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Design of
Compression
Members
Design of Compression Members Design of Steel Structures to EC3
Page(2) Dr.Mamoun Alqedra Eng.Mohammed AbuRahma Eng. Haya Baker
2.1 Classification of cross sections
Classifying cross-sections may mainly depend on four critical factors:
1- Width to thickness (c/t) ratio.
2- Support condition.
3- Yield strength of the material.
4- Stress distribution across the width of the plate element.
There are basically four different cross-section classes and they are defined
as:
1- Class 1 cross-sections: are those which can form a plastic hinge with the rotation
capacity required from plastic analysis without reduction of the resistance.
2- Class 2 cross-sections: are those which can develop their plastic moment
resistance, but have limited rotation capacity because of local buckling.
3- Class3 cross-sections: are those in which the stress in the extreme compression
fiber of the steel member assuming an elastic distribution of stresses can reach the
yield strength, but local buckling is liable to prevent development of the plastic
moment resistance.
4- Class 4 cross-sections: are those in which local buckling will occur before the
attainment of yield stress in one or more parts of the cross section.
Design of Compression Members Design of Steel Structures to EC3
Page(3) Dr.Mamoun Alqedra Eng.Mohammed AbuRahma Eng. Haya Baker
Design of Compression Members Design of Steel Structures to EC3
Page(4) Dr.Mamoun Alqedra Eng.Mohammed AbuRahma Eng. Haya Baker
Design of Compression Members Design of Steel Structures to EC3
Page(5) Dr.Mamoun Alqedra Eng.Mohammed AbuRahma Eng. Haya Baker
Design of Compression Members Design of Steel Structures to EC3
Page(6) Dr.Mamoun Alqedra Eng.Mohammed AbuRahma Eng. Haya Baker
2.2 Support Conditions
Note:
1- If buckling occurs, it will take place in a plane perpendicular to the crossponding
principal axis of inertia.
2- The slenderness to be used, λ, is the larger of those calculated in the y and z
directions, that is λ = max (λy, λz). For example, it is possible to make reference to
the compression member in the below Figure, restrained in different ways in the x–
y and x–z planes, where the x-axis is the one along the length of the member. The
effective length in the x–y plane has to be taken as L/4 (i.e. LZ = 2.25 m), whereas
in the x–z plane it is 0.7L(Ly = 6.3 m).
Design of Compression Members Design of Steel Structures to EC3
Page(7) Dr.Mamoun Alqedra Eng.Mohammed AbuRahma Eng. Haya Baker
2.3 Design for compression
1- Plastic Resistance:
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
2- Buckling Resistance:
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.
Design of Compression Members Design of Steel Structures to EC3
Page(8) Dr.Mamoun Alqedra Eng.Mohammed AbuRahma Eng. Haya Baker
2.4 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
Page(9) Dr.Mamoun Alqedra Eng.Mohammed AbuRahma Eng. Haya Baker
Design of Compression Members Design of Steel Structures to EC3
Page(10) Dr.Mamoun Alqedra Eng.Mohammed AbuRahma Eng. Haya Baker
Values of the reduction factor χ for the appropriate non-dimensional slenderness �� may be
obtained from Figure 6.4.
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
Page(11) Dr.Mamoun Alqedra Eng.Mohammed AbuRahma Eng. Haya Baker
2.5 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√210×109
275×106 = 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
Design of Compression Members Design of Steel Structures to EC3
Page(12) Dr.Mamoun Alqedra Eng.Mohammed AbuRahma Eng. Haya Baker
-Safety verification
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√210×109
275×106 = 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
Page(13) Dr.Mamoun Alqedra Eng.Mohammed AbuRahma Eng. Haya Baker
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.
buckling in X–Y plan buckling in X–Z plane
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
∴ 𝑇ℎ𝑒 𝑐𝑟𝑜𝑠𝑠 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 𝑖𝑠 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
Design of Compression Members Design of Steel Structures to EC3
Page(14) Dr.Mamoun Alqedra Eng.Mohammed AbuRahma Eng. Haya Baker
Determination of the slenderness coefficients
𝜆1 = 𝜋√210×109
355×106 = 76.4
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
Page(15) Dr.Mamoun Alqedra Eng.Mohammed AbuRahma Eng. Haya Baker
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
Page(16) Dr.Mamoun Alqedra Eng.Mohammed AbuRahma Eng. Haya Baker
Determination of the slenderness coefficients
𝜆1 = 𝜋√210×109
275×106 = 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
Page(17) Dr.Mamoun Alqedra Eng.Mohammed AbuRahma Eng. Haya Baker
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√210×109
275×106= 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
Page(18) Dr.Mamoun Alqedra Eng.Mohammed AbuRahma Eng. Haya Baker
∅ = 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
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