Lecture_22

Lecture_22 - Chapter Six

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Unformatted text preview: Chapter Six _____________________________________________________________________ Design of Reinforced Concrete Slabs 6.1 Introduction Fig. 6.1-1 Slab-beam floor plan 6.1.1 Principal slab types • One-way slab (One-way spanning slab) ly / l x > 2 ly lx Loading distribution of one-way slabs to supporting beams Load on beam AB A B Fig. 6.1-2 One-way slab. (a) Loading transfer (ly – long side; lx = short side) 202 Fig. 6.1-2 (cont’d) (b) One-way slabs on beam and girder • Two-way slab (Two-way spanning slab) ly / lx ≤ 2 ly Load on beam BC lx C Yield line A Load on beam AB Loading distribution of two-way slabs on supporting beams B (a) (b) Fig. 6.1-3 Two-way slab. (a) Loading transfer; (b) two-way slab and beam 203 • Flat slabs (without or with drops) (a) Flat slab without drops (flat plate) (b) Flat slab with drops Fig. 6.1-4 Flat slabs • Staircases • Others – such as ribbed floors, waffle slabs, etc. Fig. 6.1-5 Waffle slab 204 6.1.2 Preliminary sizing • As for beams, the dimensions of slabs may generally be governed by fire-resistance requirements and/or the cover needed to ensure adequate resistance to corrosion (durability condition). • In practice, general minimum dimensions and cover to main reinforcement for the preliminary design of solid (one-way or twoway slabs) and ribbed slabs are as follows: h = 120 ~ 200 mm (normally take h = 150 mm in buildings); d = h – 35 mm 6.1.3 Methods of analysis • Elastic methods (from general structural analysis, finite element method, etc.) • Method of design coefficients (from design manuals) • The yield-line and Hillerborg strip methods (plastic methods) 6.1.4 Design considerations • Concrete slabs behave primarily as flexural members and the design is similar to that for beams, though in general it is somewhat simpler because o The slab breadth is fixed and a unit value of 1 m is normally used in calculations. o Shear stresses are usually low, except where are heavy concentrated loads. 205 o Compression reinforcement is seldom needed. o Reinforcement detailing: main steel + distribution steel Fig. 6.1-6 Reinforcement detailing. (a) Simply supported one-way slab; (b) continuous one-way slab The purposes of arranging the distribution reinforcement are for Tying the slab together; Distributing non-uniform loads through slabs; Taking the possible bending moments in the long span. 6.1-1 206 • Design loads o Design for the single-load case of maximum design load on all spans or panels: 1.4G k + 1.6Qk ; o All support moments (except at a cantilever) should be reduced 20% and span moments increased accordingly. • Curtailment and anchorage Fig. 6.1-7 Curtailment and anchorage of reinforcement in slabs. (a) Simply supported span; (b) cantilever; (c) continuous beam 207 6.2 One-way Solid Slabs (Solid slabs spanning in one direction) 6.2.1 Ultimate bending moments and shear forces for continuous slabs Table 6.2-1 Bending moments and shear forces for continuous slabs (the coefficients in the table include an allowance for 20% reduction in support moments due to the effects of moment redistribution) 6.2.2 Moment reinforcement and distribution reinforcement • Main moment steel ρmin = 0.13% (for high-yield steel), 0.24% (for mild steel) • Distribution steel o Area ≥ the minimum area of main reinforcement (ρmin) o Bar spacing is normally not greater than 300 mm in practice. 6.2.3 Shear reinforcement • Under normal loading conditions, shear stresses are not critical and shear reinforcement is in general not required. • It is required that o v ≤ vc for a slab thickness less than 200 mm; o If v ≥ vc , shear reinforcement must be provided in heavily thick slabs, but should not be used in slabs less than 200 mm thick. 208 6.2.4 Reinforcement detailing (a) Fig. 6.2-1 Reinforcement of one-way slabs. (a) Part of floor plan; (b) section A-A; (c) section B-B 6.2.5 Concrete cover • In practice, for grade 30 concrete the cover is generally 25 mm for mild exposure. 209 6.2.6 Deflection and crack control • Deflection control Checking the span-to-depth ratio of slabs only • Crack control o No case should the clear spacing of main steel > the lesser of 3d and 750 mm. o No further check is needed in the following normal cases: 1. If mild steel is used and d ≤ 250 mm; 2. If high-yield steel is used and d ≤ 200 mm; or 3. If the reinforcement ratio As /(bd) < 0.3%. ■ Examples 1. Text Examples 8.3 (simply supported slab) and 8.4 (continuous slab) (pp.187-192). 