Step 1

We are given the uniform load on the beam is w.

We are asked to the draw the shear and bending moment for each of the two segments on compound beam.

We have the distance between point A and B is ${d_1} = L$.

We have the distance between point B and E is ${d_2} = \frac{L}{3}$.

We have the distance between point E and F is ${d_3} = \frac{L}{3}$.

We have the distance between point F and C is ${d_4} = \frac{L}{3}$.

We have the distance between point C and D is ${d_5} = L$.

 
Step 2

The free body diagram of the beam EF is shown as:

Here, ${E_x}$ is the reaction force at point E, ${E_y}$ is the reaction force at point E,${F_x}$ is the reaction force at point F and ${F_y}$ is the reaction force at point F.

 
Step 3

Applying the moment of force of equation along x-axis at point E:

\[{F_y}{d_3} - w{d_3}\left( {\frac{{{d_3}}}{2}} \right) = 0\]
 
Step 4

Substitute the values in the above equation:

\[\begin{array}{c} {F_y}\left( {\frac{L}{3}} \right) - w\left( {\frac{L}{3}} \right)\left( {\frac{{\frac{L}{3}}}{2}} \right) = 0\\ {F_y}\left( {\frac{L}{3}} \right) = w\left( {\frac{L}{3}} \right)\left( {\frac{L}{6}} \right)\\ {F_y} = \frac{{wL}}{6} \end{array}\]
 
Step 5

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

\[\begin{array}{c} \Sigma {F_y} = 0\\ {E_y} + {F_y} - w{d_3} = 0 \end{array}\]
 
Step 6

Substitute the known values in the above equation:

\[\begin{array}{c} {E_y} + \frac{{wL}}{6} - \frac{{wL}}{3} = 0\\ {E_y} = \frac{{wL}}{6} \end{array}\]
 
Step 7

The free body diagram of the beam FD is shown as:

Here, ${F_x}$ is the reaction force at point F, ${F_y}$ is the reaction force at point F, ${C_x}$ is the reaction force at point C,${C_y}$ is the reaction force at point C, ${D_x}$ is the reaction force at point D and${D_y}$ is the reaction force at point D.

 
Step 8

Applying the moment of force of equation along x-axis at point C:

\[{F_y}{d_4} + {D_y}{d_5} - w\left( {{d_4} + {d_5}} \right)\left( {{d_4}} \right) = 0\]
 
Step 9

Substitute the values in the above equation:

\[\begin{array}{c} \left( {\frac{{wL}}{6}} \right)\left( {\frac{L}{3}} \right) + {D_y}L - w\left( {L + \frac{L}{3}} \right)\left( {\frac{L}{3}} \right) = 0\\ {D_y}L = \frac{{4w{L^2}}}{9} - \frac{{w{L^2}}}{{18}}\\ {D_y}L = \frac{{7w{L^2}}}{{18}}\\ {D_y} = \frac{{7wL}}{{18}} \end{array}\]
 
Step 10

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

\[\begin{array}{c} \Sigma {F_y} = 0\\ {C_y} - {F_y} + {D_y} - w\left( {{d_4} + {d_5}} \right) = 0 \end{array}\]
 
Step 11

Substitute the known values in the above equation:

\[\begin{array}{c} {C_y} - \frac{{wL}}{6} + \frac{{7wL}}{{18}} - w\left( {L + \frac{L}{3}} \right) = 0\\ {C_y} = \frac{{wL}}{6} - \frac{{7wL}}{{18}} + \frac{{4wL}}{3}\\ = \frac{{3wL - 7wL + 24wL}}{{18}}\\ = \frac{{10wL}}{9} \end{array}\]
 
Step 12

The free body diagram of beam AE is shown as:

 
Step 13

Applying the moment of force of equation at point B:

\[ - {A_y}{d_1} + w\left( {{d_1} + {d_2}} \right)\left( {{d_3}} \right) - {E_y}{d_3} = 0\]
 
Step 14

Substitute the values in the above equation:

\[\begin{array}{c} - {A_y}L + w\left( {L + \frac{L}{3}} \right)\left( {\frac{L}{3}} \right) - \left( {\frac{{wL}}{6}} \right)\left( {\frac{L}{3}} \right) = 0\\ {A_y}L = \frac{{4w{L^2}}}{9} - \frac{{w{L^2}}}{{18}}\\ {A_y}L = \frac{{8w{L^2} - w{L^2}}}{{18}}\\ {A_y} = \frac{{7wL}}{{18}} \end{array}\]
 
