Step 1

We are given the following data:

A beam with uniform distributed load ${W_{AB}} = 4{\rm{ kip/ft}}$ between roller support A and pinned support B.

The distance between the two supports A and B is ${l_{AB}} = 30{\rm{ ft}}$.

The beam also contains a point load ${W_C} = 20{\rm{ kip}}$ at a distance of ${l_{CA}} = 15{\rm{ ft}}$ from A.

A point load ${W_D} = 20{\rm{ kip}}$ is at a distance of ${l_{BD}} = 15{\rm{ ft}}$ from point B.

We are required to draw the shear and moment diagrams for the beam.
 
Step 2

The free body diagram of the beam is shown below:

 
Step 3

To find the force we will equate all the forces in vertical direction by using relation,

\[{R_A} + {R_B} = {W_C} + \left( {{W_{AB}} \times {l_{AB}}} \right) + {W_D}\]
 
Step 4

On plugging the values in the above relation, we get,

\[\begin{array}{l} {R_A} + {R_B} = 20{\rm{ kip}} + \left( {4{\rm{ kip/ft}} \times 30{\rm{ ft}}} \right) + 20{\rm{ kip}}\\ {R_A} + {R_B} = 20{\rm{ kip}} + 120{\rm{ kip}} + 20{\rm{ kip}}\\ {R_A} + {R_B} = 160{\rm{ kip}} \end{array}\]
 
Step 5

The supports are separated at an equal distance. So the reaction forces are equal. Substitute ${R_A} = {R_B}$ in the above equation.

\[\begin{array}{c} {R_A} + {R_A} = 160{\rm{ kip}}\\ 2{R_A} = 160{\rm{ kip}}\\ {R_A} = \frac{{160{\rm{ kip}}}}{2}\\ {R_A} = 80{\rm{ kip}} \end{array}\]

Similarly, the reaction at support B will be,

\[{R_B} = 80{\rm{ kip}}\]
 
Step 6

To find the shear force at $x = 0{\rm{ ft}}$ from C, we will use the following relation.

\[{S_C} = - 20{\rm{ kip}}\]
 
Step 7

To find the shear force at$x = 15{\rm{ ft}}$ from C, we will use the following relation.

\[\begin{array}{c} {S_A} = - 20{\rm{ kip}} + 80{\rm{ kip}}\\ = 60{\rm{ kip}} \end{array}\]
 
Step 8

To find the shear force at$x = 30{\rm{ ft}}$ from C, we will use the following relation.

\[\begin{array}{c} {S_B} = - 20{\rm{ kip}} + 80{\rm{ kip}} - 4{\rm{ kip/ft}} \times 30{\rm{ ft}}\\ = - 20{\rm{ kip}} + 80{\rm{ kip}} - 120{\rm{ kip}}\\ = - 60{\rm{ kip}} \end{array}\]
 
Step 9

To find the shear force at$x = 45{\rm{ ft}}$ from C, we will use the following relation.

\[\begin{array}{c} {S_B} = - 20{\rm{ kip}} + 80{\rm{ kip}} - 4{\rm{ kip}}/{\rm{ft}} \times 30{\rm{ ft}} + 80{\rm{ kip}}\\ = - 20{\rm{ kip}} + 80{\rm{ kip}} - 120{\rm{ kip}} + 80{\rm{ kip}}\\ = 20{\rm{ kip}} \end{array}\]
 
Step 10

To find the shear force at $x = 60{\rm{ ft}}$ from C, we will use the following relation.

\[\begin{array}{c} {S_E} = - 20{\rm{ kip}} + 80{\rm{ kip}} - 4{\rm{ kip/ft}} \times 15{\rm{ ft}}\\ = - 20{\rm{ kip}} + 80{\rm{ kip}} - 60{\rm{ kip}}\\ = 0{\rm{ kip}} \end{array}\]
 
Step 11

The bending moment at C will be ${M_C} = 0$.

To find the bending moment at A from C, we will use the following relation,

\[{M_A} = - {R_A} \times {l_{CA}}\]

The negative sign indicates that the load is acting in the clockwise direction. Substitute the given values in the above formula.

\[\begin{array}{c} {M_A} = - 20{\rm{ kip}} \times 15{\rm{ ft}}\\ = - 300{\rm{ kip}} \cdot {\rm{ft}} \end{array}\]
 
Step 12

To find the bending moment between B and E from point C, we will use the following relation,

\[{M_x} = - 20{\rm{ kip}} \times x{\rm{ ft + 80 kip}} \times \left( {x - 15} \right){\rm{ ft}} + 4{\rm{ kip/ft}} \times \left( {x - 15} \right){\rm{ ft}} \times \frac{{\left( {x - 15} \right)}}{2}{\rm{ ft}}\]

To find the bending moment at$x = 45{\rm{ ft}}$we will use the following relation.

\[\begin{array}{c} {M_B} = - 20{\rm{ kip}} \times 45{\rm{ ft + 80 kip}} \times \left( {45 - 15} \right){\rm{ ft}} - 4{\rm{ kip/ft}} \times \left( {45 - 15} \right){\rm{ ft}} \times \frac{{\left( {45 - 15} \right)}}{2}{\rm{ ft}}\\ = - 900{\rm{ kip}} \cdot {\rm{ft}} + {\rm{80 kip}} \times 30{\rm{ ft}} - 4{\rm{ kip/ft}} \times 30{\rm{ ft}} \times \frac{{30}}{2}{\rm{ ft}}\\ = - 900{\rm{ kip}} \cdot {\rm{ft}} + {\rm{80 kip}} \times 30{\rm{ ft}} - 4{\rm{ kip/ft}} \times 30{\rm{ ft}} \times 15{\rm{ ft}}\\ = - 900{\rm{ kip}} \cdot {\rm{ft}} + {\rm{2400 kip}} \cdot {\rm{ft}} - 1800{\rm{ kip}} \cdot {\rm{ft}}\\ = - 300{\rm{ kip}} \cdot {\rm{ft}} \end{array}\]
 
Step 13

To find the bending moment at $x = 30{\rm{ ft}}$ we will use the following relation.

\[\begin{array}{c} {M_E} = - 20{\rm{ kip}} \times 30{\rm{ ft}} + {\rm{80 kip}} \times \left( {30 - 15} \right){\rm{ ft}} - 4{\rm{ kip/ft}} \times \left( {30 - 15} \right){\rm{ ft}} \times \frac{{\left( {30 - 15} \right)}}{2}{\rm{ ft}}\\ = - 600{\rm{ kip}} \cdot {\rm{ft}} + {\rm{80 kip}} \times 15{\rm{ ft}} - 4{\rm{ kip/ft}} \times 15{\rm{ ft}} \times \frac{{15}}{2}{\rm{ ft}}\\ = - 600{\rm{ kip}} \cdot {\rm{ft}} + {\rm{80 kip}} \times 15{\rm{ ft}} - 4{\rm{ kip/ft}} \times 15{\rm{ ft}} \times 7.5{\rm{ ft}}\\ = - 600{\rm{ kip}} \cdot {\rm{ft}} + {\rm{1200}} \cdot {\rm{ft}} - 450{\rm{ kip}} \cdot {\rm{ft}}\\ = 150{\rm{ kip}} \cdot {\rm{ft}} \end{array}\]
 
Step 14

The bending moment at D will be${M_D} = 0$.

The shear force and bending moment diagram will be,