Example- 1

The column B – E on the Figure shown below is under the action of

Design axial load NSd = 2800 kN Both sides are pinned. Check the
resistance (adequacy) of the column. Steel grade S - 275 is used.

1
Solution
Step 1: Axial load NSd = 2800 kN (given)

Step 2: Buckling length L = 4000 mm (pinned end both sides. (given).

Step 3: The section is given.

Step 4: Determine the class of the cross-section and check for a local buckling. \

➢ The section is subjected to uniform compression. For the section to be classified as

at least class 3, in order to avoid any modification to the full cross sectional area due
to local buckling, the limiting width to thickness ratio for class 3 section are (See
Table 5.2 ES EN-3).

2
• For S – 275 steel grade fy = 275 N / mm2. Thus  = 235 275 = 0.92
• Outstand element of compression flange: c/tf ≤ 14 ε.
• Internal element: c/tw ≤ 42 ε.
This gives the following limiting value for at least Class 3 :
• Outstand element of compression flange:
c / tf = ((254-9.1)/2)-33 / 16.3 = 5.49 < 14 x 0.92 = 12.88 OK!
• Internal element:
d/ tw = (310-2 (33)-2*16.3) / 9.1 = 23.23 < 42 x 0.92 = 38.64 OK!
• Therefore, the section belongs to at least Class 3.
Thus, βA = 1.0 3
Step 5: Check for uniform compression Resistance

The design resistance of the cross section for compression

NC,Rd = Np,Rd = (A.Fy)/γmo = 11000*275/1.0 = 3025KN

2800KN
≤1 OK!
3025KN

4
Step 6: Buckling Resistance
Determine the non-dimensional slenderness ratio.
▪ For S - 275 steel grade, λ1 = 93.9 ε = 93.9 x 0.92 = 86.39

▪ Slenderness ratio about y-axis: λy = L / 𝑖𝑦 = 4000 / 135 = 29.63

▪ Slenderness ratio about z-axis: λz = L / 𝑖𝑧 = 4000/63.6 = 62.89

▪ Hence, the non-dimensional slenderness ratio is determined as:

it is slander column, so check the column for buckling !

5
Step 7: Determine the appropriate buckling curves (from Table 6.2).

▪ Use curve a for buckling about y-axis and curve b for buckling about z-axis.

Step 8: Determine value of χ. Using Fig.6.4 :

▪ For y-axis: curve a

▪ For z-axis: curve b

Therefore, buckling about the z-axis becomes critical. From the fig. 6.4

6
Step 9: Calculate the design buckling resistance

𝜒𝛽𝐴 𝐴𝑓𝑦 0.77 × 1 × 11000 × 275

𝑵𝒃,𝑹𝒅 = = = 2329.3𝐾𝑁
𝛾𝑀1 1.0

2800 KN > 2329.3 KN, the column do not resist.

Solution . Add an additional hinged support at mid-height to increase the

resistance about the minor axis

7
.

y-axis: χy = 0.97 don’t varies

z-axis: Curve b for

• Hence buckling about the z-axis becomes critical

𝜒𝛽𝐴 𝐴𝑓𝑦 0.94 × 1 × 11000 × 275

𝑵𝒃,𝑹𝒅 = = = 2843.5𝐾𝑁 > 2800 𝑂𝐾!
𝛾𝑀1 1.0 8
Example- 2
The 457 x 152 x 52 UB used for a pin-ended column. Check the adequacy of
the member for a factored axial compressive load corresponding to a nominal
dead load of 160 kN and a nominal imposed load of 230kN. If the column is
3.00 m long and its steel grade is S 235.

