Consider a simply supported beam carrying a uniformly distributed load of intensity w per unit length over its whole span as shown in Fig. 1.1 (a).
Taking moment about R2, we gat
R1 × l = wl ×1/2 R1 = wl/2
R2 = wl/2
Consider a section at distance x from R1. Shear force at this cross-section is,
This is represents the equation of a straight line. At x = l/2, the shear force is zero and is positive in nature upto this length. At x = l, the shear force F = – wl/2,
i.e., it becomes negative. Hence the shear force diagram is as shown in Fig. 1.1.(b).
The bending moment at a section x frim R1 is,
M = R1 x – wx. x/2 = R1x – (wx²/2)
= (wl/2) x – (wx²/2) = w/2 (lx – x²)
This represents the equation of parabola. Hence the variation of bending moment is parabolic in nature.
At. x = 0
M = 0
At. x = l/2
Mmax = wl²/8 .........(1.1.)
and is the maximum bending moment. At x = l, M = 0. The Bending moment remains positive on the whole span of the beam. Hence the bending moment is shown in Fig. 1.1 (c).
This is represents the equation of a straight line. At x = l/2, the shear force is zero and is positive in nature upto this length. At x = l, the shear force F = – wl/2, 
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