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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, Ξ», is the larger of those calculated in the y and z
directions, that is Ξ» = max (Ξ»y, Ξ»z). For example, it is possible to make reference to
the compression member in the below Figure, restrained in different ways in the xβ
y and xβz planes, where the x-axis is the one along the length of the member. The
effective length in the xβy plane has to be taken as L/4 (i.e. LZ = 2.25 m), whereas
in the xβz 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 ππΈπ at each cross-section shall satisfy:
ππΈπ
ππ,π πβ€ 1.0
The design resistance of the cross-section for uniform compression ππ,π π should be
determined as follows:
ππ,π π = π΄ ππ¦ πΎπ0β for class 1,2 or 3 cross-sections
ππ,π π = π΄πππ ππ¦ πΎπ0β For class 4 cross-sections
2- Buckling Resistance:
A compression member should be verified against buckling as follows:
ππΈπ
ππ,π πβ€ 1.0
The design buckling resistance of a compression member should be taken as:
ππ,π π = Ο π΄ ππ¦ πΎπ1β for class 1,2 or 3 cross-sections
ππ,π π = Ο π΄πππ ππ¦ πΎπ1β for class 4 cross-sections
Where Ο 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 Ο for the appropriate non-dimensional
slenderness οΏ½οΏ½ should be determined from the relevant buckling curve according to:
Ο =1
β + ββ 2 β οΏ½οΏ½2 , ππ’π‘ Ο β€ 1.0
where β = 0.5[1 + πΌ(οΏ½οΏ½ β 0.2) + οΏ½οΏ½2]
οΏ½οΏ½ = βπ΄ ππ¦ πππβ =πΏππ
π
1
π1 for class 1,2 or 3 cross-sections
οΏ½οΏ½ = βπ΄πππ ππ¦
πππ=
πΏππ
π
βπ΄πππ /π΄
π1 for class 4 cross-sections
Ξ± is an imperfection factor.
πππ is the elastic critical force for the relevant buckling mode based on the gross
cross-sectional properties.
The imperfection factor Ξ± 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 Ο for the appropriate non-dimensional slenderness οΏ½οΏ½ may be
obtained from Figure 6.4.
For slenderness οΏ½οΏ½ β€ 0.2 or for ππΈπ
πππβ€ 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 β Assuming class 1, 2 or 3 cross sections, yields:
ππΈπ = 30 πΎπ β€ ππ,π π = π΄ππ¦ πΎπ0β = π΄ Γ 275 Γ 103/1.0
β π΄ β₯ 1.09 Γ 10β4π2 = 1.09 ππ2
Use πͺπ―πΊ ππ. π Γ π. π 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,π = β235/275 = 0.92 βπ
π‘< 50π2 β 8.41 < 50(0.92)2 β 8.41 <
42.32 ββ πΆπππ π 1
Buckling lengths β According to the support conditions, the buckling lengths are equal
in both planes, given by:
Buckling in the plane of the structure - πΏπΈ = 2 Γ 3 = 6 π
Determination of the slenderness coefficients
π1 = πβ210Γ109
275Γ106 = 86.81
π =πΏ
π=
6Γ102
0.846= 709.21
οΏ½οΏ½ =π
π1=
709.21
86.81= 8.16 > 1 β πΏπππ ππππ’ππ
Calculation of the reduction factor π
ππ’ππ£π π β πΌ = 0.21
π₯ =1
β + ββ 2 β οΏ½οΏ½2 , ππ’π‘ π₯ β€ 1.0
β = 0.5[1 + πΌ(οΏ½οΏ½ β 0.2) + οΏ½οΏ½2]
β = 0.5[1 + 0.21 Γ (8.16 β 0.2) + 8.162] =34.62
π₯ =1
34.62 + β34.622 β 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
ππ,π π = π₯π΄ππ¦ πΎπ1β = 0.0146 Γ 1.09 Γ 10β4 Γ 275 Γ 103 1.0β = 0.4376 πΎπ
As NEd = 30 kN > Nb,Rd = 0.4376 kN Safety is not verified
Try heavier section such πͺπ―πΊ πππ. π Γ ππ. π 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,π = β235/275 = 0.92 βπ
π‘< 50π2 β 17.5 < 50(0.92)2 β 8.41 <
42.32 ββ πΆπππ π 1
Buckling lengths β According to the support conditions, the buckling lengths are equal
in both planes, given by:
Buckling in the plane of the structure - πΏπΈ = 2 Γ 3 = 6 π
Determination of the slenderness coefficients
π1 = πβ210Γ109
275Γ106 = 86.81
π =πΏ
π=
6Γ102
7.32= 81.96
οΏ½οΏ½ =π
π1=
81.96
86.81= 0.944 < 1 β π βπππ‘ ππππ’ππ
Calculation of the reduction factor π
ππ’ππ£π π β πΌ = 0.21
β = 0.5[1 + 0.21 Γ (0.944 β 0.2) + 0.9442] =1.02
π₯ =1
1.02 + β1.022 β 0.9442= 0.711
-Safety verification
ππ,π π = π₯π΄ππ¦ πΎπ1β = 0.711 Γ 81.1 Γ 10β4 Γ 275 Γ 103 1.0β = 1585.70 πΎπ
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 Γ
254 Γ 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 XβY plan buckling in XβZ 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, π = β235
355= 0.81 β
π
π‘< 9π β 3.48 < 9 β 0.81 β
3.48 < 7.29 ββ πΆπππ π 1
Web: (c/t) =10.4, π = β235
355= 0.81 β
π
π‘< 33π β 10.4 < 33 β 0.81 β
10.4 < 26.73 ββ πΆπππ π 1
β΄ πβπ ππππ π π πππ‘πππ ππ πΆπππ π 1
ππ,π π = π΄ππ¦ πΎπ0β = 213 Γ 10β4 Γ 355 Γ103
1.0= 7561.5 > 6000 β ππππ
Buckling lengths β 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) - πΏπΈπ¦ = 1.0 Γ 5.0 = 5.0 π
Buckling in the plane of the structure (plane x-y) - πΏπΈπ§ = 1.0 Γ 3.0 = 3.0 π
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
π1 = πβ210Γ109
355Γ106 = 76.4
ππ¦ =πΏπΈπ¦
ππ¦=
5Γ102
11.9= 42.01, ππ¦
=ππ¦
π1=
42.01
76.4= 0.55 < 1 β πβπππ‘ ππππ’ππ
ππ§ =πΏπΈπ§
ππ§=
3Γ102
6.81= 44.05, ππ§
=ππ§
π1=
44.05
76.4= 0.57 < 1 β πβπππ‘ ππππ’ππ
Calculation of the reduction factor π
h/b = 289.1/265.2 = 1.09 < 1.2, tf=31.7 mm < 100 mm
πππππππ ππππ’ππ π§ β ππ’ππ£π π β πΌ = 0.49
β = 0.5[1 + 0.49 Γ (0.57 β 0.2) + 0.572] = 0.75
π₯ =1
0.75 + β0.752 β 0.572= 0.808
-Safety verification
ππ,π π = π₯π΄ππ¦ πΎπ1β = 0.808 Γ 213 Γ 10β4 Γ 355 Γ 103 1.0β = 6110.58 πΎπ
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 Γ
210 Γ 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, π = β235
275= 0.92 β
π
π‘< 9π β 6.58 < 9 β 0.92 β
6.58 < 8.28 ββ πΆπππ π 1
Web: (c/t) =49.6, π = β235
275= 0.92 β
π
π‘< 42π β 49.6 < 42 β 0.92 β
49.6 < 38.64 ββ πΆπππ π 4
From bluebook π΄πππ = 96.4 ππ2
ππ,π π = π΄πππ ππ¦ πΎπ0β = 96.4 Γ 10β4 Γ 275 Γ103
1.0= 2651 > 2500 β ππππ
Buckling lengths β 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) - πΏπΈπ¦ = 1.0 Γ 6.0 = 6.0 π
Buckling in the plane of the structure (plane x-y) - πΏπΈπ§ = 1.0 Γ 2.0 = 2.0 π
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
π1 = πβ210Γ109
275Γ106 = 86.81
ππ¦ =
πΏππ
ππ¦
βπ΄πππ /π΄
π1 =
6Γ100Γβ96.4/105
21.3Γ86.81= 0.31 < 1 β πβπππ‘ ππππ’ππ
ππ§ =
πΏππ
ππ§
βπ΄πππ /π΄
π1 =
2Γ100Γβ96.4/105
4.38Γ86.81= 0.50 < 1 β πβπππ‘ ππππ’ππ
Calculation of the reduction factor π
h/b = 528.3/208.8 = 2.53 > 1.2, tf=13.2 mm < 40 mm
πππππππ ππππ’ππ π§ β ππ’ππ£π π β πΌ = 0.34
β = 0.5[1 + 0.34 Γ (0.50 β 0.2) + 0.502] = 0.676
π₯ =1
0.676 + β0.6762 β 0.502= 0.88
-Safety verification
ππ,π π = π₯ π΄πππ ππ¦ πΎπ1β = 0.88 Γ 96.4 Γ 10β4 Γ 275 Γ 103 1.0β = 2332.88 πΎπ
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 β Assuming class 1, 2 or 3 cross sections, yields:
Upper cord:
ππΈπ = 742.6 πΎπ β€ ππ,π π = π΄ππ¦ πΎπ0β = π΄ Γ 275 Γ 103/1.0
β π΄ β₯ 27 Γ 10β4π2 = 27 ππ2
Use πΊπ―πΊ πππ Γ πππ Γ π for upper cord with A=35.5 cm2, I=738cm4, i=4.56cm
Classification of the section
Upper cord: c/t=12,π = β235/275 = 0.92 βπ
π‘< 33π β 12 < 33(0.92) β 12 <
30.4 ββ πΆπππ π 1
Determination of the slenderness coefficients
π1 = πβ210Γ109
275Γ106= 86.81 , LE=3.0 m
π =πΏπΈ
π=
3Γ102
4.56= 65.78 οΏ½οΏ½ =
π
π1=
65.78
86.81= 0.757 < 1 β πβπππ‘ ππππ’ππ
Calculation of the reduction factor π
ππ’ππ£π π β πΌ = 0.21
Design of Compression Members Design of Steel Structures to EC3
Page(18) Dr.Mamoun Alqedra Eng.Mohammed AbuRahma Eng. Haya Baker
β = 0.5[1 + 0.21 Γ (0.757 β 0.2) + 0.7572] =0.84
π₯ =1
0.84 + β0.842 β 0.7572= 0.83
-Safety verification
ππ,π π = π₯π΄ππ¦ πΎπ1β = 0.83 Γ 35.5 Γ 10β4 Γ 275 Γ 103 1.0β = 810.28 πΎπ
As NEd = 742.6 kN < Nb,Rd = 810.28 kN Safety is verified