Step 1

We are given the weight of the beam is $W = 280\;{\rm{lb/ft}}$ and the length of the beam is $L = 3\;{\rm{ft}} + 7\;{\rm{ft}} = 10\;{\rm{ft}}$.

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

 
Step 2

The free body diagram of the system is:

We have the vertical distance between points A and B is $y = 8\;{\rm{ft}}$.

We have the horizontal distance between points A and B is $x = 6\;{\rm{ft}}$.

 
Step 3

The formula to calculate the angle $\theta $ is,

\[\tan \theta = \frac{y}{x}\]
 
Step 4

Substitute the values in the above expression.

\[\begin{array}{c} \tan \theta = \frac{{8\;{\rm{ft}}}}{{6\;{\rm{ft}}}}\\ \theta = 53.13^\circ \end{array}\]
 
Step 5

The expression to calculate the moment about point A is,

\[\begin{array}{c} \Sigma {M_A} = 0\\ - \left( {WL} \right)\left( {\frac{L}{2}\cos \theta } \right) + {B_y}\left( y \right) = 0 \end{array}\]
 
Step 6

Substitute the values in the above expression.

\[\begin{array}{c} - \left( {280\;{\rm{lb/ft}}} \right)\left( {10\;{\rm{ft}}} \right) \times \left( {\frac{{1\;{\rm{kip}}}}{{1000\;{\rm{lb}}}}} \right)\left( {\frac{{10\;{\rm{ft}}}}{2}\cos 53.13^\circ } \right) + {B_x}\left( {8\;{\rm{ft}}} \right) = 0\\ {B_x} = 1.05\;{\rm{kip}} \end{array}\]
 
Step 7

The expression to calculate horizontal forces by equilibrium condition is,

\[\begin{array}{c} \Sigma {F_x} = 0\\ {A_x} - {B_x} = 0 \end{array}\]
 
Step 8

Substitute the values in the above expression.

\[\begin{array}{c} {A_x} - 1.05\;{\rm{kip}} = 0\\ {A_x} = 1.05\;{\rm{kip}} \end{array}\]
 
Step 9

The expression to calculate vertical forces by equilibrium condition is,

\[\begin{array}{c} \Sigma {F_y} = 0\\ {A_y} - WL = 0 \end{array}\]
 
Step 10

Substitute the values in the above expression.

\[\begin{array}{c} {A_y} - \left( {280\;{\rm{lb/ft}}} \right)\left( {10\;{\rm{ft}}} \right) \times \left( {\frac{{1\;{\rm{kip}}}}{{1000\;{\rm{lb}}}}} \right) = 0\\ {A_y} = 2.8\;{\rm{kip}} \end{array}\]
 
Step 11

The free body diagram of a segment AC as shown below:

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

 
Step 12

The expression to calculate horizontal forces by equilibrium condition is,

\[\begin{array}{c} \Sigma {F_x} = 0\\ {N_C} + {A_x}\cos \theta + {A_y}\sin \theta - Wd\sin \theta = 0 \end{array}\]
 
Step 13

Substitute the values in the above expression.

\[\begin{array}{c} \left[ \begin{array}{c} {N_C} + \left( {1.05\;{\rm{kip}}} \right)\cos 53.13^\circ + \left( {2.8\;{\rm{kip}}} \right)\sin 53.13^\circ \\ - \left( {280\;{\rm{lb/ft}}} \right)\left( {3{\rm{ft}}} \right) \times \left( {\frac{{1\;{\rm{kip}}}}{{1000\;{\rm{lb}}}}} \right)\sin 53.13^\circ \end{array} \right] = 0\\ {N_C} = - 2.2\;{\rm{kip}} \end{array}\]
 
Step 14

The expression to calculate vertical forces by equilibrium condition is,

\[\begin{array}{c} \Sigma {F_y} = 0\\ - {V_C} - Wd\cos \theta + {A_y}\cos \theta - {A_x}\sin \theta = 0 \end{array}\]
 
Step 15

Substitute the values in the above expression.

\[\begin{array}{c} \left[ \begin{array}{c} - {V_C} - \left( {280\;{\rm{lb/ft}}} \right)\left( {3\;{\rm{ft}}} \right) \times \left( {\frac{{1\;{\rm{kip}}}}{{1000\;{\rm{lb}}}}} \right)\cos 53.13^\circ + \left( {2.8\;{\rm{kip}}} \right)\cos 53.13^\circ \\ - \left( {1.05\;{\rm{kip}}} \right)\sin 53.13^\circ \end{array} \right] = 0\\ {V_C} = 0.33\;{\rm{kip}} \end{array}\]
 
Step 16

The expression for the moment of all the forces about point C is,

\[\begin{array}{c} \Sigma {M_C} = 0\\ - {A_y}\cos \theta \left( d \right) + {A_x}\sin \theta \left( d \right) + Wd\cos \theta \left( {\frac{d}{2}} \right) + {M_C} = 0 \end{array}\]
 
Step 17

Substitute the values in the above expression.

\[\begin{array}{c} \left[ \begin{array}{c} - \left( {2.8\;{\rm{kip}}} \right)\cos 53.13^\circ \left( {3\;{\rm{ft}}} \right)\\ + \left( {1.05\;{\rm{kip}}} \right)\sin 53.13^\circ \left( {3\;{\rm{ft}}} \right)\\ + \left( {280\;{\rm{lb/ft}}} \right)\left( {3\;{\rm{ft}}} \right) \times \left( {\frac{{1\;{\rm{kip}}}}{{1000\;{\rm{lb}}}}} \right)\cos 53.13^\circ \left( {\frac{{3\;{\rm{ft}}}}{2}} \right)\\ + {M_C} \end{array} \right] = 0\\ {M_C} = 1.76\;{\rm{kip}} \cdot {\rm{ft}}\;\left( {{\rm{Counter clockwise}}} \right) \end{array}\]