EUROCODES Spreadsheets Structural Design

F. A. Clignett Photography Delft - Copyright 2006

User's Guide

EUROCODESS P R E A D S H E E T SS t r u c t u r a l D e s i g n

C a r l o S i g m u n d

Edited and published by:Carlo Sigmund

Copyright 2013 Carlo Sigmund

to Excel spreadsheet fileVerification testsEN 1991-1-2: Eurocode 1

This document is made available only to allow propervalidation of the mathematical calculations carried out

on spreadsheets. Therefore, this document is notintended to replace or interpret the parts of the

standard to which it refers, and which are mentionedherein, nor does it constitute a stand-alone document.

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Copyright 2013 http://www.sigmundcarlo.netAll rights reserved. No part of this work may be reproduced, stored in a retrieval system, or transmitted by any means, electronic,mechanical, photocopying, recording, or otherwise, without prior written permission from the publisher.

------------------------------------------------

First Edition: January 2013

Sigmund, Carlo

Eurocodes - Structural Design

--------------------------------

The sponsoring editor for this document and the production supervisor was Carlo Sigmund.

Electronic mail: [email protected]

________________________________________________________

Cover Art from: F. A. Clignett Photography Delft - Copyright 2006.The Cover Art (optimized electronically) is a mirror image of the original picture.

Have not been able to contact the owner of the photograph to give full consent to the publication. The author is at the disposal of the beneficiaries.

Bridge: Erasmus Bridge Location: Rotterdam, NetherlandsLength/ main span: 802 m/284 mPylon: 139 mDesigner: Architects Ben van Berkel, Freek Loos, UN Studio.

________________________________________________________

Note: The pages of this document were created electronically using Inkscape 0.48

Copyright 1989, 1991 Free Software Foundation, Inc. 59 Temple Place, Suite 330, Boston, MA 02111-1307 USA.

www.inkscape.org

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Contents

Eurocode 1 EN 1991-1-2...........................................................................................................5

1.1 General .................................................................................................................................................... 5

1.2 Terms relating to thermal actions............................................................................................................. 5

1.3 Structural Fire design procedure.............................................................................................................. 7

1.4 Design fire scenario, design fire............................................................................................................... 7

1.5 Temperature Analysis .............................................................................................................................. 7

1.6 Thermal actions for temperature analysis (Section 3).............................................................................. 8

1.7 Nominal temperature-time curves ............................................................................................................ 9

1.8 Verification tests....................................................................................................................................... 10

1.9 References [Section 1]............................................................................................................................. 18

Eurocode 1 EN 1991-1-2Annex B ................................................................................................................19

2.1 Thermal actions for external members - Simplified calculation method................................................... 19

2.2 Verification tests....................................................................................................................................... 25

2.3 References [Section 2]............................................................................................................................. 33

Eurocode 1 EN 1991-1-2Annex C, Annex E................................................................................................35

3.1 ANNEX C: Localised fires ........................................................................................................................ 35

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EN 1991-1-2 - Eurocode 1: Actions on structures Part 1-2: General actions Actions on structures exposed to fire

CONTENTS - page iv

3.2 ANNEX E: fire load densities ....................................................................................................................38

3.3 Verification tests .......................................................................................................................................41

3.4 References [Section 3] .............................................................................................................................47

Eurocode 1 EN 1991-1-2Annex F, Annex G, Sec. B.5 Annex B................................................................................................. 49

4.1 ANNEX F: Equivalent time of fire exposure..............................................................................................49

4.2 ANNEX G: configuration factor .................................................................................................................51

4.3 ANNEX B, Section B.5: Overall configuration factors...............................................................................53

4.4 Verification tests .......................................................................................................................................53

4.5 Reference [Section 4] ...............................................................................................................................60

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Topic: Users Manual/Verification tests - EN1991-1-2_(a).xls page 5

Section 1 Eurocode 1 EN 1991-1-2

1.1 General

he methods given in this Part 1-2 of EN 1991 are applicable to buildings, with a fire load related to the building and its occupancy. This Part 1-2 of EN 1991

deals with thermal and mechanical actions on structures exposed to fire. It is intended to be used in conjunction with the fire design Parts of prEN 1992 to prEN 1996 and prEN 1999 which give rules for designing structures for fire resistance. This Part 1-2 of EN 1991 contains thermal actions related to nominal and physically based thermal actions. More data and models for physically based thermal actions are given in annexes.

In addition to the general assumptions of EN 1990 the following assumptions apply:

any active and passive fire protection systems taken into account in the design will be adequately maintained

the choice of the relevant design fire scenario is made by appropriate qualified and experienced personnel, or is given by the relevant national regulation.

The rules given in EN 1990:2002, 1.4 apply. For the purposes of this European Standard, the terms and definitions given in EN 1990:2002, 1.5 and the following apply.

1.2 Terms relating to thermal actions

FIRECOMPARTMENT.Space within a building, extending over one or several floors, which is enclosed by separating elements such that fire spread beyond the compartment is prevented during the relevant fire exposure.

FIRERESISTANCE.Ability of a structure, a part of a structure or a member to fulfil its required functions (load bearing function and/or fire separating function) for a specified load level, for a specified fire exposure and for a specified period of time.

EQUIVALENTTIMEOFFIREEXPOSURE.time of exposure to the standard temperature-time curve supposed to have the same heating effect as a real fire in the compartment.

T

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EUROCODES SPREADSHEETS STRUCTURAL DESIGN SECTION 1 EUROCODE 1 EN 1991-1-2

page 6 Topic: Users Manual/Verification tests - EN1991-1-2_(a).xls

EXTERNALMEMBER.Structural member located outside the building that may be exposed to fire through openings in the building enclosure.

GLOBALSTRUCTURALANALYSIS(FORFIRE).Structural analysis of the entire structure, when either the entire structure, or only a part of it, are exposed to fire. Indirect fire actions are considered throughout the structure.

MEMBER.Basic part of a structure (such as beam, column, but also assembly such as stud wall, truss,...) considered as isolated with appropriate boundary and support conditions.

DESIGNFIRESCENARIO.Specific fire scenario on which an analysis will be conducted.EXTERNALFIRECURVE.Nominal temperature-time curve intended for the outside of separating external walls which can be exposed to fire from different parts of the facade, i.e. directly from the inside of the respective fire compartment or from a compartment situated below or adjacent to the respective external wall.

FIRELOADDENSITY.Fire load per unit area related to the floor area , or related to the surface area of the total enclosure, including openings, .

FIRELOAD.Sum of thermal energies which are released by combustion of all combustible materials in a space (building contents and construction elements).

HYDROCARBONFIRECURVE.Nominal temperature-time curve for representing effects of an hydrocarbon type fire.

OPENINGFACTOR.Factor representing the amount of ventilation depending on the area of openings in the compartment walls, on the height of these openings and on the total area of the enclosure surfaces.

STANDARDTEMPERATURETIMECURVE.Nominal curve defined in prEN 13501-2 for representing a model of a fully developed fire in a compartment.

TEMPERATURETIMECURVES.Gas temperature in the environment of member surfaces as a function of time. They may be:

nominal: conventional curves, adopted for classification or verification of fire resistance, e.g. the standard temperature-time curve, external fire curve, hydrocarbon fire curve

parametric: determined on the basis of fire models and the specific physical parameters defining the conditions in the fire compartment.

CONVECTIVEHEATTRANSFERCOEFFICIENT.Convective heat flux to the member related to the difference between the bulk temperature of gas bordering the relevant surface of the member and the temperature of that surface.

EMISSIVITY.Equal to absorptivity of a surface, i.e. the ratio between the radiative heat absorbed by a given surface and that of a black body surface.

FLASHOVER.Simultaneous ignition of all the fire loads in a compartment.

qfqt

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EUROCODES SPREADSHEETS STRUCTURAL DESIGNSECTION 1 EUROCODE 1 EN 1991-1-2

1.3 Structural Fire design procedure

A structural fire design analysis should take into account the following steps as relevant:

selection of the relevant design fire scenarios

determination of the corresponding design fires

calculation of temperature evolution within the structural members

calculation of the mechanical behaviour of the structure exposed to fire.

Mechanical behaviour of a structure is depending on thermal actions and their thermal effect on material properties and indirect mechanical actions, as well as on the direct effect of mechanical actions.

Structural fire design involves applying actions for temperature analysis and actions for mechanical analysis according to this Part and other Parts of EN 1991. Actions on structures from fire exposure are classified as accidental actions, see EN 1990:2002, 6.4.3.3(4).

1.4 Design fire scenario, design fire

To identify the accidental design situation, the relevant design fire scenarios and the associated design fires should be determined on the basis of a fire risk assessment.

(2) For structures where particular risks of fire arise as a consequence of other accidental actions, this risk should be considered when determining the overall safety concept. Time- and load-dependent structural behaviour prior to the accidental situation needs not be considered, unless (2) applies.

For each design fire scenario, a design fire, in a fire compartment, should be estimated according to section 3 of this Part. The design fire should be applied only to one fire compartment of the building at a time, unless otherwise specified in the design fire scenario.

(3) For structures, where the national authorities specify structural fire resistance requirements, it may be assumed that the relevant design fire is given by the standard fire, unless specified otherwise.

1.5 Temperature Analysis

When performing temperature analysis of a member, the position of the design fire in relation to the member shall be taken into account. For external members, fire exposure through openings in facades and roofs should be considered.

(3) For separating external walls fire exposure from inside (from the respective fire compartment) and alternatively from outside (from other fire compartments) should be considered when required.

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Depending on the design fire chosen in section 3, the following procedures should be used:

with a nominal temperature-time curve, the temperature analysis of the structural members is made for a specified period of time, without any cooling phase;

Note ThespecifiedperiodoftimemaybegiveninthenationalregulationsorobtainedfromannexFfollowingthespecificationsofthenationalannex.

with a fire model, the temperature analysis of the structural members is made for the full duration of the fire, including the cooling phase.

Note Limitedperiodsoffireresistancemaybesetinthenationalannex.

1.6 Thermal actions for temperature analysis (Section 3)

Thermal actions are given by the net heat flux to the surface of the member. On the fire exposed surfaces the net heat flux should be determined by considering heat transfer by convection and radiation as:

(Eq.11)

where is the net convective heat flux component and is the net radiative heat flux component. The net convective heat flux component should be determined by:

(Eq.12)

where: is the coefficient of heat transfer by convection

is the gas temperature in the vicinity of the fire exposed member [C]

is the surface temperature of the member [C].

On the unexposed side of separating members, the net heat flux should be determined by using equation 1-1, with = 4 . The coefficient of heat transfer by convection should be taken as = 9 , when assuming it contains the effects of heat transfer by radiation.

The net radiative heat flux component per unit surface area is determined by:

(Eq.13)

where: is the configuration factor

is the surface emissivity of the member

is the emissivity of the fire

hnet W m2 hnet

hnet hnet c hnet r+=

hnet c hnet r

hnet c W m2 c g m =

c W m2K gm

hnetc W m2K c W m2K

hnet r W m2 m f r 273+ 4 m 273+ 4 =

mf

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is the Stephan Boltzmann constant

is the effective radiation temperature of the fire environment [C]

is the surface temperature of the member [C].

Note UnlessgiveninthematerialrelatedfiredesignPartsofprEN1992toprEN1996andprEN1999, maybeused.Theemissivityofthefireistakeningeneralas .

Where this Part or the fire design Parts of prEN 1992 to prEN 1996 and prEN 1999 give no specific data, the configuration factor should be taken as . A lower value may be chosen to take account of so called position and shadow effects.

Note Forthecalculationoftheconfigurationfactor amethodisgiveninannexG.In case of fully fire engulfed members, the radiation temperature may be represented by the gas temperature around that member. The surface temperature results from the temperature analysis of the member according to the fire design Parts 1-2 of prEN 1992 to prEN 1996 and prEN 1999, as relevant.

Gas temperatures may be adopted as nominal temperature-time curves according to 3.2, or adopted according to the fire models given in 3.3.

Note Theuseofthenominaltemperaturetimecurvesaccordingto3.2or,asanalternative,theuseofthenaturalfiremodelsaccordingto3.3maybespecifiedinthenationalannex.

1.7 Nominal temperature-time curves

STANDARDTEMPERATURETIMECURVE.The standard temperature-time curve is given by:

(Eq.14)

where: is the gas temperature in the fire compartment [C]

is the time [min].

The coefficient of heat transfer by convection is .

EXTERNALFIRECURVE.The external fire curve is given by:

(Eq.15)

where: is the gas temperature near the member [C]

is the time [min].

5 67 810 W m2K4 rm

m 0 8=f 1 0=

1=

rg

m

g

g 20 345 log10 8t 1+ +=

gt

c 25 W m2K=

g 20 660 1 0 687 e 0 32t 0 313e 3 8t +=

gt

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HYDROCARBONCURVE.The hydrocarbon temperature-time curve is given by:

(Eq.16)

where: is the gas temperature in the fire compartment [C]

is the time [min].


1.8 Verification tests

EN199112_(A).XLS.6.4 MB. Created: 3 February 2013. Last/Rel.-date: 3 May 2013. Sheets:

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CodeSec3

Annex A.

EXAMPLE 1-ASection3.1Thermalactionsfortemperatureanalysistest1Given: Determinethenetheatfluxonafireexposedsurface,incaseoffullyfireengulfed

members( aroundthemember).Suppose: (coefficientofheattransferbyconvection); (gastemperatureinthevicinityofthefireexposedmember); (surfacetemperatureofthemember); (configurationfactor);

(surfaceemissivityofthemember); (emissivityofthefire); (effectiveradiationtemperatureofthefireenvironment).[Referencesheet:CodeSec3][CellRange:A1:O1A64:O64].

Solution: Thenetconvectiveheatfluxcomponentisgivenby:.

Thenetradiativeheatfluxcomponentisdeterminedby:

Therefore,thenetheatflux(consideringheattransferbyconvectionandradiation)isgivenby:

.Now,considering with(say) , ,weget:

c 25 W m2K=

g 20 1080 1 0 325 e 0 167t 0 675e 2 5t +=

gt

c 50 W m2K=

r g c 4 00 W m2K=g 700C=m 70C= 1=

m 0 8= f 1 0= r 700C=

hnet c c g m 4 00 700 70 2520 W m2 2 52 kW m2= = = =

hnet r m f r 273+ 4 m 273+ 4 1 0 8 1 5 67108------------ 700 273+ 4 70 273+ 4 = =

hnet r 1 0 8 1 5 67108------------ 8 963 1011 0 138 1011 401011108---------- 40000 W m2 40 kW m2= = = =

hnet hnet c hnet r+ 2 52 40+ 42 52 kW m2= = =r g m 500C= g 720C=

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example-end

Hence,wefind: (seeplotabove).

EXAMPLE 1-BSection3.2Nominaltemperaturetimecurvestest2Given: Determinethestandardtemperaturetimecurveat (timeoftheexposure),

theexternalfirecurveandthehydrocarbontemperaturetimecurveat .[Referencesheet:CodeSec3][CellRange:A68:O68A190:O190].

Solution: Thestandardtemperaturetimecurveisgivenby(gastemperatureinthefirecompartment): .Sobstituting ,weget:

hnet c c g m 4 00 720 500 880 W m2 0 88 kW m2= = = =hnet r m f r 273+ 4 m 273+ 4 1 0 8 1 5 67108------------ 720 273+ 4 500 273+ 4 = =

hnet r 1 0 8 1 5 67108------------ 9 723 1011 3 570 1011 27 911011108---------- 27910 W m2 27 91 kW m2= = = =

Figure 1.1 View Plot (from input). See cells Range H63:J65 - Sheet: CodeSec3.

hnet hnet c hnet r+ 0 88 27 91+ 28 79 kW m2= = =

t 120 min=t 15 min=

g 20 345 log10 8t 1+ +=t 120 min=

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.

The external fire curve is given by (gas temperature near the member):

g 20 345 log10 8t 1+ + 20 345 log10 8 120 1+ + 20 345 2 983+ 1049C= = = =

Figure 1.2 Standard temperature-time curve.

Figure 1.3 External fire curve.

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.Sobstituting ,weget:

.

Wefindthat: for approximately.Thehydrocarbontemperaturetimecurveisgivenby(gastemperatureinthefirecompartment): .Sobstituting ,weget:

.

example-end

Wefindthat: for approximately.

EXAMPLE 1-CAnnexAParametrictemperaturetimecurvestest3Given: Forinternalmembersoffirecompartments,calculatethegastemperatureinthe

compartmentusingthemethodgivenininformativeAnnexAofEC1Part12.Thetheoryassumesthattemperatureriseisindependentoffireload.

g 20 660 1 0 687 e 0 32t 0 313e 3 8t += t 15 min=g 20 660 1 0 687 e 0 32 15 0 313e 3 8 15 + 20 660 1 0 687 e 4 8 0 313e 57 += =g 20 660 1 0 687 0 00823 0 + 676 3C=

g 680C cost= = t 40 min

g 20 1080 1 0 325 e 0 167t 0 675e 2 5 t +=t 15 min=

g 20 1080 1 0 325 e 0 167 15 0 675e 2 5 15 + 20 1080 1 0 325 e 2 505 0 675e 37 5 += =g 20 1080 1 0 325 e 2 505 0 + 20 1080 1 0 325 0 0817 0 + 1071 3C= =

Figure 1.4 Hydrocarbon curve.

g 1100C cost= = t 65 min

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Thetemperaturewithinthecompartmentisassumedtovaryasasimpleexponentialfunctionofmodifiedtimedependentonthevariationintheventilationareaandthepropertiesofthecompartmentliningsfromthisstandardcompartment.

[Referencesheet:AnnexA][CellRange:A1:O1A152:O152].Solution: Dimensionofthecompartment:

width=6,50m;lenght=15,00m;heigth=3,60m.Dimensionofwindows:numberofwindows=4;width=2,30m(meanvalue);heigth= =1,70m(weightedaverageofwindowheightsonallwalls).

Weassume:ceiling ;wallsandfloor .

Figure 1.5 Plan of fire compartment (height = 3,60 m).

[kg/m3] [J/kgK] [W/mK] [J/m2s0,5K]

CEILING 2400 1506 1,50 (a)

(a). b (thermal absorptivity) with the following limits .

WALLS 900 1250 0,24

FLOOR 900 1250 0,24

Table 1.1 Thermal properties of enclosure surfaces.

heq

c b c=

2400 1506 1 50 2328=

100 b 2200

900 1250 0 24 519 6=519 6

b 2200 J m2s0 5 K= b 520 J m2s0 5 K=

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Totalareaofverticalopeningsonallwalls:.

Totalareaofenclosure(walls,ceilingandfloor,includingopenings):.

Openingfactor:

withthefollowinglimits:0,02



With

,weget:,

.

For(say) ,wefind:

.

example-end

Roundingerror:100x(703699,7)/699,7=0,5%.

EXAMPLE 1-DAnnexAParametrictemperaturetimecurvestest4Given: Maintainingthesameassumptionsinthepreviousexampleandassuming

,calculatethecoolingphase.[Referencesheet:AnnexA][CellRange:A107:O107A152:O152].

t**max 0 2 310 qt dO-------- 0 2 310 195 110 0583------------------

2 802 0 669 2 802 1 88 h= = = =

0 5 h 0 2 310 qt dO

-------- 1 88 h 2 h=g max 250 3 t**max t* t**max x =g max 250 3 1 88 t* 1 88 1039 250 3 1 88 t* 1 88 = =

t 1 10 h t* t 1 10 2 802 3 08 h= = = =

Figure 1.6 Parametric curve: heating, cooling.

g 1039 250 3 1 88 t* 1 88 1039 250 3 1 88 3 08 1 88 703C= = =

qf d 200 MJ m2=

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Solution: Wefind:

Timefactorfunction(A.2b):,with

.

