Step 1

We are given the following data:

The magnitude of the right angle triangular loading is ${W_1} = 900\;{\rm{N/m}}$.

The magnitude of the triangular loading is ${W_2} = 900\;{\rm{N/m}}$.

The horizontal distance between the point $A$ and the point $C$ is $AC = 6\;{\rm{m}}$.

The horizontal distance between the point $C$ and the point $D$ is $CD = 4\;{\rm{m}}$.

The horizontal distance between the point $D$ and the point $E$ is $DE = 3\;{\rm{m}}$.

The horizontal distance between the point $E$ and the point $B$ is $EB = 3\;{\rm{m}}$.

We are asked to determine the reactions at the supports $A$ , $E$ and $B$ of the compound beam.

 
Step 2

We will draw a free body diagram of the part $DEB$ of the beam.

Here, ${F_{{\rm{CD}}}}$ is the horizontal force in the member $CD$, ${E_y}$ and ${B_{\rm{y}}}$ are the vertical reaction force, ${F_1}$ is the total resultant force of the right triangular loading.

The formula to calculate the resultant force of the right angle triangular loading is given by,

\[{F_1} = \frac{1}{2} \times \left( {{W_2}} \right) \times \left( {DE + EB} \right)\]

Substitute all the known values in the above formula.

\[\begin{array}{c} {F_1} = \frac{1}{2} \times \left( {900\;{\rm{N/m}}} \right) \times \left( {3\;{\rm{m}}\;{\rm{ + }}\;{\rm{3}}\;{\rm{m}}} \right)\\ = \frac{{\left( {5400\;{\rm{N}}} \right)}}{2}\\ = 2700\;{\rm{N}} \end{array}\]
 
Step 3

The formula to calculate the distance of the resultant force of the right angle triangular loading from the point $B$ is given by,

\[{x_1} = \frac{2}{3} \times \left( {DE + EB} \right)\]

Substitute all the known values in the above formula.

\[\begin{array}{c} {x_1} = \frac{2}{3} \times \left( {3\;{\rm{m}}\;{\rm{ + }}\;{\rm{3}}\;{\rm{m}}} \right)\\ = \frac{{\left( {12\;{\rm{m}}} \right)}}{3}\\ = 4\;{\rm{m}} \end{array}\]
 
Step 4

We will take the sum of the moments about point $B$ equal to zero to determine the vertical force ${E_{\rm{y}}}$.

\[\begin{array}{c} \sum {{M_{\rm{B}}}} = 0\\ \left[ {\left( {{E_{\rm{y}}}} \right)\left( {EB} \right) - \left( {{F_1}} \right)\left( {{x_1}} \right)} \right] = 0 \end{array}\]

Substitute the known value in the above equation.

\[\begin{array}{c} \left[ {\left( {{E_{\rm{y}}}} \right)\left( {3\;{\rm{m}}} \right) - \left( {2700\;{\rm{N}}} \right)\left( {4\;{\rm{m}}} \right)} \right] = 0\\ \left( {{E_{\rm{y}}}} \right)\left( {3\;{\rm{m}}} \right) = \left( {10800\;{\rm{N}} \cdot {\rm{m}}} \right)\\ {E_{\rm{y}}} = 3600\;{\rm{N}} \end{array}\]
 
Step 5

We will take the sum of the force with $x$ components equal to zero to determine the horizontal force ${F_{{\rm{CD}}}}$.

\[\begin{array}{c} \sum {{F_{\rm{x}}}} = 0\\ {F_{{\rm{CD}}}} = 0 \end{array}\]
 
Step 6

We will take the sum of the force with $y$ components equal to zero to determine the vertical force ${B_{\rm{y}}}$.

\[\begin{array}{c} \sum {{F_{\rm{y}}}} = 0\\ \left[ { - {F_1} + {E_{\rm{y}}} - {B_{\rm{y}}}} \right] = 0 \end{array}\]

Substitute all the known values in the above formula.

\[\begin{array}{c} \left[ {\left( { - 2700\;{\rm{N}}} \right) + \left( {3600\;{\rm{N}}} \right) - \left( {{B_{\rm{y}}}} \right)} \right] = 0\\ \left[ { - {B_{\rm{y}}} + \left( {900\;{\rm{N}}} \right)} \right] = 0\\ {B_{\rm{y}}} = \;900\;{\rm{N}} \end{array}\]
 
Step 7

We will draw the free-body diagram of the part $AC$ of the beam.

The formula to calculate the resultant load of the triangular loading is given by,

\[{F_2} = \frac{1}{2} \times \left( {{W_2}} \right)\left( {AC} \right)\]

Here,${F_2}$ represent the resultant effective load of the triangular loading.

Substitute all the known values in the above formula.

\[\begin{array}{c} {F_2} = \frac{1}{2} \times \left( {900\;{\rm{N/m}}} \right)\left( {6\;{\rm{m}}} \right)\\ = \frac{{\left( {5400\;{\rm{N}}} \right)}}{2}\\ = 2700\;{\rm{N}} \end{array}\]
 
Step 8

The formula to calculate the effective distance of the resultant point load from the end $A$ is given by,

\[{x_2} = \frac{1}{2}\left( {AC} \right)\]

Substitute all the known values in the above formula.

\[\begin{array}{c} {x_2} = \frac{1}{2}\left( {6\;{\rm{m}}} \right)\\ = 3\;{\rm{m}} \end{array}\]
 
Step 9

We will take the sum of the moments about point $A$ equal to zero to determine the couple ${M_{\rm{A}}}$.

\[\begin{array}{c} \sum {{M_{\rm{A}}}} = 0\\ \left[ {\left( { - {M_{\rm{A}}}} \right) + \left( {{F_2}} \right)\left( {{x_2}} \right)} \right] = 0 \end{array}\]

Substitute the known value in the above equation.

\[\begin{array}{c} \left[ {\left( { - {M_{\rm{A}}}} \right) + \left( {2700\;{\rm{N}}} \right)\left( {3\;{\rm{m}}} \right)} \right] = 0\\ \left[ {\left( { - {M_{\rm{A}}}} \right) + \left( {8100\;{\rm{Nm}}} \right)} \right] = 0\\ {M_{\rm{A}}} = 8100\;{\rm{N}} \cdot {\rm{m}} \end{array}\]
 
Step 10

We will take the sum of the forces in $x$ direction equal to zero to determine the force ${A_{\rm{x}}}$.

\[\begin{array}{c} \sum {{F_{\rm{x}}}} = 0\\ {A_{\rm{x}}} = 0 \end{array}\]
 
Step 11

We will take the sum of the forces in $y$ direction to zero to determine the vertical force ${A_{\rm{y}}}$.

\[\begin{array}{c} \sum {{F_{\rm{y}}}} = 0\\ \left[ {{A_{\rm{y}}} - {F_2}} \right] = 0 \end{array}\]

Substitute the known value in the above equation.

\[\begin{array}{c} \left[ {{A_{\rm{y}}} - \left( {2700\;{\rm{N}}} \right)} \right] = 0\\ {A_{\rm{y}}} = 2700\;{\rm{N}} \end{array}\]