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reinforced concrete
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23/7/2013
1
LESSON 3
Flat Slab
Introduction
• A reinforced concrete slab supported directly by concrete columns without the use of intermediary beams.
• Constant thickness or may be thickened in the area of column drop panel.
• The column of constant section or may be flared column head or capital.
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Introduction(cont.)
• Drop panels effective in reducing the shearing stress why??
• Drop panels increase moment resistance.
• Advantages over solid slab: – Simplified formwork.
– Reduce storey height more economical.
– Windows can extend underside the slab.
– No beams to obstruct the light and the circulation of air.
Introduction(cont.)
• Deflection requirements govern slab thickness normally not less than 180mm for fire resistance.
• The analysis divide the structure into a series of equivalent frames.
• Moment in these frames may be determined by:– Frame analysis moment distribution.
– Simplified method outlined in Table 3.12 BS8110
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Analysis
• The analysis divide the structure into a series of equivalent frames.
• Moment in these frames may be determined by:
– Frame analysis moment distribution.
– Simplified method outlined in Table 3.12 BS8110
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Analysis(cont.)
• Column and middle strips.
• Drop panels ignored smaller dimension < lx/3
• Moments from Table 3.12 are distributed between the strips such that the negative and positive moments resisted by the column and middle strips total 100% in each case.
Design
• Design as flexural members beam procedures.
• Important features: to calculate the punching shear at the head of columns and at the change in depth of the slab, if drop panels are used.
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Design(cont.)
• The design for shear should follow EC2 that requires the design shear force to be increased above the calculated value by 15% for internal columns, up to 40% for edge columns and 50% for corner columns to allow for moment transfer effects.
Design(cont.)
• Only apply to braced structures where adjacent spans do not differ by more than 25%.
• Punching shear additional requirement by EC2 on the amount and distribution of reinforcement around column heads to ensure that full punching shear capacity is developed.
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Design(cont.)
• The usual basic span‐effective depth ratios may be used but where the greater span exceeds 8.5m the basic ratio should be multiplied by 8.5/span.
• Span‐effective depth calculation should be based on the longer span.
Detailing
• Reinforcement designed to resist these slab moments may be detailed according to the simplified rules for slabs to satisfy normal spacing limits.
• Should be spread across the respective strip, but in solid slabs without drops, top steel to resist negative moments in column strips should have ½ of the area located in the central quarter‐strip width.
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Detailing(cont.)
• If column strip is narrower because of drops, the moments resisted by the column and middle strips should be adjusted proportionally.
• The reinforcement should pass through internal columns to enhance robustness.
Example
The columns are 6.5m centres in each direction and the slab supports the variable load of 5kN/m2. The characteristic material strengths are fck=25N/mm2
for the concrete, and fyk=500N/mm2 for the reinforcement.
It is decided to use a floor slab as shown in figure below with 250mm overall depth of slab, and drop panels 2.5m square by 100mm deep. The column heads are to be made 1.2m diameter.
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Solution
Permanent load
Weight of slab = 0.25x25x6.52 = 264.1kN
Weight of drop = 0.1x25x2.52 = 15.6kN
Total = 279.7kN
Variable load = 5x6.52
= 211.3kN
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Solution(cont.)
Therefore, ultimate load on the floor, F
= 1.35Gk+1.5Qk
= 1.35x279.7+1.5x211.3
= 695kN per panel
and equivalent distributed load, n
= 695/6.52
= 16.4kN/m2
The effective span,
L = clear span between column head +
at
either end
= (6.5‐1.2) + x 2 x 10‐3 = 5.65m
A concrete cover of 25mm has been allowed, and there are two equal layers of reinforcement the effective depth has been taken as the mean depth of the two layers in calculating the reinforcement areas. (d=205mm in span and 305mm in supports).
Solution(cont.)
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The drop dimension is greater than one‐third of the panel dimension, therefore the column strip is taken as the width of the drop panel (2.5m).
Bending reinforcement
Since the variable load is less than the permanent load and bay size = 6.5x6.5= 42.25m2( 30m2), from Table 3.12 BS8110:
Solution(cont.)
1. Centre of interior spanPositive moment = 0.063FL
= 0.063x695x5.65 = 247kNm
The width of the middle strip is (6.5‐2.5)=4m which is greater than half the panel dimension, therefore the proportion o this moment taken by the middle strip can be taken as 0.45 from table 3.18 adjusted as shown:
0.45x( . ⁄0.55
Solution(cont.)