If(O>0,04andqt,d



1.9 References [Section 1]

BS EN 1991-1-2. Eurocode 1: Actions on structures Part 1-2: General actions Actions on structures exposed to fire. 26 November 2002

EN 1991-1-2:2002/AC:2013. Eurocode 1: Actions on structures - Part 1-2: General actions - Actions on structures exposed to fire. CEN Brussels, February 2013.

Manual for the design of building structures to Eurocode 1 and Basis of Structural Design - April 2010. 2010 The Institution of Structural Engineers.

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Topic: Users Manual/Verification tests - EN1991-1-2_(b).xls page 19

Section 2 Eurocode 1 EN 1991-1-2Annex B

2.1 Thermal actions for external members - Simplified calculation method

his method considers steady-state conditions for the various parameters. The method is valid only for fire loads . This method allows the

determination of: the maximum temperatures of a compartment fire

the size and temperatures of the flame from openings

radiation and convection parameters.

CONDITIONSOFUSE.When there is more than one window in the relevant fire compartment, the weighted average height of windows , the total area of vertical openings and the sum of windows widths are used.

When there are windows in only wall 1, the ratio D/W is given by:

. (Eq.27)

When there are windows on more than one wall, the ratio D/W has to be obtained as follows:

, (Eq.28)

where: is the width of the wall 1, assumed to contain the greatest window

area

is the sum of window areas on wall 1

is the width of the wall perpendicular to wall 1 in the fire compartment.

When there is a core in the fire compartment, the ratio D/W has to be obtained as follows:

T qf d 200 MJ m2

heqAv

D W W2 W1=

D W W2W1-------

Av1Av--------=

W1

Av1W2

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EUROCODES SPREADSHEETS STRUCTURAL DESIGN SECTION 2 EUROCODE 1 EN 1991-1-2 ANNEX B

page 20 Topic: Users Manual/Verification tests - EN1991-1-2_(b).xls

, (Eq.29)

where: and are the length and width of the core

and are the length and width of the 'fire compartment

the size of the fire compartment should not exceed 70 m in length, 18 m in width and 5 m in height.

(5) All parts of an external wall that do not have the fire resistance (REI) required for the stability of the building should be classified as window areas. The total area of windows in an external wall is:

the total area, according to (5), if it is less than 50% of the area of the relevant external wall of the compartment

firstly the total area and secondly 50% of the area of the relevant external wall of the compartment if, according to (5), the area is more than 50%. These two situations should be considered for calculation. When using 50% of the area of the external wall, the location and geometry of the open surfaces should be chosen so that the most severe case is considered.

The flame temperature should be taken as uniform across the width and the thickness of the flame.

EFFECTOFWINDMODEOFVENTILATION,DEFLECTIONBYWIND.If there are windows on opposite sides of the fire compartment or if additional air is being fed to the fire from another source (other than windows), the calculation shall be done with forced draught conditions. Otherwise, the calculation is done with no forced draught conditions.

Flames from an opening should be assumed to be leaving the fire compartment (see figure below):

perpendicular to the facade

with a deflection of 45 due to wind effects.

D W W2 Lc Av1W1 Wc Av---------------------------------=

Lc WcW1 W2

Figure 2.7 Deflection of flame by wind (from fig. B.1).

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EUROCODES SPREADSHEETS STRUCTURAL DESIGNSECTION 2 EUROCODE 1 EN 1991-1-2 ANNEX B

CHARACTERISTICOFFIREANDFLAMES:NOFORCEDDRAUGHT.The rate of burning or the rate of heat release is given by [MW]:

. (Eq.210)

The temperature of the fire compartment is given by [K]:

. (Eq.211)

The flame height (see Figure B.2) is given by:

(Eq.212)

where: is the total area of vertical openings on all walls

is the weighted average of windows on all walls

is the total area of enclosure (walls, ceiling and floor, including openings)

is the design fire load density related to the floor area

is the floor area of the fire compartment

is the opening factor of the fire compartment

is the free burning duration (in seconds)

Figure 2.8 Flame dimensions, no through draught (from fig. B.2).

Q minAf qf d

F------------------ 3 15 1 e

0 036O

---------------

Av heqD W--------------

1 2/

;

=

Tf 6000 1 e0 1 O

O 1 e 0 00286 T0+=

LL max 0 heq 2 37Q

Avg heqg---------------------------

2 3/ 1;

=

Av Av Av ii=

heq Av i hii Av=

At

qf d MJ m2 AfAfO Av heq At =F 1200 s=

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i the ratio (see section B.2 Conditions of use)

is the initial temperature

is the internal gas density

.

The flame width is the window width (see Figure B.2). The flame depth is 2/3 of the window height: 2/3 heq (see Figure B.2).

(6) The horizontal projection of flames: in case of a wall existing above the window, is given by: if

; if and distance to any other window > ; in other cases, (with = sum of window widths on all walls)

in case of a wall not existing above the window, is given by: .

The flame length along axis is given by: when , if wall exists above window or if ;

if no wall exists above window or if

when , then .

The flame temperature at the window is given by [K]:

(Eq.213)

with . The emissivity of flames at the window may be taken as . The flame temperature along the axis is given by [K]:

(Eq.214)

with and is the axis length from the window to the point where the calculation is made. The emissivity of flames may be taken as:

(Eq.215)

where is the flame thickness [m]. The convective heat transfer coefficient is given by [W/m2K]:

. (Eq.216)

D W Af qf d AvAt=T0 273 K 20C= =g kg m3 g 9 81 m s2=

LH heq 3=heq 1 25wt LH 0 3 heq heq wt 0 54= heq 1 25wt

4wt LH 0 454 heq heq 2wt 0 54= wt

LH 0 6 heq LL heq 1 3/=

LL 0 Lf LL heq 2+= heq 1 25wtLf LL2 LH heq 3 2+ heq 2+=heq 1 25wt

LL 0= Lf 0=

Tw520

1 0 4725Lf wtQ

--------------- ---------------------------------------------------------- T0+=

Lfwt Q 1 f 1 0=

Tz Tw T0 1 0 4725 Lx wtQ----------------

T0+=

Lxwt Q 1 Lx

f 1 e 0 3df=

df

c 4 67 1 deq 0 4 Q Av 0 6 =

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(13) If an awning or balcony is located at the level of the top of the window on its whole width for the wall above the window and , the height and horizontal projection of the flame should be modified as follows:

the flame height given in eq. 2-12 is decreased by

the horizontal projection of the flame given in (6), is increased by .

With the same conditions for awning or balcony as mentioned in (13), in the case of no wall above the window or , the height and horizontal projection of the flame should be modified as follows:

the flame height given in eq. 2-12 is decreased by

the horizontal projection of the flame , obtained in (6) with the above mentioned value of , is increased by .

FORCEDDRAUGHT.The rate of burning or the rate of heat release is given by [MW]:

. (Eq.217)

The temperature of the fire compartment is given by [K]:(1)

. (Eq.218)

(1) There were errors in the equation B.19 of Annex B of the English version of the standard. These have been corrected

in the BS EN 1991-1-2:2002.

heq 1 25wt

LL Wa 1 2+ Wa

Figure 2.9 Deflection of flame by balcony (from fig. B.3).

heq 1 25wt

LL WaLH

LL Wa

QAf qf d

F------------------

=

Tf 1200 1 e0 00288

T0+=

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The flame height (see Figure B.4) is given by:

(Eq.219)

where is the wind speed, moisture content. The horizontal projection of flames is given by:

. (Eq.220)

The flame width is given by . The flame length along axis is given by .

The flame temperature at the window is given by [K]:

(Eq.221)

with . The emissivity of flames at the window may be taken as . The flame temperature along the axis is given by [K]:

(Eq.222)

where is the axis length from the window to the point where the calculation is made. The emissivity of flames may be taken as:

(Eq.223)

LL 1 366 1 u 0 43 QAv---------- heq=

u m s

LH 0 605 u2 heq 0 22 LL heq+ =

wf wt 0 4LH+=Lf LL2 LH2+ 0 5=

Figure 2.10 Flame dimensions, through or forced draught (from fig. B.4).

Tw520

1 0 3325Lf AvQ

-------------------- --------------------------------------------------------------- T0+=

Lf Av Q 1f 1 0=

Tz Tw T0 1 0 3325 Lx AvQ---------------------

T0+=

Lx

f 1 e 0 3df=

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where is the flame thickness [m]. The convective heat transfer coefficient is given by [W/m2K]:

. (Eq.224)

Regarding the effects of balconies or awnings, see Figure B.5 (below), the flame trajectory, after being deflected horizontally by a balcony or awning, is the same as before, i.e. displaced outwards by the depth of the balcony, but with a flame length unchanged.


EN199112_(B).XLS.6.85 MB. Created: 6 February 2013. Last/Rel.-date: 3 June 2013. Sheets:

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Annex B.

EXAMPLE 2-ESectionB.2Conditionsofusetest1Given: FindtheratioD/Wwhen:

Case1)therearewindowsinonlyonewallCase2)therearewindowsonmorethanonewallCase3)thereisacoreinthefirecompartment.

df

c 9 8 1 deq 0 4 Q17 5Av------------------u1 6---------+

0 6 =

Lf

Figure 2.11 Deflection of flame by awning (from fig. B.5).

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WhenCase1)or2)applies,assumethat:thewidth ofthewall1(assumedtocontainthegreatestwindowarea)isequalto0,40m

thewidth ofthewallperpendiculartowall1inthefirecompartmentisequalto0,25m

thesum ofwindowsareasonwall1isequalto4,20m2thetotalarea ofverticalopeningsonallwallsisequalto6,80m2.WhenCase3)applies,assumethat:thelength andwidth ofthecoreareequalto5,00mand3,50mrespectivelythelength andthewidth ofthefirecompartmentareequalto6,00mand6,50mrespectively.

[Referencesheet:AnnexB][CellRange:A75:Q75CommandButton].Solution: Case1).Fromeq.(B.1):

.

Case2).Fromeq.(B.2):.

Case3).Fromeq.(B.3):.

Note Allpartsofanexternalwallthatdonothavethefireresistance(REI)requiredforthestabilityofthebuildingshouldbeclassifiedaswindowareas.

W1

W2

Av1Av

Lc WcW1 W2

D W W2W1-------

0 250 40------------ 0 625= = =

Figure 2.12 PreCalculus Excel form: procedure for a quick pre-calculation: Case 1).

D W W2W1-------Av1Av--------

0 250 40------------

4 20 6 80 ---------------- 0 386= = =

D W W2 Lc Av1W1 Wc Av---------------------------------

6 50 5 00 4 20 6 00 3 50 6 80 ---------------------------------------------------- 0 371= = =

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Using the PreCalculus Excel form, for the Case 2) we find:

Finally, for the case 3), we find:

Note Thesizeofthefirecompartmentshouldnotexceed70minlength,18minwidthand5minheight.Theflametemperatureshouldbetakenasuniformacrossthewidthandthethicknessoftheflame.

example-end



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EXAMPLE 2-FSec.B.4.1,Characteristicoffireandflames:noawningorbalconytest2Given: Assumethat: ; (takenfromAnnexE); ;

; andD/W=0,625.Assumingnoawningorbalconyislocatedatthelevelofthetopofthewindows,find:therateofburningthetemperatureofthefirecompartmenttheflameheight,widthanddepththehorizontalprojectionofflamestheflametemperatureatthewindowtheflametemperaturealongtheaxistheemissivityofflamestheconvectiveheattransfercoefficient.[Referencesheet:AnnexB][CellRange:A59:O59A264:O264andA343:O343A502:O502].

Solution: Noforceddraught.Opening factor of the fire compartment:

.

Rateofburningorrateofheatrelease(eq.210):

.

Factor : .Temperatureofthefirecompartment(eq.211):

.

Internalgasdensity,say .Flameheight(eq.212):

.

Theflamewidthisthewindowwidth:say .Flamedepth:.

Af 30 00 m2= qf d 500 MJ m2= heq 1 70 m=At 112 00 m2= Av 6 80 m2=

O Av heq At 6 80 1 70 112 0 0792 m1 2/= = =

Q minAf qf d

F------------------ 3 15 1 e

0 036O

---------------

Av heqD W--------------

1 2/

;

=

min 30 5001200------------------- 3 15 1 e

0 0360 0791------------------

6 80 1 700 625---------------

1 2/

;

min 12 5 12 9; 12 5 MW= =

Af qf d AvAt 30 500 6 80 112 543 5 MJ m2= =

Tf 6000 1 e0 1 O

O 1 e 0 00286 T0+=Tf 6000 1 e

0 1 0 0792 0 0791 1 e 0 00286 543 5 293K+=

Tf 6000 0 7171 0 2812 0 7887 273+ 955K 293K+ 1248K= =Tf 1248 273 975C= =

0 50 kg m3=

LL max 0 heq 2 37Q

Avg heqg---------------------------

2 3/ 1;

=

LL max 0 1 70 2 37 12 56 80 0 50 1 70 9 81 ------------------------------------------------------------------------ 2 3/ 1;

2 056 m= =

wt 1 00 m=2heq 3 2 1 70 3 1 13 m= =

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Horizontalprojectionofflames:Casea)Wallabovethewindowand :if , .If anddistancetoanyotherwindow> ,

.

Inothercases: .Flamelengthalongaxis(when ):

.Check:

Meanvalue: (satisfactory).Theflametemperatureatthewindowisgivenby(eq.213):

.

with (caseapplicable).

Caseb)Nowallabovethewindowor :.

.

.

Theflametemperatureatthewindowisgivenby(eq.213):

.

with (caseapplicable).Theemissivityofflamesatthewindowmaybetakenas .

heq 1 25wtheq 1 25wt LH heq 3 1 70 3 0 57 m= = =heq 1 25wt 4wt

LH 0 3heq heq wt 0 54 0 3 1 70 1 70 1 00 0 54 0 68 m= = =LH 0 454heq heq 2wt 0 54 0 454 1 70 1 70 2 1 00 0 54 0 71 m= = =

LL 0Lf LL 0 5heq+ 2 056 0 5 1 70+ 2 91 m= =

Lf LL L1+ LL LH2heq2

9-------++

2 056 0 57 2 1 70 29

-------------------++

2 056 0 68 2 1 70 29

-------------------++

2 056 0 71 2 1 70 29

-------------------++

2 86 m2 94 m2 96 m

= = = =

Lf 2 86 2 94 2 96++ 3 2 92 m LL 0 5heq+ 2 91 m= = =

Tw520

1 0 4725Lf wtQ

--------------- ---------------------------------------------------------- T0+=

Tw520

1 0 4725 2 91 1 0012 5----------------------------

----------------------------------------------------------------------- 293K+ 877K 877 273 604C= = = =

Lfwt Q 2 91 1 00 12 5 0 23 1= =

heq 1 25wtLH 0 6heq LL heq 1 3/ 0 6 1 70 2 056 1 70 1 3/ 1 09 m= = =L1 0 5heq 0 5 1 70 0 85 m= =Lf LL

2 LH heq 3 2+ 0 5heq+ 2 056 2 1 09 1 70 3 2+ 0 5 1 70 + 2 97 m= = =

Tw520

1 0 4725Lf wtQ

--------------- ---------------------------------------------------------- T0+=

Tw520

1 0 4725 2 97 1 0012 5----------------------------

----------------------------------------------------------------------- 293K+ 879K 879 273 606C= = = =

Lfwt Q 2 91 1 00 12 5 0 23 1= =f 1 0=

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Theflametemperaturealongtheaxisisgivenby(eq.214):

.

Forcasea: .

Tz Tw T0 1 0 4725 Lx wtQ----------------

T0+=

Figure 2.15 Plots eq. (B.15).

Tz 877 293 1 0 4725 Lx 1 0012 5----------------------

293+=

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For(say) ,weget(casea):

.

Forcaseb: .For(say) ,weget(caseb):

.

Flamethickness(say): ,geometricalcharacteristicofanexternalstructuralelement(diameterorside): .Emissivityofflames(eq.215): .Convectiveheattransfercoefficient(eq.216):

.

Forceddraught.Rateofburningorrateofheatrelease(eq.217):

.

Temperatureofthefirecompartment(eq.218):

.

Flameheight(eq.219),withwindspeedequalto(say) :.

Thehorizontalprojectionofflamesisgivenby(eq.220):.

Theflamewidthisgivenby: .Theflamelengthalongaxisisgivenby: .Theflametemperatureatthewindowisgivenby(eq.221):

,with (caseapplicable).Theemissivityofflamesatthewindowmaybetakenas .Theflametemperaturealongtheaxisisgivenby(eq.222):

Lx 1 45 m=

Tz 877 293 1 0 4725 1 45 1 0012 5----------------------------

293+ 845K 845 273 572C= = = =

Tz 877 293 1 0 4725 Lx 1 0012 5----------------------

293+=

Lx 1 49 m=

Tz 879 293 1 0 4725 1 49 1 0012 5----------------------------

293+ 846K 846 273 573C= = = =

df 1 00 m=deq 0 70 m=

f 1 e 0 3df 1 e 0 3 1 00 0 26= = =

c 4 67 1 deq 0 4 Q Av 0 6 4 67 1 0 70 0 4 12 5 6 80 0 6 7 8 W m2K= = =

QAf qf d

F------------------ 30 00 500

1200---------------------------- 12 50 MW= = =

Tf 1200 1 e0 00288

T0+ 1200 1 e 0 00288 543 5 293+ 1242K= = =Tf 1242 273 969C= =

u 6 00 m s=LL 1 366 1 u 0 43 QAv

---------- heq 1 366 1 6 00 0 4312 56 80----------------

1 70 1 33 m= = =

LH 0 605 u2 heq 0 22 LL heq+ 0 605 6 00 2 1 70 0 22 1 33 1 70+ 3 59 m= = =wf wt 0 4LH+ 1 00 0 4 3 59 + 2 44 m= = =

Lf LL2 LH2+ 1 33 2 3 59 2+ 3 83 m= = =

Tw520

1 0 3325Lf AvQ

-------------------- --------------------------------------------------------------- T0+

520

1 0 3325 3 83 6 8012 5--------------------------------

--------------------------------------------------------------------------- 293+ 1001K= = =

Tw 1001 273 728C= = Lf Av Q 3 83 6 80 12 5 0 8 1= =f 1 00=

Tz Tw T0 1 0 3325 Lx AvQ---------------------

T0+=

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,

with(say) . Theconvectiveheattransfercoefficientisgivenby(eq.224):

example-end

.

EXAMPLE 2-GSec.B.4.1,Characteristicoffireandflames:withawningorbalconytest3Given: Considerthesameassumptionsintheexampleabove.Findtheflameheight andthe

horizontalprojection oftheflameifanawningorbalcony(withhorizontalprojection:)islocatedatthelevelofthetopofthewindowonitswholewidth.

[Referencesheet:AnnexB][CellRange:A267:O267A340:O340].Solution: Casea),wallaboveand .

Theflameheight givenineq.212isdecreasedby :.

Thehorizontalprojectionoftheflame givenin(6),isincreasedby :

Tz 1001 293 1 0 3325 2 50 6 8012 5--------------------------------

293+ 878K 878 273 605C= = = =

Lx 2 50 m=

Figure 2.16 Plots eq. (B.25).

c 9 8 1 deq 0 4 Q17 5Av------------------u1 6---------+

0 6 =

c 9 8 1 0 70 0 4 12 517 5 6 80 ---------------------------------6 001 6------------+

0 6 25 4 W m2K= =

LLLH

Wa 0 50 m=

heq 1 25wtLL Wa 1 2+

LL* LL Wa 1 2+ 2 06 0 50 1 2+ 0 85 m= = =

LH Wa

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Caseb),nowallaboveor .Theflameheight givenin(3)isdecreasedby :

.

Thehorizontalprojectionoftheflame givenin(6),withtheabovementionedvalueof,isincreasedby :

.

example-end

.


BS EN 1991-1-2. Eurocode 1: Actions on structures Part 1-2: General actions Actions on structures exposed to fire. 26 November 2002.


