A shaft of span 1 m and diameter 25 mm is simply supported at the ends. It carries a 1.5 kN concentrated load at mid-span. If E is 200 GPa, its fundamental frequency will be nearly

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  1. 3.5 Hz
  2. 4.2 Hz
  3. 4.8 Hz
  4. 5.5 Hz

Answer (Detailed Solution Below)

Option 4 : 5.5 Hz
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Detailed Solution

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Concept:

Angular frequency is given by:

\({\omega _n} = \sqrt {\frac{g}{δ }}\)

where δ = deflection of the beam

Deflection of a simply supported beam when a concentrated load is placed at midspan is given by:

δ = \(\frac{{P{L^3}}}{{48EI}}\)

Calculation:

Given:

L = 1 m, d = 25 mm, E = 200 × 109 Pa, P = 1.5 × 103 N

Deflection of the beam is:

δ = \(\frac{{P{L^3}}}{{48EI}}\)

δ = \(\frac{{1.5\; \times \;{{10}^3} \times {{1000}^3}}}{{48 \times 200 \times {{10}^9} \times \frac{\pi }{{64}} \times {{25}^4}}}\) = 8.14 mm

Angular frequency is:

\(∵ {\omega _n} = \sqrt {\frac{g}{δ }}\)

ωn = \(\sqrt {\frac{{9.81}}{{8.14\; \times \;{{10}^{ - 3}}}}} \) = 34.69 rad/s

Angular frequency can also be calculated by:

ωn = 2πf

where f = frequency in Hz

34.69 = 2πf

\(f =\frac{{34.65}}{{2\pi }}\)

∴ f = 5.52 Hz

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