MEG6007: Advanced Dynamics -Principles and Computational Methods (Fall, 2017) Lecture DOF Modeling of Shock Absorbers. This lecture covers:

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1 MEG6007: Advanced Dynamics -Principles and Computational Methods (Fall, 207) Lecture 4. 2-DOF Modeling of Shock Absorbers This lecture covers: Revisit 2-DOF Spring-Mass System Understand the den Hartog Invariant points as a design application Learn a particular optimization for minimizing vibrations There is a plenty of room for 2-DOF system to learn and apply!

2 x x 2 m 2 M f2 f C 2 k 2 /2 K /2 Two DOF Spring-Mass-Damper Suspension Model Fig.. 2-DOF Shock Isolation Model 4. Invariant Points of the Den Hartog- Ormondroyd Model Consider a Two-DOF system as shown Figure. The governing equations for the system is given by M ẍ + K x + k 2 (x x 2 ) + c 2 (ẋ ẋ 2 ) = f (t) m 2 ẍ 2 + k 2 (x 2 x ) + c 2 (ẋ 2 ẋ ) = f 2 (t) (4.) Assume that the excitation acts directly to M in the form f (t) = F e jωt, f 2 (t) = 0 (4.2) The solution of (4.) assumes x (t) = X e jωt, x 2 (t) = X 2 e jωt (4.3) Substituting these into (4.) yields the coupled equations in the frequency domain as 2

3 ( M ω 2 + K + k 2 + icω)x (k 2 + jcω)x 2 = F (k 2 + jcω)x + ( m 2 ω 2 + k 2 + jcω)x 2 = 0 (4.4) X obtained from the above equation reads: X (ω) F = ( m 2 ω 2 + k 2 + jcω) [( M ω 2 + K )( m 2 ω 2 + k 2 ) m 2 ω 2 k 2 ] + jωc[k (M + m 2 )ω 2 ] (4.5) Let us parameterize the model as Ω = K /M, ω 2 = k 2 /m 2, c = 2ζm 2 Ω, µ = m 2 /M, s = ω/ω, = F /K, f = ω 2 /Ω, c c = 2m 2 Ω (4.6) The ratio of X (ω) to is the amplification factor from the static to the dynamic response, which can be expressed as X (ω) = [ (2 c c c s) 2 + (s 2 f 2 ) 2 (2 c c c s) 2 (s 2 + µs 2 ) 2 + [µf 2 s 2 (s 2 )(s 2 f 2 )] 2 ] 2 (4.7) Figure 2 plots { X (ω) } vs. the driving (or exciting) frequency, ω, for various damping coefficients (ζ), i.e., by varying ζ while keeping other parameters constants. Note that there are two points, marked by red dots in the figures, the curves pass through for all damping ratios. We will call these two points as den Hartog s invariant points and will exploit its properties in two ways: one is to minimize the maximum amplitude resulting in good shock absorbers; and, the other is to maximize its peak amplitude so that the resulting resonators possesses a minimum loss or high-q performance. 4.2 Shock Absorbers A good shock absorber should have the following characteristics: 3

4 Frequency Response Functions for Different Damping Ratios of a 2-DOF Suspension Model 0 Invariant points Amplitude Frequency (Hz) Fig. 2. Frequency Response Function at Mass (a) The peak amplitude should be insensitive over a wide range of excitation frequencies; (b) The cost for implementing damping should be affordable; (c) The auxiliary mass (m 2 ) should be as light as possible; (d) The response time duration to reach its steady state should be short. We now proceeds with the task of designing a good shock absorber Determination of Undamped Natural Frequencies The natural frequencies are determined by setting c = 0 from the denominator of (4.7): 4

5 [ µf 2 s 2 (s 2 )(s 2 f 2 ) ] = 0 s 4 [ + ( + µ)f 2 ]s 2 + f 2 = 0 (4.8) The two distinct roots of this equation will be designated as s 2 n and s 2 2n Determination of Invariant Points Since the two invariant points would play a key role, it is necessary to find the two coordinates. Of several approaches, we utilize the fact that they are independent of system damping, c. To this end, we rearrange the frequency response function or the magnification factor (4.7) as X (ω) = H(s, c, f, µ) = [ (2 c c c s) 2 A + B (2 c c c s) 2 C + D ] 2, s = ω/ω A = B = (s 2 f 2 ) 2 C = (s 2 + µs 2 ) 2 D = [µf 2 s 2 (s 2 )(s 2 f 2 )] 2 (4.9) The damping-independency of the two invariant points implies that at the invariant points the magnification factor satisfies: A B = C D AD = BC (4.0) for which the amplification factor becomes: X (ω) = H(s, c, f, µ) = ( A C ) 2 (4.) The equation for determining the invariant points yields two 5

6 equations: (s 2 f 2 ) = (s 2 + µs 2 ) [µf 2 s 2 (s 2 )(s 2 f 2 )] (s 2 f 2 ) = (s 2 + µs 2 ) [µf 2 s 2 (s 2 )(s 2 f 2 )] (4.2) The first of the above relation yields the trivial solution, viz., s = ω/ω = 0, (4.3) a static case which is not of interest for the present discussion. The second of (4.2) can be rearranged as (2 + µ)s 4 2( + f 2 + µf 2 )s 2 + 2f 2 = 0 (4.4) The two invariant points, s and s 2 can be shown to satisfy s 2 < + µ < s2 2 (4.5) and the corresponding amplification factors are given by X (ω) = ( + µ)s 2 (,2) (4.6) Determination of Amplification Factor at the Invariant Points The foregoing analysis now supplies the coordinates of the two invariant points as P = (x, y ) = (s, ( + µ)s 2 ), Q = (x 2, y 2 ) = (s 2, (4.7) In order to see succinctly the magnification factor, we arrange as ( + µ)s 2 2 ) µ 2 + µ z2 2 ( + µ)2 f 2 (2 + µ) z = 0, z = s 2 ( + µ) (4.8) 6

