Question

Determine the reactions at A and B of the overhanging beam as shown in fig.

Answer 1

First change UDL in to point load. 

Resolved all the forces in horizontal and vertical direction. Since hinged at point A (one vertical and one horizontal reaction). 

Let reaction at hinged i.e., point A is R_AH and R_AV, Let ∑H & ∑V is the sum of horizontal and vertical component of the forces, The supported beam is in equilibrium, hence Draw the FBD of the beam as shown in fig 8.42, Since beam is in equilibrium, i.e.,

∑H = 0;

RAH = 30 cos 45° = 21.2 KN

RAH = 21.21 KN

∑V = 0;

R_AV –30 sin 45 – 40 + R_BV = 0

R_AV + R_BV = 61.2 KN ...(i) 

Taking moment about point B, 

R_AV × 6 + 40 – 30 sin 45 × 1 + 40 × 1 = 0

R_AV = – 9.8 KN

Putting the value of R_AV in equation (i), we get

R_BV = 71K N

Hence reaction at support A i.e., 

R_AV = –9.8KN, R_AH = 21.2KN, R_BV = 71KN.