Airy stress function

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The Airy stress function (\varphi):

  • Scalar potential function that can be used to find the stress.
  • Satisfies equilibrium in the absence of body forces.
  • Only for two-dimensional problems (plane stress/plane strain).

Airy stress function in rectangular Cartesian coordinates[edit]

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

     \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}

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

     \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}

Airy stress function in polar coordinates[edit]

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

    \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) 

Something to think about ...

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

Stress equation of compatibility in 2-D[edit]

In the absence of body forces,

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


     \sigma_{11,11} + \sigma_{11,22} + \sigma_{22,11} + \sigma_{22,22} = 0 \,
  • Note that the stress field is independent of material properties in the absence of body forces (or homogeneous body forces).
  • Therefore, the plane strain and plane stress solutions are the same if the boundary conditions are expressed as traction BCS.

In terms of the Airy stress function

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


     \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


     \nabla^4\varphi = 0 \,
  • The stress function (\varphi) is biharmonic.
  • Any polynomial in x_1 and x_2 of degree less than four is biharmonic.
  • Stress fields that are derived from an Airy stress function which satisfies the biharmonic equation will satisfy equilibrium and correspond to compatible strain fields.

Some biharmonic Airy stress functions[edit]

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

    \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) \\


    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

Displacements in terms of scalar potentials[edit]

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}


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

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

    \sigma_{\alpha\beta} & = 2\mu\left[\varepsilon_{\alpha\beta} + 
        \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] 

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

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

Related Content[edit]

Introduction to Elasticity