MT-1620 al.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 Unit 23 Vibration of Continuous Systems Paul A. Lagace, Ph.D. Professor of Aeronautics & Astronautics and Engineering Systems Paul A. Lagace © 2001
MT-16.20 al.2002 The logical extension of discrete mass systems is one of an infinite number of masses In the limit this is a continuous syStem T ake the generalized beam-column as a generic representation d dw El F dx 23-1) Figure 23.1 Representation of generalized beam-column dF dx 和x 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 (232) where: m(x =mass/unit length Paul A Lagace @2001 Unit 23-2
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 dx2 dx2 − d F dw = pz (23-1) dx dx Figure 23.1 Representation of generalized beam-column dF = − px 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
MT-16.20 a.2002 Use per unit length since entire equation is of this form. Thus d/d EI dx dx2 dx dx p -mw or ddw El m=p:(233) x Beam Bending Equation often f=0 and this becomes E mw =p dx dx 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 MIT - 16.20 Fall, 2002 Use per unit length since entire equation is of this form. Thus: 2 d 2 EI dw dx2 dx2 − d F dw = pz − m ˙˙ dx dx or: 2 d 2 EI dw ˙˙ dx2 dx2 − d F dw + mw = pz (23-3) dx dx Beam Bending Equation often, F = 0 and this becomes: 2 d 2 2 EI dw + mw = pz dx dx2 ˙˙ --> 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
MT-1620 Fall 2002 Notes Could also get via simple beam equations Change occurs in p- mw If consider dynamics along x, must include mu in px term: (P, -mi 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) w(x, t)=w(x)e separable solution spatially(x) and temporally (t) Consider the homogeneous case(p, =0) and let there be no axial forces (px=0→F=0) Paul A Lagace @2001 Unit 23-4
w i t MIT - 16.20 Fall, 2002 Notes: • Could also get via simple beam equations. Change occurs in: dS = pz − m ˙˙ dx • If consider dynamics along x, must include mu˙˙ in px term: ( px − mu˙˙) 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 w x(, t) = w (x) e separable solution spatially (x) and temporally (t) Consider the homogeneous case (pz = 0) and let there be no axial forces (px = 0 ⇒ F = 0) Paul A. Lagace © 2001 Unit 23 - 4
MT-1620 al.2002 El d x d2/+m=0 Also assume that El does not vary with X E (23-5) dx Placing the assumed mode in the governing equation dw E 100 I 4e mo we This gives a w E m02=0 (23-6) d x 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
it it MIT - 16.20 Fall, 2002 So: 2 d 2 EI dw + mw = 0 dx2 dx2 ˙˙ 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 − mω2 w e ω = 0 dx 4 This gives: 4 EI d w − mω2 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
MT-1620 al.2002 Can recast equation(23-6) to be w ma W (237) El The solution to this homogeneous equation is of the form Putting this into (23-7)yields 0 E m10 El So this is an eigenvalue problem (spatially). The four roots are p=+λ,-λ,+i,-i Where m10 E This yields Paul A. Lagace @2001 Unit 23-6
MIT - 16.20 Fall, 2002 Can recast equation (23-6) to be: 4 d w mω2 dx 4 − w = 0 (23-7) EI The solution to this homogeneous equation is of the form: w x() = e px Putting this into (23-7) yields 4 px mω2 pe − e px = 0 EI 4 mω2 ⇒ p = EI So this is an eigenvalue problem (spatially). The four roots are: p = + λ, -λ, + iλ, - iλ where: 14 mω2 / λ = EI This yields: Paul A. Lagace © 2001 Unit 23 - 6
MT-1620 Fall 2002 w(x)= Ae+ Be Ce/+ De or w(x)=Csinh/x C, coshnx+ C3 sin/x+ C4 cosx(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 2 EI. m= constant with x d w E m Paul A Lagace @2001 Unit 23-7
MIT - 16.20 Fall, 2002 λ w x() = Ae λ x + Be −λ x + Ce i λ x + De −i x or: w x() = C1 sinhλx + C2 coshλx + C3 sinλx + C4 cosλx (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 EI, m = constant with x 4 2 EI dw + m dw = pz dx 4 dt 2 Paul A. Lagace © 2001 Unit 23 - 7
MT-1620 al.2002 Boundary conditions. @X=0W=0 @x=e M=EI n=0 With w(x)=C sinh/x C, cosh/x+ C]sin/x+ C4 cos/x Put the resulting four equations in matrix form 0)=0 d 0 dx ()=0 sinh] cosh sin2 coS2 C30 sinh] cosh]I -sin/I -COS2/ IC4 0 Solution of determinant matrix generally yields values of n 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 Boundary conditions: @ x = 0 w = 0 2 @ x = l M = EI dw = 0 dx2 with: w x() = C1 sinhλx + C2 coshλx + C3 sinλx + C4 cosλx Put the resulting four equations in matrix form w 0 2 () = 0 0 1 0 1 C1 0 dw dx2 () = 0 0 1 0 −1 0 C2 0 = () = 0 sinhλl coshλl sinλl cosλl C3 0 w l dw sinhλl coshλl −sinλl −cosλl C4 2 0 l dx2 () = 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
MT-1620 a.2002 In this case, the determinant of the matrix yields C sin 2= 0 Note: Equations(1& 2)giVe C2=C4=0 Equations( 3&4)give 2C sin/ =0 nontrivial:λC The nontrivial solution is 入C=na (eigenvalue problem!) Recalling that m2) El me (change n to r to be consistent with El previous notation) 2_2 EI r U natural frequency Paul A Lagace @2001 Unit 23-9
MIT - 16.20 Fall, 2002 In this case, the determinant of the matrix yields: C sinλl = 0 3 Note: Equations (1 & 2) give C2 = C4 = 0 Equations (3 & 4) give 2 C3 sinλl = 0 ⇒ nontrivial: λl = nπ The nontrivial solution is: λl = nπ (eigenvalue problem!) Recalling that: 14 mω2 / λ = EI 4 mω2 n π4 ⇒ (change n to r to be consistent with = EI l 4 previous notation) 22 ⇒ ωr = r π EI ml 4 <-- natural frequency Paul A. Lagace © 2001 Unit 23 - 9
MT-1620 al.2002 As before find associated mode(eigenvector), by putting this back in the governing matrIX equation Here(setting C3=1.one""magnitude rJX sin mode shape(normal mode for:r=1,2.3 Note: A continuous system has an infinite number of modes So total solution is FX El W sina.t sin Snr兀 - Vibration modes and frequencies are Paul A Lagace @2001 Unit 23-10
rx rx MIT - 16.20 Fall, 2002 As before, find associated mode (eigenvector), by putting this back in the governing matrix equation. Here (setting C3 = 1…..one “arbitrary” magnitude): π w x() = φr = sin Vibration modes and frequencies are: Paul A. Lagace © 2001 Unit 23 - 10