1、1 : 1 Mechanical Vibration全册全册 配套完整教学课件配套完整教学课件 1 : 2 4th Year Mechanical Design Lecturer: Stephen Tiernan 2 hours lecture 2 hours lab. Lectures: Section 1: Free Vibration (wk 1 & 2) Section 2: Forced Vibration (wk 3 & 4) Section 3: Two Degrees of Freedom (wk 5 & 6) Section 4: Vibration Absorbers (w
2、k 10 & 11) Lab Time Free Vibration: week 1 & 2 Section 5: Statics, Failure Theories weeks 3 to 12 Thanks to Dr P Young of DCU for some of the course material Grading Exam 70%: (5 questions do 4) Continuous Assessment 30% Labs: 1.Free Vibration: Group Report 2.Statics Case Study Individual Report Lab
3、 Reports Write-up due one week completion of lab session. Results are most important part Clear presentation Understanding of results Accuracy of observations No waffle 1 : 3 Free Vibrations Un Damped Simplest system Mass and spring Consider a mass, m, on frictionless rollers connected to a fixed po
4、int by a spring, k: m k x = 0 x SI Units: Mass.kg StiffnessN/m Displacementm 1 : 4 Hookes Law Free Body Diagrams for Static Displacements Hookes Law: Spring Force = Stiffness x Extension (or Compression) m k x = 0 x m k x = 0 x m k.(-x) m k.(-x) Compression x has a negative value Extension x has a p
5、ositive value FBDs Spring force accelerates the body in the positive direction Spring force accelerates the body in the negative direction Spring force is proportional to and opposes displacement 1 : 5 Free Body Diagram Including Inertial Forces k.x x = 0 x m.x . For Equilibrium: 0 0 x m k x kxxm kx
6、xm 1 : 6 Solution to Equation then0sincos wheresolution trivial the Ignoring 0sincossincos Hence sincos constants are &, where sincos 2 2 tBtA tBtA m k tBtA tBtAx BA tBtAx m k 2 The solution to differential equations of this form is Simple Harmonic Motion where x is a sum of cosine and sine componen
7、ts with distinct magnitudes and the same angular velocity. 1 : 7 Natural Frequency rad/s n m k The displacement of the mass will oscillate about the equilibrium position with a frequency determined by the mass and stiffness. The solution now has the form: tBtAx m k m k sincos The constants A and B m
8、ust be determined from the initial conditions. 1 : 8 Gravity m k x x = 0 k d y A spring of natural length, y, extends by a small distance, d, when mass m is suspended from it. m.g k.d = m.g m.x . k.x Natural length of spring m.g Force due to gravity k.d New length of spring is y + d where Static Def
9、lection, d = m.g/k Selecting the static equilibrium position as x = 0 means that the static spring and gravitational forces cancel out of the equation of motion Static Mass 1 : 9 Torsional System q Kq I Inertia of Rotor Torsional Stiffness of Shaft Rotation of Rotor Similar to translational system I
10、nertial force is I.q (c.f. m.x) Stiffness force is Kq.q (c.f. k.x) . . q Kqq Iq . t I K Bt I K A KI KI q qq qq qq q q sincos :equation ntranslatio to Similar 0 1 : 10 Some Properties of Undamped Systems System is conservative no energy is lost from the system Existence of a static load changes the e
11、quilibrium position, but not the oscillation The frequency of oscillation is a function of the stiffness and mass of the components Two independent initial conditions are required to determine the exact solution Almost impossible to achieve in mechanical systems 1 : 11 Non-conservative Systems A sys
12、tem is conservative if the total energy of the system remains constant Non-conservative systems loose energy with every oscillation cycle While electronic systems may be designed to be conservative (or almost so), mechanical systems are in general non-conservative Energy may be lost from the system
13、in a variety of ways Friction in the components Internal strain losses Viscous losses as some part moves through a fluid Engineers often add such elements to a system to control the movement and response of the system to external stimulii 1 : 12 Viscous Damping m k x = 0 x c Viscous damping is gener
14、ated when an object is moved through a fluid. The greater the viscosity of the fluid the greater the energy loss. The effect on the body is a force which is proportional to the velocity. Most viscous dampers are configured as dashpots in which a plate is forced to move through a fluid with a constri
15、cted flow. This increases the viscosity of the fluid. 1 : 13 Dashpots Fluid flow around piston is required to allow motion up or down. No movement = No Force Movement = Force is proportional to velocity The constant of proportionality is termed c and has the units of Ns/m. The force acts to oppose t
16、he movement. Damping Force: F = -c.x . l c D vF 3 3 c cD l 1 : 14 c D l v F Damping Force: F = -c.x Dashpots 1 : 15 m k x = 0 x c m k x = 0 x c Damping Force m k.(-x) Negative Velocity Damping Force is Positive Zero velocity Damping Force is zero Velocity Positive Damping Force is negative FBDs Stif
17、fness and Damping forces have the same sense Stiffness and Damping forces have opposite sense Damping force is proportional to and opposes velocity m k x = 0 x c v v m k.(-x) m k.(-x) c.(-x) . c.(-x) . 1 : 16 General Case Equation of Motion x = 0 x k.x m.x . c.x . For equilibrium: 0 0 x m k x m c x
18、kxxcxm kxxcxm A second order differential equation solution has the same form as before except that the amplitude is not constant. For this derivation use x(t) = Xest 1 : 17 Solution to Viscous Damping 0 :Otherwise 0 :Solution Trivial 0 0 )( )( )( 2 2 2 22 kcsms X Xekcsms kxsxcxsm xsXestx sxsXetx Xe
19、tx st st st st Quadratic in s with two roots, so solution is: The exact form of x depends on the solution to this equation. In general: m mkcc s 2 4 2 2, 1 tsts eXeXtx 21 21 1 : 18 Critical Damping Case 1: c2 4mk, s1 and s2 are distinct real roots There is a sharp boundary between oscillating system
20、s and monotonic systems (case 2). This value of damping is termed critical damping, ccr Oscillating Monotonic ncr mmkmkc224 The damping ratio, z, is defined as the ratio between the amount of damping in the system, c, and the critical damping, ccr ncr m c mk c c c z 24 z is dimensionless 1 : 19 Tran
21、sient Solution for Free Vibration with Damping (under damped) 2 d )( )( )( 1 constants are andA )( OR constants are B andA )()( z z z n d t t dd t t at t whereand tSinAex tBSintACosex btBSinbtACosex n n 1 : 20 Transient Solution for Free Vibration with Damping (critically damped) constants are B and
22、A form thehavemust solution a so phase and amplitudefor allow toconstants twobemust There )( t t n eBtAx 1 : 21 Free Vibration of Damped Systems -1.5 -1 -0.5 0 0.5 1 1.5 0100200300400500600700 Time Displacement Free Vibration of Damped Systems -1.5 -1 -0.5 0 0.5 1 1.5 0100200300400500600700 Time Dis
23、placement Free Vibration of Damped Systems -1.5 -1 -0.5 0 0.5 1 1.5 0100200300400500600700 Time Displacement Free Vibration of Damped Systems -1.5 -1 -0.5 0 0.5 1 1.5 0100200300400500600700 Time Displacement Free Vibration of Damped Systems -1.5 -1 -0.5 0 0.5 1 1.5 0100200300400500600700 Time Displa
24、cement Comparison of Different Levels of Viscous Damping Undamped Underdamped Critically damped Overdamped 1 : 22 Free Damped Vibration Transient Response Sum of forces at any time =0 0kxxcxm Forces Initial Displacement = 0, Initial Velocity = 4.2 m/s 1 : 23 Free Damped Vibration Transient Response
