7.6 Energy Method for the Bending of Thin Plates 247
ItfollowsimmediatelythatthepotentialenergyoftheNxyloadsis
Vxy=−
1
2
∫a
0
∫b
0
2 Nxy
∂w
∂x
∂w
∂y
dxdy (7.44)
andforthecompletein-planeloadingsystemwehave,fromEqs.(7.42),(7.43),and(7.44),apotential
energyof
V=−
1
2
∫a
0
∫b
0
[
Nx
(
∂w
∂x
) 2
+Ny
(
∂w
∂y
) 2
+ 2 Nxy
∂w
∂x
∂w
∂y
]
dxdy (7.45)
Wearenowinapositiontosolveawiderangeofthin-plateproblemsprovidedthatthedeflectionsare
small,obtainingexactsolutionsifthedeflectedformisknownorapproximatesolutionsifthedeflected
shapehastobe“guessed.”
ConsideringtherectangularplateofSection7.3,simplysupportedalongallfouredgesandsubjected
toauniformlydistributedtransverseloadofintensityq 0 ,weknowthatitsdeflectedshapeisgivenby
Eq.(7.27),namely,
w=
∑∞
m= 1
∑∞
n= 1
Amnsin
mπx
a
sin
nπy
b
Thetotalpotentialenergyoftheplateis,fromEqs.(7.37)and(7.39),
U+V=
∫a
0
∫b
0
{
D
2
[(
∂^2 w
∂x^2
+
∂^2 w
∂y^2
) 2
− 2 ( 1 −ν)
{
∂^2 w
∂x^2
∂^2 w
∂y^2
−
(
∂^2 w
∂x∂y
) 2 }]
−wq 0
}
dxdy
(7.46)
SubstitutinginEq.(7.46)forwandrealizingthat“cross-product”termsintegratetozero,wehave
U+V=
∫a
0
∫b
0
{
D
2
∑∞
m= 1
∑∞
n= 1
A^2 mn
[
π^4
(
m^2
a^2
+
n^2
b^2
) 2
sin^2
mπx
a
sin^2
nπy
b
− 2 ( 1 −ν)
m^2 n^2 π^4
a^2 b^2
(
sin^2
mπx
a
sin^2
nπy
b
−cos^2
mπx
a
cos^2
nπy
b
)]
−q 0
∑∞
m= 1
∑∞
n= 1
Amnsin
mπx
a
sin
nπy
b
}
dxdy