Design of a steel column under NEd
1
Load Effects
Axial force NEd
Structural Resistances
Ed  Rd
NEd
NEd=n[(1.35gk+1.5qk )(xy)]
NEd=1.35Gk+1.5Qk
NEd=n(RL +RR)
n=number of a building storey
x=Space between vertical beams
y= Space between horizontal beams Lz Ly
RL =Reaction of left beam
RR=Reaction of right beam
Axial Nc,Rd and Nb,Rd
NEd≤ Nc,Rd= Afy /M0
(Compression resistance of a
stocky or restrained column)
NEd≤ Nb,Rd=Afy /M0
(Buckling resistance of a slender or
unrestrained column)
A = Sectional Area (mm2)
fy= Steel Strength, 275 or 265 N/mm2.
M0 =Safety factor =1.00
 = Reduction factor of flexural buckling
y
x
=min[y, z]
Nb,Rd=min[Nb,Rd,y, Nb,Rd,z]
y=(Afy/Ncr,y)0.5 , y=(Afy/Ncr,Z)0.5
Ncr,y=2EIy/L2y,
Ncr,z=2EIz/L2z
1
Buckling resistance by calculation
2
Design Equation: NEd ≤ Nb,Rd=Afy/Mo

Reduction factor:
  1    ( 2   ) 2   1
2


Value for reduction factor;   0.5[1   (  0.2)   2 ]
Slenderness ratio:
 Ly=0.85L
 Lz=0.7L
 Buckling
curves a, b,
c, d and
Imperfection
factor 

2

   1 
f y A N cr  ( L i ) / 86.6
Buckling resistance Nb,Rd using SCI table
3
Example: A 5m long column with 254x254x73UC
partially and effectively restrain about y-y and z-z.
Lz=0.70x5=3.50m
Solution: Ly=0.85x5=4.25m,
4.0<Ly=0.85x5=4.25m<5.0,
3.0<Lz=0.7x5=3.5m<4.0,
Nb,y,Rd= (4.25-5)x(2360-2240)/(4-5)
+2240=2330kN
Nb,z,Rd=(3.5- 4)x(2110-1840)/(3 - 4)
+1840=1975kN
2360kN> Nb,y,Rd >2240kN,
2110kN> Nb,z,Rd >1840kN,
3
Design of stock/restrained column by Nc,Rd
4
For a steel column in S275 that has effective lengths Lz, and Ly and
the calculated axial force NEd, if its buckling is prevented from,
2
 Take NEd≤Nc,Rd=Afy/Mo and fy=265N/mm ,
2
 The required area of the steel column: Areq≥ NEdM0/fy=267cm
 Use the SCI section table to choose a 356x406x235UKC with
Aprov=299cm2≥ Areq=267cm2
4
Design of unrestrained column by Nb,Rd
5
For a steel column in S275 that has effective lengths Ly=5m, Lz=4m and
the calculated axial force NEd, =2000kN, if the buckling can occur,
from NEd≤Nb,Rd=min {Nb,Rd,y; Nb,Rd,z},

Take Nb,Rd,y≥ NEd=2000kN and Nb,Rd,z ≥ NEd =2000kN
Use the SCI resistance table to chose a 203x203x100UKC,
For Ly=5m, Nb,Rd,y=2820kN ≥ NEd=2000kN
For Lz=4m, Nb,Rd,z =2150kN≥ NEd =2000kN


5
Assessment of a steel column under NEd
6
https://www.dailymail.co.uk/news/article-8502645/Massive-crane-collapses-homes-Bow-east-London-leaving-oneman-seriously-injured.html
8th July 2020, at
2:40pm, East
London
Ed > Rd
•
•
•
•
6
Which kind of limit state?
Possible/potential failure modes?
How to investigate the issue independently?
Measures to be taken to prevent from it?
A woman dies
and race is on to
rescue any others
trapped as 65foot crane
COLLAPSES onto
homes in east
London – with
neighbor whose
house was hit
describing impact
as 'like an
earthquake'
A steel column under NEd+MEd
7
Steel Frame
Beam
Brace
Column
 In addition to the axial force NEd, a column, especially a
column also carries bending moment MEd.
7
MOD010255
Structure and Construction Project
MOD006111
Structural Method and Design
Lecturers:
Dr Yingang Du
Room:
MAR317
Email: yingang.du@analia.ac.uk
Telephone:
01245 683945
W8-1– Design of Steel Column under NEd+MEd
8
Aim of the lecture
9


To explain the performance of a column
under an axial force NEd and bending
moment MEd.
To carry out the design and assessment of
steel column to Structural Eurocodes safely
and economically.
9
Outline of the lecture (NEd≤NRd, MEd≤MRd)
10
10

Introduction to a column under NEd and MEd

Design axial force NEd and bending moment MEd.

Behaviour of a column under NEd and MEd

Resistances of a steel column under NEd and MEd

Design method of a steel column under NEd and MEd

Examples, Tutorials and Coursework (BEng group)
Reading list
11

William M C McKenzie, Design of Structural
Elements to Eurocode (2nd Edition), Palgrave
Macmillan, Pages 431- 445.
3/2/2025
11
Introduction to a column under NEd & MEd
12
NEd
MEd
http://www.chdist.com/images/products/42-002A_BK.jpg
12
Introduction to a column under NEd and MEd
e
13
e
MEd is caused by
an eccentricity
between the
reaction of a
beam and the
centre of a
column
eL=h/2
+100
e
MEd=eNEd
eR=h/2+100
100
h
100
13
Introduction to a column under NEd & MEd
14
NEd
MED
2NEd
MED
Caused by Rigid Connection
between a column and a beam
14
NED
NED
e
Eccentric load
NED
MED
Introduction to a column under NEd & MEd
15
NEd
2NEd
NED
MN/2
ED
N/2
M
ED
e
NED
NED
N
M
1.0M
ED
0.5M
Caused by Rigid Connection
between a column and a beam
or a foundation
Eccentric load
15
Introduction – Type of columns
16

For the members of truss in compression
Pw
Pw
Pw
PP+Pw
16
Pw
Pw
Informative assessment 1
17



Explain how a bending moment is caused or
applied to a column.
Explain the terms of “rigid” and “pinned”
Describe how the type of a connection affects the
bending moment diagram of a frame.
17
Calculation of NEd and MEd on a column
18
Calculation of NEd
and MEd
eR=h/2+100
RR
eR=h/2+100
eL=h/2+100
RL
100
NEd=RL+RR
18
NEd=RR
MEd=eNEd
RR
h
100
MEd=RReR-RLeL=(RR-RL)e
Calculation of NEd and MEd on a column
Drawings taken from DESIGN OF COLUMN by Dr.Mongkol JIRAVACHARADET
1. Method of Tributary Area(half way to next column)
REd,T = Zone Area AT (m2) x Floor udl wd (kN/m2)
REd,B = Zone Area AB (m2) x Floor udl wd (kN/m2)
AT
A
T
×
×
AT
×
×
×
AB
×
AB
NEd = REd,T+REd,B,
MEd =REd,TeT - REd,BeB=(REd,T - REd,B)e
19
19
Calculation of NEd and MEd on a column
Drawings taken from DESIGN OF COLUMN by Dr.Mongkol JIRAVACHARADET
2. Method of Beam Reaction
NEd = RB1+RB2
MEd,y = RB1eB1-RB2eB2= (RB1-RB2)e
MEd,Z = RB3eB3-RB4eB4= (RB3-RB4)e
eB1
eB2
20
20
Calculation of NEd and MEd on a column
Drawings taken from DESIGN OF COLUMN by Dr.Mongkol JIRAVACHARADET
NEd =Sum of the reactions of all the upper beams and columns
MEd =Sum of those caused by the floor beam reactions only
RR,2
RL,2
NEd,2 = RL,2+RR,2
MEd,2,= RL,2eL,2- RR,2 eR,2
RL,1
RR,1
NEd,1 = NEd,2+RL,1+RR,1
MEd,2= RL,1 eL,1- RR,1 eR,1
RL,0
NEd =NEd,1+RL,0+RR,0
RR,0
,0
M
Ed,0,= RL,0eL,0- R21R,0eR,0
21
NEd and MEd on a column -Example
Figure shows a structural grid of five-storey building with the spans of AB and BC
of 5.0m and 7.0m. It carries a 200mm thick concrete slab and a characteristic
variable action of 4.5kN/m2 on the floor. The columns use 152x152x23UKC with
h=152.4mm . Calculate the design NEd and MEd of the column B2
C
3.0m
eAB (mm)
3.0m
7.0m
eBC (mm)
Reaction
of the
beam
3.5m
B
Reaction
of the
beam
RAB
2.5m
5.0m
6.0 m
6.0 m
2
h of UKC
100
mm
MEd,y =RBCeBC - RABeAB
A
1
RBC
100
mm
3
=(0.1+0.5h)(RBC-RAB)
22
22
NEd and MEd on a column – Solution
Permanent action:
Variable action
Design action on the floor:
gk=25x200/1000
=5.00 kN/m2
qk=4.5
=4.50 kN/m2
wd=(1.35x5+1.5x4.5) =13.5 kN/m2
Reaction of beam BC
C
RBC=wdsBCLBC/2 =13.5x6(7/2)=283.5 kN
7.0m
3.0m
3.0m
3.5m
Reaction of beam AB
RAB=wdsABLAB/2 =13.5x6(5/2)=202.5 kN
Total axial force in the ground floor
NEd=n(RBC+RAB)=5x(283.5+202.5)=2430kN
B
2.5m
5.0m
A
6.0 m
6.0 m
3
2
Design moment of the column in the ground floor 1
152x152x23UKC
MEd = (0.1+0.5h)(RBC- RAB)
=(0.1+0.5x0.1524)x(283.5-202.5)=14.27kNm
23
23
NEd and MEd on a column– Tutorial
Figure shows a structural grid of six-storey building with the spans of AB and BC
of 6.0m and 8.0m. It carries a 160mm thick concrete slab and a characteristic
variable action of 2.5kN/m2 simultaneously. The columns use 152x152x23UKC
with h=152.4mm . Calculate the design NEd and MEd of the column B2 and A2
eAB (mm)
C
8.0m
Reaction
of the
beam
Reaction
of the
beam
RAB
B
RBC
100
mm
6.0m
h
of UKC 100
mm
MEd,y = RBCeBC-RABeAB
A
7.0 m
7.0 m
1
24
eBC (mm)
2
3
=(0.1+0.5h)(RBC-RAB)
Load Effects of NEd and MEd – Stresses
25
NEd
ez
z
y
y
y
Mz,Ed=NEdez A = bh
y
Wz = b2h/6
Axial = NEd (kN)
Mz,Ed = NEd ez
NEd
z
z
h
b
z ez
h
b
NEd NEd A

ez
A
A Wz
 min 
NEd 
e 
1 z 

A  b / 6
z
y
y
N
M
N
+
max 
h
M
y
y
N
M
N
+
h
For full
compression,
i.e., min >0,
i.e., ez <b/6
M
25
Load Effects of NEd and MEd – Stresses
26
ey
NEd
y
y
NEd My,Ed=NEdey
y
y
z
z
b
N M
N
z
M
Axial = NEd (kN)
zb
max 
h
NEd NEd A

ey
A
A Wy
e 
N 
N M
 min  Ed 1  y 
y
y
+
A = bh
Wy = bh2/6
My,Ed = NEdey
h
y
b
26
y
z
b
ey
z
N
+
A 
h / 6
For full
M compression, i.e.,
min >0, i.e.,
ey <h/6
Load Effects of NEd and MEd – Stresses
27
NEd
e
y
Area A = bh, Axial= NEd (kN)
y
Due to ez, My,Ed= NEdey, Wy=bh2/6
z
Due to ey, Mz,Ed= NEdez, Wz=b2h/6,
b
z
D
A

 max
h
ez
y
C
y
h
ey
B
z
b
ey
N Ed 
e 
 z 
1 
A 
h /6 b 6
B 
ey
N Ed 
e 
 z 
1 
A 
h /6 b 6
C 
ey
N Ed 
e 
 z 
1 
A 
h /6 b 6
D
 min

ey
N Ed 
e 
 z 
1 
A 
h / 6 b 6
A
27
Load Effects of NEd and MEd – Stresses
28
From
D 
min
 Core area of
ey
N Ed 
ez 



1
0


section= a central
A  h / 6 b 6
ey
h/6

ez
1
b/6
[ez]<b/3
[ey]<h/3 y
y h
b
28
z
part of a section in
which any eccentrical
force can ONLY cause
either full
compressive or full
tensile stress over the
cross section.
 For a rectangular
section, it is one third
of its each side
lengths, respectively.
Load Effects of NEd and MEd – Stresses
29
Stress status of a column under NEd and MEd
NEd
NEd
MEd
min
NEd
MEd
min
min
max
Large Ned, Small MEd
MEd
max
max
Moderate NEd, & MEd
Small NEd, Large MEd
29
Informative assessment 2
30




30
Explain how to determine an axial load and
bending moment on a column.
Critically compare the axial force and bending
moment that act on a column in the group floor of a
multi-storey building.
Describe the distribution of stresses over the section
of a column under axial force and bending moment.
With an aid of sketch, explain the terms of “core
area of a section” and “one third rule”
Behaviour of Stocky or Restrained Column
31
NEd
MEd
If a stocky column or a column with its buckling
prevented from using partition walls.

Compression under NEd with Nc,Rd=Afy/M0

Bending under My,Ed with Mc,y,Rd=Wpl,yfy/M0

Bending under Mz,Ed with Mc,z,Rd=Wpl,zfy/M0

Combined maximum stress:
 max 

NEd M y, Ed M z ,Ed


 fy
A
Wy
Wz
Partition walls
Interaction between axial force NEd and
bending moments My,Ed and Mz,Rd
31
Behaviour of Slender unrestrained column
32
If a slender column or a column that has not got partition walls
or lateral restraints to get its buckling prevented from.
NEd
MEd
+
MEd
=
under Axial NEd under end moment MEd
Flexural
Lateral Torsional
Buckling
Buckling
32
NEd
Under NEd+MEd
Flexural-Torsional
Buckling
Informative assessment 3
33



Describe the terms “stocky column” and “slender
column”, “restrained column” and “unrestrained
column”
Explain the performance of stocky column under an
axial force and bending moment.
Decibel the behaviour of a slender column under
axial force and bending moment.
33
Resistance of Stocky or Restrained Column
34
If an axial load NEd is small, the plastic neutral axis NA
is in the web with z=NEd/(2fytw)≤(h-2tf)/2
=fy
C
Centre of
Section
NEd
MEd
NA
z
NM ,Rd  C  T  2 f ytw z
2
M N , y,Rd  f ybt f (h  t f )  f ytw 0.25  h  2t f   z 2 


N M , Rd  0
If z=0, pure bending only
34
T
NA
Ultimate
limit
state
=fy
M
M=fy
2
M y , Rd  f y bt f (h  t f )  f y tw 0.25  h  2t f    M N , y , Rd M


M=fy
Resistance of Stocky or restrained column
35
If an axial load NEd is large, the plastic neutral axis NA
is in the flange with z=NEd/(2fytw)>(h-2tf)/2
=fy
Centre of
Section
NEd
MEd
C
z
NA
NA
=fy
T
Ultimate
limit
state
N M , Rd  C  T  f y tw  h  2t f   2b  t f  0.5h  z  
M N , y , Rd  f y b  0.25 h 2  z 2 
M N , y ,Rd  0
If z=h/2, full compression
N Rd  C  T  f y  t w  h  2 t f   2 bt f   N M , Rd
N=fy
N=fy
35
Resistance of Stocky or restrained column
36
NEd/NM,Rd
Interaction between NEd and MEd
MEd/MN,Rd
36
Resistance of Stocky or restrained column
37
To account for the interaction between axial force NEd and
bending moment My,Ed , the reduced moment resistance Mpl,y,Rd due
to the simultaneously applied axial force:
MN,y,Rd=Mpl,y,Rd(1-n)/(1-0.5a) (≤Mpl,y,Rd)

n=NEd/Npl,Rd -Ratio of axial force to compression resistance

a=(A-2btf)/A (<0.5) – Ratio of web area to the whole area
If n=NEd/Npl,Rd≤0.25 (i.e., 25% of the overall compression resistance)
and nw=NEd/(hwtwfy/M0) ≤0.5 (i.e., 50% of the web resistance)
Then the interaction can be ignored and take: MN,y,Rd=Mpl,Rd
37
Resistance of Stocky or restrained column
38
To account for the interaction between axial force NEd and
bending moment Mz,Ed, the reduced moment resistance Mpl,z,Rd
due to the simultaneously applied axial force
MN,z,Rd=Mpl,z,Rd[1-((n-a)/(1-a))2] (≤Mpl,y,Rd)

n=NEd/Npl,Rd -Ratio of axial force to compression resistance

a=(A-2btf)/A (<0.5) – Ratio of web area to the whole area

n >a,

38
However, If n ≤a, then the interaction can be ignored amd
hence take: MN,z,Rd=Mpl,z,Rd
Resistance of Stocky or restrained column
39
39
Resistance of Slender or unrestrained column
40
MEd
+
Flexural
Buckling
Nb,Rd=Afy/M0
40
MEd
MEd
NEd
MEd
NEd
=
Lateral Torsional
Buckling(C1=1.7)
Mb,Rd=LTWpl,yfy/M0
Flexural-Torsional
Buckling
N
Myy/,EdM0
NEdb,Rd=Af
1.0
Mb,Rd=
LTWpl,yfy/
M0
Nmin,b,Rd Mb, y,Rd
Informative assessment 4
41


Describe all the types of the resistances of a column under
an axial force and bending moment.
Explain the ultimate limit states of stocky column and
slender column under an axial force and bending moment
Define the term of “interaction between axial force and
bending moment”
 Explain how to account for the effect of an axial force on
the moment resistance of a steel column
 How to account for simultaneous effects of axial force and
bending moment on safety of steel column

41
Design of Column under NEd & MEd
42
For a stocky column or a column with its buckling
prevented from using partition walls, check if the given
UKC section is adequate for the bending+compression,
i.e.,
 Under axial only:
NEd ≤ Nc,Rd = Afy/Mo
 Under uniaxial bending only:
My,Ed ≤ Mc,y,Rd = Wy,plfy/Mo
 Under uniaxial bending only:
Mz,Ed ≤ Mc,z,Rd = Wz,plfy/Mo
 Simultaneous effects
42
M y , Ed
M z , Ed
N Ed


 1.0
N c , Rd M c , y , Rd M c , z , Rd
Design of Column under NEd & MEd
43
For a stocky column or a column with its buckling
prevented from using partition walls, check if the given
UKC section is adequate for its reduced moment
resistance due to interaction with the axial force NEd
 Under uniaxial bending only:
My,Ed≤MN,y,Rd=Mpl,y,Rd(1-n)/(1-0.5a)
 Under uniaxial bending only:
Mz,Ed≤MN,z,Rd=Mpl,z,Rd[1-((n-a)/(1-a))2]
 Simultaneous effects :


where, =2, =5n,
 M y , Ed   M z , Ed 
 

  1.0 n=NEd/Npl,Rd
 M N , y , Rd   M N , z , Rd 
43
Design of Column under NEd & MEd
44
For a slender column or a column without getting its buckling
prevented from using partition walls, check if the given UKC
section is adequate for the flexural and LT buckling.
 Under axial only:
NEd ≤ Nb,Rd = Afy/Mo
 Under uniaxial bending only:
My,Ed ≤ Mb,y,Rd = LTWy,plfy/Mo
 Under uniaxial bending only:
Mz,Ed ≤ Mc,z,Rd = Wz,plfy/Mo
 Simultaneous effects:
44
My,Ed
M
NEd

1.5 z,Ed 1.0
Nmin,b,Rd Mb, y,Rd
Mc,z,Rd
Design of Column under NEd & MEd
45
To prevent both flexural bucking (in –plane) and lateral torsional
buckling (out of plane about major axis y-y) of a column under
axial force and bending moment, EC3 requires that, for a column
with a hot rolled I or H section and in simple construction, which
laterally restrained at both ends but unrestrained between floors:
SCI P362, Page 59-60
M
M
Simplified from (6.61)
NEd
 y,Ed 1.5 z,Ed 1.0 in Page (3-56) by
SN048b-EN-GB
Nmin,b,Rd Mb, y,Rd
Mcb,z,Rd
 NEd - Design value of an axial force on column
 My,Ed - Design value of a bending moment about major y-y axis
 Mz,Ed - Design value of a bending moment about minor z- z axis
45
Design of Column under NEd & MEd
46
NEd
Nmin,b, Rd

