Unformatted text preview: Chapter Six
_____________________________________________________________________ Design of Reinforced Concrete Slabs 6.1 Introduction Fig. 6.11 Slabbeam floor plan 6.1.1 Principal slab types
• Oneway slab (Oneway spanning slab)
ly / l x > 2
ly
lx Loading distribution
of oneway slabs to
supporting beams Load on
beam AB A B Fig. 6.12 Oneway slab. (a) Loading transfer (ly – long side; lx = short side) 202 Fig. 6.12 (cont’d) (b) Oneway slabs on beam and girder • Twoway slab (Twoway spanning slab)
ly / lx ≤ 2
ly Load on
beam BC lx C Yield line A Load on
beam AB Loading distribution
of twoway slabs on
supporting beams B (a) (b)
Fig. 6.13 Twoway slab. (a) Loading transfer; (b) twoway slab and beam 203 • Flat slabs (without or with drops) (a) Flat slab without drops (flat plate) (b) Flat slab with drops
Fig. 6.14 Flat slabs • Staircases
• Others – such as ribbed floors, waffle slabs, etc. Fig. 6.15 Waffle slab 204 6.1.2 Preliminary sizing
• As for beams, the dimensions of slabs may generally be governed
by fireresistance 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 (oneway 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 yieldline 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. 6.16 Reinforcement detailing. (a) Simply supported oneway slab;
(b) continuous oneway slab The purposes of arranging the distribution reinforcement are for
Tying the slab together;
Distributing nonuniform loads through slabs;
Taking the possible bending moments in the long span.
6.11 206 • Design loads
o Design for the singleload 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. 6.17 Curtailment and anchorage of reinforcement in slabs. (a)
Simply supported span; (b) cantilever; (c) continuous beam
207 6.2 Oneway Solid Slabs (Solid slabs spanning in one direction) 6.2.1 Ultimate bending moments and shear forces for continuous slabs
Table 6.21 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 highyield 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. 6.21 Reinforcement of oneway slabs. (a) Part of floor plan; (b)
section AA; (c) section BB 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 spantodepth 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 highyield 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.187192). 2. Example: Continuous oneway slab
(1) Specification
A continuous oneway 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: selfweight + 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
crosssection.
(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 2.41 (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 ([email protected]) 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 ([email protected]) 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 onehalf 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 onehalf 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 ([email protected]) 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 ([email protected]) 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 [email protected] (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.166167). 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 crosssection of slab
A sketch of the crosssection with steel reinforcement is shown below. Distribution steel 215 ...
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 Spring '06
 JSKUANG
 Shear Stress, Rebar

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