Example Composite Floor Slab

Document Ref: SX009a-EN-EU Sheet 1 of 12 Title

CALCULATION SHEET

Example: Composite floor slab

Eurocode Ref EN 1994-1-1, EN 1993-1-3, EN 1992-1-1 & EN 1993-1-1 Made by Jonas Gozzi Date March 2005 Checked by Bernt Johansson Date April 2005

Example: Composite floor slab This example shows the design of a composite floor slab for both construction stage and composite stage. The composite slab is checked at both ultimate limit state and serviceability limit state.

In many markets there are benefits in avoiding propped construction but in this example props are used for pedagogical reasons. The slab considered is propped, as shown below

1800

P PP

[mm]

18001800180018001800

P are temporary props during casting of the concrete

Sheeting data:The characteristic values for the chosen sheeting are as follows:

Yield strength fyp,k = 320 N/mm2

Thickness ts = 0,778 mm

Effective steel area Ap = 955 mm2/m

Second moment of area of steel core Ip = 33,0 × 104 mm4/m

Plastic bending resistance Mpa,Rk = 5,29 kNm/m

Sagging bending resistance = 3,41 kNm/m +Rka,M

Hogging bending resistance = 2,86 kNm/m −Rka,M

Resistance to support reaction Rw,k = 34,0 kN/m

Resistance to horizontal shear τu,Rk = 0,306 N/mm2

From trade literature

Example: Composite floor slabC

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CALCULATION SHEET



Slab data:Slab depth ht = 120 mm

Slab mean depth hred = 103,5 mm

Slab depth above sheeting hc = 75 mm

Effective depth dp = 101 mm

Concrete C25/30 fck = 25 N/mm2

Ecm = 31000 N/mm2

Sheeting geometry and slab:

10

45

30

60 15050o

27

900

[mm]

htdp

19

Partial safety factors:

• γG = 1,35 (permanent loads)

• γQ = 1,5 (variable loads)

• γM0 = 1,0

• γM1 = 1,0

• γC = 1,5

• γVS = 1,25

EN 1990

EN 1993-1-1

EN 1993-1-1

EN 1992-1-1

EN 1994-1-1

Loads:The slab is designed for both the construction stage and the composite stage. In the construction stage, the steel sheeting acts as shuttering and has to carry its own weight, the wet concrete and the construction loads. In the composite stage the slab has to carry its own weight, floor finishes and the live load. The following loads are considered in this example:


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CALCULATION SHEET



Construction stage:

Self weight of the sheeting gp = 0,09 kN/m2

Self weight of the wet concrete gc = 2,6 kN/m2

Distributed construction load q1 = 0,75 kN/m2

Concentrated construction load q2 = 1,5 kN/m2

Composite stage:

Self weight of the slab g1 = 2,5 + 0,09 = 2,6 kN/m2

Floor finishes g2 = 1,2 kN/m2

Live load (hotel) q = 5,0 kN/m2

Verification of the sheeting as shutteringThe sheeting resistance needs to be verified in the construction stage in both ultimate and serviceability limit state according to EN 1993-1-3.

Ultimate limit state: Maximum sagging bending moment:

Ed G g Q qM M M+ += ⋅ + ⋅γ γ +

= 0,81 kNm/m p c

2 20 078 0 09 1 8 0 094 2 6 1 8, , , , , ,g g gM M M+ + += + = ⋅ ⋅ + ⋅ ⋅

= 0,46 kNm/m 20 094 1 5 1 8, , ,qM + = ⋅ ⋅

= 1,78 kNm/m Ed 1 35 0 81 1 5 0 46, , , ,M + = ⋅ + ⋅


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CALCULATION SHEET



Maximum hogging bending moment:

3000

q2

gp

gc

q1

= 2,18 kNm/m Ed 1 35 1 01 1 5 0 55, , , ,G g Q qM M M− − −= ⋅ + ⋅ = ⋅ + ⋅γ γ

15,35,175,535,1qQGGEd ⋅+⋅=⋅+⋅= FFF γγ = 12,5 kN/m

gM − , qM − , Fg and Fq are calculated by computer.

Design check:Positive bending

RkRd

M0

MMγ

++ = = =

0,141,3 3,41 kNm/m > EdM + =1,78 kNm/m OK

Negative bending

RkRd

M0

MMγ

−− = = =

0,186,2 2,86 kNm/m > EdM − = 2,18 kNm/m OK

Support reaction

RkRd

M1

RRγ

= = =0,10,34 34,0 kN/m > FEd= 12,5 kN/m OK

Interaction, moment and support reaction

25,1Rdw,

Ed

Rd

Ed ≤+−

RF

MM

2 18 12 5 113 1 252 86 34 0, , , ,, ,

+ = < OK

All design checks are OK at the ultimate limit state.

EN 1993-1-3 §6.1.11 (6.28)


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http://www.access-steel.com/discovery/linklookup.aspx?id=EC163


CALCULATION SHEET



Serviceability limit state:

The deflection, δs, due to the wet concrete and the self weight of the sheeting should not, unless otherwise noted in the National Annex, exceed δs,max = L/180.

4

p cs

p

2 65 3 4384

( , , )g gEI

⋅ + ⋅ ⋅=

⋅δ

L

EN 1994-1-1 §9.6 (2)

Check if the sheeting is fully effective, i.e. does Ip need to be recalculated due to local buckling?

Maximum positive moment in serviceability limit state:

= 0,81 kNm/m 2 2sls 0 078 0 09 1 8 0 094 2 6 1 8, , , , , ,M = ⋅ ⋅ + ⋅ ⋅

Maximum compression stress in top flange:

6

slscom 4

p

0 81 10 45 1933 0 10, (,

M zI

σ ⋅= ⋅ = ⋅ −

⋅)= 63,8 N/mm2

yp

cr 28 4/

,f b t

kσλ

σ ε= =

com

235 23563 8,

εσ

= = = 1,9

with ψ = 1, Table 4.1 gives kσ = 4

p30 0 778

28 4 1 9 4 0/ ,

, , ,λ =

⋅= 0,36 1 0,ρ =

Since the reduction factor, 1 0,ρ = , no reduction of the top flange is necessary, i.e. Ip is fully effective.

4

s 42 65 0 09 3 4 2 6 1800

384 210000 33 0 10( , , , , )

,δ ⋅ + ⋅ ⋅

=⋅ ⋅ ⋅

= 3,6 mm

s,max1800

180 180Lδ = = =10 mm > 3,6 mm = δs OK

The deflection, δs, is less than 1/10 of the slab depth, therefore no ponding effects need to be taken into account. Hence, the serviceability limit state is also OK. The sheeting can work as shuttering in the construction stage.

EN 1993-1-5 § 4.4

EN 1993-1-5 Table 4.1

EN 1994-1-1 §9.3.2 (2)


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CALCULATION SHEET



Verification of the composite slab: Ultimate limit state: The continuous slab will be designed as a series of simply supported spans.

EN 1994-1-1 §9.4.2 (5)

qg1+g2

2

G 1 2 QEd 8

[ ( ) ]g g q LM

⋅ + + ⋅ ⋅=γ γ

2

Ed1 35 2 6 1 2 1 5 5 0 3 6

8[ , ( , , ) , , ] ,M ⋅ + + ⋅ ⋅

= = 20,5 kNm/m

Design bending resistance:

The sagging bending moment resistance should be calculated from the stress distribution in the figure below, if the neutral axis is above the sheeting.

dp

xpl -

+

fyp,d

0,85 fcd

z Mpl,Rd

Np

Nc,f

EN 1994-1-1 Figure 9.5

centroidal axis of the profiled steel sheeting

p yp,d

plcd0 85,

A fx

b f⋅

=⋅ ⋅

yp,kyp,d

M0

3201 0,

ff

γ= = = 320 N/mm2

ckcd

C

251 5,

ffγ

= = = 16,7 N/mm2

7,16100085,0

320955pl ⋅⋅

⋅=χ = 21,6 mm


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CALCULATION SHEET



For full shear connection:

( )pl,Rd p yd p pl 2/M A f d x= ⋅ ⋅ −

( ) 3pl,Rd 955 320 101 21 6 2 10M −= ⋅ ⋅ − ⋅, / = 27,5 kNm/m > 20,5 = MEd

Longitudinal shear by partial connection method:

Shear span required for full shear connection

c u,Rd x cN b L fNτ= ⋅ ⋅ ≤

EN 1994-1-1 §9.7.3 (8)

The distance to the nearest support, Lx, required for full shear connection can be determined by

p ydcfx

u,Rd u,Rd

A fNLb bτ τ

⋅= =

⋅ ⋅

u,Rku,Rd

Vs

0 3061 25,,

ττ

γ= = = 0,245 N/mm2

245,01000

320955x ⋅

⋅=L = 1247 mm

Hence, at a distance of 1247 mm from the support a full shear connection is fulfilled.

