Dynamics of Aerospace Structures
Dynamics of Aerospace Structures ASEN 5022
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Date Created: 10/30/15
15 2DOF Modeling of Shock Absorbers 15 3 151 INVARIANT POINTS OF THE DEN HARTOG ORMONDROYD MODEL Substituting these into 151 yields the coupled equations in the frequency domain as lez K1 k2 icwX1 k2 jcwX2 2 F1 2 154 kg jcwX1 m2w k2 jcwX2 0 X1 obtained from the above equation reads X1w 771sz 162 1060 15 5 F1 111102 KIM 7712602 162 7712602162 jwclK1 M1 7712le I Let us parameterize the model as 21VK1M w2vk2m2 C2 m2 217Mm2M17 156 sZwle xSIZFlKla fw2Q1Cc2m2S21 The ratio of X1 0 to x is the ampli cation factor from the static to the dynamic response which can be expressed as X1a 2302 s2 f22 1 JCSI 2 s2s2 1 Mszgt2 Mfzsz s2 1s2 f22 12 157 15 3 15 5 152 SHOCK ABSORBERS 152 SHOCK ABSORBERS A good shock absorber should have the following characteristics 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 1521 Determination of Undamped Natural Frequencies The natural frequencies are determined by setting 0 0 from the denominator of 157 MQ4 mf n0 it 158 Uam W 0 The two distinct roots of this equation will be designated as s12 and s22 1522 Determination of Invariant Points Since the two invariant points would play a key role it is necessary to nd 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 magni cation factor 157 as 15 5 Chapter 15 2 DOF MODELING OF SHOCK ABSORBERS 15 6 mZHltSCMZW1szwQl x3 7 7 7 2 s2CD 7 A21 159 Bs2f22 C s2 1 1 1s22 D W2 2 s2 1s2 f22 The dampingindependency of the two invariant points implies that at the invariant points the magni cation factor satis es A C B D iL 1510 AD 2 BC for which the ampli cation factor becomes X A 1 Hs c f M 2 1511 JCS C The equation for determining the invariant points 1510 yields two equations 1 s2 1 usz 2 2 2 2 2 2 2 s f Mfs s 1s f 1512 1 s2 1 usz s2 f2 Hf2s2 s2 1S2 f2l The rst of the above relation yields the trivial solution viz s 2609120 1513 15 6 15 7 152 SHOCK ABSORBERS a static case which is not of interest for the present discussion The second of 1512 can be rearranged as 21s4 21f2uf2s22f2 0 1514 The two invariant points s1 and s2 can be shown to satisfy s2 lt L lt s2 1515 1 1 M 2 and the corresponding ampli cation factors are given by XIW 4 1516 xSl Ms2152 1523 Determination of Ampli cation Factor at the Invariant Points The foregoing analysis now supplies the coordinates of the two invariant points as P X17y131 Q x27y2 32 1517 ll 1M312 ll 1MS In order to see succinctly the magni cation factor we arrange 1514 as 1 1 2 2 1 M 22 2Mz 120 22 1518 2M 2M s21M 1 so that the product of the two roots of the above equation 21 gt1lt 22 is given by 2 2 2122 1 1519 M 15 7 Chapter 15 2 DOF MODELING OF SHOCK ABSORBERS 15 8 which is independent of the frequency ratio f 12 21 It should be noted that for a shock absorber the magni cation factor should remain small for a wide range of frequencies which can be realized if we set the ampli cation 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 coef cient associated with zterms in 1518 vanishes a 1 1 1M2f20 fQ 21 um2M1 1520 1 M The ampli cation factor X 1 w xst for this particular choice at the two invariant points becomes X10 1 2 2 2 1 1521 xsl 391 1 Hs12 M 1524 