Talat Lecture 2301: Design Of Members Example 10.1: Transverse Bending Of Unsymmetrical Flange

TALAT Lecture 2301

Design of Members Deviation of linear stress distribution Example 10.1 : Transverse bending of unsymmetrical flange 4 pages Advanced Level prepared by Torsten Höglund, Royal Institute of Technology, Stockholm

Date of Issue: 1999  EAA - European Aluminium Association

TALAT 2301 – Example 10.1

1

Example 10.1. Transverse bending of unsymmetical flange Check transverse bending of the web due to shear force gradient in the flanges

Section height:

hw

400 . mm

Flange widths:

bo

300 . mm

bu

120 . mm

Flange thickness:

to

16 . mm

tu

16 . mm

Web thickness:

tw

6 . mm

Overall length:

L

6 .m

Load

q

10 . kN . m

O

0 . mm

1

Cross section, half profile Co-ordinates bu O y

tu

O O z

O bo

hw hw

2

t

tu tw to

400

300

200

kN 1000 . newton

100

6 MPa 10 . Pa

E 70000 . MPa

0 100

[1] ENV 1999-1-1.

0

100

Eurocode 9 - Design of aluminium structures - Part 1-1: General rules. 1997

[2] StBK-N5Regulations for cold-formed steel and aluminium structures. Svensk Byggtjänst Stockholm 1979 [3] Hetenyi, M. Beams on elastic foundation. University of Michigan, 1958

TALAT 2301 – Example 10.1

2

Nodes

1 .. rows ( y )

i

1 rows ( y )

Area of half cross section

dAi

ti .

yi

2

yi 1

zi

zi 1

2

1 dAi

A i =1

rows ( y )

First moment of area, y axis, gravity centre

Sy

Second moment of area, y axis, section modulus

Iy

1

dAi zi 1 . 2

zi i =1 rows ( y ) 1 zi i =1 rows ( y )

First moment of area, z axis

1

Sz

2

zi 1

Sy

z gc

z gc = 214.3 mm

A

dAi zi . zi 1 . 3

2

Iy Wy

dAi yi 1 . 2

yi

Lateral force

rows ( y )

2 . yi 1 . zi 1

2 . yi . zi

yi 1 . zi

dAi yi . zi 1 . 6

i =1 k h .q

Lateral force constant [2] Second moment of area of bottom flange + 0.27 . h w

I yz

b u .t u

k

E .t w

3

S y .S z

I yz

A 4

k h = 0.143

2 .I y

t u .b u

2 0.5 . b u . t u

0.27 . h

.t

I u.z

w w

3

Au

[2] Modulus of foundation [3]

I yz

7

3

Au

z gc

I yz = 5.811 . 10 mm

on bottom flange kh

Iy 5

1

I yz

2

W y = 9.493 . 10 mm

i =1 Second moment of area with respect to y and z axes

A . z gc

Iy

2

yu

2 0.5 . b u . t u

Au

3

4 .h w

k = 59.062 kN . m

3

2

(Transverse bending of top flange neglected)

1

Parameter λ [3]

Lateral deflection of the bottom flange at the middle of the span [3]

k 4 .E .I

λ

vc

Transverse bending moment and stresses m z in the web Lateral moment in bottom flange [3]

Mc

4

λ = 0.478 m

λ . L = 2.867

1

u.z

k h .q . k

L . L cos λ . 2 2 . . cosh ( λ L ) cos ( λ L )

2 . cosh λ . 1

k .v c .h w

m z = 0.527 kN

L L sinh λ . . sin λ . k h .q 2 2 . 2 cosh ( λ . L ) cos ( λ . L ) λ

v c = 22.3 mm

σ transv

m z.6 tw

2

M c = 1.56 kN . m

σ transv= 87.9 MPa

σ c

M c .y u I u.z

σ c= 17.3 MPa

TALAT 2301 – Example 10.1

3

Comment : If the web resistk h . q by itself then k h .q .h w m z.6 σ transv 2 tw mz

Main axis bending

My

q.

L

2

8

TALAT 2301 – Example 10.1

m z = 0.571 kN

σ transv= 95.2 MPa

My σ x Wy

4

vc

3 k h .q .h w

3 .E .

tw

3

v c = 24.184 mm

12

σ x = 47.4 MPa

σ x σ c = 64.7 MPa