Question

A massless cantilever beam, with a tip mass \( m \) of 10 kg, is modeled as an equivalent spring-mass system as shown in the figure. The beam is of length \( L = 1 \, {m} \), with a circular cross-section of diameter \( d = 20 \, {mm} \). The Young’s modulus of the beam material is 200 GPa.
The natural frequency of the spring-mass system is ............ Hz (rounded off to two decimal places).

Hint:

For calculating the natural frequency of a cantilever beam, determine the spring constant using the beam's Young's modulus and moment of inertia, and then use the spring-mass system formula for frequency.

Answer 1

Correct Answer: 3.43

We are asked to find the natural frequency of a spring-mass system representing a cantilever beam. The natural frequency \( f \) of the system is given by: \[ f = \frac{1}{2 \pi} \sqrt{\frac{k}{m}}, \] where:
- \( k \) is the spring constant,
- \( m \) is the mass.
Step 1: Calculate the Spring Constant \( k \) of the Beam
For a cantilever beam with a point mass at the end, the spring constant \( k \) can be calculated using the following formula: \[ k = \frac{3EI}{L^3}, \] where:
- \( E \) is the Young’s modulus of the beam material,
- \( I \) is the second moment of area (area moment of inertia) of the beam cross-section,
- \( L \) is the length of the beam.
The second moment of area \( I \) for a circular cross-section is given by: \[ I = \frac{\pi d^4}{64}, \] where \( d \) is the diameter of the beam. Substituting the given values: \[ d = 20 \, {mm} = 0.02 \, {m}, E = 200 \, {GPa} = 200 \times 10^9 \, {N/m}^2, L = 1 \, {m}. \] First, calculate \( I \): \[ I = \frac{\pi (0.02)^4}{64} = 7.85 \times 10^{-9} \, {m}^4. \] Now, calculate \( k \): \[ k = \frac{3 \times (200 \times 10^9) \times (7.85 \times 10^{-9})}{(1)^3} = 4710 \, {N/m}. \] Step 2: Calculate the Natural Frequency
Now that we have \( k \), we can use the formula for the natural frequency: \[ f = \frac{1}{2 \pi} \sqrt{\frac{k}{m}} = \frac{1}{2 \pi} \sqrt{\frac{4710}{10}}. \] Solving for \( f \): \[ f = \frac{1}{2 \pi} \sqrt{471} \approx \frac{1}{2 \pi} \times 21.7 \approx 3.46 \, {Hz}. \] Thus, the natural frequency lies between 3.43 Hz and 3.47 Hz, which is the correct range.