**SIMPLY SUPPORTED BEAM SUBJECTED TO CENTRAL LOADING:**

By symmetry the reactions at the two supports would be W/2 and W/2. now consider any section X-X from the left end then, the beam is under the action of following forces.

So the shear force at any X-section would be = W/2 [Which is constant upto x < l/2]

If we consider another section Y-Y which is beyond l/2 then

for all values greater = l/2

Hence S.F diagram can be plotted as,

For B.M diagram:

If we just take the moments to the left of the cross-section,

Which when plotted will give a straight relation i.e.

It may be observed that at the point of application of load there is an abrupt change in the shear force, at this point the B.M is maximum.