User:Strettle/Stress, Strain and Displacement Functions

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Stress, Strain and Displacement Functions in Linear Elasticity new article content ...

SUMMARY

Stress, strain and displacement (SSD) functions, so called because of their representation of the respective field quantities in three-dimensional linear elastic continua undergoing deformation, could be regarded as a little anachronistic in the digital age where numerical solutions to complex problems of elastic continua are routinely solved, often with just a passing understanding of the underlying mechanics. These functions are now regarded as vestiges of a century of effort that went into unlocking the equations of linear elasticity such that engineers and others could come to understand the stresses, strains and displacements that relate to particular boundary value problems of the real world. Much of the research into SSD functions for engineering purposes has now been scaled back with the advent of the emerging digital technologies. Of course mathematicians take a different view, examining the theory of linear elasticity in the broader and deeper context of the mechanics of solids which continues to the present day, in a notation (tensor indicial) that few non-mathematicians could hope to understand in any meaningful way. Although an excursion into the applications of SSD functions, in pragmatic terms, may be of limited value, they offer an illuminating pathway to an intellectual grasp of the underlying mechanics.

As mentioned, the vast majority of interested readers of this article (which includes students of engineering) would be unfamiliar with indicial notation, therefore, the multi-dimensional extension of vector notation and algebra, namely dyadic notation and algebra, whilst acknowledging its limitations, is employed for greater transparency for those readers.

INTRODUCTION

In their purest form SSD functions represent a class of functions independent of any constitutive (stress-strain) relation. Ultimately however they serve to resolve the complexity of the coupled and inter-dependent sets of equations of equilibrium and compatibility which emerge from the formulations of linear elasticity.

In determining a useful characterisation of the field quantities for three-dimensional homogeneous isotropic elastic continua it is desirable to seek an explicit function or functions in which the scalar components can be defined over the elastic region by a tractable independent set of governing equations for each of the scalar functions. That is, for the function or functions to be of any use, their ability to decouple the sets of interrelated equations of equilibrium or compatibility leading to one or more independent equations is essential. A simple example of this concept is that of the Galerkin vector. which is employed later to illustrate the point.

As a second theme to this article it is shown that the known SSD functions of linear elasticity can be unified into a common format using dyadic representation. The Beltrami-Schaeffer stress functions are recognised extensions to those of Maxwell and Morera for non-self-equilibrated stress fields in the linear three-dimensional elastic theory. What has generally not appeared in the literature is the notion of unification that these functions enable in respect of many other explicit functions such as Galerkin, Papkovich-Neuber and others. The Helmholtz characterization of the displacement vector for three-dimensional continua also emerges as a significant part of the unified formulation. The explicit functions can be shown to be solutions to the Beltrami-Schaeffer formulation and a conceptual hierarchical model is presented here. It is postulated that any derivative which may be described as an SSD function would likely conform to this model. Accordingly, this article is a practical attempt to describe the nature of SSD functions in the context of a unified theory within homogeneous isotropic linear elasticity.

Firstly the fundamental equations of elasticity which characterize the field quantities are required.

NOTATION AND THE BASIC EQUATIONS OF LINEAR ELASTICITY

The dyadic representation of stress and strain at a point in three-dimensional elastic space is given respectively


 * $$\tilde\tilde\Tau = \begin{bmatrix}

\text{ii}\sigma_{x} & \text{ij}\sigma_{xy} & \text{ik}\sigma_{xz}\\ \text{ji}\sigma_{yx} & \text{jj}\sigma_{y} & \text{jk}\sigma_{yz}\\ \text{ki}\sigma_{zx} & \text{kj}\sigma_{zy} & \text{kk}\sigma_{z} \end{bmatrix} $$          $$\tilde\tilde\epsilon = \begin{bmatrix} \text{ii}\epsilon_{x} & \text{ij}\epsilon_{xy} & \text{ik}\epsilon_{xz}\\ \text{ji}\epsilon_{yx} & \text{jj}\epsilon_{y} & \text{jk}\epsilon_{yz}\\ \text{ki}\epsilon_{zx} & \text{kj}\epsilon_{zy} & \text{kk}\epsilon_{z} \end{bmatrix} $$

where $$\text{i}$$, $$\text{j}$$ and $$\text{k}$$ are the usual unit vectors and $$\text{ii}$$, $$\text{ij} $$, .........$$\text{kk}$$ are unit dyads resulting from the extension into second-order formulation.

Note that we assume both tensors are symmetric. Intrinsic material properties relate these two tensors as shown in the next section.

The Constitutive Equations of Linear Elasticity

The relationship between the stress and strain tensors for an isotropic linear elastic material can be expressed in the dyadic form


 * $$\tilde\tilde\Tau = 2G \left [ \tilde\tilde\epsilon + \frac{\nu} {(1-2\nu)} \tilde\tilde\text{I} tr \tilde\tilde\epsilon \right ]$$

where


 * $$tr$$ is the trace or first invariant operator $$tr\tilde\tilde\epsilon = \epsilon_{x}+ \epsilon_{y}+   \epsilon_{z} $$
 * $$\tilde\tilde \text{I} $$ is the idemfactor or unit tensor $$\tilde\tilde\text{I} = \text{ii} + \text{jj} +   \text{kk} $$
 * $$\nu$$ in practical terms is Poisson's ratio, and
 * $$G$$ is Kirchoff's modulus (shear modulus)

The dyadic form above resolves to the six scalar equations


 * {| class="collapsible collapsed" width="30%" style="text-align:left"

!Constitutive equations
 * $$\sigma_x = 2G[\epsilon_x + \frac{\nu} {(1-2\nu)} tr \tilde\tilde\epsilon],$$             $$\sigma_{xy} = 2G\epsilon_{xy}$$
 * $$\sigma_y = 2G[\epsilon_y + \frac{\nu} {(1-2\nu)} tr \tilde\tilde\epsilon],$$             $$\sigma_{xz} = 2G\epsilon_{xz}$$
 * $$\sigma_z = 2G[\epsilon_z + \frac{\nu} {(1-2\nu)} tr \tilde\tilde\epsilon],$$             $$\sigma_{yz} = 2G\epsilon_{yz}$$
 * }
 * $$\sigma_z = 2G[\epsilon_z + \frac{\nu} {(1-2\nu)} tr \tilde\tilde\epsilon],$$             $$\sigma_{yz} = 2G\epsilon_{yz}$$
 * }
 * }

As an adjunct the alternative form of the constitutive equation above can be expressed


 * $$2G\tilde\tilde\epsilon = \left [ \tilde\tilde\Tau - \frac{\nu} {(1+\nu)} \tilde\tilde\text{I} tr \tilde\tilde\Tau \right ]$$

Both forms of the constitutive equation are required in the development of the SSD functions below.

The strain tensor can further be defined in terms of the displacement vector as follows


 * $$\tilde\tilde\epsilon = \frac{1} {2}(\nabla\tilde\rho+\tilde\rho\nabla) = \text{def } \tilde\rho$$

where


 * $$\tilde\rho$$ is the displacement vector ($$\tilde\rho = \text{i} u_x + \text{j} u_y + \text{k} u_z$$)
 * $$\nabla$$ is the nabla or del operator $$\left(\text{i} \frac {\partial {}} {\partial {x}} + \text{j} \frac {\partial {}} {\partial {y}} + \text{k} \frac {\partial {}} {\partial {z}} \right )$$, and
 * $$\text {def} $$ represents the deformator operator $$\frac{1} {2}[\nabla + \nabla]$$

In algebraic form the above strain-displacement relation is


 * {| class="collapsible collapsed" width="30%" style="text-align:left"

!Strain-displacement equations
 * $$\epsilon_x = \frac{\partial u_x} {\partial x} $$,        $$  \epsilon_{xy} = \frac{1} {2}(\frac{\partial u_x} {\partial y}+ \frac{\partial u_y} {\partial x})$$
 * $$\epsilon_y = \frac{\partial u_y} {\partial y} $$,        $$\epsilon_{xz} = \frac{1} {2}(\frac{\partial u_x} {\partial z}+ \frac{\partial u_z} {\partial x})$$
 * $$\epsilon_z = \frac{\partial u_z} {\partial z} $$,        $$\epsilon_{yz} = \frac{1} {2}(\frac{\partial u_y} {\partial z}+ \frac{\partial u_z} {\partial y})$$
 * }
 * $$\epsilon_z = \frac{\partial u_z} {\partial z} $$,        $$\epsilon_{yz} = \frac{1} {2}(\frac{\partial u_y} {\partial z}+ \frac{\partial u_z} {\partial y})$$
 * }
 * }

The immediate corollary of defining the strain tensor via a displacement vector is that the stress tensor can also be characterized in terms of the displacement vector giving the important stress-displacement relation. In doing so a first-order reduction is achieved in the number of scalar components from six to three (strains to displacements) in which a potential solution to the boundary value problems becomes more tractable. The relation becomes


 * $$\tilde\tilde\Tau = G \left [ \nabla\tilde\rho +\tilde\rho\nabla +\frac{2\nu}{(1-2\nu)} \nabla\cdot\tilde\rho\right ]$$

We defer further analysis of this for now.

The Equations of Equilibrium

Following the dyadic form the equations of equilibrium can be expressed as


 * $$\nabla\cdot\tilde\tilde\Tau + \tilde\text{b}= 0$$

where


 * $$\tilde\text{b}$$ is the body force vector $$\left (\tilde\text{b} = \text{i}b_x+\text{j}b_y+\text{k}b_z \right )$$

The above embodies the following scalar equilibrium equations in the three-dimensional domain


 * {| class="collapsible collapsed" width="30%" style="text-align:left"

!Equilibrium equations
 * $$\frac{\partial\sigma_x} {\partial x}+\frac{\partial\sigma_{xy}} {\partial y} +\frac{\partial\sigma_{xz}} {\partial z} +b_x= 0$$
 * $$\frac{\partial\sigma_{xy}} {\partial x}+\frac{\partial\sigma_{y}} {\partial y} +\frac{\partial\sigma_{yz}} {\partial z} +b_y= 0$$
 * $$\frac{\partial\sigma_{xz}} {\partial x}+\frac{\partial\sigma_{yz}} {\partial y} +\frac{\partial\sigma_{z}} {\partial z} +b_z= 0$$
 * }
 * $$\frac{\partial\sigma_{xz}} {\partial x}+\frac{\partial\sigma_{yz}} {\partial y} +\frac{\partial\sigma_{z}} {\partial z} +b_z= 0$$
 * }
 * }

Substitution of the constitutive relation for stress into the stress-equilibrium equation yields (homogeneity of G not assumed)


 * $$2\nabla\cdot G \left [ \tilde\tilde\epsilon + \frac{\nu} {(1-2\nu)} \tilde\tilde\text{I}tr\tilde\tilde\epsilon \right ]+\tilde\text{b}=0$$

In tensorial form we have the equivalent of three independent (but coupled) governing scalar equations embodying six scalar components of strain.

With further inclusion of the strain-displacement relation this leads to the important displacement-equilibrium equation


 * $$\nabla\cdot G \left [ \nabla\tilde\rho + \tilde\rho\nabla +\frac{2\nu}{(1-2\nu)} \tilde\tilde\text{I}\nabla\cdot\tilde\rho\right ]+\tilde\text{b}=0$$

For homogeneous isotropic linear elastic domains the above resolves to Navier's equation


 * $$\left [ \nabla^2 +\frac{1}{(1-2\nu)} \nabla\nabla\cdot\right ]\tilde\rho+\tilde\text{b°}=0$$

with the constant material property $$G$$ absorbed into $$\tilde\text{b°}$$

The Equations of Compatibility

The strain equations of compatibility are generally referred to as the compatibility conditions and may be written in the dyadic form


 * $$\nabla\times\tilde\tilde\epsilon\times\nabla = 0$$

In scalar form these may be written


 * {| class="collapsible collapsed" width="30%" style="text-align:left"

!Compatibility equations 2\frac{\partial^2\epsilon_{xy}}{\partial x \partial y} - \frac{\partial^2\epsilon_x}{\partial y^2} - \frac{\partial^2\epsilon_y}{\partial x^2} = 0$$ 2\frac{\partial^2\epsilon_{xz}}{\partial x \partial z} - \frac{\partial^2\epsilon_x}{\partial z^2} - \frac{\partial^2\epsilon_z}{\partial x^2} = 0$$ 2\frac{\partial^2\epsilon_{yz}}{\partial y \partial z} - \frac{\partial^2\epsilon_y}{\partial z^2} - \frac{\partial^2\epsilon_z}{\partial y^2} = 0$$ +\frac{\partial\epsilon_{xz}}{\partial y} +\frac{\partial\epsilon_{xy}}{\partial z}) = 0$$ -\frac{\partial\epsilon_{xz}}{\partial y} +\frac{\partial\epsilon_{xy}}{\partial z}) = 0$$ +\frac{\partial\epsilon_{xz}}{\partial y} -\frac{\partial\epsilon_{xy}}{\partial z}) = 0$$
 * $$\frac{\partial^2\epsilon_x}{\partial y \partial z} - \frac\partial {\partial x}(-\frac{\partial\epsilon_{yz}}{\partial x}
 * $$\frac{\partial^2\epsilon_x}{\partial y \partial z} - \frac\partial {\partial x}(-\frac{\partial\epsilon_{yz}}{\partial x}
 * $$\frac{\partial^2\epsilon_y}{\partial x \partial z} - \frac\partial {\partial y}(\frac{\partial\epsilon_{yz}}{\partial x}
 * $$\frac{\partial^2\epsilon_y}{\partial x \partial z} - \frac\partial {\partial y}(\frac{\partial\epsilon_{yz}}{\partial x}
 * $$\frac{\partial^2\epsilon_z}{\partial x \partial y} - \frac\partial {\partial z}(\frac{\partial\epsilon_{yz}}{\partial x}
 * $$\frac{\partial^2\epsilon_z}{\partial x \partial y} - \frac\partial {\partial z}(\frac{\partial\epsilon_{yz}}{\partial x}
 * }

The strain-compatibility equation is immediately satisfied by substitution of the strain-displacement relation


 * $$\nabla\times\tilde\tilde\epsilon\times\nabla = \nabla\times\frac{1} {2}\left(\nabla\tilde\rho+\tilde\rho\nabla\right)\times\nabla = 0$$

The curl of the grad of the displacement vector immediately resolves the above by definition. This brings us to the notion that the displacement vector as defined is in fact a special strain function which, together with a deformator operator, identically satisfies the compatibility conditions. The displacement vector is thus one of the few functions of elastic theory having a physical as well as mathematical property. This is a precursor to still higher-order functions. For example, the substitution of the strain-displacement relation into the formulation enables a reduction in the number of unknown variables from six (strains) to three (displacements). By extension, higher-order functions can further reduce the number of unknowns from three to one (stress or strain function).

If the constitutive relation is now introduced into the strain-compatibility equation without assuming homogeneity of the elastic domain we get the stress-compatibility equation


 * $$\nabla\times\frac{1}{2G}\left [\tilde\tilde\Tau - \frac{\nu} {(1+\nu)}\tilde\tilde\text{I} tr \tilde\tilde\Tau \right ]\times\nabla = 0$$

Noting that the stress-compatibilty equation is an identity then it's trace must also be zero, that is (ignoring constants)


 * $$tr\nabla\times\left [\tilde\tilde\Tau - \frac{\nu} {(1+\nu)}\tilde\tilde\text{I} tr \tilde\tilde\Tau \right ]\times\nabla = \nabla\cdot\left [\tilde\tilde\Tau - \frac{(1-\nu)} {(1+\nu)}\tilde\tilde\text{I} tr \tilde\tilde\Tau \right ]\cdot\nabla = 0$$

Since $$\nabla\cdot\tilde\tilde\Tau = 0 $$, this gives the interesting result that the trace of the stress tensor is harmonic, that is $$\nabla^2 tr\tilde\tilde\Tau = 0$$.

Employing the general identity (work it out long hand for proof!)


 * $$\nabla\times\tilde\tilde\Tau\times\nabla = \nabla^2\tilde\tilde\Tau^T - \nabla(\tilde\tilde\Tau\cdot\nabla)-(\nabla\cdot\tilde\tilde\Tau)\nabla + \tilde\tilde\text{I}\nabla\cdot\tilde\tilde\Tau\cdot\nabla + \nabla\nabla tr \tilde\tilde\Tau - \tilde\tilde\text{I}\nabla^2 tr \tilde\tilde\Tau$$

then, again since $$\nabla\cdot\tilde\tilde\Tau = 0 $$, and assuming $$\tilde\tilde\Tau$$ is symmetric, the stress-compatibility equation, when homogeneity is imposed, reduces to


 * $$\nabla^2(\tilde\tilde\Tau)+ \frac{1} {(1+\nu)}\nabla\nabla tr \tilde\tilde\Tau + 2\text{def}(\tilde\text{b})-\frac{2\nu} {(1+\nu)} \tilde\tilde\text{I}\nabla\cdot\tilde\text{b} = 0$$

In this form the equation is known as the Beltrami-Michell compatibility equation. See for example,Nadeau

For completeness the tensorial form of the Beltrami-Michell equation resolves in scalar form to


 * {| class="collapsible collapsed" width="30%" style="text-align:left"

!Beltrami-Michell equations
 * $$\nabla^2\sigma_{x}+ \frac{1} {(1+\nu)} \frac{\partial^2 }{\partial {x^2}}tr\tilde\tilde\Tau + 2\frac {\partial{b_x}}{\partial{x}} -\frac{2\nu} {(1+\nu)} \nabla\cdot\tilde\text{b} = 0$$
 * $$\nabla^2\sigma_{y}+ \frac{1} {(1+\nu)} \frac{\partial^2 }{\partial {y^2}}tr\tilde\tilde\Tau + 2\frac {\partial{b_y}}{\partial{y}} -\frac{2\nu} {(1+\nu)} \nabla\cdot\tilde\text{b} = 0$$
 * $$\nabla^2\sigma_{z}+ \frac{1} {(1+\nu)} \frac{\partial^2 }{\partial {z^2}}tr\tilde\tilde\Tau + 2\frac {\partial{b_z}}{\partial{z}} -\frac{2\nu} {(1+\nu)} \nabla\cdot\tilde\text{b} = 0$$
 * $$\nabla^2\sigma_{xy}+ \frac{1} {(1+\nu)} \frac{\partial^2 }{\partial {x}\partial{y}}tr\tilde\tilde\Tau + \frac {\partial{b_y}}{\partial{x}} + \frac {\partial{b_x}}{\partial{y}} = 0$$
 * $$\nabla^2\sigma_{xz}+ \frac{1} {(1+\nu)} \frac{\partial^2 }{\partial {x}\partial{z}}tr\tilde\tilde\Tau + \frac {\partial{b_z}}{\partial{x}} + \frac {\partial{b_x}}{\partial{z}} = 0$$
 * $$\nabla^2\sigma_{yz}+ \frac{1} {(1+\nu)} \frac{\partial^2 }{\partial {y}\partial{z}}tr\tilde\tilde\Tau + \frac {\partial{b_z}}{\partial{y}} + \frac {\partial{b_y}}{\partial{z}} = 0$$
 * }
 * $$\nabla^2\sigma_{xy}+ \frac{1} {(1+\nu)} \frac{\partial^2 }{\partial {x}\partial{y}}tr\tilde\tilde\Tau + \frac {\partial{b_y}}{\partial{x}} + \frac {\partial{b_x}}{\partial{y}} = 0$$
 * $$\nabla^2\sigma_{xz}+ \frac{1} {(1+\nu)} \frac{\partial^2 }{\partial {x}\partial{z}}tr\tilde\tilde\Tau + \frac {\partial{b_z}}{\partial{x}} + \frac {\partial{b_x}}{\partial{z}} = 0$$
 * $$\nabla^2\sigma_{yz}+ \frac{1} {(1+\nu)} \frac{\partial^2 }{\partial {y}\partial{z}}tr\tilde\tilde\Tau + \frac {\partial{b_z}}{\partial{y}} + \frac {\partial{b_y}}{\partial{z}} = 0$$
 * }
 * $$\nabla^2\sigma_{yz}+ \frac{1} {(1+\nu)} \frac{\partial^2 }{\partial {y}\partial{z}}tr\tilde\tilde\Tau + \frac {\partial{b_z}}{\partial{y}} + \frac {\partial{b_y}}{\partial{z}} = 0$$
 * }

In scalar form it is clear that managing these equations is impractical and confirms the need for the tensorial or dyadic form to gain any traction in finding solutions to them. Also, at this juncture, it is assumed that the inclusion of the body force vector leads to unnecessary complexity in furthering the underlying theory of SSD functions and it is ignored in the subsequent formulations.

Coupling and Dependency in the Equilibrium and Compatibility Equations

Before delving further into possible solutions to Navier's displacement-equilibrium equations or the Beltrami-Michell stress-compatibility equations it is worth visiting the notions of coupling and dependency within the equations which lead us to require higher-order functions to enable us to find tractability. That is, coupling and dependency within the equations force us to seek higher-order functions to enable a resolution of boundary value problem formulations.

A distinction is made between coupling and dependency. For example the six scalar components of the Beltami-Michell equation which define the six components of stress over the elastic domain exhibit both coupling and dependency. By contrast there are only three scalar components of Navier's displacement-equilibrium equations. These equations exhibit coupling but not dependency.

Coupling between equations is demonstrated when one scalar equation contains more than one scalar component of stress or displacement. This results in the 'simultaneous' nature of the equations. Naviers equations though coupled are independent since each represents a unique condition to be satisfied. With the Beltami-Michell equations from any two a third can be derived. This is the result of dependency. No such manipulation can be applied to Navier's equations.

For the Beltrami-Michell equations there are effectively the equivalent of three independent equations only to define the six scalar components of stress. The addition of the stress-equilibrium equations to these gives us a total of six coupled but mutually independent equations which are necessary for the complete description of the six components of stress. However if one is to find a solution to the elastic boundary value problem based on for example the Beltrami-Michell equations only (a sufficient condition) then the equivalent of three scalar equations only, together with the six components of stress, requires that higher-order functions be introduced to reduce the number of independent scalar variables to three. This means that the stress components must be characterized through higher order functions. This is the principle justification for the introduction of the SSD functions as a means of obtaining the general solutions to the basic equations of elasticity.

The simplest example of this concept is briefly examined below in relation to the so-called Galerkin vector which is able to decouple Navier's displacement-equilibrium equations.

Westergaard after Galerkin showed that the displacement vector could be represented as follows


 * $$2G\tilde\rho = \left [(2-2\nu)\nabla^2 + \nabla\nabla\cdot \right ]\tilde\text{F}$$

where $$\tilde\text{F}$$ is an explicit displacement vector function otherwise know as the Galerkin vector $$\tilde\text{F} = (\text{i} F_x +\text{j} F_y + \text{k} F_z)$$

Substituting the above into Navier's equation $$\left [\nabla^2 + \frac {1}{1-2\nu}\nabla\nabla\cdot\right ]\tilde\rho = 0$$ yields a tractable set of independent equations in biharmonic form for each of the three scalar components of $$\tilde\text{F}$$, namely


 * $$\nabla^2\nabla^2\tilde\text{F} = 0$$

Presto! the equilibrium equations are decoupled. The trade-off for this advantage is the requirement to operate with fourth-order functions and to formulate any boundary conditions in terms of the Galerkin functions. However the point is made and we now examine these and other types of higher-order functions in more detail.

THE BELTRAMI-SCHAEFFER FUNCTIONS

The Beltrami-Schaeffer functions are an extension to the generalized form of the complementary sets of Maxwell and Morera functions. The latter are defined in tensorial form as
 * $$\tilde\tilde\Phi_{Max} = \begin{bmatrix}

\text{ii}\phi_{x}\ & \text{ij}\ & \text{ik}\\ \text{ji}\ & \text{jj}\phi_{y} & \text{jk}\\ \text{ki}\ & \text{kj}\ & \text{kk}\phi_{z} \end{bmatrix} $$         $$\tilde\tilde\Phi_{Mor} = \begin{bmatrix} \text{ii}\ & \text{ij}\phi_{xy} & \text{ik}\phi_{zx}\\ \text{ji}\phi_{xy} & \text{jj}\ & \text{jk}\phi_{yz}\\ \text{ki}\phi_{zx} & \text{kj}\phi_{yz} & \text{kk} \end{bmatrix} $$

The stress tensor may be characterized respectively by Maxwell's and Morera's functions respectively as


 * $$\tilde\tilde\Tau = \nabla\times\tilde\tilde\Phi_{Max}\times\nabla$$  and   $$\tilde\tilde\Tau = \nabla\times\tilde\tilde\Phi_{Mor}\times\nabla$$

and in scalar form


 * {| class="collapsible collapsed" width="30%" style="text-align:left"

!Maxwell functions - \frac{\partial^2\phi_z}{\partial y^2} $$,     $$\sigma_{xy} = \frac{\partial^2\phi_z}{\partial x \partial y} $$ - \frac{\partial^2\phi_z}{\partial x^2} $$,     $$\sigma_{xz} = \frac{\partial^2\phi_y}{\partial x \partial z} $$ - \frac{\partial^2\phi_y}{\partial x^2} $$,     $$\sigma_{yz} = \frac{\partial^2\phi_x}{\partial y \partial z} $$
 * $$\sigma_x = -\frac{\partial^2\phi_y}{\partial z^2}
 * $$\sigma_x = -\frac{\partial^2\phi_y}{\partial z^2}
 * $$\sigma_y = -\frac{\partial^2\phi_x}{\partial z^2}
 * $$\sigma_y = -\frac{\partial^2\phi_x}{\partial z^2}
 * $$\sigma_z = -\frac{\partial^2\phi_x}{\partial y^2}
 * $$\sigma_z = -\frac{\partial^2\phi_x}{\partial y^2}
 * }
 * }


 * {| class="collapsible collapsed" width="40%" style="text-align:left"

!Morera functions +\frac{\partial\phi_{zx}}{\partial y} -\frac{\partial\phi_{xy}}{\partial z})$$ -\frac{\partial\phi_{zx}}{\partial y} +\frac{\partial\phi_{xy}}{\partial z})$$ +\frac{\partial\phi_{zx}}{\partial y} +\frac{\partial\phi_{xy}}{\partial z})$$
 * $$\sigma_x = 2\frac{\partial^2\phi_{yz}}{\partial y \partial z}$$,     $$\sigma_{xy} = - \frac\partial {\partial z}(\frac{\partial\phi_{yz}}{\partial x}
 * $$\sigma_x = 2\frac{\partial^2\phi_{yz}}{\partial y \partial z}$$,     $$\sigma_{xy} = - \frac\partial {\partial z}(\frac{\partial\phi_{yz}}{\partial x}
 * $$\sigma_y = 2\frac{\partial^2\phi_{zx}}{\partial x \partial z}$$,     $$\sigma_{xz} = - \frac\partial {\partial y}(\frac{\partial\phi_{yz}}{\partial x}
 * $$\sigma_y = 2\frac{\partial^2\phi_{zx}}{\partial x \partial z}$$,     $$\sigma_{xz} = - \frac\partial {\partial y}(\frac{\partial\phi_{yz}}{\partial x}
 * $$\sigma_z = 2\frac{\partial^2\phi_{xy}}{\partial x \partial y}$$,     $$\sigma_{yz} = - \frac\partial {\partial x}(-\frac{\partial\phi_{yz}}{\partial x}
 * $$\sigma_z = 2\frac{\partial^2\phi_{xy}}{\partial x \partial y}$$,     $$\sigma_{yz} = - \frac\partial {\partial x}(-\frac{\partial\phi_{yz}}{\partial x}
 * }

Beltrami's generalization of the stress tensor in terms of these functions is


 * $$\tilde\tilde \Tau =\nabla \times \tilde\tilde \Phi \times \nabla$$

where $$\tilde \tilde\Phi $$ is the combined Maxwell-Morera stress function tensor. This expression serves as a complete description of $$\tilde \tilde\Tau $$ for totally self-equilibrated stress fields.

The embodied scalar equations become


 * {| class="collapsible collapsed" width="30%" style="text-align:left"

!Beltrami's generalisation - \frac{\partial^2\phi_z}{\partial y^2} $$ - \frac{\partial^2\phi_z}{\partial x^2} $$ - \frac{\partial^2\phi_y}{\partial x^2} $$ - \frac\partial {\partial z}(\frac{\partial\phi_{yz}}{\partial x} +\frac{\partial\phi_{xz}}{\partial y} -\frac{\partial\phi_{xy}}{\partial z})$$ - \frac\partial {\partial y}(\frac{\partial\phi_{yz}}{\partial x} -\frac{\partial\phi_{xz}}{\partial y} +\frac{\partial\phi_{xy}}{\partial z})$$ - \frac\partial {\partial x}(-\frac{\partial\phi_{yz}}{\partial x} +\frac{\partial\phi_{xz}}{\partial y} +\frac{\partial\phi_{xy}}{\partial z})$$
 * $$\sigma_x = 2\frac{\partial^2\phi_{yz}}{\partial y \partial z} - \frac{\partial^2\phi_y}{\partial z^2}
 * $$\sigma_x = 2\frac{\partial^2\phi_{yz}}{\partial y \partial z} - \frac{\partial^2\phi_y}{\partial z^2}
 * $$\sigma_y = 2\frac{\partial^2\phi_{xz}}{\partial x \partial z} - \frac{\partial^2\phi_x}{\partial z^2}
 * $$\sigma_y = 2\frac{\partial^2\phi_{xz}}{\partial x \partial z} - \frac{\partial^2\phi_x}{\partial z^2}
 * $$\sigma_z = 2\frac{\partial^2\phi_{xy}}{\partial x \partial y} - \frac{\partial^2\phi_x}{\partial y^2}
 * $$\sigma_z = 2\frac{\partial^2\phi_{xy}}{\partial x \partial y} - \frac{\partial^2\phi_x}{\partial y^2}
 * $$\sigma_{xy} = \frac{\partial^2\phi_z}{\partial x \partial y}
 * $$\sigma_{xy} = \frac{\partial^2\phi_z}{\partial x \partial y}
 * $$\sigma_{xz} = \frac{\partial^2\phi_y}{\partial x \partial z}
 * $$\sigma_{xz} = \frac{\partial^2\phi_y}{\partial x \partial z}
 * $$\sigma_{yz} = \frac{\partial^2\phi_x}{\partial y \partial z}
 * $$\sigma_{yz} = \frac{\partial^2\phi_x}{\partial y \partial z}
 * }

Schaeffer introduced a modification to the functions in the form of an additional harmonic vector and arrived at what has become the Beltrami-Schaeffer functions. These may be written


 * $$\tilde\tilde\Tau =\nabla\times\tilde\tilde\Phi\times\nabla + \nabla\tilde\text{h} + \tilde\text{h}\nabla - \tilde\tilde\text{I}\nabla\cdot\tilde\text{h}$$

where $$\tilde\text{h} $$ is a vector which accounts for the non-self-equilibrated component of the stress tensor and is harmonic since $$\nabla\cdot\tilde\tilde\Tau = \nabla^2\tilde\text{h} =0$$.

At this point we introduce a special vectorial operator in shorthand notation as follows $$\text{ℝ} = \left[\nabla + \nabla - \tilde\tilde\text{I}\nabla\cdot\right]$$. Note that $$\nabla\cdot\text{ℝ} = \nabla^2$$.


 * $$\tilde \tilde\Tau =\nabla \times \tilde\tilde\Phi \times \nabla + \text{ℝ}\left(\tilde\text{h}\right)$$

The expanded form of the above may be written


 * $$\tilde\tilde\Tau =\nabla^2\tilde\tilde\Phi^T -\nabla\nabla tr \tilde\tilde\Phi - \text{ℝ}\left [\nabla\cdot\left(\tilde\tilde\Phi - \tilde\tilde \text{I} tr \tilde\tilde\Phi \right ) - \tilde\text{h}\right ] $$

and the trace of $$\tilde\tilde\Tau$$ becomes


 * $$tr\tilde\tilde\Tau = \nabla\cdot\left [\nabla\cdot\left(\tilde\tilde\Phi - \tilde\tilde \text{I} tr \tilde\tilde\Phi \right )- \tilde\text{h}\right ]$$

Noting that the trace of the constitutive equation for strain, $$\left(2G\tilde\tilde\epsilon = \tilde\tilde\Tau - \frac{\nu} {1+\nu}\tilde\tilde\text{I} tr \tilde\tilde\Tau \right)$$ is


 * $$G tr\tilde\tilde\epsilon = \frac{1-2\nu} {2+2\nu} tr\tilde\tilde\Tau$$

then through the equivalence $$tr\tilde\tilde\epsilon = \nabla\cdot\tilde\rho$$


 * $$G\nabla\cdot\tilde\rho = \frac {1-2\nu}{2+2\nu}\nabla\cdot\left[\nabla\cdot\left(\tilde\tilde\Phi - \tilde\tilde \text{I} tr \tilde\tilde\Phi \right ) - \tilde\text{h}\right]$$

This expression lends itself to integration which finally defines the displacement vector in terms of the Beltrami-Schaeffer functions as follows


 * $$G\tilde\rho = \frac {1-2\nu}{2+2\nu}\left[\nabla\cdot\left(\tilde\tilde\Phi - \tilde\tilde \text{I} tr \tilde\tilde\Phi \right ) - \tilde\text{h} + \nabla\times\tilde\text{c}\right] $$

where the vector constant of integration, $$\nabla\times\tilde\text{c}$$, has been independently maintained (not absorbed into $$\tilde\text{h}$$) and scaled to better fit the formulation. This is important. As will be demonstrated below, the vector of integration is maintained to ensure that the explicit functions of Galerkin, Papkovich-Neuber and others conform to Beltrami-Schaeffer format. This is particularly so with the Helmholtz functions where $$\nabla\times\tilde\text{c}$$ directly becomes one of the functions.

To obtain a complete formulation of the field quantities through these functions it remains to deduce the governing tensorial equation for $$\tilde\tilde\Phi$$.

Employing the stress-displacement equation


 * $$\tilde\tilde\Tau = G \left [ \nabla\tilde\rho +\tilde\rho\nabla +\frac{2\nu}{(1-2\nu)} \tilde\tilde\text{I}\nabla\cdot\tilde\rho\right ]$$

substitution for $$\tilde\rho$$ above yields
 * $$\tilde\tilde\Tau = \frac {1-2\nu}{2+2\nu} \left [ \text{ℝ}\left [\nabla\cdot\left(\tilde\tilde\Phi - \tilde\tilde \text{I} tr \tilde\tilde\Phi \right ) - \tilde\text{h} + \nabla\times\tilde\text{c}\right ] + \frac {1}{1-2\nu}\tilde\tilde\text{I} \nabla\cdot\left [\nabla\cdot\left(\tilde\tilde\Phi - \tilde\tilde \text{I} tr \tilde\tilde\Phi \right ) - \tilde\text{h} + \nabla\times\tilde\text{c}\right ]\right ] $$

If this is now equated to the initial format for the stress tensor, namely


 * $$\tilde\tilde\Tau =\nabla^2\tilde\tilde\Phi -\nabla\nabla tr \tilde\tilde\Phi - \text{ℝ}\left [\nabla\cdot\left(\tilde\tilde\Phi - \tilde\tilde \text{I} tr \tilde\tilde\Phi \right ) - \tilde\text{h}\right ] $$

we can generate the underlying tensorial equation which embodies $$\tilde\tilde\Phi$$ and $$\tilde\text{h}$$ as follows


 * $$3\text{ℝ}\left [\nabla\cdot\left(\tilde\tilde\Phi - \tilde\tilde \text{I} tr \tilde\tilde\Phi \right ) - \tilde\text{h} + \nabla\times\tilde\text{c}\right ] + \tilde\tilde\text{I} \nabla\cdot\left [\nabla\cdot\left(\tilde\tilde\Phi - \tilde\tilde \text{I} tr\tilde \tilde\Phi \right ) - \tilde\text{h} \right ] - (2+2\nu)\left(\nabla^2\tilde\tilde\Phi -\nabla\nabla tr \tilde\tilde\Phi + \text{ℝ}\nabla\times\tilde\text{c}\right)= 0 $$

A derived solution to this equation for the boundary value problems of linear elasticity requires great insight into the algebra of tensor equations, nevertheless it will be shown that the functions of Galerkin, Papkovich-Neuber and others are clearly explicit solutions which, upon substitution, yield the requisite identities. In this respect the above equation is the generalized form which embodies the characteristics of all equations defining the known functions.

For completeness an independent equation for $$\tilde\tilde\Phi$$ can be generated by twice applying the divergence operator to the above, yielding (remembering that $$\tilde\text{h}$$ is harmonic)

$$1^{st}$$ divergence


 * $$(1-2\nu)\nabla^2\left[\nabla\cdot\left(\tilde\tilde\Phi - \tilde\tilde\text{I}tr \tilde\tilde\Phi\right) + \nabla\times\tilde\text{c}\right ] + \nabla\nabla\cdot\left [\nabla\cdot\left(\tilde\tilde\Phi - \tilde\tilde\text{I} tr\tilde\tilde\Phi\right ) - \tilde\text{h}\right ]= 0 $$

$$2^{nd}$$ divergence


 * $$\nabla^2\nabla\cdot\left (\tilde\tilde\Phi - \tilde\tilde\text{I}tr\tilde\tilde\Phi\right)\cdot\nabla = 0$$

Accordingly, $$\tilde\tilde\Phi$$ is neither harmonic nor biharmonic, rendering the function itself quite useless in practical terms. The underlying tensorial equation cannot be manipulated to obtain a simple governing equation for $$\tilde\tilde\Phi$$ as one can for $$\tilde\text{h}$$. What $$\tilde\tilde\Phi$$ does do is represent, with $$\tilde\text{h}$$, a pair of functions that encode all other SSD functions. It is worth noting that the generality of the Beltrami-Schaeffer functions  is apparent with the displacement vector being characterized by the divergence of a tensor, $$\nabla\cdot\tilde\tilde\Phi$$, a vector $$\tilde\text {h}$$, the curl of a vector, $$\nabla\times\tilde\text{c}$$ and the grad of a scalar $$\nabla tr \tilde\tilde\Phi$$. That is, all classes of variable up to and including second-order tensors are represented in the displacement vector. This may be a little spurious but it allows various classes of variables to be brought in as explicit solutions to the Beltrami-Schaeffer functions.


 * {| class="collapsible collapsed" width="100%" style="text-align:left"

!Airy's stress function
 * The well-known Airy stress function, a two-dimensional degenerate form of the Maxwell stress functions, used for characterizing the stresses in the two-dimensional linear elastic theory is briefly examined below for interest.
 * The well-known Airy stress function, a two-dimensional degenerate form of the Maxwell stress functions, used for characterizing the stresses in the two-dimensional linear elastic theory is briefly examined below for interest.
 * The well-known Airy stress function, a two-dimensional degenerate form of the Maxwell stress functions, used for characterizing the stresses in the two-dimensional linear elastic theory is briefly examined below for interest.

Airy's stress function is simply the $$\text{kk}$$ component of the Maxwell stress tensor, $$\phi_z$$. In which case


 * $$\tilde\tilde\Tau = \nabla\times\text{kk}\phi_z\times\nabla = \left (\Delta\Delta - \tilde\tilde\text{I}\Delta^2\right )\phi_z$$

where


 * $$\tilde\tilde\Tau$$ in this instance defines only the stresses in the x,y (or r, $$\theta$$) plane


 * $$\tilde\tilde\text{I}$$ in this instance is the two-dimensional idemfactor $$(\text{ii} + \text{jj})$$, and


 * $$\Delta$$ is the two-dimensional nabla operator, namely $$\left(\text{i} \frac {\partial {}} {\partial {x}} + \text{j} \frac {\partial {}} {\partial {y}} \right )$$

Expansion of the above leads to the scalar equations for stress as follows


 * $$\sigma_x = -\frac{\partial^2\phi_z}{\partial y^2}

$$     $$\sigma_y = -\frac{\partial^2\phi_z}{\partial x^2} $$     $$\sigma_{xy} = \sigma_{yx} = \frac{\partial^2\phi_z}{\partial x \partial y} $$

Unlike it's three-dimensional Maxwell counterpart the Airy stress function is biharmonic. The restriction of $$\tilde\tilde\Phi$$ to $$(\text{kk}\phi_z)$$ renders the fourth-order governing equation for $$\tilde\tilde\Phi$$ to a biharmonic form as follows


 * $$\nabla^2\nabla\cdot\left (\tilde\tilde\Phi - \tilde\tilde\text{I}tr\tilde\tilde\Phi\right)\cdot\nabla = 0$$


 * $$\nabla^2\nabla\cdot\left (\text{kk}\phi_z - \tilde\tilde\text{I}tr\text{kk}\phi_z\right)\cdot\nabla => -\Delta^2\Delta^2\phi_z = 0$$

Note that the $$\nabla^2$$ reverts to $$\Delta^2$$ as the Airy stress function describes a planar state of stress which is invariant with respect to the z-coordinate.

This concludes the review of Airy's stress function.
 * }


 * {| class="collapsible collapsed" width="100%" style="text-align:left"

!Alternative Beltrami-Schaeffer stress tensors
 * As an aside an interesting development ensues from a little tensorial algebra, giving rise to alternative interpretations of the Beltrami-Schaeffer stress tensor. Beginning with a fresh representation of the stress tensor by an aribitrary admissible tensor function
 * As an aside an interesting development ensues from a little tensorial algebra, giving rise to alternative interpretations of the Beltrami-Schaeffer stress tensor. Beginning with a fresh representation of the stress tensor by an aribitrary admissible tensor function
 * As an aside an interesting development ensues from a little tensorial algebra, giving rise to alternative interpretations of the Beltrami-Schaeffer stress tensor. Beginning with a fresh representation of the stress tensor by an aribitrary admissible tensor function


 * $$\tilde\tilde\Tau = \nabla^2\tilde\tilde\Alpha $$

and employing the identity $$\nabla\times\tilde\tilde\Alpha\times\nabla = \nabla^2\tilde\tilde\Alpha^T - \nabla(\tilde\tilde\Alpha\cdot\nabla)-(\nabla\cdot\tilde\tilde\Alpha)\nabla + \tilde\tilde\text{I}\nabla\cdot\tilde\tilde\Alpha\cdot\nabla + \nabla\nabla tr \tilde\tilde\Alpha - \tilde\tilde\text{I}\nabla^2 tr \tilde\tilde\Alpha$$

an expression can be found following Little for the symmetric stress tensor as


 * $$\tilde\tilde\Tau =\nabla \times \left (\tilde\tilde\Alpha - \tilde\tilde\text{I}tr\tilde\tilde\Alpha\right ) \times \nabla + 2\text{ def }\left(\nabla\cdot \tilde\tilde\Alpha\right) -  \tilde\tilde\text{I}\nabla\cdot \left (\tilde\tilde\Alpha\right )\cdot\nabla $$

and with the substitutions $$\left(\tilde\tilde\Phi = \tilde\tilde\Alpha - \tilde\tilde\text{I} tr \tilde\tilde\Alpha\right )$$ and $$\left(\tilde\text{h} = \nabla\cdot\tilde\tilde\Alpha\right )$$ we confirm an instance in which the Beltrami-Schaeffer functions can encode another function.

Further as an adjunct of the above (after transposing) $$\tilde\tilde\Alpha = \tilde\tilde\Phi - \frac {1}{2}\tilde\tilde\text{I} tr \tilde\tilde\Phi$$


 * $$\tilde\tilde\Tau = \nabla^2\left (\tilde\tilde\Phi - \frac {1} {2}\tilde\tilde\text{I}tr\tilde\tilde\Phi\right )$$

which expanding according to the curl curl identity becomes


 * $$\tilde\tilde\Tau =\nabla \times \tilde\tilde\Phi \times \nabla + \text{ℝ}\left[\nabla\cdot\left( \tilde\tilde\Phi - \frac {1} {2}\tilde\tilde\text{I}tr\tilde\tilde\Phi\right ) \right]$$

and whilst it may be a little circular it is clear that if we assume that $$\tilde\text{h} = \nabla\cdot \left(\tilde\tilde\Phi - \frac {1} {2}\tilde\tilde\text{I}tr\tilde\tilde\Phi\right )$$ then the Beltrami-Schaeffer format arises once more.

It has been mentioned that $$\tilde\tilde\Phi$$ is not a simple function and in this particular instance enforcing the equilibrium condition yields


 * $$\nabla\cdot\tilde\tilde\Tau = \nabla^2\nabla\cdot\left (\tilde\tilde\Phi - \frac {1} {2}\tilde\tilde\text{I}tr\tilde\tilde\Phi\right ) = 0$$

Again the Beltrami-Schaeffer functions are intractably neither harmonic nor biharmonic which the above expression indicates.
 * }

EXPLICIT REPRESENTATIONS OF THE BELTRAMI-SCHAEFFER FUNCTIONS

Helmholtz Functions

The application of the Helmholtz theorem resolves the displacement vector into an irrotational and a solenoidal component for a displacement field which vanishes at infinity. That is


 * $$\tilde\rho = \tilde\rho_{irrotational} + \tilde\rho_{solenoidal}$$

Accordingly, the displacement vector may be given by


 * $$\tilde\rho = \nabla\chi^* + \nabla\times\tilde\eta^* $$

where


 * $$\nabla\chi^* $$ is the irrotational component of the displacement vector ($$\chi^*$$, a scalar function), and


 * $$\nabla\times\tilde\eta^*$$ is the solenoidal component ($$\tilde\eta^*$$, a vector function, the divergence of which is zero, $$\nabla \cdot\tilde\eta^* = 0$$)

The above functions come with a superscript since in the algebra below it will be convenient to scale them with constant material properties such that the above becomes


 * $$2G\tilde\rho = (1-2\nu)\nabla\chi + (2-2\nu)\nabla\times\tilde\eta$$

An elegant proof of Helmholtz theorem for homogeneous linear elastic continua may be demonstrated by considering the stress tensor in terms of an arbitrary admissible tensor function. Let


 * $$\tilde\tilde\Tau = (1+\nu)\nabla^2\tilde\tilde\Alpha$$

noting that the coefficient in $$(1+\nu)\nabla^2\tilde\tilde\Alpha$$ is chosen in order to obtain the prescribed Helmholtz representation above for $$\tilde\rho$$. The trace of $$\tilde\tilde\Tau$$ becomes


 * $$tr\tilde\tilde\Tau = (1+\nu)\nabla^2 tr\tilde\tilde\Alpha$$

Continuing, from the constitutive relation for strain


 * $$2G\tilde\tilde\epsilon = G(\nabla\tilde\rho+\tilde\rho\nabla) = \tilde\tilde\Tau - \frac{\nu} {(1+\nu)} \tilde\tilde\text{I} tr \tilde\tilde\Tau $$

we get the useful expression


 * $$G(\nabla\tilde\rho + \tilde\rho\nabla) = (1+\nu)\nabla^2 \left[\tilde\tilde\Alpha - \frac {\nu} {(1+\nu}\tilde\tilde\text{I}tr\tilde\tilde\Alpha \right]$$

the trace of which becomes


 * $$2G\nabla\cdot\tilde\rho = (1-2\nu)\nabla^2 tr\tilde\tilde\Alpha$$

and importantly from which integration yields


 * $$2G\tilde\rho = (1-2\nu)\nabla tr\tilde\tilde\Alpha + \nabla\times\tilde\eta^*$$

where $$ \nabla \times\tilde\eta^{*}$$ is the vector of integration

We suppose that $$tr\tilde\tilde\Alpha = \chi$$ and also that $$\tilde\eta^* = (2-2\nu)\tilde\eta $$ as before, then we get


 * $$2G\tilde\rho = (1-2\nu)\nabla\chi + (2-2\nu)\nabla\times\tilde\eta $$

Thus the desired outcome is achieved, that is, that the Helmholtz characterization of the displacement vector is confirmed through the linear elastic constitutive equation coupled with the integration of the strain tensor. Again, there is good reason for introducing the scaled functions for $$\chi $$ and $$\tilde\eta$$. It happens that the Galerkin vector and other functions below yield quite transparently to unification with this format.

It remains to investigate the Helmholtz functions a little further. We can determine a set of independent and uncoupled biharmonic functions for $$\chi$$ and $$\tilde\eta$$ by firstly setting up the underlying governing tensorial equation for $$\tilde\tilde\Alpha$$. Firstly, the Helmholtz displacement vector (in terms of $$\chi$$ and $$\tilde\eta$$) are substituted into the stress-displacement expression


 * $$\tilde\tilde\Tau = G\left [ \nabla\tilde\rho + \tilde\rho\nabla +\frac{2\nu}{(1-2\nu)} \tilde\tilde\text{I}tr\nabla\cdot\tilde\rho\right ]$$

to yield


 * $$\tilde\tilde\Tau = (1-2\nu)\nabla\nabla \chi + \nu\tilde\tilde\text{I}\nabla^2 \chi + (1-\nu)\text{def} \left(\nabla\times\tilde\eta \right)$$

The divergence of this equation leads us to the equilibrium condition $$\nabla\cdot\tilde\tilde\Tau = 0$$, in terms of the Helmholz functions, that is


 * $$(1-\nu)\nabla^2\left[\nabla \chi + \nabla\times\tilde\eta \right] = 0$$

Applying the divergence operator and to this expression yields the governing biharmonic equation for $$\chi$$. Applying the curl operator independently also demonstrates the biharmonic form of $$\tilde\eta$$.


 * $$\nabla^2\nabla^2\chi=0$$   and     $$\nabla^2\nabla^2\tilde\eta=0$$

The latter arises by applying the Helmholtz condition $$\nabla\cdot\tilde\eta = 0$$. So to be clear, the governing biharmonic equations arise from the equilibrium condition. That is, substitution of the Helmholtz displacement functions into the displacement-equilibrium equation yields the defining equations for $$\chi$$ and $$\tilde\eta$$. Since the compatibility condition is identically satisfied by the substitution of the displacement vector for the strain tensor, by extension, all displacement functions identically satisfy compatibility.

Extending the Helmholtz Functions

Now for something quite interesting. Consider the following expansion for an arbitrary symmetric tensor $$\tilde\tilde\Alpha$$


 * $$\nabla^2\tilde\tilde\Alpha = \nabla\times\left(\tilde\tilde\Alpha - \tilde\tilde\text{I}tr\tilde\tilde\Alpha\right)\times\nabla +\text{ℝ}\nabla\cdot\tilde\tilde\Alpha$$

where, as a reminder, $$\text{ℝ} = \nabla + \nabla - \tilde\tilde\text{I}\nabla\cdot $$

Applying the same relationship as before we can get an expression for $$\tilde\tilde\Tau$$ as follows


 * $$\tilde\tilde\Tau = (1+\nu)\nabla^2\tilde\tilde\Alpha = (1+\nu)\left[\nabla\times\left(\tilde\tilde\Alpha - \tilde\tilde\text{I}tr\tilde\tilde\Alpha\right)\times\nabla +\text{ℝ}\nabla\cdot\tilde\tilde\Alpha\right]$$

The latter form is that of the Beltrami-Schaeffer format and it is clear that


 * $$\tilde\tilde\Phi = (1+\nu)\left(\tilde\tilde\Alpha - \tilde\tilde\text{I}tr\tilde\tilde\Alpha\right)$$    and     $$\tilde\text{h} = (1+\nu)\nabla\cdot\tilde\tilde\Alpha$$

whereupon since $$ tr \tilde\tilde\Phi = -2(1+\nu)tr\tilde\tilde\Alpha $$ then


 * $$\tilde\tilde\Phi - \tilde\tilde\text{I}tr\tilde\tilde\Phi = (1+\nu)\left(\tilde\tilde\Alpha + \tilde\tilde\text{I}tr\tilde\tilde\Alpha\right)$$

and the displacement vector in the Beltrami-Schaeffer format, $$2G\tilde\rho = \frac {1-2\nu}{1+\nu}\left[\nabla\cdot\left(\tilde\tilde\Phi - \tilde\tilde \text{I} tr \tilde\tilde\Phi \right ) - \tilde\text{h} + \nabla\times\tilde\text{c}\right] $$ becomes


 * $$2G\tilde\rho = \frac {1-2\nu}{1+\nu}\left[(1+\nu)\nabla tr \tilde\tilde\Alpha + \nabla\times\tilde\text{c}\right] $$

It is a straitforward assumption that $$tr\tilde\tilde\Alpha = \chi$$ and $$\tilde\text{c} = \frac {1+\nu}{1-2\nu}(2-2\nu)\tilde\eta$$ to yield


 * $$2G\tilde\rho = (1-2\nu)\nabla\chi + (2-2\nu)\nabla\times\tilde\eta $$

as before.

A further interesting observation is that after some manipulation the Helmholtz stress tensor may be written directly from the substitution of the Helmholtz functions into the stress-displacement equation as


 * $$\tilde\tilde\Tau = \nabla\times\left (-\tilde\tilde\text{I}\chi\right )\times\nabla + (1-\nu)\text{ℝ}\left (\nabla\chi+\nabla\times\tilde\eta\right )$$

or more fully


 * $$\tilde\tilde\Tau = -\nabla\times\left (\tilde\tilde\text{I}\chi + \tilde\tilde\text{I}\times\tilde\eta\right )\times\nabla + (1-\nu)\text{ℝ}\left[\nabla\cdot\left (\tilde\tilde\text{I}\chi+\tilde\tilde\text{I}\times\tilde\eta\right )\right]$$

since $$\left(\tilde\tilde\text {I}\times\tilde\eta\right)$$ component vanishes under the curl curl operation.

As before the Beltrami-Schaeffer format $$\left(\tilde\tilde\Tau =\nabla\times\tilde\tilde\Phi\times\nabla + \text{ℝ}(\tilde\text{h}) \right)$$ compares to the Helmholtz stress equation above and thus


 * $$\tilde\tilde\Phi = -\left(\tilde\tilde\text{I}\chi+\tilde\tilde\text{I}\times\tilde\eta\right )$$   and    $$\tilde\text {h} = (1-\nu)\nabla\cdot\left ( \tilde\tilde\text{I}\chi+\tilde\tilde\text{I}\times\tilde\eta\right ) = -(1-\nu)\nabla\cdot\tilde\tilde\Phi $$

Substitution for $$\tilde\tilde\Phi$$ and $$\text{h}$$ into the Beltrami-Schaeffer displacement equation is interesting, giving us


 * $$2G\tilde\rho = \frac {1-2\nu}{1+\nu}\left[(2-\nu)\nabla\cdot\tilde\tilde\Phi - \nabla tr \tilde\tilde\Phi + \nabla\times\tilde\text{c}\right]$$

where substitution for the Helmholtz equivalences above yields


 * $$2G\tilde\rho = (1-2\nu)\nabla\chi + \frac {1-2\nu}{1+\nu}\left[-(2-\nu)\nabla\times\tilde\eta + \nabla\times\tilde\text{c}\right]$$

If  $$\tilde\text{c} = \frac{4-5\nu}{1-2\nu}\tilde\eta $$    then $$2G\tilde\rho = (1-2\nu)\nabla\chi + (2-2\nu)\nabla\times\tilde\eta $$

again, as before.

It is worth noting here that $$\left(\tilde\tilde\text{I}\chi+\tilde\tilde\text{I}\times\tilde\eta\right )$$ is asymmetric, however the operator terms on the right hand side of the stress equation result in symmetry for the stress tensor as you would expect.

Finally and most interestingly we can eliminate the need for the constant vector of integration in the process of determining a Helmholtz basis for the field quantities.

From above we determined a Beltrami-Schaeffer format of the Helmholtz functions to be


 * $$\tilde\tilde\Tau = -\nabla\times\left (\tilde\tilde\text{I}\chi + \tilde\tilde\text{I}\times\tilde\eta\right )\times\nabla + (1-\nu)\text{ℝ}\left[\nabla\cdot\left (\tilde\tilde\text{I}\chi+\tilde\tilde\text{I}\times\tilde\eta\right )\right]$$

Let us revise this expression such that we introduce a new constant multipier $$k_{\nu}$$ on $$\nabla\times\tilde\eta$$, giving


 * $$\tilde\tilde\Eta = -\left(\tilde\tilde\text{I}\chi+k_{\nu}\tilde\tilde\text{I}\times\tilde\eta\right )$$   and    $$\tilde\text {h} = (1-\nu)\nabla\cdot\left ( \tilde\tilde\text{I}\chi+k_{\nu}\tilde\tilde\text{I}\times\tilde\eta\right ) = -(1-\nu)\nabla\cdot\tilde\tilde\Eta $$

where $$\tilde\tilde\Eta$$ is a special version of $$\tilde\tilde\Phi$$ for this and only this configuration of the Helmholtz stress tensor.

Thus


 * $$\tilde\tilde\Tau = -\nabla\times\left (\tilde\tilde\text{I}\chi + k_{\nu}\tilde\tilde\text{I}\times\tilde\eta\right )\times\nabla + (1-\nu)\text{ℝ}\nabla\cdot\left (\tilde\tilde\text{I}\chi+k_{\nu}\tilde\tilde\text{I}\times\tilde\eta\right )$$

Picking up from the equations above, substitution for $$\tilde\tilde\Eta$$ and $$\tilde\text{h}$$ into the Beltrami-Schaeffer displacement equation is again


 * $$2G\tilde\rho = \frac {1-2\nu}{1+\nu}\left[(2-\nu)\nabla\cdot\tilde\tilde\Eta - \nabla tr \tilde\tilde\Eta + \nabla\times\tilde\text{c}\right]$$

whereupon substitution for the Helmholtz functions yields


 * $$2G\tilde\rho = (1-2\nu)\nabla\chi + \frac {1-2\nu}{1+\nu}\left[-(2-\nu)k_{\nu}\nabla\times\tilde\eta + \nabla\times\tilde\text{c}\right]$$

If $$k_{\nu}$$ is carefully chosen, in this case such that $$k_{\nu}= -\frac {1+\nu}{1-2\nu}\cdot\frac{2-2\nu}{2-\nu}$$ then the equation for $$2G\tilde\rho$$ succumbs to a simple substitution such that the vector of integration $$\nabla\times\tilde\text{c}$$ is dispensed with, giving for the fourth time


 * $$2G\tilde\rho = (1-2\nu)\nabla\chi + (2-2\nu)\nabla\times\tilde\eta $$

This leads to a relatively important characterization of all of the field quantities of linear elasticity in terms of the Helmholtz functions in the Beltrami-Schaeffer format, which also removes any awkward vectors of integration from the formulations. In summary

$$\tilde\tilde\Tau =\nabla\times\tilde\tilde\Eta\times\nabla - (1-\nu)\text{ℝ}\nabla\cdot\tilde\tilde\Eta $$      and      $$2G\tilde\rho = \frac {1-2\nu}{1+\nu}\nabla\cdot\left[(2-\nu)\tilde\tilde\Eta - \tilde\tilde \text{I} tr \tilde\tilde\Eta \right] $$

where

$$\tilde\tilde\Eta = -\left(\tilde\tilde\text{I}\chi+k_{\nu}\tilde\tilde\text{I}\times\tilde\eta\right )$$      and        $$k_{\nu} = -\frac {1+\nu}{1-2\nu}\frac {2-2\nu}{2-\nu}$$

Since both the Beltrami-Schaeffer and Helmholtz functions are fundamental to characterizing the field equations of homogeneous linear elasticity it will be interesting to see how the more explicit functions of Galerkin and others unfold in this format.

Papkovich-Neuber Functions

It is known that the Papkovich_Neuber functions may be simply derived from the Helmholtz functions. Firstly with reference to the Helmholtz stress tensor above


 * $$\tilde\tilde\Tau = \nabla\times\left (-\tilde\tilde\text{I}\chi\right )\times\nabla + (1-\nu)\text{ℝ}\left (\nabla\chi+\nabla\times\tilde\eta\right )$$

Introducing a new vector function such that


 * $$\tilde\text{B}=\nabla\chi+\nabla\times\tilde\eta$$

where $$\tilde\text{B}$$ is harmonic since from above $$\nabla^2\left[\nabla \chi+ \nabla\times\tilde\eta \right] = 0$$.

then taking the divergence of $$\tilde\text{B}$$ we get


 * $$\nabla\cdot\tilde\text{B} = \nabla^2\chi$$

To proceed we define a position vector $$\tilde\text{P} = \text{i}x + \text{j}y + \text{k}z $$

Employing the identity $$2\nabla\cdot\tilde\text{B} = \nabla^2 \left(\tilde\text{P}\cdot\tilde\text{B}\right )$$ the above can be expressed in the form


 * $$\nabla^2\left(2\chi - \tilde\text{P}\cdot\tilde\text{B}\right) = 0$$

A second function $$B_o $$ is now introduced such that


 * $$B_o = 2\chi-\tilde\text{P}\cdot\tilde\text{B}$$

leading to the harmonic nature of both of the Papkovich-Neuber functions, that is


 * $$\nabla^2\tilde\text{B} = \nabla^2B_o = 0$$

In reality $$\tilde\text{B}$$ and $$B_o$$ are just simple substitutes for the underlying combination of Helmholtz functions.

For completeness the Helmholtz functions become


 * $$\nabla\chi=\frac{1}{2}\nabla\left[B_o+\tilde\text{P}\cdot\tilde\text{B}\right]$$, and


 * $$\nabla\times\tilde\eta=\tilde\text{B}-\frac{1}{2}\nabla\left[B_o+\tilde\text{P}\cdot\tilde\text{B}\right]$$

and thus from the Helmholtz displacement vector $$2G\tilde\rho = (1-2\nu)\nabla\chi + (2-2\nu)\nabla\times\tilde\eta$$, an expression for Papkovich-Neuber displacement vector ensues, namely


 * $$G\tilde\rho = (1-\nu)\tilde\text{B} - \frac{1}{4}\nabla \left(B_o+\tilde\text{P}\cdot\tilde\text{B} \right )$$

From the stress-displacement relation the stress tensor directly becomes


 * $$\tilde\tilde\Tau=(1-\nu)\left(\nabla\tilde\text{B}+\tilde\text{B}\nabla\right)-\frac{1}{2}\nabla\nabla\left(B_o+\tilde\text{P}\cdot\tilde\text{B}\right)+\frac{\nu}{(1-2\nu)}\tilde\tilde\text{I}\nabla\cdot\tilde\text{B}$$

which can be manipulated into the format


 * $$\tilde\tilde\Tau=\nabla\times\tilde\tilde\text{I}\left[-\frac{1}{2}\left(B_o+\tilde\text{P}\cdot\tilde\text{B}\right)\right]\times\nabla + (1-\nu)\text{ℝ}\left(\tilde\text{B}\right)$$

which is unsurprising since


 * $$\tilde\tilde\Tau = \nabla\times\left (-\tilde\tilde\text{I}\chi\right )\times\nabla + (1-\nu)\text{ℝ}\left (\nabla\chi+\nabla\times\tilde\eta\right )$$

Ergo, by observation the Beltrami-Schaeffer stress functions can also encode the Papkovich-Neuber functions by putting


 * $$\tilde\tilde\Phi = -\frac{1}{2}\tilde\text{I}\left ( B_o + \tilde\text{P}\cdot\tilde\text{B}\right )$$    and    $$\tilde\text{h}=(1-\nu)\tilde\text{B}$$

Thus, along with the Helmholtz functions the Papkovich-Neuber functions are simple variants of the same generalized Beltrami-Schaeffer stress functions.

The Galerkin Vector 

The Galerkin vector can also be shown to be a variant of the Beltrami-Schaeffer functions.

From above the Helmholtz displacement vector is


 * $$2G\tilde\rho = (1-2\nu)\nabla\chi + (2-2\nu)\nabla\times\tilde\eta$$

Putting $$\chi = \nabla\cdot\tilde\text{F}$$ and $$\tilde\eta=\tilde\text{F}\times\nabla$$ then the known Galerkin form, or a variant thereof, of the displacement vector ensues


 * $$2G\tilde\rho = (1-2\nu)\nabla\nabla\cdot\tilde\text{F} + (2-2\nu)\nabla\times\tilde\text{F}\times\nabla $$

Remembering that $$\tilde\tilde\Tau = \nabla\times\left (-\tilde\tilde\text{I}\chi\right )\times\nabla + (1-\nu)\text{ℝ}\left (\nabla\chi+\nabla\times\tilde\eta\right )$$

then the stress tensor in the Galerkin form becomes


 * $$\tilde\tilde\Tau=\nabla\times\left(-\tilde\tilde\text{I}\nabla\cdot\tilde\text{F}\right)\times\nabla + \left ( 1-\nu\right ) \text{ℝ}\left ( \nabla^2\tilde\text{F}\right )$$

which upon taking the divergence results in the governing biharmonic equation for $$\tilde\text{F}$$ as before,


 * $$\nabla\cdot\tilde\tilde\text{T} = \nabla^2\nabla^2\tilde\text{F}=0$$

The Beltrami-Schaeffer functions thus take on the Galerkin form


 * $$\tilde\tilde\Phi = -\tilde\tilde\text{I}\nabla\cdot\tilde\text{F}$$   and     $$\tilde\text{h}=(1-\nu)\nabla^2\tilde\text{F}$$

In the alternative mode using the 'Helmholtz tensor'


 * $$\tilde\tilde\text{H} = \left(\tilde\tilde\text{I}\chi + \tilde\tilde\text{I}\times\tilde\eta\right) = \left(\tilde\tilde\text{I}\nabla\cdot\tilde\text{F} + \tilde\tilde\text{I}\times\tilde\text{F}\times\nabla \right) = \left(\tilde\tilde\text{I}\nabla\cdot\tilde\text{F} + \nabla\tilde\text{F} - \tilde\text{F}\nabla \right)$$

which results in the equivalent Galerkin stress function format


 * $$\tilde\tilde\Tau=\nabla\times\left(-\tilde\tilde\text{I}\nabla\cdot\tilde\text{F}\right)\times\nabla + \left ( 1-\nu\right ) \text{ℝ}\nabla\cdot\left (\tilde\tilde\text{I}\nabla\cdot\tilde\text{F}+ \nabla\tilde\text{F} -\tilde\text{F}\nabla \right )$$

A very unique expression can now be found for the Galerkin stress tensor since a null deformator field can be added to the first term and the second can be rationalized as follows


 * $$\tilde\tilde\Tau=\nabla\times\left(\nabla\tilde\text{F} + \tilde\text{F}\nabla-\tilde\tilde\text{I}\nabla\cdot\tilde\text{F}\right)\times\nabla + \left ( 1-\nu\right ) \text{ℝ}\nabla\cdot\left (\nabla\tilde\text{F} +\tilde\text{F}\nabla - \tilde\tilde\text{I}\nabla\cdot\tilde\text{F} \right )$$

which further becomes the interesting expression


 * $$\tilde\tilde\Tau=\nabla\times\text{ℝ}\left(\tilde\text{F}\right)\times\nabla + (1-\nu )\text{ℝ}\nabla\cdot\text{ℝ}\left(\tilde\text{F}\right)$$

or the most compact form of the Galerkin representation for stress


 * $$\tilde\tilde\Tau=\nabla\times\tilde\tilde\text{G}\times\nabla + (1-\nu )\text{ℝ}\nabla\cdot\tilde\tilde\text{G}$$

where, not to be confused with $$G$$, Kirchoff's modulus $$\tilde\tilde\text{G} = \text{ℝ}\left(\tilde\text{F}\right) = \left(\nabla\tilde\text{F} + \tilde\text{F}\nabla-\tilde\tilde\text{I}\nabla\cdot\tilde\text{F}\right)$$

Clearly, $$\tilde\tilde\Phi=\tilde\tilde\text{G}$$    and    $$\tilde\text{h}=(1-\nu)\nabla\cdot\tilde\tilde\text{G}$$,   whilst the elemental function within $$\tilde\tilde\text{G}$$, that is $$\tilde\text{F}$$, remains biharmonic as before.

The associated displacement vector takes the form


 * $$2G\tilde\rho = (2-2\nu)\nabla\cdot\tilde\tilde\text{G}+ \nabla tr \tilde\tilde\text{G}$$

As a reminder, Navier's equation is similar $$\left [(1-2\nu)\nabla^2 + \nabla\nabla\cdot\right ]\tilde\rho = 0$$

The resulting substitution is not very helpful unless $$\tilde\text{F}$$ is brought into the equation via $$\tilde\tilde\text{G}$$ however.

 Generalised Cylinder Stress Function

A new stress function is currently being researched which has a particular property of being applicable to uni-directional non-homogeneous continua. That is for an elastic field where G, Kirchoff's modulus, is functionally defined along one axis. This will be added to the article as the work is completed.

Additional reading

 * Morse, P.M. and Feshbach, H. (1953). Methods of Theoretical Physics.   McGraw-Hill, New York.
 * Youngdahl, C.K. (1969). On the completeness of a set of stress functions appropriate to the solution of elasticity problems in general cylindrical coordinates. Int. J. Enging. Sci., vol. 7, pp. 61-79.
 * Gurtin, M.E. (1972). The linear theory of elasticity. S. Flugge, ed., Handbuch der Physik, col. VI1/2, Springer-Verlag, Berlin.
 * Lure, A.I. (1964). Three-Dimensional Problems of the Theory of Elasticity.  Interscience, New York.
 * Beasley, A.J. (1986). A New Pair of Galerkin Based Stress Functions in Homogeneous Linear Elasticity.  Univ. of Tas., Dept of Civ and Mech. Engng, Research Report No CM 86/4.
 * Beasley, A.J. (1995). An interim unified model of stress functions for homogeneous linear elasticity, Proc. Fourteenth Australasian Conference on the Mechanics of Structures and Materials, Hobart.
 * Sternberg, E. (1960). On some recent developments in the linear theory of elasticity, Structural Mechanics, Pergamon, New York.
 * Sternberg, E., Eubanks, R.A. and Sadowsky, W.A. (1951). On the stress-function approaches of Boussinesq and Timpe to the axisymmetric problem of elasticity theory, J. App. Phys. vol. 22, pp 1121-1124.