Document Ref: SX029a-EN-EU Sheet 1 of 28 Title Example: Elastic analysis of a single bay portal frame Eurocode Ref Made by Valérie Lemaire Date April 2006 CALCULATION SHEET Checked by Alain Bureau Date April 2006 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 5,988 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 Created on Monday, January 19, 2009 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Access Steel Licence Agreement
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Document Ref: SX029a-EN-EU Sheet 1 of 28 Title
Example: Elastic analysis of a single bay portal frame
Eurocode Ref Made by Valérie Lemaire Date April 2006
CALCULATION SHEET
Checked by Alain Bureau Date April 2006
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 frameC
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Document Ref: SX029a-EN-EU Sheet 2 of 28 Title
Example: Elastic analysis of a single bay portal frame
Eurocode Ref Made by Valérie Lemaire Date April 2006
CALCULATION SHEET
Checked by Alain Bureau Date April 2006
2 Loads
EN 1991-1-12.1 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/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]
Example: Elastic analysis of a single bay portal frameC
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y, J
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9, 2
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pyrig
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all
right
s re
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ed. U
se o
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the
term
s an
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of th
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Document Ref: SX029a-EN-EU Sheet 3 of 28 Title
Example: Elastic analysis of a single bay portal frame
Eurocode Ref Made by Valérie Lemaire Date April 2006
CALCULATION SHEET
Checked by Alain Bureau Date April 2006
2.3 Wind loads EN 1991-1-4Characteristic values for wind loading in kN/m for an internal frame
Example: Elastic analysis of a single bay portal frame
Eurocode Ref Made by Valérie Lemaire Date April 2006
CALCULATION SHEET
Checked by Alain Bureau Date April 2006
5.1 Buckling amplification factor α EN 1993-1-1cr § 5.2.1In 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: γ G + γmax Q QS (combination 101).
For this combination, the amplification factor is: αcr = 14,57
The first buckling mode is shown hereafter.
EN 1993-1-1 So : αcr = 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)
310204,3866,0740,0
2001 −⋅=×× φ = φ0 αh α = m
where φ0 = 1/200
740,030,7
22==
h αh =
866,0)11(5,0 =+m
m = 2 (number of columns) =α m
Sway imperfections may be disregarded where H ≥ 0,15 V EN 1993-1-1 Ed Ed.
§ 5.3.2The effects of initial sway imperfection may be replaced by equivalent horizontal forces:
(4)
H = φ V in the combination where H < 0,15 ⎢Veq Ed Ed Ed ⎢
Example: Elastic analysis of a single bay portal frameC
Example: Elastic analysis of a single bay portal frame
Eurocode Ref Made by Valérie Lemaire Date April 2006
CALCULATION SHEET
Checked by Alain Bureau Date April 2006
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:
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)
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
Example: Elastic analysis of a single bay portal frameC
Example: Elastic analysis of a single bay portal frame
Eurocode Ref Made by Valérie Lemaire Date April 2006
CALCULATION SHEET
Checked by Alain Bureau Date April 2006
5.3.2 Ultimate limit states Moment diagram in kNm
Combination 101:
Combination 102:
Combination 103:
Combination 104:
Example: Elastic analysis of a single bay portal frameC
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Document Ref: SX029a-EN-EU Sheet 9 of 28 Title
Example: Elastic analysis of a single bay portal frame
Eurocode Ref Made by Valérie Lemaire Date April 2006
CALCULATION SHEET
Checked by Alain Bureau Date April 2006
Combination 105:
Combination 106:
6 Column verification
Profile IPE 600 - S275 (ε = 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 Classification of the cross section • Web: The web slenderness is c / tw = 42,83 EN 1993-1-1
§ 5.5mm94,48
27512161500
yw
EdN =
×==
ftN
d
548,05142
94,485142 w
Nw =×+
=+
=d
ddα > 0,50
49,591548,013
92,0396=
−×× The limit for Class 1 is : 396ε / (13α − 1) =
Then : c / tw = 42,83 < 59,49 The web is class 1.
Example: Elastic analysis of a single bay portal frameC
The effect of the shear force on the moment resistance may be neglected.
Verification to axial force EN 1993-1-1 § 6.2.4-3 N = A fpl,Rd y / γ = (15600 × 275/1,0).10M0 N = 4290 kN pl,Rd N = 161,5 kN < 0,25 NEd pl,Rd = 4290 x 0,25 = 1073 kN EN 1993-1-1
and NEd = 161,5 kN < 3,92710001
275125625,05,0
M0
yww =×
×××=
γfth
§ 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,y fy / γ = (3512 × 275/1,0).10 M0
M = 965,8 kNm pl,y,Rd
My,Ed = 755 kNm < M = 965,8 kNm pl,y,Rd
Example: Elastic analysis of a single bay portal frameC
Example: Elastic analysis of a single bay portal frame
Eurocode Ref Made by Valérie Lemaire Date April 2006
CALCULATION SHEET
Checked by Alain Bureau Date April 2006
6.3 Buckling resistance The buckling resistance of the column is sufficient if the following conditions are fulfilled (no bending about the weak axis, M
EN 1993-1-1z,Ed = 0):
§ 6.3.3
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
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 (αcr = 14,57 > 10). Then the buckling 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. 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 :
L = 5,99 m for buckling about the weak axis cr,z L = 5,99 m for torsional buckling cr,T and L = 5,99 m for lateral torsional buckling cr,LT
• Buckling about yy Lcr,y = 5,99 m
EN 1993-1-1 Buckling curve : a (αy = 0,21) § 6.3.1.2
10005990
10000920802100002
22
ycr,
y2ycr, ×
××== ππ
LEI
N (2)
=53190kN Table 6.1
284,0
10.5319027515600
3ycr,
yy =
×==
NAf
λ EN 1993-1-1 § 6.3.1.3 (1)
Example: Elastic analysis of a single bay portal frameC
Example: Elastic analysis of a single bay portal frame
Eurocode Ref Made by Valérie Lemaire Date April 2006
CALCULATION SHEET
Checked by Alain Bureau Date April 2006
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 kN Ed V = 118,50 kN Ed M = 349,10 kNm y,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 § 6.2 A = A - 2btv f + (tw+2r)t η = 1 f
598516)2122,10(16200211550 =××++××−=vA mm2
EN 1993-1-1 Av > η.hw.tw = 468×10,2 = 4774mm2
§ 6.2.8 (2)
3 3Vpl,Rd = Av (f / ) /γ = 5985×275/ /1000 = 950,3 kN y 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 kN Rd N = 136 kN < 0,25 NEd pl, = 3176 × 0,25 = 794,1 kN Rd and EN 1993-1-1
kN4,65610001
2752,104685,05,0
M0
yww =×
×××=
γfth § 6.2.8NEd = 136 kN < (2)
⇒ its effect on the moment resistance may be neglected!
Example: Elastic analysis of a single bay portal frameC
Example: Elastic analysis of a single bay portal frame
Eurocode Ref Made by Valérie Lemaire Date April 2006
CALCULATION SHEET
Checked by Alain Bureau Date April 2006
Bending EN 1993-1-1 § 6.2.5 M = 2194 × 275/1000 = 603,4 kNm pl,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 ≤+
γχ
γχ M
Mk
NN
and 1
M1
Rky,LT
Edy,zy
M1
Rkz
Ed ≤+
γχγ
χ MM
kN
N
• Buckling about yy:
For the determination of the buckling length about yy, a buckling analysis is performed to calculate the buckling amplification factor αcr for 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 6.1 Combination 101 ⇒ αcr = 37,37
EN 1993-1-1 § 6.3.1.3 (1)
EN 1993-1-1 Buckling curve : a (h/b>2) ⇒ αy = 0,21 § 6.3.1.2
kNNN 508213637,37Edcrycr, =×=α= (2)
Table 6.1
7906,010.508227511550
3ycr,
yy =
×==λ
NAf
Example: Elastic analysis of a single bay portal frameC
Example: Elastic analysis of a single bay portal frame
Eurocode Ref Made by Valérie Lemaire Date April 2006
CALCULATION SHEET
Checked by Alain Bureau Date April 2006
( ) ⎥⎦⎤
⎢⎣⎡ +−+=
2yyy 2,015,0 λλαφ
= 0,8745 ( )[ ]2y 7906,02,07906,021,015,0 +−×+×=φ
8011,07906,08745,08745,0
11222
y2
yy
y =−+
=−+
=λφφ
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
L = 6,00 m § 6.3.1.3cr,z
10006000100002141210000
22
2zcr,
z2zcr, ×
××== ππ
LEIN
= 1233kN
NCCI Torsional buckling
Lcr,T = 6,00 m SN003
)( 2Tcr,
w2
t0
Tcr, LEIGI
IAN π
+=
with yo = 0 and zo = 0 (doubly symmetrical section)
cm450340214148199)( 20
20zy0 =+=+++= AzyIII
)
600010.124937021000010.29,8980770(
100010.5034011550
2
624
4Tcr,×
+×××
= πN
Ncr,T = 3305 kN
Ncr = min ( N ; Ncr,z cr,T ) = 1233 kN EN 1993-1-1 § 6.3.1.3
605,110.123327511550
3cr
yz =
×==
NAf
λ (1)
Example: Elastic analysis of a single bay portal frameC
Example: Elastic analysis of a single bay portal frame
Eurocode Ref Made by Valérie Lemaire Date April 2006
CALCULATION SHEET
Checked by Alain Bureau Date April 2006
z
y
zy
ycr,
Ed
zmLTmyzy 6,01
1 ww
CNNCCk
−=
μ
5859,050,1
138,16,09011,01
50821361
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 ≤+
γχ
γχ M
MkN
N (6.61)
8131,0
127510.21948503,0
10.1,349116,1
1275115508011,0
1360003
6=
××
×+×
× < 1 OK
1
M1
Rky,LT
Edy,zy
M1
Rkz
Ed ≤+
γχγ
χ MM
kNN
(6.62)
5385,0
127510.21948503,0
10.1,3495859,0
1275115503063,0
1360003
6=
××
×+×
×< 1 OK
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:
N = 139,2 kN EdV = 151,3 kN EdM = 755 kNm Ed
Example: Elastic analysis of a single bay portal frameC