16-Civ-B2 Advanced Structural Design · May 2018

Question 4 of 7: Composite Steel-Concrete Bridge Floor

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

Notes on this paper

Paper format. 16-Civ-B2 Advanced Structural Design, May 2018 — a three-hour examination. The cover page allows design handbooks and textbooks but no notes, states that any five questions constitute a complete paper, that all seven questions carry equal value, and that all loads shown on the figures are unfactored. The mark split printed on page 1 is Q1 (12+6+2), Q2 (12+8), Q3 (12+4+4), Q4 (14+6), Q5 (10+5+5), Q6 (10+4+6) and Q7 (12+8). All seven questions are worked below, because the set is a study resource rather than an answer book.

Design data (page 1, reproduced verbatim). These values are used throughout and are not repeated in every question.

Design data supplied with the paper
Material	Property	Value
Concrete	Cylinder strength $f'_{c}$	30 MPa
Structural steel	$F_{y}$	350 MPa
Reinforcing steel	$f_{y}$	400 MPa
Prestressed concrete	$f_{ci}$ at transfer	35 MPa
	$f'_{c}$	50 MPa
	Modular ratio $n$	6
	$f_{ult}$	1750 MPa
	$f_{y}$ (strand)	1450 MPa
	$f_{initial}$	1200 MPa
	Loss of prestress	240 MPa

Reference texts. The answers are written to the Canadian limit-states codes that Engineers Canada lists for this examination:

CSA S16:19, Design of Steel Structures (with the CISC Handbook of Steel Construction, 11th ed.).
CSA A23.3:19, Design of Concrete Structures (with the Cement Association of Canada Concrete Design Handbook, 4th ed.).
National Building Code of Canada 2020, Part 4, for load combinations.
CSA S6:19, Canadian Highway Bridge Design Code, for the bridge floor of Question 4.
Kulak and Grondin, Limit States Design in Structural Steel, 9th ed.; MacGregor and Bartlett, Reinforced Concrete: Mechanics and Design (Canadian edition); Collins and Mitchell, Prestressed Concrete Structures.

Check — load factors. Page 1 says only that the loads shown are unfactored; it gives no dead-to-live split. Every question below therefore applies a single $\alpha = 1.5$ to the loads printed on the figures (the NBCC live-load factor, which is the conservative reading of an unidentified load) and $\alpha_{D} = 1.25$ to any self weight the solution itself introduces. Mechanisms, section classifications and all interaction equations are independent of that choice; only the magnitudes move with it. State this assumption on the answer paper, as the cover page invites.

Question 4: Composite Steel-Concrete Bridge Floor (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 14 m simple span, 12 m wide, with a 240 mm deck on steel beams at 3 m centres. Four spacings fit across the width, so the floor has five beams and an interior beam carries a 3 m tributary width.

Loading and materials
Item	Value
Span / width	14 m / 12 m
Deck slab	240 mm, $f'_{c} = 30$ MPa
Beam spacing	3.0 m (5 beams)
Deck dead load	$0.240 \times 24 = 5.76$ kPa, $w_{D} = 17.28$ kN/m
Live load	15 kPa, $w_{L} = 45.0$ kN/m
Steel	$F_{y} = 350$ MPa
Construction	unshored, 100 per cent interaction
Steel beam self weight	ignored, as the question directs

Find. The steel section for an interior beam and the number of headed shear studs needed to develop full interaction.

Approach. Because the construction is unshored, the bare steel beam carries the wet deck alone and the composite section carries only the live load; both stages must be checked. Take the effective slab width from Clause 17.4, locate the plastic neutral axis, and take $M_{r}$ from the plastic stress block. Part (b) follows from the horizontal shear that has to cross the interface between the point of maximum moment and the support.

