15.2 Unsymmetrical Bending 437
The moment resultants of the internal direct stress distribution have the same sense as the applied
momentsMxandMy.Therefore,
Mx=
∫
A
σzydA, My=
∫
A
σzxdA (15.17)
SubstitutingforσzfromEq.(15.16)in(15.17)anddefiningthesecondmomentsofareaofthesection
abouttheaxesCx,Cyas
Ixx=
∫
A
y^2 dA, Iyy=
∫
A
x^2 dA, Ixy=
∫
A
xydA
gives
Mx=
Esinα
ρ
Ixy+
Ecosα
ρ
Ixx, My=
Esinα
ρ
Iyy+
Ecosα
ρ
Ixy
or,inmatrixform
{
Mx
My
}
=
E
ρ
[
Ixy Ixx
Iyy Ixy
]{
sinα
cosα
}
fromwhich
E
ρ
{
sinα
cosα
}
=
[
Ixy Ixx
Iyy Ixy
]− 1 {
Mx
My
}
thatis,
E
ρ
{
sinα
cosα
}
=
1
IxxIyy−Ixy^2
[
−Ixy Ixx
Iyy −Ixy
]{
Mx
My
}
sothat,fromEq.(15.16),
σz=
(
MyIxx−MxIxy
IxxIyy−Ixy^2
)
x+
(
MxIyy−MyIxy
IxxIyy−Ixy^2
)
y (15.18)
Alternatively,Eq.(15.18)mayberearrangedintheform
σz=
Mx(Iyyy−Ixyx)
IxxIyy−Ixy^2
+
My(Ixxx−Ixyy)
IxxIyy−Ixy^2
(15.19)
FromEq.(15.19)itcanbeseenthatif,say,My=0,themomentMxproducesastresswhichvarieswith
bothxandy;similarlyforMyifMx=0.
Inthecasewherethebeamcrosssectionhaseither(orboth)CxorCyasanaxisofsymmetry,the
productsecondmomentofareaIxyiszeroandCxyareprincipalaxes.Equation(15.19)thenreducesto
σz=
Mx
Ixx
y+
My
Iyy
x (15.20)