3.1Prandtl Stress Function Solution 73whichgives
T=
πR^4
2G
dθ
dzthatis,
T=GJ
dθ
dz(iii)inwhichJ=πR^4 / 2 =πD^4 /32(Disthediameter),thepolarsecondmomentofareaofthebar’scross
section.
SubstitutingforG(dθ/dz)inEq.(ii)from(iii)
φ=−T
2 J
(x^2 +y^2 −R^2 )andfromEqs.(3.2)
τzy=−∂φ
∂x=
Tx
Jτzx=∂φ
∂y=−
T
J
yTheresultantshearstressatanypointonthesurfaceofthebaristhengivenbyτ=√
τzy^2 +τzx^2thatis,
τ=T
J
√
x^2 +y^2thatis,
τ=TR
J
(iv)Theprecedingargumentmaybeappliedtoanyannulusofradiusrwithinthecrosssectionofthe
barsothatthestressdistributionisgivenby
τ=Tr
JandthereforeincreaseslinearlyfromzeroatthecenterofthebartoamaximumTR/Jatthesurface.
Example 3.2
A uniform bar has the elliptical cross section that is shown in Fig. 3.6 and is subjected to equal and
oppositetorquesTateachofitsfreeends.Deriveexpressionsfortherateoftwistinthebar,theshear
stressdistribution,andthewarpingdisplacementofitscrosssection.