Chapter 5 Design Acceptable vibration levels (ISO) Vibration isolation Vibration absorbers Effects of damping in absorbers Optimization Viscoelastic damping treatments Critical Speeds Design for vibration suppression 1/51 Mechanical Engineering at Virginia Tech
5.1 Acceptable levels of vibration Each part or system in a dynamic setting is required to pass vibration muster Military and ISO provide a regulation and standards Individual companies provide their own standards Usually stated in terms of amplitude, frequency and duration of test /51 Mechanical Engineering at Virginia Tech
Example 5.1. Dissimilar devices with the same frequency m=1000 kg k=400,000 N/m c=8000 Ns/m m=1 kg k=400 N/m c=8 Ns/m car ω n = 400,000 = 0 rad/s 1000 8000 ζ = (1000)(0) = 0. ω d = 0 1 0. = 19.5959 rad/s CD drive ω n = 400 = 0 rad/s 1 8 ζ = (1)(0) = 0. ω d = 0 1 0. = 19.5959 rad/s 3/51 Mechanical Engineering at Virginia Tech
But: response magnitudes different Magnitude plot will have the same shape Time responses will have the same form for similar (but scaled) disturbances BUT WITH DIFFERENT MAGNITUDES different Xk 1 = { F (1 r) + ( ζ r) 14444443 Fig 5.3 same 4/51 Mechanical Engineering at Virginia Tech
Section 5. Isolation A major job of vibration engineers is to isolate systems from vibration disturbances or visa versa. Uses heavily material from Sections.4 on Base Excitation Important class of vibration analysis Preventing excitations from passing from a vibrating base through its mount into a structure Vibration isolation Shocks on your car Satellite launch and operation Disk drives 5/51 Mechanical Engineering at Virginia Tech
Recall from Section.4 that the FBD of SDOF for base excitation is x(t) m m k c y(t) base k( x y) c( x& y) & F= - k( x - y) - c( x& - y& ) = mx && mx && + cx& + kx = cy& + ky (1) 6/51 Mechanical Engineering at Virginia Tech
SDOF Base Excitation assumes the input motion at the base has the form y( t) = Y sin( ωt) and plug into Equation(1) mx && + cx& + kx = cωy cos( ωt) + ky sin( ωt) () 1444444443 harmonic forcing functions The steady-state solution is just the superposition of the two individual particular solutions f 0 c 64748 f } 0 s && x + ζω x& + ω x = ζω ωy ωt + ω Y ωt 144443 1443 n n n cos( ) n sin( ) (3) x ( t) x ( t) pc ps 7/51 Mechanical Engineering at Virginia Tech
Particular Solutions (sine input) With a sine for the forcing function, && x + ζω x& + ω x = f ωt n n 0s sin x = A cosωt + B sinωt = X sin( ωt φ ) ps s s s s where A B s s = = ζω ω f n 0s ( n ) + ω ω ζω ω ( ωn ω ) ( n ) + ( ) ( ) 8/51 Mechanical Engineering at Virginia Tech f 0s ω ω ζω ω n n
Particular Solutions (cosine input) With a cosine for the forcing function, we showed && x + ζω x& + ω x = f ωt n n 0c cos x = A cosωt + B sinωt = X cos( ωt φ ) pc c c c c where A B s s = = ω ω ( n ) ( n ) + ζω ω f n 0c ( n ) + ( ) ( ) 9/51 Mechanical Engineering at Virginia Tech f 0c ω ω ζω ω ω ω ζω ω n n
Magnitude X/Y Magnitude of the full particular solution X p = X pc + X ps = ( A s + B s )+ ( A c + B ) c = f 0c + f 0s (ω n ω ) + ζω n ω = ω where f 0c = ζω n ωy and f 0s = ω n Y (ζω) +ω n ( ) = ω Y (ζω) +ω n n (ω n ω ) + ( ζω n ω) if we define r = ω ω n this becomes X p = Y (ζr) +1 (1 r ) + ( ζr) 10/51 Mechanical Engineering at Virginia Tech
The magnitude plot of X/Y 40 30 0 ζ =0.01 ζ =0.1 ζ =0.3 ζ =0.7 X/Y (db) 10 0-10 -0 0 0.5 1 1.5.5 3 Frequency ratio r 11/51 Mechanical Engineering at Virginia Tech
Notes on Displacement Transmissibility Potentially severe amplification at resonance Attenuation only for r > sqrt() If r< sqrt() transmissibility decreases with damping ratio If r>>1 then transmissibility increases with damping ratio X p =Yζ/r 1/51 Mechanical Engineering at Virginia Tech
