P139, 3-1: 求图示摆的柔度系数.
d 11
d 21 d 31
1
d11 = d 21 = d31 =
l1 (m1 + m2 + m3 ) g
d 22
1
d 22 = d32 =
d 32
l1 l2 + (m1 + m2 + m3 ) g (m2 + m3 ) g
d 33
1
d 22 = d32 =
l l1 l2 + + 3 (m1 + m2 + m3 ) g (m2 + m3 ) g m3 g
m1l 2 3 M = 0
0 2 7 m2l 48
9l 2 k1 16 K = 9l 2 k1 16
l 2k K= 16
9l 2 k1 16 2 2 9l k1 l k2 + 16 4
ml 2 M= 3
0 1 0 7 /16
9 9 9 13
r
n
r
Φ CΦ = diag[Mi ] ∑αr diag[ωi2 ] 1≤i≤N r =0 1≤i≤N
T
n
r
= Φ diag[Mi ] ∑αr diag[ωi2 ] Φ1 1≤i≤N r =0 1≤i≤N
T
n
r
对角矩阵
P140,3-12 : 考察无刚体运动自由度的比例阻尼系统,证明计算其响应的模态加速度法为 u(t ) = K f (t ) ∑
m 0 u1 3k 0 m u + k 2 k u1 2ku0 = 3k u2 0
2k 4k ω1 = , ω2 = m m
u1 1 1 q1 u = 1 1 q 2 2
1 1 φ1 = , φ2 = 1 1
1 1 m 0 1 1 q1 1 1 3k 1 1 0 m 1 1 q + 1 1 k 2
=0
系统的固有频率: ω1 = 0
ω2 = ω3 =
k m
ω4 =
( M + 3m)k mM
2 容易确定ω12和ω4的特征矢量:
1 1 φ1 = 1 1
3m M 1 φ4 = 1 1
对于ω2 = ω3 =
k , 有 m
1 1 1 0 0 0 φi = 0,(i = 2,3) 0 0 0 0 0 0
m2
k1 + k2 K = k2 0
m M = 2m m
ω1 =
k m
ω2 =
2k m
ω3 =
3k m
1 φ1 = 1 1
1 φ2 = 0 1
1 φ2 = 1 1
P140,3-4: 图示带集中质量的自由梁是飞机的最简单的模型,梁的抗弯刚度EI , 质量不计, 集中质量的比值为 = m / M. 求系统的固有频率和固有振型.
ω2 = ω3 =
3EI = ml 3EI Ml
3EI (2m + M ) 3EI (2 + 1) = Mml Ml
1 φ2 = 0 1 1 φ3 = 2 1
1 φ1 = 1 1
P140,3-5: 图示系统中各质量只能沿ui,i = 1, 4方向运动, 试分析其固有模态.
0 u1 k m u1 c c u + c 3c 2c u + k 3m 2 2 m u3 0 2c 2c u3 0 k 3k 2k 0 u1 0 2k u2 = 0 2k u3 f 0
k 2k ω1 = 0, ω2 = , ω3 = m m
模态坐标系下的运动方程:
1 1 1 φ1 = 1 , φ2 = 0 , φ3 = 1 1 1/ 2 1
1 1 q1 u1 1 u = 1 0 1 q2 2 u3 1 1/ 2 1 q3
P139,3-2: 求图示系统的的刚度矩阵和柔度矩阵, 并求m1 = m2 = m, k1 = k2 = k时系统的固有频率.
1 m1l 2 l 2 2 1 m2l 2 l T= ( + m1 ( ) )θ1 + ( + m2 ( ) 2 )θ 2 2 2 12 2 2 12 4
U=
1 3l 3l 1 l k1 ( θ1 θ 2 ) 2 + k2 ( θ 2 ) 2 2 4 4 2 2
q1 (0) q1 (0) 0 q (0) = q (0) = 0 2 2 q3 (0) q3 (0) 0
0 0 q1 0 0 0 q1 1 6m q1 0 q + 0 3c / 2 0 q + 0 3k / 2 0 q = 1/ 2 f 3m / 2 2 2 2 0 6m q3 0 0 12c q3 0 0 12k q3 1
13 = 0 23 + 33 + 43 = 0
正交条件:φ3T M φ2 = 0
223 +33 +43 = 0
0 0 φ3 = 1 1
1 1 φ1 = 1 1
0 2 φ2 = 1 1
0 0 φ3 = 1 1
3m M φ4 = 1 1 1
q1 (t ) = f0 2 t 12m
t 0
q2 (t ) = ∫
f 0 1 ζ 2ω2 (t τ ) e sin ωd 2 (t τ )dτ 3m ωd 2
f0 e ζ 2ω2 t = 1 cos(ωd 2t 2 ) 2 3k 1 ζ 2
ωd 2
ζ c 1 2 = 1 ζ ω2 , ζ 2 = , 2 = tan 1 ζ 2 2mω2 2
r r
n
Kφr = ω r2 M φr , ( r = 1... N )
M 1 Kφr = ωr2 φr , (r = 1...N )
M 1 KΦ = Φ diag[ωi2 ]
1≤i ≤ N
其中:Φ = [φ1 φN ]
M 1 K = Φ diag[ωi2 ]ห้องสมุดไป่ตู้ 1
1≤i ≤ N
代入
C = M ∑αr ( M K ) = M ∑αr (Φ diag[ωi2 ]Φ1 )r
m 0 u1 3k 0 m u + k 2
2
k u1 2ku0 sin ωt = 3k u2 0 3k ω 2 m k
3k ω 2 m K ω M = k
3k 3k ω 2 m H11 (ω ) = (ω )
H 21 (ω ) =
r =1 r =1
n
n
n
1
f ( t ) K M ∑ φ r q r ( t ) K 1C ∑ φ r q r ( t )
1 r =1 r =1
1 r T
n
K
M
∑ φ q ( t ) = Φ diag[ K
r =1 r r 1≤ r ≤ N 1≤ r ≤ N
]Φ M
∑ φ q (t )
1 r =1 n
2ζ r
ωr
φr qr (t ) ∑
n
1
2 r =1 ωr
φr qr (t )
证明: Mu(t ) + Cu(t ) + Ku(t ) = f (t )
u(t ) = K f (t ) K Mu(t ) K Cu(t )
u (t ) = K
1 n
1
1
1
模态截断:u(t ) = ∑ φr qr (t ), u(t ) = ∑ φr qr (t )
P140,3-7: 图示系统左端基础作简谐激励u0 (t ) = u0 sin ωt , 试求两集中质量的稳态位移响应并讨论其 反共振现象.
mu1 (t ) = k (u1 (t ) u2 (t )) 2k (u1 (t ) u0 (t )) mu2 (t ) = 2ku2 (t ) k (u2 (t ) u1 (t ))
T T
k 1 1 q1 1 1 2ku0 = 3k 1 1 q2 1 1 0
T
1 0 q1 ω12 0 q1 ku0 / m = 0 1 q + 2 2 0 ω2 q2 ku0 / m
q1 (t ) = ∫ ku0 1 u sin ω1 (t τ )dτ = 0 (1 cos ω1t ) 0 m ω 2 1
M 3 m k ( K M )φi = k 1 m 1 1
任取32 = 42 = 1
1i = 0 2i +3i + 4i = 0 (i = 2,3)
正交条件:φiT M φ1 = φiT M φ4 = 0,(i = 2,3)
2i +3i +4i = 0, (i = 2,3)
0 2 φ2 = 1 1
系统的动能:T = 系统的势能:U =
m M M = m
1 1 1 2 2 mu12 + Mu2 + mu3 2 2 2 1 3EI 1 3EI (u3 u2 ) 2 + (u1 u2 ) 2 2 l 2 l
1 1 0 3EI K= 1 2 1 l 0 1 1
ω1 = 0
k (ω )
u1 (t ) H11 (ω ) u (t ) = H (ω ) 2ku0 sin ωt 2 21
ω=
3k 为反共振频率 m
P140,3-9: 图示系统初始静止,求左端基础产生阶跃位移u0后系统的响应.
mu1 (t ) = k (u1 (t ) u2 (t )) 2k (u1 (t ) u0 (t )) mu2 (t ) = 2ku2 (t ) k (u2 (t ) u1 (t ))
r =1 r r
n
= Φ diag[ K r 1 ][ M 1 q1 ( t ) M n q n ( t ) 0 0]T = Φ [ q1 ( t ) / ω q n ( t ) / ω
| K ω 2 M |= 0