Engineering Mechanics

(Joyce) #1

(^228)  A Textbook of Engineering Mechanics
We know that vertical component of the reaction RA
= [5 + (1.5 × 2) + 2.83] – 5.05 = 5.78 kN
∴Reaction at A,
(^) RA=+=(5.78)^22 (2.83) 6.44 kN Ans.
Let θ= Angle, which the reaction at A makes with vertical.

2.83
tan 0.4896
5.78
θ= = or θ = 26.1° Ans.
Graphical method
(a) Equivalent space diagram (b) Vector diagram
Fig. 12.20.

1. First of all, draw the *equivalent space diagram of the beam, and name all the loads and
reactions according to Bow’s notations.

2. Select some suitable point p and draw pq, qr and rs parallel and equal to the loads 5 kN,
3 kN (equivalent load) and 4 kN to some scale.

3. Select any point o and join op, oq, or and os.

4. Now extend the lines of the loads PQ, QR, RS and the reaction RB. Through A draw Ap 1
parallel to op intersecting the line of action of 5 kN load at p 1.

5. Similarly, draw p 1 p 2 , p 2 p 3 and p 3 p 4 parallel to oq, or and os respectively. Join A and p 4.
Through o, draw a line parallel to this line. Now through s, draw a vertical line (as the
reaction RB is vertical) meeting the line through o at t. Join tp.

6. Now the lengths tp and st, in the vector diagram, give the magnitudes and direction of the
reaction RA and RB respectively to the scale as shown in Fig. 12.20 (a) and (b). By
measurement, we find that RA = tp = 6.5 kN ; RB = st = 5.0 kN and θ = 26° Ans.

``````* In this case, the uniformly distributed load is assumed as an equivalent load of 1.5 × 2 = 3 kN acting at
the centre of gravity of the load i.e., at the mid point of C and D.``````