Design of Members
Rui Simões
Department of Civil Engineering
University of Coimbra
Eurocodes ‐ Design of steel buildings with worked examples
Contents

Introduction

Design of columns

Design of beams

Design of beam‐columns
Brussels, 16 ‐ 17 October 2014
Eurocodes ‐ Design of steel buildings with worked examples
Brussels, 16 ‐ 17 October 2014
INTRODUCTION
Main internal forces and
combinations
Bending+Shear
Compression+Bending+Shear
Tension/Compression
Torsion – less common
Building – master example (Cardington - UK)
Eurocodes ‐ Design of steel buildings with worked examples
Brussels, 16 ‐ 17 October 2014
INTRODUCTION
Member design:
i) resistance of cross sections;
ii) member buckling resistance.
RESISTANCE OF CROSS SECTIONS
 Cross section classification - Class 1; Class 2; Class 3 and Class 4.
 Clause 6.2 of Eurocode 3, part 1.1 provides different
Vz
My
G
approaches, depending of cross section shape, cross section
class and type of internal forces (N, M+V, N+M+V,….):
– elastic criteria (clause 6.2.1(5));
 
x , Ed

 fy  M 0





2
 
z , Ed
 
 fy  M 0





2
NEd
My,Ed
 
x , Ed
 
 fy  M 0

 
z , Ed

 f y  M 0



  3   Ed

 fy  M 0






2
 1
Eurocodes ‐ Design of steel buildings with worked examples
Brussels, 16 ‐ 17 October 2014
INTRODUCTION
– linear summation of the utilization ratios – class 1/2/3 (clause 6.2.1(7));
M y , Ed
M z , Ed
N Ed


1
N Rd
M y , Rd
M z , Rd
– nonlinear interaction formulas – class 1/2 (clause 6.2.1(6)).
 Section properties – gross section, net section
(deduction for holes) or effective section (class 4 or shear
lag effects) (clause 6.2.2 of EC3-1-1).
MEMBER BUCKLING RESISTANCE
 Buckling resistance (clause 6.3 of Eurocode 3,
part 1.1) must be checked in all members
submitted to compressive stresses, which are:
– members under axial compression N;
– members under bending moment M;
– or under a combination of both (M+N).
Eurocodes ‐ Design of steel buildings with worked examples
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DESIGN OF COLUMNS
Column cross sections and applications
 Rolled open or closed
sections, welded sections or built-up sections – The
objective is to maximize the second moment of area in the relevant buckling
plan in order to maximize the buckling resistance.
Eurocodes ‐ Design of steel buildings with worked examples
Brussels, 16 ‐ 17 October 2014
DESIGN OF COLUMNS
NEd
Compression resistance (clause 6.2.4 of EC3-1-1)
NEd
 1.0
Nc , Rd
NEd is the design value of the axial compression;
fy
Nc,Rd is the design resistance to axial compression,
given by the minimum of:
i) Plastic resistance
A
N c , Rd  A f y  M 0
(class 1, 2 or 3)
N c , Rd  Aeff f y  M 0
(class 4)
Aeff - effective area
ii) Buckling resistance – Nb,Rd, in general the flexural buckling
resistance, which is analysed hereafter.
Eurocodes ‐ Design of steel buildings with worked examples
Brussels, 16 ‐ 17 October 2014
DESIGN OF COLUMNS
Column Buckling
 Flexural buckling is in general the buckling mode, which govern the design of
a member in pure compression. For this mode in a pinned column, the elastic
critical load Ncr, defined as the maximum load supported by the column, free
from any type of imperfections, is given by the well known Euler’s formula:
N
EI
N
d 2y
dx 2
Ny 0
Ncr
L
x
Ncr 
y(x)
y(x)
y
(z)
0
N
Buckling in a bending mode
2 E I
L2
E I – Bending stiffness
L – Buckling length
(LE for other support conditions)
 In specific cases (e.g. members with cruciform cross sections) buckling may
occur in other modes: torsional buckling or flexural-torsional buckling.
Eurocodes ‐ Design of steel buildings with worked examples
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DESIGN OF COLUMNS
Column Buckling
Ncr 
2 E I
LE
 cr 
2
2 E I
A LE
2
2 E
 2


Slenderness
Critical stress
2 E

 f y  1  
12
 cr
E
fy
LE
i

fy

N
A


1
Af y
Ncr
Non-dimensional
slenderness
1.0
Euler’s curve
I
A
Radius of gyration
 
fy

i 
Euler´s curve

E fy
1
  
2 E
2
1.0
L
 E
i

Imperfections or real columns (geometrical
imperfections and material imperfections).
Eurocodes ‐ Design of steel buildings with worked examples
Brussels, 16 ‐ 17 October 2014
DESIGN OF COLUMNS
Buckling Resistance
(clause 6.3.1 of EC3-1-1)
N b.Rd   A f y  M1
(Class 1, 2 or 3)
Theoretical behaviour
N b.Rd   Aeff f y  M1
(Class 4)
 is the reduction factor for the
relevant buckling mode
 
