Depression of the midpoint of a weightless beam supported at two ends and loaded in the middle:

Let, MN represents a uniform rectangular beam of length l, which is supported near the ends of two knife edges held at the same horizontal plane, as shown in the figure. A weight \( W \) is applied at the midpoint \( O \) of the beam. So the reaction at each knife-edge is \( \frac{W}{2} \) in the vertically upward direction.

The beam may be considered as equivalent to two inverted cantilevers because the middle part of the beam is horizontal. The bending in each cantilever of length \( \frac{l}{2} \) is produced by the load \( \frac{W}{2} \), acting vertically upwards at the points M and N.

Let us consider a very small section AB at a distance \( x \) from the mid-point \( O \). Now the external bending moment is \( \displaystyle{\frac{W}{2}(\frac{l}{2}-x)} \). Again we know that the internal bending moment is \( \displaystyle{\frac{YI}{R}} \), [Read In Detail] where \( Y \) is Young’s modulus, \( I \) is the geometrical moment of inertia of the beam, \( R \) is the radius of curvature of the neutral axis of \( AB \).
In the equilibrium of the beam, the internal moment is balanced by the external bending moment, so we can write
\( \displaystyle{\frac{YI}{R}=\frac{W}{2}(\frac{l}{2}-x)} \)
or, \( \displaystyle{\frac{1}{R}=\frac{W}{2YI}(\frac{l}{2}-x)} \)
or, \( \displaystyle{\frac{d\theta}{dx}=\frac{W}{2YI}(\frac{l}{2}-x)} \)
or, \( \displaystyle{d\theta=\frac{W}{2YI}(\frac{l}{2}-x)dx} \).
where, \( \frac{1}{R}=\frac{d\theta}{dx} \), \( dx \) is the length of AB and \( d\theta \) is the angle subtended by AB at the centre of curvature.

Now the depression of B below A is
\( \displaystyle{dy=(\frac{l}{2}-x)d\theta\\=\frac{W}{2YI}{(\frac{l}{2}-x)}^2{dx}} \)
So the total depression of the midpoint \( O \) is
\( \displaystyle{y=\frac{W}{2YI}\int_0^{\frac{l}{2}}{(\frac{l}{2}-x)}^2{dx}\\=\frac{W}{2YI}\int_0^{\frac{l}{2}}[\frac{l^2}{4}+x^2-2\frac{l}{2}x]dx\\=\frac{W}{2YI}{[\frac{l^2}{4}x+\frac{x^3}{3}-l\frac{x^2}{2}]}_0^{\frac{l}{2}}\\=\frac{W}{2YI}(\frac{l^3}{8}+\frac{l^3}{24}-\frac{l^3}{8})\\=\frac{W}{2YI}\frac{l^3}{24} }\)

\( \displaystyle{y=\frac{Wl^3}{48YI} } \)

Since the beam has a rectangular cross-sectional area of depth \( d \) and breadth \( b \), so the geometrical moment of inertia of the beam is \( I=\frac{bd^3}{12} \).
Therefore the depression of the beam is,
\( \displaystyle{y=\frac{Wl^3}{48Y}\frac{12}{bd^3}\\=\frac{Wl^3}{4Ybd^3}} \)

If the beam has a circular cross-sectional area of radius \( r \), then the geometrical moment of inertia of the beam is \( I=\frac{\pi{r}^4}{4} \).
Therefore the depression of the beam is,
\( \displaystyle{y=\frac{Wl^3}{48Y}\frac{4}{\pi{r}^4}\\=\frac{Wl^3}{12Y\pi{r^4}}} \)