Step 15

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

\[\begin{array}{c} \Sigma {F_y} = 0\\ {A_y} + {B_y} - {E_y} - w\left( {{d_1} + {d_2}} \right) = 0 \end{array}\]
 
Step 16

Substitute the known values in the above equation:

\[\begin{array}{c} \frac{{7wL}}{{18}} + {B_y} - \frac{{wL}}{6} - w\left( {L + \frac{L}{3}} \right) = 0\\ {B_y} = \frac{{wL}}{6} - \frac{{7wL}}{{18}} + \frac{{4wL}}{3}\\ = \frac{{3wL - 7wL + 24wL}}{{18}}\\ = \frac{{10wL}}{9} \end{array}\]
 
Step 17

The free body diagram of the beam AB is shown as:

Here, $M$ is the moment of force and $V$ is the shear force and x is distance where uniform load is acting.

 
Step 18

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

\[\begin{array}{c} \Sigma {F_y} = 0\\ {A_y} - wx - V = 0 \end{array}\]
 
Step 19

Substitute the known values in the above equation:

\[\begin{array}{c} \frac{{7wL}}{{18}} - wx - V = 0\\ V = \frac{{7wL}}{{18}} - wx\\ = \frac{w}{{18}}(7L - 18x) \end{array}\]…… (1)
 
Step 20

At $x = 0$, the shear force at point A is calculated as:

\[\begin{array}{c} {V_A} = \frac{w}{{18}}\left( {7L - 18\left( 0 \right)} \right)\\ = \frac{w}{{18}}\left( {7L - 0} \right)\\ = \frac{{7wL}}{{18}} \end{array}\]
 
Step 21

At $x = L$, the shear force at point B is calculated as:

\[\begin{array}{c} {V_B} = \frac{w}{{18}}\left( {7L - 18L} \right)\\ = \frac{w}{{18}}\left( { - 11L} \right)\\ = - \frac{{11wL}}{{18}} \end{array}\]
 
Step 22

Assume that at $x = d$, the shear force of the beam is 0. Substituting the known values in equation (1):

\[\begin{array}{c} 0 = \frac{w}{{18}}\left( {7L - 18\left( d \right)} \right)\\ 7L - 18d = 0\\ 18d = 7L\\ d = \frac{{7L}}{{18}} \end{array}\]
 
Step 23

Applying the moment of force of equation along x-axis at section AB:

\[M + wx\left( {\frac{x}{2}} \right) - {A_y}x = 0\]
 
Step 24

Substitute the values in the above expression.

\[\begin{array}{c} M + wx\left( {\frac{x}{2}} \right) - \left( {\frac{{7wL}}{{18}}} \right)x = 0\\ M = \frac{{7wLx}}{{18}} - \frac{{w{x^2}}}{2}\\ M = \frac{w}{{18}}\left( {7Lx - 9{x^2}} \right) \end{array}\]…… (2)
Step 25

At $x = 0$, the moment of force at point A is calculated as:

\[\begin{array}{c} {M_A} = \frac{w}{{18}}\left( {7L\left( 0 \right) - 9{{\left( 0 \right)}^2}} \right)\\ = \frac{w}{{18}}\left( {0 - 0} \right)\\ = 0 \end{array}\]
 
Step 26

At $x = L$, the moment of force at point B is calculated as:

\[\begin{array}{c} {M_B} = \frac{w}{{18}}\left( {7L\left( L \right) - 9{{\left( L \right)}^2}} \right)\\ = \frac{w}{{18}}\left( {7{L^2} - 9{L^2}} \right)\\ = - \frac{{2w{L^2}}}{{18}}\\ = - \frac{{w{L^2}}}{9} \end{array}\]
 
Step 27

Substitute the known value of $x = \frac{{7L}}{{18}}$ in equation (2):

\[\begin{array}{c} {M_{AB}} = \frac{w}{{18}}\left( {7L\left( {\frac{{7L}}{{18}}} \right) - 9{{\left( {\frac{{7L}}{{18}}} \right)}^2}} \right)\\ = \frac{w}{{18}}\left( {\frac{{49{L^2}}}{{18}} - \frac{{441{L^2}}}{{324}}} \right)\\ = \frac{w}{{18}}\left( {\frac{{441{L^2}}}{{324}}} \right)\\ = \frac{{49w{L^2}}}{{648}} \end{array}\]
 
Step 28

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

Here, $M$ is the moment of force and $V$ is the shear force and x is distance where uniform load is acting.

