the Bucking of Plates

8/11/2019 the Bucking of Plates

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6 The Bucking of plates

6.1 Introduction

one dimensional members

beams and columns,

ordinary differential equations;

critical load=the failure load

two dimensional members

plates (local buckling)

partial differential equations;

failure load>the critical load;

(post-buckling)



Middle-surface

Located on the plane of half thickness, paralleled with the surface

of plate (Figure 6-1) .

the coordinate surface XOY on the middle-surface.The axis Z obeys the corkscrew rule of right hand.

x

y

z

t/2t/2

Middle-surface

x

y

z

Figure 6-1



x

y

z

Nx

Nx

Nxy

Nxy

Ny

Ny

Nyx

Nyx

forces distribute on the unit width

Figure 6-2

⎭⎬

⎫

==

==

t N t N

t N t N

yx yx xy xy

y y x x

τ τ

σ σ

（

6.2）

σ x

σ x

σ y

σ y

τ xyτ yx

middle-surface forces



Fig 6.4 bending moments, twisting moments andshears

x

y

z

Mx

Mx

My

My

MxyMxy

Myx

Myx

Qx

Qx



dx

dy

t

σ x

σ yτ

xyτ yx

τ xz

τ yz

Fig. the bending stress in thin plates



The relationship between the bending forces and the

bending stresses is:

⎪⎪

⎪

⎭

⎪⎪

⎪

⎬

⎫

⋅=⋅=

⋅=⋅=

⋅=⋅=

∫∫

∫∫

∫∫

−−

−−

−−

2

2

2

2

2

2

2

2

2

2

2

2

11

11

11

t

t yz y

t

t xz x

t

t yx yx

t

t xy xy

t

t y y

t

t x x

dzQdzQ

dz z M dz z M

dz z M dz z M

τ τ

τ τ

σ σ

(6.1)



The first denotes the plane on which it acts and thesecond the direction of the stress.

For stress, if the normal line of the plane is the sameas the coordinate, the positive when the second is thesame as the coordinate. versa.

For strain, tension is positive for normal stress andchanging to smaller angle is positive for shear strain.



Thick plates / 1/5t a > Shear stresses aren’t negligible.

Thin plates / 1/5t a ≤ Shear stresses are negligible and

some assumptions made.

rigid plates / 1/5w t ≤ small deflection problem,

membrane strain are negligible.

no tension or compression or

shear and deformation.

Flexural plates 1/5 / 5w a≤ <large deflection problem,

membrane strain aren’t

negligible

membranes / 5w a ≥ membrane strain

bending strain> >



following assumptions:

• The shear strains are negligible and lines normal to the middle

surface prior to bending remain straight and normal to the middle

surface during bending.

• The normal stress and the corresponding strain are negligible,

and therefore the transverse deflection at any point is equal to that ofthe corresponding point(x,y,0) along the middle surface.

• The transverse deflections of the plates are small compared to the

thickness of the plate. Thus middle surface stretching caused by bending can be neglected (membrane action is negligible).

Rigid plates.

• Homogeneous, isotropic, and obeys Hooke’s law.

Small deflection of uniform thickness thin plates

with rigid and flexure

z



6.2 Differential equation of plate bucking

To determine the critical loading by neutral equilibrium, it is

necessary to have the equation of equilibrium in a slight bent

configuration.

constant biaxial compression forces

Loading constant in-plane shears;

in-plane forces=externally applied loads

Two sets of forces moments and shears=transverse bending



Equilibrium of in-plane forces

y

dx

dyx

y

a

b c

d

w

a′

b′

d′

c′

wdy

y

w

∂

∂

w dx xw

∂

∂

x

w

∂

∂

y

w

∂

∂

Nx

x

w

∂

∂

Nx

dx xw

x xw )(

∂∂

∂∂+

∂∂

Ny

y

w

∂

∂

Ny

dy

y

w

y y

w)(

∂

∂

∂

∂+

∂

∂

Nxy

Nxy

Nyx

Nyx dx

y

w

x y

w)(

∂

∂

∂

∂+

∂

∂

dy

x

w

y x

w)(

∂

∂

∂

∂+

∂

∂

x

y

z

NxNx

Nxy

Nxy

Ny

Ny

Nyx

Nyx





the resultant of the middle-surface forces in the Z

direction :

dxd y

w N y x

w N x

w N y xy x ⎟⎟

⎞

⎜⎜⎝

⎛

∂

∂+∂∂

∂+∂

∂2

22

2

2

2 (6.3)



Q x

x

w

∂

∂

Q y

y

w

∂

∂

Q x+(∂Q x / ∂x)dx

dx x

w

x

w2

2

∂

∂+

∂

∂

dy y

w

y

w2

2

∂

∂+

∂

∂

Q y+(∂Q y / ∂y)dy

x

y

The sum of components of Qx and Qy in the Z direction is:

dxdy y

Q

x

Q y x ⎟⎟ ⎠

⎞⎜⎜⎝

⎛

∂

∂

+∂

∂(6.4)

Equilibrium of bending moments, twisting moments and

shears



Addition of these terms to the Z components of the middle-

surface force, given by equation 6.3 to 6.4, leads to the

equation of equilibrium in Z direction:

022

22

2

2

=∂

∂+

∂∂

∂+

∂

∂+

∂

∂+

∂

∂

y

w N

y x

w N

x

w N

y

Q

x

Q y xy x

y x (6.5)





If the higher-order terms are neglected, this relation

reduces to:

Similarly, moment equilibrium about the Y axis leads to:

0=−∂

∂+∂

∂ y xy y Q

x M

y M

（6.6）

0=−∂

∂+

∂

∂ x

yx x Q y

M

x

M （6.7）

Equation（

6.5），

（

6.6）

and（

6.7）

are equilibrium equations

of plate bucking.





Moment-displacement relations

The relationships between the force moments of Mx, My, Mxy and

stresses are given in the equation 6.1.

(1) The stress-strain relations are:

( )

( )

( ) ⎪

⎪⎪

⎭

⎪

⎪⎪

⎬

⎫

+

=

+−

=

+−

=

xy xy

x y y

y x x

E

E

E

γ

µ

τ

µε ε µ

σ

µε ε µ

σ

12

1

1

2

2

（

6.10）



(2)

The bending strains-deflection relations

··m′

A′

w

xw

∂

∂

u

xw

∂∂

z

x

z

o ··

m

A

According to the hypothesis 1 and hypothesis 3, gives;

x

w zu

∂

∂−= (6.11)



2

2

x

w z x

∂

∂−=ε （6.13）

2

2

y

w z y

∂

∂−=ε （6.14）

y xw z xy∂∂∂−=

2

2γ （6.15）



(3) The bending stress-deflection relations

The equation (6.13)-(6.15) express the relationship betweenthe bending strains and the deflections. Substituting them into

equation (6.10), gives:

⎪⎪

⎪⎪

⎭

⎪

⎪⎪⎪

⎬

⎫

∂∂

∂

+

−=

⎟⎟ ⎠

⎞⎜⎜⎝

⎛

∂

∂+∂

∂

−−=

⎟⎟ ⎠ ⎞⎜⎜

⎝ ⎛

∂∂+

∂∂

−−=

y x

w Ez

xw

yw Ez

yw

xw Ez

xy

y

x

2

2

2

2

2

2

2

2

2

2

2

1

1

1

µ τ

µ µ

σ

µ µ

σ

（6.16）



⎟⎟ ⎞

⎜⎜⎝

⎛

∂

∂+

∂

∂−=

2

2

2

2

y

w

x

w D M x µ （6.17）

⎟⎟ ⎠ ⎞⎜⎜

⎝ ⎛

∂∂+

∂∂−=

2

2

2

2

x

w

y

w D M y µ （6.18）

( ) y x

w D M xy

∂∂

∂−−=

2

1 µ （6.19）

（6.20）( )2

3

112 µ −=

Et D

in which

(4)

The relationship between the moments and deflections



The quantity D is the flexural rigidity per unit width

of plate, and it called the flexural rigidity of plate as well.

It corresponds to the bending stiffness EI of a beam.

Comparison of the beam rigidity with that of the plate,

indicates that a strip of plate is stiffer than a beam of

similar width and depth by a factor.

The difference in stiffness exists because the beam is free

to deform laterally, whereas the plate strip is constrained

from deforming in this manner by the adjacent material.



Differential equation of plate buckling

Substitution of (6.17)-(6.19) into equation (6.8), gives:

⎟⎟ ⎞

⎜⎜⎝

⎛

∂

∂+

∂∂

∂+

∂

∂4

4

22

4

4

4

2 y

w

y x

w

x

w D

y x

w

N y

w

N x

w

N xy y x ∂∂

∂+

∂

∂+

∂

∂=

2

2

2

2

2

2 （6.21）



The critical load for uniaxial compression, pure shear or due

to a combination of compression and shear can be determined.

