Limit state design and verification
Joost Walraven
25 October 2011
1
Flat slab on beams
To be considered:
beam axis 2
25 October 2011
2
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
beff
beff,2
beff,1
bw
bw
b1
b2
b1
b2
b
l0 = 0,85 l1
l1
25 October 2011
l0 =
0,15(l1 + l2 )
l0 = 0,7 l2
l2
l0 = 0,15 l2 + l3
l3
3
Cross-section of beam with slab
beff,i = 0,2bi + 0,1l0  0,2l0 and beff,I  bi
beff,i = 0,22875 +0,1(0,857125) = 1180 mm (<2875mm)
beff = S beff,i + bw = 21180 + 250 = 2610mm
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4
Beam with effective width
Cross-section at mid-span
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5
Beam with effective width
Cross-section at intermediate support
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6
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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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)
25 October 2011
8
Determination of bending reinforcement using method
with simplified concrete design stress block (3.1.7)
cu3
Ac
fcd
Fc
x
x
d
As
Fs
s
 = 0,8 for fck  50 MPa
(f ck  50)
400
for 50 < fck  90 MPa
= 1,0 – (fck – 50)/200
for fck  50 MPa
for 50 < fck  90 MPa
 0,8 
 = 1,0
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9
Simplified factors for flexure (1)
Factors for NA depth (n) and lever arm (=z) for concrete grade  50 MPa
1.20
1.00
lever arm
Factor
0.80
0.60
0.40
NA depth
0.20
0.00
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
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
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M/bd 2fck
10
Simplified factors for flexure (2)
Factors for NA depth (=n) and lever arm (=z) for concrete grade 70 MPa
1.20
lever arm
1.00
Factor
0.80
0.60
0.40
NA depth
0.20
0.00
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
0.09
0.10
0.11
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
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0.12
0.13
0.14
0.15
0.16
M/bd 2fck
0.17
11
Determination of bending reinforcement (span AB)
Example: largest bending moment in span AB: Med = 89,3
kNm
M Ed
89,3 106

 0,001
bd 2 f ck 2610  3722  25
Read in diagram: lever arm
factor = 0,99, so:
Asl ,req
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M Ed
89,3 106


 563mm2
z  f yd 0,98  372  435
12
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 =
0,02180 = 4 mm. So height of
compression zone < flange
thickness (180 mm), OK
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13
Determination of bending reinforcement (intermediate
support B
Bending moment at support B: Med = 132,9 kNm
M Ed
132,0 106

 0,154
bd 2 f ck 250  372 2  25
Read: lever arm factor 0,81
M Ed
132,9 106
Asl 

 1014mm2
z  f yd 0,81 372  435
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14
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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15
Design of beams for shear (6.2.2)
First check (6.2.2): if VEd ≥ VRd,c then shear reinforcement is
required:
VRd ,c  (0,18 /  c )k (100l f ck )1/ 3 bd
where:
fck
in Mpa
k
= 1
l
=
200
 2,0
d
with d in mm
Asl
 0,02
bw d
with d = 372mm, bw = 250mm, l = 0,61%, fck = 25MPa
VRd ,c  (0,18 / 1,5) 1,73  (0,61 25)1/ 3  250  372 103  47,8kN  115kN
so shear reinforcement is required
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16
Expressions for shear capacity at stirrup
yielding (VRd,s) and web crushing (VRd,max)
s

Vu,3
Aswfyw
z cot 
s
A
fyw
sw
zco
t
V
u,3
z

z

c = fc1
= fc
Vu,2
cc
=f1
=
fc
V
u,2
For yielding shear reinforcement:
VRd,s = (Asw/s) z fywd cot
At web crushing:
with  between 450 and 21,80
(1  cot   2,5)
with  between 450 and 21,80
(1 cot  2,5)
VRd,max = bw z  fcd /(cot + tan)
 = 0.6 (1- fck/250)
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17
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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18
Stirrup configuration near to support A
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19
Transverse shear in web-flange interface
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20
Shear between web and flanges of T-sections
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21
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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22
Check necessity of transverse reinforcement
vEd 
bf
beff

