The theory of bent beams shows that at any point in the beam the 'bending moment' is given by $K / \rho$, where $K$ is a constant (that depends upon the beam material and cross-sectional shape) and $\rho$ is the radius of curvature at that point. Consider a light beam of length $L$ whose ends, $x=0$ and $x=L$, are supported at the same vertical height and which has a weight $W$ suspended from its centre. Verify that at any point $x(0 \leq x \leq L / 2$ for definiteness) the net magnitude of the bending moments, (bending moment $=$ force $\times$ perpendicular distance) due to the weight and support reactions, evaluated on either side of $x$, is $W x / 2$.
If the beam is only slightly bent, so that $(d y / d x)^{2} \ll 1$, where $y=y(x)$ is the downward displacement of the beam at $x$, show that the beam profile satisfies the approximate equation
$$
\frac{d^{2} y}{d x^{2}}=-\frac{W x}{2 K}
$$
By integrating this equation twice and using physically imposed conditions on your solution at $x=0$ and $x=L / 2$, show that the downward displacement at the centre of the beam is $W L^{3} /(48 K)$.