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and so
∂σ
x = V z z
I y .
(5.66)
Substituting this into equation 5.65 leads to
z 2
τ v =− V z
I y b 2
bz d z .
(5.13)
z T
5.10.2 Thin-walled open cross-sections
Athin-walledopencross-sectionbeamwithashearforce V z paralleltothe z prin-
cipal axis is shown in Figure 5.11a. It is assumed that the resulting shear stresses
act parallel to the walls of the section and are constant across the wall thickness.
The corresponding horizontal shear stresses τ v are in equilibrium with the hori-
zontalbendingstresses σ .Fortheelement t × δ s × δ x showninFigure5.11b,this
equilibrium condition reduces to
∂σ
x δ xt δ s + ∂(τ v t )
δ s δ x = 0,
x
and substitution of equation 5.66 and integration leads to
s
τ v t =− V z
I y
zt d s ,
(5.16)
0
whichsatisfiestheconditionthattheshearstressiszeroattheunstressedboundary
s = 0.
The direction of the shear stress can be determined from the sign of the right-
handsideofequation5.16.Ifthisispositive,thenthedirectionoftheshearstress
isthesameasthepositivedirectionof s ,andviceversa.Thedirectionoftheshear
stress may also be determined from the direction of the corresponding horizontal
shear force required to keep the out-of-balance bending force in equilibrium, as
shownforexampleinFigure5.13a.Hereahorizontalforce τ v t f δ x inthepositive x
directionisrequiredtobalancetheresultantbendingforce δ σ δ A (duetotheshear
force V z = d M y / d x ) , and so the direction of the corresponding flange shear stress
is from the tip of the flange towards its centre.
5.10.3 Multi-cell thin-walled closed sections
The shear stress distributions in multi-cell closed sections can be determined by
extendingthemethoddiscussedinSection5.4.4forsingle-cellclosedsections.If
thesectionconsistsof n junctionsconnectedby m walls,thenthereare( m n + 1)
independent cells. In each wall there is an unknown circulating shear flow τ vc t .
 
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