Page 246 - Intro to Tensor Calculus
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              I 11. Use the results from problem 9 to verify that in cylindrical coordinates


                                                         1  ∂(ru θ )  ∂u r
                                                   ω θr =          −
                                                         2r   ∂r      ∂θ

                                                         1 ∂u r   ∂u z
                                                   ω rz =      −
                                                         2  ∂z    ∂r

                                                         1 1 ∂u z  ∂u θ
                                                   ω zθ =        −
                                                         2 r ∂θ     ∂z
              I 12. Use the results from problem 9 to verify that in spherical coordinates

                                                      1  ∂(ρu θ )  ∂u ρ
                                               ω θρ =          −
                                                     2ρ    ∂ρ     ∂θ

                                                      1   1 ∂u ρ   ∂(ρu φ)
                                               ω ρφ =            −
                                                     2ρ sin θ ∂φ     ∂ρ

                                                        1    ∂(u φ sin θ)  ∂u θ
                                               ω φθ =                  −
                                                     2ρ sin θ   ∂θ       ∂φ
              I 13. The conditions for static equilibrium in a linear elastic material are determined from the conservation
               law
                                                   j
                                                 σ i,j  + %b i =0,  i, j =1, 2, 3,
                      i
               where σ are the stress tensor components, b i are the external body forces per unit mass and % is the density
                      j
               of the material. Assume an orthogonal coordinate system and verify the following results.
                (a) Show that
                                                        1   ∂
                                                  j            √   j         mj
                                                 σ   = √      ( gσ ) − [ij, m]σ
                                                  i,j        j     i
                                                         g ∂x
                (b) Use the substitutions
                                                      j  h j
                                              σ(ij)= σ      no summation on i or j
                                                      i
                                                        h i
                                                      b i
                                               b(i)=     no summation on i
                                                     h i
                                                      ij
                                              σ(ij)= σ h i h j  no summation on i or j
                   and express the equilibrium equations in terms of physical components and verify the relations
                                       3         √                3         2
                                      X   1  ∂     gh i σ(ij)  1  X  σ(jj) ∂(h )
                                                                            j
                                         √     j             −        2     i  + h i %b(i)=0,
                                           g ∂x      h j       2     h    ∂x
                                      j=1                        j=1  j
                   where there is no summation on i.
              I 14. Use the results from problem 13 and verify that the equilibrium equations in Cartesian coordinates
               can be expressed
                                                 ∂σ xx  ∂σ xy  ∂σ xz
                                                     +       +      + %b x =0
                                                  ∂x     ∂y     ∂z
                                                 ∂σ yx  ∂σ yy  ∂σ yz
                                                      +      +      + %b y =0
                                                  ∂x     ∂y     ∂z
                                                 ∂σ zx  ∂σ zy  ∂σ zz
                                                      +      +      + %b z =0
                                                  ∂x     ∂y     ∂z
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