Wind Loads UK- Portal Frame WE

Example: Elastic analysis of a single bay portal frame (GB)

CALCULATION SHEET

Document Ref:

SX029a-EN-GB

Title

Example: Elastic analysis of a single bay portal frame

Sheet

1

of

30

Eurocode Ref Made by

Laurent Narboux

Date

Nov 2006

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.

,0 72

7,30

5,988

Created on Saturday, December 03, 2011 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Access Steel Licence Agreement

α

0 7,2

[m] 30,00

1 Basic data • • • • •

Total length : Spacing: Bay width : Height (max): Roof slope: 3,00

3,00

b = 72.00 m s = 7.20 m d = 30.00 m h = 7.30 m α = 5.0° 3,00

3,00

1

1 : Torsional restraints

3,00

0


CALCULATION SHEET

Document Ref:

SX029a-EN-GB

Title


Sheet

2

of

30


Laurent Narboux

Date

Nov 2006

Checked by

C M King

Date

Dec 2006

2 Loads EN 1991-1-1

2.1 Permanent loads •

self-weight of the beam

•

roofing with purlins

G = 0.30 kN/m2 G = 0.30 × 7.20 = 2.16 kN/m

2.2 Snow loads

EN 1991-1-3

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 s = 4,45 kN/m

α 7,30


for an internal frame:

30,00

[m]


CALCULATION SHEET

Document Ref:

SX029a-EN-GB

3

Title


Sheet

of

30


Laurent Narboux

Date

Nov 2006

Checked by

C M King

Date

Dec 2006

2.3 Wind loads

EN 1991-1-4

Characteristic values for wind loading in kN/m for an internal frame Zone G: w = 9,18

wind direction

Zone J: w = 5,25

Zone H: w = 5,25

Zone I: w = 5,25

Zone D: w = 4,59 Zone E: w = 3,28

1,46

e/10 = 1,46

30,00


3 Load combinations

EN 1990

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

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 Table A1.1

EN 1990


CALCULATION SHEET

Document Ref:

SX029a-EN-GB

4

Title


Sheet

of

30


Laurent Narboux

Date

Nov 2006

Checked by

C M King

Date

Dec 2006

3.3 SLS Combinations

EN 1990

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

4 Sections z

tf

4.1 Column Try UKB 610x229x125 – Steel grade S275 Depth

h = 612.2 mm

Web Depth

hw = 573.0 mm

tw y

y

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

Section area

A = 159 cm2

hw

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

z b

h


CALCULATION SHEET

Document Ref:

SX029a-EN-GB

Title


Sheet

5

of

30


Laurent Narboux

Date

Nov 2006

Checked by

C M King

Date

Dec 2006

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

Second moment of area /yy Iy = 41000 cm4 Second moment of area /zz Iz = 2090 cm4 Torsion constant

It = 90.7 cm4

Warping constant

Iw = 1.04 x 106 cm6

Elastic modulus /yy

Wel,y = 1770 cm3

Plastic modulus /yy

Wpl,y = 2014 cm3

Elastic modulus /zz

Wel,z = 218 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.

The frame has been modelled using the EFFEL program.

EN 1993-1-1 § 5.2


CALCULATION SHEET

Document Ref:

SX029a-EN-GB

Title


Sheet

6

of

30


Laurent Narboux

Date

Nov 2006

Checked by

C M King

Date

Dec 2006

5.1 Buckling amplification factor αcr

EN 1993-1-1 In order to evaluate the sensitivity of the frame to 2 order effects, a buckling § 5.2.1 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). nd

For this combination: αcr = 12.74


The first buckling mode is shown hereafter.

So :

αcr = 12.74 > 10

First order elastic analysis may be used.

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

φ = φ0 αh αm = where

1 × 0.740 × 0.866 = 3.204 ⋅10 −3 200

φ0 = 1/200 αh =

2 2 = = 0.740 h 7.30

αm = 0.5(1 +

1 ) = 0.866 m

m = 2 (number of columns)

EN 1993-1-1 § 5.2.1 (3)

EN 1993-1-1 § 5.3.2 (3)


CALCULATION SHEET

Document Ref:

SX029a-EN-GB

7

Title


Sheet

of

30


Laurent Narboux

Date

Nov 2006

Checked by

C M King

Date

Dec 2006

Sway imperfections may be disregarded where HEd ≥ 0.15 VEd.

