Find the tractions on the surfaces of the block and illustrate the results on a sketch of the block.

We wish to use this solution to solve the corresponding problem in which the surfaces $x_{3}=\pm c$ are traction-free. Determine an approximate corrective solution for this problem by offloading the unwanted force and moment results using elementary bending theory.

Find the maximum error in the stress $\sigma _{33}$ in the corrected solution and compare it with the maximum tensile stress in the plane strain solution.

These tractions are illustrated in the following figure

Tractions on the beam

To unload the tractions on the faces $x_{3}=\pm c$, we have to superpose the solution to a problem with equal and opposite tractions and moments. In order to use elementary bending theory, we have set up a problem with simple boundary conditions.

Let us first consider the force distribution required for the superposed problem. Since the loading is antisymmetric at $x_{3}=\pm c$, there is no net force is the $x_{3}$ direction. Similarly, there is no net moment about the $x_{2}$ axis.

However, there is a net moment about the $x_{1}$ axis. Hence, the problem to be superposed should have a bending stress distribution $\sigma _{33}=Cx_{2}$, where $C$ is a constant that is chosen so as to make the total bending moment (original problem + superposed problem) equal to zero. (Note: Think of a beam in the $x_{3}-x_{2}$ plane subjected to bending moments at the ends.)

The bending moment for a cross-section of the beam in the
$x_{3}-x_{2}$ plane about the $x_{1}$ axis is $M=-\sigma _{33}I/x_{2}\,$, where
$I=\int _{0}^{a}\int _{-b}^{b}x_{2}^{2}dx_{2}dx_{1}$.

Since $\sigma _{33}$ varies with
$x_{1}$, the total bending moment for the beam is given by