07-Str-A5 · May 2014

Question 7 of 7: Composite steel–concrete pedestrian bridge

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, May 2014 — 3 hours, closed book (handbooks and textbooks permitted, no notes on them; Casio or Sharp approved calculator). Seven questions of equal value; any five constitute a complete paper and only the first five presented are marked. 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, $f_{pi} = 1200$ MPa, losses $= 240$ MPa, hence $f_{pe} = 1200 - 240 = 960$ MPa.

Reference texts. CSA S16 Design of Steel Structures with the CISC Handbook of Steel Construction — plastic design (Cl. 8.6, 13.7), lateral–torsional buckling (Cl. 13.6), beam-columns (Cl. 13.8), plate girders (Cl. 14), composite beams (Cl. 17), welded connections (Cl. 13.13, 21); CSA A23.3 Design of Concrete Structures — flexure and shear (Cl. 10, 11), slenderness (Cl. 10.13–10.16), deflection (Cl. 9.8), 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 taken from the trial section and factored at $1.25$. A candidate assuming a different split obtains proportionally different sizes; the method is what is examined.

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

Check: readings taken from the examination drawing. Three details are shown on the drawing but not stated in the question text. In Figure 1 the bases at A and D carry the fixed-support hatching and the question text says "the frame is fixed at the bases A and D", so both bases are fixed and the frame is three times statically indeterminate. In Figure 4 the 100 kN arrow at D points towards the column, i.e. to the left; A and E both carry pin symbols, making that frame indeterminate to the first degree. In Figure 5 the deck is 3 + 3 = 6 m wide over three beams spaced at 2.5 m, leaving 0.5 m overhangs — the slab-thickness annotation is not legible on the printed figure, so the deck thickness is designed in Question 7 rather than read off.

[Figure not reproduced: Figure 1 as printed on the examination paper: rectangular steel portal, fixed at A and D, beam capacity 1.5 M₊ and column capacity M₊. All loads shown are unfactored. See the official exam paper.]

Question 7: Composite steel–concrete pedestrian bridge (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. Simply supported span 18 m. From Figure 5 the deck is $3 + 3 = 6$ m wide, carried on three steel beams at 2.5 m centres with 0.5 m overhangs and a railing at each edge. Live load 15 kPa; $f'_c = 30$ MPa, $F_y = 350$ MPa. Construction is unshored, so the bare steel carries the wet concrete.

Find. The deck thickness and steel section for flexure with full interaction, and the number of stud connectors.

Question 7: 6 m pedestrian deck on three welded beams at 2.5 m with 0.5 m overhangs; 18 m simple span, unshored.

Check: deck thickness. A 175 mm slab is adopted here, which is the usual minimum for a bridge deck spanning 2.5 m between girders and is thick enough to keep the plastic neutral axis inside the slab; part (a) therefore designs the deck rather than reading it off the drawing.

Approach. Design the interior beam, which has the larger tributary width. Take the loads on a per-beam basis, check the bare steel section during construction, then compute the composite plastic moment resistance with the neutral axis in the slab, verify deflections, and finally size the connectors from the smaller of the two horizontal-shear resistances.

Part (a) — loads on an interior beam. Tributary width 2.5 m (the exterior beams take only $0.5 + 1.25 = 1.75$ m, so the interior beam governs). Effective flange width is $b_e = \min(L/4,\ \text{spacing}) = \min(4500,\ 2500) = 2500$ mm.

Component	Intensity	Per beam
175 mm deck slab	4.2 kPa	10.5 kN/m
steel beam	—	1.54 kN/m
railings, surfacing allowance	0.2 kPa	0.50 kN/m
dead total $D$	—	12.54 kN/m
live load $L$	15 kPa	37.5 kN/m
factored $1.25D + 1.5L$	—	71.9 kN/m

Design actions. $$M_f = \frac{w_fL^2}{8} = \frac{71.9(18)^2}{8} = 2913\ \text{kN}\cdot\text{m}, \qquad V_f = \frac{w_fL}{2} = 647\ \text{kN}$$
Trial steel section. Take a welded I with flanges $300 \times 20$ and web $800 \times 10$, so $d = 840$ mm, $A_s = 20\,000$ mm², $Z = 6520\times10^3$ mm³ and $I_s = 2444\times10^6$ mm⁴. The flange, $b/t = 7.25 \le 7.75$, is Class 1; the web, $h/w = 80 \le 1700/\sqrt{F_y} = 90.9$, is Class 2, so the full plastic moment may be used.
Locate the plastic neutral axis. The steel in tension can deliver $$T_r = \phi A_sF_y = 0.9(20\,000)(350) = 6300\ \text{kN}$$ while a slab depth $a$ delivers $\alpha_1\phi_cf'_cb_ea = 0.805(0.65)(30)(2500)a = 39.24a$ N. Equating, $$a = \frac{6300\times10^3}{39\,244} = 161\ \text{mm} \;<\; t_s = 175\ \text{mm}$$ so the neutral axis lies inside the slab and the whole steel section yields in tension — the most efficient composite arrangement.
Composite moment resistance. The steel force acts at the mid-depth of the section, 420 mm above its soffit; the concrete force acts $a/2 = 80$ mm below the top of the slab, i.e. $840 + 175 - 80 = 935$ mm above the same datum: $$\boxed{M_{rc} = T_r\left(d + t_s - \frac{a}{2} - \frac{d}{2}\right) = 6300(0.515) = 3243\ \text{kN}\cdot\text{m} \ \ge\ M_f = 2913\ \text{kN}\cdot\text{m}}$$
Construction stage — the check unshored construction forces. Before the deck hardens the bare steel carries the wet concrete and its own weight plus a 1 kPa construction live load: $$w = 1.25(10.5 + 1.54) + 1.5(2.5) = 18.8\ \text{kN/m}, \qquad M = \frac{18.8(18)^2}{8} = 761\ \text{kN}\cdot\text{m}$$ With temporary bracing every 3 m, Cl. 13.6 returns $M_u = 8273$ kN·m, so the bare section develops its full $M_r = \phi ZF_y = 2054$ kN·m, comfortably above 761 kN·m. Its deflection under slab plus steel weight is $$\Delta = \frac{5wL^4}{384E I_s} = 33.7\ \text{mm} = \frac{L}{535}$$ which should be cambered out so that the finished deck is level and of uniform thickness.
Live-load deflection on the composite section. With $n = E_s/E_c = 8.11$ the slab transforms to a width of $2500/8.11 = 308$ mm. The transformed centroid lies 790 mm above the steel soffit and $$I_{tr} = 6339\times10^6\ \text{mm}^4, \qquad \Delta_L = \frac{5w_LL^4}{384EI_{tr}} = 40.4\ \text{mm} = \frac{L}{445}$$ inside the $L/400$ limit customary for pedestrian bridges. On a footbridge the vibration check would normally be run as well, since a 40 mm live deflection implies a fundamental frequency near the pedestrian pacing range.
Part (b) — horizontal shear to be transferred. For full interaction the connectors between the point of maximum moment and the support must carry the smaller of the two forces the interface can develop: $$V_h = \min\left(\phi A_sF_y,\ \alpha_1\phi_cf'_cb_et_s\right) = \min(6300,\ 6868) = 6300\ \text{kN}$$
Stud resistance. For 19 mm diameter headed studs, $A_{sc} = 283.5$ mm², $F_u = 450$ MPa and $\phi_{sc} = 0.80$: $$q_r = \min\left(0.5\phi_{sc}A_{sc}\sqrt{f'_cE_c},\ \phi_{sc}A_{sc}F_u\right) = \min(97.5,\ 102.1) = 97.5\ \text{kN}$$ Concrete crushing around the shank governs, as it usually does with normal-density 30 MPa concrete.
Number and layout. $$N = \frac{V_h}{q_r} = \frac{6300}{97.5} = 64.6 \Rightarrow 65\ \text{studs per half span}$$ $$\boxed{\text{2 studs per row at 265 mm} \Rightarrow 34\ \text{rows} = 68\ \text{studs per half span, 136 per beam}}$$ The pitch satisfies the limits $6d = 114$ mm minimum and $\min(8t_s,\,900) = 900$ mm maximum, and the transverse gauge of 100 mm exceeds the $4d = 76$ mm minimum. Because the moment diagram is parabolic and the shear flow is largest near the supports, the uniform spacing is conservative at mid-span and exact overall — which is what CSA S16 Cl. 17.9.5 permits for a beam with no concentrated loads.

Quantity	Result
Deck slab	175 mm, $b_e = 2500$ mm for an interior beam
Factored actions	$M_f = 2913$ kN·m, $V_f = 647$ kN
(a) Steel section	welded I, flanges $300 \times 20$, web $800 \times 10$, $d = 840$ mm
(a) Plastic neutral axis	$a = 161$ mm, inside the 175 mm slab
(a) Composite resistance	$M_{rc} = 3243$ kN·m
Construction stage	$M = 761$ kN·m vs $M_r = 2054$ kN·m; camber 34 mm
Live-load deflection	40.4 mm $= L/445$
(b) Horizontal shear	$V_h = 6300$ kN; $q_r = 97.5$ kN per 19 mm stud
(b) Connectors	2 × 19 mm studs per row at 265 mm — 68 per half span, 136 per beam

Topic. Composite steel–concrete beams with full shear interaction, unshored construction, and stud connector design.

Key relations. $b_e = \min(L/4,\ \text{beam spacing})$; plastic neutral axis from $\phi A_sF_y = \alpha_1\phi_cf'_cb_ea$; $M_{rc} = T_r \times$ lever arm; CSA S16 Cl. 17.7.2.1 $q_r = 0.5\phi_{sc}A_{sc}\sqrt{f'_cE_c} \le \phi_{sc}A_{sc}F_u$; $N = V_h/q_r$ per half span; transformed section with $n = E_s/E_c$ for deflection.

Why this works. Composite action moves the neutral axis up into the slab, so almost the entire steel section works in tension at its yield stress while the concrete, which is cheap and strong in compression, supplies the compression block. The lever arm becomes roughly the distance from the middle of the steel to the middle of the compression block — here 515 mm on an 840 mm beam — which is why the composite resistance, 3243 kN·m, is 58 % above the bare steel's 2054 kN·m for no extra steel at all. Unshored construction splits the problem into two states: the bare steel alone resists everything applied before the concrete gains strength, and only the loads applied afterwards are carried compositely. For strength this distinction can be ignored at the ultimate limit state, because a Class 1 or 2 section redistributes internally as it plastifies, but for deflection it cannot — the 34 mm of dead-load sag is locked in and must be cambered out.

Common pitfall. Reading the 2.5 m dimensions on Figure 5 as slab overhangs rather than as beam spacing, which turns a 6 m deck on three girders into an 11 m deck on two and roughly doubles the design moment. The second is computing the connector count from the factored shear rather than from the horizontal force at the interface: $V_h$ is set by what the section can develop, $\min(\phi A_sF_y,\ \alpha_1\phi_cf'_cb_et_s)$, not by $V_f$.

Source. CSA S16 Cl. 17 (composite beams) and Cl. 13.6; CHBDC CSA S6 Cl. 3.8 for pedestrian loading and deflection limits; Salmon, Johnson & Malhas, Steel Structures, Ch. 16.

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