fib Model Code 2010
Punching of flat slabs:
Design example
Stefan Lips, Aurelio Muttoni, Miguel Fernández Ruiz
Ecole Polytechnique Fédérale de Lausanne, Switzerland,
16.12.2011
Lips / Muttoni / Fernández Ruiz / Ecole Polytechnique Fédérale de Lausanne, Switzerland
1
1
Basic data
1.1
Geometry (dimensions in [m])
Plan view
Spans: Lx = 6.00 m and Ly = 5.60 m
Slab thickness h:
25 cm
Cover concrete c:
3 cm
1.2
The material properties can be found in chapter 5 of model code 2010.
Section trough slab
and column
Material
Concrete C30
Steel B500S (flexural and transverse reinforcement)
fck
30 MPa
fyd
435 MPa
γc
1.5
Es
200 GPa
dg
32 mm
Ductility class
B
1.3
Loads
Self-weight of concrete slab:
6.25 kN/m2
Superimposed dead load:
2 kN/m2
Live load:
3 kN/m2
g d + qd = 1.35(6.25 + 2) + 1.5 ⋅ 3 = 15.6 kN/m 2
Lips / Muttoni / Fernández Ruiz / Ecole Polytechnique Fédérale de Lausanne, Switzerland
2
2
Level I of approximation (preliminary design)
Reaction forces
The goal of the preliminary design is to check if the dimensions of the
structure are reasonable with respect to the punching shear strength and if
punching shear reinforcement is needed.
Inner (C5):
Vd ≈ 692 kN
Corner (C1, C3): Vd ≈ 93 kN
Edge (C2):
Vd ≈ 265 kN
The reaction forces in the columns are estimated by using contributive areas.
(C4 and C6 are not governing
Vd ≈ 244 kN)
Control perimeter
The effective depth dv is assumed to be 200 mm.
Eccentricity coefficient (ke) are adopted from the commentary of §7.3.5.2
Inner column
(C5)
ke=0.9
Corner Column
(C1, C3)
ke=0.65
Inner:
b0 = ke ⋅ ( 4 ⋅ bc + d v ⋅ π ) = 0.90 ⋅ ( 4 ⋅ 260 + 200 ⋅ π ) = 1501mm
Corner:
d ⋅π
⎛
b0 = ke ⋅ ⎜ 2 ⋅ bc + v
4
⎝
200 ⋅ π
⎞
⎛
⎟ = 0.65 ⋅ ⎜ 2 ⋅ 260 + 4
⎝
⎠
⎞
⎟ = 440 mm
⎠
Edge:
d ⋅π
⎛
b0 = ke ⋅ ⎜ 3 ⋅ bc + v
2
⎝
200 ⋅ π
⎞
⎛
⎟ = 0.70 ⋅ ⎜ 3 ⋅ 260 + 2
⎝
⎠
⎞
⎟ = 766 mm
⎠
Edge Column
(C2, C4, C6)
ke=0.7
Rotations
According to the commentary, the distance to the point where the radial
moment is zero rs can be estimated based on the spans.
rs ,x = 0.22 Lx = 0.22 ⋅ 6.0 = 1.32 m
By using the Level I approach, one can estimate the rotations.
ψ x = 1.5 ⋅
The maximum aggregate size of 32 mm leads to a factor kdg of
ψ y = 1.5 ⋅
k dg
32
32
=
=
= 0.67 < 0.75 → k dg = 0.75
16 + d g 16 + 32
Lips / Muttoni / Fernández Ruiz / Ecole Polytechnique Fédérale de Lausanne, Switzerland
kψ =
rs ,x f yd
d Es
rs , y f yd
d Es
rs , y = 0.22 Ly = 0.22 ⋅ 5.6 = 1.23m
= 1.5 ⋅
1.32 435
= 0.0215 Ö governing
0.200 200000
= 1.5 ⋅
1.23 435
= 0.0200
0.200 200000
1
1
=
= 0.227 ≤ 0.6
1.5 + 0.9 ⋅ψ ⋅ d ⋅ k dg 1.5 + 0.9 ⋅ 0.0215 ⋅ 200 ⋅ 0.75
3
Shear strength without shear reinforcement
Inner:
Corner: VRd ,c = kψ
Edge:
f ck
V Rd ,c = kψ
γc
f ck
γc
V Rd ,c = kψ
f ck
γc
b0 d v = 0.227
30
1501 ⋅ 200 ⋅ 10 −3 = 249 kN ≤ Vd = 692 kN
1.5
b0 d v = 0.227
30
440 ⋅ 200 ⋅ 10 −3 = 73 kN ≤ Vd = 93 kN
1.5
b0 d v = 0.227
30
766 ⋅ 200 ⋅ 10 −3 = 127 kN ≤ Vd = 265 kN
1.5
The thickness of the slab has to be increased or the slab has to be shear
reinforced.
