Talat Lecture 2301: Design Of Members Example 5.7: Axial Force Resistance Of Orthotropic Double-skin Plate

TALAT Lecture 2301

Design of Members Axial Force Example 5.7 : Axial force resistance of orthotropic double-skin plate 9 pages Advanced Level prepared by Torsten Höglund, Royal Institute of Technology, Stockholm

Date of Issue: 1999  EAA - European Aluminium Association

TALAT 2301 – Example 5.7

1

Example 5.7. Axial force resistance of orthotropic double-skin plate

Input

N newton (highlighted)

kN 1000 . N

Plate thickness

t

5 . mm

Plate width

b

Plate length "Pitch" (2a or d)

6 MPa 10 . Pa

fo

240 . MPa

300000 . mm

fu

260 . MPa

L

5000 . mm

E

70000 . MPa

w

160 . mm

t1

a

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

t

w 2 cf

0

γ M1 1.1 ht 1

a) Profiles with groove and tongue

Half bottom flange

a2

40 . mm

Thickness of bottom flange

t2

5 . mm

Profile depth

h

Web thickness

t3

5 . mm

Half width of trapezoidal stiffener at the top

a1

80 . mm

Number of webs

nw

4

Width of web

a3

TALAT 2301 – Example 5.7

70 . mm

a1

a2

2

h

2

2

a 3 = 80.6 mm

Local buckling Internal elements

0

β i 1 β

[1] Tab. 5.1

a1

ε

2 .a 2

β i 2

t1

β

250 . newton fo

β i 3

t2

max β i

mm

a3 t3

16.125

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

class i

16

= 16.1

β 1 9 .ε β 2 13 . ε

2

16

β i=

No reduction for local buckling

β 3= 18.371 class i = 3

50

0

50

100

100

50

0

50

100

Overall buckling, uniform compression Cross sectional area

A

2 .t 1 .a

2 .t 2 .a 2

e

Second moment of area

IL

nw

A = 2.812 . 10 mm 3

2

h nw 2 .t 3 .a 3 . . 2 2

2 .t 2 .a 2 .h

Gravity centre

2 .t 3 .a 3 .

e = 30.022 mm

A 2 .t 2 .a 2 .h

2

2 .t 3 .a 3 .

2 h .n w 3 2

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

2

A .e

2

I L = 2.059 . 10 mm

4

I T = 3.517 . 10 mm

4

6

2

Approx. torsional constant

IT

6

Rigidities of orthotropic plate Table 5.10

Table 5.10

Table 5.10

Bx By H

E .I L 2 .a 0.001 . N . mm

ν

G

E 2 .( 1

G .I T

ν)

B y = 1 . 10

3

8

N . mm mm

N . mm

3

H = 5.918 . 10

2 .a

TALAT 2301 – Example 5.7

B x = 9.007 . 10

0.3

2

mm 8

N . mm mm

2

2

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 = 1.067 . 10 kN 5

b 2 .π

(5.78)

2

.

b

B x .B y

H

otherwise

Buckling resistance A ef (5.69)

(5.33)

A

A ef = 2.812 . 10 mm 3

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

2

λ 0

λ c

1

χ c φ

(5.68)

λ c= 0.08

N cr

φ

2

A ef . χ .c γ

TALAT 2301 – Example 5.7

α = 0.2 λ 0= 0.1 φ = 0.501 χ c = 1.004

2

λ c fo

for one stiffener

M1

4

N cRd = 616.164 kN

2

b) Truss cross section

Half bottom flange

a2

a 2 70 . mm

Stiffener depth

h

Thickness of bottom flange

t2

5 . mm

Web thickness

t3

5 . mm

Width of web

a3

a1

Local buckling Internal elements

2

a = 80 mm a 1 = 40 mm a 2 = 40 mm

2

a 3 = 80.623 mm

h

0 2 .a 1

β i 1 β

[1] Tab. 5.1

2

a

a1

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=

t3

16 16.125

= 16.1

β 1 9 .ε β 2 13 . ε

2

16

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 50

0

50

100

TALAT 2301 – Example 5.7

5

100

50

0

50

100

Overall buckling, uniform compression 2 .t 1 .a 1

Cross sectional area A

2 .t 2 .a 2

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

e

2 .t 3 .a 3

2 .t 3 .a 3 .

A = 1.606 . 10 mm 3

h 2

e = 35 mm

A 2 .t 2 .a 2 .h

2

IL

2

2 .t 3 .a 3 .

