Step 1

We are given the distributed loading over the beam.

We are asked an equivalent resultant force and couple moment acting at point A.

 
Step 2

The following is the diagram of the beam:

We have the length of the beam between point A and B is $l = 6\;{\rm{m}}$.

We have the magnitude of force per unit length acting at point A is ${F_1} = 6\;{\rm{kN/m}}$.

We have the magnitude of force per unit length acting at point B is ${F_2} = 6\;{\rm{kN/m}}$.

We have the magnitude of force per unit length acting at midpoint of the beam is ${F_3} = 3\;{\rm{kN}}$.

 
Step 3

The equivalent resultant force acting over the beam is,

\[F = \frac{1}{2}\left[ {\left( {{F_1} - {F_3}} \right)\;\left( {\frac{l}{2}} \right)} \right] + \left( {{F_3}} \right)\left( l \right) + \frac{1}{2}\left[ {\left( {{F_2} - {F_3}} \right)\;\left( {\frac{l}{2}} \right)} \right]\]
 
Step 4

Substitute the values in the above expression.

\begin{array}{l} F = \frac{1}{2}\left[ {\left( {6 - 3} \right)\;{\rm{kN/m}}\left( {\frac{{6\;{\rm{m}}}}{2}} \right)} \right] + \left( {3\;{\rm{kN/m}}} \right)\left( {6\;{\rm{m}}} \right) + \frac{1}{2}\left[ {\left( {6 - 3} \right)\;{\rm{kN/m}}\left( {\frac{{6\;{\rm{m}}}}{2}} \right)} \right]\\ F = \frac{1}{2}\left[ {\left( {6 - 3} \right)\;{\rm{kN/m}}\left( {3\;{\rm{m}}} \right)} \right] + \left( {3\;{\rm{kN/m}}} \right)\left( {6\;{\rm{m}}} \right) + \frac{1}{2}\left[ {\left( {6 - 3} \right)\;{\rm{kN/m}}\left( {3\;{\rm{m}}} \right)} \right]\\ F = 27\;{\rm{kN}}\;\left( {{\rm{down}}} \right) \end{array}
 
Step 5

The couple moment acting at point A is,

\[M = \frac{1}{2}\left[ {\left( {{F_1} - {F_3}} \right)\;\left( {\frac{l}{2}} \right)\left( {\frac{1}{3} \times \frac{l}{2}} \right)} \right] + \left( {{F_3}} \right)\left( l \right)\left( {\frac{l}{2}} \right) + \frac{1}{2}\left[ {\left( {{F_2} - {F_3}} \right)\;\left( {\frac{l}{2}} \right)\left( {l - \frac{1}{3} \times \frac{l}{2}} \right)} \right]\]
 
Step 6

Substitute the values in the above expression.

\[\begin{array}{l} M = \left\{ \begin{array}{l} \frac{1}{2}\left[ {\left( {6 - 3} \right)\;{\rm{kN/m}}\left( {\frac{{6\;{\rm{m}}}}{2}} \right)\left( {\frac{1}{3} \times \frac{{6\;{\rm{m}}}}{2}} \right)} \right] + \left( {3\;{\rm{kN/m}}} \right)\left( {6\;{\rm{m}}} \right)\left( {\frac{{6\;{\rm{m}}}}{2}} \right)\\ + \frac{1}{2}\left[ {\left( {6 - 3} \right)\;{\rm{kN/m}}\left( {\frac{{6\;{\rm{m}}}}{2}} \right)\left( {6\;{\rm{m}} - \frac{1}{3} \times \frac{{6\;{\rm{m}}}}{2}} \right)} \right] \end{array} \right\}\\ M = 4.5\;{\rm{kN}} \cdot {\rm{m}} + 54\;{\rm{kN}} \cdot {\rm{m}} + 22.5\;{\rm{kN}} \cdot {\rm{m}}\\ M = 81\;{\rm{kN}} \cdot {\rm{m}} \end{array}\]