Shear reinforcement
To check if shear reinforcement and which system can be used, one can
calculate the minimal needed value of factor ksys.
VRd ,max = ksys kψ
f ck
γc
Vd
b0 d v ≥ Vd → ksys ≥
kψ
f ck
γc
=
b0 d v
Vd
VRd ,c
ksys depends on the performance of the used shear reinforcement system. The
model code proposes a value of ksys = 2.0 for system compliant with model
code detailing rules (§7.13.5.3). Higher values (up to ksys = 2.8) can be used if
more restrictive detailing rules are adopted and if the placing of the transverse
reinforcement is checked at the construction site.
Inner:
k sys ≥
Vd
692
=
= 2.78
VRd ,c 249
Corner:
k sys ≥
Vd
93
=
= 1.27
VRd ,c 73
Edge:
k sys ≥
Vd
265
=
= 2.09
VRd ,c 127
Conclusions
Inner column: Shear reinforcement is necessary and sufficient (accounting for
the values of ksys) to ensure punching shear strength
Corner columns: Shear reinforcement might probably not be necessary. This has
to be confirmed by a higher level of approximation.
Edge columns: Shear reinforcement might probably be necessary.
The thickness of the slab is sufficient if shear reinforcement is used.
Lips / Muttoni / Fernández Ruiz / Ecole Polytechnique Fédérale de Lausanne, Switzerland
4
The moments and the reaction forces have been calculated with a finite
element software. For the analysis, a linear-elastic model has been used.
The moment Md is the vector addition of the moments in x- and y-direction.
Md = M
2
d ,x
+M
2
d ,y
3
Level II of approximation (typical design)
3.4
Structural analysis and flexural design
Summary of the column reactions
Column
Rd [kN]
Md,x, [kNm]
Md,y [kNm]
111
266
112
252
664
246
25
42
25
3
8
5
22
0
22
36
1
34
C1
C2
C3
C4
C5
C6
For a level II approximation, one has to know the flexural reinforcement. It
was designed on the basis of the previous finite element analysis.
The flexural strength can be calculated according to the Model Code. In this
example, however, the flexural strength has been calculated assuming a rigidplastic behavior of concrete and steel:
⎛ ρ ⋅ f yd ⎞
mRd = ρ ⋅ d 2 ⋅ f yd ⎜ 1 −
⎟
2 f cd ⎠
⎝
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Md [kNm]
33
42
33
36
8
34
Reinforcement sketch
Flexural strength
ø10 @200 mm
mRd = 35 kNm/m
d = 210 mm
ø10 @100 mm
mRd = 69 kNm/m
d = 210 mm
ø10 @200 mm / ø16 @200 mm
mRd = 115 kNm/m
d = 204 mm
5
3.5
Shear design inner column C5
Design shear force
The design shear force Vd is equal to the column
reaction force Nd minus the applied load within the
control perimeter (gd + qd)·Ac.
Ac = bc2 + 2 ⋅ bc ⋅ d v +
dv 2
0.2042
π = 0.262 + 2 ⋅ 0.26 ⋅ 0.204 +
π = 0.21 m 2
4
4
Vd = N d − ( g d + qd ) ⋅ Ac = 664 − 15.6 ⋅ 0.21 = 661 kN
Control perimeter
In case of inner columns, the centroid of the column corresponds to the
centroid of the control perimeter. Therefore,
eu =
Md
8 ⋅106
− Δe =
− 0 = 12 mm
661 ⋅103
Vd
ke =
1
1
=
= 0.98
1 + eu bu 1 + 12 513
Δe = 0
bu =
4
π
Ac =
4
π
206 ⋅103 = 513mm
b0 = ke ⋅ b1 = ke ⋅ ( 4 ⋅ bc + d v ⋅ π ) = 0.98 ⋅ ( 4 ⋅ 260 + 204 ⋅ π ) = 1642 mm
Rotations
The distances rs,x and rs,y are the same as for the Level I approximation.
rs ,x = 1.32 m
rs , y = 1.23m
bs = 1.5 ⋅ rs , x ⋅ rs , y = 1.5 ⋅ 1.32 ⋅1.23 = 1.91m
kdg is calculated at Level I.