EN 1993-1-1 § 5.3.2 (4)

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. Left column 1

Right column 2

Total

ULS


0.15 ⎢VEd ⎢

HEd,1

VEd,1

HEd,2

VEd,2

HEd

VEd

kN

kN

Kn

kN

kN

kN

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

Comb.

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

⇒

VEd

Heq = φ.VEd

kN

kN

172.4

0.552

Modelling with Heq for the combination 101

EN 1993-1-1 § 5.3.2 (7)


CALCULATION SHEET

Document Ref:

SX029a-EN-GB

8

Title


Sheet

of

30


Laurent Narboux

Date

Nov 2006

Checked by

C M King

Date

Dec 2006

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: 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:

EN 1993-1-1 § 7 and EN 1990


CALCULATION SHEET

Document Ref:

SX029a-EN-GB

Title


Sheet

9

of

30

Eurocode Ref

Combination 103:


Combination 104:

Combination 105:

Combination 106:

Made by

Laurent Narboux

Date

Nov 2006

Checked by

C M King

Date

Dec 2006


CALCULATION SHEET

Document Ref:

SX029a-EN-GB

Title


Sheet

10

of

30


Laurent Narboux

Date

Nov 2006

Checked by

C M King

Date

Dec 2006

6 Column verification Profile UKB 610x229x125 - S275 (ε = 0.92)

The verification of the member is carried out for the combination 101 : NEd = VEd = MEd =

161.5 kN (assumed to be constant along the column) 122.4 kN (assumed to be constant along the column) 755 kNm (at the top of the column)

6.1 Classification of the cross section •

The web slenderness is c / tw = 46.02

Web:


dN =

α=

N Ed 161500 = = 49.35 mm t w f y 11.9 × 275

d w + d N 547.6 + 49.35 = = 0.545 > 0.50 2d w 2 × 547.6

The limit for Class 1 is : 396ε / (13α -1) = Then : c / tw = 46.02 < 59.87 •

Flange:

396 × 0.92 = 59.87 13 × 0.545 − 1

The web is class 1.

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.

EN 1993-1-1 § 5.5


CALCULATION SHEET

Document Ref:

SX029a-EN-GB

11

Title


Sheet

of

30


Laurent Narboux

Date

Nov 2006

Checked by

C M King

Date

Dec 2006

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

Av = 15900 − 2 × 229 × 19.6 + (11.9 + 2 × 12.7) × 19.6 = 7654 mm

2

Av > η.hw.tw = 6819 mm2 (η = 1)


Vpl,Rd = Av (fy /

EN 1993-1-1 § 6.2.8 (2)

OK

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. Verification to axial force

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

EN 1993-1-1 § 6.2.4

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 <

0.5hw t w f y

γ M0

0.5 × 573 × 11.9 × 275 = = 937.6 kN 1 × 103

EN 1993-1-1 § 6.2.8 (2)

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


CALCULATION SHEET

Document Ref:

SX029a-EN-GB

Title


Sheet

12

of

30


Laurent Narboux

Date

Nov 2006

Checked by

C M King

Date

Dec 2006

6.3 Buckling resistance The buckling resistance of the column is sufficient if the following conditions EN 1993-1-1 are fulfilled (no bending about the weak axis, Mz,Ed = 0): § 6.3.3 N Ed + k yy χ y N Rk

γ M1

M y,Ed ≤1 M y,Rk

χ LT

γ M1

M y,Ed ≤1 M y,Rk

N Ed + k zy χ y N Rk

χ LT

γ M1

γ M1


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 EN 1993-1-1 buckling length for in-plane buckling may be taken equal to the system length. § 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. 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 Lcr,LT = 5.99 m

and •

Buckling about yy

Lcr,y = 5.99 m

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

N cr,y = π 2

λy =

EI y L2cr,y

Af y N cr,y

=

210000 × 98600 × 10 4 =π2 =56956 kN 59902 × 103

15900 × 275 = 0.277 56956 × 103

EN 1993-1-1 § 6.3.1.2 (2) Table 6.1 EN 1993-1-1 § 6.3.1.3 (1)


CALCULATION SHEET

Document Ref:

SX029a-EN-GB

Title


Sheet

13

of

30

Eurocode Ref

[

(

Made by

Laurent Narboux

Date

Nov 2006

Checked by

C M King

Date

Dec 2006

)

]

φy = 0.5 1 + α λ y − 0.2 + λ y = 0.5 × [1 + 0.21(0.277 − 0.2) + 0.2772 ]= 0.546 χy = •

2

1

φy + φy − λ 2

2 y

=

1 0.546 + 0.546 2 − 0.277 2

= 0.984

Lcr,z = 5.99 m

Buckling about zz

EN 1993-1-1 § 6.3.1.2 (2)

Buckling curve : b (αz = 0.34)

Af y


λz =

N cr,z

[

Table 6.1

4 EI z 2 210000 × 3930 × 10 π = 2270 kN = L2cr,z 5990 2 ×103

N cr,z = π 2

EN 1993-1-1 § 6.3.1.3 (1)

15900 × 275 = 1.388 2270 ×10 3

=

(

)

]

φz = 0.5 1 + α λ z − 0.2 + λ z = 0.5 × [1 + 0.34(1.388 − 0.2 ) + 1.3882 ] = 1.665

χz =

•

1

φz + φz2 − λ

=

2 z

2

1 1.665 + 1.6652 − 1.3882

= 0.387

Buckling curve : c (αLT = 0.49) Moment diagram with linear variation : ψ = 0 M cr = C1 M cr = 1.77 ×

Lcr,LT

2

5990 × 10 2

Table 6.5

6

4

3.45 × 10 5990 × 80770 × 154 × 10 + 2 4 3930 × 10 π × 210000 × 3930 × 10 4 12

Mcr = 1517 kNm

Wpl,y f y M cr

C1 = 1.77

2

π × 210000 × 3930 × 10

λ LT =

EN 1993-1-1 § 6.3.2.3

I W Lcr,LT GI t + IZ π 2 EI Z

2

=

EN 1993-1-1 § 6.3.1.2 (1)

Lcr,LT = 5.99 m

Lateral torsional buckling

π 2 EI Z

EN 1993-1-1 § 6.3.1.2 (1)

3680 × 103 × 275 = 0.817 1517 × 106

2

4

NCCI SN003


CALCULATION SHEET

Document Ref:

SX029a-EN-GB

Title


Sheet

14

of

30

Eurocode Ref

[

Made by

Laurent Narboux

Date

Nov 2006

Checked by

C M King

Date

Dec 2006

(

)

2

φLT = 0.5 1 + α LT λ LT − λ LT,0 + βλ LT

]

EN 1993-1-1 § 6.3.2.3 (1)

with λ LT,0 = 0.40 and β =0.75

φLT = 0.5 × [1 + 0.49(0.817 − 0.4) + 0.75 × 0.8172 ] = 0.852 χ LT =

1 2

2 − βλ LT φLT + φLT

=

1 0.852 + 0.8522 − 0.75 × 0.817 2

= 0.754

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


[

(

f = 1 − 0.5 × (1 − k c ) 1 − 2 λ LT − 0.8 1 = 0.752 where: kc = 1.33 − 0.33ψ

)]

EN 1993-1-1 § 6.3.2.3 (2) Table 6.6

2

(ψ = 0)

[

]

f = 1 − 0.5 × (1 − 0.752) 1 − 2(0.817 − 0.8) = 0.876 < 1 0.754 χ = 0.861 < 1 χLT,mod = LT = f 0.876 2

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

N Ed 161.5 1− N cr, y 56956 μy = = = 0.999 N Ed 161.5 1 − χy 1 − 0.984 × N cr, y 56956 1−

EN 1993-1-1 Annex A

N Ed 161.5 1− N cr,z 2270 μz = = 0.955 = 161.5 N Ed 1− χz 1 − 0.387 × 2270 N cr,z 1−

wy =

Wpl,y

wz =

Wpl,z

Wel,y

Wel,z

=

3680 = 1.143 < 1.5 3220

=

535 = 1.56 > 1.5 343

EN 1993-1-1 Annex A ⇒

wz = 1.5


CALCULATION SHEET

Document Ref:

SX029a-EN-GB

Title


Sheet

15

of

30

Eurocode Ref

M cr,0 = C1

Made by

Laurent Narboux

Date

Nov 2006

Checked by

C M King

Date

Dec 2006

π 2 EI Z Lcr,LT

2


2

NCCI SN003

Mcr,0 is the critical moment for the calculation of λ0 for uniform bending moment as specified in Annex A. ⇒ C1 = 1 M cr,0 = 1×

