If the stress components can be expressed in terms of a single scalar function ϕ(x,y), then the corresponding strain and displacement components can likewise be written in terms of the same function. Consequently, the number of unknowns in the two-dimensional elasticity problem is reduced to one. This approach, introduced by George Biddell Airy in 1862, makes use of a function ϕ(x,y) known as the Airy stress function.
Definition and the Equations of Equilibrium
The Airy stress function, denoted by \(\phi(x, y)\), is defined such that the stress components are derived from its second partial derivatives : \[ \sigma_x = \frac{\partial^2 \phi}{\partial y^2} \quad , \quad \sigma_y = \frac{\partial^2 \phi}{\partial x^2} \quad , \quad \tau_{xy} = -\frac{\partial^2 \phi}{\partial x \partial y} \] if there are no body forces.
Rationale Behind the Form of the Airy Stress Function
If we have a pair of functions \(f(x,y)\) and \(g(x,y)\) related by \[ \frac{\partial f}{\partial x} = \frac{\partial g}{\partial y}, \] then if we find a scalar function \(U(x,y)\) such that
\[ f = \frac{\partial U}{\partial y}, \qquad g = \frac{\partial U}{\partial x}. \] Because the mixed partial derivatives of \(U\) are equal (\(\dfrac{\partial ^2 U}{\partial x \partial y}=\dfrac{\partial ^2 U}{\partial y \partial x}\)), the above relation is automatically satisfied. This observation provides a useful analogy for formulating elasticity problems in terms of potential functions.
For a two-dimensional problem without body forces, the equations of equilibrium are
\[ \frac{\partial \sigma_x}{\partial x} + \frac{\partial \tau_{xy}}{\partial y} = 0, \qquad \frac{\partial \sigma_y}{\partial y} + \frac{\partial \tau_{xy}}{\partial x} = 0. \] From the first of these, we introduce a function \(A(x,y)\) such that
\[ \sigma_x = \frac{\partial A}{\partial y}, \qquad \tau_{xy} = -\frac{\partial A}{\partial x}, \]
and from the second equation, another function \(B(x,y)\) such that
\[ \sigma_y = \frac{\partial B}{\partial x}, \qquad \tau_{xy} = -\frac{\partial B}{\partial y}. \]
Because
\[ -\tau_{xy} = \frac{\partial A}{\partial x} = \frac{\partial B}{\partial y}, \]
the functions \(A\) and \(B\) can be related through another scalar function \(\phi(x,y)\) such that
\[ A = \frac{\partial \phi}{\partial y}, \qquad B = \frac{\partial \phi}{\partial x}. \]
Substituting these into the previous definitions gives the stress components directly in terms of \(\phi\):
\[ \sigma_x = \frac{\partial^2 \phi}{\partial y^2}, \qquad \sigma_y = \frac{\partial^2 \phi}{\partial x^2}, \qquad \tau_{xy} = -,\frac{\partial^2 \phi}{\partial x,\partial y}. \]
The scalar function ( (x,y) ), known as the Airy stress function, thus generates all the in-plane stress components through its second derivatives.
The brilliance of this definition is that the equilibrium equations are automatically satisfied for any function \(\phi\) that has continuous second derivatives.
Substituting these definitions into the first equilibrium equation demonstrates this:
\[ \frac{\partial}{\partial x} \left( \frac{\partial^2 \phi}{\partial y^2} \right) + \frac{\partial}{\partial y} \left( -\frac{\partial^2 \phi}{\partial x \partial y} \right) = \frac{\partial^3 \phi}{\partial x \partial y^2} - \frac{\partial^3 \phi}{\partial y \partial x \partial y} = 0 \]
The second equilibrium equation is satisfied in the same manner. This is a significant advantage: any stress field derived from an Airy function is guaranteed to be in equilibrium.
Compatibility, Constitutive Laws, and the Biharmonic Equation
While equilibrium is satisfied, the resulting strain field must also be compatible, meaning it must correspond to a continuous physical deformation. This physical requirement is captured by the strain compatibility equation:
\[ \frac{\partial^2 \epsilon_x}{\partial y^2} + \frac{\partial^2 \epsilon_y}{\partial x^2} = \frac{\partial^2 \gamma_{xy}}{\partial x \partial y} \]
To proceed, we must connect the strains to the stresses using the material’s constitutive law. For a linear elastic, isotropic material, this is Hooke’s Law. Here we must distinguish between two types of two dimensional analysis.
Derivation of the Biharmonic Equation
Let us derive the governing equation for \(\phi\) using the plane stress.1 We start by substituting Hooke’s Law into the strain compatibility equation: \[ \frac{\partial^2}{\partial y^2} \left[ \frac{1}{E} (\sigma_x - \nu \sigma_y) \right] + \frac{\partial^2}{\partial x^2} \left[ \frac{1}{E} (\sigma_y - \nu \sigma_x) \right] = \frac{\partial^2}{\partial x \partial y} \left[ \frac{\tau_{xy}}{G} \right] \] Multiplying by \(E\) and rearranging terms gives: \[ \left( \frac{\partial^2 \sigma_x}{\partial y^2} + \frac{\partial^2 \sigma_y}{\partial x^2} \right) - \nu \left( \frac{\partial^2 \sigma_y}{\partial y^2} + \frac{\partial^2 \sigma_x}{\partial x^2} \right) = \frac{E}{G} \frac{\partial^2 \tau_{xy}}{\partial x \partial y} = 2(1+\nu) \frac{\partial^2 \tau_{xy}}{\partial x \partial y} \] Now, substitute the Airy stress function definitions (\(\sigma_x = \dfrac{\partial^2 \phi}{\partial y^2}\), \(\sigma_y = \dfrac{\partial^2 \phi}{\partial x^2}\), \(\tau_{xy} = -\dfrac{\partial^2 \phi}{\partial x \partial y}\)): \[ \left( \frac{\partial^4 \phi}{\partial y^4} + \frac{\partial^4 \phi}{\partial x^4} \right) - \nu \left( \frac{\partial^4 \phi}{\partial y^2 \partial x^2} + \frac{\partial^4 \phi}{\partial x^2 \partial y^2} \right) = -2(1+\nu) \frac{\partial^4 \phi}{\partial x^2 \partial y^2} \] Simplifying the expression: \[ \frac{\partial^4 \phi}{\partial x^4} + \frac{\partial^4 \phi}{\partial y^4} - 2\nu \frac{\partial^4 \phi}{\partial x^2 \partial y^2} = -2 \frac{\partial^4 \phi}{\partial x^2 \partial y^2} - 2\nu \frac{\partial^4 \phi}{\partial x^2 \partial y^2} \] The terms involving \(\nu\) on both sides cancel out, leaving: \[ \frac{\partial^4 \phi}{\partial x^4} + \frac{\partial^4 \phi}{\partial y^4} = -2 \frac{\partial^4 \phi}{\partial x^2 \partial y^2} \] Rearranging this gives the celebrated biharmonic equation: \[ \frac{\partial^4 \phi}{\partial x^4} + 2 \frac{\partial^4 \phi}{\partial x^2 \partial y^2} + \frac{\partial^4 \phi}{\partial y^4} = 0 \] This equation can be written compactly as \(\nabla^4 \phi = 0\). Remarkably, if the same derivation is performed using the plane strain constitutive laws, the result is the exact same biharmonic equation.
Incorporating Body Forces
1. The Specific Case of Gravity
Consider a body where its own weight is the only acting body force. With the y axis pointing vertically upward, the body force components are \(B_x = 0\) and \(B_y = -\rho g\). To satisfy the modified equilibrium equations, we adjust the stress definitions: \[ \sigma_x = \frac{\partial^2 \phi}{\partial y^2} + \rho g y \quad , \quad \sigma_y = \frac{\partial^2 \phi}{\partial x^2} + \rho g y \quad , \quad \tau_{xy} = -\frac{\partial^2 \phi}{\partial x \partial y} \] When these definitions are used in the compatibility derivation, the additional \(\rho g y\) terms cancel out. The governing equation for \(\phi\) remains the homogeneous biharmonic equation, \(\nabla^4 \phi = 0\). The effect of gravity is simply added to the final stress calculation after \(\phi\) is found.
2. The General Case Using a Potential Function
Any conservative body force field can be described by a potential function \(V(x, y)\) where \(B_x = -\dfrac{\partial V}{\partial x}\) and \(B_y = -\dfrac{\partial V}{\partial y}\). The general stress definitions become: \[ \sigma_x = \frac{\partial^2 \phi}{\partial y^2} + V \quad , \quad \sigma_y = \frac{\partial^2 \phi}{\partial x^2} + V \] In this general case, the governing equation becomes the non homogeneous biharmonic equation: \[ \nabla^4 \phi = -(1-\nu) \nabla^2 V \quad \text{(for plane stress)} \] For plane strain, the leading constant is \(-(1-2\nu)\). This general equation confirms our result for gravity, since for \(V=\rho g y\), the Laplacian \(\nabla^2 V\) is zero.
The Polynomial Solution Method
A versatile strategy for solving the biharmonic equation is to assume the solution is a polynomial in \(x\) and \(y\). A general form of the solution is \[ \phi(x,y)=\sum_{m=0}^\infty \sum_{n=0}^\infty a_{mn} x^m y^n. \]
- Degree 0 and 1 (\(m+n \le 1\)): These terms produce zero stress and can be ignored.
- Degree 2 and 3 (\(m+n = 2\) or \(3\)): Any polynomial of degree three or less automatically satisfies the biharmonic equation because all its fourth derivatives are zero. Their coefficients are determined by the problem’s boundary conditions. A second degree polynomial produces a constant stress state, while a third degree polynomial produces a linearly varying stress field.
- Degree 4 and higher (\(m+n \ge 4\)): For these terms, the coefficients are no longer independent. They must be chosen to satisfy the biharmonic equation. For example, for a term \(\phi_4 = a_{40} x^4 + a_{22} x^2y^2 + a_{04} y^4\), the coefficients must satisfy the constraint \(3a_{40} + a_{22} + 3c_{04} = 0\).
Examples
1. Uniaxial Tension For a bar under uniform tensile stress \(\sigma_x = S\), we need \(\frac{\partial^2 \phi}{\partial y^2} = S\). Integrating twice gives: \[ \phi = \frac{S}{2} y^2 \] This second degree polynomial automatically satisfies \(\nabla^4 \phi = 0\).
2. Pure Bending of a Beam The stress in a beam under pure bending is \(\sigma_x = -\frac{My}{I}\). To achieve this linearly varying stress, \(\phi\) must be a cubic function. We propose \(\phi = C y^3\). This gives \(\sigma_x = 6Cy\). Comparing this to the known formula, we find the constant is \(C = -M/(6I)\), so the solution is: \[ \phi = -\frac{M}{6I} y^3 \]
- For the plane strain condition, the derivation is similar.↩︎