16-Civ-B2 Advanced Structural Design · December 2013
Question 7 of 7: Composite steel–concrete floor system
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Examinations, December 2013 — 98-Civ-B2 Advanced Structural Design, 3 hours, closed book (handbooks and textbooks permitted). Seven design questions of equal value; any five constitute a complete paper.
Reference texts. CSA S16:19 Design of Steel Structures; CISC Handbook of Steel Construction; CSA A23.3:19 Design of Concrete Structures (Clause 18 for prestressed concrete, Clause 17 of S16 for composite members); National Building Code of Canada 2020 for load combinations; Hibbeler, Structural Analysis, for the frame and continuous-beam analysis.
Design data printed on page 1. 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, $f_c' = 50\ \text{MPa}$, modular ratio $n = 6$, $f_{ult} = 1750\ \text{MPa}$, $f_y = 1450\ \text{MPa}$, $f_{initial} = 1200\ \text{MPa}$, losses $= 240\ \text{MPa}$. Design in SI. All loads shown are unfactored.
Check: load factors. Page 1 states only that the printed loads are unfactored; it gives no dead/live split, so no code load combination can be assembled uniquely. Every load printed on Figures 1–4 is therefore factored by 1.5, and any self weight the solver itself introduces (the concrete frame, the prestressed girder, the steel sections) by 1.25, consistent with NBCC 2020 Table 4.1.3.2. State this assumption on the answer paper as page-1 Note 1 invites. The choice scales the magnitudes only — the collapse mechanisms, section classifications and interaction equations below are unaffected.
Check: section properties. Rolled-shape properties cannot be reproduced from first principles, so every steel member selected here is a welded three-plate I-section whose area, $I$, $Z$, $J$ and $C_w$ are computed from its plate dimensions and are therefore fully checkable. A candidate with the CISC Handbook would instead take the lightest rolled W-shape whose $Z_x$ (or $M_r$, $C_r$) exceeds the value derived in each question; the design logic is identical.
Question 7: Composite steel–concrete floor system (15 + 5 marks)
220 mm R.C. slab on four beams at 3.5 m centres, 10.5 m overall
Span
15 m, simply supported, one-way
Design live load
$14\ \text{kPa}$
Materials
$f_c' = 30\ \text{MPa}$, $F_y = 350\ \text{MPa}$
Interaction
complete (full shear connection)
Trial beam
welded I, web $606 \times 12$, flanges $260 \times 22$
Find. The composite moment resistance of a typical interior beam against the factored demand, the web shear check, the number and spacing of shear studs, and the live-load deflection.
Figure 4. A typical interior beam carries a 3.5 m tributary width.
Approach. Take a typical interior beam with a 3.5 m tributary width, compute the factored load and moment, locate the plastic neutral axis, form the composite moment resistance, then size the connectors from the total horizontal shear and check deflection on the transformed section.
Part (a) — loads on a typical interior beam. The slab weighs $0.220(24) = 5.28\ \text{kPa}$, so over a 3.5 m tributary width $18.48\ \text{kN/m}$; the steel beam adds $1.44\ \text{kN/m}$. The live load is $14(3.5) = 49.0\ \text{kN/m}$. Hence$$w_f = 1.25(18.48 + 1.44) + 1.5(49.0) = 98.4\ \text{kN/m}$$$$\boxed{M_f = \frac{w_fL^2}{8} = 2767.5\ \text{kN}\cdot\text{m}, \qquad V_f = \frac{w_fL}{2} = 738.0\ \text{kN}}$$Note how completely the live load dominates: $14\ \text{kPa}$ is a heavy industrial loading and contributes three quarters of the factored moment.
Effective slab width. S16 Clause 17.4.1 limits the effective flange to the least of one quarter of the span and the beam spacing:$$b_{eff} = \min\left(\frac{15\,000}{4},\ 3500\right) = 3500\ \text{mm}$$The beam spacing governs, so the entire tributary slab is effective — a consequence of the generous 15 m span relative to the 3.5 m spacing.
