Equivalent Mass of a Fixed-End Beam Figure 4.3.6a shows a motor mounted on a beam with two fixed-end supports. An imbalance in the motor’s rotating mass will produce a vertical force f that oscillates at the same frequency as the motor’s rotational speed. The resulting beam motion can be excessive

Question

Equivalent Mass of a Fixed-End Beam
Figure 4.3.6a shows a motor mounted on a beam with two fixed-end supports. An imbalance in the motor’s rotating mass will produce a vertical force f that oscillates at the same frequency as the motor’s rotational speed. The resulting beam motion can be excessive if the frequency is near the natural frequency of the beam, as we will see in a later chapter, and excessive beam motion can eventually cause beam failure. Determine the natural frequency of the beam-motor system.

Accepted Answer

Treating this system as if it were a single mass located at the middle of the beam results in the equivalent system is shown in Figure 4.3.6b, where x is the displacement of the motor from its equilibrium position. The equivalent mass of the system is the motor mass (treated as a concentrated mass [latex]m_{c}[/latex]) plus the equivalent mass of the beam. From Table 4.3.1, the beam’s equivalent mass is [latex]0.38 m_{d}[/latex] . Thus the system’s equivalent mass is [latex]m_{e} = m_{c} + 0.38 m_{d}[/latex] .

Table 4.3.1 Equivalent masses and inertias of common elements.

Translational systems
Nomenclature:	Equivalent system
[latex]m_{c}[/latex] = concentrated mass [latex]m_{d}[/latex] = distributed mass [latex]m_{e}[/latex] = equivalent lumped mass System model: [latex]m_{e} \ddot{x} + kx = 0 [/latex]
Helical spring, or rod in tension/compression	Cantilever beam

[latex]m_{e} = m_{c} + m_{d} /3 [/latex]	[latex]m_{e} = m_{c} + 0.23 m_{d}[/latex]
Simply supported beam	Fixed-end beam

[latex]m_{e} = m_{c} +0.50 m_{d}[/latex]	[latex]m_{e} = m_{c} +0.38 m_{d}[/latex]
Rotational systems
Nomenclature:	Equivalent system
[latex]I_{c}[/latex] = concentrated inertia [latex]I_{d}[/latex] = distributed inertia [latex]I_{e}[/latex] = equivalent lumped inertia System model: [latex]I_{e} \ddot{θ} + kθ = 0[/latex]
Helical spring	Rod in torsion

[latex]I_{e} = I_{c} + I_{d} /3[/latex]	[latex]I_{e} = I_{c} + I_{d} /3[/latex]

The equivalent spring constant of the beam is found from Table 4.1.1. It is

Table 4.1.1 Spring constants of common elements.

Coil spring
	[latex]k = \frac{G d^{4}}{64n R^{3}}[/latex] d = wire diameter n = number of coils
Solid rod
	[latex] k= \frac{E A}{L}[/latex]
Simply supported beam
	[latex]k = \frac{4Ewh^{3}}{L^{3}} [/latex]
Cantilever beam
	[latex]k = \frac{Ewh^{3}}{4L^{3}} [/latex] w = beam width h = beam thickness
Fixed-end beam
	[latex]k = \frac{16Ewh^{3}}{L^{3}} [/latex]
Parabolic leaf spring
	[latex]k = \frac{2Ewh^{3}}{L^{3}} [/latex]

[latex]k = \frac{16 E w h^{3}}{L^{3}} [/latex]
where h is the beam’s thickness (height) and w is its width (into the page). Thus the system model is [latex]m_{e} \ddot{x} + kx = f[/latex] where x is the vertical displacement of the beam end from its equilibrium position. The natural frequency is [latex]ω_{n} = \sqrt{k/m_{e}} [/latex], which gives
[latex]ω_{n} = \sqrt{\frac{k}{m_{e}}} = \sqrt{\frac{16 E w h^{3}/L^{3}}{m_{c} + 0.38 m_{d}}} [/latex]

Question 4.3.5: Equivalent Mass of a Fixed-End Beam Figure 4.3.6a shows a mo...

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