Talat Lecture 2301: Design Of Members Example 6.7: Shear Force Resistance Of Orthotropic Plate. Open Or Closed Stiffeners

TALAT Lecture 2301

Design of Members Shear Force Example 6.7 : Shear force resistance of orthotropic plate. Open or closed stiffeners 6 pages Advanced Level prepared by Torsten Höglund, Royal Institute of Technology, Stockholm

Date of Issue: 1999  EAA - European Aluminium Association TALAT 2301 – Example 6.7

1

Example 6.7. Shear 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

a

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

t

w

6 MPa 10 . Pa

γ M1 1.1 ht 1

2

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 L = 2.2 . 10 mm 3

Length of web panel (5.97)

if

kτ

(5.96)

λ w

Table 5.12

ρ v

Table 5.12

η

L

a = 150 mm

L > 1.00 , 5.34 2 .a

4.00 .

2 .a L

2

, 4.00

5.34 .

2 .a 2 .a L

= 7.333

2

k τ = 5.414

0.81 . 2 . a . f o t E kτ 0.48

λ w= 0.371

λ w

ρ v = 1.295

0.4

0.2 .

fu

η = 0.617

fo

Table 5.12

ρ v if ρ v > η , η , ρ v

(5.95)

V w.Rd

TALAT 2301 – Example 6.7

ρ .vb . t 1 .

ρ v= 0.617

fo

V w.Rd = 3.33 . 10 kN 3

γ M1

2

Overall buckling, shear force

5.11.6

Cross sectional area A

Gravity centre

2 .t 1 .a

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

2

h

I L = 2.751 . 10 mm

2

3 . 2 a1

3

A .e

a2

7

, 2 .a

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

E .t

By

B x = 6.42 . 10

9

E . 2 (1 ν)

3

12 . 1 3 G .t

H

0.3

B y = 2.88 . 10

7

ν

2

H = 2.016 . 10

6

7

N . mm mm N . mm

2

mm N . mm

2

mm

Elastic buckling load 1

(5.83)

(5.84)

(5.82)

φ

L. B y

4

H

η

kτ

3.25

0.567 . φ

1.92 . φ

2

φ = 0.38

B x .B y

b Bx

0.1 . φ

1.95

η = 0.047 2.75 . φ

2

. ηk

τ = 3.423

1

(5.81)

2 k τ .π 4 . B .B 3 x y b

V o.cr

V o.cr = 2.506 . 10 kN 3

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= 1.54

V o.cr

χ o

2

χ .ob . t 1 .

if χ o > 0.6 , 0.6 , χ o χ o= 0.189

fo

V o.Rd = 1.022 . 10 kN 3

γ M1

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

TALAT 2301 – Example 6.7

3

2

V Rd = 1.022 . 10 kN 3

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

a2

2

2

a 3 = 162.8 mm

h

a 4 = 70 mm

a1

Local buckling L = 2.2 . 10 mm 3

Length of web panel

(5.97)

kτ

if

(5.96)

λ w

Table 5.12

ρ v

Table 5.12

η

L am

> 1.00 , 5.34

am 4.00 .

am L

2 .a

2 .a 1

2

, 4.00

5.34 .

a m = 140 mm am L

L am

= 15.714

2

k τ = 5.356

0.81 . a m . f o t E kτ 0.48

λ w= 0.174

λ w

ρ v = 2.76

0.4

0.2 .

fu

η = 0.617

fo

Table 5.12

ρ v if ρ v > η , η , ρ v

(5.95)

V w.Rd

ρ .vb . t 1 .

TALAT 2301 – Example 6.7

ρ v= 0.617

fo

V w.Rd = 3.33 . 10 kN 3

γ M1

4

Overall buckling, shear force

5.11.6

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

2 .t 3 .a 3 .

e

IL

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

A .e

2

3

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

Torsion constant

IT

(5.79c)

C1

B

(5.79d)

4. 1

2 ν . a2

E .t

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

t2

C 1 = 3.076 . 10 mm 9

3 .a .t 1

3

3

12 . 1

4

B = 2.88 . 10 N . mm 7

ν

4. a 1 (5.79e)

7

2

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

2 .a 4

mm

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

a 3 .t 2

a 3 .t 2

3

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

3

G .I T H

1.6 . G . I T . a 4

4 .a 3 .t 2

3

B y = 3.929 . 10

7

6 .a

2 .B

2

1

2 L .a .B

1

. 1 4 π .

C1 L

H = 7.305 . 10 C2

4

8

N . mm mm N . mm mm

Elastic buckling load 1

(5.83) (5.82)

(5.81)

(5.84)

φ

kτ

2

2 . B. a

By

(5.79b)

N . mm

B x = 7.72 . 10

L. B y

4

η

b Bx 3.25

V o.cr

0.567 . φ

1.92 . φ

2

1 2 k τ .π 4 . B .B 3 x y b

TALAT 2301 – Example 6.7

H

φ = 0.392

B x .B y 1.95

0.1 . φ

η = 1.326 2.75 . φ

2

. ηk

τ = 6.52

V o.cr = 6.311 . 10 kN 3

5

2

2

TALAT 2301 – Example 6.7

6

Buckling resistance

(5.120)

(5.119)

b .t 1 .f o

λ ow

0.6

χ o

λ ow

0.8 (5.118)

λ ow= 0.97

V o.cr

V o.Rd

χ .ob . t 1 .

TALAT 2301 – Example 6.7

χ o

2

fo

if χ o > 0.6 , 0.6 , χ o

χ o= 0.345 V o.Rd = 1.861 . 10 kN 3

γ M1

7