Mathematical Methods for Physics and Engineering - Matematica.NET

TENSORSFurther, Poisson’s ratio is defined as σ = −e 22 /e 11 (or −e 33 /e 11 ) and is thus( ) ( )( ) 1 λθ 1 λσ =e 11 2µ = Ee11e 11 2µ 3λ +2µ = λ2(λ + µ) . (26.49)Solving (26.48) and (26.49) for λ and µ gives finallyp ij =σE(1 + σ)(1 − 2σ) e kkδ ij + E(1 + σ) e ij. ◭26.13 Integral theorems for tensorsIn chapter 11, we discussed various integral theorems involving vector and scalarfields. Most notably, we considered the divergence theorem, which states that, forany vector field a, ∫∮∇ · a dV = a · ˆn dS, (26.50)VSwhere S is the surface enclosing the volume V and ˆn is the outward-pointing unitnormal to S at each point.Writing (26.50) in subscript notation, we have∫∂a kdV = a k ˆn k dS. (26.51)V ∂x k SAlthough we shall not prove it rigorously, (26.51) can be extended in an obviousmanner to relate integrals of tensor fields, rather than just vector fields, overvolumes and surfaces, with the result∫∂T ij···k···mdV = T ij···k···m ˆn k dS.V ∂x k SThis form of the divergence theorem for general tensors can be very useful invector calculus manipulations.◮A vector field a satisfies ∇ · a =0inside some volume V and a · ˆn =0on the boundary∫ surface S. By considering the divergence theorem applied to T ij = x i a j , show thata dV = 0.VApplying the divergence theorem to T ij = x i a j we find∫∫∮∂T ij ∂(x i a j )dV = dV = x i a j ˆn j dS =0,V ∂x j V ∂x j Ssince a j ˆn j = 0. By expanding the volume integral we obtain∫V∫∫∂(x i a j ) ∂x idV = a j dV +∂x j V ∂x j∫= δ ij a j dVV∫= a i dV =0,V∮∮Vx i∂a j∂x jdVwhere in going from the first to the second line we used ∂x i /∂x j = δ ij and ∂a j /∂x j =0.◭954