Page 280 - Intro to Tensor Calculus
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              I 14 . For a linear elastic, homogeneous, isotropic material assume there exists a state of plane strain in
               polar coordinates. Verify the equilibrium equations are

                                             ∂σ rr  1 ∂σ rθ  1
                                                 +        + (σ rr − σ θθ )+ %b r =0
                                              ∂r    r ∂θ    r
                                                                   2
                                                    ∂σ rθ  1 ∂σ θθ
                                                        +        + σ rθ + %b θ =0
                                                     ∂r    r ∂θ    r
                                                                   ∂σ zz
                                                                       + %b z =0
                                                                    ∂z
               Hint: See problem 15, Exercise 2.3.

              I 15. For a linear elastic, homogeneous, isotropic material assume there exists a state of plane stress in
               Cartesian coordinates. Verify the equilibrium equations are

                                                    ∂σ xx  ∂σ xy
                                                         +      + %b x =0
                                                     ∂x     ∂y
                                                    ∂σ yx  ∂σ yy
                                                         +      + %b y =0
                                                     ∂x     ∂y
              I 16. Determine the compatibility equations in terms of the Airy stress function φ when there exists a state
               of plane stress. Assume the body forces are derivable from a potential function V.

              I 17. For a linear elastic, homogeneous, isotropic material assume there exists a state of plane stress in
               polar coordinates. Verify the equilibrium equations are

                                                            1
                                             ∂σ rr  1 ∂σ rθ
                                                 +        + (σ rr − σ θθ )+ %b r =0
                                              ∂r    r ∂θ    r
                                                    ∂σ rθ  1 ∂σ θθ  2
                                                        +        + σ rθ + %b θ =0
                                                     ∂r    r ∂θ    r
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