A primary difference between plates and columns is the

existence of two independent variables in the former as

opposed to a single independent variable in the latter.



6.3 Critical load of a rectangular plate uniformly

compressed in one directiona simply supported rectangular plate with sides a and b units

long.

coordinates system is shown as Figure 6-8.

The plate is acted on by a compression force per unit length,

distributed uniformly along the edges X=0 and X=a.

a

bσ xσ x

x

y



1 The differential equation of plate bucking

0122

2

4

4

22

4

4

4

=∂∂+

∂∂+

∂∂∂+

∂∂

xwt

D yw

y xw

xw

xσ (6.22)

t N

N N

x x

xy y

σ −=

== 0

The differential equation of plate bending reduces to:



2 The boundary conditions

simply supported, the boundary conditions are:

when x=0 a，w=0 Mx=0

when y=0

b，

w=0

My=0 a

b

x

y

when y=0 b ,w=0， 02

2

2

2

=∂

∂+

∂

∂

x

w

y

wµ

02

2

2

2

=

∂

∂+

∂

∂

y

w

x

wµ

when x=0

a ,w=0，





3 To determine the critical stresses by solving

the equationLet us assume that the solution to equation ( 6.22) is of the form

∑∑

∞

=

∞

== 1 1 sinsinm nmn b

yn

a

xm

Aw

π π

（a)

The coefficient that Amn is undetermined in the type , m and n

are the number of half-waves that plate buckles into in the x

and y directions, respectively.

m=1 2 3…

n=1

2

3…

The assumed solution already satisfies the boundary conditions

already.



Substitution (a ) into equation (6.22 ), gives:

∑∑∞

=

∞

= ⎢⎢⎣

⎡

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ +

1 1

2

2

2

2

24

m n

mnb

n

a

m A π 0sinsin

2

22

=⎥⎦

⎤−

b

yn

a

xm

a

m

D

t x π π π σ （ b)

The left-hand side of equation (b) consists of the sum of an infinite

number of independent functions, The only way such a sum can

vanish is if the coefficient of every one of the terms is equal to zero.

02

222

2

2

2

24 =

⎥

⎥

⎦

⎤

⎢

⎢

⎣

⎡−⎟⎟

⎠

⎞⎜⎜

⎝

⎛ +

a

m

D

t a

b

n

a

m A x

mn

π π （c)



（

1）

Amn

=0

Substituting (1) into equation (a), we get:

w=0

solve is zero .

（

2）

The term in the square brackets vanishes

02

222

2

2

2

24 =−⎟⎟

⎞⎜⎜

⎝

⎛ +

a

m

D

t

b

n

a

m x π σ π

2

2

2

2

2

2

22

⎟⎟ ⎞

⎜⎜⎝

⎛ +=

b

n

a

m

t m

Da x

π σ

or22

2

2

⎟⎟

⎠

⎞⎜⎜

⎝

⎛ +=

mb

an

a

mb

t b

D x

π σ （d ）



The critical value ofσ x is the smallest value.

The balancecurved state

the dimension,

physics properties of plate,the number of half-waves that

the plate buckles into.

n=1 results in a minimum value of x that is the plate buckles



2

2

2

⎟ ⎞

⎜⎝

⎛ +=

mb

a

a

mb

t b

D x

π σ (6.25)

n=1 results in a minimum value ofσ

x , that is, the plate buckles

in a single half-wave in the y direction

0=dm

d xσ

02 22

2

=⎟ ⎞⎜⎝ ⎛ −⎟ ⎞⎜⎝ ⎛ + bma

ab

mba

amb

t b Aπ

from which (6.26)

Substitution (6.26）

into equation（

6.25）

b

am =

( )t b

D

cr x 2

24 π σ = (6.27)



m must be an integer. The critical load given by（

6.27）

is

thus valid only when a/b is a whole number.For plates in this category, the bucking pattern consists of a

single half-wave in y direction and a/b half-waves in x direction.

But a/b is seldom an integer.

The more general case, where a/b is not an integer:

t b

Dk cr 2

2π σ = （6.28）

2

⎟ ⎞

⎜⎝

⎛ +=

mb

a

a

mbk （6.29）

The factor k depends on the aspect ratio a/b and on m and n,

the number of half-waves that the plate buckles into.

The minimal value of k is called bulking stress coefficient.



0

2

4

6

8

k

1 2 3 4a/b

2 6

m=1

m=2 m=3

m=4

Fig 6-9 k varies with a/b



The critical stress of the column thus depends on its length,

whereas that of the plate depends on the width of the plate and itsindependent of the length.

2

2 212(1 )( / )cr

k E

b t

π

σ

µ

=−

（6.51）

2

2

( / )

cr

C E

l r

π

σ = （6.52）



6.4 Strain energy of bending in a plate

1

Strain energy of the plate in tiny bending state

In small bending state, the plate has two kinds of force:

middle-surface force and bending force. Because a tiny bendingdeformation can not generate middle surface strain, so the work

done by middle surface force are negligible, the corresponding

strain energy is zero. Only bending force need to be considered.

x

y

z

Nx

Nx

Nxy

Nxy

Ny

Ny

Nyx

Nyxσ x

τ xy





work done by bending force or strain energy of the plate

gives:

( )∫∫∫ ++ dxdydz xy xy y y x x γ τ ε σ ε σ 2

1

[∫ ∫ ∫−−+= 2

20 0

22 22

1 l

l

b a

y x y x E

U σ µσ σ σ

( ) dxdydz xy

2

12 τ µ ++ （6.32）

substituting the relationship between bending stress and

curvature given by ( 6.16) :





2. The potential energy of external load

The load acted on the plate is the middle surface

load. Nx , Ny , Nxy . Base on tiny deformationhypothesis, the changes of middle surface load

produced by bending deformation are neglected.



when it buckles, the slab bends and the end A has a

translation along the X axis, it has been assumed that the

middle surface can not deform, so the length of AB is always

the same.

a

A A′ BN

x

Nx

dx dx

ds

w

dx x

w

∂

∂

B A B A ′−′=∆ (a)

2dx



2

2⎟ ⎠

⎞⎜⎝

⎛

∂

∂+= dx

x

wdxds

dx x

wds

⎥⎦

⎤

⎢⎣

⎡

⎟ ⎠

⎞

⎜⎝

⎛

∂

∂+=

2

2

11 (b)

∫ ∫∫ ′ ′′ ⎟

⎞⎜⎝

⎛ ∂

∂+==

B

A

B

A

B

Adx

x

wdxds

2

2

1A′B

(c)∫ ′ ⎟

⎞⎜⎝

⎛ ∂

∂+′= B

Adx

x

w B A

2

2

1

Substitution（

c）

into（

a），

gives:

∫ ⎟ ⎞

⎜

⎝

⎛

∂

∂=∆ dx

x

w2

2

1(6.35)

ds

w

dx x

w

∂∂



Take out a right-angle AOB (Fig 6-12) from middle-

surface of the plate along x and y direction, after

bending AOB turn into A1O1B1, and the shear

strain γ xy is:

1112

BO A xy ∠−= π

γ o1 w

dx x

w

∂

∂

dy y

w

∂

∂

o A

B

dx

dy

A1

B1

dx

dy

x

yz

is a minim，

so)2

( 111 BO A∠−π

)

2

sin( 111 BO A xy ∠−= π

γ

Fig 6—12

111cos BOA∠=



111cos BO A∠

And l 1 m1 n1—cosine in three axes of lineO1A1

l 2

m2

n2

—cosine in three axes of lineO1

BI

from Fig6-12，

212121 nnmmll xy ++=

γ （

a）

⎪

⎪⎪⎪⎪⎪

⎪

⎭

⎪⎪⎪⎪

⎪⎪⎪

⎬

⎫

∂

∂=

⎟ ⎠

⎞⎜⎝

⎛

∂

∂−≈

=

∂

∂=∂

∂

=

=

⎟ ⎠

⎞⎜⎝

⎛ ∂

∂−≈

⎟ ⎠

⎞⎜⎝

⎛ ∂

∂−

=

y

wn

y

wm

l

x

w

dx

dx x

w

n

m

x

w

dx

dx x

wdx

l

2

2

2

1

1

22

1

2

11

0

0

2

11

（ b）

substitution b into a gives



substitution（b）into（a）gives

During bending，the work done by in-plane force in

strip AB is：

yw

xw xy

∂∂⋅∂∂=γ （6.36）

∫ ⎟ ⎞

⎜⎝

⎛ ∂

∂=∆ dxdy

x

wdy N dy N x x

2

2

1

For the entire plate， the work done by in-plane force Nx

is

∫∫ ⎟ ⎞

⎜⎝

⎛ ∂

∂= dxdy

x

w N T x

2

12

1

Similarly, for the entire plate, the work done by in-plane force Ny is

∫∫ ⎟ ⎞

⎜

⎝

⎛

∂

∂= dxdy

y

w N T y

2

2

2

1

F th ti l t th k d b i l f N i



For the entire plate， the work done by in-plane force Nxy is

∫∫ ∂∂

∂∂= dxdy

y

w

x

w N T xy3

∫∫⎢⎢⎣

⎡⎟ ⎠

⎞⎜⎝

⎛

∂

∂+⎟

⎠

⎞⎜⎝

⎛ ∂

∂=

22

2

1

y

w N

x

w N V y x dxdy

y

w

x

w N xy ⎥

⎦

⎤

∂

∂

∂

∂+ 2 （6.37）

The total potential energy for the bending plate under the

action of Nx , Ny , Nxy are:



6.5 Critical load of uniaxially compressed plate, fixed

along all edges, by the energy method

1

boundary conditionwhen x=0 a，

when y=0 a，

0=w 0=∂

∂

x

w

0=w 0=∂

∂

y

w

In accordance with these conditions the plate is prevented from

moving in the z direction or rotating at the boundaries. The edges

of the plate are however, free to move in the xy plane.

a

a

x

y z

2 Boundary conditions are satisfied if w is assumed



2 Boundary conditions are satisfied if w is assumed

to be of the form：

⎟ ⎞

⎜⎝

⎛ −⎟

⎞⎜⎝

⎛ −=

a

y

a

x Aw

π π 2cos1

2cos1

（

6.39）

It is easy to prove，

formula（

6.39）

can satisfy all theboundary conditions .

3

Calculate the strain energy due to bending and thepotential energy of the external loads

∫ ∫ ⎢⎢⎣

⎡

∂∂

∂∂+⎟⎟

⎠ ⎞⎜⎜

⎝ ⎛

∂∂+⎟⎟

⎠ ⎞⎜⎜

⎝ ⎛

∂∂= a a

y xw

yw

xw DU

0 0 2

2

2

22

2

22

2

2

22

µ ( ) dxd y xw

⎥⎥⎦

⎤

⎟⎟ ⎠

⎞⎜⎜⎝

⎛

∂∂

∂−+

22

12 µ （6.40）

To evaluate this expression, the following derivatives of w are

needed:

⎫⎞⎛⎞⎛∂ A 222



⎪

⎪⎪⎪⎪

⎪

⎭

⎪⎪

⎪⎪⎪⎪

⎬

⎫

⎟ ⎠

⎞⎜⎝

⎛ ⎟ ⎠

⎞⎜⎝

⎛ =∂∂

∂

⎟ ⎠ ⎞⎜

⎝ ⎛ −⎟

⎠ ⎞⎜

⎝ ⎛ =

∂∂

⎟ ⎠

⎞⎜⎝

⎛ −⎟

⎠

⎞⎜⎝

⎛ =

∂

∂

⎟ ⎠

⎞⎜⎝

⎛ −⎟

⎠

⎞⎜⎝

⎛ =

∂

∂

⎟ ⎠

⎞⎜⎝

⎛ −⎟

⎠

⎞⎜⎝

⎛ =

∂

∂

a

y

a

x

a

A

y x

w

a x

a y

a A

yw

a

y

a

x

a

A

x

w

a

x

a

y

a

A

y

w

a

y

a

x

a

A

x

w

π π π

π π π

π π π

π π π

π π π

2sin

2sin

4

2cos12cos4

2cos1

2cos

4

2cos1

2sin

2

2cos1

2sin

2

2

22

2

2

2

2

2

2

2

2

（a）

Substitution（a）into（6.40），gives:



