490 CHAPTER 16 Shear of Beams
Fig.16.11
(a) Determination ofqs,0; (b) equivalent loading on “open” section beam.
We obtainqbby supposing that the closed beam section is “cut” at some convenient point, thereby
producingan“open”section(seeFig.16.11(b)).Theshearflowdistribution(qb)aroundthis“open”
sectionisgivenby
qb=−
(
SxIxx−SyIxy
IxxIyy−I^2 xy
)∫s
0
txds−
(
SyIyy−SxIxy
IxxIxy−Ixy^2
)∫s
0
tyds
as in Section 16.2. The value of shear flow at the cut (s=0) is then found by equating applied and
internalmomentstakenaboutsomeconvenientmomentcenter.Then,fromFig.16.11(a),
Sxη 0 −Syξ 0 =
∮
pqds=
∮
pqbds+qs,0
∮
pds,
where
∮
denotesintegrationcompletelyaroundthecrosssection.InFig.16.11(a),
δA=
1
2
δsp
sothat
∮
dA=
1
2
∮
pds
Hence,
∮
pds= 2 A
whereAistheareaenclosedbythemidlineofthebeamsectionwall.Hence,
Sxη 0 −Syξ 0 =
∮
pqbds+ 2 Aqs,0 (16.17)