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Thus middle strip positive moment = 0.55x247=136kNm
The column strip positive moment = 0.45x247=111kNm
(a) For the middle stripM/bd2fck= 136x10
6/4000x2052x25 = 0.032
As = M/0.87fykz = 136x106/0.87x500x0.95x205
= 1605mm2 bottom steel
Thus provide 16H12 bars (As=1809mm2) each way in the span, distributed evenly across the 4m width of the middle strip (spacing=250mm=max allowable for a slab)
Solution(cont.)
(b) For the column strip
will require 1310mm2 bottom steel which can be provided as 12H12 bars (As=1356mm2) in the span distributed evenly across the 2.5m width of the column strip (spacing approx. 210mm)
2. Negative moment = ‐0.063Fl
= ‐0.063x695x5.65
= 247kNm
Solution(cont.)
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and this can also be divided into
middle strip = 0.25x[4/(6.5/2)]x247
= 77kNm
column strip = (1‐0.31)x247
= 0.69x247= 170kNm
Solution(cont.)
(a) For the middle stripM/bd2fck= 77x10
6/4000x2052x25 = 0.018
As = M/0.87fykz = 77x106/0.87x500x0.95x205
= 909mm2 top steel
Thus provide 11H12 bars (As=1243mm2) to satisfy 400mm max limit.
(b) For the column strip M/bd2fck= 170x10
6/2500x3052x25 = 0.029
As = M/0.87fykz = 170x106/0.87x500x0.95x305
= 1349mm2 top steel
Thus provide 14H12 bars (As=1582mm2) at 200 centres over the full 2.5m width of the column strip.
Solution(cont.)
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Punching Shear
1. At the column head
Perimeter uo = x diameter of column head
= x 1200 = 3770mm
Shear force, Ved = Fx1.22 = 695‐
x1.22x16.4
= 676.4kN
Solution(cont.)
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To allow for the effects of moment transfer, V is increased by 15% for an internal column, thus
Ved,eff = 1.15x676.4 = 778kN
Maximum permissible shear force,
VRd,max = 0.5uod 0.6 1.
= 0.5x3770x305x 0.6 1.
= 5174kN
Solution(cont.)
Thus Ved,eff is significantly less than VRd,max
2. The first critical sectioncritical section = 2.0 x effective depth from the face of the column head, that is, a section of diameter 1.2+2x2.0x0.305=2.42m(within the drop panel)
Thus the length of the perimeter u1= x2420 =7602mm
Ultimate shear force, Ved =695‐x2.422x16.4 = 620kN
Ved,eff =1.15x620 = 713kN
Solution(cont.)
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For the unreinforced section,
VRd,cu1d = Rd,cx7602x305
with 1= y= z= =0.21%
thus from Table 7, Rd,c = 0.47N/mm2, therefore
VRd,c = 0.47x7602x305x10‐3=1090kN
As Ved,eff is less than VRd,c the section is adequate, and shear reinforcement is not needed.
Solution(cont.)
0.21%
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3. At the dropped panel the critical section is 2.0x205=410mm from the panel with a parameter given by
u = (2a+2b+2x2d)= (4x2500 +2x410) = 12576mm
the area within the perimeter is given by
(2.5+3d)2‐(4‐)(2.0x0.205)2
= (2.5+3x0.205)2‐(4‐)(0.410)2
= 9.559m2
Solution(cont.)
Ultimate shear force,Ved = 695‐9.559x16.4 = 538kN
Ved,eff = 1.15x538 = 619kN
VRd,c = Rd,cud where u=12576mm and d=205mm
1 = 0.31%, thus from Table 7,
Rd,c 0.55N/mm2
VRd,c = 0.55x12576x205 = 1418kN
As Ved,eff is less than VRd,c , the section is adequate.
Solution(cont.)
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Span‐effective depth ratiosat the centre of the span
,= = 0.20
the basic span‐effective depth ratio = 32x1.2(k factor for a flat slab) = 32x1.2 = 38.4
actual span‐effective ratio = l/d = 6500/205 = 31.7
Hence the slab‐effective depth is acceptable.
Solution(cont.)
0.20%