It is also important to look at the Force Transmissibility: F = k( x y) + c( x& y& ) = mx && T we know that x( t) = X cos( t ), so && x =- X cos( t ) T ω ω φ ω F = m X = kr X ω x(t) k φ m c F T y(t) base 13/51 Mechanical Engineering at Virginia Tech
Plot of Force Transmissibility F/kY (db) 40 30 0 10 0 ζ =0.01 ζ =0.1 ζ =0.3 ζ =0.7-10 -0 0 0.5 1 1.5.5 3 Frequency ratio r 14/51 Mechanical Engineering at Virginia Tech
Isolation is a sdof concept; Two types: moving base and fixed base Three magnitude plots to consider TR= transmissibility ratio X Y = 1+ (ζr) (1 r ) + (ζr) F T ky = r 1+ (ζr) (1 r ) + (ζr) Moving base displacement Moving base force F T F 0 = 1 + (ζr) (1 r ) + (ζr) Fixed base force 15/51 Mechanical Engineering at Virginia Tech
For displacement transmissibility, isolation occurs as a function of stiffness For stiffness such that the frequency ration is larger the root, isolation occurs, but increasing damping reduces the effect For less then root, increased damping reduces the magnitude. 16/51 Mechanical Engineering at Virginia Tech
Example 5..1 Design an isolation mount Fig 5.6 Design an isolator (chose k, c) to hold a 3 kg electronics module to less then 0.005 m deflection if the base is moving at y(t)=(0.01)sin(35t) Calculate the force transmitted through the isolator 17/51 Mechanical Engineering at Virginia Tech
T.R. Plot for moving base displacement 1.5 T.R. 1 0.5 ζ =0.01 ζ =0.05 ζ =0.1 ζ =0. ζ =0.5 ζ =1. 0 0 0.5 1 1.5.5 3 Frequency ratio r ζ For T.R. =0.5 ζ 0.01 0.05 0.1 0. 0.5 1. r 1.73 1.74 1.76 1.84.35 4.41 18/51 Mechanical Engineering at Virginia Tech
From the plot, note that T.R. = X Y = max Response max input = 0.005 0.01 = 0.5 from the plot r =1.73, ζ=0.0 works r = ω ω n = 1.73 k 3 = 35 1.73 k = 18 N/m c = ζmω n = (0.0)(3)(0.3) =.48 kg/s 19/51 Mechanical Engineering at Virginia Tech
Rattle Space δ = mg k = g X Y < 0.5 ω = 9.81 (0.3) = 0.04 m n Choice of k and c must also be reasonable As must force transmitted: 0/51 Mechanical Engineering at Virginia Tech
The transmitted force is F T = kyr 1+ (ζr) (1 r ) + (ζr) = kyr T.R. = (18)(0.01)(1.73) (0.5) = 18.376 N Transmitted force, T.R., static deflection, damping and stiffness values must all be reasonable for the application. 1/51 Mechanical Engineering at Virginia Tech
Shock Isolation π Y sinω pt 0 t t1 = ω p y( t) = (5.8) π 0 t > t1 = ω p Plot: max && x( t) ω X ω ω t max && y( t) && y( t) ω π = n versus n = n 1 p Shock pulse Pulse duration Increased isolation with increasing k /51 Mechanical Engineering at Virginia Tech
Figure 5.8 Shock Response 3/51 Mechanical Engineering at Virginia Tech
Shock versus Vibration Isolation In figure 5.8 for ζ = 0.5 requires ω n t 1 π < 1.0 k < mπ for shock isolation to occur. Thus shock isolation requires a bound on the stiffness Also from Figure 5, high damping is desirable for shock attenuation. t 1 4/51 Mechanical Engineering at Virginia Tech
Example 5..3 Design a system that is good for both shock and vibration isolation. The design constraints are that we have the choice of 3 off the shelve isolation mounts: 5 Hz, 6 Hz and 7 Hz each with 8% damping The shock input is a 15 g half sine at 40 ms The vibration source is a sine at 15 Hz The response should be limited to 15 g s and 76. mm, and 0 db of vibration isolation 5/51 Mechanical Engineering at Virginia Tech
Simulation of the response to the shock input for all three mounts choices From these numerical simulations, only the 7 Hz mount satisfies all of the shock isolation goals: Less then 15 g s Less then 3 in deflections Fig 5.11 Fig 5.1 6/51 Mechanical Engineering at Virginia Tech simulated relative displacement: z(t) simulated absolute acceleration: &x(t)
