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 W_{1} and W_{2} at a distance *L _{1} *and

*L*from the end

_{2}*A.*

Let *R _{A} *= Reaction

*at A*

*and R _{B} *= Reaction at

*B*

As the beam is in equilibrium, the equations of the equilibrium *i.e., * *F _{x} *= 0, F

*= 0 and*

_{y}*M*= 0 should be satisfied. In this case there is no horizontal force, hence the equations of equilibrium are

*F*= 0 and

_{y}*M=O.*

For* * *y *= 0, we have

R_{A }+ R_{B }= W_{1}+W_{2}

For *M *= 0, the moments about any point of all the forces should be zero.

Taking the moments about point *A, *we get

W_{1} × L_{1} + W_{2} × L_{2} – R_{B} × L = 0

or W_{1} × L_{1 }+ W_{2} × L_{2} = R_{B} × L

As W_{1} W_{2}, L_{1}, L_{2} and L are given, hence value of RB can be calculated.

Now from equation (i), we have R_{A} = (W_{1} + W_{2}) – R_{B}

**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 *R*_{A}* *and *R*_{B.}* *This step is known as construction of space diagram.

The different loads and forces *i.e., *reactions *R*_{A}* *and R_{B}) 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 W_{1} is named by *PQ, *W_{2} by *QR, *reaction *R _{B} *by

*SR*and reaction

*R*

_{A}*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., **W _{1}) *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 R_{A.} Through 1, draw the line 1-2 parallel *pO, *intersecting the line of action of load W_{1} at 2.

v. Now from point 2, draw line 2-3 parallel to *qO, *intersecting the line of action of load W_{2}* *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) 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 *R _{A} *to the same scale. Similarly, the length

*rs*represents the magnitude of reaction

*R*to the same scale.

_{B}

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