机械振动与噪声学答案

1、 cLLLmLmLm232433222211)(0)(22243gLmkLL 111312122221rmgmfrbakbafrkf1121122grmrkbak2222113rrmJrffrf 0221222222211rkbarkmrrmJ f2f1f321cosk222a kb22221ecosbkakk 22323211()()3311(1)()33mmJLnLLnLnLnLLLJmnLmnL222213JmLmnLmn L3231( 1)3JmLnnore211(1)3mmnn222222222222222221311122213eeeJmLmnLmn LnLxJm xm n LmL

2、mnLmn LJmn Ln Lf1f212fk R21fk xR12fk R21k R Rk xRRJ21fk xR21210Jk Rxkk R21fk xR1k xRmx 110mxk xk R00)(002121110 xkkRkRkRkxJm 110mxk xk R21210Jk Rxkk R211111212100 xxccccxxmm 00214333321xxkkkkkkk002112100221221xxLmTxx 2112TL1122221200120103xxm LxxT 21123LT00002123222222212121rkkrkrkrkkJJ 12, ,k r2J1

3、,-1 1, ,3 k r2J1 , 1 100002123222222212121rkkrkrkrkkJJ 12, ,k r2J1 , 1 1, ,3 k r2J1 ,-1 1)114141(4214321nkkkkmf详细推导详细推导1122123344141412212xkxkxxxkxk1122123344141412212xkxkxxxkxk12341111161644exkkkktmmkkgmtmmkkmmhgmtx21221212cossin)(2)(tctctxn2n1sincos)(n02/xc01xc 0.15 99.81.89 mm7000fFNmgkkk xx00250

4、mm, mm5674. .xx00250mm, 00nnn0() cossin0 xxxxttx 00nnnnn0()sincos0 xxxxttx 0220nnnnn00nnn0()cossin000()cossinxxxxttxxxxxtt xx00250mm, max0n7000()250.1899xx s/mm5644max. .x 00nnnnn0()sincos0 xxxxttx 4210.8 mm/snmm461. .xxx00250mm, mm5674. .0n() cosxxt 0nn()sinxxt n0,xt0()xx xx00250mm, )0(1)0(20)0()0(

5、21Jk22221Jk22222 2211utt2121cos207. 0cos207. 12211 1111u 111121mu)()(2121212121xxxxyy := Mm00m := K2 kkk2 kkm3. km := 1 := -1 0 xxMH EILEILEILEILH2232234002mRmM1.702053777EIL3m16.28198431EIL3m := K12 EIL36 EIL26 EIL24 EIL详细推导详细推导00211210022121xxmLTxx ,3232TL m3232TL m212.366025404 Tm L0.6339745960 T

6、m LRayleigh := X1, 1 12 T3 m L9 T14 m LRayleigh := X1, 1 12 T3 m L9 T14 m L4 m L3 Tm L3 T2 m L3 T2 m L3 TDunkerly T2 m L0.6339745960Tm LRayleigh 2 T3 m L9 T14 m LDunkerly T2 m L1.05156684711R12R1D1.0140108880.78867513500kxyxcxm 21tan2mkc222)()(cmkcAXtAckxxcxmcos xtsakAamgLkcamLco)(222 0cos ()FtknLgm

7、Lka22n)(222mgLkamLca212arctan2222cos ()()(1)(2)k A atkamgLi20eei tmxkxFi()0( )etx tX2i2i000eemXkXF2i00cos 2sin 2ekmikXF00222cos 2sin 2FXkmk2sin 2arctancos 2kkm tLvkYkxxgW2sin 0Wxk xyg2sinvyYtL0nn( )sinsinFx tttkWkgLv2k g L24 2v2W0( )x t2 sinkg tWk Yg L vWk g L24 2v2Wk Y g L2sin2 v tLk g L24 2v2WtLvk

8、YkxxgW2sin WkgLv2ktFx)/cos1 (n0mknplot(subs(F0=1,m=1,omegan=2,X1),t=0.20)1m nd( )F ()sin n()t ttmFtx0nn0dsintkFttkFn0n0cos10cos 00n0nsindtFx ttm000nn0ncoscoscos0tFFttttkk2 F0sin12nt0kF0t0()sin ntm nktFn 002121122222223223i()iii()kkmcckcZkckkmcc1221221111022322322220cos00ccckkkmxxxFtccckkkmxxx 102cos

9、Re0 xFtHx 121121122222223223i()iii()kkmcckcHZkckkmcck2k32m2 c2I c3Ik2 c3I3c1m2I c1k2I c2k3Ik1 c3I()(3m1c2I3c2m2I2c1c22c1c33m1c3I c1k3I2c2c3k12m2k1k2k1k3k2k3k1 c2Ik22m22m1k22m1k34m1m2) 1,1H121212,mmmccckkk1110222022cos0220 xxxmcckkFtxxxmcckk 102cosRe0 xFtHx0.12ccccmk22d1119.9 rad/snkm220.6281 n2d2211

10、ee1.88nnxx02020. .0032150. .rad/s7244d. .Ns/m441. .c39.8143.94/5.08 10ngrad s1200 /302.86,043.942sinMxkxmet该系统振动微分方程为该系统振动微分方程为22220.23082.860.579()453.4 1 2.861meXmmM223453.4 43.940.579 10507NTnFkXMX222201212TFSF或或2112202211212000.2308507N11 2.8660TFSFme22220.23082.860.579()453.4 1 2.861meXmmM(a) n 23.9 rpm2n222212120.2S(b) n = 43.45 rpm 210.151S2.771600 260.51/2.7760 2.77nrad s229812.667()60.51stngmm1122122220sin00 xkkkxMtxkkxm211222222sin0 xtkkMkxkkm2222212122122222221222sin0() ()() sin() ()kmktkkkMXkkMkmkmtkkMkm