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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.

5,988 7,30
,0 72 0

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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 = 5,0

3,00 3,00 3,00

1 : Torsional restraints

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2 Loads
2.1 Permanent loads
self-weight of the beam roofing with purlins for an internal frame: G = 0,30 kN/m2 G = 0,30 7,20 = 2,16 kN/ml EN 1991-1-1

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

EN 1991-1-3

7,30 30,00 [m]

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2.3 Wind loads

Characteristic values for wind loading in kN/m for an internal frame
wind direction Zone G: w = 9,18 Zone D: w = 4,59 Zone E: w = 3,28 Zone H: w = 5,25 Zone J: w = 5,25 Zone I: w = 5,25

EN 1991-1-4

e/10 = 1,46

1,46 30,00

3 Load combinations
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EN 1990

3.1 Partial safety factor

Gmax = 1,35 Gmin = 1,0 Q = 1,50 0 = 0,50 0 = 0,60 M0 = 1,0 M1 = 1,0

(permanent loads) (permanent loads) (variable loads) (snow) (wind) EN 1990 Table A1.1

3.2 ULS Combinations

Combination 101 : Combination 102 : Combination 103 : Combination 104 : Combination 105 : Combination 106 :

EN 1990

Gmax G + Q Qs Gmin G + Q Qw Gmax G + Q Qs + Q 0 Gmin G + Q Qs + Q 0 Qw Gmax G + Q 0 Qs + Q Qw Gmin G + Q 0 Qs + Q Qw

EN 1990

3.3 SLS Combinations

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

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4 Sections
4.1 Column
Try IPE 600 Steel grade S275 Depth Web Depth h = 600 mm hw = 562 mm dw = 514 mm Width Web thickness Flange thickness Fillet
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tf

tw y hw y

Depth of straight portion of the web b = 220 mm tw = 12 mm tf = 19 mm r = 24 mm 122,4 kg/m A = 156 cm2

z b

Mass Section area

Second moment of area /yy Iy = 92080 cm4 Second moment of area /zz Iz = 3386cm4 Torsion constant Warping constant Elastic modulus /yy Plastic modulus /yy Elastic modulus /zz Plastic modulus /zz It = 165,4 cm4 Iw = 2845500 cm6 Wel,y = 3069 cm3 Wpl,y = 3512 cm3 Wel,z = 307,80 cm3 Wpl,z = 485,60 cm3

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4.2 Rafter
Try IPE 500 Steel grade S275 Depth Web Depth h = 500 mm hw = 468 mm dw = 426 mm Width Web thickness Flange thickness Fillet Mass
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Depth of straight portion of the web b = 200 mm tw = 10,2 mm tf = 16 mm r = 21 mm 90,7 kg/m A = 115,50 cm2

Section area

Second moment of area /yy Iy = 48200 cm4 Second moment of area /zz Iz = 2141 cm4 Torsion constant Warping constant Elastic modulus /yy Plastic modulus /yy Elastic modulus /zz Plastic modulus /zz It = 89,29 cm4 Iw = 1249400 cm6 Wel,y = 1928 cm3 Wpl,y = 2194 cm3 Wel,z = 214,1 cm3 Wpl,z = 335,90 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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EN 1993-1-1 5.2.1 In order to evaluate the sensitivity of the frame to 2 order effects, a buckling analysis is performed to calculate the buckling amplification factor cr for the load combination giving the highest vertical load: max G + Q QS (combination 101).
nd

5.1 Buckling amplification factor cr

For this combination, the amplification factor is: cr = 14,57 The first buckling mode is shown hereafter.

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So :

cr = 14,57 > 10

First order elastic analysis may be used.

EN 1993-1-1 5.2.1 (3) EN 1993-1-1 5.3.2 (3)

5.2 Effects of imperfections

The global initial sway imperfection may be determined from

= 0 h m =
where

1 0,740 0,866 = 3,204 10 3 200

0 = 1/200 h =
2 2 = = 0,740 h 7,30

m = 0,5(1 +

1 ) = 0,866 m

m = 2 (number of columns) EN 1993-1-1 5.3.2 (4)

Sway imperfections may be disregarded where HEd 0,15 VEd. The effects of initial sway imperfection may be replaced by equivalent horizontal forces: Heq = VEd in the combination where HEd < 0,15 VEd

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The following table gives the reactions at supports.

Left column 1 ULS Comb. HEd,1 kN 101 102 103 104 105
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Right column 2 HEd,2 Kn 125,5 -24,47 89,48 77,01 26,72 14,25 VEd,2 kN -172,4 58,19 -105,3 -86,56 -10,57 8,17 HEd kN 0 70,69 42,42 42,42 70,69 70,69

Total VEd kN -344,70 138,9 -197,1 -159,6 1,40 38,88 51,70 20,83 29,56 23,93 0,21 5,83

VEd,1 kN -172,4 80,74 -91,77 -73,03 11,97 30,71

0,15 VEd

-125,5 95,16 -47,06 -34,59 43,97 56,44

106

The sway imperfection has only to be taken into for the combination 101: VEd kN 172,4 Heq = .VEd kN 0,552

EN 1993-1-1 5.3.2 (7)

Modelling with Heq for the combination 101

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: Snow only: Dy = 124 mm = L/241 Dy = 73 mm = L/408 EN 1993-1-1 7 and EN 1990

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 states

Moment diagram in kNm Combination 101:

Combination 102:
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Combination 103:

Combination 104:

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Combination 105:

Combination 106:

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6 Column verification
Profile IPE 600 - S275 ( = 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

Web: The web slenderness is c / tw = 42,83

dN =

N Ed 161500 = = 48,94mm t w f y 12 275

EN 1993-1-1 5.5

d w + d N 514 + 48,94 = = 0,548 > 0,50 2d w 2 514

396 0,92 = 59,49 13 0,548 1

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 = 4,74 The flange is Class 1

The limit for Class 1 is : 9 = 9 0,92 = 8,28 Then : c / tf = 4,74 < 8,28

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 : Av = A - 2btf + (tw+2r)tf > .hw.tw EN 1993-1-1 6.2.6 (3)

may be conservatively taken equal to 1

Av = 15600 2 220 19 + (12 + 2 24) 19 = 8380 mm2 > .hw.tw = 6744 mm2
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Vpl,Rd = Av (fy / Vpl,Rd = 1330 kN

3 ) /M0 = (8380275/ 3 ).10-3

VEd / Vpl,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 Npl,Rd = A fy / M0 = (15600 275/1,0).10-3 Npl,Rd = 4290 kN NEd = 161,5 kN < 0,25 Npl,Rd = 4290 x 0,25 = 1073 kN and NEd = 161,5 kN < EN 1993-1-1 6.2.4

0,5hw t w f y

M0

0,5 562 12 275 = = 927,3 kN 11000

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 = Wpl,y fy / M0 = (3512 275/1,0).10 Mpl,y,Rd = 965,8 kNm My,Ed = 755 kNm < Mpl,y,Rd = 965,8 kNm
-3

EN 1993-1-1 6.2.5

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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 y N Rk

M1
N Ed + k zy y N Rk

LT

M y,Ed 1 M y,Rk

M1

M1

LT

M y,Ed 1 M y,Rk

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 (cr = 14,57 > 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 for buckling about the weak axis Lcr,T = 5,99 m for torsional buckling and Lcr,LT = 5,99 m for lateral torsional buckling
Buckling about yy

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Lcr,y = 5,99 m EN 1993-1-1 6.3.1.2 (2) Table 6.1 EN 1993-1-1 6.3.1.3 (1)

Buckling curve : a (y = 0,21) EI y 210000 92080 10000 =53190kN N cr, y = 2 2 = 2 Lcr, y 5990 2 1000

y =

Af y N cr, y

15600 275 = 0,284 53190.10 3

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2 y = 0,5 1 + y 0,2 + y = 0,5 [1 + 0,21(0,284 0,2) + 0,2842 ]= 0,5491 EN 1993-1-1

6.3.1.2 (1)

y =

y + y
2

2 y

1 0,5491 + 0,54912 0,284 2

= 0,9813

Buckling about zz

Lcr,z = 5,99 m

N cr,z

Buckling curve : b (z = 0,34) EI 210000 3386 10000 =2 2 z =2 = 1956 kN Lcr,z 5990 2 1000
Af y N cr, z = 15600 275 = 1,481 1956.103

