ECCS Fire Des Kurs 20130429

5/22/2018 ECCS Fire Des Kurs 20130429

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A course onFire Design of Steel Structures

Eurocode 1: Actions on StructuresPart 1-2 General actions Actions on

structures exposed to fire

Eurocode 3: Design of Steel StructuresPart 1-2 General rules Structural fire design

Professor Paulo Vila Real

Stockholm, 29 of April 2013


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Introduction

Thermal Actions

Mechanical Actions

Thermal Analysis

Mechanical Analysis

Case Study

Scope

In all these topics the software for fire design of steel structural

members, Elefir-EN, will be used


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Introduction

Used Eurocodes in this course

Ambient temperature design

EN 1990 Basis of structural design

EN 1991-1-1-3, 4, 5 Actions on structures - General actions - Densities,

self-weight, imposed loads for buildings, Snow loads,

Wind loads, Thermal actions

EN 1993-1-1 Design of steel structures

General rules and rules for buildings

EN 1993-1-4 Design of steel structures

Supplementary rules for stainless steels

EN 1993-1-5 Design of steel structuresPlated structural elements

Fire design

EN 1990 Basis of structural design

EN 1991-1-2 Actions on structures exposed to fire

EN 1993-1-2 Design of steel structures Structural fire design


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Each country has its own regulations for fire safety of buildings

where the requirements for fire resistance are given

Standards for checking the structural fire resistance of the

buildings - in Europe the structural EUROCODES

Introduction

Two type of regulations or standards


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Introduction - Fire Resistance

Criteria R, E and I Example: UK Approved document B

R

E

I


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Introduction

Fire Resistance

Classif ication criteria

RER Load REILoad Load

heat hea

t

flames

flames

hotgases

hotgases

- Load bearing only: mechanical resistance (criterion R)

- Load bearing and seprarting: criteria R, E and when requested, I

R Load bearing criterion; E Integrity criterion; I Insulation criterion


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Introduction

Standard Fire Resistance Criteria R, E and I

Standard fire curve

Fire resistance is the time since the begining of the standard fire curve ISO 834

until the moment that the element doesnt fulfi ll the functions for what it has been

designed (Load bearing and/or separating functions)

2018log345 10 tT

Curva ISO 834

0

200

400

600

800

1000

1200

0 20 40 60 80 100 120

min

C

ISO 834 curve


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Introduction

Regulations for f ire safety of buildings

Normally the risk factors are:

Height of the last occupied storey in the building (h) over the reference

plane

Number of storeys below the reference plane (n)

Total gross floor area

Number of occupants (effective)

h

n

Reference plane

R30, R60, R90, ...

or

REI30, REI60. REI90, ...


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9

Introduction.

Regulations for fire safety Example: UK Approved document B


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10

Introduction.

Regulations for fire safety Example: UK Approved document B


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Introduction

The software Elefir-EN supplied with the ECCS Book


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Introduction


800 kN

800 kN

50 kNm

- 50 kNm

Nfi,Ed

M1fi,Ed

M2fi,Ed

Nfi,Ed

Member analysis and continuous beams

AB C

D

qfi,Ed

AB

qfi,Ed


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Mechanicalresponse

+

Thermal

response

Thermal

response

Introduction



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- The load-bearing function is ensured when collapse is prevented duringthe complete duration of the fire including the decay phase or

alternatively during the required period of time under standard fire

exposure.

Introduction

Prescriptive or performance-based approach

or

t

Sandard fire ISO 834

Prescriptive approacht

Natural fire

Performance-based approach


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1. Definition of the thermal loading - EC1

2. Definition of the mechanical loading - EC0 +EC1

3. Calculation of temperature evolution within the structuralmembers - EC3

4. Calculation of the mechanical behaviour of the structure

exposed to fire - EC3

Fire Design of Steel Structures

Four steps


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16

S

G

Q

Fire

W

AC

TIONS

Actions for temperature analysisThermal Action

FIRE

Actions for structural analysis

Mechanical Action

Dead Load GImposed Load QSnow SWind W

Eurocode 1:

Actions on Structures


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17

S

G

Q

Fire

W

AC

TIONS


FIRE

Eurocode 1:



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18

Thermal actions

Heat transfer at surface of building elements

rnetcnetdnet hhh,,,

Total net heat flux

dneth

,

dneth

,

dneth

,

dneth

,


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19

Thermal actions

Heat transfer at surface of building elements

Temperature of the fire

compartment

])273()273[( 44, mrmfrneth

)(, mgccneth Convective heat flux

Radiative heat flux

rg

t

or

t

Nominalfire

Naturalfire

Total net heat fluxrnetcnetdnet hhh ,,,


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20

Prescriptive Rules

(Thermal Actions given

by Nominal Fire)

Performance-Based Code

(Physically based Thermal Actions)

Design Procedures

Actions on Structures Exposed to Fire

EN 1991-1-2 - Prescript ive rules or performance-based approach

t

t


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21

Actions on Structures Exposed to Fire

EN 1991-1-2 - Prescript ive rules or performance-based approach

Simplified fire models

Nominal fires

Standard fire ISO 834External fire

Hydrocarbon fire

Natural fires

Parametric fire

Localised fire


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22

Simplified fire models

Nominal Temperature-Time Curve

200

400

600

800

1000

1200

0 1200 2400 3600

Time (s)

Gas temperature (C)

External Fire

Standard Fire

Hydrocarbon Fire

EC3 and EC9 do not use

this external fire curve.

A special Annex B on

both Eurocodes gives

a method forevaluating the heat

transfer to external

steelwork


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Boundary properties

Ceiling height

Opening Area

Fire area

Rate of heat release

Fire load density

Geometry

Fire

List of Physical Parameters needed for

Natural Fire Model


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24

Fire resistant enclosuresdefining the fire compartment

according to the national regulations

Material properties of

enclosures: c

Definition of openings

Characteristics of the Fire Compartment


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25

OccupancyFire Growth

Rate

RHRf

[kW/m]

Fire Load qf,k80% fractile

[MJ/m]

Dwelling Medium 250 948

Hospital (room) Medium 250 280

Hotel (room) Medium 250 377

Library Fast 500 1824

Office Medium 250 511

School Medium 250 347

Shopping Centre Fast 250 730

Theatre (movie/cinema) Fast 500 365

Transport (public space) Slow 250 122

Characteristics of the Fire Load from EN 1991-1-2


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26

Design value of the fire load density

[MJ/m2

]nqqk,fd,f mqq 21

10921

10

1

nnnn

i

nin

m Combustion factor. Its value is between 0 and 1. For mainly cellulosic

materials a value of 0.8 may be taken. Conservatively a value of 1 can be

used

q1 factor taking into account the fire activation risk due to the size of thecompartment

q2 factor taking into account the fire activation risk due to the type ofoccupancy

n factor taking into account the different fire fighting measures


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Fire Load DensityFire Load Density

1,90

2,00

2,13

Danger ofFire Act ivation

Compartment

floor area Af [m]

