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EN 3-2005 Example 002
WIDE FLANGE SECTION UNDER BENDING
EXAMPLE DESCRIPTION
The frame object axial strengths are tested in this example.
A beam is subjected to factored load N = 1050 kN. This example was tested
using the Eurocode 3-2005 steel frame design code. The design capacities are
compared with independent hand calculated results.
GEOMETRY, PROPERTIES AND LOADING
NEd
A
A
L/2 L/2 Section A-A
L = 1.4 m
Material Properties Loading Design Properties
E = 210x103 MPa N = 1050 kN fy = 275 MPa
v = 0.3 Section: 406x178x74 UB
G = 80770 MPa
TECHNICAL FEATURES TESTED
Section compactness (beam)
Section shear capacity
Section bending capacity with shear reduction
EN 3-2005 Example 002 - 1
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RESULTS COMPARISON
Independent results are taken from hand calculations based on the CSI steel
design documentation contained in the file “SFD-EC-3-2005.pdf,” which is
available through the program “Help” menu. Examples were taken from Example
6.5 on pp. 53-55 from the book “Designers’ Guide to EN1993-1-1” by R.S.
Narayanan & A. Beeby.
Percent
Output Parameter SAP2000 Independent Difference
Compactness Class 1 Class 1 0.00%
Section Shear Resistance,
689.2 689.2 0.00%
Vpl,Rd (kN)
Section Bending Resistance,
412.8 412.8 0.00%
Mc,y,Rd (kN-m)
Section Shear-Reduced Bending
386.8 386.8 0.00%
Resistance, Mv,y,Rd (kN-m)
COMPUTER FILE: EN 3-2005 EX002
CONCLUSION
The results show an exact comparison with the independent results.
EN 3-2005 Example 002 - 2
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HAND CALCULATION
Properties:
Material: S275 Steel
fy = 275 MPa
E = 210000 MPa
Section: 406x178x74 UB
A = 9450 mm2
b = 179.5 mm, tf = 16 mm, h = 412.8 mm, tw = 9.5 mm, r = 10.2 mm
hw = h − 2t f = 412.8 − 2 • 16 = 380.8 mm
d = h − 2 ( t f + r ) = 412.8 − 2 • (16 + 10.2 ) = 360.4 mm
b − tw − 2r 179.5 − 9.5 − 2 • 10.2
=c = = 74.8 mm
2 2
Wpl,y = 501,000 mm3
Other:
γ M 0 = 1.0
η = 1.2
Loadings:
N Ed = 0 kN
N = 1050 kN @ mid-span
Results in the following internal forces:
VEd = 525 kN
M Ed = 367.5 kN-m
EN 3-2005 Example 002 - 3
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Section Compactness:
235 235
=ε = = 0.924
fy 275
Localized Buckling for Flange:
λ cl .1 = 9ε = 9 • 0.924 = 8.32 for pure compression
c 74.8
λe = = = 4.68
tf 16
=
λ e 4.68 < λ=
cl .1 8.32
So Flange is Class 1 in pure compression
Localized Buckling for Web:
λ cl .1= 72ε= 72 • 0.924= 66.56 for pure bending
d 360.4
λe = = = 37.94
tw 9.5
=
λ e 37.94 < =
λ cl .1 66.56
So Web is Class 1 in pure bending
Since Flange & Web are Class 1, Section is Class 1.
Section Shear Capacity
Av − min = η h wtw = 1.2 • 380.8 • 9.5 = 4341mm 2
Av = A − 2bt f + (tw + 2r )t f = 9450 − 2 • 179.5 • 16 + ( 9.5 + 2 • 10.2 ) • 16
=Av 4021.2 mm 2 < Av − min
So Av = 4341mm 2
EN 3-2005 Example 002 - 4
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Av f y 4341 275
= = = 689, 245 N
γ M 0 3 1.0 3
V pl , Rd
V pl , Rd = 689.2 kN
Section Bending Capacity
W pl , y f y 1501, 000 • 275
=
M = = 412, 775, 000 N-mm
c , y , Rd
γM0 1
M c , y , Rd = 412.8 kN-m
With Shear Reduction:
2
2VEd 2 • 525 2
=
ρ − 1= − 1= 0.27
V pl , Rd 689.2
Aw = hwtw = 380.8 • 9.5 = 3617.6 mm 2
fy ρ Aw 2 275 0.27 • 3617.62
=
M v , y , Rd W − = 1,501, 000 −
γ M 0 pl , y 4tw 1.0 4 • 9.5
M v , y , Rd = 386,829, 246 N-mm
M v , y , Rd = 386.8 kN-m
EN 3-2005 Example 002 - 5
