Plates - Mechanics of Structures: Variational and Computational Methods - page 789

It is readily observed that integration of this equation for constant p z gives

p z

C 3 r 2 C 4 r 2 ln r

64 K r 4

w = C 1 + C 2 ln r + + + (2)

Then

p z r 3
16 K

d dr =− C 2

r

θ =− − 2 C 3 r − C 4 r ( 2ln r + 1 ) −

K d 3 K C 4 4

d 2

w
dr 3 +

1
r

w
dr 2 −

1
r 2

d dr p z r

2 K

V =− =−

r +

(3)

K d 2 K C 2

dr 2 + r

w d dr

m =− =−

r 2 (ν −

1 ) + C 4 [ ( 1 + ν)( 2ln r + 1 ) + 2]

p z r 2
16 K (

+ 2 C 3 ( 1 + ν) + 3 + ν)

or in matrix form,

r 2 r 2 ln r

w
V
m

1 ln r C 1

C 2
C 3
C 4

=

0 − 1 / r − 2 r − r ( 2ln r + 1 )

0 0 0 − 4 K / r

r 2

0 − K (ν − 1 )/ − 2 K ( 1 + ν) − K [ ( 1 + ν)( 2ln r + 1 ) + 2]

p z r 4

/( 64 K )

p z r 3

− /( 16 K )

+ (4)

− p z r / 2

p z r 2

− ( 3 + ν)/( 16 )

In order to convert this to transfer matrix form it is necessary to express the integration
constants C i ,i

= ... w θ = .

1 , 2 , , 4 , in terms of the state variables , , V , and m at r 0 Note

that all of these variables must have finite magnitudes at r = 0 . From this condition, it can

be reasoned that C 4 = C 2 = 0. It is of interest to note that since the deflection is symmetric,

θ = 0at r = 0 . In general, the total shear force along the perimeter of a circle of radius r is

2 π r r

2 π rV = p z ( r ) rdrd θ = 2 π p z ( r ) rdr (5)

0 0 0

so that

r

1
r

V = p z ( r ) rdr (6)

0

It is apparent that as r → 0, V → 0, i.e., V 0 = 0 . Imposition of the conditions that θ and V

are zero at r = 0 , leads again to the conclusion that C 4 = C 2 = 0 .

With C 4 = C 2 = 0, (4) at r 0 becomes

w(

=

1 C 1

C 3

0 ) 0

= (7)

m ( 0 ) 0 − 2 K ( 1 + ν)

or

C 1 = w( 0 ) = w

0 ,

=− m ( 0 )/ [2 K ( 1 + ν) ] =− m 0 / [2 K ( 1 + ν) ]

3