EN 1991-1-2 (2002) (English): Eurocode 1: Actions on structures - Part 1-2: General actions - Actions on structures exposed to fire [Authority: The European Union Per Regulation 305/2011, Directive 98/34/EC, Directive 2004/18/EC]. European Committee for Standardisation.

LH* LH Wa+ 0 57 0 50+ 1 07 m= = =LH

* LH Wa+ 0 68 0 50+ 1 18 m= = =LH

* LH Wa+ 0 71 0 50+ 1 21 m= = =

heq 1 25wtLL Wa

LL* LL Wa 2 06 0 50 1 56 m= = =

LHLL

* Wa

LH 0 6heqLL* heq 1 3/=LL

* 1 56 m=

LH 0 6heq LL* heq 1 3/ 0 6 1 70 1 56 1 70 1 3/ 0 99 m= = =

LH* Wa LH+ 0 50 0 99+ 1 49 m= = =

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Topic: Users Manual/Verification tests - EN1991-1-2_(c).xls page 35

Section 3 Eurocode 1 EN 1991-1-2Annex C, Annex E

3.1 ANNEX C: Localised fires

he thermal action of a localised fire can be assessed by using the expression given in this annex. Differences have to be made regarding the relative height

of the flame to the ceiling. The heat flux from a localised fire to a structural element should be calculated with expression (3.1):

(Eq.325)

and based on a configuration factor established according to annex G. The flame lengths of a localised fire (see Figure C.1) is given by:

. (Eq.326)

T

hnet hnet c hnet r+ c g m mf r 273+ 4 m 273+ 4 += =

Lf

Figure 3.17 From figure C.1.

Lf 1 02D 0 0148Q2 5/+=

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EUROCODES SPREADSHEETS STRUCTURAL DESIGN SECTION 3 EUROCODE 1 EN 1991-1-2 ANNEX C, ANNEX E

page 36 Topic: Users Manual/Verification tests - EN1991-1-2_(c).xls

When the flame is not impacting the ceiling of a compartment ( ; see Figure C.1) or in case of fire in open air, the temperature in the plume along the symmetrical vertical flame axis is given by:

(Eq.327)

where: is the diameter of the fire [m], see Figure C.1

is the rate of heat release [W] of the fire according to E.4

is the convective part of the rate of heat release [W], with

is the height [m] along the flame axis, see Figure C.1

is the distance [m] between the fire source and the ceiling, see Figure C.1.

The virtual origin of the axis is given by:

. (Eq.328)

When the flame is impacting the ceiling ( ; see Figure C.2) the heat flux received by the fire exposed unit surface area at the level of the ceiling

is given by:

if

if (Eq.329) if

where:

Lf H z

z 20 0 25Qc2 3/ z z0 5 3/+ 900C=

D

Q

Qc Qc 0 8Q=z

H

z0

Figure 3.18 From figure C.2.

z0 1 02D 0 00524Q2 5/+=

Lf Hh W m2

h 100000= y 0 30h 136300 to 121000y= 0 30 y 1 0 h 15000y 3 7= y 1 0

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EUROCODES SPREADSHEETS STRUCTURAL DESIGNSECTION 3 EUROCODE 1 EN 1991-1-2 ANNEX C, ANNEX E

is a parameter given by:

is horizontal distance [m] between the vertical axis of the fire and the point along the ceiling where the thermal flux is calculated, see Figure C.2

is the distance [m] between the fire source and the ceiling, see Figure C.2.

is the horizontal flame length (see Figure C.2) given by the following relation:

(Eq.330)

is a non-dimensional rate of heat release given by:

. (Eq.331)

is the vertical position of the virtual heat source [m] and is given by:

when

when . (Eq.332)

where:

.

The net heat flux received by the fire exposed unit surface area at the level of the ceiling, is given by:

(Eq.333)

(see expressions 3.2, 3.3).

In case of several separate localised fires, expression (C.4, see eq. 3-29) may be used in order to get the different individual heat fluxes , ,... received by the fire exposed unit surface area at the level of the ceiling. The total heat flux may be taken as:

. (Eq.334)

The rules set up to this point are valid if the following conditions are met: the diameter of the fire is limited by D < 10 m; the rate of heat release of the fire is

limited by Q < 50 MW.

y y r H z+ +Lh H z+ +---------------------------=

r

H

Lh

Lh 2 9H QH* 0 33 H=

QH*

QH* Q

1 11 610 H2 5--------------------------------------=

z

z 2 4D QD* 2 5/ QD* 2 3/ = QD* 1 0z 2 4D 1 0 QD* 2 3/ = QD* 1 0

QD*Q

1 11 610 D2 5--------------------------------------=

hnet

hnet h c m 20 mf m 273+ 4 293 4 =

h1 h2

h tot h1 h2 105 W m2 + +=

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3.2 ANNEX E: fire load densities

The fire load density used in calculations should be a design value, either based on measurements or in special cases based on fire resistance requirements given in national regulations.

The design value may be determined: from a national fire load classification of occupancies; and/or

specific for an individual project by performing a fire load survey.

The design value of the fire load is defined as:

(Eq.335)

where: is the characteristic fire load density per unit floor area [MJ/m] (see

f.i. Table E.4)

is the combustion factor equal to 0,8

is a factor taking into account the fire activation risk due to the size of the compartment (see Table E.1)

is a factor taking into account the fire activation risk due to the type of occupancy (see Table E.1)

is a factor taking into account the different active fire fighting measures i (sprinkler, detection, automatic alarm transmission, firemen,). These active measures are generally imposed for life safety reason (see Table E.2 and clauses (4) and (5)):

.

For the normal fire fighting measures, which should almost always be present, such as the safe access routes, fire fighting devices, and smoke exhaust systems in staircases, the values of Table E.2 should be taken as 1,0. However, if

Compartment floor area Af [m2]

Danger of Fire Activation q1

Danger of Fire Activation q2 Examples of Occupancies

25 1,10 0,78 Art gallery, museum, swimming pool

250 1,50 1,00 Offices, residence, hotel, paper industry

2500 1,90 1,22 Manufactory for machinery & engines

5000 2,00 1,44 Chemical laboratory, painting workshop

10000 2,13 1,66 Manufactory of fireworks or paints

Table 3.2 From table E.1 - Factors , .

qf d

qf d MJ m2 qf k m q1 q2 n =

qf k

m

q1

q2

q1 q2

n

n nii 1=

10=

ni

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these fire fighting measures have not been foreseen, the corresponding value should be taken as 1,5. If staircases are put under overpressure in case of fire alarm, the factor of Table E.2 may be taken as 0,9.

CHARACTERISTICFIRELOAD.It is defined as:

(Eq.336)

where: is the amount of combustible material [kg], according to (3) and (4)

is the net calorific value [MJ/kg], see (E.2.4)

is the optional factor for assessing protected fire loads, see (E.2.3).

CHARACTERISTICFIRELOADDENSITY.It is defined as:

(Eq.337)

where is the floor area of the fire compartment or reference space.

NETCALORIFICVALUES.Net calorific values should be determined according to EN ISO 1716:2002. The moisture content of materials may be taken into account as follows:

(Eq.338)

where: is the moisture content expressed as percentage of dry weight

is the net calorific value of dry materials.

nin8

Qfi k Mk j Hui i Qfi k j = =

Mk jHuii

qf kQfi kAf

-----------=

Af

Hu Hu0 1 0 01u 0 025u=

u

Hu0

Material Hu Material Hu Material Hu

wood 17,5 Alcohols 30 Polyvinylchloride, PVC (plastic) 20

cellulosicmaterials

20 Fuels 45 Bitumen, asphalt 40

carbon 30 Pure hydrocar-bons plastics

40 Leather 20

Paraffinseries

50 ABS (plastic) 35 Linoleum 20

Olefinseries

45 Polyester (plastic) 30 Rubber tyre 30

Aromaticseries

40 Polyisocyaneratand polyurethane

(plastics)

25

Table 3.3 Net calorific values [MJ/kg] of combustible materials for calculation of fire loads.Hu

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RATEOFHEATRELEASEQ.The growing phase may be defined by the expression:

(Eq.339)

where: is the rate of heat release [W]

is the time [s]

is the time needed to reach a rate of heat release of 1 MW.

The parameter and the maximum rate of heat release , for different occupancies, are given in Table E.5. These values according to Table E.5 are valid in case of a factor (see table E.1).

For an ultra-fast fire spread, corresponds to 75 s. The growing phase is limited by an horizontal plateau corresponding to the stationary state and to a value of

given by where: is the maximum area of the fire which is the fire compartment in

case of uniformly distributed fire load but which may be smaller in case of a localised fire

is the maximum rate of heat release produced by of fire in case of fuel controlled conditions (see Table E.5).

The horizontal plateau is limited by the decay phase which starts when 70% of the total fire load has been consumed. The decay phase may be assumed to be a linear decrease starting when 70% of the fire load has been burnt and completed when the fire load has been completely burnt. If the fire is ventilation controlled, this plateau level has to be reduced following the available oxygen content, either automatically in case of the use of a computer program based on one zone model or by the simplified expression:

Occupancy Fire growth rate t [s] RHRf [kW/m2]

Dwelling Medium 300 250

Hospital (room) Medium 300 250

Hotel (room) Medium 300 250

Library Fast 150 500

Office Medium 300 250

Classroom of a school Medium 300 250

Shopping centre Fast 150 250

Theatre (cinema) Fast 150 500

Transport (public space) Slow 600 250

Table 3.4 Fire growth rate and RHRf for different occupancies (from Table E.5).

Q 106 tt----

2=

Q

t

t

t RHRf

q2 1 0=

t

Q RHRf AfiAfi m2

RHRf 1 m2kW m2

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(Eq.340)

where: is the combustion factor

is the net calorific value of wood with

the opening area

is the mean height of the openings [m].

When the maximum level of the rate of heat release is reduced in case of ventilation controlled condition, the curve of the rate of heat release has to be extended to correspond to the available energy given by the fire load. If the curve is not extended, it is then assumed that there is external burning, which induces a lower gas temperature in the compartment.


EN199112_(C).XLS.7.20 MB. Created: 12 February 2013. Last/Rel.-date: 12 February 2013. Sheets:

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Annex C

Annex E.

EXAMPLE 3-HLocalisedfirestest1Given: Thesteeltemperatureofabeam(IPE550)hastobeverified.Itispartofanunderground

carparkbelowabuilding.Thebeamsofthecarparkareaccomplishedwithoutanyuseoffireprotectionmaterial.Themostseverefirescenarioisaburningcarinthemiddleofthebeam.Findthenetheatflux(vs @generictime ):

Assumptions:diameterofthefire:D=4,00m;rateofheatreleaseQ=Q(t)=5MW(meanreferencevalueforthecalculations@generictime );distancebetweenthefiresourceandtheceiling(beam):H=3,10m;configurationfactor ;surfaceemissivityofthemember: ;emissivityofthefire: ;surfacetemperature(1)(meanvalue)ofthemember(beam),say: ;coefficientofheattransferbyconvection:

.[Referencesheet:AnnexC][CellRange:A1:O1A81:O81].

Qmax MW 0 10 m Hu Av heq =

m 0 8=Hu Hu 17 5 MJ kg=Av m2 heq

(1) The surface temperature results from the temperature analysis of the member according to the fire design Parts

1-2 of prEN 1992 to prEN 1992 to prEN 1996 and prEN 1999, as relevant.

m thnet hnet Q m; =

Q Q t .=

t 1=

m 0 70= f 1 0=

m

m 150C=c 25 W m2K=

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Solution: Thenaturalfiremodelofalocalisedfireisused.Flamelength:,

theflameisnotimpactingtheceiling(lowerflangeofasteelbeam).Virtualoriginoftheaxis:

.

Convectivepartoftherateofheatrelease: .Height[m]alongtheflameaxis,seeFigureC.1(immediatelybelowthelowerflangeofthebeam): .Temperature intheplumealongthesymmetricalverticalflameaxis:

.

Thenetheatflux(vs @generictime )iscalculatedaccordingtoSection3.1ofEN199112(theflameisnotimpactingthebeam):

,

.

For(say) ,weget:

example-end

.

EXAMPLE 3-ILocalisedfirestest2Given: Samefirescenarioofthefirstexample.Thenaturalfiremodelofalocalisedfireisused.

Findthenetheatflux(vs @generictime ):

Assumptions:diameterofthefire:D=2,00m;rateofheatreleaseQ=Q(t)=5MW(meanreferencevalueforthecalculations@generictime );distancebetweenthefiresourceandtheceiling(beam):H=2,70m;configurationfactor ;surfaceemissivityofthemember: ;emissivityofthefire: ;surfacetemperature(meanvalue)ofthemember(beam): ;coefficientofheattransferbyconvection:

.[Referencesheet:AnnexC][CellRange:A85:O85A173:O173].

Solution: Flamelength:,

theflameisimpactingtheceiling(seeFigureC.2).

Lf 1 02D 0 0148Q2 5/+ 1 02 4 00 0 0148 5 610 2 5/+ 2 997 m H= = =

z0 1 02D 0 00524Q2 5/+ 1 02 4 00 0 00524 5 610 2 5/+ 1 57 m= = =Qc 0 8Q 0 8 5 610 4 610 W= = =

z 2 98 m= z

z 20 0 25Qc2 3/ z z0 5 3/+ 20 0 25 4 610 2 3/ 2 98 1 57+ 5 3/+ 524C 900C= = =m t

hnet c z m mf z 273+ 4 m 273+ 4 +=

hnet m t; 25 524 m 1 0 70 1 0 5 67108------------ 524 273+ 4 m 273+ 4 +=m 150C=

hnet t 25 524 150 1 0 70 1 0 5 67108------------ 524 273+ 4 150 273+ 4 + 24094 W m2= =

m thnet hnet Q m; =

Q Q t .=

t 1=

m 0 70= f 1 0=m 650C=

c 25 W m2K=

Lf 1 02D 0 0148Q2 5/+ 1 02 2 00 0 0148 5 610 2 5/+ 5 037 m H= = =

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Horizontaldistancebetweentheverticalaxisofthefireandthepointalongtheceiling(steelbeam)wherethethermalfluxiscalculated,seeFigureC.2: (assumption).Nondimensionalrateofheatrelease:

.

Horizontalflamelength(seeFigureC.2):.

Verticalposition ofthevirtualheatsource( ):

.Parameter:

,with : .Therefore: .Netheatfluxreceived(vs @generictime )bythefireexposedunitsurfaceareaattheleveloftheceiling(steelbeam):

.

.

Forsay ,weget:

example-end

.

EXAMPLE 3-JFireloaddensitiestest3Given: Thegastemperatureofafullyengulfedfireinanofficehastobedetermined.Anatural

firemodelischosenforthecalculationofthegastemperature.Forfireswithaflashover,themethodofthecompartmentfirescanbeused.AsimplecalculationmethodforaparametrictemperaturetimecurveisgiveninAnnexAofEN199112.Findthedesignvalue ofthefireloaddensity.Assumptions:floorarea ;totalareaofverticalopenings: .Thefireloadconsistedof20%plastics,11%paperand69%wood,soitconsistedmainlyofcellulosicmaterial.Thereforethecombustionfactoris: .[Referencesheet:AnnexE][CellRange:A1:O1A169:O169].

z 0=

QH* Q

1 11 610 H2 5--------------------------------------

50000001 11 610 2 70 2 5--------------------------------------------------- 0 376= = =

Lh 2 9H QH* 0 33 H 2 9 2 70 0 376 0 33 2 70 2 97 m= = =z QD* 1

QD*Q

1 11 610 D2 5--------------------------------------

50000001 11 610 2 00 2 5--------------------------------------------------- 0 796 1= = =

z 2 4D QD* 2 5/ QD* 2 3/ 2 4 2 00 0 796 2 5/ 0 796 2 3/ 0 26 m= = =

y r H z+ +Lh H z+ +---------------------------

0 2 70 0 26+ +2 97 2 70 0 26+ +------------------------------------------------ 0 50= = 0 30 y 1 h

136300 to 121000y=

h 136300 W m2 to 121000 0 50 60500 W m2 = =m t

hnet h c m 20 mf m 273+ 4 293 4 =hnet m t; h 25 m 20 1 0 7 1 0 5 67108------------ m 273+ 4 293 4 =

m 650C=hnet t h t 25 650 20 1 0 7 1 0 5 67108------------ 650 273+ 4 293 4 h

44264 W m2= =hnet t 136300 to 60500 44264 92036 to 16236 W m2= =

qf dAf 135 m2= Av 27 m2=

m 0 8=

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Solution: ForthedeterminationofthefireloaddensitytheAnnexEofEN199112offersacalculationmodel.Thedesignvalueoftheloaddensitymayeitherbegivenfromanationalfireloadclassificationofoccupanciesand/orspecificforanindividualprojectbyperformingafireloadevaluation.Atthisexample,thesecondmethodischosen.FromTableE.1forcompartmentfloorareaupto : , .FromtableE.1andclauses(4)and(5): , , , ,

, .Therefore,weget:

.

Amountofcombustiblematerial[kg],accordingto(3)and(4):

Netcalorificvalueswithu=moisturecontent(%ofdryweight):

Characteristicfireload:

.

Actualcompartmentfloorarea: .Characteristicfireloaddensity perunitarea:

.

Designvalueofthefireload:

example-end

.

EXAMPLE 3-KRateofheatreleaseQtest4Given: Plotthegrowingphaseoftherate oftheheatreleaseincaseofuniformlydistributed

fireload.Occupancy:officewithamediumfiregrowthrateandmaxrateofheatrelease

Material No. Mki [kg] Huoi [MJ/kg] ui [%] ii = 1 2000 19,5 9 1,0

i = 2 3200 20 5 1,0

i = 3 1000 20 5 1,0

Table 3.5 Amount and characteristics of combustible materials.

250 m2 q1 1 50= q2 1 00=n1 n2 n3 1 0= = = n4 0 73= n5 0 87= n6 1 0=n7 0 78= n8 n9 n10 1 0= = =

n nii 1=

10 1 0 1 0 1 0 0 73 0 87 1 0 0 78 1 0 1 0 1 0 0 495= = =

Hu1 Huo1 1 0 01u1 0 025u1 19 5 1 0 01 9 0 025 9 17 5= =Hu2 Huo2 1 0 01u2 0 025u2 20 1 0 01 5 0 025 5 18 9= =Hu3 Huo3 1 0 01u3 0 025u3 20 1 0 01 5 0 025 5 18 9= =

Qfi k Mk i Hui i Qfi k j 2000 17 5 1 0 3200 18 9 1 0 1000 18 9 1 0++= = =Qfi k Qfi k j 35000 60480 18900+ + 114380 MJ= = =

Af 135 m2=

qf kqf k Qfi k Af 114380 135 847 MJ m2= =

qf d qf k m q1 q2 n 847 0 8 1 50 1 00 0 495 503 MJ m2= = =

Q

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(factor ).Maximumareaofthefirecompartmentequalto.Openingarea ;meanheightoftheopenings: .Find

thetimeatwhichthegrowingphaseislimitedbythehorizontalplateaucorrespondingtothestationarystate.[Referencesheet:AnnexE][CellRange:A174:O174A309:O309].

Solution: FromTableE.5(FiregrowthrateandRHRffordifferentoccupancies):, .

Thegrowingphasemaybedefinedbytheexpression:.

Casea):thefireisfuelcontrolled.ThegrowingphaseislimitedbyanhorizontalplateaucorrespondingtothestationarystateandtoavalueofQgivenby:

.

Horizontalplateaustartsat(seeplotabove):

.

RHRf 250 kW m2= q2 1 0=Afi 70 m2= Av 10 m2= heq 1 80 m=

t 300 s= RHRf 250 kW m2=

Q 106 tt----

2=

Figure 3.19 Growing phase limited by the horizontal plateau: the fire is fuel controlled.

Qmax RHRf Afi 250 70 17500 kW 17 5 MW= = = =

t t 1000 RHRf Afi

106------------------------------------------------ 300 1000 250 70

106----------------------------------------- 1255 s 21 min= = = =

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Thehorizontalplateauislimitedbythedecayphasewhichstartswhen70%ofthetotalfireloadhasbeenconsumed.Thedecayphasemaybeassumedtobealineardecreasestartingwhen70%ofthefireloadhasbeenburntandcompletedwhenthefireloadhasbeencompletelyburnt.

Caseb):thefireisventilationcontrolled.Inthiscasetheplateaulevelhastobereducedfollowingtheavailableoxygencontent,eitherautomaticallyincaseoftheuseofacomputerprogrambasedononezonemodelorbythesimplifiedexpressionE.6giveninAnnexAofEN199112.Atthisexample,thesecondmethodischosen:

.

Horizontalplateaustartsat(seeplotabove):

.

Whenthemaximumleveloftherateofheatreleaseisreducedincaseofventilationcontrolledcondition,thecurveoftherateofheatreleasehastobeextendedtocorrespondtotheavailableenergygivenbythefireload.Ifthecurveisnotextended,itisthen

Figure 3.20 Growing phase limited by the horizontal plateau: the fire is ventilation controlled.

Qmax MW 0 10 m Hu Av heq =Qmax MW 0 10 0 8 17 5 10 1 80 18 78 MW= =

t t 1000 Qmax kW

106---------------------------------------------- 300 1000 18780

106----------------------------------- 1300 s 22 min= = = =

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assumedthatthereisexternalburning,whichinducesalowergastemperatureinthe

example-end

compartment.


BS EN 1991-1-2. Eurocode 1: Actions on structures Part 1-2: General actions Actions on structures exposed to fire. 26 November 2002.

ECSC Project, Development of design rules for steel structures subjected to natural fires in CLOSED CAR PARKS, CEC agreement 210-AA/211/318/518/620/933, Brussels, June 1996.


EN 1991-1-2 (2002) (English): Eurocode 1: Actions on structures - Part 1-2: General actions - Actions on structures exposed to fire [Authority: The European Union Per Regulation 305/2011, Directive 98/34/EC, Directive 2004/18/EC]. European Committee for Standardisation.

Example to EN 1991 Part 1-2: Compartment fire. PART 5a: Worked examples, 2005. P. Schaumann, T. Trautmann. University of Hannover Institute for Steel Construction, Hannover, Germany.

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Topic: Users Manual/Verification tests - EN1991-1-2_(d).xls page 49

Section 4 Eurocode 1 EN 1991-1-2Annex F, Annex G, Sec. B.5 Annex B

4.1 ANNEX F: Equivalent time of fire exposure

he following approach may be used where the design of members is based on tabulated data or other simplified rules, related to the standard fire exposure.

The method given in this annex is material dependent. It is not applicable to composite steel and concrete or timber constructions.

The concept of an equivalent period of fire exposure has been used for many years to quantify a real fire exposure (based on physical parameters) with an equivalent period of heating in the standard furnace test. The basic concept relates the maximum temperature in a natural fire to the time taken to reach the same temperature in a standard furnace test. The concept has been derived largely from tests on protected steel members although it has been extended to incorporate reinforced concrete members. If fire load densities are specified without specific consideration of the combustion behaviour (see annex E) then this approach should be limited to fire compartments with mainly cellulosic type fire loads. The equivalent time of standard fire exposure is defined by :

(Eq.441)

where: is the design fire load density according to annex E

is the conversion factor according to (4)

is the ventilation factor according to (5)

is the correction factor function of the material composing structural cross-sections and defined in Table F.1.

Where no detailed assessment of the thermal properties of the enclosure is made, the conversion factor may be taken as:

T

te d min

te d qf d kb wf kc =

qf dkbwfkc

kb

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EUROCODES SPREADSHEETS STRUCTURAL DESIGN SECTION 4 EUROCODE 1 EN 1991-1-2 ANNEX F, ANNEX G, SEC. B.5 ANNEX B

page 50 Topic: Users Manual/Verification tests - EN1991-1-2_(d).xls

when is given in .

otherwise may be related to the thermal property of the enclosure according to Table F.2.

For determining b for multiple layers of material or different materials in walls, floor, ceiling, see annex A (5) and (6).

The ventilation factor [-] may be calculated as:

(Eq.442)

where: is the area of vertical openings in the facade related to

the floor area of the compartment where the limit should be observed

is the area of horizontal openings in the roof related to the floor area of the compartment

the height of the fire compartment [m].

It shall be verified that:

where is the design value of the standard fire resistance of the members, assessed according to the fire Parts of prEN 1992 to prEN 1996 and prEN 1999.

Cross-section material Correction factor kc

Reinforced concrete 1,0

Protected steel 1,0

Not protected steel 1,37 x O

Table 4.6 From Table F.1 - Correction factor in order to cover various materials (O is the opening factor defined in Annex A).

[J/m2s1/2K] kb [min.m2/MJ]

b > 2500 0,04

720 < b < 2500 0,055

b < 720 0,07

Table 4.7 From Table F.2 - Conversion factor depending on the thermal properties of the enclosure.

kb 0 07 min m2 MJ = qf d MJ m2

kb b c=

kc

b c=

kb

wf

wf6 0H

--------- 0 3

0 62 90 0 4 v 41 bvh+ --------------------------------------+ 0 5=

v Av Af= Av Af 0 025 v 0 25

h Ah Af= Ah Af

bv 12 5 1 10v v2+ 10 0=H

te d tfi d

tfi d

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EUROCODES SPREADSHEETS STRUCTURAL DESIGNSECTION 4 EUROCODE 1 EN 1991-1-2 ANNEX F, ANNEX G, SEC. B.5 ANNEX B

4.2 ANNEX G: configuration factor

The configuration factor is defined in EN 1991-1-2 Section 1.5.4.1 and measures the fraction of the total radiative heat leaving a given radiating surface that arrives at a given receiving surface. Its value depends on the size of the radiating surface, on the distance from the radiating surface to the receiving surface and on their relative orientation.

EXTERNALMEMBERS.For the calculation of temperatures in external members, all radiating surfaces may be assumed to be rectangular in shape. They comprise the windows and other openings in fire compartment walls and the equivalent rectangular surfaces of flames, see annex B. In calculating the configuration factor for a given situation, a rectangular envelope should first be drawn around the cross-section of the member receiving the radiative heat transfer, as indicated in Figure G.2 (this accounts for the shadow effect in an approximate way). The value of should then be determined for the mid-point P of each face of this rectangle.

The configuration factor for each receiving surface should be determined as the sum of the contributions from each of the zones on the radiating surface (normally four) that are visible from the point P on the receiving surface, as indicated in Figures G.3 and G.4. These zones should be defined relative to the point X where a horizontal line perpendicular to the receiving surface meets the plane containing the radiating surface. No contribution should be taken from zones that are not visible from the point P. The contribution of each zone should be determined as follows:

receiving surface parallel to radiating surface

receiving surface perpendicular to radiating surface

Figure 4.21 From Figure G.2 - Envelope of receiving surfaces.

f,1

f,3

f,2

f,4

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receiving surface in a plane at an angle to the radiating surface.

Receiving surface parallel to radiating surface:

. (Eq.443)

Receiving surface perpendicular to radiating surface:

. (Eq.444)

Receiving surface in a plane at an angle to the radiating surface:

. (Eq.445)

Figure 4.22 From Figures G.3 and G.4 - Receiving surface in a plane parallel (perpendicular) to that of the radiating surface.

f,1

f,3

12------a

1 a2+ 0 5------------------------- arcb

1 a2+ 0 5-------------------------tanb

1 b2+ 0 5------------------------- arca

1 b2+ 0 5-------------------------tan+

=

12------ arc a tan1

1 b2+ 0 5------------------------- arca

1 b2+ 0 5-------------------------tan

=

12------ arc a tan1 b cos

1 b2 2b cos+ 0 5--------------------------------------------------- arca

1 b2 2b cos+ 0 5---------------------------------------------------tan

+=

+ 12------acos

a2 sin2+ 0 5------------------------------------ arcb cos

a2 sin2+ 0 5------------------------------------ tan arc cos

a2 sin2+ 0 5------------------------------------ tan+

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4.3 ANNEX B, Section B.5: Overall configuration factors

The overall configuration factor of a member for radiative heat transfer from an opening should be determined from:

(Eq.446)

where: is the config. factor of member face i for that opening, see annex G

is the cross-sectional dimension of member face i

is the protection coefficient of member face i as follows: for a protected face; for an unprotected face.

The configuration factor for a member face from which the opening is not visible should be taken as zero. The overall configuration factor of a member for radiative heat transfer from a flame should be determined from:

(Eq.447)

where is the configuration factor of member face i for that flame, see annex G. The configuration factors of individual member faces for radiative heat transfer from flames may be based on equivalent rectangular flame dimensions. The dimensions and locations of equivalent rectangles representing the front and sides of a flame for this purpose should be determined as given in annex G. For all other purposes, the flame dimensions given in B.4 of this annex should be used.


EN199112_(D).XLS.6.40 MB. Created: 22 February 2013. Last/Rel.-date: 22 February 2013. Sheets:

Splash

Annex F

Annex G

B.5 (Annex B).

EXAMPLE 4-LEquivalenttimeoffireexposuretest1Given: Supposeafirecompartmentwithmainlycellulosictypefireloads.Thedimensionsofthe

compartmentare:width=7,04m;length=17,72m;height=3,60m.Thedimensionsofwindowsare:width=2,20m;height=1,60m;numberofwindows=3.Totalareaof

f C1f 1 C2f 2+ d1 C3f 3 C4f 4+ d2+C1 C2+ d1 C3 C4+ d2+-------------------------------------------------------------------------------------------------------------------=

f idiCi Ci 0=

Ci 1=

f iz

z C1z 1 C2z 2+ d1 C3z 3 C4z 4+ d2+C1 C2+ d1 C3 C4+ d2+---------------------------------------------------------------------------------------------------------------------=

z iz i

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horizontalopeningsonroof .Totalareaofenclosure .Designfireloaddensity ;Thermalabsorptivity ;Totalareaofverticalopeningsonallwalls .Assumeacorrectionfactor forreinforcedconcreteorprotectedsteel(EC1Part12,tableF.1,page53).Findtheequivalenttimeofstandardfireexposure.[Referencesheet:AnnexF][CellRange:A1:O1A84:O84].

Solution: Correctionfactorforreinforcedconcreteorprotectedsteelis: (seeTableF.1Correctionfactor inordertocovervariousmaterials).Conversionfactor(EC1Part12,tableF.2,page54):for is

.

Note Ifnodetailedinformationisavailableonthethermalpropertiesofthecompartmentliningsorifthereareuncertaintiesaboutthefinalconstructionorchangesmaybemadeoverthecourseofthebuildingsdesignlifethenthedefaultvalueof shouldbeused.

Actualvalueassumedforthecalculations: .Factor .Factor .Factor .Ventilationfactor(with heightofthefirecompartment):

.

.

Designfireloaddensity(accordingtoannexE):.

Equivalenttimeofstandardfireexposure:

example-end

.

EXAMPLE 4-MEquivalenttimeoffireexposuretest1bGiven: Thepreviouscalculationhasassumedasingleopeningfactorcorrespondingtoremoval

ofallglazingwithinthefirecompartmentattheonsetofflashover.Inrealitythebreakingofglassisatimedependentvariableandsensitivitystudiesshouldbeundertakentoascertaintheeffectofonlypartoftheglazingfailingduringthefire.Inthisexampleassumethatonlyhalfoftheopeningsareavailabletocontributetothecombustionprocess.Findtheequivalenttimeofstandardfireexposure.[Referencesheet:AnnexF][CellRange:A1:O1A84:O84].

Ah 0 m2= At 427 56 m2=qf d 570 MJ m2= b 1060 J m2s1 2/ K=

Av 10 56 m2= kc

kc 1 0=kc

720 b 2500 J m2s1 2/ K kb 0 055 min m2 MJ=

kb 0 07=kb 0 07=

v Av Af 10 56 124 66 0 085 0 025 0 25; = = =h Ah Af 0 124 66 0= = =bv 12 5 1 10v v2+ 12 5 1 10 0 085 0 085 2+ 23 0 10 0= = =

H 3 60 m=

wf6 0H

--------- 0 3

0 62 90 0 4 v 41 bvh+ --------------------------------------+ 0 5=

wf6 03 60------------

0 3 0 62 90 0 4 0 085 41 23 0 0 + -----------------------------------------------+ 1 76 0 5= =

qf d 570 MJ m2=

te d qf d kb wf kc 570 0 07 1 76 1 0 70 2 min= = =

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Solution: Letusassume .Therefore:factor .Factor .Ventilationfactor:

.

Equivalenttimeofstandardfireexposure:.

Ifitisassumedthatonlyhalfoftheopeningsareavailabletocontributetothecombustion

example-end

processthenthevalueoftimeequivalentincreasesby38,7%.

EXAMPLE 4-NConfigurationfactor:externalmemberstest2Given: Inthefollowinganalysis,theradiationisassumedtobeproducedby(through)awindow

rectangularinshape(width=3,00mxheight=2,00m).Thereceiver(target)isaconcretecolumn(crosssection300mmx1000mm).

Arectangularenvelopeisdrawnaroundthecrosssectionofthecolumnreceivingtheradiatingheattransfer(seeFigure4.21).

Av 0 5 10 56 m2 5 28 m2= =v Av Af 5 28 124 66 0 042 0 025 0 25; = = =bv 12 5 1 10v v2+ 12 5 1 10 0 042 0 042 2+ 17 8 10 0= = =

wf6 03 60------------

0 3 0 62 90 0 4 0 042 41 23 0 0 + -----------------------------------------------+ 2 44 0 5= =

te d qf d kb wf kc 570 0 07 2 44 1 0 97 4 min= = =

Figure 4.23 Size and geometry of the window.

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Assumethattheentireradiatingsurfaceisinaplaneparalleltothememberfacef,1andpartoftheradiatingsurface(rectangles5and6)isperpendiculartotheplaneofthememberfacef,3(seedetailsinfigure4.22).Notethatthezones(rectangles)1234arenotvisiblefromthepointP(greenone)whichisonthefacef,3(andsymmetricallyonthefacef,4).Forthecalculationsassumethat (seedetailsinfigure4.22):

Findtheconfigurationfactorforthememberfacesf,1;f,3;f,4(seefigure4.21).Note Thememberfacef,2isnotvisiblefromthepointP:nocontributionshouldbe

taken.

[Referencesheet:AnnexG][CellRange:A1:O1A183:O183].Solution: Thecolumnwidthisequalto300mm.Therefore:

(seefigure4.21).Fromtable4.8,weget:

Casea):receivingsurfacef,1paralleltoradiantsurfacewithrectangles162345.Caseb):receivingsurfacef,3(orf,4)perpendiculartoradiantsurfacewithrectangles56.

Rectangle 1 Rectangle 2 Rectangle 3 Rectangle 4 Rectangle 5 Rectangle 6

w1 = 50 cm w2 = 150 cm w3 = 150 cm w4 = 50 cm w5 = 100 cm w6 = 100 cm

h1 = 100 cm h2 = 100 cm h3 = 100 cm h4 = 100 cm h5 = 100 cm h6 = 100 cm

Table 4.8 Radiating surfaces: geometry.

s// 100 cm=

Rectangle 16 Rectangle 45

w1 = 150 cm w2 = 150 cm

h1 = 100 cm h2 = 100 cm

Table 4.9 Radiating surfaces 5+6: geometry.

Ratio Rect. 1 Rect. 2 Rect. 3 Rect. 4 Rect. 5 Rect.6 Rect. 16 Rect. 45

100/100 100/100 100/100 100/100 - - 100/100 100/100

50/100 150/100 150/100 50/100 - - 150/100 150/100

- - - - 100/115 100/115 - -

- - - - 100/115 100/115 - -

Table 4.10 Geometrical parameters [-].

s s// 0 5b+ 100 0 5 30+ 115 cm= = =

a hs//----=

b ws//----=

a hs-----=

b ws-----=

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Casea: (receivingsurfaceparalleltoradiatingsurface):

.

Caseb: (receivingsurfaceperpendiculartoradiatingsurface):

.

Casea,rectanglesno.234516:(a=1,00;b=1,50)

.

Therefore: .

Caseb,rectanglesno.56:(a=0,87;b=0,87)

, .

example-end

Therefore: .

EXAMPLE 4-OConfigurationfactor:externalmemberstest2bGiven: Consideraconcretecolumn(receivingsurfacewithcrosssection300mmx1000mm)and

averticalradiatingsurfacerectangularinshape(width=3,00mxheight=2,00m).Eachreceivingsurfaceiisinaplaneatanangle tothatoftheradiatingsurface.Findtheconfigurationfactor foreachreceivingfaceofthecolumn.Nocontributionshouldbe

f 1 2 3 45 16+ + +=

i 12------a

1 a2+ 0 5------------------------- arcb

1 a2+ 0 5-------------------------tanb

1 b2+ 0 5------------------------- arca

1 b2+ 0 5-------------------------tan+

=

f 3 f 4 5 6+= =

i 12------ arc a tan1

1 b2+ 0 5------------------------- arca

1 b2+ 0 5-------------------------tan

=

i 12------a

1 a2+ 0 5------------------------- arcb

1 a2+ 0 5-------------------------tanb

1 b2+ 0 5------------------------- arca

1 b2+ 0 5-------------------------tan+

=

i 12------1

1 12+ 0 5------------------------- arc1 5

1 12+ 0 5-------------------------tan1 5

1 1 5 2+ 0 5------------------------------------ arc1

1 1 5 2+ 0 5------------------------------------tan+

=

i 12------ 0 707 arc 1 061 tan 0 832 arc 0 555 tan+ =

i 12------ 0 707 0 815 rad 0 832 0 507 rad+ 0 159= =f 1 2 3 45 16+ + + 0 159 0 159 0 159 0 159+ + + 0 636 1= = =

i 12------ arc a tan1

1 b2+ 0 5------------------------- arca

1 b2+ 0 5-------------------------tan

=

i 12------ arc 0 87 tan1

1 0 87 2+ 0 5--------------------------------------- arc0 87

1 0 87 2+ 0 5---------------------------------------tan

=

i 12------ 0 716 0 754 arc 0 656 tan = i12------ 0 716 0 754 0 581 0 044= =

f 3 f 4 5 6+ 0 044 0 044+ 0 088 1= = = =

if i

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takenfromzonef,4thatisnotvisiblefromtheradiatingsurfaceBA.Assume:receivingsurfacef,1,radiatingsurface(AA)rectangles(23)with(input)w=79,3cm,h=200cm,s=65,0cm.Receivingsurfacef,2,radiatingsurface(BB)rectangles(65)with(input)w=79,3cm,h=200cm,s=165,0cm.Receivingsurfacef,3,radiatingsurface(BC)rectangles(65142*3*)withinputw=(79,3+141,1+79,30,86)cm=298,84cm,h=200cm,s=100cm.Seedetailsonthefigureabove.[Referencesheet:AnnexG][CellRange:A1:O1A183:O183].

Solution: Receivingsurfacef,1,radiatingsurface(AA,inputwith ):, , . ;

; ; . ,, , , .

.

Figure 4.24 Geometry for the calculations (dimensions are given in millimetres).

45=a h s 200 65 3 077= = = b w s 79 3 65 1 220= = = 4= cos 0 707=

sin 0 707= 2 cos 1 414= sin 2 0 500= 1 b2 2b cos+ 0 5 0 874=a2 sin2+ 0 5 3 157= 1 b cos 0 137= acos 2 175= b cos 0 513=

f 1 12------ arc a tan1 b cos

1 b2 2b cos+ 0 5--------------------------------------------------- arca

1 b2 2b cos+ 0 5---------------------------------------------------tan

+=

+ 12------acos

a2 sin2+ 0 5------------------------------------ arcb cos

a2 sin2+ 0 5------------------------------------ tan a

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