东南大学：《弹性力学》课程教学课件（英文讲稿）09 Two-Dimensional Problems in Polar Coordinates

Two-Dimensional Problemsin Polar Coordinates

Two-Dimensional Problems in Polar Coordinates

Outline.Polar Coordinate FormulationAxisymmetric Solutions toBiharmonicEquationsCylinders under Boundary PressuresHole in Infinite MediaPure Bending of Curved BeamsRotating Disk/Cylinder ProblemGeneral Solutions to Biharmonic equationStress Concentration around a HoleTransverse Bending of Curved BeamsWedge ProblemsQuarter-Plane ProblemsHalf-Plane Problems2

Outline • Polar Coordinate Formulation • Axisymmetric Solutions to Biharmonic Equations • Cylinders under Boundary Pressures • Hole in Infinite Media • Pure Bending of Curved Beams • Rotating Disk/Cylinder Problem • General Solutions to Biharmonic equation • Stress Concentration around a Hole • Transverse Bending of Curved Beams • Wedge Problems • Quarter-Plane Problems • Half-Plane Problems 2

Polar Coordinate Formulation - ReviewStrain-Displacementougou1 u.ue2a0a02OrarrrHooke's Law3-K13-Kd0ap = 2GSGapOOapaβαβaβry2G42(1-x)(1+x)(1 + x)13- x13- x1oO60Gr000Oro2G2G2G441+ K1+KGG+x)。 +(3-)c,),Tro= 2G8ro)8, +(3-x)8(1-3-VFor plane strain: K = 3- 4v; For plane stress: 1+v3

Polar Coordinate Formulation – Review • Strain-Displacement • Hooke’s Law 1 1 1 , . 2 r r r rr u u u u u u r r r rr θ θ θ θ θ εε ε θ θ ∂ ∂    ∂ ∂ = = + = −+    ∂ ∂ ∂∂    3 3 For plane strain: 3 4 ; For plane stress: . 1 ν κ ν κ ν − = − = + ( ) ( ) ( ) ( ) (( ) ( ) ) ( ) (( ) ( ) ) 1 3 3 2 2 4 2 1 13 13 1 1 1 , , . 24 1 24 1 2 1 3 , 1 3 , 2. 1 1 r r rr r r r rr r G G GG G G G G αβ αβ γγ αβ αβ αβ γγ αβ θ θ θ θ θ θ θ θ θ θ κ κ ε σ σδ σ ε εδ κ κ κ κ κ ε σ σε σ σε τ κ κ σ κε κε σ κε κε τ ε κ κ − −     = − =−       −   + +   − − = −= − =     + + =− + + − =− + + − = − −

Polar Coordinate Formulation - ReviewEquilibrium equations00,+10tre +,-0e+F, = 0atre100.2tre + E= 0Orr a0Orr a0rrBeltrami-Michell equation4aFF1 aFV?(, +0。)r 00Or1+K2Navier's equation2GV(V·u)+F = 0.GV?u1-Ka'u.1 'u,22Ga1ou,OueOurur10ueu,-F=0r2Or2002r2r200rrOr001-x rrr1 0"ueo'ue2a10ueour2G 1our1ougUgur=0002r2r2Or?r2a0Orara0r1-xr001r4

Polar Coordinate Formulation – Review 1 1 2 0, 0. r rr r r F F r rr r rr r θ θ θ θθ θ σ τ σσ τ σ τ θ θ ∂∂ − ∂ ∂ + + += + + += ∂ ∂ ∂ ∂ ( ) 2 4 1 . 1 r r r F F F r rr θ σ σθ κ θ   ∂ ∂ ∇ + =− + +   +∂ ∂   4 • Equilibrium equations • Beltrami-Michell equation ( ) 2 2 2 2 22 2 2 2 2 2 22 2 2 2 0. 1 11 2 2 1 0, 1 1 1 2 21 1 0. 1 r r r r r r r r r r G G u u u u uu u u G G F r rr r r r r r r r uu u u u G u u u G F r rr r r r r r r r θ θ θθ θ θ θ θ κ θθ κ θ θ θ κθ θ ∇ − ∇ ∇⋅ + = −    ∂∂∂ ∂ ∂ ∂   ∂    + + − − − ++ +=      ∂ ∂ ∂ ∂ −∂ ∂ ∂   ⇒     ∂∂∂ ∂ ∂ ∂   ∂   + + + − − ++ +=       ∂ ∂ ∂ ∂ − ∂∂ ∂   u uF • Navier’s equation

Polar Coordinate Formulation - Review?Airy Stress Function representationa2022a211 a12(1-x)a21 01aOr200200212Or2r.2aer orr Or1+xOra'ya'y1a1ay1ayaOOOr?P6a02r2Orr orr0. Traction boundary conditionsCRf,(r,0) = T(") =0,n, + Tronefo(r,0)=T(") =Tron, +Ogne. Without body forces, the plane problem is then reducedto a single governing biharmonic equation5

( ) 2 2 2 2 2 22 2 22 2 2 2 2 2 2 2 22 2 11 11 1 1 1 , . 2 1 1 1 . 1 r r r rr r r rr r r r r r r r V r rr r V V θ θ ψ θ θ ψψ ψ ψ σ σσ θ θ κ κ θ −   ∂∂∂ + +   ∂ ∂ ∂∂ ∂ ∂     ++ ++ =     ∂ ∂ ∂∂ ∂ ∂   ∂ ∂ ∂ ∂ ∂   =   +∂ ∂ ∂ + == −   ∂ ∂ ∂ ∂∂   +   + R S x y θ r  • Airy Stress Function representation ( ) ( ) (, ) (, ) r r rr r r r fr T n n fr T n n θ θ θ θ θ θθ θ στ θ τσ = = + = = + n n • Traction boundary conditions • Without body forces, the plane problem is then reduced to a single governing biharmonic equation. 5 Polar Coordinate Formulation – Review

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