Part (a) — factored actions on an interior beam. With $1.25D + 1.5L$ on a 3 m tributary width, $$w_{f} = 1.25(17.28) + 1.5(45.0) = 21.6 + 67.5 = 89.1\ \text{kN/m}$$ $$M_{f} = \frac{w_{f}L^{2}}{8} = \frac{89.1(14)^{2}}{8} = 2183\ \text{kN}\cdot\text{m}, \qquad V_{f} = \frac{w_{f}L}{2} = 623.7\ \text{kN}$$ For the construction stage the bare beam carries only the factored wet deck, $M_{f} = 1.25(17.28)(14)^{2}/8 = 529$ kN·m.
Effective slab width. Clause 17.4.1 limits the effective width of a simply supported interior beam to the smaller of a quarter of the span and the beam spacing: $$b_{e} = \min\!\left(\frac{14\,000}{4},\ 3000\right) = 3000\ \text{mm}$$ so the whole 3 m tributary slab is effective.
Trial section and the plastic neutral axis. Try W610×125, $A = 15\,793\ \text{mm}^{2}$. The steel can deliver $$T_{r} = \phi A F_{y} = 0.90(15\,793)(350) = 4975\ \text{kN}$$ while the slab in full compression could deliver $C_{r} = 0.85\phi_{c}f'_{c}b_{e}t_{s} = 0.85(0.65)(30)(3000)(240) = 11\,934$ kN. Steel governs, so the neutral axis lies inside the slab at $$a = \frac{T_{r}}{0.85\phi_{c}f'_{c}b_{e}} = \frac{4975 \times 10^{3}}{0.85(0.65)(30)(3000)} = 100\ \text{mm} \lt 240\ \text{mm}$$ The entire steel section is in tension — the classic, and most efficient, composite arrangement.
Composite moment resistance. The lever arm runs from the centroid of the steel section to the centroid of the compressed slab block: $$M_{r} = T_{r}\left(\frac{d}{2} + t_{s} - \frac{a}{2}\right) = 4975\left(\frac{612}{2} + 240 - 50\right)\times 10^{-3} = \boxed{2467\ \text{kN}\cdot\text{m}}$$ against $M_{f} = 2183$ kN·m, a utilisation of 0.885.
Construction stage and shear. On the bare steel, $Z_{x} = 3.635 \times 10^{6}\ \text{mm}^{3}$ gives $M_{r} = \phi Z_{x} F_{y} = 1145$ kN·m against 529 kN·m, a ratio of 0.46, so the beam carries the wet deck without shoring. For shear, $h/w = 48.1$ is below $439\sqrt{k_{v}/F_{y}} = 54.2$, so the web yields: $$V_{r} = \phi A_{w}(0.66F_{y}) = 0.90(612 \times 11.9)(231) = 1514\ \text{kN} \ge 623.7\ \text{kN}$$ a ratio of 0.41.
Serviceability, and the decision it forces. With $n = E/E_{c} = 200\,000/24\,648 = 8.11$, the transformed uncracked section has $I_{tr} = 3.833 \times 10^{9}\ \text{mm}^{4}$, so the live-load deflection is $$\Delta_{L} = \frac{5 w_{L} L^{4}}{384 E I_{tr}} = 29.4\ \text{mm} = \frac{L}{477}$$ That clears the ordinary floor criterion of $L/360 = 38.9$ mm, but it does not meet the $L/800 = 17.5$ mm that CSA S6 offers for vehicular bridge superstructures. Strength alone gives W610×125; if the bridge deflection criterion is enforced the beam must grow to W840×176, which returns 14.1 mm ($L/991$). The bare-steel deflection under the wet deck is 44 mm on W610×125 and should be cambered out either way.

Check — which section to ship. Part (a) asks for a section that carries the 15 kPa live load, so W610×125 is the answer to the question as set and is carried into part (b). The deflection calculation is recorded because a bridge floor is where $L/800$ normally applies, and a candidate who checks it and states the consequence earns the marks that a candidate who stops at strength does not.

Part (b) — horizontal shear to be transferred. For full interaction Clause 17.9.4 requires the connectors between the point of maximum moment and the support to carry the smaller of the two forces computed in Step 3: $$V_{h} = \min(T_{r},\, C_{r}) = \min(4975,\ 11\,934) = 4975\ \text{kN}$$ per half span.
Resistance of one stud. Take 19 mm diameter headed studs, $A_{sc} = 284\ \text{mm}^{2}$, $F_{u} = 450$ MPa, with $E_{c} = 4500\sqrt{30} = 24\,648$ MPa. Clause 17.7.2.2 gives $$q_{r} = 0.5\,\phi_{sc}A_{sc}\sqrt{f'_{c}E_{c}} = 0.5(0.80)(284)\sqrt{30 \times 24\,648} = 97.6\ \text{kN}$$ capped by $\phi_{sc}A_{sc}F_{u} = 102.2$ kN, so $q_{r} = 97.5$ kN governs.
Number and spacing of studs. $$n = \frac{V_{h}}{q_{r}} = \frac{4975}{97.5} = 51.0 \;\Rightarrow\; \boxed{52\ \text{studs per half span, 104 per beam}}$$ Arranged in transverse pairs that is 26 pairs in each half, at a uniform pitch of $7000/26 = 269$ mm, say 265 mm. The pitch satisfies the Clause 17.7 limits comfortably — minimum $6d = 114$ mm along the beam, transverse spacing $4d = 76$ mm on a 229 mm flange, and maximum spacing the lesser of $8t_{s} = 1920$ mm and 600 mm. Uniform spacing is permissible because the section is Class 1 and the connectors are ductile; a shear-proportional layout would bunch them near the supports to no advantage. Across the whole floor, five beams need 520 studs.

Composite deck-and-beam cross-section for the 14 m bridge floor.

Question 4 — final results
Quantity	Value
Design actions	$M_{f} = 2183$ kN·m, $V_{f} = 624$ kN
Construction-stage moment	529 kN·m on the bare steel
Effective slab width	3000 mm
Steel section (strength)	W610×125, $F_{y} = 350$ MPa
Depth of compression block	$a = 100$ mm, inside the slab
$M_{r}$ composite	2467 kN·m, ratio 0.885
$M_{r}$ bare steel	1145 kN·m, ratio 0.46
$V_{r}$	1514 kN, ratio 0.41
Live-load deflection	29.4 mm ($L/477$); W840×176 gives 14.1 mm ($L/991$) if $L/800$ is enforced
Shear connectors	19 mm studs, $q_{r} = 97.5$ kN
Number of studs	52 per half span, 104 per beam, 520 for the floor
Stud pitch	26 pairs per half span at 265 mm

Concept & Theory

Topic. Plastic design of an unshored composite floor beam and its shear connection.

Key relations. Effective width $b_{e} = \min(L/4,\ \text{spacing})$; $T_{r} = \phi A F_{y}$; $C_{r} = 0.85\phi_{c}f'_{c}b_{e}t_{s}$; $a = T_{r}/(0.85\phi_{c}f'_{c}b_{e})$; $M_{r} = T_{r}(d/2 + t_{s} - a/2)$ when the neutral axis is in the slab; $q_{r} = 0.5\phi_{sc}A_{sc}\sqrt{f'_{c}E_{c}} \le \phi_{sc}A_{sc}F_{u}$; $n = V_{h}/q_{r}$.

Why this works. Composite action works because the studs stop the slab and the beam sliding past one another, so the pair bends about a single neutral axis far above the steel centroid. The lever arm roughly doubles and the resistance rises with it, which is why the composite $M_{r}$ here is 2.2 times the bare-steel value for the same tonnage. Unshored construction splits the job in two: the wet deck is carried by the bare beam and locked into it as a permanent stress, while only the loads applied after the concrete has set enjoy the composite section. Ductile studs let the interface shear redistribute, so a uniform spacing is enough.

Common pitfall. Applying the full factored load to the composite section and forgetting that unshored construction leaves the wet-deck moment locked into the bare steel — a section can pass the composite check and still fail during the pour. A second trap is computing $V_{h}$ from $M_{f}$ instead of from the smaller of the two full-section forces; full interaction is defined by the section, not by the applied moment.

Source. CSA S16:19, Clause 17 (17.4 effective width, 17.7 connectors, 17.9 flexural resistance); CSA S6:19, Clause 3.4.4 for the deflection criterion; Kulak and Grondin, Chapter 9.

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