25、Sum of forces at any time =0 0kxxcxm Forces Initial Displacement = .01m, Initial Velocity = 0 m/s 1 : 24 Free Vibration of Damped Systems -1.5 -1 -0.5 0 0.5 1 1.5 0100200300400500600700 Time Displacement Properties of Viscous Damping Response Ae-znt -Ae-znt td Ae-zntSin(dt + ) A0 A1 )( )( 11 00 1 0
26、z z tSinAeA tSinAeA d t d t n n t0 t1 1 : 25 Logarthmic Decrement, d d 22 22 2 1 0 4 hence 1 2 1 2 1 22 but ln d d z z z z zd z t tzd n n n d d dn A A n n d d dn A A n A A 0 2 1 0 ln 1 cycles ofnumber n :cylces ofnumber aOver 1 22 but ln d z t tzd 1 : 26 Summary of Free Damped Equations conditions i
27、nitial from found constants are and )( OR conditions initial from found constants are B andA SolutionTransient 2C Damping Critical Ratio Damping 2 Time Periodic 1Frequency Damped Frequency Natural )( )( c 2 d z z z z CtSinCex tBSintACosex m C C T m k d t t dd t t n cd n n n n 22 0 4 ln 1 Decrement L
28、og d d z d n A A n 1 : 27 Sample Problem A mass of 38 kg is suspended by a spring that deflects by 1 mm when the mass is attached. If a viscous damper is attached in parallel to the mass spring system. The damping ratio is 0.2. a. Calculate the damping value for critical damping and the damped frequ
29、ency of oscillation and the logarithmic decrement. b. Draw a sketch of the system and calculate the natural frequency of oscillation. If the mass is pulled down by 5 mm and held until released at time t = 0. c. How long does it take for the system to oscillate within 1% of maximum displacement from
30、the equilibrium position? d. Sketch the response of the system, indicating the period of oscillation and the amplitude envelope. e. Write down the equation of motion for the response. 1 : 28 Notation System Translation Rotation Displacement x metres q radians Force F Newtons T Newton-Metres Mass M k
31、ilograms I Kilogram-metre2 Stiffness K Newtons/metre Kq Newton-metres/radian Viscous Damping C Newton-seconds /metre Cq Newton-metre-seconds /radian 1 : 29 Harmonic Excitation of Single Degree of Freedom Systems m k c f(t) = Fsin(t) x(t) Phase lag Equation: Solution: x(t) = XSin(t - ) )sin()(ttfkxxc
32、xmF X F 1 : 30 Solution: General = Particular + Homogeneous + = Steady State Forced Response at same frequency as excitation x(t) = XSin(et - ) Free Vibration Response to change in forcing function at damped frequency x(t) = Ae-zntSin(dt - ) 1 : 31 Response of System to Sinusoidal Input Combined Mag
33、nitude & Phase Chart 1 : 32 General Solution to Forced Vibration )( )( )( )sin(F)( 2 tSinXx tCosXx tXSinx ttfkxxcxm )sin(F)()sin(X)cos(X)sin(X 2 ttftktctm These are vectors and therefore must be added as such (i.e. each element has a direction) 1 : 33 Phasor Diagrams For instance the combination of
34、a Sine and Cosine wave may be determined: y = A.sinq + B.cosq y = R.sin(q + a) A q A.sinq B B.cosq q a R Where a solution is comprised of orthogonal components the resultant may be understood through the use of Phasor diagrams. A B 22 BAR arctan a 1 : 34 Phasor Diagrams for Forced Vibration kX t cX
35、)sin()()sin()cos()sin( 2 ttftktctmFXXX 2mX F kX- 2mX cX F 1 : 35 Dynamic Magnification Factor 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 22 21 1 21 )( )( nn nn s X X k F X cmk F X cmk F X XcXmkXF z z 2 1 2 1 2 1 2 tan tan tan n n mk c XmkX Xc z Using Pythagorus 1 : 36 Response of System to Sinusoidal Input - M
36、agnitude Dimensionless graph representative of all spring-mass-dashpot systems. Shape of curve changes with damping ratio. Natural Frequency and Spring Stiffness determine unit values on the axes. 1 : 37 Response of System to Sinusoidal Input - Phase Dimensionless graph representative of all spring-
37、mass-dashpot systems. Shape of curve changes with damping ratio. Natural Frequency determined by stiffness and mass values. 1 : 38 Phasor Diagrams for Forced Vibration kX t cX F )sin()()sin()cos()sin( 2 ttftktctmFXXX 2mX n F m 2 F X Phasor Diagrams for Forced Vibration Therefore resultant amplitude
38、dominated by mass 1 : 41 Zones on Frequency Response k F X m 2 F X Mass Damping Stiffness n Slow Oscillation Fast Oscillation c F X 1 : 42 Zones on Frequency Response Mass Damping Stiffness To make the graph easier to read log scales are often used for both frequency and response axes. 1 : 43 Dynami
39、c Magnification Factor (DM Factor) z 2 1 Factor QMax resonanceAt s n X X m k c f(t) = Fsin(t) x(t) It is difficult to measure Xs as it can be very small 2 2 2 21 1 Factor DM nn s X X z z2 Xmax s X Note: At Resonance 1 : 44 Derivation What frequency is amp pointspower half at the width band theis 1 2
40、 1 Factor DMMax resonanceAt Re 242 2 12 12 1212 12 pp DM pp Thus X X member pppp so pp but n s n nn n z z z nn n pp p pppppp Thus P P PP PP Then Let nn n z zzzz zzz zzz zz zz 1212 2 2 1 2 2 2 22 22 2222 222 22 2 2 2 4 small is if 414 12)21 ( 2 )81 (4)24() 12(2 0)81 ()24( 0821 P 12 2 2 2 2 max 2 2 2
41、p 821 222 21 X zz z z nn nn pp s pp s XXX 2 max X p X 1 : 45 A method to determine damping and the DM factor is the Half Power Point Method z 2 112 DM n pp Thus By drawing a graph of amp vs frequency n, p1 and p2 may be measured and damping and DM calculated. pointspower half at the width band theis
42、 1 p 2 p 1 : 46 Measurement of DM Factor For light damping, the damping ratio, and hence the DM factor associated with any mode of vibration can be found from the amplitude frequency measurements at resonance and the half power points. Care must be taken that the exciting device does not load the sy
43、stem and alter its response. In real systems Xmax can be difficult to measure and also p1 and p2 may be very close. The table below shows the relationship between DM factor and the bandwidth. DM Factor 500 50 5 Resonance Freq (Hz) Frequency Bandwidth (Hz) 10 0.02 .2 2 100 0.2 2 20 1000 2 20 200 May
44、be difficult to measure as p1 and p2 very close 1 : 47 Excitation Due to Eccentric Rotating Mass m k x = 0 x c me Excitation force is proportional to the speed of rotation of the eccentricity: F = mere2 f(t) = Fsint A rotating element within a machine (e.g. motor in a lathe or vehicle or shafts, pul
45、leys & gears in a housing) will tend to include some form of eccentricity. If the machine is mounted on some form of suspension then it will oscillate at the frequency of the rotating component. 1 : 48 Reciprocating/Rotating Imbalance Force These produce forces which are proportional to 2. For an un
46、balanced rotating component f = mere2sint statically applied wereforce thisif deflection theis This 21 1 Factor DM 2 2 2 2 k rm k F X X X ee s s nn z X is the resulting deflection of the machine 1 : 49 Response of Single DOF System to Sinusoidal Excitation 1 : 50 Resonance Peak - Frequency 2 1, 0z n X X n s Resonant frequency decreases as damping increases Peak occurs when the slope of the graph is zero 2 1 z n d Resonance peak occurs below the damped frequency 1 : 51 Resonance Peak - Amplitude 2 max 12 1 zz s X X Resonant amplitude decreases as damping increases Substituting in at maximum n