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

z =

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z = 0,5 1 + z 0,2 + z = 0,5 [1 + 0,34(1,481 0,2 ) + 1,4812 ] = 1,814

2

z =

z +
2 z

2 z

1 1,814 + 1,814 2 1,4812

= 0,3495

Lateral torsional buckling

Lcr,LT = 5,99 m C1 = 1,77

Buckling curve : c (LT = 0,49) Moment diagram with linear variation : = 0

M cr = C1
M cr = 1,77
2

EN 1993-1-1 6.3.2.3 Table 6.5 NCCI SN003

2 EI Z
Lcr,LT
2

I W Lcr,LT GI t + IZ 2 EI Z
6 2 6

210000 3386 10000 2845500.10

5990 10 3386.10
4

5990 80770 165,4.10 2 210000 3386.10 4

2

Mcr = 1351 kNm

LT =

Wpl,y f y M cr

3512.103 275 = 0,8455 1351.106

LT = 0,5 1 + LT LT LT,0 + LT
with LT,0 = 0,40 and =0,75

EN 1993-1-1 6.3.2.3 (1)

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LT = 0,5 1 + 0, 49 (0 ,8455 0 , 4 ) + 0, 75 0,8455 2 = 0,8772

LT =

LT +
kc =

2 LT

2 LT

1 0,8772 + 0,8772 2 0, 75 0,8455 2

( = 0)

= 0, 7352

1 = 0,7519 1,33 0,33

f = 1 0,5 (1 0,7519) 1 2(0,8455 0,8) = 0,8765 < 1 0,7352 LT,mod = LT = = 0,8388 < 1 0,8765 f
2

f = 1 0,5 (1 k c ) 1 2 LT 0,8

)]
2

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

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N Ed 161,5 1 N cr,y 53190 = = 0,9999 y = N Ed 161,5 1 0,9813 1 y 53190 N cr,y 1

N Ed 161,5 1 N cr,z 1956 = = 0,9447 z = N Ed 161,5 1 0,3495 1 z 1956 N cr,z 1

EN 1993-1-1 Annex A

wy = wz =

Wpl,y Wel,y Wpl,z Wel,z

= =

3512 = 1,144 < 1,5 3069 485,6 = 1,578 > 1,5 307,8 wz = 1,5

EN 1993-1-1 Annex A

Critical axial force in the torsional buckling mode

N cr,T =

NCCI SN003

2 EI w A (GI t + ) I0 Lcr,T 2

For a doubly symmetrical section,

2 2 I 0 = I y + I z + ( y 0 + z 0 ) A = 92080 + 3386 = 95466 cm4

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N cr,T =

15600 210000 2845500 .10 6 80770 165, 4.10 4 + 2 95466.10 4 1000 5990 2

Ncr,T = 4869 kN

M cr,0 = C1

2 EI Z
Lcr,LT
2

I W Lcr,LT GI t + IZ 2 EI Z

NCCI SN003

Mcr,0 is the critical moment for the calculation of 0 for uniform bending moment as specified in Annex A. C1 = 1
M cr,0 = 1
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2 210000 3386.10 4
5990 2 10 6

2845500.10 6 5990 2 80770 165,4.10 4 + 3386.10 4 2 210000 3386.10 4

M cr,o = 763,3kNm

0 =

Wpl,y f y M cr,o

3512.10 3 275 = 1,125 763,3.10 6

N Ed N )(1 Ed ) N cr,z N cr,TF
(doubly

EN 1993-1-1 Annex A

0 lim = 0,2 C1 4 (1
with Ncr,TF = Ncr,T

symmetrical section)

0 lim = 0,2 1,77 4 (1

0 > 0 lim

161,5 161,5 )(1 ) = 0,2582 1956 4869

C my = C my,0 + (1 C my,0 )

y a LT
1 + y a LT

with

y =

M y,Ed N Ed

I A = 23,76 (class 1) and a LT = 1 t = 0,9982 Iy Wel, y

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Calculation of the factor Cmy,0

C my,0 = 0,79 + 0,21 y + 0,36( y 0,33)

N Ed N cr, y

EN 1993-1-1 Annex A Table A2

y =0

C my,0

N = 0,79 0,1188 Ed = 0,7896 N cr,y

Calculation of the factors Cmy and Cm,LT :

Cmy = Cmy,0 + (1 Cmy,0 ) y aLT 1 + y aLT
23,76 0,9982 1 + 23,76 0,9982
= 0,9641

C my = 0,7896 + (1 0,7896)
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2 C mLT = C my

a LT (1 N Ed N )(1 Ed ) N cr,z N cr,T

0,9982

EN 1993-1-1 Annex A

C mLT = 0,96412

161,5 161,5 (1 )(1 ) 1956 4869

= 0,9843 < 1

CmLT = 1
Calculation of the factors Cyy and Czy :

Cyy = 1 + ( wy 1)[(2
npl =

W 1,6 2 1,6 2 2 Cmy max Cmy max )npl bLT ] el, y wy wy Wpl, y

EN 1993-1-1 Annex A

N Ed 161500 = = 0,03765 N Rk / M1 15600 275 / 1

max = z = 1,4810

Mz,Ed = 0 bLT = 0 and d LT = 0

Cyy = 1 + (1,144 1) [(2

1,6 1,6 0,96412 1,481 0,96412 1,4812 ) 0,03765] 1,144 1,144 W 3069 = 0,8739 Cyy = 0,9849 > el, y = Wpl, y 3512

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Czy = 1 + ( wy 1)[(2

w W 14 2 2 Cmy max )npl d LT ] 0,6 y el, y 5 wy wz Wpl, y

14 0 ,9641 2 1, 481 2 ) 0 ,03765 ] = 0 ,9318 1,144 5

C zy = 1 + (1,144 1)[( 2 C zy = 0 ,9318 > 0,6

wy Wel,y wz Wpl,y

= 0,6

1,144 3069 = 0,4579 1,5 3512

Calculation of the factors kyy and kzy :

k yy = CmyCmLT y 1 N Ed C yy 1 N cr, y

EN 1993-1-1 Annex A

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k yy = 0,9641 1

0,9999 1 = 0,9818 161,5 0,9849 1 53190

k zy = CmyCmLT

z
1 N Ed N cr, y

wy 1 0,6 Czy wz

k zy = 0,96411

0,9447 1 1,144 0,6 = 0,5138 161,5 0,9318 1,50 1 53190

Verification with interaction formulae

N Ed + k yy y N Rk

M1

LT

M y,Ed 1 M y,Rk

EN 1993-1-1 6.3.3

M1

161500 755.10 6 = 0,9534 <1 OK + 0,9818 15600 275 3512.10 3 275 0,9813 0,8388 1 1

N Ed + k zy z N Rk

M1

LT

M y,Ed 1 M y,Rk

M1

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161500 755.10 6 + 0,5138 = 0,5867 <1 OK 15600 275 3512.10 3 275 0,3495 0,8388 1 1

So the buckling resistance of the member is satisfactory.

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

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Web:

hw = 468 mm tw = 10,2 mm c = 426 mm

NEd= 136 kN
dN = N Ed 136000 = = 48,5mm t w f y 10,2 275

c / tw = 41,76

=
c / tw = 41,76 <

d + d N 426 + 48,5 = = 0,557 > 0,5 2d 2 426

EN 1993-1-1 5.5

396 396 0,92 = 58,38 = 13 1 13 0,557 1

class 1

Flanges b = 200 mm tf = 16 mm r = 21 mm c = 71 mm

c / tf = 4,44

part to compression

c / tf < 9 = 8,28
c / tf = 4,44

(S275 = 0,92 )
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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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: NEd = 136,00 kN VEd = 118,50 kN My,Ed = 349,10 kNm
754,98 kNm 349,10 kNm

305,23 kNm

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Combination 101: Bending moment diagram along the rafter

