Question

A vertical shaft of 100 mm diameter and 1 m length has its upper end fixed at the top. The other end carries a disc of 5000 N and the modulus of elasticity of the shaft material is 2 × 10⁵ N/mm². Neglecting the weight of the shaft, the frequency of the longitudinal vibrations will be nearly
1. 279.5 Hz
2. 266.5 Hz
3. 253.5 Hz
4. 241.5 Hz

Answer 1

Correct Answer - Option 1 : 279.5 Hz

Concept:

Elongation due to axial load is

\(\delta l = \frac{{PL}}{{AE}}\)      ------ (1)

Let, k be the stiffness of the material then

P = kδl      ------ (2)

Now, substituting equation (2) in equation (1);

\(\therefore k = \frac{{AE}}{L}\)

Frequency (f) \( = \frac{{{\omega _n}}}{{2\pi }}\)

Where, ω_n \(= \sqrt {\frac{k}{m}}\)

Calculation:

Given:

d = 100mm; L = 1000mm; W = 5000N; E = 2 ×10⁵ N/mm² = 2 × 10¹¹ N/m²

\(k = \frac{{AE}}{L}\)

\(k = \frac{{\frac{\pi }{4}{{\left( {0.1} \right)}^2} \times 2 \times {{10}^{11}}}}{1}\)

∴ k = 1570796327 N/m.

Now,

\({\omega _n} = \sqrt {\frac{k}{m}} = \sqrt {\frac{{1570796327}}{{\frac{{5000}}{{9.81}}}}}\)

ω_n = 1755.535 rad/s

\({f_n} = \frac{{{\omega _n}}}{{2\pi }} = \frac{{1755.535}}{{2\pi }}\)

∴ f_n = 279.402 Hz.