Composite Structures 48 (2000) 265±273
The eect of rotatory inertia on the dynamic response of laminated composite plate Y.Y. Wang *, K.Y. Lam, G.R. Liu Institute of High Performance Computing, 89C Science Park Drive, #02-11/12 The Rutherford 118261, Singapore
Abstract The strip element method (SEM) is extended to include rotatory inertia for analysing the dynamic response of laminated composite plates. The transient responses of rectangular symmetric laminated plates are computed for various loading using the newly developed SEM program. The eect of the rotatory inertia is investigated for plates of dierent thickness. It is found that the rotatory inertia has less eect on thin plates whose thickness±length ratio is less than 1/20, but signi®cant eect on thicker plates. The eects of other parameters such as elastic constants, material density and ®ber orientation on the responses of plate are also studied and discussed in detail. Ó 2000 Elsevier Science Ltd. All rights reserved. Keywords: Strip element method; Composite plate; Rotatory inertia; Dynamics analysis; Transient analysis; Numerical method
1. Introduction High performance of composite structures can only be achieved by a proper design based on eective and accurate dynamic and static analysis of the composite material. Plate theories have been used in such analysis, whose theories include classical laminated plate theory (CLPT) [1,17], the ®rst-order shear deformation theory (FSDT) [2±4,18] and the third-order laminated theory (TLT) [5,19]. Researches on application and comparison of these theories in the dynamic analysis of laminated composite plates have also been carried out. Khdeir and Reddy [6] presented an exact solution for the dynamic response of simply supported symmetrically laminated cross-ply plates. The exact solution had been compared with CLPT and FSDT, and it was shown that the difference between the center de¯ections predicted by various theories decreases as the thickness decreases. Mallikarjuna and Kant [7] presented an isoparametric ®nite element formulation based on a high-order displacement model for dynamic analysis of a multi-layered symmetric composite plate. Reddy [8] investigated the eect of lamination angle, layers, shear deformation, aspect ratio and material orthotropy on the solution of the forced motion of a rectangular composite plate. Lu *
Corresponding author.
[9] employed the Rayleigh±Ritz method and the method of superposition of normal modes to solve the dynamic response of laminated angle-ply plates with clamped boundary conditions subjected to explosive loadings. It has been shown that symmetrically laminated plates produce the least transverse deformation compared to the anti-symmetrically and non-symmetrically stacked plates. The strip element method (SEM) developed recently by Liu and Achenbach [10,11] and Wang et al. [12,13] provided a good alternative for static and dynamic analysis of composite laminates. Generally, the rotatory inertia is not considered in the CLPT which is applicable to thin plates only. However, the eect of the rotatory inertia cannot be ignored for thicker plates, especially for the transient analysis. The purpose of this paper is to investigate the eect of rotatory inertial of laminated plate on the transient analysis. A new SEM formulation is derived to include the inertia terms for dynamic analysis. The dynamic response of the plate in the frequency domain is then investigated. The Fourier transform technique is used to obtain the time domain responses and an exponential window method is introduced to avoid singularities in the Fourier integration. The transient responses of a rectangular symmetric laminated plate are presented for various loading and boundary conditions. The eects of rotatory inertia on the responses of plate are discussed in detail. The eects of other parameters such as material constants, ®ber orientation have also been investigated.