Shear

VEd = 118,50 kN Av = A - 2btf + (tw+2r)tf

=1

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

Av = 11550 2 200 16 + (10,2 + 2 21) 16 = 5985 mm2

Av > .hw.tw = 46810,2 = 4774mm2 Vpl,Rd = Av (fy /

3 ) /M0 = 5985275/ 3 /1000 = 950,3 kN

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

its effect on the moment resistance may be neglected!
Compression

Npl,Rd = 11550 x 275/1000 = 3176 kN NEd = 136 kN < 0,25 Npl,Rd= 3176 0,25 = 794,1 kN
and

EN 1993-1-1 6.2.4

NEd = 136 kN <

0,5 hw t w f y

M0

0,5 468 10, 2 275 = = 656, 4 kN 1 1000

EN 1993-1-1 6.2.8 (2)

its effect on the moment resistance may be neglected!

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Bending

Mpl,y,Rd = 2194 275/1000 = 603,4 kNm My,Ed = 349,10 kNm < Mpl,y,Rd = 603,4 kNm

EN 1993-1-1 6.2.5

7.3 Buckling resistance

Uniform members in bending and axial compression

Verification with interaction formulae

N Ed + k yy y N Rk M y,Ed 1 M y,Rk

EN 1993-1-1 6.3.3
M y,Ed 1 M y,Rk

and

M1
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LT

N Ed + k zy z N Rk

M1

M1

LT

M1

Buckling about yy:

For the determination of the buckling length about yy, a buckling analysis is performed to calculate the buckling amplification factor cr for the load EN 1993-1-1 combination giving the highest vertical load, with a fictitious restraint at 6.3.1.2 (2) top of column: Table 6.1 Combination 101 cr = 37,37

EN 1993-1-1 6.3.1.3 (1)

Buckling curve : a (h/b>2)

y = 0,21

N cr, y = cr N Ed = 37,37 136 = 5082 kN

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

y =

Af y N cr, y

11550 275 = 0,7906 5082.103

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y = 0,5 1 + y 0, 2 + y

y = 0, 5 1 + 0, 21 (0, 7906 0, 2 ) + 0, 7906 2 = 0,8745

y =

y + y
2

2 y

1 0,8745 + 0,87452 0,7906 2

= 0,8011

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 Lcr,z = 6,00 m EN 1993-1-1 6.3.1.3

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N cr,z = 2

EI z L2 z cr,

=2

210000 2141 10000 = 1233kN 6000 2 1000

NCCI SN003

Torsional buckling Lcr,T = 6,00 m

N cr,T = A 2 EI w (GI t + ) 2 I0 Lcr,T

with yo = 0 and zo = 0

(doubly symmetrical section)

2 2 I 0 = I y + I z + ( y0 + z0 ) A = 48199 + 2141 = 50340 cm4

N cr,T =

11550 210000 1249370 .10 6 ) (80770 89,29.10 4 + 2 50340.10 4 1000 6000 2

Ncr,T = 3305 kN Ncr = min ( Ncr,z ; Ncr,T ) = 1233 kN

EN 1993-1-1 6.3.1.3 (1)

z =

Af y N cr

11550 275 = 1,605 1233.103

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Buckling curve : b z = 0,34

EN 1993-1-1 6.3.1.2 (1)

z = 0,5 [1 + 0,34 (1,605 0,2) + 1,6052 ] =2,027

z =

z = 0,5 1 + z 0,2 + z

Table 6.1

z +
2 z

2 z

1 2,027 + 2,027 2 1,6052

= 0,3063

Lateral torsional buckling :

Lcr,LT = 6,00 m
Buckling curve : c
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EN 1993-1-1 6.3.1.3 LT = 0,49 Table 6.5

Moment diagram along the part of rafter between restraints: Combination 101 Calculation of the critical moment: NCCI SN003
qL2 = = - 0,123 8M

= - 0,487
q = - 9,56 kN/m
C1 = 2,75

M cr = C1
M cr = 2,75

2 EI Z
Lcr,LT
2

I W Lcr,LT GI t + IZ 2 EI Z
2141.10 4 + 6000 2 80770 89,29.10 4 2 210000 2141.10 4

NCCI

2 210000 2141 10 4 1249400.10 6

6000 2 10 6

Mcr = 1159 kNm

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LT =

Wpl, y f y

LT = 0,5 1 + LT LT LT,0 + LT
with LT,0 = 0,40 and =0,75

M cr

2195.103 275 = 0,7215 1159.106

EN 1993-1-1 6.3.2.3 (1)

