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8/10/2019 Lecture 6 Post Buckling
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Post-Buckling5.2.4
Post-buckled composite stiffened panel under shear
8/10/2019 Lecture 6 Post Buckling
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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
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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
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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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Derivation of von Karman
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
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Derivation of von Karman
equations (large deflections)
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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Derivation of von Karman
equations (large deflections)
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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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Analysis model versus FE (45-dominated
laminate)
Results from R. Kroese MS Thesis TUDelft 2013
Excellent agreement between FE and Analysis
end displa-
cement
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Analysis model versus FE (Quasi-isotropic
laminate)
Results from R. Kroese MS Thesis TUDelft 2013
Excellent agreement between FE and Analysis
end displa-
cement
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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
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