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25 October 2011
1
Limit state design and verification
Joost Walraven
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25 October 2011 2
Flat slab on beams
To be considered:
beam axis 2
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25 October 2011 3
b
b1
b1
b2 b2
bw
bw
beff,1
beff,2
beff
Determination of effective width (5.3.2.1)
beff = S beff,i + bw b
where beff,i = 0,2bi + 0,1l0 0,2l0 and beff,I bi
l3
l1 l2
0,15(l1 + l2 )l =0
l0 = 0,7 l2 l0 = 0,15 l2 + l3l0 = 0,85 l1
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25 October 2011 4
Cross-section of beam with slab
beff,i = 0,2bi + 0,1l0 0,2l0 and beff,I bi
beff,i = 0,22875 +0,1(0,857125) = 1180 mm (
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25 October 2011 5
Beam with effective width
Cross-section at mid-span
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25 October 2011 6
Beam with effective width
Cross-section at intermediate support
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25 October 2011 7
Maximum design bending moments and shear forces
Maximum design moments
Med in kNm (values for
different load cases)
Maximum shear forces Ved in
kN (values for different load
cases)
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25 October 2011 8
Maximum design bending moments and shear forces
Maximum design moments
Med in kNm (values for
different load cases)
Maximum shear forces Ved in
kN (values for different load
cases)
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25 October 2011 9
Determination of bending reinforcement using method with simplified concrete design stress block (3.1.7)
As
d
fcd
Fs
x
s
x
cu3
Fc Ac
400
)508,0 ck
(ffor 50 < fck 90 MPa
= 0,8 for fck 50 MPa
= 1,0 for fck 50 MPa
= 1,0 – (fck – 50)/200 for 50 < fck 90 MPa
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25 October 2011 10
Factors for NA depth (n) and lever arm (=z) for concrete grade 50 MPa
0.00
0.20
0.40
0.60
0.80
1.00
1.20
M/bd 2fck
Fac
tor
n 0.02 0.04 0.07 0.09 0.12 0.14 0.17 0.19 0.22 0.24 0.27 0.30 0.33 0.36 0.39 0.43 0.46
z 0.99 0.98 0.97 0.96 0.95 0.94 0.93 0.92 0.91 0.90 0.89 0.88 0.87 0.86 0.84 0.83 0.82
0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10 0.11 0.12 0.13 0.14 0.15 0.16 0.17
lever arm
NA depth
Simplified factors for flexure (1)
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25 October 2011 11
Factors for NA depth (=n) and lever arm (=z) for concrete grade 70 MPa
0.00
0.20
0.40
0.60
0.80
1.00
1.20
M/bd 2fck
Fa
cto
r
n 0.03 0.05 0.08 0.11 0.14 0.17 0.20 0.23 0.26 0.29 0.33
z 0.99 0.98 0.97 0.96 0.95 0.94 0.93 0.91 0.90 0.89 0.88
0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10 0.11 0.12 0.13 0.14 0.15 0.16 0.17
Simplified factors for flexure (2)
lever arm
NA depth
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25 October 2011 12
Determination of bending reinforcement (span AB)
Example: largest bending moment in span AB: Med = 89,3
kNm
001,0253722610
103,892
6
2
ck
Ed
fbd
M
Read in diagram: lever arm factor = 0,99, so:
26
, 56343537298,0
103,89mm
fz
MA
yd
Edreqsl
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25 October 2011 13
Determination of bending reinforcement (span AB)
Example: largest bending moment in span AB: Med = 89,3 kNm
Moreover, from diagram: neutral axis depth factor is 0,02, so xu = 0,02180 = 4 mm. So height of compression zone < flange thickness (180 mm), OK
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25 October 2011 14
Determination of bending reinforcement (intermediate support B
Bending moment at support B: Med = 132,9 kNm
154,025372250
100,1322
6
2
ck
Ed
fbd
M
Read: lever arm factor 0,81
26
101443537281,0
109,132mm
fz
MA
yd
Edsl
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25 October 2011 15
Maximum design bending moments and shear forces
Maximum design moments
Med in kNm (values for
different load cases)
Maximum shear forces Ved in
kN (values for different load
cases)
Shear may be determined at distance d from support, so Ved 115 kN
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25 October 2011 16
Design of beams for shear (6.2.2)
First check (6.2.2): if VEd ≥ VRd,c then shear reinforcement is
required:
where: fck in Mpa
k = with d in mm
l =
with d = 372mm, bw = 250mm, l = 0,61%, fck = 25MPa
so shear reinforcement is required
bdfkV cklccRd3/1
, )100()/18,0(
0,2200
1 d
02,0db
A
w
sl
kNkNV cRd 1158,4710372250)2561,0(73,1)5,1/18,0(33/1
,
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25 October 2011 17
