Step 1

We are given the uniform load on the beam $w = 30{\rm{ lb/ft}}$ and the moment of force for the beam is $M = 180\;{\rm{lb}} \cdot {\rm{ft}}$.

We are asked the draw shear and bending moment of the diagram for the beam.

We have the distance between point A and B is ${d_1} = 9{\rm{ ft}}$.

We have the distance between B and C is ${d_2} = 4.5{\rm{ ft}}$.

 
Step 2

The free body diagram of the 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 at point A:

\[\begin{array}{c} \Sigma {F_x} = 0\\ {A_x} = 0 \end{array}\]
 
Step 4

Applying the moment of force of equation at point A:

\[{B_y}{d_1} - \left( {\frac{1}{2}w{d_1}} \right)\left( {\frac{2}{3}{d_1}} \right) - M = 0\]
 
Step 5

Substitute the known values in the above equation:

\[\begin{array}{c} \left\{ {{B_y}\left( {9{\rm{ ft}}} \right) - \left( {\frac{1}{2}\left( {30\;{\rm{lb/ft}}} \right)\left( {9\;{\rm{ft}}} \right)} \right)\left( {\frac{2}{3}\left( {9\;{\rm{ft}}} \right)} \right) - 180\;{\rm{lb}} \cdot {\rm{ft}}} \right\} = 0\\ {B_y}\left( {9\;{\rm{ft}}} \right) = 810\;{\rm{lb}} \cdot {\rm{ft}} + 180\;{\rm{lb}} \cdot {\rm{ft}}\\ {B_y} = \frac{{\left( {990\;{\rm{lb}} \cdot {\rm{ft}}} \right)}}{{\left( {9\;{\rm{ft}}} \right)}}\\ = 110\;{\rm{lb}} \end{array}\]
 
Step 6

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

\[\begin{array}{c} \Sigma {F_y} = 0\\ {A_y} + {B_y} - \frac{1}{2}w{d_1} = 0 \end{array}\]
 
Step 7

Substitute the known values in the above equation:

\[\begin{array}{c} \left\{ \begin{array}{l} {A_y} + 110{\rm{ lb}} - \\ \frac{1}{2}\left( {30\;{\rm{lb/ft}}} \right)\left( {9\;{\rm{ft}}} \right) \end{array} \right\} = 0\\ {A_y} = 135\;{\rm{lb}} - 110\;{\rm{lb}}\\ = 25\;{\rm{lb}} \end{array}\]
 
Step 8

The free body diagram of section AB is shown as:

Here, V is the shear force, M is the moment force and x is the distance at which moment force acts.

 
Step 9

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

\[\begin{array}{c} \sum {F_y} = 0\\ {A_y} - \frac{1}{2}\left( {\frac{{wx}}{{{d_1}}}} \right)x - V = 0 \end{array}\]
 
Step 10

Substitute the known values in the equation:

\[\begin{array}{c} 25\;{\rm{lb}} - \left( {\frac{1}{2}} \right)\left( {\frac{{\left( {{\rm{30}}\;{\rm{lb/ft}}} \right)x}}{{9\;{\rm{ft}}}}} \right)x - V = 0\\ V = 25\;{\rm{lb}} - \left( {{\rm{1}}{\rm{.667}}\;{\rm{lb/f}}{{\rm{t}}^2}} \right){x^2} \end{array}\]   ...... (1)
 
Step 11

At $x = 0\;{\rm{ft}}$, the shear force at point A is calculated as:

\[\begin{array}{c} {V_A} = 25\;{\rm{lb}} - \left( {{\rm{1}}{\rm{.667}}\;{\rm{lb/f}}{{\rm{t}}^2}} \right){\left( {0\;{\rm{ft}}} \right)^2}\\ = 25\;{\rm{lb}} - 0\;{\rm{lb}}\\ = 25\;{\rm{lb}} \end{array}\]
 
Step 12

At $x = 9\;{\rm{ft}}$, the shear force at point B is calculated as:

\[\begin{array}{c} {V_B} = 25\;{\rm{lb}} - \left( {{\rm{1}}{\rm{.667}}\;{\rm{lb/f}}{{\rm{t}}^2}} \right){\left( {9\;{\rm{ft}}} \right)^2}\\ = 25\;{\rm{lb}} - 135\;{\rm{lb}}\\ = - 110\;{\rm{lb}} \end{array}\]
 
Step 13

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

\[\begin{array}{c} 0 = 25\;{\rm{lb}} - \left( {1.667\;{\rm{lb/f}}{{\rm{t}}^2}} \right){d^2}\\ \left( {1.667\;{\rm{lb/f}}{{\rm{t}}^2}} \right){d^2} = 25\;{\rm{lb}}\\ {d^2} = \frac{{\left( {25\;{\rm{lb}}} \right)}}{{\left( {1.667\;{\rm{lb/f}}{{\rm{t}}^2}} \right)}}\\ d = 3.87\;{\rm{ft}} \end{array}\]
 
Step 14

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

\[ - {A_y}x + \frac{1}{2}\left( {\frac{{wx}}{{{d_1}}}} \right)x\left( {\frac{x}{3}} \right) + M = 0\]
 
