16.2 Shear of Open Section Beams 483and
∂vt
∂z=pdθ
dz−xRsinψdθ
dz+yRcosψdθ
dz(16.9)
AlsofromEq.(16.7)
∂vt
∂z=pdθ
dz+
du
dzcosψ+dv
dzsinψ (16.10)ComparingthecoefficientsofEqs.(16.9)and(16.10),weseethat
xR=−dv/dz
dθ/dzyR=du/dz
dθ/dz(16.11)
16.2 ShearofOpenSectionBeams......................................................................
TheopensectionbeamofarbitrarysectionshowninFig.16.5supportsshearloadsSxandSysuchthat
thereisnotwistingofthebeamcrosssection.Forthisconditiontobevalid,theshearloadsmustboth
passthroughaparticularpointinthecrosssectionknownastheshearcenter.
Sincetherearenohoopstressesinthebeam,theshearflowsanddirectstressesactingonanelement
ofthebeamwallarerelatedbyEq.(16.2)—thatis,
∂q
∂s+t∂σz
∂z= 0
Weassumethatthedirectstressesareobtainedwithsufficientaccuracyfrombasicbendingtheoryso
thatfromEq.(15.18)
∂σz
∂z=
[(∂My/∂z)Ixx−(∂Mx/∂z)Ixy]
IxxIyy−Ixy^2x+[(∂Mx/∂z)Iyy−(∂My/∂z)Ixy]
IxxIyy−Ixy^2yFig.16.5
Shear loading of open section beam.