Expressions for shear capacity at stirrup yielding (VRd,s) and web crushing (VRd,max)
Vu,2
cc1 = f
= fc
Vu,3
s
z
z cot
Afswyw
Vu,2
c c1 = f
= f c
Vu,3
s
z
z cot
A fsw yw
For yielding shear reinforcement: VRd,s = (Asw/s) z fywd cot with between 450 and 21,80
(1 cot 2,5)
At web crushing: VRd,max = bw z fcd /(cot + tan) with between 450 and 21,80
(1 cot 2,5) = 0.6 (1- fck/250)
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25 October 2011 18
Design of beams for shear
Basic equation for determination of shear reinforcement:
VEd,s = (Asw/s) z fywd cot
With Ved,s = 115000 N, fywd = 435 Mpa, z = 0,9d, d = 372 mm and cot = 2,5 it is
found that
Asw/s ≥ 0,32 e.g. stirrups 6mm – 175mm
Check upper value of shear capacity (web crushing criterion)
VRd,max = bw z fcd /(cot + tan) with bw = 250mm, d = 372mm, z = 0,9d, = 0,6(1-fck/250) = 0,54, fcd = 25/1,5 = 13,3 Mpa and cot = 2,5 it is found that VRd,max = 1774 kN which is much larger than the design shear force of 115 kN
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25 October 2011 19
Stirrup configuration near to support A
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25 October 2011 20
Transverse shear in web-flange interface
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25 October 2011 21
Shear between web and flanges of T-sections
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25 October 2011 22
Shear between web and flanges of T-sections
Strut angle : 1,0 ≤ cot f ≤ 2,0 for compression flanges (450 f 26,50 1,0 ≤ cot f ≤ 1,25 for tension flanges (450 f 38,60) No transverse tension ties required if shear stress in interface vEd = Fd/(hf·x) ≤ kfctd (recommended k = 0,4)
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25 October 2011 23
Check necessity of transverse reinforcement
MPahz
V
b
bv
f
Ed
eff
f
Ed 86,01803729,0
115000
2610
1180
No transverse reinforcement required if vEd 0,4fctd For C25/30 fctd = fctk/c =1,8/1,5 = 1,38 Mpa, so the limit value for interface shear is 0,4fctk = 0,41,38 = 0,55 MPa. Transverse shear reinforcement is required at the end of the beam.
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25 October 2011 24
Maximum design bending moments and shear forces
Maximum design moments
Med in kNm (values for
different load cases)
Maximum shear forces Ved in
kN (values for different load
cases)
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25 October 2011 25
Areas in beam axis 2 where transverse reinforcement is required
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25 October 2011 26
Areas in beam axis 2 where transverse reinforcement is required
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25 October 2011 27
Example: transverse reinforcement near to support A
Required transverse reinforcement for Ved = 115 kN
e.g. 8 – 250 (=0,20 mm2/mm)
mmmmzf
V
b
b
s
A
fyd
Ed
eff
fst /18,00,2
1
435335
115000
2610
1180
cot
1 2
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25 October 2011 28
Design of slabs supported by beams
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25 October 2011 29
Design of slabs supported by beams
Load transmission from slabs to beams
Simplified load transmission model Dead load G1 = 0,1825 = 4,5 kN/m
2 Partitions, etc. G2 = 3,0 kN/m
2 Variable load Q = 2,0 kN/m2 Ged = 1,3(4,5 + 3,0) = 9,75 kN/m
2 Qed = 1,52,0 = 3,0 kN/m
2
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25 October 2011 30
Load transfer from slabs to beams
Loading cases on arbitrary strip (dashed in left figure)
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25 October 2011 31
Longitudinal reinforcement in slabs on beams
Examples of reinforced areas
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25 October 2011 32
Floor type 2: flat slab d = 210 mm
From floor on beams to flat slab: replace beams by strips with
the same bearing capacity
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25 October 2011 33
From slab on beams to flat slab
hidden strong strip
-Strips with small width and large reinforcement ratio favourable for punching
resistance
- Strips not so small that compression reinforcement is necessary
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25 October 2011 34
Methods of analysis: Equivalent Frame
Analysis – Annex I
(Informative)
lx (> ly)
ly
ly/4 ly/4
ly/4
ly/4
= lx - ly/2
= ly/2
= ly/2 A
B
B
A – Column strip
B – Middle strip
Negative moments Positive moments
Column Strip
60 - 80%
50 - 70%
Middle Strip
40 - 20%
50 - 30% Note: Total negative and positive moments to be resisted by the column and
middle strips together should always add up to 100%.
