DERIVATION OF MILES EQUATION FOR AN APPLIED FORCE Revision C By Tom Irvine Email: tomirvine@aol.com August 6, 8 Introduction The obective is to derive a Miles equation which gives the overall response of a single-degreeof-freedom system to an applied force, where the excitation is in the form of a random vibration acceleration power spectral density. Derivation Consider a single-degree-of-freedom system. f(t) m & x& c where m c x f(t) is the mass is the viscous damping coefficient is the stiffness is the displacement of the mass is the applied force Note that the double-dot denotes acceleration.
The free-body diagram is f(t) & x& m x c& x Summation of forces in the vertical direction F m& x () m & x cx& x f (t) () m & x cx& x f (t) () & x (c / m)x& ( / m)x (/ m)f (t) (4) By convention, (c / m) ( / m) ωn ω n where ω n is the natural frequency in (radians/sec), and is the damping ratio. & x ωn x& ωn x (/ m)f (t) (5) Let & x ωn x& ωn x ( ωn / )f (t) (6) f (t) x(t) Fˆ exp(ωt) (7) Xˆ exp(ωt) (8)
ω ωωn ω n Xˆ exp(ωt) ( ωn / )Fˆ exp(ωt) (9) Tae the Fourier transform of each side. ω ωω ω n n X n ( ω) ( ω / ) F( ω) () ( ω / )F( ω) X ( ω) n () ω ω ωω n n (f / )F() f X () f n () f f f n fn ( f ) (/ )F X () f () ( f ) (/ )F X () f (4) Multiply each side by its complex conugate. (/ ) (/ ) (5) () X * () f F() f F* () f X f
(/ ) (/ ) (6) () X * () f F() f F* () f X f The transfer function H( ) times its complex conugate is / / H ( ) H *( ) (7) Solve for the roots R and R of the first denominator. ( ) ± 4 R,R (8) ± 4 4 R,R (9) R,R ± () Solve for the roots R and R4 of the second denominator. ± 4 R,R4 () ± 4 4 R,R4 () R,R4 ± () Summary, (4) R (5) 4
R R R4 (6) (7) (8) Note R -R* (9) R R* () R4 -R* () Now substitute into the denominators. H ( ) H *( ) / () / H ( ) H *( ) () ( R)( R)( R)( R4) / H ( ) H *( ) (4) ( R)( R* )( R* )( R) 5
Expand into partial fractions. ( R)( R* )( R* )( R) ( R) α ( R* ) λ ( R* ) σ ( R) (5) Multiply through by the denominator on the left-hand side of equation (5). α ( R)( R* )( R* )( R) ( R) ( R)( R* )( R* )( R) ( R* ) λ ( R)( R* )( R* )( R) ( R* ) σ ( ) ( R) R R* R* R (6) α λ σ ( R* )( R* )( R) ( R)( R* )( R) ( R)( R* )( R) ( R)( R* )( R* ) (7) 6
α λ σ α λ σ ( R* )( R) ( ( R R* ) RR* )( R) ( ( R R* ) RR* )( R) ( ( R R* ) RR* )( R* ) (8) ( R R* RR* ) ( ( R R R* ) ( RR* R RR* ) R R* ) ( ( R R R* ) ( RR* R RR* ) R R* ) ( ( R* R R* ) ( RR* RR* R* ) RR* ) α λ σ ( R R* RR* ) ( R* R R R* ) ( R* R R R* ) ( R R* RR* ) (9) [ α λ σ] [ Rα R* R* λ Rσ] [ R* α R R λ R* σ] [ RR* α R R* R R* λ RR* σ] (4) (4) 7
[ α λ σ] [ Rα R* R* λ Rσ] [ R* α R R λ R* σ] [ R* α R Rλ R* σ] RR * (4) Equation (4) can be broen up into four separate equations, α λ σ (4) [ R R* R* λ Rσ ] α (44) R * α R R λ R* σ (45) [ R * α R Rλ R* σ] RR* (46) The four equations are assembled into matrix form. Recall R R* R* R R* R R R* R R α R R* λ R* σ / RR* (47) (48) R R* (49) 8
9 R R* (5) σ λ α R* R R R* R* R R R* R R* R* R (5) Multiply the first row by * R and add to the third row. σ λ α R* R R R* R* R R* R R R* R* R (5) Scale the third row. σ λ α R* R R R* R R* R* R (5a) Multiply the third row by - and add to the first row. σ λ α R* R R R* R R* R* R (54b)
Multiply the first row by -R and add to the second row. Also multiply the first row by R* and add to the fourth row. R* R R* R α R λ R* σ (55) Multiply the third row by -R* and add to the second row. Also, multiply the third row by R and add to the fourth row. R* R α R λ R* σ (56) The first row equation yields α σ (57) The third row equation yields λ (58) Equation (56) thus reduces to R* R R λ R* σ (59) R* λ Rσ (6) R * λ Rσ (6) λ R R* σ (6) Rλ R* σ (6)
R R σ R* σ R* σ R R R* σ R* R R R* R* (64) (65) (66) R* σ (67) [ R R* ] λ R R* σ (68) R λ (69) [ R R* ] Recall, α σ (7) λ (7) The complete solution set is thus α λ σ [ R R* R* R ] R R* (7)
Note that 4 R* R (7) σ λ α 4 (74) σ λ α 8 (75) σ λ α 8 (76) Let ψ (77)
α ψ ψ λ 8 ψ σ ψ (78) R (79) ( R)( R* )( R* )( R) ψ 8 ψ 8 ψ 8 ψ 8 (8) The overall displacement is found by integration. [ ( f ) ] n, H( ) H * ( ) x RMS Fˆ PSD (f ) df (8) [ ( f ) ] n, H( ) H * ( ) x RMS f n Fˆ PSD (f ) d (8) Assume that the force PSD is constant Fˆ PSD (f ) A (8)
[ ( f ) ] n, H( ) H *( ) x RMS A f n d (84) { ( fn, ) } x RMS 8 A fn ψ d ψ d ψ d ψ d (85) 8 A f n { ( f n, ) } ( ψ ) x RMS ( ψ ) ( ψ ) ( ψ ) ln ln ln ln (86) Note that 4
y ln [ x y ] ln x y (87) x { ( f n, ) } x RMS 8 A f n ( ψ ) ln ( ψ ) ln ( ψ ) ln ( ψ ) ln (88) Both the real and imaginary components of the natural log terms cancel out at the lower integration limit. The imaginary components of the natural log terms cancel out at the upper limit. The sum of the real components of the natural log terms approaches zero as the upper limit approaches infinity. 5
6 { } ψ ψ ψ ψ A f n 8 f n, x RMS (89)
7 { } ψ ψ ψ ψ A f n 8 f n, x RMS (9)
8 { } ψ ψ ψ ψ A fn 8 fn, x RMS (9)
9 { } ψ π ψ π ψ π ψ A fn 8 fn, x RMS (9) { } π A fn 8 fn, x RMS (9) { } A fn 4 fn, x RMS π (94)
x RMS ( fn, ) πa fn (95) 4 πa x m RMS π (96) 4 The RMS displacement is finally / / 4 / 4 A x RMS 8 m (97) As a review, m A is the mass is the stiffness is viscous damping ratio is the amplitude of the force PSD in dimensions of [force^ / Hz] at the natural frequency
APPENDIX A Velocity Response Recall ( f ) (/ )F X () f (A-4) The corresponding velocity V(f) is ( f ) (π f / )F V () f (A-4) ( f ) (πf / () fn )F V f fn (A-4) ( f ) π F V () f fn (A-4) Multiply each side by its complex conugate. πfn (A-5) () V * () f F() f F* () f V f
πfn (A-6) () V * () f F() f F* () f V f The transfer function H( ) times its complex conugate is πf H ( ) H *( ) n (A-7) Solve for the roots R and R of the first denominator. ( ) ± 4 R,R (A-8) ± 4 4 R,R (A-9) R,R ± (A-) Solve for the roots R and R4 of the second denominator. ± 4 R,R4 (A-) ± 4 4 R,R4 (A-) R,R4 ± (A-) Summary, (A-4)
R (A-5) R R R4 (A-6) (A-7) (A-8) Note R -R* R R* R4 -R* (A-9) (A-) (A-) Now substitute into the denominators. H ( ) H *( ) π fn (A-) π H ( ) H *( ) fn (A-) ( R)( R)( R)( R4)
π H ( ) H *( ) fn (A-4) ( R)( R* )( R* )( R) 4
Expand into partial fractions. ( R)( R* )( R* )( R) ( R) α ( R* ) λ ( R* ) σ ( R) (A-5) Multiply through by the denominator on the left-hand side of equation (A-5). α ( R)( R* )( R* )( R) ( R) ( R)( R* )( R* )( R) ( R* ) λ ( R)( R* )( R* )( R) ( R* ) σ ( ) ( R) R R* R* R (A-6) α λ σ ( R* )( R* )( R) ( R)( R* )( R) ( R)( R* )( R) ( R)( R* )( R* ) (A-7) 5
α R* λ σ ( R) ( R R* ) RR* ( R) ( R R* ) RR* ( R) ( R R* ) RR* ( R* ) λ σ α R R* ( R R R* ) ( R R R* ) RR* RR* R RR* R R* RR* R RR* (A-8) ( R* R R* ) RR* RR* R* RR* R R* (A-9) α R R* RR* R* R R R* λ σ R* R R R* R R* RR* (A-4) 6
[ α λ σ] [ Rα R* R* λ Rσ] R* α R R λ R* σ RR* α R R* R R* λ RR* σ (A-4) [ α λ σ] [ Rα R* R* λ Rσ] R* α R R λ R* σ [ R* α R Rλ R* σ] RR * (A-4) Equation (4) can be broen up into four separate equations, α λ σ (A-4) [ R R* R* λ Rσ] α (A-44) R * α R R λ R* σ (A-45) [ R * α R Rλ R* σ] RR* (A-46) 7
