A centiliver beam of negligible mass 0.6m .it has rectangular cross-se...
Calculation of Young's modulus of a Centiliver beam
Given
- Length of the beam, L = 0.6m
- Width of the beam, b = 8mm = 0.008m
- Height of the beam, h = 6mm = 0.006m
- Mass of the beam, m = 1.4kg
- Natural frequency of the system, f = 10 rad/s
Formula
The natural frequency of a beam is given by:
f = (1/2π) √(EI/ml^3)
where E is the Young's modulus of the material, I is the moment of inertia of the cross-section of the beam, and m is the mass of the beam.
The moment of inertia of a rectangular cross-section is given by:
I = bh^3/12
Young's modulus is given by:
E = (mL^3f^2)/(4b^3h^3π^2)
Solution
Substituting the given values in the formula:
I = (0.008 x 0.006^3)/12 = 8 x 10^-11 m^4
E = (1.4 x 0.6^3 x 10^2)/(4 x 0.008^3 x 0.006^3 x π^2) = 1.2 x 10^11 Pa
Therefore, the Young's modulus of the material of the beam is 1.2 x 10^11 Pa.
Explanation
A Centiliver beam is a type of beam that is clamped at one end and free at the other end. The natural frequency of such a beam depends on the length, mass, and Young's modulus of the material. The formula for calculating the natural frequency involves the moment of inertia of the cross-section, which depends on the width and height of the beam. The Young's modulus is a measure of the stiffness of the material and is given by the ratio of stress to strain. In this problem, we use the formula for the natural frequency of a beam and the moment of inertia of a rectangular cross-section to calculate the Young's modulus of the material of the beam.