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25 October 2011 35
Flat slab with “hidden strong strips”
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25 October 2011 36
Punching shear control column B2
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25 October 2011 37
Punching column B2
Junction column to slab
Vertical load from slab to
column Ved = 705 kN
Simplified assumptions for eccentricity factor according to EN 1992-1-1 Cl. 6.4.3
= 1,4
= 1,5
= 1,15
C
B A
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25 October 2011 38
How to take account of eccentricity (simplified case)
du
Vv
i
EdEd Or, how to determine in equation
= 1,4
= 1,5
= 1,15
C
B A
Only for structures where
lateral stability does not
depend on frame action and
where adjacent spans do
not differ by more than 25%
the approximate values for
shown left may be used:
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25 October 2011 39
Upper limit value for design punching shear stress in design
cdRdEd
Ed fvdu
Vv
4,0max,
0
At the perimeter of the loaded area the maximum punching shear stress should satisfy the following criterion:
where: u0 = perimeter of loaded area = 0,6[1 – fck/250]
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25 October 2011 40
Punching shear column B2
1. Check of upper limit value of punching shear capacity
Further data: dy = 210 – 30 – 16/2 = 172mm
dz = 210 – 30 – 16 – 16/2 = 156 mm
Mean effective depth 0,5(172 + 156) = 164mm
= 0,6(1 + fck/250) = 0,54
vRd,max = 0,4fcd = 0,40,54(25/1,5) = 3,60 Mpa
vEd = Ved/(u0d) = 1,15705000/(4500164)
= 2,47 Mpa < 3,60 Mpa
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25 October 2011 41
Definition of control perimeters
The basic control perimeter u1 is taken at a distance 2,0d from the loaded area and should be constructed as to minimise its length
Length of control perimeter of column 500x500mm: u = 4500 + 22164 = 4060 mm
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25 October 2011 42
Punching shear capacity column B2
Punching shear stress at perimeter:
No punching shear reinforcement required if:
MPadu
Vv EdEd 22,1
1644060
70500015,1
1
cRdEd vv ,
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25 October 2011 43
Limit values for design punching shear stress in design
cRdEd vv ,
The following limit values for the punching shear stress are used in design: If no punching shear reinforcement required
)()100( 1min13/1
,, cpcpcklcRdcRd kvkfkCv
where:
where: k1 = 0,10 (advisory value)
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25 October 2011 44
Punching shear capacity of column B2
No punching shear reinforcement required if vEd < vRd,c
With CRd,c = 0,12
k = 1 + (200/d) = 1 + (200/164) < 2, so k = 2,0
= (xy) = (0,860,87) = 0,865%
fck = 25 Mpa
It is found that vRd,c = 0,67 Mpa
Since vEd = 1,22 MPa> 0,67 MPa punching
shear reinforcement should be applied.
3/1
,, )100( cklcRdcRd fkCv
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25 October 2011 45
Punching shear reinforcement
Capacity with punching shear reinforcement
Vu = 0,75VRd,c + VS
Shear reinforcement within 1,5d from column is accounted for with fy,red = 250 + 0,25d(mm) fywd
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25 October 2011 46
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
Punching shear reinforcement
The outer control perimeter at which
shear reinforcement is not required,
should be calculated from:
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.
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47
Design of punching shear reinforcement
The necessary punching shear reinforcement per perimeter is found from:
1 ,
,
( 0,75 )
1,5
r Ed Rd c
sw
ywd ef
u s v vA
f
with:
vEd = 1,22 N/mm2
vRd,c= 0,67 N/mm2
u1 = 4060 mm
fyd,ef = 250 + 0,25 164 = 291 N/mm2
sr = 0,75 164 = 123 mm 120 mm
It is found that: Asw = 800 mm2 per
reinforcement perimeter
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25 October 2011 48
Design of column B2 for punching shear
Determination of the outer perimeter for which vEd = vRd,c The distance from this perimeter to the edge of the column follows from: The outer punching shera reinforcement should be at a distance of not more than 1,5d from the outer perimeter. This is at a distance 5,22d – 1,5d = 3,72d = 610 mm. The distance between the punching shear reinforcement perimeters should not be larger than 0,75d = 0,75164 = 123mm.
mmdvVu cRdEdout 737816467,0/()70500015,1()/( ,
dmmhua out 22,5856)2/()50047378(2/)4(
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25 October 2011 49
Punching shear design of slab at column B2
Perimeters of shear reinforcement
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25 October 2011 50
Design of column B2
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25 October 2011 51
General background: Second order effects at axial loading (EC2, 5.8.2, 5.8.3.1 & 5.8.3.3
- Second-order effects may be ignored if they are smaller than 10% of the corresponding 1th order effects
- “Slenderness”: is defined as = l0/i where i = (l/A) so for rectangular cross-section = 3,46 l0/h and for circular cross section = 4l0/h - Second order effects may be ignored if the slenderness is smaller than the limit value lim
- In case of biaxial bending the slenderness should be calculated for any direction; second order effects need only to be considered in the direction(s) in which lim is exceeded.
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25 October 2011 52
General background: “Slender” versus “short” columns
Definition of slenderness
)/(
00
AI
l
i
l
l0 effective height of the column i radius of gyration of the uncracked concrete section I moment of inertia around the axis considered A cross-sectional area of column
Basic cases EC2 fig. 5.7
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25 October 2011 53
General background: when is a column slender?
Relative flexibilities of rotation-springs at the column ends 1 en 2 k = (/M)(EI/l) where = rotation of restraining members for a bending moment M EI = bending stiffness of compression member l = height of column between rotation-springs
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25 October 2011 54
General background: when is a column slender?
)45,0
1)(45,0
1(5,02
2
1
10
k
k
k
kll
)101(21
210
kk
kkll
Determination of effective column height in a frame
For unbraced frames: the largest value of:
and
where k1 and k2 are the relative spring stiffnesses at the ends of the column, and l is the clear height of the column between the end restraints
For braced frames: Failing column
Non failing
column
End 1
End 2
Non-failing
column
1 2
0
1 2
1 11 1
k kl l
k k
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25 October 2011 55
General background: determination of effective column length (1) (5.8, 5.8.3.2)
Failing
column
Non failing
column
End 1
End 2
Non failing
column Simplifying assumption: * The contribution of the adjacent “non failing ” columns to the spring stiffness is ignored (if this contributes in a positive sense to the restraint) * for beams for /M the value l/2EI may be assumed (taking account of loss of beam stiffness due to cracking)
Assuming that the beams are symmetric with regard to the column and that their dimensions are the same for the two stories, the following relations are found: k1 = k2 = [EI/l]column / [SEI/l]beams = [EI/l]column / [22EI/l]beams = 0,25 where: = [EI/l]column / [EI/l]beams
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25 October 2011 56
General background: Determination of effective column length (2) (5.8, 5.8.3.2)
or
0 (fixed end)
0.25
0.5
1.0
2.0
(pinned end)
k1 = k2 0 0.0625 0.125 0.25 0.50 1.0
l0 for braced
column
0.5 l
0.56 l
0.61 l
0.68 l
0.76 l
1.0 l
l0 for
unbraced column:
Larger of the values in the
two rows
1.0 l
1.14 l
1.27 l
1.50 l
1.87 l
∞
1.0 l
1.12 l
1.13 l
1.44 l
1.78 l
∞
The effective column length l0 can, for this situation
be read from the table as a function of )/(
00
AI
l
i
l
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25 October 2011 57
General background: when is a column slender ?
nCBA /20lim
)2,01/(1 efA
21B
A column is qualified as “slender”, which implies that second order effects should be taken into account, if lim. The limit value is defined as:
where:
mrC 7,1
ef = effective creep factor: if unknown it can be assumed that A = 0,7 = Asfyd/(Acfcd): mech. reinforcement ratio, if unknown B = 1,1 can be adopted n = NEd/(Acfcd);
rm = M01/M02: ratio between end-moments in column, with M02 M01
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25 October 2011 58
Design of column B2
Configuration of variable load on slab
B2
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25 October 2011 59
Determination of columns slenderness
First step: determination of rotational spring stiffness at end of column:
Column: EI/l = 0,043106 kNm2
Beam: EI/l = 0,052106 kNm2
K1 = [EI/l]col/[SEI]beams = 0,043/(20,052) = 0,41
If the beam would be cracked a value of 1,5 k1 would be more realistic. This would result
in l0 = 0,80l = 3,2m.
llk
k
k
kll 70,0)
02,1
41,01(5,0)
45,01)(
45,01(5,0 2
2
2
1
10
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25 October 2011 60
Verification of column slenderness
Actual slenderness of column:
Limit slenderness according to
EC2, Cl. 5.8.3.1:
With the default values A = 0,7 B=1,1 C = 0,7 whereas the value n follows from
n= Ned/(Acfcd) = 438400/(500220) = 0,88, the value of lim becomes:
Because the actual slenderness of the column is larger than the limit slenderness second
order effects have to be taken into account.
