Design of Compression Members Design of Steel …site.iugaza.edu.ps/hbaker/files/2018/02/Design-of...Design of Compression Members Design of Steel Structures to EC3 Page(6) Dr.Mamoun

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.









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).



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.



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.





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.



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



-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.


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 in the plane of the structure - 𝐿𝐸 = 2 × 3 = 6 𝑚


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


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


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



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


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 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




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


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


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.



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


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 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




𝜆1 = 𝜋√210×109

275×106 = 86.81

𝜆𝑦 =

𝐿𝑐𝑟

𝑖𝑦


𝜆1 =

6×100×√96.4/105

21.3×86.81= 0.31 < 1 → 𝑆ℎ𝑜𝑟𝑡 𝑐𝑜𝑙𝑢𝑚𝑛

𝜆𝑧 =

𝐿𝑐𝑟

𝑖𝑧


𝜆1 =

2×100×√96.4/105

4.38×86.81= 0.50 < 1 → 𝑆ℎ𝑜𝑟𝑡 𝑐𝑜𝑙𝑢𝑚𝑛


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


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.



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


Upper cord: c/t=12,𝜀 = √235/275 = 0.92 →𝑐

𝑡< 33𝜀 → 12 < 33(0.92) → 12 <

30.4 →→ 𝐶𝑙𝑎𝑠𝑠 1


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


𝑐𝑢𝑟𝑣𝑒 𝑎 → 𝛼 = 0.21



∅ = 0.5[1 + 0.21 × (0.757 − 0.2) + 0.7572] =0.84

𝑥 =1

0.84 + √0.842 − 0.7572= 0.83


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

Documents

Design of Compression Members Design of Steel …site.iugaza.edu.ps/hbaker/files/2018/02/Design-of...Design of Compression Members Design of Steel Structures to EC3 Page(6) Dr.Mamoun