A vibrating system having a mass of 30 kg is supported by 4 springs, each having a stiffness of k. The system runs at 420 rpm. If only 10% of the shaking force is allowed to be transmitted to the supporting structure, then the value of stiffness k will be close to:
1.32 N/mm
This problem involves understanding the concepts of mechanical vibrations, specifically focusing on a vibrating system with springs and mass, and calculating the required stiffness based on a given force transmissibility. We need to determine the stiffness of each spring ('k') when the system's mass, operating speed, and the allowable force transmitted to the supporting structure are known.
Understanding Vibrating System Parameters
A vibrating system typically consists of a mass supported by elastic elements like springs. When an external force or unbalance causes vibrations, these are transmitted to the supporting structure. The goal is often to minimize this transmitted force, which is quantified by transmissibility.
- Mass (m): The total mass of the vibrating system is given as 30 kg.
- Springs: The system is supported by 4 springs, each having an individual stiffness 'k'. Therefore, the equivalent stiffness ($K_{eq}$) of the entire spring system will be the sum of individual stiffnesses, since they are in parallel: $K_{eq} = 4k$.
- Operating Speed (N): The system operates at a speed of 420 revolutions per minute (rpm). This speed needs to be converted into angular frequency ($\omega$) in radians per second.
- Transmissibility (TR): Only 10% of the shaking force is allowed to be transmitted to the supporting structure. This means the transmissibility ($TR$) is 0.10. Transmissibility is a dimensionless ratio that describes how much of the exciting force is transmitted to the base or support.
Key Formulas for Vibration Analysis
To solve this problem, we will use the fundamental formulas related to mechanical vibrations:
- Angular Frequency of Excitation ($\omega$): This is derived from the operating speed. $$ \omega = \frac{2\pi N}{60} $$ where N is in rpm.
- Natural Frequency ($\omega_n$): This is an inherent property of the system, dependent on its mass and stiffness. $$ \omega_n = \sqrt{\frac{K_{eq}}{m}} $$ where $K_{eq}$ is the equivalent stiffness and $m$ is the mass.
- Frequency Ratio (r): This is the ratio of the excitation frequency to the natural frequency. $$ r = \frac{\omega}{\omega_n} $$
- Transmissibility (TR): For an undamped system, the transmissibility is given by: $$ TR = \frac{1}{\left|1 - r^2\right|} $$ Since the transmissibility (0.10) is less than 1, it implies that the system is operating above its resonant frequency (i.e., $r > \sqrt{2}$). In this case, the formula simplifies to: $$ TR = \frac{1}{r^2 - 1} $$
Step-by-Step Stiffness Calculation
Let's calculate the required stiffness 'k' step by step using the given information and formulas.
1. Excitation Frequency Calculation
First, convert the operating speed from rpm to angular frequency in radians per second.
$$ N = 420 \text{ rpm} $$ $$ \omega = \frac{2\pi N}{60} = \frac{2\pi \times 420}{60} $$ $$ \omega = 14\pi \text{ rad/s} $$ $$ \omega \approx 43.98 \text{ rad/s} $$2. Frequency Ratio Determination
Next, use the given transmissibility to find the frequency ratio 'r'.
$$ TR = 0.10 $$ $$ TR = \frac{1}{r^2 - 1} $$ $$ 0.10 = \frac{1}{r^2 - 1} $$Rearranging the equation:
$$ r^2 - 1 = \frac{1}{0.10} $$ $$ r^2 - 1 = 10 $$ $$ r^2 = 11 $$ $$ r = \sqrt{11} $$ $$ r \approx 3.3166 $$3. Natural Frequency Calculation
Now that we have the excitation frequency ($\omega$) and the frequency ratio (r), we can calculate the natural frequency ($\omega_n$).
$$ r = \frac{\omega}{\omega_n} $$ $$ \omega_n = \frac{\omega}{r} $$ $$ \omega_n = \frac{14\pi}{\sqrt{11}} $$ $$ \omega_n \approx \frac{43.98}{3.3166} $$ $$ \omega_n \approx 13.262 \text{ rad/s} $$4. Equivalent Stiffness Determination
Using the natural frequency ($\omega_n$) and the mass (m), we can find the equivalent stiffness ($K_{eq}$) of the entire spring system.
$$ m = 30 \text{ kg} $$ $$ \omega_n = \sqrt{\frac{K_{eq}}{m}} $$Squaring both sides:
$$ \omega_n^2 = \frac{K_{eq}}{m} $$ $$ K_{eq} = m \cdot \omega_n^2 $$ $$ K_{eq} = 30 \text{ kg} \times (13.262 \text{ rad/s})^2 $$ $$ K_{eq} = 30 \times 175.879 $$ $$ K_{eq} \approx 5276.37 \text{ N/m} $$5. Individual Spring Stiffness (k)
The system is supported by 4 springs, each with stiffness 'k'. Therefore, the total equivalent stiffness is $K_{eq} = 4k$.
$$ K_{eq} = 4k $$ $$ k = \frac{K_{eq}}{4} $$ $$ k = \frac{5276.37 \text{ N/m}}{4} $$ $$ k \approx 1319.09 \text{ N/m} $$6. Convert Stiffness to N/mm
The options are given in N/mm, so we need to convert our calculated stiffness.
$$ k \approx 1319.09 \text{ N/m} $$Since 1 m = 1000 mm:
$$ k \approx \frac{1319.09 \text{ N}}{1000 \text{ mm}} $$ $$ k \approx 1.31909 \text{ N/mm} $$Result and Conclusion
The calculated value for the stiffness 'k' of each spring is approximately 1.31909 N/mm.
Comparing this value with the given options:
| Option | Value (N/mm) |
|---|---|
| 1 | 1.09 |
| 2 | 3.96 |
| 3 | 2.18 |
| 4 | 1.32 |
The calculated value of 1.31909 N/mm is very close to 1.32 N/mm. This means that for only 10% of the shaking force to be transmitted, each spring needs to have a stiffness close to 1.32 N/mm.
Important Questions from Transmissibility and Magnification Factor
When ω/ωn = 1, the transmissibility is
where ω = frequency of excitation, rad/s
ωn = Natural frequency, rad/s
The Transmissibility of a vibrating system, for the values of (ω/ωn) >√2, will be
For a single degree of freedom viscous damped system, if the frequency ratio is greater than √2, it implies that:
Dynamic Magnification Factor is computed as:
A single degree of freedom system having mass 1 kg and stiffness 10 kN/m initially at rest is subjected to an impulse force of magnitude 5 kN for 10-4 seconds. The amplitude in mm of the resulting free vibration is