1,225,0
2,346,346,3 0
h
l
n
CBA
20lim
5,1188,0
7,01,17,020lim
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25 October 2011 61
General : Method based on nominal curvature
Mt = NEd (e0 + ei + e2) Different first order eccentricities e01 en e02 At the end of the column can be replaced by an equivalent eccentricity e0 defined as:
e0 = 0,6e02 + 0,4e01 0,4e02 e01 and e02 have the same sign if they lead to tension at the same side, otherwise different signs. Moreover e02 e01
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25 October 2011 62
General : Method based on the nominal curvature
2
0lvei
Mt = NEd (e0 + ei + e2) The eccentricity ei by imperfection follows from (5.2(7)):
where l0 = effective column height around the axis regarded
200
1
100
1
lv
where l = the height of the column in meters
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25 October 2011 63
General: Method based on nominal curvature
1)150200
35,0(1 efckfK
Mt = NEd (e0 + ei + e2) The second order eccentricity e2 follows from:
where
0,1
balud
Edudr
NN
NNKand
20
2 2 0, 45
yd
r
le K K
d
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25 October 2011 64
Calculation of bending moment including second order effects
The bending moment on the column follows from:
e0 = Med/Ned = 42/4384 = 0,010m = 10mm . However, at least the maximum value of
{l0/20, b/20 or 20mm} should be taken. So, e0 =b/20 = 500/20 = 25mm.
ei = i(l0/2) where i = 0hm 0 = 1/200 rad, h =2/l = 1 and
so that ei =(1/200)(4000/2) = 10mm
where and
and finaly
)( 20 eeeMM iEdt
1)1
11(5,0)
11(5,0
mm
20
2 2 0, 45
yd
r
le K K
d
effckfK
)
15020035,0(1
tEdEqpeff MM ,00 )/( balu
Edur
nn
nnK
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25 October 2011 65
Calculation of bending moment including second order effects
where (estimated value = 0,03)
so Kr = 0,62 and finaly:
4,0225,1
23,0
0
0
Ed
Eqp
effM
M
14,14,0)150
9,22
200
3035,0(1)
15020035,0(1 eff
ckfK
balu
Edur
nn
nnK
65,1
20
43503,011
cd
yd
uf
fn
88,0cdc
EdEd
fA
Nn 4,0baln
mmd
lKKe
yd
r 1445425,0
1017,2320062,015,1
45,0
3
2
2
2
2
02
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25 October 2011 66
Calculation of bending moment including second order effects and reinforcement
Determination of reinforcement
kNmeeeNM Edtot 21510)141025(4384)(3
210
58,030500
43840002
ck
Ed
bhf
N
06,030500
21500032
cd
Ed
fbh
M
From diagram: So: (1,4%)
15,0ck
yks
bhf
fA
22
3448435
3050020,0mmAs
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25 October 2011 67
Design of shear wall
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25 October 2011 68
Design of shear wall
The stability of the building is ensured by two shear walls (one at any end of the
building) and one central core
shear wall 1 core shear wall 2 I = 0,133 m4 I = 0,514 m4 I = 0,133 m4 Contribution of shear wall 1: 0,133/(20,133 + 0,514) = 0,17 (17%)
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25 October 2011 69
Second order effects to be regarded?
“If second order effects are smaller than 10% of the first
order moments they can be neglected”.
Moment magnification factor:
]1/
1[0
EdB
EdEdNN
MM
2
2
)12,1( l
EINB
lqN vEd
NB is the buckling load of the system sketched, l = height of building, qv = uniformely distributed load in vertical direction, contributing to 2nd order deformation.
qv
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25 October 2011 70
Second order effects to be regarded?
The moment magnification factor is:
where n = NB/NEd
Requiring f < 1,1 and substituting the corresponding
values in the equation above gives the condition:
(Eq.1)
Assuming 30% of the variable load as permanent, the
load per story is 3014,2510,65 = 4553 kN. Since the
storey height is 3m, this corresponds with qv=1553
kN/m’ height.
With l = 19m, E = 33.000/1,2 = 27.500 MPA, I = 0,78 m4
1
n
nf
84,0EI
lql vEd
84,070,078,05,27
191518
10
193
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25 October 2011 71
Second order effects to be regarded?
However, in the calculation it was assumed that the stabilizing
elements were not cracked. In that case a lower stiffness
should be used.
For the shear wall the following actions apply:
Max My = 66,59 kNm = 0,0666 MNm
Corresp. N = -2392,6 kN = 2,392 MN/m2
So the shear wall remains indeed uncracked and 2nd order
effects may be ignored.
2/78,425,02
2392mMNN
2/99,301667,0
0666,0mMN
W
MM
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25 October 2011 72
Alternative check by Eq. 5.18 in EC2
According to Cl. 5.8.3.3 of EC-22nd order effects may be ignored if:
Where
FV,Ed total vertical load (both on braced and unbraced elements)
ns number of storeys
L total height of building above fixed foundation
Ecd design E-modulus of the concrete
Ic moment of intertia of stabilizing elements
The advisory value of the factor k1 is 0,31. If it can be shown that the
stabilizing elements remain uncracked k1 may be taken 0,62
21, 6,1 L
IE
n
nkF ccd
s
sEdV
S
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25 October 2011 73
Alternative check by Eq. 5.18 in EC2
Verification for the building considered:
Condition:
or: 27.318 29.084
so the condition is indeed fullfilled
21, 6,1 L
IE
n
nkF ccd
s
sEdV
S
2
6
19
78,0105,27
6,16
662,045536
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25 October 2011 74
Monodirectional slab with embedded lighting elements
-
25 October 2011 75
Bearing beams in floor with embedded elements
-
25 October 2011 76
Design for bending of main bearing beam in span 1-2
Med = 177,2 kNm
Effective width:
Midspan: beff = 2695 mm
from diagram z = 0,98d = 365mm
bbbb wieffeff S , 0, 1,02,0 lbb iieff
02,0253722695
102,1722
6
2
ck
Ed
fbd
M
26
1367435365
102,172mm
fz
MA
yd
Edsl
-
25 October 2011 77
Design for bending of main bearing beam in span 1-2 (intermediate support)
Med = 266 kNm
Effective width:
Internal support: beff = 926 mm
At intermediate support: !?
bbbb wieffeff S , 0, 1,02,0 lbb iieff
31,025372250
102662
6
2
ck
Ed
fbd
M
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25 October 2011 78
Factors for NA depth (n) and lever arm (=z) for concrete grade 50 MPa
0.00
0.20
0.40
0.60
0.80
1.00
1.20
M/bd 2fck
Fac
tor
n 0.02 0.04 0.07 0.09 0.12 0.14 0.17 0.19 0.22 0.24 0.27 0.30 0.33 0.36 0.39 0.43 0.46
z 0.99 0.98 0.97 0.96 0.95 0.94 0.93 0.92 0.91 0.90 0.89 0.88 0.87 0.86 0.84 0.83 0.82
0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10 0.11 0.12 0.13 0.14 0.15 0.16 0.17
lever arm
NA depth
Simplified factors for flexure (1)
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25 October 2011 79
Design for bending of main bearing beam in span 1-2 (intermediate support)
Med = 266 kNm
Effective width:
Internal support: beff = 926 mm
At intermediate support compression reinforcement required:
e.g. 320
bbbb wieffeff S , 0, 1,02,0 lbb iieff
222
826)35372(435
37225025)167,031,0(
)'(
)'(mm
ddf
bdfKKA
yd
cksc
-
25 October 2011 80
Design for bending of main bearing beam in span 1-2 (intermediate support)
Med = 266 kNm
Effective width:
Internal support: beff = 926 mm
Calculation of tensile reinforcement:
For K = 0,167 z = 0,81372=301 mm
e.g. 720 = 2198 mm2
bbbb wieffeff S , 0, 1,02,0 lbb iieff
26
2031435301
10266mm
fz
MA
yd
Edsl
-
25 October 2011 81
Design of one-way beams with embedded elements
Loads:
G1 = 2,33 kN
G2 = 3,0
Q = 2,0
Qed = 1,3(2,33+3,0) + 1,52,0=9,93 kN/m2
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25 October 2011 82
Beams with embedded elements: design for bending at intermediate support
Compression reinforcement required
In any rib 203 mm2
167,0294,025189240
10632
6
2
ck
Ed
fbd
Mk
222
406)35189(435
18924025)167,0294,0(
)'(
)'(mm
ddf
bdfKKA
yd
cksc
-
25 October 2011 83
Beams with embedded elements: design for bending at intermediate support
Tensile reinforcement: for K = 0,167 z = 0,8189=151 mm
e.g. 12-100 = 1130 mm2 or
26
959435151
1063mm
fz
MA
yd
Edsl
10-75 = 1040 mm2
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25 October 2011 84
Beams with embedded elements: design for bending at midspan
From diagram z = 0,95d = 0,95189 = 180 mm
044,0251891000
102,392
6
2
ck
Ed
fbd
MK
26
501435180
102,39mm
fz
MA
yd
Edsl
251 mm2 per rib
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25 October 2011 85
Deflection control by slenderness limitation
23
0ck
0ck 12,35,111
ffK
d
l if 0 (7.16.a)
0
ck0
ck
'
12
1
'5,111
ffK
d
lif > 0 (7.16.b)
l/d is the limit span/depth
K is the factor to take into account the different structural systems
0 is the reference reinforcement ratio = fck 10-3
is the required tension reinforcement ratio at mid-span to resist the moment
due to the design loads (at support for cantilevers)
’ is the required compression reinforcement ratio at mid-span to resist the
moment due to design loads (at support for cantilevers)
For span-depth ratios below the following limits no further checks is needed
-
Deflection control by slenderness limitation
)(
500
310
,
,
provs
reqsyk
s
A
Af
The expressions given before (Eq. 7.6.a/b) are derived based on many different
assumptions (age of loading, time of removal of formwork, temperature and humidity
effects) and represent a conservative approach.
