494 CHAPTER 16 Shear of Beams
centerofasectionproduceszerotwist.Itfollowsthatdθ/dzinEq.(16.22)iszerosothat
0 =
∮
qs
Gt
ds
or
0 =
∮
1
Gt
(qb+qs,0)ds
whichgives
qs,0=−
∮
(qb/Gt)ds
∮
ds/Gt
(16.27)
IfGt=constant,thenEq.(16.27)simplifiesto
qs,0=−
∮
qbds
∮
ds
(16.28)
ThecoordinateηSisfoundinasimilarmannerbyapplyingSxthroughS.
Example 16.3
Athin-walledclosedsectionbeamhasthesinglysymmetricalcrosssectionshowninFig.16.13.Each
wallofthesectionisflatandhasthesamethicknesstandshearmodulusG.Calculatethedistanceof
theshearcenterfrompoint4.
Theshearcenterclearlyliesonthehorizontalaxisofsymmetrysothatitisonlynecessarytoapply
ashearloadSythroughSandtodetermineξS.Ifwetakethexreferenceaxistocoincidewiththeaxis
Fig.16.13
Closed section beam of Example 16.3.