Now consider the vibration isolation by plotting shock isolator design s transmissibility: For the 7 Hz shock isolator design, the reduction in Transmissibility is only 9.4 db. From this plot, and recalling Fig 5.7 less damping is required. However, less damping is not possible Fig 5.13 7/51 Mechanical Engineering at Virginia Tech
5.3 Vibration Absorbers Consider a harmonic disturbance to a singledegree-of freedom system Suppose the disturbance causes large amplitude vibration of the mass in steady state A vibration absorber is a second spring mass system added to this primary mass, designed to absorb the input disturbance by shifting the motion to the new added mass (called the absorber mass). 8/51 Mechanical Engineering at Virginia Tech
Absorber concept Primary mass (optical table) m F(t) = F 0 sinωt x absorber k / k a x a k/ m a Primary system experiences resonance Add absorber system as indicated Look at equations of motion (now dof) 9/51 Mechanical Engineering at Virginia Tech
The equations of motion become: m 0 && x k + ka ka x F0 sinωt 0 m a x + = a ka k a x a 0 && To solve assume a harmonic displacement: x( t) = X sinωt assumed solution: xa ( t) = X a sinωt ω X k + ka m ka 0 k X = a 0 a ka maω F 30/51 Mechanical Engineering at Virginia Tech
The form of the response magnitude suggests a design condition allowing the motion of the primary mass to become zero: X X a = k + k a mω k a k a k a m a ω 1 F 0 0 X = (k m ω )F pick m a and k a a 0 a (k + k a mω )(k a m a ω to make zero ) k a X a = k a F 0 (k + k a mω )(k a m a ω ) k a All the system motion goes into the absorber motion 31/51 Mechanical Engineering at Virginia Tech
Choose the absorber mass and stiffness from: ω = k a m a This causes the primary mass to be fixed and the absorber mass to oscillate at: x a (t) = F 0 sinωt, a force magnitude of k a k a k a x a = k a F 0 k a = F 0 As in the case of the isolator, static deflection, rattle space and force magnitudes need to be checked in each design 3/51 Mechanical Engineering at Virginia Tech
Other pitfalls in absorber design Depends on knowing ω exactly Single frequency device If ω shifts it could end up exciting a system natural frequency (resonance) Damping, which always exists to some degree, spoils the absorption let s examine these: 33/51 Mechanical Engineering at Virginia Tech
Avoiding resonance (robustness) µ = ω ω p a = = k m a m k m a a = = k ω p β = ω ω a p k m a a Mass ratio ω µ µβ Original natural frequency of primary system before absorber is attached Natural frequency of absorber before it is attached to primary mass frequency ratio Stiffness ratio 34/51 Mechanical Engineering at Virginia Tech
r a = ω ω a, r p = ω ω p Define a dimensionless amplitude of the primary mass Xk F 0 = 1+ µβ r p 1 r a 1 r a µβ 35/51 Mechanical Engineering at Virginia Tech
Normalized Magnitude of Primary Xk ω 1 1 µ. β ω ωp ω ωa. 1 ω ωa µ. β Xk ω 1 0 0.5 1 1.5 ω 36/51 Mechanical Engineering at Virginia Tech ωa Absorber zone Fig 5.15 ω/ω 1 ω/ω
Robustness to driving frequency shifts If ω hits ω 1 or ω resonance occurs Using Xk/F 0 <1, defines useful operating range of absorber In this range some absorption still occurs The characteristic equation is: ω n ω a = 1+ β (1+ µ) β ± 1 β β 4 (1+ µ) β (1 µ) + 1 = Frequency dependence on mass and frequency ratio 37/51 Mechanical Engineering at Virginia Tech
Mass ratio versus frequency Referring to fig. 5.16, as µ increases, frequencies split farther apart for fixed β thus if µ is too small, system will not tolerate much fluctuation in driving frequency indicating a poor design Rule of thumb 0.05< µ <0.5 Fig 5.16 38/51 Mechanical Engineering at Virginia Tech
