From the beam loading diagram shown in Figure 13.9, we can easily find out support reaction at fixed end O of the beam as follows:

\begin{array}{ll} R_{ O _y}=+P \sin \theta(\uparrow) & M_z=+P L \sin \theta \\ R_{ O _z}=-P \cos \theta & M_y=+P L \cos \theta \end{array}

Let us consider a section ‘X−X’ at a distance x from O and draw its free-body diagram (Figure 13.10) to determine the bending moments at that section as follows:

At section X–X,

M_y=-P(L-x) \cos \theta \text {  and  } M_z=-P(L-x) \sin \theta

are the bending moments about y and z axes, respectively. For a rectangular section, they are the principal axes. Thus,

\bar{I}_{y z}=0             (refer to Figure 13.11)

Hence, from Eq. (13.21) of generalised flexure formula, we get the bending stress at any point P on the beam cross-section as

\sigma_{x x}=\left\lgroup \frac{z \bar{I}_{z z}-y \bar{I}_{y z}}{\bar{I}_{y y} \bar{I}_{z z}-\bar{I}_{y z}^2} \right\rgroup M_y+\left\lgroup \frac{z \bar{I}_{y z}-y \bar{I}_{y y}}{\bar{I}_{y y} \bar{I}_{z z}-\bar{I}_{y z}^2} \right\rgroup M_z               (13.21)

\sigma_{x x}=\left\lgroup \frac{M_y z}{\bar{I}_{y y}}-\frac{y M_z}{\bar{I}_{z z}} \right\rgroup               (1)

For neutral axis, we must have, \sigma_{x x}=0 . So, from Eq. (1)

\frac{y}{z}=\tan \phi=\left\lgroup \frac{M_y}{M_z} \right\rgroup\left\lgroup \frac{\bar{I}_{z z}}{\bar{I}_{y y}} \right\rgroup

\tan \phi=\cot \theta \frac{b h^3}{h b^3}=\cot \theta\left\lgroup \frac{h}{b} \right\rgroup^2=\cot \theta \cdot \tan ^2 \theta

\tan \phi=\tan \theta

or              \phi=\theta

Therefore, neutral axis will be along the other diagonal of the rectangle.