9
Step 1: Determine the class of the cross-section and check for local
buckling.
For, 𝑡𝑓 = 10.9 < 40 and S 235 steel grade ( see table 3.1)

𝑓𝑦 = 235 𝑁 / 𝑚𝑚2. Thus, 𝜀 = 235

ൗ𝑓𝑦 = 1

These limiting values are:

• Outstand element of compression flange: 𝑐/𝑡𝑓 ≤ 14𝜀 = 14
• Internal compression element: c/𝑡𝑤 ≤ 42 𝜀 = 42

10
For the 457 x 152 x 52 UB profile, the actual values are given:
• Outstand element of compression flange:

𝑐
= (152.4 − 7.6/2 − 10.2) / 10.9 = 5.7 < 14 𝑂𝐾.
𝑡𝑓
• Internal compression element (web) :
d / 𝑡𝑤 = (449.8 – 2 𝑥 10.9 – 2 𝑥 10.2) / 7.6 = 53.63 > 42

• Therefore, the flange satisfies the Class 3 requirement, but the web is
Class 4 section. Consequently, there must be a reduction in the strength of the
section to allow for the load buckling which will take place in the web. Therefore, the
effective area, 𝑨𝒆𝒇𝒇 must be determined for the web. 11
Explanation for the effect.

12
• The effective width is
ഥ𝒆𝒇𝒇 = 𝑟𝑒𝑑𝑢𝑐𝑡𝑖𝑜𝑛 𝑓𝑎𝑐𝑡𝑜𝑟 × 𝑏ത = 𝜌 × 𝑏ത
𝒃

The method to calculate the effective area (𝑨𝒆𝒇𝒇) is explained in section

4.3.4 of ES-3.

• To calculate the reduction factor 𝝆 is as follow for internal element

𝑎) 𝜌 = 1, 𝑖𝑓 𝜆ҧ𝑝 ≤ 0.673
ഥ 𝑝 − 0.055(3 + Ψ)ൗ𝜆ҧ2𝑝 ≤ 1.0, 𝑖𝑓 𝜆ҧ𝑝 > 0.673
𝑏) 𝜌 = (𝜆

𝑓𝑦 𝑏ത
Where; 𝜆ҧ𝑝 = 𝜎𝑐𝑟
= 𝑡
ൗ 28.4𝜀 𝑘𝜎
13
• In our example, since the column is axially loaded the stress
distribution is uniform, i.e. 𝜎1 = 𝜎2. (Table 4.1 in this slid,page-22), for
internal element is used to calculate the effective width. Thus, Ψ =
𝜎1/ 𝜎2 = 1, and 𝑘𝜎 = 4.0

𝑏ത = 𝑏𝑤 = 449.8 − 2 ∗ 10.9 − 2 ∗ 10.2 = 407.6𝑚𝑚

𝑏തൗ = 407.6Τ7.6 = 53.6
𝑡𝑤

𝜆ҧ𝑝 = 53.6൘ = 0.944 > 0.673

(28.4 × 1 × 4 )

(𝜆ҧ𝑝 − 0.055(3 + 1) (0.944 − 0.22)ൗ

∴𝜌= ൘ 2ҧ = = 0.812
𝜆𝑝 0.9442

𝐴𝑛𝑑 𝑏ത𝑒𝑓𝑓 = 𝜌𝑏ത = 0.812 × 407.6 = 331.2𝑚𝑚 14

 ▪ Therefore the area that should be ignored at the center of the web is:
∆A = 407.7 − 331.2 × 7.6 = 581.4𝑚𝑚 2 ,

Aeff = A - ∆A = 6650 - 581.4 = 6068.6mm2 Then βA = AAeff = 6068.6Τ6650 = 0.913

Step 2: Check for uniform compression Resistance

Design compression force
NEd = 1.35 x160 +1.5x230 = 561 KN

The design resistance of the c – 4 cross section for compression

NC,Rd = (Aeff.Fy)/γmo = 6068.6*235/1.0 = 1426.12KN
561KN /1426.12KN = 0.39 ≤ 1 OK!

15
Step 3: Buckling resistance

Step 3.1: Appropriate column buckling curve.

• For ℎ / 𝑏 = 449.8 / 152.4 = 2.95 > 1.2; and

3000
𝜆𝑦 = 𝐿𝑒/𝑖 = = 16.76
▪ 𝑡𝑓 = 10.9 < 40 𝑚𝑚; 179

3000
✓ Use curve a for y-y axis and 𝜆𝑧 = 𝐿𝑒/𝑖 = = 96.46
31.1

✓ Use curve 𝒃 for buckling about z-z axis. 𝜆1 = 93.9Ɛ = 93.9*1 = 93.9 λത

ഥλ𝑦 = 16.76
0.913= 0.171< 0.2
93.9
Step 3.2: Determine the value of χ. Hence buckling resistance not need to calculate

ഥλ𝑧 = 96.46/93.9√0.913 = 0.982> 0.2

Hence buckling resistance need to be calculated
16
The appropriate buckling curve depends on h/ b and The appropriate buckling curve depends on h/ b and
steel grade steel grade

h 449.8 h 449.8
= = 2.95 > 1.2 and t < 40 mm = = 2.95 > 1.2 and t < 40 mm
b 152.4 b 152.4
therefore, refer buckling curve a for y-y axis αy = 0.21 therefore, refer buckling curve b for z-z axis αz = 0.34
𝜙𝑦 = 0.5 x 1 + 0.21 0.171 − 0.2 + 0.1712 = ϕz = 0.5 x 1 + 0.34 0.982 − 0.2 + 0.9822 = 1.12

𝜙𝑦 = 0.512 1
χz = = 0.603 < 1.0
1.12 1.122 − 0.9822
1
χy = = 1.005 ≅ 1.0
0.512 + 0.5122 − 0.1712 Therefore, 𝝌 = 𝟎. 𝟔𝟎𝟑 (least value)

χ𝑧 βA Aeff fy 0.603 X 0.913𝑋6068.6 X 235

Nb,Rd = = = 785,057.6N = 785.06KN > 160 kN
ϑM1 1

Therefore, the section is adequate to resist for bucking

17
Example-3
Design the column BD of the steel structure represented in the figure below,
using a UB cross section in S 355 steel, according to EC1993-1-1. The column
is fixed at the base and hinged at section B (with respect to the two principal
axis of the cross section). Cross section B is fixed in both horizontal directions,
in the plane of the structure (due to the beam itself) and in the perpendicular
plane (because of secondary bracing members). Loading already factored for
ULS.

18
Solution:

Step – 1: Compute the design applied compressive axial force 𝑁𝐸𝐷

• Load on column 𝑁𝐸𝐷 ∗ 10= 80*12*6+800*10

80∗12∗6
• ⇒ 𝑁𝐸𝐷 = +800 = 1376KN
10

19
Step – 2: Select a preliminary cross section

Assume class1, 2 or 3 cross section, and considering minimum cross section resistance.

𝐴𝑓𝑦 𝑁𝐸𝐷∗𝛾𝑚𝑜 1376∗103 𝑁∗1.0

𝑁𝐸𝐷 = 1376KN ≤ 𝑁𝐶𝑅𝐷 = , A≥ = = 0.003876𝑚2
𝛾𝑚𝑜 𝑓𝑦 355𝑁/𝑚𝑚2

Trial-1: As it is expected that buckling resistance will govern the member design, section
of 254*254*85 steel profile with the following properties (geometrical and
mechanical).

20
Properties of 254 × 254 x 85 UKB section

h = 254.3 mm. d = ℎ𝑤 = 200.3 mm b = 260.4 mm.

𝑡𝑓 = 14.3 mm. 𝑡𝑤 = 14.4 mm. r = 12.7 mm

𝑖𝑦𝑦 = 10.6 cm 𝑖𝑧𝑧 = 6.24 cm. A = 108 cm2

𝐶𝑤 𝐶𝑓
= 7.71 = 7.71 =13.9 𝐼𝑍𝑍 = 4215 cm4
𝑡𝑤 𝑡𝑓

𝐼𝑦𝑦 = 12284 cm4

21
 235 235
Stress factor (ε) = = = 0.81
𝑓𝑦 355

Step – 3: Check the class of Classification of the section

𝐶𝑓 𝐶𝑤
From table: = 7.71 and = 13.9
𝑡𝑓 𝑡𝑤

Outstand element of flange Internal element of web

𝐶𝑓 𝐶𝑤
For Class -1 limit: ≤ 9ϵ For Class -1 limit: ≤ 33 ϵ
𝑡𝑤
𝑡𝑓

𝐶𝑓 𝐶𝑤
≤ 9 x 0.81= 7.71 ≤ 7.29---Ok! ≤ 13.9 ≤ 33 X 0.81 =>13.9 ≤ 26.73
𝑡𝑤
𝑡𝑓
-- Ok! 22
Step – 2: For uniform compression Resistance

𝑁𝐸𝑑
≤1
𝑁𝐶,𝑅𝑑

The design resistance of the cross section for compression

𝐴𝑓𝑦 10800∗355
𝑁𝐶𝑅𝐷 = = = 3465666.2 KN = 3465.66 KN > 1532.25 KN
𝛾𝑚𝑜 1.0
NEd 1532.25
= = 0.44 < 1.0 ------------ Ok!
NC,Rd 3465.66

23
Step – 3: Check for the buckling resistance (Check for instability)

Step 3.1: Identify the Buckling length in both directions.

According to the support conditions, the buckling lengths are equal in both
planes, given by:
𝜒 𝐴 fy
Nb,Rd = => for class 1
ϑM1

NEd
≤1 λ1 = 93.9 ε = 93.9 x 0.81 = 76.059
Nb,Rd
1
𝜒= ≤ 1.0 ; 𝜙 = 0.5 1 + 𝛼 𝜆ҧ − 0.2 + 𝜆2
𝜙+ 𝜙2 −𝜆2 )

24
 𝐴𝑓𝑦 𝑙𝑒 1
𝜆ҧ𝑖 = = ∗
𝑁𝑐𝑟 𝑖 𝜆1

As the beam and column are flexible connected, we shall assume it as pin
connected and column at the base is pin connected therefore k = 0.7L =>
𝐿𝑒𝑦 = 𝐿𝑒𝑧 = 0.7L = 0.7*8m = 5.6m = 5600mm

25
 Buckling about major axis (y-y) Buckling about minor axis (z-z)
5600 1
𝜆ҧ𝑦 = = 0.695 ≥ 0.2 5600 1
106 76.059 𝜆ҧ𝑧 = = 1.8 ≥ 0.2
62.4 76.059
Hence buckling resistance need to be calculated
The appropriate buckling curve depends on h/ b and Hence buckling resistance need to be calculated

steel grade The appropriate buckling curve depends on h/ b and

steel grade
h 254.3
= = 0.98 ≤ 1.2 and t ≤ 100 mm
b 260.4 h 254.3
= = 0.98 < 1.2 and t ≤ 100 mm
Therefore, refer buckling curve b for y-y axis αy = b 260.4
Therefore, refer buckling curve c for z-z axis αz =
0.34
2 0.49
𝜙𝑦 = 0.5 x 1 + 0.34 0.695 − 0.2 + 0.695 = 0.83
ϕz = 0.5 x 1 + 0.49 1.8 − 0.2 + 1.82 = 2.512
1 1
χy = = 0.78 < 1.0 χz = = 0.235 < 1.0
0.83 + 0.832 − 0.6952 2.512 2.5122 − 1.82

26
 χ𝑧 A fy 0.235 X 10800 X 355
Nb,Rd = = = 797103.226 N
ϑM1 1

Nb,Rd < NEd ⇒ 900.990KN < 1376KN`

Hence, the section is not safe.

Therefore, the section is not adequate for the resistance for bucking

Go to step 2: Select a preliminary cross section

Trai-2: Section of 305x305x97 with gross area of A = 123cm2 from UKB

steel table

27