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Detailing of Members
Beams
Slabs, one-way and two-way spanning
Flat slabs
Columns
Walls
Deep beams
Foundations: Pile caps, Column & wall footings
Tying systems
Detailing rules for particular situations
Detailing comparisons EC2 v BS 8110
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As,min = 0,26 (fctm/fyk)btd but 0,0013btd
As,max = 0,04Ac
Section at supports should be designed for a
hogging moment 0,25 max. span moment
Any design compression reinforcement () should beheld by transverse reinforcement with spacing 15
BeamsEC2: Cl. 9.2
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Tension reinforcement in a flanged beam at
supports should be spread over the effective width
(see 5.3.2.1)
BeamsEC2: Cl. 9.2
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Shear Design: Links
Variable strut methodallows a shallower strut angle
hence activating more links.
As strut angle reduces concrete stress increases
Angle = 45 V carried on 3 links Angle = 21.8 V carried on 6 links
d
V
z
x
d
x
V
z
s
EC2: Cl. 6.2.3
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Where av 2dthe applied shear force, VEd, for a point load(eg, corbel, pile cap etc) may be reduced by a factor av/2d
where 0.5 av 2d provided:
dd
av av
The longitudinal reinforcement is fully anchored at the support. Only that shear reinforcement provided within the central 0.75av is
included in the resistance.
Short Shear Spans with Direct
Strut ActionEC2: Cl. 6.2.3 (8)
Note: see PD6687-1:2010 Cl 2.14 for more information.
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Shear reinforcement
Minimum shear reinforcement, w,min = (0,08(fck)/fyk
Maximum longitudinal spacing, sl,max = 0,75d (1+cot)
Maximum transverse spacing, st,max = 0,75d 600 mm
EC2: Cl. 9.2.2
For vertical links sl,max = 0,75d
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(1) Sufficient reinforcement should be provided at all sections to resist the
envelope of the acting tensile force, including the effect of inclined cracks
in webs and flanges.
(2) For members with shear reinforcement the additional tensile force, Ftd,
should be calculated according to 6.2.3 (7). For members without shear
reinforcement Ftd
may be estimated by shifting the moment curve a
distance al = d according to 6.2.2 (5). This "shift rule may also be used
as an alternative for members with shear reinforcement, where:
al = z (cot - cot )/2 = 0.5 z cot for vertical shear links
z= lever arm, = angle of compression strut
al = 1.125 d when cot = 2.5 and 0.45 d when cot = 1
CurtailmentEC2: Cl. 9.2.1.3
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For members without shear reinforcement this is satisfied with al = d
a lFtd
a l
Envelope of (MEd /z+NEd)
Acting tensile force
Resisting tensile force
lbd
lbd
lbd
lbd
lbd lbd
lbd
lbdFtd
Shift rule
Curtailment of reinforcementEC2: Cl. 9.2.1.3, Fig 9.2 Concise: 12.2.2
For members with shear reinforcement: al = 0.5 zCot
But it is always conservative to use al = 1.125d
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lbd is required from the line of contact of the support.
Simple support (indirect) Simple support (direct)
As bottom steel at support 0.25As provided in the span
Transverse pressure may only be taken into account with
a direct support.
Shear shift rule
al
Tensile Force Envelope
Anchorage of Bottom
Reinforcement at End SupportsEC2: Cl. 9.2.1.4
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Simplified Detailing Rules for
Beams
How to.EC2
Detailing section
Concise: Cl 12.2.4
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h /31
h /21
B
A
h /32h /22
supporting beam with height h1
supported beam with height h2 (h1 h2)
The supporting reinforcement is in
addition to that required for other
reasons
A
B
The supporting links may be placed in a zone beyond
the intersection of beams
Supporting Reinforcement at
Indirect Supports
Plan view
EC2: Cl. 9.2.5Concise: Cl 12.2.8
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Solid Slabs
One-way and two-way spanning
EC2: Cl 9.3
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Curtailment as beams except for the Shift rule al = d
may be used
Flexural Reinforcement min and max areas as beam
Secondary transverse steel not less than 20% main
reinforcement
Reinforcement at Free Edges
Solid slabsEC2: Cl. 9.3
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Where partial fixity exists, not taken into account in design: Internalsupports:As,top 0,25As of Mmax in adjacent spanEnd supports: As,top 0,15As of Mmax in adjacent span
This top reinforcement should extend 0,2 adjacent span
Solid slabsEC2: Cl. 9.3
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Flat slabs
EC2: Cl 9.4
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What are flat slabs? Solid concrete floors of constant thickness
They have flat soffits
Flat Slabs - Introduction
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Distribution of momentsEC2: Figure I.1 Concise Figure 5.11
Particular rules for flat slabs
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Distribution of moments
EC2: Table I.1 Concise: Table 5.2
Particular rules for flat slabs
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Arrangement of reinforcement should reflect behaviour
under working conditions.
At internal columns top reinforcement of area 0.5Atshould be placed in a width = 0.25 panel width.
At least two bottom bars should pass through internal
columns in each orthogonal direction.
Particular rules for flat slabs
EC2: Cl. 9.4 Concise: 12.4.1
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Design reinforcement at edge and corner reinforcement
should be placed within the becz
cy
y
be = cz + y
A
cz
cyy
A
be = z + y/2
z
A
Particular rules for flat slabsEC2: Figure 9.9 Concise Figure 5.12
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Flat slabs Top steel example
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8m
Flat slabs Top steel example
Equivalent frame method
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TR 64 Figure 14
Reduction in maximum hogging moment
at columns
Flat slabs Top steel example
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Flat slabs Top steel example
Flat slab example design information
Area of reinforcement to resist full negative moment, At = 8000mm
2
Panel width = 8.0m
Column strip width = 4.0m Middle strip width = 4.0m
0.5 At in a width of 0.125 panel width either side of the column ie:
4000mm2 in width of 2m
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Distribution of top reinforcement
Distance (m)
600
100
200
300
400
500
0
Bendingmom
ent(kNm/m
)
0 1 2 3 4 5 6 7 8
Centre column strip:4000 mm2 or 2000 mm2/m
Outer column strip:0.75 x 8000 4000
= 2000 mm2 or 1000 mm2/m
Middle strip:0.25 x 8000 = 2000 mm2
or 500 mm2/m
Assume a total area of steel, At = 8000 mm2
Distribution 75% Column strip & 25% Middle strip6000 mm2 2000 mm2
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Punching shear
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Punching shear does not use the Variable Strut inclination method
and is similar to BS 8110 methods
The basic control perimeter is set at 2dfrom the loaded area
The shape of control perimeters have rounded corners
bz
by
2d 2d 2d
2du1
u1 u1
Punching Shear
Cl. 6.4 Figure 8.3
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kd
Outer control
perimeter
Outer perimeter of shear
reinforcement
1.5d (2d if > 2d from
column)
0.75d
0.5dA A
Section A - A
0.75d
0.5d
Outer control
perimeter
kd
The outer control perimeter at
which shear reinforcement is not
required, should be calculatedfrom:
uout,ef = VEd / (vRd,c d)
The outermost perimeter of
shear reinforcement should be
placed at a distance not
greater than kd( k = 1.5)
within the outer control
perimeter.
Punching Shear Reinforcement (1)EC2: Cl. 6.4.5 Figures 12.5 & 12.6
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1,5d
2d
d
d
> 2d
1,5d
uout
uout,ef
Where proprietary systems are used the control perimeter at which
shear reinforcement is not required, uout or uout,ef(see Figure) should be
calculated from the following expression:
uout,ef = VEd / (vRd,c d)
Punching Shear Reinforcement (2)EC2: Cl. 6.4.5 Figure 8.10
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Columns
EC2: Cl 9.5
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h 4b
min 12
As,min = 0,10NEd/fyd but 0,002 Ac
As,max= 0.04 Ac (0,08Ac at laps)
Minimum number of bars in a circular column is 4.
Where direction of longitudinal bars changes more than1:12 the spacing of transverse reinforcement should be
calculated.
Columns
EC2: Cl. 9.5.2
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scl,tmax = min {20 min; b ; 400mm}
150mm
150mm
scl,tmax
scl,tmax should be reduced by a factor 0,6:
in sections within h above or below a beamor slab
near lapped joints where > 14.