∫ ∫ ⎢⎣

⎡

⎜⎝

⎛

−⎟ ⎠

⎞

⎜⎝

⎛

=

a a

a

y

a

x

a

A D

U 0 0

2

4

242

cos21

2

cos

16

2

π π π

( )

dxdya

y

a

x

a

y

a

y

a

x

a

x

a

x

a

x

a

y

a

y

⎥⎦

⎤⎟

⎠

⎞⎜⎝

⎛ ⎟

⎠

⎞⎜⎝

⎛ ⋅

−+⎟ ⎠

⎞⎜⎝

⎛ −⋅

⎟ ⎠

⎞⎜⎝

⎛ −+⎟

⎠

⎞+

⎜⎝

⎛ −⎟

⎠

⎞⎜⎝

⎛ +⎟ ⎠

⎞+

π π

µ π π

π π µ

π

π π π

2sin

2sin

122

cos2

cos

2cos

2cos2

2cos

2cos21

2cos

2cos

22

2

22

22

（ b）

Making use of the following definite integrals：

⎫a ax2π



⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

=

=

=

∫

∫∫

a

a

a

dxa

x

adx

a

x

adx

a

x

0

0

2

0

2

02

cos

2

2cos

2

2sin

π

π

π

（c）

Equation（

b）

reduces into：

⎢⎣

⎡⎟ ⎠

⎞⎜⎝

⎛ ++⎟

⎠

⎞⎜⎝

⎛ +=

2222

84

24a

aaa

aa

a

A DU

π

( ) ⎥⎦

⎤−+⎟⎟

⎠

⎞⎜⎜⎝

⎛ +

412

42

22aa

µ µ

The strain energy of the bending plate is equals



substitution into（

a），

==>

gy g p q

to：2

24

16a

A DU π = （6.41）

The potential energy of the external loads derive

from（

6.37）：

∫ ∫ ⎟ ⎞

⎜⎝

⎛ ∂∂−=

a a x dxdy

x

w N V

0 0

2

2（a）

x

w

∂

∂

∫ ∫ ⎜⎝

⎛ −−=

a a x

a

y

a

x

a

N AV

0 0

2

2

22 2cos21

2sin

2

4 π π π dxdy

a

y⎟

⎞+

π 2cos2

⎥⎦

⎤⎢⎣

⎡⎟ ⎠

⎞⎜⎝

⎛ +−=

22

22

22a

aa

a

N AV xπ

（6.42）

or 3 22

x N AV

π −=



2V −=

The total potential energy of the system is obtained by adding

the expressions in（6.41） and （6.42）：

2316

22

2

24

x N Aa

A DV U π π −=+ （6.43）

( )

( ) 0332

0

2

2

4

=−

=+

cr x N A

a

A D

dA

V U d

π π

The critical loading and critical stress can be determined by

finding the slightly bent configuration for which the total

potential energy has a stationary value.



( )2

2

2

2

67.103

32

a

D

a

D N

cr x

π π == （6.44

）

Critical stress：

( ) ( )

t a

D

t

N cr x

cr x 2

2

67.10 π

σ == （6.45）

Exact solution :

( ) 2

2

07.10a

D N cr x

π

=

Critical load:

6.7 CRITICAL STRESSS OF A RECTANGULAR



6.7 CRITICAL STRESSS OF A RECTANGULAR

PLATE IN SHEARThis shows for bucking to take place, it is not necessary

that a member be loaded in axial compression. All that is

necessary is that compression stresses exist in some part ofthe member.

x

y

a

a xy

We solve this problem with Galerkin method introducing of 2 4



We solve this problem with Galerkin method introducing of 2.4.

1, the brief introduction of Galerkin equation

Consulting 2.4, general form of Galerkin equation is：

( ) ( )∫ ∫ =a a

i dxdy xgwQ0 0

0 （6.59）

Q (w ) is the right type of the differential equation about flexion；

gi(x) is nember i function of flexion modal

( ) y x

w N

y

w

y x

w

x

wwQ xy

∂∂

∂+

∂

∂+

∂∂

∂+

∂

∂=

2

4

2

22

4

4

4

22 （6.60）

To determine critical loading and critical stress by solving theGalerkin equation