0263-8223/00/$ - see front matter Ó 2000 Elsevier Science Ltd. All rights reserved. PII: S 0 2 6 3 - 8 2 2 3 ( 9 9 ) 0 0 1 1 4 - 2
266
Y.Y. Wang et al. / Composite Structures 48 (2000) 265±273
2. Formulations A ®ber reinforced symmetrically laminated composite plate shown in Fig. 1 is considered in this paper. The plate consists of K layers of orthotropic plies lying in the x±y plane, the thickness of the plate in z direction is H. The reference plane z 0 is located at the undeformed mid-plane of the plate. The rectangular plate is bounded by ÿa=2 6 x 6 a=2, 0 6 y 6 b and ÿH =2 6 z 6 H =2. The boundaries are denoted by S1 , S2 , S3 and S4 , as shown in Fig. 2. The kth layer is located between the points z zkÿ1 and z zk in the thickness direction and its principal material coordinate oriented at an angle ak to the laminate coordinate, x, as shown in Fig. 3. Let w denote the mid-plane displacement in the directions of z, the lateral mid-surface de¯ection of the plate is governed by the following dierential equation: o4 w o4 w o4 w 4D16 3 2
D12 2D66 2 2 4 ox ox oy ox oy 4 4 2 ow ow ow 4D26 D22 4 I0 2 oxoy 3 oy ot 2 2 2 o ow ow ÿ q 0; ÿ I2 2 ot ox2 oy 2
Fig. 2. An in®nite plate is divided to strip elements in y-direction, while the region bounded by S1 , S2 , S3 and S4 is the problem domain.
D11
1
where Dij are the coecients of the bending stiness, I0 the mass moment and I2 are the rotatory inertia which are given as Z H =2 Z H =2 q dz; I2 qz2 dz:
2 I0 ÿH =2
ÿH =2
When the rotatory inertia is considered, the shear forces will be in the form of
Fig. 3. Geometry and coordination system of the kth layer of a rectangular plate in the x±y plane with ®ber orientation of ak .
oMx oMxy o3 w I2 2 ; ox oy ot ox oMxy oMy o3 w I2 2 : Qy ox oy ot oy
Qx
3
To apply the concept of the SEM, an in®nite length rectangular plate is ®rst considered as shown in Fig. 2. The plate occupies the region of ÿ1 6 x 6 1 and 0 6 y 6 b, and is divided in the y direction into N strip elements. The displacement ®eld in an element is assumed to be W N
yV e
x; t;
4
where matrix N(y) and vector Ve (x) are given by: N
y n1
y n2
y n3
y n4
y n5
y n6
y;
5
V e
x w1
x; t h1
x; t w2
x; t h2
x; t w3
x; t h3
x; t
T
T
v1
x; t v2
x; t v3
x; t v4
x; t v5
x; t v6
x; t
6
Fig. 1. The coordination system of the laminated composite plate.
and wi (i 1; 2; 3) are the lateral de¯ections on the node lines, hi (i 1; 2; 3) are the rotation angles on the node lines given by
Y.Y. Wang et al. / Composite Structures 48 (2000) 265±273
hi
ow : oy yyi
7
The components in matrix N(y) are given in the paper by Wang et al. [12]. Applying the principle of minimum potential energy to the strip element in the in®nite plate, the system of approximate ordinary dierential equation for an element can be obtained [13,14]. e 3 e 2 e o4 V e eo V eo V e oV Be5 V e B B B 2 3 4 ox4 ox3 ox2 ox o2 V e o2 o2 V Pe T e ;
B e6 ÿ Be7 2 ÿ B e8 2 ot ot ox2
8
where coecient matrix B ei
i 1; . . . ; 5, force vector Pe and traction vector on the boundary lines T e for an element have been given by Wang et al. [13]. The components of the matrix B ei
i 6; . . . ; 8 can be obtained from the following integration: Z be e nk nl dy; B6kl I0 Be7kl I2 Be712 I2 Be756 I2 Be8kl
