Module 4 Boundary value problems in linear elasticity

1 Module 4 Boundary value problems in linearelasticityLearning Objectives formulate the general Boundary value problem of linear elasticity in three dimensions understand the stress and displacement formulations as alternative solution approachesto reduce the dimensionality of the general elasticity problem solve uniform states of strain and stress in three dimensions specialize the general problem to planar states of strain and stress understand the stress function formulation as a means to reduce the general problemto a single differential equation. solve aerospace-relevant problems in plane strain and plane stress in cartesian andcylindrical Summary of field equationsReadings: BC 3 Intro, Sadd Equations of equilibrium ( 3 equations, 6 unknowns ): ji,j+fi= 0( ) Compatibility ( 6 equations, 9 unknowns): ij=12( ui xj+ uj xi)( )7778 Module 4.

2 Boundary value problems IN linear ELASTICITYe1e2e3 Bbfb Buut BtbuFigure : Schematic of generic problem in linear elasticityor alternatively the equations of strain compatibility (6 equations, 6 unknowns), seeModule ij,kl+ kl,ij= ik,jl+ jl,ik( ) Constitutive Law (6 equations, 0 unknowns) : ij=Cijkl kl( ) ij=Sijkl kl( )In the specific case of linear isotropic elasticity : ij= kk ij+ 2 ij( ) ij=1E[(1 + ) ij kk ij]( ) Boundary conditions of two types: Traction or natural Boundary conditions: For tractions timposed on the portionof the surface of the body Bt:ni ij=tj= tj( ) Displacement or essential Boundary conditions: For displacements uimposed onthe portion of the surface of the body Bu, this includes the supports for whichwe have u=0:ui= ui( ) SOLUTION PROCEDURES79We observe that the general elasticity problem contains 15 unknown fields: displacements(3), strains (6) and stresses (6); and 15 governing equations: equilibrium (3), pointwisecompatibility (6), and constitutive (6), in addition to suitable displacement and tractionboundary conditions.

3 One can prove existence and uniqueness of the solution ( the fields:ui(xj), ij(xk), ij(xk)) in B assuming the convexity of the strain energy function or thepositive definiteness of the stiffness can be shown that the system of equations has a solution (existence) which is unique(uniqueness) providing that the bulk and shear moduli are positive:K=E3(1 2 )>0, G=E2(1 + )>0which poses the following restrictions on the Poisson ratio: 1< < Solution ProceduresThe general problem of 3D elasticity is very difficult to solve analytically in general. Thefirst step in trying to tackle the solution of the general elasticity problem is to reduce thesystem to fewer equations and unknowns by a process of elimination.

4 Depending on theprimary unknown of the resulting equations we have Displacement formulationReadings: BC , Sadd this case, we try to eliminate the strains and stresses from the general problem andseek a reduced set of equations involving only displacements as the primary unknowns. Thisis useful when the displacements are specified everywhere on the Boundary . The formula-tion can be readily derived by first replacing the constitutive law, Equations ( ) in theequilibrium equations, Equations ( ):(Cijkl kl),j+fi= 0and then replacing the strains in terms of the displacements using the stress-strain relations,Equations ( ):(Cijkluk,l),j+fi= 0( )These are the so-called Navier equations.

5 Once the displacement field is found, the strainsfollow from equation ( ), and the stresses from equation ( ).Concept Question s 4. Boundary value problems IN linear ELASTICITYS pecialize the general Navier equations to the case of isotropic elasticitySo-lution:The strategy is to replace the strain-displacement relations in the constitutive lawfor isotropic elasticity ij= kk ij+ 2 ij= uk,k ij+ (ui,j+uj,i),and then insert this in the equilibrium equations:0 = ij,j+fi= uk,kj ij+ (ui,jj+uj,ij) +fi= uk,ki+ (ui,jj+uj,ij) + can be slightly simplified to finally obtain:( + )uj,ji+ ui,jj+fi= 0It can be observed that this is the component form of the vector equation:( + ) ( u) + 2u+f= Question volumetric that in the case that the body forces are uniform or vanish the volumetric defor-matione= kk=uk,kis harmonic, its Laplacian vanishes identically: 2e=e,ii= 0.

6 (Hint: apply the divergence operator ()i,ito Navier s equations)Solution:[( + )uj,ji+ ui,jj+fi],i= 0[( + )e,i+ ui,jj+fi],i= 0( + )e,ii+ ui,jji+fi,i= 0( + )e,ii+ e,jj+fi,i= 0( + 2 )e,ii+fi,i= zero or constant,fi,i= 0 and we obtain the sought resulte,ii= 2e xi2= SOLUTION PROCEDURES81 Concept Question solution to Navier s a problem with body forces given byf= f1f2f3 = 6Gx2x32Gx1x310Gx1x2 ,whereG=E2(1+ )and = 1 displacements given byu= u1u2u3 = C1x21x2x3C2x1x22x3C3x1x2x23 .Determine the constants,C1,C2, andC3allowing the displacement fielduto be solutionof the Navier :By applying the Navier s equationsE2(1 + )(1 2 )uj,ji+E2(1 + )ui,jj+fi= 0E2(1 + )(1 2 )e,i+E2(1 + )ui,jj+fi= 0,wheree= kk=uk, this specific problem , the Navier equations can be simplified to2 Guj,ji+Gui,jj+fi= 02Ge,i+Gui,jj+fi= the displacement field we can calculateu1=C1x21x2x3, u1,1= 11= 2C1x1x2x3, u1,11= 2C1x2x3, u1,22=u1,33= 0,u2=C2x1x22x3, u2,2= 22= 2C2x1x2x3, u2,22= 2C2x1x3, u2,11=u2,33= 0,u3=C3x1x2x23, u3,3= 33= 2C3x1x2x3, u3,33= 2C3x1x2, u3,11=u3,22= 0,e= 11+ 22+ 33= 2(C1+C2+C3) the above expressions into the Navier s equations, we obtain(6C1+ 4C2+ 4C3 6)Gx2x3= 0(4C1+ 6C2+ 4C3+ 2)Gx1x3= 0(4C1+ 4C2+ 6C3+ 10)