Design check using the simplified partial interaction diagram:

For any cross section along the span it has to be shown that the corresponding design bending moment, MEd, does not exceed the design bending resistance, MRd. In the figure x is the distance from the support.


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CALCULATION SHEET



MRd, MEd

[kNm/m]

x [m] 0 0.4 0.8 1.2 1.6 2

0

10

20

30

MRd

G 1 2 QEd 2

[ ( ) ]( )

g g q xM L x

⋅ + + ⋅ ⋅= ⋅ −

γ γ

Lx

Mpl,Rd

Mpa

Ed RdM M≤ for all cross sections

Vertical shear:

G 1 2 QEd 2

g g q LV

γ γ⋅ + + ⋅ ⋅=

[ ( ) ]

Ed1 35 2 6 1 2 1 5 5 0 3 6

2V ⋅ + + ⋅ ⋅

=[ , ( , , ) , , ] , = 22,7 kN/m

Design vertical shear resistance:

1 3v,Rd Rd,c I ck 1 cp w100 /( ) pV C k f k bρ σ⎡= ⋅ ⋅ ⋅ ⋅ + ⋅ ⋅ ⋅⎣ d⎤⎦

d

with a minimum of

v,Rd,min min 1 cp( ) w pV v k bσ= + ⋅ ⋅ ⋅

Rd,cC

0 18 0 18 0 121 5

, , ,,

Cγ

= = =

p

200 2001 1101

,kd

= + = + = 2 4

EN 1992-1-1 §6.2.2

See Note in EN 1992-1-1 §6.2.2


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CALCULATION SHEET



02,0pw

sll ≤

⋅=

dbAρ

Asl is the area of the tension reinforcement in [mm], i.e Asl = Ap

bw = 400 mm/m, i.e. the smallest width in [mm] of the section in the tension area.

02,0024,0101400

955l >=

⋅=ρ 02,0l =ρ

Edcp

c

0NA

σ = = , since NEd = 0, i.e. no axial forces or prestress.

k1 = 0,15

1 3v,Rd 0 12 2 4 100 0 02 25 0 15 0 400 101/, , ( , ) ,V ⎡ ⎤= ⋅ ⋅ ⋅ ⋅ + ⋅ ⋅ ⋅⎣ ⎦

Vv,Rd = 42,8 kN/m

See Note in EN 1992-1-1 §6.2.2

Minimum value

= 0,65 3 2 1/2 3 2 1 2min ck0 035 0 035 2 4 25/ /, , ,v k f= ⋅ ⋅ = ⋅ ⋅

/

= 26,3 kN/m v,Rd,min 0 65 0 15 0 400 101V = + ⋅ ⋅ ⋅( , , )

Vv,Rd = 42,8 kN/m > 22,7 kNm/m = VEd OK

All design checks of the composite slab in the ultimate limit state are OK.

Serviceability limit state: Cracking of concrete:

As the slab is designed as simply supported, only anti-crack reinforcement is needed. The cross-sectional area of the reinforcement above the ribs should be not less than 0,4% of the cross-sectional area of the concrete above the ribs.

min = 300 mms c0 004 0 004 1000 75, ,A b h= ⋅ ⋅ = ⋅ ⋅ 2/m

φ8 s160 mm will be enough for this purpose.

EN 1994-1-1 §9.8.1 (2)


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CALCULATION SHEET



Deflection:

For the calculations of the deflections of the slab, the slab is considered to be continuous. The following approximations apply:

• the second moment of area may be taken as the average of the values for the cracked and un-cracked section;

• for concrete, an average value of the modular ratio, n, for both long- and short-term effects may be used.

