Q.A Simply Supported beam of length 4 m carries two point loads of 50 kN and 100 kN are placed at a distance of 1 m and 3 m from the left hand support. Find the reactions.

 

SOLUTION :

Total load  =50+100= 150kN

Taking moments about A

RB x 4 = 50 x 1 + 100 x 3

RB x 4 = 350

RB= 350/4 = 87.5kN

RA= Total load – RB

RA= 150 – 87.5 = 62.5kN

( or)

Taking moments about B

RA x 4 = 100×1 +50×3

RA x 4 = 250

RA= 250/4 = 62.5kN

RB = Total load – 62.5 

RB= 150-62.5 = 87.5kN