2 Calculate critical shearing stress

Choosing an expression for the deflected shape of the member

that satisfies the boundary conditions



t at sat s es t e bou da y co d t o s：

in Galerkin equation：

i=1，2

a y

a x A

a y

a x Aw π π π π 2sin2sinsinsin 21 += （6.58）

( )a

y

a

x xg

π π sinsin1 = （6.61）

( )a y

a x xg π π 2sin2sin2 = （6.62）

Substitution（6.58）into（6.60）and according to i=1

and i=2, using（6.61）and（6.62）separately，get two

Galerkin equations：

A44



∫ ∫ +

a a

a

y

a

x

a

A

0 0

22

4

4

1

sinsin

4 π π π

a

y

a

x

a

x

a

A

a

y

a

y

a

x

a

x

a

A

D

N a

y

a

y

a

x

a

x

a

A

xy

π π π π π

π π π π

π π π π π

2cossin

2cos

4sin

cossincos2

sin2

sinsin2

sin64

2

2

2

2

2

1

4

4

2

+⋅

⎜⎜

⎝

⎛ +

0sin =⎥⎦

⎤⎟ ⎠

⎞⋅ dxdya

yπ

（6.63）

⎡ A 4 224



∫ ∫ ⎢⎣

⎡a a

a

y

a

y

a

x

a

x

a

A

0 0 4

41 2

sinsin

2

sinsin

4 π π π π π

a

x

a

x

a

A

a

y

a

y

A

x

a

x

a

A

D

N a

x

a

x

a

A

xy

π π π π π

π π π

π π π

2sin

2cos

42sincos

2sincos

2

2sin

2sin

64

2

22

2

21

22

4

42

+⋅

⎜⎜⎝

⎛ +

+

02sin2cos =⎥⎦

⎤⎟ ⎠ ⎞⋅ dxdy

a

y

a

y π π

（6.64）

The definite integrals appearing in（

6.64）

and（

6.64）

have

the following value：



∫ =

a a

dxa

x

0

2

2sin

π

∫

∫

=

=

a

a

dxa x

a x

dxa

x

a

x

0

0

0sincos

0sin2

sin

π π

π π

∫

∫

=

−=

a

a

adxa x

a x

adx

a

x

a

x

0

0

34cos2sin

3

2sin

2cos

π π π

π

π π

Hence the equation (6.63) and (6.64) can be reduced to

03

42

2

64

03242

24

2

2

21

2

4

42

2

2

2

2

2

4

4

1

=⎟ ⎞

⎜⎝

⎛ +⎟

⎞⎜⎝

⎛

=⎟ ⎠ ⎞⎜

⎝ ⎛ −+⎟

⎠ ⎞⎜

⎝ ⎛

π

π π

π π π

a

a

A

D

N a

a

A

aa

A D N a

a A

xy

xy

After simplifying ,gives



⎪⎪⎭

⎪⎪⎬

⎫

=+

=+

016

9

32

09

32

22

4

1

212

4

Aa

A D

N

A D

N Aa

xy

xy

π

π

（

6.65）

The indifferent equilibrium can become possible , only

and isn’t zero completely 。 So, the coefficient

determinant of the equation (6.65 ) must be zero.

016

9

329

32

2

4

2

4

=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

a D

N D

N

a

xy

xy

π

π

（6.66）

1 A

2 A

( )2

2

1.11a

D N

cr xy

π =



a

Critical shear stress：

( )t a

Dcr xy 2

2

1.11 π

τ =

Bending bucking coefficient： k =11.1

using double trigonometric series with energy law to

have former 18 terms , obtained an exacter solution：

k =9.34

Standardize formulae in Germany's steel construction：4



（

6.67）

2

4

5.34k = +1

a

b = ≥

（ 6.68）2

5.344k

= +

result in fig

6-3Germany

standardizes

formulae

α

1

k

5.34

9.34

1.0

Fig. bending bucking coefficients varies with

1a

b = <





6.10 Inelastic bucking of plates



A theory of the inelastic behavior is extremely complex and

beyond the scope of this book.

However, the conclusions that have been drawn from inelastic

plate-buckling studies are simple and straightforward and will

be briefly considered here.

Young’ modulus is replaced by a reduced modulus.

( )

2

2

2

2

2

112⎟ ⎠

⎞⎜⎝

⎛

−==

b

t E k

t b

Dk cr

µ

η π η π σ （6.92）

Where is a plasticity reduction factor. .η 1<η



No generally applicable expression, like the tangent modulus in the

columns, therefore exists for the reduced modulus of a plate.

η

the shape of the stress-strain curve,

the type of loading,

the length-to-width ratio of the plate,

the boundary conditions

A relation for considerably more suited for design purposes has

been derived by Bleich. Using an approximate theory, he obtained

η



E

E t =η

The advantage of this relation is that it leads to conservative results

for any long plate, regardless of the boundary conditions, and that it

can be used for shear as well as axial compression.

But remember: k should be decided by

4 η

α α =

y g pp y,

the simple expression:

6.11 finite deflection theory of plates



Plates, unlike columns, do not collapse when the critical load isreach.it is thus obvious that the postbuckling behavior should be

considered in formulating design criteria for plates. However, if one

desires to study the behavior of the member, it is necessary to

consider deformations of finite magnitude.

