LONGITUDINAL NATURAL FREQUENCIES OF RODS AND RESPONSE TO INITIAL CONDITIONS Revision B

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1 By Tom Irvine ONGITUDINA NATURA FREQUENCIES OF RODS AND RESPONSE TO INITIA CONDITIONS Revision B Marh 4, 009 Consider a thin rod. E, A, m E is the modulus of elastiity. A is the ross-setion area. m is the mass per unit length. The longitudinal displaement u(x, t) is governed by the equation x EA x u u ( ) mx ( ) x t (1) This equation is taken from Referene 1. For a uniform ross-setion, the governing equation simplifies to u u EA( x) mx ( ) () x t Consider a beam with uniform mass density. The governing equation simplifies to u x u (3) E t where is the mass per unit volume. 1

2 et E (4) Note that is the longitudinal wave veloity. Substitute equation (4) into (3). u 1 u x (5) t Separate the variables. et uxt (, ) UxT(t ( ) ) (6) Substitute equation (6) into (5). 1 UxT(t ( ) ) UxT(t ( ) ) x (7) t Perform the partial differentiation. 1 U ( x) T(t) UxT ( ) ( t) (8) Divide through by U(x)T(t). U( x) Ux ( ) 1 T() t T(t) (9) U ( x) Ux ( ) T () t T(t) (10) Eah side of equation (10) must equal a onstant. et be a onstant. U ( x ) T t () (11) Ux ( ) T(t)

3 The time equation is T() t T(t) (1) T() t T(t) (13) T() t T(t) 0 (14) Propose a solution T(t) a sint b os t (15) T() t a ost b sin t (16) T() t a sintb ost (17) Verify the proposed solution. Substitute into equation (14). a sin t b ost sint os t 0 (18) Equation (15) is thus verified as a solution. 0 = 0 (19) There is not a unique, however, in equation (11). This is demonstrated later in the derivation. Thus a subsript n must be added as follows. The spatial equation is Tn() t an sinnt bn os nt (0) U ( x ) (1) Ux ( ) U( x) U( x) () U x ( ) U( x) 0 (3) 3

4 U( x) Ux ( ) 0 (4) Equation (4) is similar to equation (14). Thus, a solution an be found by inspetion. x x Ux ( ) dsin eos (5) The slope equation is x x U ( x) d os e sin (6) Now onsider three boundary ondition ases as shown in the following appendies. REFERENCE 1.. Meirovith, Analytial Methods in Vibrations, Mamillan, New York,

5 APPENDIX A Case I. Fixed-Fixed The left boundary ondition is u( 0, t) 0 (zero displaement) (A-1) U( 0) T( t) 0 (A-) U( 0 ) = 0 (A-3) The right boundary ondition is ut (, ) 0 (zero displaement) (A-4) UT(t ( ) ) 0 (A-5) U ( ) 0 (A-6) Substitute equation (A-3) into (5). e = 0 (A-7) Thus, the displaement equation beomes x Ux ( ) dsin (A-8) Substitute equation (A-6) into (A-8). d sin 0 (A-9) The onstant d must be non-zero for a non-trivial solution. Thus, n n, n 1,, 3,... (A-10) The term is given a subsript n beause there are multiple roots. n n, n 1,, 3,... (A-11) 5

6 The displaement funtion the fixed-fixed rod is nx Un( x) dn sin Un( x) dn sin n x (A-1) (A-13) Substitute the natural frequeny term into the time equation. n t n t Tn() t an sin b n os (A-14) The displaement funtion is thus nx nt nt uxt (, ) dn sin a b n sin n os n 1 (A-15) The oeffiients an be simplified as follows An dn an (A-16) Bn dn bn (A-17) By substitution, the displaement equation is nx nt nt uxt (, ) sin A B n sin n os n 1 (A-18) 6

7 APPENDIX B Case II. Fixed-Free The left boundary onditions is u( 0, t) 0 (zero displaement) (B-1) U( 0) T( t) 0 (B-) U( 0 ) = 0 (B-3) The right boundary ondition is x uxt (, ) x 0 (zero stress) (B-4) U( ) T(t) 0 (B-5) U( ) 0 (B-6) Substitute equation (B-3) into (5). e = 0 (B-7) Thus, the displaement equation beomes x Ux ( ) dsin x U ( x) d os (B-8) (B-9) 7

