punching of flat slabs: design example

fib Model Code 2010

Punching of flat slabs: Design example Stefan Lips, Aurelio Muttoni, Miguel Fernández Ruiz Ecole Polytechnique Fédérale de Lausanne, Switzerland, 04.11.2010

Lips / Muttoni / Fernández Ruiz / Ecole Polytechnique Fédérale de Lausanne, Switzerland 1

1 Basic data 1.1 Geometry (dimensions in [m]) Spans: Lx = 6.00 m and Ly = 5.60 m

Plan view Section trough slab and column

Slab thickness h: 25 cm Cover concrete c: 3 cm

1.2 Material The material properties can be found in chapter 5 of model code 2010. 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

21.35(6.25 2) 1.5 3 15.6kN/md dg q+ = + + ⋅ =


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. The reaction forces in the columns are estimated by using contributive areas.

Inner (C5): Vd ≈ 692 kN Corner (C1, C3): Vd ≈ 93 kN Edge (C2): Vd ≈ 265 kN (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 Corner Column Edge Column (C5) (C1, C3) (C2, C4, C6) ke=0.9 ke=0.65 ke=0.7

Inner: ( ) ( )0 4 0.90 4 260 200 1501mme c vb k b d π π= ⋅ ⋅ + ⋅ = ⋅ ⋅ + ⋅ =

Corner: 0

2002 0.65 2 260 440mm4 4

ve c

db k b π π⋅ ⋅⎛ ⎞ ⎛ ⎞= ⋅ ⋅ + = ⋅ ⋅ + =⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

Edge: 0

2003 0.70 3 260 766mm2 2

ve c

db k b π π⋅ ⋅⎛ ⎞ ⎛ ⎞= ⋅ ⋅ + = ⋅ ⋅ + =⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

Rotations According to the commentary, the distance to the point where the radial moment is zero rs can be estimated based on the spans. By using the Level I approach, one can estimate the rotations. The maximum aggregate size of 32 mm leads to a factor kdg of

48 48 1.016 16 32dg

g

kd

= = =+ +

, 0.22 0.22 6.0 1.32ms x xr L= = ⋅ = , 0.22 0.22 5.6 1.23ms y yr L= = ⋅ =

, 1.32 4351.5 1.5 0.02150.200 200000

yds xx

s

frd E

ψ = ⋅ = ⋅ = governing

, 1.23 4351.5 1.5 0.02000.200 200000

s y ydy

s

r fd E

ψ = ⋅ = ⋅ =

1 1 0.25 0.61.5 0.6 1.5 0.6 0.0215 200 1.0dg

kd kψ ψ

= = = ≤+ ⋅ ⋅ ⋅ + ⋅ ⋅ ⋅


Shear strength without shear reinforcement

Inner: 3

, 0

300.25 1501 200 10 274 kN 692kN1.5

ckRd c v d

c

fV k b d Vψ γ

−= = ⋅ ⋅ = < =

Corner: 3, 0

300.25 440 200 10 80 kN 93kN1.5

ckRd c v d

c

fV k b d Vψ γ

−= = ⋅ ⋅ = < =

Edge: 3, 0

300.25 766 200 10 140 kN 265kN1.5

ckRd c v d

c

fV k b d Vψ γ

−= = ⋅ ⋅ = < =

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.

,max 0,

0

ck d dRd sys v d sys

c Rd cckv

c

f V VV k k b d V kVf

k b dψ

ψ

γγ

= ≥ → ≥ =

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 may be used if they are experimentally verified.

Inner: ,

692 2.5274

dsys

Rd c

VkV

≥ = =

Corner: ,

93 1.280

dsys

Rd c

VkV

≥ = =

Edge: ,

265 1.9140

dsys

Rd c

VkV

≥ = =

Conclusions Inner column: Shear reinforcement is certainly necessary. It might be that a

special shear reinforcement system has to be used since factor ksys needs to be larger than the value proposed in the model code (ksys = 2.0). This has to be checked by using a higher level of approximation. 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.


3 Level II of approximation (typical design) 3.4 Structural analysis and flexural design 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.

2 2, ,d d x d yM M M= +

Summary of the column reactions Column Rd [kN] Md,x, [kNm] Md,y [kNm] Md [kNm] C1 111 25 22 33 C2 266 42 0 42 C3 112 25 22 33 C4 252 3 36 36 C5 664 8 1 8 C6 246 5 34 34

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 rigid-plastic behavior of concrete and steel:

2 12

ydRd yd

cd

fm d f

fρ

ρ⋅⎛ ⎞

= ⋅ ⋅ −⎜ ⎟⎝ ⎠

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


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.

2 22 2 20.2042 0.26 2 0.26 0.204 0.21

4 4v

c c c v

dA b b d m π π= + ⋅ ⋅ + = + ⋅ ⋅ + =

( ) 664 15.6 0.21 661 kNd d d d cV N g q A= − + ⋅ = − ⋅ =

Control perimeter In case of inner columns, the centroid of the column corresponds to the centroid of the control perimeter. Therefore,

0eΔ =

6

3

8 10 0 12 mm661 10

du

d

Me e

V⋅= − Δ = − =⋅

34 4 206 10 513mmu cb Aπ π

= = ⋅ =

1 1 0.981 1 12 513e

u u

ke b

= = =+ +

( ) ( )0 1 4 0.98 4 260 204 1642 mme e c vb k b k b d π π= ⋅ = ⋅ ⋅ + ⋅ = ⋅ ⋅ + ⋅ =

Rotations The distances rs,x and rs,y are the same as for the Level I approximation.

, 1.32ms xr = , 1.23ms yr =

, ,1.5s s xb r= ⋅ 1.5 1.32s y = ⋅ 1.23 1.91mr⋅ ⋅ =

,,

661 8 85kNm/m8 2 8 2 1.91

d x d xdsd x

s

M V eVmb

− ⋅ Δ= + = + =

⋅ ⋅

,,

661 1 83kNm/m8 2 8 2 1.91

d y d ydsd y

s

M V eVmb

− ⋅ Δ= + = + =

⋅ ⋅

kdg is calculated at Level I.

1.5 1.5,

,

1.32 435 851.5 1.5 0.01330.204 200000 115

yds x sdx

s Rd x

fr md E m

ψ⎛ ⎞ ⎛ ⎞= ⋅ ⋅ = ⋅ ⋅ =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

governing

1.5 1.5,

,

1.23 435 831.5 1.5 0.01210.204 200000 115

s y yd sdy

s Rd y

r f md E m

ψ⎛ ⎞ ⎛ ⎞= ⋅ ⋅ = ⋅ ⋅ =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

1 1 0.32 0.61.5 0.6 1.5 0.6 0.0133 204 1.0dg

kd kψ ψ

= = = <+ ⋅ ⋅ ⋅ + ⋅ ⋅ ⋅


Punching strength without shear reinforcement The punching shear strength of the concrete is not sufficient. Consequently, shear reinforcement is necessary.

3

, 0

300.32 1642 204 10 390 kN 661kN1.5

ckRd c v d

c

fV k b d Vψ γ

−= = ⋅ ⋅ = < =

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. 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.

,max , 02 390 781kN 1223kNck

Rd sys Rd c vc

fV k V b d

γ= = ⋅ = ≤ =

,max 781kN 661kNRd dV V= ≥ =

200000 0.0133 444MPa 435MPa6 6s

swd ywd

E fψσ ⋅= = = > =

( )( )

3, 2661 390 10

636mmsin 0.98 435 sin 90

d Rd csw

e swd

V VA

k σ α− ⋅−

= = =⋅ ⋅ °

( ) ( )2

,min

0.5 0.5 661 774mmsin 0.98 435 sin 90

dsw

e ywd

VAk f α

⋅ ⋅= = =⋅ ⋅ ⋅ ⋅ °

governing

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. 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.

