11

FIRE DESIGN OF HOLLOW CORE SLABS

IPHA SeminarGothenburg, 6-7 November 2007

André de ChefdebienCERIBConstruction Products Division

Yahia MsaadCERIBConstruction Products DivisionFire section

22

Plan

calculation of the temperature field on supports and at mid-span

shear resistance modelo Interaction model

o Simplified shear-flexure model

o Validation and results

33

Temperature along the strand: midspan and support

Fire - ISO 834

Air circulation 20°C

Fire test in CERIB(up side down)

FEM thermal model

44

Boundary conditions for thermal model derived from EC2-1-2

The heat capacity of air is neglected

The convective heattransfer coefficient in the cavity is the same that for non exposed surface

55

Thermal calculation

3D-temperature (°K) after 2h of Iso-834 fire curve

660

100

200

300

400

500

600

700

0 20 40 60 80 100 120 140 160 180 200

Voie n°6

Voie n°15

M03_C2

M14_C2

num

Strands temperatures

77

After 1h

0

50

100

150

200

250

300

350

-200 -150 -100 -50 0 50 100 150 200

numexpexp2

Temperature along the strandOn supportFar from support

88

Calculation method

for shear strength at high temperature

99

Interaction model

∫∫= cvdR fV max,

Calculation of a « upper bound » value of the shear strength :Integration of the shear strength along the assumed failure surface

Shear strength –normal stress interaction curvegiven by EC2-1-1 §12

)(2

2

lim,

2lim,2

lim,

2lim,

cdctdctdcdc

ccpctdcpctdcvdccp

ctdcpctdcvdccp

ffff

fffif

fffif

+−=

⎟⎟⎠

⎞⎜⎜⎝

⎛ −−+=⇒>

+=⇒≤

σ

σσσσσ

σσσ

1010

Interaction model

•Assume a 45° failure surface•Calculate the bond/anchorage capacity of reinforcement•Calculate the longitudinal stress field equilibrating MEd

and thermal strain (1st step with a reduced load, include shift rule)•Calculate VR,fi by integration of the interaction curve•Adjustment of the load level: VEdfi = VR,fi

•Iterations until convergence

l

d a laVEd = p (l/2 - la/2 - a - d)MEd = p (la/2 + a + d)(l/2 - la/4 - a/2- d/2)

1111

Compatibility of shear strain : Kinematic factor

glissement γ

τ (M

pa)

50°C

300°C

500°C

Interaction model

∫∫= iicvd

icvdR ds

kf

V,

,max,

min

,

min, γγ

γi

icvd

Gf

iicvdk ==

∫∫= iicvd

R dsf

V5.1

,max,

Assumption: shear stress / shear strain relationship at high temperature proportional to axial stress / strain relationship

1212

yksfibpdfibpd

p

fiyksfirap

sfiaRpfiaRfiaR

fAfxfl

A

fAA

FFF

+⎟⎟⎠

⎞⎜⎜⎝

⎛+=

+=

+=

φαφα

σ

2

,

2

,

,,

,,,,,,,,

.''.

φ12 fully anchoredand Ks(θ)=1

fic

CCtcCctm

pfibpd

TTkff

,

2560,50,

12,2

7,0'

γηη

⎟⎠⎞

⎜⎝⎛ +

= ( )fic

CtcCctmpfibpd

Tkff

,

25,25,12,

7,0γ

ηη=

Interaction model

Bond and achorage of reinforcement

1313

A proposal of simplified

calculation method

for shear strength

1414

Shear flexure simplified model

( )( ) dbkfkCV wficpficfilcRdficRd ,1

3/1

,,,,, 100( σρ +=

dbF

w

fiaRfil

1500

,,,

∑=ρ wb web minimal width

ratio of longitudinal reinforcements brought back to the minimal section

ficf ,mean compressive concrete strength on the total section, i.e. temperature at mid heigth of the slab

average concrete stress due to prestressing on the considered section

);)(min( ,,,20,,

c

fipaRCcppficp A

Fk °= σθσ

1515

Comparison between shear-flexure simplified model and interaction model

0

5

10

15

20

25

0 5 10 15 20 25

Vr-simplified(kN/cell)

Vr-intrinsic curve(kN/cell)

y=x

1616

comparison between shear load and shear strength given by simplified method

0

5

10

15

20

25

0 5 10 15 20 25

shear load applied (kN/cell)

shear strength(kN/cell)

shear failure observedno failure observedy=x

NO SHEAR FAILURELINE OF SHEAR FAILURE VEd,fi = VRd,fi

Data base (shear failure tests):-Arnold Van Acker: 5-Tauno Hietanen: 4

Comparison with available test results

1717

Test report.Test report. YearYear OriginOrigin Slab Slab heightheight

mmmm

Prestressing Prestressing strandsstrands

SpanSpanx widthx width

mm²²

HA12HA12 Self Self WeightWeightkN/mkN/m²²

VtestVtest//dalledallekNkN

Test Test durationduration

VrVr/dalle /dalle kNkN

DkDk--BetonelemBetonelem

ententforeningenforeningen 20052005 DenmarkDenmark 265265

2T12.5 2T12.5 a=40a=40

2,94 x 2,94 x 2,42,4 yesyes 3,653,65 89.4889.48 45'45' 90.5590.55

FEBEFEBE--RUG RUG 91589158--aa 19981998 BelgiumBelgium

265265+30+30

2T12.5 2T12.5 a=50a=50 2,9 x 2,42,9 x 2,4 yesyes 4,554,55 85.5485.54 120'120' 83.183.1

