Eurocodes Print

8/12/2019 Eurocodes Print

1/123

F. A. Clignett Photography Delft - Copyright 2006

User's Guide

EUROCODESS P R E A D S H E E T S

S 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 SigmundCopyright 2014 Carlo Sigmund

to Excelspreadsheet file

Prints from spreadsheets


2/123

Copyright 2014http://www.sigmundcarlo.net

All 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 writtenpermission from the publisher.

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

First Edition: January 2014

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 fullconsent to the publication. The author is at the disposal of thebeneficiaries.

Bridge: Erasmus Bridge

Location: Rotterdam, Netherlands

Length/ main span: 802 m/284 m

Pylon: 139 m

Designer: 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


3/123

http://www.sigmundcarlo.net

EUROCODES Spreadsheets Structural Design

PROJECT: ???

CLIENT: ???

JOB NO.: ??

DESIGN BY: ??

REVIEW BY: ??

[Table 2.1-EN1990] Indicative design working life

Design life working category: Indicative design working life: Examples:

4 50 years

[Table A1.1-EN1990] Recommended values of factors for buildings

Action: 0 1 2

Imposed loads in buildings, category (see EN 1991-1-1)

Category B : office areas 0,7 0,5 0,3

Action: 0 1 2

Snow loads on buildings (see EN 1991-1-3)*

a) Finland, Iceland, Norway, Sweden 0,7 0,5 0,2

[Table B1-EN1990] Definition of consequences classes

Consequences Class Description Examples of buildings and civil engineering works

CC2

[Table B2-EN1990] Recommended minimum values for reliability index (ultimate limit states)

Reliability Class Minimum values for : 1 year refer. period Minimum values for: 50 years refer. period

RC2 4,7 3,8

[Table B3-EN1990] KFIfactor for actions

Reliability class KFI

RC2 1,0

Building structures and other common structures.

NOTE The values may be set by the National

annex.

* For countries not mentioned below, see relevant

local conditions.

Residential and office buildings, public buildings where

consequences of failure are medium (e.g. an office building).

Medium consequence for loss of human

life, economic, social or environmental

consequences considerable.

Page A1/2 EN1990.xls (rel. xx.x) - Sheet: Annex A1-B - 06/04/2013 --

- The header and logo can be customized -


4/123



PROJECT: ???

CLIENT: ???

JOB NO.: ??

DESIGN BY: ??

REVIEW BY: ??

[Table B4-EN1990] Design supervision levels (DSL)

Design Superv. Levels Characteristics Minimum recommended requirements for checking of

calculations, drawings and specifications

DSL2 relating to RC2

[Table B5-EN1990] Inspection levels (IL)

Inspection Levels Characteristics Requirements

IL2 Relating to RC2 Inspection in accordance with the procedures of theorganisation.

Checking by different persons than those originally

responsible and in accordance with the procedure of the

organisation.

Normal supervision

Normal inspection

Page A2/2 EN1990.xls (rel. xx.x) - Sheet: Annex A1-B - 06/04/2013 --



5/123



PROJECT: ???

CLIENT: ???

JOB NO.: ??

DESIGN BY: ??

REVIEW BY: ??

[Table C1-EN1990] Relation between and Pf

Pf (reliability index)

0,00001 = 0,01 4,3

[Table C.2-EN1990] Target reliability index for Class RC2 structural members ( See Annex B)

Limit state Target reliability index (1 year) Target reliability index (50 years)

Ultimate 1= 4,7 (n = 50) = 3,8

The actual frequency of failure is significantly dependent upon human error, which are not considered in partial factor design

(See Annex B). Thus does not necessarily provide an indication of the actual frequency of structural failure.

Note: Target reliability index: n= [(n)]-1

= {[(1)]n}-1

= 3,7 (n = 100 years)

(3)

EN1990 - Section C7: Approach for calibration of design values

The design values of action effects E dand resistances Rdshould be defined such that the probability of having a more

unfavourable value is as follows:

(C.6a) P(E > Ed) = (+ E)

(C.6b) P(R Rd) = (- R)

(C7) Standard deviations of the action effect and resistance, respectively: E= 5,00 R= 5,00

Case 1 0,16 < E/R< 7,6 [Case Applicable] E/R= 1,00 Active value: = 4,3

Case 2 "Case 1 Not Applicable" [Not Applicable]

(C.8a) P(E > Ed) = (+ E) = (- 0,7) = (- 0,7(4,30)) = (- 3,01) =

(C.8b) P(R Rd) = (- R) = (- 0,8) = (- 0,8(4,30)) = (- 3,44) =

(4) Where condition (C.7) is not satisfied = 1,0 should be used for the variable with the larger standard deviation, and= 0,4 for the variable with the smaller standard deviation. [ See points (C.8a), (C.8b) below].

Note: is negative for unfavourable actions and action effects, and positive for resistances.

larger standard deviation: P(E > Ed) = (+ E,max) = (- 1,0) =

(C.8a) P(E > Ed) = (+ E) =

(4) smaller standard deviation : P(E > Ed) = (+ E,min) = (- 0,4) =

larger standard deviation: P(R Rd) = (- R,max) = (- 1,0) =

(C.8b) P(R Rd) = (- R) =

(4) smaller standard deviation : P(R Rd) = (- R,min) = (- 0,4) =

(5) When the action model contains several basic variables, expression (C.8a) should be used for the leading variable only.

For the accompanying actions the design values may be defined by :

(C.9) P(E > Ed) = (+ E0,4) = (- 0,70,4(4,30)) = (- 1,204) = 1,14E-01

[Not Applicable]

[Not Applicable]

[Not Applicable]

[Not Applicable]

1,31E-03

2,91E-04

(- 0,7(0,4)) =

Page A1/3 EN1990.xls (rel. xx.x) - Sheet: Annex C - 06/04/2013 --



6/123



PROJECT: ???

CLIENT: ???

JOB NO.: ??

DESIGN BY: ??

REVIEW BY: ??

(C.9) P(E > Ed) = larger standard deviation: P(E > Ed) = (- 1,0(0,4)) = (- 0,4(4,30)) =

(4) = (+ E0,4) =

smaller standard deviation: P(E > Ed) = (- 0,4(0,4)) = (- 0,16(4,30)) =

Note: For = 3,8 the values defined by expression (C.9) correspond approximately to the 0,90 fractile.

[Table C3-EN1990] Design values for various distribution functions

Distribution Design value

Gumbel u - (1/a)ln{ - ln[(- )]} where: u = - 0,577/a a = /[6]

Note: In the expression above , and V are, respectively, the mean value, the standard deviation and the coefficient of variation

of a given variable. For variable actions, these should be based on the same reference period as for .

Mean value and standard deviation (active value) of the action effect: (E) E= 30,00 E= 5,00

Mean value and standard deviation (active value) of the resistance: (R) R= 30,00 R= 5,00

Reliability index: = 3,8

max{E; R} = 5,00

Coefficient of variation V = /of a given variable: (E) E/E= 0,16667 = VE

(R) R/R= 0,16667 = VR

(5) When the action model contains several basic variables, expression (C.8a) should be used for the leading variable only.

For the the leading/accompanying actions the design values may be defined by :

Basic variable: leading variable action: * = 1,0 Reliability index * = 1,0= 1,0= 3,8

E - E E/E R - R R/RC.8a/b -0,7 2,66 0,1667 0,8 -3,0400 0,1667

E - E E/E R - R R/RC.8a/b 1,00 -3,80 0,1667 -0,40 1,52 0,1667 [Not Applicable]

(4)

Distribution: Design value (leading variable):

C.8a/b Gumbel u - (1/a)ln{ - ln[(- )]} a = /[6] u = - 0,577/a

E= -0,7

(E) E= 5,00 Xdi,E= Xdi,E/Ei= 1,65 0,25651 27,7506

- E= 2,66 [Case Applicable]

(- E) =

R= 0,8

(R) R= 5,00 Xdi,R= Ri/Xdi,R = 1,48 0,25651 27,7506

- R= -3,0400 [Case Applicable]

(- R) =

(- E)

9,96E-01 [Case Applicable]

[Not Applicable]

1,18E-03

(- R)

[Not Applicable]

(- E) (- R)

7,23E-05 9,36E-01

Ratio

9,96E-01

1,18E-03

4,94E+01

2,03E+01




7/123



PROJECT: ???

CLIENT: ???

JOB NO.: ??

DESIGN BY: ??

REVIEW BY: ??

(4) Where condition (C.7) is not satisfied = 1,0 should be used for the variable with the larger standard deviation, and

= 0,4 for the variable with the smaller standard deviation. [ See points (C.8a), (C.8b) below].

Distribution: Design value (leading variable):

C.8a/b Gumbel u - (1/a)ln{ - ln[(- )]} a = /[6] u = - 0,577/a

(4) E= 1,00

(E) E= 5,00 Xdi,E= Xdi,E/Ei= 0,63 0,25651 27,7506

- E= -3,80 [Not Applicable]

(- E) =

R= -0,40

(R) R= 5,00 Xdi,R= Ri/Xdi,R = 0,78 0,25651 27,7506

- R= 1,5200 [Not Applicable]

(- R) =

EN1990 - Section C10: 0factors

(1) Table C4 gives expressions for obtaining the 0factors (see Section 6) in the case of two variable actions.

(2) The expressions in Table C4 have been derived by using the following assumptions and conditions:

- the two actions to be combined are independent of each other; the basic period (T 1or T2) for each action is constant;

- T1is the greater basic period; the action values within respective basic periods are constant; the intensities of an

action within basic periods are uncorrelated; the two actions belong to ergodic processes.

(3) The distribution functions in Table C4 refer to the maxima within the reference period T. These distribution functions

are total functions which consider the probability that an action value is zero during certain periods.

Reference period: T = 50 years

Ratio: N1= T/T1= 7

Greater of the basic periods: T1= 7 years (approximated to the nearest integer).

(for actions to be combined)Reliability index: = 3,8[not to be confused with rate parameter ] (shape parameter)

Coefficient of variation of the accompanying 11,1 = action for the reference period: V = 0,30

11,1 = 1/(scale parameter)

[Table C4-EN1990] - Expressions for ofor the case of two variable actionsa)General:

0,97 / 1,65 = 0,59 with: 3,3

= 3,8

Note:Fs(x; , ) "Gamma distribution" with shape and rate . [not to be confused with reliability index!].

b)Approximation for very large N1:

0,99 / 1,98 = 0,50 with: 3,3

= 3,8

Note:Fs(x; , ) "Gamma distribution" with shape and rate . [not to be confused with reliability index!].

c)Normal (approximation):

0,91 / 1,80 = 0,51

d)Gumbel (approximation):

0,85 / 2,16 = 0,39

3,83E+01

9,36E-01

Ratio

1,90E+01

7,23E-05

1

1

N1

saccompanying

0 N1leading

s

F 0,4F

F F 0,7

1 10,7 / N

1

s 1accompanying

0 1leading s

F exp N 0,4F

F F 0,7

accompanying 10

leading

F 1 0,28 0,7 lnN V

F 1 0,7 V

2 2

21k VV

2

2 2 2V

V

1 10,7 / N

1accompanying

0

leading

1 0,78V 0,58 ln ln 0,28 lnNF

F 1 0,78V 0,58 ln ln 0,7




8/123



PROJECT: ???

CLIENT: ???

JOB NO.: ??

DESIGN BY: ??

REVIEW BY: ??

D7 STATISTICAL DETERMINATION OF A SINGLE PROPERTY

D7.2 Assessment via the characteristic value

(1) The design value of a property X should be found by using:

(D.1)

where dis the design value of the conversion factor. The assessment of the relevant conversion factor is strongly

dependent on the type of test and the type of material.

The value of kncan be found from Table D1.

[Table D1-EN1990] Values of knfor the 5% characteristic value

Number of experiments:(numerical test results)

n = 30 kn= 1,67 (VXknown) kn= 1,73 (VXunknown)

Note: This table is based on the Normal distribution.

(2) When using table D1, one of two cases should be considered as follows.

- The row "VXknown" should be used if the coefficient of variation, VX, or a realistic upper bound of it, is known

from prior knowledge.

- The row "VXunknown" should be used if the coeff icient of variation VXis not known from prior knowledge and so

needs to be estimated from the sample as:

(D.2)

(D.3)

(3) The partial factor mshould be selected according to the field of application of the test results.

Note: With a log-normal distribution expression (D.1) becomes:

If VXis known from prior knowledge:

If VXis unknown from prior knowledge:

NUMERICAL TEST RESULTS, X = {x1, x2, x3,xi}

[Input max 75 rows]. Number of tests active in input = 30

n xi n xi n xi n xi n xi

1 19,3 16 17,3 31 46 61

2 19,8 17 19,2 32 47 623 20,1 18 22,4 33 48 63

4 20,4 19 16 34 49 64

5 20,3 20 15 35 50 65

6 19,3 21 15,6 36 51 66

7 18 22 18,2 37 52 67

8 17,4 23 17,4 38 53 68

9 21,3 24 19,2 39 54 69

10 19,4 25 16,3 40 55 70

11 20,2 26 15,3 41 56 71

12 20,5 27 14 42 57 72

13 21 28 13 43 58 73

14 22,3 29 15,3 44 59 74

15 18,5 30 16,5 45 60 75

k(n) dd d X n Xm m

XX m 1 k V

n

22

X i X

i 1

1s x m

n 1

X X XV s / m

dd y n y

m

X exp m k s

n

y i

i 1

1m ln x

n

2y X Xs ln V 1 V

n

2

y i y

i 1

1s ln x m

n 1

Page B1/7 EN1990.xls (rel. xx.x) - Annex D - 06/04/2013 --



9/123



PROJECT: ???

CLIENT: ???

JOB NO.: ??

DESIGN BY: ??

REVIEW BY: ??

mX = 18,283 Mean of the n = 30 sample results

(D.2) = 6,011 Variance of the term X

(D.3) = 0,134 Coefficient of variation

sX= = 2,452 Standard deviation.

kn= 1,67 (VXknown)

Values of knfor the 5% characteristic value:

kn= 1,73 (VXunknown)

Design value of a property X:

14,1889 (VXknown)

(D.1) =

14,0418 (VXunknown) [See next section D7.3].

NUMERICAL TEST RESULTS, = {1, 2, 3,i}. i0

[Input max 75 rows]. Number of tests active in input = 30

n i n i n i n i n i

1 19,3 16 17,3 31 46 61

2 19,8 17 19,2 32 47 62

3 20,1 18 22,4 33 48 63

4 20,4 19 16 34 49 64

5 20,3 20 15 35 50 65

6 19,3 21 15,6 36 51 66

7 18 22 18,2 37 52 67

8 17,4 23 17,4 38 53 68

9 21,3 24 19,2 39 54 69

10 19,4 25 16,3 40 55 70

11 20,2 26 15,3 41 56 71

12 20,5 27 14 42 57 72

13 21 28 13 43 58 73

14 22,3 29 15,3 44 59 74

15 18,5 30 16,5 45 60 75

Note: Log-normal distribution (i0)

= 2,897 Estimated value mYfor E()

(of the n = 30 sample results).

0,09 Estimated value sfor . (If Vis known from prior knowledge).

[coefficient of variation V of the i terms ].

= 0,139 Estimated value sfor .(If Vis unknown from prior knowledge).

kn= 1,67 (Vknown)

Value of knfor the 5% characteristic value:

kn = 1,73 (V unknown)

With a log-normal distribution expression (D.1) becomes:

15,5891 If Vis known from prior knowledge.

=

14,2415 If Vis unknown from prior knowledge. [See next section D7.3].

n

22

X i Xi 1

1s x m

n 1

X X XV s / m

dd y n y

m

X exp m k s

d

m

k(n) dd d X n Xm m

XX m 1 k V

d

m

2ys s ln V 1 V

n2

y i y

i 1

1s s mn 1

n n

y i i

i 1 i 1

1 1m ln

n n




10/123



PROJECT: ???

CLIENT: ???

JOB NO.: ??

DESIGN BY: ??

REVIEW BY: ??

D7.3 Direct assessment of the design value for ULS verifications

(1) The design value Xdfor X should be found by using:

Note: In this case, dshould cover all uncertainties not covered by the tests.

(2) kd,nshould be obtained from table D2:

[Table D2-EN1990] Values of kd,nfor the ULS design value. [Leading value].

Number of experiments:

(numerical test results)

n = 30 kd,n= 3,13 (VXknown) kd,n= 3,44 (VXunknown)

Note: This table is based on the assumption that the design value corresponds to a product R= 0,8 x 3,8 = 3,04 (see

annex C) and that X is Normally distributed. This gives a probability of observing a lower value of about 0,1 %: P{X < Xd} = 0,1%

mX= 18,283

[See NUMERICAL TEST RESULTS, X = {x1, x2, x3,xi}]

VX= 0,134

Design value of a property X (leading variable ):

10,61 If Vxis known from prior knowledge.

(1)

9,86 If Vxis unknown from prior knowledge.

Note: With a log-normal distribution expression (D.4) becomes:

mY= 2,897

[See NUMERICAL TEST RESULTS, = {1, 2, 3,i}. i0 ] 0,09 (If Vis known).

sY=

0,139 (If Vis unknown).

Design value of a property X (leading variable ):

13,67 (If Vis known).

11,23 (If Vis unknown).

D8 STATISTICAL DETERMINATION OF RESISTANCE MODELS

D8.2 Standard evaluation procedure (Method (a))

(1) For the standard evaluation procedure the following assumptions are made:

- the resistance function is a function of a number of independent variables X

- a sufficient number of test results is available

- all relevant geometrical and material properties are measured

- there is no correlation (statistical dependence) between the variables in the resistance function

- all variables follow either a Normal or a log-normal distribution.

Note: Adopting a log-normal distribution for a variable has the advantage that no negative values can occur.

D8.2.2.1 Step 1: Develop a design model: say: rti= AiBiCiDiHiLiMiNiQiTi. [See numerical test result below].

d d X d,n xX m 1 k V

d d X d,n xX m 1 k V d

d d y d,n yX exp m k s d




11/123



PROJECT: ???

CLIENT: ???

JOB NO.: ??

DESIGN BY: ??

REVIEW BY: ??

NUMERICAL TEST RESULTS, re= {re1, re2, re3,re1}. rei0

(rtitheoretical values; reiexperimental values from the tests).

[Input max: 75 pairs of values]. Number of tests active in input = 30

rti rei rti rei rti rei rti rei rti rei

10,5 10,9 18,9 18,4

12,6 12,3 19,4 18,9

14,7 14,9 19,7 19,5

14,9 14,2 20,4 20,8

15,1 14,8 20,8 21

15,3 14,7 21,4 21,7

15,8 15,2 21,9 22

16,1 15,6 22,5 22,8

16,5 15,5 22,9 23,2

16,9 15 23,6 23,9

17,2 16,5 23,9 24,1

17,4 16,9 24,7 25

17,8 17,5 25,2 25,5

18,1 18,5 25,9 26,218,5 18,3 26,4 25

D8.2.2.2 Step 2: Compare experimental and theoretical values [See numerical test result below].

Note: (2) The points representing pairs of corresponding values ( rti, rei) are plotted on a diagram, as indicated in figure D1.

(3) If the resistance function is exact and complete, then all of the points will lie on the line= /4 (equation: re = rt).

In practice the points will show some scatter, but the causes of any systematic deviation from that line should be

investigated to check whether this indicates errors in the test procedures or in the resistance function.