Determination of Absorber Damping In the previous section we imposed the condition on the ampli cation factor at the two invariant points to be the same However it is possible that the ampli cation factor may be larger at other exciting frequencies Of several possible choices we impose that the maximum ampli cation 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 159 with respect to s 0 521 set the resulting expression to zero 3 X lt gt 1a O 1522 as x3 15 8 15 9 for the two invariant points obtained from 1514 Fors s1 2 2 Cc Fors 2 s2 2 52 cc 152 SHOCK ABSORBERS 1 3 M 81 103 1 M 81 103 1523 In practice we must have one darnping value for all the frequency range To this end we take the average value to be 3M 81 103 1524 1525 Applications First in order to utilize Matlab capability we express 155 in its transfer function forrn X1 s quot12s2 k2 CS 15 25 F1 111132 K0071st k2 771232162 SCK1 M1 m2SZl I where s is now the Laplace Transform variable The preceding equation can be arranged to read X1S 2 61032 6113 02 JCS 1 b0S4 b1S3 l72S2 b3S 74 2 1526 6101 012291 0202 170 1 b121M 91 172 15 9 911Mw b3 252 Chapter 15 2 DOF MODELING OF SHOCK ABSORBERS we express Ci from 1523 as c 2 B y2D 0 6 y2C A A 4s2 B 2 s2 f22 C 4s2s2 1 usz2 Dw f w Dw w where yuaz is the magnitude of the two invariant points P Q 1 27 Q x27 1 2 s27 1 1Ls1 Px1y1317 m 2 and s1 and s2 are the two roots of am 2afkwdefz 0 u 1 1 2 a 2Mw h aa w bv4 am74 Observe that at s12 and s3 we have 2 B 2 2 y B gt yC AB yDO which means we have from 1528 c2B fD 0 KR mm 15 14 15 14 5 5 0mm 0mm 0mm 15 15 152 SHOCK ABSORBERS Hence we can apply L Hopital s rule to determine 52 B y2D Cc yZC A 8A 35172 A m etc 1533 If one used 1522 Ci2 would have to be computed by seeking the following quadratic equation AC AC 54 AD BC AD BC 2 31 BB 2 0 1534 CC Cc We now apply 1533 to compute the optimum damping ratio for the den Hartog oscillator First as we set the magnitude ofy X1sxst to be the same at the two inavriant points Viz ys12 ys we have from 1529 1 11s1211s 1 1535 Substituting s12 and s3 from 1530 yields Za1l f2 f2lt gt2 f 1 1Ms121 1M2i a 1536 11 mm X1 somam 27 A Mathematica program to compute the optimum damping ratio is given below 15 15 Chapter 15 2 DOF MODELING OF SHOCK ABSORBERS 15 16 compute the optimum damping ratio for den Hartog shock isolation system Clear12 h2 pi pr mu 1 X squot2 in the lecture notes four quantities A1 4X B1 X f2quot2 C1 4X 1 mux 1quot2 D1 mu12X x 1x f2 2 s1 root values pr 1 1 mu122 mu pi Sqrt 1 2f2 1 muquot212 22 mu A1d DAl X Differentiate A by X squot2 etc B1d DB1 x C1d DC1 x D1d DD1 x parameter set h2 1 2mu optimum magnituded squared prv Simplify pr 12 gt 11 muquot2 piv Simplify pi 12 gt 11 muquot2 p Simplifyprv piv A1d SimplifyA1d f2 gt 11 mu 2 B1d SimplifyB1d f2 gt 11 mu 2 C1d SimplifyC1d f2 gt 11 mu 2 D1d SimplifyD1d f2 gt 11 mu 2 nomin Simplify B1 h2D1 x gt p denom Simplifyh2C1 A1 x gt p damping Simplifyvalnominvaldenom ClearAll 15 16 15717 The results reads 152 SHOCK ABSORBERS smph eauem after canymg outhand smph eauons 1Lbecomes 3 J1 27 M Z saw 3 7 3 1 Fi c5 H 81 m3 1115 the avaage ofthe precedmg two wh1ch we have used 10 obtam For x pomL a s1m1lar computahon leads 27 311 Z Wow wh1ch 1 the same aipresswn obmmed m 15 24 1517 15 37 15 38 15 39