Calculating Transverse Vibration Frequency of a Shaft with Multiple Loads

This problem asks us to determine the frequency of transverse vibration for a simply supported shaft carrying multiple point loads. The natural frequency of vibration is related to the stiffness and mass distribution of the system. For beams with multiple point loads, methods like Dunkerley's method or Rayleigh's method based on static deflections can be used to estimate the fundamental frequency.

Given Information:

• Shaft Diameter, \(d = 50 \text{ mm} = 0.05 \text{ m}\)
• Shaft Length, \(L = 3 \text{ m}\)
• Loads and their distances from the left support:
  • \(W_1 = 1000 \text{ N}\) at \(a_1 = 1 \text{ m}\)
  • \(W_2 = 1500 \text{ N}\) at \(a_2 = 2 \text{ m}\)
  • \(W_3 = 750 \text{ N}\) at \(a_3 = 2.5 \text{ m}\)
• Young's Modulus, \(E = 200 \frac{\text{GN}}{\text{m}^2} = 200 \times 10^9 \frac{\text{N}}{\text{m}^2}\)
• Acceleration due to gravity, \(g \approx 9.81 \frac{\text{m}}{\text{s}^2}\)

Step 1: Calculate the Area Moment of Inertia (\(I\))

For a solid circular shaft, the area moment of inertia is given by:

\(I = \frac{\pi d^4}{64}\)

Substituting the diameter \(d = 0.05 \text{ m}\):

\(I = \frac{\pi (0.05)^4}{64} = \frac{\pi \times 0.00000625}{64}\)

\(I \approx 3.0679 \times 10^{-7} \text{ m}^4\)

Step 2: Calculate \(EI\)

The flexural rigidity of the shaft is \(EI\):

\(EI = (200 \times 10^9 \frac{\text{N}}{\text{m}^2}) \times (3.0679 \times 10^{-7} \text{ m}^4)\)

\(EI \approx 61358 \text{ N m}^2\)

Step 3: Calculate Static Deflection at Each Load Point due to Each Load

We will use Rayleigh's method, which requires the total static deflection at each load point due to all loads combined. The deflection \(\delta(x)\) at a point \(x\) due to a load \(W\) at a point \(c\) on a simply supported beam of length \(L\) is given by:

• If \(x \le c\): \(\delta(x) = \frac{W (L-c) x}{6 E I L} (L^2 - x^2 - (L-c)^2)\)
• If \(x \ge c\): \(\delta(x) = \frac{W c (L-x)}{6 E I L} (L^2 - (L-x)^2 - c^2)\)

Note that for the case when \(x=c\), the deflection at the load point itself is also given by the simpler formula \(\delta(c) = \frac{W c^2 (L-c)^2}{3 E I L}\).

Let's calculate the total deflection \(\delta_i^{total}\) at the position of load \(W_i\) (i.e., at \(x = a_i\)) due to all loads \(W_j\) at positions \(a_j\).

\(6 E I L = 6 \times 61358 \times 3 = 1104444 \text{ N m}^3\)

Deflection at Point 1 (\(x = 1 \text{ m}\)) due to all loads: \(\delta_1^{total}\)

• Due to \(W_1 = 1000 \text{ N}\) at \(c_1 = 1 \text{ m}\) (\(x=c\)):
  \(\delta_{11} = \frac{W_1 a_1^2 (L-a_1)^2}{3 E I L} = \frac{1000 \times 1^2 \times (3-1)^2}{3 \times 61358 \times 3} = \frac{1000 \times 1 \times 4}{552222} = \frac{4000}{552222} \approx 0.007244 \text{ m}\)
• Due to \(W_2 = 1500 \text{ N}\) at \(c_2 = 2 \text{ m}\) (\(x=1 \le c=2\)):
  \(\delta_{12} = \frac{W_2 (L-c_2) x_1}{6 E I L} (L^2 - x_1^2 - (L-c_2)^2) = \frac{1500 (3-2) 1}{1104444} (3^2 - 1^2 - (3-2)^2) = \frac{1500 \times 1 \times 1}{1104444} (9 - 1 - 1) = \frac{1500 \times 7}{1104444} = \frac{10500}{1104444} \approx 0.009507 \text{ m}\)
• Due to \(W_3 = 750 \text{ N}\) at \(c_3 = 2.5 \text{ m}\) (\(x=1 \le c=2.5\)):
  \(\delta_{13} = \frac{W_3 (L-c_3) x_1}{6 E I L} (L^2 - x_1^2 - (L-c_3)^2) = \frac{750 (3-2.5) 1}{1104444} (3^2 - 1^2 - (3-2.5)^2) = \frac{750 \times 0.5 \times 1}{1104444} (9 - 1 - 0.5^2) = \frac{375}{1104444} (8 - 0.25) = \frac{375 \times 7.75}{1104444} \approx 0.002631 \text{ m}\)

\(\delta_1^{total} = \delta_{11} + \delta_{12} + \delta_{13} \approx 0.007244 + 0.009507 + 0.002631 \approx 0.019382 \text{ m}\)

Deflection at Point 2 (\(x = 2 \text{ m}\)) due to all loads: \(\delta_2^{total}\)

• Due to \(W_1 = 1000 \text{ N}\) at \(c_1 = 1 \text{ m}\) (\(x=2 \ge c=1\)):
  \(\delta_{21} = \frac{W_1 c_1 (L-x_2)}{6 E I L} (L^2 - (L-x_2)^2 - c_1^2) = \frac{1000 \times 1 \times (3-2)}{1104444} (3^2 - (3-2)^2 - 1^2) = \frac{1000 \times 1 \times 1}{1104444} (9 - 1 - 1) = \frac{1000 \times 7}{1104444} \approx 0.006338 \text{ m}\)
• Due to \(W_2 = 1500 \text{ N}\) at \(c_2 = 2 \text{ m}\) (\(x=c\)):
  \(\delta_{22} = \frac{W_2 a_2^2 (L-a_2)^2}{3 E I L} = \frac{1500 \times 2^2 \times (3-2)^2}{3 \times 61358 \times 3} = \frac{1500 \times 4 \times 1}{552222} = \frac{6000}{552222} \approx 0.010865 \text{ m}\)
• Due to \(W_3 = 750 \text{ N}\) at \(c_3 = 2.5 \text{ m}\) (\(x=2 \le c=2.5\)):
  \(\delta_{23} = \frac{W_3 (L-c_3) x_2}{6 E I L} (L^2 - x_2^2 - (L-c_3)^2) = \frac{750 (3-2.5) 2}{1104444} (3^2 - 2^2 - (3-2.5)^2) = \frac{750 \times 0.5 \times 2}{1104444} (9 - 4 - 0.5^2) = \frac{750}{1104444} (5 - 0.25) = \frac{750 \times 4.75}{1104444} \approx 0.003226 \text{ m}\)

\(\delta_2^{total} = \delta_{21} + \delta_{22} + \delta_{23} \approx 0.006338 + 0.010865 + 0.003226 \approx 0.020429 \text{ m}\)

Deflection at Point 3 (\(x = 2.5 \text{ m}\)) due to all loads: \(\delta_3^{total}\)

• Due to \(W_1 = 1000 \text{ N}\) at \(c_1 = 1 \text{ m}\) (\(x=2.5 \ge c=1\)):
  \(\delta_{31} = \frac{W_1 c_1 (L-x_3)}{6 E I L} (L^2 - (L-x_3)^2 - c_1^2) = \frac{1000 \times 1 \times (3-2.5)}{1104444} (3^2 - (3-2.5)^2 - 1^2) = \frac{1000 \times 1 \times 0.5}{1104444} (9 - 0.5^2 - 1^2) = \frac{500}{1104444} (9 - 0.25 - 1) = \frac{500 \times 7.75}{1104444} \approx 0.003509 \text{ m}\)
• Due to \(W_2 = 1500 \text{ N}\) at \(c_2 = 2 \text{ m}\) (\(x=2.5 \ge c=2\)):
  \(\delta_{32} = \frac{W_2 c_2 (L-x_3)}{6 E I L} (L^2 - (L-x_3)^2 - c_2^2) = \frac{1500 \times 2 \times (3-2.5)}{1104444} (3^2 - (3-2.5)^2 - 2^2) = \frac{1500 \times 2 \times 0.5}{1104444} (9 - 0.25 - 4) = \frac{1500 \times 4.75}{1104444} \approx 0.006451 \text{ m}\)
• Due to \(W_3 = 750 \text{ N}\) at \(c_3 = 2.5 \text{ m}\) (\(x=c\)):
  \(\delta_{33} = \frac{W_3 a_3^2 (L-a_3)^2}{3 E I L} = \frac{750 \times 2.5^2 \times (3-2.5)^2}{3 \times 61358 \times 3} = \frac{750 \times 6.25 \times 0.25}{552222} = \frac{1171.875}{552222} \approx 0.002122 \text{ m}\)

\(\delta_3^{total} = \delta_{31} + \delta_{32} + \delta_{33} \approx 0.003509 + 0.006451 + 0.002122 \approx 0.012082 \text{ m}\)

Load \(W_i\) (N)	Position \(a_i\) (m)	Total Static Deflection \(\delta_i^{total}\) at \(a_i\) (m)
1000	1	0.019382
1500	2	0.020429
750	2.5	0.012082

Step 4: Apply Rayleigh's Method Formula

Rayleigh's method for estimating the fundamental natural frequency \(f\) for a beam with multiple loads is given by:

\(f \approx \frac{1}{2\pi} \sqrt{\frac{g \sum_{i=1}^n W_i \delta_i^{total}}{\sum_{i=1}^n W_i (\delta_i^{total})^2}}\)

Calculate the sums:

\(\sum W_i \delta_i^{total} = (1000 \times 0.019382) + (1500 \times 0.020429) + (750 \times 0.012082)\)

\(\sum W_i \delta_i^{total} \approx 19.382 + 30.6435 + 9.0615 = 59.087 \text{ N m}\)

\(\sum W_i (\delta_i^{total})^2 = (1000 \times 0.019382^2) + (1500 \times 0.020429^2) + (750 \times 0.012082^2)\)

\(\sum W_i (\delta_i^{total})^2 \approx (1000 \times 0.0003756) + (1500 \times 0.0004173) + (750 \times 0.0001460)\)

\(\sum W_i (\delta_i^{total})^2 \approx 0.3756 + 0.6260 + 0.1095 = 1.1111 \text{ N m}^2\)

Now, substitute these values into Rayleigh's formula (using \(g = 9.81 \text{ m/s}^2\)):

\(f \approx \frac{1}{2\pi} \sqrt{\frac{9.81 \times 59.087}{1.1111}}\)

\(f \approx \frac{1}{2\pi} \sqrt{\frac{579.74}{1.1111}}\)

\(f \approx \frac{1}{2\pi} \sqrt{521.76}\)

\(f \approx \frac{1}{2\pi} \times 22.842\)

\(f \approx 3.636 \text{ Hz}\)

Conclusion

The calculated frequency of transverse vibration using Rayleigh's method is approximately 3.636 Hz, which is closest to 3.5 Hz among the given options.

The final answer is \(\boxed{3.5 Hz}\).