, 204 30 174v outd d c m= − = − = m

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:

3

0

,

661 10 3258mm300.32 174

1.5

d

ckv out

c

Vbf

k dψ γ

⋅= = =

0 3258 519mm2 2out

brπ π

= = = ( )1 1 0.99

1 2 1 12 2 519eu out

ke r

= = = + + ⋅

0 3258 3296mm0.99out

e

bbk

= = =

8@100@100 0.50%swρ = ∅ =

( )224 4 0.35 0.35sw w c v v c v vA b d d b d dρ π π⎡ ⎤= ⋅ ⋅ + ⋅ − ⋅ ⋅ − ⋅⎣ ⎦

( )220.005 4 260 204 204 4 260 0.35 204 0.35 204swA π π⎡ ⎤= ⋅ ⋅ + ⋅ − ⋅ ⋅ ⋅ − ⋅ ⋅⎣ ⎦2 21263mm 774mmswA = >

4 700 174 3347 mm 3296 mmoutb π= ⋅ + ⋅ = >


3.6 Shear design corner column C1 and C3 Design shear force


2 22 2 20.2102 0.26 0.26 0.210 0.13

2 16 16v v

c c c

d dA b b m π π= + ⋅ ⋅ + = + ⋅ + =

( ) 112 15.6 0.13 110 kNd d d d cV N g q A= − + ⋅ = − ⋅ =

Control perimeter

3 1 34 2 2 4 2

v cx y c c v d be e b b d⎛ ⎞ ⎛ ⎞Δ = Δ = + − = +⎜ ⎟ ⎜ ⎟

⎝ ⎠⎝ ⎠

1 3 1 3260 210 144 mm4 2 4 2x y c ve e b d⎛ ⎞ ⎛ ⎞Δ = Δ = + = + =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

2 144 2 203mmxe eΔ = Δ ⋅ = ⋅ =

6

3

33 10 203 97 mm110 10

du

d

Me e

V⋅= − Δ = − =⋅

34 4 131 10 408mmu cb Aπ π

= = ⋅ =

1 1 0.811 1 97 408e

u u

ke b

= = =+ +

0 1 2 0.81 2 260 210 554 mm4 4

ve e c

db k b k b π π⋅⎛ ⎞ ⎛ ⎞= ⋅ = ⋅ ⋅ + = ⋅ ⋅ + ⋅ =⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

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.

, 1.32ms xr = , 1.23ms yr =

, ,1.5s s xb r= ⋅ 1.5 1.32 1.= ⋅ ⋅ 23 1.91ms yr⋅ =

2 2 0.26 0.52 msr cb b= ⋅ = ⋅ = governing

,,

110 25 110 0.144 11031kN 55kN8 8 0.52 2 2

d x d xd dsd x

s

M V eV Vm

b− ⋅ Δ − ⋅= + = + = < = =

,,

110 22 110 0.144 11025kN 55kN8 8 0.52 2 2

d y d yd dsd y

s

M V eV Vm

b− ⋅ Δ − ⋅= + = + = < = =

1.5 1.5, ,

,

1.32 435 551.5 1.5 0.01460.21 200000 69

yds x sd xx

s Rd x

fr md E m

ψ⎛ ⎞ ⎛ ⎞= ⋅ ⋅ = ⋅ ⋅ =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

governing



1.5 1.5, ,

,

1.23 435 551.5 1.5 0.01360.21 200000 69

s y yd sd yy

s Rd y

r f md E m

ψ⎛ ⎞ ⎛ ⎞= ⋅ ⋅ = ⋅ ⋅ =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

1 1 0.30 0.61.5 0.6 1.5 0.6 0.0146 210 1dg

kd kψ ψ

= = = <+ ⋅ ⋅ ⋅ + ⋅ ⋅ ⋅

Punching strength without shear reinforcement The punching shear strength of the concrete is sufficient. Thus, no shear reinforcement will be necessary

3

0

300.30 554 210 10 127 kN 110 kN1.5

ckRd v

c

fV k b dψ γ

−= = ⋅ ⋅ = >

Integrity reinforcement Since no shear reinforcement has been used and msd < mRd, integrity reinforcement needs to be provided. For the design of the integrity reinforcement, the accidental load case can be used. Thus, the design load can be reduced.