If a plate is bent into a nondevelopable surface or if its edges are

restrained from approaching one another during bending, membrane

strains will be induced in the middle surface of the plate.once the

transverse deflections become of the order of magnitude of the plate

thickness, stretching of the middle surface is no longer negligible.

The main difference between that in this article and theinfinitesimal deformation theory is that middle-surface strains due

to bending that were neglected previously will now be considered.

Derivation of equations



Equilibrium

No change occurs. On the other hand there are in-plane forces due to

membrane action in addition to the forces applied along the edges of the

plate when large deflections are considered.Forces in the x direction of in-plane to zero:

For the y direction of in-plane

In the z direction::

0=∂

∂+

∂

∂

y

N

x

N xy x

0=∂

∂+

∂

∂

x

N

y

N xy y

⎟⎟ ⎠

⎞⎜⎜⎝

⎛

∂

∂+

∂∂

∂+

∂

∂4

4

22

4

4

4

2 y

w

y x

w

x

w D

y x

w N

y

w N

x

w N xy y x

∂∂

∂+

∂

∂+

∂

∂=

2

2

2

2

2

2 （6.113）

Compatibility

The displacements and of a point in the plate consist of t o parts:



The displacements u and v of a point in the plate consist of two parts:

(1)The displacement in the middle plane

(2)the bending displacements

2)(2

1

x

w

x

u o xo

∂

∂+

∂

∂=ε

y

w

x

w

x

v

y

u oo xyo

∂

∂

∂

∂

+∂

∂

+∂

∂

=γ

2)(21

yw

yv o

yo∂∂+

∂∂=ε

Relationship



)(1

y x xo N N

Eh

µ ε −=

)(1

x y yo N N Eh

µ ε −=

xy xyo N Eh

)1(2 µ γ

+=

To reduce the number of equations that must be solved

simultaneously a stress function is introduced



simultaneously, a stress function is introduced.

2

2

y

F h N x

∂

∂=

2

2

x

F h N y

∂

∂=

y x

F h N xy

∂∂

∂−=

2

Von Karman large deflection plate equations:



])[(

2

2

2

2

22

2

4

4

22

4

4

4

y

w

x

w

y x

w E

y

F

y x

F

x

F

∂

∂

∂

∂−

∂∂

∂

=∂

∂+∂∂

∂+∂

∂

)2(

2

22

2

2

2

2

2

2

2

2

4

4

22

4

4

4

y x

w

y x

F

y

w

x

F

x

w

y

F

D

h

yw

y xw

xw

∂∂

∂

∂∂

∂−

∂

∂

∂

∂+

∂

∂

∂

∂

=∂∂+∂∂ ∂+∂∂

6.12 postbuckling behavior of axially compressed plates



Simply supported square plate subjected to a uniaxial compressionforce:

2)

2cos

2(cos

32

22 y

a

y

a

x Ef F xaσ π π

−+=

2

22

8a

f E cr xa

π σ σ +=

a

ycr xa xa x

π σ σ σ σ 2cos)( −+=

a x

cr xa yπ σ σ σ 2cos)( −=



Conclusions:



1.plates can continue to carry increasing load

subsequent to reaching the critical stress; this is, they

exhibit postbuckling strength.

2.transverse tensile stresses that arise subsequent to thestart of buckling are primarily responsible for the

presence of postbuckling strength in plates.

3. The material near the longitudinal edges of the plateresists most of the increase in load that occurs in the

postbuckling range.

6.13 ultimate strength of axially compressed plates



Where is the average longitudinal stress at failure.

)1/1(

21 +=

ycr cr

fa

σ σ σ σ

fa



yeu t bP σ =



6.14 design provisions for local buckling



In designing, it is common to proportion the member so thatoverall failure occurs prior to local buckling.

ycr F F >

For I section axial column, web:

flange: y

f

t

b/235)1.010( λ +≤

y

w

o f t h /235)5.025( λ +≤

For I section eccentric column, web:

flange: y f

t

b/235151 ≤

yo

w

o f t

h/235)255.016( ++≤ λ α

yo

w

o f t

h/235)2.265.048( −+≤ λ α



For beam, it is the most complex in this class.

Disadvantages:

1. Elastic analysis;

2. No postbuckling is calculated but considered by using

high buckling stress coefficient k, no initial

imperfections.





Documents

the Bucking of Plates