Microsoft Word - FireSafety Hom

7/27/2019 Microsoft Word - FireSafety Hom

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Politecnico di Milano

Facoltà di Ingegneria Civile, Ambientale e Territoriale

Master of Science in Civil Engineering

Fire Safety of Material and Structure

Lecturers: Prof. Gambarova

Prof. Felicetti

Dr. P. Bamonte

Student: Yixiang Huang



CONTENT

Basic Information…………………………………………………………………….1

Evaluation of Fire Scenario…………………………………………………………2

Design of Beam and Evaluation in Fire…………………………………………..10

Analysis of T-Beam………………………………………………………………...24

Design of Column…………………………………………………………………..30

Annex: Excel working sheet..……………………………………………… ………..37



Fire Resistance of Material and Structure Homework Yixiang Huang 737534

BASIC INFORMATION

A possible fire scenario occurs in the library room, heating the beam located on the top of this

floor, which bearing the load of upper layer. We are required to evaluate the behavior of the given

beam under fire conditions.

F=the third letter of first name “X” stands for 24

L=the third letter of surname “A” stands for 1

So, in my case

[ ]1 14 3.99100

L m= − =

[ ]2

14 4.01

100 L m= + =

Thermal characteristics of walls, floor and ceiling are

Mass per unit volume324

1000 1 1096250

kg m ρ

= ⋅ + =

Specific heat [ ]1

900 1 896.4250

c J kg

= ⋅ − =

℃

Thermal conductivity [ ]0.3 W mλ =℃

Derived thermal properties:

Thermal diffusivity7 20.3

3.0536 101096 896.4

m sc

λ α

ρ

− = = = × ⋅

Thermal inertia5 2 4 2

2.94736 10c W s mλρ = × ℃




EVALUATION OF FIRE SCENARIO

Determination of fire load densities

According to EN 1991-1-2:2002(E) Annex E:

In our case, the design value of fire load is determined from classification of occupancies

Which is defined as

, , 1 2 f d f k q q nq q m δ δ δ = ⋅ ⋅ ⋅ ⋅

2 MJ m

Where,

m the combustion factor [ ]−

1qδ the factor taking into account the fire activation risk due to the size of the compartment,

according to “Table E.1” in Annex E [ ]−

2qδ the factor taking into account the fire activation risk due to the type of occupancy, according

to “Table E.1” in Annex E [ ]−

nδ 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, according to “Table E.2” and “clauses (4) and (5)” in Annex E.

10

1

n ni

i

δ δ =

= ∏ [ ]−

, f k q the characteristic fire load density per unit floor area, according to “Table E.4” in Annex E

Calculation:

According to Annex E.3 combustion behavior:

For mainly cellulosic materials, the combustion factor may be assumed as m = 0.8.

·

Computing 1qδ




Using interpolation method, there is a relation between 1qδ and compartment floor area

f A

δδδδq1 interpolation

y = 0.1713Ln(x) + 0.5514

0

0.5

1

1.5

2

2.5

10 100 1000 10000

Which 1qδ is a function of

f A ,

2

1 2 12 8 96 f

A l l m = ⋅ = × =

( ) ( )1 0.1713 ln 0.5514 0.1713 ln 96 0.5514 1.33327q f

Aδ = × + = × + =

·

Computing 2qδ

In our case of a library room, type of occupancy is 2: (offices, residence, hotel, paper industry)

So, 2 1.0qδ =

· Computingn

δ




WITH Safe Access Routes8 1.0nδ =

WITH Fire Fighting Devices 9 1.0nδ =

NO Smoke Exhaust system10 1.5nδ =

So,10

1

1.17n ni

i

δ δ =

= =∏

·

Computing , f k q

We choose library type Occupancy, 80% Fractile as Gumbel distribution assumes, thus:

2

, 1824 f k

q MJ m =

So, finally

2

, , 1 2 1824 0.8 1.33327 1.0 1.17 2276.24873 f d f k q q n

q q m MJ mδ δ δ = ⋅ ⋅ ⋅ ⋅ = × × × × =

Parametric temperature time curve in the heating phase:




Γ time factor function of the opening factor O and the thermal absorptivity b

( )( )

2

20.04 1160

O bΓ = 1160ref

b = , 0.04ref

O = [ ]−

b thermal absorptivity for total enclosure

b c ρ λ = with the following limits 100 2200b≤ ≤ 2 0.5

J m s ℃

ρ density of boundary of enclosure 1096= 3

kg m

c specific heat of boundary of enclosure 896.4= [ ] J kg ℃

λ thermal conductivity of boundary of enclosure 0.3= [ ]W m℃

O opening factor

v eq

t

A h

O A= with the following limits 0.02 0.2O≤ ≤ 0.5

m

v A total area of vertical openings on the wall

2m

eqh weighted average of windows height on all walls [ ]m

t A total area of enclosure (walls, ceiling and floor, including openings)

2

m

Calculation:

Consider the enclosure surface just has one layer of material and b value ( , ,c ρ λ ) for floor, walls,

ceiling are the same.