LT = 0 , 5 1 + 0 , 49 (0 , 7215 0 , 4 ) + 0 , 75 0 , 7215 2

= 0 , 7740

LT =

LT +
2 LT

2 LT

1 0,7740 + 0,7740 2 0,75 0,72152

= 0,8125

kc = 0,91
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f = 1 0,5 (1 0,91) 1 2 (0,7215 0,8) = 0,9556 < 1

2

f = 1 0,5 (1 k c ) 1 2 LT 0,8

)]
2

EN 1993-1-1 6.3.2.3 (2) Table 6.6

LT,mod =

LT
f

0,8125 = 0,8503 < 1 0,9556

Combination 101

NEd = 136 kN compression My,Ed = 349,10 kNm Mz,Ed = 0

Section class1
N Ed + k yy y N Rk

M y,Ed 1 M y,Rk

My,Ed = 0 et Mz,Ed = 0
N Ed + k zy z N Rk M y,Ed 1 M y,Rk

EN 1993-1-1 6.3.3

M1

LT

M1

M1

LT

M1

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N Ed 136 1 N cr, y 5082 = = 0,9946 y = N 136 1 y Ed 1 0,8011 N cr, y 5082 1

N Ed 136 1 N cr,z 1233 z = = = 0,9208 N Ed 136 1 0,3063 1 z 1233 N cr,z 1

EN 1993-1-1 Annex A

wy = wz =

Wpl,y Wel, y Wpl,z Wel,z

= =

2194 = 1,138 < 1,50 1928 335,9 = 1,569 > 1,50 214,1
I W Lcr,LT GI t + IZ 2 EI Z
2

EN 1993-1-1 Annex A wz = 1,5

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M cr,0 = C1

2 EI Z
Lcr,LT
2

NCCI SN003

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

2 210000 2141.10 4 1249400.10 6

2141.10 4

M cr,o

6000 2 10 6 = 421,5kNm

6000 2 80770 89 , 29.10 4 2 210000 2141.10 4

0 =

Wpl, y f y M cr,o

2195.10 3 275 = 1,196 421,5.10 6

EN 1993-1-1 Annex A with C1 = 2,75

0 lim = 0,2 C1 4 (1
with Ncr,TF = Ncr,T

N Ed N )(1 Ed ) N cr,z N cr,TF

(doubly

symmetrical section)

0 lim = 0,2 2,75 4 (1

136 136 )(1 ) = 0,3187 1233 3305

0 =1,196 > 0 lim =0,3187

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Cmy = Cmy,0 + (1 Cmy,0 )

y aLT
1 + y aLT

EN 1993-1-1 Annex A

with y = and

M y,Ed N Ed

A 349,10.106 11550 =15,38 (class 1) = Wel,y 136000 1928 103

89,29 It = 0,9981 = 1 48200 Iy

aLT = 1

Calculation of the factor Cmy,0

Moment diagram along the rafter: My,Ed = maximum moment along the rafter = 755kNm 30m 2 EI y x N = 1+ 2 1 Ed L M y,Ed N cr, y
2 210000 48200 10 4 179 136 =0,9803 1 = 1+ 30000 2 755 106 5082

EN 1993-1-1 Annex A Table A2

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= maximum displacement along the rafter = 179mm

Cmy,0

Cmy,0

Calculation of the factors Cmy and Cm,LT :

Cmy = Cmy,0 + (1 Cmy,0 )

y aLT
1 + y aLT

Cmy = 0,9803 + (1 0,9803)

15,38 0,9982 = 0,996 1 + 15,38 0,9982

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2 CmLT = Cmy

aLT (1 N Ed N )(1 Ed ) N cr,z N cr,T

EN 1993-1-1 Annex A

CmLT = 0,996 2

0,9981 = 1,072 > 1 136 136 ) )(1 (1 3305 1233

Calculation of the factors Cyy and Czy

C yy = 1 + ( wy 1)[(2

W 1,6 2 1,6 2 2 Cmy max Cmy max )npl bLT ] el, y wy wy Wpl,y

EN 1993-1-1 Annex A

npl =
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N Ed 136000 = = 0,0428 N Rk / M1 11550 275 / 1