1,50

1,1025

250

2500

5000

10000

q10,78

1,00

1,22

1,44

1,66

Danger ofFire Act ivationq2

Examples

ofOccupancies

Art gallery, museum,

swimming pool

Residence, hotel, office

Manufactory for machinery& engines

Chemical laboratory,

Painting workshopManufactory of f ireworks

or paints

Automatic

Water

Extinguishing

System

Independent

Water

Supplies

Automatic fi re

Detection

& Alarm

by

Heat

by

Smoke

Automatic

Alarm

Transmission

to

Fire Brigade

Function of Act ive Fire Safety Measuresni

0 1 2

Automatic Fire Suppression Automatic Fire Detect ion

n1 n2 n3 n4 n5

0,61 0,87 or 0,73 0,871,0 0,87 0,7

Work

Fire

Brigade

Off Site

Fire

Brigade

Safe

Access

Routes

Fire

Fighting

Devices

Smoke

Exhaust

System

n10

Manual Fire Suppression

n6

n7

n8

n9

0,61 or 0,780,9 or 1

1,5

1,0

1,5

1,0

1,5

Characteristics of the Fire Load from EN 1991-1-2

k,fni2q1qd,f q.m...q


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RHR [MW]

Time [min]0 tdecay

70% (qf,d Afi)Decay phase

Rate of Heat Release Curve from EN 1991-1-2

Characteristics of the Fire Load

2

tt)t(Q

Qmax = Afi x RHRf


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Demonstration of real fire tests in car parks and high buildings Contract no. 7215 PP 025, Projecto Europeu

Car of class 3

Rate of Heat Release of a class 3 car.

Experimental evaluation: Natural Fire Model


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30

An idealized Rate of Heat Release Curve for a car burning

Characteristics of the Fire Load

30

Curve of the rate of heat release of one car

from ECSC Project: Demonstration of real fire tests in car parks and

high buildings.

Note: This curve can be introduced in Elefir-EN as an user defined RHR curve


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31

Annex C of EN 1991-1-2:

Flame is not impacting the ceiling of a compartment (Lf< H) Fires in open air

The flame length Lfof a

localised fire is given by :

Flame axis

L

z D

f

H

(z) = 20 + 0,25 (0,8 Qc)2/3 (z-z0)-5/3 900C

Lf = -1,02 D + 0,0148 Q2/5

Localised Fire:

HESKESTAD Method


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Annex C of EN 1991-1-2: Flame is impacting the ceiling (Lf> H)

Localised Fire:

HASEMI Method


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Fire load density -

Opening factor -

Wall factor -

- area of vertical openings; - total area of enclosure

Limitations :

Afloor 500 m

No horizontal openings

H 4 m

Wall factor from 1000to 2200

Fire load density, qt,d from 50 to 1000 MJ/m

dfq ,

tv AhAO /

cb

tAvA

Parametric fire

Needed parameters

Temperature = (t)


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34

Annex A of EN 1991-1-2

Parametric fire curves function of - O

For a given qf,d, b, At and Af

0

200

400

600

800

1000

0 10 20 30 40 50 60 70 80 90 100 110

[min]

[ C]

2/102.0 mO

2/106.0 mO

2/11.0 mO

2/114.0 mO

2/120.0 mO

2/105.0 mO

2/107.0 mO

Ventilation controlled fires Fuel controlled fires

Fully Engulfed Compartment

Parametric Fire


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Office

Af = 45,0 m2

O = 0,08 m1/2

qf,k = 511 MJ/m2

m = 0,8

Fully Engulfed Compartment

Parametric Fire - Influence of the Actives Fire Safety Measures

Time [min]

No Fire Active Measures

Off Site Fire Brigade

Safe access routes

Automatic Fi re Detect ion & Alarm by Smoke

Fire fighting devices

Automat ic water ext inguishing system - Spr inklers

Automatic alarm transmission to fi re br igade

qf,d (MJ/m2) 1567 815 397 210

1567 = 511x0,8x1,14x1x

1x1x1x1x1x1x1x1,5x1,5x1,5

815 = 511x0,8x1,14x1x

1x1x1x1x1x1x0,78x1x1,5x1,5

397 = 511x0,8x1,14x1x

1x1x1x0,73x1x1x0,78x1x1x1,5

210 = 511x0,8x1,14x1x

0,61x1x1x0,73x0,87x1x0,78x1x1,5x1,5

Influncia das protec es activas

0

200

400

600

800

1000

1200

1400

0 10 20 30 40 50 60 70 80

Tempo (min)

Temperatura(C)

k,fi

niq1d,f qmq = q2


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Time-temperature curves

Worked examples with the software Elefir-EN


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Compartment fire

Nominal curves

- Standard temperature-time curve ISO 834

- Hydrocarbon temperature-time curve

Parametric fire curve

Localised fire

Hasemi method

Heskestad method

User defined curve

Time-temperature curves



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Nominal Time-temperature curves



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Gas temperature. Examples with the software Elefir-EN

Standard f ire curve ISO 834 and Hydrocarbon Fire

Standard fire ISO 834

Hydrocarbon fire


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Example 3.3 from the ECCS Book: calculate and plot the

parametric time temperature curve for a bedroom in a hotel. Theplan view is rectangular with dimensions of 3.2 m by 6.4 m. The

floor to ceiling distance is 2.60 m. The floor and the ceiling are

made of normal weight concrete while the walls are made of

normal weight concrete covered by a 12.6 mm thick layer ofgypsum. The only opening is a door of size 1.10 m by 2.20 m.


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41



Normal weight concreteUnit mass = 2300 Kg/m3

Conductivity = 1.6 W/mK

Specific heat = 1000 J/KgK

b = 1918 J/(m2s1/2K)

Gypsum

Unit mass = = 1150 Kg/m3Conductivity = = 0.485 W/mKSpecific heat = c = 1000 J/KgKb = 749 J/(m2s1/2K)

Material of boundaries:

cb


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42

Hotel room

q f,d = 377 MJ/m (no active fire fighting measures)

Medium (20 min.)

Af = 3.2 x 6.4 = 20.48 m

At = 90.88 m2

H = 2.6 m

h x b = 1.1 x 2.2 m2

heq = 1.1 m

Av = 2.42 mO = 0.0395 m1/2

b = 1290 J/m2s0.5K

Building type:

Fire load:

Fire growth rate:

Floor area:

Total area:Height:

One opening:

Average window height:

Area of vertical openings:Vertical opening factor:

Wall factor:


Parametric fire curve - parameters

teqv A/hAO

AvhAhi

ivieq

iiiA/Abb


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43

t,d f,d f tq q A A

where

0,43 h < tlim

= 0,333 h fuel controlled

0,43 h > tlim = 0,333 h ventilation controlled

0,2 10-3 qt,d / O = 0,43 h


Fuel or venti lation controlled?


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44

0.2 t* 1.7 t* 19 t*g 20 1325 1 0.324 e 0.204 e 0.472 e

t* t

2

2O b

0.04 1160

Calculation of the heating curve:

where:


Heating curve

heating curve

tmax

max


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45

3

t,d

max

lim

0.2 10 q O

t t max t

The maximum temperature is needed for determining

the cooling curve:


Maximum temperature

heating curve

tmax

max


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46

t* t

3max t,dt * 0.2 10 q O

g max max625 t * t * x

where

If fire is ventilation controlled: x = 1.0

If fire is fuel controlled: x = tlim / t*max

Calculation of the cooling curve:


Cooling curve

cooling curve

tmax

max


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53

Determination of the gas temperatures at the level of the

ceiling exposed to a localized fire in a library. Natural firemodel for localised fires

EN 1991-1-2: Annex C

Localised fire

Task

Heskestad Method Hasemi Method


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54

Library

H = 3.0 m

Afi = 72 m2

m = 0.8

Building

Type:

Height:Fire area:

Combustion factor:

Localised fire

Parameters


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Localised fire

Example with the software Elefir-EN


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56

Localised fireExample with the software Elefir-EN


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Localised fireResults from Elefir-EN


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RHR

Flame legth

Temperature



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Development of the fire area

Development of the diameter



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60

Example B.1 from ECCS Book: calculate the evolution of the

rate of heat release and flame length of a localised fire in a

ticket counter that has an area of 36 m. It is assumed that theflame does not touch the ceiling.

Consider that :

- the ticket counter is considered as a library

- the fire load is is mainly cellulosic (m = 0.8)

- the height of the compartment is H = 6 m

Active fire fighting measures:

- automatic fire detection and alarm by smoke- off site fire brigade but no work fire brigade

- safe access routes

- normal fire fighting devices

- no staircases and, thus, no requirement of an air exhaust system



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RHR

Flame legth

Temperature



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Development of the fire area

Development of the diameter



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66


Example B.2 from ECCS Book: modify the RHR curve of previous

example if the fire described in Example B.1 occurs in a compartment

with five windows of 2 meters wide and 1 meter high.

21025 mAv

21mAvhAh

i

ivieq


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User defined curveExample with the software Elefir-EN

Suppose that a temperature-time curve was obtained from another software or

that the data have been obtained from a real fire. It is possible to use this

curve with the software Elefir-EN.In this example the temperature-time curve was obtained with the software OZone

V2.2 and it was copied to a text file to be read by Elefir-EN:


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User defined curveExample with the software Elefir-EN


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User defined curveResults from Elefir-EN


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4. Calculation of the mechanical behaviour of the structureexposed to fire - EC3

Fire Design of Steel StructuresFour steps


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72

S

G

Q

Fire

W

ACT

IONS


FIRE

Actions for structural analysisMechanical Action

Dead Load GImposed Load QSnow S

Wind W



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73

S

G

Q

Fire

W

ACT

IONS


FIRE

Actions for structural analysisMechanical Action

Dead Load GImposed Load QSnow S

Wind W



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At room temperature (20 C)

1. Fire is an accidental action.2. The simultaneous occurrence of other independent

accidental actions need not be considered

1,1 Qk,1 Frequent value of the representative value of the variable action Q12,1 Qk,1 Quasi-permanent value of the representative value of the variable act ion Q1A

d

Indirect thermal action due to fire induced by the restrained thermal expansion

may be neglected for member analysis

In fire situation

1

,,01,1,1,1

,,i

ikiQkQj

jkjG QQG

di

ikikj

k AQQG 1,,21,1,21,1

11, )ou(

Combination Rules for Mechanical ActionsEN 1990: Basis of Structural Design


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Action 1 2Imposed loads in buildings, category

(see EN 1991-1-1)0.5 0.3

Imposed loads in congregation

areas and shopping areas0.7 0.6

Imposed loads in storage areas 0.9 0.8

vehicle weight 30 kN 0.7 0.630 kN vehicle weight 160 kN 0.5 0.3Imposed loads in roofs 0.0 0.0

Snow (Norway, Sweden ) 0.2 0.0

Wind loads on bui ldings 0.2 0.0

The Norwegian Annex

recommends 1,Q1,only when the wind isthe dominant action and

the Swedish Annex

recommends 1,Q1 forall dominant variable

actions. In bothCountries wind is

always considered and

so horizontal actions are

always taken into

account

di

ikikj

k AQQG

1,,21,1,21,1

11,

)ou(



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76

0.2W1.0G + 0.3Q

1.0G + 0.3Q

1.0G + 0.3Q

1.0G + 0.3Q

1.0G + 0.3Q

1.0G + 0.3Q

1.0G + 0.3Q

1.0G + 0.3Q

0N0N

1.0G + 0.3Q

1.0G + 0.3Q

1.0G + 0.3Q

1.0G + 0.3Q

1.0G + 0.3Q

1.0G + 0.3Q

1.0G + 0.3Q

1.0G + 0.3Q

Fire situation


ii212,1dfi, QQGE

ii211,1dfi, QQGE

Efi,d = G + 0,2 W + 0,3 Q Efi,d = G + 0,0 W + 0,3 Q

Representative value of the variable action : wind

or

ii212,11,1dfi, Q)Qor(GE


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77

The effect of actions

can be obtained from:

dfidfi EE

,

whereLoad combination in fire situation

Load combination at 20 C

Design value of the

effect of actionsat normal temperature

1,1,

1,1,1

kQkG

kkGAfi

QGQG

di ikikj k

AQQG

1 ,,21,1,21,11 1,

)ou(

Instead of using:

as a simplification can be taken as 0.65

Member analysisSimplified rule


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78

5m

Characteristic loading (kN/m):

Permanent Gk = 11.82

Variablel Qk,1= 22.8

- At normal temperature

G Gk + Q.1 Q1 = 1.35x11.82 + 1.5x22.8 = 50.16 kN/mDesign value of the moment at 20 C:

MEd = 50.16x52/8

= 156.75 kNm

- In fire situation

Gk +1.1 Q1 = 11.82 + 0.5x22.8 = 23.22 kN/mDesign value of the moment in fi re situation:

Mfi,Ed = 23.22x52/8

= 72.6 kNmMfi.Ed = fi MEd

=0.463 x 156.75 = 72.6 kNm

fi =23.22 / 50.16 = 0.463

EC3fi =0.65

or

Actions on StructuresExample: Office building

On the

safe

side


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Fire Design of Steel StructuresFour steps


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80

Heat conduction equation

tcQ

yyxx p

)( cc hq

)()())(()(2244

r ara

h

aaa hqr

convection

radiation

Boundary conditions

Thermal response

Note: this equation can be simplified for the case of current steel profiles


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Thermal properties of carbon steel and stainless steel

0

10

20

30

40

50

60

0 200 400 600 800 1000 1200

(C)

a (W/mK)Stainless steel

Carbon s teel

0

1000

2000

3000

4000

5000

0 200 400 600 800 1000 1200

(C)

Ca (J/kgK)

Carbon s teel

Stainless s teel

Thermal conductivity (W/mK) Specific Heat (J/kgK)


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Temperature increase in time step t:

448

27327310675

mrmfr,net x,h

Heat flux has 2 parts:

Radiation:

mgcc,net

h

Convection:

Steel

temperature

Steel

Fire

temperature

Temperature increase of unprotected steelSimplified equation of EC3

thc

VAk d,netaa

msht.a

d,neth


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perimeter

c/s area

exposed perimeter

c/s area

h

b

2(b+h)

c/s area

thc

VAk d,netaa

msht.a

Section factor Am/VUnprotected steel members


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c/s area

b

2(b+h)

h

c/s area

h

2h+b

b

bm V][ A - Section factor as the profile has a hollow encasement fire protection

For I-sections under nominal fire: ksh = 0.9 [Am/V]b/[Am/V]

In all other cases: ksh = [Am/V]b/[Am/V]

For cross-sections convex shape: ksh = 1

Correction factor for theShadow effect ksh


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Structural fire protection

Passive Protection

Insulating BoardGypsum, Mineral fibre, Vermiculite.

Easy to apply, aesthetically acceptable.

Difficulties with complex details.

Cementitious Sprays

Mineral fibre or vermiculite in cement binder.

Cheap to apply, but messy; clean-up may be expensive.

Poor aesthetics; normally used behind suspended ceilings.

Intumescent Paints

Decorative finish under normal conditions.

Expands on heating to produce insulating layer.

Can be done off-site.


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Structural fire protection

Columns:

Beams:

St t l fi t ti


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Structural fire protectionIntumescent paint

St t l fi t ti


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Structural fire protectionCementitious Sprays

St t l fi t ti


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Structural fire protectionInsulating Board

Temperature increase of protected steel


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Steeltemperature

Steel

Protection

Fire

temperature

dp

Some heat stored in

protection layer.

V

Ad

c

cpp

aa

pp

Heat stored in protection layer

relative to heat stored in steel

t.g/t.at.gp

aa

ppt.a et

/V

A

c

d/

1

31

1 10

Temperature rise of steel in t ime

increment t

Temperature increase of protected steelSimplified equation of EC3

Section factor Ap/V


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h

b

t.g/

t.at.g

p

aa

pp

t.aet

/V

A

c

d/

1

31

1 10

Section factor Ap/VProtected steel members

Steel perimeter

steel c/s area

2(b+h)

c/s area

inner perimeter

of board

steel c/s area

Moist fire protection materials


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Moist fire protection materialsEffect of the moisture content

ISO834

Temp (C)

0Time (min)

crit

30 60 90

Time delay tv in the rise of

steel temperature when itreaches 100 C.

p

pp

v

dpt

5

2

p = moisture content in %

With moisture content

tv

100 C

Temperature development in protected and unprotected steel


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Temperature development in protected and unprotected steelWorked examples with the software Elefir-EN



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Temperature development in protected and unprotected steelWorked examples

Example 4.1 from ECCS Book: What is the temperature of an

unprotected rectangular bar with a cross-section of 200 x 50 mm2after 30 minutes of standard fire exposure on four sides?

15005020

0502022

m..

)..(

tb

)tb(V/Am

1shk

1-m505001 .]/[ VAk msh

Answer:As rectangular solid sections are not in the data base of the

software Elefir-EN the section factor must be evaluated first:



95/303

95


Other



96/303

96


Temperature development in protected and unprotected steelW k d l


97/303

97


Other



98/303

98

e pe a u e de e op e p o ec ed a d u p o ec ed s eeWorked examples



99/303

99

p p p pWorked examples


unprotected circular hollow section with a diameter of d = 220 mmand a thickness of t = 5 mm after 60 minutes of standard fire curve

exposure?

d

t



100/303

100


Other



101/303

101




102/303

102

Worked examples



103/303

103

Worked examples


unprotected HE 200 A profile after 30 minutes of standard firecurve exposure on four sides?



104/303

104

Worked examples



105/303

105

Worked examples



106/303

106

Worked examples

Example 4.4 from ECCS Book: What would be the thickness of

fibre-cement board encasement for a IPE 300 heated on threesides to be classified as R90 if the critical temperature is 654 C?



107/303

107

p



108/303

108

p



109/303

109



110/303

110

Example 4.5 from ECCS Book: How much time is needed for a column

of HE220B protected with gypsum board encasement with 20 mm ofthickness to reach the temperature of 559 C when heated by the

standard fire curve on four sides?



111/303

111



112/303

112



113/303

113



114/303

114



115/303

115



116/303

116



117/303

117

Right -clicking on the chart the

temperature-time curves can be

copied to be used in another

program, for instance with theExcel



118/303

118

Data obtained with the program Elefir-EN and chart made with Excel

0

200

400

600

800

1000

1200

0 20 40 60 80 100 120

Time [min.]

Temperat

ure[C]

ISO 834

Without moisture

content

With moisture content

Critical temp. 559 C

Effect of the moisture content in the temperature of the HE 220 B



119/303

119

Example from section 4.11.1 from ECCS Book: What is the

temperature of the unprotected circular hollow section of Example 4.2after 60 minutes of standard fire curve exposure if it is made of

stainless steel?



120/303

120



121/303

121



122/303

122

Localised fireTask


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123

Determination of the gas temperatures at the level of the

ceiling exposed to fire by burning cars and the temperature

of an IPE 500 using natural fire model for localised fires

EN 1991-1-2: Annex C

Heskestad Method Hasemi Method

Localised fires in a car parkFive fire scenarios


124/303

124

H = 2.7 mD = 2.0 m

IPE 500

Height:Diameter of flame:

Steel Beams:

Fire Scenario 1

Fire Scenario 2 Fire Scenario 4

20 x 2.40 m

16.00 m

START

Fire Scenario 3

START

START

Fire Scenario 5

START

Most severe scenario

Localised fire

Two fire scenarios. Horizontal distance


125/303

125r = 0.0 m

Horizontal distance

from flame axis to beam: r1 = r2 = 5.63 m

r3 = 6.57 m

Horizontal distance

from flame axis to beam:

Scenario 1 Scenario 2

2

2

2

1 ddr

i

r1r2r3

XX16 m

5 m

3rd 1st2nd

d1

d2ir

Localised fire

Rate of heat release


126/303

126

Curve of the rate of heat release of one car

from ECSC Project: Demonstration of real fire tests in car parks and

high buildings.

Localised fire

Rate of heat release of three cars


127/303

127

Curve of the rate of heat release of each car

1st 2nd 3rd

from ECSC Project: Demostration of real fire tests in car parks and

high buildings.

