Note: Please use the actual date you accessed this material in your citation.

Transcription

1 MIT OpenCourseWare Electromagnetic Fields, Forces, and Motion, Spring 005 Please use the following citation format: Markus Zahn, Electromagnetic Fields, Forces, and Motion, Spring 005. (Massachusetts Institute of Technology: MIT OpenCourseWare). (accessed MM DD, YYYY). License: Creative Commons Attribution-Noncommercial-Share Alike. Note: Please use the actual date you accessed this material in your citation. For more information about citing these materials or our Terms of Use, visit:

2 6.641, Electromagnetic Fields, Forces, and Motion Prof. Markus Zahn Lecture 16: Elastic Waves On a Thin Rod I. Governing Equations A δ x x x 1 ρ = T 11 x 1 + T 11 1 x 1 A + 1 t 1 FA x ρ δ = T 11 + F (F is body force density) t x 1 T 11 = E δ x 1 Hooke s Law (stress-strain relation) Young s modulus (modulus of elasticity) δ δ ρ = E + F t x 1 Prof. Markus Zahn Page 1 of 17

3 II. Example Case Study - Change in Rod Length Prof. Markus Zahn Page of 17

4 δ Steady State: = 0, F = 0 t δ = 0 δ = ax + b 1 x 1 Boundary Condition at x 1 = l: A T 11 (x 1 = l) = AE δ x 1 x 1 = l = AEa = f δ (x 1 = 0) = 0 = b δ (x 1 ) = fx AE δ (x 1 = l) = fl AE 11 E = N m (Aluminum) A = m, f = 445 N, l = 0.1m ( 1 ) δ x = l = ( ) ( ) = m = 6.36 µm III. Elastic Waves : F = 0 A. Wave Equation δ = E δ, v p = E v p = E t ρ x ρ ρ δ δ = v t p x δ=δ ( ) +δ x v t (x + vp t ) + p Prof. Markus Zahn Page 3 of 17

5 Superposition: δ=δ + (x vp t ) =δ + (α), α= x vp t δ δ α δ = = v p t α t α v p δ δ = v p α δ = v p t α α t α v p δ δ α δ = = x α x α 1 δ δ δ α δ x = x α = α α x = α 1 δ δ δ δ = v = v = v Q.E.D t p α p x p α For δ= δ (x + v p t ) let v p v p B. Velocity and Stress Variables v = δ, T = E t δ x δ v T ρ = ρ = t t x T δ = E = E t x t v x v T ρ = t t x T v = E x t x T v v = ρ = E x t t x v = v v, v = E ρ t p x p Prof. Markus Zahn Page 4 of 17

6 v 1 T = x t ρ x 1 T 1 T = ρ x E t v 1 T = t x E t T = T vp, t x v p = E ρ v = v+ ( x vp t ) + v ( x + vp t ) T = T+ ( x vp t ) + T ( x + vp t ) C. Method of Characteristics Lagrange Multipliers v v T T λ 1 ρ = 0 t x t x v v T T λ 0 E = 0 t x t t v λ ρλ E v λ T λ 1 T 1 + = 0 t ρλ 1 x ρλ 1 t ρλ x 1 dv dt λ dt ρλ 1 dt v v dv ( x, t ) = dt + dx t x dt (x, t ) = T dt + T dx t x dx λ E λ 1 = = dt ρλ 1 λ λ ρ λ ρ = = ± λ 1 E λ 1 E Prof. Markus Zahn Page 5 of 17

7 dv λ dt 1 dt + ρλ dt = 0 dv ± dt 1 ρe = 0 1 dx dv + dt = 0 on = E ρ = v p ρe dt 1 dx dv dt = 0 on =+ E ρ =+ v p ρ E dt 1 dx v + ρe T = C on dt = v p 1 dx v T = C + on =+v p ρe dt C + + C v = T C C = + ρ E D. Cook-Book Method dv ( x, t ) = v dx + v dt, ρ v = T x t t x ( ) = T dx + T dt, dt x, t T v = E x t t x ρ v t 0 0 E 1 0 v x 0 = dt dx 0 0 T t dv 0 0 dt dx T x dt Prof. Markus Zahn Page 6 of 17

8 E 1 0 det dv dx 0 0 v dt 0 dt dx set each determinant to = t zero for general solution ρ E 1 0 det dt dx dt dx 1 st Characteristic equation ρ E 1 0 det = 0 dt dx dt dx E ρ dx dt E dt dx 0 dt dx E 1 E 1 =ρdx dt dx 0 0 dt = ( dx ) ρ+ E ( dt ) dx E dx = = ± dt ρ dt E ρ = ± v p Prof. Markus Zahn Page 7 of 17

9 nd Characteristic equation E 1 0 E 1 0 det = dv dx 0 dv dx 0 0 dt 0 dt dt 0 dt dx dv 0 dv dx = E dt dt + dt 0 = Edvdt dx dt = 0 dx E Edv dt = 0 Edv dt = 0 dt ρ dt dv = 0 v ρe T ρe = constant T dx v = C + on = + v ρe dt p T dx v + = C on = v p ρ E dt IV. Case Study : Region of Initial Uniform Stress T m t = 0: T (x,t = 0) = 0 x < a x > a Prof. Markus Zahn Page 8 of 17

10 Prof. Markus Zahn Page 9 of 17

11 Prof. Markus Zahn Page 10 of 17

12 VIII. Boundary Conditions Prof. Markus Zahn Page 11 of 17

13 Prof. Markus Zahn Page 1 of 17

14 IX. Sinusoidal Steady State A. Velocity Driven Thin Rod v( x,t ) = Re v x e j t ( ) ω B.C: v ( x = 0) = jv 0 Prof. Markus Zahn Page 13 of 17

15 v ( x = l) = 0 v v d v = v ω v(x) = v t p x dx p dv dx + k v x ( ) = 0, k = ω v = ρ ω p E v ( x ) = A sinkx + B coskx v ( x = 0) = B = jv 0 v ( x = l) = 0 = Asinkl jv 0 coskl A = jv 0 cotkl ( ) = jv coskl v x 0 sinkl sinkx coskx = = jv 0 (coskl sinkx coskx sinkl ) sinkl jv 0 sink(x l) sinkl T E v ω ( ) = j v 0 E k cosk (x l) = j T x t x sinkl T ( x ) = v 0 E k cosk (x l) ω sinkl = v E ρ E 0 cosk (x l) sinkl = v 0 ρe cosk (x l) sinkl ( ) Re ( ) ω sink (x l) v x e j t = v 0 v x,t = sinkl sin ωt Prof. Markus Zahn Page 14 of 17

16 ω v 0 ρe ( ) j t T x,t = Re T ( x) e = cosk ( x l ) cos ωt sinkl ω Resonance: sinkl = 0 kl = nπ k = = v p nπ l X. Electromechanical Coupling ε 0 A c 1 dc 1 v ε 0 A c C =, f x = v = s ds s f has steady part & transient part x 1 V ε A = 0 0 c ( ( l, t)) g + δ 1 ε A V δ=δ x + δ ' x t = 0 c 0 ( ) ( ) 0, ( l, t ) δ g 1 + g 1 ε A V Steady part δ = 0 0 c 0 0 t δ( l, t) g 1 + g 6.641, Electromagnetic Fields, Forces, and Motion Le cture 16 Prof. Markus Zahn Page 15 of 17

17 1 ε A V 0 c 0 g 1 δ g ( l, t) δ 0 = 0 δ x 0 = ax + b δ 0 (x = 0) = b = 0 δ 0 = ax δ 0 1 ε A V 0 c 0 al EA = + g 1 + r x g x= l = EA a r ε + 1 A V l ε A V 1 ε A V 0 c 0 0 c 0 0 c 0 a EA = + a = r 3 g g ε A V l 0 c 0 g EA r 3 g ( ) ( ) ω Transient Part: δ' xt, = Re δ x e j t δ ' E δ' ρ = δ ' ( x ) = A sinkx + A cos kx,k = ω t x 1 v p δ (x = 0) = 0 = A δ ' ( x ) = A 1 sink x B.C. at x = l δ' ε A V = 0 c 0 δ 3 g EA r '( l,t ) x x= l Prof. Markus Zahn Page 16 of 17

18 ε A V EA r kcoskl = + 0 c 0 sinkl 3 g tankl = 3 EA r g k l ε A V l 0 c 0 ( ) EA r g 3 < l k = jk i ε A V l 0 c 0 r tankl = tan ( jk l ) = j tanh ( k l ) = jk l EA g 3 i i i ε A V l 0 c 0 ω = kv i i p Prof. Markus Zahn Page 17 of 17