东南大学：《弹性力学》课程教学课件（英文讲稿）11 Three-Dimensional Problems Dimensional Problems

Three-Dimensional Problems

Three-Dimensional Problems Dimensional Problems

OutlineDisplacement Formulation ReviewHalf-Space under Uniform Pressure and GravitySpherical ShellGeneral Solution - Helmholtz RepresentationParticular Case-Lame Strain PotentialGalerkin Vector PotentialLove Strain Potential -Axi-symmetryCompleteness of Displacement PotentialsHarmonic and Bi-harmonic FunctionsKelvin's ProblemBoussinesq's ProblemCerruti's ProblemDistributed Pressure onHalf-SpaceHertz Contact Problem2

Outline • Displacement Formulation Review • Half-Space under Uniform Pressure and Gravity Space under Uniform Pressure and Gravity • Spherical Shell • General Solution General Solution – Helmholtz Representation Helmholtz Representation • Particular Case – Lamé Strain Potential • Galerkin Vector Potential Galerkin Vector Potential • Love Strain Potential – Axi-symmetry • Comp p leteness of Displacement Potentials • Harmonic and Bi-harmonic Functions • Kelvin’s Problem • Boussinesq’s Problem • Cerruti’s Problem • Distributed Pressure on Half-Space • Hertz Contact Problem 2

Review of Displacement Formulation - RCCNavier's equationGv?u+(a+G)v(V·u)+F= 0Gui,kh +(a+G)uk,ki + F, = 0Displacement-strain relation:+Vu)8CiiHooke'slaw' = aTr()I +2G8,0,=180,+2GEEv2=G=(1+v)(1-2v)2(1+v)3

Review of Displacement Formulation – RCC     2 • Navier’s equation       2 0 0 G G G GF          u uF • Displacement-strain relation: , ,   0 G i kk k ki i u     G u F Displacement strain relation:     1 1 ε     u u  u u   ,  , ,  2 2 ij i j j i ε     u u  u u • Hooke s’ law: Tr 2   , 2 σ ε     G G     ij kk ij ij   I ε      , , ij kk ij ij E E G     3      , 1 1 2 21  G     

Review of Displacement Formulation - CylindricalNavier's equation210ugou.oueourduru,+F=0a+2.200ar00ozarrr2ou,oueaOu,ug1ur(α +F=0Oz0000raorraOu,ouOug1.W+F.=0ara0OzOzLrα?a2α21a1V2022Or.200?r2LarDisplacement-strain relation:augou1ou1oue(1ou,ugu88.Oz2a0a0Orarr1Ou.Ouroue1 ou.a80Oza02arOz2rHooke's law.4

Review of Displacement Formulation – Cylindrical • Navier’s equation        2 2 2 2 1 0 2 11 r rr z r r u u u uu u Gu G F r r r r rr z                                       2 2 2 2 11 0 1 r rr z u uu u u u Gu G F r r r r rr z uu u u                                        2 2 2 1 0 1 1 rr z z z uu u u Gu G F z r rr z                      2 2  2 2 1 1 r rr r       2 2  z     • Displacement-strain relation: 1 11 , , 2 r zr r r zr u u u uu u u r r z r rr                              Displacement strain relation: 2 11 1 , 2 2 z zr z zr r r z r rr u u uu zr r z                                • Hooke’s law. 4

Displacement Formulation - Axi-symmetricNavier's equationaOu.ou,u,v?u, -+G)+F=0T2arOrOzIaaurou.u,GVF=0OzOrOzra2a21a2Or2022arr Displacement-strain relationauOu.1Ou.ouu,6.C-2OzOrOzOrrHooke's lawOuauauauou,u2G0,=2(s, +8+8.)+2G6,=2(+6+6)+2G20azararOzarrrOuouOuauauu,2G0,=2(8,+8g+8:)+2G8,==2G6razazararOzI*EEv23(1+v)(1-2v)"2(1+v)5

Displacement Formulation – Axi-symmetric • Navier’s equation   2 2 0 r rr z r r u uu u Gu G F r rrr z                           2 2 2 0 rr z z z uu u Gu G F zrr z                2 2 2 2 2 1 r rr z          • Displacement-strain relation 1 , , 2 r r z zr r z zr u u u uu r r z rz                     Displacement strain relation    r r z rz 2   • Hooke’s law   22 22   rr z r rr z r uu u u uu u u   GG GG              2 2 , 2 2 2 2 ,2 rr z r rr z r rr z r r z rr z z z r z r z z rz rz GG GG rr z r rr z r uu u u u u G G GG rr z z r z                                                                          , 1 1 2 21 rr z z r z E E G                5