Introduction to Aircraft Structural Analysis (Elsevier Aerospace Engineering)

3.1Prandtl Stress Function Solution 69

Fig.3.3

Derivation of torque on cross section of bar.

Therefore,weareinapositiontoobtainanexactsolutiontoatorsionproblemifastressfunction
φ(x,y)can be found to satisfy Eq. (3.4) at all points within the bar and that vanishes on the surface
of the bar, provided that the external torques are distributed over the ends of the bar in an identical
mannertothedistributionofinternalstressoverthecrosssection.Althoughthelastprovisoisgenerally
impracticable,weknowfromSt.Venant’sprinciplethatonlystressesintheendregionsareaffected;
therefore,thesolutionisapplicabletosectionsatdistancesfromtheendsusuallytakentobegreaterthan
thelargestcross-sectionaldimension.Wehavenowsatisfiedalltheconditionsoftheproblemwithout
theuseofstressesotherthanτzyandτzx,demonstratingthatouroriginalassumptionswerejustified.
Usually, in addition to the stress distribution in the bar, we must know the angle of twist and the
warpingdisplacementofthecrosssection.First,however,weshallinvestigatethemodeofdisplacement
ofthecrosssection.Wehaveseenthatasaresultofourassumedvaluesofstress,

εx=εy=εz=γxy= 0

Itfollows,fromEqs.(1.18)andthesecondofEqs.(1.20),that

∂u

∂x

=

∂v

∂y

=

∂w

∂z

=

∂v

∂x

+

∂u

∂y

= 0

which result leads to the conclusions that each cross section rotates as a rigid body in its own plane
aboutacenterofrotationortwist,andthatalthoughcrosssectionssufferwarpingdisplacementsnormal
totheirplanes,thevaluesofthisdisplacementatpointshavingthesamecoordinatesalongthelength
ofthebarareequal.Therefore,eachlongitudinalfiberofthebarremainsunstrained,aswehaveinfact
assumed.
Letussupposethatacrosssectionofthebarrotatesthroughasmallangleθaboutitscenteroftwist
assumedcoincidentwiththeoriginoftheaxesOxy(seeFig.3.4).SomepointP(r,α)willbedisplaced
toP′(r,α+θ),thecomponentsofitsdisplacementbeing

u=−rθsinα v=rθcosα