The four equations are assembled into matrix form. Recall R R R* R* R* R R R* R R α R R* λ R* σ (A-47) (A-48) R R* (A-49) R R* (A-5) R R* R* R* R R R* R R α R R* λ R* σ (A-5) Multiply the first row by R * and add to the third row. R R* R* R R* R R* R R* R α R λ R* σ Scale the third row. (A-5) 8
9 σ λ α R* R R R* R R* R* R (A-5a) Multiply the third row by - and add to the first row. σ λ α R* R R R* R R* R* R (A-54b) Multiply the first row by -R and add to the second row. Also multiply the first row by R* and add to the fourth row. σ λ α R* R R R R* R* (A-55) Multiply the third row by -R* and add to the second row. Also, multiply the third row by R and add to the fourth row. σ λ α R* R R R* (A-56) The first row equation yields σ α (A-57)
The third row equation yields λ (A-58) Equation (56) thus reduces to R* R R λ R* σ (A-59) Note that the determinant of the coefficient matrix from Reference is 4 R* 4R 6 (A-94) Apply Cramer s rule. λ 6 det R R* (A-67) λ R* 6 (A-67) λ 6 (A-67) λ 8 (A-67)
λ (A-67) 8 σ 6 R* det R (A-67) σ R 6 (A-67) σ 8 (A-67) σ (A-67) 8 σ (A-67) 8 Recall, α σ (A-7) λ (A-7)
The complete solution set is thus σ λ α 8 (A-74) Let ψ (A-77) ψ ψ ψ ψ σ λ α 8 (A-78) ψ ψ ψ ψ 8 8 8 8 R R* R* R (A-8)
The overall velocity is found by integration. [ ( f ) ] n, H( ) H * ( ) x& RMS Fˆ PSD (f )df (A-8) [ ( f ) ] n, H( ) H * ( ) x& RMS f n Fˆ PSD (f )d (A-8) Assume that the force PSD is constant Fˆ PSD (f ) A (A-8) x& Af RMS n d (A-84) [ ( f n, ) ] H( ) H * ( ) { & ( fn, ) } x RMS 8 πfn A fn ψ d ψ d ψ d ψ d (A-85)
8 πfn A fn { & ( fn, ) } ( ψ ) x RMS ( ψ ) ( ψ ) ( ψ ) ln ln ln ln (A-86) Note that y ln [ x y ] ln x y (A-87) x 4
{ & ( f n, ) } x RMS 8 πf n A f n ( ψ ) ln ( ψ ) ln ( ψ ) ln ( ψ ) ln (A-88) Both the real and imaginary components of the natural log terms cancel out at the lower integration limit. The imaginary components of the natural log terms cancel out at the upper limit. The sum of the real components of the natural log terms approaches zero as the upper limit approaches infinity. 5
6 { } ψ ψ ψ ψ π A f n 8 f n f n, x RMS & (A-89)
7 { } ψ ψ ψ ψ π A f n 8 f n fn, x RMS & (A-9)
8 { } ψ ψ ψ ψ π A fn 8 fn fn, x RMS & (A-9)
9 { } ψ π ψ π ψ π ψ π A fn 8 fn fn, x RMS & (A-9) { } π π A f n 8 f n f n, x RMS & (A-9)
π x& RMS (A-94) 8 πfn A fn { ( fn, ) } π Af x& n RMS (A-94) { ( f n, ) } A / x& RMS (A-94) 8 m / { ( f n, ) } A x& RMS (A-94) 8 / m / { ( fn, ) } The RMS velocity is finally / A x& RMS (A-94) 8 / 4 m / 4 Recall that the RMS displacement is x RMS / 4 / 4 / A m (A-97) 8 x& x RMS RMS / A / 4 / 4 m 8 / 4 / 4 / A m 8 (A-94) 4
x& RMS x RMS / m / (A-94) x& x RMS RMS ωn (A-94) 4
APPENDIX B Acceleration Response ACCELERANCE MAGNITUDE ( ACCELERATION / FORCE ) SDOF SYSTEM: mass g fn Hz Damp.5 ACCELERANCE ( m/sec / N )... EXCITATION FREQUENCY (Hz) Figure B-. An accelerance FRF curve is shown for a sample system in Figure B-. Note that the normalized accelerance converges to as the excitation frequency becomes much larger than the natural frequency. As a result, the acceleration response would be infinitely high for a white noise force excitation which extended up to an infinitely high frequency. Thus, a Miles equation for the acceleration response to a white noise applied force cannot be derived. 4
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