07-Str-A5 · December 2015

Question 3 of 7: Composite steel–concrete warehouse floor

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

Notes on this paper

Paper format. National Exams, December 2015 — 07-Str-A5 Advanced Structural Design. Three hours, "closed book" (textbooks and design handbooks permitted, no notes). Seven questions; any five constitute a complete paper and all are of equal value, the printed mark split being 20 marks each. All loads shown on Figures 1–4 are unfactored. All seven questions are answered here.

Design data printed on page 1 and used throughout. Concrete $f'_c = 30\ \text{MPa}$; structural steel $F_y = 350\ \text{MPa}$; reinforcing steel $f_y = 400\ \text{MPa}$; prestressed concrete $f'_{ci} = 35\ \text{MPa}$ at transfer and $f'_c = 50\ \text{MPa}$ in service, $n = 6$; $f_{pu} = 1750\ \text{MPa}$, $f_{py} = 1450\ \text{MPa}$, $f_{p,\text{initial}} = 1200\ \text{MPa}$, losses $= 240\ \text{MPa}$, hence $f_{pe} = 960\ \text{MPa}$ and the loss ratio $\eta = 0.80$.

Check: load factors and figure data. The paper states only that "all loads shown are unfactored" and gives no load classification, so every printed load is treated as a specified live load and factored at 1.5, while member and element self weight is treated as dead and factored at 1.25 (NBCC 2020 Case 2, $1.25D + 1.5L$). Resistance factors are $\phi = 0.90$ (steel), $\phi_w = 0.67$ with the additional 0.67 of CSA S16 Cl. 13.13.2.2 (welds), $\phi_{sc} = 0.80$ (shear studs), $\phi_c = 0.65$, $\phi_s = 0.85$, $\phi_p = 0.90$ (CSA A23.3). Figure 2 has a free tip at A, a roller at B and a pin at C; Figure 4 has a hatched fixed base at A and a hatched fixed end at C, with the 200 kN arrow pointing towards the frame at mid-height of the column.

Reference texts for 07-Str-A5.

CSA S16:19, Design of Steel Structures, with the CISC Handbook of Steel Construction (11th ed.) — Cl. 13.3 (compression), 13.4 (shear), 13.5–13.6 (flexure and lateral–torsional buckling), 13.8 (beam–columns), 13.13 (welds), 14.3–14.6 (plate girders), Cl. 17 (composite beams).
CSA A23.3:19, Design of Concrete Structures, with the CAC Concrete Design Handbook (4th ed.) — Cl. 9 (deflections), Cl. 10 (flexure, columns, slenderness), Cl. 11 (shear), Cl. 13 (two-way shear), Cl. 18 (prestressed concrete).
Salmon, Johnson and Malhas, Steel Structures: Design and Behavior, 5th ed. — plastic analysis (Ch. 10), plate girders (Ch. 11), composite construction (Ch. 16).
MacGregor and Bartlett, Reinforced Concrete: Mechanics and Design, Canadian edition — frames, beam–columns, slender columns, detailing.
Collins and Mitchell, Prestressed Concrete Structures — cable zone, transfer and service stress checks, harped profiles.
National Building Code of Canada 2020 for load combinations and deflection limits.

Question 3: Composite steel–concrete warehouse floor (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. From Figure 3, a 9.0 m wide floor carried on four steel beams at 2.5 m centres with 0.75 m overhangs each side ($0.75 + 3 \times 2.5 + 0.75 = 9.0$ m), a reinforced-concrete slab on top, a design span of 14 m and a design live load of 16 kPa.

Given data (Figure 3 and its note)
Quantity	Value
Floor width / beam spacing / overhang	9.0 m / 2.5 m / 0.75 m
Design span	14.0 m
Design live load	16 kPa
Slab (adopted)	150 mm, $f'_c = 30\ \text{MPa}$, $n = 6$
Construction	Unshored, 100 % interaction

Find. (a) the slab thickness, the effective flange width and the steel section for the governing interior beam; (b) the number of shear connectors.

Figure 3 floor cross-section, and the composite design section of one interior beam with its 2500 mm effective slab width.

Approach. Design the interior beam, whose 2.5 m tributary width governs over the 2.0 m of the edge beams; check the bare steel section for the wet-concrete construction stage because the floor is unshored, then check the composite section at ultimate and the composite stiffness for live-load deflection; finally transfer the full horizontal shear with studs.

