NMC Example 111215

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Lips / Muttoni / Fernndez Ruiz / Ecole Polytechnique Fdrale de Lausanne, Switzerland 2

1 Basic data1.1 Geometry (dimensions in [m])

Spans:Lx = 6.00 m andLy = 5.60 m

Plan view Section trough slab

and column

Slab thickness h: 25 cm

Cover concrete c: 3 cm

1.2 MaterialThe 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

c 1.5 Es 200 GPa

dg 32 mm Ductility class B

1.3 LoadsSelf-weight of concrete slab: 6.25 kN/m

2

Superimposed dead load: 2 kN/m2

Live load: 3 kN/m2

21.35(6.25 2) 1.5 3 15.6 kN/md d

g q+ = + + =

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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 kNCorner (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 = + = + =

Corner:0

2002 0.65 2 260 440mm

4 4

v

e c

db k b

= + = + =

Edge:0

2003 0.70 3 260 766mm

2 2

v

e c

db k b

= + = + =

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 factorkdgof

75.075.067.03216

32

16

32=

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Shear strength without shear reinforcement

Inner: kN692kN2491020015015.1

30227.0 3

0,====

dv

c

ck

cRdVdb

fkV

Corner: kN93kN73102004405.1

30227.0 30,

==== dv

c

ck

cRdVdbfkV

Edge: kN265kN127102007665.1

30227.0 3

0,====

dv

c

ck

cRdVdb

fkV

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 factorksys.

,max 0

,

0

ck d d

Rd sys v d sys

c Rd cck

v

c

f V VV k k b d V k

Vfk b d

= =

ksys depends on the performance of the used shear reinforcement system. Themodel code proposes a value ofksys = 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: 78.2249

692

,

==cRd

d

sysV

Vk

Corner: 27.173

93

,

==cRd

d

sysV

Vk

Edge: 09.2127

265

, == cRd

d

sys V

V

k

Conclusions

Inner column: Shear reinforcement is necessary and sufficient (accounting for

the values ofksys) to ensure punching shear strengthCorner 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.

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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 momentMd 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

yd

Rd yd

cd

fm d f

f

=

Reinforcement sketch

Flexural strength

10 @200 mm mRd= 35 kNm/m d= 210 mm

10 @100 mm mRd= 69 kNm/m d= 210 mm

10 @200 mm / 16 @200 mm mRd= 115 kNm/m d= 204 mm

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3.5 Shear design inner column C5Design shear force

The design shear force Vd is equal to the column

reaction force Nd minus the applied load within thecontrol perimeter (gd+ qd)Ac.

2 2

2 2 20.2042 0.26 2 0.26 0.204 0.21

4 4

v

c c c v

dA b b d m = + + = + + =

( ) 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,

0e =

6

3

8 100 12mm

661 10

d

u

d

Me e

V

= = =

3

4 4206 10 513mm

u cb A

= = =

1 1 0.981 1 12 513

e

u u

ke b

= = =+ +

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

Rotations

The distances rs,x and rs,y are the same as for the Level I approximation. , 1.32ms xr = , 1.23ms yr =

, ,1.5 1.5 1.32 1.23 1.91m

s s x s yb r r= = =

,

,

661 885kNm/m

8 2 8 2 1.91

d x d xd

sd x

s

M V eVm

b

= + = + =

,

,

661 183kNm/m

8 2 8 2 1.91

d y d yd

sd y

s

M V eVm

b

= + = + =

kdg is calculated at Level I.

1.5 1.5

,

,

1.32 435 851.5 1.5 0.0133

0.204 200000 115

yds x sd

x

s Rd x

fr m

d E m

= = =

governing

1.51.5

,

,

1.23 435 831.5 1.5 0.0121

0.204 200000 115

s y yd sd

y

s Rd y

r f m

d E m

= = =

6.0300.075.02040133.09.05.1

1

9.05.1

1=

+=

+=

dgkd

k

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Punching strength without shear reinforcement

The punching shear strength of the concrete is not sufficient. Consequently,

shear reinforcement is necessary. kN661kN3671020416425.1

30300.0 3

0,====

dv

c

ck

cRdVdb

fkV

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.

The bond strength is taken as fbd = 3 MPa (according to MC 2010 for

corrugated bars).

kN1223kN73436720,

====v

c

ck

cRdsysRd,maxdb

fVkV

kN661kN734 ==dRd,max

VV

+

=

+=

8

204

435

31

6

0133.02000001

6wywd

bds

swd

d

f

fE

MPa435MPa521 =>=ywdswd

f

( )( )

2, mm69090sin43598.0

367661

sin=

=

=

swde

cRdd

swk

VVA

( )2

,mm779

90sin43598.0

6615.0

sin

5.0=

=

=

swde

d

minswk

VA 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 dminus the concrete coverc on the bottom surface of the

slab:,

204 30 174v out

d d c mm= = =

Assuming a circular control perimeter for the estimation of the eccentricity,

factorke can be estimated as detailed in the right hand side column.

Possible shear reinforcement layout:

mm3468

1745.1

30300.0

10661

3

,

0 =

==

outv

c

ck

d

df

k

Vb

mm5522

3468

2

0 ===

br

out

( )99.0

5522121

1

21

1=

+=

+=

outu

ere

k

mm350599.0

34680 ===e

outk

bb

8@100@100 0.50%sw

= =

( )224 4 0.35 0.35

sw w c v v c v vA b d d b d d = +

( )220.005 4 260 204 204 4 260 0.35 204 0.35 204

swA = +

2 21263 mm 774 mm

swA = >

mm3505mm37461748004 >=+= out

b

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3.6 Shear design corner column C1 and C3Design shear force

The design shear force Vd is equal to the column reaction

forceNdminus the applied load within the control perimeter(gd+ qd)Ac.

2 2

2 2 20.2102 0.26 0.26 0.210 0.13

2 16 16

v v

c c c

d dA b b m = + + = + + =

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

Control perimeter

3 1 3

4 2 2 4 2

v c

x y c c v

d be e b b d

= = + = +

1 3 1 3260 210 144 mm

4 2 4 2x y c v

e e b d

= = + = + =

2 144 2 203mmx

e e = = =

6

3

33 10203 97 mm

110 10

d

u

d

Me e

V

= = =

3

4 4131 10 408mm

u cb A

= = =

1 10.81

1 1 97 408e

u u

ke b

= = =+ +

0 12 0.81 2 260 210 554mm

4 4

v

e e c

db k b k b

= = + = + =

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.

,1.32m

s xr =

,1.23m

s yr =

, ,1.5 1.5 1.32 1.23 1.91m

s s x s yb r r= = =

2 2 0.26 0.52msr c

b b= = = governing

,

,110 25 110 0.144 11031kN 55kN

8 8 0.52 2 2

d x d xd d

sd x

s

M V eV Vmb

= + = + = < = =

,

,

110 22 110 0.144 11025kN 55kN

8 8 0.52 2 2

d y d yd d

sd y

s

M V eV Vm

b

= + = + = < = =

1.5 1.5

, ,

,

1.32 435 551.5 1.5 0.0146

0.21 200000 69

yds x sd x

x

s Rd x

fr m

d E m

= = =

governing

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1.51.5

, ,

,

1.23 435 551.5 1.5 0.0136

0.21 200000 69

s y yd sd y

y

s Rd y

r f m

d E m

= = =

6.0280.075.02100146.09.05.1

1

9.05.1

1=

+=

+=

dgkd

k


The punching shear strength of the concrete is sufficient. Thus, no shear

reinforcement will be necessary kN110kN119102105545.1

30280.0 3

0,=>===

dv

c

ck

cRdVdb

fkV

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+ = + + =

( ),

10.1110 71kN

15.6

d d acc

d 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

It is assumed that only straight bars will be used, thus ult= 20.

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 int has to be chosen such that

int 0.12 dres

( ) ( )2

3

,

mm44220sin08.1435

1071

sin =

==ultkytyd

accd

sfff

V

A

kN2851684645.1

30,

==resint

c

ck

accddb

fV

2x2 12As= 452 mm2

(2 in each direction with a spacing of 100 mm)

mm16812103022502 ===bottomtopres chd

mm4641682

10022

2int

=+=+=

resintdsb

mm2016812.012.0mm12int

===res

d

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3.7 Shear design edge column C2Design shear force

The design shear force Vd is equal to the column reaction

forceNd minus the applied load within the control perimeter(gd+ qd)Ac.