VEd
1180
115000


 0,86MPa
z  h f 2610 0,9  372 180
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 = 0,41,38 = 0,55 MPa.
Transverse shear reinforcement is required at the end of the beam.
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23
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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24
Areas in beam axis 2 where
transverse reinforcement is required
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25
Areas in beam axis 2 where transverse
reinforcement is required
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26
Example: transverse reinforcement near to
support A
Required transverse reinforcement for Ved = 115 kN
Ast b f VEd
1
1180 115000 1






 0,18mm2 / mm
s
beff zf yd cot  f 2610 335  435 2,0
e.g. 8 – 250 (=0,20 mm2/mm)
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27
Design of slabs supported by beams
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28
Design of slabs supported by beams
Load transmission from slabs to beams
Simplified load transmission model
Dead load G1 = 0,1825 = 4,5 kN/m2
Partitions, etc. G2 = 3,0 kN/m2
Variable load Q = 2,0 kN/m2
Ged = 1,3(4,5 + 3,0) = 9,75 kN/m2
Qed = 1,52,0 = 3,0 kN/m2
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29
Load transfer from slabs to beams
Loading cases on arbitrary strip
(dashed in left figure)
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30
Longitudinal reinforcement in slabs on beams
Examples of reinforced areas
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31
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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32
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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33
Methods of analysis: Equivalent Frame
l (> l )
Analysis – Annex I
l /4 l /4
B = l - l /2
(Informative)
x
y
y
y
x
y
ly/4
A – Column strip
ly/4
B = ly/2
ly
B – Middle strip
A = ly/2
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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34
Flat slab with “hidden strong strips”
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35
Punching shear control column B2
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36
Punching column B2
C
 = 1,5
B
 = 1,4
Junction column to slab
Vertical load from slab to
column Ved = 705 kN
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A
 = 1,15
Simplified assumptions for
eccentricity factor  according to
EN 1992-1-1 Cl. 6.4.3
37
How to take account of eccentricity
(simplified case)
VEd
vEd  
Or, how to determine  in equation
ui d
C
 = 1,5
B
 = 1,4
25 October 2011
A
 = 1,15
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:
38
Upper limit value for design punching
shear stress in design
At the perimeter of the loaded area the maximum punching shear
stress should satisfy the following criterion:
vEd 
VEd
u0 d
 vRd ,max  0,4f cd
where:
u0 = perimeter of loaded area
 = 0,6[1 – fck/250]
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39
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 = 0,4fcd = 0,40,54(25/1,5) = 3,60 Mpa
vEd = Ved/(u0d) = 1,15705000/(4500164)
= 2,47 Mpa < 3,60 Mpa
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40
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 = 4500 + 22164 = 4060 mm
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41
Punching shear capacity column B2
Punching shear stress at perimeter:
vEd 
VEd
u1d

1,15  705000
 1,22MPa
4060 164
No punching shear reinforcement required if:
vEd  vRd ,c
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42
Limit values for design punching
shear stress in design
The following limit values for the punching shear stress are used in
design:
If
vEd  vRd ,c
no punching shear reinforcement required
where:
vRd ,c  CRd ,c k (100l f ck )1/ 3  k1 cp  (vmin  k1 cp )
where: k1 = 0,10 (advisory value)
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43
Punching shear capacity of column B2
No punching shear reinforcement required if vEd < vRd,c
vRd ,c  CRd ,c k (100l f ck )1/ 3
With CRd,c = 0,12
k = 1 + (200/d) = 1 + (200/164) < 2, so k = 2,0
 = (xy) = (0,860,87) = 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.
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44
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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45
Punching shear reinforcement
Outer control
perimeter
Outer perimeter of shear
reinforcement
The outer control perimeter at which
shear reinforcement is not required,
1.5d (2d if > 2d from should be calculated from:
column)
uout,ef = VEd / (vRd,c d)
A
0.75d
kd
A
0.5d
0.75d
0.5d
Outer control
perimeter
kd
25 October 2011
Section A - A
The outermost perimeter of shear
reinforcement should be placed at a
distance not greater than kd (k =
1.5) within the outer control
perimeter.
46
Design of punching shear reinforcement
The necessary punching shear reinforcement per perimeter is found from:
Asw 
u1sr (vEd  0, 75vRd ,c )
1,5 f ywd ,ef
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
47
Design of column B2 for punching shear
Determination of the outer perimeter for which vEd = vRd,c
uout  VEd /(vRd ,c  d )  (1,15  705000) /(0,67 164  7378mm
The distance from this perimeter to the edge of the column follows from:
a  (uout  4h) / 2  (7378  4  500) /( 2 )  856mm  5,22d
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 = 0,75164 = 123mm.
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48
Punching shear design of slab at column B2
Perimeters of shear reinforcement
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49
Design of column B2
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50
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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51
General background: “Slender” versus “short” columns
Definition of slenderness
l0
l0
 
i
( I / A)
l0
i
I
A
effective height of
the column
radius of gyration of the
uncracked concrete section
moment of inertia around the axis
considered
cross-sectional area of column
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EC2 fig. 5.7
Basic cases
52
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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53
General background: when is a column slender?
Non-failing
column
Determination of effective column height
in a frame
End 1
For braced frames:
l0  0,5l (1 
Failing
column
k1
k2
)(1 
)
0,45  k1
0,45  k2
End 2
For unbraced frames: the largest value of:
k k
l0  l (1  10 1 2 )
k1  k2
and
Non failing
column

k1 
k2 
l0  l  1 
1 

1

k
1

k

1 
2 
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
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54
General background: determination of effective
column length (1)
(5.8, 5.8.3.2)
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)
End 1
Failing
column
End 2
Non failing
column
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 / [22EI/l]beams = 0,25 
where:
 = [EI/l]column / [EI/l]beams
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55
General background: Determination of effective
column length (2) (5.8, 5.8.3.2)
The effective column length l0 can, for this situation
be read from the table as a function of 

l0
l0
 
i
( I / A)
0
(fixed end)
0.25
0.5
1.0
2.0

(pinned end)
0
0.0625
0.125
0.25
0.50
1.0
0.5 l
0.56 l
0.61 l
0.68 l
0.76 l
1.0 l
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
∞
or
k1 = k2
l0
for braced
column
l0 for
unbraced
column:
Larger of the
values in the
two rows
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56
General background: when is a column slender ?
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:
lim  20  A  B  C / n
where:
A  1 /(1  0,2ef )
B  1 2
C  1,7  rm
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);
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rm = M01/M02: ratio between endmoments in column, with
M02 M01
57
Design of column B2
B2
Configuration of variable load on slab
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58
Determination of columns slenderness 
First step: determination of rotational spring stiffness at end of column:
Column: EI/l = 0,043106 kNm2
Beam: EI/l = 0,052106 kNm2
K1 = [EI/l]col/[SEI]beams = 0,043/(20,052) = 0,41
l0  0,5l (1 
k1
k2
0,41 2
)(1 
)  0,5l (1 
)  0,70l
0,45  k1
0,45  k2
1,02
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.
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59
Verification of column slenderness
Actual slenderness of column:
Limit slenderness according to
EC2, Cl. 5.8.3.1:

3,46l0 3,46  3,2

 22,1
h
0,5
lim 
20 A  B  C
n
With the default values A = 0,7 B=1,1 C = 0,7 whereas the value n follows from
n= Ned/(Acfcd) = 438400/(500220) = 0,88, the value of lim becomes:
lim 
20  0,7 1,1 0,7
 11,5
0,88
Because the actual slenderness of the column is larger than the limit slenderness second
order effects have to be taken into account.
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60
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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61
General : Method based on the nominal curvature
Mt = NEd (e0 + ei + e2)
The eccentricity ei by imperfection follows from (5.2(7)):
l0
ei  v
2
where l0 = effective column height around the axis regarded
v
1
1

100 l 200
where l = the height of the column in meters
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62
General: Method based on nominal curvature
Mt = NEd (e0 + ei + e2)
The second order eccentricity e2 follows from:
e2  K Kr
l02
 yd
 2 0, 45d
where
f ck

