Axisymmetric Solutions. Navier's Displacement Formulation (without body forces)(r)er, F=0.u=u."u,2G12ougaou10uuuO-r2002ao1-x OrararaeAO1 g2 Qu,au2G 11 OugauuOue+F=0.Or20000r200ar1-kr00arrrrDisplacement fielddudduduuuur.=2C,→u.=CdrdrdrdrYrA Strain and stress fieldOu1-arOug1+K)c+(3-K)6aeouaue1Tr=2G8ro=0Ue1naeOrr6
( ) 2 2 2 22 2 , . 11 2 r r rr r u r uu u u G r rr r r θ θ θ = = ∂∂ ∂ ∂ ++ − ∂∂∂ ∂ u e F0 2 2 1 1 r r r u uu G u r rrrr θ κ θ ∂ ∂ ∂ − − ++ −∂ ∂ ∂ Fr + 2 2 2 22 2 2 0, 1 1 2 21 1 0. 1 r r r uu u u u G u u u G F r rr r r r r r r r θθ θ θ θ θ θ θ κθ θ = ∂∂∂ ∂ ∂ ∂ ∂ + + + − − ++ += ∂ ∂ ∂ ∂ − ∂∂ ∂ Axisymmetric Solutions • Navier’s Displacement Formulation (without body forces) 1 1 12 1 0 2 r r r r r r r d du u du u du u C C u Cr C dr dr r dr r dr r r ⇒ + =⇒ + = ⇒ + = ⇒ = + • Displacement field • Strain and stress field ( ) (( ) ( ) ) ( ) (( ) ( ) ) 1 2 2 1 2 2 1 2 2 2 1 2 2 1 1 13 2 , , 1 1 1 1 2 1 , 13 2 , 1 1 1 1 2 0. 0. 2 r r r r r r r r r r r u G C C G CC r r r u G u CC G CC r r r u u u G r rr θ θ θ θ θ θ θ θ θ θ σ κε κε ε σ κ κ ε σ κε κε θκ κ τ ε ε θ ∂ = = − =− + + − =− + = ∂ − − ∂ = + = + ⇒ =− + + − =− − ⇒ ∂− − ∂ ∂ = = = −+ = ∂ ∂ 2 2 , , 0. r A B r A B r θ θ σ τ + =− + = 6
Axisymmetric Solutions? Airy Stress Formulation (without body forces02d?11a1 da21dd72+Or.2dr200?r orr drr drdr1 d()]1-0r drd1 ddydy1-4 = 死[(au=二1Alnr+Br drr drdrdrdrdrddydy= Ar? nr+B,r? +CArlnr+Br =drdrdrdyArlnr+Br+= = y= Ar? Inr+B,r? +Clnr+Ddry=a+a lnr+a,r?+a,r?Inr7
Axisymmetric Solutions 2 2 2 2 2 2 1 1 r rr r θ ∂∂ ∂ ∇= + + ∂∂∂ 2 2 4 2 2 1 1 1 1 1 0 1 11 ln ln ln 1 d d dd r dr r dr r dr dr d d r r dr dr d dd d dd A dd r rA d r r ArB dr r dr dr dr r dr dr r r dr dr d d d r Ar r Br r A r r B r C dr d d r r dr r dr dr ψ ψ ψψ ψ ψ ψ =+ = ⇒∇ = = ⇒ =⇒ =⇒ = + ⇒ = +⇒ = + + ⇒ 2 2 1 1 2 2 0 2 2 1 23 ln ln ln ln ln d C Ar r Br Ar r B r C r D dr r a a r ar ar r ψ ψ ψ = + +⇒= + + + + + + ⇒ = • Airy Stress Formulation (without body forces) 7
Axisymmetric Solutions. Stress and strain fieldy=a+a,lnr+ar? +ar?Inr11ay[0, =号+2a, +a,(1+21nr)rr不a24+2a, +a,(3+2lnr)-O2Tro=0O00(4(2-K)(1+x)(1+x)4114(13-KaKoaa42G42G1+x r?1+x1+ K1+ K1+K41 (1+x)(1(1+x)4x3x4(1-x)a,4(1-C+a2G42G41+x r21+K1+K1+K1+x1=0.GreTro2G8
2 2 01 2 3 2 2 2 ln ln 1 1 r a a r ar ar r rr r ψ ψ ψ σ θ =+ + + ∂ ∂ = + ∂ ∂ 2 2 , , 1 r r r r θ θ ψ σ ψ τ θ ∂ = ∂ ∂ ∂ = − ∂ ∂ ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 1 2 2 3 1 2 2 3 1 2 2 3 1 2 1 2ln 2 3 2ln 0 . 1 314 1 1 41 42 41 ln , 2 4 1 2 41 1 1 1 1 31 4 1 1 24 1 24 1 r r r r r a aa r r a aa r r a a a r G G r a G G θ θ θ θ θ σ σ τ κ κ κ κ κκ ε σσ κ κ κ κκ κ κ κ ε σσ κ κ =+ + + ⇒ =− + + + = + − + − −− = −= − − + + + + ++ + + − ⇒= − = − + + ( ) ( ) 2 2 3 4 1 4 4 1 ln , 1 11 1 0. 2 r r a a r r G θ θ κ κ κ κ κκ ε τ − − − +− + ++ = = Axisymmetric Solutions • Stress and strain field 8
Axisvmmetric Solutions. Displacement field+(1-x)a,r+a,(r+(1-x)rlnr))+f(0)u, =[s,dr =2G1+K-- () + ()u。= [(reg-u,)deIde=r-f(0))142G(1ouaugue0=8= f'(0)+[ f(0)d0-g(r)+rg'(r)=02r00Or= f'(0)+[f(0)d0=g(r)-rg'(r)=K = g(r)=0r+K, f(0)=u cos+v。sin1(+(1-x)a,r+a, (r+(1-x)rlnr)+u, cos0 +v, sinou.2G1+ xagr-u, sino+v。cos0+o.r+K,K=0e2G. Identification of rigid-body displacements in polar coordinatescosesin[ coso(-o.rsino+u.)+sino(o.rcoso+v.) 1)u.cos+ysino-oy+u-sino coso[-sino(-の.rsing+u.)+coso(orcos+v.))[-u sin+v。cosの+or0x+v9
( ) ( ( ) ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 1 2 3 3 3 1 1 1 ln 2 1 1 2 2 1 1 0 0 2 , co s s in 1 r r r r r o o o r a u dr a r a r r r f G r u r u d ar f d ar f d g r G G u u u f f d g r rg r r rr f f d g r rg r g r r f u v K K u θ θ θ θ θ ε κ κθ κ κ ε θ θ θ θ θθ ε θ θθ θ θ θθ ω θ θ θ = =− + − + + − + + + =− = − = − + ∂ ∂ = = −+ ⇒ + − + = ′ ′ ∂ ∂ ⇒ + =− ⇒ = = ′ ′ = + = − + ⇒ ∫ ∫ ∫ ∫ ∫ ∫ ( ) ( ( ) ) 1 2 3 3 1 1 ln cos sin 2 1 sin cos 2 o o oo o a ar a r r r u v G r u ar u G θ v r K κ κ θθ κ θ θ θω +− + +− + + + =− + + + , K ≡ 0 • Displacement field Axisymmetric Solutions ( ) ( ) ( ) ( ) cos sin cos sin sin cos cos sin sin cos sin sin cos cos sin cos o o oo o o o o o o oo o o o o o y u ru r v u v x v ru r v u v r θ θ ω θ ω θ θω θ θ θ θ θ ω θ ω θ θω θ θ θω − + − ++ + + = = − + − − ++ + −+ + • Identification of rigid-body displacements in polar coordinates 9
Axisymmetric Solutions. Stress formulation.Displacementformulation4+2a, +a,(1+2lnr),一a=u,=C4 +2a, +a,(3+2 lnr)u。= 0. (assumed)+(1-x)a,r+u.coso+vsine2G+a, (r+(1-x)rlnr)1+Karo-u。sino+vcoso+o.rue2G The displacement formulation does not contain the logarithmicterms. Thus, these terms are not consistent with single-valueddisplacements. The compatibility condition is not sufficientThe a, term leads to multivalued behavior, and is not foundfollowing the displacement formulation approach.. The candidacy of individual terms depend on domain singularity10
Axisymmetric Solutions • Displacement formulation • Stress formulation ( ) 1 2 1 2 2 1 2 2 0. assumed 1 , 2 1 2 , 1 2 1 2 . 1 r r u Cr C r G CC r C r u θ G C θ σ κ σ κ = + =− + − =− − = − ( ) ( ) ( ) ( ( ) ) 1 2 2 3 1 2 2 3 1 2 3 3 2 1 2ln , 2 3 2ln . 1 1 cos sin , 2 1 ln 1 sin cos . 2 r r o o oo o a aa r r a aa r r a a r u r u v G ar r r u ar u v r G θ θ σ σ κ θ θ κ κ θ θ θω =+ + + =− + + + + − = − + + + +− + = −++ • The displacement formulation does not contain the logarithmic terms. Thus, these terms are not consistent with single-valued displacements. The compatibility condition is not sufficient. • The a3 term leads to multivalued behavior, and is not found following the displacement formulation approach. • The candidacy of individual terms depend on domain singularity. 10