7.3 Plates Subjected to a Distributed Transverse Load 235
SubstitutionforwfromEq.(i)intotheexpressionsforbendingmoment,Eqs.(7.7)and(7.8),yields
Mx=
16 q 0
π^4
∑∞
m=1,3,5
∑∞
n=1,3,5
[(m^2 /a^2 )+ν(n^2 /b^2 )]
mn[(m^2 /a^2 )+(n^2 /b^2 )]^2
sin
mπx
a
sin
nπy
b
(iii)
My=
16 q 0
π^4
∑∞
m=1,3,5
∑∞
n=1,3,5
[ν(m^2 /a^2 )+(n^2 /b^2 )]
mn[(m^2 /a^2 )+(n^2 /b^2 )]^2
sin
mπx
a
sin
nπy
b
(iv)
Maximumvaluesoccuratthecenteroftheplate.Forasquareplatea=b,andthefirstfivetermsgive
Mx,max=My,max=0.0479q 0 a^2
ComparingEqs.(7.3)withEqs.(7.5)and(7.6),weobservethat
σx=
12 Mxz
t^3
σy=
12 Myz
t^3
Again,themaximumvaluesofthesestressesoccuratthecenteroftheplateatz=±t/2sothat
σx,max=
6 Mx
t^2
σy,max=
6 My
t^2
Forthesquareplate,
σx,max=σy,max=0.287q 0
a^2
t^2
Thetwistingmomentandshearstressdistributionsfollowinasimilarmanner.
The infinite series (Eq. (7.27)) assumed for the deflected shape of a plate gives an exact solution
fordisplacementsandstresses.However,amorerapid,butapproximate,solutionmaybeobtainedby
assumingadisplacementfunctionintheformofapolynomial.Thepolynomialmust,ofcourse,satisfy
thegoverningdifferentialequation(Eq.(7.20))andtheboundaryconditionsofthespecificproblem.
The “guessed” form of the deflected shape of a plate is the basis for the energy method of solution
describedinSection7.6.
Example 7.2
Showthatthedeflectionfunction
w=A(x^2 y^2 −bx^2 y−axy^2 +abxy)
isvalidforarectangularplateofsidesaandb,builtinonallfouredgesandsubjectedtoauniformly
distributedloadofintensityq.IfthematerialoftheplatehasaYoung’smodulusEandisofthickness
t,determinethedistributionsofbendingmomentalongtheedgesoftheplate.