12.1 Introduction to column buckling• Buckling: “Buckling can be defined as the

sudden large deformation of structure due to a slight increase of an existing load under which the structure had exhibited little, if any, deformation before the load was increased.” No failure implied!!!

Reinforced concrete steel

Stability of equilibrium condition

• Easy to visualize for a ball on a surface

• Note that we use curvature to decide stability, but the surface can curve up with zero curvature

Euler formula• For simply supported column

( )

2 22

22 or using , /

/ : Slenderness ratio

cr crEI EAP I Ar PL L r

L r

π π= = =

Large displacements• Slenderness ratio and yield stress govern type of

post-buckling response

Dashed lines represent behavior without yielding

Governing equation for beam column

⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟⎠

⎞⎜⎜⎝

⎛=

=

+−=

+=

=+

−==

xEIPBx

EIPAxy

EIPODE

xBxAdxyd

xBxAxyTry

yEIP

dxyd

yEIP

EIM

dxyd

sincos)(

:into Substitute

))sin()cos((

)sin()cos()(:

0

22

2

2

2

2

2

ω

ωωω

ωω

Apply boundary conditions

2

2

2

22

2

2

2

2

22

)/(:load buckling criticalat stress (normal) eCompressiv

ratio sSlendernes:/)/(

, usingor

load buckling critical thedefines 1n pinned, endsboth h column wit aFor

gyration. of radius theis

mode. buckling thedefininginteger theis

,..3,2,10sin

0)(0)0(

rLE

APσ

rLrLEIPArI

LEIP

AIrwhereArI

nLEInP

nnLEIPL

EIPB

Lyandy

crcr

crcr

π

ππ

π

π

==

===

=

==

=

==→=⎟⎟⎠

⎞⎜⎜⎝

⎛

==

Example 1A 3m column with the cross section shown is constructed from two pieces of timber, that act as a unit. If the modulus of elasticity of timber is E=13 GPa, determine a) The slenderness ratio b) Critical buckling load c) Axial stress in the column when the critical load is applied

mmrr

mmAI

mmAI

mmI

mmI

mmy

mmA

yx

y

x

c

3.32

3.32rand51.59r

inertia of Radii

10x625.1512

)50(15012

)150(50

10x13.53)50)(50(15012

)150(50)50)(50(15012

)50(150section cross theof centroid about the inertia of Moments

bottom from75000,15

)150)(50)(7550()150)(50(25000,15)50)(150(2

section cross theof Properties

min

yx

4633

4623

23

2

==⇒

====

•

=+=

=+++=

•

=++

=

==

•

MParLE

APσc

kNrLEAPb

a

crcr

cr

85.14)/(

Stress Buckling Critical)

75.222)/(

LoadBuckling Critical)

933.32

3000rLratio sSlendernes)

2

2

2

2

===

==

==

π

π

Example 2C229 x 30 structural steel channels with E=200 GPa are used for a 12 m column. Determine the total compressive load required to buckle the column if a) One channel is used b) Two channels are laced 150 mm back to back as shown

4646

2

10x 01.110x3.25

8.143795channel onefor propertiesSection

mmImmI

mmxmmA

ycxc

c

==

==

Solution for single channel

.

kNrLEAP

rL

mmrr

mmAI

rmmAIr

cr

yy

xx

82.13)/(

:load Buckling Critical

2.7363.16

12000:ratio sSlendernes

3.16

3.16and1.81

channel single a of inertia of Radii

2

2

min

==

==

==⇒

====

π

Two laced channels.

[ ]

kNrLEAP

rL

mmrrmmAI

rmmAIr

mmAII

mmII

mmA

cr

yy

xx

ycy

cxx

3.694)/(

:load Buckling Critical

9.1467.81

12000:ratio sSlendernes

7.813.91and7.81

:inertia of Radii

10x23.63)8.1475(2

10x6.502:section cross theof centroid about the inertia of Moments

7590)3975(2

2

2

min

462

46

3

==

==

==→====

=++=

==

==

π

Higher buckling modes

• With appropriate boundary conditions can get higher modes

2

2

)2(

2

2

)/(4

44

rLE

APσ

PLEIP

cr

cr

π

π

==

==

Buckling modes : n = 1,2,3.

Imperfection approach

• Put in imperfection in the form of eccentricity

• Moment equation

LxPePyxM −−=)(

Eccentrically loaded pinned-end columns

Solution of eccentric load• Differential equation of equilibrium

• General solution

• With boundary conditions

LxforyEIPk

Lxekyk

dxyd

,00

, 22

22

2

==

=−=+

LexxkBxkAy −+= cossin

⎟⎟⎠

⎞⎜⎜⎝

⎛−=Lx

Lkxkey

sinsin

Reading assignmentSections 12.3-4: Question: Why equation 12.26 does not really

guarantee buckling? What does it guarantee?

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