Elasticity/Prandtl stress function

Prandtl Stress Function φ

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 $\varphi (x_1,x_2)$ using

${\sigma }_{13}={\varphi }_{,2};{\sigma }_{23}=-{\varphi }_{,1}$

You can easily check that this definition satisfies equilibrium.

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

$\frac{d\varphi }{ds}=0\mathrm{\forall }(x_1,x_2)\in \partial \text{S}$

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:

$\varphi =0\mathrm{\forall }(x_1,x_2)\in \partial \text{S}$

From the compatibility condition, we get a restriction on $\varphi$

${\mathrm{\nabla }}^2\varphi =C\mathrm{\forall }(x_1,x_2)\in \text{S}$

where $C$ is a constant.

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

${\mathrm{\nabla }}^2\varphi =-2\mu \alpha \mathrm{\forall }(x_1,x_2)\in \text{S}$

Therefore, the twist per unit length is

$\alpha =-\frac{1}{2\mu }{\mathrm{\nabla }}^2\varphi$

The applied torque is given by

$T=-\underset{S}{\int }(x_1{\varphi }_{,1}+x_2{\varphi }_{,2})dA$

For a simply connected cylinder,

$T=2\underset{S}{\int }\varphi dA$

The projected shear traction is given by

$\tau =\sqrt{({\varphi }_{,1}{)}^2+({\varphi }_{,2}{)}^2}$

The projected shear traction at any point on the cross-section is tangent to the contour of constant $\varphi$ at that point.

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

${\psi }_{,1}=\frac{1}{\mu \alpha }{\varphi }_{,2}+x2;{\psi }_{,2}=-\frac{1}{\mu \alpha }{\varphi }_{,1}-x1$

Membrane Analogy

The equations

${\mathrm{\nabla }}^2\varphi =-2\mu \alpha \mathrm{\forall }(x_1,x_2)\in \text{S};\varphi =0\mathrm{\forall }(x_1,x_2)\in \partial \text{S}$

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.