EN1994-1-1 §9.8.2 (5)

p p

cmcmcm

210000 1021 3100032 3

'

E En

EE E= = = ≈

⎛ ⎞ ⋅⋅ +⎜ ⎟⎝ ⎠

Second moment of area for the cracked section

3

2cbc p p c3

( )b x

pI A d x In

⋅= + ⋅ − +

⋅

p pi ic

i p

21 1

n A b dA zxA b n A

∑

∑

⎛ ⎞⋅ ⋅ ⋅⋅⎜ ⎟= = ⋅ +⎜ ⎟⋅⎝ ⎠

−

c10 955 2 1000 1011 1

1000 10 955x

⎛ ⎞⋅ ⋅ ⋅= ⋅ + −⎜ ⎟⎜ ⎟⋅⎝ ⎠

= 35,4 mm

3

2 4bc

1000 35 4 955 101 35 4 33 0 10 5 92 103 10

, ( , ) , ,I ⋅= + ⋅ − + ⋅ =

⋅6⋅ mm4/m

Second moment of area for the un-cracked section

22 330 p 0 p pc c c

bu u t u

2p p u p

12 2 12 2

( )

b h b h hb h b h hI x h xn n n n

A d x I

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

+

2pc

0 p t p p

c 0 p p

2 2u

hhb b h h n Ax

b h b h n A

⎛ ⎞⋅ + ⋅ ⋅ − + ⋅ ⋅⎜ ⎟

⎝ ⎠=⋅ + ⋅ + ⋅

d


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CALCULATION SHEET



275 451000 650 45 120 10 955 1012 2

1000 75 650 45 10 955ux

⎛ ⎞⋅ + ⋅ ⋅ − + ⋅ ⋅⎜ ⎟⎝ ⎠=

⋅ + ⋅ + ⋅= 58,3 mm

23 3

bu

22

4 6 4

1000 75 1000 75 75 610 4558 312 10 10 2 12 10

610 45 45120 58 3 955 101 58 310 2

33 0 10 13 5 10 mm /m

,

, (

, ,

I ⋅ ⋅ ⋅⎛ ⎞= + ⋅ − +⎜ ⎟⋅ ⋅⎝ ⎠

⋅ ⎛ ⎞

, )

+

⋅ − − + ⋅ −⎜ ⎟⎝ ⎠

⋅ = ⋅

+

Average Ib of the cracked and un-cracked section

6 6bc bub

5 92 13 5 10 9 7 10 mm /m2 2

, , ,I II + += = ⋅ = ⋅ 4

Deflections

The total deflection under the worst load case should not exceed L/250.

Weight of floor finishes:

2

4 42

c 6b

0 0068 0 0068 1 2 3600210000 9 7 10,

, , ,,g

g LE I⋅ ⋅ ⋅ ⋅

= =⋅ ⋅ ⋅

δ = 0,67 mm

EN1992-1-1 §7.4.1(4)

Live load, worst case:

q q

4 41

c 6b

0 0099 0 0099 0 7 5 0 3600210000 9 7 10,

, , , ,,q

q LE I⋅ ⋅ ⋅ ⋅ ⋅ ⋅

= =⋅ ⋅ ⋅ψδ = 2,86 mm

Removal of the props:

G1' G1' G1'

1 13 62 6

2 2LG g′ = ⋅ = ⋅

,, = 4,68 kN/m

1

3 31

6cb

0 01146 0 01146 4680 3600210000 9 7 10,

, ,,G

G LE I

δ ′

′⋅ ⋅ ⋅ ⋅= =

⋅ ⋅ ⋅= 1,23 mm


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CALCULATION SHEET



Total deflection:

21

c c cc1 23 0 67 2 86, ,,, , ,g qG ′= + + = + +δ δ δ δ = 4,76 mm

mm4,14250

3600250

mm76,4 ==<=L

cδ OK

E x a m p l e : C o m p o s i t e f l o o r s l a bC

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Example: Composite floor slab SX009a-EN-EU

Quality Record

RESOURCE TITLE Example: Composite floor slab

Reference(s)

ORIGINAL DOCUMENT

Name Company Date

Created by Jonas Gozzi SBI 10/03/2005

Technical content checked by Bernt Johansson SBI 08/04/2005

Editorial content checked by

Technical content endorsed by the following STEEL Partners:

1. UK G W Owens SCI 7/7/05

2. France A Bureau CTICM 17/8/05

3. Sweden A Olsson SBI 8/8/05

4. Germany C Muller RWTH 10/8/05

5. Spain J Chica Labein 12/8/05

Resource approved by Technical Coordinator

G W Owens SCI 06/7/06

TRANSLATED DOCUMENT

This Translation made and checked by:

Translated resource approved by:


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Documents

Example Composite Floor Slab