π 2 × 210000 × 3930 × 10 4 5990 2 × 106

3.45 × 1012 5990 2 × 80770 × 154 × 10 4 + 3930 × 10 4 π 2 × 210000 × 3930 × 10 4


M cr,o = 857.2 kNm Wpl,y f y

λ0 =

M cr,o

3680 ×103 × 275 = = 1.087 857.2 × 106

λ 0 lim = 0.2 C1 4 (1 −

N Ed N )(1 − Ed ) N cr,z N cr,TF

with Ncr,TF = Ncr,T

(doubly

EN 1993-1-1 Annex A

symmetrical section)

Critical axial force in the torsional buckling mode

N cr,T

π 2 EI w A = (GI t + ) I0 h2

For a doubly symmetrical section,

I 0 = I y + I z + ( y02 + z02 ) A = 98600 + 3930 = 102530 cm4 N cr,T =

12 15900 4 2 210000 × 3.45 × 10 × ( 80770 × 154 × 10 + π ) × 10 −3 4 2 102530 × 10 5990

Ncr,T = 5019 kN So: λ 0 lim = 0.2 1.77 4 (1 −

161.5 161.5 )(1 − ) = 0.259 2270 5019

λ 0 > λ 0 lim → Section susceptible to torsional deformations

NCCI SN003


CALCULATION SHEET

Document Ref:

SX029a-EN-GB

Title


Sheet

16

of

30


Laurent Narboux

Date

Nov 2006

Checked by

C M King

Date

Dec 2006

C my = C my,0 + (1 − C my,0 )

with

εy =

and aLT = 1 −

ε y a LT 1 + ε y a LT

M y,Ed A 755 × 103 × 159 × 10 2 = 23.08 (class 1) = N Ed Wel,y 161.5 × 3220 × 103 154 It = 0.998 = 1− Iy 98600

Calculation of the factor Cmy,0


Cmy,0 = 0.79 + 0.21ψ y + 0.36(ψ y − 0.33)

ψy =0

Cmy,0

N Ed N cr,y

161.5 = 0.79 − 0.1188 = 0.790 56956

EN 1993-1-1 Annex A Table A2

Calculation of the factors Cmy and Cm,LT : C my = Cmy,0 + (1 − C my,0 )

ε y aLT 1 + ε y aLT

Cmy = 0.790 + (1 − 0.790)

23.08 × 0.998 = 0.964 1 + 23.08 × 0.998

a LT

2 C mLT = C my

(1 − CmLT = 0.964 2 ×

N Ed N )(1 − Ed ) N cr,z N cr,T

≥1

0.998 = 0.978 < 1 161.5 161.5 (1 − )(1 − ) 2270 5019

⇒ CmLT = 1

EN 1993-1-1 Annex A


CALCULATION SHEET

Document Ref:

SX029a-EN-GB

17

Title


Sheet

of

30


Laurent Narboux

Date

Nov 2006

Checked by

C M King

Date

Dec 2006

Calculation of the factors Cyy and Czy :

C yy = 1 + ( wy − 1)[(2 −

npl =

W 1.6 2 1.6 2 2 Cmy λ max − Cmy λ max )npl − bLT ] ≥ el,y wy wy Wpl,y

EN 1993-1-1 Annex A

N Ed 161.5 ×103 = 0.037 = N Rk / γ M1 159 ×102 × 275 / 1

Mz,Ed = 0 ⇒ bLT = 0 and d LT = 0 Cyy = 1 + (1.143 − 1) × [(2 −

λ max = λ z = 1.388

1.6 1.6 × 0.9642 ×1.388 − × 0.9642 ×1.3882 ) × 0.037] 1.143 1.143

Cyy = 0.988


Cyy >

Wel,y Wpl,y

=

3220 = 0.875 3680

Czy = 1 + ( wy − 1)[(2 −

w W 14 2 2 Cmy λ max ) npl − d LT ] ≥ 0.6 y el,y 5 wy wz Wpl,y

C zy = 1 + (1 .143 − 1)[( 2 −

Czy > 0.6

wy Wel,y wz Wpl,y

OK

= 0 .6

14 × 0 .964 2 × 1 .388 2 ) × 0 .037 ] = 0 .943 5 1 .143 1.143 3220 × = 0.458 1.5 3680

Calculation of the factors kyy and kzy : k yy = CmyCmLT

μy 1 N 1 − Ed C yy N cr, y

k yy = 0.964 × 1×

1 0.999 × = 0.978 161.5 0.988 1− 56956

OK

EN 1993-1-1 Annex A


CALCULATION SHEET

Document Ref:

