Airy stress function

Definition

The Airy stress function (\varphi):

If the coordinate basis is rectangular Cartesian (\mathbf{e}_1,~\mathbf{e}_2) with coordinates denoted by (x_1, ~x_2) then the Airy stress function (\varphi) is related to the components of the Cauchy stress tensor (\boldsymbol{\sigma}) by

\begin{align} \sigma_{11} &= \varphi_{,22} = \cfrac{\partial^2 \varphi}{\partial x_2^2} \\ \sigma_{22} &= \varphi_{,11} = \cfrac{\partial^2 \varphi}{\partial x_1^2} \\ \sigma_{12} &= -\varphi_{,12}= -\cfrac{\partial^2 \varphi}{\partial x_1 \partial x_2} \end{align}

Alternatively, if we write the basis as (\mathbf{e}_x, \mathbf{e}_y) and the coordinates as (x, y)\,, then the Cauchy stress components are related to the Airy stress function by

\begin{align} \sigma_{xx} &= \cfrac{\partial^2 \varphi}{\partial y^2} \\ \sigma_{yy} &= \cfrac{\partial^2 \varphi}{\partial x^2} \\ \sigma_{xy} &= -\cfrac{\partial^2 \varphi}{\partial x \partial y} \end{align}

In polar basis (\mathbf{e}_r, \mathbf{e}_\theta) with co-ordinates (r, \theta)\,, the Airy stress function is related to the components of the Cauchy stress via

\begin{align} \sigma_{rr} & = \cfrac{1}{r}\cfrac{\partial\varphi}{\partial r} + \cfrac{1}{r^2}\cfrac{\partial^2\varphi}{\partial \theta^2} \\ \sigma_{\theta\theta} & = \cfrac{\partial^2\varphi}{\partial r^2} \\ \sigma_{r\theta} & = -\cfrac{\partial}{\partial r} \left(\cfrac{1}{r}\cfrac{\partial\varphi}{\partial \theta}\right) \end{align}

Something to think about ...

Do you think the Airy stress function can be extended to three dimensions?

Stress equation of compatibility in 2-D

In the absence of body forces,

\nabla^2{(\sigma_{11} + \sigma_{22})} = 0

or,

\sigma_{11,11} + \sigma_{11,22} + \sigma_{22,11} + \sigma_{22,22} = 0 \,

In terms of the Airy stress function

\varphi_{,1122} + \varphi_{,2222} + \varphi_{,1111} + \varphi_{,1122} = 0 \,

or,

\cfrac{\partial^4 \varphi}{\partial x_1^4} + 2\cfrac{\partial^4 \varphi}{\partial x_1^2 \partial x_2^2} + \cfrac{\partial^4 \varphi}{\partial x_2^4} = 0

or,

\nabla^4\varphi = 0 \,

Some biharmonic Airy stress functions

In cylindrical co-ordinates, some biharmonic functions that may be used as Airy stress functions are

\begin{align} \varphi &= C\theta \\ \varphi &= Cr^2\theta \\ \varphi &= Cr\theta\cos\theta \\ \varphi &= Cr\theta\sin\theta \\ \varphi &= f_n(r)\cos(n\theta) \\ \varphi &= f_n(r)\sin(n\theta) \\ \end{align}

where

\begin{align} f_0(r) & = a_0 r^2 + b_0 r^2 \ln r + c_0 + d_0 \ln r \\ f_1(r) & = a_1 r^3 + b_1 r + c_1 r \ln r + d_1 r^{-1} \\ f_n(r) & = a_n r^{n+2} + b_n r^n + c_n r^{-n+2} + d_n r^{-n} ~,~~n > 1 \end{align}

Displacements in terms of scalar potentials

If the body force is negligible, then the displacements components in 2-D can be expressed as

2\mu u_1 = -\varphi_{,1} + \alpha \psi_{,2} ~,~~ 2\mu u_2 = -\varphi_{,2} + \alpha \psi_{,1}

where,

\alpha = \begin{cases} 1-\nu & \text{for plane strain} \\ \cfrac{1}{1+\nu} & \text{for plane stress} \end{cases}

and \psi(x_1,x_2) is a scalar displacement potential function that satisfies the conditions

\nabla^4 {\psi} = 0 ~,~~ \psi_{,12} = \nabla^4 {\varphi}

To prove the above, you have to use the plane strain/stress constitutive relations

\begin{matrix} \sigma_{\alpha\beta} & = 2\mu\left[\varepsilon_{\alpha\beta} + \left(\cfrac{1-\alpha}{2\alpha-1}\right) \varepsilon_{\gamma\gamma}\delta_{\alpha\beta}\right] \\ \varepsilon_{\alpha\beta} & = \cfrac{1}{2\mu}\left[\sigma_{\alpha\beta} + \left(1-\alpha\right) \sigma_{\gamma\gamma}\delta_{\alpha\beta}\right] \end{matrix}

Note also that the plane stress/strain compatibility equations can be written as

\nabla^2{\sigma_{\gamma\gamma}} = -\cfrac{1}{\alpha} f_{\gamma,\gamma}

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