D8.2.2.3 Step 3: Estimate the mean value correction factor b:

reirti= 11400,6

b = (reirti)/(r2ti) = 0,99127

r2ti= 11501

Figure D1 - re- rtdiagram

re = rt

0

5

10

15

20

25

30

0 5 10 15 20 25 30rt

re




12/123



PROJECT: ???

CLIENT: ???

JOB NO.: ??

DESIGN BY: ??

REVIEW BY: ??

(1) Probabilistic model of the resistance r r = brt

(2) The mean value of the theoretical resistance function, calculated using the mean values Xmof the basic variables,

can be obtained from:

From the design model: rti= AiBiCiDiHiLiMiNiQiTi. [See step 1 ]

D8.2.2.4 Step 4 : Estimate the coefficient of variation of the errors.

(1) The error term ifor each experimental value reishould be determined from expression (D9):

NUMERICAL TEST RESULTS, = {1, 2, 3,1}. i0

(ierror term for each test; i= ln(i) estimated value for V).

i i i i i i i i i i

1,047 0,046 0,982 -0,018 0,000 0,000 0,000 0,000 0,000 0,000

0,985 -0,015 0,983 -0,017 0,000 0,000 0,000 0,000 0,000 0,000

1,023 0,022 0,999 -0,001 0,000 0,000 0,000 0,000 0,000 0,000

0,961 -0,039 1,029 0,028 0,000 0,000 0,000 0,000 0,000 0,000

0,989 -0,011 1,019 0,018 0,000 0,000 0,000 0,000 0,000 0,000

0,969 -0,031 1,023 0,023 0,000 0,000 0,000 0,000 0,000 0,000

0,970 -0,030 1,013 0,013 0,000 0,000 0,000 0,000 0,000 0,000

0,977 -0,023 1,022 0,022 0,000 0,000 0,000 0,000 0,000 0,000

0,948 -0,054 1,022 0,022 0,000 0,000 0,000 0,000 0,000 0,000

0,895 -0,110 1,022 0,021 0,000 0,000 0,000 0,000 0,000 0,000

0,968 -0,033 1,017 0,017 0,000 0,000 0,000 0,000 0,000 0,000

0,980 -0,020 1,021 0,021 0,000 0,000 0,000 0,000 0,000 0,000

0,992 -0,008 1,021 0,021 0,000 0,000 0,000 0,000 0,000 0,000

1,031 0,031 1,020 0,020 0,000 0,000 0,000 0,000 0,000 0,000

0,998 -0,002 0,955 -0,046 0,000 0,000 0,000 0,000 0,000 0,000

-0,135

n = 30 Numerical test results: pairs of values.

(3) -0,005 Estimated value for E().

0,031

(4) 0,001 Estimated value s2for

2.

(5) 0,032 Coefficient of variation Vof the ierror terms.

D8.2.2.5 Step 5: Analyse compatibility

(1) The compatibility of the test population with the assumptions made in the resistance function should be analysed.

(2) If the scatter of the (rei, rti) values is too high to give economical design resistance functions, this scatter may be reduced.

(3) To determine which parameters have most influence on the scatter, the test results may be split into subsets with

respect to these parameters.

(4) When determining the fractile factors kn (see step 7), the kn value for the sub-sets may be determined on the basis

of the total number of the tests in the original series.

m t m rt mr br X bg X

eii

ti

r

br

n n

i ii 1 i 1

1 1ln

n n

n

22

ii 1

1s

n 1

n

2

i

i 1

2V exp s 1

n n

i ii 1 i 1

ln




13/123



PROJECT: ???

CLIENT: ???

JOB NO.: ??

DESIGN BY: ??

REVIEW BY: ??

D8.2.2.6 Step 6: Determine the coefficients of variation VXiof the basic variables

(1) The coefficients of variation VXiwill normally need to be determined on the basis of some prior knowledge.

From the design model: rti= AiBiCiDiHiLiMiNiQiTi [See step 1 ]

Coefficient of variation [VXi0]:

[Input max: 10 values]. Number of tests active in input = 4

VA VB VC VD VH VL VM VN VQ VT

0 0,04 0,05 0,07

Note: Input VA = 0 means A = cost (number).

Resistence function: r = cost B C D

(see next step).

D8.2.2.7 Step 7: Determine the characteristic value rkof the resistance

(1) The resistance function for j basic variables is a product function of the form:

Coefficient of variation:

= 0,010 with: 0,032 0,009

(4) If the number of tests is limited (say n < 100) allowance should be made in the distribution offor statistical uncertainties.

The distribution should be considered as a central t -distribution with the parameters , Vand n.

(5) In this case the characteristic resistance rkshould be obtained from:

with:

Qrt, rtversus coefficients of variation [Vr, V0]:

V2r= 0,010 V

2rt= 0,009 V

2= 0,001

0,100 Q 0,095 Qrt

0,032 Q

0,950 rt (weighting factor for Qrt). 0,320 (weighting factor for Q).

Values of knfor the 5% characteristic value: kn= 1,73 (for the case Vunknown)

k= 1,64 (for n ).

(D.17) exp ( 1,64 x 0,950 x 0,095 1,73 x 0,320 x 0,032 0,5 x 0,100)

0,843 [characteristic value of the resistance].

j

2 2 2

r Xii 1

V V 1 V 1 1

2V exp s 1

2k rt m rt rt nr bg X exp k Q k Q 0,5Q

2

rt ln(rt ) rt

2

ln( )

2

ln(r ) r

rtrt

Q ln V 1

Q ln V 1

Q ln V 1

Q

Q

Q

Q

n2 2

rt Xii 1

V V

2k rt m rt rt n mr bg X exp k Q k Q 0,5Q r

2k rt m rt rt n mr bg X exp k Q k Q 0,5Q r

t 1 2 jr br b X X ...X




14/123



PROJECT: ???

CLIENT: ???

JOB NO.: ??

DESIGN BY: ??

REVIEW BY: ??

D8.3 Standard evaluation procedure (Method (b))

(1) In this case the procedure is the same as in D8.2, excepted that step 7 is adapted by replacing the characteristic fractile

factor knby the design fractile factor kd,nequal to the product Rassessed at 0,8 X 3,8 = 3,04 as commonly accepted

(see Annex C) to obtain the design value rdof the resistance.

(2) For the case of a limited number of tests the design value rdshould be obtained from:

(D.21)

where:

- kd,n is the design fractile factor from table D2 for the case VXunknown

- kd,is the value of kd,nfor n [kd,= 3,04].

Values of kd,n

for the ULS design value for leading value. [Active value]: kd,n

=3,13

(VXknown)

kd,n= 3,44 (VXunknown).

(D.21) exp ( 3,04 x 0,950 x 0,095 3,44 x 0,320 x 0,032 0,5 x 0,100)

0,730 [design value of the resistance].

Dividing the characteristic value by the design value we obtain:

(0,843/0,730) = 1,15

D8.4 Use of additional prior knowledge

(1) If the validity of the resistance function rtand an upper bound (conservative estimate) for the coefficient of variation Vr

are already known from a significant number of previous tests, the following simplified procedure may be adopted when

further tests are carried out.

b) two or three further tests are carried out

Maximum coefficient of variation observed in previous tests: Vr= 0,09

provided that each extreme (maximum or minimum) value ree satisfies the condition: |ree - rem| 0,10rem

Reduction factor applicable in the case of pr ior knowledge:

(D.26) exp ( 2,0 x 0,09 0,5 x 0,090) = 0,832

c1= 1,0

c2= 2,0

Characteristic value rk:

re re

rk= 0,832

rem rem [Case Applicable]

where reis the result of the test for case a) and remis the mean value of the test results (case b).

2d rt m d, rt rt d,nr bg X exp k Q k Q 0,5Q

2d rt m d, rt rt d,n mr bg X exp k Q k Q 0,5Q r

2d rt m d, rt rt d,n mr bg X exp k Q k Q 0,5Q r

kR

d

r

r

2

k 1 2 r r c exp c V 0,5V

k kr




15/123



PROJECT: ???

CLIENT: ???

JOB NO.: ??

DESIGN BY: ??

REVIEW BY: ??

SECTION 6 - IMPOSED LOADS ON BUILDINGS

6.3 Characteristic values of Imposed Loads6.3.1 Residential, social, commercial and administration areas - 6.3.1.1 Categorie

[Table 6.1-EN1991-1-1] Categories of use

Category Specific Use

C

Example

NOTE 1 Depending on their anticipated uses, areas likely to be categorised as C2, C3, C4 may be categorised

as C5 by decision of the client and/or National annex.

NOTE 2 The National annex may provide sub categories to A, B, C1 to C5, D1 and D2.

NOTE 3 See 6.3.2 for storage or industrial activity.

6.3 Characteristic values of Imposed Loads

6.3.1 Residential, social, commercial and administration areas - 6.3.1.2 Values of action

(1)P The categories of loaded areas, as specified in Table 6.1, shall be designed by using characteristic values qk(uniformly

distributed load) and Qk(concentrated load).

Note Values for kand Q kare given in Table 6.2 below. Where a range is given in this table, the value

may be set by the National annex. The recommended values, intended for separate application, areunderlined. q kis intended for determination of general effects and Q k for local effects. The National annex

may define different conditions of use of this Table.

[Table 6.2-EN1991-1-1] Imposed loads on floors, balconies and stairs in buildings

Categories of loaded areas qk[kN/m2] Qk[kN]

Category A- Floors 1,5 to 2,0 2,0 to 3,0

- Stairs 2,0 to 4,0 2,0 to 4,0

- Balconies 2,5 to 4,0 2,0 to 3,0

Category B 2,0 to 3,0 1,5 to 4,5

Category C

- C1 2,0 to 3,0 3,0 to 4,0

- C2 3,0 to 4,0 2,5 to 7,0

- C3 3,0 to 5,0 4,0 to 7,0

- C4 4,5 to 5,0 3,5 to 7,0

- C5 5,0 to 7,5 3,5 to 4,5

Category D

- D1 4,0 to 5,0 3,5 to 7,0

- D2 4,0 to 5,0 3,5 to 7,0

Areas where people may congregate (with the exception of areas defined under category A,

B, and D).

