Step 1

We are given the following data:

The magnitude of the uniformly varying load is $W = 4\;{\rm{kip/ft}}$.

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

The distance between the point $C$ and point $B$ is $BC = 6\;{\rm{ft}}$.

The distance between the point $A$ and point $B$ is $AB = 12\;{\rm{ft}}$.

We are asked to determine the internal shear force and moment acting at point $C$ in the beam.

 
Step 2

The equation to calculate the resultant force of the uniform varying loading is given by,

\[F = \frac{1}{2} \times \left( W \right) \times \left( {AC} \right)\]

Here, $F$ represent the resultant point force of the uniform variable load.

Substitute all the known values in the above formula.

\[\begin{array}{c} F = \frac{1}{2} \times \left( {4\;{\rm{kip/ft}}} \right)\left( {6\;{\rm{ft}}} \right)\\ = \frac{{\left( {24\;{\rm{kip}}} \right)}}{2}\\ = 12\;{\rm{kip}} \end{array}\]
 
Step 3

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

\[{x_1} = \frac{2}{3} \times \left( {AC} \right)\]

Here, ${x_1}$ represent the distance of the resultant point force from the point $A$.

Substitute all the known values in the above formula.

\[\begin{array}{c} {x_1} = \frac{2}{3} \times \left( {6\;{\rm{ft}}} \right)\\ = \frac{{\left( {12\;{\rm{ft}}} \right)}}{3}\\ = 4\;{\rm{ft}} \end{array}\]
 
Step 4

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

Here, ${A_y}$ and ${B_{\rm{y}}}$ are the vertical reaction force, and ${A_{\rm{x}}}$ is the horizontal force.

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

\[\begin{array}{c} \sum {{M_{\rm{A}}}} = 0\\ \left[ {\left( {{B_{\rm{y}}}} \right)\left( {AB} \right) - \left( F \right)\left( {{x_1}} \right)} \right] = 0 \end{array}\]

Substitute the known value in the above equation.

\[\begin{array}{c} \left[ {\left( {{B_{\rm{y}}}} \right)\left( {12\;{\rm{ft}}} \right) - \left( {12\;{\rm{kip}}} \right)\left( {4\;{\rm{ft}}} \right)} \right] = 0\\ \left[ {\left( {{B_{\rm{y}}}} \right)\left( {12\;{\rm{ft}}} \right) - \left( {48\;{\rm{kip}} \cdot {\rm{ft}}} \right)} \right] = 0\\ {B_{\rm{y}}} = 4\;{\rm{kip}} \end{array}\]
 
Step 5

We will draw the free-body diagram of the member $BC$ of the frame.

Here, ${V_{\rm{C}}}$ is the vertical shear force at point $C$ of the beam and ${M_{\rm{C}}}$ represent the moment at the point $C$.

We will take the sum of the forces along the $y$ axis and equate to zero to determine the shear force ${V_{\rm{C}}}$.

\[\begin{array}{c} \sum {{F_{\rm{y}}}} = 0\\ \left[ {{V_{\rm{C}}} + {B_{\rm{y}}}} \right] = 0 \end{array}\]

Substitute the known value in the above equation.

\[\begin{array}{c} \left[ {{V_{\rm{C}}} + \left( {4\;{\rm{kip}}} \right)} \right] = 0\\ {V_{\rm{C}}} = - \left( {4\;{\rm{kip}}} \right) \end{array}\]
 
Step 6

We will take the sum of the moment about the point $C$ and equate to zero to determine the moment acting at point $C$ .

\[\begin{array}{c} \sum {{M_{\rm{C}}}} = 0\\ \left[ {\left( {{B_{\rm{y}}}} \right)\left( {BC} \right) - {M_{\rm{C}}}} \right] = 0 \end{array}\]

Substitute the known value in the above equation.

\[\begin{array}{c} \left[ {\left( {4\;{\rm{kip}}} \right)\left( {6\;{\rm{ft}}} \right) - {M_{\rm{C}}}} \right] = 0\\ \left[ {\left( {24\;{\rm{kip}} \cdot {\rm{ft}}} \right) - {M_{\rm{C}}}} \right] = 0\\ {M_{\rm{C}}} = 24\;{\rm{kip}} \cdot {\rm{ft}} \end{array}\]