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)∫s0txds−(
SyIyy−SxIxy
IxxIxy−Ixy^2)∫s0tydsas 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
δspsothat
∮
dA=
1
2
∮
pdsHence,
∮
pds= 2 A
whereAistheareaenclosedbythemidlineofthebeamsectionwall.Hence,
Sxη 0 −Syξ 0 =∮
pqbds+ 2 Aqs,0 (16.17)