Elasticity/Compatibility example 1

Example 1

Given:

The compatibility equations in terms of the strains are (in index notation)

e_{i k r} e_{j l s} ε_{i j, k l} = 0

The stress-strain relations for a linear elastic material are

ε_{i j} = \frac{1}{E} [(1 + ν) σ_{i j} - ν σ_{m m} δ_{i j}]

Show:

Substituting the stress-strain relations into the compatibility equations, show that the compatibility equation of stress can be expressed as

σ_{i i, j j} δ_{r s} - σ_{i i, r s} - (1 + ν) (σ_{i j, i j} δ_{r s} + σ_{r s, i i} - σ_{i s, i r} - σ_{i r, i s}) = 0

Substituting the stress-strain relations into the left hand side of the compatibility equations and multiplying both sides by $E$ , we get,

\begin{matrix} E e_{i k r} e_{j l s} ε_{i j, k l} & = e_{i k r} e_{j l s} [(1 + ν) σ_{i j, k l} - ν σ_{m m, k l} δ_{i j}] \\ = (1 + ν) e_{i k r} e_{j l s} σ_{i j, k l} - ν e_{i k r} e_{j l s} δ_{i j} σ_{m m, k l} \\ = (1 + ν) e_{i k r} e_{j l s} σ_{i j, k l} - ν e_{n k r} e_{n l s} σ_{m m, k l} \\ = (1 + ν) e_{i k r} e_{j l s} σ_{i j, k l} - ν e_{k r n} e_{l s n} σ_{m m, k l} \end{matrix}

Now, the $δ - e$ rule states that

e_{i j k} e_{p q k} = δ_{i p} δ_{j q} - δ_{i q} δ_{j p}

Therefore,

\begin{matrix} E e_{i k r} e_{j l s} ε_{i j, k l} & = (1 + ν) e_{i k r} e_{j l s} σ_{i j, k l} - ν (δ_{k l} δ_{r s} - δ_{k s} δ_{r l}) σ_{m m, k l} \\ = (1 + ν) e_{i k r} e_{j l s} σ_{i j, k l} - ν (σ_{m m, n n} δ_{r s} - σ_{m m, s r}) \end{matrix}

Recall that,

\begin{matrix} e_{i k r} e_{j l s} & = det [\begin{matrix} δ_{i j} & δ_{i l} & δ_{i s} \\ δ_{k j} & δ_{k l} & δ_{k s} \\ δ_{r j} & δ_{r l} & δ_{r s} \end{matrix}] \\ = δ_{i j} δ_{k l} δ_{r s} - δ_{i j} δ_{k s} δ_{r l} - δ_{i l} δ_{k j} δ_{r s} + δ_{i l} δ_{k s} δ_{r j} + δ_{i s} δ_{k j} δ_{r l} - δ_{i s} δ_{k l} δ_{r j} \end{matrix}

Therefore,

\begin{matrix} δ_{i j} δ_{k l} δ_{r s} σ_{i j, k l} & = σ_{i i, j j} δ_{r s} \\ δ_{i j} δ_{k s} δ_{r l} σ_{i j, k l} & = σ_{i i, r s} \\ δ_{i l} δ_{k j} δ_{r s} σ_{i j, k l} & = σ_{i j, i j} δ_{r s} \\ δ_{i l} δ_{k s} δ_{r j} σ_{i j, k l} & = σ_{i r, i s} \\ δ_{i s} δ_{k j} δ_{r l} σ_{i j, k l} & = σ_{i s, i r} \\ δ_{i s} δ_{k l} δ_{r j} σ_{i j, k l} & = σ_{s r, j j} \end{matrix}

Hence,

\begin{matrix} E e_{i k r} e_{j l s} ε_{i j, k l} & = (1 + ν) (σ_{i i, j j} δ_{r s} - σ_{i i, r s} - σ_{i j, i j} δ_{r s} + σ_{i r, i s} + σ_{i s, i r} - σ_{s r, j j}) - ν (σ_{i i, j j} δ_{r s} - σ_{i i, r s}) \\ = σ_{i i, j j} δ_{r s} - σ_{i i, r s} - (1 + ν) (σ_{i j, i j} δ_{r s} + σ_{r s, i i} - σ_{i s, i r} - σ_{i r, i s}) = 0 \end{matrix}

Hence shown.