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Document Ref: SX029a-EN-EU Sheet 1 of 28Title Example: Elastic analysis of a single bay portal frame

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Made by Valrie Lemaire Date April 2006

CALCULATION SHEET

Checked by Alain Bureau Date April 2006

Example: Elastic analysis of a single bay portalframe

A single bay portal frame made of rolled profiles is designed according to

EN 1993-1-1. This worked example includes the elastic analysis of the

frame using first order theory, and all the verifications of the members

under ULS combinations.

30,00

5,

988

[m]

7,20

7,

30

72,00

1 Basic data Total length : b = 72,00 m

Spacing: s = 7,20 m

Bay width : d= 30,00 m

Height (max): h = 7,30 m

Roof slope: = 5,0

1

3,00 3,00 3,00 3,00 3,00

1 : Torsional restraints

Example: Elastic analysis of a single bay portal frame

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2 LoadsEN 1991-1-1

2.1 Permanent loads

self-weight of the beam

roofing with purlins G = 0,30 kN/m2

for an internal frame: G = 0,30 7,20 = 2,16 kN/ml

2.2 Snow loads EN 1991-1-3Characteristic values for snow loading on the roof in [kN/m]

S= 0,8 1,0 1,0 0,772 = 0,618 kN/m

for an internal frame: S= 0,618 7,20 = 4,45 kN/m

7,

30

30,00

s = 4,45 kN/m

[m]


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2.3 Wind loads EN 1991-1-4Characteristic values for wind loading in kN/m for an internal frame

30,00

e/10 = 1,46 1,46

Zone D:w = 4,59

Zone G:w = 9,18

Zone H:w = 5,25

Zone J:w = 5,25 Zone I:

w = 5,25

Zone E:w = 3,28

wind direction

3 Load combinations EN 1990

3.1 Partial safety factor

Gmax = 1,35 (permanent loads)

Gmin = 1,0 (permanent loads) EN 1990

Table A1.1 Q = 1,50 (variable loads) 0 = 0,50 (snow)

0 = 0,60 (wind)

= 1,0 M0

= 1,0 M1

3.2 ULS Combinations EN 1990

Combination 101 : Gmax G + QsQ

Combination 102 : Gmin G + Q Qw

Combination 103 : Gmax G + QQ s + Q0

Combination 104 : Gmin G + Q Qs + Q0 Qw

Combination 105 : Gmax G + Q 0 Qs + Q Qw

Combination 106 : Gmin G + Q0 Qs + Q Qw

3.3 SLS Combinations EN 1990Combinations and limits should be specified for each project or by National

Annex.


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4 Sections

hw

y y

z

z

tf

tw

b

h

4.1 Column

Try IPE 600 Steel grade S275

Depth h = 600 mm

Web Depth hw = 562 mm

Depth of straight portion of the web

dw = 514 mm

Width b = 220 mm

Web thickness tw= 12 mm

Flange thickness t = 19 mmf

Fillet r= 24 mm

Mass 122,4 kg/m

Section area A = 156 cm2

Second moment of area /yy Iy = 92080 cm4

Second moment of area /zz Iz = 3386cm4

Torsion constant I = 165,4 cm4t

Warping constant Iw = 2845500 cm6

Elastic modulus /yy Wel,y = 3069 cm3

Plastic modulus /yy Wpl,y = 3512 cm3

Elastic modulus /zz W = 307,80 cm3el,z

Plastic modulus /zz Wpl,z = 485,60 cm3


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4.2 RafterTry IPE 500 Steel grade S275

Depth h = 500 mm

Web Depth hw = 468 mm

Depth of straight portion of the web

dw = 426 mm

Width b = 200 mm

Web thickness tw = 10,2 mm

Flange thickness tf= 16 mm

Fillet r= 21 mm

Mass 90,7 kg/m

Section area A = 115,50 cm2


Second moment of area /zz Iz = 2141 cm4

Torsion constant It = 89,29 cm4

Warping constant Iw = 1249400 cm6


Plastic modulus /yy Wpl,y = 2194 cm3

Elastic modulus /zz Wel,z = 214,1 cm3

Plastic modulus /zz Wpl,z = 335,90 cm3

5 Global analysisThe joints are assumed to be:

pinned for column bases

rigid for beam to column.

EN 1993-1-1

5.2

The frame has been modelled using the EFFEL program.


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5.1 Buckling amplification factor EN 1993-1-1cr 5.2.1

In order to evaluate the sensitivity of the frame to 2nd order effects, a buckling

analysis is performed to calculate the buckling amplification factorcrfor the

load combination giving the highest vertical load: G + max QQS

(combination 101).

For this combination, the amplification factor is: cr = 14,57

The first buckling mode is shown hereafter.

EN 1993-1-1So : cr = 14,57 > 10 5.2.1First order elastic analysis may be used.

(3)

5.2 Effects of imperfections EN 1993-1-1 5.3.2The global initial sway imperfection may be determined from (3)

310204,3866,0740,0200

1 == 0h =m

where 0 = 1/200

740,030,7

22 ==h

h =

866,0)1

1(5,0 =+m

m = 2 (number of columns)=m

Sway imperfections may be disregarded whereH 0,15 V EN 1993-1-1Ed Ed.

5.3.2The effects of initial sway imperfection may be replaced by equivalent

horizontal forces:

(4)

H = V in the combination whereH < 0,15 Veq Ed Ed Ed


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The following table gives the reactions at supports.

Left column 1 Right column 2 Total

ULS

Comb.HEd,1

kN

VEd,1

kN

HEd,2

Kn

VEd,2

kN

HEd

kN

VEd

kN

0,15 VEd

101 -125,5 -172,4 125,5 -172,4 0 -344,70 51,70

102 95,16 80,74 -24,47 58,19 70,69 138,9 20,83

103 -47,06 -91,77 89,48 -105,3 42,42 -197,1 29,56

104 -34,59 -73,03 77,01 -86,56 42,42 -159,6 23,93

105 43,97 11,97 26,72 -10,57 70,69 1,40 0,21

106 56,44 30,71 14,25 8,17 70,69 38,88 5,83

The sway imperfection has only to be taken into for the combination 101:

VEdkN

Heq = .VEdkN

172,4 0,552

Modelling withHeq for the combination 101

EN 1993-1-1

5.3.2 (7)

5.3 Results of the elastic analysis

5.3.1 Serviceability limit states

Combinations and limits should be specified for each project or in National

Annex.

For this example, the deflections obtained by modeling are as follows:

EN 1993-1-1

7 and

EN 1990

Vertical deflections:

G + Snow: Dy = 124 mm = L/241

Snow only: Dy = 73 mm = L/408

Horizontal deflections:

Deflection at the top of column by wind only

Dx = 28 mm = h/214


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5.3.2 Ultimate limit statesMoment diagram in kNm

Combination 101:

Combination 102:

Combination 103:

Combination 104:


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Combination 105:

Combination 106:

6 Column verification

Profile IPE 600 - S275 (= 0,92)

The verification of the member is carried out for the combination 101 :

N = 161,5 kN (assumed to be constant along the column)Ed= 122,4 kN (assumed to be constant along the column)VEd

M = 755 kNm (at the top of the column)Ed

6.1 Classif ication of the cross section

Web: The web slenderness is c / tw = 42,83 EN 1993-1-1 5.5

mm94,4827512

161500

yw

EdN =

==ft

Nd

548,05142

94,48514

2 w

Nw =+

=+

=d

dd > 0,50

49,591548,013

92,0396=

The limit for Class 1 is : 396 / (13 1) =

Then : c / tw = 42,83 < 59,49 The web is class 1.


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Flange: The flange slenderness is c / tf= 80 / 19 = 4,74The limit for Class 1 is : 9 = 9 0,92 = 8,28

Then : c / t = 4,74 < 8,28 The flange is Class 1f

So the section is Class 1. The verification of the member will be based on

the plastic resistance of the cross-section.

6.2 Resistance of cross section

Verification for shear force

Shear area : A =A - 2btv f+ (tw+2r)t > .hf w.t EN 1993-1-1w 6.2.6

may be conservatively taken equal to 1(3)

838019)24212(19220215600 =++=vA mm2 > .hw.tw = 6744 mm

2

Vpl,Rd =Av (fy / 3 ) /M0 = (8380275/ 3 ).10-3

V = 1330 kNpl,Rd

V / VEd pl,Rd = 122,40/1330 = 0,092 < 0,50

The effect of the shear force on the moment resistance may be neglected.

Verification to axial force EN 1993-1-1

6.2.4-3N =Afpl,Rd y / = (15600 275/1,0).10M0

N = 4290 kNpl,Rd

N = 161,5 kN < 0,25NEd pl,Rd = 4290 x 0,25 = 1073 kNEN 1993-1-1

and NEd = 161,5 kN < 3,92710001

275125625,05,0

M0

yww =

=

fth

6.2.8kN (2)

The effect of the axial force on the moment resistance may be neglected.

Verification to bending moment EN 1993-1-1

6.2.5-3 M = Wpl,y,Rd pl,yfy / = (3512 275/1,0).10M0

M = 965,8 kNmpl,y,Rd

My,Ed = 755 kNm < M = 965,8 kNmpl,y,Rd


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6.3 Buckling resistanceThe buckling resistance of the column is sufficient if the following conditions

are fulfilled (no bending about the weak axis,MEN 1993-1-1

z,Ed = 0): 6.3.3

1

M1

Rky,

LT

Edy,

yy

M1

Rky

Ed +

M

Mk

N

N

1

M1

Rky,

LT

Edy,

zy

M1

Rky

Ed +

M

Mk

N

N

The k and kyy zy factors will be calculated using the Annex A of EN 1993-1-1.

The frame is not sensitive to second order effects (cr= 14,57 > 10). Then thebuckling length for in-plane buckling may be taken equal to the system length.

EN 1993-1-1

5.2.2

(7)

Lcr,y = 5,99 m

Note: For a single bay symmetrical frame that is not sensitive to second order

effects, the check for in-plane buckling is generally not relevant. Theverification of the cross-sectional resistance at the top of the column will bedeterminant for the design.

Regarding the out-of-plane buckling, the member is laterally restrained at bothends only. Then :

L = 5,99 m for buckling about the weak axiscr,zL = 5,99 m for torsional bucklingcr,T

and L = 5,99 m for lateral torsional bucklingcr,LT

Buckling about yy Lcr,y = 5,99 m

EN 1993-1-1Buckling curve : a (y = 0,21) 6.3.1.2

10005990

10000920802100002

2

2

ycr,

y2

ycr,

== L

EIN

(2)=53190kN

Table 6.1

284,010.53190

275156003

ycr,

y

y =

==N

Af EN 1993-1-1

6.3.1.3 (1)


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( ) ( )[ ]22

yyy 284,02,0284,021,015,02,015,0 ++=++= EN 1993-1-1= 0,5491 6.3.1.2 (1)

9813,0284,05491,05491,0

11

222

y2

yy

y =+

=+

=

Buckling about zz L = 5,99 m EN 1993-1-1cr,z 6.3.1.2

Buckling curve : b (z= 0,34)(2)

Table 6.1

10005990

1000033862100002

2

2

zcr,

z2

zcr,

==

L

EIN = 1956 kN

EN 1993-1-1481,1

10.1956

275156003

zcr,

y

z =

==N

Af 6.3.1.3 (1)

( ) ( )[ ]22zzz 481,12,0481,134,015,02,015,0 ++=++= EN 1993-1-1= 1,814 6.3.1.2 (1)

3495,0481,1814,1814,1

11

222

z2

zz

z =+

=+

=

Lateral torsional buckling Lcr,LT = 5,99 m

EN 1993-1-1Buckling curve : c (LT = 0,49) 6.3.2.3Moment diagram with linear variation : = 0 C1 = 1,77

