Plates - Mechanics of Structures: Variational and Computational Methods - page 790

Thus, the state variables in terms of their initial values have the form

r 2 p z r 4

64 K

w = w −

+ ν) +

m 0

0

2 K ( 1

p z r 3
16 K

r

θ = m 0

+ ν) −

(8)

K ( 1

p z r
2

V =−

p z r 2
16

= − ( + ν)

m m 0 3

or in matrix notation,

= +

r 2 p z r 4 w

V
m

100 − / [2 K ( 1 + ν) ] w 0

θ 0
V 0
m 0

/( 64 K )

p z r 3 000 r / [ K ( 1 + ν) ] − /( 16 K )

(9)

000 0 − p z r / 2

p z r 2 000 1 − ( 3 + ν)/ 16

U i z i

= +

z z

where U i is the transfer matrix.

EXAMPLE 13.5 Responses of a Circular Plate Without a Center Hole
Calculate the responses for a circular plate with no center hole. Suppose the symmetrical
loading is uniformly distributed. The plate is clamped on the boundary r .

The responses can be calculated using Eq. (9) of Example 13.4. The initial parameters

= b

w 0 ,

θ 0 , V 0 , and M 0 have to be determined first. It is already known from Example 13.4 that

θ = V 0 = 0 (1)

0

At the fixed outer boundary where r = b,

w | = 0 , θ | = 0 (2)

r = b r = b

Substitution of (1) and (2) into Eq. (9) of Example 13.4 with r = b leads to

m 0 b 2 p z b 4

w 0 − / [2 K ( 1 + ν) ] + /( 64 K ) = 0

(3)

p z b 3

m 0 b / [ K ( 1 + ν) ] − /( 16 K ) = 0

Solve these relationships to find

p z b 4
64 K

p z b 2
16

w = m 0 = ( 1 + ν) (4)

0

Thus, the responses are

r 2 p z r 4

64 K =

p z
64 K (

r 2 b 2

2

w = w 0 − m 0 + − )

2 K ( 1 + ν)

r p z r 3

16 K =

p z r
16 K (

b 2 r 2

θ = m 0 − − )

K ( 1 + ν)