Discussion
A 5 kg mass is suspended at the free end of an overhanging massless beam, having a pin support and a roller support, as shown in the figure below. Young's modulus of the material of the beam is 200 GPa and area moment of inertia of the beam is \(10^{-8}\ \text{m}^{4}\). The natural frequency of the beam in rad/s is
Answer: A
The natural frequency is 10 rad/s.
Treat the beam as a spring under the mass. For an overhanging beam with span \(L = 1\) m and overhang \(a = 2\) m, a tip load \(F\) deflects the tip by
\[ \begin{aligned} \delta &= \frac{Fa^{2}(L + a)}{3EI} = \frac{F(4)(3)}{3EI} = \frac{4F}{EI} \\ k &= \frac{EI}{4} = \frac{200\times 10^{9}\times 10^{-8}}{4} = 500\ \text{N/m} \\ \omega_{n} &= \sqrt{k/m} = \sqrt{500/5} = 10\ \text{rad/s} \end{aligned} \]
The first term of the deflection comes from the overhang bending as a cantilever and the rest from rotation over the roller; dropping either part gives a stiffer spring and a higher, wrong frequency. 5 and the fractional values do not follow from \(k = 500\) N/m.
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