18.48 + 3.50 + 1.48) + 1.5(49.0) = 102.8\ \text{kN/m},$$ so $M_f = w_fL^2/8 = 2892$ kN·m and $V_f = w_fL/2 = 771$ kN.
Fix the effective slab width. For an interior beam, $b_{eff}$ is the lesser of the beam spacing and one quarter of the span: $\min(3500,\ 15\,000/4 = 3750) = 3500$ mm.
Select the steel section and locate the plastic neutral axis. Try a welded I with flanges $260 \times 22$ mm and a web $650 \times 12$ mm, $d = 694$ mm, $A = 19\,240$ mm². The two interface forces are $$T_r = \phi A_sF_y = 6061\ \text{kN}, \qquad C_r = 0.85\phi_cf'_cb_{eff}t = 12\,763\ \text{kN}.$$ Steel governs, so the neutral axis lies in the slab at a compression-block depth $$a = \frac{T_r}{0.85\phi_cf'_cb_{eff}} = \frac{6.061 \times 10^6}{58\,013} = 104.5\ \text{mm} \; < \; 220\ \text{mm}.$$ The entire steel section is in tension, so its local plate slendernesses cannot govern.
Check flexure. The steel force acts at mid-depth of the section and the concrete force at mid-depth of the block, so the lever arm is $d/2 + t - a/2 = 347 + 220 - 52.2 = 514.8$ mm and $$M_r = T_r\left(\frac{d}{2} + t - \frac{a}{2}\right) = 6061(0.5148) = \boxed{3120\ \text{kN}\cdot\text{m}} \; > \; 2892\ \text{kN}\cdot\text{m}.$$
Check shear. Shear is taken by the steel web alone. With $h/w = 650/12 = 54.2$, which is at the limit $439\sqrt{k_v/F_y} = 54.2$ for an unstiffened web, $F_s = 0.66F_y = 231$ MPa and $$V_r = \phi\,d\,w\,F_s = 0.90(694)(12)(231) = 1731\ \text{kN} \; > \; 771\ \text{kN}.$$ Provide bearing stiffeners at the two ends only.
Check the construction stage. If the floor is built unshored, the bare steel beam carries the wet concrete and the steel itself plus a construction live load: $w_f = 27.6$ kN/m, $M_f = 776$ kN·m, against a bare-steel resistance $\phi ZF_y = 1610$ kN·m. Satisfactory, with the formwork providing lateral restraint. The dead-load deflection on the bare steel is 42 mm, so specify 40 mm of camber.
Check deflection. With $n = E_s/E_c = 8.11$ the transformed slab width is $3500/8.11 = 431$ mm; the elastic neutral axis lies 187 mm below the top of the slab and $I_{tr} = 5.29 \times 10^9$ mm⁴. The live-load deflection is then $$\Delta_L = \frac{5w_LL^4}{384EI_{tr}} = 30.5\ \text{mm} \; < \; \frac{L}{360} = 41.7\ \text{mm}.$$ Note how much the composite action buys: $I_{tr}$ is 3.4 times the bare-steel $I_x$.
(b) Size the shear connectors. Full interaction requires the smaller of the two interface forces, $V_h = 6061$ kN, to be transferred between the point of maximum moment and each support. For a 19 mm headed stud, $A_{sc} = 283.5$ mm² and $E_c = 24\,648$ MPa, so $$q_r = 0.5\phi_{sc}A_{sc}\sqrt{f'_cE_c} = 0.5(0.8)(283.5)(859.9) = 97.5\ \text{kN} \; \le \; \phi_{sc}A_{sc}F_u = 102.1\ \text{kN},$$ so $q_r = 97.5$ kN governs. The number required in each half-span is $$n = \frac{V_h}{q_r} = \frac{6061}{97.5} = 62.2 \; \Rightarrow \; \boxed{63\ \text{studs per half-span, 126 in total.}}$$
Space the studs. Place them in pairs across the 260 mm flange at a 100 mm transverse gauge (above the $4d = 76$ mm minimum, with 80 mm edge distance). Thirty-two pairs over each 7.5 m half-span give a pitch of 230 mm, which comfortably exceeds the $6d = 114$ mm longitudinal minimum and is far below the maximum of eight slab thicknesses. Uniform spacing is permissible because the section is compact and the studs are ductile enough to redistribute; no closer spacing is needed near the supports.

Question 7 — final results
Item	Result
Design loading (interior beam)	$w_f = 102.8$ kN/m; $M_f = 2892$ kN·m, $V_f = 771$ kN
Effective slab width	3500 mm (beam spacing governs)
Steel section	Welded I: flanges $260 \times 22$, web $650 \times 12$, $d = 694$ mm (151 kg/m)
Compression block	$a = 104.5$ mm, wholly within the 220 mm slab
Flexure	$M_r = 3120$ kN·m $> 2892$ kN·m ✓
Shear	$V_r = 1731$ kN $> 771$ kN ✓
Construction stage	$M_f = 776$ kN·m vs bare-steel $M_r = 1610$ kN·m; camber 40 mm
Live-load deflection	30.5 mm $< L/360 = 41.7$ mm ($I_{tr} = 5.29 \times 10^9$ mm⁴)
Shear connectors	126 studs, 19 mm dia.; 63 per half-span, in pairs at 230 mm pitch

Topic. Full-interaction composite beam design — plastic moment resistance of the composite section and the shear connection that makes it real.

Key relations. $b_{eff} = \min(s,\ L/4)$ for an interior beam; interface force $V_h = \min(\phi A_sF_y,\ 0.85\phi_cf'_cb_{eff}t)$; $a = V_h/(0.85\phi_cf'_cb_{eff})$; $M_r = V_h(d/2 + t - a/2)$ when the neutral axis is in the slab; $q_r = 0.5\phi_{sc}A_{sc}\sqrt{f'_cE_c} \le \phi_{sc}A_{sc}F_u$; $n = V_h/q_r$ per half-span.

Why this works. Composite action moves the whole steel section into tension and puts the compression into concrete that is there anyway. The lever arm grows from $d/2 \approx 347$ mm for the bare beam to 515 mm, and the moment resistance nearly doubles for no extra steel. The price is the interface: every newton of the 6061 kN steel force has to cross from slab to beam through the studs, which is why part (b) is not an afterthought but the condition on which part (a) depends. The deflection check shows the same effect in service, where the transformed inertia is 3.4 times the bare-steel value — and it is the reason a 15 m span at 14 kPa is feasible at all with a 694 mm beam.

Common pitfall. Sizing the studs from the factored shear $V_f$ rather than from the interface force $V_h$. The two are unrelated quantities: $V_h$ is the axial force in the slab at mid-span, roughly eight times $V_f$ here, and using $V_f$ would give about a sixth of the studs actually needed. A second pitfall is forgetting the unshored construction case, in which the bare steel beam alone carries the wet concrete — the stage at which composite floors are most often overstressed.

Source. CSA S16 Cl. 17.3, 17.4, 17.7 and 17.9; Salmon, Johnson & Malhas, Ch. 16 (composite steel–concrete construction).

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