7.3 Plates Subjected to a Distributed Transverse Load 227Multiplyingthenumeratoranddenominatorofthisequationbythefactor(1−ν)yields
Mxy=D( 1 −ν)∂^2 w
∂x∂y(7.14)
Equations (7.7), (7.8), and (7.14) relatethebending and twisting moments to theplatedeflection
andareanalogoustothebendingmoment–curvaturerelationshipforasimplebeam.
7.3 PlatesSubjectedtoaDistributedTransverseLoad...............................................
The relationships between bending and twisting moments and plate deflection are now employed in
establishing the general differential equation for the solution of a thin rectangular plate, supporting
a distributed transverse load of intensityqper unit area (see Fig. 7.8). The distributed load may, in
general, vary over the surface of the plate and is, therefore, a function ofxandy. We assume, as in
theprecedinganalysis,thatthemiddleplaneoftheplateistheneutralplaneandthattheplatedeforms
such that plane sections remain plane after bending. This latter assumption introduces an apparent
inconsistencyinthetheory.Forplanesectionstoremainplane,theshearstrainsγxzandγyzmustbe
zero.However,thetransverseloadproducestransverseshearforces(andthereforestresses)asshownin
Fig.7.9.Wethereforeassumethatalthoughγxz=τxz/Gandγyz=τyz/Garenegligible,thecorresponding
shearforcesareofthesameorderofmagnitudeastheappliedloadqandthemomentsMx,My,and
Mxy. This assumption is analogous to that made in a slender beam theory in which shear strains are
ignored.
The element of plate shown in Fig. 7.9 supports bending and twisting moments as previously
describedand,inaddition,verticalshearforcesQxandQyperunitlengthonfacesperpendiculartothe
xandyaxes,respectively.Thevariationofshearstressesτxzandτyzalongthesmalledgesδx,δyof
theelementisneglected,andtheresultantshearforcesQxδyandQyδxareassumedtoactthroughthe
Fig.7.8
Plate supporting a distributed transverse load.