Doubt in derivation of bending of beam, It's about derivatives and intergration

Question

Radius of curvature of the beam in above picture is given as:

$$ \frac{1}{R} = \frac{d^2 y}{dx^2}$$

Please help me two points used as steps of a derivation in my book:

1. How was the radius of curvature of rod found?
2. How did eqtn 4-27 turn to eqtn 4-28? I see no dx, so I am confused how an integration could have been done.

Answer 1

1. Recall that radius of curvature is defined as:

$$ R= \frac{(1+y'^2)^{\frac32}}{y''}$$

In the particular case in the picture at $x=0$, $y'=0$ (due to choice of coordinate axes. Hence, $$ R = \frac{1}{y''}$$

2. They multiplied both side with $dx$ and integrated.

$$ W(l-x) = (YI_g) \frac{d^2 y}{dx^2}$$

Multiply both side by dx:

$$ W(l-x) dx = YI_g \frac{d^2 y }{dx^2} dx$$

And integrate.