Lecture 8: Mechanical Vibration

1 1 ENE 5400 ( ) ( ) ( ) ( ) , Spring 20041 Lecture 8: Mechanical Vibration Discrete systems Energy method Lumped-parameter analysis 1 (Eigenvalue analysis) Continuous systems Direct solving of partial differential equations Rayleigh s method (the energy approach) Example: a laterally-driven folded-flexure comb-drive resonatorReference: Singiresu S. Rao, Mechanical Vibrations, 2nd Ed., Addison-W esleyPublishing Company, Inc., 1990 ENE 5400 ( ) ( ) ( ) ( ) , Spring 20042 Energy Method Conservation of energy; the maximum kinetic energy is equal to the maximum potential energy: Tmax= Vmax Also known as Rayleigh s energy method Example: Effect of spring mass ms on the resonant frequency n=sdTmydylkKinetic energy of spring length dy:=TTotal kinetic energy:x(t)2 ENE 5400 ( ) ( ) ( ) ( ) , Spring 20043 Cont d The total potential energy: By assuming a harmonic motion x(t) = X cos nt, By equating Tmax= Vmax,221kxU====2max22max21)3(21kXUXmmTns ====++++==== 3/snmmk++++==== ENE 5400 ( ) ( ) ( ) ( ) , Spring 20044 Lumped-Parameter Model Simplified description of 3D physical model using minimum required number of variables (coordinates) Do we have mass-less spring?

2 A valid assumption? Can consist of a set of ordinary differential equations depending on the number of variables In Linear Control Systems , we call them the state-space equationsL-shape springmkkkkx= ?3 ENE 5400 ( ) ( ) ( ) ( ) , Spring 20045 Degree of Freedom The minimum number of independent coordinatesrequired to determine completely the positions of all parts of a system at any instant of time defines the degree of freedom of the systemk3x2m2m1k1x1k2m1k1x1k21 degree of freedom system2 degree of freedom system ENE 5400 ( ) ( ) ( ) ( ) , Spring 20046 Equations of Motion for a 2 Systemk3m2k1x1(t)k2x2(t)F1(t)F2(t)m1b1b2 b3==2211xmxm&&&&4 ENE 5400 ( ) ( ) ( ) ( )

3 , Spring 20047 Equations of Motion for a 2 System ==== ++++ ++++==== ++++ ++++==== ========++++++++2132222132222121][,][,00 ][][][][FFFkkkkkkkbbbbbbbmmmFxkxbxmrr&r& &rIn addition to the free-body diagram, equation of motion can also bederived through the Lagrange s equation from the energy perspective ENE 5400 ( ) ( ) ( ) ( ) , Spring 20048 Solving of Dynamic Equation Gives complete transient response under Free Vibration : without external applied force How can a structure move without a force? Natural frequency and damped natural frequency can be obtained Forced Vibration : with external applied force Motion Types: Underdamped Critical damped Overdamped Remember how to solve a set of linear ordinary differential equations for multiple systems?

4 5 ENE 5400 ( ) ( ) ( ) ( ) , Spring 20049 Determine Resonant Frequency Design of micromechanical devices needs to know natural frequency and damping To many performance indexes of the transient response, such as rise time, overshoot, and settling time Resonant frequencies of a lumped-parameter Mechanical system can be obtained by Solving the eigenvalue problem (exact solution) Rayleigh s Method (approximate solution) etc ENE 5400 ( ) ( ) ( ) ( ) , Spring 200410 Eigenvalue Problem Under free Vibration and no damping, natural frequencies of a system are solutions of the eigenvalue problem The roots i= m i2/k, so ican be solved The eigenvector corresponding to the individual eigenvalue is the mode shape of the system (){00000212122= = = = = = ]]D[km]I[[,]]M[]K[]I[[]K[]]M[]K[[,xx]]M[ ]K[[then,tsinxxLetrr6 ENE 5400 ( ) ( ) ( ) ( ) , Spring 200411 Examplek1m1m2x1x2k2 From the free-body diagram:m3x3k3 ENE 5400 ( ) ( ) ( ) ( ) , Spring 200412 Cont d Let m1= m2= m3= m, k1= k2= k3= k, and = (k/m).}

