Wind Loads UK- Portal Frame WE

Document Ref: SX029a-EN-GB Sheet 1 of 30 Title

Example: Elastic analysis of a single bay portal frame

Eurocode Ref Made by Laurent Narboux Date Nov 2006

CALCULATION SHEET

Checked by C M King Date Dec 2006

Localized resource for UK

Example: Elastic analysis of a single bay portal frame 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

5,98

8

α

[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: α = 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 (GB)C

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CALCULATION SHEET


2 Loads 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/m

EN 1991-1-1

2.2 Snow loads Characteristic values for snow loading on the roof in [kN/m] S = 0.8 × 1.0 × 1.0 × 0.772 = 0.618 kN/m²

⇒ for an internal frame: S = 0.618 × 7.20 = 4.45 kN/m

α

7,30

30,00

s = 4,45 kN/m

[m]

EN 1991-1-3


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CALCULATION SHEET


2.3 Wind loads Characteristic values for wind loading in kN/m for an internal frame

30,00e/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

EN 1991-1-4

3 Load combinations 3.1 Partial safety factor

• γGmax = 1.35 (permanent loads)

• γGmin = 1.0 (permanent loads)

• γQ = 1.50 (variable loads)

• ψ0 = 0.50 (snow)

• ψ0 = 0.60 (wind)

• γM0 = 1.0

• γM1 = 1.0

EN 1990

EN 1990

Table A1.1

3.2 ULS Combinations Combination 101 : γGmax G + γQ Qs

Combination 102 : γGmin G + γQ Qw

Combination 103 : γGmax G + γQ Qs + γQ ψ0 Qw

Combination 104 : γGmin G + γQ Qs + γQ ψ0 Qw

Combination 105 : γGmax G + γQ ψ0 Qs + γQ Qw

Combination 106 : γGmin G + γQ ψ0 Qs + γQ Qw

EN 1990


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CALCULATION SHEET


3.3 SLS Combinations Combinations and limits should be specified for each project or by National Annex.

EN 1990

4 Sections 4.1 Column Try UKB 610x229x125 – Steel grade S275

Depth h = 612.2 mm

Web Depth hw = 573.0 mm

Depth of straight portion of the web

dw = 547.6 mm

Width b = 229.0 mm

Web thickness tw = 11.9 mm

Flange thickness tf = 19.6 mm

Fillet r = 12.7 mm

Mass 125.1 kg/m

hw

y y

z

z

tf

tw

b

h

Section area A = 159 cm2

Second moment of area /yy Iy = 98600 cm4

Second moment of area /zz Iz = 3930 cm4

Torsion constant It = 154 cm4

Warping constant Iw = 3.45 x 106 cm6

Elastic modulus /yy Wel,y = 3220 cm3

Plastic modulus /yy Wpl,y = 3680 cm3

Elastic modulus /zz Wel,z = 343 cm3

Plastic modulus /zz Wpl,z = 535 cm3


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CALCULATION SHEET


4.2 Rafter Try UKB 457x191x89 – Steel grade S275

Depth h = 463.4 mm

Web Depth hw = 428.0 mm

Depth of straight portion of the web

dw = 407.6 mm

Width b = 191.9 mm

Web thickness tw = 10.5 mm

Flange thickness tf = 17.7 mm

Fillet r = 10.2 mm

Mass 89.3 kg/m

Section area A = 114 cm2



Torsion constant It = 90.7 cm4

Warping constant Iw = 1.04 x 106 cm6


Plastic modulus /yy Wpl,y = 2014 cm3


Plastic modulus /zz Wpl,z = 338 cm3

5 Global analysis The 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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http://www.access-steel.com/discovery/linklookup.aspx?id=EC002





CALCULATION SHEET


5.1 Buckling amplification factor αcr In order to evaluate the sensitivity of the frame to 2nd order effects, a buckling analysis is performed to calculate the buckling amplification factor αcr for the load combination giving the highest vertical load: γmax G + γQ QS (combination 101).

For this combination: αcr = 12.74

The first buckling mode is shown hereafter.

EN 1993-1-1§ 5.2.1

So : αcr = 12.74 > 10

First order elastic analysis may be used.

EN 1993-1-1§ 5.2.1 (3)

5.2 Effects of imperfections The global initial sway imperfection may be determined from

φ = φ0 αh αm = 310204.3866.0740.02001 −⋅=××

EN 1993-1-1§ 5.3.2 (3)

where φ0 = 1/200

αh = 740.030.7

22==

h

αm = 866.0)11(5.0 =+m

m = 2 (number of columns)


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CALCULATION SHEET


Sway imperfections may be disregarded where HEd ≥ 0.15 VEd.

The effects of initial sway imperfection may be replaced by equivalent horizontal forces:

Heq = φ VEd in the combination where HEd < 0.15 ⎢VEd ⎢

The following table gives the reactions at supports.

EN 1993-1-1§ 5.3.2 (4)

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 -161.5 122.4 -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:

VEd

kN Heq = φ.VEd

kN

172.4 0.552

⇒ Modelling with Heq for the combination 101

EN 1993-1-1§ 5.3.2 (7)


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CALCULATION SHEET


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 = 164 mm = L/183

Snow only: Dy = 105 mm = L/286

Horizontal deflections:

Deflection at the top of column by wind only

Dx = 38 mm = h/157

5.3.2 Ultimate limit states Moment diagram in kNm

Combination 101:

Combination 102:


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CALCULATION SHEET


Combination 103:

Combination 104:

Combination 105:

Combination 106:


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CALCULATION SHEET


6 Column verification Profile UKB 610x229x125 - S275 (ε = 0.92) The verification of the member is carried out for the combination 101 : NEd = 161.5 kN (assumed to be constant along the column) VEd = 122.4 kN (assumed to be constant along the column) MEd = 755 kNm (at the top of the column)

6.1 Classification of the cross section • Web: The web slenderness is c / tw = 46.02

mm35.492759.11

161500

yw

EdN =

×==

ftNd

545.06.547235.496.547

d2 w

N =×+

=+

=ddwα > 0.50

The limit for Class 1 is : 396ε / (13α -1) = 87.591545.013

92.0396=

−××

Then : c / tw = 46.02 < 59.87 The web is class 1.

EN 1993-1-1§ 5.5

• Flange: The flange slenderness is c / tf = 95.9/ 19.6= 4.89

The limit for Class 1 is : 9 ε = 9 × 0.92 = 8.28

Then : c / tf = 4.89 < 8.28 The flange is Class 1

So the section is Class 1. The verification of the member will be based on the plastic resistance of the cross-section.


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CALCULATION SHEET


6.2 Resistance of cross section NOTE: This example uses a value for yield strength of 275 N/mm2, irrespective of flange thickness. The UK National Annex requires yield strength to be taken from the product standard, which will reduce the value for thicknesses over 16 mm. (UK NA still to be confirmed at April 2007)

Verification to shear force Shear area : Av = A - 2btf + (tw+2r)tf

76546.19)7.1229.11(6.19229215900 =××++××−=vA mm2

vA > η.hw.tw = 6819 mm2 (η = 1) OK

Vpl,Rd = Av (fy / 3 ) /γM0 = (7654×275/ 3 )×10-3

Vpl,Rd = 1215 kN

VEd / Vpl,Rd = 122.4 / 1215 = 0.10 < 0.50

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

EN 1993-1-1§ 6.2.8 (2)

Verification to axial force

Npl,Rd = A fy / γM0 = (15900 × 275/1.0) ×10-3

Npl,Rd = 4372.5 kN

NEd = 161.5 kN < 0.25 Npl,Rd = 0.25 x 4372.5 = 1093.1 kN

and NEd = 161.5 kN < 6.937101

2759.115735.05.03

M0

yww =×

×××=

γfth

kN

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

EN 1993-1-1§ 6.2.4

EN 1993-1-1§ 6.2.8 (2)

