Step 1

We are given the force $F = 200\;{\rm{N}}$, the radius of the rod as $R = 0.5\;{\rm{m}}$, the angle $\alpha $ as $\alpha = 30^\circ $, and the angle $\beta $ as $\beta = 45^\circ $.

We are asked to determine the internal normal force, shear force, and moment acting at point $B$ and $C$.

 
Step 2

We will draw a free body diagram of the bottom segment.

Here, ${N_C}$ is the normal force at point $C$, ${M_C}$ is the moment about point $C$, ${V_C}$ is the shear force at point $C$, and $\theta $ is the angle of force.

 
Step 3

We will find the angle $\theta $ from the figure.

\[\begin{array}{c} \tan \theta = \frac{3}{4}\\ \theta = {\tan ^{ - 1}}\left( {\frac{3}{4}} \right)\\ \approx 37^\circ \end{array}\]
 
Step 4

We will resolve the forces in the horizontal direction to find the internal normal force at point $C$.

\[\begin{array}{c} \sum {{F_x}} = 0\\ {N_C} - F \times \sin \left( {\theta + \alpha } \right) = 0\\ {N_C} = F \times \sin \left( {\theta + \alpha } \right) \end{array}\]

Substitute the given value in the above equation.

\[\begin{array}{c} {N_C} = \left( {{\rm{200}}\;{\rm{N}}} \right) \times \sin \left( {37^\circ + 30^\circ } \right)\\ \approx 184.1\;{\rm{N}} \end{array}\]
 
Step 5

We will resolve the forces in the vertical direction to find the shear force at point $C$.

\[\begin{array}{c} \sum {{F_y}} = 0\\ - {V_C} - F\cos \left( {\theta + \alpha } \right) = 0\\ {V_C} = - F\cos \left( {\theta + \alpha } \right) \end{array}\]

Substitute the given value in the above equation.

\[\begin{array}{c} {V_C} = - \left( {{\rm{200}}\;{\rm{N}}} \right) \times \cos \left( {37^\circ + 30^\circ } \right)\\ \approx - 78.1\;{\rm{N}} \end{array}\]
 
Step 6

We will take the moment about point $C$ to find the moment about point $C$.

\[\begin{array}{c} \sum {{M_C}} = 0\\ \left[ \begin{array}{l} F\cos \theta \times \left[ {\left( {0.5\sin 30^\circ } \right){\rm{m}}} \right] - \\ F\sin \theta \times \left[ {0.5\left( {1 - \cos 30^\circ } \right){\rm{m}}} \right] + {M_C} \end{array} \right] = 0\\ {M_C} = \left\{ \begin{array}{l} F\sin \theta \times \left[ {0.5\left( {1 - \cos 30^\circ } \right){\rm{m}}} \right]\\ - F\cos \theta \times \left[ {\left( {0.5\sin 30^\circ } \right){\rm{m}}} \right] \end{array} \right\} \end{array}\]

Substitute the given value in the above equation.

\[\begin{array}{c} {M_C} = \left( {200\;{\rm{N}}} \right)\left( {\sin 37^\circ } \right) \times \left[ {0.5\left( {1 - \cos 30^\circ } \right){\rm{m}}} \right] - \left( {200\;{\rm{N}}} \right)\left( {\cos 37^\circ } \right) \times \left[ {\left( {0.5\sin 30^\circ } \right){\rm{m}}} \right]\\ = \left( {200\;{\rm{N}}} \right)\left[ {\left( {\sin 37^\circ } \right) \times \left[ {0.5\left( {1 - \cos 30^\circ } \right){\rm{m}}} \right] - \left( {\cos 37^\circ } \right) \times \left[ {\left( {0.5\sin 30^\circ } \right){\rm{m}}} \right]} \right]\\ \approx - 32\;{\rm{N}} \cdot {\rm{m}} \end{array}\]
 
Step 7

We will draw a free body diagram of the upper segment.

Here, ${N_B}$ is the normal force at point $B$, ${M_B}$ is the moment about point $B$, and ${V_B}$ is the shear force at point $B$.

 
Step 8

We will resolve the forces in the horizontal direction to find the internal normal force at point $B$.

\[\begin{array}{c} \sum {{F_x}} = 0\\ {N_B} - F \times \sin \left( {\beta - \theta } \right) = 0\\ {N_B} = F \times \sin \left( {\beta - \theta } \right) \end{array}\]

Substitute the given value in the above equation.

\[\begin{array}{c} {N_B} = \left( {{\rm{200}}\;{\rm{N}}} \right) \times \sin \left( {45^\circ - 37^\circ } \right)\\ \approx 28\;{\rm{N}} \end{array}\]
 
Step 9

We will resolve the forces in the vertical direction to find the shear force at point $B$.

\[\begin{array}{c} \sum {{F_y}} = 0\\ - {V_B} + F\cos \left( {\beta - \theta } \right) = 0\\ {V_B} = - F\cos \left( {\beta - \theta } \right) \end{array}\]

Substitute the given value in the above equation.

\[\begin{array}{c} {V_B} = - \left( {{\rm{200}}\;{\rm{N}}} \right) \times \cos \left( {45^\circ - 37^\circ } \right)\\ \approx - 198\;{\rm{N}} \end{array}\]
 
Step 10

We will take the moment about point $B$ to find the moment about point $B$.

\[\begin{array}{c} \sum {{M_B}} = 0\\ \left[ \begin{array}{l} F\cos \theta \times \left[ {\left( {0.5\sin 45^\circ } \right){\rm{m}}} \right] - \\ F\sin \theta \times \left[ {0.5\left( {1 + \cos 45^\circ } \right){\rm{m}}} \right] + {M_B} \end{array} \right] = 0\\ {M_B} = \left\{ \begin{array}{l} F\sin \theta \times \left[ {0.5\left( {1 + \cos 45^\circ } \right){\rm{m}}} \right]\\ - F\cos \theta \times \left[ {\left( {0.5\sin 45^\circ } \right){\rm{m}}} \right] \end{array} \right\} \end{array}\]

Substitute the given value in the above equation.

\[\begin{array}{c} {M_B} = \left( {200\;{\rm{N}}} \right)\left( {\sin 37^\circ } \right) \times \left[ {0.5\left( {1 + \cos 45^\circ } \right){\rm{m}}} \right] - \left( {200\;{\rm{N}}} \right)\left( {\cos 37^\circ } \right) \times \left[ {\left( {0.5\sin 45^\circ } \right){\rm{m}}} \right]\\ = \left( {200\;{\rm{N}}} \right)\left[ {\left( {\sin 37^\circ } \right)\left[ {0.5\left( {1 + \cos 45^\circ } \right){\rm{m}}} \right] - \left( {\cos 37^\circ } \right) \times \left[ {\left( {0.5\sin 45^\circ } \right){\rm{m}}} \right]} \right]\\ \approx 46.3\;{\rm{N}} \cdot {\rm{m}} \end{array}\]