Z
0
be 0
Z
d2 nl dy dy 2
n1
d2 n2 dy I2 ; dy 2
n5
d2 n6 dy ÿ I2 ; dy 2
be
0
Z
be
0
Z I2
nk
be 0
except Be712 and Be756 ;
9
1 ÿ1
P
x; t eÿixt dt;
12
where x is the angular frequency and `' stands for a variable in the transformed domain. The application of the Fourier transform to Eq. (10) leads to the following ordinary dierential equation: ~ ~ ~ d4 V d3 V d2 V dV~ 2 B
B x B B 2 3 8 4 dx4 dx3 dx2 dx 2 ~ ~ ~ B5 V ÿ x
B6 ÿ B 7 V P:
B1
13
The general solution of Eq. (13) consists of two parts, one is the complementary solution which satis®es the homogeneous equation corresponding to Eq. (13), and another is the particular solution which satis®es Eq. (13). 3.1. Complementary solution in the frequency domain
~ d 0 exp
ikx V
nk nl dy;
o4 V o3 V o2 V oV B B B4 B5 V 2 3 ox4 ox3 ox2 ox o2 V o2 o2 V P:
B 6 ÿ B7 2 ÿ B 8 2 ot ot ox2
ÿ1
Z
The complementary solution can be obtained by solving the homogeneous equation corresponding to Eq. ~ 0). Assuming (13) (P
where be is the width of the strip element (see Fig. 2). The details of matrices B ei
i 6; . . . ; 8 are given in Appendix A. Assembling all the strip elements of the domain, T e will vanish except on the boundary lines of y 0, y b and the node lines where there are concentrated loads, a system of approximate dierential equations for the whole domain can be obtained. B1
The variables V and P in Eq. (10) are functions of coordinate x and time t. The integral transform to one parameter will deduce the partial dierential equations to ordinary dierential equations. We introduce the Fourier transformation with respect to the time t as Z 1 ~ x V
x; V
x; t eÿixt dt;
11 ~ x P
x;
B e1
267
10
The matrices B i
i 1; . . . ; 8 and vector P can be obtained by assembling the corresponding matrices and vectors of adjacent elements just like is done in the ®nite element method. If the plate is divided to N strip elements, then Bi (i 1; . . . ; 8) will be M M
M 4N 2 matrices.
and substituting it into the homogeneous equation of Eq. (13), we obtain the following eigenvalue equation with respect to k: k 4 B 1 ÿ ik 3 B 2 ÿ k 2
B 3 x2 B 8 ikB 4 B5 ÿ x2
B 6 ÿ B7 d 0 0:
15
For a given frequency, this equation can be changed to a standard eigenvalue equation as Ad kBd;
16 where 2 6 6 A6 4
0
I
0
0
0
0
I
0
0
0
0
2
ÿB 5 x
B6 ÿ B 7 ÿiB4 3 I 0 0 0 60 I 0 0 7 6 7 B6 7; 40 0 I 0 5 2
0
3. Solution of dierential equations The set of partial dierential equations given in Eq. (10) can be solved using the integral transform method.
14
0
0
I 2
B3 x B8
3 7 7 7; 5
iB 2
B1
17
d T d 0 kd 0 k 2 d 0 k 3 d 0 :
18
268
Y.Y. Wang et al. / Composite Structures 48 (2000) 265±273
Solving Eq. (16), 4M eigenvalues kj (j 1; 2; . . . ; 4M) and eigenvectors which are the functions of x can be obtained. The ®rst M components in the eigenvectors are corresponding to d0 . If the part of the jth eigenvector corresponding to d0 is denoted by Uj
x UTj
x /j1 /j2 . . . /jM
19 the complementary solution can then be written by superposition of these eigenvectors as ~ c
x; x V
4M X j1
G
x; xC
x;
20
where the superscript c indicates the complementary solution and
/1M exp
ik1 x
/21 exp
ik2 x /22 exp
ik2 x .. .
/2M exp
ik2 x
.. .
/L1 exp
ikL x /L2 exp
ikL x .. .