1
2
  

2
but   1.0

2
  0.5 1     0.2   


Neglect BUCKLING if:
  0.2
or
NEd Ncr  0.04
Eurocodes ‐ Design of steel buildings with worked examples
Brussels, 16 ‐ 17 October 2014
DESIGN OF COLUMNS
Buckling Resistance
(clause 6.3.1 of EC3-1-1)
Flexural buckling
  A fy Ncr 
Lcr 1
i 1
  Aeff fy Ncr 
1  
Lcr
i
(Class 1, 2 or 3)
Aeff A
1
E f y  93 .9 
(Class 4)
ε  235 f y
Torsional or flexural-torsional buckling
T 
A f y Ncr
T 
Aeff fy Ncr
(Class 1, 2 or 3)
(Class 4)
 - buckling in flexural buckling mode about z
axis
Eurocodes ‐ Design of steel buildings with worked examples
Brussels, 16 ‐ 17 October 2014
DESIGN OF COLUMNS
EXAMPLE 1
Safety verification of a column member of the building represented in the
figure.
A’
A
E’
C’
B
C
D
E
F
4
6.00 m
3
4.50 m
2b
2.50 m
2a
2.00 m
2
6.00 m
1
4.00 m
4.50 m 4.50 m
4.00 m
Building – master example
i) The inner column E-3 represented in the figure, at base level, is selected. This member has
a length of 4.335 m and is composed by a section HEB 340 in steel S 355.
In this column the bending moments (and the shear force) may be neglected; the design
axial force (compression) obtained from the previous analysis is given by NEd = 3326.0 kN.
Eurocodes ‐ Design of steel buildings with worked examples
Brussels, 16 ‐ 17 October 2014
DESIGN OF COLUMNS
EXAMPLE 1
ii) Cross section classification – section HEB 340 in pure compression.
Geometric characteristics: A = 170.9 cm2, b = 300 mm, h = 340 mm, tf = 21.5 mm,
tw = 12 mm, r = 27 mm, Iy = 36660 cm4, iy = 14.65 cm, Iz = 9690 cm4, iz = 7.53 cm.
Mechanical properties of the steel: fy = 355 MPa and E = 210 GPa.
Web in compression (Table 5.2 of EC3-1-1)
c (340  2  21.5  2  27)

 20.25  33 
t
12
(class 1)
 33  0.81  26.73
Flange in compression (Table 5.2 of EC3-1-1)
c 300 2  12 2  27

 5.44  9   9  0.81  7.29
21.5
t
(class 1)
HEB 340 cross section, steel S 355, in pure compression is class 1.
c
Eurocodes ‐ Design of steel buildings with worked examples
Brussels, 16 ‐ 17 October 2014
DESIGN OF COLUMNS
EXAMPLE 1
iii) Cross section verification - class 1 in pure compression.
N Ed  3326 . 0 kN  N c , Rd 
A fy
 M0

170 . 9  10  4  355  10 3
 6067 . 0 kN .
1 .0
iv) Buckling resistance.
Buckling lengths – Assuming that the design forces were obtained by a second order
structural analysis, the buckling lengths are considered (conservatively) equal to the real
lengths (mid-distance between floors), given by:
Buckling in the plan x-z (around y)
-
LEy  4.335 m
Buckling in the plan x-y (around z)
-
LEz  4.335 m
Determination of the slenderness
coefficients
1  
210  106
3
355  10
 76.41
y 
z 
LEy
iy

4.335
14.65  102
 29.59
LEz
4.335

 57.57
iz
7.53  10  2
y 
y
 0.39
1
z 
z
 0.75
1
Eurocodes ‐ Design of steel buildings with worked examples
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DESIGN OF COLUMNS
EXAMPLE 1
z
Calculation of the reduction factor min
h 340

 1.13  1.2
b 300

and
flexural buckling around y  curve b (  0.34)
flexural buckling around z  curve c (  0.49).
As  z  0.75  y  0.39
and

tf  21.5 mm  100mm
curve c

 min  z
curve b
340
y
300
HEB 340
Eurocodes ‐ Design of steel buildings with worked examples
Brussels, 16 ‐ 17 October 2014
DESIGN OF COLUMNS
EXAMPLE 1



