Step 1

We are given the load $w = 9\;{\rm{kN/m}}$.

We are asked to draw the shear and moment diagrams.

 
Step 2

The free body diagram of the beam is shown below:

The length of between the reaction force and point $E$ is ${d_2} = 2\,{\rm{m}}$.

The reaction force on the load is calculated as:

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

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

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

\[\begin{array}{l} R = \left( {9\;{\rm{kN/m}}} \right) \times \left( {4\;{\rm{m}}} \right)\\ R = 36\;{\rm{kN}} \end{array}\]

To find the moment we will use the relation,

\[\begin{array}{c} \Sigma {M_E} = 0\\ R \times {d_2} - {Y_F} \times {d_1} = 0 \end{array}\]

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

\[\begin{array}{c} 36\;{\rm{kN}} \times \left( {2\;{\rm{m}}} \right) - {Y_F} \times \left( {4\;{\rm{m}}} \right) = 0\\ {Y_F} = 18\;{\rm{kN}} \end{array}\]

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

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

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

\[\begin{array}{c} \Sigma {F_y} = 0\\ {Y_E} - R + {Y_F} = 0 \end{array}\]

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

\[\begin{array}{c} {Y_E} - 36\;{\rm{kN}} + 18\;{\rm{kN}} = 0\\ {Y_E} = 18\;{\rm{kN}} \end{array}\]
 
Step 3

The free body diagram of the beam is shown below:

The length between force at $B$ and point $A$ is ${d_3} = 4.5\;{\rm{m}}$.

To find the moment we will use the relation,

\[\begin{array}{c} \Sigma {M_A} = 0\\ \frac{1}{2}w{d_3}^2 \times \left( {\frac{2}{3}} \right) - {Y_B} \times {d_3} = 0 \end{array}\]

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

\[\begin{array}{c} \frac{1}{2}\left( {9\;{\rm{kN/m}}} \right){\left( {4.5\;{\rm{m}}} \right)^2} \times \left( {\frac{2}{3}} \right) - {Y_B} \times \left( {4.5\;{\rm{m}}} \right) = 0\\ {Y_B} = 13.5\,{\rm{kN}} \end{array}\]

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

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

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

\[\begin{array}{c} \Sigma {F_y} = 0\\ {Y_A} - \frac{1}{2} \times w \times {d_3} + {Y_B} = 0 \end{array}\]

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

\[\begin{array}{c} {Y_A} - \left( {\frac{1}{2}} \right)\left( {9\;{\rm{kN/m}}} \right)\left( {4.5\;{\rm{m}}} \right) + 13.5\,{\rm{kN}} = 0\\ {Y_A} = 6,75\;{\rm{kN}} \end{array}\]
 
Step 4

The free body diagram of the beam is shown below:

The length between force at $B$ and point $C$ is ${d_4} = 2\;{\rm{m}}$.

The length on which load is acting is equal to${d_5} = 6\;{\rm{m}}$.

The length between reaction force and point $C$ is ${d_6} = 1\;{\rm{m}}$.

The length between force at $D$ and point $C$ is ${d_7} = 2\;{\rm{m}}$.

The length between force at $E$ and point $C$ is ${d_8} = 4\;{\rm{m}}$.

To find the moment we will use the relation,

\[\begin{array}{c} \Sigma {M_C} = 0\\ - {Y_B} \times {d_4} + w \times {d_5} \times {d_6} - {Y_D} \times {d_7} + {Y_E} \times {d_8} = 0 \end{array}\]

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

\[\begin{array}{c} - \left( {13.5\;{\rm{kN}}} \right) \times \left( {2\;{\rm{m}}} \right) + \left( {9\;{\rm{kN/m}}} \right) \times \left( {6\;{\rm{m}}} \right) \times \left( {1\;{\rm{m}}} \right) - {Y_D} \times \left( {2\;{\rm{m}}} \right) + \left( {18\;{\rm{kN}}} \right) \times \left( {4\;{\rm{m}}} \right) = 0\\ {Y_D} = 49.5\;{\rm{kN}} \end{array}\]

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

\[\begin{array}{c} \Sigma {F_y} = 0\\ {Y_C} + {Y_D} - {Y_B} - \omega \times {d_5} - {Y_E} = 0 \end{array}\]

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

\[\begin{array}{c} {Y_C} + \left( {49.5\;{\rm{kN}}} \right) - \left( {13.5\;{\rm{kN}}} \right) - \left( {9\;{\rm{kN/m}}} \right) \times \left( {6\;{\rm{m}}} \right) - \left( {18\;{\rm{kN}}} \right) = 0\\ {Y_C} = 36\,{\rm{kN}} \end{array}\]
 
Step 5

The free body diagram of the entire system is shown below:

To find the shear force we will use the relation,

\[\begin{array}{l} {F_{x = 0}} = {Y_A}\\ {F_{x = 0}} = 6.75\;{\rm{kN}} \end{array}\]

To find the shear force at $4.5\;{\rm{m}}$we will use the relation,

\[\begin{array}{c} {F_{x = 4.5}} = {F_{x = 0}} - \left( {\frac{1}{2}w{d_3}} \right)\\ {F_{x = 4.5}} = 6.75\;{\rm{kN}} - \left( {\frac{1}{2} \times \left( {9\;{\rm{kN/m}}} \right) \times \left( {4.5\;{\rm{m}}} \right)} \right)\\ {F_{x = 4.5}} = - 13.5\;{\rm{kN}} \end{array}\]

To find the shear force at $6.5\,{\rm{m}}$ we will use the relation,

\[\begin{array}{c} {F_{x = 6.5}} = {F_{x = 4.5}} - w \times {d_2}\\ {F_{x = 6.5}} = - 13.5\;{\rm{kN}} - \left( {9\;{\rm{kN/m}}} \right) \times \left( {2\,{\rm{m}}} \right)\\ {F_{x = 6.5}} = - 31.5\;{\rm{kN}} \end{array}\]

To find the shear force at $6.5\,{\rm{m}}$ we will use the relation,

\[\begin{array}{c} {F_{x = 6.5}}' = {F_{x = 6.5}} + {Y_C}\\ {F_{x = 6.5}}' = - 31.5\;{\rm{kN}} + \left( {36\;{\rm{kN}}} \right)\\ {F_{x = 6.5}}' = 4.5\;{\rm{kN}} \end{array}\]

To find the shear force at $8.5\,{\rm{m}}$ we will use the relation,

\[\begin{array}{l} {F_{x = 8.5}} = {F_{x = 6.5}}' - w \times {d_2}\\ {F_{x = 8.5}} = 4.5\;{\rm{kN}} - \left( {9\;{\rm{kN/m}}} \right) \times \left( {2\;{\rm{m}}} \right)\\ {F_{x = 8.5}} = - 13.5\;{\rm{kN}} \end{array}\]

To find the shear force at $8.5\,{\rm{m}}$ we will use the relation,

\[\begin{array}{l} {F_{x = 8.5}}' = {F_{x = 8.5}} + {Y_D}\\ {F_{x = 8.5}}' = \left( { - 13.5\;{\rm{kN}}} \right) + \left( {49.5\;{\rm{kN}}} \right)\\ {F_{x = 8.5}}' = 36\;{\rm{kN}} \end{array}\]

To find the shear force at $10.5\,{\rm{m}}$ we will use the relation,

\[\begin{array}{l} {F_{x = 10.5}} = {F_{x = 8.5}} - w \times {d_2}\\ {F_{x = 10.5}} = \left( {36\;{\rm{kN}}} \right) - \left( {9\;{\rm{kN/m}}} \right) \times \left( {2\;{\rm{m}}} \right)\\ {F_{x = 10.5}} = 18\;{\rm{kN}} \end{array}\]

To find the shear force at $14.5\,{\rm{m}}$ we will use the relation,

\[\begin{array}{l} {F_{x = 14.5}} = {F_{x = 10.5}}' - w \times {d_8}\\ {F_{x = 14.5}} = \left( {18\;{\rm{kN}}} \right) - \left( {9\;{\rm{kN/m}}} \right) \times \left( {4\;{\rm{m}}} \right)\\ {F_{x = 14.5}} = - 18\;{\rm{kN}} \end{array}\]

To find the shear force at $14.5\,{\rm{m}}$ we will use the relation,

\[\begin{array}{l} {F_{x = 14.5}}' = {F_{x = 14.5}} + {Y_E}\\ {F_{x = 14.5}}' = \left( { - 18\;{\rm{kN}}} \right) + \left( {18\;{\rm{kN}}} \right)\\ {F_{x = 14.5}}' = 0\;{\rm{kN}} \end{array}\]

The shear force diagram is shown below:

 
Step 6

To find the bending moment we will use the relation,

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

To find the bending moment at $2.598\;{\rm{m}}$ we will use the relation,

\[{M_{x = 2.598}} = 11.69\;{\rm{kN}} \cdot {\rm{m}}\]

To find the bending moment $4.5\;{\rm{m}}$ we will use the relation,

\[{M_{x = 4.5}} = 0\]

To find the bending moment at $6.5\;{\rm{m}}$ we will use the relation,

\[{M_{x = 6.5}} = {M_{x = 4.5}} - \left( {\frac{1}{2}} \right)\left( {{F_{x = 8.5}} + {F_{x = 6.5}}} \right) \times {d_2}\]

On plugging the values in the above relation we get,

\[\begin{array}{l} {M_{x = 6.5}} = 0 - \left( {\frac{1}{2}} \right)\left( {13.5\;{\rm{kN}} + 31.5\;{\rm{kN}}} \right) \times \left( {2\;{\rm{m}}} \right)\\ {M_{x = 6.5}} = - 45\;{\rm{kN}} \cdot {\rm{m}} \end{array}\]

To find the bending moment at $7\;{\rm{m}}$ we will use the relation,

\[{M_{x = 7}} = {M_{x = 6.5}} + \frac{1}{2} \times {F_{x = 6.5}} \times {d_0}\]

Here, ${d_0}$ is the horizontal distance obtained from shear force diagram.

On plugging the values in the above relation we get,

\[\begin{array}{l} {M_{x = 7}} = \left( { - 45\;{\rm{kN}} \cdot {\rm{m}}} \right) + \frac{1}{2} \times \left( {4.5\;{\rm{kN}}} \right) \times \left( {0.5\;{\rm{m}}} \right)\\ {M_{x = 7}} = - 43.875\;{\rm{kN}} \cdot {\rm{m}} \end{array}\]

To find the bending moment at $8.5\;{\rm{m}}$ we will use the relation,

\[{M_{x = 8.5}} = {M_{x = 7}} - \frac{1}{2} \times {F_{x = 8.5}} \times \left( {{d_2} - {d_0}} \right)\]

On plugging the values in the above relation we get,

\[\begin{array}{l} {M_{x = 8.5}} = - 43.875\;{\rm{kN}} \cdot {\rm{m}} - \frac{1}{2} \times \left( {13.5\;{\rm{kN}}} \right) \times \left( {2\;{\rm{m}} - 0.5\;{\rm{m}}} \right)\\ {M_{x = 8.5}} = - 54\;{\rm{kN}} \cdot {\rm{m}} \end{array}\]

To find the bending moment at $12.5\;{\rm{m}}$ we will use the relation,

\[{M_{x = 12.5}} = {M_{x = 8.5}} + \frac{1}{2} \times {F_{x = 8.5'}} \times {d_8}\]

On plugging the values in the above relation we get,

\[\begin{array}{l} {M_{x = 12.5}} = - 54\;{\rm{kN}} \cdot {\rm{m}} + \frac{1}{2} \times \left( {36\;{\rm{kN}}} \right) \times \left( {4\;{\rm{m}}} \right)\\ {M_{x = 12.5}} = 18\;{\rm{kN}} \cdot {\rm{m}} \end{array}\]

To find the bending moment at $14.5\;{\rm{m}}$ we will use the relation,

\[{M_{x = 14.5}} = {M_{x = 12.5}} - \frac{1}{2} \times {F_{x = 10.5}} \times \left( {{d_2}} \right)\]

On plugging the values in the above relation we get,

\[\begin{array}{l} {M_{x = 14.5}} = 18\;{\rm{kN}} \cdot {\rm{m}} - \frac{1}{2} \times \left( {18\;{\rm{kN}}} \right) \times \left( {2\;{\rm{m}}} \right)\\ {M_{x = 14.5}} = 0\;{\rm{kN}} \cdot {\rm{m}} \end{array}\]

The bending moment diagram is shown below: