Post on 15-Mar-2018
transcript
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