442 CHAPTER 15 Bending of Open and Closed, Thin-Walled Beams
Thecomponentsuandvofζareinthenegativedirectionsofthexandyaxes,respectively,sothat
u=−ζsinα, v=−ζcosα (15.26)Differentiating Eqs. (15.26) twice with respect tozand then substituting forζfrom Eq. (15.25), we
obtain
sinα
ρ=−
d^2 u
dz^2,
cosα
ρ=−
d^2 v
dz^2(15.27)
InthederivationofEq.(15.18),weseethat
1
ρ{
sinα
cosα}
=
1
E(IxxIyy−Ixy^2 )[
−Ixy Ixx
Iyy −Ixy]{
Mx
My}
(15.28)
Substituting in Eqs. (15.28) for sinα/ρand cosα/ρfrom Eqs. (15.27) and writingu′′=d^2 u/dz^2 ,
v′′=d^2 v/dz^2 ,wehave
{
u′′
v′′
}
=
− 1
E(IxxIyy−Ixy^2 )[
−Ixy Ixx
Iyy −Ixy]{
Mx
My}
(15.29)
ItisinstructivetorearrangeEq.(15.29)asfollows
{
Mx
My}
=−E
[
Ixy Ixx
Iyy Ixy]{
u′′
v′′}
(seederivationofEq.(15.18)) (15.30)thatis,
Mx=−EIxyu′′−EIxxv′′
My=−EIyyu′′−EIxyv′′}
(15.31)
The first of Eqs. (15.31) shows thatMxproduces curvatures—that is, deflections—in both thexz
andyzplaneseventhoughMy=0;similarlyforMywhenMx=0.Thus,forexample,anunsymmetrical
beamwilldeflectbothverticallyandhorizontallyeventhoughtheloadingisentirelyinaverticalplane.
Similarly,verticalandhorizontalcomponentsofdeflectioninanunsymmetricalbeamareproducedby
horizontalloads.
For a beam having either Cxor Cy(or both) as an axis of symmetry,Ixy=0 and Eqs. (15.29)
reduceto
u′′=−My
EIyy, v′′=−Mx
EIxx(15.32)
Example 15.5
DeterminethedeflectioncurveandthedeflectionofthefreeendofthecantilevershowninFig.15.16(a);
theflexuralrigidityofthecantileverisEIanditssectionisdoublysymmetrical.