7 Gx1x2= 0,which allows us to write the following system of equations 6 4 44 6 44 4 6 C1C2C3 = 6 2 10 .82 Module 4. Boundary value problems IN linear ELASTICITYThe solution of this algebraic system is (C1,C2,C3) =17(27, 1, 29), which leads to thefollowing displacement fieldu= u1u2u3 =17 27x21x2x3 x1x22x3 29x1x2x23 .Practical solutions of Navier s equations can be obtained for fairly complex elasticityproblems via the introduction of displacement potential functions to further simplify Stress formulationReadings: Sadd this case we attempt to eliminate the displacements and strains and obtain equationswhere the stress components are the only unknowns.

8 This is useful when the tractionsare specified on the Boundary . To eliminate the displacements, instead of using the strain-displacement conditions to enforce compatibility, it is more convenient to use the SaintVenant compatibility equations ( ) The derivation is based on replacing the constitutivelaw, Equations ( ) into these equations, and then use the equilibrium equations ( ), first step involves doing:(Sijmn mn ij),kl+ (Sklmn mn kl),ij= (Sikmn mn ik),jl+ (Silmn mn jl),ik( )Concept Question s the Beltrami-Michell equations corresponding to isotropic elasticity (Hint: asa first step, use the compliance form of the isotropic elasticity constitutive relations andreplace them into the Saint-Venant strain compatibility equations, if you take it this far, Iwill explain some additional simplifications)Solution.

9 The general expression for thestress compatibility conditions is given by ij,kl+ kl,ij= ik,jl+ jl,ik,but it is possible to find the six meaningful relationships by settingk=l, ij,kk+ kk,ij= ik,jk+ jk, the compliance expression for isotropic materials mn=1E[(1 + ) mn pp mn] SOLUTION PROCEDURES83into the compatibility expressions, we obtain that1E[(1 + ) ij,kk pp,kk ij] +1E[(1 + ) kk,ij pp,ij kk] =1E[(1 + ) ik,jk pp,jk ik] +1E[(1 + ) jk,ik pp,ik jk],and after a little algebra we can write ij,kk+ kk,ij ik,jk jk,ik= 1 + [ pp,jk ik+ pp,ik jk pp,kk ij pp,ij kk] ij,kk+ kk,ij ik,jk jk,ik= 1 + [ pp,ij+ pp,ij pp,kk ij 3 pp,ij] ij,kk+ kk,ij ik,jk jk,ik= 1 + kk,ij+ 1 + pp,kk the equilibrium equations ij,j+fi= 0, the above expresion can be simplified to ij,kk+11 + kk,ij= 1 + pp,kk ij fi,j fj,i.

10 ( )For the casei=j, the previous equation reduces to ii,kk+11 + kk,ii=3 1 + pp,kk 2fi,i ii,kk+11 + ii,kk=3 1 + ii,kk 2fi,i ii,kk= pp,kk= 1 + 1 fi,i= 1 + 1 fk,k,which allows us to write the Equation as ij,kk+11 + kk,ij= 1 fk,k ij fi,j fj,iConcept Question s equations the case of constant body forces. Expand the general Beltrami-Michell equationswritten in index form into the six independent scalar equationsSolution:TheBeltrami-Michell s equations in index notation are written as ij,kk+11 + kk,ij= 1 fk,k ij fi,j fj,i,and for constant body forces they reduce to(1 + ) ij,kk+ kk,ij= 4. Boundary value problems IN linear elasticity (1 + ) 2 11+ 2 x21( 11+ 22+ 33)= (1 + ) 2 11+ 2I1 x21= 0(1 + ) 2 22+ 2 x22( 11+ 22+ 33)= (1 + ) 2 22+ 2I1 x22= 0(1 + ) 2 33+ 2 x23( 11+ 22+ 33)= (1 + ) 2 33+ 2I1 x23= 0(1 + ) 2 12+ 2 x1 x2( 11+ 22+ 33)= (1 + ) 2 12+ 2I1 x1 x2= 0(1 + ) 2 13+ 2 x1 x3( 11+ 22+ 33)= (1 + ) 2 13+ 2I1 x1 x3= 0(1 + ) 2 23+ 2 x2 x3( 11+ 22+ 33)= (1 + ) 2 23+ 2I1 x2 x3= 0,where 2= 2 x21+ 2 x22+ 2 x23andI1= 11+ 22+ 33is the first invariant of the stress the six non-vanishing equations obtained, only three represent independent equations(just as with Saint-Venant strain compatibility equations).