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Linear vibration of a coupled string–rigid bar system
Journal of Sound and Vibration (1995) 183(3), 383–399
LINEAR VIBRATION OF A COUPLED STRING–RIGID BAR SYSTEM B. Y Department of Mechanical Engineering, University of Southern California, Los Angeles, California 90089-1453, U.S.A. (Received 30 August 1993, and in final form 11 April 1994) Linear vibration of a coupled string – rigid bar system is investigated. Owing to dynamic interaction between the strings and the rigid bar, the coupled system is not self-adjoint in its original equations of motion. Consequently, the classical modal analysis cannot be applied. This study develops an augmented formulation for the eigenvalue and boundary value problems of the coupled system. In this formulation, orthogonal eigenfunctions are obtained, and the general response of the coupled string–bar system is represented by an eigenfunction series. Green’s function formula for the coupled system is derived by two methods: modal analysis and Laplace transform. The analytical predictions about the system eigensolutions and response are illustrated in numerical simulation.
1. INTRODUCTION
Coupled flexible–rigid body systems that are assemblages of multiple flexible and rigid bodies are commonly seen in structures and machines. Applications are diverse, including vehicles, airplanes, spacecraft, robots, computer storage systems, rotating machinery, highway structures, buildings, bridges and pipelines conveying fluids. Accurate modelling is essential to analysis, control and optimal design of such systems. Flexible–rigid body coupling mechanisms can be divided into two classes: (1) type I, a flexible body is joined to several rigid bodies at some interior points (see Figure 1(a))—in many cases, the rigid bodies can be idealized as dimensionless, lumped masses; (2) type II, a number of flexible bodies are interconnected to rigid bodies in some regions or at some points of their boundaries (see Figure 1(b))—the sizes of the rigid bodies are always considered. The dynamic coupling of flexible and rigid bodies renders the system eigenfunctions non-orthogonal in the original equations of motion. As a result, the classical modal analysis for conventional flexible systems is not applicable to coupled flexible–rigid body systems. Dynamics of type I systems have been extensively studied in the literature; for instance, see References [1–7] and the papers cited therein. Previous research on type II systems focused on finite element analysis and other approximate methods; analytical prediction of the system response to arbitrary external and initial excitations is rarely addressed. Recently, in a vibration analysis of stepped beams and plates, Lee and Bergman [8] developed a dynamic flexibility method, which represents the interconnecting forces between the beam or plate segments by Green’s functions of the flexible subsystems, and delta functions [5, 6]. This method yields analytical solutions of various dynamic problems for stepped beams and plates, and can be directly applied to type II coupled systems. The current work is concerned with an alternative approach for the analysis of type II coupled systems. The linear vibration of a coupled string–rigid bar system, which is shown 383 0022–460X/95/230383 + 17 $08.00/0
7 1995 Academic Press Limited
384
.
Figure 1. Two types of flexible–rigid body coupling mechanisms: (a) type I—a flexible body joined to rigid bodies at its interior points; (b) type II—flexible bodies interconnected with rigid bodies at their boundaries.
in Figure 2, is investigated. The coupled system is probably the simplest type II system, yet its dynamic characteristics are representative of generic systems. The coupled system itself is a model for certain structural systems such as the secondary mirror of a telescope suspended by tensioned vanes, and cable-supporting systems for bridges and television towers. A self-adjoint formulation for the eigenvalue and boundary value problems of the coupled string–bar system is developed. In this formulation, orthogonal eigenfunctions are obtained, and the response of the coupled system to arbitrary external and initial excitations is represented by eigenfunction expansion. The Green’s function formula for the coupled system is derived by two methods: modal analysis and Laplace transform. The analytical predictions about the system eigensolutions and response are illustrated in several numerical examples.
2. EQUATIONS OF MOTION
In Figure 2, a rigid bar of length 2a is attached to two uniform strings of length l1 and l2 , respectively. The rigid bar has mass m and moment of inertia I. The strings have the
Figure 2. A schematic of a coupled string–rigid bar system.
–
385
same linear density r, and are under tension T. The strings are subject to the external force f(x, t); the rigid bar is subject to the force F0 (t) and torque t0 (t) applied at its center of mass. Let the left and right strings be defined in the regions V− = {x=−a − l1 Q x Q −a} and V+ = {x=a Q x Q a + l2 }, respectively. The transverse displacement w(x, t) of the strings is governed by the differential equation rwtt (x, t) − Twxx (x, t) = f(x, t)
x $ V = V − * V+ ,
(1)
with the boundary conditions w(−a − l1 , t) = 0,
w(a + l2 , t) = 0,
(2)
where ( )t = 1( )/1t and ( )x = 1( )/1x. According to the free body diagram in Figure 3, the equations of motion of the rigid bar are my¨ = T(sin u+ − sin u− ) + F0 ,
Iu = Ta [sin(u+ − u) + sin (u− − u)] + t0 ,
(3)
where the overdot denotes temporal differentiation, y(t) is the vertical displacement of the mass center and u(t) is the rotational displacement of the bar. Note that y(t) = 12 [w(a, t) + w(−a, t)],
(4)
and for small oscillations sin u21u2 1 wx (2a, t),
sin (u2−u) 1 u2−u,
u(t) 1 (1/2a)[w(a, t) − w(−a, t)].
Inserting the above in equation (3) and rearranging yields the matching conditions for the strings awtt (−a, t) + bwtt (a, t) = −Twx (−a, t) − (T/2a)[w(−a, t) − w(a, t)] + F− (t), bwtt (−a, t) + awtt (a, t) = Twx (a, t) + (T/2a)[w(−a, t) − w(a, t)] + F+ (t),
(5)
where a = 14 (m + I/a 2), F− (t) = 12 [F0 (t) − t0 (t)/a],
b = 14 (m − I/a 2), F+ (t) = 12 [F0 (t) + t0 (t)/a].
(6)
Note that b is always positive because for any rigid body ma 2 q I. The above equations of motion can also be derived using the generalized Hamilton principle
g
t2
(dKE − dPE + dW) dt = 0.
t1
Figure 3. A free body diagram of the rigid bar.
.
386
Here, for the coupled string–bar system, the kinetic energy, the potential energy and the virtual work done by the external forces are given by KE =
g
1 2
rwt2 dx + 12 my˙ 2 + 12 Iu 2,
PE =
V
g
1 2
Twx2 dx + aTu 2,
V
dW =
g
fdw dx + F0 dy + t0 du,
V
respectively. Assume that the initial conditions of the strings are wt (x, 0) = n0 (x),
w(x, 0) = u0 (x),
x $ V.
(7)
The initial conditions of the rigid bar are dependent on the given functions u0 (x) and n0 (x); i.e., y(0) = 12 [u0 (a) + u0 (−a)], y˙ (0) = 12 [n0 (a) + n0 (−a)], u(0) = (1/2a)(u0 (a) − u0 (−a)),
u (0) = (1/2a)(n0 (a) − n0 (−a)).
Equations (1), (2), (5) and (7) form a boundary value problem for the coupled string–bar system. The fundamental problem is to determine the system response w(x, t) for given external and initial disturbances. In what follows, the exact solution will be derived in terms of eigenfunction expansion. 3. EIGENVALUE PROBLEM
3.1. Letting w(x, t) = n(x) eivt, i = z−1, and f(x, t), F0 (t), t0 (t) = 0 in the equations of motion leads to the eigenvalue problem for the coupled system: n0(x) + g 2n(x) = 0, n(−a − l1 ) = 0,
g = vzr/T,
x $ V,
(8a)
w(a + l2 ) = 0,
(8b)
v 2(an(−a) + bn(a)) = Tn'(−a) + (T/2a)(n(−a) − n(a)),
(8c)
v 2(bn(−a) + an(a)) = −Tn'(a) − (T/2a)(n(−a) − n(a)),
(8d)
where n' = dn/dx and v is the natural frequency. Let the eigenfunction be in the form n(x) =
6
A sin g(x + a + l1 ), B sin g(−x + a + l2 ),
x $ V− , x $ V+ ,
(9)
satisfying equations (8a) and (8b), where A and B are constants. Substituting equation (9) into equations (8c) and (8d) gives
$
mg sin gl1 − 2r cos gl1 (Ig 2 − 2ra) sin gl1 − 2ra 2g cos gl1
%0 1
mg sin gl2 − 2r cos gl2 −(Ig 2 − 2ra) sin gl2 + 2ra 2g cos gl2
A = 0. (10) B
The characteristic equation is obtained by letting the determinant of the matrix in equation (10) vanish. Solution of the characteristic equation gives the natural frequencies of the coupled string–bar system. With the known natural frequencies, the associate eigenfunctions (mode shapes) can be determined using equations (9) and (10).
–
387
3.2. When the two strings in Figure 2, have the same length, i.e., l1 = l2 = l, equation (10) becomes two decoupled equations (mg sin gl − 2r cos gl)(A + B) = 0,
(11a)
((Ig 2 − 2ra) sin gl − 2ra 2g cos gl)(A − B) = 0.
(11b)
This implies that the translation and rotation of the rigid bar in each mode of vibration are not coupled, which is discussed as follows. 3.2.1. Translational modes Every point of the rigid body has the same transverse motion; i.e., A = B. By equation (11), the characteristic equation is tan m = 1/(hM m),
m = gl,
(12)
where the mass ratio hM = m/(2rl). The natural frequencies of translational modes are vkr =
mkt zT/r, l
k = 1, 2, . . . ,
(13)
where mkt are the roots of equation (12). The associate mode shapes are given by equation (9) with A = B. 3.2.2. Rotational modes The rigid bar rotates about its mass center which does not move; i.e., A = −B. By equation (11), the characteristic equation becomes tan m = m/[hI m 2 − (l/a)],
m = gl,
(14)
where the mass ratio hI = I/(2rla 2). The natural frequencies of rotational modes are vkr = (mkr /l)zT/r,
k = 1, 2, . . . ,
(15)
where m are the roots of equation (14). The associate mode shapes are given by equation (9) with A = −B. The vkt and vkr shall be called the translational and rotational frequencies, respectively. The system eigenvalues are the intersections of tan m and functions 1/hM m (for mkt ) and m/(hI m 2 − l/a) (for mkr ); see Figure 4. Because ma 2 q I, the two mass ratios satisfy the inequality h M q hI . (16) r k
Figure 4. Frequency interlacing: w, mkt = vkt lzr/T, eigenvalues of translational modes; W, mkr = vkr lzr/T, eigenvalues of rotational modes.
388 . The following observations about the natural frequencies of the coupled string–bar system can be drawn from Figure 4; first, the translational (rotational) frequencies decrease as the mass ratio hM (hI ) increases; second, the translational frequencies are interlaced with the rotational frequencies; i.e., k = 1, 2, . . . ; (17) vkt Q vkr Q vkt + 1 , and third by equation (17), the coupled string–bar system always has distinctive natural frequencies. However, when the mode number k is very large, vkt , vkr 1 kp(1/l)zT/r, indicating that translational and rotational modes tend to be coupled at high frequencies. 4. ORTHOGONALITY OF EIGENFUNCTIONS
The dynamic coupling between the strings and rigid bar renders the eigenvalue problem (8) non-self adjoint; the eigenfunctions given in equations (9) are not orthogonal, i.e., fV nj (x)nk (x) dx $ 0 for j $ k. As a result, the classical modal analysis cannot be applied to determine the response of the coupled system. In this section, an augmentation formalism proposed in References [9] and [10] is generalized to define an equivalent eigenvalue problem which admits orthogonal eigenfunctions. Consider a function space H = H(V)$R 2, where H(V) is a Hilbert space, and V is the domain for the strings (see equation (1)). Any element U of H has the form U = (u(x) u1 u2 )T, where u(x) is a function defined on V, and u1 and u2 are two scalars. The inner product on H is U, V =
g
U, V $ H.
u(x)v(x) dx + u1 v1 + u2 v2 ,
(18)
V
Define the augmented operators M=
K −T 1 2 1x G G 0 K=G G G k 0 2
&
'
r=x $ V 0 0 0 a=x = −a b=x = a , 0 b=x = −a a=x = a
b
0
0
x$V
0
T
1 1 + 1x 2a −
T 2a
b
1b
− x = −a
x = −a
0
T 2a
b
x=a
1 1 + 1x 2a
T −
1b
L G G G, G G l
(19)
x=a
where the parameters a and b are given in equation (6). The domain of the operators is a subspace of H: D 0 {U = U = (u(x) u(−a) u(a))T, u $ H(V); u(−a − l1 ) = u(a + l2 ) = 0}. (20) By equations (19) and (20), the original eigenvalue problem (8) is equivalent to v 2M[V] = K[V], V $ D. (21) The eigensolutions of equation (21) are represented by those of the original problem (equation (8)); namely, vk = gk zT/r, Vk (x) = [vk (x) vk (−a) vk (−a)T, k = 1, 2, . . . , (22) where gk are roots of the characteristic equation deduced from equation (10), and vk (x) are the scalar eigenfunctions given in equation (9).
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389
It can be shown that both M and K are self-adjoint, and positive definite; namely, for any elements U, V $ D, M[U], V = U, M[V], K[U], V = U, K[V]
(23a)
and M[U], U e 0,
K[U], U e 0,
(23b)
where the equalities in equation (23b) hold if and only if U = 0. Thus, in its augmented form, the eigenvalue problem (21) admits orthogonal eigenfunctions. The statement of the orthogonality for the eigenfunctions is M[Vj ], Vk = djk , K[Vj ], Vk = djk vk2 ,
j, k = 1, 2, . . . .
(24a, b)
Equation (24a) requires normalization of the eigenfunctions Vk , i.e., M[Vk ], Vk =
g
vk2 (x) dx + a[vk2 (−a) + vk2 (a)] + 2bvk (−a)vk (a) = 1.
(25)
V
By equations (9) and (10), the kth scalar eigenfunction is written as vk (x) =
6
Ak Dk sin gk (x + a + l1 ), Ak sin gk (−x + a + l2 ),
x $ V− , x $ V+ ,
(26)
where mgk sin gk l2 − 2r cos gk l2 mgk sin gk l1 − 2r cos gk l1
Dk = −
or
(Igk2 − 2ra) sin gk l2 − 2ra 2gk cos gl2 . (Igk2 − 2ra) sin gk l1 − 2ra 2gk cos gl1
It follows from equations (25) and (26) that the normalization coefficient for the kth mode is Ak =
6
r 2 r D (l − s1 c1 /gk ) + (l2 − s2 c2 /gk ) + aDk2 (s12 + s22 ) + 2bDk s1 s2 2 k 1 2
7
−1/2
,
(27)
where sj = sin gk lj and cj = cos gk lj , j = 1, 2. As an example, consider the bar coupled to two identical strings (l1 = l2 = l); see section 3.2. For a translational mode, Dk = 1, and the normalization coefficient is
60
Ak = rl 1 −
1
1 sin 2mkt + m sin2 mkt 2mkt
7
−1/2
.
(28a)
For a rotational mode, Dk = −1, and the normalization coefficient is
60
Ak = rl 1 −
1
1 I sin 2mkr + 2 sin2 mkr 2mkr a
7
−1/2
.
(28b)
5. SOLUTION OF BOUNDARY VALUE PROBLEM
With the augmented formulation, the original boundary value problem described in section 2 is restated as M[Wtt (x, t)] + K[W(x, t)] = F(x, t), with the initial conditions u0 (x) v0 (x) W(x, 0) = U0 (x) = u0 (−a) , Wt (x, 0) = V0 (x) = v0 (−a) . u0 (a) v0 (a)
2 3
2 3
(29a)
(29b)
.
390 Here,
W(x, t) = (w(x, t) w(−a, t) w(a, t))T $ D
for any t e 0,
F(x, t) = ( f(x, t) F− (t) F+ (t))T,
(30a) (30b)
and u0 (x) and v0 (x) are given in equation (7). The solution of the boundary value problem (29) is obtained by expanding W in an eigenfunction series a
W(x, t) = s qk (t)Vk (x),
(31)
k=1
where qk (t) are the general co-ordinates, and Vk (x) are the eigenfunctions given in equation (22). By equation (31) and the orthogonality relations (24), equations (29a, b) are decoupled into an infinite set of the second order different equations q¨k (t) + vk2 qk (t) = pk (t) = Vk , F,
qk (0) = Vk , M[U0 ],
q˙k (0) = Vk , M[V0 ].
k = 1, 2, . . . ,
(32)
The solution of the above is qk (t) = qk (0) cos vk t + q˙k (0)
1 sin vk t + vk
By equations (31) and (33), W(x, t) =
g
t
0
1 sin vk (t − t)pk (t) dt. vk
g
{Gt (x, j, t)FM[U0 (j)] + G(x, j, t)FM[V0 (j)]} dj
+
gg
(33)
V t
G(x, j, t − t)FF(j, t) dj dt,
(34)
V
0
where F = diag{1 d(x + a) d(x − a)}, and the 3 × 3 matrix Green’s function is a
G(x, j, t) = s k=1
1 sin vk tVk (x)VkT (j), vk
x, j $ V,
t e 0.
(35)
With the solution (34) to the augmented boundary value problem, the response w(x, t) of the strings can be presented in an eigenfunction series. To this end, define the function a
1 sin vk tvk (x)vk (j), v k k=1
g(x, j, t) = s
x, j $ V,
where vk (x) are the normalized eigenfunctions given in equations (35) and (36), g(x, j, t) g(x, −a, t) G(x, j, t) = g(−a, j, t) g(−a, −a, t) g(a, j, t) g(a, −a, t)
&
t e 0,
(36)
equations (26) and (27). By
'
g(x, a, t) g(−a, a, t) . g(a, a, t)
(37)
Substituting equation (37) into equation (34) and using equations (29b) and (30b) gives the string response w(x, t) =
g
{gt (x, j, t)ru0 (j) + g(x, j, t)rv0 (j)} dj
V
+ gt (x, −a, t){au0 (−a) + bu0 (a)} + g(x, −a, t){av0 (−a) + bv0 (a)} + gt (x, a, t){bu0 (−a) + au0 (a)} + g(x, a, t){bv0 (−a) + av0 (a)}
–
gg g t
+
g(x, j, t − t)f(j, t) dj dt +
V
0
g
391
t
g(x, −a, t − t)F− (t) dt
0
t
g(x, a, t − t)F+ (t) dt,
+
(38)
0
for x $ V. It is seen that g(x, j, t) is the Green’s function of the coupled string-bar system. For a vibrating continuum without coupling to a rigid body, its Green’s function is usually represented by a series of orthogonal eigenfunctions. Because the eigenfunctions vk (x) of the coupled–string bar system are non-orthogonal, the system Green’s function cannot be directly obtained by the classical modal analysis. Nevertheless, through introduction of an augmented formulation, we are able to express the Green’s function of the coupled system by an eigenfunction series. The method presented here is applicable to general coupled flexible–rigid body systems. 6. GREEN’S FUNCTION FORMULA
Equation (38) is the Green’s function formula for the coupled string–bar system. Although g(x, j, t) has been given in the eigenfunction series (35), the Green’s function formula does not depend on either modal analysis or the augmented formalism; it can be obtained by other methods. In this section, for the sake of theoretical development, the Green’s function formula is re-derived using the Laplace transform method. The Laplace transformation of equations (1), (2) and (5) gives rs 2w¯ (x, s) − Tw¯xx (x, s) = feI (x, s), w¯ (−a − l1 , s) = 0,
x $ V,
w¯ (a + l2 , s) = 0,
(39a) (39b)
s (aw¯ (−a, s) + bw¯ (a, s)) + Tw¯xx (−a, s) + (T/2a)(w¯ (−a, s) − w¯ (a, s)) = g− (s),
(39c)
s (bw¯ (−a, s) + aw¯ (a, s)) − Tw¯xx (a, s) − (T/2a)(w¯ (−a, s) − w¯ (a, s)) = g+ (s).
(39d)
2
2
Here, feI (x, s) = f (x, s) + r(su0 (x) + v0 (x)), g− (s) = F− (s) + s(au0 (−a) + bu0 (a)) + av0 (−a) + bv0 (a),
(40)
g+ (s) = F+ (s) + s(bu0 (−a) + au0 (a)) + bv0 (−a) + av0 (a). The overbar denotes the Laplace transform with respect to time and s is the complex Laplace transform parameter. By superposition, the solution of equation (39) can be written as w¯ (x, s) = w¯ f(x, s) + w¯ b(x, s), (41) where w¯ f(x, s) satisfies f rs 2w¯ f(x, s) − Tw¯ xx (x, s) = feI (x, s),
x $ V,
(42)
subject to associate homogeneous boundary conditions, i.e., equations (39b–d) with g− (s) = g+ (s) = 0, and w¯ b(x, s) satisfies b rs 2w¯ b(x, s) − Tw¯xx (x, s) = 0,
x $ V,
(43)
subject to the inhomogeneous boundary conditions (39b–d). The solution w¯ f(x, s) of equation (42) can be represented by an integral form w¯ f(x, s) =
g
V
g¯ (x, j, s)feI (j, s) dj,
x $ V,
(44)
.
392
where the integral kernel g¯ (x, j, s) satisfies the differential equation rs 2g¯ (x, j, s) − Tg¯jj (x, j, s) = d(x − j),
x, j $ V,
(45)
with the associate homogeneous boundary conditions. The complex function g¯ (x, j, s) is the distributed transfer function of the coupled string–bar system [11]. To evaluate the solution w¯ b(x, s) of equation (43), consider the integral
g
b g¯ (x, j, s)(rs 2 w¯ b(j, s) − Tw¯jj (j, s)) dj = 0.
(46)
V
Integration by part and use of the boundary conditions of w¯ b(x, s) and g¯ (x, j, s) leads to
g
w¯ b(j, s)(rs 2g¯ (x, j, s) − Tg¯jj (x, j, s)) dj − g¯ (x, −a, s)g− (s) − g¯ (x, a, s)g+ (s) = 0.
V
According to equation (45), the above becomes w¯ b(x, s) = g¯ (x, −a, s)g− (s) + g¯ (x, a, s)g+ (s).
(47)
Thus, by equation (41), the solution to equation (39) is w¯ (x, s) =
g
g¯ (x, j, s)feI (j, s) dj + g¯ (x, −a, s)g− (s) + g¯ (x, a, s)g+ (s).
(48)
V
The inverse Laplace transformation of equation (48) yields equation (38), with the Green’s function g(x, j, t) given by the inverse Laplace transform of the distributed transfer function g¯ (x, j, s). Although the current study is limited to the coupled string–bar, the above derivation of the system Green’s function formula is valid for general coupled flexible–rigid body systems. With the Green’s function formula, the transient response of the coupled system subject to arbitrary external and initial disturbances can be conveniently estimated, as shall be shown in the next section. 7. EXAMPLES
The analytical results obtained are illustrated on a rigid bar coupled to two identical strings (l1 = l2 = l). In the following numerical simulation, the string and bar parameters are chosen as l = 1 m,
r = 1 kg/m,
T = 20 N,
a = 0·5 m,
m = 0·2 kg,
I = 1/60 kg m2.
(49)
The system transient response is evaluated based on Green’s function formula (38), where 100 terms are taken to compute Green’s function. 7.1. According to section 3.2, the translation and rotation of the rigid bar are not coupled in each mode of vibration. The first ten translational and rotational frequencies of the coupled string–bar are calculated based on equations (12) and (14), and listed in Table 1. As predicted by equation (17), interlacing of the translational and rotational frequencies is observed. In addition, the mode shapes of the first three translational and rotational modes (vkt (x) and vkr (x)) are plotted in Figure 5, where the solid lines represent the transverse displacement of the strings.
–
393
T 1 Natural frequencies of the coupled string–bar system (rad/s) k
vkt
vkr
1 2 3 4 5 6 7 8 9 10
6·3901 10·2561 32·3251 45·6170 59·0956 71·7141 86·4331 100·2244 114·0680 127·9504
10·0763 22·1251 35·0941 48·3950 61·8528 75·4064 89·0290 102·7063 116·4290 130·1901
7.2. Let the external disturbances be zero. Without loss of generality, let the coupled system have non-zero initial displacement and zero initial velocity (i.e., u0 (x) $ 0 and v0 (x) = 0, x $ V). By equation (38), the system response is w(x, t) =
g
gt (x, j, t)ru0 (j) dj + gt (x, a, t){bu0 (−a) + au0 (a)}
V
+ gt (x, −a, t){au0 (−a) + bu0 (a)}.
(50)
Assume that the initial displacement of the strings is of the form w(x, 0) = u0 (x) =
6
u−(x + l + a)/l, u+(−x + l + a)/l,
x $ V− , x $ V+ ,
(51)
Figure 5. The translational (left) and rotational (right) mode shapes of the coupled string–bar system: the solid lines in each subfigure represent the transverse displacement of the strings, vkt (x) and vkr (x). (a) First mode; (b) second mode; (c) third mode.
394
.
Figure 6. The initial and external disturbances considered in the numerical simulation: (a) the coupled string–bar subject to initial displacement, u2=w(2a, 0); (b) an impulse f0 d(t) applied to the right string; (c) an impulsive force Q0 d(t) and an impulsive torque G0 d(t) applied to the rigid bar; (d) a transverse force f0 travelling on the left string at a speed cf .
where u2 = w(2a, 0); see Figure 6(a). The initial displacements of the rigid bar depend on u2 : y(0) = 12 (u+ + u− ), u(0) = (1/2a)(u+ − u− ). (52) Choose y(0) = 0·1 m and u(0) = 0·1 rad in the simulation. In Figure 7 is shown the free response w(x, t) of the strings, where the solid lines represent the string displacement. The translation y(t) and rotation u(t) of the rigid bar are plotted in Figure 8. It is seen that the first translational mode (v1t = 6·3901 rad/s) is dominant in the bar translation while more modes are involved in the bar rotation. 7.3. 7.3.1. Impact applied to the strings An impulse f0 d(t) is applied to the right string at a distance b from the rigid bar; see Figure 6(b). Assume zero initial conditions. The system response is w(x, t) = g(x, a + b, t)f0
(53)
where x = a + b is the location of the impulse. Let f0 = 1 kg m/s, and b = 0·3 m. The transient response of the coupled system is shown in Figure 9, where the horizontal axis is for the spatial co-ordinate x and the solid lines represent the string transverse displacement w(x, t). The dynamic behavior of the coupled system under the impulse is explained as follows.
–
395
Figure 7. The free response of the strings: the solid lines in each subfigure represent the transverse displacement w(x, t) of the strings at a specific time t. (a) t = 0; (b) t = 0·1 s; (c) t = 0·2 s; (d) t = 0·3 s; (e) t = 0·4 s; (f) t = 0·5 s; (g) t = 0·6 s; (h) t = 0·7 s.
The strings are initially at rest. Shortly after the impulse is applied, a disturbance (a peak) is generated at the impulse location; see the response at time t = 0·002 s in Figure 9. This disturbance then spreads along the string to both the left and right, at a speed c = zT/r = 4·47 m/s (Figures 9(c) and 9(d)). At the moment, the left string has no motion because the travelling disturbance has not yet reached the rigid bar. At t = b/c = 0·067 s, the disturbance hits the rigid bar, causing the bar to vibrate. Owing to the string–bar coupling, a wave is transmitted to the left string (Figure 9(e)), and a wave is reflected back to the right string (Figure 9(f)). In the meantime, the disturbance continues to propagate to the right, approaching the fixed end of the right string at t = (l − b)/c = 0·16 s (Figure 9(g)). Because of its step-pulse shape, the disturbance cancels its reflected wave, rendering zero string displacement near the boundary x = 1·5 m (Figure 9(f)). The transmitted wave in the left string arrives at the fixed end at t = (l + b)/c = 0·029 s, and is reflected back with a sign change, as indicated by the response at t = 0·26 and 0·34 s.
Figure 8. The free response of the rigid bar: ——, y(t); – – – , u(t).
.
396
Figure 9. The transient response of the coupled system subject to an impulse applied to the right string: the solid lines in each subfigure represent the transverse displacement w(x, t) of the strings at a specific time t. (a) t = 0; (b) t = 0·002 s; (c) t = 0·01 s; (d) t = 0·04 s; (e) t = 0·08 s; (f) f = 0·12 s; (g) t = 0·16 s; (h) t = 0·2 s; (i) t = 0·26 s; (j) t = 0·34 s.
7.3.2. Impact applied to the rigid bar An impulsive force Q0 d(t) and an impulsive torque G0 d(t) are applied to the rigid bar at the mass center; see Figure 6(c). The system response, by equation (38), is w(x, t) = 12 (Q0 − G0 /a)g(x, −a, t) + 12 (Q0 + G0 /a)g(x, a, t),
x $ V.
(54)
In Figure 10 is shown the transient displacement (solid lines) of the coupled system for Q0 = 0 and G0 = 0·5 N m. Because the external disturbance is a torque, only rotational modes are excited. As a result, the string vibration is anti-symmetric with respect to the mass center of the rigid bar, which does not move at all. Again, wave reflection and transmission at the string boundaries and string–bar interconnecting points are seen. 7.4. In Figure 6(d), a transverse force f0 is travelling on the left string at a speed cf . Let the force initially be at the fixed end of the left string. The system response then is w(x, t) = −f0
g
t
g(x, cf t − a − l, t − t) dt,
for 0 E t E l/cf .
(55)
0
Choose f0 = 1 N and cf = 1 m/s. The system transient response for 0 E t E l/cf = 1 s is shown in Figure 11. When the load starts to move, it incurs a wave (denoted by $ ) Y
–
397
propagating along the left string to the right. The wave is travelling ahead of the moving load because c q cf . The right string remains undisturbed until the wave hits the rigid bar (at t = l/c = 0·22 s). At the moment, a wave ( ') is reflected back to the left string and a wave is transmitted to the right string; see the response at t = 0·3 s. This transmitted wave later is reflected at the fixed end of the right string (see the response at t = 0·4 and 0·5 s). As the load moves on, more transmission and reflection of waves occur at the string fixed ends and string–bar interconnecting points. The load arrives at the string–bar interconnecting point at t = 1·0 s. U
8. CONCLUSIONS
The linear vibration of a coupled string–rigid bar system has been studied. Owing to dynamic interaction between the strings and rigid bar, the coupled system is not self-adjoint in its original equations of motion. As a result, the classical modal analysis cannot be directly used. This paper introduces an augmented formulation under which orthogonal eigenfunctions are obtained. Through use of both eigenfunction expansion and Laplace transform, a Green’s function formula for the coupled system is derived, with which the system response to arbitrary initial and external disturbances can be conveniently evaluated. As shown in the numerical examples, the analytical method developed can accurately capture the dynamic characteristics of the coupled system. Although the current study is on a simple system, the basic concepts presented here can be extended to general coupled flexible–rigid body systems.
Figure 10. The transient response of the coupled system subject to an impulsive torque applied to the rigid bar: the solid lines in each subfigure represent the transverse displacement w(x, t) of the strings at a specific time t. (a) t = 0; (b) t = 0·02 s; (c) t = 0·1 s; (d) t = 0·2 s; (e) t = 0·3 s; (f) t = 0·4 s; (g) t = 0·5 s; (h) t = 0·6 s; (i) t = 0·7 s; (j) t = 0·8 s.
398
.
Figure 11. The transient response of the coupled system subject to a moving load: f0 = 1 N; the solid lines in each subfigure represent the transverse displacement w(x, t) of the strings at a specific time t. (a) t = 0; (b) t = 0·1 s; (c) t = 0·2 s; (d) t = 0·3 s; (e) t = 0·4 s; (f) t = 0·5 s; (g) t = 0·7 s; (h) t = 1·0 s.
ACKNOWLEDGMENT
This work was supported by the US Army Research Office under Grant No. DAAL03-93-G-0066 with Dr Gary L. Anderson as the technical monitor.
REFERENCES 1. S. T and D. H. Y 1959 Vibration Problems in Engineering. Princeton, New Jersey: Van Nostrand. 2. E. C. P and F. A. L 1963 Matrix Methods in Elastomechanics. New York: McGraw-Hill. 3. J. C. S 1968 Vibration and Shock in Damped Mechanical Systems. New York: John Wiley. 4. R. S. S and D. A. C 1983 Vehicle System Dynamics 16, 167–192. Road vehicle suspension system design – a review.
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5. L. A. B and J. W. N 1985 ASME Journal of Vibration, Acoustics, Stress, and Reliability in Design 107, 275–281. Forced vibration of a damped combined linear system. 6. J. W. N and L. A. B 1986 Journal of Engineering Mechanics 112, 1–13. Free vibration of combined dynamical systems. 7. B. Y 1992 Journal of Sound and Vibration 156, 425–443. Transfer functions of constrained/ combined one-dimensional continuous dynamic systems. 8. J. L and L. A. B 1991 Technical Report No. UILU ENG 91-0510, University of Illinois, Urbana. Free vibration of combined dynamical systems. 9. B. F 1956 Principles and Techniques of Applied Mathematics. New York: John Wiley. 10. D. R and N. R. A 1974 Transactions of the American Society of Mechanical Engineers, Journal of Applied Mechanics 41, 1106–1112. On vibration problems with discretely distributed loads—a rigorous formalism. 11. A. G. B 1982 Green’s Functions and Transfer Functions Handbook. New York: John Wiley.
LINEAR VIBRATION OF A COUPLED STRING–RIGID BAR SYSTEM B. Y Department of Mechanical Engineering, University of Southern California, Los Angeles, California 90089-1453, U.S.A. (Received 30 August 1993, and in final form 11 April 1994) Linear vibration of a coupled string – rigid bar system is investigated. Owing to dynamic interaction between the strings and the rigid bar, the coupled system is not self-adjoint in its original equations of motion. Consequently, the classical modal analysis cannot be applied. This study develops an augmented formulation for the eigenvalue and boundary value problems of the coupled system. In this formulation, orthogonal eigenfunctions are obtained, and the general response of the coupled string–bar system is represented by an eigenfunction series. Green’s function formula for the coupled system is derived by two methods: modal analysis and Laplace transform. The analytical predictions about the system eigensolutions and response are illustrated in numerical simulation.
1. INTRODUCTION
Coupled flexible–rigid body systems that are assemblages of multiple flexible and rigid bodies are commonly seen in structures and machines. Applications are diverse, including vehicles, airplanes, spacecraft, robots, computer storage systems, rotating machinery, highway structures, buildings, bridges and pipelines conveying fluids. Accurate modelling is essential to analysis, control and optimal design of such systems. Flexible–rigid body coupling mechanisms can be divided into two classes: (1) type I, a flexible body is joined to several rigid bodies at some interior points (see Figure 1(a))—in many cases, the rigid bodies can be idealized as dimensionless, lumped masses; (2) type II, a number of flexible bodies are interconnected to rigid bodies in some regions or at some points of their boundaries (see Figure 1(b))—the sizes of the rigid bodies are always considered. The dynamic coupling of flexible and rigid bodies renders the system eigenfunctions non-orthogonal in the original equations of motion. As a result, the classical modal analysis for conventional flexible systems is not applicable to coupled flexible–rigid body systems. Dynamics of type I systems have been extensively studied in the literature; for instance, see References [1–7] and the papers cited therein. Previous research on type II systems focused on finite element analysis and other approximate methods; analytical prediction of the system response to arbitrary external and initial excitations is rarely addressed. Recently, in a vibration analysis of stepped beams and plates, Lee and Bergman [8] developed a dynamic flexibility method, which represents the interconnecting forces between the beam or plate segments by Green’s functions of the flexible subsystems, and delta functions [5, 6]. This method yields analytical solutions of various dynamic problems for stepped beams and plates, and can be directly applied to type II coupled systems. The current work is concerned with an alternative approach for the analysis of type II coupled systems. The linear vibration of a coupled string–rigid bar system, which is shown 383 0022–460X/95/230383 + 17 $08.00/0
7 1995 Academic Press Limited
384
.
Figure 1. Two types of flexible–rigid body coupling mechanisms: (a) type I—a flexible body joined to rigid bodies at its interior points; (b) type II—flexible bodies interconnected with rigid bodies at their boundaries.
in Figure 2, is investigated. The coupled system is probably the simplest type II system, yet its dynamic characteristics are representative of generic systems. The coupled system itself is a model for certain structural systems such as the secondary mirror of a telescope suspended by tensioned vanes, and cable-supporting systems for bridges and television towers. A self-adjoint formulation for the eigenvalue and boundary value problems of the coupled string–bar system is developed. In this formulation, orthogonal eigenfunctions are obtained, and the response of the coupled system to arbitrary external and initial excitations is represented by eigenfunction expansion. The Green’s function formula for the coupled system is derived by two methods: modal analysis and Laplace transform. The analytical predictions about the system eigensolutions and response are illustrated in several numerical examples.
2. EQUATIONS OF MOTION
In Figure 2, a rigid bar of length 2a is attached to two uniform strings of length l1 and l2 , respectively. The rigid bar has mass m and moment of inertia I. The strings have the
Figure 2. A schematic of a coupled string–rigid bar system.
–
385
same linear density r, and are under tension T. The strings are subject to the external force f(x, t); the rigid bar is subject to the force F0 (t) and torque t0 (t) applied at its center of mass. Let the left and right strings be defined in the regions V− = {x=−a − l1 Q x Q −a} and V+ = {x=a Q x Q a + l2 }, respectively. The transverse displacement w(x, t) of the strings is governed by the differential equation rwtt (x, t) − Twxx (x, t) = f(x, t)
x $ V = V − * V+ ,
(1)
with the boundary conditions w(−a − l1 , t) = 0,
w(a + l2 , t) = 0,
(2)
where ( )t = 1( )/1t and ( )x = 1( )/1x. According to the free body diagram in Figure 3, the equations of motion of the rigid bar are my¨ = T(sin u+ − sin u− ) + F0 ,
Iu = Ta [sin(u+ − u) + sin (u− − u)] + t0 ,
(3)
where the overdot denotes temporal differentiation, y(t) is the vertical displacement of the mass center and u(t) is the rotational displacement of the bar. Note that y(t) = 12 [w(a, t) + w(−a, t)],
(4)
and for small oscillations sin u21u2 1 wx (2a, t),
sin (u2−u) 1 u2−u,
u(t) 1 (1/2a)[w(a, t) − w(−a, t)].
Inserting the above in equation (3) and rearranging yields the matching conditions for the strings awtt (−a, t) + bwtt (a, t) = −Twx (−a, t) − (T/2a)[w(−a, t) − w(a, t)] + F− (t), bwtt (−a, t) + awtt (a, t) = Twx (a, t) + (T/2a)[w(−a, t) − w(a, t)] + F+ (t),
(5)
where a = 14 (m + I/a 2), F− (t) = 12 [F0 (t) − t0 (t)/a],
b = 14 (m − I/a 2), F+ (t) = 12 [F0 (t) + t0 (t)/a].
(6)
Note that b is always positive because for any rigid body ma 2 q I. The above equations of motion can also be derived using the generalized Hamilton principle
g
t2
(dKE − dPE + dW) dt = 0.
t1
Figure 3. A free body diagram of the rigid bar.
.
386
Here, for the coupled string–bar system, the kinetic energy, the potential energy and the virtual work done by the external forces are given by KE =
g
1 2
rwt2 dx + 12 my˙ 2 + 12 Iu 2,
PE =
V
g
1 2
Twx2 dx + aTu 2,
V
dW =
g
fdw dx + F0 dy + t0 du,
V
respectively. Assume that the initial conditions of the strings are wt (x, 0) = n0 (x),
w(x, 0) = u0 (x),
x $ V.
(7)
The initial conditions of the rigid bar are dependent on the given functions u0 (x) and n0 (x); i.e., y(0) = 12 [u0 (a) + u0 (−a)], y˙ (0) = 12 [n0 (a) + n0 (−a)], u(0) = (1/2a)(u0 (a) − u0 (−a)),
u (0) = (1/2a)(n0 (a) − n0 (−a)).
Equations (1), (2), (5) and (7) form a boundary value problem for the coupled string–bar system. The fundamental problem is to determine the system response w(x, t) for given external and initial disturbances. In what follows, the exact solution will be derived in terms of eigenfunction expansion. 3. EIGENVALUE PROBLEM
3.1. Letting w(x, t) = n(x) eivt, i = z−1, and f(x, t), F0 (t), t0 (t) = 0 in the equations of motion leads to the eigenvalue problem for the coupled system: n0(x) + g 2n(x) = 0, n(−a − l1 ) = 0,
g = vzr/T,
x $ V,
(8a)
w(a + l2 ) = 0,
(8b)
v 2(an(−a) + bn(a)) = Tn'(−a) + (T/2a)(n(−a) − n(a)),
(8c)
v 2(bn(−a) + an(a)) = −Tn'(a) − (T/2a)(n(−a) − n(a)),
(8d)
where n' = dn/dx and v is the natural frequency. Let the eigenfunction be in the form n(x) =
6
A sin g(x + a + l1 ), B sin g(−x + a + l2 ),
x $ V− , x $ V+ ,
(9)
satisfying equations (8a) and (8b), where A and B are constants. Substituting equation (9) into equations (8c) and (8d) gives
$
mg sin gl1 − 2r cos gl1 (Ig 2 − 2ra) sin gl1 − 2ra 2g cos gl1
%0 1
mg sin gl2 − 2r cos gl2 −(Ig 2 − 2ra) sin gl2 + 2ra 2g cos gl2
A = 0. (10) B
The characteristic equation is obtained by letting the determinant of the matrix in equation (10) vanish. Solution of the characteristic equation gives the natural frequencies of the coupled string–bar system. With the known natural frequencies, the associate eigenfunctions (mode shapes) can be determined using equations (9) and (10).
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387
3.2. When the two strings in Figure 2, have the same length, i.e., l1 = l2 = l, equation (10) becomes two decoupled equations (mg sin gl − 2r cos gl)(A + B) = 0,
(11a)
((Ig 2 − 2ra) sin gl − 2ra 2g cos gl)(A − B) = 0.
(11b)
This implies that the translation and rotation of the rigid bar in each mode of vibration are not coupled, which is discussed as follows. 3.2.1. Translational modes Every point of the rigid body has the same transverse motion; i.e., A = B. By equation (11), the characteristic equation is tan m = 1/(hM m),
m = gl,
(12)
where the mass ratio hM = m/(2rl). The natural frequencies of translational modes are vkr =
mkt zT/r, l
k = 1, 2, . . . ,
(13)
where mkt are the roots of equation (12). The associate mode shapes are given by equation (9) with A = B. 3.2.2. Rotational modes The rigid bar rotates about its mass center which does not move; i.e., A = −B. By equation (11), the characteristic equation becomes tan m = m/[hI m 2 − (l/a)],
m = gl,
(14)
where the mass ratio hI = I/(2rla 2). The natural frequencies of rotational modes are vkr = (mkr /l)zT/r,
k = 1, 2, . . . ,
(15)
where m are the roots of equation (14). The associate mode shapes are given by equation (9) with A = −B. The vkt and vkr shall be called the translational and rotational frequencies, respectively. The system eigenvalues are the intersections of tan m and functions 1/hM m (for mkt ) and m/(hI m 2 − l/a) (for mkr ); see Figure 4. Because ma 2 q I, the two mass ratios satisfy the inequality h M q hI . (16) r k
Figure 4. Frequency interlacing: w, mkt = vkt lzr/T, eigenvalues of translational modes; W, mkr = vkr lzr/T, eigenvalues of rotational modes.
388 . The following observations about the natural frequencies of the coupled string–bar system can be drawn from Figure 4; first, the translational (rotational) frequencies decrease as the mass ratio hM (hI ) increases; second, the translational frequencies are interlaced with the rotational frequencies; i.e., k = 1, 2, . . . ; (17) vkt Q vkr Q vkt + 1 , and third by equation (17), the coupled string–bar system always has distinctive natural frequencies. However, when the mode number k is very large, vkt , vkr 1 kp(1/l)zT/r, indicating that translational and rotational modes tend to be coupled at high frequencies. 4. ORTHOGONALITY OF EIGENFUNCTIONS
The dynamic coupling between the strings and rigid bar renders the eigenvalue problem (8) non-self adjoint; the eigenfunctions given in equations (9) are not orthogonal, i.e., fV nj (x)nk (x) dx $ 0 for j $ k. As a result, the classical modal analysis cannot be applied to determine the response of the coupled system. In this section, an augmentation formalism proposed in References [9] and [10] is generalized to define an equivalent eigenvalue problem which admits orthogonal eigenfunctions. Consider a function space H = H(V)$R 2, where H(V) is a Hilbert space, and V is the domain for the strings (see equation (1)). Any element U of H has the form U = (u(x) u1 u2 )T, where u(x) is a function defined on V, and u1 and u2 are two scalars. The inner product on H is U, V =
g
U, V $ H.
u(x)v(x) dx + u1 v1 + u2 v2 ,
(18)
V
Define the augmented operators M=
K −T 1 2 1x G G 0 K=G G G k 0 2
&
'
r=x $ V 0 0 0 a=x = −a b=x = a , 0 b=x = −a a=x = a
b
0
0
x$V
0
T
1 1 + 1x 2a −
T 2a
b
1b
− x = −a
x = −a
0
T 2a
b
x=a
1 1 + 1x 2a
T −
1b
L G G G, G G l
(19)
x=a
where the parameters a and b are given in equation (6). The domain of the operators is a subspace of H: D 0 {U = U = (u(x) u(−a) u(a))T, u $ H(V); u(−a − l1 ) = u(a + l2 ) = 0}. (20) By equations (19) and (20), the original eigenvalue problem (8) is equivalent to v 2M[V] = K[V], V $ D. (21) The eigensolutions of equation (21) are represented by those of the original problem (equation (8)); namely, vk = gk zT/r, Vk (x) = [vk (x) vk (−a) vk (−a)T, k = 1, 2, . . . , (22) where gk are roots of the characteristic equation deduced from equation (10), and vk (x) are the scalar eigenfunctions given in equation (9).
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389
It can be shown that both M and K are self-adjoint, and positive definite; namely, for any elements U, V $ D, M[U], V = U, M[V], K[U], V = U, K[V]
(23a)
and M[U], U e 0,
K[U], U e 0,
(23b)
where the equalities in equation (23b) hold if and only if U = 0. Thus, in its augmented form, the eigenvalue problem (21) admits orthogonal eigenfunctions. The statement of the orthogonality for the eigenfunctions is M[Vj ], Vk = djk , K[Vj ], Vk = djk vk2 ,
j, k = 1, 2, . . . .
(24a, b)
Equation (24a) requires normalization of the eigenfunctions Vk , i.e., M[Vk ], Vk =
g
vk2 (x) dx + a[vk2 (−a) + vk2 (a)] + 2bvk (−a)vk (a) = 1.
(25)
V
By equations (9) and (10), the kth scalar eigenfunction is written as vk (x) =
6
Ak Dk sin gk (x + a + l1 ), Ak sin gk (−x + a + l2 ),
x $ V− , x $ V+ ,
(26)
where mgk sin gk l2 − 2r cos gk l2 mgk sin gk l1 − 2r cos gk l1
Dk = −
or
(Igk2 − 2ra) sin gk l2 − 2ra 2gk cos gl2 . (Igk2 − 2ra) sin gk l1 − 2ra 2gk cos gl1
It follows from equations (25) and (26) that the normalization coefficient for the kth mode is Ak =
6
r 2 r D (l − s1 c1 /gk ) + (l2 − s2 c2 /gk ) + aDk2 (s12 + s22 ) + 2bDk s1 s2 2 k 1 2
7
−1/2
,
(27)
where sj = sin gk lj and cj = cos gk lj , j = 1, 2. As an example, consider the bar coupled to two identical strings (l1 = l2 = l); see section 3.2. For a translational mode, Dk = 1, and the normalization coefficient is
60
Ak = rl 1 −
1
1 sin 2mkt + m sin2 mkt 2mkt
7
−1/2
.
(28a)
For a rotational mode, Dk = −1, and the normalization coefficient is
60
Ak = rl 1 −
1
1 I sin 2mkr + 2 sin2 mkr 2mkr a
7
−1/2
.
(28b)
5. SOLUTION OF BOUNDARY VALUE PROBLEM
With the augmented formulation, the original boundary value problem described in section 2 is restated as M[Wtt (x, t)] + K[W(x, t)] = F(x, t), with the initial conditions u0 (x) v0 (x) W(x, 0) = U0 (x) = u0 (−a) , Wt (x, 0) = V0 (x) = v0 (−a) . u0 (a) v0 (a)
2 3
2 3
(29a)
(29b)
.
390 Here,
W(x, t) = (w(x, t) w(−a, t) w(a, t))T $ D
for any t e 0,
F(x, t) = ( f(x, t) F− (t) F+ (t))T,
(30a) (30b)
and u0 (x) and v0 (x) are given in equation (7). The solution of the boundary value problem (29) is obtained by expanding W in an eigenfunction series a
W(x, t) = s qk (t)Vk (x),
(31)
k=1
where qk (t) are the general co-ordinates, and Vk (x) are the eigenfunctions given in equation (22). By equation (31) and the orthogonality relations (24), equations (29a, b) are decoupled into an infinite set of the second order different equations q¨k (t) + vk2 qk (t) = pk (t) = Vk , F,
qk (0) = Vk , M[U0 ],
q˙k (0) = Vk , M[V0 ].
k = 1, 2, . . . ,
(32)
The solution of the above is qk (t) = qk (0) cos vk t + q˙k (0)
1 sin vk t + vk
By equations (31) and (33), W(x, t) =
g
t
0
1 sin vk (t − t)pk (t) dt. vk
g
{Gt (x, j, t)FM[U0 (j)] + G(x, j, t)FM[V0 (j)]} dj
+
gg
(33)
V t
G(x, j, t − t)FF(j, t) dj dt,
(34)
V
0
where F = diag{1 d(x + a) d(x − a)}, and the 3 × 3 matrix Green’s function is a
G(x, j, t) = s k=1
1 sin vk tVk (x)VkT (j), vk
x, j $ V,
t e 0.
(35)
With the solution (34) to the augmented boundary value problem, the response w(x, t) of the strings can be presented in an eigenfunction series. To this end, define the function a
1 sin vk tvk (x)vk (j), v k k=1
g(x, j, t) = s
x, j $ V,
where vk (x) are the normalized eigenfunctions given in equations (35) and (36), g(x, j, t) g(x, −a, t) G(x, j, t) = g(−a, j, t) g(−a, −a, t) g(a, j, t) g(a, −a, t)
&
t e 0,
(36)
equations (26) and (27). By
'
g(x, a, t) g(−a, a, t) . g(a, a, t)
(37)
Substituting equation (37) into equation (34) and using equations (29b) and (30b) gives the string response w(x, t) =
g
{gt (x, j, t)ru0 (j) + g(x, j, t)rv0 (j)} dj
V
+ gt (x, −a, t){au0 (−a) + bu0 (a)} + g(x, −a, t){av0 (−a) + bv0 (a)} + gt (x, a, t){bu0 (−a) + au0 (a)} + g(x, a, t){bv0 (−a) + av0 (a)}
–
gg g t
+
g(x, j, t − t)f(j, t) dj dt +
V
0
g
391
t
g(x, −a, t − t)F− (t) dt
0
t
g(x, a, t − t)F+ (t) dt,
+
(38)
0
for x $ V. It is seen that g(x, j, t) is the Green’s function of the coupled string-bar system. For a vibrating continuum without coupling to a rigid body, its Green’s function is usually represented by a series of orthogonal eigenfunctions. Because the eigenfunctions vk (x) of the coupled–string bar system are non-orthogonal, the system Green’s function cannot be directly obtained by the classical modal analysis. Nevertheless, through introduction of an augmented formulation, we are able to express the Green’s function of the coupled system by an eigenfunction series. The method presented here is applicable to general coupled flexible–rigid body systems. 6. GREEN’S FUNCTION FORMULA
Equation (38) is the Green’s function formula for the coupled string–bar system. Although g(x, j, t) has been given in the eigenfunction series (35), the Green’s function formula does not depend on either modal analysis or the augmented formalism; it can be obtained by other methods. In this section, for the sake of theoretical development, the Green’s function formula is re-derived using the Laplace transform method. The Laplace transformation of equations (1), (2) and (5) gives rs 2w¯ (x, s) − Tw¯xx (x, s) = feI (x, s), w¯ (−a − l1 , s) = 0,
x $ V,
w¯ (a + l2 , s) = 0,
(39a) (39b)
s (aw¯ (−a, s) + bw¯ (a, s)) + Tw¯xx (−a, s) + (T/2a)(w¯ (−a, s) − w¯ (a, s)) = g− (s),
(39c)
s (bw¯ (−a, s) + aw¯ (a, s)) − Tw¯xx (a, s) − (T/2a)(w¯ (−a, s) − w¯ (a, s)) = g+ (s).
(39d)
2
2
Here, feI (x, s) = f (x, s) + r(su0 (x) + v0 (x)), g− (s) = F− (s) + s(au0 (−a) + bu0 (a)) + av0 (−a) + bv0 (a),
(40)
g+ (s) = F+ (s) + s(bu0 (−a) + au0 (a)) + bv0 (−a) + av0 (a). The overbar denotes the Laplace transform with respect to time and s is the complex Laplace transform parameter. By superposition, the solution of equation (39) can be written as w¯ (x, s) = w¯ f(x, s) + w¯ b(x, s), (41) where w¯ f(x, s) satisfies f rs 2w¯ f(x, s) − Tw¯ xx (x, s) = feI (x, s),
x $ V,
(42)
subject to associate homogeneous boundary conditions, i.e., equations (39b–d) with g− (s) = g+ (s) = 0, and w¯ b(x, s) satisfies b rs 2w¯ b(x, s) − Tw¯xx (x, s) = 0,
x $ V,
(43)
subject to the inhomogeneous boundary conditions (39b–d). The solution w¯ f(x, s) of equation (42) can be represented by an integral form w¯ f(x, s) =
g
V
g¯ (x, j, s)feI (j, s) dj,
x $ V,
(44)
.
392
where the integral kernel g¯ (x, j, s) satisfies the differential equation rs 2g¯ (x, j, s) − Tg¯jj (x, j, s) = d(x − j),
x, j $ V,
(45)
with the associate homogeneous boundary conditions. The complex function g¯ (x, j, s) is the distributed transfer function of the coupled string–bar system [11]. To evaluate the solution w¯ b(x, s) of equation (43), consider the integral
g
b g¯ (x, j, s)(rs 2 w¯ b(j, s) − Tw¯jj (j, s)) dj = 0.
(46)
V
Integration by part and use of the boundary conditions of w¯ b(x, s) and g¯ (x, j, s) leads to
g
w¯ b(j, s)(rs 2g¯ (x, j, s) − Tg¯jj (x, j, s)) dj − g¯ (x, −a, s)g− (s) − g¯ (x, a, s)g+ (s) = 0.
V
According to equation (45), the above becomes w¯ b(x, s) = g¯ (x, −a, s)g− (s) + g¯ (x, a, s)g+ (s).
(47)
Thus, by equation (41), the solution to equation (39) is w¯ (x, s) =
g
g¯ (x, j, s)feI (j, s) dj + g¯ (x, −a, s)g− (s) + g¯ (x, a, s)g+ (s).
(48)
V
The inverse Laplace transformation of equation (48) yields equation (38), with the Green’s function g(x, j, t) given by the inverse Laplace transform of the distributed transfer function g¯ (x, j, s). Although the current study is limited to the coupled string–bar, the above derivation of the system Green’s function formula is valid for general coupled flexible–rigid body systems. With the Green’s function formula, the transient response of the coupled system subject to arbitrary external and initial disturbances can be conveniently estimated, as shall be shown in the next section. 7. EXAMPLES
The analytical results obtained are illustrated on a rigid bar coupled to two identical strings (l1 = l2 = l). In the following numerical simulation, the string and bar parameters are chosen as l = 1 m,
r = 1 kg/m,
T = 20 N,
a = 0·5 m,
m = 0·2 kg,
I = 1/60 kg m2.
(49)
The system transient response is evaluated based on Green’s function formula (38), where 100 terms are taken to compute Green’s function. 7.1. According to section 3.2, the translation and rotation of the rigid bar are not coupled in each mode of vibration. The first ten translational and rotational frequencies of the coupled string–bar are calculated based on equations (12) and (14), and listed in Table 1. As predicted by equation (17), interlacing of the translational and rotational frequencies is observed. In addition, the mode shapes of the first three translational and rotational modes (vkt (x) and vkr (x)) are plotted in Figure 5, where the solid lines represent the transverse displacement of the strings.
–
393
T 1 Natural frequencies of the coupled string–bar system (rad/s) k
vkt
vkr
1 2 3 4 5 6 7 8 9 10
6·3901 10·2561 32·3251 45·6170 59·0956 71·7141 86·4331 100·2244 114·0680 127·9504
10·0763 22·1251 35·0941 48·3950 61·8528 75·4064 89·0290 102·7063 116·4290 130·1901
7.2. Let the external disturbances be zero. Without loss of generality, let the coupled system have non-zero initial displacement and zero initial velocity (i.e., u0 (x) $ 0 and v0 (x) = 0, x $ V). By equation (38), the system response is w(x, t) =
g
gt (x, j, t)ru0 (j) dj + gt (x, a, t){bu0 (−a) + au0 (a)}
V
+ gt (x, −a, t){au0 (−a) + bu0 (a)}.
(50)
Assume that the initial displacement of the strings is of the form w(x, 0) = u0 (x) =
6
u−(x + l + a)/l, u+(−x + l + a)/l,
x $ V− , x $ V+ ,
(51)
Figure 5. The translational (left) and rotational (right) mode shapes of the coupled string–bar system: the solid lines in each subfigure represent the transverse displacement of the strings, vkt (x) and vkr (x). (a) First mode; (b) second mode; (c) third mode.
394
.
Figure 6. The initial and external disturbances considered in the numerical simulation: (a) the coupled string–bar subject to initial displacement, u2=w(2a, 0); (b) an impulse f0 d(t) applied to the right string; (c) an impulsive force Q0 d(t) and an impulsive torque G0 d(t) applied to the rigid bar; (d) a transverse force f0 travelling on the left string at a speed cf .
where u2 = w(2a, 0); see Figure 6(a). The initial displacements of the rigid bar depend on u2 : y(0) = 12 (u+ + u− ), u(0) = (1/2a)(u+ − u− ). (52) Choose y(0) = 0·1 m and u(0) = 0·1 rad in the simulation. In Figure 7 is shown the free response w(x, t) of the strings, where the solid lines represent the string displacement. The translation y(t) and rotation u(t) of the rigid bar are plotted in Figure 8. It is seen that the first translational mode (v1t = 6·3901 rad/s) is dominant in the bar translation while more modes are involved in the bar rotation. 7.3. 7.3.1. Impact applied to the strings An impulse f0 d(t) is applied to the right string at a distance b from the rigid bar; see Figure 6(b). Assume zero initial conditions. The system response is w(x, t) = g(x, a + b, t)f0
(53)
where x = a + b is the location of the impulse. Let f0 = 1 kg m/s, and b = 0·3 m. The transient response of the coupled system is shown in Figure 9, where the horizontal axis is for the spatial co-ordinate x and the solid lines represent the string transverse displacement w(x, t). The dynamic behavior of the coupled system under the impulse is explained as follows.
–
395
Figure 7. The free response of the strings: the solid lines in each subfigure represent the transverse displacement w(x, t) of the strings at a specific time t. (a) t = 0; (b) t = 0·1 s; (c) t = 0·2 s; (d) t = 0·3 s; (e) t = 0·4 s; (f) t = 0·5 s; (g) t = 0·6 s; (h) t = 0·7 s.
The strings are initially at rest. Shortly after the impulse is applied, a disturbance (a peak) is generated at the impulse location; see the response at time t = 0·002 s in Figure 9. This disturbance then spreads along the string to both the left and right, at a speed c = zT/r = 4·47 m/s (Figures 9(c) and 9(d)). At the moment, the left string has no motion because the travelling disturbance has not yet reached the rigid bar. At t = b/c = 0·067 s, the disturbance hits the rigid bar, causing the bar to vibrate. Owing to the string–bar coupling, a wave is transmitted to the left string (Figure 9(e)), and a wave is reflected back to the right string (Figure 9(f)). In the meantime, the disturbance continues to propagate to the right, approaching the fixed end of the right string at t = (l − b)/c = 0·16 s (Figure 9(g)). Because of its step-pulse shape, the disturbance cancels its reflected wave, rendering zero string displacement near the boundary x = 1·5 m (Figure 9(f)). The transmitted wave in the left string arrives at the fixed end at t = (l + b)/c = 0·029 s, and is reflected back with a sign change, as indicated by the response at t = 0·26 and 0·34 s.
Figure 8. The free response of the rigid bar: ——, y(t); – – – , u(t).
.
396
Figure 9. The transient response of the coupled system subject to an impulse applied to the right string: the solid lines in each subfigure represent the transverse displacement w(x, t) of the strings at a specific time t. (a) t = 0; (b) t = 0·002 s; (c) t = 0·01 s; (d) t = 0·04 s; (e) t = 0·08 s; (f) f = 0·12 s; (g) t = 0·16 s; (h) t = 0·2 s; (i) t = 0·26 s; (j) t = 0·34 s.
7.3.2. Impact applied to the rigid bar An impulsive force Q0 d(t) and an impulsive torque G0 d(t) are applied to the rigid bar at the mass center; see Figure 6(c). The system response, by equation (38), is w(x, t) = 12 (Q0 − G0 /a)g(x, −a, t) + 12 (Q0 + G0 /a)g(x, a, t),
x $ V.
(54)
In Figure 10 is shown the transient displacement (solid lines) of the coupled system for Q0 = 0 and G0 = 0·5 N m. Because the external disturbance is a torque, only rotational modes are excited. As a result, the string vibration is anti-symmetric with respect to the mass center of the rigid bar, which does not move at all. Again, wave reflection and transmission at the string boundaries and string–bar interconnecting points are seen. 7.4. In Figure 6(d), a transverse force f0 is travelling on the left string at a speed cf . Let the force initially be at the fixed end of the left string. The system response then is w(x, t) = −f0
g
t
g(x, cf t − a − l, t − t) dt,
for 0 E t E l/cf .
(55)
0
Choose f0 = 1 N and cf = 1 m/s. The system transient response for 0 E t E l/cf = 1 s is shown in Figure 11. When the load starts to move, it incurs a wave (denoted by $ ) Y
–
397
propagating along the left string to the right. The wave is travelling ahead of the moving load because c q cf . The right string remains undisturbed until the wave hits the rigid bar (at t = l/c = 0·22 s). At the moment, a wave ( ') is reflected back to the left string and a wave is transmitted to the right string; see the response at t = 0·3 s. This transmitted wave later is reflected at the fixed end of the right string (see the response at t = 0·4 and 0·5 s). As the load moves on, more transmission and reflection of waves occur at the string fixed ends and string–bar interconnecting points. The load arrives at the string–bar interconnecting point at t = 1·0 s. U
8. CONCLUSIONS
The linear vibration of a coupled string–rigid bar system has been studied. Owing to dynamic interaction between the strings and rigid bar, the coupled system is not self-adjoint in its original equations of motion. As a result, the classical modal analysis cannot be directly used. This paper introduces an augmented formulation under which orthogonal eigenfunctions are obtained. Through use of both eigenfunction expansion and Laplace transform, a Green’s function formula for the coupled system is derived, with which the system response to arbitrary initial and external disturbances can be conveniently evaluated. As shown in the numerical examples, the analytical method developed can accurately capture the dynamic characteristics of the coupled system. Although the current study is on a simple system, the basic concepts presented here can be extended to general coupled flexible–rigid body systems.
Figure 10. The transient response of the coupled system subject to an impulsive torque applied to the rigid bar: the solid lines in each subfigure represent the transverse displacement w(x, t) of the strings at a specific time t. (a) t = 0; (b) t = 0·02 s; (c) t = 0·1 s; (d) t = 0·2 s; (e) t = 0·3 s; (f) t = 0·4 s; (g) t = 0·5 s; (h) t = 0·6 s; (i) t = 0·7 s; (j) t = 0·8 s.
398
.
Figure 11. The transient response of the coupled system subject to a moving load: f0 = 1 N; the solid lines in each subfigure represent the transverse displacement w(x, t) of the strings at a specific time t. (a) t = 0; (b) t = 0·1 s; (c) t = 0·2 s; (d) t = 0·3 s; (e) t = 0·4 s; (f) t = 0·5 s; (g) t = 0·7 s; (h) t = 1·0 s.
ACKNOWLEDGMENT
This work was supported by the US Army Research Office under Grant No. DAAL03-93-G-0066 with Dr Gary L. Anderson as the technical monitor.
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