![振动力学(第二版)](https://wfqqreader-1252317822.image.myqcloud.com/cover/176/31729176/b_31729176.jpg)
2.1 运动方程的建立
【例2.1】 已知弹簧—质量系统如图2.4所示,质量为m,弹簧刚度为k,弹簧原长为l。试确定系统的振动方程。
【解】 图2.4是最简单的单自由度系统。考察弹簧—质量系统沿铅垂方向的自由振动。设x1向下为正,由牛顿第二定律知系统的运动方程为
![](https://epubservercos.yuewen.com/B43255/17180246604477406/epubprivate/OEBPS/Images/img00014001.jpg?sign=1739531691-3ryYGKnF4WbX21kj2TGbIt8MALkPVMOE-0-074613a96b95c87bb1941bc78684daf7)
若设偏离静平衡位置的位移为x,则因x1=x+l+mg/k,故上式变为
![](https://epubservercos.yuewen.com/B43255/17180246604477406/epubprivate/OEBPS/Images/img00014002.jpg?sign=1739531691-WXCxxPbH5witcM6IMIWqtjsKk4eWExlZ-0-2fb403b49ab5e71e89a48a4a0a12aff9)
因此,当像重力一类的不变力作用时,可只考虑偏离系统静平衡位置的位移,那么运动方程中不会再出现重力这类常力,使方程形式变得简洁。现约定,以后若无特别指明,一律以系统稳定的静平衡位置作为运动(或广义)坐标的原点。
【例2.2】 如图2.5所示扭摆,已知扭轴的切变模量为G,极惯性矩为Ip,转动惯量为J,轴长为l。试求扭摆的振动方程。
【解】 如图2.5所示,相对于固定轴x发生扭动,以θ为广义坐标建立系统的转动运动方程。经分析知有两力矩作用在圆盘上,即惯性力矩和恢复力矩
。由动静法原理得
![](https://epubservercos.yuewen.com/B43255/17180246604477406/epubprivate/OEBPS/Images/img00014005.jpg?sign=1739531691-ygBQXO27Kxp2EWGV6bOJwCPHH5quC2iH-0-6d084f97a81d8752dbc52356d5f0bac5)
![](https://epubservercos.yuewen.com/B43255/17180246604477406/epubprivate/OEBPS/Images/img00014006.jpg?sign=1739531691-T5mDRlr7U5rTf6TaYyKWfaI6Zt8i6DLq-0-7a8a1d7e465558ab9a93d6a03a8988d1)
图 2.4
![](https://epubservercos.yuewen.com/B43255/17180246604477406/epubprivate/OEBPS/Images/img00014007.jpg?sign=1739531691-6qYgVXfGkJf0a14t3kk4UVEPISb9MRK0-0-df4b1f79c867ec10884cbdc94abc3cd0)
图 2.5
其中 为轴的扭转刚度,设为kt,故
![](https://epubservercos.yuewen.com/B43255/17180246604477406/epubprivate/OEBPS/Images/img00014009.jpg?sign=1739531691-PN6Oi9Hs05u3R7QTTcU02y9HdMPJe8Lu-0-c51c7e35cad5b1fe650b74c2cc9da508)
【例2.3】 一质量为m的重物附加在简支梁上,系统参数及截面尺寸如图2.6(a)、(b)所示。试将系统简化为单自由度系统,并求其振动方程。
【解】 梁的质量与m相比可略去。弹簧常数k取决于质量m在梁上的位置。对图2.6(a)所示的简支梁,由材料力学得
![](https://epubservercos.yuewen.com/B43255/17180246604477406/epubprivate/OEBPS/Images/img00014010.jpg?sign=1739531691-XrhBMIlv787paeySB7By4gRXtqy2cplo-0-62e80215d1309a5a5ebde4fdda0bad77)
![](https://epubservercos.yuewen.com/B43255/17180246604477406/epubprivate/OEBPS/Images/img00014011.jpg?sign=1739531691-5FB0wOkZi6fOrl8fuqArMumalqn9ptg0-0-e57370603480157f9615be24c5bc83cc)
因矩形横截面惯性矩 ,所以
![](https://epubservercos.yuewen.com/B43255/17180246604477406/epubprivate/OEBPS/Images/img00014013.jpg?sign=1739531691-GJl0qRoB7p8FloydQGIukH76MrKipS8c-0-e520413cabc3b2027cc508c38e23717c)
将带重物的简支梁简化为图2.6(c)所示的相当系统,惯性力与弹性恢复力相平衡,则有
![](https://epubservercos.yuewen.com/B43255/17180246604477406/epubprivate/OEBPS/Images/img00015001.jpg?sign=1739531691-Xu3J5YbqwPpwPfWD8ZN80B5LaiIHmEGj-0-3b3a531712304c89f8a953f3fd2550f9)
![](https://epubservercos.yuewen.com/B43255/17180246604477406/epubprivate/OEBPS/Images/img00015002.jpg?sign=1739531691-llcVmdVwHNZ99WLKnb8GgsquNxiNSY4N-0-dd733b661ae85e7b3908b89e873c1556)
如果梁的两端不是简支,则Δ应有不同数值。
![](https://epubservercos.yuewen.com/B43255/17180246604477406/epubprivate/OEBPS/Images/img00015003.jpg?sign=1739531691-BgYbLajHGlTTALWl8WxVjOAm99xJS8KQ-0-7adeeff31b7aed1a1525b6883a320466)
图 2.6
【例2.4】 如图2.7所示系统,相关参数已在图上标出。试求系统的振动方程。
【解】 求解时可以选择任意坐标x1、x2、θ作为变量,但它们相互关联,只有一个是独立的。现取绕固定轴O的转角θ为独立坐标,则等效转动惯量
![](https://epubservercos.yuewen.com/B43255/17180246604477406/epubprivate/OEBPS/Images/img00015004.jpg?sign=1739531691-TyUHd5p6b5DW3guK0l7t6A547UvtU06t-0-604a74828b7d2bd5441341159454d338)
图 2.7
Jc=m1a2+m2b2+m3r2
其中,r是m3的惯性半径。系统的等效角刚度
ke=k1a2+k2b2+k3c2
![](https://epubservercos.yuewen.com/B43255/17180246604477406/epubprivate/OEBPS/Images/img00015005.jpg?sign=1739531691-k1AJ7FQKW3K2NDj04J6VGfYJoopWU5Ka-0-e54bedee586e32351d98f41d468444cf)
![](https://epubservercos.yuewen.com/B43255/17180246604477406/epubprivate/OEBPS/Images/img00015006.jpg?sign=1739531691-lA1fKR7pqb4sBYdIQ6T2Yzk3BB7G699N-0-bb27b99ee78351c373aa5863d1e2a5d8)
![](https://epubservercos.yuewen.com/B43255/17180246604477406/epubprivate/OEBPS/Images/img00015007.jpg?sign=1739531691-zYCDydM3yojLjPJswb3wFhYxNJ0mXbCc-0-2d8bd5ef58f40ed1ab9820c735f68ef8)
可以取x1为独立坐标,于是
![](https://epubservercos.yuewen.com/B43255/17180246604477406/epubprivate/OEBPS/Images/img00015008.jpg?sign=1739531691-5ori2lVK3a7iEq2TWNmrWZCMoAAeziCg-0-6dc2b3a1a5c2e8ea95811557061fbd71)
![](https://epubservercos.yuewen.com/B43255/17180246604477406/epubprivate/OEBPS/Images/img00015009.jpg?sign=1739531691-QjGMWslllVKdtLbHDLMgL5h7KhaWbwtw-0-e51bd328a8e89c3d212b406efbf2dcf1)
![](https://epubservercos.yuewen.com/B43255/17180246604477406/epubprivate/OEBPS/Images/img00015010.jpg?sign=1739531691-SQWibXCrfiwkIc4cw0DIDPFzTGLh0gRA-0-a478f3d29616ba34fc14c9666555fa67)
经推导可得系统运动方程
![](https://epubservercos.yuewen.com/B43255/17180246604477406/epubprivate/OEBPS/Images/img00015011.jpg?sign=1739531691-psQaxzWN6SZSRHMVkp6bvAM6hoi3jEbh-0-6db39bc1d774730a8e9b1d099844c862)
同理以x2为独立坐标,可得
![](https://epubservercos.yuewen.com/B43255/17180246604477406/epubprivate/OEBPS/Images/img00015012.jpg?sign=1739531691-ql04tCAYiiYvazrX1CcF8pZakMY8RGkn-0-f04310d994401a531f7fdb91a69022b6)
![](https://epubservercos.yuewen.com/B43255/17180246604477406/epubprivate/OEBPS/Images/img00015013.jpg?sign=1739531691-zXdodUUVgX5Gh1idebZZk2wXYLePoWHc-0-ca88994945b8b23630028deabca77f85)
不难验证A=B=C。
可见,对结构较复杂的单自由度系统(其中有些元件作平移,另一些作转动),不管选择哪一个坐标变量作为独立坐标,其振动方程形式不变。这说明系统固有振动规律与坐标选择无关。