Buckling AnalysisBuckling Analysis

THE UNIVERSITY OF HONG KONGTHE UNIVERSITY OF HONG KONGDr. Ray SuDr. Ray Su

Department of Civil EngineeringDepartment of Civil Engineering

Buckling of TrussBuckling of Truss

Before buckling After buckling

Maximum Load = 58N(Buckling load)

A crane collapsed on a Hennessy Centre demolition site in Causeway Bay, Hong Kong in 2007. The accident killed two construction workers.

Buckling InstabilityBuckling Instability• Buckling is an instability phenomenon where a

structural system or a structural member under predominant compressive forces is unduly sensitive to changes.

• Buckling load is the maximum load that the structure can be withstood before buckling.

• The aim of this course is to use matrix method to determine the buckling load and the buckling displacements (buckling modes) of frame structures

Neutral Equilibrium Unstable

A small disturbance will lead to excessive movement

A curved surface

Equilibrium positionat local min. Stable

Concept of Stability

M

θ

M

Bending Stiffness Kθ

=M/θ

The additional axial force increases the end rotations ,

lateral deflection x

and internal moment (Madd

=Px) and reduces the bending stiffness.

MM

θPP

Under combined axial and bending loads

x

The Effect of Axial Load on the The Effect of Axial Load on the Bending StiffnessBending Stiffness

Force approach

Consider a simple structural system with two struts only

Roller

Hinge

Hinge

Spring

L

L

L = the length of each portion of the columnk = spring stiffness

S Joint equilibrium

S

N

N/S = L/xS = (x/L) N

By the geometric argument

compression

compression

Force approach

L

L

F

N = axial load, compression taken to be positiveF = the applied lateral loadX = lateral deflection

S

= (x/L) N

If F0 kx

k

2S

Horizontal force equilibrium

F= kx

– (2N/L)

x (1)

Spring force

Lateral force generated by N

Rearranging

(k-2N/L) x

= F (2)

Additional stiffness(geometric stiffness)

Notes:Notes:• The effect of the axial force N

is equivalent to an

additional stiffness of value (-2N/L) acting laterally at the pin location. We term it as geometric stiffness.

• When the axial force is tensile (-ve), the geometric stiffness is positive. The effects on the system tend to be beneficial, the lateral deflection and the restraining force in the spring are both reduced.

• When the axial force is compressive (+ve), the geometric stiffness is negative. The effects on the system tend to be detrimental, the lateral deflection and the restraining force in the spring are both increased.

when N=Ncr , the negative geometric stiffness will cancel out the spring stiffness, causing the lateral stiffness of the system to become zero.

When the compressive force is kept increasing, N > 0 and assuming that F=0 (no horizontal applied force)

When buckling happen, the lateral (or bending) stiffness of the member becomes ZERO

For an axially loaded beam, the lateral stiffness comes from the bending stiffness. Buckling implies that the bending stiffness is cancelled out by the geometric stiffness

Critical load (Ncr)

k-2Ncr /L = 0

Ncr = kL/2 (3)

If the analysis is to be accurate to within 10%, it may need to allow for the effects of buckling if N/Ncr >0.1.

Stiffness K

with geometric effects

Decreasing in the lateral stiffness K=

k

(1-N/Ncr

)

increasing in lateral displacement

Buckling equation Substituting (3) into (2)

k(1-N/Ncr ) x

= F (4)

F

Nk (1-N/Ncr

)k

==k/K

Effects of displacement amplification

10.01.110.9

5.001.250.8

3.331.430.7

2.501.670.6

2.002.000.5

1.672.500.4

1.503.330.3

1.255.000.2

1.1110.000.1

1.000

Displacement Amplification

crN/Ncr

10.01.110.9

5.001.250.8

3.331.430.7

2.501.670.6

2.002.000.5

1.672.500.4

1.503.330.3

1.255.000.2

1.1110.000.1

1.000

Displacement Amplification

crN/Ncr

Energy approachWe will use the energy approach to derive the stiffness and geometric matrices.

y

=2Δ= 2 [ L -

d ]

Ignore the higher order terms

(of the spring)

L

Δ

2

Δ=y

Energy gain +ve

Energy loss -ved

x

L

kx

W

By the theory of minimum potential energy (see page B7)

U(with geometric effect)

A curve surface

Equilibrium positionat local min. i.e. δП=0

Stable

Concept of Stability

П

x

U+W= =U

=W

δП≠0

δП≠0δП=0

N

this implies

x

can be any

value. i.e. buckling! N=kL/2

U=

=W

=U+W

U = 0П

= W

F = 0 -F x =0П

= W = 0

=WδП=0for all x

a line

No local minimum

x=F/(k-2N/L)

Notes on TomorrowNotes on Tomorrow’’s Tests Test• The test venue is at Knowles Building 419• The design part is on slab design.• The theory part is on sketch of BMD and assembly

of system matrix and vectors• Please bring along the HK code and a calculator.• Table for reinforcement areas will be provided.

Buckling Problems with Multiple Degrees of Freedom

Energy Approach

Assumed disp.

shape

Have been derived in Matrix Method for Static Load Analysis

(ai

=Nodal displacements)

Consider a beam segment of length dx

Longitudinal deformations1.

Hooke’s law (axial shortening)2.

Geometric

u’

= du/dxv’

= dv/dx

Binomial expansionSee page B5

(due to geometric deformation)

Hooke’s law Geometric

dxdu dx

dx=ds

du

dv

v

ignored

Since v = Σ

ai

Ni

4

i=1

(see page B6)

-ΔP=

044

33

22

11

aa

aa

aa

aa

Hence:

Not a function of xa function of x

Π

is a function of ai

Strain energy axial load nodal forcesθ=dv/dx=v’dθ/dx=v’’dθ=v”dxM=EI v”

dU=½Mdθ

Note:

4x4

scalar

Geometric stiffness affecting the total stiffness of the beam element by the applied axial load P

PP

=kij

=gij

These two matrices will be given in the examination.

Stiffness matrix

Geometric stiffness matrix

22

22

22

22

3

433336336

343336336

304626

6126122646612612

LLLLLL

LLLLLL

LP

LLLLLL

LLLLLL

LEIPgk

The stiffness matrix and geometric stiffness matrix

It will be provided in the Examination paper.

Consider the last two eqns.

(Buckling Analysis)

Only considering the deformation due to geometric change

There is no end shear force and moment.

2

22

vd

=0

u2

PWhen buckling just occurred, u2 =0

Detailed solution of the Detailed solution of the eigenvalueseigenvalues

043

3363046

612det 223

ll

ll

Pll

llEI

043336

3046612

det2

EI

Pl

EIPl

30

2

043336

46612

det

04436363612

det

013515612 2

222 1807.32or4860.230lEI

lEI

lEIP

0)36()44)(3612( 2

2

2

2

2

lvv

043336

3046612

det 3

l

PlEI

Set θ=lθl l l l

=0.0829 or 1.0727

By formula02 cbxax

aacbbx

242

by the first equation9.01685/l×1-

5.7514θ2

=0 or θ2

=9.01685/5.7514l=1.5677/lor by the second equation-5.7514×1

+ 3.66854lθ2

=0 or θ2

=5.7514/3.66854l=1.5677/l

2DOF only

Subdivided into more line segments /elements

Knowing the nodal displacements, the column deformation can be calculated by interpolation.a3 a4

Sketch the mode shapes

Mode 1, v2 =1 and θ2 =1.5677/l

v

x/lθ2

θ1 =0

v

x/lθ2

θ1 =0

Mode 2, v2 =1 and θ2 =-9.567/l

v2 1 v2 1

=0

XY

=0

u2

PWhen buckling just occurred, u2 =0

C

=0 Axial force=0

Stiffness matrix of a beam Stiffness matrix of a beam elementelement

Stiffness equations for element 1

212

121

42

24

ML

EI

ML

EI

v1

=0 v2

=0u1

=0 u2

=0

2

2

2

1

1

1

22

22

22

22

3

0

0

4626612612

2646612612

uv

uv

LLLLLL

IAL

IAL

LLLLLL

IAL

IAL

LEI

0

0

removed

1 2

For Element 1

NO AXIAL FORCE (P=0)

0

Node 1 is fixed and vertical displacement at node 2 =0

1

2

For Element 2

Node 1 is fixed

2

v2

The stiffness matrix of element 2 is (the same as that in Example 1)

For element 2

XY

Element from example 1

Rotate the element and the local coordinate axes by 90o clockwise

The coordinate system is the same as that of element 1 in example 1.

Eqn(3)

Involving one DOF only.

Stiffness from element 1

To get accurate results, we have to use more elements

C

System

≠0Sliding only, no rotation

System

=0By static condensation reduced 3DOFs to 2DOFs

System &

Matrix condensation 2DOF 1DOF

=0By static condensation reduced 3DOFs to 2DOFs

221

121

2

1

2

1

42

24

4224

MLEI

MLEI

MM

LEI

The effects of pinned joint i.e. M1

’=0 has been considered.

&System

No transformθ1

u1

With stiffness matrix only

Stiffness+ geometric

matrices

Ignore the axial deformations v1

=v2

=0, u1

=u2

2

1

YX

Fixed end

θ2

u1

Take away l

in the matrices of the above equation

System

With only numerical values in the matrices

(details see next slide)

Expansion of DeterminantExpansion of DeterminantFor a 2×2 determinant

For a 3×3 determinant

2bcacbba

edcb

dfdeb

bfeec

afedecbdba

= acf

+ 2bde -

ae2 -

b2f –

cd

2

It will be provided in the Examination paper.

P

>

Considering the symmetric buckling mode of the 1Considering the symmetric buckling mode of the 1stst

mode, the mode, the problem can be simplified as a cantilever with a length problem can be simplified as a cantilever with a length LL/2 subjected /2 subjected to a point load to a point load

Past examination question, year 2000

EndEnd

(Shear Building)

The flexural stiffness of the horizontal beam elements is infinitely high and the beams would not be bent and rotated. This type of buildings

is called shear building. Each story has one horizontal DOF only.

Undeformed

shape Deformed shape

2DOF=0

+

SkippedIllustration of Energy Method

≠0

k Deq

dD

(internal W.D.)+ -P dD

(external W.D.)0

Skipped

Formulation for the stiffness Formulation for the stiffness matrix of element 1matrix of element 1

No use Stiffness for element 1

removed

212211

121111

420

240

ML

EIFv’L

EAv’L

EA

ML

EIFv’L

EAv’L

EA

v’1

= -v1 v’2

= -v1

2

2

1

1

2

12

1

1

22

22

22

22

3

42

24

MFMF

v’v’

v’

LLI

ALI

ALLL

IAL

IAL

LEI

v1 v1

u1

=0 u2

=0

2

2

2

1

1

1

22

22

22

22

3

0

0

4626612612

2646612612

uv

uv

LLLLLL

IAL

IAL

LLLLLL

IAL

IAL

LEI