For the given beam section, we get

M_y=-M \sin 45^{\circ} \quad \text { and } \quad M_z=-M \cos 45^{\circ}

For the semicircular section, by parallel-axis theorem,

\bar{I}_{y y}=\left(I_{y y}\right)_0-\left\lgroup \frac{\pi r^2}{2} \right\rgroup\left\lgroup \frac{4 r}{3 \pi} \right\rgroup^2

=\frac{\pi r^4}{8}-\frac{8 r^4}{9 \pi}=\left\lgroup \frac{\pi}{8}-\frac{8}{9 \pi} \right\rgroup r^4=0.1098 r^4

and            \bar{I}_{z z}=\frac{\pi r^4}{8}=0.393 r^4

So the NA is oriented at an angle of \phi with z-axis, where

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

=(1) \frac{\frac{\pi}{8}}{\left\lgroup \frac{\pi}{8}-\frac{8}{9 \pi} \right\rgroup}=\frac{1}{1-\frac{64}{9 \pi^2}}

Thus, \phi=74.4^{\circ} . Hence, we sketch the neutral axis as shown in Figure 13.13:

We identify that points D and E are at extreme positions from the neutral axis. Now, to get the coordinates of E, we note from the geometry as shown in Figure 13.13(b) that

y=r \sin 15.6^{\circ}=0.269 r

and        z = −(r cos15.6° − 0.424r )

= −0.539r

Therefore,

\sigma_{ E }=\frac{M_y z}{\bar{I}_{y y}}-\frac{M_z y}{\bar{I}_{z z}}

\text { where } M_y=-M \sin 45^{\circ} \text { and } M_z=-M \cos 45^{\circ} . Substituting, we get

\sigma_{ E }=M\left[\frac{\left(-\sin 45^{\circ}\right)(-0.539 r)}{0.1098 r^4}+\frac{\left(\cos 45^{\circ}\right)(0.269 r)}{0.393 r^4}\right]

=3.956 \frac{M}{r^3}

Similarly, we would calculate stress at D as

\sigma_{ D }=\frac{M\left(-\sin 45^{\circ}\right)(0.424 r)}{0.1098 r^4}-\frac{M\left(-\cos 45^{\circ}\right)(-r)}{0.393 r^4}

=-4.53 \frac{M}{r^3}

From the magnitude point of view, the maximum bending stress is 4.53 M/r³ (compressive) developed at point D. And the maximum tensile stress developed is at point E where OE is perpendicular to NA and

\left(\sigma_{ t }\right)_{\max }=3.956 \frac{M}{r^3}