Step 1

We are given the distributed load on section BC is $w = 1.5\;{\rm{kN/m}}$.

We are asked the shear and 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 $AB = 2\;{\rm{m}}$.

We have the distance between points B and C is $BC = 2\;{\rm{m}}$.

We have the distance between points A and C is $AC = 4\;{\rm{m}}$.

 
Step 3

The forces acting on the system in horizontal direction by equilibrium condition is,

\[{A_x} = 0\]
 
Step 4

The moment about point A is,

\[\begin{array}{c} \Sigma {M_A} = 0\\ {C_y}\left( {AC} \right) - w\left( {BC} \right)\left( {AB + \frac{{BC}}{2}} \right) = 0 \end{array}\]
 
Step 5

Substitute the values in the above expression.

\[\begin{array}{c} {C_y}\left( {4\;{\rm{m}}} \right) - \left( {1.5\;{\rm{kN/m}}} \right)\left( {2\;{\rm{m}}} \right)\left( {2\;{\rm{m}} + \frac{{2\;{\rm{m}}}}{2}} \right) = 0\\ {C_y} = 2.25\;{\rm{kN}} \end{array}\]
 
Step 6

The forces acting on the system in vertical direction is,

\[{A_y} + {C_y} - w\left( {BC} \right) = 0\]
 
Step 7

Substitute the values in the above expression.

\[\begin{array}{c} {A_y} + \left( {2.25\;{\rm{kN}}} \right) - \left( {1.5\;{\rm{kN/m}}} \right)\left( {2\;{\rm{m}}} \right) = 0\\ {A_y} = 0.75\;{\rm{kN}} \end{array}\]
 
Step 8

Consider A section between AB and at A distance x from point A. For $0 \le x < 2\;{\rm{m}}$.

 
Step 9

The forces acting on the section in vertical direction is,

\[{A_y} - V = 0\]
 
Step 10

Substitute the values in the above expression.

\[\begin{array}{c} 0.75\;{\rm{kN}} - V = 0\\ V = 0.75\;{\rm{kN}} \end{array}\]
 
Step 11

The shear force at point A and B is,

\[{V_A} = {V_B} = V = 0.75\;{\rm{kN}}\]
 
Step 12

The moment about the section is,

\[\begin{array}{c} \Sigma M = 0\\ M - {A_y}x = 0 \end{array}\]
 
Step 13

Substitute the values in above expression.

\[M = \left( {0.75\;{\rm{kN}}} \right)x\]
 
Step 14

Substitute $x = 0$ for moment at point A is,

\[\begin{array}{l} {M_A} = \left( {0.75\;{\rm{kN}}} \right)\left( 0 \right)\\ {M_A} = 0 \end{array}\]
 
Step 15

Substitute $x = 2\;{\rm{m}}$ for moment at point B is,

\[\begin{array}{l} {M_B} = \left( {0.75\;{\rm{kN}}} \right)\left( {2\;{\rm{m}}} \right)\\ {M_B} = 1.5\;{\rm{kN}} \cdot {\rm{m}} \end{array}\]
 
Step 16

Consider A section at A distance x in segment BC. For $\left( {2\;{\rm{m}} < x \le 4\;{\rm{m}}} \right)$.

 
Step 17

The force acting on the segment along vertical direction is,

\[\begin{array}{c} {A_y} - w\left( {x - AB} \right) - V = 0\\ V = {A_y} - w\left( {x - AB} \right) \end{array}\]
 
Step 18

Substitute the values in the above expression.

\[\begin{array}{c} V = 0.75\;{\rm{kN}} - \left( {1.5\;{\rm{kN}}} \right)\left( {x - 2\;{\rm{m}}} \right)\\ V = \left( { - 1.5x + 3.75} \right)\;{\rm{kN}} \end{array}\]
 
Step 19

Substitute $x = 4\;{\rm{m}}$for shear force at point C is,

\[\begin{array}{l} {V_C} = \left[ { - 1.5\left( {4\;{\rm{m}}} \right) + 3.75} \right]\;{\rm{kN}}\\ {V_C} = - 2.25\;{\rm{kN}} \end{array}\]
 
Step 20

The moment about the section is,

\[\begin{array}{c} \Sigma M = 0\\ \left[ \begin{array}{l} - {A_y}x + \left[ {w\left( {x - AB} \right)\left( {\frac{{x - AB}}{2}} \right)} \right]\\ + M \end{array} \right] = 0\\ M = {A_y}x - \left[ {w\left( {x - AB} \right)\left( {\frac{{x - AB}}{2}} \right)} \right] \end{array}\]
 
Step 21

Substitute the values in the above expression.

\[\begin{array}{l} M = \left( {0.75\;{\rm{kN}}} \right)x - \;\left[ {\left( {1.5\;{\rm{kN}}} \right)\left( {x - 2\;{\rm{m}}} \right)\left( {\frac{{x - 2\;{\rm{m}}}}{2}} \right)} \right]\\ M = \left[ {0.75x - 0.75{{\left( {x - 2} \right)}^2}} \right]\;{\rm{kN}} \cdot {\rm{m}} \end{array}\]
 
Step 22

The distance at which shear force becomes zero is,

\[V = {A_y} - w\left( {x - AB} \right)\]
 
Step 23

Substitute the values in the above expression.

\[\begin{array}{c} 0 = 0.75\;{\rm{kN}} - \left( {1.5\;{\rm{kN}}} \right)\left( {x - 2\;{\rm{m}}} \right)\\ x = 2.5\;{\rm{m}} \end{array}\]
 
Step 24

The maximum magnitude of bending moment is,

\[M = \left[ {0.75x - 0.75{{\left( {x - 2} \right)}^2}} \right]\;{\rm{kN}} \cdot {\rm{m}}\]
 
Step 25

Substitute $x = 2.5\;{\rm{m}}$ in the above expression.

\[\begin{array}{l} M = \left[ {0.75\left( {2.5} \right) - 0.75{{\left( {2.5 - 2} \right)}^2}} \right]\;{\rm{kN}} \cdot {\rm{m}}\\ M = 1.68\;{\rm{kN}} \cdot {\rm{m}} \end{array}\]
 
Step 26

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