Aircraft Design Day8

DAY8

BUCKLING OF AIRCRAFT STRUCTURES

BUCKLING Buckling 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.

π 2 EI Pcr = k L2

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

Load P

Pcr

Perfect structures

p im

ct u tr s ct e f er

e ur

s

Lateral Deflection

COLUMN CLASSIFICATION Type

Short Intermediate Long

Material Structural Steel

Aluminium alloy (6000)

Aluminium alloy (2000)

Wood

SR<40

SR<9.5

SR<12

SR<11

40<SR<150

9.5<SR<66

12<SR<55

SR>150

SR>66

SR>55

Slenderness Ratio (SR) =Leff /ρ

Long Intermediate

Short

Column Theory Johnson

11<SR<30 Inelastic SR>30

Euler

BUCKLING SHAPES

BEAM BENDING EQUATION From Flexure formula

M 1 = ... (1) EI R

Radius of curvature

d2 y

2 1 dx = ... (2) 2 3/ 2 R   dy    1 +  dx      

Ignoring higher order terms 1 d2 y = 2 ... (3) R dx

From (1) & (3)

d2 y EI 2 = M... (4) dx

EULER BUCKLING FORMULA d 2v

M = dx 2 EI

Beam deflection equation

d 2v

Applying the column load

dx 2

Actual solution Substituting in equation

P v=0 EI

v = 0 at

x = 0 ⇒ 0 = C2

v = 0 at

x = L ⇒ 0 = C1 sin(λL)

x v = C1 sin(nπ ) L

P − λ2C1 sin(λx) + C1 sin(λx ) = 0 EI

(

P − λ2 )C1 sin(λx) = 0 EI

P

B

B Spring

L

A

v = C1 sin(λ x) + C 2 cos(λ x)

Solution to the above equation Boundary conditions

+

P

λ=

A

Fully Aligned

nπ L

P

P

B

v

M= -Pv

x

A

P P

B

P = EIλ 2

Pcr =

n 2π 2 EI L2

A n=1

n=2

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

Euler Engesser

Pcr =

Reduced modulus theory Pcr =

π2 Er I

(L ) eff

2

Er =

π2 Et A  Leff    ρ  

(

2

4 EE t E + Et

)

2

REDUCED MODULUS FORMULA

d1

As load remains constant

∫σ 0

Moment equilibrium

d1

∫σ ( y x

0

1

x

dA = ∫ σ v dA

………… (1)

0

d2

+ e )dA + ∫ σ v ( y 2 − e ) dA = − Pv

σ1 σx = y1 d1 Change in slope

d2

………… (2)

0

σv =

σ2 y2 d2

σ1 σ2 d 2v = = dz 2 Ed 1 E t d 2

………… (3) ………… (4)

REDUCED MODULUS FORMULA d 2v E 2 dz

Equation (1) becomes

d1

d 2v ∫0 y1dA − E t dz 2

d2

∫ y dA = 0 2

0

………… (5)

Equation (2) becomes d d2 d1 d2 2    d 2 v  1 2 d v 2    = − Pv E y dA + E y dA + e E y dA + E y dA 1 t ∫ 2 1 t ∫ 2 2  2  ∫ ∫   dz  0 dz  0 0 0  

………… (6)

Equation (5) in (6) gives d 2v ( EI1 + E t I 2 ) = − Pv dz 2

………… (7)

Equation (7) can be rewritten as

d 2v E r I 2 + Pv = 0 dz

………… (8)

Solving (8) we get

π2 Er I Pcr = l e2

………… (9)

EULER ENGESSER FORMULA Equation (9) gives

But

π2 Er I Pcr = l e2

E r I = EI1 + E t I 2

………… (10) ………… (11)

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

E r I = E t I1 + E t I 2 = E t I

………… (12)

Now Equation (10) becomes

π2 Et I Pcr = l e2

………… (13)

EULER ENGESSER CURVE Euler Engesser curve is divided in to three regions 1) Block compression 2) Short column range ( Plasticity effects) 3) Long column range (Euler buckling)

TANGENT & SECANT MODULUS Tangent modulus

Et =

Et =

E  3  F  1 + n    7  F0.7  

n −1

   

n = 1+

E   0.002 En  σ  n−1  1 +       σ  σ   cy   cy   

Secant modulus

Es =

E n     0 . 002 E σ  1 +        σ  σ    cy   

log e ( 17 / 7 )  F0.7     F0.85 

COLUMNS ON ELASTIC FOUNDATION

The stiffness of elastic foundation increases the buckling load and reduces the buckling length Critical column load

π 2 EI Pcr = 2 L

 βL4   2 4   m π EI 

4 β L m 2 ( m + 1 )2 = 4 π EI

µ β= a

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 CRIPPLING The 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 CRIPPLING The 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 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

Unrestrained

Restrained

WEB CRIPPLING STRESS DISTRIBUTION

Straight unloaded edges

Unloaded edges free to warp in the plane of plate

POST BUCKLING OF PLATES Thick 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

Fcs

(F E )

1

cy

= n

ce

( t) b′

0.75

c e = 0.316

(for two edges free)

c e = 0.342

(for one edge free)

c e = 0.366

(for no edge free)

b′ =

a+b 2

Crippling load

For other sections

for an angle section

Pcs = Fcs A Fcs =

∑ crippling loads of angles ∑ Area of angles

PREDICTION OF CRIPPLING STRESS (Contd…) 2

Gerard method

For angles, V-groove plates, stiffened panels with distorted unloaded edges 0.85

 2  Fcs gt = 0.56   A Fcy 

 E    Fcy

   

1

   

2

For T, H, cruciform, plates with undistorted edges

 2  Fcs gt = 0.67   A Fcy 

 E    Fcy

   

1

2

   

0.40

For Z,J and channel

 2  t  E Fcs  = 3.2    A  Fcy Fcy  

   

1

3   

0.75

CORRECTION FOR CLADDING Maximum crippling stress

  σ cl    f  1 + 3   σ cr   η= (1 + 3 f ) f=Cladding thickness / Total thickness =0.1 for 2024-T3 = 0.08 for 7075-T3

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

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

IL

i.e

 Dmin   t  f

4

AL 2.73 − ≥5 3 bf t bf t

   − 1.6 D min   t   f

bL b ≤ 0.328 f tL tf 3

   − 0.374 Dmin   t   f

2

For lip

 b  = 7.44 f  t   f

   

For Bulb

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 If t > t other  t other b=b + 2 If t other > t  t b=b + 2

Extruded / Machined

Formed section

 b = b + 0.57 R

SECTION DETAILS

t 0 + t1 t= 2 Tapered section Stepped section

SECTION DETAILS

t thick ,cor

 bthick = 3t thin   bthin

  

Thin / thick section

3 1  t f b f Rθ = 20  b w t w3

  Rα  

tf 2 Rα = cos α +  b  f

2

  sin 2 α  

Adjoining flange

SECTION DETAILS  Dmin   t  f

4

Bulb section

   − 1.6 D min   t   f

3

   − 0.374 D min   t   f

2

 b  = 7.44 f  t   f

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 analysis 2) 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 condition 3) 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

kc π 2 E  t  Fcr = 2   12 1 − ν  b 

(

t = 3.6E  b

)

2

2

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

 E   w = 1.9t  F   cy  NASA Structures manual 2w e = Kt

( E s ) skin ( E s ) stiff

 1     f   stiff 

 Es   2w e = Kt   f   stiff 

f stiff =

P Astiff

For skin & stiffener different material

For skin & stiffener same material

K = 1.3 1.7

for one edge free for no edge free

CURVED PANEL

Pcr = P flat + Pcurved

Pcr = ( Fc ) stiff ( Ast + 4t s w e ) + ( Fc ) curved ( b − 2 w e ) t s

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 cutoff 2) Plastic plate buckling 3) Empirical crippling curve

JOHNSON-EULER FORMULA

COLUMN BUCKLING CURVE

INCREASING CRIPPLING STRESS • Select a material with higher E and yield stress • Reduce b/t ratio • Add appropriate lip or bulb to change the edge conditions

PLATE STRUCTURES 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 :  δ < 0.5  Load resisted by bending t  δ 0 . 5 < < 5 Load is resisted by bending b) Thin plate : and inplane actiont c) Membrane : δ > 5 t

Load is resisted by tension

PLATE THEORIES • KIRCHHOFF PLATE THEORY – Shear deformation is ignored w = w( x, y ) ∂w u = −z ∂x ∂w v = −z ∂y

ε x = −z

∂θ y

∂x ∂θ ε y = −z x ∂y  ∂θ ∂θ  γ xy = z  y − x  ∂x   ∂y

γ yz = γ xz = 0

• MINDLIN PLATE THEORY – Shear deformation is considered ∂θ w = w( x, y ) εx = z y ∂x ∂w ∂θ x u=z ε y = −z ∂x ∂y ∂w  ∂θ y ∂θ x  v = −z  γ xy = z  − ∂y ∂y ∂x 



∂w −θx ∂y ∂w γ xz = −θ y ∂x

γ yz =

PLATE BENDING Qyz

Myx

My Qxz Mxy

Mx dy

Mxy

z Mx

y

Qxz

dx dz

x Myx

My Qyz

PLATE BENDING (Contd….) •

Plate bending equation is derived based on the following a) Strain-displacement relation  ∂ 2 w0   2   ∂2x  εx   θx   ε y  = z  θ y  = − z  ∂ w20   ∂y  γ  2θ   xy   xy   ∂ 2 w0    ∂ x ∂ y  

………….. (1)

b) Stress-strain relation εx  1 1  ε y  = − ν γ  E  0  xy 

−ν 1 0

0  σ x  0  σ y  2(1 + ν ) τ xy 

………….. (2)

PLATE BENDING (Contd….) c) Moment & Force resultants  M x  t / 2 σ x   M y  = ∫ z σ y  dz  M  −t / 2 τ   xy   xy  Qxz  = t / 2 τ xz dz Q yz  −t∫/ 2 τ yz 

………….. (3)

………….. (4)

d) Equilibrium equations ∂M xy ∂M y Q yz = + ∂x ∂y ∂M x ∂M yx Qxz = + ∂x ∂y

∂Qxz ∂Q yz + = −p ∂x ∂y

………….. (5) ………….. (6)

………….. (7)

PLATE BENDING (Contd….) (5) & (6) in (7) gives

∂ 2 M xy ∂ 2 M y ∂2M x +2 + = −p 2 2 ∂x ∂x∂y ∂y

………….. (8)

(3) in (8) gives

∂ 2σ xy ∂ 2σ y   ∂ 2σ x dz = − p + ∫ z  2 + 2 2  −t / 2 ∂x∂y ∂y   ∂x t/2

………….. (9)

Rearranging (2), we get

 σ x  1 ν E σ y  = ν 1 2 τ  (1 − ν )   xy  0 0

 0  ε x  0  ε y  (1 − ν )  γ xy     2 

 ∂ 2 w0   2   ∂2x  σ x  1 ν 0   ∂ w σ y  = − Ez 2 ν 1 0   20  ( 1 − ν ) 0 0 (1 − ν )  ∂y  τ  xy    ∂ 2 w0     ∂x∂y 

………….. (10)

(1) in (10) gives

………….. (11)

PLATE BENDING (Contd….) (11) in (9) gives

 ∂ 4 w0 ∂ 4 w0  ∂ 4 w0 ∂ 4 w0   Ez 2   ∂ 4 w0   4 + ν 2 2  + 2(1 − ν ) 2 2 +  4 + ν 2 2  dz = p ∫ 2  −t / 2 1 − ν ∂x ∂y  ∂x ∂y  ∂y ∂x ∂y     ∂x t/2

 ∂ 4 w0 ∂ 4 w0 ∂ 4 w0  Ez 2 dz 4 + 2 2 2 + = p ∫ 2 4  −t / 2 1 − ν ∂x ∂y ∂y   ∂x t/2

But

3 E Et 2 D= ∫ z dz = 2 −t / 2 1 − ν 12(1 − ν 2 )

………….. (12)

t/2

………….. (13)

(13) in (12) gives

 ∂4 ∂4 ∂4  D 4 + 2 2 2 + 4  w0 = p ∂x ∂y ∂y   ∂x

p ∇ w= D 4

………….. (14)

PLATE SOLUTION Plate equation is given as

 ∂4w ∂4w ∂ 4 w  p ( x, y )  4 + 2 2 2 + 4  = ∂x ∂y ∂y  D  ∂x

………….. (14)

Assume

 mπx   nπy  w( x, y ) = ∑ ∑ Amn sin   sin   m =1 n =1  a   b  ∞ ∞  mπx   nπy  q ( x, y ) = ∑ ∑ amn sin   sin   m =1 n =1  a   b  ∞

∞

………….. (15) ………….. (16)

4

∂4w ∞ ∞  mπ   mπx   nπy  = A sin ∑ ∑ mn     sin   4 m =1 n =1 ∂x a a b       4

∂ w  nπ   mπx   nπy  = A sin ∑ ∑    sin  mn  4 m =1 n =1 ∂y b a b       4

∞

∞

………….. (17)

2

………….. (18)

2

∞ ∞ ∂4w  mπ   nπ   mπx   nπy  = ∑ ∑ Amn     sin   sin   2 2 m = 1 n = 1 ∂x ∂y  a   b   a   b 

………….. (19)

PLATE SOLUTION (17), (18) & (19) in (14) gives 2 2 4 ∞ ∞    mπ  4  a mn   mπx   nπy  m π n π n π         sin  ∑ ∑  Amn    + 2    +   −  sin  = 0 ………….. (20)  m =1 n =1  a   b   b   D   a   b    a  From (20), we get 2 2 4   mπ  4  amn m π n π n π        Amn   =0  + 2   +   − ………….. (21) a a b b D          

Amn =

amn

(

2 m π D 4

1 ∞ ∞ w( x, y ) = 4 ∑ ∑ 2 π D m =1 n =1 m

(

a

2 n 2 +

a

) b

amn 2 n + 2

………….. (22)

2

2

b2

)

 mπx   nπy  sin   sin  2 a b    

For uniformly distributed load (q ( x, y ) = q0 )  mπ   nπ  sin   sin   ∞ 16 q0 ∞ 2 2     w( x, y ) = 6 ∑ ∑ 2 π D m =1, 3, 5,... n =1, 3, 5,...   m 2  n 2  mn    +     a   b   For concentrated load  mπa   nπb   mπx   nπy   sin     sin  sin   L  sin  L    L L 4P  x   y   x   y  w( x, y ) = 4 ∑ ∑ 2 π DL x L y m =1 n =1   m 2  n 2    +      Lx   L y       ∞

………….. (23)

………….. (24)

∞

………….. (25)

PLATE SOLUTION Moment

2   m 2  n     +ν     a  ∞  b   16q0 ∞  mπx   nπy   Mx = 4 ∑ sin ∑   sin   π m =1,3, 5,...n =1,3, 5,...   m  2  n  2  2  a   b  mn   +     a   b     m 2  n 2  ν   +      a  b  ∞ 16q0 ∞   sin  mπx  sin  nπy  My = 4 ∑ ∑     π m =1, 3, 5,...n =1, 3, 5,...   m  2  n  2  2  a   b  mn   +     a   b  

Stress

12M x z σx = t3

12M y z σy = 3 t

………….. (26)

………….. (27)

………….. (28)

………….. (29)

BUCKLING OF PLATES Nx

Nx

b

a Displacement

 mπx   nπy  w( x, y ) = ∑ ∑ Amn sin   sin   m =1 n =1  a   b  ∞

∞

………….. (1)

Potential energy 2 2   ∂ 2 w ∂ 2 w  ∂ 2 w 2   1a b  ∂w   ∂ 2 w ∂ 2 w    U + V = ∫ ∫  D  2 + 2  − 2(1 −ν )  2 −     − N x   dxdy  2 0 0 2 ∂x ∂y   ∂x      ∂x∂y     ∂x ∂y   ………….. (2) 2

∞ ∞ π abD ∞ ∞ 2  m   n   π 2b 2 U +V = N x ∑ ∑ m 2 Amn ∑ ∑ Amn   +    − m =1 n =1 8 m=1 n=1  a   b   8 4

2

2

………….. (3)

BUCKLING OF PLATES (Contd..) Differentiating 2

 m   n   π 2b ∂ (U + V ) π abD = Amn   +    − N x m 2 Amn ∂Amn 4 4a  a   b   ………….. (4) 2

4

2

Critical buckling load

π a D  m   n   N Cr =   +    2 m  a   b   2

kπ 2 D N Cr = b2 Critical buckling stress

σ Cr

2

2

2

2

 mb   a  k =  +   a   mb 

where

t = kπ E   b 2

………….. (5)

2

………….. (6)

2

………….. (7)

END FIXITY COEFFICIENT

BUCKLING OF PLATES (from column formula) π 2 EI Pcr = k L2 σy σ εx = x − ν E E σy σ εy = −ν x E E

• Euler column formula • Stress-strain relation

………….. (1)

………….. (2)

• Assuming that the plate has no curvature in y direction ε y = 0 σ y = ν σx gives εx =

σx 1− ν 2 E

(

)

………….. (3)

• Flat plate has a smaller elongation compared to a column 2 Pcr =

π EI

(1 − ν )( L ) 2

eff

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

bt 3 L = a, I = 12

π2 E  t  σ cr =   2 12 1 − ν  a 

(

)

2

………….. (4)

and σ cr = Pcr tb ………….. (5) 2

………….. (6)

SHEAR BUCKLING OF PLATES

SHEAR BUCKLING OF PLATES • Shear buckling formula

Fcs π t =k E  2 η 12 1 − ν b 2

(

2

)

METHODOLOGY • Calculate a/b from plate dimensions measured between panel supports • Determine edge restraint fixity 3) 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 Fcs π t =k E  2 η 12 1 − ν b 2

(

2

)

There are three primary effective ways to increase the buckling load of a panel: 1) Decrease the "b" dimension of the panel 2) Increase the thickness of the panel 3) Increase the fixity of the panel supports

LINEAR BUCKLING ANALYSIS

[ K ]{ X } = λ [ K G ]{ X } [K] – Stiffness matrix [KG] – Geometric Stiffness matrix {X} – Buckling shape λ - Buckling load factor

Solution method Lanczos method Subspace iteration Backward iteration

NONLINEAR STATIC ANALYSIS

[ K L + K NL ]{ X } = [ P] [KL] – Linear Stiffness matrix [KNL ] – Nonlinear Stiffness matrix {X} – Deflection vector [P} - Load vector

Solution method Newton-Raphson Method