Attention is drawn to 6.3.1.1(2), in particular for C4 and C5. See EN 1990 when dynamic effects need to be considered.

For Category E, see Table 6.3.

C1: Areas with tables, etc. e.g. areas in schools, cafs, restaurants, dining halls, reading rooms, receptions.

C2: Areas with fixed seats, e.g. areas in churches, theatres or cinemas, conference rooms, lecture halls, assembly halls,

waiting rooms, railway waiting rooms.

C3: Areas without obstacles for moving people, e.g. areas in museums, exhibition rooms, etc. and access areas in public

and administration buildings, hotels, hospitals, railway station forecourts.

C4: Areas with possible physical activities, e.g. dance halls, gymnastic rooms, stages.

C5: Areas susceptible to large crowds, e.g. in buildings for public events like concert halls, sports halls including stands,

terraces and access areas and railway platforms.

Page A1/6 EN1991-1-1.xls (rel. xx.x) - Sheet: CodeSec6 - 06/04/2013 --



16/123



PROJECT: ???

CLIENT: ???

JOB NO.: ??

DESIGN BY: ??

REVIEW BY: ??

(2) Where necessary qkand Qkshould be increased in the design (e.g. for stairs and balconies depending on the occupancy

and on dimensions).

(3) For local verifications a concentrated load Qkacting alone should be taken into account.

(5)P The concentrated load shall be considered to act at any point on the floor, balcony or stairs over an area with a shape

which is appropriate to the use and form of the floor.

Note: The shape may normally be assumed as a square with a width of 50 mm. See also 6.3.4.2(4)

(8) Self-weight of the partitions:

c) movable partitions with a self-weight > 2 3,0 kN/m

Note: [6.2.1 Floors, beams and roofs:Imposed loads from a single category may be reduced according to the areas supported

by the appropriate member, by a reduction factor A according to 6.3.1.2(10).

(10) In accordance with 6.2.1(4) imposed loads from a single category may be reduced according to the areas supported by the

appropriate member, by a reduction factor.

(See Table 6.2 and subclauses (8) and (9)) and for accessible roofs, Category I (see Table 6.9).

Note 1: The recommended value for the reduction factor Afor categories A to D is determined as follows:

Factor according to EN 1990 Annex A1 Table A1.1: 0= 0,7

Categories of loaded areas (see Table 6.2): Category C 0,6 [Restriction required]

A = 75,00 m2

= 0,63

Loaded area A:

(influence area) A0= 10 m2

= (5/7) x 0,7 + 10/75 = 0,63

Note: The National Annex may give an alternative method.

(11) In accordance with 6.2.2(2) and provided that the area is classified according to table 6.1 into the categories A to D,

for columns and walls the total imposed loads from several storeys may be multiplied by the reduction factor n.

6.2.2(2) Where imposed loads from several storeys act on columns and walls, the total imposed loads may be reduced by a

factor naccording to 6.3.1.2(11) and 3.3.1(2)P.

Note 1: The recommended values for nare given below:

[2 + (4 - 2) x 0,7]/4 = 0,85

Number of storey (> 2) above the loaded structural element from the same category: n = 4

Factor according to EN 1990 Annex A1 Table A1.1 [Active value]: 0= 0,7


6.3.2 Areas for storage and industrial activities - 6.3.2.1 Categories

[Table 6.3-EN1991-1-1 Categories of storage and industrial use

Category Specific use

E1 Areas susceptible to accumulation of goods, including access areas.

Example

Areas for storage use including storage of books and other documents.

0A 0

A51,0

7 A

0n

2 n 2

n




17/123



PROJECT: ???

CLIENT: ???

JOB NO.: ??

DESIGN BY: ??

REVIEW BY: ??

6.3.2 Areas for storage and industrial activities - 6.3.2.2 Values for Actions

[Table 6.4-EN1991-1-1 Imposed loads on floors due to storage

Categories of loaded areas qk[kN/m2] Qk[kN]

E1 7,5 7,0

Note: Recommended values for q kand Q kare given in Table 6.4 above. The values may be changed if necessary according

to the usage (see Table 6.3 and Annex A) for the particular project or by the National annex. q kis intended for determination

of general effects and Q kfor local effects. The National annex may define different conditions of use of Table 6.4.


6.3.2 Areas for storage and industrial activities - 6.3.2.3 Actions induced by forklifts

[Table 6.5-EN1991-1-1 Dimensions of forklift according to classes FL

Class of Forkilt

FL2 31 15 0,95 1,10 3,00

(2) The static vertical axle load Qkof a forklift depends on the forklift classes FL1 to FL6 and should be obtained from Table 6.6.

[Table 6.6-EN1991-1-1 Axle loads of forklifts

Class of Forkilt Axle load Qk[kN]

FL2 40

(3) The static vertical axle load Qkshould be increased by the dynamic factor using expression (6.3):

(6.3) Qk,dyn= Qk= 1,40 x 40 = 56,0 kN

(4)

Dynamic factor: = 1,40

a) for pneumatic tyres

(5) For forklifts having a net weight greater than 110 kN the loads should be defined by a more accurate analysis.

(6) The vertical axle load Qkand Qk,dynof a forklift should be arranged according to Figure 6.1.

Net weight [kN] Hoisting load [kN] Width of axle a [m] Overall width b

[m]

Overall length l

[m]




18/123



PROJECT: ???

CLIENT: ???

JOB NO.: ??

DESIGN BY: ??

REVIEW BY: ??

(7) Horizontal loads due to acceleration or deceleration of forklifts may be taken as 30 % of the vertical axle loads Qk:

Hk,dyn= 0,30 Qk= 0,30 x 40 = 12,0 kN (Dynamic factors need not be applied).


6.3.3 Garages and vehicle traffic areas (excluding bridges) - 6.3.3.1 Categories

[Table 6.7-EN1991-1-1] Traffic and parking areas in buildings

Category of Specific Use

traffic areas

G

Examples

Access routes; delivery zones; zones accessible to fire engines (160 kN gross vehicle weight).

Note: Access to areas designed to category F should be limited by physical means built into the structure.

Areas designed to categories F and G should be posted with the appropriate warning signs.


6.3.3 Garages and vehicle traffic areas (excluding bridges) - 6.3.3.2 Values of actions

(1) The load model which should be used is a single axle with a load Qkwith dimensions according to Figure 6.2 and a uniformly

distributed load qk. The characteristic values for qkand Qkare given in Table 6.8.

[Figure 2 - Dimension of axle load]

Note: qkis intended for determination of general effects and Qkfor local effects. The National annex may define different

conditions of use of this Table.

Note: For category F (see Table 6.8) the width of the square surface is 100 mm and for category G (see Table 6.8) the width of a

square surface is 200 mm [see next (2)].

[Table 6.8-EN1991-1-1 Imposed loads on garages and vehicle traffic areas

Category of qk[kN/m2] Qk[kN]

traffic areas

F: Gross vehicle weight: 30 kN 1,5 to 2,5 10,0 to 20,0

G: 30 kN < gross vehicle weight 160 kN 5,0 40,0 to 90,0

Traffic and parking areas for medium vehicles (>30 kN, 160 kN gross vehicle weight, on 2axles).




19/123



PROJECT: ???

CLIENT: ???

JOB NO.: ??

DESIGN BY: ??

REVIEW BY: ??

(2) The axle load should be applied on two square surfaces with a 100 mm side for category F and a 200 mm side for Category G

in the possible positions which will produce the most adverse effects of the action.


6.3.4 Roofs - 6.3.4.1 Categories

[Table 6.9-EN1991-1-1 Categorization of roofs

Catogories of Specific use

loaded area

I Roofs accessible with occupancy according to categories A to G.

(2) Imposed loads for roofs of category H should be those given in Table 6.10. Imposed loads for roofs of category I are given

in Tables 6.2, 6.4 and 6.8 according to the specific use.

(3) The loads for roofs of category K which provide areas for helicopter landing areas should be for the helicopter classes HC,

see Table 6.11.


6.3.4 Roofs - 6.3.4.2 Values of actions

[Table 6.10-EN1991-1-1 Imposed loads on roofs of category H

Category H qk[kN/m2] Qk[kN]

Range: 0,0 to 1,0 0,9 to 1,5

Recommended values: 0,4 1,0

Note: Where a range is given the values may be set by the National Annex. q kmay be varied by the National Annex dependent

upon the roof slope. q kmay be assumed to act on an area A which may be set by the National Annex. The recommended

value for A is 10 m2, within the range of zero to the whole area of the roof. See also 3.3.2(1).

(2) The minimum values given in Table 6.10 do not take into account uncontrolled accumulations of construction materials that

may occur during maintenance.

(4) Roofs, other than those with roof sheeting, should be designed to resist 1,5 kN on an area based on a 50 mm sided square.Roof elements with a profiled or discontinuously laid surface, should be designed so that the concentrated load Qkacts over

the effective area provided by load spreading arrangements.

[Table 6.11-EN1991-1-1 Imposed loads on roofs of category K for helicopters

Class of Helicopter

HC2 20 kN < Q 60 kN 60 kN 0,3 x 0,3

Qk,dyn= 1,40 x (60 kN) = 84 kN

Take-off load Q of

helicopter

Take-off load Qk

=

Dimension of the loaded

area [m x m]




20/123



PROJECT: ???

CLIENT: ???

JOB NO.: ??

DESIGN BY: ??

REVIEW BY: ??

(7) Access ladders and walkways should be assumed to be loaded according to Table 6.10 for a roof slope < 20.

For walkways which are part of a designated escape route, qkshould be according to Table 6.2. For walkways for service

a minimum characteristic value Qkof 1,5 kN should be taken.

(8) The following loads should be used for the design of frames and coverings of access hatches (other than glazing),

the supports of ceilings and similar structures:

a) without access: no imposed load;

b) with access: 0,25 kN/m2distributed over the whole area or the area supported, and the concentrated load of 0,9 kN

so placed so as to produce maximum stresses in the affected member.

6.4 Horizontal loads on parapets and partition walls acting as barriers

(1) The characteristic values of the line load qkacting at the height of the partition wall or parapets but not higher than 1,20 m

should be taken from Table 6.12.

[Table 6.12-EN1991-1-1 Horizontal loads on partition walls and parapets

Loaded areas qk[kN/m] qk[kN/m]

(range) (recommended values)

Category A 0,2 to 1,0 0,5

Category B and C1 0,2 to 1,0 0,5

Categories C2 to C4 and D 0,8 to 1,0 1,0

Category C5 3,0 to 5,0 3,0

Category E 0,8 to 2,0 2,0

Category F

Category G

Note: For areas of category E the horizontal loads depend on the occupancy. Therefore the value of q kis defined as a minimum

value and should be checked for the specific occupancy.

(2) For areas susceptible to significant overcrowding associated with public events e.g. for sports stadia, stands, stages,

assembly halls or conference rooms, the line load should be taken according to category C5.

See Annex B

See Annex B




21/123



PROJECT: ???

CLIENT: ???

JOB NO.: ??

DESIGN BY: ??

REVIEW BY: ??

ANNEX A - Informative

[Table A.1-EN1991-1-1] Construction materials-concrete and mortar

Materials

Concrete:

8) heavy weight 24,0 to > 24,0 ) )

) Increase by 1kN/m3for normal percentage of reinforcing and pre-stressing steel.

) Increase by 1kN/m3for unhardened concrete.

Mortar:

3) lime-cement mortar 18,0 to 20,0

Note: See Section 4.

[Table A.2-EN1991-1-1] Construction materials-masonry

Materials See:

Masonry units:

5) manufactured stone masonry units - to - EN 771-5

Natural stones, see prEN 771-6:

h) volcanic tuff 20,0 to 20,0 -


[Table A.3-EN1991-1-1] Construction materials-wood

Materials

Wood: (see EN 338 for timber strength

timber strength class D70 10,8 classes)

Glued laminated timber: (see EN 1194 for Timber strength

8) combined glulam GL36c 4,2 classes)

Plywood:

2) birch plywood 7,0

Particle boards:

1) chipboard 7,0 to 8,0

cont'd

Density [kN/m3]

Density [kN/m3]

Density [kN/m3]

[TableA.3-EN1991-1-

Page B1/7 EN1991-1-1.xls (rel. xx.x) - Sheet: Annex A - 06/04/2013 --



22/123



PROJECT: ???

CLIENT: ???

JOB NO.: ??

DESIGN BY: ??

REVIEW BY: ??

cont'd

2) medium density fibreboard 8,0


[Table A.4-EN1991-1-1] Construction materials-metals

Materials

Metals:

9) zinc 71,0 to 72,0

[Table A.5-EN1991-1-1] Construction materials- other materials

Materials

Other materials:

2) glass, in sheets 25,0

Plastics:

2) polystyrene, expanded, granules 0,3

[Table A.6-EN1991-1-1] Bridge materials

Materials

Pavement of road bridges:

1) gussasphalt and asphaltic concrete 24,0 to 25,0

Infills for bridges:

3) hardcore 18,5 to 19,5

Pavement of rail bridges:

2) normal ballast (e.g. granite, gneiss, etc.) 20,0

Weight per unit bed length

gk[kN/m]

Structures with ballasted bed:

b) prestressed concrete sleeper with track fastenings 4,8

cont'd

Density [kN/m3]

Density [kN/m3]

Density [kN/m3]

Density [kN/m3]

[TableA.6-EN1991-1-

[TableA.3

]




23/123



PROJECT: ???

CLIENT: ???

JOB NO.: ??

DESIGN BY: ??

REVIEW BY: ??

cont'd

Weight per unit bed length

Materials gk[kN/m] ) )

Structures without ballasted bed:

a) 2 rails UIC 60 with track fastenings 1,7

Note: ) Given in other tables as stored materials.

) Excludes an allowance for ballast.

) Assumes a spacing of 600 mm.

Note 1 The values for track are also applicable outside railway bridges.

Note 2 See Section 4.

[Table A.7-EN1991-1-1] Stored materials - building and construction

Materials

Aggregates (see EN 206):

1) lightweight 9,0 to 20,0 30

Gravel and sand, bulked 15,0 to 20,0 35

Sand 14,0 to 19,0 30

Blast furnace slag:

2) granules 12,0 30

Brick sand, crushed brick, broken bricks 15,0 35

Vermiculite:

2) crude 6,0 to 9,0 -

Bentonite:

loose 8,0 40

Cement:

1) in bulk 16,0 28

Fly ash 10,0 to 14,0 25

Glass, in sheets 25,0 -

Gypsum, ground 15,0 25

Lignite filter ash 15,0 20

Lime 13,0 25

Limestone, powder 13,0 25 to 27

Magnesite, ground 12,0 -

cont'd

Angle of repose []Density [kN/m3]

[TableA.6-EN1991-

1-

[TableA.7-EN1991-1-




24/123



PROJECT: ???

CLIENT: ???

JOB NO.: ??

DESIGN BY: ??

REVIEW BY: ??

cont'd

Materials

Plastics:

1) polyethylene, polystyrol granulated 6,4 30

Water, fresh 10,0 -

[Table A.8-EN1991-1-1] Stored products - agricultural

Farmyard:

3) dry chicken manure 6,9 45

Fertiliser, artificial:

2) basic slag, crushed 13,7 to 13,7 35

Fodder, green, loosely stacked 3,5 to 4,5 -

Grain whole:

whole (14 % moisture content unless indic. otherwise)

j) wheat in bulk 7,8 30

Grass cubes 7,8 40

Hay:

1) baled 1,0 to 3,0 -

Hides and skins 8,0 to 9,0 -

Hops 1,0 to 2,0 25

Malt 4,0 to 6,0 20

Meal:

1) ground 7,0 45

Peat:

2) dry, compressed in bales 5,0 -

Silage 5,0 to 10,0 -

Straw:

1) in bulk (dry) 0,7 -

Tobacco in bales 3,5 to 5,0 -

Wool:

1) in bulk 3,0 to 3,0 -

Note: ) see table A.1 for density classes of lightweight concrete

See Section 4.

Density [kN/m3] Angle of repose []

[TableA.7-EN1991-1-




25/123



PROJECT: ???

CLIENT: ???

JOB NO.: ??

DESIGN BY: ??

REVIEW BY: ??

[Table A.9-EN1991-1-1] Stored products - foodstuffs

Materials

Eggs, in stands 4,0 to 5,0 -

Flour:

1) bulk 6,0 25

Fruit:

3) cherries 7,8 -

Sugar:1) loose, piled 7,5 to 10,0 35

Vegetable, green:

2) lettuce 5,0 -

Vegetable, legumes:

1) beans: general 8,1 30

Vegetable, root:

3) carrots 7,8 35

Patatoes:1) in bulk 7,6 35

Sugarbeet:

2) raw 7,6 -


[Table A.10-EN1991-1-1] Stored products - liquids

Materials

Beverages:

1) beer 10,0

Natural oils:

2) glycerol (glycerine) 12,3

Organic liquids and acids:

8) turpentine, white spirit 8,3

cont'd

Density [kN/m3]


[TableA.1

0-EN1991-1-




26/123



PROJECT: ???

CLIENT: ???

JOB NO.: ??

DESIGN BY: ??

REVIEW BY: ??

cont'd

Hydrocarbons:

n) liquid gas (butane)

Other liquids:

3) white lead, in oil 38


[Table A.11-EN1991-1-1] Stored products - solid fuels

Materials

Charcoal:

1) air-filled 4 -

Coal:

j) all other kinds of coal 30 to 35

Firewood 5,4 45

Lignite/brown coal:

3) damp 9,8 30 to 40

Peat:

1) black, dried, firmly packed 6 to 9 -


[Table A.12-EN1991-1-1] Stored products - industrial and general

Materials

Books and documents:

1) books and documents, 6,0 -

Filing racks and cabinets 6,0 -

Garments and rags, bundled 11,0 -

Ice, lumps 8,5 -

Leather, piled 10,0 -

cont'd

8,3


5,7


[TableA.1

0-EN1991-1

-

[TableA.1

2-EN1991-1-




27/123



PROJECT: ???

CLIENT: ???

JOB NO.: ??

DESIGN BY: ??

REVIEW BY: ??

cont'd

Paper:

2) piled 11,0 -

Rubber 10 to 17 -

Rock salt 22,0 45

Salt 12,0 40

Sawdust

2) dry, loose 2,5 45

Tar, bitumen 14,0 -

[TableA.1

2-EN1991-1

-




28/123



PROJECT: ???

CLIENT: ???

JOB NO.: ??

DESIGN BY: ??

REVIEW BY: ??

ANNEX B - Informative

VEHICLE BARRIERS AND PARAPETS FOR CAR PARKS

B(1) Barriers and parapets in car parking areas should be designed to resist the horizontal loads given in B(2).

B(2) The horizontal characteristic force F (in kN), normal to and uniformly distributed over any length of 1,5 m of a barrier for a car

park, required to withstand the impact of a vehicle is given by:

(characteristic value).

B(3) The car park has been designed on the basis that the gross mass of the vehicles using it will not exceed 2500 kg:

Deformations of the vehicle (positive in mm): c= 100 mm (unless better evidence is available).

Deformations of the barrier (positive in mm): b= 0 mm

Velocity of the vehicle (in m/s) normal to the barrier: v = 4,5 m/s

Gross mass of the vehicle in (kg): m = 1500 kg (case m 2500 kg).

(The mass of 1500 kg is taken as being more representative

of the vehicle population than the extreme value of 2500 kg).

Case:

B(3) F = 0,5 x (1500 kg) x (4,5 m/s)/[(100 mm + 0 mm)/(10 mm/m)] = 151875 N = 151,88 kN [Rigid Barrier]

Design value: Fd= Fx F = 1,5 x 151,88 kN = 227,82 kN Bending moment (design value):

Md= Fdx hd= (227,82 kN) x (0,375 m)

B(5) Bumper eight above finish floor level (FFL): hd= 375 mm (design height). = 85,43 kNm

Case:

B(4) The car park has been designed for vehicles whose gross mass exceeds 2500 kg:

F = 0,5 x (3000 kg) x (4,5 m/s)/[(100 mm + 0 mm)/(10 mm/m)] = 303750 N = 303,75 kN [Rigid Barrier]

Actual mass of the vehicle for which the car park is designed: m = 3000 kg

(case m > 2500 kg).

B(5) Bumper eight (actual) above finish floor level (FFL): hd= 550 mm

Design value: Fd= Fx F = 1,5 x 303,75 kN = 455,63 kN Bending moment (design value):

Md= Fdx hac= (455,63 kN) x (0,550 m)

Bumper eight above finish floor level (FFL): hac= 550 mm (actual height). = 250,60 kNm

B(6) Barriers to access ramps of car parks have to withstand one half of the force determined in B(3) or B(4) acting at a height

of 610 mm above the ramp:

Rif. Case:

B(3) Design val.: Fd= Fx 0,5F = 1,5 x 0,5 x 151,88 kN = 113,91 kN Bending moment (design value):

Md= Fdx hd= (113,91 kN) x (0,610 m)

Bumper eight above finish floor level (FFL): hd= 610 mm (design height). = 69,49 kNm

Rif. Case:

B(4) Design val.: Fd= Fx 0,5F = 1,5 x 0,5 x 303,75 kN = 227,81 kN Bending moment (design value):

Md= Fdx hd= (227,81 kN) x (0,610 m)


cont'd

2

k

c b

1 mvF

2

Page C1/2 EN1991-1-1.xls (rel. xx.x) - Sheet: Annex B - 06/04/2013 --



29/123



PROJECT: ???

CLIENT: ???

JOB NO.: ??

DESIGN BY: ??

REVIEW BY: ??

cont'd

B(7) Opposite the ends of straight ramps intended for downward travel which exceed 20 m in length the barrier has to withstand

twice the force determined in B(3) acting at a height of 610 mm above the ramp:

Rif. Case:

B(3) Design value: Fd= Fx 2F = 1,5 x 2 x 151,88 kN = 455,64 kN Bending moment (design value):

Md= Fdx hd= (455,64 kN) x (0,610 m)


Page C2/2 EN1991-1-1.xls (rel. xx.x) - Sheet: Annex B - 06/04/2013 --



30/123



PROJECT: ???

CLIENT: ???

JOB NO.: ??

DESIGN BY: ??

REVIEW BY: ??

SECTION 3 - THERMAL ACTIONS FOR TEMPERATURE ANALYSIS

3.1 General rules

1(P) Thermal actions are given by the net heat flux to the surface of the member.

(2) On the fire exposed surfaces the net heat flux should be determined by considering heat transfer by convection and

radiation as:

(3.1) (net convective heat flux component + net radiative heat flux component).

Coefficient of heat transfer by convection: c= 4,00 W/m2K

(4) For the coefficient of heat transfer by convection crelevant for nominal temperature-time curves, see 3.2.

(5) On the unexposed side of separating members, the net heat flux should be determined by using equation (3.1), with

c= 4 W/m2K. The coefficient of heat transfer by convection should be taken as c= 9 W/m

2K when assuming it contains

the effects of heat transfer by radiation.

Gas temperature in the vicinity of the fire exposed member: g= 700 C

(10) Gas temperatures gmay be adopted as nominal temperature-time curves according to 3.2, or adopted according to the fire

models given in 3.3.

Surface temperature of the member: m= 70 C

= (4,00 W/mK) x (700 C - 70 C) = 2520 W/m = 2,52 kW/m

Configuration factor: = 1,0

Note For the calculation of the configuration factor a method is given in annex G.

Surface emissivity of the member: m= 0,8

Note 1 Unless given in the material related fire design Parts of prEN 1992 to prEN 1996 and prEN 1999, m= 0,8 may be used.

Emissivity of the fire: f= 1,0

Note 2 The emissivity of the fire is taken in general as f= 1,0.

Effective radiation temperature of the fire environment: r= 700 C

(8) In case of fully fire engulfed members, the radiation temperature rmay be represented by the gas temperature garound

that member.

Surface temperature of the member: m= 70,0 C

The surface temperature mresults 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.

1,0 x 0,8 x 1 x x {[(700 + 273)] [(70 + 273)]} = 40.028 W/m

= 40,03 kW/m

where is the Stephan Boltzmann constant (= 5,67 x 10-8

W/m2K

4).

2

neth [W / m ]

neth

2

net net,c net,r h h h [W / m ]

neth

net,c c g mh

4 4

net,r m f r mh 273 273

Page A1/5 EN1991-1-2_(a).xls (rel. xx.x) - Sheet: CodeSec3 - 06/04/2013 --



31/123



PROJECT: ???

CLIENT: ???

JOB NO.: ??

DESIGN BY: ??

REVIEW BY: ??

(3.1) (2,52 + 40,03) = 42,55 kW/m

3.2 Nominal temperature-time curves

3.2.1 Standard temperature-time curve

(1) The standard temperature-time curve is given by (gas temperature in the fire compartment):

g= 20 + 345log10(8t + 1) = 20 + 345 x log[(8 x 120) + 1] = 1049 C

Time of the exposure: t = 120 min

(2) The coefficient of heat transfer by convection is: = 25 W/m2K.


3.2.2 External fire curve (see figure below)

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

g= 20 + 660(1 0,687e0,32t

0,313e3,8t

) = 20 + 660 x [1 0,687exp(0,32 x 15) 0,313exp(3,8 x 15)] = 676C

Time of the exposure: t = 15 min [See next page ].


net net,c net,r h h h

1049

0

200

400

600

800

1000

1200

0 50 100 150 200 250

t [min]

g

[C]

Standard temperature-time curve




32/123



PROJECT: ???

CLIENT: ???

JOB NO.: ??

DESIGN BY: ??

REVIEW BY: ??



676

0

100

200

300

400

500

600

700

800

0 50 100 150 200 250 300 350 400

t [min]

g

[C]

1071

0

200

400

600

800

1000

1200

0 50 100 150 200 250 300 350 400

t [min]

g

[C]

External fire curve

Hydrocarbon curve




33/123



PROJECT: ???

CLIENT: ???

JOB NO.: ??

DESIGN BY: ??

REVIEW BY: ??


3.2.3 Hydrocarbon curve (see figure above)

(1) The hydrocarbon temperature-time curve is given by (gas temperature in the fire compartment):

g= 20 + 1080(1 0,325e0,167t

0,675e2,5t

) = 20 + 1080 x [1 0,325exp(0,167 x 15) 0,675exp(2,5 x 15)] = 1071C

Time of the exposure: t = 15 min [See previous page ].


3.3 Natural fire models3.3.1 Simplified fire models - 3.3.1 General

(1) Simple fire models are based on specific physical parameters with a limited field of application.

Note For the calculation of the design fire load density qf,da method is given in annex E.

(2) A uniform temperature distribution as a function of time is assumed for compartment fires. A nonuniform temperature

distribution as a function of time is assumed in case of localised fires.

(3) When simple fire models are used, the coefficient of heat transfer by convection should be taken as c= 35 W/m2K.

3.3 Natural fire models

3.3.1 Simplified fire models - 3.3.1.2 Compartment fires

(1) Gas temperatures should be determined on the basis of physical parameters considering at least the fire load density and the

ventilation conditions.

Note 1 The national annex may specify the procedure for calculating the heating conditions.

Note 2 For internal members of fire compartments, a method for the calculation of the gas temperature in the compartment is given

in annex A.

(2) For external members, the radiative heat flux component should be calculated as the sum of the contributions of the fire

compartment and of the flames emerging from the openings.

Note For external members exposed to fire through openings in the facade, a method for the calculation of the heating conditions is

given in annex B.


3.3.1 Simplified fire models - 3.3.1.3 Localised fires

(1) Where flash-over is unlikely to occur, thermal actions of a localised fire should be taken into account.

Note The national annex may specify the procedure for calculating the heating conditions. A method for the calculation of thermal

actions from localised fires is given in annex C.


3.3.2 Advanced fire models

(1) Advanced fire models should take into account the following: - gas properties, - mass exchange, - energy exchange.Available calculation methods normally include iterative procedures. For the calculation of the design f ire load density qf,d

a method is given in annex E. For the calculation of the rate of heat release Q a method is given in annex E.

Notes

1,2,3




34/123



PROJECT: ???

CLIENT: ???

JOB NO.: ??

DESIGN BY: ??

REVIEW BY: ??

(2) One of the following models should be used:

one-zone models assuming a uniform, time dependent temperature distribution in the compartment;

two-zone models assuming an upper layer with time dependent thickness and with time dependent uniform temperature, as

well as a lower layer with a time dependent uniform and lower temperature;

Computational Fluid Dynamic models giving the temperature evolution in the compartment in a completely time dependent

and space dependent manner.

Note The national annex may specify the procedure for calculating the heating conditions. A method for the calculation of thermal

actions in case of one-zone, two-zone or computational fluid dynamic models is given in annex D.

(3) The coefficient of heat transfer by convection should be taken as c= 35 [W/m2K], unless more detailed information is available.

(4) In order to calculate more accurately the temperature distribution along a member, in case of a localised fire, a combination of

results obtained with a two-zone model and a localised fire approach may be considered.

Note The temperature field in the member may be obtained by considering the maximum effect at each location given by the two

fire models.




35/123



PROJECT: ???

CLIENT: ???

JOB NO.: ??

DESIGN BY: ??

REVIEW BY: ??

ANNEX A - Informative

PARAMETRIC TEMPERATURE-TIME CURVES

(1) The following temperature-time curves are valid for fire compartments up to 500 m2of floor area, without openings in the roof

and for a maximum compartment height of 4 m. It is assumed that the fire load of the compartment is completely burnt out.

(2) If fire load densities are specified without specific consideration to the combustion behaviour (see annex E), then this approach

should be limited to fire compartments with mainly cellulosic type fire loads.

(3) The temperature-time curves in the heating phase are given by:

(A.1) g= 20 + 1325(1 0,324e0,2t*

0,204e1,7t*0,472e19t*)

where:

gis the gas temperature in the fire compartment;

t* = tthe time [h].

Dimensions of the compartment (see figure above): Dimensions of windows (mean values):

Width = 6,50 m Number of windows: 4

Length = 15,00 m Width = 2,30 m

Height = 3,60 m Height = 1,70 m (heq)

(weighted average of window heights on all walls).

Thermal properties of enclosure surfaces

c bj= [c] Aj(**)

[kg/m3] [J/kgK] [W/mK] [J/m

2s

1/2K] [m

2]

Note: ,c,may be taken at ambient temperature.

CEILING 2400 1506 1,50 2200,0 (*) 97,5

WALLS 900 1250 0,24 519,6 139,2

FLOOR 900 1250 0,24 519,6 97,5

(*) b (thermal absorptivity) with the following limits: 100 b 2200.

(**) Ajis the area of enclosure surface j, openings not included.

PLAN

Page B1/3 EN1991-1-2_(a).xls (rel. xx.x) - Sheet: Annex A - 06/04/2013 --



36/123



PROJECT: ???

CLIENT: ???

JOB NO.: ??

DESIGN BY: ??

REVIEW BY: ??

Total area of vertical openings on all walls: Av= 15,64 m

Total area of enclosure: At= 349,8 m

(walls, ceiling and floor, including openings)

Opening factor: O = Av[heq]/At= 0,0583 m1/2

(with the following limits: 0,02 O 0,2).

(6) To account for different b factors in walls, ceiling and floor, b = [c] should be introduced as:

b = ((bjAj)/(AtAv) = [2.200 x 97,5 + 520 x 139,16 + 520 x 97,5]/(349,80 15,64) = 1.010

(*) b (thermal absorptivity) with the following limits: 100 b 2200.

Time factor function: = [O/b]/(0,04/1160) = [0,058/1.010]/(0,04/1160) = 2,802 [-]

Design value of the fire load density related

to the surface area A fof the floor: qf,d= 700 MJ/m (taken from Annex E).

Floor area of the fire compartment: Af= 97,50 m2Aj (FLOOR)

Design value of the fire load density related

to the total surface area A t of the enclosure: q t,d= qf,dx (Af/At) = 700 x (97,5/349,8) = 195,11 MJ/m

(with the following limits: 50q t,d1000).

Fire growth rate: 2) medium fire growth rate tlim= 20 min = 0,33 h

0,2103

qt,d/O = (0,2/1000) x 195,11/0,058 = 0,67 h

tmax= max [(0,2103qt,d/O); 0,33 h] = 0,67 h tmax> tlim [TRUE]

The fire is ventilation controlled.

Case a: the fire is ventilation controlled [Case Applicable].

t*max= tmax= 0,67 x 2,802 = 1,876 h

(7) The maximum temperature maxin the heating phase happens for t* = t*max

Case b: the fire is fuel controlled [Case Not Applicable]

(A.8) Time factor funct.: lim= [Olim/b]/(0,04/1160) = [0,0585/1.010]/(0,04/1160) = 2,825 [-]

Olim= 0,1103

qt,d/tlim= (0,1/1000) x 195,11/0,33 = 0,0585

(9) If (O > 0,04 and qt,d< 75 and b < 1160), limin (A.8) has to be multiplied by k given by:

= 1,00 [Case Not Applicable]

t* = t klim= t x 2,825 (see eq. (A.1))

t*max= tmaxklim= 0,669 x 1,00 x 2,825 = 1,891 h

(7) The maximum temperature maxin the heating phase happens for t* = t*maxSee eq. (A.1) below:

t,dq 75O 0,04 1160 bk 10,04 75 1160

[J/m2s

1/2K]




37/123



PROJECT: ???

CLIENT: ???

JOB NO.: ??

DESIGN BY: ??

REVIEW BY: ??

Maximum temperature (heating phase):

(A.1) max= 20 + 1325(1 0,324e0,2t*

0,204e1,7t*0,472e19t*) = 1.038,9 C

t*max= 1,876 h tmax= 0,669 h

Cooling phase (t t*max)

t*max= 1,876 h Note: the maximum temperature maxin the heating phase happens for t* = t*max

tmax> tlim [TRUE] x = 1 [Case Applicable]

tmax= tlim x = tlim/t**max= 0,333 x 2,802/1,876 = 0,50 [Case Not Applicable]

(A.12) t**max= (0,2103

qt,d/O)= 1,876

(11) The temperature-time curves in the cooling phase are given by:

(A.11a) g= max625(t* t**maxx) = max 625 x (t* 1,876x) [Case Not Applicable]

for t**max0,5

(A.11b) g= max250(3 t**max)(t* t**maxx) = max 250(3 1,876)(t* 1,876x) [Case Applicable]

for 0,5 < t**max< 2

(A.11c) g= max250(t* t**maxx) = max 250 x (t* 1,876x) [Case Not Applicable]

for t**max2

t* = t= t x 2,802 (see eq. A.2a).

1038,9

(Heating)

Cooling

Cooling

0

200

400

600

800

1000

1200

1400

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

t [h]

g

[C]

Parametric curve




38/123



PROJECT: ???

CLIENT: ???

JOB NO.: ??

DESIGN BY: ??

REVIEW BY: ??

ANNEX B - Informative

THERMAL ACTIONS FOR EXTERNAL MEMBERS - SIMPLIFIED CALCULATION METHOD

B.1 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.

This method considers steady-state conditions for the various parameters. The method is valid only for fire loads qf,dhigher

than 200 MJ/m2.

B.2 When there is more than one window in the relevant fire compartment, the weighted average height of windows heq, the total

area of vertical openings Avand the sum of window widths (wt= wi) are used.

B.3 Effect of wind

B.3.1 Mode of ventilation

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 onditions.

B.3.2 Flame deflection by wind

Flames from an opening should be assumed to be leaving the fire compartment (see Figure B.1):

perpendicular to the facade;

with a deflection of 45 due to wind effects.

Figure B.1 - Deflection of flame by wind.

B.4 Characteristics of fire and flames

B.4.1 No forced draught

(1) The rate of burning or the rate of heat release is given by:

Floor area of the fire compartment: Af= 30,00 m2

Design fire load density related to the surface area Af: qf,d= 500 MJ/m2

(taken from Annex E).

Free burning fire duration: F= 1200 s

1/20,036

f f ,d eqOv

F

A q hQ m in ; 3,15 1 e A

D / W

Page A1/9 EN1991-1-2_(b).xls (rel. xx.x) - Sheet: Annex B - 06/04/2013 --



39/123


40/123



PROJECT: ???

CLIENT: ???

JOB NO.: ??

DESIGN BY: ??

REVIEW BY: ??

(3) LL= max {0; 1,7 x [2,37 x (12,50/(6,8 x 0,5 x [1,7 x 9,81]))^(2/3) 1]} = 2,06 m

Width of window (see figure B.2 above): wt= 1,00 m

Wall above and heq1,25wt

Case a): if heq1,25wt 1) LH= heq/3 = (1,70/3) = 0,57 m [Case Not Applicable]

if heq> 1,25wtand distance to any

other window > 4wt 2) LH= 0,3heq(heq/wt)0,54

= 0,3 x 1,7 x (1,7/1,00)^0,54 = 0,68 m

in other cases: 3) LH= 0,454heq(heq/2wt)0,54

= 0,454 x 1,7 x (1,7/(2 x 1,00))^0,54 = 0,71 m

LfLL+ heq/2 = 2,06 + (1,7/2) = 2,91 m Lf= 0 (Lf= flame length along axis).

[Case Not Applicable]

LL> 0 [Case Applicable] if LL= 0

(Lf= flame length along axis).

[0,57 + 1,7/9] = 1) L1= 0,80 m Lf= LL+ L1= (2,06 + 0,80) = 2,86 m

[0,68 + 1,7/9] = 2) L1= 0,88 m Lf= LL+ L1= (2,06 + 0,88) = 2,94 m

[0,71 + 1,7/9] = 3) L1= 0,91 m Lf= LL+ L1= (2,06 + 0,91) = 2,96 m

Mean value: (2,86 + 2,94 + 2,96)/3 = 2,92 m.

(8) The flame temperature at the window is given by:

Case a):

Tw= 520/(1 0,4725(Lfwt/Q)) + T0= 520/(1 0,4725 x (2,91 x 1,00/12,50)) + 293 K = 877 K (604C)

Lfwt/Q < 1 [Case Applicable] Lf= 2,91 m

(9) The emissivity of flames at the window may be taken as f= 1,0.

No wall above or heq> 1,25wt

Case b): LH= 0,6heq(LL/heq)1/3

= 0,6 x 1,7 x (2,06/1,7)^(1/3) = 1,09 m

Lf= (LL2+ (LHheq/3)

2)1/2

+ heq/2 = [2,06 + (1,09 1,7/3)]^(1/2) + 1,7/2 = 2,97 m

LL> 0 [Case Applicable]

Lf= 0

[Case Not Applicable]

if LL= 0

(Lf= flame length along axis).

L1heq / 2 = 1,7/2 = 0,85 m.


Case b):

Tw= 520/(1 0,4725(Lfwt/Q)) + T0= 520/(1 0,4725 x (2,97 x 1/12,50)) + 293 K = 879 K (606C)

Lfwt/Q < 1 [Case Applicable] Lf= 2,97 m

(9) The emissivity of flames at the window may be taken as f= 1,0.

2

eq2

1 H

hL L

9




41/123



PROJECT: ???

CLIENT: ???

JOB NO.: ??

DESIGN BY: ??

REVIEW BY: ??

(10) The flame temperature along the axis is given by (see plots below):

Tz= (TwT0)(1 0,4725(Lxwt/Q)) + T0 case a) [Case Applicable]

Lxwt/Q < 1

Lxwt/Q < 1 case b) [Case Applicable]

Case a) 0 Lx 2,91 Lx= 1,45 m Tz= 845 K = 572C

Lx= 1,49 m Tz= 846 K = 573C

Case b) 0 Lx 2,97 Lxis the axis length from the window to the point where the

calculation is made.

Note:

Lxwt/Q < 1

[Case Applicable]

Note:

Lxwt/Q < 1

[Case Applicable]

845,1Tz[K] - Case a)

0

100

200

300

400

500

600

700

800

900

1000

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

Lx[m]

Tz

[K]

845,9Tz[K] - Case b)

0

100

200

300

400

500

600

700

800

900

1000

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

Lx[m]

Tz

[K]

[m]

[m]




42/123



PROJECT: ???

CLIENT: ???

JOB NO.: ??

DESIGN BY: ??

REVIEW BY: ??

Flame thickness: df= 1,00 m

Geometrical characteristic of an external structural

element (diameter or side): deq= 0,70 m

(11) Emissivity of flames: f= 1e0,3df

= 1 exp(0,3 x 1,00) = 0,26 [-]

(12) Convective heat transfer coefficient: c= 4,67(1/deq)0,4

(Q/Av)0,6

= 4,67 x (1/0,70)^(0,4) x (12,50/6,80)^(0,6) = 7,8

Q = 12,50 MW (see input pages A1-A2)

Av= 6,80 m2

(see input pages A1-A2)

c= 4,67(1/deq)0,4

(Q/Av)0,6

= 4,67 x (1/0,7)^(0,4) x (1,84)^(0,6) = 7,8

Q/Av= 1,84 MW/m2

(13) If an awning or balcony (with horizontal projection: Wa) is located at the level of the top of the window on its whole width (see

figure B.3), for the wall above the window and heq1,25wt, the height and horizontal projection of the flame should be modified

as follows:

Awning or balcony with orizontal projection equal to: Wa= 0,50 m (see figure below):

Figure B.3 - Deflection of flame by balcony.

Wall above and heq1,25wt

Case a):

The flame height LLgiven in (3) is decreased by Wa(1 + 2)

LL* = LLWa(1 + 2) = 2,06 [0,50 x (1 + 2)] = 0,85 m

LL= 2,06 m

cont'd

W/m2K

W/m2K




43/123



PROJECT: ???

CLIENT: ???

JOB NO.: ??

DESIGN BY: ??

REVIEW BY: ??

cont'd

Wall above or heq1,25wt

Case a):

The horizontal projection of the flame LHgiven in (6), is increased by Wa:

if heq1,25wt 1) LH= 0,57 m LH* = Wa + LH= 0,50 + 0,57 = 1,07 m

if heq> 1,25wtand distance to any

other window > 4wt 2) LH= 0,68 m LH* = Wa + LH= 0,50 + 0,68 = 1,18 m

in other cases: 3) LH= 0,71 m LH* = Wa + LH= 0,50 + 0,71 = 1,21 m

No wall above and heq> 1,25wt

Case b):

(14) The flame height LLgiven in (3) is decreased by Wa:

LL* = LLWa= 2,06 0,50 = 1,56 m

LL= 2,06 m.

The horizontal projection of the flame LHgiven in (6), with the above mentioned value of LL*, is increased by Wa:

LH= 0,6heq(LL*/heq)1/3

= 0,6 x 1,70 x (1,56/1,70)^(1/3) = 0,99 m

LL* = 1,56 m

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

B.4 Characteristics of fire and flames

B.4.2 Forced draught

(1) The rate of burning or the rate of heat release is given by:

Q = (Afqf,d)/F= (30 x 500)/1200 = 12,50 MW

Af= 30,00 m2

qf,d= 500 MJ/m2

(taken from Annex E).

F= 1200 s

(3) The flame height (see Figure B.4, next page) is given by:

LL= (1,366(1/u)0,43

Q/[Av]) heq= (1,366 x [(1/6,00)^(0,43)] x 12,50/[6,8]) 1,7 = 1,33 m

heq= 1,70 m

Av= 6,80 m2

Wind speed, moisture content: u = 6,00 m/s

(4) The horizontal projection of flames is given by:

LH= 0,605(u2/heq)

0,22(LL+ heq) = 0,6054 x [(6,00/1,70)^(0,22)] x (1,33 + 1,70) = 3,59 m

LL = 1,33 m




44/123



PROJECT: ???

CLIENT: ???

JOB NO.: ??

DESIGN BY: ??

REVIEW BY: ??

(5) The flame width is given by:

wf= wt+ 0,4LH= 1,00 + 0,4 x 3,59 = 2,44 m

Width of window (see figure B.4 below): wt= 1,00 m

(6) The flame length along axis is given by:

Lf= (LL2+ LH

2)1/2

= [1,33 + 3,59] = 3,83 m

Figure B.4 - Flame dimensions, through or forced draught


Tw= 520/(1 0,3325Lf[Av]/Q) + T0= 520/(1 0,3325 x 3,83 x [6,80]/12,50) + 293 K = 1.001 K = 728C

Q = (Afqf,d)/F= (30 x 500)/1200 = 12,50 MW

Av= 6,80 m2

T0= 293 K (initial temperature)

with Lf(Av)1/2

/Q < 1 [Case Applicable]

(8) The emissivity of flames at the window may be taken as f= 1,0

(9) The flame temperature along the axis is given by (see plot, next page ):

(1 0,3325 x 2,5 x [6,80]/12,50) x (1.001 293) + 293 = 878 K = 605C

Axis length from the window to the point where

the calculation is made: 0 Lx 3,83 m Lx= 2,50 m

(10) The emissivity of flames may be taken as: f= 1e0,3df

Flame thickness: df= 1,50 m

Geometrical characteristic of an external structural

element (diameter or side): deq= 0,70 m

(10) Emissivity of flames: f= 1e0,3df

= 1 exp(0,3 x 1,50) = 0,36 [-]

x vz w 0 0L A

T 1 0,3325 T T TQ




45/123



PROJECT: ???

CLIENT: ???

JOB NO.: ??

DESIGN BY: ??

REVIEW BY: ??

(11) The convective heat transfer coefficient is given by:

c= 9,8(1/deq)0,4

[Q/(17,5Av) + u/1,6]0,6

= 9,8 x [(1/0,70)^(0,4)] x [12,50/(17,5 x 6,80) + 6,00/1,6]^(0,6) = 25,4

Wind speed, moisture content: u = 6,00 m/s

Note:

Lxwt/Q < 1

[Case Applicable]

(12) Regarding the effects of balconies or awnings, see Figure B.5, 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 Lf

unchanged.

Figure B.5 - Deflection of flame by awning.

W/m2K

878,3

Tz[K] - Eq. (B.25)

0

200

400

600

800

1000

1200

0,0 0,5 1,0 1,5 2,0 2,5 3,0 3,5 4,0 4,5

Lx[m]

Tz

[K]




46/123



PROJECT: ???

CLIENT: ???

JOB NO.: ??

DESIGN BY: ??

REVIEW BY: ??

(2) The temperature of the fire compartment is given by (forced draught):

(B.19) Tf= 1200(1 e0,00288

) + T0= 1200 x (1 exp(0,00288 x 543,54)) + 293 K = 1.242 K = 969C

= (Af qf,d)/(Av At)1/2

= 543,54 MJ/m2

T0= 293 K = 20C (initial temperature).




47/123



PROJECT: ???

CLIENT: ???

JOB NO.: ??

DESIGN BY: ??

REVIEW BY: ??

ANNEX C - Informative

LOCALISED FIRES

(1) The 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.

(2) The heat flux from a localised fire to a structural element should be calculated with expression (3.1), and based on a

configuration factor established according to annex G.

(3) The flame length Lfof a localised fire (see Figure C.1 - Model A) is given by:

Lf= 1,02D + 0,00148Q2/5

LEGENDA

Lf= flame length of the localised

fire.

D= diameter of the fire/flame.

H= distance between the fire source

and the ceiling (beam).

Model A: Lf< H

[Case Applicable].

Figure C.1 - Model A.

Diameter of the fire (see figg. C.1 and C.2): D = 4,00 m 10 m [Input ok]

Rate of heat release [W] of the fire according to E.4: 5,0E+06 W 50 MW [Input ok]

(@ generic time t). = 5 MW

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

= 1,02 x 4,00 + 0,0148 x 5000000^(2/5) = 2,997 m

[Model A applies].

Distance between the fire source and the cei ling (beam): H = 3,10 m

(see Figure C.1 ).

(4) When the flame is not impacting the ceiling of a compartment (L f< H; see Figure C.1) or in case of fire in open air, the

temperature (z)in the plume along the symetrical vertical flame axis is given by:

(z)= 20 + 0,25Qc2/3

(z z0)5/3

900C [Case Applicable].

(5) The virtual origin z0of the axis is given by: [Case Applicable].

z0= 1,02D + 0,00524Q2/5

= 1,02 x 4,00 + 0,00524 x 5000000^(2/5) = -1,57 m < 0.

Convective part of the rate of heat release: Qc= 0,8Q = 0,8 x 5000000 = 400

Eurocodes Print

Documents

Transcript of Eurocodes Print