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ESE Mechanical 2019: Official Paper

Option 1 : 279.5 Hz

CT 3: Building Materials

2962

10 Questions
20 Marks
12 Mins

__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 ×10^{5} N/mm^{2} = 2 × 10^{11} N/m^{2}

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

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

∴ 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.**