Consider a cantilever beam, having negligible mass and uniform flexural rigidity, with length 0.01 m. The frequency of vibration of the beam, with a 0.5 kg mass attached at the free tip, is 100 Hz. The flexural rigidity (in N. ?2) of the beam is::
Question
Consider a cantilever beam, having negligible mass and uniform flexural rigidity, with length 0.01 m. The frequency of vibration of the beam, with a 0.5 kg mass attached at the free tip, is 100 Hz. The flexural rigidity (in N. ?2) of the beam is::
Solution
The frequency of vibration of a cantilever beam with a mass at the free end is given by the formula:
f = (1/2π) * √(3EI/(mL^3))
where: f is the frequency of vibration, E is the modulus of elasticity, I is the moment of inertia, m is the mass at the free end, L is the length of the beam.
We are asked to find the flexural rigidity (EI) of the beam. The flexural rigidity is the product of the modulus of elasticity (E) and the moment of inertia (I).
Rearranging the formula to solve for EI gives:
EI = m * L^3 * (2πf)^2 / 3
Substituting the given values:
m = 0.5 kg, L = 0.01 m, f = 100 Hz,
we get:
EI = 0.5 kg * (0.01 m)^3 * (2π * 100 Hz)^2 / 3 = 0.5 kg * 1e-6 m^3 * (200π rad/s)^2 / 3 = 0.5 kg * 1e-6 m^3 * (40000π^2 rad^2/s^2) / 3 = 0.5 * 1e-6 * 40000π^2 / 3 N.m^2 = 20π^2 / 3 N.m^2
So, the flexural rigidity of the beam is approximately 20π^2 / 3 N.m^2.
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