K  1  (0,35 

)ef  1
200 150
25 October 2011
and
N ud  N Ed
Kr 
 1,0
N ud  N bal
63
Calculation of bending moment including
second order effects
The bending moment on the column follows from:
M t  M Ed (e0  ei  e2 )
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 = 0hm 0 = 1/200 rad, h =2/l = 1 and
 m  0,5(1 
e2  K Kr
l02
1
1
)  0,5(1  )  1 so that ei =(1/200)(4000/2) = 10mm
m
1
 yd
 2 0, 45d
where
eff  (M 0 Eqp / M 0 Ed ),t
25 October 2011
f ck


)eff and
200 150
n  nEd
Kr  u
nu  nbal
K  1  (0,35 
and finaly
64
Calculation of bending moment including
second order effects
eff 
M 0 Eqp
M 0 Ed
 
0,3  2
 2  0,4
1,5  2
K  1  (0,35 
Kr 
nu  nEd
nu  nbal
nEd 
f ck

30 22,9

)eff  1  (0,35 

)0,4  1,14
200 150
200 150
where
N Ed
 0,88
Ac f cd
nu  1 
f yd
f cd
 1
nbal  0,4
0,03  435
 1,65 (estimated value  = 0,03)
20
so Kr = 0,62 and finaly:
 yd
32002 2,17 103
e2  K K r 2
 1,15  0,62 
 14mm
 0,45d
 2 0,25  454
l02
25 October 2011
65
Calculation of bending moment including
second order effects and reinforcement
M tot  N Ed (e0  e1  e2 )  4384  (25  10  14) 103  215kNm
Determination of reinforcement
N Ed
4384000

 0,58
bhf ck 5002  30
M Ed
215000

 0,06
2
3
bh f cd 500  30
From diagram: As f yk  0,15
bhf ck
0,20  5002  30
As 
 3448mm2
435
(1,4%)
So:
25 October 2011
66
Design of shear wall
25 October 2011
67
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
I = 0,133 m4
core
I = 0,514 m4
Contribution of shear wall 1: 0,133/(20,133 + 0,514) = 0,17
25 October 2011
shear wall 2
I = 0,133 m4
(17%)
68
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:
qv
M Ed  M 0 Ed [1 
NB 

]
N B / N Ed  1
 2 EI
(1,12l ) 2
N Ed  qvl
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.
25 October 2011
69
Second order effects to be regarded?
The moment magnification factor is:
f 
n
n 1
where n = NB/NEd
Requiring f < 1,1 and substituting the corresponding
values in the equation above gives the condition:
l
qvEd l
 0,84
EI
(Eq.1)
Assuming 30% of the variable load as permanent, the
load per story is 3014,2510,65 = 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
19
103
25 October 2011
1518 19
 0,70  0,84
27,5  0,78
70
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
N 
 2392
 4,78MN / m 2
2  0,25
M 
M
0,0666

 3,99MN / m 2
W 0,01667
So the shear wall remains indeed uncracked and 2nd order
effects may be ignored.
25 October 2011
71
Alternative check by Eq. 5.18 in EC2
According to Cl. 5.8.3.3 of EC-22nd order effects may be ignored if:
FV , Ed  k1 
Where
FV,Ed
ns
L
Ecd
Ic
ns SEcd I c
ns  1,6 L2
total vertical load (both on braced and unbraced elements)
number of storeys
total height of building above fixed foundation
design E-modulus of the concrete
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
25 October 2011
72
Alternative check by Eq. 5.18 in EC2
Verification for the building considered:
FV , Ed  k1 
ns SEcd I c
ns  1,6 L2
Condition:
6
27,5 106  0,78
6  4553  0,62 

6  1,6
192
or:
27.318  29.084
so the condition is indeed fullfilled
25 October 2011
73
Monodirectional slab with embedded lighting
elements
25 October 2011
74
Bearing beams in floor with embedded elements
25 October 2011
75
Design for bending of main bearing
beam in span 1-2
Med = 177,2 kNm
beff  Sbeff ,i  bw  b
Effective width:
Midspan: beff = 2695 mm
M Ed
172,2 106