Step 15

Substitute the known values in the above equation:

\[\begin{array}{c} - \left( {25\;{\rm{lb}}} \right)x + \left( {\frac{1}{2}} \right)\left( {\frac{{\left( {30\;{\rm{lb/ft}}} \right)x}}{{9\;{\rm{ft}}}}} \right)x\left( {\frac{x}{3}} \right) + M = 0\\ M = \left( {25\;{\rm{lb}}} \right)x - \left( {0.556\;{\rm{lb/f}}{{\rm{t}}^2}} \right){x^3} \end{array}\]   ...... (2)
 
Step 16

At $x = 0\;{\rm{ft}}$, the moment of force at point A is calculated as:

\[\begin{array}{c} {M_A} = \left( {25\;{\rm{lb}}} \right)\left( {0\;{\rm{ft}}} \right) - \left( {0.556\,{\rm{lb/f}}{{\rm{t}}^2}} \right){\left( {0\;{\rm{ft}}} \right)^3}\\ = 0 \end{array}\]
 
Step 17

At $x = 9\;{\rm{ft}}$, the moment of force at point B is calculated as:

\[\begin{array}{c} {M_B} = \left( {25\;{\rm{lb}}} \right)\left( {9\;{\rm{ft}}} \right) - \left( {0.556\,{\rm{lb/f}}{{\rm{t}}^2}} \right){\left( {9\;{\rm{ft}}} \right)^3}\\ = 225\;{\rm{lb}} \cdot {\rm{ft}} - 405\;{\rm{lb}} \cdot {\rm{ft}}\\ = - 180\;{\rm{lb}} \cdot {\rm{ft}} \end{array}\]
 
Step 18

Substitute the known value of $x = 3.87\;{\rm{ft}}$ in equation (2):

\[\begin{array}{c} {M_{AB}} = \left( {25\;{\rm{lb}}} \right)\left( {3.87\;{\rm{ft}}} \right) - \left( {0.556\,{\rm{lb/f}}{{\rm{t}}^2}} \right){\left( {3.87\;{\rm{ft}}} \right)^3}\\ = 96.7\;{\rm{lb}} \cdot {\rm{ft}} - 32.2\;{\rm{lb}} \cdot {\rm{ft}}\\ = 64.5\;{\rm{lb}} \cdot {\rm{ft}} \end{array}\]
 
Step 19

The free body diagram of section AC is shown as:

Here, V is the shear force, M is the moment force, ${B_y}$ is the reaction force at B and x is the distance at which moment force acts.

 
Step 20

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

\[\begin{array}{c} \sum {F_y} = 0\\ {A_y} + {B_y} - \frac{1}{2}w{d_1} - V = 0 \end{array}\]
 
Step 21

Substitute the known values in the equation:

\[\begin{array}{c} \left\{ \begin{array}{l} 25\;{\rm{lb}} + 110\;{\rm{lb}} - \\ \frac{1}{2}\left( {{\rm{30}}\;{\rm{lb/ft}}} \right)\left( {9\;{\rm{ft}}} \right) - V \end{array} \right\} = 0\\ V = 25\;{\rm{lb}} + 110\;{\rm{lb}} - 135\;{\rm{lb}}\\ = 0\;{\rm{lb}} \end{array}\]
 
Step 22

At $x = 9\;{\rm{ft}}$, the shear force at point B is calculated as:

\[{V_B} = 0\;{\rm{lb}}\]
 
Step 23

At $x = 13.5\;{\rm{ft}}$, the shear force at point C is calculated as:

\[{V_C} = 0\;{\rm{lb}}\]
 
Step 24

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

\[ - {A_y}x - {B_y}\left( {x - {d_1}} \right) + \frac{1}{2}w{d_1}\left( {x - \frac{2}{3}{d_1}} \right) + M = 0\]
 
Step 25

Substitute the known values in the above equation:

\[\begin{array}{c} \left\{ \begin{array}{l} - \left( {25\;{\rm{lb}}} \right)x - \left( {110\;{\rm{lb}}} \right)\left( {x - 9\;{\rm{ft}}} \right)\\ + \frac{1}{2}\left( {30\;{\rm{lb/ft}}} \right)\left( {9\;{\rm{ft}}} \right)\left( {x - \frac{2}{3}\left( {9\;{\rm{ft}}} \right)} \right) + M \end{array} \right\} = 0\\ M = \left\{ \begin{array}{l} \left( {25\;{\rm{lb}}} \right)x + \\ \left( {110\;{\rm{lb}}} \right)\left( {x - 9\;{\rm{ft}}} \right) - \\ \left( {135\;{\rm{lb}}} \right)\left( {x - 6\;{\rm{ft}}} \right) \end{array} \right\}\\ = \left\{ \begin{array}{l} \left( {25\;{\rm{lb}}} \right)x + \\ \left( {110\;{\rm{lb}}} \right)x - \left( {990\;{\rm{lb}} \cdot {\rm{ft}}} \right) - \\ \left( {135\;{\rm{lb/ft}}} \right)x - 810\;{\rm{lb}} \cdot {\rm{ft}} \end{array} \right\}\\ = - 180\;{\rm{lb}} \cdot {\rm{ft}} \end{array}\]
 
Step 26

At $x = 9\;{\rm{ft}}$, the moment of force at point B is calculated as:

\[{M_B} = - 180\;{\rm{lb}} \cdot {\rm{ft}}\]
 
Step 27

At $x = 13.5\;{\rm{ft}}$, the moment of force at point C is calculated as:

\[{M_C} = - 180\;{\rm{lb}} \cdot {\rm{ft}}\]
 
Step 28

The shear and bending diagram of the first beam is shown as: