Introduction to Elasticity/Airy example 1

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Example 1 - Beltrami solution

Given:

Beltrami's solution for the equations of equilibrium states that if


           \boldsymbol{\sigma} = \boldsymbol{\nabla} \times \boldsymbol{\nabla} \times \mathbf{A}

where \mathbf{A} is a stress function, then


           \boldsymbol{\nabla} \bullet \boldsymbol{\sigma} = 0~;~~\boldsymbol{\sigma} = \boldsymbol{\sigma}^{T}

Airy's stress function is a special form of \mathbf{A}, given by (in 3\times3 matrix notation)


           \left[A\right] = \begin{bmatrix}
                  0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & \varphi
                 \end{bmatrix}

Show:

Verify that the stresses when expressed in terms of Airy's stress function satisfy equilibrium.

Solution

In index notation, Beltrami's solution can be written as


   \sigma_{ij} = e_{imn}~e_{jpq}~A_{mp,~nq}

For the Airy's stress function, the only non-zero terms of A_{mp,~nq}\, are A_{33,~nq} = \varphi_{,nq}\, which can have nine values. Therefore,

\begin{align}
   \sigma_{11} & = e_{13n}~e_{13q}~\varphi_{,~nq} \\
   \sigma_{22} & = e_{23n}~e_{23q}~\varphi_{,~nq} \\
   \sigma_{33} & = e_{33n}~e_{33q}~\varphi_{,~nq} \\
   \sigma_{23} & = e_{23n}~e_{33q}~\varphi_{,~nq} \\
   \sigma_{31} & = e_{33n}~e_{13q}~\varphi_{,~nq} \\
   \sigma_{12} & = e_{13n}~e_{23q}~\varphi_{,~nq}
\end{align}

Since e_{33k} = 0\, for k=1,2,3, the above set of equations reduces to

\begin{align}
   \sigma_{11} & = e_{13n}~e_{13q}~\varphi_{,~nq} \\
   \sigma_{22} & = e_{23n}~e_{23q}~\varphi_{,~nq} \\
   \sigma_{33} & = 0 \\
   \sigma_{23} & = 0 \\
   \sigma_{31} & = 0 \\
   \sigma_{12} & = e_{13n}~e_{23q}~\varphi_{,~nq}
\end{align}

Now, e_{13k}\, is non-zero only if k = 2, and e_{23k}\, is non-zero only if k=1. Therefore, the above equations further reduce to

\begin{align}
   \sigma_{11} & = e_{132}~e_{132}~\varphi_{,~22} \\
   \sigma_{22} & = e_{231}~e_{231}~\varphi_{,~11} \\
   \sigma_{33} & = 0 \\
   \sigma_{23} & = 0 \\
   \sigma_{31} & = 0 \\
   \sigma_{12} & = e_{132}~e_{231}~\varphi_{,~21}
\end{align}

Therefore, (using the values of e_{132}\,, e_{231}\, and the fact that the order of differentiation does not change the final result), we get

\begin{align}
   \sigma_{11} & = \varphi_{,~22} \\
   \sigma_{22} & = \varphi_{,~11} \\
   \sigma_{33} & = 0 \\
   \sigma_{23} & = 0 \\
   \sigma_{31} & = 0 \\
   \sigma_{12} & = -\varphi_{,~12}
\end{align}

The equations of equilibrium (in the absence of body forces) are given by


  \sigma_{ji,j} = 0

or,

\begin{align}
  \sigma_{11,1} + \sigma_{21,2} + \sigma_{31,3} & = 0 \\
  \sigma_{12,1} + \sigma_{22,2} + \sigma_{32,3} & = 0 \\
  \sigma_{13,1} + \sigma_{23,2} + \sigma_{33,3} & = 0 
\end{align}

Plugging the stresses in terms of \varphi into the above equations gives,

\begin{align}
  \varphi_{,221} - \varphi_{,122} + 0 & = 0 \\
  -\varphi_{,121} + \varphi_{,112} + 0 & = 0 \\
  0 + 0 + 0 & = 0 
\end{align}

Noting that the order of differentiation is irrelevant, we see that equilibrium is satisfied by the Airy stress function.

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