Localised fire

Flame length

Heskestad Method


128/303

128

if Lr H Hasemi method has to be used

if Lr< H Heskestad method has to be used

0

1

2

3

4

5

6

7

Height[m]

0 10 20 30 40 50 60 70

Time[min]

Flame Length

Ceiling (H)

Hasemi Method

Localised fire



129/303

129

Localised fire


Scenario 1


130/303

130

Localised fire


Scenario 1


131/303

131

Localised fire

Results from Elefir-EN

Scenario 1


132/303

132

Localised fire


Scenario 2


133/303

133

Scenario 2

Localised fire



134/303

134

Scenario 2

Localised fire



135/303

135

Scenario 2

Localised fire



136/303

136

Scenario 2

Localised fire



137/303

137

Fire Design of Steel Structures

Four Steps


138/303



3. Calculation of temperature evolution within the structural

members - EC3


Degree of simplif ication of the structure


139/303

139

Analysis of: a) Global structure; b) Parts of the structure; c) Members

a) b)

c)

a) b)

c)

Stress (N/mm2)

Mechanical properties of carbon steel

Stress-strain relationship at elevated temperatures


140/303

140Strain (%)

0.5 1.0 1.5 2.00

300

250

200

150

100

50

20C

200C300C

400C

500C

600C

700C

800C

Steel softens progressively

from 100-200C up.

Only 23% of ambient-

temperature strength

remains at 700C.

At 800C strength reduced

to 11% and at 900C to

6%.

Melts at about 1500C.




141/303

141

Strain

Stress

E = tan a,

y,p, u,

fy,

fp,

t,

= 2% = 15% = 20%

St th/ tiff d ti

Stress (N/mm2)

30020C




142/303

142

Strength/stiffness reduction

factors for elastic modulusand yield strength (2% total

strain).

Strain (%)0.5 1.0 1.5 2.00

300

250

200

150

100

50

20C

200C300C

400C

500C

600C

700C

800C

Elastic modulus at 600Creduced by about 70%.

Yield strength at 600C

reduced by over 50%.

f f

% of the value at 20 C

Reduction factors for stress-strain relationship of carbon steel

at elevated temperatures


143/303

143

yyy ffk /,,

Yield Strength

Reduction factor for

yield strength andYoung modulus of

carbon steel

0 300 600 900 1200

1

.8

.6

.4

.2

Temperature (C)

Young Modulus

aaE EEk /,,

Reduction factors for stress-strain relationship of carbon steel


Reduction factors at temperature a relative to the value of fy or Ea

at 20C

Steel Reduction factor Reduction factor Reduction factor


144/303

144

Temperature

a

(relative tofy)for effective yield

strength

ky, = fy,/fy

(relative tofy)for proportional limit

kp, = fp,/fy

(relative toEa)for the slope of thelinear elastic range

kE, = Ea,/Ea

20C 1,000 1,000 1,000

100C 1,000 1,000 1,000

200C 1,000 0,807 0,900

300C 1,000 0,613 0,800

400C 1,000 0,420 0,700

500C 0,780 0,360 0,600

600C 0,470 0,180 0,310

700C 0,230 0,075 0,130

800C 0,110 0,050 0,090

900C 0,060 0,0375 0,0675

1000C 0,040 0,0250 0,0450

1100C 0,020 0,0125 0,0225

1200C 0,000 0,0000 0,0000

*

* This opt ion is never used in the calculations with Elefir-EN. It wasincluded just as an aid for the users.

Annex C of EN 1993-1-2

Mechanical properties of stainless steel



145/303

145

20 C

300

(MPa) Stainless steelCarbon steel

20 C

300


S235 vs. 1.4301 (or 304)

Mechanical properties of carbon and stainless steel

Stress-strain relationship at room and at elevated temperature


146/303

146

0

50

100

150

200

250

0 0.002 0.004 0.006 0.008 0.01 0.012 0.014 0.016 0.018 0.02

235

210

(0.2%)

0

50

100

150

200

250

0 0.002 0.004 0.006 0.008 0.01 0.012 0.014 0.016 0.018 0.02

235

210

(0.2%)

600 C

0

20

40

60

80

100

120

140

160

0 0.002 0.004 0.006 0.008 0.01 0.012 0.014 0.016 0.018 0.02


146.6

110.5

(2.0%)

600 C

0

20

40

60

80

100

120

140

160

0 0.002 0.004 0.006 0.008 0.01 0.012 0.014 0.016 0.018 0.02


146.6

110.5

(2.0%)

( )

Reduction factors of yield strength

Reduction factors of carbon steel and stainless steel



147/303

147

0.0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

0 200 400 600 800 1000 1200

(C)

ky,Grade 1.4301

Grade 1.4401/1.4404

Grade 1.4003Grade 1.4571

Grade 1.4462

Carbon Steel

0.0

0.2

0.4

0.6

0.8

1.0

0 200 400 600 800 1000 1200

(C)

kE,

Stainless steel

Carbon steel

Reduction factors of Young

modulus

Checking Fire Resistance:

Strategies

Procedure usually adopted

with advanced calculationEurocodes allow fire


148/303

148

with advanced calculation

models but also used with

simple EC3 method. Find

design time (time for

collapse) and compare with

required time.

Find reduced resistance at

design temperature for the

required time and compare

with the effect of the actions.

Most usual simple EC3

method. Find critical

temperature for loading,

compare with design

temperature at required time.

resistance to be establishedin any of 3 domains :

Time: tfi,d > tfi,requ

Load resistance: Rfi,d,t > Efi,d,t

Temperature: cr,d >d

R E,


Strategies with nominal fires


149/303

149

1. Time:

tfi,d > t fi,requ

2. Load resistance:

Rfi,d,t > Efi,d

3. Temperature:d cr,d

2

1

3

Rfi,dEfi,d

d

cr,d

tfi,requ tfi,dt

t

1

3

2

R E,1. Load resistance:

Rfi,d,t > Efi,d


Strategies with natural fires


150/303

150

Rfi,d

Efi,d

d

cr,d

t

t

2

1

2. Temperature:

d cr,dcollapse is prevented during the

complete duration of the fire

including the decay phase or

during a required period of time.

collapse is prevented during the

complete duration of the fireincluding the decay phase or

during a required period of time.

Note: With the agreement of authorities,

verification in the time domain can be

performed. The required periodo of t ime

defining the fire resistance must be accepted

by the authorities.

The Load-bearing function is ensured if collapse is prevented during the


Strategies with natural fires


151/303

151

complete duration of the fire including the decay phase, or during a required

period of time.

Rfi,d, t

Efi,d

t

Rfi,d, t

Efi,d

t

R E, R E,

t fi,requ t fi,d1 t fi,requ

2

Collapse is prevented during the

complete duration of the fire

including the decay phase.

Collapse is prevented during a

required period of t ime, t1fi,req.

Checking Fire Resistance using Elefir-EN:

Strategies


152/303

152

Time: tfi,d > t fi,requ

Resistance: Rfi,d,t > Efi,d

Temperature: cr,d >d

Time: t fi,d > t fi,requ

Temperature: cr,d

>d

Checking Fire Resistance using Elefir-EN:

Strategies

Gives cr,d to be used inExample for elements submited to tension


153/303

153

the temperature domain:

cr,d >d

Gives Rfi,d,t

to be used in

the resistance domain:

Rfi,d,t > Efi,d

Gives tfi,d to be used in

the time domain:

tfi,d > t fi,requ

These four options are available for all the load cases

Temperature

Unprotectedd

Checking Fire Resistance in the temperature domain:

Strategy for nominal fires.


