Talat Lecture 2301: Design Of Members Example 9.2: Beam-column With Rectangular Hollow Section

TALAT Lecture 2301

Design of Members Axial force and bending moment Example 9.2 : Beam-column with rectangular hollow section 4 pages Advanced Level prepared by Torsten Höglund, Royal Institute of Technology, Stockholm

Date of Issue: 1999  EAA - European Aluminium Association TALAT 2301 – Example 9.2

1

Example 9.2 Beam-column with rectangular hollow section

Dimensions and material properties Length

6 MPa 10 . Pa

3800 . mm

lc

Thickness

tw

Width

hy

180 . mm

hi

hy

2 .t f

h i = 168 mm

by

120 . mm

bi

by

2 .t w

b i = 108 mm

[1] Table 3.2b

Alloy: EN AW-6060 T6 EP t < 15 mm

[1] (5.4), (5.5)

fo

f 0.2

fa

fu

E

tf

6 . mm

kN 1000 . newton

6 . mm

f 0.2

70000 . MPa

140 . MPa

γ M1 1.1

fu

170 . MPa

γ M2

1.25

Forces and moment Axial force

N Ed

110 . kN

Transverse force

F Ed

Bending moment

M y.Ed

8 . kN F Ed . l c 4

M y.Ed = 7.6 kN . m

Classification of the cross section in axial compression 2 .t f

hy

Web

β w

[1] Tab. 5.1

β 1w 11 . ε

tw

class c [1] 5.8.4.1

Ae

β w= 28 β 2w 16 . ε

ε

250 . MPa fo

β 3w 22 . ε

if β w > β 1w , if β w > β 2w , if β w > β 3w , 4 , 3 , 2 , 1

A

TALAT 2301 – Example 9.2

η

1.0

2

ε = 1.336

class c = 3

Classification of the cross section in y-y axis bending 2 .t f

hy

Web

β w 0.4 .

[1] Tab. 5.1

β 1w 11 . ε

[1] 5.4.5

class w

Flange

β f

[1] Tab. 5.1

β 1f 11 . ε

[1] 5.4.5

class f

β w= 11.2

tw

β 2w 16 . ε

β 3w 22 . ε

if β w > β 1w , if β w > β 2w , if β w > β 3w , 4 , 3 , 2 , 1 2 .t w

by

class w = 1

β =f 18

tf

β 2f 16 . ε

β 3f 22 . ε

if β >f β 1w , if β f > β 2f , if β f > β 3f , 4 , 3 , 2 , 1

class f = 2

Classification of the total cross-section iny-y axis bending: if class f > class w , class f , class w

class y

class y = 2

Classification of the cross section in z-z axis bending 2 .t w

by

Web

β w 0.4 .

[1] Tab. 5.1

β 1w 11 . ε

[1] 5.4.5

class w

Flange

β f

[1] Tab. 5.1

β 1f 11 . ε

[1] 5.4.5

class f

β w= 7.2

tf

β 2w 16 . ε

β 3w 22 . ε

if β w > β 1w , if β w > β 2w , if β w > β 3w , 4 , 3 , 2 , 1 2 .t f

hy

class w = 1

β =f 28

tw

β 2f 16 . ε

β 3f 22 . ε

if β >f β 1w , if β f > β 2f , if β f > β 3f , 4 , 3 , 2 , 1

class f = 3

Classification of the total cross-section inz-z axis bending: class z

if class f > class w , class f , class w

class z = 3

Cross section constants A

b y .h y

b i.h i b i.h i

12

12

W el.y W pl.y

TALAT 2301 –

iy

I y = 1.565 . 10 mm 5

b y .h y

b i.h i

4

4

Iy

ExampleA 9.2

4

Iz

W el.y = 1.738 . 10 mm

hy

h y .b y

h i.b i

12

12

3

7

I y .2

class y = 2

2

3

2

[1] 5.6.2.1

3

b y .h y

3

Iy

A = 3.456 . 10 mm

3

I z.2

W el.z

by

3

I z = 8.28 . 10 mm 6

W el.z = 1.4 . 10 mm 5

2

W pl.y = 2.1 . 10 mm 5

W pl.y α y W el.y

α y = 1.208

i y = 67.3 mm

3

3

class z = 3 iz

Iz A

α z 1 i z = 49 mm

4

3

Flexural buckling TALAT (5.6)

[1] 5.8.4.1

lc

λ y

π .i y

.

η .f o

φ y 0.5 . 1

0.20 . λ y

0.1

1

χ y φ y [1] Table 5.5

λ y= 0.804

E

k1

1

N Rd

2

λ y

χ y = 0.779 2

φ y

k2

λ z

lc

π .i z

.

η .f o

λ z= 1.105

E

φ z 0.5 . 1

0.20 . λ z

χ z

1

2

λ y

2

φ z

λ z

1

A .f o

N Rd = 439.9 kN

γ M1

Exponents in interaction formula [1] 5.9.4.2 (4)

ψ

α z.α y

ψ

ψ c χ z.ψ

if ( ψ > 2 , 2 , ψ )

ψ = 1.208

ψ c if ψ c < 0.8 , 0.8 , ψ c

ψ c= 0.8

Cross weld in mid section [1] Tab. 5.2

HAZ softening factor

[1] 5.9.4.5

ω o

ρ haz 0.65

ρ haz . f u . γ M1 f o . γ M2 fo M y.Rd α y . W el.y. γ M1 fo M z.Rd α .zW el.z. γ M1 χ min χ y

ω o= 0.695

ω x ω o

M y.Rd = 26.721 kN . m

M y.Ed = 7.6 kN . m

M z.Rd = 17.572 kN . m

M z.Ed

0 . kN . m

χ min = 0.779

Flexural buckling check [1] 5.9.4.2 (4)

N Ed .ω .N χ min x Rd

TALAT 2301 – Example 9.2

ψc

1 . M y.Ed ω o M y.Rd

4

1.7

M z.Ed M z.Rd

1.7 0.6

= 0.939

2

λ z

χ z = 0.586 2

φ z

0.1

< 1,0 OK !