3 7 7 7; 5
/LM exp
ikL x
21
where L 4M. In Eq. (20), C is the constant vector which will be determined using the boundary conditions on S2 and S4 after the particular solution is obtained. 3.2. Particular solution in the frequency domain The particular solution of Eq. (13) can be obtained by a seminumerical method, in which the Fourier transformation with respect to coordinate x is introduced as following: Z 1 ~ ~ p
x; xeÿikx dx; ~ p
k; x V
22 V ~~ x P
k;
Z
ÿ1
1
ÿ1
~ xeÿikx dx; P
x;
23
where superscript p indicates the particular solution. The application of the Fourier transform to Eq. (13) leads to the following equation in the transform domain: ~~ k 4 B ÿ ik 3 B ÿ k 2
B x2 B ikB B P 1 2 3 8 4 5 ~ ~ p: ÿ x2
B6 ÿ B 7 V
where B m /Lm B/Rm km , /Lm
28 /Rm
and are the mth eigenvalues and left and and right eigenvectors which are obtained for a given frequency x from /Lm A ÿ km B 0;
Cj
xUj
x exp
ikj x
G
x; x 2 /11 exp
ik1 x 6 /12 exp
ik1 x 6 6 .. 4 .
Applying the modal analysis technique [15]. we obtain 4M X /Lm p/Rm d ;
27
km ÿ kBm m1
24
This equation can be rewritten as: p A ÿ kBd;
25
where n o ~~ ; pT 0 0 0 ÿ P n o ~ pT k 3 V ~ pT : ~ pT k V ~ pT k 2 V dT V
26
A ÿ km B/Rm 0:
29
It can be seen that km , /Lm and /Rm are the same as that obtained in the previous sub-section. The right eigenvectors /Rm and left eigenvectors /Lm can be written in the form of sub-vectors as 8 R 9 / > > > > > m1 > > > > > > > > R > > = < /m2 > ; /Lm /Lm1 /Lm2 /Lm3 /Lm4 ;
30 /Rm > R > > > > > / > > m3 > > > > > > > > : R ; /m4 where /Rmi
i 1; 2; 3; 4 and /Lmj
j 1; 2; 3; 4 have the same dimension. From Eqs. (26), (27) and (30) we obtain that
4M L R ~~ p
k; x ÿX /m4 P/m1 : V
km ÿ kB m m1
31
Once the external load is speci®ed, the vector of the ~~ can be obtained. Apload Fourier transformation P plying inverse Fourier transformation to Eq. (31), the particular solution of Eq. (13) can be obtained. Z 1 1 ~~ p
k; xeikx dk V V~ p
x; x 2p ÿ1 Z 1X 4M ~~ R 1 /Lm4 P/ m1 eikx dk:
32 2p ÿ1 m1
k ÿ km B m For a line load in the x direction acting at x x0 , the force vector in Eq. (13) can be written as P P0
td
x ÿ x0
33
and the Fourier transformations are given as below. ~ P ~ 0
xd
x ÿ x0 ; P
34
~~ P ~ 0
xeÿikx0 : P
35
Hence, from CauchyÕs theorem, the integration in Eq. (32) can be carried out: 8 2M L X/ P ~ 0 /R > > > i m4 m1 eikm
xÿx0 ; x P x0 ; > < Bm m1 p ~
36 V
x; x 2M Lÿ ~ Rÿ > X > P / ÿ
xÿx 0 /m1 ikm > m4 0 > ÿi e ; x < x0 ; : Bÿ m m1
Y.Y. Wang et al. / Composite Structures 48 (2000) 265±273
where `+' denotes variables evaluated for the cases that the eigenvalues are positive real numbers or the imaginary parts of the eigenvalues are positive, while `ÿ' denotes variables evaluated for the other cases. For a distributed load in the x-direction, the solution can be obtained in the form of superposition integral over the solution of the line load. Such as, for the sinusoidally distributed load in the x-direction, 1 x ÿ p
y; t
37 p
x; y; t sin p 2 a the particular solution can be had from the following integration: 2M L ~ R Z x 1 x X / P0 / p ~ sin p i m4 m1 eikm
xÿx0 dx ÿ V 2 a m1 Bm ÿa=2 X Z a=2 2M ~ 0 /Rÿ ÿ 1 x /Lÿ P sin p ÿ i m4 ÿ m1 eikm
xÿx0 dx ÿ Bm 2 a m1 x
38a which gives that ÿ 2M X ~ 0 /Rÿ /Lm4 P p m1 ~ i V Bÿ m m1
p a
sin
2M X ~ 0 /R /Lm4 P m1 i B m m1
p a
px a
:
38b 3.3. The general solution in the frequency domain
