484 CHAPTER 16 Shear of Beams
UsingtherelationshipsofEqs.(15.23)and(15.24)—thatis,∂My/∂z=Sx,andsoon—thisexpression
becomes
∂σz
∂z
=
(SxIxx−SyIxy)
IxxIyy−Ixy^2
x+
(SyIyy−SxIxy)
IxxIyy−Ixy^2
y
Substitutingfor∂σz/∂zinEq.(16.2)gives
∂q
∂s
=−
(SxIxx−SyIxy)
IxxIyy−Ixy^2
tx−
(SyIyy−SxIxy)
IxxIyy−Ixy^2
ty (16.12)
IntegratingEq.(16.12)withrespecttosfromsomeoriginforstoanypointaroundthecrosssection,
weobtain
∫s
0
∂q
∂s
ds=−
(
SxIxx−SyIxy
IxxIyy−Ixy^2
)∫s
0
txds−
(
SyIyy−SxIxy
IxxIyy−Ixy^2
)∫s
0
tyds (16.13)
Iftheoriginforsistakenattheopenedgeofthecrosssection,thenq=0whens=0,andEq.(16.13)
becomes
qs=−
(
SxIxx−SyIxy
IxxIyy−Ixy^2
)∫s
0
txds−
(
SyIyy−SxIxy
IxxIyy−I^2 xy
)∫s
0
tyds (16.14)
ForasectionhavingeitherCxorCyasanaxisofsymmetryIxy=0andEq.(16.14)reducesto
qs=−
Sx
Iyy
∫s
0
txds−
Sy
Ixx
∫s
0
tyds
Example 16.1
Determine the shear flow distribution in the thin-walled Z-section shown in Fig. 16.6 due to a shear
loadSyappliedthroughtheshearcenterofthesection.
Theoriginforoursystemofreferenceaxescoincideswiththecentroidofthesectionatthemidpoint
oftheweb.Fromantisymmetry,wealsodeducebyinspectionthattheshearcenteroccupiesthesame
position.SinceSyisappliedthroughtheshearcenter,notorsionexistsandtheshearflowdistribution
isgivenbyEq.(16.14)inwhichSx=0;thatis,
qs=
SyIxy
IxxIyy−Ixy^2
∫s
0
txds−
SyIyy
IxxIyy−I^2 xy
∫s
0
tyds
or
qs=
Sy
IxxIyy−Ixy^2
⎛
⎝Ixy
∫s
0
txds−Iyy
∫s
0
tyds
⎞
⎠ (i)