SX029a-EN-GB

Title


Sheet

18

of

30

Eurocode Ref

k zy = CmyCmLT

Made by

Laurent Narboux

Date

Nov 2006

Checked by

C M King

Date

Dec 2006

μz 1−

k zy = 0.964 × 1×

N Ed N cr,y

w 1 0.6 y Czy wz

1 1.143 0.955 × × 0.6 × = 0.513 161.5 0.943 1 . 50 1− 56956

Verification with interaction formulae

N Ed + k yy χ y N Rk

γ M1

M y,Ed ≤1 M y,Rk

χ LT

γ M1

161.5 × 10 755.106 + 0.978 × = 0.88 <1 OK 3 15900 × 275 3680 × 10 × 275 0.984 × 0.861 × 1 1


3

N Ed + k zy χ z N Rk

γ M1

M y,Ed ≤1 M y,Rk

χ LT

γ M1

161.5 × 103 755.106 + 0.513 × = 0.54 <1 OK 3 15900 × 275 3680 × 10 × 275 0.387 × 0.861 × 1 1 So the buckling resistance of the member is satisfactory.

EN 1993-1-1 § 6.3.3


CALCULATION SHEET

Document Ref:

SX029a-EN-GB

Title


Sheet

19

of

30


Laurent Narboux

Date

Nov 2006

Checked by

C M King

Date

Dec 2006

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 dN =


NEd= 136 kN

α= c / tw = 38.81 <

•

N Ed 136 ×103 = = 47.10 mm t w f y 10.5 × 275

d w + d N 407.6 + 47.10 = = 0.558 > 0.5 2d w 2 × 407.6

396ε 396 × 0.92 = = 58.25 13α − 1 13 × 0.558 − 1

Flanges b = 191.9 mm tf = 17.7 mm r = 10.2 mm c = 80.5 mm

⇒

class 1

c / tf = 4.55

(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.

EN 1993-1-1 § 5.5


CALCULATION SHEET

Document Ref:

SX029a-EN-GB

20

Title


Sheet

of

30


Laurent Narboux

Date

Nov 2006

Checked by

C M King

Date

Dec 2006

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 754,98 kNm 349,10 kNm


305,23 kNm

Combination 101: Bending moment diagram along the rafter

Verification to shear force

VEd = 118.50 kN η=1

Av = A - 2btf + (tw+2r)tf

Av = 11400 − 2 × 191.9 × 17.7 + (10.5 + 2 × 10.2) × 17.7 = 5154 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

Verification to axial force -3

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

EN 1993-1-1 § 6.2 EN 1993-1-1 § 6.2.8 (2)

EN 1993-1-1 § 6.2.4

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

0.5hw t w f y

γ M0

0.5 × 428 ×10.5 × 275 = ×10−3 = 617.9kN 1

EN 1993-1-1 § 6.2.8 (2)


CALCULATION SHEET

Document Ref:

SX029a-EN-GB

Title


Sheet

21

of

30


Laurent Narboux

Date

Nov 2006

Checked by

C M King

Date

Dec 2006

EN 1993-1-1 § 6.2.5

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

7.3 Buckling resistance Uniform members in bending and axial compression

Verification with interaction formulae N Ed + k yy χ y N Rk


γ M1

•

M y,Ed ≤1 M y,Rk

χ LT

and

γ M1

EN 1993-1-1 § 6.3.3

M y,Ed N Ed ≤1 + k zy χ z N Rk M y,Rk χ LT γ M1 γ M1

Buckling about yy:

For the determination of the buckling length about yy, a buckling analysis is performed to calculate αcr for the load combination giving the highest EN 1993-1-1 vertical load, with a fictitious restraint at top of column: § 6.3.1.2 (2) Combination 101 ⇒ αcr = 32.40 Table 6.1

EN 1993-1-1 § 6.3.1.3 (1)

Buckling curve : a (h/b>2)

⇒ αy = 0.21

N cr,y = α cr N Ed = 32.40 × 136 = 4406 kN λy =

Af y N cr,y

=

11400 × 275 = 0.844 4406.103

EN 1993-1-1 § 6.3.1.2 (2) Table 6.1


CALCULATION SHEET

Document Ref:

SX029a-EN-GB

Title


Sheet

22

of

30

Eurocode Ref

[

Made by

Laurent Narboux

Date

Nov 2006

Checked by

C M King

Date

Dec 2006

(

)

2

φ y = 0.5 1 + α λ y − 0.2 + λ y

[

]

]

φ y = 0.5 × 1 + 0.21× (0.844 − 0.2) + 0.8442 = 0.924 χy =

•

1 φy + φy − λ 2

2 y

=

1 0.924 + 0.924 2 − 0.844 2

= 0.769

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 N cr,z = π 2

4 EI z 2 210000 × 2090 × 10 = π = 1203 kN L2cr,z 6000 2 × 10 3

Torsional buckling

NCCI

Lcr,T = 6.00 m N cr,T =

SN003

π 2 EI w A (GI t + ) 2 I0 Lcr,T

with yo = 0 and zo = 0

(doubly symmetrical section)

I 0 = I y + I z + ( y 02 + z 02 )A = 41000 + 2090 = 43090 cm4 N cr,T =

12 11400 4 2 210000 × 1.04.10 −3 × × × × + π 10 ( 80770 90 . 7 10 ) 43090.10 4 6000 2

Ncr,T = 3522 kN


CALCULATION SHEET

Document Ref:

SX029a-EN-GB

Title


Sheet

23

of

30


Laurent Narboux

Date

Nov 2006

Checked by

C M King

Date

Dec 2006

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

λz =

N cr

EN 1993-1-1 § 6.3.1.3 (1)

11400 × 275 = 1.614 1203 × 103

=

Buckling curve : b

EN 1993-1-1 § 6.3.1.2 (1)

αz = 0.34

[

(

)

2

φ z = 0 . 5 1 + α λ z − 0 .2 + λ z

[

]

Table 6.1

]

φ z = 0.5 × 1 + 0.34 × (1.614 − 0.2 ) + 1.614 2 =2.043


χz = •

1 φz + φ2z − λ

2 z

=

1 2.043 + 2.0432 − 1.6142

= 0.303

Lateral torsional buckling :

EN 1993-1-1 § 6.3.1.3

Lcr,LT = 6.00 m Buckling curve : c

αLT = 0.49

Table 6.5

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

Calculation of the critical moment:

NCCI

ψ = - 0.487 q = - 9.56 kN/m ⇒

C1 = 2.75

SN003

qL2 = - 0.123 μ= 8M


CALCULATION SHEET

Document Ref:

SX029a-EN-GB

Title


Sheet

24

of

30

Eurocode Ref

M cr = C1 M cr = 2.75 ×

π 2 EI Z Lcr,LT

Made by

Laurent Narboux

Date

Nov 2006

Checked by

C M King

Date

Dec 2006

NCCI

2


2

π2 × 210000 × 2090 ×104 60002 ×106

1.04 ×1012 60002 × 80770 × 90.7 ×104 + 2090 ×104 π 2 × 210000 × 2090 ×104

Mcr = 1100 kNm λ LT =

Wpl, y f y M cr

[

2010 × 103 × 275 = 0.709 1100 × 106

=

(

)

2

φLT = 0.5 1 + α LT λ LT − λ LT,0 + βλ LT

]

EN 1993-1-1 § 6.3.2.3 (1)


with λ LT,0 = 0.40 and β =0.75

[

]

φLT = 0.5 × 1 + 0.49 × (0.709 − 0.4) + 0.75 × 0.7092 = 0.764 χ LT =

1 φLT + φ2LT − βλ

kc = 0.91

2 LT

1

=

[

0.764 + 0.7642 − 0.75 × 0.7092

(

f = 1 − 0.5 × (1 − k c ) 1 − 2 λ LT − 0.8

[

)]

= 0.821

2

]

f = 1 − 0.5 × (1 − 0.91) × 1 − 2 × (0.709 − 0.8) = 0.956 < 1

χLT,mod =

2

EN 1993-1-1 § 6.3.2.3 (2) Table 6.6

χ LT 0.821 = = 0.859 < 1 f 0.956

Combination 101

NEd = 136 kN compression My,Ed = 349.1 kNm Mz,Ed = 0 ⇒

Section class1


γ M1

M y,Ed ≤1 M y,Rk

χ LT

γ M1

ΔMy,Ed = 0 and ΔMz,Ed = 0 N Ed + k zy χ z N Rk

γ M1

M y,Ed ≤1 M y,Rk

χ LT

γ M1

EN 1993-1-1 § 6.3.3


CALCULATION SHEET

Document Ref:

SX029a-EN-GB

Title


Sheet

25

of

30


Laurent Narboux

Date

Nov 2006

Checked by

C M King

Date

Dec 2006

N Ed 136 1− N cr, y 4406 μy = = = 0.993 136 N Ed 1 − 0.769 × 1 − χy 4406 N cr, y 1−

EN 1993-1-1 Annex A

N Ed 136 1− N cr,z 1203 = 0.918 μz = = N Ed 136 1 − χz 1 − 0.303 × N cr,z 1203


1−

wy =

Wpl, y

wz =

Wpl, z 338 = = 1.550 > 1.50 Wel,z 218

Wel, y

M cr,0 = C1

=

EN 1993-1-1 Annex A

2010 = 1.136 < 1.50 1770

π 2 EI Z Lcr,LT

2

⇒

wz = 1.50

NCCI

2


SN003

Mcr,0 is the critical moment for the calculation of λ0 for uniform bending moment as specified in Annex A. ⇒ C1 = 1 π 2 × 210000 × 2090 × 10 4 6000 2 × 106 = 400.2 kNm

M cr,0 = 1 ×

M cr,o

λ0 =

Wpl,y f y M cr,o

=

1.04 × 1012 6000 2 × 80770 × 90.7 × 10 4 + 2 π × 210000 × 2090 × 10 4 2090 × 10 4

EN 1993-1-1 Annex A

2010 ×103 × 275 = 1.175 400.2 ×106

λ 0 lim = 0.2 C1 4 (1 −

N N Ed )(1 − Ed ) N cr, TF N cr, z

with Ncr,TF = Ncr,T

(doubly

λ 0 lim = 0.2 2.75 4 (1 −

with C1 = 2.75

symmetrical section)

136 136 )(1 − ) = 0.319 1203 3522

λ 0 =1.175 > λ 0 lim =0.319


CALCULATION SHEET

Document Ref:

SX029a-EN-GB

26

Title


Sheet

of

30


Laurent Narboux

Date

Nov 2006

Checked by

C M King

Date

Dec 2006

Cmy = Cmy,0 + (1 − Cmy,0 )

with ε y = and

ε y aLT

EN 1993-1-1 Annex A

1 + ε y aLT

M y,Ed A 349.1×106 11400 = × =16.53 (class 1) 3 N Ed Wel,y 136 × 10 1770 × 103

a LT = 1 −

It 90.7 = 0.998 =1− Iy 41000

EN 1993-1-1 Annex A

Calculation of the factor Cmy,0

Moment diagram along the rafter:


Table A2 My,Ed = maximum moment along the rafter = 755kNm

δ = maximum displacement along the rafter = 164 mm

30m

⎡ π 2 EI y δ x ⎤ N Cmy,0 = 1 + ⎢ 2 − 1⎥ Ed ⎢⎣ L M y,Ed ⎥⎦ N cr,y ⎡ π2 × 210000 × 41000 × 104 × 164 ⎤ 136 C my,0 = 1 + ⎢ − 1⎥ =0.978 300002 × 755 × 106 ⎣ ⎦ 4406 Calculation of the factors Cmy and Cm,LT : Cmy = Cmy,0 + (1 − Cmy,0 )

ε y aLT 1 + ε y aLT

C my = 0.978 + (1 − 0.978)

16.53 × 0.998 = 0.996 1 + 16.53 × 0.998


CALCULATION SHEET

Document Ref:

SX029a-EN-GB

27

Title


Sheet

of

30


Laurent Narboux

Date

Nov 2006

Checked by

C M King

Date

Dec 2006

aLT

2 CmLT = Cmy

(1 −

N Ed N )(1 − Ed ) N cr,z N cr,T

EN 1993-1-1 Annex A

≥1

0.998 = 1.072 > 1 136 136 ) )(1 − (1 − 3522 1203

C mLT = 0.996 2 ×

Calculation of the factors Cyy and Czy

C yy = 1 + ( w y − 1)[(2 −


n pl =

W 1.6 2 2 1.6 2 Cmy λ max )n pl − b LT ] ≥ el, y Cmy λ max − wy Wpl, y wy

N Ed 136 × 103 = 0.043 = N Rk / γ M1 11400 × 275 / 1

Mz,Ed = 0 ⇒ bLT = 0 and d LT = 0 C yy = 1 + (1.136 − 1)[( 2 −

C zy = 1 + ( w y − 1)[(2 −

Wel, y 1770 = = 0.88 OK Wpl, y 2010

w y Wel, y 14 2 2 C my λ max ) n pl − d LT ] ≥ 0.6 5 wy w z Wpl, y

C zy = 1 + (1.136 − 1)[(2 − w y Wel, y w z Wpl, y

λ max = λ z = 1.614

1 .6 1. 6 × 0.996 2 × 1.614 − × 0.996 2 × 1.614 2 ) × 0.043] 1.136 1.136

Cyy = 0.977 ≥

C zy ≥ 0.6

EN 1993-1-1 Annex A

14 × 0.996 2 × 1.614 2 ) × 0.043] = 0.900 5 1.136

= 0 .6 ×

1.136 1770 = 0.460 OK 1.50 2010

Calculation of the factors kyy and kzy : k yy = CmyCmLT

μy 1−

N Ed N cr, y

k yy = 0.996 × 1.072 ×

1 C yy

1 0.993 × = 1.120 136 0.977 1− 4406

EN 1993-1-1 Annex A


CALCULATION SHEET

Document Ref:

SX029a-EN-GB

Title


Sheet

28

of

30

Eurocode Ref

k zy = C myCmLT

Made by

Laurent Narboux

Date

Nov 2006

Checked by

C M King

Date

Dec 2006

wy μz 1 0.6 N wz 1 − Ed Czy N cr, y

k zy = 0.996 × 1.072 ×

1 1.136 0.918 × × 0.6 × = 0.587 136 0.900 1 . 50 1− 4406

Verification with interaction formulae


γ M1

M y,Ed ≤1 M y,Rk

χ LT

EN 1993-1-1 § 6.3.3 (6.61)

γ M1

136 × 10 349.1 × 106 + 1.120 × = 0.88 < 1 OK 3 11400 × 275 2010 × 10 × 275 0.769 × 0.859 × 1 1


3

N Ed + k zy χ z N Rk

γ M1

M y,Ed ≤1 M y,Rk

χ LT

γ M1

136 × 103 349.1 × 106 + 0.587 × = 0.57 < 1 OK 3 11400 × 275 2010 × 10 × 275 0.303 × 0.859 × 1 1

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 = VEd = MEd =

139.2 kN 151.3 kN 755 kNm

(6.62)


CALCULATION SHEET

Document Ref:

SX029a-EN-GB

29

Title


Sheet

of

30


Laurent Narboux

Date

Nov 2006

Checked by

C M King

Date

Dec 2006

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

Second moment of area /yy

Iy = 230520 cm4

Second moment of area /zz

Iz = 2141 cm4

Elastic modulus /yy

Wel,y = 4610 cm3

Elastic modulus /zz

Wel,z = 214 cm3

1000 mm

200 mm


Properties of the compression part:

Section at the mid-length of the haunch including 1/6th of the web depth A = 44 cm2

Section area Second moment of area /yy

Iy = 554 cm

Second moment of area /zz

Iz =1068 cm4

⇒ iz =

λz =

120 mm

4

1068 = 4.93cm 44

200 mm

L fz 3000 = = 0.704 i z λ1 49.30 × 86.39

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

[

⇒ α = 0.76

(

)

]

[

]

φz = 0.5 1 + α λ z − 0.2 + λ z = 0.5 × 1 + 0.76 × (0.704 − 0.2 ) + 0.704 2 = 0.939

χz =

2

1 φz + φ − λ 2 z

2 z

=

1 0.939 + 0.9392 − 0.7042

= 0.641


CALCULATION SHEET

Document Ref:

SX029a-EN-GB

Title


Sheet

30

of

30


Laurent Narboux

Date

Nov 2006

Checked by

C M King

Date

Dec 2006

Compression in the bottom flange:

N Ed, f = 139.24 ×

4400 755000 × 1000 + × 4400 = 760kN 16080 4610.103 × 1000

Verification of buckling resistance of the bottom flange:


N Ed,f 760 × 103 = 0.980 < 1 OK = χ z N Rk 0.641× 4400 × 275


SX029a-EN-UK

Quality Record RESOURCE TITLE


Reference(s)

SX029a-EN-GB


LOCALISED RESOURCE DOCUMENT Name

Company

Date

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

Page 31

Wind Loads UK- Portal Frame WE

Recommend Documents