Question 2 of 7: Composite Steel-Concrete Pedestrian Bridge
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. 98-Civ-B2 Advanced Structural Design, National Exams May 2016 — 3 hours, closed book (design handbooks permitted). Seven design questions of equal value; any five constitute a complete paper, and all seven are solved here.
Given. A simply supported two-girder composite deck carrying a heavy crowd load over an 18 m span.
Quantity
Value
Span, simply supported
L = 18 000 mm
Deck width / slab thickness
5000 mm / 220 mm
Beam spacing
4000 mm (tributary width 2500 mm each)
Live load
14 kPa over the full deck width
Concrete / steel
f'c = 30 MPa, Fy = 350 MPa
Interaction
100 per cent (full shear connection), as stated
Find. (a) an adequate rolled steel section for each girder acting compositely with the slab, and (b) the number of shear connectors needed to develop that composite action.
Figure 2 — the pedestrian-bridge cross-section. Each girder carries a 2.5 m strip of deck, which is also what caps the effective flange width used in the composite calculation.
Approach. Work on one girder and its 2.5 m strip of deck. Build the factored uniformly distributed load, size the section from the plastic composite resistance with the neutral axis located first, then check shear and live-load deflection; finally take the full horizontal shear between the point of zero moment and mid-span and divide by the resistance of one stud.
Part (a) — assemble the loading on one girder. Each girder carries a 2.5 m strip. The slab contributes $0.220\times24 = 5.28$ kPa, so $w_{D,\text{slab}} = 5.28\times2.5 = 13.20$ kN/m, and the trial W760 by 147 adds $1.42$ kN/m. The live load gives $w_{L} = 14.0\times2.5 = 35.0$ kN/m. Hence $$w_{f} = 1.25(13.20 + 1.42) + 1.5(35.0) = 18.28 + 52.50 = 70.78\ \text{kN/m}$$
Get the design actions. For a simply supported span, $$M_{f} = \frac{w_{f}L^{2}}{8} = \frac{70.78\times18^{2}}{8} = \boxed{2867\ \text{kN.m}}\qquad V_{f} = \frac{w_{f}L}{2} = 637\ \text{kN}$$ The live load alone is three quarters of the total, which is what makes the deflection check in step 6 the real constraint rather than an afterthought.
Fix the effective flange width. S16 Cl 17.4.1 takes the least of one quarter of the span, the centre-to-centre beam spacing, and the slab actually available. Here that is $\min(18\,000/4,\ 4000,\ 2500) = 2500$ mm — the physical 2.5 m strip governs, because a 5 m deck shared between two girders leaves nothing more to mobilise.
Locate the plastic neutral axis. Compare the tensile resistance of the whole steel section with the compressive resistance of the slab. For W760 by 147, $A_{s} = 18\,501$ mm$^{2}$ from its nominal plate dimensions (753 deep, 265 wide flanges 17.0 thick, 13.2 web), so $$T_{r} = \phi A_{s}F_{y} = 0.9\times18\,501\times350 = 5828\ \text{kN}$$ $$C_{r,\text{slab}} = 0.85\phi_{c}f'_{c}b_{e}t_{s} = 0.85\times0.65\times30\times2500\times220 = 9116\ \text{kN}$$ Since $T_{r} \lt C_{r,\text{slab}}$ the steel governs, the neutral axis lies inside the slab, and the depth of the stress block is $a = 5\,828\,000/(0.85\times0.65\times30\times2500) = 140.6$ mm, less than the 220 mm slab.
Compute the composite moment resistance. With the whole steel section yielding in tension at its own centroid, the lever arm runs from there to the centroid of the concrete block: $$M_{rc} = T_{r}\left(\frac{d}{2} + t_{s} - \frac{a}{2}\right) = 5828\left(376.5 + 220 - 70.3\right)\times10^{-3} = \boxed{3066\ \text{kN.m}}$$ so $M_{f}/M_{rc} = 2867/3066 = 0.935$, adequate. Shear is carried by the steel web alone: $h/w = 719/13.2 = 54.5$, so $F_{s}\approx0.66F_{y}$ and $V_{r} = 0.9\times(753\times13.2)\times231 = 2066$ kN, only 31 per cent used.
Check live-load deflection. With $E_{c} = 3300\sqrt{30} + 6900 = 24\,975$ MPa the modular ratio is $n = 8.01$, so the slab transforms to a width of $2500/8.01 = 312$ mm. Taking first moments about the steel soffit puts the composite centroid 760 mm up, and the transformed second moment of area is $I_{tr} = 5.356\times10^{9}$ mm$^{4}$. Then $$\Delta_{L} = \frac{5w_{L}L^{4}}{384EI_{tr}} = \frac{5\times35\times18\,000^{4}}{384\times200\,000\times5.356\times10^{9}} = 44.7\ \text{mm}$$ which is $L/403$, just inside the $L/400$ live-load limit of S16 Appendix D. A lighter section would satisfy strength but fail this, so deflection is what fixes the girder depth.
Part (b) — find the horizontal shear to be transferred. Full interaction means the connectors must carry the entire compressive force developed in the slab between the support and mid-span, which equals the governing resistance found in step 4: $V_{h} = \min(T_{r}, C_{r,\text{slab}}) = 5828$ kN.
Size one connector and count them. For a 19 mm diameter headed stud, $A_{sc} = \pi(19)^{2}/4 = 284$ mm$^{2}$ and S16 Cl 17.7.2.2 gives $$q_{r} = 0.5\phi_{sc}A_{sc}\sqrt{f'_{c}E_{c}} = 0.5(0.8)(284)\sqrt{30\times24\,975} = 98.2\ \text{kN}$$ capped by $\phi_{sc}A_{sc}F_{u} = 0.8\times284\times450 = 102$ kN, so $q_{r} = 98.2$ kN governs. The number required in each half span is $$n = \frac{V_{h}}{q_{r}} = \frac{5828}{98.2} = 59.4 \rightarrow \boxed{60\ \text{studs per half span, 120 per girder}}$$
Lay the studs out. Placing them in pairs across the 265 mm flange gives 30 rows over 9 m, a uniform pitch of 300 mm. That satisfies the longitudinal minimum of $4d = 76$ mm and the maximum of the lesser of $8t_{s} = 1760$ mm and 600 mm, and the transverse gauge of 100 mm clears the $4d$ minimum with edge distance to spare. Uniform spacing is acceptable because the section is compact and fully plastic at ultimate, so the connectors redistribute.
Full-interaction plastic stress block. The whole steel section yields in tension and the neutral axis falls inside the 220 mm slab, so the lever arm is measured from the steel centroid to the centroid of the concrete block.
Check — construction sequence. The question does not say whether the deck is shored. Checked both ways: unshored, the bare steel section carries the wet concrete, $M = (13.20+1.42)\times18^{2}/8 = 592$ kN.m against $\phi Z_{x}F_{y} = 1582$ kN.m for W760 by 147, so the girder is adequate either way and the composite result above stands.
