DAY8

BUCKLING OF AIRCRAFT STRUCTURES

BUCKLINGBuckling is a failure mode characterized by a sudden failure of a structural member subjected to high compressive stresses, where the actual compressive stresses at failure are smaller than the ultimate compressive stresses that the material is capable of withstanding

Buckling is also described as failure due to elastic instability

BUCKLING TYPES

Stable or Gentle Buckling is a buckling in which the displacements increase in a controlled fashion as loads are increased, ie. the structure's ability to sustain loads is maintained

Unstable or violent Buckling is a buckling in which the displacements increase instantaneously, the load carrying capacity nose- dives and the structure collapses catastrophically

STRUCTURAL MEMBERS• Column

– A structural member which transmits the load of the structure above it through compression to other members

• Strut– A structural member designed to resist longitudinal

compression• Plate / Panel

– A structural member whose third dimension is small compared to the other two dimensions

• Shell– A thin shell is defined as a shell with a thickness which is

relatively small compared to its other dimensions and in which deformations are not large compared to thickness. A primary difference between a shell structure and a plate structure is that, in the unstressed state, the shell structure has curvature as opposed to plates structures which are flat

COLUMN BUCKLING• Column buckling

– Buckling is defined as an instance of lateral bending or bowing of the column shape due to a compressive load on a column.

2

2

Lkcr

EIP

π=

S.No Type k

1 Pinned 1

2 Fixed-fixed 4

3 cantilever 1/4

COLUMN BUCKLING (Contd..)• Column buckling can be classified as

– Primary instability• Cross sections are translated or rotated but not distorted

– Secondary instability• Cross sections are distorted but not translated or rotated

LOAD VS DEFLECTION

imperfe

ct stru

ctures

Perfect structures

Lateral Deflection

Lo

ad P

Pcr

COLUMN CLASSIFICATION

Slenderness Ratio (SR) =Leff /ρ

Type Material ColumnTheoryStructural

SteelAluminium alloy (6000)

Aluminiumalloy (2000)

Wood

Short SR<40 SR<9.5 SR<12 SR<11 Johnson

Intermediate 40<SR<150 9.5<SR<66 12<SR<55 11<SR<30 Inelastic

Long SR>150 SR>66 SR>55 SR>30 Euler

Long ShortIntermediate

BUCKLING SHAPES

BEAM BENDING EQUATION

(1) ... REI

M 1=

(2) ...

1

d

2

2/32

21

+=

dxdy

dxy

R

From Flexure formula

Radius of curvature

(3) ...d

2

2

1

dx

y

R=

Ignoring higher order terms

From (1) & (3)(4) M...

dEI

2

=2dx

y

EULER BUCKLING FORMULA

Spring

A

B

P

Fully Aligned

A

B

P

L

A

B

P

P

P

v

x P

M= -Pv

EI

M

dx

vd =2

2

02

2

=+ vEI

P

dx

vd

Beam deflection equation

)cos()sin( 21 xCxCv λλ +=

Applying the column load

Boundary conditions)sin(00

000

1

2

LCLxatv

Cxatv

λ=⇒===⇒==

L

nπλ =

Solution to the above equation

Actual solution )sin(1 L

xnCv π=

A

B

n=1 n=2

0)sin()sin( 112 =+− xC

EI

PxC λλλSubstituting in equation

0)sin()( 12 =− xC

EI

P λλ

2λEIP =

2

22

L

EInPcr

π=

PLASTICITY REDUCTION FACTOR

• If the elastic buckling stress is more than the yield stress, plasticity reduction factor has to be applied. For columns, plasticity reduction factor is applied through tangent modulus

INELASTIC BUCKLING• For a column with intermediate length, buckling occurs after the

stress in the column exceeds the proportional limit of the column material and before the stress reaches the ultimate strength. This kind of situation is called inelastic buckling

2

2

ρ

π=eff

tcr

L

AEPEuler Engesser

Reduced modulus theory

( ) 2

2

eff

rcr

L

IEP

π= ( ) 2

4

t

tr

EE

EEE

+=

REDUCED MODULUS FORMULA

dAdAd

v

d

x ∫∫ σ=σ21

00

As load remains constant ………… (1)

Moment equilibrium

11

1 ydx

σ=σ 2

2

2 ydv

σ=σ

………… (2)( ) ( ) PvdAeydAeyd

v

d

x −=−σ++σ ∫∫ 2

00

1

21

Change in slope

2

2

1

12

2

dEEddz

vd

t

σ=

σ=

………… (3)

………… (4)

REDUCED MODULUS FORMULA0

21

0

22

2

0

12

2

=− ∫∫ dAydz

vdEdAy

dz

vdE

d

t

d

………… (5)Equation (1) becomes

Equation (2) becomes

PvdAyEdAyEdz

vdedAyEdAyE

dz

vdd

t

dd

t

d

−=

++

+ ∫∫∫∫

2121

0

2

0

12

2

0

22

0

212

2

( ) PvIEEIdz

vdt −=+ 212

2

………… (6)

Equation (5) in (6) gives

Equation (7) can be rewritten as

………… (7)

………… (8)

Solving (8) we get

………… (9)

02

2

=+ Pvdz

vdIEr

2

2

e

rcr

l

IEP

π=

EULER ENGESSER FORMULA………… (10)

Equation (9) gives

But ………… (11)

If there is no strain reversal, then entire region gets compressive stress

Now Equation (10) becomes

………… (12)

………… (13)

2

2

e

rcr

l

IEP

π=

21 IEEIIE tr +=

IEIEIEIE tttr =+= 21

2

2

e

tcr

l

IEP

π=

EULER ENGESSER CURVEEuler Engesser curve is divided in to three regions

1) Block compression2) Short column range ( Plasticity effects)3) Long column range (Euler buckling)

TANGENT & SECANT MODULUS

Tangent modulus

Secant modulus

+

=−1

707

31

n

.

t

F

Fn

EE

+=

850

70

7171

.

.

e

F

F

)/(logn

σσ

σ+

=n

cy

s

E.

EE

00201

σσ

σ

+

=−1

00201

n

cycy

t

En.

EE

COLUMNS ON ELASTIC FOUNDATION

The stiffness of elastic foundation increases the buckling load andreduces the buckling length

EI

L)m(m

4

422 1

πβ=+

Critical column load

Pcr

π

βπ=EIm

L

L

EI42

4

2

2

a

µ=β

CRIPPLING•Crippling is defined as the post-buckling failure of a axial section that is comprised of plate elements joined together at their boundaries•All the members subjected to axial load are to be checked for crippling•When local buckling takes around 0.7 - 0.8 Fcy , The crippling stress will be same as buckling stress

Web and Flange elements

CRIPPLING -ASSUMPTIONS

•Material is isotropic

•Material is ductile

•b/t ratio is less than 3.0

•Transverse shear is ignored

• Webs are assumed to have constant thickness

Crippling stress

STRESS DISTRIBUTION•As the buckling takes place, the increasing load is transferred to the corners. •Stress build up at the corner after the buckling is not well understood•Boundary restraint between flange and plate element is unknown

FLANGE CRIPPLINGThe crippling stress is defined by dividing the failure load at which the flange collapses by the area of the flange.

Pre-Buckled Post-Buckled

Stress distribution in a flange

WEB CRIPPLINGThe crippling of a web is similar to flange. The stress is uniform before buckling and increases near the edge after buckling

Stress distribution in a Post-Buckled web

INPLANE WARPING

Unrestrained Restrained

The post-buckled stress distribution in a flange or web is affected by the presence of restraints for in-plane lateral deflections at the un-loaded edges

WEB CRIPPLING STRESS DISTRIBUTION

Straight unloaded edgesUnloaded edges free to warp in the plane of plate

POST BUCKLING OF PLATESThick plates crippling will take place in plastic range and for thin plates in elastic range

Stress distribution

PREDICTION OF CRIPPLING STRESS

1 Angle method (Needham method) • Member is divided into number of angles• Crippling strength is obtained by summation of

individual crippling strength

( ) ( ) 7501 .

e

n

cs

tb

c

E

F

′=

cyF

free) edge no(for 0.366

free) edge one(for 0.342

free) edges two(for 0.316

===

e

e

e

c

c

c

section angle anfor 2

bab

+=′

AFP cscs =Crippling load

For other sections ∑∑=

anglesofArea

anglesofloadscripplingFcs

PREDICTION OF CRIPPLING STRESS (Contd…)

4002

12

670

.

cy

cs

F

E

A

gt.

F

=

cyF

8502

12

560

.

cy

cs

F

E

A

gt.

F

=

cyF

For angles, V-groove plates, stiffened panels with distorted unloaded edges

For T, H, cruciform, plates with undistorted edges

For Z,J and channel750

31

2

23

.

cy

cs

F

E

A

t.

F

=

cyF

2 Gerard method

CORRECTION FOR CLADDING

f=Cladding thickness / Total thickness

=0.1 for 2024-T3

= 0.08 for 7075-T3

( )f

fcr

cl

31

31

+

σσ

+

=η

S.No Section Value

1 Angles 0.7Fcy

2 V-groove plates Fcy

3 Multi-corner sections

0.8Fcy

4 Stiffened panels Fcy

5 H, T, Cruciform 0.8Fcy

6 Z,J, Channel 0.9Fcy

Maximum crippling stress

RESTRAINT BY LIPS & BULB

• Compressive buckling coefficient of a element can be increased the presence of lip or bulb

• Compressive buckling coefficient for a

– Plate element is 4

– Flange is 0.43

• To provide a simple support the lip and bulb dimensions should satisfy

53

≥−tt f

L

f

L

b

A

b

I2.73

f

f

L

L

t

b

t

b 3280.≤

i.e

For lip

For Bulb

=

−

−

f

f

f

min

f

min

f

min

t

b.

t

D.

t

D.

t

D447374061

234

PREDICTION OF CRIPPLING STRESS (BOEING)

• Divide the section into segments. • For each segment, determine the ratio of the width to

the thickness (b/t) • For each segment, determine the boundary

conditions (1EF or NEF)• For each segment, determine the crippling stress and

crippling load based on the method of analysis appropriate for the b/t region

• Add the contribution of the crippling load from each segment to obtain the total crippling load

• The crippling stress is obtained by dividing the crippling load by the calculated area

SECTION DETAILS

Formed section

Extruded / Machined

2

2

tbb

If

tbb

If

other

+=

>

+=

>

tt

tt

other

other

R.bb 570+=

SECTION DETAILS

Stepped section

Tapered section

210 tt

t+

=

SECTION DETAILS

Thin / thick section

Adjoining flange

=

thin

thickthincor,thick b

btt 3

αθ

= R

tb

btR

ww

ff

3

3

20

1

α

+α=α

2

2

2 sinb

tcosR

f

f

SECTION DETAILSBulb section

=

−

−

f

f

f

min

f

min

f

min

t

b.

t

D.

t

D.

t

D447374061

234

The bulb will fall into one of the following three categories:

1) Case 1: Diameter large enough (D > Dmin )The flange may be considered a web (NEF) for

the purposes of the crippling analysis2) Case 2: Intermediate diameter (2 tf < D < Dmin )The flange may be considered as a web (NEF), but

the crippling stress for this segment (including the bulb) will be adjusted to 70% of the stress

calculated assuming the no-edge-free condition3) Case 3: Diameter too small (D < 2 tf )Consider the flange as one edge-free. The area of the

bulb should be added to the flange area, and bf

should be measured to the tip of the bulb

SHEET EFFECTIVE WIDTH• Aircraft structures consists of

sheet and stringers together

• Sheet and stringer deform together. Hence, the effective sheet width has to be taken into account in calculating the crippling stress

• Ignoring the sheet will be over conservative design

( )2

2

2

2

112

=

ν−π=

b

t

b

tEkc

3.6E

F cr

SHEET EFFECTIVE WIDTH (Contd..)Von-Karman Sechler method

=

cyF

Et.91w

NASA Structures manual

For skin & stiffener different material

For skin & stiffener same material

stiffstiff A

Pf =

free edge nofor

free edge onefor

71

31

.

.K =

( )( )

=

stiffstiffs

skins

fE

EKt

1e2w

=

stiff

s

f

EKte2w

CURVED PANEL

curvedflatcr PPP +=

( ) ( ) ( ) ( ) securvedcesststiffccr twbFwtAFP 24 −++=

REGIONS OF CRIPPLING CURVE

For calculating the crippling stress, the crippling curve is divided into three regions based on the value of b/t:1) Stress cutoff2) Plastic plate buckling3) Empirical crippling curve

JOHNSON-EULER FORMULA

COLUMN BUCKLING CURVE

• Select a material with higher E and yield stress

• Reduce b/t ratio• Add appropriate lip or bulb to change

the edge conditions

INCREASING CRIPPLING STRESS

PLATE STRUCTURES

55.0 <<tδ

5>t

δ

< 5.0

tδ

DEFINITION : PLATE IS A STRUCTURAL MEMBER WHOSE THIRD DIMENSION IS COMPARATIVELY SMALLER TO THE OTHER TWO DIMENSIONS AND SUBJECTED TO NORMAL LOAD / INPLANE LOAD

CLASSIFICATION OF PLATES:

a) Thick plate : Load resisted by bending

b) Thin plate : Load is resisted by bending and inplane action

c) Membrane : Load is resisted by tension

PLATE THEORIES• KIRCHHOFF PLATE THEORY

– Shear deformation is ignored

• MINDLIN PLATE THEORY– Shear deformation is considered

y

wzv

x

wzu

yxww

∂∂−=

∂∂=

= ),(

∂

∂−∂

∂=

∂∂−=

∂∂

=

xyz

yz

xz

xy

xy

xy

y

x

θθγ

θε

θε

∂

∂−∂

∂=

∂∂−=∂

∂−=

xyz

yz

xz

xy

xy

xy

y

x

θθγ

θε

θε

y

wzv

xw

zu

yxww

∂∂−=∂∂−=

= ),(

yxz

xyz

x

wy

w

θγ

θγ

−∂∂=

−∂∂=

0== xzyz γγ

PLATE BENDING

y

x

z

dz

dy

dx

Mxy

Mx

Mx

Qxz

Qxz

Mxy

Qyz

QyzMyx

Myx

My

My

PLATE BENDING (Contd….)• Plate bending equation is derived based

on the following

a) Strain-displacement relation

b) Stress-strain relation

+−

−=

xy

y

x

xy

y

x

E τσσ

νν

ν

γεε

)1(20001011

………….. (1)

………….. (2)

∂∂∂∂

∂∂

∂

−=

=

yx

wy

wx

w

zzxy

y

x

xy

y

x

0

2

2

0

2

2

0

2

2θθθ

γεε

PLATE BENDING (Contd….)c) Moment & Force resultants

d) Equilibrium equations

dzzMMM

xy

y

xt

txy

y

x

∫=

− τσσ2/

2/

dzQQ t

t yz

xz

yz

xz ∫

=

−

2/

2/ττ

y

M

x

MQ yxx

xz ∂∂

+∂

∂=

y

M

x

MQ yxy

yz ∂∂

+∂

∂= ………….. (5)

………….. (4)

………….. (3)

………….. (6)

………….. (7)p

y

Q

x

Q yzxz −=∂

∂+

∂∂

PLATE BENDING (Contd….)

( )

−−=

xy

y

x

xy

y

x E

γεε

νν

ν

ντσσ

2)1(

00

0101

1 2

………….. (10)

………….. (9)

………….. (8)

………….. (11)

(5) & (6) in (7) gives

(3) in (8) gives

Rearranging (2), we get

(1) in (10) gives

( )

∂∂∂∂

∂∂

∂

−−−=

yx

wy

wx

w

Ez

xy

y

x

0

2

2

0

2

2

0

2

2)1(00

0101

1 νν

ν

ντσσ

py

M

yx

M

x

M yxyx −=∂

∂+

∂∂∂

+∂

∂2

22

2

2

2

pdzyyxx

zt

t

yxyx −=∫

∂

∂+

∂∂∂

+∂

∂−

2/

2/2

22

2

2

2σσσ

PLATE BENDING (Contd….)

………….. (13)

………….. (12)

(11) in (9) gives

But

( )2

32/

2/

2

2 1121 νν −=∫ −

=−

Etdzz

ED

t

t

pwyyxx

D =

∂∂+

∂∂∂+

∂∂

04

4

22

4

4

4

2

( ) pdzyx

w

y

w

yx

w

yx

w

x

wEzt

t

=∫

∂∂∂+

∂∂+

∂∂∂−+

∂∂∂+

∂∂

−−

2/

2/22

0

4

4

0

4

22

0

4

22

0

4

4

0

4

2

2

121

νννν

py

w

yx

w

x

wdz

Ezt

t

=∫

∂∂+

∂∂∂+

∂∂

−−

2/

2/4

0

4

22

0

4

4

0

4

2

2

21 ν

………….. (14)

(13) in (12) gives

D

pw =∇ 4

PLATE SOLUTION Plate equation is given as

Assume

Dyxp

yw

yxw

xw ),(

24

4

22

4

4

4

=

∂∂+

∂∂∂+

∂∂

………….. (14)

………….. (15)

∑

∑=

∞

=

∞

= 11sinsin),(

nmn

m byn

axm

ayxqππ

∑

∑=

∂∂ ∞

=

∞

= 1

4

14

4

sinsinn

mnm b

yn

a

xm

a

mA

x

w πππ

∑

∑=

∂∂ ∞

=

∞

= 1

4

14

4

sinsinn

mnm b

yna

xmb

nA

yw πππ

………….. (16)

………….. (17)

∑

∑=

∂∂∂ ∞

=

∞

= 1

22

122

4

sinsinn

mnm b

yn

a

xm

b

n

a

mA

yx

w ππππ

………….. (18)

………….. (19)

∑

∑=

∞

=

∞

= 11sinsin),(

nmn

m b

yn

a

xmAyxw

ππ

PLATE SOLUTION

………….. (20)

………….. (21)

………….. (22)

………….. (23)

………….. (24)

0sinsin21

4224

1=∑

−

+

+

∑

∞

=

∞

= n

mnmn

m byn

axm

D

a

bn

bn

am

am

Aππππππ

024224

=−

+

+

D

a

b

n

b

n

a

m

a

mA mn

mn

ππππ

(17), (18) & (19) in (14) gives

( )2

2

2

2

24

bn

amD

aA mn

mn

+=

π

( )∑

+∑=

∞

=

∞

= 12

2

2

2

214

sinsin1

),(n

mn

m b

yn

a

xm

bn

am

a

Dyxw

πππ

From (20), we get

)),(( 0qyxq =For uniformly distributed load

For concentrated load

∑

+

∑=∞

=

∞

= ,...5,3,1222,...5,3,1

6

0 2sin

2sin

16),(

nm

b

n

a

mmn

nm

D

qyxw

ππ

π

………….. (25)∑

+

∑=∞

=

∞

= 12221

4

sinsinsinsin4

),(n

yx

yxyx

myx

L

n

L

m

Lyn

Lxm

Lbn

Lam

LDL

Pyxw

ππππ

π

PLATE SOLUTION

∑

+

+

∑=∞

=

∞

= byn

axm

b

n

a

mmn

bn

am

qM

nmx

ππν

πsinsin

16,...5,3,1

222

22

,...5,3,14

0

∑

+

+

∑=∞

=

∞

= byn

axm

bn

am

mn

bn

am

qM

nmy

ππν

πsinsin

16,...5,3,1

222

22

,...5,3,14

0

Moment

Stress

3

12

t

zM xx =σ

3

12

t

zM y

y =σ

………….. (26)

………….. (27)

………….. (28)

………….. (29)

BUCKLING OF PLATES

∑

∑=

∞

=

∞

= 11sinsin),(

nmn

m byn

axm

Ayxwππ

Displacement

Potential energy

NxNx

a

b

( )∫

∂∂−

∂∂∂−

∂∂

∂∂−−

∂∂+

∂∂

∫=+b

x

a

dxdyx

wN

yx

w

y

w

x

w

y

w

x

wDVU

0

222

2

2

2

22

2

2

2

2

0

122

1 ν

∑∑∑ −

+

∑=+

∞

=

∞

=

∞

=

∞

= 1

22

1

2

1

222

2

1

4

88 nmn

mx

nmn

mAmN

b

b

n

a

mA

abDVU

ππ

………….. (1)

………….. (2)

………….. (3)

BUCKLING OF PLATES (Contd..)

mnxmn

mn

AmNa

b

b

n

a

mA

abD

A

VU 22

2224

44

)( ππ −

+

=

∂+∂

Differentiating

222

2

22

+

=

b

n

a

m

m

DaNCr

π

2

2

b

DkNCr

π= where2

+

=

mb

a

a

mbk

2

2

=

b

tEkCr πσ

Critical buckling load

Critical buckling stress

………….. (4)

………….. (5)

………….. (6)

………….. (7)

END FIXITY COEFFICIENT

BUCKLING OF PLATES(from column formula)

• Euler column formula

• Stress-strain relation

• Assuming that the plate has no curvature in y direction gives

• Flat plate has a smaller elongation compared to a column

• For a rectangular plate• (5) in (4) gives

2

2

Lkcr

EIP

π=

EE

EE

xyy

yxx

σν−

σ=ε

σν−

σ=ε

0=ε y

………….. (1)

………….. (2)

( )21 ν−σ

=ε

ν σ=σ

Ex

x

xy

( )( ) 22

2

1 eff

crL

EIP

ν−π=

………….. (3)

tbPbt

I,aL cr=σ== cr and 12

3

………….. (4)

( )2

2

2

112

ν−π=σ

a

tEcr

………….. (5)

………….. (6)

SHEAR BUCKLING OF PLATES

SHEAR BUCKLING OF PLATES• Shear buckling formula

( )2

2

2

112

ν−π=

η b

tEk

Fcs

METHODOLOGY• Calculate a/b from plate dimensions measured between panel

supports • Determine edge restraint fixity3) Select the buckling coefficient curve for the edge condition most nearly

representing the support conditions existing, enter curve with a/b from (1) and obtain "K" (or "k"). If the support condition is believed to be between two conditions represented by curves, obtain "K" for both, calculate average value or interpolate as desired.

4) Determine buckling stress from equation 12. If this stress is in the elastic range, η = 1.0 (skip to step (5))

5) If the stress is in the plastic range, obtain the proper plasticity reduction factor η

6) If the material is Alclad material, calculate the cladding reduction factor

INCREASING THE BUCKLING LOAD OF PANELS

There are three primary effective ways to increase the buckling load of a panel:1) Decrease the "b" dimension of the panel2) Increase the thickness of the panel3) Increase the fixity of the panel supports

( )2

2

2

112

ν−π=

η b

tEk

Fcs

LINEAR BUCKLING ANALYSIS

[ ]{ } [ ]{ }XKXK Gλ=

Solution methodLanczos methodSubspace iterationBackward iteration

[K] – Stiffness matrix[KG] – Geometric Stiffness matrix

{X} – Buckling shapeλ - Buckling load factor

NONLINEAR STATIC ANALYSIS

Solution methodNewton-Raphson Method

[KL] – Linear Stiffness matrix

[KNL ] – Nonlinear Stiffness matrix

{X} – Deflection vector[P} - Load vector

[ ]{ } [ ] P =+ XKK NLL