16.2 Shear of Open Section Beams 483
and
∂vt
∂z
=p
dθ
dz
−xRsinψ
dθ
dz
+yRcosψ
dθ
dz
(16.9)
AlsofromEq.(16.7)
∂vt
∂z
=p
dθ
dz
+
du
dz
cosψ+
dv
dz
sinψ (16.10)
ComparingthecoefficientsofEqs.(16.9)and(16.10),weseethat
xR=−
dv/dz
dθ/dz
yR=
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^2
x+
[(∂Mx/∂z)Iyy−(∂My/∂z)Ixy]
IxxIyy−Ixy^2
y
Fig.16.5
Shear loading of open section beam.