The coefficient K follows from the static system:
The expressions have been derived for an assumed stress of 310 Mpa under the quasi
permanent load. If another stress level applies, or if more reinforcement than required
is provided, the values obtained by Eq. 7.16a/b can be multiplied with the factor
where s is the stress in the reinforcing steel at mid-span
-
Rules for large spans
For beams and slabs (no flat slabs) with spans larger than 7m, which support partitions liable to damage by excessive deflections, the values l/d given by Eq. (7.16) should be multiplied by 7/leff (leff in meters).
For flat slabs where the greater span exceeds 8,5m, and which support partitions to be damaged by excessive deflections, the values l/d given by expression (7.16) should be multiplied by 8,5/ leff.
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25 October 2011 88
0
10
20
30
40
50
60
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2
Reinforcement percentage (As/bd)
lim
itin
g s
pan
/dep
th r
ati
o
fck =30 40 50 60 70 80 90
Eq. 7.16 as a graphical representation, assuming K = 1 and s = 310 MPa
-
25 October 2011 89
Tabulated values for l/d calculated from Eq. 7.16a/b The table below gives the values of K (Eq.7.16), corresponding to the structural system. The table furthermore gives limit l/d values
for a relatively high (=1,5%) and low (=0,5%) longitudinal reinforcement ratio. These values are calculated for concrete C30 and s = 310 MPa and satisfy the deflection limits given in 7.4.1 (4) and (5).
Structural system K = 1,5% = 0,5%
Simply supported slab/beam
End span
Interior span
Flat slab
Cantilever
1,0
1,3
1,5
1,2
0,4
l/d=14
l/d=18
l/d=20
l/d=17
l/d= 6
l/d=20
l/d=26
l/d=30
l/d=24
l/d=8
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25 October 2011 90
Beams with embedded elements: design for bending at midspan
From diagram z = 0,95d = 0,95189 = 180 mm
044,0251891000
102,392
6
2
ck
Ed
fbd
MK
26
501435180
102,39mm
fz
MA
yd
Edsl
251 mm2 per rib (e.g. 214 = 308 mm2)
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25 October 2011 91
Control of deflection slab with embedded elements
Reinforcement ratio at midspan = Asl/bed = 501/(1000189) = 0,265%
According to Cl. 7.4.2(2) no detailed calculation is necessary if the l/d ratio of
the slab is smaller than the limit value:
So:
23
0ck
0ck 12,35,111
ffK
d
l
49)1265,0
5,0(52,3
256,0
5,0255,111[3,1 2/3
d
l
-
25 October 2011 92
Control of deflection slab with embedded elements
Moreover correction for real steel stress versus 310 N/mm2 as default value:
Quasi permanent load: Qqp=2,33 + 3,0 + 0,32,0 = 5,93
Ultimate design load: Qed = 9,93
Steel stress under quasi permanent load 2 = (5,93/9,93)435 = 260 Mpa
Corrected value of l/d is:
Actual value is l/d = 7,125/189 = 38 so OK
4,5849260
310)(
310
,
d
l
d
l
qps
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25 October 2011 93
Theory of crack width control
sr
se
steel stress
concrete stress
ctmf
t t
w
The crack width is the difference between the steel deformation and the concrete deformation over the length 2lt, where lt is the “transmission length”, necessary to build-up the concrete stength from 0 to the tensile strength fctm. Then the maximum distance between two cracks is 2lt (otherwise a new crack could occur in-between). It can be found that the transmission length is equal to:
bm
ctmt
fl
4
1
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25 October 2011 94
EC-formula’s for crack width control
For the calculation of the maximum (or characteristic) crack width, the difference between steel and concrete deformation has to be calculated for the largest crack distance, which is sr,max = 2lt. So ( )
cm sm k w
r s
max ,
where sr,max is the maximum crack distance and (sm - cm) is the difference in deformation between steel and concrete over the maximum crack distance. Accurate formulations for sr,max and (sm - cm) will be given
sr
se
steel stress
concrete stress
ctmf
t t
w
Eq. (7.8)
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25 October 2011 95
EC-2 formula’s for crack width control
where: s is the stress in the steel assuming a cracked section e is the ratio Es/Ecm p,eff = (As + Ap)/Ac,eff (effective reinforcement ratio including eventual prestressing steel Ap is bond factor for prestressing strands or wires kt is a factor depending on the duration of loading (0,6 for short and 0,4 for long term loading)
(Eq. 7.9)
s
s
s
effpe
effp
effct
ts
cmsmEE
fk
6,0
)1( ,,
,
-
25 October 2011 96
EC-2 formulae for crack width control
For the crack spacing sr,max a modified expression has been derived, including the concrete cover. This is inspired by the experimental observation that the crack at the outer concrete surface is wider than at the reinforcing steel. Moreover, cracks are always measured at the outside of the structure (!)