In the present study, zero initial conditions were assumed. A three-layer cross-ply
0°=90°=0° square laminated plate is considered. The length of the plate is a b 0:762 m. All the layers are assumed to be of the same thickness and material properties are given as
~ p
x; x ~ x V ~ c
x; x V V
x;
39
Eq. (39) gives the general solution with an unknown constant vector C of 4M elements. To obtain a special solution for the problem domain, the boundary conditions on S2 and S4 have to be used to determine the constant vector C. Boundary conditions needed to determine C have been discussed in detail by Wang et al. [12,13]. Once the constant vector C is determined, the responses in the frequency domain can be obtained. 3.4. Solution in the time domain The solution in the time domain is given by the inverse Fourier transformation to Eq. (39). Z 1 1 ~ xeixt dx: V
x;
40 V
x; t 2p ÿ1 The integration of Eq. (40) usually has to be carried out numerically. For the undamped plates considered in this
E2 6:895 GPa; m12 0:25;
3
q 1603:03 kg=m :
The load acting on the plate is sinusoidally distributed on the whole surface of the plate and varies with time according to one of the expressions given as follows: 1 x py ÿ sin F
t;
41 q
x; y; t q0 sin p 2 a b
The general solution of Eq. (13) can be given by adding the complementary solution given by Eq. (20) and the particular solution given by Eq. (32). ~ p
x; x: G
x; xC
x V
4. Numerical examples
G12 3:448 GPa;
sin pxa ÿ ikm cos pxa pa eikm
xa=2 ÿ
km 2
pa2
paper, diculties with the integration result from sin~ t
x; x at x 0 and at the cut-o fregularities of V quencies
k 0, as discussed by Vasudeven and Mal [16]. To avoid the singularities, the exponential window method (EWM) is used [10]. In the EWM, a complex frequency, x ÿ ig, is introduced, where x is real and g is positive which does not depend on x. This method is good for computing the results in the time domain, because the contamination in the solution resulting from the introduction of g can be eliminated when the solution is transformed back to the time domain [16].
E1 172:369 GPa;
ÿ ÿ ikmÿ cos pxa ÿ pa eikm
xÿa=2 2 2 ÿ
kmÿ
pa
269
0 6 t 6 t1 ; sine loading; t > t1 ; 1; 0 6 t 6 t1 ; step loading; F
t 0; t > t1 ;
1 ÿ t=t1 ; 0 6 t 6 t1 ; triangular loading; 0; t > t1 sin
pt=t1 ; 0;
eÿct
explosive blast loading;
42 ÿ1
in which t1 0:006 s and c 330 s . The intensity of the transverse load is taken to be q0 68:9476 MPa when H a=5, and q0 689:476 KPa when H a=10 and H a=20. These data are taken from Khdeir and Reddy [6] for comparison purposes. Fig. 4 shows the time history of transverse de¯ection at the center of the plate for various loads. The SEM results are compared with exact solutions which do not consider the rotatory inertia and the exact solutions given by Khdeir and Reddy [6] who considered the rotatory inertia. From Fig. 4, it is found that the SEM solutions have very good agreement with the exact solutions with and without the eect of the rotatory inertia. The stress responses in the time domain can be obtained from the displacement distribution. The
270
Y.Y. Wang et al. / Composite Structures 48 (2000) 265±273
Fig. 4. Time history of the centre de¯ection for simple supported symmetric cross-ply laminated plate subjected to various loading. Comparison of the SEM solution and exact solution with and without rotatory inertia (a 0:762 m, H a=5).
