First of all, let?s assume length between end supports be ?l? maximum bending moment in a simply supported beam with point load at its centre is wl/4. We know that in simply supported beam maximum bending moment occurs at centre only.

A point at which bending moment changes its sign from positive to negative and vice versa. Such point is termed as point of contraflexure. At this point, value of bending moment is zero (0).

According to BIS, standard symbols used for sketching SFD are
Point load = ????
Udl load = \

A Beam extended beyond its support. And position of extension is called as over hung portion.

A fixed support offers resistance against horizontal and vertical movement and against rotation of member and that in turn developers support moment.

Consider a simply supported beam subjected to udl for entire span considered a free body diagram of small portion of elemental length dx.

Moment at B = 0
Moment at C = ? (10 × 3) × (3/2)
= ? 45 kNm
Moment at A = ? (10 × 3) × (1.5 + 2 )
Maximum BM at A = ? 105 kNm
= 105 Nm (hogging).

Explanation:

Cantilever beam carries a uniformly distributed load:

Shear force:

SFxx = wx

Shear force diagram is a triangle with zero at free end and WL at fixed end, âˆ´ it is linear.

Bending Moment:

$B{M}_{xx}=w\frac{x^2}{2}$

Concept:

Slope of shear force diagram = Load intensity at that section

$\frac{\mathrm{δ}F}{\mathrm{δ}x}=−w$

Slope of bending moment diagram = Shear force at that section

$\frac{\mathrm{δ}M}{\mathrm{δ}x}=F$

$w=−\frac{\mathrm{δ}F}{\mathrm{δ}x}=\frac{{\mathrm{δ}}^{2}M}{\mathrm{δ}{x}^2}$

$F=−∫wdx;M=∫Fdx$

If w is n degree curve, V will be (n+1) degree curve and M will be (n + 2) degree curve.

In case of uniformly distributed load, w is uniform, V is linear, and M is of second order parabola.

Figure 1 : Simply supported beam

Figure 2 : Cantilever beam

Figure 3 : Fixed beam

The bending moment at a section is considered to be positive when it causes convexity downwards such bending moment is called sagging bending moment positive bending moment.