15.3 Deflections due to Bending 455
Fig.15.24
Determination of the deflection of a cantilever.
Example 15.12
Determine the horizontal and vertical components of the tip deflection of the cantilever shown in
Fig.15.24.ThesecondmomentsofareaofitsunsymmetricalsectionareIxx,Iyy,andIxy.
FromEqs.(15.29)
u′′=
MxIxy−MyIxx
E(IxxIyy−Ixy^2 )
(i)
Inthiscase,Mx=W(L−z),My=0sothatEq.(i)simplifiesto
u′′=
WIxy
E(IxxIyy−Ixy^2 )
(L−z) (ii)
IntegratingEq.(ii)withrespecttoz,
u′=
WIxy
E(IxxIyy−Ixy^2 )
(
Lz−
z^2
2
+A
)
(iii)
and
u=
WIxy
E(IxxIyy−Ixy^2 )
(
L
z^2
2
−
z^3
6
+Az+B
)
(iv)
in whichu′denotesdu/dzand the constants of integrationAandBare found from the boundary
conditions;thatis,u′=0andu=0whenz=0.FromthefirstoftheseandEq.(iii),A=0,whilefrom
thesecondandEq.(iv),B=0.Hence,thedeflectedshapeofthebeaminthexzplaneisgivenby
u=
WIxy
E(IxxIyy−Ixy^2 )
(
L
z^2
2
−
z^3
6
)
(v)