BelgiumBelgium19711971 19711971 BelgiumBelgium 265265 T12.5 a=31T12.5 a=31 2,9 x 1,22,9 x 1,2 nono 3,863,86 54.7254.72 33'33' 52.1652.16

DenmarkDenmark19981998 19981998 DenmarkDenmark 185185 T9.3 a=31T9.3 a=31 6,2 x 2,46,2 x 2,4 nono 2,622,62 43.543.5 22'22' 47.847.8

VTT/PAL VTT/PAL 43504350 19841984 FinlandFinland 265265 T9.3 a=60T9.3 a=60

5,08 x 5,08 x 2,42,4 nono 3,863,86 25.1125.11 130'130' 27.4327.43

VTT/PAL VTT/PAL 2480/822480/82 19821982 FinlandFinland 265265 T12.5 a=65T12.5 a=65 3,9 x 2,43,9 x 2,4 nono 3,63,6 44.2844.28 63'63' 43.6843.68

VTT/PAL VTT/PAL 4248/844248/84 19841984 FinlandFinland 265265 T12.5 a=64T12.5 a=64

5,185 x 5,185 x 2,42,4 nono 3,63,6 48.6548.65 49'49' 48.5448.54

VTT/PAL VTT/PAL 566d/85566d/85 19851985 FinlandFinland 265265 T12.5 a=57T12.5 a=57

5,165 x 5,165 x 2,42,4 yesyes 3,63,6 55.5655.56 77'77' 54.8654.86

VTT/PAL VTT/PAL 90228/8990228/89 19891989 FinlandFinland 265265

T9.3 a=62 T9.3 a=62 T12.5 a=92T12.5 a=92

5,165 x 5,165 x 2,42,4 yesyes 3,63,6 77.4477.44 27'27' 79.3279.32

Characteristics of tests with shear failure

1818

Influence of longitudinal bars and protruding strands

ConfigurationConfiguration VrVr ((kNkN//cellcell) ) intrinsicintrinsiccurvecurve

VrVr((kNkN//cellcell))SimplifiedSimplified methodmethod

No No φφ1212L=0 L=0

11.211.2 12.112.1

φφ1212L=0L=0

16.716.7 15.115.1

φφ1212L=0.1L=0.1

19.819.8 17.817.8

l=0.07l=0.07

HC : 270 mm

Span : 12.5 m

R120

1919

Concluding remarks on the shear resistance models

Both formulas seems to give a correct answer compared to test results of HC slabs

The “interaction method” allows to take into account the inter-action between the slabs ant supports (compression for floors with low slenderness ratio –tension for floors with cable effect), this approach could be developed for a global study (fire development, behaviour of the structure)

It is obvious that “shear flexure” simplified method is too conservative for short time fire exposure, and probably for thick HC

2020

Tentative review of of tabulated dataShear capacityVRd,fi/VRd,cold

( %)Slab thickness (mm)

Standard fire resistance

160 200 265 320 400

R30

73 70 71 76R60

66 68 56 / 62 55 / 63 52 / 70R90

60* 63 53 50 48 / 68R120

55* * 50 47 46R180

44* * * 42

* Does not satisfy the insulation criterion

In blue : « shear-flexure » in red : « interaction »

2121

Shear by prestressing Shear by thermal strain

2222

Dk-Betonelement (Denmark)

cellkNVmkNV cellfiEdfiEd /56.16/3.73 /,, =⇒=⇒

fiyksfibpdfibpd

p

fiyksfirapsfiaRpfiaRfiaR

fAfxfl

A

fAAFFF

,2

,

2

,

,,,,,,,,,,

.''.+⎟⎟

⎠

⎞⎜⎜⎝

⎛+=

+=+=

φαφα

σ

7.0)(25011070

40;0'4.460

=⇒°==+=

===⇒=

scts

ctmck

TkCTmmammx

mmalMPafMPaf

MPafTkff

MPa

Tkff

ykfis

sykfiyk

fic

stcctmpfibpd

500)()12(

59.21

7.04.47.02.1

)(7,0

,

12,

,

,12,

==×

=

=××

=

=

γφ

γηη

φ

kNFmmA

mmATfira

s

p 5.67²4.90

²186;19.0 ;5.122,

2 =⇒⎭⎬⎫

⎩⎨⎧

=

==× α

Effort anchored by reinforcement

2323

Dk-Betonelement (Denmark)

%46.122540

41 500,

,

==⇒⎭⎬⎫

⎩⎨⎧

=−==

dbmmhdmmb firaF

filω

ω ρ

MPaA

dxF

c

pfiaRficp 2.2

)(,,,, =

+=σ

( )( ) kNdbkfkCV wficpficfilcRdficRd 77.16100( ,1

3/1

,,,,, =+= σρ

MPakfffic

cckfic 8.5898.060)(

,, =×==

γθ

94.12001 =+=d

k18.0, =cRdC 15.01 =k

ratio of longitudinal reinforcements brought back to the minimal section

average concrete stress at x+d (shift rule):

Mean compressive concrete strength