2. Example: Continuous one-way slab (1) Specification A continuous one-way slab has three equal spans 3.5 m each. The slab depth is assumed to be 140 mm. The loads on the slabs are as follows. Dead loads: self-weight + screed + finish + partitions + ceiling = 5.2 kN/m2 Imposed load: 3.0 kN/m2 210 The construction materials are grade 30 concrete and grade 460 reinforcement. The condition of exposure is mild and the cover required is 25 mm. Design the slab and show the reinforcement on a sketch of the cross-section. (2) Design load Consider a strip 1 m wide. Design load = (1.4 × 5.2) + (1.6 × 3) = 12 .08 kN/m Design load per span = 12.08 × 3.5 = 42.28 kN The single load case of maximum design loads on all spans is shown in the following figure. (3) Bending moments and shear forces in slabs The bending moments shear forces in slab are calculated using BS8110: Part 1, Table 3.12 (the table in Notes p.208). The values are shown in the following table. The moment redistribution is 20%. Design moments and shears (Assuming that end supports are simply supported) Supports Moment (kNm/m) Shear force (kN/m) A P B Q 0 +0.086×42.28×3.5 = 12.73 −0.086×42.28×3.5 = −12.73 +0.063×42.28×3.5 = 9.32 0.46 × 42.28 = 16.91 0 0.60 × 42.28 = 25.37 0 211 (4) Design of moment steel Assume 10 mm diameter bars with 25 mm cover. The effective depth d = 140 – 25 – 5 = 110 mm The calculations for steel areas are set out below. i) First interior support B: MB = 12.73 kNm/m; when Eq. (2.41) or Table 2.4-1 (Notes p.75) is used with βb = 0.8, K ' = 0.402 × 0.4 − 0.18 × 0.4 2 = 0.132 K= 12.73 × 106 = 0.035 < K ' 1000 × 1102 × 30 ( z = 110 0.5 + 0.25 − 0.035 / 0.9 ) = 105.5 mm ≈ 0.95d = 104 mm 12.73 × 106 As = = 304 mm 2 /m 0.95 × 460 × 105 Provide 8 mm bars at 160 mm centres (T8@160) to give an area of 314 mm2/m. Provide the same reinforcement at the end span P. ii) Interior span Q: MQ = 9.32 kNm/m. From i), z = 0.95d = 104 mm As = 223 mm2/m Provide 8 mm bars at 220 mm centres (T8@220) to give an area of 228 mm2/m. 212 iii) The moment reinforcement is shown in the following figure. Curtailment of bars has not been made because one-half of the calculated steel areas would fall below the minimum area of steel permitted. The minimum area of reinforcement As,min = 0.13 × 1000 × 140/1000 = 182 mm2/m At the end support A, the top steel is equal in area to one-half the midspan steel, i.e., 152 mm2/m, but not less than the minimum area of 182 mm2/m has to be provided. Provide 8 mm bars at 250 mm centres (T8@250) to give 201 mm2/m, as shown in the figure above. The tension bars in the bottom of support A are stopped off at the line of support. (5) Distribution steel The minimum area of reinforcement (182 mm2/m) has to be provided. The clear spacing between bars is not to exceed 3d = 330 mm. Provide 8 mm bars at 250 mm centres (T8@250) to give 201 mm2/m, which is same as that provided at the end support A. 213 (6) Shear resistance Enhancement in design strength close to the support has not been taken into account. i) End Support. The shear resistance is based on the top bars T8@250 (As = 201 mm2/m). 100 As 100 × 201 = = 0.183 bd 1000 × 110 vc = 0.79 3 0.183 4 v= 400 3 30 /1.25 = 0.53 N / mm 2 100 25 16.91 × 103 = 0.15 N / mm 2 1000 × 110 v < vc; thus no shear reinforcement is required. ii) Interior Support. 100 As 100 × 314 = = 0.285 1000 ×110 bd 25.37 × 10 3 2 vc = 0.61 N / mm , v = = 0.23 N / mm 2 1000 ×110 v < vc; thus no shear reinforcement is required. (7) Deflection The slab is checked for deflection using the rules from Clause 3.4.6 of the code (Notes pp.166-167). The end span is checked. The basic span/effective depth ration is 26 for the continuous slab. M/bd2 = 1.05 fs = 5 × 460 × 304.4 = 278.7 N / mm 2 8 × 314 214 The modification factor is 0.55 + 477 − 278.7 = 1.39 120(0.9 + 1.05) Allowable span/depth ratio = 1.39 × 26 = 36.1 Actual span/depth ratio = 3500/110 = 31.8 < 36.1 The slab is therefore satisfactory with respect to deflection. (8) Crack control Because the steel grade is 460, the slab depth is less than 200 mm and the clear spacing does not exceed 3d = 330 mm, the slab is satisfactory with respect to cracking. Refer to BS 8110: Part 1, clause 3.12.11.2.7. (9) Sketch of cross-section of slab A sketch of the cross-section with steel reinforcement is shown below. Distribution steel 215 ...
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This note was uploaded on 12/24/2011 for the course CIVL 232 taught by Professor Jskuang during the Spring '06 term at HKUST.

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