 z  0.5 1   z  0.2   z
2



 z  0.5  1  0.49  0.75  0.2  0.752  0.92
z 
1
2
2
0.92  0.92  0.75
 z  0.69
 0.69
 min   z  0.69
v) Safety verification
λz  0.75
Nb,Rd   z A fy  M1  0.69  170.9  104  355  103 1.0  4186.2 kN
As, NEd  3326.0 kN  Nb,Rd  4186.2 kN
safety is verified with the cross section HEB 340 in S 355 steel.
Eurocodes ‐ Design of steel buildings with worked examples
Brussels, 16 ‐ 17 October 2014
DESIGN OF BEAMS
Beam cross sections and applications
 A beam may be defined as a member subjected
essentially to bending and shear force.
Castellated beams
Hot-rolled sections (IPE, HEA or HEB, RHS,….)
Welded sections
Welded sections in non-uniform beams
Eurocodes ‐ Design of steel buildings with worked examples
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DESIGN OF BEAMS
Cross section resistance
Uniaxial bending (clause 6.2.5 of EC3-1-1)
M Ed
 1 .0
M c.Rd
 Class 1 or 2
Mc.Rd  Wpl fy  M 0
 Class 3
Mc.Rd  Wel. min fy  M0
 Class 4
Mc.Rd  Weff .min fy  M0
Bi-axial bending (clause 6.2.9 of EC3.1.1)
 M y , Ed

 M pl , y.Rd


 M z , Ed




 M pl , z.Rd




 1 .0
I or H   2;   5 n
CHS
  2
RHS
  
but   1
1.66
2
1  1.13 n
but
  6
n  NEd N pl , Rd
Eurocodes ‐ Design of steel buildings with worked examples
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DESIGN OF BEAMS
Cross section resistance
VEd
 1.0
Vc,Rd
Shear (clause 6.2.6 of EC3-1-1)
PLASTIC RESISTANCE Vpl.Rd

Vpl.Rd  Av fy
ELASTIC RESISTANCE

3  M0
 Ed
VEd
Av – Shear area
fy
 3 
G
y
6.2.6 (3) of EC3-1-1 or
from tables of profiles).
f
y
3

M0
Vz
Av
(obtained from clause
 1.0
My
G
 Ed 
VEd S
It


e. n. a.
f
z
y
Shear stresses - 
 Shear buckling for webs without stiffeners should be verified in accordance with EC3-1-5, if:
hw

 72
tw

 
235 / fy
hw and tw are the height and thickness of the web and  is in
accordance with EC3-1-5.
3

Eurocodes ‐ Design of steel buildings with worked examples
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DESIGN OF BEAMS
My,V.Rd
Cross section resistance
z
Bending and Shear Interaction
Vz
(clause 6.2.8 of EC3-1-1)
My
VEd  50% Vpl , Rd
NO REDUCTION
VEd  50 % Vpl , Rd
REDUCED MOMENT
fyr  1    fy

fy
y
hm
tw
fyr
fyr
 (M y)
+
fy
(Vz )
  2 VEd Vpl.Rd  12
For I and H cross sections of equal flanges, with bending about the major axis y, the
bending moment resistance My,V,Rd is given by (clause 6.2.8 of EC3-1-1):
M y ,V .Rd

 Aw 2

 Wpl , y 

4 tw

 f
 y M
y , c , Rd

M0

AW  hw tw
Eurocodes ‐ Design of steel buildings with worked examples
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DESIGN OF BEAMS
Lateral-Torsional Buckling
 Instability phenomenon characterized by the occurrence of large transversal
displacements and rotation about the member axis, under bending moment
about the major axis (y axis).
 This instability phenomenon involves lateral bending (about z axis) and torsion of
cross section.
z
My
y
Eurocodes ‐ Design of steel buildings with worked examples
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DESIGN OF BEAMS
Lateral-Torsional Buckling
 In the study of lateral-torsional buckling of beams, the Elastic Critical Moment Mcr
plays a fundamental role; this quantity is defined as the maximum value of bending
moment supported by a beam, free from any type of imperfections.
 For a simple supported beam with a double symmetric section, with supports prevent
lateral displacements and rotation around member axis (twist rotations), but allowing
warping and rotations around cross section axis (y and z), submitted to a uniform
bending moment My (“standard case”), the elastic critical moment is given by:
L
My
My
C
B
A
z
x´
z´
a) Elevation
E

Mcr
x

 2 E IW
G IT E I z 1  2

L
L G IT






Which depend mainly of:
Loading and support conditions;
Length between lateral braced sections (L);
Lateral bending stiffness (E Iz);
Torsional stiffness (G IT);
Warping stiffness (E Iw).
Eurocodes ‐ Design of steel buildings with worked examples
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DESIGN OF BEAMS
z
Lateral-Torsional Buckling
z
y
CG
G
Elastic critical moment
Mcr

 E I z  k z

 C1
2  k
k z L  w

2
C
 IW k z L2 G IT


 C2 z g  C3 z j
 2 E Iz
 Iz
2

z g  z a  z s 

2




0.5


 C2 z g  C3 z j 




Mcr,1>Mcr
Mcr
Mcr,2<Mcr
P


z j  z s   0 .5 y 2  z 2 z dA  I y


A




C
P
P
C
C
- applicable to member with symmetric and mono-symmetric cross sections,
- include the effects of the loading applied below or above the shear centre;
- several degrees of restriction to lateral bending (kz) and warping (kw);
- several shapes of bending moment diagram (C1, C2 and C3 in the next tables).
y
Eurocodes ‐ Design of steel buildings with worked examples
DESIGN OF BEAMS
Lateral-Torsional Buckling
Elastic critical moment
- Publication nº 119 do ECCS
(Boissonnade et al. 2006).
- LTBeam software
http://www.cticm.com
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Eurocodes ‐ Design of steel buildings with worked examples
Brussels, 16 ‐ 17 October 2014
DESIGN OF BEAMS
Lateral-Torsional Buckling
Lateral-torsional buckling resistance (clause 6.3.2 of EC3-1-1)
M Ed
 1 .0
M b.Rd
Mb.Rd   LT Wy fy  M1
Wy = Wpl.y Class 1 and 2;
Wy = Wel.y Class 3;
Wy = Weff.y Class 4.
LT is the reduction factor for lateral-torsional buckling, which can be calculated
by one of two methods, depending of member cross section.
Eurocodes ‐ Design of steel buildings with worked examples
Brussels, 16 ‐ 17 October 2014
DESIGN OF BEAMS
Lateral-Torsional Buckling
i) General method
 LT 

1
2
 LT   LT   LT


2 0.5


 LT  1.0
 LT  0 .5 1   LT  LT  0 .2   LT
 LT  Wy fy Mcr 0.5
Mcr - Elastic critical moment
2

Table 6.4 -
Eurocodes ‐ Design of steel buildings with worked examples
Brussels, 16 ‐ 17 October 2014
DESIGN OF BEAMS
Lateral-Torsional Buckling
ii) Alternative method (rolled sections or equivalent welded sections)
 LT 

1


 LT   LT    LT


2
2 0.5
 LT  1.0
 LT  1  LT 2
 LT  0 .5 1   LT  LT   LT ,0    LT
 LT  Wy fy Mcr 0.5
Mcr - Elastic critical moment
2

Table 6.5 -
 LT ,0  0.4
  0.75
(may be specified in National
Annexes of Eurocode 3)
Eurocodes ‐ Design of steel buildings with worked examples
Brussels, 16 ‐ 17 October 2014
DESIGN OF BEAMS
Lateral-Torsional Buckling
 LT ,mod 
 LT
 LT ,mod  1.0
f


2
f  1  0.5 1  k c  1  2.0  LT  0.8 


f  1.0
Neglect LTB if:
 LT   LT ,0
M Ed M cr   LT ,0
2
Eurocodes ‐ Design of steel buildings with worked examples
Brussels, 16 ‐ 17 October 2014
DESIGN OF BEAMS
EXAMPLE 2
Safety check of a beam of the building illustrated in the figure (along line E). The beam is composed by a
IPE 600 with 9 m length at the central span; the lateral spans with 6 m length (the governing spans) are
composed by a section IPE 400 in steel S 355. For the lateral buckling check, two cases are considered:
a) a beam with 6 m length, laterally braced only at the end support sections;
b) a beam with 6 m length, laterally braced at the end support sections and at mid-span section.
A’
A
The geometrical and mechanical properties of
C
D
E
F
6.00 m
3
A = 84.46 cm2, b = 180 mm, h = 400 mm,
iy = 16.55 cm, Iz = 1318 cm4, iz = 3.95 cm,
B
4
the section IPE 400 in S 355 steel are:
tf = 13.5 mm, tw = 8.6 mm, Iy = 23130 cm4,
E’
C’
4.50 m
2b
2.50 m
2a
2.00 m
2
IT = 51.08 cm4 ; Iw = 490x103 cm6;
fy = 355 MPa and E = 210 GPa.
6.00 m
1
4.00 m
4.50 m 4.50 m
Building plan – master example
4.00 m
Eurocodes ‐ Design of steel buildings with worked examples
Brussels, 16 ‐ 17 October 2014
DESIGN OF BEAMS
EXAMPLE 2
139.1 kN
a) Beam laterally braced at supports
i) The internal forces (neglecting the
axial force) are represented in the figure.
Vz,Ed
70.7 kN
The design values are MEd = 114.3 kNm
71.6 kN
140.1 kN
and VEd = 75.9 kN.
255.7 kNm
93.7 kNm
ii) Cross section classification
246.3 kNm
111.4 kNm
99.2 kNm
109.7 kNm
My,Ed
Web (an internal part) in bending:
c
331

 38 .49  72   72  0 .81  58 .32
t
8 .6
Flange (outstand part) in compression:
(180  2  21  8 .6) 2
c

 4 .79  9   9  0 .81  7 .29
13 .5
t
The cross section is class 1
75.2 kN
75.9 kN
114.3 kNm
113.6 kNm
163.0 kNm
Eurocodes ‐ Design of steel buildings with worked examples
Brussels, 16 ‐ 17 October 2014
DESIGN OF BEAMS
EXAMPLE 2
iii) Cross section verification
Bending resistance:
MEd  114.3 kNm  Wpl,y fy  M0  1307  106  355  103 1.0  464.0 kNm
Shear resistance:
VEd  75.9 kN  Vpl , Rd 
Av fy
 M0 3

42.69  10 4  355  103
1.0  3
 875.0 kN
hw

373 .0
0 .81

 43 .4  72  72 
 58 .3
8 .6
1 .0
tw

So, it is not necessary to verify the shear buckling resistance.
Bending + Shear:
VEd  75.9 kN  0.50  Vpl,Rd  0.50  875.0  437.5 kN
So, it is not necessary to reduce the bending resistance to account for the shear force.
Cross section resistance is verified.
Eurocodes ‐ Design of steel buildings with worked examples
Brussels, 16 ‐ 17 October 2014
DESIGN OF BEAMS
EXAMPLE 2
iv) Lateral buckling resistance
Assuming the support conditions of the “standard case” and the loading
applied at the upper flange level, the elastic critical moment can be
zg=200 mm
400
obtained from the following equation, with L = 6.00 m, kz = kw = 1.0,
CG
IPE 400
C1 ≈ 1.80 and C2 ≈ 1.60 (Boissonnade et al., 2006) and zg = 200 mm.
Mcr

 E I z  k z
 C1

k z L2  kw

2
2
 IW

 Iz
k L G IT  C z  C z 2 
 z2
2 g
3 j

 E Iz

2


0.5
180


 C2 z g  C3 z j 




C
A
3m
Mcr  164.7 kNm
93.7 kNm
(Using LTBeam
–> Mcr = 175.64 kNm)
3m
6m
111.4 kNm
My,Ed
114.3 kNm
  93.7 111.4  0.84
Eurocodes ‐ Design of steel buildings with worked examples
Brussels, 16 ‐ 17 October 2014
DESIGN OF BEAMS
EXAMPLE 2
Mcr  164.7 kNm ; Wy  Wpl ,y  1307 cm3

 LT  Wy fy Mcr 0.5
 LT  1.68
Table 6.4 -
General method:
Rolled cross section IPE 400 with
h/b=400/180=2.2>2

- Curve b
 LT  0.34
LT  2.16
 LT  0.28
M b, Rd  0 .28  1307  10  6 



 LT  0 .5 1   LT  LT  0 .2   LT
355  10 3
 129 .9 kNm  114.3 kNm
1 .0
So, the safety is verified (utilization ratio = 114.3/129.9=0.88).
 LT 

1
2
 LT   LT   LT

2
2 0.5

Eurocodes ‐ Design of steel buildings with worked examples
Brussels, 16 ‐ 17 October 2014
DESIGN OF BEAMS
EXAMPLE 2
B
A
C
b) Beam laterally braced at supports and mid-span
i) Cross section verifications are not changed.
3m
3m
6m
ii) Lateral buckling check:
93.7 kNm
As the beam is laterally braced at mid span cross
section, the critical moment can be evaluated
111.4 kN
  93.7 114.3  0.82
My,Ed
with L = 3.00 m and a conservative hypothesis of
kz = kw = 1.0. For the given bending moment
114.3 kNm
shape between lateral braced cross sections,
M
C1 = 2.6 (Boissonnade et al., 2006) .





zj
C3

zg
C2




2

5
.
0


zj
C3

zg
C2


IT
G Iz
2 E
L
2
kz
2




IWIz
z





kzkw
I 2
E Lz
2 k
C1
r
Mc


M


Mcr  1778.8 kNm

(Using LTBeam – Mcr = 1967.7 kNm)
1.0
2.06
1.000
0.850
0.5
2.15
1.000
0.650
1.0
2.35
1.000
1.3  1.2 f
0.5
2.42
0.950
0.77  f
1.0
2.60
1.000
0.55   f
0.5
2.45
0.850
0.35   f
1.0
2.60
 f
 f
0.5
2.45
 0.125  0.7  f
 0.125  0.7 f
Eurocodes ‐ Design of steel buildings with worked examples
Brussels, 16 ‐ 17 October 2014
DESIGN OF BEAMS
EXAMPLE 2
Mcr  1778.8 kNm ; Wy  Wpl ,y  1307 cm3

 LT  Wy fy Mcr 0.5
 LT  0.51
Table 6.4 -
General method:
Rolled cross section IPE 400 with
h/b=400/180=2.2>2

- Curve b
 LT  0.34
LT  0.68


LT  0.89
M b, Rd  0 .89  1307  10


 LT  0 .5 1   LT  LT  0 .2   LT
6
355  10 3

 412 .9 kNm  114.3 kNm
1 .0
So, the safety is verified (utilization ratio = 114.3/412.9=0.28).
 LT 

1
2
 LT   LT   LT

2
2 0.5

Eurocodes ‐ Design of steel buildings with worked examples
Brussels, 16 ‐ 17 October 2014
DESIGN OF BEAM‐COLUMNS
Cross section resistance (clause 6.2.9 of EC3-1-1)
 Class 1 or 2 – Uniaxial bending
MEd  MN,Rd
Double-symmetric I or H sections
M N , y , Rd  M pl , y , Rd
1n
1  0 .5  a
but
M N , z , Rd  M pl , z , Rd
M N , z , Rd
if
2

n  a 
 M pl , z , Rd 1  
 

 1  a  
n  NEd N pl.Rd
if
MN , y , Rd  M pl , y , Rd
n a
N
N pl
Eixo
de about
menor
inércia
Bending
minor
axis -- zz
1.0
Eixo deabout
maior
inércia
Bending
major
axis - -y y
n a
a  A  2 b t f  A  0.50
No reduction if
N Ed  0.25 N pl , Rd
NEd  0.5 hw tw fy  M 0
My
HEA
NEd  hw tw fy  M0 (z axis)
(y axis)
0
0
M pl , y
1.0
, Mz
M pl , z
Eurocodes ‐ Design of steel buildings with worked examples
Brussels, 16 ‐ 17 October 2014
DESIGN OF BEAM‐COLUMNS
Cross section resistance (clause 6.2.9 of EC3-1-1)
M
z,Ed
 Class 1 or 2 – Bi-axial bending
 M y , Ed

 M N , y.Rd

 M z , Ed



 M N , z.Rd




I or H

 1 .0
N
 Class 3 or 4
  2;   5 n but   1
Circular hollow sections
  2
Rectangular hollow sections
d
R
,
l
p
d
NE
n


 x , Ed 
fy
 M0
M y,Ed
NEd
 x , Ed 
  
N Ed
A

1.66
1  1.13 n2
M y , Ed
Iy
z
6
M z , Ed
Iz
y
Bending, shear and axial force (clause 6.2.10 of EC3-1-1) – Similar to bending
and shear interaction.
Eurocodes ‐ Design of steel buildings with worked examples
Brussels, 16 ‐ 17 October 2014
DESIGN OF BEAM‐COLUMNS
Member stability
Members with high slenderness subjected to bending and compression, may fail
by flexural buckling or lateral-torsional buckling.
Flexural buckling
cross-section):
and
lateral-torsional
buckling
(doubly-symmetric
My.Ed  My , Ed
M z.Ed  M z , Ed
NEd
 k yy
 k yz
 1.0 (Eq. 6.61 of EC3-1-1)
 y NRk  M1
 LT My , Rk  M1
M z , Rk  M1
eN,y NEd (class 4)
M y.Ed  M y , Ed
M z.Ed  M z , Ed
N Ed
 k zy
 k zz
 1.0 (Eq. 6.62 of EC3-1-1)
 z N Rk  M1
 LT M y , Rk  M1
M z , Rk  M1
kyy, kyz, kzy and kzz - interaction factors, which are dependent of instability phenomena
and plasticity – Annex A of EC3-1-1 (Method 1) or Annex B (Method 2).
Eurocodes ‐ Design of steel buildings with worked examples
Brussels, 16 ‐ 17 October 2014
DESIGN OF BEAM‐COLUMNS
Member stability
i) Members with closed hollow sections or open sections restrained to torsion are
not susceptible to torsional deformation.
ii) Members with open sections (I or H sections) are susceptible to torsional
deformation.
Members not susceptible to torsional deformation – checking of flexural
buckling against y-axis and z-axis, considering eqs. (6.61) and (6.62) with
LT = 1.0 and interaction factors kyy, kyz, kzy and kzz in members not susceptible to
torsional deformation.
Members susceptible to torsional deformation – checking of lateral-torsional
buckling, considering eqs (6.61) and (6.62) with LT according to 6.3.2 of EC3-1-1
and interaction factors kyy, kyz, kzy and kzz in members susceptible to torsional
deformation.
Eurocodes ‐ Design of steel buildings with worked examples
DESIGN OF BEAM‐COLUMNS
Member stability
Method 2
(Annex B of EC3-1-1)
Interaction factors for members not
susceptible to torsional deformations
(Table B.1 of EC3-1-1).
Brussels, 16 ‐ 17 October 2014
Eurocodes ‐ Design of steel buildings with worked examples
DESIGN OF BEAM‐COLUMNS
Member stability
Method 2 (Annex B
of EC3-1-1)
Interaction factors for
members susceptible to
torsional deformations
(Table B.2 of EC3-1-1).
Brussels, 16 ‐ 17 October 2014
Eurocodes ‐ Design of steel buildings with worked examples
DESIGN OF
BEAM‐COLUMNS
Member stability
Method 2
(Annex B of EC3-1-1)
Equivalent factors of
uniform moment Cmi
(Table B.3 of EC3-1-1)
Brussels, 16 ‐ 17 October 2014
Eurocodes ‐ Design of steel buildings with worked examples
Brussels, 16 ‐ 17 October 2014
DESIGN OF BEAM‐COLUMNS
EXAMPLE 3
Safety check of a beam-column of the first storey of the building illustrated in the figure. The
member, composed by a HEB 320 cross section in steel S 355, has a length of 4.335.
The relevant geometric characteristics of HEB 320 cross
section are: A = 161.3 cm2; Wpl,y = 2149 cm3,
Iy = 30820 cm4, iy = 13.82 cm; Iz = 9239 cm4,
iz = 7.57 cm; IT = 225.1 cm4 and IW = 2069 x 103 cm6.
The mechanical characteristics of the material are:
fy = 355 MPa, E = 210 GPa and G = 81 GPa.
Eurocodes ‐ Design of steel buildings with worked examples
Brussels, 16 ‐ 17 October 2014
DESIGN OF BEAM‐COLUMNS
EXAMPLE 3
The design internal forces obtained through the structure
analysis
(second
combinations
are
order)
for
illustrated
the
in
the
various
figure.
load
z
65.5 kNm
y
Two
simplification are assumed for the subsequent design
verifications: i) the shear force is sufficient small so can
53.0 kN
201 kN
417 kN
72.3 kNm
29.8 kNm
41.1 kN
58.6 kNm
32.0 kNm
41.1 kN
57.1 kNm
29.3 kNm
40.2 kN
55.9 kNm
27.9 kNm
39.3 kN
53.8 kNm
29.0 kNm
be neglected; ii) the shape of the bending moment
diagram is linear.
Design values are: NEd = 1704 kN; My,Ed = 24.8 kNm
630 kN
841 kN
at the base cross section.
1053 kN
i) Cross section classification
43.2 kN
1262 kN
54.5 kNm
68.6 kNm
As the compression force is high, the cross section is
As the section HEB 320 is a stocky section, even under this
load condition, is class 1.
50.5 kN
1496 kN
classified under compression only (conservative approach).
29.4 kN
1704 kN
NEd
Vz,Ed
10.6 kNm
73.5 kNm
24.8 kNm
My,Ed
Eurocodes ‐ Design of steel buildings with worked examples
Brussels, 16 ‐ 17 October 2014
DESIGN OF BEAM‐COLUMNS
EXAMPLE 3
ii) Cross section resistance
The design internal forces: My,Ed = 24.6 kNm and NEd = 1704 kN (compression).
N pl ,Rd  A fy  M0  161.3  104  355  103 1.0  5726.2 kN
NEd
My,Ed
As, NEd  1704 kN  N pl ,Rd  5726.2 kN, the axial force resistance is verified.
Since, NEd  1704kN  0.25 Npl, Rd  1431.5 kN ,
in accordance with clause 6.2.9.1(4) of EC3-1-1, it is necessary to
reduce the plastic bending resistance (to MN,y,Rd):
M pl ,y ,Rd 
Wpl ,y fy
 Mo
MN,y , Rd  M pl , y , Rd
As,
2149  106  355  103
A  2 b tf
161.3  2  30  2.05

 762.9 kNm
a

 0.24
1.0
A
161.3
1n
1  0.30
 762.9 
 606.9 kNm
NEd
1704
n

 0.30
1  0.5 a
1  0.5  0.24
N pl ,Rd
5726.2
M y , Ed  24 .8 kNm  M N , y , Rd  606 .9 kNm , the bending resistance is verified.
Eurocodes ‐ Design of steel buildings with worked examples
Brussels, 16 ‐ 17 October 2014
DESIGN OF BEAM‐COLUMNS
EXAMPLE 3
iii) Verification of the stability of the member
In this example the Method 2 is applied. As the member is susceptible to torsional
deformations (thin-walled open cross section), it is assumed that lateral-torsional buckling
constitutes the relevant instability mode. Since Mz,Ed = 0, the following conditions must be
verified:
My , Ed
NEd
 k yy
 1.0
 y NRk  M1
 LT My ,Rk  M1
M y , Ed
N Ed
 k zy
 1 .0
 z N Rk  M1
 LT M y , Rk  M1
Step 1: characteristic resistance of the cross section
NRk  A fy  161.3  10 4  355  103  5726.2 kN
M y , Rk  Wpl , y f y  2149  10 6  355  10 3  762 .9 kNm
Eurocodes ‐ Design of steel buildings with worked examples
Brussels, 16 ‐ 17 October 2014
DESIGN OF BEAM‐COLUMNS
EXAMPLE 3
z
Step 2: reduction coefficients due to flexural buckling, y and z
h 320

 1.07  1.2
b 300

and
y
320
tf  20.5 mm  100mm
300
flexural buckling around y  curve b (  0.34)
flexural buckling around z  curve c (  0.49).
HEB 320
Plane xz - LE,y = 4.335 m.
y 
LE,y 1
4.335
1


 0.41
iy 1 13.82  102 93.9  0.81
 y  0.62

 y  0.92
Plane xy - LE,z = 4.335 m
z 
LE , z 1
4.335
1


 0.75

2
93.9  0.81
i z 1 7.57  10
 z  0.92

 z  0.69


  0 .5 1     0 .2   


2
 
1
2
    2   


0.5
Eurocodes ‐ Design of steel buildings with worked examples
Brussels, 16 ‐ 17 October 2014
DESIGN OF BEAM‐COLUMNS
EXAMPLE 3
Step 3: calculation of the LT using the alternative method applicable to rolled or equivalent
welded sections (clause 6.3.2.3 of EC3-1-1)
The length between braced sections is L = 4.335 m. The critical
5 kNm
10.6 kNm
moment Mcr assuming a linear diagram, in this example obtained
just by LTBeam software, is given by:
Mcr  5045.1 kNm


 LT  2149  10
6
3
Rolled I or H sections with h b  320 300  1.07  2 
curve b, and
 LT  0.34
Taking LT ,0  0.4
and
Table 6.4 -
  0.75

 LT  0 .5  1  0 .34  0 .39  0 .4   0 .75  0 .39 2
 0 .56
 LT 

1
2
0.56  0.56  0.75  0.39

2 0.5

 355  10 5045.1
 0.99

0.5
 0.39
24.8 kNm
My,Ed
Eurocodes ‐ Design of steel buildings with worked examples
Brussels, 16 ‐ 17 October 2014
DESIGN OF BEAM‐COLUMNS
EXAMPLE 3
Step 3: calculation of the LT using the alternative method applicable to rolled or equivalent
welded sections (clause 6.3.2.3 of EC3-1-1)
The correction factor kc, according to Table 6.6 of EC3-1-1, with  = 10.6/(-24.8) = - 0.43, is
given by:
kc 
1
1

 0.68
1.33  0.33  1.33  0.33  (0.43)


2
f  1  0.5  1  kc   1  2.0   LT  0.8 




 1  0.5  1  0.68  1  2.0  0.39  0.8  0.89
2
The modified lateral-torsional buckling reduction factor is
given by:
 LT ,mod  0.99 0.89  1.11  1.00
So,  LT ,mod  1.00
must be adopted.
Eurocodes ‐ Design of steel buildings with worked examples
Brussels, 16 ‐ 17 October 2014
DESIGN OF BEAM‐COLUMNS
EXAMPLE 3
Step 4: interaction factors kyy and kzy.
The equivalent factors of uniform moment Cmy and CmLT are obtained based on the bending
moment diagram, between braced sections according to the z direction in case of Cmy and
laterally in case of CmLT. Assuming the member braced in z direction and laterally just at the
base and top cross sections, the factors Cmy and CmLT must be calculated based on the bending
moment diagram along the total length of the member.
Since the bending moment diagram
is assumed linear, defined by:
My,Ed,base= -24.8 kNm;
My,Ed,top = 10.4 kNm, from Table B.3
of EC3-1-1, is obtained:
  My,Ed,top My,Ed,base  10.6  24.8  0.43
5 kNm
10.6 kNm

C my  C mLT  0.60  0.4   0.43  0.43 ( 0.40)
24.8 kNm
My,Ed
Eurocodes ‐ Design of steel buildings with worked examples
Brussels, 16 ‐ 17 October 2014
DESIGN OF BEAM‐COLUMNS
EXAMPLE 3
Because the member is susceptible to torsional deformations, the interaction factors kyy and kzy
are obtained from Table B.2 of EC3-1-1, through the following calculations:




NEd
1704
k yy  C my 1  y  0.2
  0.43  1  0.41  0.2 
  0.46;
0.92  5726.2 1.0 
 y NRk  M1 




NEd
k yy  0.46
  0.54 , then
As k yy  C my 1  0.8


 y NRk  M1 





NEd
0.1z

k zy   1 
CmLT  0.25  z NRk  M1 



0.1  0.75
1704
 1
 0.82
0.43  0.25 0.69  5726.2 1.0 


As k zy  1 


NEd
0.1
 0.76
C mLT  0.25  z NRk  M1 
then k zy  0.82
Step 5: Finally, the verification of equations 6.61
and 6.62 of EC3-1-1 yields:
1704
24.8
 0.46 
 0.34  1.0
0.92  5726.2 1.0
1.00  762.9 1.0
1704
24.8
 0.82 
 0.46  1.0
0.69  5726.2 1.0
1.00  762.1 1.0
Section O.K.
Eurocodes ‐ Design of steel buildings with worked examples
Brussels, 16 ‐ 17 October 2014
 Free software for design of steel members in accordance
with EC3-1-1.
Beam-columns design
Design of cellular beams
http://www.constructalia.com
http://www.steelconstruct.com
Eurocodes ‐ Design of steel buildings with worked examples
Brussels, 16 ‐ 17 October 2014
Thank you for your attention
rads@dec.uc.pt