The section is antisymmetric with C being its centroid. Therefore,

\bar{I}_{y y}=2\left[\frac{1}{12}(20)(100)^3+(100)(20)(40)^2\right]+\frac{1}{12}(160)(20)^3  mm ^4

=9.84 \times 10^6  mm ^4

Again,

\bar{I}_{z z}=2\left[\frac{1}{12}(100)(20)^3+(100)(20)(90)^2\right]+\frac{1}{12}(20)(160)^3  mm ^4

=39.36 \times 10^6  mm ^4

and                  \bar{I}_{y z}=2[0+(100)(20)(-40)(90)]  mm ^4

=-14.4 \times 10^6  mm ^4

Obviously, the fixed end of the beam is subjected to following bending moments:

M_z=-14\left(10^3\right) \sin 71.6^{\circ} \times 2.0  Nm

=-26.57 \times 10^6  Nmm

and        M_y=-(14)\left(10^3\right) \cos 71.6^{\circ} \times 2.0  Nm

=-8.84 \times 10^6  Nmm

From generalised flexure equation (refer to Eq. 13.21), we get

\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)

For the neutral axis orientation with respect to z-axis, we set \sigma_{x x}=0 \text {, so }

\frac{y}{z}=\frac{\bar{I}_{z z} M_y+M_z \bar{I}_{y z}}{\bar{I}_{y z} M_y+M_z \bar{I}_{y y}}=\tan \phi           (Let)

or            \tan \phi=\frac{(-8.84)(39.36)+(-26.57)(-14.4)}{(-14.4)(-8.84)+(-26.57)(9.84)}=-0.258

or        \phi=165.5^{\circ}

Let us draw the neutral axis in the section as shown in Figure 13.17:

By simple geometry, we note points A and B are furthest from neutral axis and note their ( y, z) coordinates. Thus,

We see that identical stress but compressive in nature is developed at point B.