eksempel - momentbelastet søjle

8/9/2019 Eksempel - Momentbelastet søjle

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Aarhus UniversitySchool of Engineering

Department of Mechanical Engineering 59

Example – axial loaded member with bending moment.

The steel column shown below is free at the upper end and supported by a fixed support at the lower

end. The column is free to deflect in both y- and z-direction.

The design load at the column is:

Axial force NEd = 100 kN

Transverse force: FEd = 20 kN (in z-direction)

The cross section is a rolled I- section type IPE400, see cross-section below. Steel material is S235,

normal control class. IPE400 is a class 3 cross-section.

t =13.5f

t

=8.6

y

zx

b = 1 8 0

h=400

w = 3 9 0 0

FEd

Section A-A

M =Ed F l Ed

l

NEd

y,

A A

Bending moment diagram

Question:

Check the resistance of the column for the attached load.

Solution:

Cross-section values for IPE400:

mmi

mmW mm I

mmi

mmW mm I

mm A

z

z z

y

y y

5.39

10146 and 102.13

165

101160 and 103.231

1045.8

:table-IPE400From

3326

3346

23

=

⋅=⋅=

=

⋅=⋅=

⋅=

43

69

10514

10490

mm I

mm I

v

w

⋅=

⋅=

The column length:

mml Lcr 7800390022 =⋅=⋅= (concerning both the y-axis and the z-axis).


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For IPE400: 2.2180

400==

b

h

Slenderness:

6.1)and6.2Tablesee(0.21a""curvebuckling50.019.93

165

7800

9.93.=⇒→=

⋅=

⋅=⋅= y

y

cr

ycr

y

yi

L

N f A α

ε λ

6.1)and6.2Tablesee(0.34b""curvebuckling10.219.93

5.39

7800

9.93.=⇒→=

⋅=

⋅=

⋅= z

z

cr

zcr

y

z

i

L

N

f Aα

ε λ

Buckling reduction factor:

( )[ ] ( )[ ] 66.050.02.050.021.015.02.015.0

6.4)Figurefrom(or92.050.066.066.0

11

22

2222

=+−⋅+=+−+=

=

−+

=

−+

=

y y y y

y y y

y

λ λ α φ

λ φ φ

χ

( )[ ] ( )[ ] 03.310.22.010.234.015.02.015.0

6.4)Figurefrom(or19.010.203.303.3

11

22

2222

=+−⋅+=+−+=

=

−+

=

−+

=

z z z z

z z z

z

λ λ α φ

λ φ φ χ

Design resistance for the axial force:

07.0104.1522

10100

104.152220.1

2351045.892.0

3

3

,,

33

1

,,

=⋅

⋅==

⋅=⋅⋅⋅

=⋅⋅

=

y Rd b

Ed

y

M

y y

y Rd b

N N n

N f A

N γ

χ

32.0104.314

10100

104.31420.1

2351045.819.0

3

3

,,

33

1

,,

=⋅

⋅==

⋅=⋅⋅⋅

=⋅⋅

=

z Rd b

Ed

z

M

y z

z Rd b

N

N n

N f A

N γ

χ

Interaction factors (see Table B1, B2 and B3):

( ) ( ) 92.06.0192.007.050.06.0190.0)0 and 0 :B3(Table90.0

6.016.01

h

=⇒⋅+≤=⋅⋅+⋅=

===

⋅+≤⋅⋅+=

yy ymy yy

hmy

ymy y ymy yy

k nC k

M C

nC nC k

α

λ

0

789.320

98.098.032.025.090.0

05.0194.032.0

25.090.0

10.205.01

)0 and 0 :B1(Table90.0

25.0

05.01

25.0

05.01

,

,

h

=

=⋅=⋅=

⋅⋅=

=⇒=⋅−

−≥=⋅−

⋅−=

===

⋅−

−≥⋅−

⋅−=

z

Ed Ed y

yd y LT

Ed y

y

zy zy

hmLT

z

mLT

z

mLT

z zy

m

kNmlF M

f W

M m

k k

M C

nC

nC

k

χ

α

λ


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The lateral-torsional buckling reduction factor χ LT must be found (same procedure as in Example 2 on

page 55):

48.2104906.2

390010514

9

232

=⋅⋅⋅

⋅⋅⋅=

⋅

⋅⋅=⋅

E

E

I E

l I Glk

w

v

Nmmhl

I E lk M t

z E

6

2

622

2

22

109.4425.3863900

102.1321000048.21

21

2⋅=⋅

⋅⋅⋅

+⋅=⋅

⋅⋅

⋅+⋅=

π

π

π

π

( ) ( )

Nmm M c M

c

cc

lk

cccc

E LT cr

LT

LT

66

2

21

2

2

2

2

2

2

21

103.354109.44280.0

80.049.24

28.1

49.24

28.1130.1

52pageseeflange,upperat theload- 1-

6.6Tablesee- 28.16.6;Tablesee-30.1

49.248.24

4

44

1

⋅=⋅⋅=⋅=

=

+

−+

+⋅=

=

==

=⋅=⋅⋅=

+

⋅++

+⋅=

β

π π κ

κ β

κ

88.0 103.354

235101160

6

3

=⋅

⋅⋅=

⋅=

cr

y y

LT M

f W λ

( )[ ] ( )[ ]

51.023510116068.0

20.11078

)71.088.075.091.091.0

1 :case(alternat. 68.0

88.00.10.1

1

6.3Tablesee- 34.06.4)(Tablebcurvebucklingonimperfecti22.25.13

400

0.188.02.088.034.015.02.015.0

1

3

6,

2222

22

22

=⋅⋅⋅

⋅⋅=

⋅⋅=

=

⋅−+

==

−+

=

=⇒⇒>==

=+−+=+−+=

−+

=

yd y LT

Ed y

y

LT LT

LT

LT LT LT LT

LT LT LT

LT

f W

M m

b

h

χ

χ χ

α

λ λ α φ

λ φ φ χ

Criteria, which must be satisfied:

O.K. 154.051.092.007.01

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