Question

Determine the reaction at support $A$ and $B$ of loaded beam shown in figure below.

Answer 1

As we know that we can use a concept of equilibrium in this question, as given the beam is balancing the load applied in a downward direction with supports on both ends, the supports of the same magnitude but in opposite directions.
Given First downward load, ${L_1} = 60KN$
Second downward load ${L_2} = 20KN/m$
Distance where second downward load occur, $D = 2m$
Now, Second downward load will be, $20KN/m \times 2m = 40KN$ .

Complete step by step solution:
A loaded beam has two loads in downward direction, and it is still stationary means a force of the same magnitude but in the opposite direction helps this beam to be stationary.
Firstly we will calculate total downward load given on the beam.
${L_{Total}} = {L_1} + {L_2}$
Put values of ${L_1}$ and ${L_2}$ in above equation
${L_{Total}} = 60 + 40 = 100KN$
Here, we got the total load on the beam in a downward direction that is $60KN$ , means $60KN$ force is also applied in an upward direction.
So, let the first support in upward direction $= {U_1}$
And second support in upward direction $= {U_2}$
Total support in upward direction $= {U_{Total}}$
$$ ${U_{Total}} = {U_1} + {U_2}......1$
As per question we can give two supports on $A$ and $B$ to keep this beam stationary, and the magnitude of that force we have already calculated.
Hence, ${U_{Total}} = {L_{Total}}$
Both upward support should be equal.
So, ${U_1} = {U_2}$
Now, equation 1 can be written as
${U_{Total}} = {U_1} + {U_1}$
${U_{Total}} = 2{U_1} = 100KN$
$\Rightarrow {U_1} = {U_2} = \dfrac{{100KN}}{2} = 50KN$
Here, two supports of $50KN$ strength will act upward on $A$ and $B$ to balance the beam.

Note:
We should know the concept of equilibrium to solve the above question. In equilibrium, a beam is supported by force at both ends, all the download loads are balanced by equal and opposite upward forces and the beam gets stationary.