Chapter 6 less HW solution.doc

7/28/2019 Chapter 6 less HW solution.doc

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Chapter 6 Buckling of Plate Elements

1. Differential Equation of Plate Buckling, a Linear Theory

Consider an isolated freebody of a plate element in the deformed configuration (necessary for

stability problems examining equilibrium in the deformed configuration, neighboring

equilibrium). The plate material is assumed to be isotropic, homogeneous and obey Hookes law.

The plate is assumed to be prismatic (constant thickness) and forces expressed per unit width of

the plate are assumed constant.

Fig. 6-1 In-plane forces on plate element

zF for xN :2

20x x

w w wN dx dy N dy

x x x

+ =

x

y

dy

dx

yN w

y

w

x

w

y

w

x

xN

yN

xNxyN

yxN

yxN

xyN

t

w wdx

x x x

+

w wdx

y x y

+

w wdy

x y x +

w wdy

y y y

+

1


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for yN :

2

20y y

w w wN dy dx N dx

y y y

+ =

for xyN :

2

0xy xyw w w

N dx dy N dy

y x y y

+ =

for yxN :

2

0yx yxw w w

N dy dx N dxx x y x

+ =

Since xy yxN N= (this can be readily proved by taking the in-plane moment at a corner), it follows

Fig. 6-2 Bending shear

zF :2 2 2

2 22x y xy

w w wN N N dxdy

x y x y

+ +

(a)

yxQQ

dxdyx y

+

(b)

(b)

From (a) and (b), one obtains

2 2 2

2 22 0

yxx y xy

QQ w w wN N N

x y x y x y

+ + + + =

(6-1)

yQ

xxQ

xx

QQ dx

x

+

z

yy

y

QQ dy

y

+

2


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Fig. 6-3 Moment components

It should be noted that the positive direction of xyM in Fig. 6-3 is reversed from that given by

Timoshenko and Woinowsky-Krieger (2nd edition, McGraw-Hill, 1959). xy yxM M =

0xM = (The condition that the sum of moments about the x-axis must vanish yields.)

02

y xy yxy

M M QQ dxdydydydx dxdy Q dxdy dxdydy

y x x y

+ =

Neglecting the higher order terms, the above equation reduces to

0y xy

yM M Qy x

+ =

(6-2)

Similarly, moments about the y-axis leads to

0yxx

x

MMQ

x y

+ =

(6-3)

x

y

z

yM

yxM

xyM

xM xx

MM dx

x

+

yx

yx

MM dy

y

+

xy

xy

MM dx

x

+

y

y

MM dy

y

+

3


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Differentiating Eqs. (6-2) and (6-3) and substituting the results into Eq. (6-1), yields

2 22 2 2 2

2 2 2 22 2 0

xy yxx y xy

M MM w w wN N N

x x y y x y x y

+ + + + + =

(6-4)

If one considers (at least temporarily) x y xyN , N , and N are known, then Eq. (6-4) contains four

unknowns, x y xyM , M , M , and w . In order to determine these quantities uniquely, one needs

three additional relationships. These three additional equations may be obtained from the

geometric compatibility conditions.

Moment-Displacement Relationship

2

2

t /

x xt /

M zdz

= (6-5)

2

2

t /

y yt /

M zdz

= (6-6)

2

2

t /

xy xyt /

M zdz

= (6-7)

andx yM M are positive when ( )x y and z are positive. Note the in-plane shear stress, xy ,

due to twisting moment, xyM , is positive if it directs toward positive axis at positive z.

From Hookes law of plane stress problems, it follows immediately

( )

( )( )2 1

x x y

y y x

xy xy xy

/ E

/ E

/ G / E

= = = = +

(6-8)

Solving for stresses, yields

( ) ( )

( ) ( )

( ) ( )

2

2

1

1

2 1

x x y

y y x

xy xy

E /

E /

E /

= + = +

= +

(6-9)

4


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a

b

x

y

xN

4 4 4 2 2 2

4 2 2 4 2 22 2x y xy

w w w w w wD t

x x y y x y x y

+ + = + +

Recall x x y y xy xyN t , N t , N t = = = .

Eq. (6-13) is the differential equation of a rectangular plate element under the action of in-plane

forces (linear problems).

2. Critical Load of a Plate Uniformly Compressed in One Direction

Fig. 6-4 Simply supported plate subjected to uniaxial force

For the problem depicted in the Fig. 6-4, 0y xyN N= = . The governing equation becomes

4 40 0x ,xx x ,xxD w N w or w N w / D + = + =

If the plate is simply supported at 0 andx x a= = (conservative assumption), then

, 0 at 0,xxw w x a= = =

Assume the solution to be of a form

( ) ( ), sin with 1,2,3 . . .n nn x

w x y Y y na

= =

This is a standard procedure of separating variable to transform a partial differential equation

into ordinary differential equation, which will reduce the computational efforts significantly.

( )nY y is a function of the independent variable y only.

Taking appropriate derivatives and substituting into the governing equation above, gives

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2 4 2

2 sin 0iv xn n nNn n n n x

Y Y Ya a D a a

+ =

(6-14)

Since sin 0n x

a

for all values of x, the expression inside the brace must vanish. Let

2

2 xN auD n

=

, then ( )2 4

22 1 0ivn n nn n

Y Y u Y a a

+ =

(6-15)

Assume the homogeneous solution of Eq. (6-15) to be of a formmy

nY e= . Taking successive

derivatives, substituting back to Eq. (6-15), and solving the resulting characteristic equation,

gives

1 1 2 1 3 2 4 2cosh sinh cos sinnY c k y c k y c k y c k y= + + +

where

1 21 and 1n n

k u k ua a

= + =

Assume the rectangular plate shown in Fig. 6-5 is simply supported at / 2x a= and elastically

restrained at / 2y b= . Then, the deflection corresponding to the smallest xN is a symmetric

function ofy based on the coordinate system given. Hence, nY must be an even function and

2 4 0c c= = . The deflection surface becomes

( ) ( )1 1 3 2, cosh cos cosn x

w x y c k y c k ya

= + (6-16)

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Fig. 6-5 Elastically restrained rectangular plate

Fig. 6-6 Elastically restrained boundary condition

The elastically restrained boundary conditions shown in Fig. 6-6 are

at / 2yw

M k y by

= = and at / 2yw

M k y by

= + =

where

k = rotational spring constant

where

( ), ,y yy xxM D w w= + and /2 ,y b yyM Dw= = since /2 ,0 0y b xxw w= =

2 2

1 1 1 3 2 2cosh cosny c k k y c k k y =

/2 1 1 3 20 cosh cos 02 2

y b

b bw c k c k = = + = (a)

xN

xN

y

x

yMyMy

,z w

wy

w

y

+

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9/53

2 2

1 1 1 3 2 2

2 2

1 1 1 3 2 2

cosh cos cos2 2

sinh sin cos2 2

y b by y

w b b n xM k D c k k c k k

y a

b b n xk c k k c k k

a

= =

= =

=

(b)

2 2

1 1 1 3 2 2

2 2

1 1 1 3 2 2

cosh cos cos2 2

sinh sin cos2 2

y b by y

w b b n xM k D c k k c k k

y a

b b n xk c k k c k k

a

= =

= =

=

(c)

It is noted that equation (c) is identical to equation (b). Let2D

bk = . Then equation (b) becomes

2 21 1 1 3 2 2 1 1 1 3 2 2cosh cos sinh sin

2 2 2 2

D b b b bc k k c k k c k k c k k k =

2 2

1 1 1 1 1 3 2 2 2 2sinh cosh sin cos 02 2 2 2 2 2

b b b b b bc k k k k c k k k k

=

(d)

Setting the coefficient determinant of equations (a) and (d) for the constants 1c and 3c , yields

1 2

2 2

1 1 1 1 2 2 2 2

cosh cos2 2

0sinh cosh sin cos

2 2 2 2 2 2

b bk k

b b b b b bk k k k k k k k = +

( )2 22 1 1 2 1 21 2 2 11

cos cosh cosh sin sinh cos2 2 2 2 2 2 2

k b k b k b k b k b k bb k k k k + = +

( )2 22 2 1 21 2 2 11

cos sin tanh cos2 2 2 2 2

k b k b k b k bb k k k k + = +

( )2 2 2 1

1 2 2 1

1

tan tanh2 2 2

k b k b

b k k k k + = + , ( )2 21 2

1 2 1 2

1

tanh tan 02 2 2

k b k b

k k b k k + + =

Let a / b = , the aspect ratio of the rectangular plate. Then

1 12 2

b nk u

= + and 2 1

2 2

b nk u

=

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1 tanh 1 1 tan 1 02 2

n n nu u u u u

+ + + =

(e)

Equation (e) is the general buckling condition equation. If the plate is simply supported along

the boundary at / 2y b= , then 0k = B . Then, equation (e) becomes

1 tan 1 02

nu u

=

as 1 tanh 12

nu u

+ +

is a finite value.

tan 1 1 1 12 2 2

n n nu u u

= = =

( )

22 22 2

2

2 2

1 1 1 1 xNn a

u u un n D n

= = = + =

( )

2 22 2 3 2

2 22

2 2 22 2 22

2 2 2 2

1 112 1

xcr

n Et a nN D

n a n b a

na n D a nb D nD

n ab a b nb a b n

= + = +

= + = + = +

( )

2 22

2

/

12 1cr xcr

t E a nbN t

b nb a

= = +

Let

2

'n

kn

= +

. Then

( )

2

2

2

'

12 1

cr

k E

b

t

=

(6-17)

For the smallest xcrN ,2 2 2

2 22 10 0xcrdN D n n n

d b n n

= = + = =

If 1n = , then( )

2 2

22

2

4 '1 ' 4 '=4

12 1

xcr cr

D k Ek N k

b b

t

= = = =

10


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If 2n = , then 2 ' 4k = =

If 3n = , then 3 ' 4k = =

Fig. 6-7 Plate buckling coefficient

3. Longitudinally Stiffened Plates

Fig. 6-8 Longitudinally stiffened plate strip

Consider a rectangular plate simply supported on all four edges with a longitudinal

stiffener at the center of the plate as shown in Fig. 6-8. From Eqs. (6-14) and (6-16), the

deflection surfaces can be written

n= 4n= 3n= 2

n= 1

4

1 3 422 6

k

A

A

y

z

1w 2w

2

b+

2

ba

2

b

2

b

xx

y

x

S.S.

S.S.

11

A-A


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( )1 1 1 2 1 3 2 4 2sin cosh sinh cos sin for 0n x

w c k y c k y c k y c k y ya

= + + +

( )2 1 1 2 1 3 2 4 2sin cosh sinh cos sin for 0n x

w c k y c k y c k y c k y ya

= + + +

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Chapter 6 less HW solution.doc