Q. A beam of length 4 m. It is supported at 1 m and 3 m from left end of the beam. A udl 2 kN/m is acting on complete length Find the reactions.

Q. A beam of length 4 m. It is supported at 1 m and 3 m from left end of the beam. A udl 2 kN/m is acting on complete length Find the reactions.

SOLUTION:
Total load = 2 * 4 = 8kN
As beam and loading is symmetrical
Hence R A =R B = Total load 2 = 8/2 = 4kN . Reactions R_{A} = R_{B} = 4kN

Q. Calculate the reactions at supports for a double over hanging beam as shown in figure

SOLUTION:
Total load= 60 + 40 + 20 * 6 = 220kN
To find reactions
Taking moment about A
R_{B} * 6 + 60 * 3 = 20 * 6 * 0/2 + 40 * 8
= 360 + 320 = 680
R{B} * 6 = 680 – 180 = 500
R{B} = 500/6 = 83.33kN
R{A} = 220 – 83.33 = 136.67kN
(OR)
Taking moments about B
R{A} * 6 + 40 * 2 = (20 * 6) * 0/2 + 60 * 9
= 360 + 540 = 900
R{A} * 6 = 900 – 80 = 820
R{A} = 820/6 = 136.67kN