72 CHAPTER 3 Torsion of Solid Sections
whichmaybewritten,fromFig.3.5,as
τzx=−
∂φ
∂x
dx
dn
−
∂φ
∂y
dy
dn
=−
∂φ
∂n
(3.16)
where, in this case, the direction cosineslandmare defined in terms of an elemental normal of
lengthδn.
Therefore,wehaveshownthattheresultantshearstressatanypointistangentialtothelineofshear
stressthroughthepointandhasavalueequaltominusthederivativeofφinadirectionnormaltotheline.
Example 3.1
Determine the rate of twist and the stress distribution in a circular section bar of radiusRwhich is
subjectedtoequalandoppositetorquesTateachofitsfreeends.
Ifweassumeanoriginofaxesatthecenterofthebar,theequationofitssurfaceisgivenby
x^2 +y^2 =R^2
Ifwenowchooseastressfunctionoftheform
φ=C(x^2 +y^2 −R^2 ) (i)
theboundaryconditionφ=0issatisfiedateverypointontheboundaryofthebarandtheconstantC
maybechosentofulfilltheremainingrequirementofcompatibility.Therefore,fromEqs.(3.11)and(i)
4 C=− 2 G
dθ
dz
sothat
C=−
G
2
dθ
dz
and
φ=−G
dθ
dz
(x^2 +y^2 −R^2 )|2(ii)
SubstitutingforφinEq.(3.8)
T=−G
dθ
dz
(∫∫
x^2 dxdy+
∫∫
y^2 dxdy−R^2
∫∫
dxdy
)
ThefirstandsecondintegralsinthisequationbothhavethevalueπR^4 /4, whereasthethirdintegralis
equaltoπR^2 ,theareaofcrosssectionofthebar.Then,
T=−G
dθ
dz
(
πR^4
4
+
πR^4
4
−πR^4