Engineering Mechanics

(^230) „„„„„ A Textbook of Engineering Mechanics
∴ RA=+=(6.205)^22 (0.995) 6.28 kN
Let θ = Angle, which the reaction at A makes with the vertical.
∴
0.995
tan 0.1604
6.205
θ= = or θ = 9.1°
Example 12.9. A beam has hinged support at A and roller support at B as shown in Fig. 12.23.
Fig. 12.23.
The beam is subjected to loads as shown. Determine analytically the reactions at A and B.
Solution. Given: Span = 9 m
Let RA = Reaction at A, and
RB = Reaction at B.
The reaction at B, supported on rollers and inclined at an angle of 30°
with the vertical is shown in Fig. 12.24. We know that as the beam is hinged at
A, therefore the reaction at this end will be the resultant of vertical and hori-
zontal forces, and thus will be inclined with the vertical.
Resolving the 12 kN load at E vertically
= 12 sin 30° = 12 × 0.5 = 6 kN
and now resolving it horizontally
= 12 cos 30° = 12 × 0.866 = 10.4 kN
We know that vertical component of reaction RB.
= RB cos 30° = RB × 0.866 = 0.866 RB
and anticlockwise moment due to vertical component of reaction RB about A
= 0.866 RB × 9 = 7.794 RB ...(i)
We also know that sum of clockwise moments due to loads about A
= (6 × 3) + (9 × 6) + (3 × 3) = 81 kN-m ...(ii)
Now equating the anticlockwise and clockwise moments given in (i) and (ii),
7.794 RB = 81 or^81 10.4 kN
7.794
RB== Ans.
We know that vertical component of the reaction RB
= 0.866 RB = 0.866 × 10.4 = 9.0 kN
and horizontal component of reaction RB
= RB sin 30° = 10.4 × 0.5 = 5.2 kN
Vertical component of reaction RA
= (6 + 9) – 9 = 6 kN
Fig. 12.24.