Step 1

We are given the uniformly varying load over the beam is $w = 6\;{\rm{kN/m}}$.

We are asked the internal normal force, shear force, and moment at point C.

 
Step 2

The free body diagram of the system is:

We have the distance between the points A and B is $AB = 6\;{\rm{m}}$.

We have the distance between the points A and C is $AC = 3\;{\rm{m}}$.

We have the distance between the points B and C is $BC = 3\;{\rm{m}}$.

 
Step 3

The moment about point A is,

\[\begin{array}{c} \Sigma {M_A} = 0\\ {B_y}\left( {AB} \right) - \frac{1}{2}\left( w \right)\left( {AB} \right)\left( {\frac{{AB}}{3}} \right) = 0 \end{array}\]
 
Step 4

Substitute the values in the above expression.

\[\begin{array}{c} {B_y}\left( {6\;{\rm{m}}} \right) - \frac{1}{2}\left( {6\;{\rm{kN/m}}} \right)\left( {6\;{\rm{m}}} \right)\left( {\frac{{6\;{\rm{m}}}}{3}} \right) = 0\\ {B_y} = 6\;{\rm{kN}} \end{array}\]
 
Step 5

The forces acting in horizontal direction by equilibrium condition is,

\[{B_x} = 0\]
 
Step 6

The free body diagram of the segment BC is:

 
Step 7

The forces acting in horizontal direction by equilibrium condition is,

\[{N_C} = 0\]
 
Step 8

The forces acting in vertical direction by equilibrium condition is,

\[{V_C} + {B_y} - \frac{1}{2}\left( {\frac{w}{2}} \right)\left( {BC} \right) = 0\]
 
Step 9

Substitute the values in the above expression.

\[\begin{array}{l} {V_C} + \left( {6\;{\rm{kN}}} \right) - \frac{1}{2}\left( {\frac{{6\;{\rm{kN/m}}}}{2}} \right)\left( {3\;{\rm{m}}} \right) = 0\\ {V_C} = - 1.5\;{\rm{kN}} \end{array}\]
 
Step 10

The moment about point C is,

\[\begin{array}{c} \Sigma {M_C} = 0\\ {B_y}\left( {BC} \right) - \frac{1}{2}\left( {\frac{w}{2}} \right)\left( {BC} \right)\left( {\frac{{BC}}{3}} \right) - {M_C} = 0 \end{array}\]
 
Step 11

Substitute the values in the above expression.

\[\begin{array}{c} \left[ \begin{array}{l} \left( {6\;{\rm{kN}}} \right)\left( {3\;{\rm{m}}} \right) - \frac{1}{2}\left( {\frac{{6\;{\rm{kN/m}}}}{2}} \right)\left( {3\;{\rm{m}}} \right)\left( {\frac{{3\;{\rm{m}}}}{3}} \right)\\ - {M_C} \end{array} \right] = 0\\ {M_C} = 13.5\;{\rm{kN}} \cdot {\rm{m}} \end{array}\]