MIT - 16.20 Fall, 2002 Unit 23 Vibration of Continuous Systems Paul A. Lagace, Ph.D. Professor of Aeronautics & Astronautics and Engineering Systems Paul A. Lagace ? 2001 MIT - 16.20 Fall, 2002 The logical extension of discrete mass systems is one of an infinite number of masses. In the limit, this is a continuous system . Take the generalized beam-column as a generic representation: 2 d 2 ? EI dw ? dx 2 ?? dx 2 ? ? ? d ?? F dw ? ? = p z (23-1) dx ? dx ? Figure 23.1 Representation of generalized beam-column dF = ? p x dx This considers only static loads. Must add the inertial load(s). Since the concern is in the z-displacement (w): ˙˙ Inertial load unit length = mw (23-2) where: m(x) = mass/unit length Paul A. Lagace ? 2001 Unit 23 - 2 w MIT - 16.20 Fall, 2002 Use per unit length since entire equation is of this form. Thus: 2 d 2 ? EI dw ? dx 2 ?? dx 2 ? ? ? d ?? F dw ? ? = p z ? m ˙˙ dx ? dx ? or: 2 d 2 ? EI dw ? ˙˙ dx 2 ?? dx 2 ? ? ? d ?? F dw ? ? + mw = p z (23-3) dx ? dx ? Beam Bending Equation often, F = 0 and this becomes: 2 d 2 2 ?? EI dw ? + mw = p z dx ? dx 2 ?? ˙˙ --> This is a fourth order differential equation in x --> Need four boundary conditions --> This is a second order differential equation in time --> Need two initial conditions Paul A. Lagace ? 2001 Unit 23 - 3 w it MIT - 16.20 Fall, 2002 Notes : ? Could also get via simple beam equations. Change occurs in: dS = p z ? m ˙˙ dx ? If consider dynamics along x, must include mu ˙˙ in p x term: ( p x ? m u ˙˙ ) Use the same approach as in the discrete spring-mass systems: Free Vibration Again assume harmonic motion. In a continuous system, there are an infinite number of natural frequencies ( eigenvalues ) and associated modes (eigenvectors) so: ω wx (, t ) = w ( x ) e separable solution spatially (x) and temporally (t) Consider the homogeneous case ( p z = 0) and let there be no axial forces (p x = 0 ? F = 0) Paul A. Lagace ? 2001 Unit 23 - 4 it it MIT - 16.20 Fall, 2002 So: 2 d 2 ? EI dw ? + mw = 0 dx 2 ?? dx 2 ?? ˙˙ Also assume that EI does not vary with x: 4 ˙˙ EI dw + mw = 0 (23-5) dx 4 Placing the assumed mode in the governing equation: 4 EI d w e ω ? m ω 2 w e ω = 0 dx 4 This gives: 4 EI d w ? m ω 2 w = 0 (23-6) dx 4 which is now an equation solely in the spatial variable (successful separation of t and x dependencies) _ Must now find a solution for w(x) which satisfies the differential equations and the boundary conditions. Note : the shape and frequency are intimately linked (through equation 23-6) Paul A. Lagace ? 2001 Unit 23 - 5 MIT - 16.20 Fall, 2002 Can recast equation (23-6) to be: 4 d w m ω 2 dx 4 ? w = 0 (23-7) EI The solution to this homogeneous equation is of the form: wx () = e px Putting this into (23-7) yields 4 px m ω 2 pe ? e px = 0 EI 4 m ω 2 ? p = EI So this is an eigenvalue problem (spatially). The four roots are: p = + λ , - λ , + i λ , - i λ where: 14 ? m ω 2 ? / λ = ? ? ? EI ? This yields: Paul A. Lagace ? 2001 Unit 23 - 6 MIT - 16.20 Fall, 2002 λ wx () = A e λ x + B e ? λ x + C e i λ x + D e ? ix or : wx () = C 1 sinh λ x + C 2 cosh λ x + C 3 sin λ x + C 4 cos λ x (23-8) The constants are found by applying the boundary conditions (4 constants ? 4 boundary conditions) Example : Simply-supported beam Figure 23.2 Representation of simply-supported beam E I , m = constant with x 4 2 EI dw + m dw = p z dx 4 dt 2 Paul A. Lagace ? 2001 Unit 23 - 7 MIT - 16.20 Fall, 2002 Boundary conditions: @ x = 0 w = 0 2 @ x = l M = E I dw = 0 dx 2 with: wx () = C 1 sinh λ x + C 2 cosh λ x + C 3 sin λ x + C 4 cos λ x Put the resulting four equations in matrix form w 0 2 () = 0 ? 0 1 0 1 ? ? C 1 ? ? 0 ? dw dx 2 () = 0 ?? 0 1 0 ? 1 ? ? ? ? ? 0 ? C 2 ? ? 0 ? ? ? ? = ? ? () = 0 ? sinh λ l cosh λ l sin λ l cos λ l ? ? C 3 ? ? 0 ? wl dw ? ? sinh λ l cosh λ l ? sin λ l ? cos λ l ? ? ? C 4 ? ? 0 ? 2 ? ? ? ? l dx 2 () = 0 Solution of determinant matrix generally yields values of λ which then yield frequencies and associated modes (as was done for multiple mass systems in a somewhat similar fashion) Paul A. Lagace ? 2001 Unit 23 - 8 MIT - 16.20 Fall, 2002 In this case, the determinant of the matrix yields: C sin λ l = 0 3 Note : Equations (1 & 2) give C 2 = C 4 = 0 Equations (3 & 4) give 2 C 3 sin λ l = 0 ? nontrivial: λ l = n π The nontrivial solution is: λ l = n π (eigenvalue problem!) Recalling that: 14 ? m ω 2 ? / λ = ? ? ? EI ? 4 m ω 2 n π 4 (change n to r to be consistent with ? = EI l 4 previous notation) 22 ? ω r = r π EI ml 4 <-- natural frequency Paul A. Lagace ? 2001 Unit 23 - 9 rx rx MIT - 16.20 Fall, 2002 As before, find associated mode (eigenvector), by putting this back in the governing matrix equation. Here (setting C 3 = 1 …..one “ arbitrary ” magnitude): π wx () = φ r = sin <-- mode shape (normal mode) l for: r = 1, 2, 3,…… ∞ Note : A continuous system has an infinite number of modes So total solution is: π ? 22 EI ml 4 ? wx (, t ) = φ r sin ω r t = sin sin ? r π t ? l ? ? --> Vibration modes and frequencies are: Paul A. Lagace ? 2001 Unit 23 - 1 0 MIT - 16.20 Fall, 2002 Figure 23.3 Representation of vibration modes of simply-supported beam 1st mode ω = π 2 1 4 EI ml 2nd mode ω = 4 π 2 2 4 EI ml 3rd mode ω = 9 π 2 3 4 EI ml etc. Same for other cases Paul A. Lagace ? 2001 Unit 23 - 1 1 MIT - 16.20 Fall, 2002 Continue to see the similarity in results between continuous and multi-mass (degree-of-freedom) systems. Multi-mass systems have predetermined modes since discretization constrains system to deform accordingly. The extension is also valid for … Orthogonality Relations They take basically the same form except now have continuous functions integrated spatially over the regime of interest rather than vectors : l ∫ 0 mx x x () φ r () φ s () dx = M r δ r s (23-9) = 1 for r = s where: δ rs = kronecker delta = 0 for r ≠ s l () φ r 2 () d x M r = ∫ 0 m x x generalized mass of the rth mode Paul A. Lagace ? 2001 Unit 23 - 1 2 MIT - 16.20 Fall, 2002 So: l ∫ 0 m φφ s d x = 0 r ≠ s r l ∫ 0 m φφ r d x = M r r Also can show (similar to multi degree-of-freedom case): 2 l d 2 ? ? EI d dx φ 2 r ?? ? φ s dx = δ rs M r ω r 2 (23-10) ∫ 0 dx 2 ? This again, leads to the ability to transform the equation based on the normal modes to get the … Normal Equations of Motion Let: ∞ wx (, t ) = ∑ φ r ( x ) ξ r () t (23-11) r = 1 normal mode normal coordinates Paul A. Lagace ? 2001 Unit 23 - 1 3 px px MIT - 16.20 Fall, 2002 Place into governing equation: 2 2 d 2 2 ?? EI dw ? + m dw = z () dx ? dx 2 ?? dt 2 l multiply by φ s and integrate ∫ 0 dx to get: ∞ ˙˙ ∞ d 2 ? 2 ∑ ξ r ∫ 0 l m φ r φ s d x + ∑ ξ r ∫ 0 l φ s dx 2 ? ? EI d dx φ 2 r ??? dx = ∫ 0 l φ s f d x r = 1 r = 1 Using orthogonality conditions, this takes on the same forms as before: ˙˙ M ξ + M ω 2 ξ = Ξ (23-12) rr r r r r r = 1, 2, 3,…… ∞ l with: M r = ∫ 0 m φ r 2 d x - Generalized mass of rth mode l Ξ r = ∫ 0 φ r z (, t ) d x - Generalized force of rth mode ξ r (t) = normal coordinates Paul A. Lagace ? 2001 Unit 23 - 1 4 MIT - 16.20 Fall, 2002 Once again ? each equation can be solved independently ? allows continuous system to be treated as a series of “simple ” one degree-of-freedom systems ? superpose solutions to get total response ( Superposition of Normal Modes ) ? often only lowest modes are important ? difference from multi degree-of-freedom system: n --> ∞ --> To find Initial Conditions in normalized coordinates … same as before: (, 0 ) = ∑ φ r ( x ) ξ r () wx 0 r etc. Thus: 1 l 0 ξ r () = M ∫ 0 m φ r w 0 ( x ) d x r (23-13) ˙ r () = 1 ∫ 0 l m φ r w ˙ 0 () d x ξ 0 x M r Paul A. Lagace ? 2001 Unit 23 - 1 5 MIT - 16.20 Fall, 2002 Finally, can add the case of … Forced Vibration Again … response is made up of the natural modes ? Break up force into series of spatial impulses ? Use Duhamel ’ s (convolution) integral to get response for each normalized mode ξ ω τ τ τ r rr r t M t t ( ) = ( ) ? ( ) ∫ 1 0 Ξ sin ω r d (23-14) ? Add up responses (equation 23-11) for all normalized modes (Linear ? Superposition) What about the special case of … --> Sinusoidal Force at point x A Paul A. Lagace ? 2001 Unit 23 - 1 6 xF MIT - 16.20 Fall, 2002 Figure 23.4 Representation of force at point x A on simply-supported beam Ft () = F o sin ? t As for single degree-of-freedom system, for each normal mode get: t A ξ r () = φ r () o 2 sin ? t M r ω r 2 ? ? 1 ? ? 2 ? ? ? ω ? r for steady state response (Again, initial transient of sin ω r t dies out due to damping) Add up all responses … Paul A. Lagace ? 2001 Unit 23 - 1 7 MIT - 16.20 Fall, 2002 Note : ? Resonance can occur at any ω r ? DMF (Dynamic Magnification Factor) associated with each normal mode --> Can apply technique to any system. ? Get governing equation including inertial terms ? Determine Free Vibration Modes and frequencies ? Transform equation to uncoupled single degree-of-freedom system (normal equations) ? Solve each normal equation separately ? Total response equal to sum of individual responses Modal superposition is a very powerful technique! Paul A. Lagace ? 2001 Unit 23 - 1 8