154/303

154

time

dcrit

ISO 834

Unprotected

section

Protected

section

crit

d

d

trequ

Temperature Natural

fire

Checking Fire Resistance in the temperature domain:

Strategy for natural f ires


155/303

155mxcrit time

critmx

mx UnprotectedsectionProtected

section

*

* - or using active fire

fighting measures

Temperature

Unprotected

Natural

fire

Checking Fire Resistance in the time domain:

Strategy for natural fires if accepted by the authorit ies


156/303

156trequ t fi,dtime

crit

Unprotected

section

Protected

section

*

* - or using active fire

fighting measures

trequtfi,d t fi,d

T b l t d d t (N t i EC3)

Design procedures


157/303

Tabulated data (Not in EC3)

Simple calculation models

Advanced calculation models


Design procedures


158/303


Simple calculation models

Advanced calculation models

Cross-sections are classified based on the parameter

Classification of the cross-sections - 1


159/303

159

210000235 Efy

For the case of carbon steel at normal temperature the, Young modulus

takes the value 210 GPa:

- At normal temperature

- At elevated temperature

with fy and E in MPa

yf

235

850235850 .f

.y

210000

E

f

235

k

k

210000

Ek

fk

235

210000

E

f

235

yy

,E,E

yyy

Classification of the cross-sections - 2


160/303

160

85.0f

23585.0

f

235

k

k

yy,y

,E

y,yy,y,y

0

0,2

0,4

0,6

0,8

11,2

0 200 400 600 800 1000 1200

[C]

,

,

y

E

k

k

0,85

yf.

235850

Classification of the cross-sections - 3For carbon steel:

and tables from EN 1993-1-1


161/303

161

Element Class 1 Class 2 Class 3

Flange c / t =9 c / t=10 c / t=14

Web subjected

to

compression

d / t=33 d / t=38 d / t=42

Web subectedtobending

d / t=72 d / t=83 d / t=124

For stainless steel:

210000

235850

E

f.

y

and tables from EN 1993-1-4

Fire Resistance:

Safety factors


162/303

162

1

.8

% of the value at 20 CFactor de

reduo

Fire Resistance:

Tension members - 1


163/303

163

]/[NkN fi,MMRd,yRd,,fi 0

NRd = design resistance of the cross-section Npl,Rdfor normal temperature design

The design resistance of a tensionmember with uniform temperature ais:

0 300 600 900 1200

.8

.6

.4

.2

Temperature (C)

yyy ffk /,,

or

fi,My,yRd,,fi /AfkN

Fire Resistance:

Tension members - 2


164/303

164

y,yfiRd,,fi,b fAkN

1 Design buckling resistance of a

compression member with uniform

temperature a is

Fire Resistance:

Compression members with Class 1, 2 or 3 cross-sections - 1


165/303

165

fi,M

Bracing system

lfi=0,7L

lfi=0,5L

,, / Ey kk

Non.dimensional slenderness:

22

1

fi

2121

With

yf/23565.0 (Curves a, b, c, d, a0)

Fire Resistance:

Compression members with Class 1, 2 or 3 cross-sections - 2


166/303

166

The design moment

Fire Resistance: Laterally restrained beams with Class 1, 2

or3 cross-sections with uniform temperature - 1


167/303

167

g

resistance of a Class 1, 2 or

Class 3 cross-section with a

uniform temperature a is:

fi,M

M,yRdRd,,fi kMM

0

MRd = Mpl,Rd Class 1 or 2 cross-sections

MRd = Mel,Rd Class 3 cross-sections

Adaptation factors to take into account

the non-uniform temperature distribution


or 3 cross-sections with non-uniform temperature - 2


168/303

168

Temp

Moment Resistance:

21

0 1

fi,M

M,yRdRd,,fi kMM

1=1,0 for a beam exposed on all four sides

1=0,7 for an unprotected beam exposed on three sides

1=0,85 for a protected beam exposed on three sides

1 is an adaptation factor for non-uniformtemperature across the cross-section

Adaptation factores to take into account

the non-uniform temperature distribution




169/303

169

Temp TempTemp

21

0 1

fi,M

M,yRdRd,,fi kMM

2=0,85 at the supports of a statically indeterminate beam

2 is an adaptation factor for non-uniform temperature alongthe beam.

Moment Resistance:

2=1.0 in all other cases




170/303

170




171/303

171

Fire Resistance: Laterally unrestrained beams with Class 1, 2 or

3 cross-sections - 1Fixed


172/303

172

Lateral-torsional buckling

Unloaded position

Applied load

Buckled position

lfiLTRdfib fkWM1

Design lateral torsional buckling

resistance moment of a laterally

unrestrained beam at the max.

y,elW (if Class 3)


3 cross-sections - 2


173/303

173

fi.Mycom..yy.plfi.LTRd..fi.b fkWM temp. in the comp. flange a.com is

com..Ecom..yLTcom..LT k/k

LT.fi the reduction factor for lateral-

torsional buckling in the fire designsituation.

2

,,

2

,,,,

,

][][

1

comLTcomLTcomLT

fiLT

2,,,,,, )(12

1comLTcomLTcomLT

yf/23565.0

(Curves a, b, c, d)cr

yypl

LTM

fW ,




174/303

174




175/303

175



gg

z

tz

z

w

w

z

z

zcr zC)zC(

EI

GI)Lk(

I

I

k

k

)Lk(

EICM 2

222

22

2

2

1


176/303

176

Elefir-EN can behelpful to find the

elastic critical

moment to be used

also with EC31-1, if

needed.

Design shear resistance

MRdwebyRdtfi VkV 0

Fire Resistance: Shear Resistance

Beams with Class 1, 2 or 3 cross-sections - 1


177/303

177

fi,M

,MRdweb,,yRd,t,fi VkV 0

VRd is the shear resistance of the gross cross-section

for normal temperature design, according to EN 1993-1-1.

web is the average temperature in the web of thesection.

ky,,web is the reduction factor for the yield strength of steelat the steel temperature web.




178/303

178




179/303

179

Fire Resistance: Shear and bending


Rd,t,fiEd,fi V.V 50

The effect of the applied shear force on the moment resistance of the section

can be neglected when the shear force is less than half the plastic shear

resistance of the section


180/303

180

Otherwise use a reduced yield strength on the shear area

y y

fy(1- ) fy(1- )

2

,,

,1

2

Rdtfi

Edfi

V

V

Class 1 or 2 Class 3

Fire Resistance: Shear and bending



181/303

181

MkMkN

Without lateral-torsional buckling

Class 1 and 2

Fire Resistance: Members with Class 1, 2 or 3 cross-sections,

subject to combined bending and axial compression - 1


182/303

182

1

fi,M

y,yz,pl

Ed,fi,zz

fi,M

y,yy,pl

Ed,fi,yy

fi,M

y,yfimin,

Ed,fi

fkW

Mk

fkW

Mk

fkA

N

1

fi,M

y,yz,el

Ed,fi,zz

fi,M

y,yy,el

Ed,fi,yy

fi,M

y,yfimin,

Ed,fi

fkW

Mk

fkW

Mk

fkA

N

Class 3

Class 1 and 2

MkMkN

With lateral-torsional buckling




183/303

183

Class 3

1

fi,M

y,yz,pl

Ed,fi,zz

fi,M

y,yy,plfi,LT

Ed,fi,yLT

fi,M

y,yfi,z

Ed,fi

fkW

Mk

fkW

Mk

fkA

N

1

fi,M

y,yz,el

Ed,fi,zz

fi,M

y,yy,elfi,LT

Ed,fi,yLT

fi,M

y,yfi,z

Ed,fi

fkW

Mk

fkW

Mk

fkA

N




184/303

184




185/303

185




186/303

186




187/303

187




188/303

188

If there are moments about the

major axis, the program allows

for the possibility of considering

LTB

Fire Resistance: verifications of the fire resistance not covered

by EN 1993-1-2

Clause 1.1.2 (Scope of Part 1.2 of Eurocode 3) of EN 1993-1-2 states This

Part 1-2 of EN 1993 deals with the design of steel structures for the

accidental situation of fire exposure and is intended to be used in conjunctionwith EN 1993-1-1 and EN 1991-1-2. This part 1.2 only identifies differences


189/303

189

p y

from or supplements to normal temperature design

This means that for the cases not covered by EN 1993-1-2, the formulae

from the part 1.1 of EC3 should be used but modified to take into account the

effect of the temperature.

Fire Resistance: Cross-sectional verification of a member

subjected to bending and axial force (compression or tension) - 1

For class 1 and class 2

01.MM

MM Ed,fi,zEd,fi,y


190/303

190

MM Rd,fi,z,NRd,fi,y,N

where MN,y,fi,Rd and MN,z,fi,Rd are the the design plastic moment resistance

reduced due to the axial force.

For class 3

01.

M

M

M

M

N

N

Rd,fi,z

Ed,fi,z

Rd,fi,y

Ed,fi,y

Rd,fi

Ed,fi


subjected to bending and axial force (compression or tension) - 2


191/303

191

For class 1 and class 2

0.1MM Ed,fi,zEd,fi,y


subjected to bi-axial bending - 1


192/303

192

MM Rd,fi,zRd,fi,y

For class 3

0.1M

M

M

M

Rd,fi,z

Ed,fi,z

Rd,fi,y

Ed,fi,y Ed,fi,yM

Ed,fi,zM

Example: a purlin


subjected to bi-axial bending - 2


193/303

193

AB C D

EI = Const.

p kN/m

l

E

l l l

a) Structural scheme.

Fire Resistance:

Design of continuous beams - 1


194/303

194

A B C D

l

E

l l l

l l l l

A B C D E

b) End span mechanism.

c) Intermediate span mechanism

Typical collapse mechanisms for a continuous beam

Fire Resistance:



195/303

195

Fire Resistance:



196/303

196

Two procedures:

1. In the absence of calculation a critical temperature of 350 Cshould be considered (conservative results).

Fire Resistance:

Members with Class 4 cross-sections


197/303

197

2. Alternatively use Annex E, considering the effective area

and the effective section modulus determined inaccordance with EN 1993-1-3 and EN 1993-1-5, i.e. based on

the material properties at 20C. For the design under f ire

conditions the design strength of steel should be taken as

the 0,2 percent proof strength instead of the strength

corresponding to 2% total strain.


Tenso

fy,

f

Cross-sectional Class 1, 2 and 3

Eurocode 3: Fire Resistance

Design yield strength to be used with simple calculation models


198/303

198Extenso

E = tan a,

y,p, u,

fp,

t,0,2%

f0.2p,

Cross-sectionalClass 4

= 2%

Annex E ofEN 1993-1-2

Two procedures:

1. In the absence of calculation, a crit ical temperature of 350 Cshould be considered (conservative results).




199/303

199

2. Alternatively use Annex E, considering the effective area

and the effective section modulus determined inaccordance with EN 1993-1-3 and EN 1993-1-5, i.e. based on

the material properties at 20C.

, 20p C (Slenderness of the plates)(See next slide)


y

y

cr

y

p

f

Ek

tb

b

Etk

ff 1

)1(12)1(12 2

2

22

22

1tb




200/303

210000

235

235

210000

)1(12 2

2 E

fk

y

k

tb

k

tb

E

f

k

tb

y

4.28

1

4.28

210000

235

1

4.28

210000

235 E

fy

with yf andEin MPa


For Class 4 cross-sections and

high temperatures:

,

0.2,

235 2351.00

E

p y y

k

k f f




201/303

0

0.2

0.4

0.6

0.8

1

1.2

1.4

0 200 400 600 800 1000 1200

1.00

20C

, 20p C

Tenso

fy,

f 0 2

Cross-sectional Class 1, 2 and 3

Fire Resistance:

Design yield strength to be used with simple calculation modelsNote: for the time being

Elefir-EN does not

allow for Class 4

elements


202/303

202Extenso

E = tan a,

y,p, u,

fp,

t,0,2%

fp0.2,

Cross-sectional Class

4

= 2%

Annex E of

EN 1993-1-2

crit,1crit,2 2a,cr fi,d = 576.1C ko!


223/303

223

Nfi,Rd = 177 kN < Nfi,Ed = 780 kN ko!

Verification in the resistance domain:

Fire Resistance


Example 5.2 from the ECCS Book: Laterally restrained beam. Consider

a simply supported restrained beam 4.0 long, constructed from an IPE

300 section, in steel grade S235, supporting a concrete slab. Assuming

the steel beam does not act compositely with the concrete slab and thatthe design load in the fire situation is qfi,Ed = 33.8 kN/m, verify if it is

necessary to protect the beam for a fire resistance period of R90. If fire

protection is needed, use fibre-cement boards.


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protection is needed, use fibre cement boards.

Design value of the moment in f ire situation at

the mid span:

Mfi,Ed = 33.8x42/8

= 67.6 kNm

Design value of the shear in fire situation at the

supports:

Vfi,Ed = 33.8x4/2

= 67.6 kN

Solution:

Fire Resistance


At mid span:


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Fire Resistance



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Fire Resistance



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tfi,d = 16.7 min < trequ = 90 min ko!

Passive fire protection is needed!

Fire Resistance



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Fire Resistance



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Fire Resistance



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Fire Resistance



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Fire Resistance



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Fire Resistance


Check the shear resistance at the supports


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Fire Resistance


As the critical temperature of

729.5 is bigger than the critical



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729.5 is bigger than the critical

temperature obtained for the

bending moment (624 C),

shear is not l imitative.

Fire Resistance


Example 5.3 from the ECCS Book: Unprotected column under axial

compression

Consider a 3.5 m long HE 180 B column in S275 grade steel, located in

an intermediate storey of a braced frame and subject to a compressionload of Nfi,Ed = 495 kN in the fire situation. Assuming that the column

doesnt have any fire protection and that the required fire resistance is

R30, verify the fire resistance


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Buckling length:

lfi = 0.5x3.5 = 1.75 kNm

Solution:

Fire Resistance



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Fire Resistance



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Fire Resistance



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tfi,d = 17.47 min < t requ = 30 min ko!

Fire Resistance


Example 5.6 from the ECCS Book: Unrestrained beam.

Consider a simply supported IPE 300, S235 grade steel beam from an office building,

with fork supports. The member is 5.0 m long and is subjected to a transverse

uniform load at normal temperature qEd = 19.2 kN/m. The transverse loading is

assumed to act at the shear centre of the beam. Evaluate the critical temperatureconsidering that the beam is unrestrained. Lateral-torsional buckling is not prevent

and may therefore occur.

Solution:Design value of the uniform load in fire situation:


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Design value of the moment span in f ire

situation at the mid span:

Mfi,Ed = 12.48x52/8 = 39.0 kNm

Design value of the shear in fire situation at thesupports:

Vfi,Ed = 12.48x5/2 = 31.2 kN

Design value of the uniform load in fire situation:

qfi,Ed = fixqEd = 0.65x19.2 = 12.48 kN/m

Fire Resistance




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cr = 845.7 C

Fire Resistance


Lateral-torsional buckling:


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Fire Resistance



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cr = 517 C cr = min(845.7 C, 517 C)= 517 C

Fire Resistance


Example 5.7 from the ECCS Book: Unrestrained beam-column.

A 6.0 m long, simply supported beam-column constructed from an IPE 450 section is

subjected to a uniform load in a fire situation of qfi,Ed = 15.89 kN/m and to an axial

compression load of Nfi,Ed = 136.5 kN . Assuming that the steel grade is S235 andthat the load is applied at the shear centre of the beam, evaluate the critical

temperature in the following conditions:

a) When the beam-column is lateral restrained.


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b) When the beam-column is unrestrained. Lateral-torsional buckling is notprevent and may therefore occur.

15,89 kN/m

136.5 kN 136.5 kN

Fire Resistance


a) The beam-column is lateral restrained.


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Fire Resistance



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Fire Resistance


b) The beam-column is laterally unrestrained.


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Fire Resistance



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Fire Resistance




supports:

Vfi,Ed = 15.89x6/2 = 47.67 kN

Solution:


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Fire Resistance




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cr = min(515.4 C, 881.7 C)= 515.4 C

Fire Resistance


Example 5.8 from the ECCS Book: Beam-column with restrained lateral

displacements.

Consider a 3.0 m high, HE 200 B column in an intermediate storey of a braced frame,

subjected to a bi-triangular bending moment diagram with moments at the ends in

fire situation of My,fi,Ed = 50 kNm about major axis and to an axial compressionload in fire situation of Nfi,Ed = 800 kN. Assuming that the steel grade is S235, and

that lateral-torsional buckling is not allowed, evaluate the critical temperature of

the column.


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800 kN

800 kN

50 kNm

- 50 kNm

50 kNm

- 50 kNm

Fire Resistance



supports:

Vfi,Ed = [50 - (-50)]/3 = 33.3 kN

Solution:

Buckling length in fire situation:

lyfi = 0.5x3.0 = 1.5 m


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Fire Resistance



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Fire Resistance



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Fire Resistance




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cr = min(516.6 C, 822.5 C)= 516.6 C

Fire Resistance


Example 5.9 from the ECCS Book: Class 4 column subjected to axial

compression.

Consider a 3.2 m high HE 500 A column in steel grade S460 in an intermediate storey

of a braced frame, subjected to an axial compression load. Evaluate the critical

temperature of the column for the following axial loads in a fire situation:

a) Nfi,Ed = 5460 kN.

b) Nfi,Ed = 3520 kN.


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Solution:

Buckling length in fire situation:

lyfi = 0.5x3.2 = 1.6 m

Fire Resistance



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Note: For the time being the program Elefir-EN

does not evaluate the crit ical temperature of

profi les with Class 4 cross-sections.

Fire Resistance


Example 5.10 from the ECCS Book: Three-span continuous beam.

Consider a three-span continuous beam in an office building, with equal spans of 6.0

m, constructed from an IPE 300 section in S235 grade steel, as shown in the Fig..

The beam is subjected to a permanent uniform load of Gk = 20 kN/m (including the

weight of the profile) and a live load of Qk = 8 kN/m. Assuming that the out ofplane displacements are restrained and that the beam is unprotected and supports

a concrete slab, evaluate the critical temperature.

A D

qfi,Ed


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Solution:

Load combination:

q fi,Ed = Gk + 2Qk = 20 + 0.3x8 = 22.4 kN/m

A B C

6.0 m

D

6.0 m 6.0 m

Fire Resistance



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Fire Resistance


No shear effect.


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Fire Resistance


No shear effect.


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cr = min(665.8 C, 708.7 C)= 665.8 C

Fire Resistance


Considering shear effect.


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Fire Resistance




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Fire Resistance




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cr = min(664.0 C, 699.3 C)= 664.0 C

Fire Resistance


Example 5.11 from the ECCS Book: Member subjected to combined

bending and tension.

Consider an HE 200 A section in S355 grade steel subjected to tension and major axis

end moments as shown in the figure.a) Evaluate the critical temperature.

b) Verify the fire resistance of the member after 30 minutes of standard fire exposure

on four sides.


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Mfi,Ed= 20kNm

Nfi,Ed= 100 kNNfi,Ed

Mfi,Ed= -10kNm

Fire Resistance


a) Evaluate the critical temperature.


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Fire Resistance



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Fire Resistance


b) Fire resistance after 30 min of ISO 834 curve on 4 sides.


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Fire Resistance



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Fire Resistance


Example 5.12 from the ECCS Book: Laterally restrained stainless steel

beam.

Suppose that a welded section equivalent to the IPE 300 is made of Grade 1.4301

stainless steel. What is the critical temperature if it is supporting a concrete slaband subjected to a bending moment of Mfi,Ed = 67.6 kNm and what is the time

needed to reach the critical temperature if the profile is unprotected and exposed

to the ISO 834 fire curve in three sides?


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Fire Resistance



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Fire Resistance



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Fire Resistance



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Fire Resistance



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Fire Resistance



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Case Study

2.0 m

1.85 m

8.15 m

Case study from Chapter 9 using simple calculation models

Single storey hall for indoor sport activities

IPE 500


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For more details seeChapter 9 from

ECCS book

5.5 x 12 = 66 m

Case Study

6 x 14 = 84 m

Single storey hall for

indoor sport activities

Steel grade S355


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2.75 x 24 = 66 m For more details seeChapter 9 from

ECCS book

2.0 m1.85 m

8.15 m

Case StudySection N M

1

M2

q lfi,y

lfi,z

- [kN] [kNm] [kNm] [kN/m]

ECCS Fire Des Kurs 20130429

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Transcript of ECCS Fire Des Kurs 20130429