8 Substitute equation (B-6) into equation (B-9). d os 0 (B-10) The onstant d must be non-zero for a non-trivial solution. Thus, n n 1, n 13,,,... (B-11) The term is given a subsript n beause there are multiple roots. n n 1, n 13,,,... (B-1) The displaement funtion for the fixed-free rod is nx Un( x) dnsin n 1 Un( x) dn sin x (B-13) (B-14) Substitute the natural frequeny term into the time equation. n 1t n 1 t Tn() t an sin b n os (B-15) The displaement funtion is thus n 1x n 1t n 1t uxt (, ) dn sin a b nsin n os n1 (B-16) 8

9 Simplify the oeffiients. n 1x n 1t n 1t uxt (, ) sin A B nsin n os n1 Now determine the effetive mass of the rod for the fundamental mode. The stiffness k at free end of the fixed-free longitudinal rod is (B-17) EA k (B-18) The formula for the fundamental frequeny of a single-degree-of-freedom system is Solve for the mass m. k 1 m (B-19) k m (B-0) 1 Substitute the stiffness term from equation (B-18). EA m (B-1) 1 Add a subsript e to denote that the mass is the effetive mass. m e EA (B-) 1 Calulate the fundamental frequeny from equation (B-1). 1 1 (B-3) 1 1 E 1 (B-4) 9

10 1 E 4 (B-5) Substitute the frequeny term from equation (B-5) into (B-). m e me 4 EA (B-6) E 4A (B-7) et M be the mass of the rod. The effetive mass is 4 me M (B-8) 10

11 APPENDIX C Case III. Free-Free The left boundary onditions is x uxt (, ) x0 0 (zero stress) (C-1) U( 0) T(t) 0 (C-) U( 0) 0 (C-3) The right boundary ondition is x uxt (, ) x 0 (zero stress) (C-4) U( ) T(t) 0 (C-5) U( ) 0 (C-6) Apply equation (C-3) to (5). Thus d 0 x Ux ( ) os (C-7) (C-8) The slope equation is x U ( x) e sin (C-9) Substitute equation (C-6) into (C-9). e sin 0 (C-10) 11

12 The onstant e must be non-zero for a non-trivial solution. Thus, n n, n 1,, 3,... (C-11) The term is given a subsript n beause there are multiple roots. n n, n 13,,,... (C-1) The displaement funtion for the free-free rod is nx Un( x) en os n x Un( x) en os (C-13) (C-14) Substitute the natural frequeny term into the time equation. n t n t Tn() t an sin b n os (C-15) The displaement funtion is thus nx nt nt uxt (, ) en os a b n sin n os n 1 (C-16) Simplify the oeffiients. nx nt nt uxt (, ) os A B n sin n os n 1 (C-17) 1

13 APPENDIX D Fixed-Free Rod Subjeted to Initial Displaement and Initial Veloity Reall u (x, t) sin n x A n sinn B n osn t t (D-1) n 1 et u(x,0) f (x) (D-) u (x,0) g(x) (D-3) f (x) n Bn sin x (D-4) n 1 Premultiply by sin mx and integrate. f (x) sin m x dx B m x n x n sin sin dx (D-5) 0 n 1 0 m 1 m, m 1,,3,... (D-6) 13

14 For m n, For m = n, The integral on the right hand side of (D-5) goes to zero. The steps are omitted for brevity. f (x) sin mx dx mx Bm sin dx (D-7) f (x) sin m x dx m Bm 1 os x dx (D-8) 0 0 mx 1 f (x) sin dx x sin mx 0 Bm (D-9) m 0 1 f (x) sin mx dx m Bm sin (D-10) 0 m m 1 m, m 1,,3,... (D-11) mx 1 f (x) sin dx Bm (D-1) 0 B mx m f (x)sin dx (D-13) 0 14

15 u (x, t) n n sin x An osn t Bn sinn t (D-14) n 1 g (x) n n sin x An (D-15) n 1 g (x) sin m x dx m n m A x x n sin sin dx (D-16) 0 n 1 0 Equation (D-16) is similar to (D-5). A mx m f (x)sin dx (D-17) m 0 Again, m 1 m, m 1,,3,... (D-11) 15