Verification to bending moment

Mpl,y,Rd = (3680 × 275/1.0) ×10-3

Mpl,y,Rd = 1012 kNm

My,Ed = 755 kNm < Mpl,y,Rd = 1012 kNm OK

EN 1993-1-1§ 6.2.5


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CALCULATION SHEET


6.3 Buckling resistance The buckling resistance of the column is sufficient if the following conditions are fulfilled (no bending about the weak axis, Mz,Ed = 0):

1

M1

Rky,LT

Edy,yy

M1

Rky

Ed ≤+

γχ

γχ M

MkN

N

1

M1

Rky,LT

Edy,zy

M1

Rky

Ed ≤+

γχ

γχ M

MkN

N

EN 1993-1-1§ 6.3.3

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

The frame is not sensitive to second order effects (αcr = 12.7> 10). Then the buckling length for in-plane buckling may be taken equal to the system length. Lcr,y = 5.99 m

EN 1993-1-1§ 5.2.2 (7)

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. The verification of the cross-sectional resistance at the top of the column will be determinant for the design.

Regarding the out-of-plane buckling, the member is laterally restrained at both ends only. Then : Lcr,z = 5.99 m and Lcr,LT = 5.99 m

• Buckling about yy Lcr,y = 5.99 m

EN 1993-1-1§ 6.3.1.2 (2)

h/b = 2.67 > 1.2 and tf = 19.6 < 40 mm Buckling curve : a (αy = 0.21) Table 6.1

32

42

2ycr,

y2ycr, 105990

1098600210000×

××== ππ

LEI

N =56956 kN

EN 1993-1-1§ 6.3.1.3 (1) 277.0

105695627515900

3ycr,

yy =

××

==NAf

λ


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CALCULATION SHEET


( )[ ] ( )[ ]22yyy 277.02.0277.021.015.02.015.0 +−+×=+−+= λλαφ = 0.546

984.0277.0546.0546.0

11222

y2

yy

y =−+

=−+

=λφφ

χ

EN 1993-1-1§ 6.3.1.2 (1)

• Buckling about zz Lcr,z = 5.99 m EN 1993-1-1§ 6.3.1.2 (2)

Buckling curve : b (αz = 0.34) Table 6.1

32

42

2zcr,

z2zcr, 105990

103930210000×

××== ππ

LEIN = 2270 kN

EN 1993-1-1§ 6.3.1.3 (1)

388.110227027515900

NAf

3zcr,

yz =

××

==λ

( )[ ] ( )[ ]22zzz 388.12.0388.134.015.02.015.0 +−+×=+−+= λλαφ = 1.665

387.0388.1665.1665.1

11222

z2zz

z =−+

=−+

=λφφ

χ

EN 1993-1-1§ 6.3.1.2 (1)

• Lateral torsional buckling Lcr,LT = 5.99 m Buckling curve : c (αLT = 0.49) Moment diagram with linear variation : ψ = 0 C1 = 1.77

Z2

t2

LTcr,

Z

W2

LTcr,

Z2

1cr EIGIL

II

LEICM

ππ

+=

42

42

4

12

62

42

cr 10393021000010154807705990

1039301045.3

10599010393021000077.1

××××××

+××

××××

×=π

πM

Mcr = 1517 kNm

817.0101517

2751036806

3

cr

yypl,LT =

×××

==M

fWλ

EN 1993-1-1§ 6.3.2.3

Table 6.5

NCCI SN003


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http://www.access-steel.com/discovery/linklookup.aspx?id=SN003




CALCULATION SHEET


( )[ ]2LTLT,0LTLTLT 15.0 βλλλαφ +−+=

with 40.0LT,0 =λ and β =0.75

EN 1993-1-1§ 6.3.2.3 (1)

( )[ ] 852.0817.075.04.0817.049.015.0 2LT =×+−+×=φ

754.0817.075.0852.0852.0

11222

LT2LTLT

LT =×−+

=−+

=βλφφ

χ

The influence of the moment distribution on the design buckling resistance moment of the beam is taken into account through the f-factor :

( )[ ]2LTc 8.021)1(5.01 −−−×−= λkf

where: kc = 752.033.033.1

1=

− ψ (ψ = 0)

( )[ ] 876.08.0817.021)752.01(5.01 2 =−−−×−=f < 1

χLT,mod = 861.0876.0754.0

==fLTχ < 1

EN 1993-1-1§ 6.3.2.3 (2)

Table 6.6

Calculation of the factors kyy and kzy according to Annex A of EN 1993-1-1

999.0

569565.161984.01

569565.1611

1

1

ycr,

Edy

ycr,

Ed

y =×−

−=

−

−=

NN

NN

χμ

EN 1993-1-1Annex A

955.0

22705.161387.01

22705.1611

1

1

zcr,

Edz

zcr,

Ed

z =×−

−=

−

−=

NN

NN

χμ

EN 1993-1-1Annex A

143.132203680

yel,

ypl,y ===

WW

w < 1.5

56.1343535

zel,

zpl,z ===

WW

w > 1.5 ⇒ wz = 1.5


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CALCULATION SHEET


Z

2t

2LTcr,

Z

W2

LTcr,

Z2

1cr,0 EIGIL

II

LEICM

ππ

+=

Mcr,0 is the critical moment for the calculation of 0λ for uniform bending moment as specified in Annex A.

⇒ C1 = 1

NCCI

SN003

42

42

4

12

62

42

cr,0 10393021000010154807705990

1039301045.3

1059901039302100001

××××××

+××

××××

×=π

πM

kNmM 2.857ocr, =

6

3

ocr,

yypl,0

102.857275103680

×××

==M

fWλ = 1.087

4

TFcr,

Ed

zcr,

Ed1lim0 )1)(1(2.0

NN

NNC −−=λ

with Ncr,TF = Ncr,T (doubly symmetrical section)

Critical axial force in the torsional buckling mode

)( 2w

2

t0

Tcr, hEI

GIIAN

π+=

For a doubly symmetrical section,

cm102530393098600)( 20

20zy0 =+=+++= AzyIII 4

EN 1993-1-1Annex A

NCCI

SN003

32

1224

4cr, 10)5990

1045.32100001015480770(10102530

15900 −×××

+××××

= πTN

Ncr,T = 5019 kN

So: 4lim0 )5019

5.1611)(2270

5.1611(77.12.0 −−=λ = 0.259

0λ > →lim0λ Section susceptible to torsional deformations


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CALCULATION SHEET


LTy

LTymy,0my,0my 1

)1(a

aCCC

ε

ε

+−+=

with 3

23

yel,Ed

Edy,y 1032205.161

1015910755×××××

==W

AN

Mε = 23.08 (class 1)

and 9860015411 −=−=

y

tLT I

Ia = 0.998

Calculation of the factor Cmy,0

ycr,

Edyymy,0 )33.0(36.021.079.0

NNC −++= ψψ

0y =ψ 56956

5.1611188.079.0my,0 −=C = 0.790

EN 1993-1-1Annex A

Table A2

Calculation of the factors Cmy and Cm,LT :

LTy

LTymy,0my,0my 1

)1(a

aCCC

ε+

ε−+=

964.0998.008.231

998.008.23)790.01(790.0 =×+

×−+=myC

EN 1993-1-1Annex A

1)1)(1(

Tcr,

Ed

zcr,

Ed

LT2mymLT ≥

−−

=

NN

NN

aCC

978.0)

