Design of Compression Members
Design of
Compression
Members
Design of Compression Members Design of Steel Structures to EC3
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2.1 Classification of cross sections
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Design of Compression Members Design of Steel Structures to EC3
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2.2 Support Conditions
2.3 Design for compression
The design values of the compression force 𝑁𝐸𝑑 at each cross-section shall satisfy:
𝑁𝐸𝑑
𝑁𝑐,𝑅𝑑≤ 1.0
The design resistance of the cross-section for uniform compression 𝑁𝑐,𝑅𝑑 should be
determined as follows:
𝑁𝑐,𝑅𝑑 = 𝐴 𝑓𝑦 𝛾𝑀0⁄ for class 1,2 or 3 cross-sections
𝑁𝑐,𝑅𝑑 = 𝐴𝑒𝑓𝑓 𝑓𝑦 𝛾𝑀0⁄ for class 4 cross-sections
Design of Compression Members Design of Steel Structures to EC3
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2.4 Buckling resistance of compression members
A compression member should be verified against buckling as follows:
𝑁𝐸𝑑
𝑁𝑏,𝑅𝑑≤ 1.0
The design buckling resistance of a compression member should be taken as:
𝑁𝑐,𝑅𝑑 = χ 𝐴 𝑓𝑦 𝛾𝑀1⁄ for class 1,2 or 3 cross-sections
𝑁𝑐,𝑅𝑑 = χ 𝐴𝑒𝑓𝑓 𝑓𝑦 𝛾𝑀1⁄ for class 4 cross-sections
where χ is the reduction factor for the relevant buckling mode.
2.5 Buckling curves
For axial compression in members the value of χ for the appropriate non-dimensional
slenderness �� should be determined from the relevant buckling curve according to:
χ =1
∅ + √∅2 − ��2 , 𝑏𝑢𝑡 χ ≤ 1.0
where ∅ = 0.5[1 + 𝛼(�� − 0.2) + ��2]
�� = √𝐴 𝑓𝑦 𝑁𝑐𝑟⁄ =𝐿𝑐𝑟
𝑖
1
𝜆1 for class 1,2 or 3 cross-sections
�� = √𝐴𝑒𝑓𝑓 𝑓𝑦
𝑁𝑐𝑟=
𝐿𝑐𝑟
𝑖
√𝐴𝑒𝑓𝑓 /𝐴
𝜆1 for class 4 cross-sections
α is an imperfection factor.
𝑁𝑐𝑟 is the elastic critical force for the relevant buckling mode based on the gross
cross-sectional properties.
The imperfection factor α corresponding to the appropriate buckling curve should be
obtained from Table 6.1 and Table 6.2.
Design of Compression Members Design of Steel Structures to EC3
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Values of the reduction factor χ for the appropriate non-dimensional slenderness �� may be
obtained from Figure 6.4
Design of Compression Members Design of Steel Structures to EC3
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For slenderness �� ≤ 0,2 or for 𝑁𝐸𝑑
𝑁𝑐𝑟≤ 0,04 the buckling effects may be ignored and only cross
sectional checks apply.
Design of Compression Members Design of Steel Structures to EC3
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2.6 Solved Problems
Problem (1)
Design a lighting column subjected to an axial compression force 30 KN, using a CHS
(circular hollow section) cross section in S 275 steel, according to EC3-1-1. The
column is fixed at the base and free from the other end. With length of 3 m.
Solution:
Preliminary design – Assuming class 1, 2 or 3 cross sections, yields:
𝑁𝐸𝑑 = 30 𝐾𝑁 ≤ 𝑁𝑐,𝑅𝑑 = 𝐴𝑓𝑦 𝛾𝑀0⁄ = 𝐴 × 275 × 103/1.0
→ 𝐴 ≥ 1.09 × 10−4𝑚2 = 1.09 𝑐𝑚2
Use 𝑪𝑯𝑺 𝟐𝟔. 𝟗 × 𝟑. 𝟐 section with A=2.38 cm2, d/t=8.41, I = 1.7 cm4, i= 0.846 cm.
Classification of the section
d/t=8.41,𝜀 = √235/275 = 0.92 →𝑑
𝑡< 50𝜀2 → 8.41 < 50(0.92)2 → 8.41 <
42.32 →→ 𝐶𝑙𝑎𝑠𝑠 1
Buckling lengths – According to the support conditions, the buckling lengths are equal
in both planes, given by:
Buckling in the plane of the structure - 𝐿𝐸 = 2 × 3 = 6 𝑚
Determination of the slenderness coefficients
𝜆1 = 𝜋√210×106
275×103 = 86.81
𝜆 =𝐿
𝑖=
6×102
0.846= 709.21
�� =𝜆
𝜆1=
709.21
86.81= 8.16 > 1 → 𝐿𝑜𝑛𝑔 𝑐𝑜𝑙𝑢𝑚𝑛
Calculation of the reduction factor 𝒙
𝑐𝑢𝑟𝑣𝑒 𝑎 → 𝛼 = 0.21
𝑥 =1
∅ + √∅2 − ��2 , 𝑏𝑢𝑡 𝑥 ≤ 1.0
∅ = 0.5[1 + 𝛼(�� − 0.2) + ��2]
∅ = 0.5[1 + 0.21 × (8.16 − 0.2) + 8.162] =34.62
𝑥 =1
34.62 + √34.622 − 8.162= 0.0146
-Safety verification
Design of Compression Members Design of Steel Structures to EC3
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𝑁𝑏,𝑅𝑑 = 𝑥𝐴𝑓𝑦 𝛾𝑀1⁄ = 0.0146 × 1.09 × 10−4 × 275 × 103 1.0⁄ = 0.4376 𝐾𝑁
As NEd = 30 kN > Nb,Rd = 0.4376 kN Safety is not verified
Try heavier section such 𝑪𝑯𝑺 𝟐𝟏𝟗. 𝟏 × 𝟏𝟐. 𝟓 section with A=81.1 cm2, d/t=17.5,
I = 4350 cm4, i= 7.32 cm.
Classification of the section
d/t=17.5,𝜀 = √235/275 = 0.92 →𝑑
𝑡< 50𝜀2 → 17.5 < 50(0.92)2 → 8.41 <
42.32 →→ 𝐶𝑙𝑎𝑠𝑠 1
Buckling lengths – According to the support conditions, the buckling lengths are equal
in both planes, given by:
Buckling in the plane of the structure - 𝐿𝐸 = 2 × 3 = 6 𝑚
Determination of the slenderness coefficients
𝜆1 = 𝜋√210×106
275×103 = 86.81
𝜆 =𝐿
𝑖=
6×102
7.32= 81.96
�� =𝜆
𝜆1=
81.96
86.81= 0.944 < 1 → 𝑠ℎ𝑜𝑟𝑡 𝑐𝑜𝑙𝑢𝑚𝑛
Calculation of the reduction factor 𝒙
𝑐𝑢𝑟𝑣𝑒 𝑎 → 𝛼 = 0.21
∅ = 0.5[1 + 0.21 × (0.944 − 0.2) + 0.9442] =1.02
𝑥 =1
1.02 + √1.022 − 0.9442= 0.711
-Safety verification
𝑁𝑏,𝑅𝑑 = 𝑥𝐴𝑓𝑦 𝛾𝑀1⁄ = 0.711 × 81.1 × 10−4 × 275 × 103 1.0⁄ = 1585.70 𝐾𝑁
As NEd = 30 kN < Nb,Rd = 1585.70 kN Safety is verified
Design of Compression Members Design of Steel Structures to EC3
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Problem (2)
Check a column subjected to an axial compression force 6000 KN, using a UC
254 × 254 × 167 (universal column) cross section in S 355 steel, according to EC3-
1-1. The column is supported as shown in the figure. With length of 5 m.