Slab and loads on the interior beam. A 150 mm slab spanning 2.5 m continuously needs $A_s \approx 455\ \text{mm}^2/\text{m}$ at $M_f \approx w_fL^2/10 = 17.8\ \text{kN}\cdot\text{m/m}$; use 15M at 300 mm each way. For the beam, with a trial welded section of web $550 \times 10$ and flanges $250 \times 20$ ($A_s = 15\,500\ \text{mm}^2$, $d = 590\ \text{mm}$, 1.19 kN/m), $$w_D = 0.150(24)(2.5) + 1.19 = 10.19\ \text{kN/m}, \qquad w_L = 16(2.5) = 40.0\ \text{kN/m}$$ $$w_f = 1.25(10.19) + 1.5(40.0) = 72.74\ \text{kN/m} \;\Rightarrow\; M_f = \frac{w_fL^2}{8} = \boxed{1782\ \text{kN}\cdot\text{m}}, \quad V_f = 509\ \text{kN}$$
Construction stage (unshored). Before the slab hardens the bare steel beam alone carries the wet concrete and its own weight, $M_f = 1.25(10.19)(14)^2/8 = 312\ \text{kN}\cdot\text{m}$. The section is Class 1 ($b/t = 6.0 < 7.75$, $h/w = 55.0 < 58.8$), so $\phi Z F_y = 1136\ \text{kN}\cdot\text{m}$ — ample. This check is what makes unshored construction legitimate; it is not a formality, because on deeper spans it can size the beam.
(a) Effective width and the composite plastic capacity. CSA S16 Cl. 17.4 gives $$b_e = \min\left(\frac{L}{4},\ \text{beam spacing}\right) = \min(3500,\ 2500) = \boxed{2500\ \text{mm}}$$ The steel can deliver $T = \phi A_s F_y = 4882.5\ \text{kN}$, while the slab could resist $0.85\phi_c f'_c b_e t = 6216\ \text{kN}$. Since $T$ is the smaller, the plastic neutral axis lies inside the slab at $$a = \frac{T}{0.85\phi_c f'_c b_e} = 117.8\ \text{mm} \;<\; 150\ \text{mm}$$ and the whole steel section is in tension, so the lever arm runs from its centroid to the centroid of the slab block: $$M_{rc} = T\left(\frac{d}{2} + t - \frac{a}{2}\right) = 4882.5\left(295 + 150 - 58.9\right)\times10^{-3} = \boxed{1885\ \text{kN}\cdot\text{m}} \;>\; 1782\ \text{kN}\cdot\text{m}$$
Shear and deflection. The web takes the shear alone: with $h/w = 55.0$, $F_s = 670\sqrt{F_y}/(h/w) = 227.9\ \text{MPa}$ and $V_r = \phi A_w F_s = 1128\ \text{kN} > 509\ \text{kN}$. For deflection, transform the slab with the paper's $n = 6$, giving $b_{tr} = 416.7\ \text{mm}$, an elastic neutral axis 148.5 mm below the top (just inside the slab) and $I_{tr} = 2.769 \times 10^{9}\ \text{mm}^4$. Then $$\Delta_{LL} = \frac{5w_LL^4}{384EI_{tr}} = 36.1\ \text{mm} \;<\; \frac{L}{360} = 38.9\ \text{mm}$$ The dead-load deflection is carried by the bare steel, $26.8\ \text{mm}$, so camber the beams 25 mm.
(b) Horizontal shear to be transferred. For full interaction the connectors between the point of maximum moment and the nearer support must carry the smaller of the two capacities computed in step 3: $$V_h = \min\left(\phi A_sF_y,\ 0.85\phi_cf'_cb_et\right) = 4882.5\ \text{kN}$$
(b) Stud capacity and count. With $E_c = 4500\sqrt{30} = 24\,648\ \text{MPa}$ and 19 mm studs ($A_{sc} = 283.5\ \text{mm}^2$, $F_u = 450\ \text{MPa}$), CSA S16 Cl. 17.7.2.2 gives $$q_{rs} = 0.5\phi_{sc}A_{sc}\sqrt{f'_cE_c} = 97.5\ \text{kN} \;\le\; \phi_{sc}A_{sc}F_u = 102.2\ \text{kN}$$ so $q_{rs} = 97.5\ \text{kN}$ and $$n = \frac{V_h}{q_{rs}} = \frac{4882.5}{97.5} = 50.1 \;\Rightarrow\; \boxed{51\ \text{studs per half span}}$$ that is 102 studs per beam. Placed in pairs, 26 rows at 260 mm satisfy the Cl. 17.7.2.4 limits (minimum $4d = 76$ mm, maximum the lesser of 800 mm and $4t = 600$ mm).

Question 3 — final results
Item	Value
Slab	150 mm, 15M at 300 mm each way
Steel beam (interior, governs)	Welded I: web 550 × 10, flanges 250 × 20 (d = 590 mm)
Effective slab width $b_e$	2500 mm
$M_f$ / $M_{rc}$	1782 / 1885 kN·m (ratio 0.95)
Construction-stage $M_f$ / $\phi ZF_y$	312 / 1136 kN·m
$V_f$ / $V_r$	509 / 1128 kN
Live-load deflection	36.1 mm < L/360 = 38.9 mm; camber 25 mm
Horizontal shear $V_h$ / stud $q_{rs}$	4882.5 kN / 97.5 kN
Connectors	51 per half span = 102 per beam; 26 pairs at 260 mm

Topic. Unshored composite floor beams: effective width, plastic composite capacity and full-interaction connector design.

Key relations. $b_e = \min(L/4, \text{spacing})$; $a = \phi A_sF_y/(0.85\phi_cf'_cb_e)$; $M_{rc} = T(d/2 + t - a/2)$ for a neutral axis in the slab; $V_h = \min(\phi A_sF_y,\ 0.85\phi_cf'_cb_et)$; $q_{rs} = 0.5\phi_{sc}A_{sc}\sqrt{f'_cE_c} \le \phi_{sc}A_{sc}F_u$.

Why this works. Once studs weld the slab to the top flange, the two materials share one neutral axis and the slab supplies a compression flange far larger than any rolled shape could carry. The plastic capacity follows from the simplest possible statics: whichever of the slab or the steel runs out of force first sets the couple, and the lever arm is the distance between the two resultants. Unshored construction splits the job in time — the bare steel carries the wet concrete, and only the loads applied after the concrete gains strength are carried compositely — which is why the dead-load deflection is computed on $I_s$ and the live-load deflection on $I_{tr}$.

Common pitfall. Using the composite section for the construction stage, which silently assumes shoring the question forbids; and computing the live-load deflection on the gross steel section, which overstates it by a factor of about three here and can lead to a needlessly heavy beam. Watch also that the edge beam's tributary width (0.75 + 1.25 = 2.0 m) is smaller than the interior beam's, so the interior beam governs on this floor.

Source. CSA S16:19 Cl. 17.4, 17.7 and 17.9; Salmon, Johnson and Malhas, Steel Structures: Design and Behavior, 5th ed., Ch. 16.

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