Talat Lecture 2301: Design Of Members Example 5.1: Axial Force Resistance Of Square Hollow Section

TALAT Lecture 2301

Design of Members Axial Force Example 5.1 : Axial force resistance of square hollow section 2 pages Advanced Level prepared by Torsten Höglund, Royal Institute of Technology, Stockholm

Date of Issue: 1999  EAA - European Aluminium Association TALAT 2301 – Example 5.1

1

Example 5.1. Axial force resistance of square hollow section

5.4.3

Width

b

70 . mm

6 MPa 10 . Pa

Thickness

t

1.9 . mm

fo

200 . MPa

Column length

L

1000 . mm

E

70000 . MPa

b f = 66.2 mm

bf

β

t

= 34.8

Element classification 250 . MPa

ε

Tab. 5.1

2 .t

b

β 5.4.4

γ M1 1.0

Slenderness parameter bf

kN 1000 . newton

ε = 1.118

fo

β 1 11 . ε β 2 16 . ε

β 1= 12.3 β 2= 17.9

β 3 22 . ε β 3= 24.6 class if β > β 1 , if β > β 2 , if β > β 3 , 4 , 3 , 2 , 1 class = 4 5.4.5

Local buckling

5.4.5 (3) c) heat-treated unwelded

ρ c if

β

22 , 1.0 ,

ε

32

220

β

β

ε

4 .( b

A eff 4

r

b

(b

12

t eff = 1.52 mm

t ) . t eff 2 .t )

ρ c= 0.8

ε

if class 4 , t . ρ ,ct

t eff

I

2

A eff = 414.2 mm

4

2

I = 4.004 . 10 mm 5

12 I

4

According to 5.8.4.1 (1)

r = 31.1 mm

A eff

A .η .f o

λ

N cr

Substitute 5.8.4

Flexural buckling K .L

Table 5.7

K

1

l

Table 5.5 and 5.6

α

0.2

λ o

See to the right

λ

(5.33)

φ φ

5.8.3 (1)

N cr 0.1

k1

1

k2

l. 1 . f o r π E 0.5 . 1

α . λ

λ o

λ

2

TALAT 2301 – Example 5.1

λ 1

χ φ

χ = 0.891 fo .A χ .k 1 .k 2 . eff γ M1

2

l

A ef A

r

φ

2

λ

2

l. 1 . f o r π E

Note: λ with a bar cannot be written

N b.Rd = 73.8 kN

2

η

Result

λ = 0.547

= 0.694

N b.Rd

1

2 π .E .I

I A ef