Solution:
Section with A=213 cm2, (c/t) flange=3.48, (c/t) web=10.4, Iy-y = 30000 cm4, Iz-z = 9870
cm4, iy-y = 11.9 cm, iz-z = 6.81 cm
Classification of the section
Flange: (c/t) =3.48, 𝜀 = √235
355= 0.81 →
𝑐
𝑡< 9𝜀 → 3.48 < 9 ∗ 0.81 →
3.48 < 7.29 →→ 𝐶𝑙𝑎𝑠𝑠 1
Web: (c/t) =10.4, 𝜀 = √235
355= 0.81 →
𝑐
𝑡< 33𝜀 → 10.4 < 33 ∗ 0.81 →
10.4 < 26.73 →→ 𝐶𝑙𝑎𝑠𝑠 1
𝑁𝑐,𝑅𝑑 = 𝐴𝑓𝑦 𝛾𝑀0⁄ = 213 × 10−4 × 355 ×103
1.0= 7561.5 > 6000 → 𝑆𝑎𝑓𝑒
Buckling lengths – According to the support conditions, the buckling lengths are equal
in both planes, given by:
Buckling in the plane of the structure (plane x-z) - 𝐿𝐸𝑦 = 1.0 × 5.0 = 5.0 𝑚
Buckling in the plane of the structure (plane x-y) - 𝐿𝐸𝑧 = 1.0 × 3.0 = 3.0 𝑚
Determination of the slenderness coefficients
𝜆1 = 𝜋√210×106
355×103 = 76.4
Design of Compression Members Design of Steel Structures to EC3
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𝜆𝑦 =𝐿𝐸𝑦
𝑖𝑦=
5×102
11.9= 42.01, 𝜆𝑦
=𝜆𝑦
𝜆1=
42.01
76.4= 0.55 < 1 → 𝑆ℎ𝑜𝑟𝑡 𝑐𝑜𝑙𝑢𝑚𝑛
𝜆𝑧 =𝐿𝐸𝑧
𝑖𝑧=
3×102
6.81= 44.05, 𝜆𝑧
=𝜆𝑧
𝜆1=
44.05
76.4= 0.57 < 1 → 𝑆ℎ𝑜𝑟𝑡 𝑐𝑜𝑙𝑢𝑚𝑛
Calculation of the reduction factor 𝒙
h/b = 289.1/265.2 = 1.09 < 1.2, tf=31.7 mm < 100 mm
𝑏𝑒𝑛𝑑𝑖𝑛𝑔 𝑎𝑟𝑜𝑢𝑛𝑑 𝑧 → 𝑐𝑢𝑟𝑣𝑒 𝑐 → 𝛼 = 0.49
∅ = 0.5[1 + 0.49 × (0.57 − 0.2) + 0.572] = 0.75
𝑥 =1
0.75 + √0.752 − 0.572= 0.808
-Safety verification
𝑁𝑏,𝑅𝑑 = 𝑥𝐴𝑓𝑦 𝛾𝑀1⁄ = 0.808 × 213 × 10−4 × 355 × 103 1.0⁄ = 6110.58 𝐾𝑁
As NEd = 6000 kN < Nb,Rd = 6110.58 kN Safety is verified.
Design of Compression Members Design of Steel Structures to EC3
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Problem (3)
Check a column subjected to an axial compression force 2500 KN, using a UB
533 × 210 × 82 (universal beam) cross section in S 275 steel, according to EC3-1-1.
The column is supported as shown in the figure. With length of 6 m.
Solution:
Section with A=105 cm2, (c/t) flange=6.58, (c/t) web=49.6, Iy-y = 47500 cm4, Iz-z = 2010
cm4, iy-y = 21.3 cm, iz-z = 4.38 cm
Classification of the section
Flange: (c/t) =6.58, 𝜀 = √235
275= 0.92 →
𝑐
𝑡< 9𝜀 → 6.58 < 9 ∗ 0.92 →
6.58 < 8.28 →→ 𝐶𝑙𝑎𝑠𝑠 1
Web: (c/t) =49.6, 𝜀 = √235
275= 0.92 →
𝑐
𝑡< 42𝜀 → 49.6 < 42 ∗ 0.92 →
49.6 < 38.64 →→ 𝐶𝑙𝑎𝑠𝑠 4
From bluebook 𝐴𝑒𝑓𝑓 = 96.4 𝑐𝑚2
𝑁𝑐,𝑅𝑑 = 𝐴𝑒𝑓𝑓 𝑓𝑦 𝛾𝑀0⁄ = 96.4 × 10−4 × 275 ×103
1.0= 2651 > 2500 → 𝑆𝑎𝑓𝑒
Buckling lengths – According to the support conditions, the buckling lengths are equal
in both planes, given by:
Buckling in the plane of the structure (plane x-z) - 𝐿𝐸𝑦 = 1.0 × 6.0 = 6.0 𝑚
Buckling in the plane of the structure (plane x-y) - 𝐿𝐸𝑧 = 1.0 × 2.0 = 2.0 𝑚
Design of Compression Members Design of Steel Structures to EC3
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Determination of the slenderness coefficients
𝜆1 = 𝜋√210×106
275×103 = 86.81
𝜆𝑦 =
𝐿𝑐𝑟
𝑖𝑦
√𝐴𝑒𝑓𝑓 /𝐴
𝜆1 =
6×100×√96.4/105
21.3×86.81= 0.31 < 1 → 𝑆ℎ𝑜𝑟𝑡 𝑐𝑜𝑙𝑢𝑚𝑛
𝜆𝑧 =
𝐿𝑐𝑟
𝑖𝑧
√𝐴𝑒𝑓𝑓 /𝐴
𝜆1 =
2×100×√96.4/105
4.38×86.81= 0.50 < 1 → 𝑆ℎ𝑜𝑟𝑡 𝑐𝑜𝑙𝑢𝑚𝑛
Calculation of the reduction factor 𝒙
h/b = 528.3/208.8 = 2.53 > 1.2, tf=13.2 mm < 40 mm
𝑏𝑒𝑛𝑑𝑖𝑛𝑔 𝑎𝑟𝑜𝑢𝑛𝑑 𝑧 → 𝑐𝑢𝑟𝑣𝑒 𝑏 → 𝛼 = 0.34
∅ = 0.5[1 + 0.34 × (0.50 − 0.2) + 0.502] = 0.676
𝑥 =1
0.676 + √0.6762 − 0.502= 0.88
-Safety verification
𝑁𝑏,𝑅𝑑 = 𝑥 𝐴𝑒𝑓𝑓 𝑓𝑦 𝛾𝑀1⁄ = 0.88 × 96.4 × 10−4 × 275 × 103 1.0⁄ = 2332.88 𝐾𝑁
As NEd = 2500 kN > Nb,Rd = 2332.88 kN Safety is not verified.
Design of Compression Members Design of Steel Structures to EC3
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Problem (4)
The following truss design the upper cord members compressed members, considering
the same type of cross section, that is: Square hollow sections (SHS), with welded
connections between the members of the structure.
Solution:
Based on the axial force diagrams represented in Figure 3.53, the most
compressed chord member is under an axial force of 742.6 kN and it is
simultaneously one of the longest members, with L = 3.00 m; For the definition of the
buckling lengths of the members, it is assumed that all the nodes of the truss are braced
in the direction perpendicular to the plane of the structure.
Preliminary design – Assuming class 1, 2 or 3 cross sections, yields:
Upper cord:
𝑁𝐸𝑑 = 742.6 𝐾𝑁 ≤ 𝑁𝑐,𝑅𝑑 = 𝐴𝑓𝑦 𝛾𝑀0⁄ = 𝐴 × 275 × 103/1.0
→ 𝐴 ≥ 27 × 10−4𝑚2 = 27 𝑐𝑚2
Use 𝑺𝑯𝑺 𝟏𝟐𝟎 × 𝟏𝟐𝟎 × 𝟖 for upper cord with A=35.5 cm2, I=738cm4, i=4.56cm
Classification of the section
Upper cord: c/t=12,𝜀 = √235/275 = 0.92 →𝑐
𝑡< 33𝜀 → 12 < 33(0.92) → 12 <
30.4 →→ 𝐶𝑙𝑎𝑠𝑠 1
Determination of the slenderness coefficients
𝜆1 = 𝜋√210×106
275×103 = 86.81 , LE=3.0 m
𝜆 =𝐿𝐸
𝑖=
3×102
4.56= 65.78 �� =
𝜆
𝜆1=
65.78
86.81= 0.757 < 1 → 𝑆ℎ𝑜𝑟𝑡 𝑐𝑜𝑙𝑢𝑚𝑛
Calculation of the reduction factor 𝒙
𝑐𝑢𝑟𝑣𝑒 𝑎 → 𝛼 = 0.21
Design of Compression Members Design of Steel Structures to EC3
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∅ = 0.5[1 + 0.21 × (0.757 − 0.2) + 0.7572] =0.84
𝑥 =1
0.84 + √0.842 − 0.7572= 0.83
-Safety verification
𝑁𝑏,𝑅𝑑 = 𝑥𝐴𝑓𝑦 𝛾𝑀1⁄ = 0.83 × 35.5 × 10−4 × 275 × 103 1.0⁄ = 810.28 𝐾𝑁
As NEd = 742.6 kN < Nb,Rd = 810.28 kN Safety is verified