Talat Lecture 2301: Design Of Members Example 5.2: Axial Force Resistance Of Symmetric Hollow Extrusion

TALAT Lecture 2301

Design of Members Axial Force Example 5.2 : Axial force resistance of symmetric hollow extrusion 4 pages Advanced Level prepared by Torsten Höglund, Royal Institute of Technology, Stockholm

Date of Issue: 1999  EAA - European Aluminium Association TALAT 2301 – Example 5.2

1

Example 5.2 Axial force resistance of symmetric hollow extrusion Half width

b

50 . mm

fo

300 . MPa

Indent

c

10 . mm

E

70000 . MPa

18 . mm

γ M1 1.0

Half width of a flat parts Thickness

s

1.2 . mm

Length

L

1200 . mm

Nodes no., co-ordinates, thickness

0

b

1

b

a

s

2

a

3

a

b

s

4

a

b

s

6 MPa 10 . Pa

5

a

s

kN 1000 . newton

6

b

b

8

c

b

y b

9

c

b

b

7 i

s

c

a

s

c a

s

a

s

z

a

c

t

s

a

10

a

11

a

b

s

12

a

b

s

13

a b

14

b

s c

b c

s

c

s

a

s

15

b

a

s

16

b

a

s

50 0 z mm

0 0

50

Nodes

i

50

1 .. rows ( y )

1

0 y mm

5.4.5

Local buckling ε

5.4.3 (1) c)

5.4.5 (3) c) heat-treated, unwelded

βi

250 . MPa fo yi

if t i > 0 ,

ρ c if i

β ε

i

yi 1

2

zi

zi 1

ti > 22 , 32 .

TALAT 2301 – Example 5.2

ε β

220 . i

,1

2

ε β

2

, 1.0 i

2

50

Effective thickness

ρ .ct

t eff

βi Effective area of elements

t eff .

dAi

yi

i

2

yi 1

rows ( y )

zi

zi 1

2

1

Area of effective A eff cross section

A eff = 356.591 mm

dAi

2

i =1 rows ( y ) First moment of area. Gravity centre

1

Sy

dAi zi 1 . 2

zi

Sy

z gc

A eff

i =1 z gc = 0 m

Axial force resistance, no flexural buckling A eff .

N Rd

fo

t eff

i = βi =

ε

=

ρ c= i

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

9.129 28.402 9.129 32.863 9.129 28.402 9.129 32.863 9.129 28.402 9.129 32.863 9.129 28.402 9.129 32.863

1 0.854 1 0.77 1 0.854 1 0.77 1 0.854 1 0.77 1 0.854 1 0.77

8.333 25.927 8.333 30 8.333 25.927 8.333 30 8.333 25.927 8.333 30 8.333 25.927 8.333 30

i

mm

1.2 1.025 1.2 0.924 1.2 1.025 1.2 0.924 1.2 1.025 1.2 0.924 1.2 1.025 1.2 0.924

N Rd = 107 kN

γ M1

50 0 z mm 0 z GC

0

mm

50 50

5.8.4

0

Second moment of area of effective cross section

I

Radius of gyration

r

Table 5.7

K

1

l

Table 5.5 and 5.6

α

0.2

λ o

See to the right

λ

rows ( y )

y , mm mm

1 zi

2

50

y GC

Flexural buckling

zi 1

zi . zi 1 .

2

ti .

i =1

yi

yi 1

2

zi

zi 1

2

3 I = 4.701 . 10 mm 5

I A eff K .L

l = 1.2 . 10 mm 3

0.1

k1

1

k2

l. 1 . f o r π E

TALAT 2301 – Example 5.2

1

λ = 0.689

3

=

4

(5.33)

φ

0.5 . 1

α . λ

λ o

λ

2

1

χ φ

5.8.3 (1)

N b.Rd

χ .k 1 .k 2 .

TALAT 2301 – Example 5.2

fo

γ M1

.A

φ

φ 2

λ

= 0.796

2

N b.Rd = 89.5 kN

eff

4

χ = 0.837