Step 1

We are given that the first force acting on beam is ${F_1} = 800\;{\rm{lb}}$, the second force acting on beam is ${F_2} = 700\;{\rm{lb}}$ and the third force acting on beam is ${F_3} = 600\;{\rm{lb}}$. The angle of tilt of beam is $\theta = 30^\circ $.

We are asked to calculate the internal force, shear force and moment at point D.

We have the distance between point A and C is ${d_1} = 1.5\;{\rm{ft}}$.

We have the distance between point C and point of first is ${d_2} = 1.5\;{\rm{ft}}$.

We have the distance between two points of forces is ${d_3} = 3\;{\rm{ft}}$.

We have the distance between point of second force and point D is ${d_4} = 2\;{\rm{ft}}$.

We have the distance between point of point D and point of third force is ${d_5} = 1\;{\rm{ft}}$.

We have the distance between point of third force and point B is ${d_6} = 3\;{\rm{ft}}$.

 
Step 2

The free body diagram of beam is shown as:

Here, ${A_x}$ is the reaction force at point A, ${A_y}$ is the reaction force at point A and ${B_y}$ is the reaction force at point B.

 
Step 3

Applying the equilibrium force of equation along x-axis:

\[\begin{array}{c} \sum {F_x} = 0\\ {A_x} + {F_1}\sin \theta - {F_3}\sin \theta = 0 \end{array}\]
 
Step 4

Substitute the known values in the equation:

\[\begin{array}{c} \left\{ \begin{array}{l} {A_x} + \left( {800\;{\rm{lb}}} \right)\sin 30^\circ \\ - \left( {600\;{\rm{lb}}} \right)\sin 30^\circ \end{array} \right\} = 0\\ {A_x} = - \left( {800\;{\rm{lb}}} \right)\left( {0.5} \right) + \left( {600\;{\rm{lb}}} \right)\left( {0.5} \right)\\ = - 400\;{\rm{lb}} + 300\;{\rm{lb}}\\ = - 100\;{\rm{lb}} \end{array}\]
 
Step 5

Applying the equilibrium force of equation along y-axis:

\[\begin{array}{c} \sum {F_y} = 0\\ {A_y} + {B_y} - {F_1}\cos \theta - {F_2} - {F_3}\cos \theta = 0 \end{array}\]
 
Step 6

Substitute the known values in the equation:

\[\begin{array}{c} \left\{ \begin{array}{l} {A_y} + {B_y} - \left( {800\;{\rm{lb}}} \right)\cos 30^\circ \\ - 700\;{\rm{lb}} - \left( {600\;{\rm{lb}}} \right)\cos 30^\circ \end{array} \right\} = 0\\ {A_y} + {B_y} = \left( {800\;{\rm{lb}}} \right)\left( {0.866} \right) + 700\;{\rm{lb}} + \left( {600\;{\rm{lb}}} \right)\left( {0.866} \right)\\ = 692.8\;{\rm{lb}} + 700\;{\rm{lb}} + 519.6\;{\rm{lb}}\\ = 1912.4\;{\rm{lb}} \end{array}\]…... (1)
 
Step 7

To calculate the horizontal distance between point A and point of second force we use the formula:

\[x = \left( {{d_1} + {d_2} + {d_3}} \right)\cos \theta \]
 
Step 8

Substitute the known values in the equation:

\[\begin{array}{c} x = \left( {1.5\;{\rm{ft}} + 1.5\;{\rm{ft}} + 3\;{\rm{ft}}} \right)\cos 30^\circ \\ = \left( {6\;{\rm{ft}}} \right)\left( {0.866} \right)\\ = 5.196\;{\rm{ft}} \end{array}\]
 
Step 9

To calculate the horizontal distance between point B and point of second force we use the formula:

\[y = \left( {{d_4} + {d_5} + {d_6}} \right)\cos \theta \]
 
Step 10

Substitute the known values in the equation:

\[\begin{array}{c} y = \left( {2\;{\rm{ft}} + 1\;{\rm{ft}} + 3\;{\rm{ft}}} \right)\cos 30^\circ \\ = \left( {6\;{\rm{ft}}} \right)\left( {0.866} \right)\\ = 5.196\;{\rm{ft}} \end{array}\]
 
Step 11

To calculate the horizontal distance between point G and point of third force we use the formula:

\[x' = \left( {{d_4} + {d_5}} \right)\cos \theta \]
 
Step 12

Substitute the known values in the equation:

\[\begin{array}{c} x' = \left( {2\;{\rm{ft}} + 1\;{\rm{ft}}} \right)\cos 30^\circ \\ = \left( {3\;{\rm{ft}}} \right)\left( {0.866} \right)\\ = 2.598\;{\rm{ft}} \end{array}\]
 
Step 13

To calculate the vertical distance between point G and point of second force we use the formula:

\[y' = \left( {{d_4} + {d_5}} \right)\sin \theta \]
 
Step 14

Substitute the known values in the equation:

\[\begin{array}{c} y' = \left( {2\;{\rm{ft}} + 1\;{\rm{ft}}} \right)\sin 30^\circ \\ = \left( {3\;{\rm{ft}}} \right)\left( {0.5} \right)\\ = 1.5\;{\rm{ft}} \end{array}\]
 
Step 15

Applying the moment of force equation about point A:

\[{B_y}\left( {x + y} \right) - {F_1}\left( {{d_1} + {d_2}} \right) - {F_2}x - {F_3}\cos \theta \left( {x + x'} \right) - {F_3}\sin \theta \left( {{d_6}\sin \theta } \right) = 0\]
 
Step 16

Substitute the known values in the equation:

\[\begin{array}{c} \left\{ \begin{array}{l} {B_y}\left( {{\rm{5}}{\rm{.196}}\;{\rm{ft}} + 5.196\;{\rm{ft}}} \right) - \\ \left( {800\;{\rm{lb}}} \right)\left( {1.5\;{\rm{ft}} + {\rm{1}}{\rm{.5}}\;{\rm{ft}}} \right) - \\ \left( {700\;{\rm{lb}}} \right)\left( {5.196\;{\rm{ft}}} \right) - \\ \left( {600\;{\rm{lb}}} \right)\left( {{\rm{cos30}}^\circ } \right)\left( {5.196\;{\rm{ft}} + 2.598\;{\rm{ft}}} \right) + \\ \left( {600\;{\rm{lb}}} \right)\left( {\sin 30^\circ } \right)\left( {\left( {3\;{\rm{ft}}} \right)\sin 30^\circ } \right) \end{array} \right\} = 0\\ {B_y}\left( {10.392\;{\rm{ft}}} \right) = \left\{ \begin{array}{l} \left( {800\;{\rm{lb}}} \right)\left( {3\;{\rm{ft}}} \right) + \left( {700\;{\rm{lb}}} \right)\left( {5.196\;{\rm{ft}}} \right)\\ + \left( {600\;{\rm{lb}}} \right)\left( {0.866} \right)\left( {7.794\;{\rm{ft}}} \right)\\ - \left( {600\;{\rm{lb}}} \right)\left( {0.5} \right)\left( {3\;{\rm{ft}}} \right)\left( {0.5} \right) \end{array} \right\}\\ {B_y} = \frac{{\left\{ \begin{array}{l} 2400\;{\rm{lb}} \cdot {\rm{ft}} + 3637.2\;{\rm{lb}} \cdot {\rm{ft}}\\ + {\rm{4049}}{\rm{.8 lb}} \cdot {\rm{ft}} - 450\;{\rm{lb}} \cdot {\rm{f}} \end{array} \right\}{\rm{t}}}}{{10.392\;{\rm{ft}}}}\\ = 927.4\;{\rm{lb}} \end{array}\]
 
Step 17

Substitute the known value in equation (1):

\[\begin{array}{c} {A_y} + 927.4\;{\rm{lb}} = 1912.4\;{\rm{lb}}\\ {A_y} = 1912.4\;{\rm{lb}} - 927.4\;{\rm{lb}}\\ = 985\;{\rm{lb}} \end{array}\]
 
Step 18

The free body diagram of segment BD is shown as:

 
Step 19

Applying the equilibrium force of equation along x-axis at point D:

\[\begin{array}{c} \sum {F_x} = 0\\ {V_D} - {F_3} + {B_y}\cos \theta = 0 \end{array}\]
 
Step 20

Substitute the known values in the equation:

\[\begin{array}{c} {V_D} - 600\;{\rm{lb}} + \left( {927.4\;{\rm{lb}}} \right)\cos 30^\circ = 0\\ {V_D} = 600\;{\rm{lb}} - \left( {927.4\;{\rm{lb}}} \right)\left( {0.866} \right)\\ = 600\;{\rm{lb}} - 803.13\;{\rm{lb}}\\ = - 203.13\;{\rm{lb}} \end{array}\]
 
Step 21

Applying the equilibrium force of equation along y-axis at point D:

\[\begin{array}{c} \sum {F_y} = 0\\ {N_D} - {B_y}\sin \theta = 0 \end{array}\]
 
Step 22

Substitute the known values in the equation:

\[\begin{array}{c} {N_D} - \left( {927.4\;{\rm{lb}}} \right)\sin 30^\circ = 0\\ {N_D} = \left( {927.4\;{\rm{lb}}} \right)\left( {0.5} \right)\\ = 463.7\;{\rm{lb}} \end{array}\]
 
Step 23

Applying the moment of force equation about point D:

\[ - {M_D} - {F_3}{d_5} + {B_y}\left( {{d_5} + {d_6}} \right)\cos \theta = 0\]
 
Step 24

Substitute the known values in the equation:

\[\begin{array}{c} \left\{ \begin{array}{l} - {M_D} - \left( {600\;{\rm{lb}}} \right)\left( {1\;{\rm{ft}}} \right) + \\ \left( {927.4\;{\rm{lb}}} \right)\left( {1\,{\rm{ft}} + 3\;{\rm{ft}}} \right)\cos 30^\circ \end{array} \right\} = 0\\ {M_D} = - 600\;{\rm{lb}} \cdot {\rm{ft}} + \left( {927.4\;{\rm{lb}}} \right)\left( {4\;{\rm{ft}}} \right)\left( {0.866} \right)\\ = - 600\;{\rm{lb}} \cdot {\rm{ft}} + 3212.5\;{\rm{lb}} \cdot {\rm{ft}}\\ = 2612\;{\rm{lb}} \cdot {\rm{ft}} \end{array}\]