Step 1

We are given the load $P = 15\;{\rm{kN/m}}$, force $F = 8\,{\rm{kN}}$ and moment $M = 20\;{\rm{kN}} \cdot {\rm{m}}$.

We are asked to draw the shear and moment diagrams.

 
Step 2

The reaction force on the load is calculated as:

\[R = P \times {d_1}\]

Here, ${d_1}$ is the distance on which uniform distributed load is applied.

On plugging the values in the above relation, we get,

\[\begin{array}{l} R = \left( {15\;{\rm{kN/m}}} \right) \times \left( {3\;{\rm{m}}} \right)\\ R = 45\;{\rm{kN}} \end{array}\]

The free body diagram of the beam is shown below:

The length between reaction force and point $A$ is ${d_2} = 6.5\;{\rm{m}}$.

The length between force and point $A$ is ${d_3} = 3\;{\rm{m}}$.

The length between force at $B$ and point $A$ is ${d_4} = 5\;{\rm{m}}$.

To find the moment we will use the relation,

\[\begin{array}{c} \Sigma {M_A} = 0\\ M + R \times {d_2} + F \times {d_3} - {Y_B} \times {d_4} = 0 \end{array}\]

On plugging the values in the above relation, we get,

\[\begin{array}{c} 20\;{\rm{kN}} \cdot {\rm{m}} + \left( {45\;{\rm{kN}}} \right) \times \left( {6.5\;{\rm{m}}} \right) + \left( {8\;{\rm{kN}}} \right) \times \left( {3\;{\rm{m}}} \right) - {Y_B} \times \left( {5\;{\rm{m}}} \right) = 0\\ {Y_B} = 67.3\;{\rm{kN}} \end{array}\]

To find the force we will equate all the forces in horizontal direction by using relation,

\[\begin{array}{c} \Sigma {F_x} = 0\\ {X_A} = 0 \end{array}\]

To find the force we will equate all the forces in vertical direction by using relation,

\[\begin{array}{c} \Sigma {F_y} = 0\\ - {Y_A} - F - R + {Y_B} = 0 \end{array}\]

On plugging the values in the above relation, we get,

\[\begin{array}{c} - {Y_A} - 8\;{\rm{kN}} - 45\;{\rm{kN}} + 67.3\;{\rm{kN}} = 0\\ {Y_A} = 14.3\;{\rm{kN}} \end{array}\]
 
Step 3

To find the shear force we will use the relation,

\[\begin{array}{l} {F_{0 - 3}} = - {Y_A}\\ {F_{0 - 3}} = - 14.3\;{\rm{kN}} \end{array}\]

To find the shear force we will use the relation,

\[\begin{array}{c} {F_{3 - 5}} = {F_{0 - 3}} - F\\ {F_{3 - 5}} = - 14.3\;{\rm{kN}} - 8\;{\rm{kN}}\\ {F_{3 - 5}} = - 22.3\;{\rm{kN}} \end{array}\]

To find the shear force at $5\,{\rm{m}}$ we will use the relation,

\[\begin{array}{c} {F_{x = 5}} = {F_{3 - 5}} + {Y_B}\\ {F_{x = 5}} = - 22.3\;{\rm{kN}} + 67.3\;{\rm{kN}}\\ {F_{x = 5}} = 45\;{\rm{kN}} \end{array}\]

To find the shear force at $8\,{\rm{m}}$ we will use the relation,

\[\begin{array}{l} {F_{x = 8}} = {F_{x = 5}} - R\\ {F_{x = 8}} = 45\;{\rm{kN}} - 45\;{\rm{kN}}\\ {F_{x = 8}} = 0\;{\rm{kN}} \end{array}\]

The shear force diagram is shown below:

 
Step 4

To find the bending moment we will use the relation,

\[{M_{x = 0}} = 0\]

To find the bending moment at $2\;{\rm{m}}$ we will use the relation,

\[{M_{x = 2}} = {F_{0 - 3}} \times {d_0}\]

Here, ${d_0}$ is the horizontal distance obtained from shear force diagram between point $A$ and first shear force.

On plugging the values in the above relation we get,

\[\begin{array}{l} {M_{x = 2}} = - 14.3\;{\rm{kN}} \times 2\;{\rm{m}}\\ {M_{x = 2}} = - 28.6\;{\rm{kN}} \cdot {\rm{m}} \end{array}\]

To find the bending moment $2\;{\rm{m}}$ we will use the relation,

\[{M_{x = 2}}' = {M_{x = 2}} + M\]

On plugging the values in the above relation we get,

\[\begin{array}{l} {M_{x = 2}}' = - 28.6\;{\rm{kN}} \cdot {\rm{m}} + 20\;{\rm{kN}} \cdot {\rm{m}}\\ {M_{x = 2}}' = - 8.6\;{\rm{kN}} \cdot {\rm{m}} \end{array}\]

To find the bending moment at $3\;{\rm{m}}$ we will use the relation,

\[{M_{x = 3}} = {M_{x = 2}}' + {F_{0 - 3}} \times {d_0}'\]

Here, ${d_0}'$ is the horizontal distance obtained from shear force diagram between first shear force and second shear force.

On plugging the values in the above relation we get,

\[\begin{array}{l} {M_{x = 3}} = - 8.6\;{\rm{kN}} \cdot {\rm{m}} - 14.3\;{\rm{kN}} \times \left( {1\;{\rm{m}}} \right)\\ {M_{x = 3}} = - 22.9\;{\rm{kN}} \cdot {\rm{m}} \end{array}\]

To find the bending moment at $5\;{\rm{m}}$ we will use the relation,

\[{M_{x = 5}} = {M_{x = 3}} + {F_{3 - 5}} \times {d_0}''\]

Here, ${d_0}''$ is the horizontal distance obtained from shear force diagram between second shear force and third shear force.

On plugging the values in the above relation we get,

\[\begin{array}{l} {M_{x = 5}} = - 22.9\;{\rm{kN}} \cdot {\rm{m}} - 22.3\;{\rm{kN}} \times 2\;{\rm{m}}\\ {M_{x = 5}} = - 67.5\;{\rm{kN}} \cdot {\rm{m}} \end{array}\]

To find the bending moment at $8\;{\rm{m}}$ we will use the relation,

\[{M_{x = 8}} = {M_{x = 5}} + \frac{1}{2} \times R \times {d_0}'''\]

Here, ${d_0}'''$ is the horizontal distance obtained from shear force diagram between third shear force and last shear force.

On plugging the values in the above relation we get,

\[\begin{array}{l} {M_{x = 8}} = - 67.5\;{\rm{kN}} \cdot {\rm{m}} + \frac{1}{2} \times 45\,{\rm{kN}} \times 3\,{\rm{m}}\\ {M_{x = 8}} = 0\;{\rm{kN}} \cdot {\rm{m}} \end{array}\]

The bending moment diagram is shown below: