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The following are the methods of finding out the reactions at the two supports of a beam:

1.Analytical method, and

2.Graphical method.

2.10.1. Analytical Method. Fig. 2.37 shows a beam AB of length L and is simply supported at the ends A and B. The beam carries two point loads W1 and W2 at a distance L1 and L2 from the end A.

Let RA = Reaction at A 

and RB = Reaction at B

As the beam is in equilibrium, the equations of the equilibrium i.e., Fx = 0, Fy = 0 and M = 0 should be satisfied. In this case there is no horizontal force, hence the equations of equilibrium are Fy = 0 and M=O.

For  y = 0, we have
RA + RB = W1+W2

For M = 0, the moments about any point of all the forces should be zero.
Taking the moments about point A, we get

W1 × L1 + W2 × L2 – RB × L = 0

or  W1 × L1 + W2 × L2 = RB × L

As W1 W2, L1, L2 and L are given, hence value of RB can be calculated.

Now from equation (i), we have RA = (W1 + W2) – RB


2.10.2. Graphical Method for Finding Out the Reactions of a Beam. The graphical method consists of the following steps :

(a) Construction of space diagram;
(b) Use of Bow’s notations; and

(c) Construction of vector diagram.

The given beam is drawn to a suitable scale along with the loads and the reactions RA and RB. This step is known as construction of space diagram.

The different loads and forces i.e., reactions RA and RB) are named by two capital letters, placed on their either side of the space diagram as shown in Fig. 2.38. This step is known as Bow’s notation. The load W1 is named by PQ, W2 by QR, reaction RB by SR and reaction RA by SP.

Now the vector diagram is drawn according to the following steps:

i.            Choose a suitable scale to represent the various loads. Now take any point p and draw pq parallel and equal to the load PQ (i.e., W1) vertically. downward to the same scale.

ii.            Now through q, draw qr parallel and equal to QR vertically downward to the same scale.

    iii.            Select any suitable point O. Now join the point O to points p, q and r as shown in Fig. 2.38 (b).

    iv.            Now in Fig. 2.38 (a), extend the lines of action of the loads and the two reactions. Take any point 1, on the line of action of the reaction RA. Through 1, draw the line 1-2 parallel pO, intersecting the line of action of load W1 at 2.

v.            Now from point 2, draw line 2-3 parallel to qO, intersecting the line of action of load W2 at 3. Similarly, from point 3, draw the line 3-4 parallel to rO, intersecting the line of action of reaction RB at point 4.

Fig.2.38 (a)

Fig. 2.38 (a) Space diagram


Fig. 2.38 (b) Vector diagram

    vi.            Now join the point 1 to point 4. The line 1-4 is known as closing line. Now from point O (i.e., from vector diagram) draw line Os parallel to line 1-4.

     vii.            Now in the vector diagram the length sp represents the magnitude of reaction RA to the same scale. Similarly, the length rs represents the magnitude of reaction RB to the same scale.