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194 Energy and Variational Methods for Solution of Lateral Deflections

The maximum deflection at x  =  y  = a/ 2, using the first three terms (m = n = 1;m = 1, n = 3; m = 3, n = 1 ) of the series expression (4.2.28) is

wmax =

7.9289 p0a4

= 0.002625

 p0a4

.π7D D

A comparison with the corresponding result of a more exact solution, wmax =0.00263 p0a

4 /D, given in Ref. [2], indicates only an insignificant error. This

example illustrates the high accuracy obtainable by the Ritz method, providedthat proper shape functions are used. Considering more terms, even this rela-tively small error can be eliminated.

ILLUSTRATIVE EXAMPLE II

Determine the maximum deflection of the clamped rectangular plate shown in

Fig. 4.2.2 by the Ritz method. Assume that the plate is subjected to constant

lateral load and use a/b = 1.5 span ratio.

b

O

 X 

A A

b

aa

Y 

(a)

 p z = p0 = const

 X 

Section A− A

(b)

Figure 4.2.2 Rectangular plate with fixed edges.

To utilize the apparent symmetry of the deflected plate surface, we take the

coordinate axes through the middle of the plate parallel to the sides. In this case,

the deflection given by Eq. (2.5.30) becomes∞

∞ 

mπ  x 

 

nπ  y 

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W mnw(x, y) =  m   n 4

1 −  ( −1 )m cos a 1 −  ( −1 )

n cos b

 

for m, n = 1, 3, 5, . . ..  (4.2.30)

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Ritz’s Method 

This series expression satisfies the given boundary conditions

(w) x =±a = 0,

∂w

= 0;∂x 

 x =±a ( 4.2.3

(w)y =±b = 0,∂w

= 0.∂y 

y =±b

For the sake of simplicity, let us consider only the first term (m  =  n  =  1 )  in Eq.(4.2.30). Thus, we can write

W π  x 

π  y 

w=

11

1 + cos 1 + cos  . ( 4.2.31a )4

a b

Substitution of this expression into Eq. (4.2.6) gives the strain energy of the plate in

bending; therefore

D   a   b

Dπ4W 

23

b 3a2 . ( 4.2.32 )

Ub = ( ∇2 w)

2 d x d y =

11

+ +2

−a

32 a3 b3 ab

−b

Similarly, from Eq. (4.1.9), the potential of the external forces is computed:

V = − p0 +a

 +b

 w(x, y) d x d y = − p0W 11ab.

−a −b

Minimization of the total potential,

∂(Ub + V )

=

0,∂W 11

yields

W 11 =

16 p0a4

1

.Dπ4

3 + 3(a4 /b

4  ) + 2(a

2

 /b2 )

( 4.2.33 )

( 4.2.34 )

If a/b = 1.5 and ν = 0.3, themaximum deflection at  x   =

 y  = 0 is calculatedfrom Eqs.(4.2.31a) and

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A comparison with the

“exact” solution of the

problem [2], which is

shows that the approximate

solution is accurate enough

for most practical pur-poses.

By considering more terms

in the series representation

of the deflections, a more

accurate solution can beobtained.