notes.pdf - Engi6933 Mechanical Vibrations 1 1 Introduction...

This preview shows page 1 out of 61 pages.

You've reached the end of this preview.

Unformatted text preview: Engi6933 - Mechanical Vibrations 1 1 Introduction 1.1 Definitions • Vibration – A motion that repeats itself after a time interval or oscillation (e.g. pendulum, plucked guitar string). • Vibrating system – Consists of a mass, spring and possibly a damper. – Mass (or inertia element for rotating systems) is a rigid body that stores kinetic energy – Spring stores potential energy – Damper dissipates energy 1.2 Spring Elements • A spring element is generally assumed to have no mass and no damping. It will store potential energy due to stretch (or compression) or twist (for torsional springs). • For a linear spring the spring force is proportional to the change in length of the spring element: F = kx (1) where, x is the measured deformation of the spring from its undeformed length, and k is the spring constant. Engi6933 - Mechanical Vibrations 2 • The work done deforming a spring from x1 to x2 is: Z x2 Z x2 kx dx = F dx = U12 = x1 x1 k 2 (x − x21 ) 2 2 (2) and this work is stored as potential energy. • The spring constant of a helical spring is: Gd4 (3) 8nD3 wehere G is the shear modulus of the spring material, d is the diameter of the wire, D is the mean coil diameter, and n is the number of active turns of the coil. k= • A rod can act as a spring (or consider an elevator cable when the winding drum suddenly stops). Consider a rod being stretched by a force F : • For linear elastic behaviour Hooke’s Law applies: σ =E  Since σ = F/A and  = ∆x/L, then: FL =E A∆x or AE ∆x = k∆x L i.e. the linear spring constant for a rod (or cable in tension) is: F = k= AE L (4) Engi6933 - Mechanical Vibrations 1.3 3 Damping Elements • Damping is the mechanism by which vibrational energy is converted into heat or sound. • A damping element is assumed to have zero mass and elasticity. • There are three standard models of damping: (1) viscous damping; (2) Coulomb or dry friction damping; and (3) hysteretic damping. • Viscous Damping: – Fluid dynamic drag is used to dissipate energy (e.g. air drag on a pendulum, drag as a liquid is forced through an orifice (shock absorber)). – Viscous damping is the most common form of damping used in vibrating systems. – The damping force is proportional to the rate of change of length of the damping element or the relative velocity between the two ends of the damping element. F = cx˙ (5) – Note: Kinematics of a rigid body is used to give the position, velocity and acceleration of various components of a suspension system. The distance between the mounting points of the spring is used to give the spring force. The rate of change of the distance between the mounting points of the damper is used to give the damping force. If the spring and damper are coaxial, only one length is of interest. • Coulomb Damping: – Coulomb damping is constant in magnitude but opposite in direction to the motion. – Due to the friction between rubbing surfaces (dry or with insufficient lubrication). F = µk N (6) • Hysteretic, Solid or Material Damping: – As a material is deformed, the work done on the material is stored as strain energy or heat. – Since energy is absorbed, a body subjected to material damping shows a hysteresis loop on a stress-strain diagram. Engi6933 - Mechanical Vibrations 1.3.1 4 A Simple Viscous Damper • Consider two parallel plates separated by a distance h. The bottom plate is fixed and the top plate moves to the right at a constant speed v. The space between the plates is occupied by a fluid with dynamic viscosity µ. • Solution of the Navier-Stokes equations shows that the velocity profile is linear, i.e. Couette flow. • The shear stress in the fluid is: τ =µ ∂u v =µ ∂y h • The drag (i.e. frictional) force exerted by the fluid on the bottom plate is: F = Aτ |y=0 = µA v h where c = µA/h is the damping constant and the damping force behaves linearly wrt v. Engi6933 - Mechanical Vibrations 2 5 Free Vibration of a One Degree of Freedom System 2.1 Undamped Motion of a Single DOF System • Consider a mass, m, resting on a flat frictionless surface and connected to a rigid support by a linear spring, k: 2.1.1 Equation of Motion • We will need an equation governing the motion of the mass as it is moved from its equilibrium position. Such an equation of motion can be derived from Newton’s 2nd Law using the following steps: 1. Sketch the system and define a suitable positive co-ordinate direction to define the motion of the mass. 2. Determine the static equilibrium position of the system. It is common (and recommended) to define the positive co-ordinate from the equilibrium position. 3. Draw a free body diagram (FBD) of the mass, assuming positive displacement, velocity and acceleration are given to it. Draw the forces acting on the mass. 4. Use Newton’s 2nd Law to derive an equation of motion. +→ X Fx = max m¨ x = −kx or m¨ x + kx = 0 (7) Engi6933 - Mechanical Vibrations 6 • Equation (7) is the equation governing the free vibration of the mass. • What about a mass, m, suspended from a rigid support by a linear spring, k? +↓ X Fx = max m¨ x = −kx + (W − kδst ) But, from statics: W = kδst so the governing equation is: m¨ x + kx = 0 • Note: The same governing equation arises when x is defined from the equilibrium position (this is the advantage of defining x measured from equilibrium). 2.2 Solution of the Governing Equation • The governing equation for undamped free vibration of a 1 DOF mass-spring system is an homogeneous ordinary differential equation (ODE): m¨ x + kx = 0 (8) • Assume a solution of the form: x(t) = Cest then: x(t) ˙ = Csest x ¨(t) = Cs2 est • Substitution of the assumed solution into Eq. (8) gives: mCs2 est + kCest = 0 or ms2 + k = 0 which is called the characteristic equation. (9) Engi6933 - Mechanical Vibrations 7 • The solution of Eq. (9) is: k s=± − m  1/2 = ±iωn (10) where k 1/2 (11) m is the natural frequency of the undamped system. It is the frequency of oscillation of the undamped system.   ωn = • The two values of s are the roots or eigenvalues of the characteristic equation. Since both values of s will satisfy the governing equation, the solution for x(t) is written as: x(t) = C1 eiωn t + C2 e−iωn t (12) where C1 and C2 are constants to be determined from initial conditions. It is difficult to interpret the motion given by Eq. (12), so let’s make it look pretty. • Euler’s formula states: e±iωn t = cos ωn t ± i sin ωn t • Substitution of Euler’s formula into Eq. (12) gives: x(t) = C1 [cos ωn t + i sin ωn t] + C2 [cos ωn t − i sin ωn t] or x(t) = (C1 + C2 ) cos ωn t + i(C1 − C2 ) sin ωn t • Since x(t) is a displacement, it must be real, therefore, (C1 +C2 ) must be real and (C1 −C2 ) must be imaginary. For this to be true C1 and C2 must be complex conjugates: C1 = a + ib ; C2 = a − ib so: x(t) = 2a cos ωn t − 2b sin ωn t • Then the displacement x(t) can be written as: x(t) = A1 cos ωn t + A2 sin ωn t (13) where A1 and A2 are new constants evaluated from the initial conditions of the system. Two conditions are required (i.e. the same number as the order of the governing equation). • Specifying the initial displacement and velocity at t = 0: x(t = 0) = xo x(t ˙ = 0) = x˙ o and solving for A1 and A2 : A1 = x o x(t) ˙ = −A1 ωn sin ωn t + A2 ωn cos ωn t x˙ o = ωn A2 x˙ o A2 = ωn Engi6933 - Mechanical Vibrations 8 • So the solution to Eq. (8) is: x(t) = xo cos ωn t + x˙ o sin ωn t ωn (14) i.e. harmonic motion. Equation (14) is the motion of any system whose governing equation can be written in the form of Eq. (8). • Note: 1. If xo = 0: x(t) = x˙ o sin ωn t ωn 2. If x˙ o = 0: x(t) = xo cos ωn t • Equation (13) can be written in a more convenient form by defining: A1 = A cos φ (15) A2 = A sin φ (16) • Square and add Eqs. (15) and (16) to find A: A=  A21 + A22 1/2 x2o =  + x˙ o ωn 2 !1/2 • Divide Eq. (16) by Eq. (15) to find φ: φ = tan−1  A2 A1  = tan−1  x˙ o x o ωn  • Substitution of Eqs. (15) and (16) into Eq. (13) gives: x(t) = A cos φ cos ωn t + A sin φ sin ωn t • Using the trigonometric identity: cos(x − y) = cos x cos y + sin x sin y • Then the solution for x(t) can be written in the following convenient form: x(t) = A cos(ωn t − φ) (17) where the amplitude A and phase angle φ are:  !1/2 x˙ o 2 A = + ωn   x˙ o −1 φ = tan xo ωn x2o  (18) (19) Engi6933 - Mechanical Vibrations 9 • The motion can be plotted against time, t(s), or frequency, ωn t(rad). • Note: – The vertical displacement at t = ωn t = 0 is xo , the initial displacement of the system from equilibrium. – The slope of the x(t) curve at t = ωn t = 0 is x˙ o , the initial velocity of the system. – The amplitude, A of the motion is the maximum displacement from equilibrium. – The phase angle, φ, can be measured as the horizontal from t = ωn t = 0 to the first peak. The ωn t plot gives φ directly in radians, while the t plot gives φ in seconds, which will have to be muliplied by ωn to get the angle in radians. Engi6933 - Mechanical Vibrations 10 • The velocity and acceleration of mass m are: π ) 2 (20) x ¨(t) = −ωn2 A cos(ωn t − φ) = ωn2 A cos(ωn t − φ + π) (21) x(t) ˙ = −ωn A sin(ωn t − φ) = ωn A cos(ωn t − φ + i.e. x(t) ˙ leads x(t) by π/2 (or 90o ) and x ¨(t) leads x(t) by π (or 180o ). • For a system with ωn = π radians, i.e. τ = 2 s, with initial conditions xo = 0.02 m and x˙ o = 0 we get the following motion. Note: – Maximum velocity occurs as the system reaches the equilibrium position. – Maximum acceleration occurs when the sytem reaches maximum displacement from equilibrium. Here the system is stopping an changing direction. • The motion x(t) can also be expressed as a sine wave by defining A1 and A2 in Eq. (13) as: A1 = A sin φo A2 = A cos φo Engi6933 - Mechanical Vibrations 11 Then: A=  A21 A22 + −1 φo = tan 1/2  A1 A2 x2o =   + −1 x˙ o ωn  = tan 2 !1/2 xo ωn x˙ o  • Substitution of the definitions of A1 and A2 into Eq. (13) gives: x(t) = A sin φo cos ωn t + A cos φo sin ωn t • Using the trigonometric identity: sin(x + y) = sinx cos y + cos x sin y • Then x(t) = A sin(ωn t + φo ) (22) with the same magnitude as Eq. (17), but a new phase angle, φo : φo = tan−1 2.3  xo ωn x˙ o  (23) Free Vibration with Viscous Damping • Consider a mass, m, resting on a flat frictionless surface and connected to a rigid support by a linear spring, k, and a viscous damper, c: • We will need an equation governing the motion of the mass as it is moved from its equilibrium position. Such an equation of motion can be derived using Newton’s 2nd Law. Engi6933 - Mechanical Vibrations 2.3.1 12 Equation of Motion • The FBD for a 1 DOF mass-spring-damper system is shown below: • To derive the equation of motion assume a positive displacement, velocity and acceleration of the mass m. • Newton’s 2nd Law gives: +→ X Fx = m¨ x = −Fs − Fd = −kx − cx˙ where, for a linear spring Fs = kx, and a viscous damper Fd = cx. ˙ • The governing equation can be rearranged to give: m¨ x + cx˙ + kx = 0 (24) This equation governs the motion of all 1 DOF systems in translation with a linear spring and viscous damper. 2.3.2 Solution of the Governing Equation • The governing equation for viscously damped vibration of a 1 DOF system is an homogeneous ODE: m¨ x + cx˙ + kx = 0 • As for undamped motion, assume a solution of the form: x(t) = Cest • Substitution of the assumed solution into the governing equation, Eq. (24), gives the following characteristic equation: mCs2 est + cCsest + kCest = 0 or ms2 + cs + k = 0 i.e. a quadratic equation in s. The roots of the characteristic equation are: s √  −c ± c2 − 4mk c c 2 k =− ± − s1,2 = 2m 2m 2m m (25) (26) Engi6933 - Mechanical Vibrations 13 • So the solution to Eq. (24) is: x(t) = C1 es1 t + C2 es2 t or  x(t) = C1 e q c − 2m + ( c 2 k −m 2m   ) c − 2m − t + C2 e q ( c 2 k −m 2m )  t (27) where, C1 and C2 are constants obtained from initial conditions. This form of the solution is hard to interpret so other forms are used. • Define: 1. Critical Damping, cc , as the damping constant c which makes the radical in Eq. (26) zero:   cc 2 k − =0 2m m or s √ k = 2 km = 2mωn (28) cc = 2m m 2. Damping ratio, ζ, as the ratio of the damping constant c to the critical damping constant cc : ζ = c/cc (29) • Using Eqs. (28) and (29): c c cc = · = ζωn 2m cc 2m (30) • Using Eqs. (30) and the definition of ωn , Eq. (11), the roots s1 and s2 , Eq. (26), can be written in terms of ζ and ωn :  s1,2 = −ζ ± q ζ2  − 1 ωn (31) and solution of Eq. (24) can be rewritten in the following form:  x(t) = C1 e √ −ζ+  ζ 2 −1 ωn t  + C2 e −ζ− √  ζ 2 −1 ωn t (32) • Note: if ζ = 0, the solution reduces to that for undamped free vibration, Eq. (12). • It should be obvious that due to the ζ 2 − 1 term in the radical different types of solution will arise if ζ < 1, ζ = 1 and ζ > 1 . These three cases correspond to underdamped, critically damped, and overdamped cases, respectively. The solutions for these different cases will now be considered separately. Engi6933 - Mechanical Vibrations 2.3.3 14 Case 1: Underdamped Motion (ζ < 1, c < cc ) • If ζ < 1, then ζ 2 − 1 < 0, and written as: p p ζ 2 − 1 = i 1 − ζ 2 , therefore, the roots s1 and s2 can be  q −ζ ± i 1 − s1,2 = ζ2  ωn = −ζωn ± iωd (33) where ωd is the damped frequency of oscillation: ωd = q 1 − ζ 2 ωn (34) • The solution to Eq. (24) can be written in the following form: x(t) = C1 e(−ζωn +iωd )t + C2 e(−ζωn −iωd )t or h x(t) = e−ζωn t C1 eiωd t + C2 e−iωd t i (35) • Compare Eq. (35) with Eq. (12) for undamped motion. The solution will be harmonic with an exponentially decaying amplitude. The frequency of oscillation will be ωd = p 1 − ζ 2 ωn , i.e. less than ωn for undamped motion. Due to the similarities of Eqs. (12) and (35) the same techniques may be used to rewrite Eq. (35) in more easily interpreted forms. x(t) = e−ζωn t [(C1 + C2 ) cos ωd t + i(C1 − C2 ) sin ωd t] or x(t) = e−ζωn t C10 cos ωd t + C20 sin ωd t   (36) Then: x(t) = Xe−ζωn t sin (ωd t + φo ) (37) x(t) = Xe−ζωn t cos (ωd t − φ) (38) or Where the final two forms of the solution clearly show that the motion is harmonic with frequency ωd and exponential decay in the amplitude. • The constants C10 and C20 can be derived from the initial conditions: x(t = 0) = xo x(t ˙ = 0) = x˙ o Solving for C10 and C20 : C10 = xo (39)     x(t) ˙ = −ζωn e−ζωn t C10 cos ωd t + C20 sin ωd t + e−ζωn t −ωd C10 sin ωd t + ωd C20 cos ωd t x˙ o = −ζωn C10 + ωd C20 C20 = x˙ o + ζωn xo ωd (40) Engi6933 - Mechanical Vibrations 15 • The amplitude X and phase angles φ and φo are defined as follows: X = q (C10 )2 + (C20 )2 φo = tan−1 (C10 /C20 ) φ = tan −1 (C20 /C10 ) (41) (42) (43) The amplitude X is the maximum amplitude that would exist at t = 0, for zero phase angle. • A typical underdamped system response plotted versus ωd t is shown below. The motion is harmonic, with exponential decay of amplitude and damped frequency of oscillation, ωd , which is always less than ωn since ζ < 1. • Vehicle suspension systems are examples of underdamped systems. Engi6933 - Mechanical Vibrations 16 • Consider two cases of a system with ωn = 4π = 12.57 rad/s (τ = 0.5 s), with initial conditions xo = 0.05 m and x˙ o = 1 m/s: 1. Case 1: ζ = 0.2 → ωd = 12.31 rad/s, τd = 0.510 s, X = 0.104 m and φ = 1.07 rad. 2. Case 2: ζ = 0.4 → ωd = 11.52 rad/s, τd = 0.546 s, X = 0.120 m and φ = 1.14 rad. • Note: – Neither case reaches X displacement due to the phase angle. Case 2 has a larger X, but it reaches a lower peak displacement due to the quicker delay cause by ζ in the exponential term. – Both overshoot the equilibrium position and oscillate about it. Case 2 more quickly stabilizes at the equilibrium. – Case 2 is an example of the behaviour of modern sports car, and Case 1 is the behaviour of a land yacht from the 1960’s. A vehicle with the behaviour of Case 1 would respond more easily to bumps (have a smoother ride) but would not feel like it was stable when driven fast. The motions here are exaggerated but Case 1 is a Toyota Corolla and Case 2 is a VW Golf R...different suspension tuning for different markets. Engi6933 - Mechanical Vibrations 2.3.4 17 Case 2: Critically Damped Systems (ζ = 1, c = cc ) • If ζ = 1 the roots s1 and s2 are not distinct: cc s1 = s2 = − = −ζωn = −ωn 2m (44) • To determine the motion take the limit of Eq. (36) as ζ → 0. lim x(t) = lim e−ζωn t C10 cos ωd t + C20 sin ωd t  ζ→1 ζ→1 −ωn t  = e  (C10 + C20 ωd t Remember: ωd =  q 1 − ζ 2 ωn lim (cos ψ) = 1 ψ→0 lim (sin ψ) = ψ ψ→0 • So the solution can be written as: x(t) = (C10 + C20 ωd t)e−ωn t • Using Eqs. (39) and (40), the motion of a critically damped system is given by: x(t) = (xo + (x˙ o + ωn xo ) t) e−ωn t (45) The motion is not harmonic, it is aperiodic and decreases to zero as t increases due to the e−ωn t term. • Critical damping, cc , is the lowest value of c that will not produce oscillations. • Consider two cases of a system with ωn = 4π = 12.57 rad/s (τ = 0.5 s), with initial conditions xo = 0.05 m and x˙ o = 1 m/s. Engi6933 - Mechanical Vibrations 2.3.5 18 Case 3: Overdamped System (ζ > 1, c > cc ) • For this case, the roots s1 and s2 are both real, distinct and less than 1.   q  (46) ζ 2 − 1 ωn −ζ − s2 = − 1 ωn ζ2 −ζ + s1 =  q (47) and s2 < s1 . • The solution is:  x(t) = C1 e √ −ζ+  ζ 2 −1 ωn t  + C2 e −ζ− √  ζ 2 −1 ωn t (48) • For initial conditions: x(t = 0) = xo x(t ˙ = 0) = x˙ o the constants are:  C1 = xo ωn ζ + ζ 2 − 1 + x˙ o p 2ωn ζ 2 − 1  C2 =  p −xo ωn ζ − p p  ζ 2 − 1 − x˙ o 2ωn ζ 2 − 1 (49) (50) • Consider two cases of a system with ωn = 4π = 12.57 rad/s (τ = 0.5 s), with initial conditions xo = 0.05 m and x˙ o = 1 m/s. Engi6933 - Mechanical Vibrations 19 • The motion is aperiodic. It has a similar form to the critically damped solution, however, it undershoots the maximum amplitude of the critically damped system and takes longer to return to equilibrium. • A good example of an overdamped system is the closer on a screen door which takes forever and an age to close. 2.3.6 Summary • Consider three cases of a system with ωn = 4π = 12.57 rad/s (τ = 0.5 s), with initial conditions xo = 0.05 m and x˙ o = 1 m/s. • Underdamped motion: – is harmonic with exponential decay of amplitude. p – The frequency of oscillation of underdamped motion (ωd = 1 − ζ 2 ωn ) is always less than the frequency of oscillation of undamped motion (ωn ). – Underdamped motion will not reach the same peak amplitude as undamped motion, unless the phase angle is zero. – The underdamped case is the only solution that results in oscillatory motion. – Underdamped systems return to equilibrium quickly, but continue to oscillate. In the context of a suspension system the wheels would be in position to respond to another bump more quickly than a critically damped system. Engi6933 - Mechanical Vibrations 20 • Critically damped motion: – is aperiodic with exponential decay to equilibrium position. – Critically damped systems are quickest to return to the equilibrium postion. – A critically damped system would transmit too much force to car passengers. Vehicle suspensions are underdamped. • Overdamped systems: – take a long time to return to the equilibrium position. This is not practical for suspension systems. These systems are useful for controlled, slow return to equilibrium without oscillation. Screen door closers are overdamped. 2.3.7 Logarithmic Decrement • Consider the motion of an underdamped system. • The ratio of displacements at two times separated by one period of damped oscillation is: x1 Xe−ζωn t1 cos(ωd t1 − φ) = x2 Xe−ζωn t2 cos(ωd t2 − φ) Since t2 = t1 + τd where τd = 2π/ωd , the angle in the cosine terms would be different by 360o (2π radians or one full oscillation), therfore, the cosines would be equivalent. x1 e−ζωn t1 = −ζω (t +τ ) x2 e n 1 d • Define the logarithmic decrement as the natural log of the displacement ratios: δ = ln x1 2π 2π 2πζ = ζωn τd = ζωn = ζωn p =p 2 x2 ωd ωn 1 − ζ 1 − ζ2 (51) • Note: in the log domain the amplitude decay is constant. In the figure above the dashed line is Xe−ζωn t . Engi6933 - Mechanical Vibrati...
View Full Document

  • Fall '14
  • NHookey

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern