SX008a en EU Example Sway Stability

Document Ref: SX008a-EN-EU Sheet 1 of 9 Title

Example: Sway stability

Eurocode Ref EN 1993-1-1 Made by Jonas Gozzi Date April 2005

CALCULATION SHEET

Checked by Bernt Johansson Date July 2005

Example: Sway stability This example deals with design for global instability of frames, or sway stability. The frame considered is a non-braced two-storey building frame..

A two-storey building is considered in this example according to the figure below. The spacing between the frames in the building is s = 10 m.

A

3,5

3,5

7,07,0

[m]

All connections are designed as rigid according to the figure below.

A

Example: Sway stabilityCr

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CALCULATION SHEET


Frame data:Outer columns HEA 200 S355

Second moment of area 4y 3692 10I = mm4Inner columns HEA 220 S355

Second moment of area 4y 5410 10I = mm4Upper beams IPE 400 S355

Second moment of area 4y 23130 10I = mm4Lower beams IPE 450 S355

Second moment of area 4y 33740 10I = mm4

Loads:The following loads are considered on the structure.

Permanent loads:

Slab gslab = 4,0 kN/m2

Floor finishes gffin = 0,8 kN/m2

Steel gs = 0,3 kN/m2

Partition walls gW = 0,5 kN/m2

Suspended ceiling gsc = 0,2 kN/m2

Installations gI = 0,2 kN/m2

Roof and insulation groof = 0,4 kN/m2

Live loads:

Indoor live load q = 2,5 kN/m2

Snow qsnow = 1,0 kN/m2

Wind, windward qwind1 = 0 5 0 75 0 375, , , = kN/m2 Wind, leeward qwind2 = 0 5 0 4 0 2, , , = kN/m2

Load factors:

G = 1,35 (permanent loads) Q = 1,5 (variable loads) 0 = 0,7 (live load and snow load)

EN 1990


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CALCULATION SHEET


Design loads:

The design loads are calculated with the wind load as the principal load.

1 G roof slab s I sc Q 0 snow( ( ) )q g g g g g q s = + + + + + = 82,1 kN/m 2 G ffin W slab s I sc Q 0( ( ) )q g g g g g g q s = + + + + + + =107,3kN/m = 5,6 kN/m w1 wind11 5,q q= s

s = 3,0 kN/m w2 wind21 5,q q=

EN1990

6.4.3.2

(6.10)

In the figure below the structure is shown with these design loads.

q1

q2

qw1 qw2

Check if sway imperfections can be disregarded.

Ed Ed0 15,H V = 60,2 kN Ed w1 w 2 5 6 3 0 7( ) ( , ,H q q h= + = + ) = 2651,6 kN Ed 1 2 82 1 107 3 14( ) ( , , )V q q L= + = + Ed Ed60 2 0 15 2651 6 398 0 15, , , ,H V= < = = Sway imperfections have to be taken into account.

EN 1993-1-1

5.3.2 (4)B


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CALCULATION SHEET


Calculate global initial sway imperfections.

0 h m= 0

1200

=

h2h

= where h is the height of the structure in m

m10 5 1,m

= + where m is the number of columns in a row

31 2 10 5 1 3 09 10200 37 0

= + = , ,,

EN 1993-1-1

5.3.2 (3)

Calculate the equivalent horizontal forces, H1 and H2, due to the sway imperfections:

q1

q2

H1

H2

H1 + H2

EN 1993-1-1

5.3.2 (7)

= 3,55 kN 31 1 3 09 10 82 1 14H q L = = , , = 4,64 kN 32 2 3 09 10 107 3 14H q L = = , ,


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Determine sensitivity to sway, cr:There are several different ways to calculate the cr. One is to use the method given in EN 1993-1-1 5.2.1 (4)B described first below. Another is to perform a buckling analysis in a finite element package. This will also be described and used below. The influence from sway can be neglected if the following criterion is fulfilled.

crcrEd

10FF

=

See NCCI SN001

EN 1993-1-1

5.2.1 (3)

Check the sensitivity to sway for each storey:

EdcrEd H,Ed

H hV

=

HEd is the horizontal force. In this example only the equivalent horizontal force is used.

VEd is the total design vertical load on the structure on the bottom of the storey.

H,Ed is the horizontal displacement at the top of the storey due to the applied horizontal loads.

h is the storey height.

EN 1993-1-1

5.2.1

Eq. (5.2)

See NCCI SN001

EN 1993-1-1 Figure 5.1

The displacements on each storey due to the applied equivalent horizontal loads are calculated by means of a frame analysis software. The displacements, H,Ed1 and H,Ed2, are shown in the figure below.

H1

H2

H,Ed1 = 0,69 mm H,Ed2 = 1,23 mm

H,Ed1

H,Ed2


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CALCULATION SHEET


Total vertical load on the two storeys:

= 1149,4 kN Ed1 1 82 1 14,V q L= = = 1502,2 kN Ed2 2 107 3 14,V q L= = cr for the upper storey:

1crEd1 H,Ed1

3 55 35001149 4 0 69

,, ,

H hV

= =

=15,66 > 10

cr for the lower storey:

1 2crEd1 Ed2 H,Ed2

3 55 4 64 35001149 4 1502 2 1 23

, ,, , ,

H H hV V

+ + = = + +

= 8,79 < 10

The sway effects cannot be neglected for this frame.

The second alternative is to use a finite element package for determining cr. In this case only the vertical loads are considered as concentrated to the joints of the frame according to the figure below.

VEd1o

VEd2o VEd2o

VEd1oVEd1i

VEd2i

From a buckling analysis the cr can be directly determined. cr = 7,51 < 10 Hence, the sway effects cannot be neglected.


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CALCULATION SHEET


As can be seen the different methods gives different results. The finite element analysis should give the most correct results since the equation from EN 1993-1-1 is a simplification. In this case it gives a conservative result because the columns are uniform and the upper part has less load than the lower part. However, for large cr values the difference in the contribution from the sway on the frame is rather small. In this example the cr from the finite element analysis will be used.

Determine if local bow imperfections have to be consideredLocal bow imperfections should be considered if the following two conditions are met:

at least one moment resistant joint at one member end and

yEd

0 5,A fN

>

where

NEd is the design value of the compression force and

ycr

A fN

= is the in-plane non-dimensional slenderness calculated for the member considered as hinged at its ends

EN 1993-1-1

5.3.2(6)

EN 1993-1-1

6.3.1.2(1)

The conditions can be reformulated as follows:

y ycr Ed

0 5,A f AN N > f

Ed cr0 25,N N>

For the columns the buckling loads assuming hinged ends are:

Outer columns, HEA 200

2 2

2 2210000 3692 10

3500crE IN

l = =

4

= 6247 kN

Inner columns, HEA 220

2 2

2 2210000 5410 10

3500crE IN

l = =

4

= 9153 kN


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CALCULATION SHEET


The member that carries the highest normal force is the lower part of the outer column and the middle column.

The normal force in the outer column, lower part, according to first order analysis

NEd = 568 kN

Ed cr568 kN 1562 kN 0 25,N N= < = The normal force in the inner column, lower part, according to first order analysis

NEd =1533 kN

Ed cr1533 kN 2288 kN 0 25,N N= < = Since this two members do not meet the condition then no other member will and hence, the bow imperfections does not have to be considered.

Select method of allowing for sway effectsThe second order sway effects may be calculated according to first order theory if the horizontal loads, i.e. wind, and equivalent horizontal loads are increased by the factor:

cr

11 1

provided that cr 3 0,

EN 1993-1-1

5.2.2(5)B & (6)B

In this case cr = 7,51 and hence, this method can be used. The following loads will be considered in the first order analysis:

q1

q2

qw1a qw2aH1a

H2a

1 3

109

87

6

5

42

The numbers are the member numbers.


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CALCULATION SHEET


Where the horizontal loads marked with a are increased by the factor

cr

1 1 1151 1 1 1 7 51

,, = =

i.e.

= 4,08 kN 1a 115 3 55H = , , = 5,34 kN 2a 115 4 64H = , , = 6,44 kN/m w1a 115 5 6, ,q = = 3,45 kN/m w2a 115 3 0, ,q =

Member design forces for the frame:In the table below the design forces for all members are shown. For shear force and moment the two values corresponds to the two ends of the columns and the beams. The first value for the columns is for the lower end and the second is for the upper end and for the beams the first value is for the left end and the second is for the right end.

Member NEd [kN] VEd [kN] MEd [kNm]

1 546,9 0,95 21,6 6,0 42,2

2 238,3 42,9 65,4 86,0 103,5

3 1533,1 26,9 26,9 48,4 45,8

4 668,4 9,6 9,6 16,3 17,3

5 570,9 50,8 38,7 67,5 89,1

6 242,7 72,0 59,9 114,0 116,8

7 16,0 308,6 442,1 128,2 595,4

8 33,3 422,5 328,2 533,3 203,2

9 69,5 238,3 336,4 103,5 446,8

10 59,9 332,0 242,7 429,5

116,8

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Example: Sway stability SX008a-EN-EU.doc

Quality Record

RESOURCE TITLE Sway stability

Reference(s)

ORIGINAL DOCUMENT

Name Company Date

Created by Jonas Gozzi SBI 30/04/2005

Technical content checked by Bernt Johansson SBI 18/05/2005

Editorial content checked by

Technical content endorsed by the following STEEL Partners:

1. UK G W Owens SCI 7/7/05

2. France A Bureau CTICM 17/8/05

3. Sweden A Olsson SBI 8/8/05

4. Germany C Muller RWTH 10/8/05

5. Spain J Chica Labein 12/8/05

Resource approved by Technical Coordinator

G W Owens SCI 21.05.06

TRANSLATED DOCUMENT

This Translation made and checked by:

Translated resource approved by:


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SX008a en EU Example Sway Stability

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