M y, Ed
M b, y,Rd
 1.5
M z ,Ed
M c, z ,Rd
 1.0
SCI P362, Page 59-60
Simplified from (6.61)
in Page (3-56) by
SN048b-EN-GB
Nmin,b,R =min [yAfy/M0; zAfy/M1]- Flexural buckling
resistance under axial compression design force of NEd
Mb,y,Rd = LTWpl,yfy/M1 - Lateral buckling resistance under
bending moment of My,Ed about the y-y axis along the member
Mc,z,Rd = Wpl,zfy/M1 -Bending resistance under bending moment
of Mz,Ed about the z-z axis along the member Mz,Ed
46
Effective length of a column
47
The effective length of a column depends on its end conditions.
EC3 gives no direct guidance and use those given in BS 5950.
47
Effective length of a column
48
Distribution factors for an “isolated” column
Top End of The Column
KKc c KK11
η 1 

1
KKc  KK1  KK11 KK12
c
1
11
12
Effective Stiffness for a Element
EI
L
Fixed at far end
1.0
Pinned at far end
0.75
Rotation as at near end
(double curvature)
1.5
EI
L
Rotation equal and opposite to
that at near end (single curvature)
0.5
EI
L
EI
L
Bottom End of The Column
η2  K
2
48
K c  K
K
K22
c

K
c
2  K 21  K 22
K c  K 2  K 21  K 22
Effective length of a column
49
Taken from SN008a–EN-EU
Lcr=Effective length of a column, Lcr,y =yL0, Lcr,z =zL0
 y and y=End restrained factor

1

Lcr
 0.5  0.14(1   2 )  0.055(1   2 ) 2
L
Non-sway frame
under vertical loads
2
49
Effective length of a column
50
Taken from SN008a–EN-EU
 Lcr=Effective length of a column, Lcr,y =yL0, Lcr,z =zL0
 y and y=End restrained factor
1
 
L cr

L
1  0 . 2 ( 1   2 )  0 . 12 1 2
1  0 . 8 ( 1   2 )  0 . 6 1 2
Sway frame under horizontal load
50
2
Effective length of a column-Example
51
A steel frame has
203x203x127UC
and
457x191x82UB for
its all beams and
columns in grade
S275.
 If the frame carries
vertical loads only,
determine the
effective length of
the column BE

G
5m
A
B
C
D
E
F
5m
5m
H
8m
6m
51
Effective length of a column-Solution
52
End
Top End Members
ColumnBottom End members
BG
BC
BA
BE
ED
EF
EH
203x203 457x191 457x191 203x203 457x19 457x191x 203x203x
UKB/UKC
x127+
1x82
82
127+
x127+
x82
x82
Member
Length
L0 (m)
5
8
6
5
6
8
5
Iy (cm4)
15400
37100
37100
15400
37100
37100
15400
Far End
Fixed
Pinned
K12=
0.75I/L
34.78
Pinned
K11=
0.75I/L
46.38
Pinned Pinned
K21=
K22=
Kc=I/L
0.75I/L 0.75I/L
30.80
46.38
34.78
Fixed
Stiffness
Ki
K2= I/L
30.80
Using Chart
0.66
2
0.66
β =0.5+0.14(ƞ1+ƞ2)+0.055(ƞ1+ƞ2)
52
30.80
0.432 (Kc+K2)/(Kc+K2+K21+K22)=ƞ2
ƞ1=(Kc+K1)/(Kc+K1+K11+K12)= 0.432

K2= I/L
Effective length of column L=Ly=L0=0.66x5=3.3m
Effective length of a column-Solution
53
L=L0=
=0.66x5
=3.3m
53
Effective length of a column-Tutorial
54
A steel frame has
305x305x158UC
and 406x140x53UB
for its all beams and
columns in grade
S275.
 If the frame carries
vertical loads only,
determine the
effective length of
the column BE
(5.16m)

54
G
5m
A
B
C
D
E
F
6m
4m
H
7m
8m
Steel column – Example
A 254x254x73 UKC steel column in grade S275 carries the
characteristic permanent and variable actions of 5.0 kN/m2 and
2.5kN/m2 over an area of 100m2 with an eccentricity of e=0.05m
about its major axis y-y. It has a length of 5.0m and effectively and
partially restrained about its minor and major axis, respectively.
eAB (mm)
eBC (mm)
C
Reaction
of the
beam
11m
Reaction
of the
beam
RAB
B
RBC
100
mm
9m
h of UKC 100
mm
MEd,y =RBCeBC - RABeAB
A
10m
1
10m
2
3
=(0.1+0.5h)(RBC-RAB)
55
55
Steel column – Example
56




56
Calculate the design axial force NEd and bending moment MEd of
the column B2
If the column is prevented from buckling using partition walls, check
by calculation whether the column can economically and safely
carry the axial force NEd and bending moment MEd.
If the above partition walls have been removed, use SCI table to
check whether the column can still safely and economically carry
the axial force NEd and bending moment MEd.
If the above partition walls have been removed, use SCI resistances
table to choose an alternative UC section for the column. Confirm
by calculation the alternative section can safely and economically
carry the above axial force NEd and bending moment MEd
a. Design axial force NEd & moment MEd
57
2
Permanent action gk=
5.0
kN/m
Permanent action gk=
2.5
kN/m
Floor area Af=
100
m
Eccentricity ey about y-y major axis =
0.05
m
2
2
Eccentricity ez about z-z minor axis =
0.00
m
Axial Dean Load
GK =gkAFloor
500
KN
Axial Live Load
PK =qkAFloor
250
KN
Permanent End Moment
My,GK =Gkey=
25
KNm
Variable End Moment
My,PK =Pkey=
12.5
KNm
Safety factor for permanent
 G=
 P=
1.35
Safety factor for variable
1.5
Design Axial force
NEd =  GGk+ PPk=
1050
Design Major moment
My,Ed =  GMy,Gk+ PMy,Pk=eyNEd=
52.5
KNm
Design minor moment
Mz,Ed =
0
KNm
KN
57
b. Compression/bending resistance
58
254x254x73
Section Width
254.6
254.1
mm
Section Depth
b=
h=
Fillets Depth
Web thickness
d=
tw=
200.3
8.6
mm
mm
Flange Thickness
tf=
14.2
mm
Root Radius
r=
A=
W py=
12.7
93.1
992
mm
W pz=
Iyy=
465
11400
cm
3
cm
4
Izz =
Iw =
3910
0.562
cm
4
Sectional Area
Plastic Modulus about y-y
Plastic Modulus about z-z
Section Moment
Section Moment
Warping Constant Iw =
2
cm
3
cm
dm
6
4
Torsion Constant It =
It =
57.6
cm
Self-weigh of steel beam gk,s=
gk,b=
73.1
kg/m
fy =
275
N/mm 2
Steel Strength
58
mm
b. Compression/bending design by calculation
59
Compression resistance Npl,Rd = Afy/ M0=
> Design force NEd =
2560.3
1050
KN
KNm
0.410
Safe
Moment resistance Mpl,y,Rd=Wpl,yfy/ M0 =
273
KNm
0.192
Design moment My,Ed =
52.5
KNm
Safe
Moment resistance Mpl,z,Rd=Wpl,zfy/ M0 =
127.9
KNm
0.000
Design moment Mz,Ed =
0
KNm
Safe
NEd/Npl,Rd+M y,Ed/M pl,y,Rd+M z,Ed/M pl,z,Rd=
0.60
<
1.00
>
>
Safe
N Ed M y , Ed M z , Ed


 1 .0
N Rd M y , Rd M z , Rd
59
b. Compression/bending design by calculation
60
Compression Ratio
n=NEd/Nc,Rd=
0.41
>
0.25
Web to Gross Area Ratio
a=(A-2btf)/A=
0.223
<
0.5
Web height
Web compression resistance
hw =h-2tf=
Nw,Rd=tw hw fy/ M0=
nw =NEd/Nw,Rd=
225.7
534
1.97
mm or hw=d+2r= 225.7
KN
>
The moment resistance of a UKC section about its major axis y-y
KNm
MN,y,Rd=Mc,y,Rd(1-n)/(1-0.5a)=
181
Design moment M y,Ed =
0.290
273
KNm
Safe
The moment resistance of a UKC section about its minor axis z-z
KNm
MN,z,Rd=Mc,z,Rd(1-((n-a)/(1-a))^2)=
120
128
>
>
Design moment M z,Ed =
<
52.5
=
=5n=
0
2
2.05
Limiting ratio =
0.084
1
(My,Ed/MN,y,Rd) +(Mz,Ed/MN,z,Rd) =
60
0.5
KNm
Safe
Safe
b. Compression/bending design by SCI Table
61
61
c-1. Flexural buckling Nb,Rd by calculation
62
about Y-Y
Buckling of Column
Initial Length
Restrained Factor
Effective Length
L0=
=
L=L0=
2
2
5000
5000
0.85
0.70
4250
3500
mm
KN
Elastic Buckling
Ncr= EI/L =
13081
6615
Reference Slenderness
 = (Afy/Ncr) =
0.5
0.442
0.622
b
c
0.34
0.49
Factor


0.639
0.797
Reduction Factor
fy       
0.909
0.772
Nb,Rd=Afy/M0
Nb,Rd=min{Nb,y,Rd; Nb,z,Rd}
2327
1977
Design force NEd =
1050
Buckling Curve
Imperfection Factor
>
62
about Z-Z
mm
1977
KNm
Safe
c-1. Buckling resistance using SCI table
63
4.0<Ly=0.85x5=4.25m<5.0,
3.0<Lz=0.7x5=3.5m<4.0,
Nb,y,Rd= (4.25-5)x(2360-2240)/(4-5)
+2240=2330kN
Nb,z,Rd=(3.5- 4)x(2110-1840)/(3 - 4)
+1840=1975kN
2360kN> Nb,y,Rd >2240kN,
2110kN> Nb,z,Rd >1840kN,
63
c-2. Lateral Torsional buckling Nb,Rd by calculation
64
Free Warping at beam ends
k=
kw=
Distance from load to shear centre
zg=h/2=
127.05
Moment Pattern by udl
C1=
1.77
Free Rotation about z-z
C2=
Buckling Force about z-z
Ncrz =  2EIz/(kL)2=
GIt=
mm
0
3241567
N
46656000000
Nmm2
z
14393
mm2
z
GIt/Ncr =
14393
mm2
2
GIt/Ncr =
14373
mm2
2
(C2zg) =
C2zg=
0
0
mm2
mm
Mcr=
973.13
kNm
0.5
0.529
1.00
b
0.34
0.75
(k/kw) Iw/Iz=
Non -dimensional slenderness ratio  LT= 1=[Wpl,yf y/Mcr] =
Sectional Ratio
h/b =
Buckling curve
Imperfection Factor
αLT=
β=
64
1
1
<
c-2. Lateral Torsional buckling Nb,Rd by calculation
65
65
c-2. LTB buckling Mb,y,Rd by SCI table
66
Ly=4.25m, C1=1.7 for the case of applied moment at ONE end.
Mb,Rd=273kNm>My,Ed=52.5kNm, Safe
66
c- 4) Flexural –Torsional buckling
67
NEd= 1050
Nb,Rd= 1977
My,Ed= 53
Mb,y,Rd= 273
Mz ,Ed= 0
Mc,z,Rd= 128
KN
kNm
kNm
<1.0
Safe
<1.0
Safe
<1.0
Safe
NEd/Nb,Rd=
My,Ed/Mb,y,Rd=
Mz,Ed/Mc,z,Rd=
NEd/Nb,Rd+My,Ed/Mb,y,Rd+1.5Mz,Ed/Mc,z,Rd=
<
N Ed
N min,b , Rd

M y , Ed
M b , y , Rd
 1.5
M z , Ed
M c , z , Rd
0.53
0.19
0.00
0.72
1.00
Safe
 1.0
67
Steel column - tutorial
68
A 305x305x97 UKC steel column in grade S275 has an effective length
of 7.0m and is subjected to the characteristic permanent and variable
axial forces of 500 kN and 250kN with an eccentricity of e=0.25m. It is
assumed that the column is effectively and partially restrained about its
minor and major axis, respectively.
 Determine the design axial load and bending moment on the column.
(NEd=1050kN, MEd =262.5kNm)
If the buckling of the column is prevented from using partition walls,
check if the column is safe to carry the design actions. (Nc,Rd=3382kN,
Nc,Rd=438kNm)
If the above partition walls have been removed, check if the column is
safe to carry the applied loads. (Nb,Rd=2385kN, Mb,Rd=438kNm,
1.04>1.0, not suitable)

68