Module a Structural Analysis Suscos 2013 2014 l4 Wa

European Erasmus Mundus Master Course

Sustainable Constructions under Natural Hazards and Catastrophic Events

520121-1-2011-1-CZ-ERA MUNDUS-EMMC

Conceptual Design of Buildings(Course unit code 1C2)

Module AStructural Analysis

J.P. Jaspart (University of Liège)

L4 – Elastic critical load of framed structures - Buckling analysis


Sustainable Constructions under Natural Hazards

and Catastrophic Events

Conceptual Design of Buildings

2

Module A – Structural Analysis

In many flexural structures, such as current steel framed structures, the ultimateload tends to be governed by instability phenomena - collapse modescharacterized by large lateral displacements induced by members submitted tohigh axial compression forces.

The collapse by instability phenomena may involve the global structure (globalinstability) or only some members (partial instability).

Load

Displacement

1st order elastic analysis

1st order plastic analysis

Elastic stability analysis (buckling load)

2st order elastic analysis

2st order plastic analysis (advanced analysis) Service load

ULS load

List of contents

Buckling analysisApproximate methodsExamples

Buckling in a global mode

Buckling in a local mode






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i) Analytical evaluation (out of scope of this course).ii) Numerical calculation (used by computer programs – Robot,….).iii)Approximate methods (Horne, Wood, …).

In general, non-sway structures (which include braced structures) presenthigh buckling loads and sway structures present low buckling loads.

Buckling in a sway mode Buckling in a non-sway mode

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The elastic buckling load (also called elastic critical load) of a structuremay be determined by several processes:






4


2

2

ecr L

EIN

Ed

crcr N

N

Approximate methods – WOOD’s METHOD (can be found in thepublication P119 of ECCS or other references).

The application of this method consists of the following steps:

- determination of the equivalent length Le for the column to be studied(depending if the structure is “sway” or “non-sway”);

- determination of the critical load of the column Ncr, using equation:

- calculation of cr by:

This process must be repeated for all columns of the structure in order to findthe lowest critical load multiplier cr .

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5


N

K11

K12

K21

K22

1

N

2

Kc

K1

K2

No-sway frames

Sway frames

12111

11 KKKK

KK

c

c

22212

22 KKKK

KK

c

c

Calculation of the equivalent length Le

Kc is the column stiffness coefficient, given by I/L;K1 and K2 are the stiffness coefficients for the adjacent columns, also given by I/L;Kij represent the effective stiffness coefficients of the adjacent beams;I denotes the moment of inertia (second moment of area) and L is the length of the member.

Column bases: 2 = 0.0 for fixed bases and 2 = 1.0 for pinned bases.

N

K11

K12

K21

K22

LE

1

N

2

Kc

K1

K2

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Other situations can be found in P119 of ECCS

a) b)

Beam with single curvature

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Beam with double curvature






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List of contents

Buckling analysisApproximate methodsExamples Applicable for plane frames and one-storey frames with low inclination of the

beams ( ), unbraced and with low axial force ( ):

EdH

i

baseEd

topEdcr

hVH

,)(

)(

Ed

y

NfA

3,0º26H,Ed

VEd

HEd

hi

Approximate methods – HORNE’s METHOD (in accordance with clause5.2.1(4) of EC3-1-1).

where: HEd is the total horizontal reaction at the top of the storey,VEd is the total vertical reaction at the bottom of the storey, H,Ed is the relative horizontal displacement between the top and the bottom of a

given storey, when the frame is loaded with the design horizontal loads (included imperfections) and

hi is the height of the storey, such as illustrated in figure above.







EXAMPLE 1

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Consider the following steel frame fixed at base supports, composed by columnsHEA 260 and beams IPE 400 submitted to the shown load combination.Calculate the elastic buckling load (load factor multiplier cr) using:a) Horne’s method;b) Wood’s method;c) Computer program Robot.

33.6 kN/m

204.0 kN 204.0 kN

10 m

5 m

5 m

45.0 kN/m

18.2 kN

14.8 kN

253.5 kN 253.5 kN







EXAMPLE 1

9

List of contents


a) Horne’s method - Using a 1st order elastic analysis, the internal forcediagrams due to vertical loads and horizontal loads are obtained.

78.0 kN

36.0 kN

225.0 kN

168.0 kN

225.0 kN

168.0 kN

36.0 kN

78.0 kN

-372.0 kN

-850.5 kN

-78.0 kN

41.9 kN

-372.0 kN

-850.5 kN

Shear force diagramsAxial force diagrams

Vertical loading diagrams (which due symmetry do not induce horizontaldisplacements): axial force and shear force diagrams







EXAMPLE 1

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List of contents


309.0 kNm309.0 kNm

200.9 kNm

60.3 kNm

119.9 kNm

189.0 kNm

200.9 kNm 200.9 kNm

119.9 kNm

200.9 kNm

60.3 kNm

189.0 kNm

219.1 kNm

253.5 kNm

Bending moment diagrams

Vertical loading diagrams (which due symmetry do not induce horizontaldisplacements): bending moment diagrams.







