Question

Compute the fundamental natural frequency of the transverse vibration of a uniform beam of rectanqular cross section, with one end of the beam is free and one end is fixed. The cross section has a base of 100 mm and height of 300 mm, length of the beam is 5.9 m, E=20.5x10^10 N/m2 and density of 7830 kg/m3. Write your answer in rad/sec with 2 decimal points.

Answer 1

Answer:

The natural angular frequency of the rod is 53.56 rad/sec

Explanation:

Since the beam is free at one end and fixed at the other hence the beam is a cantilevered beam as shown in the attached figure

We know that when a unit force is placed at the end of a cantilever the displacement of the free end is given by

$\Delta x=\frac{PL^3}{3EI}$

Hence we can write

$P=\frac{3EI\cdot \Delta x}{L^3}$

Comparing with the standard spring equation $F=kx$ we find the cantilever analogous to spring with $k=\frac{3EI}{L^3}$

Now the angular frequency of a spring is given by

$\omega =\sqrt{\frac{k}{m}}$

where

'm' is the mass of the load

Thus applying values we get

$\omega _{beam}=\sqrt{\frac{\frac{3EI}{L^{3}}}{Area\times density}}$

$\omega _{beam}=\sqrt{\frac{\frac{3\times 20.5\times 10^{10}\times \frac{0.1\times 0.3^3}{12}}{5.9^{3}}}{0.3\times 0.1 \times 7830}}=53.56rad/sec$

Answer 2

Answer:

its a lil confusing

Explanation: