2.2 Stress Functions 47sothat
εx=1
E
[( 1 −ν^2 )σx−ν( 1 +ν)σy]and
εy=1
E
[( 1 −ν^2 )σy−ν( 1 +ν)σx]Also,
γxy=2 ( 1 +ν)
EτxySubstitutingasbeforeinEq.(1.21)andsimplifyingbyusingtheequationsofequilibrium,wehavethe
compatibilityequationforplanestrain
(
∂^2
∂x^2+
∂^2
∂y^2)
(σx+σy)=−1
1 −ν(
∂X
∂x+
∂Y
∂y)
(2.5)
The two equations of equilibrium together with the boundary conditions from Eq. (1.7) and one of
the compatibility equations (2.4) or (2.5) are generally sufficient for the determination of the stress
distributioninatwo-dimensionalproblem.
2.2 StressFunctions......................................................................................
Thesolutionofproblemsinelasticitypresentsdifficulties,buttheproceduremaybesimplifiedbythe
introductionofastressfunction.Foraparticulartwo-dimensionalcase,thestressesarerelatedtoasingle
function ofxandysuch that the substitution for the stresses in terms of this function automatically
satisfies the equations of equilibrium irrespective of what form the function may take. However, a
largeproportionoftheinfinitenumberoffunctionswhichfulfillthisconditionareeliminatedbythe
requirement that the form of the stress function must also satisfy the two-dimensional equations of
compatibility,(2.4)and(2.5),plustheappropriateboundaryconditions.
Forsimplicity,letusconsiderthetwo-dimensionalcaseforwhichthebodyforcesarezero.Now,the
problemistodetermineastress–stressfunctionrelationshipthatsatisfiestheequilibriumconditionsof
∂σx
∂x+
∂τxy
∂y= 0
∂σy
∂y+
∂τyx
∂x= 0
⎫
⎪⎪
⎬
⎪⎪
⎭
(2.6)
andaformforthestressfunctiongivingstresses,whichsatisfythecompatibilityequation
(
∂^2
∂x^2+
∂^2
∂y^2)
(σx+σy)= 0 (2.7)