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

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
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 = G & & . Assume separation of variables: ) ( ) ( t q X φ = to get 2 or σ = = = q q G q G q & & & & so that ˙˙ q G q += ′′ σφ 00 2 and where ω 22 = G . The clamped-inertia boundary condition is θ (0,t) = 0, and GJ ). , ( ) , ( 0 t l J t l & & = This yields that φ (0) = 0 and GJ lqt J lqt J l G qt == φφ () () () ˙˙ () 2 or Jl J l = 0 2 The solution of the spatial equation is of the form φσ ( ) sin cos xA 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 ll σσ cos sin tan )( . ) = 0 2 0 15 05 m 10kg m (2700kg/m m 4 2 3 So the frequency equation is tan l l = 675 Using the MATLAB function fsolve ; this has the solutions : 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
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
6- 28 6.31 Compare the frequencies calculated in the previous problem to the frequencies of the lumped-mass single-degree-of-freedom approximation of the same system. Solution: First calculate the equivalent torsional stiffness of the rod. k GJ l Jk == × =− += = = ( . )( ) . . ˙˙ . . 2 5 10 5 05 25 10 0 10 2 5 10 0 2 5 10 0 9 10 0 0 10 9 θθ θ or so that ω 2 = 2.5 × 10 9 , = 5 × 10 5 rad/s or about 80,000 Hz, far from the 482 Hz of problem 6.30.
Background image of page 2
6- 29 6.32 Calculate the natural frequencies and mode shapes of a shaft in torsion of shear modulus G , length l , polar inertia J, and density ρ that is free at x = 0 and connected to a disk of inertia J 0 at x = l . Solution: Assume zero initial conditions, i.e. θ ( x ,0) = ) 0 , ( x θ & = 0. From equation 6.66 = 2 2 2 2 ρ (,) xt t Gx t x (1) The boundary condition at x = l and at x = 0 is GJ lt x J t t x =− = θθ ( ,) 0 2 2 0 0 Using separation of variable in (1) of form θ ( x,t ) = Θ ( x )T( t ) yields: 2 2 ) ( ) ( 1 ) ( ) ( σ = = Θ Θ ′ t T t T c x x & & (2) where 2 2 and G c = is a separation constant. (2) can now be rewritten as 2 equations ′′ += = = ΘΘ () ˙˙ xx Tt c G
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 02/11/2010 for the course MECHANICAL ms316 taught by Professor Abduljaba during the Spring '09 term at Kalamazoo.

Page1 / 10

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

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online