Step 1

We are given two point load as $800\;{\rm{N}}$ and $1200\;{\rm{N}}$ and one distributing load of $400\;{{\rm{N}} \mathord{\left/ {\vphantom {{\rm{N}} {\rm{m}}}} \right. } {\rm{m}}}$ on the section CD.

We are asked to determine the internal normal force, shear force and the moment at point E and F of the compound beam.

 
Step 2

The free-body diagram of the member BC can be drawn as:

Here, ${B_x}$ is the horizontal component of joint B, ${B_y}$ is the vertical component of joint B and ${C_y}$ is the vertical component of joint C.

On taking moment about point B, we get:

\[\begin{array}{c} \sum {M_B} = 0\\ {C_y} \times \left( {3\;{\rm{m}}} \right) - \left( {\left( {1200\;{\rm{N}}} \right) \times \left( {\frac{4}{5}} \right) \times \left( {2\;{\rm{m}}} \right)} \right) = 0\\ {C_y} \times \left( {3\;{\rm{m}}} \right) = 1920\;{\rm{N}} \cdot {\rm{m}}\\ {C_y} = 640\;{\rm{N}} \end{array}\]

Similarly, on taking the moment about point C, we get:

\[\begin{array}{c} \sum {M_C} = 0\\ \left( {\left( {1200\;{\rm{N}}} \right) \times \left( {\frac{4}{5}} \right) \times \left( {1\;{\rm{m}}} \right)} \right) - {B_y} \times \left( {3\;{\rm{m}}} \right) = 0\\ \left( {960\;{\rm{N}} \cdot {\rm{m}}} \right) = {B_y} \times \left( {3\;{\rm{m}}} \right)\\ {B_y} = 320\;{\rm{N}} \end{array}\]

On balancing the horizontal forces, we get:

\[\begin{array}{c} \sum {F_H} = 0\\ \left( {\left( {1200\;{\rm{N}}} \right) \times \left( {\frac{3}{5}} \right)} \right) - {B_x} = 0\\ {B_x} = 720\;{\rm{N}} \end{array}\]
 
Step 3

The free-body diagram of the section AB at point E can be drawn as:

Here, ${M_E}$ is the moment at point E, ${V_E}$ is the shear force at point E and ${N_E}$ is the normal force at point E.

On balancing the horizontal forces, we get:

\[\begin{array}{c} \sum {F_H} = 0\\ {B_x} - {N_E} = 0\\ {N_E} = 720\;{\rm{N}} \end{array}\]

Similarly, on balancing the vertical forces, we get:

\[\begin{array}{c} \sum {F_V} = 0\\ {V_E} - \left( {800\;{\rm{N}}} \right) - {B_y} = 0\\ {V_E} = \left( {800\;{\rm{N}}} \right) + \left( {320\;{\rm{N}}} \right)\\ {V_E} = 1120\;{\rm{N}} \end{array}\]

Here, we consider the anti-clockwise moments to be positive and clock-wise moments to be negative. So, on taking the moment about point E, we get:

\[\begin{array}{c} \sum {M_E} = 0\\ - {M_E} - {B_y} \times \left( {1\;{\rm{m}}} \right) = 0\\ - {M_E} = \left( {320\;{\rm{N}}} \right) \times \left( {1\;{\rm{m}}} \right)\\ {M_E} = - 320\;{\rm{N}} \cdot {\rm{m}}\;\;\;\left( {{\rm{Clock - wise}}} \right) \end{array}\]
 
Step 4

The free-body diagram of the section CD at point F can be drawn as:

Here, ${M_F}$ is the moment at point F, ${V_F}$ is the shear force at point F and ${N_F}$ is the normal force at point F.

On balancing the horizontal forces, we get:

\[\begin{array}{c} \sum {F_H} = 0\\ {N_F} = 0 \end{array}\]

Similarly, on balancing the vertical forces, we get:

\[\begin{array}{c} \sum {F_V} = 0\\ - {V_F} - {C_y} - \left( {400\;{{\rm{N}} \mathord{\left/ {\vphantom {{\rm{N}} {\rm{m}}}} \right. } {\rm{m}}}} \right) \times \left( {1.5\;{\rm{m}}} \right) = 0\\ - {V_F} = \left( {640\;{\rm{N}}} \right) + \left( {600\;{\rm{N}}} \right)\\ {V_F} = - 1240\;{\rm{N}}\left( \downarrow \right) \end{array}\]

Here, we consider the anti-clockwise moments to be positive and clock-wise moments to be negative. So, on taking the moment about point F, we get:

\[\begin{array}{c} \sum {M_F} = 0\\ {M_F} + \left( {400\;{{\rm{N}} \mathord{\left/ {\vphantom {{\rm{N}} {\rm{m}}}} \right. } {\rm{m}}}} \right) \times \left( {1.5\;{\rm{m}}} \right) \times \left( {\frac{{1.5\;{\rm{m}}}}{2}} \right) + {C_y} \times \left( {1.5\;{\rm{m}}} \right) = 0\\ {M_F} = - \left( {450\;{\rm{N}} \cdot {\rm{m}}} \right) - \left( {640\;{\rm{N}}} \right) \times \left( {1.5\;{\rm{m}}} \right)\\ {M_F} = - 1410\;{\rm{N}} \cdot {\rm{m}}\;\;\;\left( {{\rm{Clock - wise}}} \right) \end{array}\]