( ) 21.0(6.25 2) 0.6 3 10.1kN/md d accg q+ = + + ⋅ =

( ),

10.1 110 71kN15.6

d d accd acc d

d d

g qV V

g q+

= ⋅ = ⋅ =+

The material properties can be found in chapter 5 of model code 2010. Ductility class B : (ft/fy)k = 1.08 and εuk = 0.05 It is assumed that only straight bars will be used, thus ° . 0α =

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

( ) ( )

3, 2

2 2

1.5 1.5

71 10 916mmcos 0cos/ 1 435 1.08 1

1 1.25 1 1.25 0.05

d accs

yd t y kuk

VA

f f f αε

⋅= = =⎛ ⎞ ⎛ ⎞

⋅ ⋅ − ⋅ ⋅ −⎜ ⎟ ⎜ ⎟⎜ ⎟+ + ⋅⎝ ⎠ ⎝ ⎠

2x2 ø20 As = 1257 mm2 (2 in each direction)

( ) ( )0.12 0.12 2 0.12 250 2 30 10 20 19mm 20mmres top bottomd h c φ φ= − ⋅ − − = − ⋅ − − = ≈


3.7 Shear design edge column C2 Design shear force


2 22 2 23 0.213 0.26 0.26 0.21 0.17 m

2 8 2 8v v

c c c

d dA b b π π= + ⋅ ⋅ + = + ⋅ + =

( ) 266 15.6 0.17 263 kNd d d d cV N g q A= − + ⋅ = − ⋅ =

Control perimeter

( )

( )

122 2 2222 2

v v vc v c c c

cx

vc v c

d d db d b b b bedb d b

⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ ⋅ + + + ⋅ +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠Δ = −

⎛ ⎞+ + +⎜ ⎟⎝ ⎠

2 22 6 314 3 2

c c v v xc v

b b d deb d

+ +Δ = ⋅

+

2 2 2 22 6 31 1 2 260 6 260 210 3 210 124 mm4 3 2 4 3 260 2 210

c c v vx

c v

b b d deb d

+ + ⋅ + ⋅ ⋅ + ⋅Δ = ⋅ = ⋅ =+ ⋅ + ⋅

0mmyeΔ = 124mmxe eΔ = Δ =

6

3

42 10 124 35mm263 10

du

d

Me e

V⋅= − Δ = − =⋅

34 4 167 10 461mmu cb Aπ π

= = ⋅ =

1 1 0.931 1 35 461e

u u

ke b

= = =+ +

0 1 3 0.93 3 260 210 1031mm2 2

ve e c

db k b k b π π⋅⎛ ⎞ ⎛ ⎞= ⋅ = ⋅ ⋅ + = ⋅ ⋅ + ⋅ =⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

Rotations The distances rs,x and rs,y are the same as for the Level I approximation. In case of edge columns, the width of the support strip may be limited by the distance bsr.

, 1.32ms xr = , 1.23ms yr =

, ,1.5 1.5 1.32 1.23 1.91ms s x s yb r r= ⋅ ⋅ = ⋅ ⋅ =

, 3 3 0.26 0.78msr x cb b= ⋅ = ⋅ = governing

,

0.260 1.91 1.09m2 2 2 2

c ssr y

b bb = + = + = governing

,,

263 42 263 0.124 45kN8 8 0.78

d x d xdsd x

s

M V eVm

b− ⋅ Δ − ⋅= + = + =

,,

263 0 26333kN 66kN8 2 8 2 1.09 4 4

d y d yd dsd y

s

M V eV Vm

b− ⋅ Δ

= + = + = < = =⋅ ⋅



1.5 1.5, ,

,

1.32 435 451.5 1.5 0.0110.21 200000 69

yds x sd xx

s Rd x

fr md E m

ψ⎛ ⎞ ⎛ ⎞= ⋅ ⋅ = ⋅ ⋅ =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

1.5 1.5, ,

,

1.23 435 661.5 1.5 0.0180.21 200000 69

s y yd sd yy

s Rd y

r f md E m

ψ⎛ ⎞ ⎛ ⎞= ⋅ ⋅ = ⋅ ⋅ =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

governing

1 1 0.27 0.61.5 0.6 1.5 0.6 0.018 210 1.0dg

kd kψ ψ

= = = <+ ⋅ ⋅ ⋅ + ⋅ ⋅ ⋅

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.