 0,02
bd 2 f ck 2695  3722 25
beff ,i  0,2bi  0,1l0
from diagram z = 0,98d = 365mm
M Ed 172,2 106
Asl 

 1367mm2
z  f yd
365  435
25 October 2011
76
Design for bending of main bearing beam in
span 1-2 (intermediate support)
Med = 266 kNm
beff  Sbeff ,i  bw  b
Effective width:
Internal support: beff = 926 mm
At intermediate support:
25 October 2011
beff ,i  0,2bi  0,1l0
M Ed
266 106

 0,31
2
2
bd f ck 250  372  25
!?
77
Simplified factors for flexure (1)
Factors for NA depth (n) and lever arm (=z) for concrete grade  50 MPa
1.20
1.00
lever arm
Factor
0.80
0.60
0.40
NA depth
0.20
0.00
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
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
25 October 2011
M/bd 2fck
78
Design for bending of main bearing beam in
span 1-2 (intermediate support)
Med = 266 kNm
beff  Sbeff ,i  bw  b
Effective width:
Internal support: beff = 926 mm
beff ,i  0,2bi  0,1l0
At intermediate support compression reinforcement required:
( K  K ' ) f ck bd 2 (0,31  0,167)25  250  3722
Asc 

 826mm2
f yd (d  d ' )
435  (372  35)
25 October 2011
e.g. 320
79
Design for bending of main bearing beam in
span 1-2 (intermediate support)
Med = 266 kNm
beff  Sbeff ,i  bw  b
Effective width:
Internal support: beff = 926 mm
beff ,i  0,2bi  0,1l0
Calculation of tensile reinforcement:
For K = 0,167 z = 0,81372=301 mm
M Ed
266 106
Asl 

 2031mm2
z  f yd 301 435
25 October 2011
e.g. 720 = 2198 mm2
80
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) + 1,52,0=9,93 kN/m2
25 October 2011
81
Beams with embedded elements: design for
bending at intermediate support
M Ed
63 106
k 2
 0,294  0,167
bd f ck 240 1892  25
Compression reinforcement required
( K  K ' ) f ck bd 2 (0,294  0,167)  25  240 1892
Asc 

 406mm2
f yd (d  d ' )
435  (189  35)
In any rib 203 mm2
25 October 2011
82
Beams with embedded elements: design for
bending at intermediate support
Tensile reinforcement: for K = 0,167 z = 0,8189=151 mm
M Ed
63 106
Asl 

 959mm2
z  f yd 151 435
25 October 2011
e.g. 12-100 = 1130 mm2 or
10-75 = 1040 mm2
83
Beams with embedded elements: design
for bending at midspan
M Ed
39,2 106
K 2

 0,044
2
bd f ck 1000 189  25
M Ed
39,2 106
Asl 

 501mm2
z  f yd 180  435
25 October 2011
From diagram z = 0,95d = 0,95189 = 180 mm
251 mm2 per rib
84
Deflection control by slenderness limitation
For span-depth ratios below the following limits no further checks is needed
3

2




l
 K 11 1,5 fck 0  3,2 fck  0  1 
d


 
 


0
l
1
 K 11 1,5 fck

fck
d
   ' 12

l/d
K
0

’
' 

0 
if   0
(7.16.a)
if  > 0
(7.16.b)
is the limit span/depth
is the factor to take into account the different structural systems
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)
25 October 2011
85
Deflection control by slenderness
limitation
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
s
310

f yk  (
500
As ,req
As , prov
)
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.
Eq. 7.16 as a graphical representation,
assuming K = 1 and s = 310 MPa
fck =30
40 50 60 70 80 90
60
limiting span/depth ratio
50
40
30
20
10
0
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
Reinforcement percentage (As/bd)
25 October 2011
88
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
Simply supported slab/beam 1,0
End span
1,3
Interior span
1,5
Flat slab
1,2
Cantilever
0,4
25 October 2011
 = 1,5%
 = 0,5%
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
89
Beams with embedded elements: design
for bending at midspan
M Ed
39,2 106
K 2

 0,044
2
bd f ck 1000 189  25
M Ed
39,2 106
Asl 

 501mm2
z  f yd 180  435
25 October 2011
From diagram z = 0,95d = 0,95189 = 180 mm
251 mm2 per rib (e.g. 214 = 308 mm2)
90
Control of deflection slab with embedded
elements
Reinforcement ratio at midspan  = Asl/bed = 501/(1000189) = 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:
25 October 2011
3

 0
 2

l
0
 K 11 1,5 fck
 3,2 fck 
 1 
d


 
 

l
0,5
0,5
 1,3[11  1,5 25 
 3,2  5  (
 1)3 / 2  49
d
0,256
0,265
91
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 + 0,32,0 = 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:
l
310 l
310

( ) 
 49  58,4
d  s ,qp d
260
Actual value is l/d = 7,125/189 = 38 so OK
25 October 2011
92
Theory of crack width control
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:
25 October 2011
sr
se
steel stress
t
t
fctm
concrete stress
w
1 f ctm 
lt 
4  bm 
93
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
wk
 s , (   )
r max sm
cm
sr
Eq. (7.8)
se
steel stress
t
t
fctm
concrete stress
where
and
w
sr,max is the maximum crack distance
(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
25 October 2011
94
EC-2 formula’s for crack width control
 s  kt
 sm   cm 
f ct ,eff
 p ,eff
(1   e  p ,eff )
Es
 0,6
s
(Eq. 7.9)
Es
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)
25 October 2011
95
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
96
EC-3 formula’s for crack width control
Maximum final crack spacing sr,max
s r ,max  3.4c  0.425 k1k2  p,eff
(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
97
EC-2 formula’s for crack width control
element loaded
in tension
t
a
eff. crosssection
h
c
d
h-xe
3
beam
2.5 (h-d) <
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:
c
gravity line
of steel
smallest value of
2.5 . (c + /2) of t/2
slab
2,5 (h-d)
(h-x)/3
h/2
25 October 2011
b
c

smallest value of
2.5 . (c + /2)
of
(h - xe)/3
98
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
XC2,XC3,XC4
0.3
XD1,XD2,XS1,XS2,
XS3
25 October 2011
0.2
Decompression
99
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:
 s ,qp 
Qqp

As ,req
QEd As , prov
 f yd  0,597  0,85  435  220MPa
Reinforcement ratio: s,eff = Asl/bd = 959/(100050) = 1,92%
Crack distance:
25 October 2011
 s ,max  k3  c  k1  k2  k4 

 s ,eff
 3,4 19  0,8 1,0  0,425 
12
 277mm
0,0192
100
Crack width control at intermediate
support of slabs with embedded elements
 s  kt
Average strain:
 sm   cm 
f ct ,eff
(1   e  p ,eff )
 p ,eff

 0,6 s
Es
Es
220  0,4
 sm   cm 
Characteristic crack width:
2,6
(1  7  0,0192)
0,0192
 0,79 103
200.000
wk  sr ,max { sm   cm }  227  0,79 103  0,18mm  0,30mm
so, OK
25 October 2011
101
Crack width at mid-span beams with
embedded elements
Height of tensile bar: smallest value of 2,5(h-d), (h-x)/3 or h/2.
Critical value 2,5(h-d) = 2,529 = 72 mm.
s,eff = Asl/bheff = 308/(12072) = 3,56%
 s ,qp 
25 October 2011
Qqp

As ,req
QEd As , prov
Cross-section of
tensile bar
 f yd  0,597  0,81 435  210MPa
102
Crack width at mid-span beams with
embedded elements
Cross-section of
tensile bar
 s ,max  k3  c  k1  k2  k4 
 s  kt
 sm   cm 
f ct ,eff
 p ,eff

 s ,eff
 3,4  29  0,8  0,5  0,425 
(1   e  p ,eff )
Es
210  0,4

2,6
(1  7  0,0356)
0,0356
 0,87 10 3
200.000
wk  sr ,max ( sm   cm )  156  0,87 103  0,14mm
25 October 2011
12
 156mm
0,0356
OK
103
Different cultures: different floors
25 October 2011
104