Talat Lecture 2301: Design Of Members Example 5.3: Resistance Of Cross Section With Radiating Outstands

TALAT Lecture 2301

Design of Members Axial Force Example 5.3 : Resistance of cross section with radiating outstands 6 pages Advanced Level prepared by Torsten Höglund, Royal Institute of Technology, Stockholm

Date of Issue: 1999  EAA - European Aluminium Association

TALAT 2301 – Example 5.3

1

Example 5.3. Resistance of cross section with Flange width

b

80 . mm

Length

Inner circle

c

18 . mm

fo

300 . MPa

Inner thickness

ti

10 . mm

E

70000 . MPa

Outer thickness

to

5 . mm

k

φk

1 , 4 .. 16

φ Nodes no., co-ordinates, thickness

i

0 .. 16

π .k 1 4 3

1

k

1

π 4

φ

k

φ

18

φ φ

k

17

zk 1 c . sin φ k 1 z18 z0

tlk 1

o

th k 1

o

tlk

to

th k

tlk 1

ti . 0.8

th k 1

tlk 1

tl18

ti

th 18

=

i = φi =

mm

1 -0.785 2 0 3 0 4 0 5 0.785 6 0.785 7 0.785 8 1.571 9 1.571 10 1.571 11 2.356 12 2.356 13 2.356 14 3.142 15 3.142 16 3.142 17 3.927 18 3.927

12.728 18 80 18 12.728 56.569 12.728 0 0 0 -12.728 -56.569 -12.728 -18 -80 -18 -12.728 -56.569

i

1 .. rows ( y )

TALAT 2301 – Example 5.3

zi mm

=

6 MPa 10 . Pa

1

c . cos φ k 1

y12

kN 1000 . newton

γ M1 1.0 G 27000 . MPa

zk 1 b . sin φ k 1 zk c . sin φ k

yi

Nodes

4

1200 . mm

L

yk 1 b . cos φ k 1 yk c . cos φ k yk 1

y18

π .k 2 4 3

π

radiating outstands

ti to

tli th i = = mm mm

-12.728 5 0 8 0 0 0 5 12.728 8 56.569 0 12.728 5 18 8 80 0 18 5 12.728 8 56.569 0 12.728 5 0 8 0 0 0 5 -12.728 8 -56.569 10

10 8 0 10 8 0 10 8 0 10 8 0 10 8 0 10 8 5

1

2

90

45

0

45

90

90

45

0

45

90

Cross-section constants, varying thickness Sectorial co-ordinates

Length of elements

ω

0

0 . mm

2

.z

ω 0 yi i

1 i

Dti

tli

li

th i

yi

ωi

t1 i

2

yi 1

rows ( y ) Area

yi . zi 1

1

zi

tli

A

Dti

tli

th i

t2 i

2

zi 1

ω

i

ω 0 i

1

tli

Dti

2

3

t3 i

Dti

3

4

2

.l

A = 3.916 . 10 mm 3

i

2

tli

2

i =1 rows ( y ) First moments of area

1

Sz

yi 1 . t1 i

yi

yi 1 . t2 i . li

S z = 1.186 . 10

zi 1 . t1 i

zi

zi 1 . t2 i . li

S y = 2.457 . 10 mm

11

mm

3

i =1 rows ( y )

1

Sy

4

3

i =1 rows ( y )

1

ω

Sω

i

1

. t1

ω

i

ω

i

i

1

. t2 . l i

S ω = 2.692 . 10 mm 6

i

4

i =1 rows ( y ) Second moments of area

1 2 yi 1 . t1 i

Iz

2 . yi 1 . yi

yi 1 . t2 i

2 yi 1 . t3 i . li

yi

i =1 rows ( y )

I z = 4.551 . 10 mm 6

1 2 zi 1 . t1 i

Iy

2 . zi 1 . zi

zi 1 . t2 i

zi

2 zi 1 . t3 i . li

i =1 rows ( y )

I y = 3.402 . 10 mm 6

ω

i

1

2.

t1 i

2 . ω i 1. ω i

ω

i

1

. t2

i

ω

ω

i

i =1 rows ( y )

4

1

Iω Mixed

4

i

1

2.

t3 i . li

I ω = 2.638 . 10 mm 9

6

1 yi 1 . zi 1 . t1 i

I yz

yi 1 . zi

zi 1 . yi

zi 1

yi 1 . t2 i

yi

yi 1 . zi

zi 1 . t3 i . li

i =1 I yz = 1.059 . 10 rows ( y )

mm

4

1 yi 1 . ω i 1 . t1 i

I yω

10

yi 1 . ω i

ω

i

1

ω

i

1

. y i

yi 1 . t2 i

yi

yi 1 . ω i

ω

i =1 I yω = 5.018 . 10 mm 7

TALAT 2301 – Example 5.3

3

5

i

1

. t3 . l i i

rows ( y )

1 zi 1 . ω i 1 . t1 i

I zω

zi 1 . ω i

ω

i

1

ω

i

1

. z i

zi 1 . t2 i

zi 1 . ω i

zi

ω

i

1

i =1 I zω = 1.689 . 10 mm 7

Iai

2 th i . tli

tli

rows ( y ) Influence of thickness

I yz

1

I yz

2

th i

l . i 48

Iai . yi

yi 1 . zi li

i =1 rows ( y ) Iz

1

Iz

Iai . zi li

rows ( y ) Iy

1

Iy

Iai . yi

rows ( y )

tli

10

I z = 4.56 . 10 mm 6

mm

4

4

2

I y = 3.413 . 10 mm 6

2

1

It

I yz = 1.061 . 10

2

2

yi 1 li

i =1

zi 1

2

zi 1

i =1

5

2 th i . tli

th i

2

l . i . 1.05 12

I t = 8.602 . 10 mm 4

4

4

i =1

Sy

z gc y gc I yz Iω

Principal axis

α

Iξ Iη

z gc = 6.274 mm

A Sz A I yz

A

if I z

2 1. 2

A Sω

Iω

1.

S y .S z

y gc = 3.028 . 10

15

I yz = 1.805 . 10

10

mm

4

Iy

A . z gc

2

I y = 3.259 . 10 mm

Iz

Iz

A . y gc

2

I z = 4.56 . 10 mm

I yω

I yω

I zω

I zω

2

I ω = 7.875 . 10 mm 8

Iy < Iy

6

2 . I yz 1 9 I z . 10 , 0 , . atan Iz Iy 2

Iy

Iz

Iz

Iy

Iy

Iz

Iz

Iy

TALAT 2301 – Example 5.3

mm

Iy

2

4 . I yz

2

4 . I yz

2

2

4

6

6

S z.S ω A S y .S ω A

4

4

I yω = 5.018 . 10 mm 7

I zω = 3.309 . 10

α.

180

π

9

mm

= 7.944 . 10

I ξ = 4.56 . 10 mm 6

15

4

I η = 3.259 . 10 mm 6

5

4

5

. t3 . l i

i

Shear centre Warping constant

Polar radius gyration

I zω . I z

y sc

I yω . I yz

I y .I z

Iw

Iy

ip

2

I yz

z sc . I yω

Iω

Iz

I y .I z

I zω . I yz

y sc = 1.625 . 10

2

I yz

y gc

2

15

mm

z sc = 11.005 mm

y sc . I zω

y sc

A

I yω . I y

z sc

I w = 2.353 . 10 mm 8

z sc

z gc

2

6

i p = 44.932 mm

0 50

Dashed line = gravity centre

0 0

Point = shear centre

50

50

0

50

Axial force resistance, flexural-torsional buckling Buckling length l Reference buckling loads

l = 1200 mm

L

N Ey

2 π .E .I y

N Ez

2

l

N Ey = 1.56 . 10 kN 3

Given Solution

N Ey

N . N Ez

N cr

N . NT

minerr( N )

TALAT 2301 – Example 5.3

2 π .E .I z

NT

2

l

N Ez = 2.19 . 10 kN 3

2 N .i p

z sc 3

5

2 π .E .I ω

t

2

l

N T = 1.34 . 10 kN

2 2 z gc . N . N Ey

N cr = 1.32 . 10 kN

G .I

3

N

y sc

. 1 2 ip Guess:

N

2 2 y gc . N . N Ez

0.2 . N Ez N

0

Cross-section entirely of radiating outstands, no welding, unsymmetical section Table 5.5 and 5.8

λ c λ bar (5.37)

α

k2

0.6

f o .A

λ c φ

λ 1

0.2

λ c= 0.945

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

2.4 . ψ

λ c

2.

1

5.8.3 (1)

N b.Rd

χ .k 1 .k 2 .

TALAT 2301 – Example 5.3

2 λ c . 1

fo

γ M1

λ c

.A

2

φ

2

φ

= 0.981

χ = 0.804

2

λ c

min z

z gc = 62.8 mm

max z

z gc = 73.7 mm

ψ = 0.08

k 1 = 0.998

N b.Rd = 942.32 kN

6