2

h

A .e

2

3

I L = 1.309 . 10 mm

4

I T = 1.952 . 10 mm

4

6

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

Torsion constant

IT

6

Orthotropic plate constant Table 5.10

E .I L

Bx

B x = 5.728 . 10

2 .a E .t 1 .t 2 .h

N . mm

8

mm

2

Table 5.10

Table 5.10

By

t1

G .I T

H

N . mm

B y = 8.575 . 10

8

t2

H = 3.285 . 10

2 .a

8

mm N . mm mm

Elastic buckling load (5.77)

2 π . Bx 2 b L

N cr

L B y. b

2 .H

4

2

if

L b

<

Bx By

N cr = 6.786 . 10 kN 4

b 2 .π

(5.78)

2

.

b

B x .B y

H

otherwise

Buckling resistance A ef (5.69)

A

A ef = 1.606 . 10 mm

A ef . f o

λ c= 0.08

λ c

3

N cr

α

if ( ht > 0 , 0.2 , 0.32 )

φ

0.5 . 1

λ 0 if ( ht > 0 , 0.1 , 0 )

α . λ c

α = 0.2 λ 0= 0.1

2

λ 0

λ c

φ = 0.5 (5.33)

1

χ c φ

(5.68)

N cRd

φ

2

χ c = 1.005

TALAT 2301 – Example 5.7

2

λ c

2 . A ef . χ .c γ

fo

for two pitches

M1

6

2

N cRd = 704.4 kN

2

2

2

c) Frame cross section

Half bottom flange

a

37.5 . mm

a1

a

Stiffener depth

h

70 . mm

a2

a

Thickness of top flange

t1

5 . mm

a 1 = 37.5 mm

Thickness of bottom flange

t2

5 . mm

a 2 = 37.5 mm

Web thickness

t3

5 . mm

Width of web

a3

h

Local buckling Internal elements

0 2 .a 1

β i 1 β

[1] Tab. 5.1

a 3 = 70 mm

ε

β i 2

t1

2 .a 2

β i 3

t2

max β i

β

250 . newton fo

class i

mm

a3

15 β i= 15

t3

14

= 15

β 1 9 .ε β 2 13 . ε

2

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 50

0

50

100

TALAT 2301 – Example 5.7

7

100

50

0

50

100

Overall buckling, uniform compression 2 .t 1 .a

Cross sectional area A

Gravity centre

Second moment of area

2 .t 2 .a 2

2 .t 2 .a 2 .h e

t 3 .a 3

t 3 .a 3 .

A = 1.1 . 10 mm 3

h 2

e = 35 mm

A 2 .t 2 .a 2 .h

t 3 .a 3 .

2

IL

2

2

h

A .e

2

3

I L = 1.062 . 10 mm

4

I T = 1.901 . 10 mm

4

6

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

Torsion constant

IT

6

Orthotropic plate constant (5.80d)

E .I L

Bx

B x = 9.909 . 10

2 .a

E .t 1

3

(5.80a

By

12 . 1

. . 10 b . 2 2 ν 32 . a

a .t 3

2

a .t 3

3

3

3.

a .t 2 t 3 t1 3

t2

mm

3

6 .h .t 2

3 2

t . 1 2 3 3 2 3 .h .t 1 .t 2 L a .t 3

3

N . mm

B y = 1.118 . 10

7

(5.80b)

3

2 .E

H

t3

3. 1

t1

. 1

2 .a

t2

6 .t 1 2 .a

1

t3

3

6 .t 2 2 .a

2

3

3

2 .h . t 1

N . mm

8

6

t3

mm N . mm

H = 8.75 . 10

2

2

mm

Elastic buckling load (5.77)

2 π . Bx 2 b L

N cr

L B y. b

2 .H

4

2

if

L b

<

Bx

N cr = 1.174 . 10 kN 5

By

b 2 .π

(5.78)

2

2

.

b Buckling resistance

A ef

(5.69)

λ c

(5.33)

B x .B y

H

otherwise

H = 7.501 kN 3

A ef . f o

α

if ( ht > 0 , 0.2 , 0.32 )

φ

0.5 . 1

λ 0 if ( ht > 0 , 0.1 , 0 )

α . λ c

2

λ 0

λ c

φ

2

TALAT 2301 – Example 5.7

χ c = 1.011

2

λ c

2 . A ef . χ .c γ

α = 0.2 λ 0= 0.1 φ = 0.496

1

χ c

2

λ c= 0.047

N cr

N cRd

B x .B y

A ef = 1.1 . 10 mm

A

φ

(5.68)

2 .π . b

fo

for two pitches

M1

8

N cRd = 485.112 kN

Summary a)

b)

c)

N Rd.a = 616.2 kN

A a = 2.812 . 10 mm

2

N Rd.b = 704.4 kN

A b = 3.212 . 10 mm

2

N Rd.c = 485.1 kN

A c = 2.2 . 10 mm

TALAT 2301 – Example 5.7

3

3

3

9

2

N Rd.a Aa N Rd.b Ab N Rd.c Ac

= 219.1

N mm

= 219.3

N mm

= 220.5

2

2

N mm

2