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Applied Mechanics - Engineering

Q1:

The beam shown in below figure is supported by a hinge at A and a roller at B. The reaction RA of the hinged support A of the beam, is

A 10.8 t

B 10.6 t

C 10.4 t

D 10.2 t.

ANS:D - 10.2 t.

Rb * 5 + 5 * 2 = 2 * 8 * 4.
Rb*5 = 64 - 10.
Rb*5 = 54.
Rb = 10.8.

So, Ra + Rb = 21,
Hence, Ra = 10.2. Compare both side moments at B,
(3*2*1.5) = (2*5*2.5) + (5*7) - (5Ra).
9 = 25+35-5Ra.
Ra=51/5.
Ra=10.2t. The resultant of the distributed load between A and B is:
2 t/m * 5 m = 10 t ... act at midway between A and B.

The resultant of the distributed load between B and the right free end of the beam is:
2 t/m * 3 m = 6 t ... act at midway.

Applying the moments=0 rule at point B:
(Clockwise is positive).

0= -5(7) + 5Ra - 10(2.5) + 6(1.5).
Solving for Ra gives:
Ra = 10.2 t.