Relation between Shear force (s) , Bending Moment (M) and Loading rate (W/m)

Now ,

$\sum F_y=0$ ---- (+ve Left to Right )

$s-W*dx-(s+ds)=0$

$s-Wdx-s-ds=0$

$-Wdx=ds$

$w=\frac{-ds}{dx}$

The rate of change of the shearing force wrt c is equal to the load.

OR

The slope of the shear diagram at a given point equals to the load at that point.

Now, $\sum M_{2-2}=0$

$M+\delta*dx-(w*dx)*\frac{dx}{2}-(M+dM)=0$

$M+\delta dx-w(\frac{dx^2}{2}-M-dM=0$

$(dx^2)$ is very small, neglecting it

$M+\delta dx-M-dm=0$

$\delta dx-dM=0$

$\delta=\frac{dM}{dx}$

The rate of change of the bending moment wrt x is equal to the shearing force.

OR

The slope of the moment diagram at a given point is the shear at that point.