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SolSec 6.4

# SolSec 6.4 - 6 27 Problems and Solutions Section 6.4(6.30...

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6- 27 Problems and Solutions Section 6.4 (6.30 through 6.39) 6.30 Calculate the first three natural frequencies of torsional vibration of a shaft of Figure 6.7 clamped at x = 0, if a disk of inertia J 0 = 10 kg m 2 /rad is attached to the end of the shaft at x = l . Assume that l = 0.5 m, J = 5 m 4 , G = 2.5 × 10 9 Pa, ρ = 2700 kg/m 3 . Solution: The equation of motion is θ ρ θ = G & & . Assume separation of variables: ) ( ) ( t q X φ θ = to get 2 or σ φ φ ρ φ ρ φ = = = q q G q G q & & & & so that ˙˙ q G q + = ′′ + = ρ σ φ σ φ 0 0 2 and where ω ρ σ 2 2 = G . The clamped-inertia boundary condition is θ (0,t) = 0, and GJ ). , ( ) , ( 0 t l J t l θ θ & & = This yields that φ (0) = 0 and GJ l q t J l q t J l G q t = = φ φ φ ρ σ ( ) ( ) ( )˙˙( ) ( ) ( ) 0 0 2 or J l J l = φ σ ρ φ ( ) ( ) 0 2 The solution of the spatial equation is of the form φ σ σ ( ) sin cos x A x B x = + but the clamped boundary condition yields B = 0. The inertia boundary condition yields JA l J A l l J J l l l σ σ σ ρ σ σ ρ σ σ cos sin tan )( . ) = = = 0 2 0 1 5 0 5 m 10kg m (2700kg/m m 4 2 3 So the frequency equation is tan σ σ l l = 675 Using the MATLAB function fsolve ; this has the solutions : σ σ σ σ σ σ 1 2 3 1 2 3 1 5685 4 7054 7 8424 3 1462 9 4388 15 7312 l l l = = = = = = . . . . . . or Thus ω 1 = 2134 rad/s f 1 = 340 Hz ω 2 = 6403 rad/s f 2 = 1019 Hz ω 3 = 10672 rad/s f 3 = 1699 Hz

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