Talat Lecture 2301: Design Of Members Example 6.8: Shear Force Resistance Of Orthotropic Double-skin Plate

TALAT Lecture 2301

Design of Members Shear Force Example 6.8 : Shear 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 6.8

1

Example 6.8. Shear 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 6.8

70 . mm

a1

a2

2

h

2

2

a 3 = 80.6 mm

Local buckling of top flange Length and width of web panel

(5.97)

L = 5 . 10 mm 3

kτ

(5.96)

λ w

Table 5.12

ρ v

if

L am

am

am

4.00 .

> 1.00 , 5.34

a m = 80 mm

a1 2

L

5.34 .

, 4.00

am L

0.81 . a m . f o t1 E kτ 0.48

λ w

= 62.5

2

k τ = 5.341

λ w= 0.328

η

0.4

Table 5.12

ρ v if ρ v > η , η , ρ v

(5.95), two flanges

V w.Rd

ρ .vb . t 1

L am

t3 .

0.2 .

fu

ρ v= 1.462 η = 0.617

fo

ρ v= 0.617 fo

V w.Rd = 4.036 . 10 kN 5

γ M1

No reduction for local buckling

50

0

50

100

100

50

0

50

100

Overall buckling, shear Cross sectional area

Gravity centre

Second moment of area

A

2 .t 1 .a

2 .t 2 .a 2

nw

A = 2.812 . 10 mm 3

2

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

2 .t 2 .a 2 .h e

IL

2 .t 3 .a 3 .

e = 30.022 mm

A 2 .t 2 .a 2 .h

2

2 .t 3 .a 3 .

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

2 h .n w 3 2

2

A .e

2

I L = 2.059 . 10 mm

4

I T = 3.517 . 10 mm

4

6

2

Approx. torsional constant

IT

TALAT 2301 – Example 6.8

3

6

Rigidities of orthotropic plate Table 5.10

Table 5.10

Table 5.10

E .I L

Bx

ν

2 .a

0.001 . N . mm

By

B x = 9.007 . 10

0.3

E

G

2 .( 1

B y = 1 . 10

ν)

G .I T

H

mm

N . mm

3

N . mm

8

mm

Elastic buckling load 1

(5.83)

(5.82)

(5.84)

φ

L. B y

4

kτ

3.25

H

η

b Bx 0.567 . φ

1.92 . φ

2

φ = 1.711 . 10

B x .B y 0.1 . φ

1.95

5

η = 6.236 . 10

5

2.75 . φ

2

. ηk

. τ = 1.216 10

6

1

(5.81)

2 k τ .π 4 . B .B 3 x y b

V o.cr

V o.cr = 0.039 kN

Buckling resistance

(5.120)

(5.119)

b .t 1 .f o

λ ow

V o.Rd V Rd

3

0.6

χ o

λ ow

0.8 (5.118)

λ ow= 3.039 . 10

V o.cr

χ o

2

χ .ob . t 1 .

if χ o > 0.6 , 0.6 , χ o χ o= 6.495 . 10

fo

if V w.Rd < V o.Rd , V w.Rd , V o.Rd

TALAT 2301 – Example 6.8

8

V o.Rd = 0.021 kN

γ M1

4

2

mm

H = 5.918 . 10

2 .a

N . mm

8

V Rd = 0.021 kN

2

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

a

a1

a = 80 mm

2

a 1 = 40 mm a 2 = 40 mm

2

2

a 3 = 80.623 mm

h

Local buckling of top flange Length and width of web panel

(5.97)

L = 5 . 10 mm 3

kτ

(5.96)

λ w

Table 5.12

ρ v

L

if

am

2 .a 1

am

4.00 .

> 1.00 , 5.34

am

L

a m = 80 mm 2

, 4.00

L

5.34 .

am

am 2

k τ = 5.341

L

0.81 . a m . f o t1 E kτ 0.48

λ w

η

λ w= 0.328 0.2 .

0.4

Table 5.12

ρ v if ρ v > η , η , ρ v

(5.95), three flanges (web)

V w.Rd

ρ .vb . t 1

= 62.5

t2

fu

ρ v= 1.462 η = 0.617

fo

ρ v= 0.617

t3 .

fo

V w.Rd = 6.055 . 10 kN 5

γ M1 50

0

50

100

TALAT 2301 – Example 6.8

5

100

50

0

50

100

Overall buckling, shear Cross sectional area A

2 .t 1 .a 1

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 .t 3 .a 3 .

2

IL

2

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 1

(5.83)

(5.82)

(5.81)

(5.84)

φ

L. B y

4

b Bx

kτ

3.25

H

η 0.567 . φ

1.92 . φ

2

0.1 . φ

1.95

1 2 k τ .π 4 3 . B .B x y b

V o.cr

φ = 0.018

B x .B y

η = 0.469 2.75 . φ

2

. ηk

τ = 4.156

V o.cr = 105.983 kN

Buckling resistance

(5.120)

(5.119)

b .t 1 .f o

λ ow

0.6

χ o

λ ow

0.8 (5.118)

V o.Rd V Rd

λ ow= 58.282

V o.cr

χ o

2

χ .ob . t 1 .

if χ o > 0.6 , 0.6 , χ o χ o= 1.766 . 10

fo

if V w.Rd < V o.Rd , V w.Rd , V o.Rd

TALAT 2301 – Example 6.8

4

V o.Rd = 57.795 kN

γ M1

6

2

V Rd = 57.795 kN

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

a 3 = 70 mm

Local buckling of top flange if t 1 < t 2 , t 1 , t 2

Thickness

tm

Length and width of web panel

L = 5 . 10 mm

(5.97)

kτ

3

(5.96)

λ w

Table 5.12

ρ v

L

if

am

t m = 5 mm 2 .a 1

am

> 1.00 , 5.34

4.00 .

am L

L

a m = 75 mm 2

, 4.00

5.34 .

am

am

2

k τ = 5.341

L

0.81 . a m . f o tm E kτ 0.48

λ w

η

λ w= 0.308 0.4

Table 5.12

ρ v if ρ v > η , η , ρ v

(5.95), two flanges

V w.Rd

ρ .vb . t 1

t2 .

= 66.667

0.2 .

fu

ρ v= 1.559 η = 0.617

fo

ρ v= 0.617 fo

V w.Rd = 4.036 . 10 kN 5

γ M1

No reduction for local buckling

50

0

50

100

TALAT 2301 – Example 6.8

7

100

50

0

50

100

Overall buckling, shear Cross sectional area A

Gravity centre

Second moment of area

2 .t 1 .a

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

(5.80b)

12 . 1

. . 10 b . 2 2 ν 32 . a 2

3

2 .a

1

t1

2 .h . t 1

3

t2

6 .t 1 2 .a

a .t 2 t 3

3

6 .h .t 2

mm

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

3

2 .a

N . mm

H = 8.75 . 10

6

t3

mm

mm

Elastic buckling load 1

(5.83)

(5.82)

(5.81)

(5.84)

φ

L. B y

4

η

b Bx

kτ

3.25

0.567 . φ

1.92 . φ

2

H

φ = 5.432 . 10

B x .B y 1.95

0.1 . φ

3

η = 0.083 2.75 . φ

1 2 k τ .π 4 . B .B 3 x y b

V o.cr

2

. ηk

τ = 3.409

V o.cr = 3.848 kN

Buckling resistance (5.120)

(5.119)

b .t 1 .f o

λ ow

0.6

λ ow

0.8 (5.118)

V o.Rd V Rd

λ ow= 305.887

V o.cr

χ o

χ o

2

χ .ob . t 1 .

if χ o > 0.6 , 0.6 , χ o

fo

if V w.Rd < V o.Rd , V w.Rd , V o.Rd

TALAT 2301 – Example 6.8

χ o= 6.412 . 10

6

V o.Rd = 2.099 kN

γ M1

8

2

3

6 .t 2

1

t3

t2

N . mm

3

3

t1

.

t3

3. 1

a .t 3

3

3

2 .E

H

a .t 3

3.

8

V Rd = 2.099 kN

2

2

Summary a)

b)

c)

V Rd.a = 0 kN

A a = 2.812 . 10 mm

2

V Rd.b = 57.8 kN

A b = 3.212 . 10 mm

2

V Rd.c = 2.1 kN

A c = 2.2 . 10 mm

TALAT 2301 – Example 6.8

3

3

3

9

2

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

= 7.6 . 10

3

N mm

= 18

N mm

=1

N mm

2

2

2