Normalized Magnitude of the primary mass with and without the absorber Adding absorber increase the number of resonances (or modes) from one to two. Smaller response of primary structure is at absorber natural frequency Effective over limited bandwidth Amplitude (X Xk/F) 60 40 0 0-0 µ=0.1 and β=0.71 39/51 Mechanical Engineering at Virginia Tech No vibration absorber With vibration absorber -40 0 0.5 1 1.5 Frequency ratio (r a )
What happens to the mass of the vibration absorber? In the operational range of the vibration absorber the absorber mass has relatively large motion Beware of deflection limits!! Amplitude (X ak/f) a 60 40 0 0-0 µ=0.1 and β=0.71-40 0 0.5 1 1.5 Frequency ratio (r a ) 40/51 Mechanical Engineering at Virginia Tech
Example 5.3.1: design an absorber given F 0 = 13 N m=73.16 kg, k=600 N/m, ω = 180 cpm, x a < 0.0 m From (5.0) : k a = m a ω X = 0 k a = F 0 X a = 13 N 0.00 m = 6500 N/m k a = m a ω m a = k a 6500 N/m = ω 180 60 π check : µ = 18.9 73.16 = 0.5 =18.9 kg 41/51 Mechanical Engineering at Virginia Tech
Example 5.3. Compute the bandwidth of the absorber design in 5.3.1 Xk F 0 =1 1 ω = 1+ µ ω a ω ω a ω p ω p 1 ω ω a ω ω a = ± 1+ µ, at µ = 0.5 ω ω a =1.1180 µ ω a ω p For Xk F 0 = 1, ω a = 6500 18.9, ω b = 600 73.16, µ = 0.5 ω ω a = 0.399, 1.138 4/51 Mechanical Engineering at Virginia Tech 3 roots
Comparing these 3 roots to the plot yields that: 0.399ω a < ω < 1.1180ω a 7.4089 < ω < 1.081 rad/s This is the range that the driving frequency can safely lie in and the absorber will still reduce the vibration of the primary mass. 43/51 Mechanical Engineering at Virginia Tech
5.4 Damped absorber system Undamped primary m 0 && x( t) ca ca x& ( t) 0 m a xa ( t) + ca c a xa ( t) && & k + ka ka x( t) F0 + sin t = k ( ) 0 ω a k a xa t X ( k m ω ) + ω c a a a F = 0 ( k m ω )( k a m aω ) m k k a aω + ( m + m a ) ω caω Cannot be zero! 44/51 Mechanical Engineering at Virginia Tech
Magnitude of primary mass for 3 levels of damping 10 Xk ω, 0.01 Xk ω, 0.1 Xk ω, 0.4 8 6 4 As damping increases, the absorber fails, but the resonance goes away 1 Region of absorption 0 0.5 1 1.5.5 3 ω ω p 45/51 Mechanical Engineering at Virginia Tech Go to mathcad example 5.16
Effect of damping on performance In the operational range of the vibration absorber decreases with damping The bandwidth increases with damping Resonances are decreased i.e. could be used to reduce resonance problems during run up See Fig 5.19 Amplitude (Xk k/f) 50 40 30 0 10 0-10 -0 µ=0.5 and β=0.8 46/51 Mechanical Engineering at Virginia Tech No VA c a =0.01 c a =0.1 c a =1-30 0 0.5 1 1.5 Frequency ratio (r a )
Three parameters effect making the amplitude small This curves show that just increasing the damping does not result in the smallest amplitude. The mass ratio and β also matter This brings us to the question of optimization µ = 0.5,β = 1.0,ζ min = 0.4 µ = 0.5,β = 0.8,ζ min = 0.7 47/51 Mechanical Engineering at Virginia Tech
With damping in the absorber: Undamped absorber has poor bandwidth Small damping extends bandwidth But, ruins complete absorption of motion Becomes a design problem to pick the most favorable µ, β, ζ. 48/51 Mechanical Engineering at Virginia Tech
Viscous Vibration Absorber Primary system Viscous absorber x(t) k m Fig 5. x a c a m a Rotating machine applications Rotational inertia, shaft stiffness and a fluid damper Often called a Houdaille damper illustrated in the following: 49/51 Mechanical Engineering at Virginia Tech
Houdaille Damper Equation of motion: Fig 5.3 J1 0 && θ 1 ca ca & θ 1 0 J + θ c a c && a & θ θ + 0 0 = θ 0 ka 0 1 M 0 e jωt 50/51 Mechanical Engineering at Virginia Tech
Frequency response of primary mass 100 X r,.5,.01 10 X r,.5, 0.1 X r,.5, 0.4 0 0.5 1 1.5 X r, µ, ζ 0.1. 4 ζ r 4. ζ r µ. r 1 r 1. r r Figure 5.4 Design on µ and ζ 51/51 Mechanical Engineering at Virginia Tech