Page 264 - Intro to Tensor Calculus
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258



                   and in spherical coordinates


                                          ∂   1 ∂   2        1   ∂             1   ∂u φ
                                   (λ + µ)         (ρ u ρ )+       (u θ sin θ)+         +
                                               2
                                         ∂ρ   ρ ∂ρ         ρ sin θ ∂θ        ρ sin θ ∂φ
                                                                                            2
                                              2      2 ∂u θ  2u θ cot θ   2   ∂u φ         ∂ u ρ
                                        2
                                     µ(∇ u ρ −  u ρ −      −         −            )+ %b ρ =%
                                                      2
                                                                         2
                                              ρ 2    ρ ∂θ       ρ 2    ρ sin θ ∂φ           ∂t 2
                                        1 ∂     1 ∂  2       1   ∂             1   ∂u φ
                                 (λ + µ)           (ρ u ρ )+       (u θ sin θ)+         +
                                               2
                                        ρ ∂θ  ρ ∂ρ         ρ sin θ ∂θ        ρ sin θ ∂φ
                                                                                            2
                                              2
                                           µ(∇ u θ +  2 ∂u ρ  −  u θ  −  2 cos θ ∂u φ )+ %b θ =%  ∂ u θ
                                                                 2
                                                                           2
                                                                        2
                                                     2
                                                    ρ ∂θ     ρ sin θ  ρ sin θ ∂φ            ∂t 2
                                                              2
                                      1   ∂     1 ∂  2       1   ∂             1   ∂u φ
                             (λ + µ)               (ρ u ρ )+       (u θ sin θ)+         +
                                               2
                                   ρ sin θ ∂φ  ρ ∂ρ        ρ sin θ ∂θ        ρ sin θ ∂φ
                                                 1          2                               2
                                        2                       ∂u ρ   2cos θ ∂u θ         ∂ u φ
                                     µ(∇ u φ −     2  u φ +  2      +      2     )+ %b φ =%   2
                                               2
                                                                        2
                                              ρ sin θ     ρ sin θ ∂φ   ρ sin θ ∂φ           ∂t
                       2
               where ∇ is determined from either equation (2.1.12) or (2.1.13).
               Boundary Conditions
                   In elasticity the body forces per unit mass (b i ,i =1, 2, 3) are assumed known. In addition one of the
               following type of boundary conditions is usually prescribed:
                 • The displacements u i ,  i =1, 2, 3 are prescribed on the boundary of the region R over which a solution
                   is desired.
                 • The stresses (surface tractions) are prescribed on the boundary of the region R over which a solution is
                   desired.
                 • The displacements u i ,i =1, 2, 3 are given over one portion of the boundary and stresses (surface
                   tractions) are specified over the remaining portion of the boundary. This type of boundary condition is
                   known as a mixed boundary condition.
               General Solution of Navier’s Equations

                   There has been derived a general solution to the Navier’s equations. It is known as the Papkovich-Neuber
               solution. In the case of a solid in equilibrium one must solve the equilibrium equations

                                                                    ~
                                                               2
                                           (λ + µ)∇ (∇· ~u)+ µ∇ ~u + %b =0 or
                                                     1             %           1                      (2.4.38)
                                              2
                                            ∇ ~u +      ∇(∇· ~u)+   ~ b =0 (ν 6=  )
                                                   1 − 2ν          µ           2
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