5 {010001000111012101201000100011101210121 22= = 44443444421 DkmImk][xxxkxxxm][xxxkkkkkkkkkxxxmmm0110 1210121000100010000000003213213213333222 21321321= + = + ++ &&&&&&&&&&&&7 ENE 5400 ( ) ( ) ( ) ( ) , Spring 200413 Cont d = m i2/k, , , , ENE 5400 ( ) ( ) ( ) ( ) , Spring 200414 Cont d: Mode Shapes For each solved i, recall that: We can solve the eigenvector xjiwith respect to each i010001000111012101210001000111012101203 213212= = = iiiiiiiiixxxkxxxm)mk(kx]]M[]K[[rr8 ENE 5400 ( ) ( ) ( ) ( ) , Spring 200415 Cont d: Mode Shapes 1st mode, 1= 2nd mode, 2= 3rd mode, 3= ==== ==== ==== ENE 5400 ( ) ( ) ( ) ( ) , Spring 200416 Vibration of Continuous Systems A system of infinite degrees of freedom The equation of motion may be described by a partial differential equation which can be solved by the method of separation of variables Many methods can be used to find approximate resonant frequencies and mode shapes ( the Rayleigh s method)9 ENE 5400 ( ) ( ) ( ) ( ) , Spring 200417 Example: Lateral Vibration of Beams What is the dynamic equation?}

6 The inertia force ( f= ma):xyy(x,t)xf(x,t): force per unit lengthM(x,t)M(x,t) + dM(x,t)V(x,t)V(x,t) + dV(x,t)dxy(x,t)OO LFree-bodydiagramf(x,t) ENE 5400 ( ) ( ) ( ) ( ) , Spring 200418 Example: the Lateral Vibration of Beams The sum of moments around the point Ois ZERO Substitute V= M/ xinto the last equation:)t,x(ft)t,x(yAx)t,x(yEIt)t,x(y) x(A)t,x(f][xt)t,x(y)x(A)t,x(fx)t,x(M= + =+ =+ 224422222222M(x,t)M(x,t) + dM(x,t)V(x,t)V(x,t) + dV(x,t)dxy(x,t)OO For a uniform beam:10 ENE 5400 ( ) ( ) ( ) ( ) , Spring 200419 Example: Lateral Vibration of Beams For free Vibration , f(x,t) = 0, we require Two initial conditions, for example: y(x, t = 0) = yo(x) = 0 y/ t|(x, t = 0)= 0 Four boundary conditions, for example.

7 Free end Bending moment = EI( 2y/ x2) = 0 Shear force = EI 3y/ x3= 0 Simply supported (pinned) end Deflection y = 0 Bending moment = EI( 2y/ x2) = 0 Clamped end Deflection y = 0 Slope y/ x= 00),(),(2244==== ++++ ttxyAxtxyEI We will use these s to solve for theFixed-pinned beam ENE 5400 ( ) ( ) ( ) ( ) , Spring 200420 Solve Lateral Vibration of Beams Use the method of separation of variables y(x,t) = Y(x) T(t)222442244)()(1)()(/0)()()()( ======== ==== ====++++adttTdtTxxYxYAEIdttTdxAYdxxYdtEI T(2)(3)11 ENE 5400 ( ) ( ) ( ) ( ) , Spring 200421 Solve the Lateral Vibration of Beams Y(x)can be solved as:xixixxeCeCeCeCxY ++++++++++++====4321)(xCxCxCxCxY sinhcoshsincos)(4321++++++++++++====422) (AlEIlAEI ========Or,The natural frequencies of the beam are (from (1)):The lproduct depends on the boundary conditions ENE 5400 ( ) ( ) ( ) ( ) , Spring 200422 Solve Lateral Vibration of a Fixed-PinnedBeam Four s for a fixed-pinned beam are substituted into Y(x).

8 0)(0)(0)0(0)0(022================43421ld xYdEIlYdxdYY()4002124421331)xsinhx(sinC) xcoshx(cosC)x(YCC)CC(CCCC + = = =+ = =+()()()7060502121= + + = + + = + )lsinhl(sin)lcoshl(coslsinhlsinlcoshlcos )lsinhl(sinC)lcoshl(cosC)lsinhl(sinC)lco shl(cosCSo,12 ENE 5400 ( ) ( ) ( ) ( ) , Spring 200423 Cont d From the last matrix, we get the determinant: The many roots of this equation, nl, will define the natural frequencies: Mode shape: Yn(x), Y(x), yn(x,y), and y(x,t):ltanhltan = 42 AlEI)l(nn = ()()shapeemodfinalThe),t,x(y)t,x(y)tsinB tcosA)(x(Y)t,x(yfrom)],xsinhx)(sinlsinhl sinlcoshlcos()xcoshx[(cosC)x(Yfrom),lsin hlsinlcoshlcos(CCnnnnnnnnnnnnnnnnnnnnnnn n == + = = =111245 ENE 5400 ( ) ( ) ( ) ( ) , Spring 200424 Results of nlfor Various Beam Constraints(1) Cantilever beam(2) Doubly-clamped beam(3) fixed-pinned beam 1l= 2l= 3l = 4l = 1l= 2l= 3l = 4l = 1l= 2l= 3l = 4l = ENE 5400 ( ) ( ) ( ) ( ) , Spring 200425 Rayleigh s Method An approximate analysis using the energy perspective to find thefundamental natural frequency of continuous systems The kinetic energy of a beam.

9 Assume a harmonic variation y(x,t) = Y(x) cos( t), the maximum kinetic energy:=TdxxAxYTl)()(2022max ==== ENE 5400 ( ) ( ) ( ) ( ) , Spring 200426 Cont d The potential energy V of a beam: (neglecting the work done by the shear forces) The maximum value of y(x,t) is Y(x), so the maximum potential energy: =lmaxdx)dx)x(Yd(EIV02222114 ENE 5400 ( ) ( ) ( ) ( ) , Spring 200427 Rayleigh s Method By equating Tmaxto Vmax, we obtain: For example, a stepped beam with various cross sections: Where is Y(x) from? You have to choose Y(x), and make sure: (1) it is a reasonable beam deflection curve; (2) Y(x) must satisfy the beam boundary conditionsLL++++++++++++++++==== 211121)()())(())((22021022222222112lllll ldxxYAdxxYAdxdxxYdIEdxdxxYdIE ====lldxxAYdxdxxYdEI0202222)())(( ENE 5400 ( ) ( ) ( ) ( ) , Spring 200428 Example: Find the Resonant Frequency Use the deflection curve Y(x) = (1 x/l)2 The cross section A(x) = hx/l The moment of inertia I(x) = 1 (hx/l)3/12 By equating Tmaxto Vmax The exact frequency is (for comparison):= 2x1lhy2/142)( ====anchored15 ENE 5400 ( ) ( ) ( ) ( ) , Spring 200429 Lateral Folded-flexure Comb-Drive Resonator What is the resonant frequency of the resonator?

10 A lumped-parameter model would be used for analysisSource: W illiam Tang, Dissertation, UC Berkeley, 1990 ENE 5400 ( ) ( ) ( ) ( ) , Spring 200430 Spring Constant kx When the resonant plate moves Xounder a given force Fo, the point B and D moves Xo/2, respectively The force acting on each beam is Fo/4 The slope at both ends of the beams are identically zeroplateanchortrussbeam16 ENE 5400 ( ) ( ) ( ) ( ) , Spring 200431 Cont d The deflection curve of beam AB is: ENE 5400 ( ) ( ) ( ) ( ) , Spring 200432 Lateral Resonant Frequency By Rayleigh s energy method: For the beam segment AB, remember that:=++=++== (((())))(((())))(((())))(((())))(((()))) ==== ======== ====32332232482/023124/LyLyXyxEILFXLxLyf oryLyEIFyxoABzxoABzxAB17 ENE 5400 ( ) ( ) ( ) ( ) , Spring 200433 Cont d So the velocity profile for segment AB (multiply ) is: The for beam AB is:() =32232 LyLyXyvoAB() = = = = 44443444421