Talat Lecture 2301: Design Of Members Example 5.6: Axial Force Resistance Of Orthotropic Plate. Open Or Closed Stiffeners

TALAT Lecture 2301

Design of Members Axial Force Example 5.6 : Axial force resistance of orthotropic plate. Open or closed stiffeners 5 pages Advanced Level prepared by Torsten Höglund, Royal Institute of Technology, Stockholm

Date of Issue: 1999  EAA - European Aluminium Association

TALAT 2301 – Example 5.6

1

Example 5.6. Axial force resistance of orthotropicplate. Open or closed stiffeners

Dimensions

(highlighted)

Plate thickness

t

16.5 . mm

Plate width

b

Plate length Stiffener pitch

fo

240 . MPa

N newton

1500 . mm

fu

260 . MPa

kN 1000 . newton

L

2200 . mm

E

70000 . MPa

w

300 . mm

t1

t

w

a

γ M1 1.1 ht 1

2

If heat-treated alloy, thenht = 1 else ht = 0

6 MPa 10 . Pa

If cold formed (trapezoidal), thencf. = 1 else cf. = 0

cf

0

a) Open stiffeners Half stiffener pitch

a = 150 mm

Half bottom flange

a2

50 . mm

Thickness of bottom flange

t2

10 . mm

Stiffener depth

h

Half web thickness

t3

4.4 . mm

Half width of trapezoidal stiffener at the top

a1

0 . mm

Width of web

a3

h

160 . mm

a 3 = 160 mm

Local buckling Internal elements

[1] Tab. 5.1

2 .a

β i 1

t1 250 . MPa

ε

fo

class i Outstand [1] Tab. 5.1

β

a3 β i 2 2 .t 3

18.182 β =i 18.182

β 1 9 .ε β 2 11 . ε

β 3 18 . ε if β > β 1 , if β > β 2 , if β > β 3 , 4 , 3 , 2 , 1

a2 t2

β =5

β 1 2.5 . ε β 2 4 .ε

β 3 5 .ε class o if β > β 1 , if β > β 2 , if β > β 3 , 4 , 3 , 2 , 1 class if class o > class i , class o , class i No reduction for local buckling

TALAT 2301 – Example 5.6

2

β

max β i

β

= 18.182

β 1= 9.186 β 2= 11.227 β 3= 18.371 class i = 3

β 1= 2.552 β 2= 4.082 β 3= 5.103 class o = 3 class = 3

5.11.6

Overall buckling, uniform compression 2 .t 1 .a

Cross sectional area A

Gravity centre

2 .t 2 .a 2

2 .t 2 .a 2 .h e

2 .t 3 .a 3

2 .t 3 .a 3 .

2 .t 1 .a 1

A = 7.358 . 10 mm 3

2

h 2

e = 37.054 mm

A

Second moment of area 2 .t 2 .a 2 .h

s

2 .t 3 .a 3 .

2

IL

if cf 1 , 2 . a

2 .a

3

2

h

A .e

I L = 2.751 . 10 mm

2

3 . 2 a1

7

, 2 .a

a2

4

s = 300 mm

Rigidities of orthotropic plate

Table 5.10

Table 5.10

Table 5.10

Bx

E .I L

ν

2 .a

G

2 .a .

By

E .t

E . 2 (1 ν)

3

s 12 . 1 3 2 .a . G .t s 6

H

0.3

N . mm

B x = 6.42 . 10

9

mm N . mm

B y = 2.88 . 10

7

ν

2

H = 2.016 . 10

7

(5.77)

N cr

2 .H

L B y. b

4

2

if

L b

<

Bx By

N . mm mm

N cr = 2.031 . 10 kN 4

b 2 .π

(5.78)

2

.

b

B x .B y

H

otherwise

Buckling resistance A ef (5.69)

Table 5.6

(5.33)

A ef = 7.358 . 10 mm 3

A A ef . f o

λ c α

if ( ht > 0 , 0.2 , 0.32 )

φ

0.5 . 1

N cRd

λ 0 if ( ht > 0 , 0.1 , 0 )

α . λ c

φ

2

A ef . χ .c γ

TALAT 2301 – Example 5.6

α = 0.2 λ 0= 0.1

2

λ 0

λ c

φ = 0.563

1

χ c φ

(5.68)

λ c= 0.295

N cr

χ c = 0.959

2

λ c fo

for one stiffener

M1

3

2

mm

Elastic buckling load 2 π . Bx 2 b L

2

N cRd = 1.54 . 10 kN 3

2

2

b) Closed stiffeners Half stiffener pitch

a = 150 mm

Plate thickness

t 1 = 16.5 mm

Half bottom flange

a 2 = 50 mm

Thickness of bottom flange

t 2 = 10 mm

Stiffener depth

h = 160 mm

Web thickness

t3

9 . mm

Half width of trapezoidal stiffener at the top

a1

80 . mm

width of web

a3

a1

a4

a

a 3 = 162.8 mm a 4 = 70 mm 9.697

2 .a 1

β i 1 β

[1] Tab. 5.1

2

h

a1

Local buckling Internal elements

2

a2

ε

β i 2

t1

2 .a 2

β i 3

t2

max β i

β

250 . newton fo

class i

mm

a3

β i 4

t3

t1

10 β =i 18.088 8.485

= 18.1

β 1 9 .ε β 2 13 . ε

2

2 .a 4

β 1= 9.186 β 2= 13.268

β 3 18 . ε if β > β 1 , if β > β 2 , if β > β 3 , 4 , 3 , 2 , 1

β 3= 18.371 class i = 3

No reduction for local buckling Cross sectional area A

Gravity centre

2 .t 1 .a

2 .t 2 .a 2

2 .t 2 .a 2 .h e

Second moment of area

IL

Torsion constant

IT

(5.79c)

C1

2 .t 3 .a 3

2 .t 3 .a 3 .

A = 8.88 . 10 mm 3

h 2

e = 44.415 mm

A 2 .t 2 .a 2 .h

2

2

2 .t 3 .a 3 .

2

h

3

A .e

2

I L = 3.309 . 10 mm

4

I T = 3.097 . 10 mm

4

7

4. h. a 1 a 2 2 .a 1 2 .a 2 a3 2. t1 t2 t3 2

(5.79d)

B

4. 1 E .t 12 . 1

TALAT 2301 – Example 5.6

2 ν . a2

7

2 2 2 a 3 .a 1 .a 4 .h .

3

t2 3 .a .t 1

3

C 1 = 3.076 . 10 mm 9

B = 2.88 . 10 N . mm 7

ν

2

4

4

4. a 1 (5.79e)

2 a 2 .a 1 .a 4 . 1

C2

3 a 2 . 3 .a 3

a1

a2

a2

a1

2

a 3 a 1 .a 3 t 2 . t1

4 .a 2

C 2 = 4.851

Rigidities of orthotropic plate Table 5.10

(5.79a)

E .I L

Bx

ν

2 .a

0.3

E

G

2 .( 1

9

ν )

3 3 2 .a 1 .a 3 .t 1 . 4 .a 2 .t 3

mm

3 3 a 3 .t 1 . 4 .a 2 .t 3

a 3 .t 2

a 3 .t 2

1.6 . G . I T . a 4

4 .a 3 .t 2

3

7

2

1

2 L .a .B

N . mm

B y = 3.929 . 10

6 .a

2 .B

H

3

3 3 a 1 . t 3 . 12 . a 2 . t 3

3

G .I T

1

. 1 4 π .

C1 L

H = 7.305 . 10

mm N . mm

8

mm

C2

4

Elastic buckling load (5.77)

2 π . Bx 2 b L

N cr

2 .H

L B y. b

4

2

if

L b

<

Bx By

N cr = 3.378 . 10 kN 4

b 2 .π

(5.78)

2

.

b

B x .B y

H

otherwise

Buckling resistance A ef (5.69)

(5.33)

A ef = 8.88 . 10 mm 3

A A ef . f o

λ c α

if ( ht > 0 , 0.2 , 0.32 )

φ

0.5 . 1

N cRd

λ 0 if ( ht > 0 , 0.1 , 0 )

α . λ c

φ

2

A ef . χ .c γ

TALAT 2301 – Example 5.6

α = 0.2 λ 0= 0.1

2

λ 0

λ c

φ = 0.547

1

χ c

2

λ c= 0.251

N cr

φ (5.68)

2

2 . B. a

By 2 .a 4

(5.79b)

N . mm

B x = 7.72 . 10

χ c = 0.969 2

λ c fo

for one stiffener

M1

5

N cRd = 1.877 . 10 kN 3

2

2