Q. A simply supported beam of span 5 m three concentrated loads 10 kN, 15 kN and 10 kN are acting at 1 m, 3 m and 4 m respectively from left hand support. Find the reactions.

SOLUTION:
Total load = 10 + 25 + 10 = 45 kN
To find the reactions
Taking moments about A
RB × 5 = 10 × 1 + 25 × 3 + 10 × 4
= 10 + 75 + 40 = 125
RB = 125/5 = 25kN
RA =Total load – RB
= 45 – 25 = 20kN
Taking moments about B
RA × 5 = 10 × 1 + 25 × 2 + 10 × 4
= 10 + 50 + 40 = 100
RA = 100/5 = 20kN
Reactions
RA = 20kN and RB = 25kN