Engineering Mechanics

Chapter 12 : Support Reactions „„„„„ 239

and reaction at Q RQ=+ =(2)^22 (1.73) 2.64 kN

Let θ = Angle, which the reaction RQ makes with the vertical.

∴

2

tan 1.160

1.73

θ= = or θ = 49.1° Ans.

Example 12.17. A truss hinged at A, and supported on rollers inclined at 45° with the horizontal
at D, is loaded as shown in Fig. 12.33.

Fig. 12.33.
Find the reaction at A and D.
Solution. Since the truss is supported on rollers at D, therefore the reaction at this support
will be normal to the support i.e. inclined at 45° with the vertical (because the support is inclined at an
angle of 45° with the horizontal) as shown in Fig. 12.34. The reaction at A will be the resultant of
vertical and horizontal forces.

Let RA= Reaction at A, and

RD= Reaction at D.

∴ Horizontal component of reaction at D,

RDH=RDV = RD cos 45° = 0.707 RD

Now taking moments about A and equating the same,

RDV × 9 – RDH × 4 = (5 × 3) + (2 × 6)

(0.707 RD × 9) – (0.707 RD × 4) = 27

or

27

7.64 kN

D 3.535

R == Ans.

∴ RDH=RDV = 7.64 × 0.707 = 5.4 kN

Now vertical component of reaction at A,

RAV= (5 + 2) – 5.4 = 1.6 kN

and horizontal component of reaction at A

RAH=RDH = 5.4 kN

∴ RA=+=(1.6)^22 (5.4) 5.63 kN Ans.

Fig. 12.34.