Introduction to Aircraft Structural Analysis (Elsevier Aerospace Engineering)

17.1 Torsion of Closed Section Beams 507

The minimum thickness of the beam corresponding to the maximum allowable shear stress of
200N/mm^2 isobtaineddirectlyusingEq.(17.1)inwhichTmax=15kNm.
Then

tmin=

15 × 106 × 4

2 ×π× 2002 × 200

=1.2mm

TherateoftwistalongthebeamisgivenbyEq.(17.4)inwhich
∮
ds
t

=

π× 200

tmin

Hence

dθ

dz

=

T

4 A^2 G

×

π× 200

tmin

(i)

TakingtheoriginforzatoneofthefixedendsandintegratingEq.(i)forhalfthelengthofthebeam,
weobtain

θ=

T

4 A^2 G

×

200 π

tmin

z+C 1 ,

whereC 1 isaconstantofintegration.Atthefixedendwherez=0,θ=0,soC 1 =0.

Hence

θ=

T

4 A^2 G

×

200 π

tmin

z

Themaximumangleoftwistoccursatthemidspanofthebeamwherez=1m.Hence

tmin=

15 × 106 × 200 ×π× 1 × 103 × 180

4 ×(π× 2002 / 4 )^2 × 25000 × 2 ×π

=2.7mm

Theminimumallowablethicknessthatsatisfiesbothconditionsistherefore2.7mm.

Example 17.2
Determinethewarpingdistributioninthedoublysymmetricalrectangular,closedsectionbeam,shown
inFig.17.4,whensubjectedtoananticlockwisetorqueT.

Fromsymmetry,thecenteroftwistRwillcoincidewiththemidpointofthecrosssection,andpoints
ofzerowarpingwilllieontheaxesofsymmetryatthemidpointsofthesides.Weshall,therefore,take
theoriginforsatthemidpointofside14andmeasuresinthepositive,anticlockwise,sensearoundthe
section.AssumingtheshearmodulusGtobeconstant,werewriteEq.(17.5)intheform

ws−w 0 =

Tδ

2 AG

(

δOs

δ

−

AOs

A

)

(i)