design of compression members design of steel structures to...

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



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



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.



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.



2.6 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



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√210×106

275×103 = 86.81

𝜆 =𝐿

𝑖=

6×102

7.32= 81.96

�� =𝜆

𝜆1=

81.96

86.81= 0.944 < 1 → 𝑠ℎ𝑜𝑟𝑡 𝑐𝑜𝑙𝑢𝑚𝑛


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.

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

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√210×106

355×103 = 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√210×106

275×103 = 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√210×106

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


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

design of compression members design of steel structures to...

Documents