Step 1

We are given the force acting at point C is $P = 150\;{\rm{lb}}$.

We are asked to determine the internal normal force, shear force, and moment at points D and E of the frame.

 
Step 2

The free body diagram of the system is:

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

We have the distance between points A and F is $AF = 8\;{\rm{ft}}$.

We have the distance between points A and B is $AB = 8\;{\rm{ft}} + 4\;{\rm{ft}} = 12\;{\rm{ft}}$.

We have the distance between points B and F is $BF = 4\;{\rm{ft}}$.

We have the vertical distance between points D and F is $DF = 1\;{\rm{ft}}$.

We have the angle of member AC from horizontal is $\theta = 30^\circ $.

 
Step 3

The expression to calculate the moment about point A is,

\[\begin{array}{c} \Sigma {M_A} = 0\\ {R_B}\left( {AB} \right) - P\left( {AF\tan \theta } \right) = 0 \end{array}\]
 
Step 4

Substitute the values in the above expression.

\[\begin{array}{c} {R_B}\left( {12\;{\rm{ft}}} \right) - \left( {150\;{\rm{lb}}} \right)\left( {8\tan 30^\circ \;{\rm{ft}}} \right) = 0\\ {R_B} = 57.73\;{\rm{lb}} \end{array}\]
 
Step 5

The expression to calculate vertical forces by equilibrium condition is,

\[\begin{array}{c} \Sigma {F_y} = 0\\ {A_y} + {R_B} = 0 \end{array}\]
 
Step 6

Substitute the values in the above expression.

\[\begin{array}{c} {A_y} + \left( {57.73\;{\rm{lb}}} \right) = 0\\ {A_y} = - 57.73\;{\rm{lb}} \end{array}\]
 
Step 7

The expression to calculate horizontal forces by equilibrium condition is,

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

Substitute the values in the above expression.

\[\begin{array}{c} {A_x} + \left( {150\;{\rm{lb}}} \right) = 0\\ {A_x} = - 150\;{\rm{lb}} \end{array}\]
 
Step 9

The free body diagram of the member AC is shown as:

 
Step 10

The expression to calculate the moment about point A is,

\[\begin{array}{c} \Sigma {M_A} = 0\\ {F_{CF}}\left( {AF} \right) - P\left( {AF\tan \theta } \right) = 0 \end{array}\]
 
Step 11

Substitute the values in the above expression.

\[\begin{array}{c} {F_{CF}}\left( {8\;{\rm{ft}}} \right) - \left( {150\;{\rm{lb}}} \right)\left( {8\tan 30^\circ \;{\rm{ft}}} \right) = 0\\ {F_{CF}} = 86.60\;{\rm{lb}} \end{array}\]
 
Step 12

Take a section at point D, and the free body of the member CD is shown as:

 
Step 13

The expression to calculate vertical forces by equilibrium condition is,

\[\begin{array}{c} \Sigma {F_y} = 0\\ - {F_{CF}} - {N_D} = 0 \end{array}\]
 
Step 14

Substitute the values in the above expression.

\[\begin{array}{c} - \left( {86.60\;{\rm{lb}}} \right) - {N_D} = 0\\ {N_D} = - 86.60\;{\rm{lb}} \end{array}\]
 
Step 15

The expression to calculate horizontal forces by equilibrium condition is,

\[\begin{array}{c} \Sigma {F_x} = 0\\ {V_D} = 0 \end{array}\]
 
Step 16

The expression to calculate the moment about the section is,

\[\begin{array}{c} \Sigma {M_{{\rm{section}}}} = 0\\ {M_D} = 0 \end{array}\]
 
Step 17

Take a section at point E, and the free body of the member EB is shown as:

We have the distance between points B and E is $BE = 5\;{\rm{ft}} + 4\;{\rm{ft}} = 9\;{\rm{ft}}$.

We have the distance between points E and F is $EF = 5\;{\rm{ft}}$.

 
Step 18

The expression to calculate vertical forces by equilibrium condition is,

\[\begin{array}{c} \Sigma {F_y} = 0\\ {V_E} - {F_{CF}} + {R_B} = 0 \end{array}\]
 
Step 19

Substitute the values in the above expression.

\[\begin{array}{c} {V_E} - \left( {86.60\;{\rm{lb}}} \right) + 57.73 = 0\\ {V_E} = 28.9\;{\rm{lb}} \end{array}\]
 
Step 20

The expression to calculate horizontal forces by equilibrium condition is,

\[\begin{array}{c} \Sigma {F_x} = 0\\ {N_E} = 0 \end{array}\]
 
Step 21

The expression to calculate the moment about the section is,

\[\begin{array}{c} \Sigma {M_{{\rm{section}}}} = 0\\ - {M_E} - {F_{CF}}\left( {EF} \right) + {R_B}\left( {BE} \right) = 0 \end{array}\]
 
Step 22

Substitute the values in the above expression.

\[\begin{array}{c} - {M_E} - \left( {86.60\;{\rm{lb}}} \right)\left( {5\;{\rm{ft}}} \right) + \left( {57.73\;{\rm{lb}}} \right)9\left( {5\;{\rm{ft}}} \right) = 0\\ {M_E} = 86.60\;{\rm{lb}} \cdot {\rm{ft}} \end{array}\]