Mz,Ed = 0 bLT = 0 and d LT = 0

C yy = 1 + (1,138 1)[( 2

max = z = 1,605

1,6 1,6 0,996 2 1,605 0,996 2 1,605 2 ) 0,0428] 1,138 1,138

Cyy = 0,9774 Czy = 1 + ( wy 1)[(2 wy Wel,y 14 2 2 Cmy max )npl d LT ] 0,6 5 wy wz Wpl, y

Czy = 1 + (1,138 1)[(2

14 0,996 2 1,6052 ) 0,0428] = 0,9011 5 1,138

EN 1993-1-1 Annex A

Calculation of the factors kyy and kzy :

k yy = CmyCmLT
1 N Ed C yy 1 N cr, y

k yy = 0,996 1,072

0,9946 1 = 1,116 136 0,9774 1 5082

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k zy = CmyCmLT

z
1 N Ed N cr, y

wy 1 0,6 Czy wz

k zy = 0,996 1,072

0,9208 1 1,138 0,6 = 0,5859 136 0,9011 1,50 1 5082

Verification with interaction formulae

N Ed + k yy y N Rk

M1
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LT

M y,Ed 1 M y,Rk

EN 1993-1-1 6.3.3 (6.61)

M1

136000 349,1.10 6 + 1,116 = 0,8131 < 1 OK 11550 275 2194.10 3 275 0,8011 0,8503 1 1

N Ed + k zy z N Rk

M1

LT

M y,Ed 1 M y,Rk

(6.62)

M1

136000 349,1.10 6 + 0,5859 = 0,5385 < 1 OK 11550 275 2194.10 3 275 0,3063 0,8503 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 = 139,2 VEd = 151,3 MEd = 755

kN kN kNm

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Properties of the whole section:

The calculation of elastic section properties for this case is approximate, ignoring the middle flange. Section area Second moment of area /yy Second moment of area /zz Elastic modulus /yy Elastic modulus /zz

A = 160,80 cm2 Iy = 230520 cm4 Iz = 2141 cm4 Wel,y = 4610 cm3 Wel,z = 214 cm3
200 mm 1000 mm

Properties of the compression part:

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Section at the mid-length of the haunch including 1/6th of the web depth Section area Second moment of area /yy Second moment of area /zz iz =

A = 44 cm2 Iy = 554 cm
4

120 mm

Iz =1068 cm4
200 mm

1068 = 4,93cm 44

z =

Lfz 3000 = = 0,7044 iz 1 49,30 86,39

= 0,76

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

z = 0,5 1 + z 0,2 + z = 0,5 [1 + 0,76 (0,7044 0,2) + 0,7044 2 ] = 0,9397

2

curve d

z =

z + z2

2 z

1 0,9397 + 0,9397 2 0,7044 2

= 0,640

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Compression in the bottom flange:

N Ed,f = 139,24
4400 755000 1000 + 4400 = 760 kN 16080 4610.10 3 1000

Verification of buckling resistance of the bottom flange:

N Ed, f

z N Rk

760000 = 0,981 < 1 OK 0,640 4400 275

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Example: Elastic analysis of a single bay portal frame

Example: Elastic analysis of a single bay portal frame SX029a-EN-EU

Quality Record
RESOURCE TITLE Reference(s) ORIGINAL DOCUMENT Name Created by Technical content checked by Editorial content checked by Technical content endorsed by the following STEEL Partners: 1. UK 2. France 3. Sweden
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Example: Elastic analysis of a single bay portal frame

T2703

Company CTICM CTICM

Date 25/10/2005 26/10/2005

Valrie LEMAIRE Alain BUREAU

G W Owens A Bureau B Uppfeldt C Muller J Chica G W Owens

SCI CTICM SBI RWTH Labein SCI

10/04/06 10/04/06 10/04/06 10/04/06 10/04/06 18/09/06

4. Germany 5. Spain Resource approved by Technical Coordinator TRANSLATED DOCUMENT This Translation made and checked by: Translated resource approved by:

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