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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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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
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
3 Load combinations EN 1990
3.1 Partial safety factor
• γGmax = 1,35 (permanent loads) • γGmin = 1,0 (permanent loads) EN 1990
Table A1.1• γQ = 1,50 (variable loads)
• ψ0 = 0,50 (snow)
• ψ0 = 0,60 (wind)
= 1,0 • γM0
= 1,0 • γM1
3.2 ULS Combinations EN 1990 Combination 101 : γGmax G + γ QsQ
Combination 102 : γGmin G + γQ Qw
Combination 103 : γGmax G + γ QQ s + γQ ψ0
Combination 104 : γGmin G + γQ Qs + γQ ψ0 Qw
Combination 105 : γGmax G + γ ψQ 0 Qs + γQ Qw
Combination 106 : γGmin G + γQ ψ0 Qs + γQ Qw
3.3 SLS Combinations EN 1990 Combinations and limits should be specified for each project or by National Annex.
Example: Elastic analysis of a single bay portal frameC
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Document Ref: SX029a-EN-EU Sheet 4 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
4 Sections
hw
y y
z
z
tf
tw
b
h
4.1 Column Try IPE 600 – Steel grade S275
Depth h = 600 mm
Web Depth hw = 562 mm
Depth of straight portion of the web
dw = 514 mm
Width b = 220 mm
Web thickness tw = 12 mm
Flange thickness t = 19 mm f
Fillet r = 24 mm
Mass 122,4 kg/m
Section area A = 156 cm2
Second moment of area /yy Iy = 92080 cm4
Second moment of area /zz Iz = 3386cm4
Torsion constant I = 165,4 cm4t
Warping constant Iw = 2845500 cm6
Elastic modulus /yy Wel,y = 3069 cm3
Plastic modulus /yy Wpl,y = 3512 cm3
Elastic modulus /zz W = 307,80 cm3el,z
Plastic modulus /zz Wpl,z = 485,60 cm3
Example: Elastic analysis of a single bay portal frameC
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Acc
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Document Ref: SX029a-EN-EU Sheet 5 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
4.2 Rafter Try IPE 500 – Steel grade S275
Depth h = 500 mm
Web Depth hw = 468 mm
Depth of straight portion of the web
dw = 426 mm
Width b = 200 mm
Web thickness tw = 10,2 mm
Flange thickness tf = 16 mm
Fillet r = 21 mm
Mass 90,7 kg/m
Section area A = 115,50 cm2
Second moment of area /yy Iy = 48200 cm4
Second moment of area /zz Iz = 2141 cm4
Torsion constant It = 89,29 cm4
Warping constant Iw = 1249400 cm6
Elastic modulus /yy Wel,y = 1928 cm3
Plastic modulus /yy Wpl,y = 2194 cm3
Elastic modulus /zz Wel,z = 214,1 cm3
Plastic modulus /zz Wpl,z = 335,90 cm3
5 Global 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.
Example: Elastic analysis of a single bay portal frameC
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Document Ref: SX029a-EN-EU Sheet 6 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
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
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Document Ref: SX029a-EN-EU Sheet 7 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
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
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Acc
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Document Ref: SX029a-EN-EU Sheet 8 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
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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men
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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
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Document Ref: SX029a-EN-EU Sheet 10 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
• Flange: The flange slenderness is c / tf = 80 / 19 = 4,74 The limit for Class 1 is : 9 ε = 9 × 0,92 = 8,28
Then : c / t = 4,74 < 8,28 The flange is Class 1 f
So the section is Class 1. The verification of the member will be based on the plastic resistance of the cross-section.
6.2 Resistance of cross section Verification for shear force Shear area : A = A - 2btv f + (tw+2r)t > η.hf w.t EN 1993-1-1w
§ 6.2.6 η may be conservatively taken equal to 1 (3)
838019)24212(19220215600 =××++××−=vA mm2 > η.hw.tw = 6744 mm2
Vpl,Rd = Av (fy / 3 ) /γM0 = (8380×275/ 3 ).10-3
V = 1330 kN pl,Rd
V / VEd pl,Rd = 122,40/1330 = 0,092 < 0,50
The effect of the shear force on the moment resistance may be neglected.
Verification to axial force EN 1993-1-1 § 6.2.4-3 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
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gree
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Document Ref: SX029a-EN-EU Sheet 11 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
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
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gree
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Document Ref: SX029a-EN-EU Sheet 12 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
( )[ ] ( )[ ]22yyy 284,02,0284,021,015,02,015,0 +−+×=+−+= λλαφ EN 1993-1-1 = 0,5491
§ 6.3.1.2 (1)
9813,0284,05491,05491,0
11222
y2
yy
y =−+
=−+
=λφφ
χ
• Buckling about zz L = 5,99 m EN 1993-1-1 cr,z § 6.3.1.2 Buckling curve : b (αz = 0,34)
(2)
Table 6.1
10005990100003386210000
22
2zcr,
z2zcr, ×
××== ππ
LEI
N = 1956 kN
EN 1993-1-1 481,1
10.195627515600
3zcr,
yz =
×==
NAf
λ § 6.3.1.3 (1)
( )[ ] ( )[ ]22zzz 481,12,0481,134,015,02,015,0 +−+×=+−+= λλαφ EN 1993-1-1 = 1,814
§ 6.3.1.2 (1)
3495,0481,1814,1814,1
11222
z2zz
z =−+
=−+
=λφφ
χ
• Lateral torsional buckling Lcr,LT = 5,99 m
EN 1993-1-1 Buckling curve : c (αLT = 0,49) § 6.3.2.3 Moment diagram with linear variation : ψ = 0 C1 = 1,77 Table 6.5
Z2
t2
LTcr,
Z
W2
LTcr,
Z2
1cr EIGIL
II
LEICM
ππ
+=
NCCI SN003
42
42
4
6
62
2
cr 10.338621000010.4,165807705990
10.338610.2845500
10599010000338621000077,1
××××
+×
××××=
ππM
Mcr = 1351 kNm
8455,010.1351
27510.35126
3
cr
yypl,LT =
×==
MfW
λ
( )[ ]2
LTLT,0LTLTLT 15,0 βλλλαφ +−+= EN 1993-1-1 § 6.3.2.3 (1)
and β =0,75 40,0LT,0 =λ with
Example: Elastic analysis of a single bay portal frameC
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Document Ref: SX029a-EN-EU Sheet 13 of 28 Title
Example: Elastic analysis of a single bay portal frame
Eurocode Ref Made by Valérie Lemaire Date April 2006
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( )[ ] 8772,08455,075,04,08455,049,015,0 2LT =×+−+×=φ
7352,08455,075,08772,08772,0
11222
LT2LTLT
LT =×−+
=−+
=βλφφ
χ
7519,033,033,1
1=
− ψEN 1993-1-1 k (ψ = 0) c = § 6.3.2.3
( )[ ]2LTc 8,021)1(5,01 −−−×−= λkf
(2) Table 6.6
( )[ ] 8765,08,08455,021)7519,01(5,01 2 =−−−×−=f < 1
8388,08765,07352,0
==fLTχ χLT,mod = < 1
and kCalculation of the factors kyy zy according to Annex A of EN 1993-1-1 EN 1993-1-1 Annex A
9999,0
531905,1619813,01
531905,1611
1
1
ycr,
Edy
ycr,
Ed
y =×−
−=
−
−
=
NN
NN
χμ
9447,0
19565,1613495,01
19565,1611
1
1
zcr,
Edz
zcr,
Ed
z =×−
−=
−
−=
NN
NN
χμ
EN 1993-1-1 144,1
30693512
yel,
ypl,y ===
WW
w < 1,5 Annex A
578,1
8,3076,485
zel,
zpl,z ===
WW
w > 1,5 ⇒ w z = 1,5
NCCI Critical axial force in the torsional buckling mode
SN003)( 2
Tcr,
w2
t0
Tcr,L
EIGIIAN π
+=
For a doubly symmetrical section,
cm4 95466338692080)( 20
20zy0 =+=+++= AzyIII
Example: Elastic analysis of a single bay portal frameC
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Document Ref: SX029a-EN-EU Sheet 14 of 28 Title
Example: Elastic analysis of a single bay portal frame
Eurocode Ref Made by Valérie Lemaire Date April 2006
CALCULATION SHEET
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⎟⎟⎠
⎞⎜⎜⎝
⎛ ×+××
×= 2
624
4cr, 599010.284550021000010.4,16580770
100010.9546615600 πTN
Ncr,T = 4869 kN
NCCI
Z2
t2
LTcr,
Z
W2
LTcr,
Z2
1cr,0 EIGIL
II
LEICM
ππ
+= SN003
0λMcr,0 is the critical moment for the calculation of for uniform bending moment as specified in Annex A.
⇒ C1 = 1
42
42
4
6
62
42
cr,0 10.338621000010.4,165807705990
10.338610.2845500
10599010.33862100001
××××
+×××
×=π
πM
kNmM 3,763ocr, =
EN 1993-1-1 6
3
ocr,
yypl,0
10.3,76327510.3512 ×
==M
fWλ = 1,125 Annex A
4
TFcr,
Ed
zcr,
Ed1lim0 )1)(1(2,0
NN
NN
C −−=λ
with N = Ncr,TF cr,T (doubly symmetrical section)
4lim0 )4869
5,1611)(1956
5,1611(77,12,0 −−=λ = 0,2582
0λ > lim0λ
LTy
LTymy,0my,0my 1
)1(a
aCCC
ε
ε
+−+=
yel,Ed
Edy,y W
AN
M=ε
y
tLT I
Ia −=1= 23,76 (class 1) and = 0,9982 with
Example: Elastic analysis of a single bay portal frameC
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uesd
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r 14
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icen
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gree
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t
Document Ref: SX029a-EN-EU Sheet 15 of 28 Title
Example: Elastic analysis of a single bay portal frame
Eurocode Ref Made by Valérie Lemaire Date April 2006
CALCULATION SHEET
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Calculation of the factor C my,0
EN 1993-1-1
ycr,
Edyymy,0 )33,0(36,021,079,0
NN
C −++= ψψ Annex A
Table A2
ycr,
Edmy,0 1188,079,0
NN
C −= = 0,7896 0y =ψ
Calculation of the factors C and Cmy m,LT :
LTy
LTymy,0my,0my 1
)1(a
aCCC
ε+
ε−+=
9641,09982,076,231
9982,076,23)7896,01(7896,0Cmy =
×+
×−+=
EN 1993-1-1 1
)1)(1(Tcr,
Ed
zcr,
Ed
LT2mymLT ≥
−−
=
NN
NN
aCC Annex A
9843,0)
48695,1611)(
19565,1611(
9982,09641,0 2mLT =
−−×=C < 1
⇒ C = 1 mLT
Calculation of the factors Cyy and C : EN 1993-1-1 zy Annex A
ypl,
yel,LTpl
2max
2my
y
max2my
yyyy ])6,16,12)[(1(1
WW
bnCw
Cw
wC ≥−λ−λ−−+=
03765,01/27515600
161500/ M1Rk
Edpl =
×==
γNNn
Mz,Ed = 0 ⇒ and 0LT =b 0LT =d 4810,1zmax =λ=λ
]03765,0)481,19641,0144,1
6,1481,19641,0144,1
6,12[()1144,1(1 222yy ×××−××−×−+=C
8739,035123069
ypl,
yel, ==>WW
Cyy = 0,9849
Example: Elastic analysis of a single bay portal frameC
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uesd
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icen
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gree
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t
Document Ref: SX029a-EN-EU Sheet 16 of 28 Title
Example: Elastic analysis of a single bay portal frame
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ypl,
yel,
z
yLTpl
2max
2my5
yyzy 6,0])142)[(1(1
WW
ww
dnCw
wC ≥−λ−−+=
9318,0]03765,0)481,19641,0144,1142)[(1144,1(1 22
5zy =×××−−+=C
4579,035123069
5,1144,16,06,0
ypl,
yel,
z
y ==WW
ww
>= 9318,0zyC
Calculation of the factors kyy and kzy : EN 1993-1-1 Annex A
yy
ycr,
Ed
ymLTmyyy
1
1 CNNCCk
−
μ=
9818,09849,01
531905,1611
9999,019641,0yy =×−
××=k
z
y
zy
ycr,
Ed
zmLTmyzy 6,01
1 ww
CNNCCk
−=
μ
5138,050,1
144,16,09318,01
531905,1611
9447,019641,0zy =×××−
××=k
Verification with interaction formulae
EN 1993-1-1 1
M1
Rky,LT
Edy,
M1
Rky
Ed ≤+
γχ
γχ M
MkN
Nyy
§ 6.3.3
9534,0
127510.35128388,0
10.7559818,0
1275156009813,0
1615003
6=
××
×+×
× <1 OK
1
M1
Rky,LT
Edy,zy
M1
Rkz
Ed ≤+
γχγ
χ MM
kNN
Example: Elastic analysis of a single bay portal frameC
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Acc
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icen
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gree
men
t
Document Ref: SX029a-EN-EU Sheet 17 of 28 Title
Example: Elastic analysis of a single bay portal frame
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5867,0
127510.35128388,0
10.7555138,0
1275156003495,0
1615003
6=
××
×+×
×<1 OK
So the buckling resistance of the member is satisfactory.
7 Rafter verification
7.1 Classification Case with maximum compression in beam: (Combination 101)
• Web: h = 468 mm w
tw = 10,2 mm c / tw = 41,76
c = 426 mm
NEd= 136 kN mmft
Nd 5,482752,10
136000
yw
EdN =
×==
557,04262
5,484262
N =×+
=+
=dddα > 0,5
c / tw = 41,76 < 38,581557,013
92,0396113
396=
−××
=−αε ⇒ class 1
EN 1993-1-1 § 5.5
• Flanges b = 200 mm tf = 16 mm r = 21 mm c / tf = 4,44 c = 71 mm
part to compression
c / tf < 9ε = 8,28 (S275 ⇒ ε = 0,92 )
c / t = 4,44 class 1 f
So the section is Class 1. The verification of the member will be based on the plastic resistance of the cross-section.
Example: Elastic analysis of a single bay portal frameC
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gree
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Document Ref: SX029a-EN-EU Sheet 18 of 28 Title
Example: Elastic analysis of a single bay portal frame
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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: 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
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Acc
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gree
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Document Ref: SX029a-EN-EU Sheet 19 of 28 Title
Example: Elastic analysis of a single bay portal frame
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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
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Acc
ess
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el L
icen
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gree
men
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Document Ref: SX029a-EN-EU Sheet 20 of 28 Title
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( ) ⎥⎦⎤
⎢⎣⎡ +−+=
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
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ed o
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uesd
ay, S
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gree
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Document Ref: SX029a-EN-EU Sheet 21 of 28 Title
Example: Elastic analysis of a single bay portal frame
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Buckling curve : b EN 1993-1-1 § 6.3.1.2αz = 0,34 (1)
( )[ ]2zzz 2,015,0 λλαφ +−+=
Table 6.1
( )[ ]2
z 605,12,0605,134,015,0 +−×+×=φ =2,027
3063,0605,1027,2027,2
11222
z2zz
z =−+
=−+
=λφφ
χ
• Lateral torsional buckling : EN 1993-1-1 § 6.3.1.3 L , = 6,00 m cr LT
Table 6.5 = 0,49 Buckling curve : c αLT
Moment diagram along the part of rafter between restraints:
Combination 101
NCCI Calculation of the critical moment:
SN003 ψ = - 0,487
MqL8
2
q = - 9,56 kN/m μ = = - 0,123
⇒ C1 = 2,75
NCCI
Z2
t2
LTcr,
Z
W2
LTcr,
Z2
1cr EIGIL
II
LEICM
ππ
+=
42
42
4
6
62
42
cr 10.214121000010.29,89807706000
10.214110.1249400
10600010214121000075,2
××××
+×
××××=
ππM
Mcr = 1159 kNm
Example: Elastic analysis of a single bay portal frameC
reat
ed o
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uesd
ay, S
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r 14
, 201
0T
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Acc
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icen
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gree
men
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Document Ref: SX029a-EN-EU Sheet 22 of 28 Title
Example: Elastic analysis of a single bay portal frame
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CALCULATION SHEET
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7215,0
10.115927510.21956
3
cr
yypl,LT =
×==
MfW
λ
( )[ ]2
LTLT,0LTLTLT 15,0 βλλλαφ +−+=
EN 1993-1-1 § 6.3.2.3 with 40,0LT,0 =λ and β =0,75 (1)
( )[ ] 7740,07215,075,04,07215,049,015,0 2LT =×+−×+×=φ
8125,07215,075,07740,07740,0
11222
LT2LTLT
LT =×−+
=−+
=λβφφ
χ
kc = 0,91
( )[ ]28,021)1(5,01 −−−×−= LTckf λ
EN 1993-1-1 § 6.3.2.3
( )[ ] 9556,08,07215,021)91,01(5,01 2 =−×−×−×−=f (2)
< 1 Table 6.6
8503,09556,08125,0
==fLTχ χLT,mod = < 1
Combination 101 N = 136 kN compression Ed
M = 349,10 kNm y,Ed
M z,Ed = 0
Section class1 ⇒ ΔM = 0 et ΔMy,Ed z,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
Example: Elastic analysis of a single bay portal frameC
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uesd
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icen
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gree
men
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Document Ref: SX029a-EN-EU Sheet 23 of 28 Title
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EN 1993-1-1
9946,0
50821368011,01
50821361
1
1
ycr,
Edy
ycr,
Ed
y =×−
−=
−
−=
NN
NN
χμ
Annex A
9208,0
12331363063,01
12331361
1
1
zcr,
Edz
zcr,
Ed
z =×−
−=
−
−=
NN
NN
χμ
EN 1993-1-1 138,1
19282194
yel,
ypl,y ===
WW
w < 1,50 Annex A
569,1
1,2149,335
zel,
zpl,z ===
WW
w > 1,50 ⇒ w = 1,5 z
NCCI
Z2
t2
LTcr,
Z
W2
LTcr,
Z2
1cr,0 EIGIL
II
LEICM
ππ
+= SN003
0λMcr,0 is the critical moment for the calculation of for uniform bending moment as specified in Annex A.
⇒ C1 = 1
42
42
4
6
62
42
cr,0 10.214121000010.29,89807706000
10.214110.1249400
10600010.21412100001
××××
+×××
×=π
πM
kNmM 5,421ocr, =
EN 1993-1-1 6
3
ocr,
yypl,0
10.5,42127510.2195 ×
==M
fWλ = 1,196
Annex A
4
TFcr,
Ed
zcr,
Ed1lim0 )1)(1(2,0
NN
NNC −−=λ with C1 = 2,75
with N = Ncr,TF cr,T (doubly symmetrical section)
4lim0 )33051361)(
12331361(75,22,0 −−=λ = 0,3187
0λ =1,196 > lim0λ =0,3187
Example: Elastic analysis of a single bay portal frameC
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uesd
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Document Ref: SX029a-EN-EU Sheet 24 of 28 Title
Example: Elastic analysis of a single bay portal frame
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EN 1993-1-1
LTy
LTymy,0my,0my 1
)1(a
aCCC
ε
ε
+−+= Annex A
with 3
6
yel,Ed
Edy,y 101928
11550136000
10.10,349×
×==W
AN
Mε =15,38 (class 1)
and 48200
29,8911y
tLT −=−=
IIa = 0,9981
Calculation of the factor C EN 1993-1-1 my,0 Annex AMoment diagram along the rafter: Table A2
My,Ed = maximum moment along the rafter = 755kNm
δ = maximum displacement along the rafter = 179mm 30m
ycr,
Ed
Edy,2
xy2
my,0 11NN
MLEI
C⎥⎥⎦
⎤
⎢⎢⎣
⎡−+=
δπ
50821361
10755300001791048200210000
1 62
42
my,0⎥⎥⎦
⎤
⎢⎢⎣
⎡−
××××××
+=π
C =0,9803
Calculation of the factors C and C my m,LT :
LTy
LTymy,0my,0my 1
)1(a
aCCC
ε
ε
+−+=
996,09982,038,151
9982,038,15)9803,01(9803,0my =×+
×−+=C
Example: Elastic analysis of a single bay portal frameC
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uesd
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gree
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Document Ref: SX029a-EN-EU Sheet 25 of 28 Title
Example: Elastic analysis of a single bay portal frame
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EN 1993-1-1 1
)1)(1(Tcr,
Ed
zcr,
Ed
LT2mymLT ≥
−−=
NN
NN
aCC Annex A
072,1)
33051361)(
12331361(
9981,0996,0 2mLT =
−−×=C > 1
Calculation of the factors Cyy and C EN 1993-1-1 zyAnnex A
ypl,
yel,LTpl
2max
2my
y
max2my
yyyy ])6,16,12)[(1(1
WW
bnCw
Cw
wC ≥−−−−+= λλ
0428,01/27511550
136000/ M1Rk
Edpl =
×==
γNNn
M 605,1max == zλλz,Ed = 0 ⇒ and 0LT =b 0LT =d
]0428,0)605,1996,0138,1
6,1605,1996,0138,1
6,12)[(1138,1(1 222yy ×××−××−−+=C
Cyy = 0,9774
ypl,
yel,
z
yLTpl
2max
2my5
yyzy 6,0])142)[(1(1
WW
ww
dnCw
wC ≥−−−+= λ
9011,0]0428,0)605,1996,0138,1142)[(1138,1(1 22
5zy =×××−−+=C
Calculation of the factors kyy and kzy : EN 1993-1-1 Annex A
yy
ycr,
Ed
ymLTmyyy
1
1 CNNCCk
−=
μ
116,19774,01
50821361
9946,0072,1996,0yy =×−
××=k
Example: Elastic analysis of a single bay portal frameC
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ect t
o th
e te
rms
and
cond
ition
s of
the
Acc
ess
Ste
el L
icen
ce A
gree
men
t
Document Ref: SX029a-EN-EU Sheet 26 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
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
reat
ed o
n T
uesd
ay, S
epte
mbe
r 14
, 201
0T
his
mat
eria
l is
copy
right
- a
ll rig
hts
rese
rved
. Use
of t
his
docu
men
t is
subj
ect t
o th
e te
rms
and
cond
ition
s of
the
Acc
ess
Ste
el L
icen
ce A
gree
men
t
Document Ref: SX029a-EN-EU Sheet 27 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
Properties of the whole section:
The calculation of elastic section properties for this case is approximate, ignoring the middle flange.
1000 mm
200 mm
Section area A = 160,80 cm2
Second moment of area /yy Iy = 230520 cm4
Second moment of area /zz Iz = 2141 cm4
Elastic modulus /yy Wel,y = 4610 cm3
Elastic modulus /zz W = 214 cm3el,z
Properties of the compression part:
Section at the mid-length of the haunch including 1/6th of the web depth
Section area A = 44 cm2 120 mm
Second moment of area /yy Iy = 554 cm4
Second moment of area /zz Iz =1068 cm4
⇒ cmi 93,444
1068z == 200 mm
7044,0
39,8630,493000
1z
fz =
×==
λλ
iL z
Buckling of welded I section with h/b > 2 :
⇒ curve d ⇒ α = 0,76
( )[ ] ( )[ ] 9397,07044,02,07044,076,015,02,015,0 22zzz =+−×+×=+−+= λλαφ
640,07044,09397,09397,0
11222
z2zz
z =−+
=−+
=λφφ
χ
Example: Elastic analysis of a single bay portal frameC
reat
ed o
n T
uesd
ay, S
epte
mbe
r 14
, 201
0T
his
mat
eria
l is
copy
right
- a
ll rig
hts
rese
rved
. Use
of t
his
docu
men
t is
subj
ect t
o th
e te
rms
and
cond
ition
s of
the
Acc
ess
Ste
el L
icen
ce A
gree
men
t
Document Ref: SX029a-EN-EU Sheet 28 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
Compression in the bottom flange:
kNN 7604400100010.4610
100075500016080440024,139 3fEd, =×
××
+×=
Verification of buckling resistance of the bottom flange:
981,02754400640,0
760000
Rkz
fEd, =××
=N
Nχ
< 1 OK
Example: Elastic analysis of a single bay portal frameC
reat
ed o
n T
uesd
ay, S
epte
mbe
r 14
, 201
0T
his
mat
eria
l is
copy
right
- a
ll rig
hts
rese
rved
. Use
of t
his
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Acc
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el L
icen
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gree
men
t
Example: Elastic analysis of a single bay portal frame SX029a-EN-EU
Quality Record
Example: Elastic analysis of a single bay portal frame RESOURCE TITLE
Reference(s) T2703
ORIGINAL DOCUMENT
Name Company Date
Created by Valérie LEMAIRE CTICM 25/10/2005
Technical content checked by Alain BUREAU CTICM 26/10/2005
Editorial content checked by
Technical content endorsed by the following STEEL Partners:
1. UK G W Owens SCI 10/04/06
2. France A Bureau CTICM 10/04/06
3. Sweden B Uppfeldt SBI 10/04/06
4. Germany C Muller RWTH 10/04/06
5. Spain J Chica Labein 10/04/06
Resource approved by Technical Coordinator
G W Owens SCI 18/09/06
TRANSLATED DOCUMENT
This Translation made and checked by:
Translated resource approved by:
Page 29
Example: Elastic analysis of a single bay portal frameC
reat
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uesd
ay, S
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r 14
, 201
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rved
. Use
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