230 CHAPTER 7 Bending of Thin Plates
and (7.14), and the shear forces per unit lengthQxandQyare found from Eqs. (7.17) and (7.18) by
substitutionforMx,My,andMxyintermsofthedeflectionwoftheplate;thus,
Qx=∂Mx
∂x−
∂Mxy
∂y=−D
∂
∂x(
∂^2 w
∂x^2+
∂^2 w
∂y^2)
(7.21)
Qy=∂My
∂y−
∂Mxy
∂x=−D
∂
∂y(
∂^2 w
∂x^2+
∂^2 w
∂y^2)
(7.22)
Direct and shear stresses are then calculated from the relevant expressions relating them toMx,My,
Mxy,Qx,andQy. Before discussing the solution of Eq. (7.20) for particular cases, we shall establish
boundaryconditionsforvarioustypesofedgesupport.
7.3.1 The Simply Supported Edge
Letussupposethattheedgex=0ofthethinplateshowninFig.7.10isfreetorotatebutnottodeflect.
Theedgeisthensaidtobesimplysupported.Thebendingmomentalongthisedgemustbezeroand
alsothedeflectionw=0.Thus,
(w)x= 0 =0and(Mx)x= 0 =−D(
∂^2 w
∂x^2+ν∂^2 w
∂y^2)
x= 0= 0
Theconditionthatw=0alongtheedgex=0alsomeansthat
∂w
∂y=
∂^2 w
∂y^2= 0
alongthisedge.Theprecedingboundaryconditions,therefore,reduceto
(w)x= 0 = 0(
∂^2 w
∂x^2)
x= 0= 0 (7.23)
Fig.7.10
Plate of dimensionsa×b.