2 2

2 2 23 0.213 0.26 0.26 0.21 0.17 m

2 8 2 8

v v

c c c

d dA b b = + + = + + =

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

Control perimeter

( )

( )

12

2 2 22

222

v v vc v c c c

c

x

vc v c

d d db d b b b

be

db d b

+ + + + + =

+ + +

2 22 6 31

4 3 2

c c v v

x

c v

b b d d e

b d

+ + =

+

2 2 2 22 6 31 1 2 260 6 260 210 3 210124mm

4 3 2 4 3 260 2 210

c c v v

x

c v

b b d d e

b d

+ + + + = = =

+ +

0mmy

e = 124mmx

e e = =

6

3

42 10124 35mm

263 10

d

u

d

Me e

V

= = =

3

4 4167 10 461mm

u cb A

= = =

1 10.93

1 1 35 461e

u u

ke b

= = =+ +

0 13 0.93 3 260 210 1031mm

2 2

v

e e c

db k b k b

= = + = + =

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.

,1.32m

s xr =

,1.23m

s yr =

, ,1.5 1.5 1.32 1.23 1.91m

s s x s yb r r= = =

,3 3 0.26 0.78m

sr x cb b= = = governing

,

0.260 1.91 1.09m2 2 2 2

c s

sr y

b bb = + = + = governing

,

,

263 42 263 0.12445kN

8 8 0.78

d x d xd

sd x

s

M V eVm

b

= + = + =

,

,

263 0 26333kN 66kN

8 2 8 2 1.09 4 4

d y d yd d

sd y

s

M V eV Vm

b

= + = + = < = =

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1.5 1.5

, ,

,

1.32 435 451.5 1.5 0.011

0.21 200000 69

yds x sd x

x

s Rd x

fr m

d E m

= = =

1.51.5

, ,

,

1.23 435 661.5 1.5 0.018

0.21 200000 69

s y yd sd y

y

s Rd y

r f m

d E m

= = =

governing

6.0247.075.0210018.09.05.1

1

9.05.1

1=

+=

+=

dgkd

k


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.

kN263kN1951021010315.1

30247.0 3

0,=

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4 Level III of approximation (detailed design or assessmentof 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 ofmsd,x

and msd,y will be maximized.

, , ,sd x d x d xym m m=

, , ,sd y d y d xym m m=

,0.64m

s xr =

,1.18m

s yr =

, ,1.5 1.5 0.64 1.18 1.30 m

s s x s yb r r= = =

,3 3 0.26 0.78m

sr x cb b= = =

,

0.26 1.300.78m

2 2 2 2

c s

sr y

b bb = + = + =

,24kNm/m

sd xm =

,43kNm/m

sd ym = (average value on support strip)

, ,

20.64m 0.52m

3

s x sr xr b= > =

, ,

21.18m 0.52m

3

s y sr yr b= > =

1.5 1.5

,

,

0.64 435 241.2 1.2 0.0016

0.21 200000 69

yds x sd

x

s Rd x

fr m

d E m

= = =

1.51.5

,

,

1.18 435 431.2 1.2 0.0073

0.21 200000 69

s y yd sd

y

s Rd y

r f m

d E m

= = =

governing

6.0395.075.02100073.09.05.1

1

9.05.1

1

=+=+=dg

kdk

The punching shear strength of the concrete is sufficient. Thus, no shear

reinforcement will be necessary kN263kN3121021010315.1

30395.0 3

0,=>===

dv

c

ck

cRdVdb

fkV

Integrity reinforcement

Since no shear reinforcement has been used and msd< mRd, integrity

reinforcement needs to be provided to prevent a progressive collapse of thestructure.

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+ = + + = ( )

,

10.1263 171kN

15.6

d d acc

d acc d

d d

g qV V

g q

+= = =

+

( ) ( )

2

3

,mm1064

20sin08.1435

10171

sin

=

=

=ultkytyd

accd

s

fff

VA

kN5221668615.1

30,

==resint

c

ck

accsddb

fV

3x3 14As= 1385 mm2

(3 in each direction with a spacing of 100 mm)

mm16614103022502 ===bottomtopres

chd

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

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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 ult= 20.

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 inthas to be chosen such that

int 0.12dres

mm8611662

20032

3int

=+=+=

resintdsb

mm2016612.012.0mm14int

===res

d

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.

Documents

NMC Example 111215