A min of 3 bars is required in lap length
scl,tmax = min {12 min; 0.6b ; 240mm}
Columns
EC2: Cl. 9.5.3
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Column starter bars - foundations
Column starter bars should be anchored, with a full compression
lap length.
Bends and hooks cannot be considered to contribute to the
compression anchorage The compression anchorage length should therefore be taken as
41 for concrete class C25/30 (smaller for higher concrete
classes)
The effect of cover cannot be considered (
2) Whittle has recently reported on some testing that shows that it
may be considered. Reduces anchorage length to 29 (class
C25/30)
Whittle, R.Are modern pad footings and pile caps too shallow?Concrete May
2011 pp 53- 55, The Concrete Society
MPA - The Concrete Centre 33
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Walls
EC2: Cl 9.6
W ll
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Walls
As,vmin = 0,002Ac (half located at each face)
As,vmax = 0.04Ac (0,08Ac at laps)
svmax = 3 wall thickness or 400mm
Vertical Reinforcement
Horizontal Reinforcement
As,hmin = 0,25 Vert. Rein. or 0,001Ac
shmax = 400mm
Transverse Reinforcement
Where total vert. rein. exceeds 0,02Ac links required as
for columns
Where main rein. placed closest to wall links are required.
(at least 4No. m2)
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Foundations
EC2: Cl 9.8
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Tying systems
EC2: Cl 9.10
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Detailing rules for particularsituations
PD 6687-1:2010 Annex B
Replacement for EC2: Annex J
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PD 6687-1:2010 Annex B
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Detailing Comparisons
d or 150 mm from main bar9.2.2 (8): 0.75 d 600 mm
9.2.1.2 (3) or 15 from main bar
st,max
0.75d9.2.2 (6): 0.75 dsl,max
0.4 b s/0.87 fyv9.2.2 (5): (0.08 b s fck)/fykAsw,min
Links
Table 3.28Table 7.3NSmax
dg + 5 mm or 8.2 (2): dg + 5 mm or or 20mmsmin
Spacing of Main Bar s
0.04 bh9.2.1.1 (3): 0.04 bdAs,max
0.002 bh--As,min
Main Bar s i n Compr essi on
0.04 bh9.2.1.1 (3): 0.04 bdAs,max
0.0013 bh9.2.1.1 (1): 0.26fctm/fykbd
0.0013 bd
As,min
ValuesClause / Val uesMain Bar s i n Tensi on
BS 8110EC2Beams
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Detailing Comparisons
places of maximum moment:
main: 2h 250 mm
secondary: 3h 400 mm
3dor 750 mmsecondary: 3.5h 450 mmSmax
dg + 5 mm or 8.2 (2): dg + 5 mm or or 20mm
9.3.1.1 (3): main 3h 400 mm
smin
Spacing of Bar s
0.04 bh9.2.1.1 (3): 0.04 bdAs,max
0.002 bh9.3.1.1 (2): 0.2As for single way
slabs
As,min
Secondar y T r ansver se Bar s
0.04 bh0.04 bdAs,max
0.0013 bh9.2.1.1 (1): 0.26fctm/fykbd
0.0013 bd
As,min
ValuesClause / Val uesMain Bar s i n Tensi on
BS 8110EC2Slabs
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Detailing Comparisons
Columns
150 mm from main bar9.5.3 (6): 150 mm from main bar
129.5.3 (3): min (12min; 0.6 b;240 mm)Scl,tmax
0.25 or 6 mm9.5.3 (1) 0.25 or 6 mmMin size
Links
0.06 bh9.5.2 (3): 0.04 bhAs,max
0.004 bh9.5.2 (2): 0.10NEd/fyk 0.002bhAs,min
Main Bar s i n Comp r ession
1.5d9.4.3 (1):
within 1st control perim.: 1.5d
outside 1st control perim.: 2d
St
0.75d9.4.3 (1): 0.75dSr
Spacing of Li nks
Total = 0.4ud/0.87fyv9.4.3 (2): Link leg = 0.053 sr st
(fck)/fyk
Asw,min
ValuesClause / Val uesLinks
BS 8110EC2Punching Shear