Plastic neutral axis and moment resistance. The two candidate forces are the steel in tension and the slab in compression:$$T_r = \phi A_sF_y = 0.9(18\,712)(350) = 5894.3\ \text{kN}, \qquad C_{r,slab} = 0.85\phi_cf_c'b_{eff}t = 12\,762.8\ \text{kN}$$Since $T_r < C_{r,slab}$ the neutral axis falls inside the slab and the steel yields throughout — the efficient case. The depth of the stress block is$$a = \frac{T_r}{0.85\phi_cf_c'b_{eff}} = 101.6\ \text{mm} < 220\ \text{mm}\ \checkmark$$and the lever arm from the steel centroid to the concrete resultant is $d/2 + (t - a/2) = 325 + 169.2 = 494.2\ \text{mm}$, so$$\boxed{M_r = T_r\left(\frac{d}{2} + t - \frac{a}{2}\right) = 2912.9\ \text{kN}\cdot\text{m} > M_f = 2767.5\ \text{kN}\cdot\text{m}}$$
Composite section of a typical interior beam.
Web shear. The steel section alone carries the shear. With $h/w = 606/12 = 50.5$ and $k_v = 5.34$ for an unstiffened web, $439\sqrt{k_v/F_y} = 54.2 > 50.5$, so the web reaches its shear yield plateau and $F_s = 0.66F_y$:$$V_r = \phi h w (0.66F_y) = 0.9(606)(12)(231) = 1511.9\ \text{kN} > V_f = 738.0\ \text{kN}$$No transverse stiffeners are needed, which is the practical reward for keeping the web just under the 54.2 threshold.
Part (b) — total horizontal shear. For full interaction, the connectors between the point of zero moment and the point of maximum moment must transfer the smaller of the two flange capacities, computed unfactored per Clause 17.9.4:$$V_h = \min\left(A_sF_y,\ 0.85f_c'b_{eff}t\right) = \min(6549.2,\ 19\,635) = 6549.2\ \text{kN}$$This is the force that must cross the steel–concrete interface over each half span.
Stud resistance and count. For a 19 mm headed stud ($A_{sc} = 283.5\ \text{mm}^2$, $F_u = 450\ \text{MPa}$) with $E_c = 4500\sqrt{30} = 24\,648\ \text{MPa}$ and $\phi_{sc} = 0.80$,$$q_r = \phi_{sc}\min\left(0.5A_{sc}\sqrt{f_c'E_c},\ A_{sc}F_u\right) = 0.80\min(121.9,\ 127.6) = 97.5\ \text{kN}$$so the number required in each half span is$$\boxed{n = \frac{V_h}{q_r} = \frac{6549.2}{97.5} = 67.2 \rightarrow 68\ \text{studs per half span}, \ 136\ \text{in total}}$$Placed two per row, that is 34 rows over 7.5 m, i.e. pairs of 19 mm studs at 220 mm centres, uniform along the span since the load is uniform and Clause 17.9.7 permits uniform spacing.
Deflection under live load. Transforming the slab by $n = E_s/E_c = 8.11$ gives an effective concrete width of $3500/8.11 = 431.4\ \text{mm}$; the composite elastic centroid then lies $688.4\ \text{mm}$ above the bottom of the steel and$$I_{comp} = 4692\times10^6\ \text{mm}^4 \quad(\text{3.5 times the bare steel value})$$$$\Delta_L = \frac{5w_LL^4}{384EI_{comp}} = 34.4\ \text{mm} < \frac{L}{360} = 41.7\ \text{mm}\ \checkmark$$The margin is only 17 %, confirming that on a 15 m composite floor with a 14 kPa live load, deflection rather than strength sets the practical beam depth. Cambering the beam for the wet-concrete dead load is the usual refinement.
