Question

The critical speed of the shaft is affected by the

• A. Diameter and the eccentricity of the shaft
• B. Span and the eccentricity of the shaft
• C. Diameter and the span of the shaft
• D. Frequency of vibrations

Hint:

Critical speed of a shaft depends entirely on its structural stiffness and mass distribution: - Higher diameter ($d \uparrow$) $\implies$ Stiffness ($I \uparrow$) $\implies$ Critical speed $\uparrow$ - Longer span ($L \uparrow$) $\implies$ Stiffness $\downarrow$ $\implies$ Critical speed $\downarrow$ - Eccentricity influences vibration amplitude, but does not alter the critical speed value itself.

Answer 1

The Correct Option is C

Concept: The critical speed (or whirling speed) of a rotating shaft is the angular velocity at which the system becomes dynamically unstable, causing violent lateral deflections. Analytically, the critical speed of a shaft matches its natural frequency of lateral vibration (\(\omega_n\)). For a simple rotor-shaft assembly, the fundamental lateral natural frequency is given by: \[ \omega_n = \sqrt{\frac{g}{\delta}} \quad \text{rad/s} \quad \text{or} \quad \omega_n = \sqrt{\frac{k}{m}} \] Where:
• \(\delta\) = Static lateral deflection of the shaft under its own weight or mounted rotor mass.
• \(k\) = Lateral stiffness of the shaft.
• \(m\) = Mass of the system. Let us examine how stiffness \(k\) and deflection \(\delta\) depend on geometric and material properties. For example, consider a simply supported shaft of length (span) \(L\) carrying a central point mass \(M\). The static deflection \(\delta\) is expressed as: \[ \delta = \frac{M g L^3}{48 E I} \] Where:
• \(E\) = Young's modulus of elasticity of the shaft material.
• \(I\) = Area moment of inertia of the shaft cross-section. For a solid circular shaft of diameter \(d\), \(I = \frac{\pi d^4}{64}\).

Step 1: Substituting the variables to see dependencies.
We substitute the expression for the area moment of inertia \(I\) into our deflection equation: \[ \delta = \frac{M g L^3}{48 E \left(\frac{\pi d^4}{64}\right)} = \frac{4 M g L^3}{3 \pi E d^4} \] Now, we substitute this deflection back into the natural frequency critical speed formula: \[ \omega_n = \sqrt{\frac{g}{\delta}} = \sqrt{\frac{g}{\left(\frac{4 M g L^3}{3 \pi E d^4}\right)}} = \sqrt{\frac{3 \pi E d^4}{4 M L^3}} = d^2 \sqrt{\frac{3 \pi E}{4 M L^3}} \]

Step 2: Evaluating geometric factors.
From the resulting relationship, the natural critical speed exhibits explicit functional dependencies: \[ \omega_n \propto d^2 \quad (\text{Square of the shaft diameter}) \] \[ \omega_n \propto \frac{1}{L^{3/2}} \quad (\text{Inverse relationship to the span length to the power } 1.5) \] Thus, the parameters that dictate the inherent critical speed are the diameter of the shaft and the span (length) of the shaft.

Step 3: Clarifying the role of eccentricity.
Eccentricity (\(e\)), which is the distance between the geometric center of the shaft and its mass center, governs the amplitude of the whirling deflection during rotation: \[ y = \frac{e}{\left(\frac{\omega_n}{\omega}\right)^2 - 1} \] While eccentricity directly affects the magnitude of vibrations and stresses, it does not alter the underlying system properties (\(k\) or \(m\)) and therefore does not shift the critical frequency \(\omega_n\) itself. Thus, Option (3) is the correct choice.