3

0

300.27 1031 210 10 211kN 263kN1.5

ckRd v

c

fV k b dψ γ

−= = ⋅ ⋅ = <


4 Level III of approximation (detailed design or assessment of existing structure)

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. 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.

, , ,sd x d x d xym m m= − , , ,sd y d y d xym m m= −

, 0.64ms xr = , 1.18ms yr =

, ,1.5 1.5 0.64 1.18 1.30 ms s x s yb r r= ⋅ ⋅ = ⋅ ⋅ =

, 3 3 0.26 0.78msr x cb b= ⋅ = ⋅ =

,

0.26 1.30 0.78m2 2 2 2

c ssr y

b bb = + = + =

, 24kNm/msd xm = (average value on support strip) , 43kNm/msd ym =

, ,

20.64m 0.52m3s x sr xr b= > = , ,

21.18m 0.52m3s y sr yr b= > =

1.5 1.5,

,

0.64 435 241.2 1.2 0.00160.21 200000 69

yds x sdx

s Rd x

fr md E m

ψ⎛ ⎞ ⎛ ⎞= ⋅ ⋅ = ⋅ ⋅ =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

1.5 1.5,

,

1.18 435 431.2 1.2 0.00730.21 200000 69

s y yd sdy

s Rd y

r f md E m

ψ⎛ ⎞ ⎛ ⎞= ⋅ ⋅ = ⋅ ⋅ =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

governing

1 1 0.41 0.61.5 0.6 1.5 0.6 0.0073 210 1.0dg

kd kψ ψ

= = = <+ ⋅ ⋅ ⋅ + ⋅ ⋅ ⋅

The punching shear strength of the concrete is sufficient. Thus, no shear reinforcement will be necessary

3

0

300.41 1047 210 10 332kN 263kN1.5

ckRd

c

fV k b dψ γ

−= = ⋅ ⋅ = >

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. For the design of the integrity reinforcement, the accidental load case can be used. Thus the design load can be reduced.

( ) 21.0(6.25 2) 0.6 3 10.1kN/md d accg q+ = + + ⋅ =

( ),

10.1 263 171kN15.6

d d accd acc d

d d

g qV V

g q+

= ⋅ = ⋅ =+

( ) ( )

3, 2

2 2

1.5 1.5

171 10 2194mmcos 0cos/ 1 435 1.08 1

1 1.25 1 1.25 0.05

d accs

yd t y kuk

VA

f f f αε

⋅= = =⎛ ⎞ ⎛ ⎞

⋅ ⋅ − ⋅ ⋅ −⎜ ⎟ ⎜ ⎟⎜ ⎟+ + ⋅⎝ ⎠ ⎝ ⎠

2x3ø20 + 3ø16 As = 2488 mm2 (3 in each direction)

( ) ( )0.12 0.12 2 0.12 250 2 30 10 20 19 mm 20 mmres top bottomd h c φ φ= − ⋅ − − = − ⋅ − − = ≈

rsy=1.18 m

my=0

x

msd,y [kNm/m]

bs,y0

-32.9-30.4

-54.7-56.0-51.9

-48.9

-36.8-42.0

-53.8y

msd,x [kNm/m]

bsr,x

-25.7-34.1-29.6

-17.4-11.0-11.8

-20.7-35.1

-31.0-17.3

2

-bsr,x2

rsx=0.64 m

mx=0


The material properties can be found in chapter 5 of model code 2010. Ductility class B : (ft/fy)k = 1.08 and εuk = 0.05 It is assumed that only straight bars will be used, thus α = 0°. 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 Luca Tassinari The authors would also like to thank Carsten Siburg (RWTH Aachen, Germany) for the independent check of the example he performed.


punching of flat slabs: design example

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