 
Step 29

Applying the equilibrium force of equation along y-axis at section AC:

\[\begin{array}{c} \Sigma {F_y} = 0\\ {A_y} + {B_y} - wx - V = 0 \end{array}\]
 
Step 30

Substitute the known values in the above equation:

\[\begin{array}{c} \frac{{7wL}}{{18}} + \frac{{10wL}}{9} - wx - V = 0\\ V = \frac{{7wL}}{{18}} + \frac{{10wL}}{9} - wx\\ = \frac{{27wL}}{{18}} - wx\\ = \frac{w}{2}(3L - 2x) \end{array}\]…… (3)
 
Step 31

At $x = L$, the shear force at point B is calculated as:

\[\begin{array}{c} {V_B} = \frac{w}{2}\left( {3L - 2\left( L \right)} \right)\\ = \frac{{wL}}{2} \end{array}\]
 
Step 32

At $x = 2L$, the shear force at point C is calculated as:

\[\begin{array}{c} {V_C} = \frac{w}{2}\left( {3L - 2\left( {2L} \right)} \right)\\ = \frac{w}{2}\left( {3L - 4L} \right)\\ = - \frac{{wL}}{2} \end{array}\]
 
Step 33

Assume that at $x = d$, the shear force of the beam is 0. Substituting the known values in equation (3):

\[\begin{array}{c} 0 = \frac{w}{2}\left( {3L - 2\left( d \right)} \right)\\ 3L - 2d = 0\\ 2d = 3L\\ d = \frac{{3L}}{2} \end{array}\]
 
Step 34

Applying the moment of force of equation along x-axis at section AC:

\[M + wx\left( {\frac{x}{2}} \right) - {B_y}\left( {x - {d_1}} \right) - {A_y}x = 0\]
 
Step 35

Substitute the values in the above expression.

\[\begin{array}{c} M + wx\left( {\frac{x}{2}} \right) - \left( {\frac{{10wL}}{9}} \right)\left( {x - L} \right) - \left( {\frac{{7wL}}{{18}}} \right)x = 0\\ M = - \frac{{w{x^2}}}{2} + \frac{{10wLx}}{9} - \frac{{10w{L^2}}}{9} + \frac{{7wLx}}{{18}}\\ = - \frac{{w{x^2}}}{2} + \frac{{27wLx}}{{18}} - \frac{{10w{L^2}}}{9}\\ = \frac{w}{{18}}\left( { - 9{x^2} + 27Lx - 20{L^2}} \right) \end{array}\]…… (4)
 
Step 36

At $x = L$, the moment of force at point B is calculated as:

\[\begin{array}{c} {M_A} = \frac{w}{{18}}\left( { - 9{{\left( L \right)}^2} + 27L\left( L \right) - 20{L^2}} \right)\\ = \frac{w}{{18}}\left( { - 2{L^2}} \right)\\ = - \frac{{w{L^2}}}{9} \end{array}\]
 
Step 37

At $x = 2L$, the moment of force at point B is calculated as:

\[\begin{array}{c} {M_B} = \frac{w}{{18}}\left( { - 9{{\left( {2L} \right)}^2} + 27L\left( {2L} \right) - 20{L^2}} \right)\\ = \frac{w}{{18}}\left( { - 36{L^2} + 54{L^2} - 20{L^2}} \right)\\ = \frac{w}{{18}}\left( { - 2{L^2}} \right)\\ = - \frac{{w{L^2}}}{9} \end{array}\]
 
Step 38

Substitute the known value of $x = \frac{{3L}}{2}$ in equation (4):

\[\begin{array}{c} {M_{AC}} = \frac{w}{{18}}\left( { - 9{{\left( {\frac{{3L}}{2}} \right)}^2} + 27L\left( {\frac{{3L}}{2}} \right) - 20{L^2}} \right)\\ = \frac{w}{{18}}\left( { - \frac{{81{L^2}}}{4} + \frac{{81{L^2}}}{2} - 20{L^2}} \right)\\ = \frac{w}{{18}}\left( {\frac{{{L^2}}}{4}} \right)\\ = \frac{{w{L^2}}}{{72}} \end{array}\]
 
Step 39

The free body diagram of the beam AD is shown as:

Here, $M$ is the moment of force and $V$ is the shear force and x is distance where uniform load is acting.

 
Step 40

Applying the equilibrium force of equation along y-axis at section AD:

\[\begin{array}{c} \Sigma {F_y} = 0\\ {A_y} + {B_y} + {C_y} - wx - V = 0 \end{array}\]
 
Step 41

Substitute the known values in the above equation:

\[\begin{array}{c} \frac{{7wL}}{{18}} + \frac{{10wL}}{9} + \frac{{10wL}}{9} - wx - V = 0\\ V = \frac{{7wL}}{{18}} + \frac{{10wL}}{9} + \frac{{10wL}}{9} - wx\\ = \frac{{47wL}}{{18}} - wx\\ = \frac{w}{{18}}(47L - 18x) \end{array}\]…… (5)
 
Step 42

At $x = 2L$, the shear force at point C is calculated as:

\[\begin{array}{c} {V_C} = \frac{w}{{18}}\left( {47L - 18\left( {2L} \right)} \right)\\ = \frac{w}{{18}}\left( {47L - 36L} \right)\\ = \frac{{11wL}}{{18}} \end{array}\]
 
Step 43

At $x = 3L$, the shear force at point D is calculated as:

\[\begin{array}{c} {V_D} = \frac{w}{{18}}\left( {47L - 18\left( {3L} \right)} \right)\\ = \frac{w}{{18}}\left( {47L - 54L} \right)\\ = - \frac{{7wL}}{{18}} \end{array}\]
 
Step 44

Assume that at $x = d$, the shear force of the beam is 0. Substituting the known values in equation (5):

\[\begin{array}{c} 0 = \frac{w}{{18}}\left( {47L - 18\left( d \right)} \right)\\ 47L - 18d = 0\\ 18d = 47L\\ d = \frac{{47L}}{{18}} \end{array}\]
 
Step 45

Applying the moment of force of equation along x-axis at section AD:

\[M + wx\left( {\frac{x}{2}} \right) - {C_y}\left( {x - {d_1} - {d_5}} \right) - {B_y}\left( {x - {d_1}} \right) - {A_y}x = 0\]
 
Step 46

Substitute the values in the above expression.

\[\begin{array}{c} \left\{ \begin{array}{l} M + wx\left( {\frac{x}{2}} \right) - \left( {\frac{{10wL}}{9}} \right)\left( {x - L - L} \right)\\ - \left( {\frac{{10wL}}{9}} \right)\left( {x - L} \right) - \left( {\frac{{7wL}}{{18}}} \right)x \end{array} \right\} = 0\\ M = \left\{ \begin{array}{l} - \frac{{w{x^2}}}{2} + \frac{{10wLx}}{9} - \frac{{10w{L^2}}}{9} - \frac{{10w{L^2}}}{9} + \\ \frac{{10wLx}}{9} - \frac{{10w{L^2}}}{9} + \frac{{7wLx}}{{18}} \end{array} \right\}\\ = - \frac{{w{x^2}}}{2} + \frac{{47wLx}}{{18}} - \frac{{30w{L^2}}}{9}\\ = \frac{w}{{18}}\left( { - 9{x^2} + 47Lx - 60{L^2}} \right) \end{array}\]…… (6)
 
Step 47

At $x = 2L$, the moment of force at point C is calculated as:

\[\begin{array}{c} {M_C} = \frac{w}{{18}}\left( { - 9{{\left( {2L} \right)}^2} + 47L\left( {2L} \right) - 60{L^2}} \right)\\ = \frac{w}{{18}}\left( { - 36{L^2} + 94{L^2} - 60{L^2}} \right)\\ = \frac{w}{{18}}\left( { - 2{L^2}} \right)\\ = - \frac{{w{L^2}}}{9} \end{array}\]
 
Step 48

At $x = 3L$, the moment of force at point D is calculated as:

\[\begin{array}{c} {M_D} = \frac{w}{{18}}\left( { - 9{{\left( {3L} \right)}^2} + 47L\left( {3L} \right) - 60{L^2}} \right)\\ = \frac{w}{{18}}\left( { - 81{L^2} + 141{L^2} - 60{L^2}} \right)\\ = \frac{w}{{18}}\left( 0 \right)\\ = 0 \end{array}\]
 
Step 49

Substitute the known value of $x = \frac{{47L}}{{18}}$ in equation (6):

\[\begin{array}{c} {M_{CD}} = \frac{w}{{18}}\left( { - 9{{\left( {\frac{{47L}}{{18}}} \right)}^2} + 47L\left( {\frac{{47L}}{{18}}} \right) - 60{L^2}} \right)\\ = \frac{w}{{18}}\left( { - \frac{{19881{L^2}}}{{324}} + \frac{{2209{L^2}}}{{18}} - 60{L^2}} \right)\\ = \frac{w}{{18}}\left( {\frac{{441{L^2}}}{{324}}} \right)\\ = \frac{{49w{L^2}}}{{648}} \end{array}\]
 
Step 50

The shear and force bending moment diagram is shown as: