21.2 Shear 579
Table 21.1
Stringer/boom y(mm) σz(N/mm^2 )
1 381.0 302.4
2,16 352.0 279.4
3,15 269.5 213.9
4,14 145.8 115.7
5,13 0 0
6,12 −145.8 −115.7
7,11 −269.5 −213.9
8,10 −352.0 −279.4
9 −381.0 −302.4
closedsectionbeam.TheshearflowdistributionisthereforegivenbyEq.(19.11),inwhichthedirect
stress-carryingcapacityoftheskinisassumedtobezero,thatis,tD=0,thus,
qs=−
(
SxIxx−SyIxy
IxxIyy−I^2 xy
) n
∑
r= 1
Bryr−
(
SyIyy−SxIxy
IxxIyy−Ixy^2
) n
∑
r= 1
Brxr+qs,0 (21.1)
Equation (21.1) is applicable to loading cases in which the shear loads are not applied through the
sectionshearcentersothattheeffectsofshearandtorsionareincludedsimultaneously.Alternatively,
ifthepositionoftheshearcenterisknown,theloadingsystemmaybereplacedbyshearloadsacting
through the shear center together with a pure torque, and the corresponding shear flow distributions
maybecalculatedseparatelyandthensuperimposedtoobtainthefinaldistribution.
Example 21.2
ThefuselageofExample21.1issubjectedtoaverticalshearloadof100kNappliedatadistanceof
150mmfromtheverticalaxisofsymmetryasshown,fortheidealizedsection,inFig.21.2.Calculate
thedistributionofshearflowinthesection.
AsinExample21.1,Ixy=0,and,sinceSx=0,Eq.(21.1)reducesto
qs=−
Sy
Ixx
∑n
r= 1
Bryr+qs,0 (i)
inwhichIxx=2.52× 108 mm^4 asbefore.Then,
qs=
− 100 × 103
2.52× 108
∑n
r= 1
Bryr+qs,0
or
qs=−3.97× 10 −^4
∑n
r= 1
Bryr+qs,0 (ii)