Talat Lecture 2301: Design Of Members Example 5.4: Axial Force Resistance Of Channel Cross Section

TALAT Lecture 2301

Design of Members Axial Force Example 5.4 : Axial force resistance of channel cross section 7 pages Advanced Level prepared by Torsten Höglund, Royal Institute of Technology, Stockholm

Date of Issue: 1999  EAA - European Aluminium Association TALAT 2301 – Example 5.4

1

Example 5.4. Axial force resistance of channel cross section Width/depth

b

80 . mm

h

100 . mm

fo

300 . MPa

Inner thickness/outstand

ti

8 . mm

c

25 . mm

E

70000 . MPa

Outer/bottom thickness

to

3.5 . mm

tu

4 . mm

Length

L

1200 . mm

Nodes no., co-ordinates, thickness

yi i = mm 0 1 2 3 4 5 6 7 8

=

25 50 50 50 0 -50 -50 -50 -25

zi mm

i

0 .. 8

k

1 .. 3

tli

to

th i

to

th 1

ti

tl8

ti

th 4

tu

tl4

tu

th 5

th 4

tl5

tl4

=

80 77.75 40 0 0 0 40 77.75 80

tli mm

=

th i mm

3.5 3.5 3.5 3.5 4 4 3.5 3.5 8

0.5 . h

yk

y0

0.5 . h

y8

y0

c

=

G

γ M1 1.0

zi

yk 4 0.5 . h y4 0 . mm

kN 1000 . newton E 2.6

6 MPa 10 . Pa

b

zk 2 0 . mm z2 0.5 . z1 z6

0.5 . z1

z1

z1

z7

z1

0.5 . ( ti

to )

3.5 8 3.5 3.5 4 4 3.5 3.5 3.5

90

65

40

Nodes

i

1 .. rows ( y )

15

1

Cross-section constants, varying thickness Sectorial co-ordinates

ω0

0 . mm

.z

1 i

Thicknesses to simplify expressions

Dti

tli

Length of elements

li

yi

A

ωi

t1 i

1

2

zi

tli

th i 2

i =1 rows ( y ) First moments of area

yi . zi 1

yi 1

rows ( y ) Area

60

30

0

30

60

2

ω 0 yi i th i

10

zi 1

t2 i

2

1

tli

Dti

2

3

ω 0 i t3 i

tli

Dti

3

4

A = 1.233 . 10 mm 3

i

2

1 yi

yi 1 . t2 i . li

i =1

TALAT 2301 – Example 5.4

i

2

.l

yi 1 . t1 i

Sz

Dti

tli

ω

2

S z = 8.47 . 10

13

mm

3

TALAT 2301 – Example 5.4

3

rows ( y )

1 zi 1 . t1 i

Sy

zi 1 . t2 i . li

zi

S y = 4.388 . 10 mm 4

3

i =1 rows ( y )

1

ω

Sω

i

1

. t1

ω

i

ω

i

i

1

. t2 . l i

S ω = 7.328 . 10 mm 6

i

4

i =1 rows ( y ) Second moments of area

1

Iz

2 yi 1 . t1 i

2 . yi 1 . yi

yi 1 . t2 i

yi

2 zi 1 . t1 i

2 . zi 1 . zi

zi 1 . t2 i

zi

2 yi 1 . t3 i . li

i =1 rows ( y )

1

Iy i =1 rows ( y )

1

ω

Iω Mixed

2 zi 1 . t3 i . li

i

1

2.

2 . ω i 1. ω i

t1 i

ω

. t2

i

ω

yi 1 . t2 i

yi

i

1

ω

i

i

1

2.

t3 i . li

i =1 rows ( y )

1 yi 1 . zi 1 . t1 i

I yz i =1 rows ( y )

yi 1 . zi

zi 1 . yi

zi 1

yi 1 . zi

zi 1 . t3 i . li

1

I yω

yi 1 . ω i 1 . t1 i

yi 1 . ω i

ω

i

1

ω

i

1

. y i

yi 1 . t2 i

zi 1 . ω i 1 . t1 i

zi 1 . ω i

ω

i

1

ω

i

1

. z i

zi 1 . t2 i

yi

yi 1 . ω i

ω

i

1

. t3 . l i i

i =1 rows ( y )

1

I zω

zi

zi 1 . ω i

ω

i

1

. t3 . l i i

i =1

Influence of thickness

Iai

2 th i . tli

tli

th i

2

rows ( y )

l . i 48

I yz

1

I yz

Iai . yi

li

i =1 rows ( y ) Iz

1

Iz

tli

rows ( y )

2

zi 1 li

i =1 rows ( y ) 1 It

Iai . zi

Iy

2

yi 1 . zi

Iy

1

zi 1

2

Iai . yi li

i =1

2 th i . tli

th i

2

l . i . 1.05 12

yi 1 2

I t = 8.425 . 10 mm 3

2

4

i =1 z gc y gc

Sy A Sz A

I yz

I yz

Iω

Iω

S y .S z A Sω A

TALAT 2301 – Example 5.4

z gc = 35.593 mm

Iy

Iy

A . z gc

2

I y = 1.325 . 10 mm

y gc = 0 mm

Iz

Iz

A . y gc

2

I z = 2.151 . 10 mm

I yz = 1.816 . 10

10

mm

4

I yω

I yω

I zω

I zω

2

I ω = 9.438 . 10 mm 9

4

6

6

6

S z.S ω A S y .S ω A

4

4

I yω = 1.057 . 10 mm 8

I zω = 5.294 . 10

8

5

mm

y 5

Shear centre

y sc

Warping constant

Iw

I zω . I z

I yω . I yz

I y .I z

2

I yz

z sc . I yω

Iω

I yω . I y

z sc

I zω . I yz

I y .I z

y sc = 4.668 . 10

2

I yz

14

mm

z sc = 49.159 mm

y sc . I zω

I w = 4.24 . 10 mm 9

6

Polar radius gyration 0

ip

Iy

Iz A

50

y sc

y gc

2

z sc

z gc

2

i p = 100 mm

0 0

50

50

0

50

Buckling of stiffened flanges 5.4.5

Local buckling, internal elements 250 . MPa

ε

5.4.5 (3) c) heat-treated unwelded

ε = 0.913

fo

ρ c if i

β

i

ε

22 , 1.0 ,

32

β

βi 220

β

i

ε

3 , 5 .. 7

i

2 i

ε

t efl

i

t efl

i

yi

ρ c. tli i ρ c. tli i

1

yi 2

t efh

i

t efh

i

2

zi

2 2 zi 2 . tli 1 th i

ρ c. th i i 1

ρ c. th i i βi =

ρ c= i

t efl

i

mm

=

t efh mm

i

=

22.214 0.943 3.302 3.302 25 0.875 3.501 3.501 22.214 0.943 3.302 3.302

5.4.5

Local buckling, outstand elements 1

t fic

i

if tli > th i ,

TALAT 2301 – Example 5.4

3 tli . th i

4

1 , 8 .. 8

i

1

,

3 th i . tli

5

4

βi

yi

yi 1

2

zi

2 1 zi 1 . t fic i

β

ρ c if i

5.4.5 (3) c) heat-treated unwelded

i

10

6 , 1.0 ,

ε

β

24

β

i

ε

ρ c. tli i

t efl

2

i

t efh

ρ c. th i i

i

i

ρ c= i

βi =

ε

3.858 3.858

1 1

t fic

i

mm

=

6.506 6.506

t efl t efh i i = = mm mm 3.5 8

8 3.5

Reference: Pekoz, T., Flanges of nonuniform thickness. Aluminium Design Workshop, Cornell University Oct. 1997

Buckling of edge stiffener t efh

Effective area

t1 i

Length of elements

li

Area

t efl

t efl

i

t efl

i

yi

yi 1

2

t efl

Ar

2

zi t efh

i

2

zi 1 i.

t efh

i

t2 i

2

i

i

2

1 .. 2 t efl

i

z2 = 40 mm

i

3

2

A r = 269 mm

li

2

i =1 2 Centre of gravity

zi 1 . t1 i

zr

zi

i =1

Second moment of area Dti Length of elements

li

i th i

tli

yi

yi 1 2

Area

t1 i

A r1

2

tli

th i 2

1 .. 2 tli

zi

1 zi 1 . t2 i . li . Ar

z2

z1

Dti

t2 i

2

zi 1

z r = 69.528 mm

15 . tl2

th 2 . 0.5

tli

Dti

2

3

t3 i

z6

z2

tli

Dti

3

4

2

.l

A r1 = 328.1 mm

i

i =1 2 Centre of gravity

yi 1 . t1 i

yr i =1

TALAT 2301 – Example 5.4

yi

1 yi 1 . t2 i . li . A r1

6

y r = 45.218 mm

2

z2 = 25.25 mm

Second moments of area

2 2 yi 1 . t1 i

Ir z2

i =1 z1 . 0.5

z6

z2

2 . yi 1 . yi

yi 1 . t2 i

2 yi 1 . t3 i . li

yi

2 y r . A r1

I r = 1.668 . 10 mm 4

6 .. 8

k

i

0 .. 2

j

0 .. 8 yr

zr

mm

mm

4

Left: Cross section for second moment of areaIr based on t Right: Effective cross section areaAr based on t ef

s1

y3

y5

b1

3 b1 . 1

Table 5.5 and 5.8

α

0.2

(5.37)

φ

0.5 . 1

b 1 = 69.528 mm

3 I r . th 3

1.05 . E .

N r.cr

zr

λ 1 α . λ c

s 1 = 100 mm

1.5 . s 1 . th 4 b 1 . th 3

3

N r.cr = 52.195 kN

3

f o .A r

λ c

0.6 2

λ 1

λ c

1

χ r φ

i

1 .. 2

t efl

i

7 .. 8

t efl

χ .rt efl i χ .rt efl

i i

i

t efh

i

t efh

i

λ c= 1.243

N r.cr

φ

2

li

yi

yi 1 8

Area

2

zi t efh

t efl

i

A eff

2

zi 1 i.

TALAT 2301 – Example 5.4

φ = 1.337

1 .. 8 t efl

i

2

mm A eff = 904.9 mm

li

Axial force resistance, flexural-torsional buckling L

λ c

χ r= 0.547

i

i =1

Buckling length l

if χ r > 1 , 1 , χ r

χ .rt efh i χ .rt efh i

Length of elements

χ r 2

l = 1200 mm

7

2

=

1.913 1.805 3.302 3.501 3.501 3.302 1.805 4.372

t efh mm

i

=

4.372 1.805 3.302 3.501 3.501 3.302 1.805 1.913

li mm

=

25.101 38.875 38.875 50 50 38.875 38.875 25.101

Reference buckling loads

2 π .E .I y

N Ey

N Ez

2

l

Solution

N Ey

N . N Ez

N cr

l

3

N . NT

2 N .i p

z sc

G .I t

NT

2

N Ez = 1.03 . 10 kN

N Ey = 635.91 kN Given

2 π .E .I z

2 π .E .I ω

l

N T = 475.37 kN

2 2 z gc . N . N Ey

. 1

2

N

ip

2

Guess:

y sc

N

2 2 y gc . N . N Ez

0.2 . N Ez N

0

N cr = 348.11 kN

minerr( N )

General cross-section Table 5.5 and 5.8

λ c λ bar (5.37)

α

0.4

k2

f o . A eff

λ c φ

λ 1

0.35

λ c= 0.883

N cr

0.5 . 1

1

α . λ c

1

2

λ 1

λ c χ φ

Table 5.5

ψ

min z

z gc

max z

z gc

min z

z gc

max z

z gc 2

Table 5.5

k1

1

λ c

2 2.4 . ψ .

1

5.8.3 (1)

N b.Rd

χ .k 1 .k 2 .

TALAT 2301 – Example 5.4

2 λ c . 1

fo

γ M1

λ c

.A

2

φ

2

φ

= 0.974

χ = 0.721

2

λ c

min z

z gc = 35.6 mm

max z

z gc = 44.4 mm

ψ = 0.11

k 1 = 0.996

N b.Rd = 265.8 kN

8