· Compute opening factor O:




Fire-resistant door not considered as opening

2 2

1 2 3 1 1 2 2 3 3 200 150 3 90000 9v A A A A B h B h B h cm m

= + + = + + = × × = =

[ ] [ ]1 1 2 2 3 3 150 1.5eq

v

A h A h A hh cm m

A

+ += = =

[ ]1 600 600 1200l cm= + = ; [ ]2 800l cm= ; the height of the roomr

h is 3.5 m

( ) ( ) 2 21 2 1 22 2 1200 800 1200 350 800 350 3320000 332t r r A l l l h l h cm m = + + = × × + × + × = =

So,

0.59 1.50.0332

332

v eq

t

A hO m

A

× = = =

Checking for the limit: 0.02 0.0332 0.2< < ok!

·

Computing thermal absorptivity for total enclosure b:

b is the square root of thermal inertia which has been calculated before

5 2 0.52.94736 10 542.896b c J m s ρ λ = = × = ℃

Checking for the limit: 100 542.896 2200< < ok!

Computing the gas temperature in the fire compartmentg

Θ

( )

( )

( )

( )

2 2

2 2

0.0332 542.8963.1453

0.04 1160 0.04 1160

O bΓ = = =

Final expression of gΘ :

( )* * *

0.2 1.7 19

20 1325 1 0.324 0.204 0.472t t t

g e e e− − −

Θ = + − − − [ ]℃

Where*

3.1453t t t = ⋅ Γ = [ ]h




( ), ,t d f d f t q q A A= ⋅ the following limits should be observed ,50 1000

t d q≤ ≤

2 MJ m

limt time for maximum gas temperature in case of fuel controlled fire [ ]h

Calculation:

0.50.0332O m = from above computation

2, 2276.24873 f d q MJ m = from ………

( ) ( ) 2

, , 2276.24873 96 332 658.1924t d f d f t

q q A A MJ m = ⋅ = × =

Checking for 50 658.1924 1000< < ok!

Choosinglim

t

According to Table E.5 in Annex E

For Occupancy type Library, fire growth rate is Fast.

In Annex A clause (10):

25min in case of slow fire growth rate




The temperature and time curve in cooling phase

(perform the worse case between EN 1991-1-2 and Buchanan’s suggestion)

It is assumed to be linearly decrease, we are required to compare with cooling rate in EC1 and

Buchanan’s formula.

·

According to EN 1991-1-2:2002(E) Annex A clause (11)

For*

max 2t ≥ ,

( )* *

max max250g

t t xΘ = Θ − − ⋅

Where

( )* * *max max max0.2 1.7 19

max 20 1325 1 0.324 0.204 0.472t t t

e e e− − −Θ = + − − −

*3.14530t t t = ⋅ Γ =

[ ]*

max 12.4708t h=

1.0 x = formax lim

t t >

So, as a result

( )max 250 3.14530 12.4708gt Θ = Θ − × − in the decay phase

So, in EC1, The cooling rate is [ ]250 3.1453 786.32568 h× =℃

·

According to Buchanan’s suggestion (textbook p76 and p78)

It is more accurate to use 1900ref b = , but he referred to ENV 1991 (1995) with lower

maxt

0.04

1160ref

OdT dT

dt dt b

= ⋅

Where

( )

*

max

* *

max max

625 0.5

250 3 0.5< 2

t dT

t t dt

≤

= − <




Conclusion:

The temperature-time curve is as follow

0

200

400

600

800

1000

1200

1400

0.00 1.00 2.00 3.00 4.00 5.00 6.00 7.00 8.00 9.00

EC1

ISO 834

EC1's cooling

Buchanan's cooli ng

For standard time intervals

real t. (min) t/tmax* t* (Γ ) T (°C)

30 0.13 1.5727 1012.9

60 0.25 3.1453 1114.9

90 0.38 4.7180 1177.8

120 0.50 6.2906 1223.0

150 0.63 7.8633 1255.9

180 0.76 9.4359 1280.0

210 0.88 11.0086 1297.5

240 in cooling phase 1297.9

Checking for the EC1 validity limits:

h C1 i lid f fi 5002

f fl

℃

h




DESIGN OF BEAM AND EVALUATION IN FIRE

The envelop of moment diagram

Given permanent load2

7.00k

G kN m = , variable load2

2.00k

Q kN m =

Determine envelop of moment diagram

The possible load combinations are as below

According to EN 1991 EC1 “actions on structures”, the load magnification factor is chosen as

1.35Gγ = (favorable), 1.0 (unfavorable); 1.5Qγ = (favorable), 0 (unfavorable)

Because of one way slab, the effective area of the beam A-B-C is showed as the shadow in the

figure above (half span on both sides), so:




Distributed load [ ]1.35 42 1.5 12 74.7G k Q k

q G Q kN mγ γ = + = × + × =

Moment Diagram of Load Combination1 [ ]kN m⋅

Load Combination2 (Maximum Positive Moment in A-B span)

QQk γ

γGGk

L2=4.01mL1=3.99m

CBA

Distributed load in A-B span [ ]1.35 42 1.5 12 74.7 AB G k Q k q G Q kN mγ γ = + = × + × =

In B-C span [ ]1.0 42 0 12 42 BC G k Q k q G Q kN mγ γ = + = × + × =

Moment Diagram of Load Combination2 [ ]kN m⋅ Shear Diagram of Load Combination2 [ ]kN

The position of maximum moment in A-B span is where shear force equal to 0: [ ]1.604 x m=




CBA

x1=1.6m x2=1.61m

-149.403

96.127 97.368

[ ][ ]

[ ]

,

,

,

96.127149.403

97.368

Ed AB

Ed B

Ed BC

M kN m M kN m

M kN m

= ⋅= − ⋅

= ⋅

Design of reinforcement for the beam sections

Firstly, we assume not to change the geometry of the sections and design for longitudinal

reinforcements

A′ ands

A .

About material resistance

f yd

sc

f ck

3.5

Reduction factor of material resistance 1.5cγ = , 1.15s

γ = for static analysis

Characteristic value of c.s of concrete and yield stress of steel [ ]30ck

f MPa= , [ ]500 yk

f MPa=

[ ]30

0.85 0.85 171.5

ck

cd

c

f f MPa

γ = = × = ; [ ]

500434.783

1.15

yk

yd

s

f f MPa

γ = = =




Due to the axial force equilibrium C=T (no axial force applied in longitudinal direction)

0.8cd yd s yd s

bxf f A f A′+ = ①

Due to the moment equilibrium about the neutral axis

1 1 10.8 0.42 2 2

Rd cd yd s yd s

h h h M bxf x f A d f A d

′ ′= − + − + −

②

The value [ ]1 24 240h cm mm= = , [ ]60 600b cm mm= =

Net cover [ ]30net c mm= , I take [ ]40c mm= initially, then [ ]240 40 200d mm= − =

For section in A-B span

Applying a calculation approach given in the class note

First assume no compressive reinforcement is needed, so equilibrium equation ① and ② become:

0.8 cd yd sbxf f A=

( ) ( )0.8 0.4 0.4 Rd cd yd s

M bxf d x f A d x= − = −

Then assume that

6, 296.127 10

12280.9 0.9 434.783 200

Ed AB

s

yd

M A mm

f d

× ≈ = ≈ × ×

Then, [ ]434.783 1228

65.430.8 0.8 600 17

yd s

cd

f A x mm

bf

×= = =× ×

( ) [ ]0.4 434.783 1228 (200 0.4 65.43) 92.809 Rd yd s M f A d x kN m= − = × × − × = ⋅

2 checks

Rd Ed M M > NO! → adding

s A

[ ]0.4 80 x d mm< = YES!

Design

For s A : 2 22 2 20Φ + Φ ; For s

A′ : 4 12Φ




5 36 360.0035 2 10 =700

s cu s

x d x x E

x x xσ ε

′− − −′ = ⋅ = × × ⋅ ⋅

Recalculate x: [ ]0.8 58.89cd s s yd sbf x A f A x mmσ ′ ′+ = → =

[ ]1 1 10.8 0.4 104.352 2 2

Rd cd s s yd s

h h h M bxf x A d f A d kN mσ

′ ′ ′= − + − + − = ⋅

3 checks

Rd Ed M M > YES!

0.4 x d < YES!

Check min

ρ , for [ ]30ck f MPa= , [ ]2.9ct

f MPa=

3 3

min

1

2.9 13880.26 0.26 1.734 10 9.638 10

434.783 600 240

ct s

yd

f A

f bh ρ − −= = × = × < = = ×

×YES!

For section in B-C span

Exactly the same approach and design:

Fors

A : 2 22 2 20Φ + Φ ; Fors

A′ : 4 12Φ

For section B

For the negative moment [ ],

149.403 Ed B

M kN m= − ⋅

It requires approximately 1 2 more the amount of reinforcement than mid-span sections, and also

needs to adds

A′ .

Design

For s A : 5 20 2 18Φ + Φ ; For s A′ : 2 16 2 14Φ + Φ

So,

2

, 2079s prov

A mm = ;2

, 710s prov

A mm′ = ; [ ]30 20 2 40c mm= + = ; [ ]30 16 2 38c mm′ = + =




min

1

s A

bh ρ < YES!

Tensile rein. Compressive rein.

section

d

[ ]mm

d ′

[ ]mm

x

[ ]mm

Ed M

[ ]kN m⋅

Rd M

[ ]kN m⋅

rebar

s A

2mm

position rebar

s A′

2mm

position

A-B span 199 36 58.89 96.127 104.35 2Φ22+2Φ20 1388 bottom 4 12Φ 452 top

B 200 38 79.12 149.403 150.54 5Φ20+2Φ18 2079 top 2Φ16+2Φ14 710 bottom

B-C span 199 36 58.89 97.368 104.35 2Φ22+2Φ20 1388 bottom 4 12Φ 452 top

A-B&B-C

600

2 4 0

22

202

2

124

3 6

4 1

B

600

2 4 0

20

182

5

162

3 8

4 0

142

500℃

isotherm method

· concrete with 500Θ ≥ ℃ has zero strength and stiffness

· concrete within 500Θ < ℃ no reduction on design value of compression strength and stiffness

· yielding stress for steel at high temperature determined from Table 3.2a in EN1992-1-2 2004(E)

First determine the position of isothermal lines at difference time intervals

The cross-section had been heated at the bottom

Using Abaqus

General procedure:

· open “retangle.cae” already been given

importing time-temperature data into amplitudes---heating curve




The yielding strength and elastic modulus of reinforcements for different cross-sections

For hot rolled

Assume the ratio ,sy yk f f θ and ,s s

E E θ vary linearly with temperature, we can obtain:

Cross-section Reinforcement Time [ ]min 30 60 90 120 150 180 210 240

Θ [ ]℃

215 366 469 548 611 664 709 747

s A

,sy f θ [ ] MPa 500 500 424 316 222 158 110 87

Θ [ ]℃ 21 24 33 46 62 80 98 116

,sy f θ [ ] MPa 500 500 500 500 500 500 500 500

A-B&B-C

s A′

,s E θ [ ]2kN mm 200 200 200 200 200 200 200 197




Design value of moment resistance varies with time

For section A-B

This is also the same for section B-C

A-B section s

A :1388s

A′ :452 d’:36 [ ],

500sy

MPa f θ

′ =

Time [ ]min [ ]F mmb [ ]F

mmd ,sy f θ [ ] MPa s

σ ′ [ ] MPa [ ]mm x x d [ ] Rd kN m M ⋅

0 600 199 500 129.11 44.14 0.22 124.78

30 600 199 500 129.11 44.14 0.22 124.78

60 600 199 500 129.11 44.14 0.22 124.78

90 600 199 424 55.79 39.12 0.20 107.39

120 600 199 316 -70.96 32.69 0.16 82.28

150 600 199 222 -205.20 27.84 0.14 60.19

180 600 199 158 -309.83 24.95 0.13 45.10

210 600 199 110 -395.23 23.01 0.12 33.77

240 600 199 87 -435.66 22.06 0.11 28.32

Notes:s

σ ′ <0 indicates tension in top rebar, that is because x smaller than d’

For section B

B section s

A :2079s

A′ :710 d’:38 [ ],

500sy

MPa f θ

=

Time [ ]min [ ]F mmb [ ]F mmd ,s

E θ [ ]2

kN mm ,s θ σ ′ or ,sy f θ

′ [ ]mm x x d [ ] Rd kN m M ⋅

0 600 200 200 254.12 59.66 0.30 180.54

30 600 184 169 353.79 54.74 0.30 168.48

60 600 172 124 354.66 54.70 0.32 159.04

90 600 163 106 351.00 54.88 0.34 151.84

120 600 155 60 229.00 60.90 0.39 140.90

150 600 148 37 152.00 64.69 0.44 131.25

180 600 142 24 102.00 67.16 0.47 123.08




, , ,

,

, , ,

cu s s sy

s

sy s sy

x d y E f

x

f f

θ θ θ θ

θ

θ θ θ

ε σ σ

σ

′− +′ ′⋅ <

′ = ′ ′ ′≥

1 1,0.8 0.4

2 2 Rd ck sy s

h h M bxf y x f A d θ θ

′′= − − + −

Where yθ is the position of 500 ℃ isothermal line measured from bottom

d ′′ is the net cover 30mm +half diameter of tensile reinforcement, which is 40mm

So:

The above curve is approximated by connecting specific points with curves (it can also be

connected with lines)

Some observations can be made from the above figure:

In the beginning, moment resistance decrease more significantly at section B than section A-B &

B-C, because there is a reduction of cross-section due to effective depth at B, and also affected by

decreasing value of elastic modulus of compression reinforcing bars




Where,

E is distributed forces in our case

[ ]kN mm

fiη the reduction factor for the design load level for the fire situation

,1

,1 ,1

k fi k

fi

G k Q k

G Q

G Q

ψ η

γ γ

+=

+in this case, simply as 0.7

For section A’ and C’

,0.7( ) 0.7G k Q k Ed fi Ed q G Q M M γ γ = + → =

So, ( ) [ ], ,' 0.7 0.7 96.127 67.29 Ed fi Ed AB M A M kN m= = × = ⋅

( ) [ ], ,' 0.7 0.7 97.368 68.16 Ed fi Ed BC

M C M kN m= = × = ⋅

2 0

3 0

4 0

5 0

6 0

7 0

8 0

9 0

1 0 0

1 1 0

1 2 0

1 3 0

1 4 0

0 3 0 6 0 9 0 1 2 0 1 5 0 1 8 0 2 1 0 2 4 0 2 7 0

M R d a t A ' $ C ' M E d a t A ' M E d a t C '

min

kN m⋅




7 0

8 0

9 0

1 0 0

1 1 0

1 2 0

1 3 0

1 4 0

1 5 0

1 6 0

1 7 0

1 8 0

1 9 0

0 3 0 6 0 9 0 1 2 0 1 5 0 1 8 0 2 1 0 2 4 0 2 7 0

M R d a t B M E d a t B

In this case, during 240 min of fire, the cross-section is still not fail

Fire endurance of the beam: { } [ ]max ' 'min , , 138 min A B C t t t t = =

Kinematic approach (Upper-bound Approach)

Case 1

CBA

1.6m 2.39m

( ), ' 1 2 , 2

1 1

i Rd A Rd B L M M

xL p L

θ θ θ

θ

= ⋅ + + ⋅

kN m⋅

min




3 0 . 0 0

4 0 . 0 0

5 0 . 0 0

6 0 . 0 0

7 0 . 0 0

8 0 . 0 0

9 0 . 0 0

1 0 0 . 0 0

1 1 0 . 0 0

1 2 0 . 0 0

0 3 0 6 0 9 0 1 2 0 1 5 0 1 8 0 2 1 0 2 4 0 2 7 0

P u p

[ ]1 170 mint =

Case 2

A B C

2.4m 1.61m

( ), ' 1 2 , 1

2 22

2

i Rd C Rd B

e u

L M M

x L p L

θ θ θ

θ

= ⋅ + + ⋅

= ⋅ ⋅

1 2 2

1.610.67

4.01 1.61θ θ θ = =

−, i e

L L= → , ' ,0.517 0.208u Rd C Rd B p M M = +

{ } [ ]0.7max 0.7 74.7 52.29 BC p q kN m= = × =

min

kN m⋅




3 0 . 0 0

4 0 . 0 0

5 0 . 0 0

6 0 . 0 0

7 0 . 0 0

8 0 . 0 0

9 0 . 0 0

1 0 0 . 0 0

1 1 0 . 0 0

1 2 0 . 0 0

0 3 0 6 0 9 0 1 2 0 1 5 0 1 8 0 2 1 0 2 4 0 2 7 0

P u p

[ ]2 165 mint =

Case 3

2.39m1.6m

A B C

2.4m 1.61m

Apparently, this can not be the failure mechanism because some of the forces do negative work

So, { } [ ]max 1 2 3min , , 165 mint t t t = =

The reason formax

t in kinematic method is much larger is that:

Because of the fire heated from the bottom the moment resistance of A’&C’ decrease much more

min

kN m⋅




What happens to ductility in different cross-section?

For central-span A’&C’, the value ( ) x d decreases with time and temperature, the cross-section

become more and more ductile.