Table 6.5

Z

2

t

2

LTcr,

Z

W

2

LTcr,

Z

2

1crEI

GIL

I

I

L

EICM

+=

NCCI

SN00342

42

4

6

62

2

cr

10.3386210000

10.4,165807705990

10.3386

10.2845500

105990

10000338621000077,1

+

=

M

Mcr= 1351 kNm

8455,010.1351

27510.35126

3

cr

yypl,LT =

==

M

fW

( ) 2LTLT,0LTLTLT 15,0 ++= EN 1993-1-1 6.3.2.3 (1)

and=0,7540,0LT,0 =with


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( )[ ] 8772,08455,075,04,08455,049,015,0 2LT =++=

7352,08455,075,08772,08772,0

11

222

LT2LTLT

LT =+

=+

=

7519,033,033,1

1=

EN 1993-1-1k (= 0)c = 6.3.2.3

( )2LTc 8,021)1(5,01 = kf(2)

Table 6.6

( ) 8765,08,08455,021)7519,01(5,01 2 ==f < 1

8388,08765,0

7352,0==

f

LT LT,mod = < 1

and kCalculation of the factors kyy zy according to Annex A of EN 1993-1-1 EN 1993-1-1

Annex A

9999,0

53190

5,1619813,01

53190

5,1611

1

1

ycr,

Edy

ycr,

Ed

y =

=

=

N

N

N

N

9447,0

1956

5,1613495,01

1956

5,1611

1

1

zcr,

Edz

zcr,

Ed

z =

=

=

N

N

N

N

EN 1993-1-1144,1

3069

3512

yel,

ypl,

y ===W

Ww < 1,5 Annex A

578,18,3076,485

zel,

zpl,z === W

Ww > 1,5 wz = 1,5

NCCICritical axial force in the torsional buckling mode

SN003)(

2Tcr,

w2

t0

Tcr,L

EIGI

I

AN

+=

For a doubly symmetrical section,

cm495466338692080)( 2020zy0 =+=+++= AzyIII


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+

=2

624

4cr, 5990

10.284550021000010.4,16580770

100010.95466

15600TN

Ncr,T = 4869 kN

NCCI

Z

2

t

2

LTcr,

Z

W

2

LTcr,

Z

2

1cr,0EI

GIL

I

I

L

EICM

+= SN003

0Mcr,0is the critical moment for the calculation of for uniform bending

moment as specified in Annex A.C1 = 1

42

42

4

6

62

42

cr,010.3386210000

10.4,165807705990

10.3386

10.2845500

105990

10.33862100001

+

=

M

kNmM 3,763ocr, =

EN 1993-1-1

6

3

ocr,

yypl,0

10.3,76327510.3512 ==

MfW = 1,125 Annex A

4

TFcr,

Ed

zcr,

Ed1lim0 )1)(1(2,0

N

N

N

NC =

withN = Ncr,TF cr,T (doubly symmetrical section)

4lim0 )4869

5,1611)(

1956

5,1611(77,12,0 = = 0,2582

0 > lim0

LTy

LTy

my,0my,0my1

)1(a

aCCC

++=

yel,Ed

Edy,

yW

A

N

M=

y

tLT

I

Ia =1= 23,76 (class 1) and = 0,9982with


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Calculation of the factor C my,0

EN 1993-1-1

ycr,

Edyymy,0 )33,0(36,021,079,0

N

NC ++= Annex A

Table A2

ycr,

Edmy,0 1188,079,0

N

NC = = 0,78960y =

Calculation of the factors C and Cmy m,LT:

LTy

LTy

my,0my,0my1

)1(a

aCCC

+

+=

9641,09982,076,231

9982,076,23)7896,01(7896,0Cmy =

+

+=

EN 1993-1-11

)1)(1(Tcr,

Ed

zcr,

Ed

LT2

mymLT

=

N

N

N

N

aCC Annex A

9843,0

)4869

5,1611)(

1956

5,1611(

9982,09641,0 2mLT =

=C < 1

C = 1mLT

Calculation of the factors Cyy and C : EN 1993-1-1zyAnnex A

ypl,

yel,

LTpl

2

max2

my

y

max2

my

y

yyy ])6,16,1

2)[(1(1W

WbnC

wC

wwC +=

03765,01/27515600

161500

/ M1Rk

Edpl =

==N

Nn

Mz,Ed = 0 and0LT =b 0LT =d 4810,1zmax ==

]03765,0)481,19641,0144,1

6,1481,19641,0

144,1

6,12[()1144,1(1 222yy +=C

8739,03512

3069

ypl,

yel, ==>W

W Cyy= 0,9849


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2011



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16/29


Eurocode Ref


CALCULATION SHEET


ypl,

yel,

z

y

LTpl

2

max2

my5

y

yzy 6,0])14

2)[(1(1W

W

w

wdnC

wwC +=

9318,0]03765,0)481,19641,0144,1

142)[(1144,1(1 22

5zy=+=C

4579,03512

3069

5,1

144,16,06,0

ypl,

yel,

z

y ==W

W

w

w>= 9318,0zyC

Calculation of the factors kyy

andkzy

: EN 1993-1-1

Annex A

yy

ycr,

Ed

y

mLTmyyy

1

1C

N

NCCk

=

9818,09849,0

1

53190

5,1611

9999,019641,0yy =

=k

z

y

zy

ycr,

Ed

zmLTmyzy 6,0

1

1w

w

C

N

NCCk=

5138,050,1

144,16,0

9318,0

1

53190

5,1611

9447,019641,0zy =

=k

Verification with interaction formulae

EN 1993-1-1 1

M1

Rky,LT

Edy,

M1

Rky

Ed +

M

Mk

N

Nyy 6.3.3

9534,0

1

27510.35128388,0

10.7559818,0

1

275156009813,0

1615003

6

=

+


17/29


Eurocode Ref


CALCULATION SHEET


5867,0

1

27510.35128388,0

10.7555138,0

1

275156003495,0

1615003

6

=

+

0,5

c / tw = 41,76 < 38,581557,013

92,0396

113

396=

=

class 1

EN 1993-1-1 5.5

Flangesb = 200 mm

tf= 16 mm

r= 21 mm c / tf= 4,44c = 71 mm

part to compression

c / tf < 9 = 8,28 (S275 = 0,92 )

c / t = 4,44 class 1f

So the section is Class 1. The verification of the member will be based on

the plastic resistance of the cross-section.


CreatedonWednesday,

December21,

2011





18/29


Eurocode Ref


CALCULATION SHEET


7.2 Resistance of cross-section

Verification with maximum moment along the member in cross-section of IPE

500:

Combination 101

Maximum force in IPE 500 at the end of the haunch:

N = 136,00 kNEdV = 118,50 kNEdM = 349,10 kNmy,Ed

305,23 kNm

349,10 kNm

754,98 kNm

Combination 101: Bending moment diagram along the rafter

Shear V = 118,50 kN EN 1993-1-1Ed 6.2A =A - 2btv f+ (tw+2r)t = 1f

598516)2122,10(16200211550 =++=vA mm2

EN 1993-1-1 Av > .hw.tw = 46810,2 = 4774mm

2

6.2.8 (2)

3 3Vpl,Rd =Av (f / ) / = 5985275/ /1000 = 950,3 kNy M0

V / VEd pl,Rd = 118,5/950,3 = 0,125 < 0,50

its effect on the moment resistance may be neglected!

Compression EN 1993-1-1

6.2.4Npl, = 11550 x 275/1000 = 3176 kNRd

N = 136 kN < 0,25 NEd pl, = 3176 0,25 = 794,1 kNRd

and

EN 1993-1-1

kN4,65610001

2752,104685,05,0

M0

yww =

=

fth 6.2.8NEd = 136 kN < (2)

its effect on the moment resistance may be neglected!


CreatedonWednesday,

December21,

2011





19/29


Eurocode Ref


CALCULATION SHEET


Bending EN 1993-1-1 6.2.5

M = 2194 275/1000 = 603,4 kNmpl,y,Rd

M = 349,10 kNm < My,Ed pl,y,Rd = 603,4 kNm

7.3 Buckling resistance

Uniform members in bending and axial compression EN 1993-1-1

6.3.3Verification with interaction formulae

1

M1

Rky,

LT

Edy,

yy

M1

Rky

Ed +

M

Mk

N

Nand 1

M1

Rky,

LT

Edy,

zy

M1

Rkz

Ed +

M

Mk

N

N

Buckling about yy:

For the determination of the buckling length about yy, a buckling analysis

is performed to calculate the buckling amplification factorcrfor the load

combination giving the highest vertical load, with a fictitious restraint at

top of column:

EN 1993-1-1

6.3.1.2 (2)

Table 6.1Combination 101 cr = 37,37

EN 1993-1-1

6.3.1.3 (1)

EN 1993-1-1Buckling curve : a (h/b>2) y= 0,21 6.3.1.2

kNNN 508213637,37Edcrycr, === (2)

Table 6.1

7906,010.5082

275115503

ycr,

y

y =

==N

Af


CreatedonWednesday,

December21,

2011



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20/29


Eurocode Ref


CALCULATION SHEET


( ) ++=2yyy 2,015,0

= 0,8745( )[ ]2y 7906,02,07906,021,015,0 ++=

8011,07906,08745,08745,0

11

222

y2

yy

y =+

=+

=

Buckling about zz:

For buckling about zzand for lateral torsional buckling, the bucklinglength is taken as the distance between torsional restraints:

Lcr= 6,00m

Note: the intermediate purlin is a lateral restraint of the upper flange only.

Its influence could be taken into account but it is conservatively neglected

in the following.

Flexural bucklingEN 1993-1-1

L = 6,00 m 6.3.1.3cr,z

10006000100002141210000 222

zcr,

z2zcr, ==

LEIN = 1233kN

NCCITorsional buckling

Lcr,T = 6,00 m SN003

)(2

Tcr,

w

2

t

0

Tcr,L

EIGI

I

AN

+=

with yo = 0 and zo = 0 (doubly symmetrical section)

cm450340214148199)( 2020zy0 =+=+++= AzyIII

)

6000

10.124937021000010.29,8980770(

100010.50340

115502

624

4Tcr,

+

= N

Ncr,T = 3305 kN

Ncr= min (N ;Ncr,z cr,T ) = 1233 kN EN 1993-1-1 6.3.1.3

605,110.1233

27511550

3cr

y

z =

==N

Af

(1)


CreatedonWednesday,

December21,

2011



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21/29


Eurocode Ref


CALCULATION SHEET


Buckling curve : b EN 1993-1-1 6.3.1.2

z= 0,34(1)

( ) 2zzz 2,015,0 ++=Table 6.1

( )[ ]2z 605,12,0605,134,015,0 ++= =2,027

3063,0605,1027,2027,2

11

222

z2

zz

z =+

=+

=

Lateral torsional buckling : EN 1993-1-1 6.3.1.3 L , = 6,00 mcr LT

Table 6.5= 0,49Buckling curve : c LT

Moment diagram along the part of rafter between restraints:

Combination 101

NCCICalculation of the critical moment:

SN003 = - 0,487

qL

8

2

q = - 9,56 kN/m = = - 0,123

C1 = 2,75

NCCI

Z

2

t

2

LTcr,

Z

W

2

LTcr,

Z

2

1crEI

GIL

I

I

L

EICM

+=

42

42

4

6

62

42

cr10.2141210000

10.29,89807706000

10.2141

10.1249400

106000

10214121000075,2

+

=

M

Mcr= 1159 kNm


CreatedonWednesday,

December21,

2011



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22/29


Eurocode Ref


CALCULATION SHEET


7215,010.1159

27510.21956

3

cr

yypl,LT ===

M

fW

( ) 2LTLT,0LTLTLT 15,0 ++=

EN 1993-1-1

6.3.2.3with 40,0LT,0 = and =0,75 (1)

( )[ ] 7740,07215,075,04,07215,049,015,0 2LT =++=

8125,0

7215,075,07740,07740,0

11

222

LT2

LTLT

LT =

+

=

+

=

kc = 0,91

( )28,021)1(5,01 = LTckf EN 1993-1-1

6.3.2.3

( ) 9556,08,07215,021)91,01(5,01 2 ==f

(2)

< 1 Table 6.6

8503,09556,0

8125,0==

f

LT LT,mod= < 1

Combination 101 N = 136 kN compressionEd

M = 349,10 kNmy,Ed

M z,Ed = 0

Section class1 M = 0 et My,Ed z,Ed = 0

EN 1993-1-1

6.3.3

1

M1

Rky,

LT

Edy,

yy

M1

Rky

Ed +

M

Mk

N

N1

M1

Rky,

LT

Edy,

zy

M1

Rkz

Ed +

M

Mk

N

N


CreatedonWednesday,

December21,

2011





23/29


Eurocode Ref


CALCULATION SHEET


EN 1993-1-1

9946,0

5082

1368011,01

5082

1361

1

1

ycr,

Edy

ycr,

Ed

y =

=

=

N

N

NN

Annex A

9208,0

1233

1363063,01

1233

1361

1

1

zcr,

Edz

zcr,

Ed

z =

=

=

N

N

N

N

EN 1993-1-1138,1

1928

2194

yel,

ypl,

y ===W

Ww < 1,50 Annex A

569,11,214

9,335

zel,

zpl,

z ===W

Ww > 1,50 w = 1,5z

NCCI

Z

2

t

2

LTcr,

Z

W

2

LTcr,

Z

2

1cr,0EI

GIL

I

I

L

EICM

+=

SN003

0Mcr,0is the critical moment for the calculation of for uniform bending

moment as specified in Annex A.

C1 = 1

42

42

4

6

62

42

cr,010.2141210000

10.29,89807706000

10.2141

10.1249400

106000

10.21412100001

+

=

M

kNmM 5,421ocr, =

EN 1993-1-16

3

ocr,

yypl,0

10.5,42127510.2195 ==

MfW = 1,196 Annex A

4

TFcr,

Ed

zcr,

Ed1lim0 )1)(1(2,0

N

N

N

NC = with C1 = 2,75

withN = Ncr,TF cr,T (doubly symmetrical section)

4lim0 )3305

1361)(

1233

1361(75,22,0 = = 0,3187

0 =1,196 > lim0 =0,3187


CreatedonWednesday,

December21,

2011



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24/29


Eurocode Ref


CALCULATION SHEET


EN 1993-1-1

LTy

LTy

my,0my,0my1

)1(a

aCCC

++= Annex A

with3

6

yel,Ed

Edy,

y101928

11550

136000

10.10,349

==

W

A

N

M =15,38 (class 1)

and48200

29,8911

y

tLT ==

I

Ia = 0,9981

Calculation of the factor C EN 1993-1-1my,0 Annex AMoment diagram along the rafter:

Table A2

My,Ed = maximum moment

along the rafter = 755kNm

= maximum displacement

along the rafter = 179mm30m

ycr,

Ed

Edy,

2xy

2

my,0 11NN

MLEIC

+=

5082

1361

1075530000

17910482002100001

62

42

my,0

+=

C =0,9803

Calculation of the factors C and C my m,LT:

LTy

LTy

my,0my,0my1

)1(a

aCCC

++=

996,09982,038,151

9982,038,15)9803,01(9803,0my =

+

+=C


CreatedonWednesday,

December21,

2011



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25/29


Eurocode Ref


CALCULATION SHEET


EN 1993-1-11

)1)(1(Tcr,

Ed

zcr,

Ed

LT2

mymLT

=

N

N

N

N

aCC Annex A

072,1

)3305

1361)(

1233

1361(

9981,0996,0 2mLT =

=C > 1

Calculation of the factors Cyy and C EN 1993-1-1zyAnnex A

ypl,

yel,

LTpl

2max

2

my

y

max2

my

y

yyy ])6,16,12)[(1(1

WWbnC

wC

wwC +=

0428,01/27511550

136000

/ M1Rk

Edpl =

==N

Nn

M 605,1max == zz,Ed = 0 and0LT =b 0LT =d

]0428,0)605,1996,0138,1

6,1605,1996,0

138,1

6,12)[(1138,1(1 222yy +=C

Cyy= 0,9774

ypl,

yel,

z

y

LTpl

2

max2

my5

y

yzy 6,0])14

2)[(1(1W

W

w

wdnC

wwC +=

9011,0]0428,0)605,1996,0138,1

142)[(1138,1(1 22

5zy=+=C

Calculation of the factors kyy andkzy : EN 1993-1-1Annex A

yy

ycr,

Ed

y

mLTmyyy

1

1C

N

NCCk=

116,19774,0

1

5082

1361

9946,0072,1996,0yy =

=k


CreatedonWednesday,

December21,

2011





26/29


Eurocode Ref


CALCULATION SHEET


z

y

zy

ycr,

Ed

zmLTmyzy 6,0

1

1w

w

C

N

NCCk

=

5859,050,1

138,16,0

9011,0

1

5082

1361

9208,0072,1996,0zy =

=k

Verification with interaction formulae EN 1993-1-1

6.3.3

1

M1

Rky,

LT

Edy,yy

M1

Rky

Ed +

MMk

NN (6.61)

8131,0

1

27510.21948503,0

10.1,349116,1

1

275115508011,0

1360003

6

=

+

< 1 OK

1

M1

Rky,

LT

Edy,

zy

M1

Rkz

Ed +

M

Mk

N

N

(6.62)

5385,0

1

27510.21948503,0

10.1,3495859,0

1

275115503063,0

1360003

6

=

+

< 1 OK

8 Haunch verification

For the verification of the haunch, the compression part of the cross-section isconsidered as alone with a length of buckling about the zz-axis equal to 3,00m

(length between the top of column and the first restraint).

Maximum forces and moments in the haunch:

N = 139,2 kNEdV = 151,3 kNEdM = 755 kNmEd


CreatedonWednesday,

December21,

2011



bjecttothetermsandconditionsoftheAccessS

teelLicenceAgreement


27/29


Eurocode Ref


CALCULATION SHEET


Properties of the whole section:

The calculation of elastic section properties for this case is approximate,

ignoring the middle flange.

1000 mm

200 mm

Section area A = 160,80 cm2


Second moment of area /zz Iz = 2141 cm4


Elastic modulus /zz W = 214 cm3el,z

Properties of the compression part:

Section at the mid-length of the haunch including 1/6th of the web depth

Section area A = 44 cm2

120 mm


Second moment of area /zz Iz =1068 cm4

cmi 93,444

1068z == 200 mm

7044,039,8630,49

3000

1z

fz =

==

i

Lz

Buckling of welded I section with h/b > 2 :

curve d = 0,76

( ) ( )[ ] 9397,07044,02,07044,076,015,02,015,0 22zzz =++=++=

640,07044,09397,09397,0

11

222

z2

zz

z =+

=+

=


CreatedonWednesday,

December21,

2011





28/29


Eurocode Ref


CALCULATION SHEET


Compression in the bottom flange:

kNN 7604400100010.4610

1000755000

16080

440024,139

3fEd,=

+=

Verification of buckling resistance of the bottom flange:

981,02754400640,0

760000

Rkz

fEd, =

=N

N

< 1 OK


CreatedonWednesday,

December21,

2011





29/29


SX029a-EN-EU

Quality Record

Example: Elastic analysis of a single bay portal frameRESOURCE TITLE

Reference(s) T2703

ORIGINAL DOCUMENT

Name Company Date

Created by Valrie LEMAIRE CTICM 25/10/2005

Technical content checked by Alain BUREAU CTICM 26/10/2005

Editorial content checked by

Technical content endorsed by the

following STEEL Partners:

1. UK G W Owens SCI 10/04/06

2. France A Bureau CTICM 10/04/06

3. Sweden B Uppfeldt SBI 10/04/06

4. Germany C Muller RWTH 10/04/06

5. Spain J Chica Labein 10/04/06

Resource approved by TechnicalCoordinator

G W Owens SCI 18/09/06

TRANSLATED DOCUMENT

This Translation made and checked by:

Translated resource approved by:


dnesday,

December21,

2011

copyright-allrightsreserved.



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