EXAMPLE 1

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List of contents


Shear force diagramsAxial force diagrams

Horizontal loading diagrams: axial force and shear force diagrams

16.5 kN

9.1 kN 5.3 kN

10.4 kN

9.1 kN

16.5 kN

5.3 kN

15.8 kN

-9.1 kN

-7.4 kN

-5.3 kN

-15.8 kN







EXAMPLE 1

12

List of contents


Bending moment diagrams

Horizontal loading diagrams: bending moment diagrams

26.6 kNm

49.2 kNm

52.2 kNm

18.9 kNm 33.4 kNm

49.1 kNm

33.3 kNm

26.6 kNm

52.3 kNm

18.9 kNm







EXAMPLE 1

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List of contents


From the previous internal force diagrams, it is obtained the displacementsalong the structures, in particular the horizontal displacements at floor levels.So according the Horne’s method we have:

23.5 mm

12.4 mm

1st floor: HEd = 33.0 kN, VEd = 1701.0 kN,h = 5.00 m and H,Ed = 12.4 mm.

82.7104.1200.5

0.17010.33

3

cr

2nd floor: HEd = 18.2 kN, VEd = 744.0 kN, h = 5.00 m and H,Ed = 11.1 mm.

02.11101.1100.5

0.7442.18

3

cr

cr is minimum value, so cr =7.82.







EXAMPLE 1

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List of contents


b) Wood’s methodConsidering the frame with lateral displacements (sway frame), the stiffnesscoefficients for the columns are given by:

The effective stiffness coefficients of theadjacent beams (double curvature) are given by:

where Ib is the in-plane second moment of area (Iy = 23130 cm4 for IPE 400) of the beam and Lb is the length of the beam.

Lowest sway buckling mode

where Ic is the in-plane second moment of area(Iy = 10450 cm4 for HEA 260) of the columns andLc is the length of the column.

9.20500

10450

c

cc L

IK

9.20500

10450︶︵ 21 c

c

LIKorK

1 2

3 4

70.3410002313050.15.12212

b

b

LIKK







EXAMPLE 1

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List of contents


b) Wood’s methodThe distribution coefficients for the upper (1) and lower (2) ends of columns1 and 2 are given by:

55.070.349.209.20

9.209.20121

11

KKKKK

c

c

02 (fixed base).

From the abacus for sway structures we obtain: mLLL

EE 30.600.526.126.1

N

K11

K12

K21

K22

1

N

2

Kc

K1

K2







EXAMPLE 1

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List of contents


b) Wood’s method

Similarly, the distribution coefficients for the upper (1) and lower (2) ends of columns 3 and 4 are given by:

38.070.349.20

9.2012

1

KK

Kc

c

55.070.349.209.20

9.209.20222

22

KKKKK

c

c

mLLL

EE 00.700.540.140.1 From the abacus for sway structures we obtain:

N

K11

K12

K21

K22

1

N

2

Kc

K1

K2







EXAMPLE 1

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b) Wood’s methodThe critical loads Ncr of columns of both levels are:

Level 1 (columns 1 and 2):

With the maximum axial forces, on each level, previous obtained NEd12 = 866.3 kNand NEd34 = 377.3 kN, comes:

kNLEINe

cr 0.54573.6

1010450102102

862

2

2

3.63.8660.5457

12, Ed

crcr N

N

Level 2 (columns 3 and 4):

kNLEINe

cr 17.44200.7

1010450102102

862

2

2

7.113.37717.4420

34, Ed

crcr N

NSo, in accordance with Wood’s method cr = 6.3.







EXAMPLE 1

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List of contents


c) Using Robot:

So, in accordance with software Robot, cr = 7.59.

Buckling mode 1cr = 7.59










EXAMPLE 2

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List of contents


Consider the 2D frame analysed in Example 1 (Lesson 3). Calculate the firstbuckling load with lateral displacements, using the Horne’s method and thecomputer program ROBOT.

Frame loading

Pinned supportsRigid jointsBeams: Levels 1 and 2 – IPE 450

Level 3 – IPE 360Columns: External – HEB 220

Internal – HEB 260Material: S 235

i) Pre-design solution 1.

Horne’s method – level 1(evaluated with 1st order elasticinternal forces and displacementsof Example 1 - Lesson 3)

41.53.8

350049.333475.42

cr

13.5crSoftware ROBOT







EXAMPLE 2

20

List of contents


Frame loading

Pinned supportsRigid jointsBeams: Levels 1 and 2 – IPE 450


Internal – HEB 360Material: S 235

ii) Pre-design solution 2.

Column cross sections increased.

27.12cr

70.102.4

350049.333475.42

cr

Horne’s method – level 1(evaluated with 1st order elasticinternal forces and displacementsof Example 1 – Lesson 3)

Software ROBOT







EXAMPLE 2

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List of contents


Frame loading

iii) Pre-design solution 3.

Initial cross sections but introducingbracing members.

Pinned supportsBeam-column joints pinnedBeams: Levels 1 and 2 – IPE 450


Internal – HEB 260Bracing members – CHS 60x5Material: S 235

68.20cr

Horne’s method (not applicable)

Software ROBOT

This lecture was prepared for the Edition 1 of SUSCOS (2012/14) by RUI SIMÕES (UC) and FLOREA DINU (UPT).

Adaptations brought by J.P. Jaspart (Ulg) for Edition 2 of SUSCOS

The SUSCOS powerpoints are covered by copyright and are for the exclusive use by the SUSCOS teachers in the framework of this Erasmus Mundus Master. They may be

improved by the various teachers throughout the different editions.

Thank you for your attention

http://steel.fsv.cvut.cz/suscos

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Module a Structural Analysis Suscos 2013 2014 l4 Wa

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