Step 1

We are given the distributed load on section AB is $w = 50\;{\rm{lb/ft}}$, and the moment at point C is $M = 200\;{\rm{lb}} \cdot {\rm{ft}}$.

We are asked the shear and bending moment diagrams for the beam.

 
Step 2

The free body diagram of the system is:

We have the distance between points A and B is $AB = 20\;{\rm{ft}}$.

We have the distance between points B and C is $BC = 10\;{\rm{ft}}$.

We have the distance between points A and C is $AC = 30\;{\rm{ft}}$.

 
Step 3

The moment about point A is,

\[\begin{array}{c} \Sigma {M_A} = 0\\ {B_y}\left( {AB} \right) - M - w\left( {AB} \right)\left( {\frac{{AB}}{2}} \right) = 0 \end{array}\]
 
Step 4

Substitute the values in the above expression.

\[\begin{array}{c} {B_y}\left( {20\;{\rm{ft}}} \right) - 200\;{\rm{lb}} \cdot {\rm{ft}} - \left( {50\;{\rm{lb/ft}}} \right)\left( {20\;{\rm{ft}}} \right)\left( {\frac{{20\;{\rm{ft}}}}{2}} \right) = 0\\ {B_y} = 510\;{\rm{lb}} \end{array}\]
 
Step 5

The forces acting on the beam in vertical direction is,

\[{A_y} + {B_y} - w\left( {AB} \right) = 0\]
 
Step 6

Substitute the values in the above expression.

\[\begin{array}{c} {A_y} + 510\;{\rm{lb}} - \left( {50\;{\rm{lb/ft}}} \right)\left( {20\;{\rm{ft}}} \right) = 0\\ {A_y} = 490\;{\rm{lb}} \end{array}\]
 
Step 7

The forces acting on the beam in horizontal direction is,

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

Consider a section between AB and at a distance x from point A. For $0 \le x < 20\;{\rm{ft}}$.

 
Step 9

The forces acting on the section in vertical direction is,

\[{A_y} - V - wx = 0\]
 
Step 10

Substitute the values in the above expression.

\[\begin{array}{c} 490\;{\rm{lb}} - V - \left( {50\;{\rm{lb/ft}}} \right)x = 0\\ V = \left( {490 - 50x} \right)\;{\rm{lb}} \end{array}\]
 
Step 11

The moment about the section is,

\[\begin{array}{c} \Sigma M = 0\\ M + wx\left( {\frac{x}{2}} \right) - {A_y}x = 0 \end{array}\]
 
Step 12

Substitute the values in the above expression.

\[\begin{array}{c} M + \left( {50\;{\rm{lb/ft}}} \right)x\left( {\frac{x}{2}} \right) - \left( {490\;{\rm{lb}}} \right)x = 0\\ M = 490x - 25{x^2} \end{array}\]
 
Step 13

The shear force at point A is,

\[{V_A} = {A_y}\]
 
Step 12

Substitute the values in the above expression.

\[{V_A} = 490\;{\rm{lb}}\]
 
Step 13

The moment at point A is,

\[{M_A} = 490x - 25{x^2} = 490\left( 0 \right) - 25{\left( 0 \right)^2} = 0\]
 
Step 14

The shear force at point B is,

\[{V_B} = {A_y} - w\left( {AB} \right)\]
 
Step 15

Substitute the values in the above expression.

\[\begin{array}{c} {V_B} = 490\;{\rm{lb}} - \left( {50\;{\rm{lb/ft}}} \right)\left( {20\;{\rm{ft}}} \right)\\ {V_B} = - 510\;{\rm{lb}} \end{array}\]
 
Step 16

The moment at point B is,

\[{M_B} = 490x - 25{x^2}\]
 
Step 17

Substitute the values in the above expression.

\[\begin{array}{c} {M_B} = 490\;{\rm{lb}}\left( {20\;{\rm{ft}}} \right) - \left( {25\;{\rm{lb/ft}}} \right){\left( {20\;{\rm{ft}}} \right)^2}\\ {M_B} = - 200\;{\rm{lb}} \cdot {\rm{ft}} \end{array}\]
 
Step 18

At maximum bending moment shear force becomes zero.

The location of bending moment is,

\[\begin{array}{c} V = 0\\ \left( {490\;{\rm{lb}}} \right) - \left( {50\;{\rm{lb/ft}}} \right)x = 0\\ x = 9.8\;{\rm{ft}} \end{array}\]
 
Step 19

The magnitude of maximum bending moment is,

\[{M_{\max }} = \left( {490x - 25{x^2}} \right)\;{\rm{lb}} \cdot {\rm{ft}}\]
 
Step 20

Substitute the values in the above expression.

\[\begin{array}{l} {M_{\max }} = \left[ {490\left( {9.81\;{\rm{ft}}} \right) - 25{{\left( {9.8\;{\rm{ft}}} \right)}^2}} \right]\;{\rm{lb}} \cdot {\rm{ft}}\\ {M_{\max }} = 2401\;{\rm{lb}} \cdot {\rm{ft}} \end{array}\]
 
Step 21

Consider a section BC and at a distance x from point A. For $20\;{\rm{ft}} \le x < 30\;{\rm{ft}}$.

 
Step 22

The forces acting on the section in vertical direction is,

\[\begin{array}{c} \Sigma {F_y} = 0\\ {A_y} - V + {B_y} - w\left( {AB} \right) = 0 \end{array}\]
 
Step 23

Substitute the values in the above expression.

\[\begin{array}{c} \left( {490\;{\rm{lb}}} \right) - V + \left( {510\;{\rm{lb}}} \right) - \left( {50\;{\rm{ft/lb}}} \right)\left( {20\;{\rm{ft}}} \right) = 0\\ V = 0 \end{array}\]
 
Step 24

The moment about the section $20\;{\rm{ft}} \le x < 30\;{\rm{ft}}$ is,

\[\begin{array}{c} \Sigma {M_x} = 0\\ M + w\left( {AB} \right)\left( {x - 10\;{\rm{ft}}} \right) - {A_y}x - {B_y}\left( {x - 20\;{\rm{ft}}} \right) = 0 \end{array}\]
 
Step 25

Substitute the values in the above expression.

\[\begin{array}{c} \left[ \begin{array}{c} M + \left( {50\;{\rm{lb/ft}}} \right)\left( {20{\rm{ ft}}} \right)\left( {x - 10\;{\rm{ft}}} \right)\\ - \left( {490\;{\rm{lb}}} \right)x - \left( {510\;{\rm{lb}}} \right)\left( {x - 20\;{\rm{ft}}} \right) \end{array} \right] = 0\\ M = \left( {490x + 510x - 10200 - 1000x + 10000} \right)\;{\rm{lb}} \cdot {\rm{ft}}\\ M = - {\rm{200}}\;{\rm{lb}} \cdot {\rm{ft}} \end{array}\]
 
Step 26

The shear force and bending moment at point B is,

\[\begin{array}{l} {V_B} = 0\\ {M_B} = - 200\;{\rm{lb}} \cdot {\rm{ft}} \end{array}\]
 
Step 27

The shear force and bending moment at point C is,

\[\begin{array}{l} {V_C} = 0\\ {M_C} = - 200\;{\rm{lb}} \cdot {\rm{ft}} \end{array}\]
 
Step 28

The shear force and bending moment diagrams of the system are shown as: