Introduction to Aircraft Structural Analysis (Elsevier Aerospace Engineering)

16.3 Shear of Closed Section Beams 491

If the moment center is chosen to coincide with the lines of action ofSxandSy, then Eq. (16.17)
reducesto

0 =

∮

pqbds+ 2 Aqs,0 (16.18)

Theunknownshearflowqs,0followsfromeitherofEqs.(16.17)or(16.18).
Itisworthwhiletoconsidersomeoftheimplicationsoftheaboveprocess.Equation(16.14)repre-
sentstheshearflowdistributioninanopensectionbeamfortheconditionofzerotwist.Therefore,by
“cutting”theclosedsectionbeamofFig.16.11(a)todetermineqb,weare,ineffect,replacingtheshear
loadsofFig.16.11(a)byshearloadsSxandSyactingthroughtheshearcenteroftheresulting“open”
sectionbeamtogetherwithatorqueTasshowninFig.16.11(b).WeshallshowinSection17.1that
theapplicationofatorquetoaclosedsectionbeamresultsinaconstantshearflow.Inthiscase,the
constantshearflowqs,0correspondstothetorquebutwillhavedifferentvaluesfordifferentpositionsof
the“cut,”sincethecorrespondingvarious“open”sectionbeamswillhavedifferentlocationsfortheir
shearcenters.Anadditionaleffectof“cutting”thebeamistoproduceastaticallydeterminatestructure,
sincetheqbshearflowsareobtainedfromstaticalequilibriumconsiderations.Itfollowsthatasingle
cellclosedsectionbeamsupportingshearloadsissinglyredundant.

16.3.1 Twist and Warping of Shear-Loaded Closed Section Beams

Shearloadswhicharenotappliedthroughtheshearcenterofaclosedsectionbeamcausecrosssections
totwistandwarp;inotherwords,inadditiontorotation,theysufferoutofplaneaxialdisplacements.
Expressionsforthesequantitiesmaybederivedintermsoftheshearflowdistributionqsasfollows.
Sinceq=τtandτ=Gγ(seeChapter1),thenwecanexpressqsintermsofthewarpingandtangential
displacementswandvtofapointinthebeamwallbyusingEq.(16.6).Thus,

qs=Gt

(

∂w

∂s

+

∂vt

∂z

)

(16.19)

Substitutingfor∂vt/∂zfromEq.(16.10),wehave

qs

Gt

=

∂w

∂s

+p

dθ

dz

+

du

dz

cosψ+

dv

dz

sinψ (16.20)

IntegratingEq.(16.20)withrespecttosfromthechosenoriginforsandnotingthatGmayalsobea
functionofs,weobtain

∫s

0

qs

Gt

ds=

∫s

0

∂w

∂s

ds+

dθ

dz

∫s

0

pds+

du

dz

∫s

0

cosψds+

dv

dz

∫s

0

sinψds

or
∫s

0

qs

Gt

ds=

∫s

0

∂w

∂s

ds+

dθ

dz

∫s

0

pds+

du

dz

∫s

0

dx+

dv

dz

∫s

0

dy