50195.1611)(

22705.1611(

998.0964.0 2mLT =

−−×=C < 1

⇒ CmLT = 1


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CALCULATION SHEET


Calculation of the factors Cyy and Czy : EN 1993-1-1Annex A

ypl,

yel,LTpl

2max

2my

y

max2my

yyyy ])6.16.12)[(1(1

WW

bnCw

Cw

wC ≥−−−−+= λλ

037.01/27510159

105.161/ 2

3

M1Rk

Edpl =

×××

==γN

Nn

Mz,Ed = 0 ⇒ and 0LT =b 0LT =d 388.1zmax == λλ

]037.0)388.1964.0143.16.1388.1964.0

143.16.12[()1143.1(1 222

yy ×××−××−×−+=C

Cyy = 0.988

Cyy > 875.036803220

ypl,

yel, ==WW

OK

ypl,

yel,

z

yLTpl

2max

2my5

yyzy 6.0])142)[(1(1

WW

ww

dnCw

wC ≥−−−+= λ

943.0]037.0)388.1964.0143.1142)[(1143.1(1 22

5zy =×××−−+=C

Czy > 458.036803220

5.1143.16.06.0

ypl,

yel, =×=WW

ww

z

y OK

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

yy

ycr,

Ed

ymLTmyyy

1

1 CNNCCk

−

μ=

978.0988.01

569565.1611

999.01964.0yy =×−

××=k


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CALCULATION SHEET


z

y

zy

ycr,

Ed

zmLTmyzy 6.01

1 ww

CNNCCk

−=

μ

513.050.1143.16.0

943.01

569565.1611

955.01964.0zy =×××−

××=k

Verification with interaction formulae

EN 1993-1-1§ 6.3.3

1

M1

Rky,LT

Edy,

M1

Rky

Ed ≤+

γχ

γχ M

MkN

Nyy

88.0

1275103680861.0

10.755978.0

127515900984.0

105.1613

63

=××

××+

××

× <1 OK

1

M1

Rky,LT

Edy,zy

M1

Rkz

Ed ≤+

γχγ

χ MM

kNN

54.0

1275103680861.0

10.755513.0

127515900387.0

105.1613

63

=××

××+

××

× <1 OK

So the buckling resistance of the member is satisfactory.


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CALCULATION SHEET


7 Rafter verification 7.1 Classification Case with maximum compression in beam: (Combination 101)

• Web:

hw = 428 mm

tw = 10.5 mm c / tw = 38.81

c = 407.6 mm

NEd= 136 kN mmft

Nd 10.472755.1010136 3

yw

EdN =

××

==

558.06.407210.476.407

2N =

×+

=+

=w

w

dddα > 0.5

c / tw = 38.81 < 25.581558.013

92.0396113

396=

−××

=−αε ⇒ class 1

EN 1993-1-1§ 5.5

• Flanges b = 191.9 mm tf = 17.7 mm r = 10.2 mm c / tf = 4.55 c = 80.5 mm

(S275 ⇒ ε = 0.92 )

c / tf = 4.55 < 9ε = 8.28 ⇒ class 1

So the section is Class 1. The verification of the member will be based on the plastic resistance of the cross-section.


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CALCULATION SHEET


7.2 Resistance of cross-section Maximum force and moment at the end of the haunch: Combination 101 NEd = 136.00 kN VEd = 118.50 kN My,Ed = 349.10 kNm

305,23 kNm

349,10 kNm 754,98 kNm

Combination 101: Bending moment diagram along the rafter

Verification to shear force VEd = 118.50 kN

Av = A - 2btf + (tw+2r)tf η = 1

51547.17)2.1025.10(7.179.191211400 =××++××−=vA mm2

Av > η.hw.tw = 428×10.5 = 4494 mm2

Vpl,Rd = Av (fy / 3 ) /γM0 = 5154×275/ 3 /1000 = 818.3 kN

VEd / Vpl,Rd = 118.5 / 818.33 = 0.145 < 0.50 OK

EN 1993-1-1§ 6.2

EN 1993-1-1§ 6.2.8 (2)

Verification to axial force

Npl,Rd = 11400 x 275 x 10-3 = 3135 kN

NEd = 136 kN < 0.25 Npl,Rd= 0.25 × 3135 = 783.8 kN and

NEd = 136 kN < kNfth

9.617101

2755.104285.05.0 3

M0

yww =××××

= −

γ

OK

EN 1993-1-1§ 6.2.4

EN 1993-1-1§ 6.2.8 (2)


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CALCULATION SHEET


Verification to bending moment

Mpl,y,Rd = 2014 × 275 × 10-3 = 553.9 kNm My,Ed = 349.1 kNm < Mpl,y,Rd = 553.9 kNm OK

EN 1993-1-1§ 6.2.5

7.3 Buckling resistance Uniform members in bending and axial compression EN 1993-1-1

§ 6.3.3


1

M1

Rky,LT

Edy,yy

M1

Rky

Ed ≤

γχ

+

γχ M

MkN

N and 1

M1

Rky,LT

Edy,zy

M1

Rkz

Ed ≤

γχ

+

γχ M

MkN

N

• Buckling about yy:

EN 1993-1-1§ 6.3.1.2 (2)

Table 6.1

EN 1993-1-1§ 6.3.1.3 (1)

For the determination of the buckling length about yy, a buckling analysis is performed to calculate αcr for the load combination giving the highest vertical load, with a fictitious restraint at top of column: Combination 101 ⇒ αcr = 32.40

EN 1993-1-1§ 6.3.1.2 (2)

Buckling curve : a (h/b>2) ⇒ αy = 0.21

kN440613640.32NN Edcrycr, =×=α= Table 6.1

844.010.440627511400

NAf

3ycr,

yy =

×==λ


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CALCULATION SHEET


( )[ ]2yyy 2.015.0 λ+−λα+=φ

( )[ ]2y 844.02.0844.021.015.0 +−×+×=φ = 0.924

769.0844.0924.0924.0

11222

y2

yy

y =−+

=λ−φ+φ

=χ

• Buckling about zz: For buckling about zz and for lateral torsional buckling, the buckling length 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 buckling EN 1993-1-1§ 6.3.1.3

Lcr,z = 6.00 m

32

42

2zcr,

z2zcr, 106000

102090210000LEIN

×××

π=π= = 1203 kN

Torsional buckling

Lcr,T = 6.00 m

)( 2Tcr,

w2

t0

Tcr, LEIGI

IAN π

+=

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

cm43090209041000A)zy(III 20

20zy0 =+=+++= 4

NCCI

SN003

)6000

10.04.1210000107.9080770(1010.43090

11400N 2

12243

4Tcr,×

π+××××= −

Ncr,T = 3522 kN


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CALCULATION SHEET


Ncr = min ( Ncr,z ; Ncr,T ) = 1203 kN

614.110120327511400

NAf

3cr

yz =

××

==λ

EN 1993-1-1§ 6.3.1.3 (1)

Buckling curve : b

αz = 0.34

( )[ ]2zzz 2.015.0 λ+−λα+=φ

( )[ ]2z 614.12.0614.134.015.0 +−×+×=φ =2.043

303.0614.1043.2043.2

11222

z2zz

z =−+

=λ−φ+φ

=χ

EN 1993-1-1§ 6.3.1.2 (1)

Table 6.1

• Lateral torsional buckling : Lcr,LT = 6.00 m

Buckling curve : c αLT = 0.49

EN 1993-1-1§ 6.3.1.3

Table 6.5

Moment diagram along the part of rafter between restraints : Combination 101

Calculation of the critical moment:

ψ = - 0.487

q = - 9.56 kN/m μ =M

qL8

2

= - 0.123

⇒ C1 = 2.75

NCCI

SN003


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CALCULATION SHEET


Z2

t2

LTcr,

Z

W2

LTcr,

Z2

1cr EIGIL

II

LEICM

ππ

+=

42

42

4

12

62

42

cr 102090210000107.90807706000

1020901004.1

10600010209021000075.2M

×××π×××

+××

××××π

×=

Mcr = 1100 kNm

NCCI

709.0101100

275102010M

fW6

3

cr

yypl,LT =

×××

==λ

( )[ ]2LTLT,0LTLTLT 15.0 βλ+λ−λα+=φ

with 40.0LT,0 =λ and β =0.75

( )[ ] 764.0709.075.04.0709.049.015.0 2LT =×+−×+×=φ

821.0709.075.0764.0764.0

11222

LT2LTLT

LT =×−+

=λβ−φ+φ

=χ

EN 1993-1-1§ 6.3.2.3 (1)

kc = 0.91 ( )[ ]2

LTc 8.021)k1(5.01f −λ−−×−= EN 1993-1-1§ 6.3.2.3 (2)

Table 6.6

( )[ ] 956.08.0709.021)91.01(5.01f 2 =−×−×−×−= < 1

χLT,mod = 859.0956.0821.0

fLT ==

χ < 1

Combination 101 NEd = 136 kN compression

My,Ed = 349.1 kNm

Mz,Ed = 0

Section class1 ⇒ ΔMy,Ed = 0 and ΔMz,Ed = 0 EN 1993-1-1§ 6.3.3

1

M1

Rky,LT

Edy,yy

M1

Rky

Ed ≤+

γχ

γχ M

MkN

N 1

M1

Rky,LT

Edy,zy

M1

Rkz

Ed ≤+

γχγ

χ MM

kNN


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CALCULATION SHEET


993.0

4406136769.01

44061361

NN1

NN1

ycr,

Edy

ycr,

Ed

y =×−

−=

χ−

−=μ

918.0

1203136303.01

12031361

NN1

NN1

zcr,

Edz

zcr,

Ed

z =×−

−=

χ−

−=μ

EN 1993-1-1Annex A

EN 1993-1-1Annex A

136.117702010

WW

wyel,

ypl,y === < 1.50

550.1218338

WW

wzel,

zpl,z === > 1.50 ⇒ wz = 1.50

Z

2t

2LTcr,

Z

W2

LTcr,

Z2

1cr,0 EIGIL

II

LEICM

ππ

+=

Mcr,0 is the critical moment for the calculation of 0λ for uniform bending moment as specified in Annex A.

⇒ C1 = 1

42

42

4

12

62

42

cr,0 102090210000107.90807706000

1020901004.1

1060001020902100001M

×××π×××

+××

××××π

×=

kNm2.400M ocr, =

NCCI

SN003

EN 1993-1-1Annex A

6

3

ocr,

yypl,0

102.400275102010

MfW

×××

==λ = 1.175

4

TFcr,

Ed

zcr,

Ed1lim0 )

NN1)(

NN1(C2.0 −−=λ with C1 = 2.75

with Ncr,TF = Ncr,T (doubly symmetrical section)

4lim0 )35221361)(

12031361(75.22.0 −−=λ = 0.319

0λ =1.175 > lim0λ =0.319


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CALCULATION SHEET


EN 1993-1-1Annex A

LTy

LTymy,0my,0my 1

)1(a

aCCC

ε

ε

+−+=

with 33

6

yel,Ed

Edy,y 101770

1140010136101.349

WA

NM

××

××

==ε =16.53 (class 1)

and 41000

7.901II1a

y

tLT −=−= = 0.998

Calculation of the factor Cmy,0

Moment diagram along the rafter:

ycr,

Ed

Edy,2

xy2

my,0 11NN

MLEI

C⎥⎥⎦

⎤

⎢⎢⎣

⎡−+=

δπ

44061361

10755300001641041000210000

1C 62

42

my,0 ⎥⎦

⎤⎢⎣

⎡−

××××××π

+= =0.978

EN 1993-1-1Annex A

Table A2

My,Ed = maximum moment along the rafter = 755kNm

δ = maximum displacement along the rafter = 164 mm 30m

Calculation of the factors Cmy and Cm,LT :

LTy

LTymy,0my,0my 1

)1(a

aCCC

ε

ε

+−+=

996.0998.053.161

998.053.16)978.01(978.0Cmy =×+

×−+=


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CALCULATION SHEET


1)1)(1(

Tcr,

Ed

zcr,

Ed

LT2mymLT ≥

−−=

NN

NN

aCC

072.1)

35221361)(

12031361(

998.0996.0C 2mLT =

−−×= > 1

EN 1993-1-1Annex A

Calculation of the factors Cyy and Czy

ypl,

yel,LTpl

2max

2my

y

max2my

yyyy W

W]bn)C

w6.1C

w6.12)[(1w(1C ≥−λ−λ−−+=

043.01/27511400

10136/N

Nn3

M1Rk

Edpl =

××

=γ

=

Mz,Ed = 0 ⇒ and 0LT =b 0LT =d 614.1zmax =λ=λ

]043.0)614.1996.0136.1

6.1614.1996.0136.1

6.12)[(1136.1(1C 222yy ×××−××−−+=

Cyy = 0.977 88.020101770

WW

ypl,

yel, ==≥ OK

EN 1993-1-1Annex A

ypl,

yel,

z

yLTpl

2max

2my5

yyzy W

Www

6.0]dn)Cw142)[(1w(1C ≥−λ−−+=

900.0]043.0)614.1996.0136.1142)[(1136.1(1C 22

5zy =×××−−+=

460.020101770

50.1136.16.0

WW

ww

6.0Cypl,

yel,

z

yzy =×=≥ OK

Calculation of the factors kyy and kzy :

yy

ycr,

Ed

ymLTmyyy

1

1 CNNCCk

−=

μ

120.1977.01

44061361

993.0072.1996.0kyy =×−

××=

EN 1993-1-1Annex A


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CALCULATION SHEET


z

y

zy

ycr,

Ed

zmLTmyzy w

w6.0

C1

NN1

CCk−

μ=

587.050.1

136.16.0900.01

44061361

918.0072.1996.0kzy =×××−

××=


1

M1

Rky,LT

Edy,yy

M1

Rky

Ed ≤+

γχ

γχ M

MkN

N

88.0

1275102010859.0

101.349120.1

127511400769.0

101363

63

=××

×

××+

××

× < 1 OK

EN 1993-1-1§ 6.3.3

(6.61)

1

M1

Rky,LT

Edy,zy

M1

Rkz

Ed ≤+

γχγ

χ MM

kNN

57.0

1275102010859.0

101.349587.0

127511400303.0

101363

63

=××

×

××+

××

× < 1 OK

(6.62)

8 Haunch verification For the verification of the haunch, the compression part of the cross-section is considered 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:

NEd = 139.2 kN VEd = 151.3 kN MEd = 755 kNm


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CALCULATION SHEET


Properties of the whole section: The calculation of elastic section properties for this case is approximate, ignoring the middle flange.

Section area A = 160.80 cm2





1000 mm

200 mm

Properties of the compression part:

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

200 mm

120 mm Section area A = 44 cm2


Second moment of area /zz Iz =1068 cm4

⇒ cm93.444

1068iz ==

704.0

39.8630.493000

iL

1z

zfz =

×=

λ=λ

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

⇒ curve d ⇒ α = 0.76

( )[ ] ( )[ ] 939.0704.02.0704.076.015.02.015.0 22zzz =+−×+×=λ+−λα+=φ

641.0704.0939.0939.0

11222

z2zz

z =−+

=λ−φ+φ

=χ


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CALCULATION SHEET


Compression in the bottom flange:

kN7604400100010.4610

100075500016080440024.139N 3fEd, =×

××

+×=

Verification of buckling resistance of the bottom flange:

980.02754400641.0

10760N

N 3

Rkz

fEd, =××

×=

χ < 1 OK


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SX029a-EN-UK

Quality Record

RESOURCE TITLE Example: Elastic analysis of a single bay portal frame

Reference(s) SX029a-EN-GB

LOCALISED RESOURCE DOCUMENT

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Created by Laurent Narboux SCI Nov 06

Technical content checked by Charles King SCI Dec 06

Editorial content checked by D C Iles SCI April 2007

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Wind Loads UK- Portal Frame WE