ENG2032-N: Structural Analysis and Design

Design of steel structures

Design of Steel Columns

Steel members under axial load (Tension, Compression)

Dr Thadshajini Suntharaligam
Lecturer in Civil Engineering
7th December 2023
 Intended Learning Outcomes

What we’ll cover in this lecture

• Design of steel members under tension.

• Design of steel members under pure compression.
• Design of steel columns under compression and bending (simple
construction).
 Sources
• Trahair NS, Bradford MA, Nethercot D, and Gardner L (2018), ‘The Behaviour and Design of
Steel Structures to EC3’, 4th Edition, Taylor and Francis Group.
• Draycott T (2019), ‘Structural Elements Design Manual: Working with Eurocodes’. CRC Press.
• Arya Chanakya, ‘Design of Structural Elements: Concrete, Steelwork, Masonry and Timber
Designs to British Standards and Eurocodes, 3rd Edition. Taylor and Francis.
• SCI Publication P360 (2011), ‘Stability of Steel Beams and Columns’.
• Davison and Owens (2012), ‘Steel Designers’ Manual’, John Wiley and Sons.
• McKenzie WMC (2015), ‘Design of Structural Elements’, Macmillan Education UK.

Useful links of organizations that are related to steel construction industry with a lot of
information and design guidelines.
(i) The Steel Construction Institute (SCI), www.steel-sci.org
(ii) The British Constructional Steelwork Association, www.steelconstruction.org
(iii)CORUS, www.corusconstruction.com
Local Buckling & Section Classification
Columns in building may be subject to pure
compression if the load is applied concentrically,
or in a combination of compression and bending
moment.

Members in steel trusses can be under

compression or tension depending on their
position.
Load transmission by structural members

Source: Trahair NS et al
DESIGN OF STEEL TENSION MEMBERS
 Perfect tension members

Source: Trahair NS et al

Concentrically loaded uniform tension members with no holes, initial crookedness

or residual stresses is a perfect tension member.
Steel members under tension

Jiang et al (2021)

Jia et al (2014)
 Design of tension members
According to EC3, there are two limit states to consider when designing a tension
member, i.e.
(1) The yield of the gross section.
(2) The fracture of net section.
Both of the limit states are covered by a single equation: 𝑁𝐸𝑑 ≤ 𝑁𝑡,𝑅𝑑

𝐴𝑓𝑦 0.90Α𝑛𝑒𝑡 𝑓𝑢
𝑁𝑡,𝑅𝑑 = 𝑚𝑖𝑛 𝑁𝑝𝑙,𝑅𝑑 , 𝑁𝑢,𝑅𝑑 = 𝑚𝑖𝑛 ,
𝛾𝑀0 𝛾𝑀2
A is the gross section area and 𝐴𝑛𝑒𝑡 the net area of cross-section.

Special requirement for capacity design.

𝐴𝑓𝑦 0.90Α𝑛𝑒𝑡 𝑓𝑢 A design method for achieving the plastic
≤
𝛾𝑀0 𝛾𝑀2 deformation capacity of a member by providing
additional strength in its connections

Source: Trahair NS et al
LOCAL BUCKLING & SECTION CLASSIFICATION
OF MEMBERS UNDER COMPRESSION
 Global and local buckling

Member buckling Local buckling

Member buckling
 Local buckling

Open
Different steel sections
section

Closed
section

Sufficiently small Insufficient ‘plate

‘plate slenderness’ slenderness’
 Slenderness limits

A column under compression

will have all plate elements
under compression

Source: Davison & Owens

𝑤ℎ𝑒𝑟𝑒 𝜀 = 235/𝑓𝑦

From Steel Designers’ Manual Width c is for the flat part ignoring the fillet or the
weld of the root.
Slenderness limits
Table from EN1993 – 1 – 1
Slenderness limits

Source: Trahair NS et al
 Section class & columns
o Under central axial compression, columns with Class 1, 2 and 3 sections
will yield without local buckling.
o Only Class 4 sections will experience local buckling.

Source: Draycott
DESIGN OF COMPRESSED STEEL
MEMBERS (STRUTS)
 Compression resistance of section
The first step in strut design must involve consideration of local buckling as it
influences axial capacity.

The compression resistance Nc,Rd of a cross – section is given as:

𝐴𝑓𝑦
𝑁𝑐,𝑅𝑑 =
𝛾Μ0
(for Class 1, 2 or 3 cross – sections) Compact sections

𝐴𝑒𝑓𝑓 𝑓𝑦
𝑁𝑐,𝑅𝑑 = (for Class 4 cross – sections)
𝛾Μ0 Slender sections

where:
A is the gross area of the cross-section.
Aeff is the effective area of the cross-section.
fy is the yield strength.
 Buckling strength of struts under axial load

The buckling resistance Nb,Rd of a compressed strut is:

𝐴𝑓𝑦
𝑁𝑏,𝑅𝑑 =𝜒 (for Class 1, 2 or 3 cross – sections) Compact sections
𝛾Μ1

𝐴𝑒𝑓𝑓 𝑓𝑦
𝑁𝑏,𝑅𝑑 =𝜒 (for Class 4 cross – sections)
𝛾Μ1 Slender sections
 Compressed members – flexural buckling
The buckling resistance 𝑵𝒃,𝑹𝒅 is given in terms of a reduction factor 𝜒 that
,
ഥ
depends on the non-dimensional slenderness 𝝀.
𝐴𝑓𝑦
𝑁𝑏,𝑅𝑑 =𝜒 Yielding strength
𝛾Μ1 (for Class 1, 2 and 3 cross – sections)
Euler strength
Plastic capacity
𝑁𝑝𝑙,𝑅𝑑
𝜆ҧ =
Ν𝑐𝑟
Elastic (Euler) buckling load
𝒍𝒆𝒇𝒇 Imperfection factor α
𝒐𝒓 𝒆𝒒𝒖𝒊𝒗𝒂𝒍𝒆𝒏𝒕𝒍𝒚 𝒂𝒔 𝝀 = Buckling curve a0 a b c d
𝒊
Imperfection factor α 0.13 0.21 0.34 0.49 0.76
𝝀 𝚬
ത𝝀𝒊 = 𝒊 𝒂𝒏𝒅 𝝀𝟏 = 𝝅
𝝀𝟏 𝒇𝒚
𝒊 = 𝒚 𝒐𝒓 𝒛 1
𝜒= ≤ 1.0 Φ = 0.5 1 + 𝛼 𝜆ҧ − 0.2 + 𝜆ҧ2
Φ + Φ2 − 𝜆ҧ2
Dimensionless load and slenderness

Source: Trahair NS et al
Compressed members – flexural buckling
STEEL COLUMNS UNDER N AND BIAXIAL M
Members under axial compression and bending

Flexural buckling Flexural buckling

mode (strong axis) mode (weak axis)

Isolated column (strut)

member under axial load
and bending moments
Torsional – flexural
buckling mode Local buckling mode
General equations for column buckling
𝑁𝐸𝑑 𝑀𝑦,𝐸𝑑 + Δ𝑀𝑦,𝐸𝑑 𝑀𝑧,𝐸𝑑 + Δ𝑀𝑧,𝐸𝑑
+ 𝑘𝑦𝑦 + 𝑘𝑦𝑧 ≤1
𝜒𝑦 𝑁𝑅𝑘 𝑀𝑦,𝑅𝑘 𝑀𝑧,𝑅𝑘
𝜒𝐿𝑇 𝛾Μ1
𝛾Μ1 𝛾Μ1

𝑁𝐸𝑑 𝑀𝑦,𝐸𝑑 + Δ𝑀𝑦,𝐸𝑑 𝑀𝑧,𝐸𝑑 + Δ𝑀𝑧,𝐸𝑑

+ 𝑘𝑧𝑦 + 𝑘𝑧𝑧 ≤1
𝜒𝑧 𝑁𝑅𝑘 𝑀𝑦,𝑅𝑘 𝑀𝑧,𝑅𝑘
𝛾Μ1 𝜒𝐿𝑇 𝛾Μ1
𝛾Μ1
COLUMN DESIGN IN ‘SIMPLE CONSTRUCTION’
 Simple construction

Structures composed of members connected by nominally pinned joints and

resistance to horizontal forces is provided by bracing.

NCCI Publication SN048, ‘Verification of columns in simple construction – a

simplified interaction criterion’.
 Criteria for the use of the simplified method

The criteria to be satisfied for the use of the simplified method are:
1. the column is hot-rolled I or H section, or rectangular hollow section.
2. the cross-section is Class 1, Class 2 or Class 3 under compression.
3. the bending moment diagrams about each axis are linear.
4. the column is restrained laterally in both the y and z directions at each floor but is
unrestrained between floors.
5. the design moments have been derived assuming ‘simple construction’ in which the
beam vertical reactions are assumed to act at a distance of 100 mm from the
web/flange face of the column − satisfied.
6. the end-moment ratio ψ is restricted as indicated in Table 7.4 (See next slide).
McKenzie WMC (2015), ‘Design of
Structural Elements’.
 Criteria for the use of the simplified method

Bending moment ratios

McKenzie WMC (2015), ‘Design of
Structural Elements’.

e.g. for I and H sections 𝑘𝑦𝑦 = 1.0 for 𝜓 ≤ −0.11 and 𝑘𝑦𝑧 = 1.5 for all cases.
* If the column is pin-ended, i.e. 𝜓 = 0, the simplified expression is still valid with 𝑘𝑦𝑦 = 1.0 if 𝑁𝐸𝑑 Τ𝑁𝑦,𝑏,𝑅𝑑
≤ 0.83
 Simplified interaction criterion
𝑵𝑬𝒅 𝑴𝒚,𝑬𝒅 𝑴𝒛,𝑬𝒅
+ + 𝟏. 𝟓 ≤𝟏
𝑵𝒃,𝑹𝒅 𝑴𝒚,𝒃,𝑹𝒅 𝑴𝒛,𝒄𝒃,𝑹𝒅

𝜒𝑦 𝑓𝑦 𝐴 𝜒𝑧 𝑓𝑦 𝐴
𝑁𝑏,𝑅𝑑 = 𝑚𝑖𝑛
𝛾Μ1
,
𝛾Μ1 Flexural buckling strength
𝑓𝑦 𝑊𝑝𝑙,𝑦
𝑀y,𝑏,𝑅𝑑 = 𝜒𝐿Τ
𝛾Μ1
Bending strength for strong axis bending

𝑓𝑦 𝑊𝑝𝑙,𝑧
𝑀𝑧,𝑐𝑏,𝑅𝑑 =
𝛾Μ0
Bending strength for weak axis bending

Nb,Rd: Design compression resistance of a column

My,b,Rd: Bending strength in strong axis y – y
Mz,cb,Rd: Bending strength in weak axis z – z
NCCI Publication SN048, ‘Verification of columns in simple construction – a simplified interaction criterion’.
 Eccentricity bending moments

𝑵𝑬𝒅 𝑴𝒚,𝑬𝒅 𝑴𝒛,𝑬𝒅

+ + 𝟏. 𝟓 ≤𝟏
𝑵𝒃,𝑹𝒅 𝑴𝒚,𝒃,𝑹𝒅 𝑴𝒛,𝒄𝒃,𝑹𝒅

SCI Publication 395, ‘Steel Building Design: Medium Rise Braced Frames’.
 Design summary
Columns under concentric load
• Calculate the ultimate axial load N applied to the column.
• Select a trial section if it is a design (size selection) problem (e.g. assuming
χ=0.5 and selecting section such as 𝐴 ≥ 𝑁𝐸𝑑 Τ𝜒).
• Determine the effective (critical) lengths 𝑙𝑐𝑟,𝑦 and 𝑙𝑐𝑟,𝑧 .
• Calculate the slenderness rations 𝜆𝑦 = 𝑙𝑐𝑟,𝑦 Τ𝑖𝑦 and 𝜆𝑧 = 𝑙𝑐𝑟,𝑧 Τ𝑖𝑧 . Ensure that
neither are greater than 180*.
• For the y and z axes separately.: Estimate the compression resistance Nb,Rd.
• The compression strength of the column will be the smallest value of the two
strengths found in the previous step.
* Draycott T (2019), ‘Structural Elements Design Manual: Working with Eurocodes’. CRC Press.
 Design summary
Columns under eccentric load
If the column is under eccentric load, e.g. in simple construction frames, then
continue with the following calculations.
• Calculate the nominal eccentricities e and the nominal moments 𝑀𝑦,𝐸𝑑 and
𝑀𝑧,𝐸𝑑 .
• Considering 𝐶 = 1 estimate the bending resistance of the member for bending
about the strong axis 𝑀𝑏,𝑦,𝑅𝑑 .
• Noting that LTB does not occur when a UC section is bent about its z-z axis,
determine the bending resistance moment for z-z bending 𝑀𝑏,𝑧,𝑅𝑑 = 𝑊𝑝𝑙 𝑓𝑦 .
• Check that the trial section satisfies the interaction formula:
𝑵𝑬𝒅 𝑴𝒚,𝑬𝒅 𝑴𝒛,𝑬𝒅
+ + 𝟏. 𝟓 ≤𝟏
𝑵𝒃,𝑹𝒅 𝑴𝒚,𝒃,𝑹𝒅 𝑴𝒛,𝒄𝒃,𝑹𝒅
COLUMN BUCKLING LENGTH (BRACED FRAMES)
Effect of lateral restraint on the effective length

Theoretical effective lengths

Source: Trahair NS et al
 Column Buckling Length (Braced frames)

Pinned/ Fixed/ Fixed/

End restraints
Pinned Pinned Fixed
Effective length
1.0 0.85 0.7
factor
SN008a-En-EU_NCCI-Buckling Lengths of
Columns – Rigorous

SCI Publication P360 (2011), ‘Stability of Steel Beams and Columns’.

Column Buckling Length (Braced frames)

SCI Publication P360 (2011), ‘Stability

of Steel Beams and Columns’.
Worked examples
 Rolled Universal Column design
Problem
Check the ability of a 203 × 203 × 52 UC in grade S275 steel to withstand a
design axial compressive load of 1150 kN over an unsupported height of 3.6 m
assuming that both ends of the member are pinned. Design to BS EN 1993-1-1.
The problem is as shown in the sketch below:
Rolled Universal Column design
Rolled Universal Column design
Rolled Universal Column design
 Pinned column with intermediate lateral restraints
Problem
A 254 × 254 × 89 UC in grade S275 steel is to be used as a
12.0 m column with pin ends and intermediate lateral braces
provided restraint against minor axis buckling at third points
along the column length. Check the adequacy of the
column, according to BS EN 1993-1-1, to carry a design
axial compressive load of 1250 kN. The problem is as shown
in the next sketch:

Source: Davison and Owens

Pinned column with intermediate lateral restraints
Pinned column with intermediate lateral restraints
Pinned column with intermediate lateral restraints
Problem: Axially Loaded Columns with Moments from Eccentric Loads
A steel column is 4.5 m high and carries the beams shown in Figure 5.29. The beam reactions
and column self-weight (SW) are also shown. The column is fixed to a substantial foundation at
the bottom and is restrained in position but not direction at the top. Choose a UC section in
S275 steel and determine whether it is satisfactory.

Pinned/ Fixed/ Fixed/

End restraints
Pinned Pinned Fixed
Effective length
1.0 0.85 0.7
factor
SN008a-En-EU_NCCI-Buckling Lengths of
Columns – Rigorous

Source: Draycott
Calculate the forces at ULS

Beam A: 1.35 X 21 + 1.50 X 27 = 68.9kN

BeamB: 1.35 X 15 + 1.50 X 18 = 47.3kN
Beam C: 1.35 X 54 + 1.50 X 48 =144.9kN
Column SW: 1.35 X 2.0 = 2.7kN
ltimate axial load NEd = 68.9 + 47.3 + 144.9 + 2.7 NEd = 263.8 kN
A ume a trial ection of 152 X 152 X 37 kg/m UC
From Table 5.8 for 152 X 152 X 37 kg/m UC h = 161.8mm
b = 154.4mm
fw = 8.0mm
ff= 11.5mm
From Table 5.9 for 152 X 152 X 37 kg/m UC Iy = 2210cm4
Iz = 706cm4
iy = 6.9cm
iz = 3.9cm
Wpt,y = 309 cm 3
Wp 1,z = 140cm 3
lw = 0.040dm6
IT= 19.2cm4
A= 47.1 cm 2
From Table 5.2 E = 210kN/mm2
G = 81kN/mm 2
British Standard provision
 2
C 1r E lz
Mer = 1
L2

X 210 X 10 X 706 X 10 4
3
0.040 X 10 12 3,830 2 X 81 X 10 3 X 19.2 X 10 4
= l.0 X
2
1r
----+-----------
3,830 2 706 X 10 4 1r X210 X 10 X 706 X 10
2 3 4

= 145.4 X 106 mm Mee = 145.4 Ill

JyWply = 275 X 309 X 10 3 = 85.0 X 106 mm fy Wpl,y = 85.0 Ill

fy Wpl,y {85.o
= = 0.765
Mer Vl45A ALT

h/bforthe ection = 161.8/154.4 = 1.05 ofromTable5.13

Fro1n Table 5.14 ,,\LT ' O = 0.4

(3 = 0.75

<J>LT = 0.781

1 1
= - = 0.837
XLT q\T + <J>tT✓ - J3AtT ✓
0.781 + 0.781 - 0.75 X 0.765 2
2 XLT
 1 1
Check that XLT i not more than 1.0 and i not more than _2 = = 1.71 Accept
,\LT 0.765 2

C 1 = LO� of= 1.0 f= 1.0

XLT,mod= XLT!f = 0.837/1.0 LTmod
, = 0.837
De ign buckling re i tance moment My,b,Rd = XLT,moctfy Wpl,y
= 0.837 X 85.0 m My,b,Rd = 71.1 k m

Bending about z-z axis

ominal eccentricity e = tw/2 + 100 = 8/2 + 100 e = 104mm
ominal moment MzEd
' = VEd Xe = 68.9 X 0.104 - 47.3 X 0.104 MzEd
, = 2.25 m
fyWplz, = 275 X 140 X 103 = 38.5 X 106 mm JyWply = 3 .5 m

Su1nmary of de ign values and re istances

De ign value Re i tance
Axial force NEd = 263.8k Nb 'Rd607 k
=

- bending My,Ed = 26.2k.Nm My b Rd = 7 1.1 111

z-z bending Mz'Ed = 2.25k.Nm JyWpl,z = 38.5k.Nm

Interaction check

NEd My,fal Mz, Ed 263·8 + 26·2 + 5 2·25

+ + 1.5 = 1. X = 0.435 + 0.368 + 0.088 = 0.891 -
< 1.0
Nb,Rd Myb,Rd fyWpl,z 607 71.l 38.5
Adopt 152 X 152 X 37 kg/m UC
Steel joints
Steel joints

Source: Trahair NS et al
Steel joints
Steel joints
Steel joints
Steel joints