Explanation:

The effective depth is not changing in this case (d is const), because the yielding stress of tensile

reinforcing bar keep reducing in fire, due to the force equilibrium the compression zone of

concrete keep reducing (x is reducing). This result in a decreasing value of ( ) x d .

For support B, the value ( ) x d increasing with time and temperature, the cross-section become

more and more brittle.

Explanation:

500 ℃ isotherm line move upwards with time, this result in a decreasing d

Because of the decreasing ,s E θ and decreasing yielding stress of compression reinforcing bar,

the proportion of compression force taken by the concrete keep increasing, while the tensile force

is constant. This result in an increasing x

As a result, ( ) x d increases.

In conclusion, under fire condition, cross-sections under positive bending moment become more

and more ductile while cross-sections under negative moment become more and more brittle.




ANALYSIS OF T-BEAM

Design of reinforcement

[ ]1.35 42 1.5 12 74.7G k Q k q G Q kN mγ γ = + = × + × =

Maximum bending moment in the center

[ ]2 21 174.7 8 597.6

8 8 B

M qL kN m= = × × = ⋅

Material properties as before

The cross-section is as below:

bf =600

3 6 0

h 2 = 6 0 0

b=400

h f =2 4 0

Because2

240.4

60

f h

h= = , to satisfy the requirement for ductility, assume the compression zone is

within the upper part body (within 24cm)

Formula in calculation

Due to the axial force equilibrium

0.8 f cd s s yd sb xf A f Aσ ′ ′+ = ①

Due to the moment equilibrium about the neutral axis




2

, 2945s prov

A mm = ;2

, 678s prov

A mm′ = ; [ ]532.5d mm≈ ; [ ]12

30+ 362

d mm′ = =

[ ]0.8 116.7 f cd s s yd sb f x A f A x mmσ ′ ′+ = → =

[ ]2 2 20.8 0.4 625.62 2 2

Rd f cd s s yd s

h h h M b xf x A d f A d kN mσ

′ ′ ′= − + − + − = ⋅

3 checks

Rd Ed M M > YES!

0.4 x d < YES!

Check min

ρ , for [ ]30ck f MPa= , [ ]2.9ct

f MPa=

3 3

min

2

2.9 29450.26 0.26 1.734 10 8.18 10

434.783 600 600

ct s

yd f

f A

f b h ρ − −= = × = × < = = ×

×YES!

500℃

isotherm method

Determinate the position of 500 ℃ isotherm line




width. Determine the value of fi

b , , f fih .

fid is not changing in this case, because the tensile

zone exposed to fire. According to EN 1992-1-2:2004 (E) Annex B.1 “500 ℃ isotherm method”,The rounded corners of isotherms can be regarded by approximating the real form of the isotherm

to a rectangle or a square.

As a result:

Time [ ]min 30 60 90 120 150 180 210 240

fib [ ]mm 370 340 320 315 300 280 260 240

, f fih [ ]mm 225 210 205 200 190 185 180 175

Determinate the temperature and mechanical properties of reinforcing bars




Reinforcement Time [ ]min 30 60 90 120 150 180 210 240

Θ [ ]℃

350 600 750 850 930 1000 1100 1200

2 25Φ corner-bottom

,sy f θ [ ] MPa 500 235 85 42.5 27 20 10 0

Θ [ ]℃

225 375 500 600 650 720 780 830

2 25Φ mid-bottom

,sy f θ [ ] MPa 500 500 390 235 175 103 74.2 47.5

Θ [ ]℃

250 400 550 650 720 800 850 900

2 25Φ upper

,sy f θ [ ] MPa 500 500 312.5 175 103 55 42.5 30

In conclusion:

B section s

A :2945s

A′ :678 d’:36 [ ],

500sy

MPa f θ

′ =

Time [ ]min [ ] f mmb [ ]mmd ,sy mean f [ ] MPa s

σ ′ [ ] MPa [ ]mm x x d [ ] Rd kN m M ⋅

0 600 532.5 500 398.23 83.51 0.16 734.22

30 600 532.5 500 398.23 83.51 0.16 734.22

60 600 532.5 411.7 332.38 68.55 0.13 610.45

90 600 532.5 262.5 156.12 46.33 0.09 395.48

120 600 532.5 150.8 -54.42 33.40 0.06 231.39

150 600 532.5 101.7 -172.02 28.90 0.05 158.88

180 600 532.5 59.3 -285.51 25.57 0.05 96.20

210 600 532.5 42.2 -334.21 24.37 0.05 70.92

240 600 532.5 25.8 -382.41 23.28 0.04 46.67

Notes: sσ ′ <0 indicates tension in top rebar, that is because x smaller than d’

f is the mean value of yielding stress of “ 2 25Φ outer bottom” “ 2 25Φ inner




0 . 0 0

1 0 0 . 0 0

2 0 0 . 0 0

3 0 0 . 0 0

4 0 0 . 0 0

5 0 0 . 0 0

6 0 0 . 0 0

7 0 0 . 0 0

8 0 0 . 0 0

9 0 0 . 0 0

0 3 0 6 0 9 0 1 2 0 1 5 0 1 8 0 2 1 0 2 4 0 2 7 0

M R d a t B ' M E d a t B '

[ ]max 87 mint =

Kinematic approach (Upper-bound Approach)

,

,22

2

8

4

i Rd B

Rd Bu

ue

L M

M p p L L L

θ

θ

= ⋅

→ ==

[ ]0.7 0.7 74.7 52.29 p q kN m= = × =

7 0 . 0 0

8 0 . 0 0

9 0 . 0 0

1 0 0 . 0 0

1 1 0 . 0 0

kN m⋅

min

kN m⋅




[ ]max 87 mint =

The value of x d is decreased as stated before.

Fi R i t f M t i l d St t H k Yi i H 737534




DESIGN OF COLUMN

QQk γ

γGGk

L2=4.01mL1=3.99m

CBA

In this mechanism, the column at B suffered from the maximum axial force

Where [ ]1.35 42 1.5 12 74.7G k Q k q G Q kN mγ γ = + = × + × =

[ ]373.5 Ed N kN =

Predimensioning of column2

300 300 90000c A mm = × =

Rd cd c yd s N f A f A= +

Verify: 17 90000 1260000 373500cd c Ed f A N = × = > = ok!

choosing 4 12Φ ,2

452s

A mm =

0.3% 0.5% 3%s

s

c

A

A ρ < = = < ok!

300

3 0 0

124





,c mk is the mean reduction factor of concrete compressive strength in the core.

According to EN 1992-1-2 (2004) Annex B.2:

w is chosen as half the smallest dimension of a 4 sided exposed column. [ ]150w mm=

the value 1 za can be obtained from figure B.5 in EC Annex





The mean reduction coefficient of the particular section

According to EN 1992-1-2 (2004) formula B.13

( )( )

11.3

1.3,

,1 1c m z

z c m c M

c M

k aa w k k

k wθ

θ

= − → = − ⋅

Time [ ]min 0 30 60 90 120 180 240

Reduction Coefficient ( )c M k θ 1.0 1.0 1.0 1.0 0.99 0.95 0.9

So, as a result

Time [ ]min 0 30 60 90 120 180 240

Mean reduction Coefficient ,c mk 1.00 0.91 0.86 0.80 0.75 0.67 0.59





Read the result from Abaqus “Visualization”, the above figure shows contour lines at 90min

The protection cover for reinforcing bars is 30mm, so the center of rebar to both edges is of 36mm

length. Red point in the figure indicates reinforcing bar.

Time [ ]min 0 30 60 90 120 180 240

temperature Θ [ ]℃ 20 400 650 800 900 1100 1200 ,sy

f θ [ ] MPa 500 500 175 55 30 10 0

Bearing capacity verification under fire

Column s

A :452 [ ]0.7 373.5=261.45Ed

N kN = ×




g g

0

5 0 0

1 0 0 0

1 5 0 0

2 0 0 0

2 5 0 0

3 0 0 0

3 5 0 0

0 5 0 1 0 0 1 5 0 2 0 0 2 5 0 3 0 0

N R d N E d

This plot also the table shows that the external load will never exceed the resistance of column

under such parabolic time-temperature curve.

-------------------------------------------------------END---------------------------------------------------------

kN

min

nnex



fire load densities according to EC1 (annex E - fire load densities)

qf,k = 1824 MJ/m2 qf,d = 2276.24873 MJ/m

2

floor area

m = 0.8 combustion factor

comp.floor area

compart.floor area = 96 m2 δδδδq1 = 1.33327 size of compartment 25 1.1

250 1.5

type of occupancy 2 δδδδq2 = 1 type of occupancy 2500 1.9

5000 2

1 artgallery, museum, swimming pool 0.78 10000 2.13

2 offices, residence, hotel, paper industry 1.003 manufactory for machinery & engines 1.22

4 chemical lab., painting workshop 1.44

5 manufact. of fireworks or paints 1.66

δδδδn = 1.17 active measures characteristic values of the fire load q f,k

Automatic Fire Suppression (yes, no) no 1.00

# Independent Water Supplies (0, 1, 2) 0 1.00

Autom Detect & Alarm by heat (yes, no) no

1.00

Autom Detect & Alarm by smoke (yes, no) no

Aut Alarm Transm to Fire Brigade (yes, no) no 1.00

On site Fire Brigade (yes, no) no

0.78Off Site Fire Brigade (yes, no) yes

Safe Access Routes (yes, no) yes 1.00

Fire Fighting Devices (yes, no) yes 1.00

Smoke Exhaust System (yes, no) no 1.50

δδδδq1 interpolation

y = 0.1713Ln(x) + 0.5514

0

0.5

1

1.5

2

2.5

10 100 1000 10000

time tmax to reach Tmax

qf,d = 2276.2487 MJ/m2 qt,d = 658.19240 MJ/m

2



minimum time for propagation - tlim

total area of vertical openings Av = 9 m2

opening factor slow 25

average height of the openings hv = 1.5 m O = 0.03320 medium 20

total area of the enclosure At = 332 m2

fast 15

tlim = 15 min

b factor - thermal inertia

area (m2) ρ (kg/m

3) c (J/kg°C) λ (W/m°C) bi bi·Ai tmax = 3.96491 hours

walls 131 1096 896.4 0.3 543 71119 tmax* = 12.47083 hours

floor 96 1096 896.4 0.3 543 52118 ventilation controlled fire

ceiling 96 1096 896.4 0.3 543 52118 Tmax = 1309.55470 °C

Total 323 542.89623 175355

(openings excluded) W s1/2

/ m2°C bref,Buch = 1900 bref,ec = 1160 W s

1/2/ m

2°C according to EC1 bref = 1160

Oref = 0.04 according to Buchanan (p.76) = 1900

but he referred to ENV 1991 (1995) with lower tmax

value of b in the case of two layers Γ ΓΓ Γ 3.14530 time scale for ventilation-controlled fire (and for cooling in both cases)

thickness (m) ρ (kg/m3) c (J/kg°C) λ (W/m°C) bi Γ ΓΓ Γ lim 197.78259 time scale for fuel-controlled fire (but not for cooling)

layer 1 0.015 0

layer 2 0.2 0 tmax* for cooling = 12.47083 hours

0 cooling rate EC1 786.32568 °C/h

slim,1 = #DIV/0! m Buchanan 332.93206 °C/h

t/tmax* t* (Γ ) real t. (h) T (°C) time (h) TISO 834 real t. (min) t/tmax* t* (Γ ) T (°C)

0.00 0.0000 0.0000 20.0 0 20.0 30 0.13 1.5727 1012.9

0.05 0.6235 0.1982 872.4 0.1 603.1 60 0.25 3.1453 1114.90.10 1.2471 0.3965 978.0 0.2 705.4 90 0.38 4.7180 1177.8

0.15 1.8706 0.5947 1038.4 0.3 765.7 120 0.50 6.2906 1223.0

0.20 2.4942 0.7930 1080.4 0.4 808.5 150 0.63 7.8633 1255.9

0.25 3.1177 0.9912 1113.5 0.5 841.8 180 0.76 9.4359 1280.0

0.30 3.7412 1.1895 1141.4 0.6 869.0 210 0.88 11.0086 1297.5

0.35 4.3648 1.3877 1165.5 0.7 892.0 240 1297.9

0.40 4.9883 1.5860 1186.6 0.8 912.0

0.45 5.6119 1.7842 1205.2 0.9 929.6

0.50 6.2354 1.9825 1221.6 1 945.3

0.55 6.8590 2.1807 1236.1 1.1 959.6

0.60 7.4825 2.3789 1248.9 1.2 972.60.65 8.1060 2.5772 1260.1 1.3 984.6

0.70 8.7296 2.7754 1270.1 1.4 995.7

0.75 9.3531 2.9737 1278.9 1.5 1006.0

0.80 9.9767 3.1719 1286.6 1.6 1015.6

0.85 10.6002 3.3702 1293.5 1.7 1024.7

0.90 11.2237 3.5684 1299.5 1.8 1033.3

0.95 11.8473 3.7667 1304.8 1.9 1041.4

1.00 12.4708 3.9649 1309.6 2 1049.0

cooling EC1 3.9649 1309.6

5.6049 20

Buchanan 3.9649 1309.6

7.8382 20

in cooling phase

0

200

400

600

800

1000

1200

1400

0.00 1.00 2.00 3.00 4.00 5.00 6.00 7.00 8.00 9.00

EC1

ISO 834

EC1's cooling

Buchanan's cooling

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