Step 1

We are given the varying load of $6\;{{{\rm{kN}}} \mathord{\left/ {\vphantom {{{\rm{kN}}} {\rm{m}}}} \right. } {\rm{m}}}$ at a distance of $3\;{\rm{m}}$ and one point load of $3\;{\rm{kN}}$ at support B.

We are asked to draw the shear and moment diagrams for the beam.

 
Step 2

The free-body diagram of the cantilever beam can be drawn as:

Here, ${A_x}$ is the horizontal reaction at support A, ${A_y}$ is the vertical reaction at support A and ${M_A}$ is the moment at support A.

On balancing the vertical forces, we get:

\[\begin{array}{c} \sum {F_V} = 0\\ {A_y} - \left( {\frac{1}{2} \times \left( {6\;{{{\rm{kN}}} \mathord{\left/ {\vphantom {{{\rm{kN}}} {\rm{m}}}} \right. } {\rm{m}}}} \right) \times \left( {3\;{\rm{m}}} \right)} \right) - \left( {3\;{\rm{kN}}} \right) = 0\\ {A_y} = \left( {3\;{\rm{kN}}} \right) + \left( {9\;{\rm{kN}}} \right)\\ {A_y} = 12\;{\rm{kN}} \end{array}\]

On balancing the horizontal forces, we get:

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

On taking the moment about point A, we get:

\[\begin{array}{c} {M_A} - \left[ {\left( {\frac{1}{2} \times \left( {6\;{{{\rm{kN}}} \mathord{\left/ {\vphantom {{{\rm{kN}}} {\rm{m}}}} \right. } {\rm{m}}}} \right) \times \left( {3\;{\rm{m}}} \right) \times \left( {1.5\;{\rm{m}}} \right)} \right)} \right] - \left[ {\left( {3\;{\rm{kN}}} \right) \times \left( {3\;{\rm{m}}} \right)} \right] = 0\\ {M_A} = \left\{ \begin{array}{l} \left( {9\;{\rm{kN}} \cdot {\rm{m}}} \right)\\ + \left( {13.5\;{\rm{kN}} \cdot {\rm{m}}} \right) \end{array} \right\}\\ {M_A} = 22.5\;{\rm{kN}} \cdot {\rm{m}} \end{array}\]
 
Step 3

The free-body diagram of the beam section from point C to support B can be drawn as:

Here, V is the shear force.

The shear force at point A can be calculated as:

\[\begin{array}{c} {V_A} = {A_y}\\ {V_A} = 12\;{\rm{kN}} \end{array}\]

The shear force at point C can be calculated as:

\[\begin{array}{c} {V_C} = {A_y} - \left( {\frac{1}{2} \times \left( {6\;{{{\rm{kN}}} \mathord{\left/ {\vphantom {{{\rm{kN}}} {\rm{m}}}} \right. } {\rm{m}}}} \right) \times \left( {1.5\;{\rm{m}}} \right)} \right)\\ {V_C} = \left( {12\;{\rm{kN}}} \right) - \left( {4.5\;{\rm{kN}}} \right)\\ {V_C} = 7.5\;{\rm{kN}} \end{array}\]

The shear force just at left of point B can be calculated as:

\[\begin{array}{c} {V_{{B_L}}} = {A_y} - \left( {\frac{1}{2} \times \left( {6\;{{{\rm{kN}}} \mathord{\left/ {\vphantom {{{\rm{kN}}} {\rm{m}}}} \right. } {\rm{m}}}} \right) \times \left( {1.5\;{\rm{m}}} \right)} \right) - \left( {\frac{1}{2} \times \left( {6\;{{{\rm{kN}}} \mathord{\left/ {\vphantom {{{\rm{kN}}} {\rm{m}}}} \right. } {\rm{m}}}} \right) \times \left( {1.5\;{\rm{m}}} \right)} \right)\\ {V_{{B_L}}} = \left( {12\;{\rm{kN}}} \right) - \left( {4.5\;{\rm{kN}}} \right) - \left( {4.5\;{\rm{kN}}} \right)\\ {V_{{B_L}}} = 3\;{\rm{kN}} \end{array}\]

The shear force just at right of point B can be calculated as:

\[\begin{array}{c} {V_{{B_R}}} = {V_{{B_L}}} - \left( {3\;{\rm{kN}}} \right)\\ {V_{{B_R}}} = \left( {3\;{\rm{kN}}} \right) - \left( {3\;{\rm{kN}}} \right)\\ {V_{{B_R}}} = 0 \end{array}\]

On taking the moment about point C, we get:

\[\begin{array}{c} \sum {M_C} = 0\\ \left[ { - {M_C} - \left( {\frac{1}{2} \times \left( {6\;{{{\rm{kN}}} \mathord{\left/ {\vphantom {{{\rm{kN}}} {\rm{m}}}} \right. } {\rm{m}}}} \right) \times \left( {1.5\;{\rm{m}}} \right) \times \left( {0.5\;{\rm{m}}} \right)} \right) - \left( {3\;{\rm{kN}}} \right) \times \left( {1.5\;{\rm{m}}} \right)} \right] = 0\\ - {M_C} = \left\{ \begin{array}{l} \left( {4.5\;{\rm{kN}} \cdot {\rm{m}}} \right)\\ + \left( {2.25\;{\rm{kN}} \cdot {\rm{m}}} \right) \end{array} \right\}\\ {M_C} = - 6.75\;{\rm{kN}} \cdot {\rm{m}} \end{array}\]
 
Step 4

The shear force and bending moment diagram for the beam can be drawn as: