Question

One end of a steel rectangular girder is embedded into a wall (figure shown above). Due to gravity it sags slightly. Find the radius of curvature of the neutral layer (see the dotted line in the figure above) in the vicinity of the point $O$ if the length of the protruding section of the girder is equal to $l=6.0m$ and the thickness of the girder equals $h=10cm$.

Answer 1

A beam clamped at one end supporting an applied load at the free end is called a cantilever. The theory of cantilevers is discussed in advanced text book on mechanics. The key result is that elastic forces in the beam generate a couple, whose, moment, called the moment of resistances, balances the external bending moment due to weight of the beam, load etc. The moment of resistance, also called internal bending moment (I.B.M) is given by
$I.B.M.=\frac{EI}{R}$
Here $R$ is the radius of curvature of the beam at the representative point $(x,y)$. $I$ is called the geometrical moment of inertia
$I=\int z^{2}dS$
of the cross section relative to the axis passing through the neutral layer which remains unstretched. (Figure shown below). The section of the beam beyond $P$ exerts the bending moment $N\left(x\right)$ and we have,
$\frac{EI}{R}=N\left(x\right)$
If there is no load other than that due to the weight of the beam, then
$N\left(x\right)=\frac{1}{2}\rho g{(l-x)}^{2}bh$
where $\rho =$ density of steel.
Hence, at $x=0$
${\left(\frac{1}{R}\right)}_0=\frac{\rho g{l}^{2}bh}{2EI}$
Here $b=$ width of the beam perpendicular to the paper.
Also, $I={\int }_{-\frac{h}{2}}^{\frac{h}{2}}{z}^{2}bdz=\frac{b{h}^3}{12}$.
Hence, ${\left(\frac{1}{R}\right)}_0=\frac{6\rho g{l}^2}{E{h}^2}={\left(0.121km\right)}^{-1}$