-
25 October 2011 97
EC-3 formula’s for crack width control
Maximum final crack spacing sr,max
eff p r k k c s
, 2 1 max , 425 . 0 4 . 3
(Eq. 7.11)
where c is the concrete cover is the bar diameter k1 bond factor (0,8 for high bond bars, 1,6 for bars with an effectively plain surface (e.g. prestressing tendons) k2 strain distribution coefficient (1,0 for tension and 0,5 for bending: intermediate values van be used)
-
25 October 2011 98
EC-2 formula’s for crack width control
In order to be able to apply the crack width formulae, basically valid for a concrete tensile bar, to a structure loaded in bending, a definition of the “effective tensile bar height” is
necessary. The effective height hc,ef is the minimum of: 2,5 (h-d) (h-x)/3 h/2
d h
gravity lineof steel
2.5
(h
-d)
<h
-xe
3
eff. cross-section
beam
slab
element loaded in tension
ct
smallest value of
2.5 . (c + /2) of t/2
c
smallest value of
2.5 . (c + /2)
of(h - x )/3
e
a
b
c
-
25 October 2011 99
EC-2 requirements for crack width control (recommended values)
Exposure class RC or unbonded PSC members
Prestressed members with bonded tendons
Quasi-permanent load
Frequent load
X0,XC1 0.3 0.2
XC2,XC3,XC4 0.3
XD1,XD2,XS1,XS2,XS3
Decompression
-
25 October 2011 100
Crack width control at intermediate support of slabs with embedded elements
Assumption: concentric tension of upper slab of 50 mm.
Steel stress s,qp under quasi permanent load:
Reinforcement ratio: s,eff = Asl/bd = 959/(100050) = 1,92%
Crack distance:
MPafA
A
Q
Qyd
provs
reqs
Ed
qp
qps 22043585,0597,0,
,
,
mmkkkckeffs
s 2770192,0
12425,00,18,0194,3
,
4213max,
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25 October 2011 101
Crack width control at intermediate support of slabs with embedded elements
Average strain:
Characteristic crack width:
so, OK
s
s
s
effpe
effp
effct
ts
cmsmEE
fk
6,0
)1( ,,
,
31079,0000.200
)0192,071(0192,0
6,24,0220
cmsm
mmmmsw cmsmrk 30,018,01079,0227}{3
max,
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25 October 2011 102
Crack width at mid-span beams with embedded elements
Cross-section of tensile bar
Height of tensile bar: smallest value of 2,5(h-d), (h-x)/3 or h/2.
Critical value 2,5(h-d) = 2,529 = 72 mm.
s,eff = Asl/bheff = 308/(12072) = 3,56%
MPaf
A
A
Q
Qyd
provs
reqs
Ed
qp
qps 21043581,0597,0,
,
,
-
25 October 2011 103
Crack width at mid-span beams with embedded elements
Cross-section of tensile bar
mmkkkckeffs
s 1560356,0
12425,05,08,0294,3
,
4213max,
3
,
,
,
1087,0000.200
)0356,071(0356,0
6,24,0210)1(
s
effpe
effp
effct
ts
cmsmE
fk
mmsw cmsmrk 14,01087,0156)(3
max, OK
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25 October 2011 104
Different cultures: different floors
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