7 so that the product of the two roots of the above equation, (z z 2 ), is given by z z 2 = 2 + µ µ = + 2 µ (4.9) which is independent of the frequency ratio f = ω 2 /Ω. It should be noted that for a shock absorber the magnification factor should remain small for a wide range of frequencies, which can be realized if we set the amplification factors to attain their maximum at the two invariant points and at the same time their peaks to be the same. The latter condition can be realized if the coefficient associated with z-terms in (4.8) vanishes: ( + µ) 2 f 2 = 0 f = ω 2 Ω = + µ, µ = m 2/M (4.20) The amplification factor, (X (ω)/ ), for this particular choice at the two invariant points becomes X (ω) = ( + µ)s 2 (,2) = + 2 µ (4.2) Determination of Absorber Damping In the previous section, we imposed the condition on the amplification factor at the two invariant points to be the same. However, it is possible that the amplification factor may be larger at other exciting frequencies. Of several possible choices, we impose that the maximum amplification factors for the frequency ranges of interest would not exceed that at the invariant points. This can be carried out as follows. First, we differentiate (4.9) with respect to s = ω/ω set the resulting expression to zero: ( s ) [X (ω) ] = 0 (4.22) 7

8 for the two invariant points obtained from (4.2.3). For s = s : ζ 2 = ( c c c ) 2 = µ 3 ( µ µ+2 ) 2 8( + µ) 3 For s = s 2 : ζ 2 = ( c c c ) 2 = µ 3 + ( µ µ+2 ) 2 8( + µ) 3 (4.23) In practice, we must have one damping value for all the frequency range. To this end, we take the average value to be: ζ = 3µ 8( + µ) 3 (4.24) Applications First, in order to utilize Matlab capability, we express (4.5) in its transfer function form: X (s) F = (m 2 s 2 + k 2 + cs) [(M s 2 + K )(m 2 s 2 + k 2 ) + m 2 s 2 k 2 ] + sc[k + (M + m 2 )s 2 ] (4.25) where s is now the Laplace Transform variable. The preceding equation can be arranged to read: X (s) = Ω 2 (a 0 s 2 + a s + a 2 ) (b 0 s 4 + b s 3 + b 2 s 2 + b 3 s + b 4 ) a 0 =, a = 2ζΩ, a 2 = ω 2 2 b 0 =, b = 2( + µ)ζω, b 2 = Ω 2 + ( + µ)ω 2 2, b 3 = 2ζΩ 3 (4.26) The amplification factor ( X (s) ) vs. the driving frequency (ω) is illustrated in Fig. 3, with (Ω = 0(Hz), ω 2 = 9.5(Hz), µ = 0.0). Observe the two invariant points at which the amplification factors are different, namely, at s and s 2 points determined by (4.2.3) and its magnitudes given by (4.7). When the amplification factor is optimized to have its peak at the invariant points by selecting the frequency ratio according to 8

9 Proof of two invariant points for den Hartog oscillator Q Amplitude Ratio (X(ω)/xst) 0 Ω =0, ω 2 = 9.5, µ =0.0 Invariant points P 0 0 s s 2 Frequency ( ω ) (Hz) 0 Fig. 3. Invariant Points of a Den Hartog Oscillator Optimized Amplitude for den Hartog oscillator ζ = Amplitude Ratio (X(ω)/xst) 0 ζ = Ω =0, ω 2 =9.5, µ =0.0 ζ = opt 0 0 Frequency ( ω ) (Hz) 0 Fig. 4. Den Hartog suspension model optimized for a wider range of frequencies 9

10 Optimized Amplitude for den Hartog oscillator Amplitude Ratio (X(ω)/xst) 0 Ω =0, ω 2 =9.5, µ = Frequency ( ω ) (Hz) 0 Fig. 5. Den Hartog suspension model optimized for a wider range of frequencies (4.20) and the damping ratio according to (4.24) (ζ = ), its amplification attains its maximum at the invariant points. This is illustrated along with two non-optimal damping coefficients in Fig. 4. Observe that the optimum amplification factor at the invariant points are about 3.5, which may not be acceptable. Figure 5 illustrates the reduction of the peak amplification factor by about half by increasing the small mass from m 2 = 0.0M to five times (m 2 = 0.05M ). In practice, a suspension system is designed to operate far smaller then the invariant points or sufficiently larger then the invariant point frequencies. Under this scenario the infrequent disturbances close to the invariant points can still be bounded. The two-dof suspension model that we have been studying so far is applicable to machinery shock isolation design. For vehicles subject to road roughness conditions, a modified model is more appropriate as shown in Fig. 6. An analysis of this model for achieving an optimum shock isolation is left for an exercise. 0

11 Two-DOF Vehicle Shock Isolation Model m 2 x C x 2 M K /2 K 2 Ground input: x (t) g Fig. 6. A Two-DOF Vehicle Suspension Model