3.1Prandtl Stress Function Solution 73
whichgives
T=
πR^4
2
G
dθ
dz
thatis,
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
J
τzx=
∂φ
∂y
=−
T
J
y
Theresultantshearstressatanypointonthesurfaceofthebaristhengivenby
τ=
√
τzy^2 +τzx^2
thatis,
τ=
T
J
√
x^2 +y^2
thatis,
τ=
TR
J
(iv)
Theprecedingargumentmaybeappliedtoanyannulusofradiusrwithinthecrosssectionofthe
barsothatthestressdistributionisgivenby
τ=
Tr
J
andthereforeincreaseslinearlyfromzeroatthecenterofthebartoamaximumTR/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.