Lips / Muttoni / Fernández Ruiz / Ecole Polytechnique Fédérale de Lausanne, Switzerland
msd ,x =
Vd M d , x − Vd ⋅ Δex
661
8
+
=
+
= 85 kNm/m
8
2 ⋅ bs
8
2 ⋅1.91
msd , y =
Vd M d , y − Vd ⋅ Δey
661
1
+
=
+
= 83kNm/m
8
2 ⋅ bs
8
2 ⋅1.91
1.5
r f ⎛ m
ψ x = 1.5 ⋅ s ,x yd ⋅ ⎜⎜ sd
d Es ⎝ mRd , x
1.5
⎞
1.32 435 ⎛ 85 ⎞
⋅⎜
= 0.0133 Ö governing
⎟⎟ = 1.5 ⋅
0.204 200000 ⎝ 115 ⎟⎠
⎠
r f ⎛ m
ψ y = 1.5 ⋅ s , y yd ⋅ ⎜ sd
d Es ⎜⎝ mRd , y
1.5
⎞
1.23 435 ⎛ 83 ⎞
⋅
= 0.0121
⎟⎟ = 1.5 ⋅
0.204 200000 ⎜⎝ 115 ⎟⎠
⎠
kψ =
1.5
1
1
=
= 0.300 ≤ 0.6
1.5 + 0.9 ⋅ψ ⋅ d ⋅ k dg 1.5 + 0.9 ⋅ 0.0133 ⋅ 204 ⋅ 0.75
6
Punching strength without shear reinforcement
The punching shear strength of the concrete is not sufficient. Consequently,
shear reinforcement is necessary.
f ck
V Rd ,c = kψ
γc
b0 d v = 0.300
30
1642 ⋅ 204 ⋅ 10 −3 = 367 kN ≤ Vd = 661 kN
1.5
Punching strength with shear reinforcement
Firstly, one has to check if the design shear force Vd is smaller than the
maximum punching strength VRd,max. This is done assuming ksys=2.
VRd,max = k sysVRd ,c = 2 ⋅ 367 = 734 kN ≤
The design shear force Vd is below the maximum punching strength VRd,max.
Therefore, the slab can be reinforced with shear reinforcement complying
with detailing rules defined in subclause §7.13.5.3.
VRd,max = 734 kN ≥ Vd = 661kN
σ swd =
The bond strength is taken as fbd = 3 MPa (according to MC 2010 for
corrugated bars).
E sψ
6
Assuming a circular control perimeter for the estimation of the eccentricity,
factor ke can be estimated as detailed in the right hand side column.
Possible shear reinforcement layout:
b0 d v = 1223 kN
⎛
f
d ⎞ 200000 ⋅ 0.0133 ⎛
3 204 ⎞
⋅ ⎜1 +
⋅
⋅ ⎜1 + bd ⋅ ⎟ =
⎟
⎟
⎜
f ywd φ w ⎠
6
⎝ 435 8 ⎠
⎝
Vd − VRd ,c
(661 − 367 ) = 690 mm 2
=
k eσ swd sin α 0.98 ⋅ 435 ⋅ sin (90°)
Asw,min =
In this example, the calculating value of the effective depth dv is equal to the
effective depth d minus the concrete cover c on the bottom surface of the
slab: d v ,out = d − c = 204 − 30 = 174 mm
γc
σ swd = 521 MPa > f ywd = 435 MPa
Asw =
To avoid a failure outside the shear reinforced area, the outer perimeter need
to have a minimal length. The design shear force can be reduced to account
for the loads applied inside the outer perimeter. This effect is neglected as a
safe estimate.
f ck
0.5 ⋅ Vd
0.5 ⋅ 661
= 779 mm 2 Ö governing
=
k eσ swd sin α 0.98 ⋅ 435 ⋅ sin (90°)
Vd
b0 =
kψ
f ck
γc
=
d v ,out
661 ⋅ 10 3
30
0.300
174
1.5
rout =
b0
3468
=
= 552 mm
2π
2π
bout =
b0 3468
=
= 3505 mm
0.99
ke
ke =
= 3468 mm
1
1 + eu 2rout
=
1
= 0.99
1 + 12 (2 ⋅ 552)
ρ sw = ∅8@100@100 = 0.50%
2
Asw = ρ w ⎡ 4 ⋅ bc ⋅ d v + d v2 ⋅ π − 4 ⋅ bc ⋅ 0.35d v − ( 0.35d v ) ⋅ π ⎤
⎣
⎦
2
Asw = 0.005 ⎡ 4 ⋅ 260 ⋅ 204 + 2042 ⋅ π − 4 ⋅ 260 ⋅ 0.35 ⋅ 204 − ( 0.35 ⋅ 204 ) ⋅ π ⎤
⎣
⎦
2
2
Asw = 1263mm > 774 mm
bout = 4 ⋅ 800 + 174 ⋅ π = 3746 mm > 3505 mm
Lips / Muttoni / Fernández Ruiz / Ecole Polytechnique Fédérale de Lausanne, Switzerland
7
3.6
Shear design corner column C1 and C3
Design shear force
The design shear force Vd is equal to the column reaction
force Nd minus the applied load within the control perimeter
(gd + qd)·Ac.
Ac = bc2 + 2 ⋅ bc ⋅
dv dv 2
0.2102
+
π = 0.262 + 0.26 ⋅ 0.210 +
π = 0.13 m2
2 16
16
Vd = N d − ( g d + qd ) ⋅ Ac = 112 − 15.6 ⋅ 0.13 = 110 kN
Control perimeter
Δex = Δey =
d
3⎛
bc + v
⎜
4⎝
2
3 ⎞
⎞ bc 1 ⎛
⎟ − 2 = 4 ⎜ bc + 2 d v ⎟
⎝
⎠
⎠
Δex = Δey =
1⎛
3 ⎞ 1⎛
3
⎞
bc + d v ⎟ = ⎜ 260 + 210 ⎟ = 144 mm
⎜
4⎝
2 ⎠ 4⎝
2
⎠
Δe = Δex ⋅ 2 = 144 ⋅ 2 = 203mm
eu =
Md
33 ⋅106
− Δe =
− 203 = 97 mm
Vd
110 ⋅103
ke =
1
1
=
= 0.81
1 + eu bu 1 + 97 408
bu =
4
π
Ac =
4
π
131 ⋅103 = 408mm
d ⋅π ⎞
π⎞
⎛
⎛
b0 = ke ⋅ b1 = ke ⋅ ⎜ 2 ⋅ bc + v ⎟ = 0.81 ⋅ ⎜ 2 ⋅ 260 + 210 ⋅ ⎟ = 554 mm
4 ⎠
4⎠
⎝
⎝
Rotations
The distances rs,x and rs,y are the same as for the Level I
approximation.
In case of corner columns, the width of the support strip
may be limited by the distance bsr.
rs ,x = 1.32 m
rs , y = 1.23m
bs = 1.5 ⋅ rs , x ⋅ rs , y = 1.5 ⋅ 1.32 ⋅1.23 = 1.91m
bsr = 2 ⋅ bc = 2 ⋅ 0.26 = 0.52 m Ö governing
msd ,x =
Vd M d , x − Vd ⋅ Δex 110 25 − 110 ⋅ 0.144
V
110
+
=
+
= 31kN < d =
= 55 kN
8
8
0.52
2
2
bs
msd , y =
Vd M d , y − Vd ⋅ Δey 110 22 − 110 ⋅ 0.144
V
110
+
=
+
= 25 kN < d =
= 55 kN
8
8
0.52
2
2
bs
r f ⎛m
ψ x = 1.5 ⋅ s ,x yd ⋅ ⎜⎜ sd ,x
d Es ⎝ mRd , x
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1.5
1.5
⎞
1.32 435 ⎛ 55 ⎞
⋅ ⎜ ⎟ = 0.0146 Ö governing
⎟⎟ = 1.5 ⋅
0.21 200000 ⎝ 69 ⎠
⎠
8
r f ⎛m
ψ y = 1.5 ⋅ s , y yd ⋅ ⎜ sd , y
d Es ⎜⎝ mRd , y
kdg is calculated at Level I.
kψ =
1.5
1.5
⎞
1.23 435 ⎛ 55 ⎞
⋅ ⎜ ⎟ = 0.0136
⎟⎟ = 1.5 ⋅
0.21 200000 ⎝ 69 ⎠
⎠
1
1
=
= 0.280 ≤ 0.6
1.5 + 0.9 ⋅ψ ⋅ d ⋅ k dg 1.5 + 0.9 ⋅ 0.0146 ⋅ 210 ⋅ 0.75
Punching strength without shear reinforcement
The punching shear strength of the concrete is sufficient. Thus, no shear
reinforcement will be necessary
f ck
VRd ,c = kψ
γc
b0 d v = 0.280
30
554 ⋅ 210 ⋅ 10 −3 = 119 kN > Vd = 110 kN
1.5
Integrity reinforcement
Since no shear reinforcement has been used and msd < mRd, integrity
reinforcement needs to be provided.
As =
For the design of the integrity reinforcement, the accidental load case can be
used. Thus, the design load can be reduced.
Vd ,acc ≤
(g
d
+ qd )acc = 1.0(6.25 + 2) + 0.6 ⋅ 3 = 10.1kN/m 2
Vd ,acc =
(g
d
+ qd )acc
g d + qd
10.1
⋅ Vd =
⋅110 = 71kN
15.6
The material properties can be found in chapter 5 of model code 2010.
Ductility class B : (ft/fy)k = 1.08
It is assumed that only straight bars will be used, thus αult = 20°.
With respect to integrity reinforcement, two restrictions should be fulfilled:
Vd ,acc
71 ⋅ 103
=
= 442 mm 2 f yd ⋅ ( f t f y )k ⋅ sin α ult 435 ⋅ 1.08 ⋅ sin (20°)
f ck
γc
bint d res =
30
⋅ 464 ⋅ 168 = 285 kN 1.5
Ö 2x2 ø12 As = 452 mm2 (2 in each direction with a spacing of 100 mm)
d res = h − 2 ⋅ c − φtop − φbottom = 250 − 2 ⋅ 30 − 10 − 12 = 168 mm bint = 2 ⋅ sint +
π
2
d res = 2 ⋅ 100 +
π
2
⋅ 168 = 464 mm φint = 12 mm ≤ 0.12d res = 0.12 ⋅ 168 = 20 mm
-the integrity reinforcement should at least be composed of four bars
-the diameter of the integrity bars øint has to be chosen such that
øint ≤ 0.12 dres
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9
3.7
Shear design edge column C2
Design shear force
The design shear force Vd is equal to the column reaction
force Nd minus the applied load within the control perimeter
(gd + qd)·Ac.
Ac = bc2 + 3 ⋅ bc ⋅
dv dv 2
3
0.212
+
π = 0.262 + 0.26 ⋅ 0.21 +
π = 0.17 m 2
2
8
2
8
Vd = N d − ( g d + qd ) ⋅ Ac = 266 − 15.6 ⋅ 0.17 = 263 kN
Control perimeter
1
d
d
d
+ d v ) ⋅ ⎛⎜ bc + v ⎞⎟ + 2 ⎛⎜ bc + v ⎞⎟ ⋅ ⎛⎜ bc + v
2
2
2
2⎝
⎝
⎠
⎝
⎠
Δex =
( bc + dv ) + 2 ⎛⎜ bc + dv 2 ⎞⎟
⎝
⎠
2
2
1 2b + 6bc d v + 3d v
Δex = ⋅ c
4
3bc + 2d v
(b
c
⎞
⎟ b
⎠− c
2
Δex =
1 2bc2 + 6bc d v + 3d v2 1 2 ⋅ 2602 + 6 ⋅ 260 ⋅ 210 + 3 ⋅ 2102
⋅
= ⋅
= 124 mm
4
3bc + 2d v
4
3 ⋅ 260 + 2 ⋅ 210
Δey = 0 mm
Δe = Δex = 124 mm
eu =
Md
42 ⋅106
− Δe =
− 124 = 35 mm
263 ⋅103
Vd
ke =
1
1
=
= 0.93
1 + eu bu 1 + 35 461
d ⋅π
⎛
b0 = ke ⋅ b1 = ke ⋅ ⎜ 3 ⋅ bc + v
2
⎝
bu =
4
π
Ac =
4
π
167 ⋅103 = 461mm
π⎞
⎞
⎛
⎟ = 0.93 ⋅ ⎜ 3 ⋅ 260 + 210 ⋅ 2 ⎟ = 1031mm
⎝
⎠
⎠
Rotations
The distances rs,x and rs,y are the same as for the Level I approximation.