comparison of the SEM solution with the exact solution of normal stress response is given in Fig. 5, in which the normal stresses are normalized as rxx rxx
a=2; b=2; H =2=q0 :
43
Also very good agreements are observed. The rotatory inertia has the eect of decreasing the frequency of the plate or increasing the wavelength. It is also found that the rotatory inertial has less eect on the de¯ection amplitude when t 6 t1 , but has signi®cant eect when t > t1 as shown in Figs. 4 and 5. This is because the plate will be in the state of free vibration after t > t1 under the initial conditions (displacement and velocity) at t t1 . A small change in natural frequency of the plate gives very dierent initial conditions at t t1 . The dierent initial conditions result in large dierence in the amplitude of the free vibration after t > t1 . Figs. 4(c) and 6 show, respectively, the displacement responses for plates with thickness±length ratios of 1/5, 1/10 and 1/20. It is observed that the smaller the
Fig. 5. The time history of the dimensionless normal stress rxx for simple supported symmetric cross-ply laminated plate under step loading. Comparison of the SEM and exact solution with and without rotatory inertia (H a=5).
Y.Y. Wang et al. / Composite Structures 48 (2000) 265±273
271
thickness±length ratio the smaller the dierence between the results with and without the eect of rotatory inertia. When the thickness±length ratio is 1/20, as shown in Fig. 6(b), there are nearly no observable dierences between the results with and without the eect of rotatory inertia. We can therefore conclude that the eect of rotatory inertia can be ignored if the thickness±length ratio is less than 1/20. The rotatory inertia may need to be considered if the thickness±length ratio is between 1/5±1/20. For plates with thickness±length ratio of larger than 1/5, the CLPT is no longer valid, and a higher order plate theory should be used. The eects of elastic constant E1 and material density q have also been investigated and results are shown in
Figs. 7±9. In Fig. 7, the thin line refers to E1 165:474 GPa (decreased about 4%) and the dash line refers to q 1550:16 kg=m3 (decreased about 3.3%) and all other parameters are unchanged. It is also found that the small change of the natural frequency results in large dierence in the amplitude of the free vibration after t > t1 . Comparing Fig. 7(a) and (b), it is found that the dierence of the amplitude after t > t1 is not so large when the rotatory inertia is considered, which shows that the rotatory inertia is very important in the analysis of thick plates. Figs. 8 and 9 show the response of the thin plate where H/a 1/20. From Figs. 7 and 8, it is found that the elastic constant aects not only the natural frequency but also the amplitude. The amplitudes
Fig. 6. The eect of the thickness on the centre of square plate under sine loading.
Fig. 7. The eect of elastic constant E1 and material density q on the centre de¯ection of plate subjected to step loading.
272
Y.Y. Wang et al. / Composite Structures 48 (2000) 265±273
Fig. 8. The eect of the elastic constant E1 on the centre de¯ection of square plate subjected to step loading.
Fig. 9. The eect of material density q on the centre de¯ection of square plate under step loading.
increase as the elastic constant decreases when t < t1 , but it is not true t > t1 in which the initial condition at t t1 will aect the response. The in¯uence of the material density can be found from Figs. 7 and 9, which show that the material density has less eect on the amplitude when t < t1 but will aect the natural frequency and the amplitude when t > t1 . The eects of the ®ber orientation on the dynamic response of symmetric laminated plate have been studied and the results are shown in Fig. 10. It is found that the ®ber orientation of the face layers aects the response of the plate signi®cantly and the ®ber orientation of the inner layer has less eect on the dynamic response.
Fig. 10. The eect of ®ber orientation on the centre de¯ection of plate under sine loading.
5. Conclusions The SEM based on CLPT has been extended for taking into account the rotatory inertia of the plate. The transient responses of a rectangular symmetric laminated plate are presented for various loading using the newly developed SEM program. The parameters which aect the dynamic responses of the composite laminated plate are investigated. It is found that the rotatory inertia has less eect on the thin plate but signi®cant eect on the thick plate. The rotatory inertia can be ignored in the analysis, if the thickness of the plate is less than 1/20 of its length. The elastic constants aect not only the
Y.Y. Wang et al. / