# Introduction to Elasticity/Prandtl stress function

## Prandtl Stress Function ($\phi$ )

The traction free BC is obviously difficult to satisfy if the cross-section is not a circle or an ellipse.

To simplify matters, we define the Prandtl stress function $\phi (x_{1},x_{2})\,$ using

${\sigma _{13}=\phi _{,2}~~;~~\sigma _{23}=-\phi _{,1}}$ You can easily check that this definition satisfies equilibrium.

It can easily be shown that the traction-free BCs are satisfied if

${{\frac {d\phi }{ds}}=0~~\forall ~(x_{1},x_{2})\in \partial {\text{S}}}$ where $s$ is a coordinate system that is tangent to the boundary.

If the cross section is simply connected, then the BCs are even simpler:

${\phi =0~~\forall ~(x_{1},x_{2})\in \partial {\text{S}}}$ From the compatibility condition, we get a restriction on $\phi$ ${\nabla ^{2}{\phi }=C~~\forall ~(x_{1},x_{2})\in {\text{S}}}$ where $C$ is a constant.

Using relations for stress in terms of the warping function $\psi$ , we get

${\nabla ^{2}{\phi }=-2\mu \alpha ~~\forall ~(x_{1},x_{2})\in {\text{S}}}$ Therefore, the twist per unit length is

${\alpha =-{\frac {1}{2\mu }}\nabla ^{2}{\phi }}$ The applied torque is given by

${T=-\int _{S}(x_{1}\phi _{,1}+x_{2}\phi _{,2})dA\,}$ For a simply connected cylinder,

${T=2\int _{S}\phi dA\,}$ The projected shear traction is given by

${\tau ={\sqrt {(\phi _{,1})^{2}+(\phi _{,2})^{2}}}}$ The projected shear traction at any point on the cross-section is tangent to the contour of constant $\phi \,$ at that point.

The relation between the warping function $\psi \,$ and the Prandtl stress function $\phi \,$ is

${\psi _{,1}={\frac {1}{\mu \alpha }}\phi _{,2}+x2~;~~\psi _{,2}=-{\frac {1}{\mu \alpha }}\phi _{,1}-x1}$ ### Membrane Analogy

The equations

$\nabla ^{2}{\phi }=-2\mu \alpha ~~\forall ~(x_{1},x_{2})\in {\text{S}}~~;~~~\phi =0~~\forall ~(x_{1},x_{2})\in \partial {\text{S}}$ are similar to the equations that govern the displacement of a membrane that is stretched between the boundaries of the cross-sectional curve and loaded by an uniform normal pressure.

This analogy can be useful in estimating the location of the maximum shear stress and the torsional rigidity of a bar.

• The stress function is proportional to the displacement of the membrane from the plane of the cross-section.
• The stiffest cross-sections are those that allow the maximum volume to be developed between the deformed membrane and the plane of the cross-section for a given pressure.
• The shear stress is proportional to the slope of the membrane.