Page 163 - Aircraft Stuctures for Engineering Student

5.6 Energy method for the bending of thin plates 147

We are now in a position to solve a wide range of thin plate problems provided that
the deflections are small, obtaining exact solutions if the deflected form is known or
approximate solutions if the deflected shape has to be 'guessed'.
Considering the rectangular plate of Section 5.3, simply supported along all four
edges and subjected to a uniformly distributed transverse load of intensity qo, we
know that its deflected shape is given by Eq. (5.27), namely
mrx my
w = 2 gAmnsin-sin-
a
m=l n=l b
The total potential energy of the plate is, from Eqs (5.37) and (5.39)

Substituting in Eq. (5.46) for IC' and realizing that 'cross-product' terms integrate to
zero, we have
coco 2m~x 2n~-v
sin -
U+V=s"s"{4c m=ln=l EA:n[~4($+$)zsin - b
a
0
0
m2n27r4 2mrx . 2nry
sin --cos
-2(1- v)- (sin - -
a2P a b a
-qo EArn,
x
x
nz=l n=l a
The term multiplied by 2( 1 - v) integrates to zero and the mean value of sin2 or cos'
over a complete number of half waves is 4, thus integration of the above expression
yields
U+V=TC EA; 03 4ab
D""
3c,
- (5.47)
nz=1.3,5 n=1:3,5 ni=1,3,5 n=1,3:5
n4ab (rnn ;z)2
From the principle of the stationary value of the total potential energy we have
4ab
a(U+V) D -qor=O
=-2AmnT -+-
aAmn 2 rmn
so that
16%
A, =
r6Dmn[(m2/d) + (n2/b2)]'
giving a deflected form
sin (mTx/a) sin( my/ b)
+ (n2/b2)12

which is the result obtained in Eq. (i) of Example 5.1.