Step 1

We are given the following data:

Abeam with a triangular load ${W_{AC}} = 12{\rm{ kN/m}}$ and a uniform distributed load ${W_{CB}} = 12{\rm{ kN/m}}$.

The triangular load is distributed at a distance of ${l_{AC}} = 3{\rm{ m}}$.

The uniform distributed load is distributed at a distance of ${l_{CB}} = 3{\rm{ m}}$.

We are required to draw the shear and moment diagram for the beam with roller joint at A and pinned joint at B.

 
Step 2

The free body diagram will be,

 
Step 3

The reactant force ${R_A}$ can be calculated by considering the equilibrium of forces in the vertical direction.

\[\begin{array}{c} \sum {{F_y}} = 0\\ {R_A} + {R_B} - {W_{AC}} \times \frac{{{l_{AC}}}}{2} - {W_{CB}} \times \frac{{{l_{CB}}}}{2} = 0 \end{array}\]
 
Step 4

Substitute the given values in the above equation.

\[\begin{array}{c} {R_A} + {R_B} - \left( {12{\rm{ kN}} \cdot {\rm{m}} \times \frac{{3{\rm{ m}}}}{2}} \right) - \left( {12{\rm{ kN}} \cdot {\rm{m}} \times 3{\rm{ m}}} \right) = 0\\ {R_A} + {R_B} = \left( {12{\rm{ kN}} \cdot {\rm{m}} \times \frac{{3{\rm{ m}}}}{2}} \right) + \left( {12{\rm{ kN}} \cdot {\rm{m}} \times 3{\rm{ m}}} \right)\\ {R_A} + {R_B} = \left( {6{\rm{ kN}} \cdot {\rm{m}} \times 3{\rm{ m}}} \right) + \left( {12{\rm{ kN}} \cdot {\rm{m}} \times 3{\rm{ m}}} \right)\\ {R_A} + {R_B} = 18{\rm{ kN}} + 36{\rm{ kN}}\\ {R_A} + {R_B} = 54{\rm{ kN}} \end{array}\]   ........ (1)
 
Step 5

The formula to find support reactions ${R_A}$ by taking moment about A is,

\[\left( {{R_B} \times {l_{AB}}} \right) - \left( {\frac{{{W_{AC}} \times l_{AC}^2}}{3}} \right) - \left( {{W_{CB}} \times {l_{CB}} \times \frac{{{l_{CB}}}}{2}} \right) - {W_{CB}} \times {l_{CB}} \times {l_{AC}} = 0\]
 
Step 6

Substitute the given values in the above formula.

\[\begin{array}{c} \left( {{R_B} \times 6{\rm{ m}}} \right) - \left( {\frac{{12{\rm{ kN/m}} \times {3^2}{\rm{ }}{{\rm{m}}^2}}}{{3{\rm{ m}}}}} \right) - \left( {12{\rm{ kN/m}} \times 3{\rm{ m}} \times \frac{{3{\rm{ m}}}}{2}} \right) - 12{\rm{ kN/m}} \times 3{\rm{ m}} \times 3{\rm{ m}} = 0\\ \left( {{R_B} \times 6{\rm{ m}}} \right) - \left( {\frac{{12{\rm{ kN/m}} \times 9{\rm{ }}{{\rm{m}}^2}}}{{3{\rm{ m}}}}} \right) - \left( {6{\rm{ kN/m}} \times 3{\rm{ m}} \times 3{\rm{ m}}} \right) - 108{\rm{ kNm}} = 0\\ \left( {{R_B} \times 6{\rm{ m}}} \right) - 36{\rm{ kN}} \cdot {\rm{m}} - 54{\rm{ kN}} \cdot {\rm{m}} - 108{\rm{ kN}} \cdot {\rm{m}} = 0\\ \left( {{R_B} \times 6{\rm{ m}}} \right) - 198{\rm{ kN}} \cdot {\rm{m}} = 0 \end{array}\]
 
Step 7

On further solving the above equation,

\[\begin{array}{c} \left( {{R_B} \times 6{\rm{ m}}} \right) = 198{\rm{ kN}} \cdot {\rm{m}}\\ {R_B} = \frac{{198{\rm{ kN}} \cdot {\rm{m}}}}{{6{\rm{ m}}}}\\ {R_B} = 33{\rm{ kN}} \end{array}\]
 
Step 8

Substitute the value of ${R_B}$ in equation (1).

\[\begin{array}{c} {R_A} + 33{\rm{ kN}} = 54{\rm{ kN}}\\ {R_A} = 54{\rm{ kN}} - 33{\rm{ kN}}\\ {R_A} = 21{\rm{ kN}} \end{array}\]
 
Step 9

The shear force at A will be,

\[\begin{array}{c} {S_A} = {R_A}\\ = 21{\rm{ kN}} \end{array}\]
 
Step 10

The shear force at C will be,

\[\begin{array}{c} {S_C} = {R_A} - {W_{CB}} \times \frac{{{l_{CB}}}}{2}\\ = 21{\rm{ kN}} - 12{\rm{ kN/m}} \times \frac{{3{\rm{ m}}}}{2}\\ = 21{\rm{ kN}} - 18\,{\rm{kN}}\\ = 3{\rm{ kN}} \end{array}\]
 
Step 11

The shear force at B will be,

\[\begin{array}{c} {S_B} = {R_B}\\ = 33{\rm{ kN}} \end{array}\]
 
Step 12

The bending moment at A will be ${M_A} = 0$.

 
Step 13

The bending moment at C will be,

\[{M_C} = {R_A} \times {l_{AC}} - \left( {\frac{{{W_{CB}} \times {l_{CB}}}}{2}} \right)\left( {\frac{{{l_{CB}}}}{3}} \right)\]
 
Step 14

Substitute the given values in the above formula.

\[\begin{array}{c} {M_C} = 21{\rm{ kN}} \times 3{\rm{ m}} - \left( {\frac{{12{\rm{ kN/m}} \times 3{\rm{ m}}}}{2}} \right)\left( {\frac{{3{\rm{ m}}}}{3}} \right)\\ = 63{\rm{ kN}} \cdot {\rm{m}} - 18{\rm{ kN}} \cdot {\rm{m}}\\ = 45{\rm{ kN}} \cdot {\rm{m}} \end{array}\]
 
Step 15

The bending moment at B will be ${M_B} = 0$.

 
Step 16

The general equation of shear force is,

\[{S_x} = {R_A} - 4x \times \frac{x}{2}\]

Substitute the given values in the above formula.

\[\begin{array}{c} {S_x} = 21{\rm{ kN}} - 4x \times \frac{x}{2}\\ = 21 - 2{x^2} \end{array}\]
 
Step 17

The bending moment will be maximum when shear force is zero.

\[\begin{array}{c} 0 = 21 - 2{x^2}\\ - 2{x^2} = - 21\\ {x^2} = \frac{{ - 21}}{{ - 2}} \end{array}\]
 
Step 18

On further solving the above equation,

\[\begin{array}{c} {x^2} = 10.5\\ x = \sqrt {10.5} \\ x = 3.24{\rm{ m}} \end{array}\]
 
Step 19

The general equation of moment to find the distance x from A at which the shear force will be zero is,

\[{R_B}\left( {{l_{AB}} - x} \right) - \frac{{{W_{AB}}{{\left( {{l_{AB}} - x} \right)}^2}}}{2} - {M_x} = 0\]
 
Step 20

Substitute the given values in the above equation.

\[33{\rm{ kN}}\left( {6{\rm{ m}} - x} \right) - \frac{{12{\rm{ kN/m}}{{\left( {6{\rm{ m}} - x} \right)}^2}}}{2} - {M_x} = 0\]

On solving the above equation,

\[\begin{array}{c} 33\left( {6 - x} \right) - 6{\left( {6 - x} \right)^2} - {M_x} = 0\\ 198 - 33x - 6\left( {{6^2} - 2\left( 6 \right)\left( x \right) + {x^2}} \right) - {M_x} = 0\\ 198 - 33x - 6\left( {36 - 12x + {x^2}} \right) - {M_x} = 0 \end{array}\]
 
Step 21

On solving the above equation,

\[\begin{array}{c} 198 - 33x - 216 + 72x - 6{x^2} - {M_x} = 0\\ - 18 + 39x - 6{x^2} - {M_x} = 0\\ {M_x} = 6{x^2} - 39x + 18 \end{array}\]
 
Step 22

The bending moment will be maximum when the shear force is zero. Differentiate the above equation with respect to x using the power rule to find the distance at which the shear force will be zero.

\[\begin{array}{c} {S_x} = 0\\ \frac{d}{{dx}}\left( {{M_x}} \right) = 0\\ \frac{d}{{dx}}\left( {6{x^2} - 39x + 18} \right) = 0\\ 6\left( 2 \right){x^{2 - 1}} - 39{x^{1 - 1}} + 0 = 0\\ 12x - 39 = 0 \end{array}\]
 
Step 23

On further solving the above equation,

\[\begin{array}{c} 12x = 39\\ x = \frac{{39}}{{12}}\\ x = 3.25{\rm{ m}} \end{array}\]

The shear force and bending moment diagram will be,