Step 1

We are given the uniformly distributed load of $2\;{{{\rm{kN}}} \mathord{\left/ {\vphantom {{{\rm{kN}}} {\rm{m}}}} \right. } {\rm{m}}}$ at a distance of $6\;{\rm{m}}$ from support B.

We are asked to determine the internal normal force, shear force and the moment at points C and D.

 
Step 2

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

Here, ${A_x}$ is the horizontal component of support A, ${A_y}$ is the vertical component of support A and ${B_y}$ is the vertical component of support B.

On taking the moment about point A, we get:

\[\begin{array}{c} \sum {M_A} = 0\\ \left[ \begin{array}{l} \left\{ {{B_y} \times \left( {6\;{\rm{m}}} \right)} \right\} + \left\{ {{B_y} \times \left( {\left( {6\;{\rm{m}}} \right)\cos 45^\circ } \right)} \right\} - \left\{ {\left( {2\;{{{\rm{kN}}} \mathord{\left/ {\vphantom {{{\rm{kN}}} {\rm{m}}}} \right. } {\rm{m}}}} \right) \times \left( {6\;{\rm{m}}} \right) \times \left( {\frac{{6\;{\rm{m}}}}{2}} \right)} \right\} - \\ \left\{ {\left( {2\;{{{\rm{kN}}} \mathord{\left/ {\vphantom {{{\rm{kN}}} {\rm{m}}}} \right. } {\rm{m}}}} \right) \times \left( {6\;{\rm{m}}} \right) \times \left( {\left( {6\;{\rm{m}}} \right)\cos 45^\circ } \right)} \right\} \end{array} \right] = 0\\ \left( {10.24\;{\rm{m}}} \right){B_y} = 86.91\;{\rm{kN}} \cdot {\rm{m}}\\ {B_y} = 8.48\;{\rm{kN}} \end{array}\]

On balancing the horizontal forces, we get:

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

Similarly, on balancing the vertical forces, we get:

\[\begin{array}{c} \sum {F_V} = 0\\ {A_y} + {B_y} - \left( {2\;{{{\rm{kN}}} \mathord{\left/ {\vphantom {{{\rm{kN}}} {\rm{m}}}} \right. } {\rm{m}}}} \right) \times \left( {6\;{\rm{m}}} \right) = 0\\ {A_y} = - \left( {8.48\;{\rm{kN}}} \right) + \left( {12\;{\rm{kN}}} \right)\\ {A_y} = 3.52\;{\rm{kN}} \end{array}\]
 
Step 3

The free-body diagram of the section AC can be drawn as:

On balancing the horizontal forces, we get:

\[\begin{array}{c} \sum {F_H} = 0 = 0\\ {V_C} + {A_y}\cos 45^\circ = 0\\ {V_C} = - \left( {3.52\;{\rm{kN}}} \right)\cos 45^\circ \\ {V_C} = 2.48\;{\rm{kN}}\left( \downarrow \right) \end{array}\]

On balancing the vertical forces, we get:

\[\begin{array}{c} \sum {F_V} = 0\\ - {N_C} + {A_y}\sin 45^\circ = 0\\ {N_C} = \left( {3.52\;{\rm{kN}}} \right) \times \sin 45^\circ \\ {N_C} = 2.48\;{\rm{kN}}\left( \uparrow \right) \end{array}\]

On taking the moment about point C, we get:

\[\begin{array}{c} \sum {M_C} = 0\\ {M_C} - \left( {{A_y}\cos 45^\circ } \right) \times \left( {2\;{{{\rm{kN}}} \mathord{\left/ {\vphantom {{{\rm{kN}}} {\rm{m}}}} \right. } {\rm{m}}}} \right) = 0\\ {M_C} = \left( {3.52\;{\rm{kN}}} \right) \times \cos 45^\circ \times \left( {2\;{{{\rm{kN}}} \mathord{\left/ {\vphantom {{{\rm{kN}}} {\rm{m}}}} \right. } {\rm{m}}}} \right)\\ {M_C} = 4.97\;{\rm{kN}} \cdot {\rm{m}} \end{array}\]

Here, we consider the anti-clockwise moments to be positive. So, the moment at point C is in anti-clockwise direction.

 
Step 4

The free-body diagram of the section BD can be drawn as:

On balancing the horizontal forces, we get:

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

On balancing the vertical forces, we get:

\[\begin{array}{c} \sum {F_V} = 0\\ - {V_D} + {B_y} - \left( {2\;{{{\rm{kN}}} \mathord{\left/ {\vphantom {{{\rm{kN}}} {\rm{m}}}} \right. } {\rm{m}}}} \right) \times \left( {3\;{\rm{m}}} \right) = 0\\ - {V_D} = - \left( {8.48\;{\rm{kN}}} \right) + \left( {6\;{\rm{kN}}} \right)\\ {V_D} = 2.48\;{\rm{kN}}\left( \downarrow \right) \end{array}\]

On taking the moment about point D, we get:

\[\begin{array}{c} \sum {M_D} = 0\\ {M_D} + {B_y} \times \left( {3\;{\rm{m}}} \right) - \left( {2\;{{{\rm{kN}}} \mathord{\left/ {\vphantom {{{\rm{kN}}} {\rm{m}}}} \right. } {\rm{m}}}} \right) \times \left( {3\;{\rm{m}}} \right) \times \left( {\frac{{3\;{\rm{m}}}}{2}} \right) = 0\\ {M_D} = - \left( {8.48\;{\rm{kN}}} \right) \times \left( {3\;{\rm{m}}} \right) + \left( {9\;{\rm{kN}} \cdot {\rm{m}}} \right)\\ {M_D} = - 16.44\;{\rm{kN}} \cdot {\rm{m}} \end{array}\]

Here, a negative sign indicates that the moment at point D is in clock-wise direction.