Determine reactions of the overhanging beam:
Determine reactions at points A and B of the overhanging beam as shown in the figure given below.
First change UDL in point load.
Figure
Resolve all the forces in horizontal direction and vertical direction. As hinged at point A (one vertical and horizontal reaction).
Let reaction at hinged that is, point A is R_{A}_{H} and R_{AV}, Let ∑H & ∑V is the sum of horizontal and vertical component of forces, The supported beam is in equilibrium, thus Draw the free body digram of the beam as shown in the given figure, As beam is in equilibrium, that is
R_{A}_{H} = 30 cos45° = 21.2KN
R_{A}_{H} = 21.21KN .......ANS
∑V = 0;
R_{AV} -30 sin45 - 40 + R_{B}_{V} = 0
R_{A}_{V} + R_{B}_{V} = 61.2KN ...(i)
By taking moment about B,
R_{A}_{V} × 6 + 40 - 30 sin45 × 1 + 40 × 1 = 0
R_{A}_{V} = - 9.8 KN .......ANS
Putting value of R_{A}_{V} in the equation (i), we get
R_{B}_{V} = 71KN .......ANS
Thus reaction at support A that is R_{A}_{V} = -9.8KN, R_{A}_{H } = 21.2KN, R_{B}_{V} = 71KN