Lecture 6 Post Buckling

8/10/2019 Lecture 6 Post Buckling

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Post-Buckling5.2.4

Post-buckled composite stiffened panel under shear


2/32

Buckling vs Post-buckling

in general buckling does not imply failure especially

for plates

PPcr

1

wcenter

beam post-buckling curve is flat and failure strains are reached at low

post-buckling loads


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Buckling vs Post-buckling

post-buckled (stiffened) skins are significantly lighter

but have additional challenges:

susceptible to skin/stiffener separation failure

where only resin holds the structure (unlessfasteners or other means are used to hold it together)

susceptible to the creation and growth of

delamination under fatigue loading especially at high

post-buckling factors

skin/stiffener

separation

skin/stiffene

r

separation

skin/stiffener

separation


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Buckling versus Post-Buckling

If a composite structure is allowed to post-buckle,

the ratio of final failure load to buckling load (post

buckling ratio PB) must be chosen carefully

(especially for shear loads)

Conservative approach (but not very efficient): Do

not allow buckling below limit load (i.e. PB=1.5)

PB>5 great design challenge both under static and

fatigue loads


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Post-buckling mode

frame

frame

stiffeners

skin

What buckles and when?

What BCs are the different components supposed to simulate?


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Post-Buckling Scenarios

Skin buckling as a whole (stiffeners onlyincrease the bending stiffness of the skin)

Skin buckling between the bays (stiffenersact as panel breakers)

Stiffeners buckle as columns or locally(crippling)

Frames do not buckle!


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Post-buckling scenarios

Most efficient: Skin between stiffeners and frames buckles

first

Stiffeners do not buckle and do not move outof plane (depending on cross-section they may

rotate => BC implications)

When required PB value is reached, skin fails

in compression and/or stiffeners fail by crippling

stiffeners impose zero deflection

and slope

stiffeners impose zero deflection

only


8/32

Post-Buckling Analysis

governing equations: Large deflection von Karmanequations

place moment equilibrium eqn into Fz equilibrium eqn:

0222

22

2

2

2

22

2

2

zyxyxyx

yxyx p

y

wp

x

wp

y

wN

yx

wN

x

wN

y

M

yx

M

x

M

px, py, pz distributed loads (force/area)

use the moment-curvature relations to substitute forthe moments

D16=D26=0

Bij=0

yx

wDM

y

wD

x

wDM

ywD

xwDM

xy

y

x

2

66

2

2

222

2

12

2

2

122

2

11

2


9/32

Derivation of von Karman

equations (large deflections) to obtain:

1st von Karman equation: Predominantly bending

zyxyxyx py

wp

x

wp

y

wN

yx

wN

x

wN

y

wD

yx

wDD

x

wD

2

22

2

2

4

4

2222

4

66124

4

11 2)2(2


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Strain Compatibility

y

w

x

w

x

v

y

u

y

w

y

v

x

w

x

u

xyo

yo

xo

2

2

2

1

2

1

2

2

2

22

2

2

3

2

3

2

3

2

32

2

32

2

2

32

2

3

2

2

2

3

2

2

2

32

2

3

2

2

2

2

1

221

y

w

x

w

yx

w

yx

w

x

w

yx

w

y

w

yx

v

yx

u

yx

yx

w

y

w

yx

w

yx

v

yx

w

y

w

xyx

v

x

yx

wxw

yxw

yx

uyxw

xw

yyx

u

y

xyo

yo

xo

...

2

2

2

2

2

yxxy

xyoyoxo


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equations (large deflections)

use (non-linear) strain compatibility

2

2

2

22

22

2

2

2

2

y

w

x

w

yx

w

yxxy

xyoyoxo

non-linear termsxo, yo, xyo mid-planestrains

invert stress-strain eqns to express strains in terms of

Nx, Ny, Nxy

xyoxy

yoxoy

yoxox

AN

AAN

AAN

66

2212

1211

xyxyo

yxyo

yxxo

NA

NAAA

AN

AAA

A

NAAA

ANAAA

A

66

2

122211

11

2

122211

12

2

122211

122

122211

22

1

A16=A26=0


12/32



substitute in strain compatibility to obtain,

2

2

2

22

22

66

2

2

122

2

112

2

122

2

222

122211

11

y

w

x

w

yx

w

yx

N

Ax

NA

x

NA

y

NA

y

NA

AAA

xyxyyx

introduce Airy stress function and potential V that identically

satisfy stress equilibrium:

yx

FN

Vx

FN

Vy

FN

xy

y

x

2

2

2

2

2

potential of distributedloads (=0 in our post-

buckling problem)


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to obtain

2nd von Karman equation: Predominantly stretching

(and non-linear)

2

2

2

22

2

22

4

664

4

1122

4

124

4

222

122211

12

1

y

w

x

w

yx

w

yx

F

Ax

FA

yx

FA

y

FA

AAA


14/32

Post-buckling of a square anisotropic plate

under compression

simply-supported with three immovable edges:

w=0 at x=y=0 and x=y=a

u=0 at x=0

v=0 at y=0 and y=a

u=-C at x=a (constant compressive displacement)

a

a

Px

(appliedforce)

rigid

x

y


15/32

Solve the two von Karman equations

approximately

assume

a

yK

a

xK

x

a

Py

a

PF

a

y

a

xww

yx

2cos

2cos

22

sinsin

0220

22

11

substitute in the 2nd von Karman equation

a

x

aw

a

y

aw

a

x

aK

AAA

A

a

y

aK

AAA

A 2cos

2

2cos

2

2cos

162cos

164

42

114

42

114

4

202

122211

11

4

4

022

122211

22

w11, K02, K20 unknown

coefficients

32

32

2

11

11

2

12221120

2

11

22

2

12221102

w

A

AAAK

w

A

AAAK

matching coefficients

of cos(2y/a) and

cos(2x/a)

and the 2nd von

Karman eqn is

identically satisfied


16/32

Solution (contd)

Py (as a function of Px) and deflection C are determinedby integrating stress-strain eqns and using average BCs

on u and v:

2

2

1

x

w

x

uxo

yxxo NAAA

AN

AAA

A2

122211

12

2

122211

22

also

non-linear strain-displacement eq

inverted stress-strain eq

Airy stress function F

dxdyx

wdxdy

x

F

AAA

Adxdy

y

F

AAA

Adxdy

x

u a aa aa aa a 2

0 00 0

2

2

2

122211

12

0 0

2

2

2

122211

22

0 0

2

1

yx

FN

Vx

FN

Vy

FN

xy

y

x

2

2

2

2

2


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Solution (contd)

=-C =0

4

22

11

w

from which:

dxdyay

a

x

awa

P

AAA

aA

a

P

AAA

aA

yuyaua

yx 2

112122211

2

12

2122211

2

22

)sincos(2

1

),0(),(

aw

a

P

AAA

aA

a

P

AAA

aAC

yx

8

22

112

122211

12

2

122211

22

similarly, from

2

2

1

y

w

y

vyo

= Poissons ratio for

isotropic

=0 for in-plane problems

11

2

122211

22

11

11

12

8 A

AAA

aw

A

APP xy


18/32

Solution (contd)

substitute now in the 1st von Karman equation

2

2

2

222

2

2

2

2

4

4

2222

4

66124

4

11 2)2(2y

w

x

F

yx

w

yx

F

x

w

y

F

y

wD

yx

wDD

x

wD

use the following:

a

yw

A

AAA

aa

P

y

F

a

xw

A

AAA

aa

P

x

F

a

y

a

x

aw

y

w

x

w

a

y

a

x

aw

y

w

yx

w

x

w

x

y

2cos

32

2

2cos

32

2

sinsin

sinsin

2

11

22

2

122211

2

2

2

211

11

2122211

2

2

2

2

112

2

2

2

4

114

4

22

4

4

4


19/32

Solution (contd)

to substitute, and match coefficients of

sin(x/a)sin(y/a) to obtain the governing eq for w11

which for w110 leads to

=Pcr, buckling load (units of force), exact for square plate

2nd von Karman eq is satisfied

approximately!

11 22 11 12 66 22

11 2 2

11 12 66 2211 22 12 11 22

12

11

16 2 21

2 23

1

xA A D D D D PwD D D DA A A A A

a A

A

22 211 22 12 11 22 3 12

11 11 12 66 22 11

11 22 11

32( 2 ) 1 0

16 x

A A A A A Aw D D D D P w

a A A a A


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Solution (end)

For out-of-plane deflections to be possible, must have

1cr

x

P

P

the applied load Px must exceed the plate buckling load


21/32

Results-Implications

use as example,

(45)/(0/90)/(45) square plate of side 25.4 cm

Material is plain weave fabric with properties:

Ex=Ey=68.94 GPa

xy=0.05

Gxy=5.17 GPa

ply thickness = 0.19 mm


22/32

Load versus center deflection

0

1

2

3

4

5

6

7

8

9

10

0 1 2 3 4 5

center defl/plate thickness (w11/h)

P/Pcr

a

a


23/32

In-plane compression load

0

0.1

0.2

0.3

0.40.5

0.6

0.7

0.8

0.91

0 1 2 3 4 5 6 7 8

Nx/(Pcr/a)

y

/

=1P

Pcr1.2 2 5

the center of the plate sees very little load!


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Implication

after buckling approximate in-plane load

beff

a

a

peak Nx value

there is an effective width beffat the edges of the

panel over which the load is concentrated


25/32

Determination of effective width

total applied force must equal the force created bythe load applied over the effective width:

effxx bNdyN max2

a

y

a

w

A

AAA

a

P

y

FN xx

2cos

2

32

22

11

22

2

122211

2

2

xx PdyN

22

11

22

2

122211max

2

32

a

w

A

AAA

a

PN xx


26/32

Solving for beff

2211

11

11

12

311212

1

AA

A

P

P

A

Aab

x

cr

eff

Note that for quasi-isotropic layup,

x

cr

eff

P

Pab

13.12

1

beff=0.303a for Px>>Pcrand QI layup

3.0

4

1

3

12

2211

11

12

11

12

AA

A

A

A


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Effective width beff

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0.5

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

Pcr/Px

beff/a

beff

a

a

A12

A11=0.05

0.30.7

weak dependence on A12/A11


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Significance for design

beffprovides an idea of where failure is expected under

compression and where reinforcement may be needed

For (conservative) failure predictions obtain the

maximum compressive stress

h

Nx maxmax (h is plate thickness)

and compare to allowable ultimate compression stress

ult max for no failure

plus one more use in crippling of stiffener flanges


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How good is the analysis compared to

reality?

For a meaningful comparison, two conditions must be

met:

Boundary conditions in the test case must be the same as in

the model

Sufficient number of terms must be included in the solution (asopposed to only one term used here for simplicity)

As it is hard to find test results with exactly these

boundary conditions, FE results are used to validate the

analysis Two different laminates: (a) 45 degree dominated

(b) Quasi-isotropic


30/32

Analysis model versus FE (45-dominated

laminate)

Results from R. Kroese MS Thesis TUDelft 2013

Excellent agreement between FE and Analysis

end displa-

cement


31/32

Analysis model versus FE (Quasi-isotropic

laminate)

Results from R. Kroese MS Thesis TUDelft 2013

Excellent agreement between FE and Analysis

end displa-

cement


32/32

Boundary conditions discussion

If the stiffeners in a stiffened panel have very high EI andvery low GJ, and one end is attached to a frame with

also very high EI and low GJ, this analytical model would

be quite accurate

If not, the model may be conservative or unconservative Must also use enough terms in the series and account

for rectangular as opposed to square panels

stiffener

frame

skin betweentiff

Documents

Lecture 6 Post Buckling