rs ,x = 1.32 m
rs , y = 1.23m
In case of edge columns, the width of the support strip may be limited by the
distance bsr.
bs = 1.5 ⋅ rs , x ⋅ rs , y = 1.5 ⋅ 1.32 ⋅1.23 = 1.91m
bsr ,x = 3 ⋅ bc = 3 ⋅ 0.26 = 0.78 m Ö governing
bsr , y =
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bc bs 0.260 1.91
+ =
+
= 1.09 m Ö governing
2 2
2
2
msd ,x =
Vd M d ,x − Vd ⋅ Δex
263 42 − 263 ⋅ 0.124
+
=
+
= 45 kN
bs
8
8
0.78
msd , y =
Vd M d , y − Vd ⋅ Δey
V
263
0
263
+
=
+
= 33kN < d =
= 66 kN
8
2 ⋅ bs
8
2 ⋅1.09
4
4
10
kdg is calculated at Level I.
1.5
r f ⎛m
ψ x = 1.5 ⋅ s ,x yd ⋅ ⎜⎜ sd ,x
d Es ⎝ mRd , x
1.5
⎞
1.32 435 ⎛ 45 ⎞
⋅
= 0.011
⎟⎟ = 1.5 ⋅
0.21 200000 ⎜⎝ 69 ⎟⎠
⎠
r f ⎛m
ψ y = 1.5 ⋅ s , y yd ⋅ ⎜ sd , y
d Es ⎜⎝ mRd , y
1.5
⎞
1.23 435 ⎛ 66 ⎞
⋅ ⎜ ⎟ = 0.018 Ö governing
⎟⎟ = 1.5 ⋅
0.21 200000 ⎝ 69 ⎠
⎠
kψ =
1.5
1
1
=
= 0.247 ≤ 0.6
1.5 + 0.9 ⋅ψ ⋅ d ⋅ k dg 1.5 + 0.9 ⋅ 0.018 ⋅ 210 ⋅ 0.75
Punching strength without shear reinforcement
The punching shear strength of the concrete is not sufficient. Since the
strength seems to be rather close to the design load, a level III approximation
will be performed.
Lips / Muttoni / Fernández Ruiz / Ecole Polytechnique Fédérale de Lausanne, Switzerland
VRd ,c = kψ
f ck
γc
b0 d v = 0.247
30
1031 ⋅ 210 ⋅ 10 −3 = 195 kN < Vd = 263 kN
1.5
11
4
The Level III calculations are based on the results of the linear-elastic finite
element analysis.
From the results of the flexural analysis, one
can obtain the distance between the center of
the column and the point, at which the bending
moments are zero.
rs ,x = 0.64 m
rs , y = 1.18 m
bs = 1.5 ⋅ rs , x ⋅ rs , y = 1.5 ⋅ 0.64 ⋅1.18 = 1.30 m
bsr ,x = 3 ⋅ bc = 3 ⋅ 0.26 = 0.78 m
my=0
mx=0
rsy=1.18 m
bsr , y =
bc bs 0.26 1.30
+ =
+
= 0.78 m
2 2
2
2
msd , x = 24 kNm/m
rsx=0.64 m
The average moment in the support strip can
be obtained by the integration of the moments
at the strip section.
Since the flexural moments md,x and md,y at the
support regions are negative, the absolute
value of the twisting moment md,x need to be
subtracted so that the absolute value of msd,x
and msd,y will be maximized.
Level III of approximation (detailed design or assessment
of existing structure)
rs ,x = 0.64m >
y
bsr,x
2
-bsr,x
2
-17.3
-31.0
-35.1
-20.7
-11.8
msd,x [kNm/m]
-11.0
-17.4
-29.6
-34.1
-25.7
msd,y [kNm/m]
-56.0-54.7-53.8 -48.9
-42.0
-51.9
-32.9
-36.8
-30.4
0
bs,y
x
msd ,x = md ,x − md , xy
msd , y = md , y − md ,xy
2
bsr ,x = 0.52 m
3
r f ⎛ m
ψ x = 1.2 ⋅ s ,x yd ⋅ ⎜⎜ sd
d Es ⎝ mRd ,x
ψ y = 1.2 ⋅
kψ =
The punching shear strength of the concrete is sufficient. Thus, no shear
reinforcement will be necessary
msd , y = 43kNm/m
rs , y f yd ⎛ msd
⋅⎜
d Es ⎜⎝ mRd , y
(average value on support strip)
rs , y = 1.18 m >
2
bsr , y = 0.52 m
3
1.5
1.5
⎞
0.64 435 ⎛ 24 ⎞
⋅ ⎜ ⎟ = 0.0016
⎟⎟ = 1.2 ⋅
0.21 200000 ⎝ 69 ⎠
⎠
1.5
1.5
⎞
1.18 435 ⎛ 43 ⎞
⋅ ⎜ ⎟ = 0.0073 Ö governing
⎟⎟ = 1.2 ⋅
0.21 200000 ⎝ 69 ⎠
⎠
1
1
=
= 0.395 ≤ 0.6
1.5 + 0.9 ⋅ψ ⋅ d ⋅ k dg 1.5 + 0.9 ⋅ 0.0073 ⋅ 210 ⋅ 0.75
VRd ,c = kψ
f ck
γc
b0 d v = 0.395
30
1031 ⋅ 210 ⋅ 10 −3 = 312kN > Vd = 263 kN
1.5
Integrity reinforcement
Since no shear reinforcement has been used and msd < mRd, integrity
reinforcement needs to be provided to prevent a progressive collapse of the
structure.
As =
Vd ,acc
171 ⋅ 103
=
= 1064 mm 2 f yd ⋅ ( f t f y )k ⋅ sin α ult 435 ⋅ 1.08 ⋅ sin (20°)
f ck
30
⋅ 861 ⋅ 166 = 522 kN 1.5
For the design of the integrity reinforcement, the accidental load case can be
used. Thus the design load can be reduced.
Vd ,accs ≤
(g
Ö 3x3 ø14 As = 1385 mm2 (3 in each direction with a spacing of 100 mm)
d
+ qd )acc = 1.0(6.25 + 2) + 0.6 ⋅ 3 = 10.1kN/m 2
Vd ,acc =
(g
d
+ qd )acc
g d + qd
⋅ Vd =
10.1
⋅ 263 = 171kN
15.6
Lips / Muttoni / Fernández Ruiz / Ecole Polytechnique Fédérale de Lausanne, Switzerland
γc
bint d res =
d res = h − 2 ⋅ c − φtop − φbottom = 250 − 2 ⋅ 30 − 10 − 14 = 166 mm 12
The material properties can be found in chapter 5 of model code 2010.
Ductility class B : (ft/fy)k = 1.08
It is assumed that only straight bars will be used, thus αult = 20°.
bint = 3 ⋅ sint +
π
2
d res = 3 ⋅ 200 +
π
2
⋅ 166 = 861 mm φint = 14 mm ≤ 0.12d res = 0.12 ⋅ 166 = 20 mm
With respect to integrity reinforcement, two restrictions should be fulfilled:
-the integrity reinforcement should at least be composed of four bars
-the diameter of the integrity bars øint has to be chosen such that
øint ≤ 0.12·dres
Corners of walls should be checked following the same methodology.
Acknowledgements:
The authors are very appreciative of the contributions of Dr. Juan Sagaseta
Albajar and Dr. Luca Tassinari
The authors would also like to thank Carsten Siburg (RWTH Aachen,
Germany) for the independent check of the example he performed.
Lips / Muttoni / Fernández Ruiz / Ecole Polytechnique Fédérale de Lausanne, Switzerland
13