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.
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 RAH and RAV, 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
RAH = 30 cos45° = 21.2KN
RAH = 21.21KN .......ANS
∑V = 0;
RAV -30 sin45 - 40 + RBV = 0
RAV + RBV = 61.2KN ...(i)
By taking moment about B,
RAV × 6 + 40 - 30 sin45 × 1 + 40 × 1 = 0
RAV = - 9.8 KN .......ANS
Putting value of RAV in the equation (i), we get
RBV = 71KN .......ANS
Thus reaction at support A that is RAV = -9.8KN, RAH = 21.2KN, RBV = 71KN