76 CHAPTER 3 Torsion of Solid Sections
theiroriginalunloadedshape,althoughtheymaysufferwarpingdisplacementsnormaltotheirplane.
Thefirstoftheseassumptionsleadstotheconclusionthatcrosssectionsrotateasrigidbodiesabout
a center of rotation or twist. This fact was also found to derive from the stress function approach of
Section3.1sothat,referringtoFig.3.4andEq.(3.9),thecomponentsofdisplacementinthexandy
directionsofapointPinthecrosssectionare
u=−θyv=θxItisalsoreasonabletoassumethatthewarpingdisplacementwisproportionaltotherateoftwistand
isthereforeconstantalongthelengthofthebar.Hence,wemaydefinewbytheequation
w=dθ
dzψ(x,y), (3.17)whereψ(x,y)isthewarpingfunction.
Theassumedformofthedisplacementsu,v,andwmustsatisfytheequilibriumandforceboundary
conditionsofthebar.Wenoteherethatitisunnecessarytoinvestigatecompatibility,asweareconcerned
with displacement forms which are single-valued functions and therefore automatically satisfy the
compatibilityrequirement.
ThecomponentsofstraincorrespondingtotheassumeddisplacementsareobtainedfromEqs.(1.18)
and(1.20)andare
εx=εy=εz=γxy= 0γzx=∂w
∂x+
∂u
∂z=
dθ
dz(
∂ψ
∂x−y)
γzy=∂w
∂y+
∂ν
∂z=
dθ
dz(
∂ψ
∂y+x)
⎫
⎪⎪
⎪⎪
⎪⎪
⎬
⎪⎪
⎪⎪
⎪⎪
⎭
(3.18)
Thecorrespondingcomponentsofstressare,fromEqs.(1.42)and(1.46)
σx=σy=σz=τxy= 0τzx=Gdθ
dz(
∂ψ
∂x−y)
τzy=Gdθ
dz(
∂ψ
∂y+x)
⎫
⎪⎪
⎪⎪
⎪⎪
⎬
⎪⎪
⎪⎪
⎪⎪
⎭
(3.19)
Ignoring body forces, we see that these equations identically satisfy the first two of the equilibrium
equations(1.5)andalsothatthethirdisfulfilledifthewarpingfunctionsatisfiestheequation
∂^2 ψ
∂x^2+
∂^2 ψ
∂y^2=∇^2 ψ= 0 (3.20)The direction cosinenis zero on the cylindrical surface of the bar, and so the first two of the
boundaryconditions(Eqs.(1.7))areidenticallysatisfiedbythestressesofEqs.(3.19).Thethirdequation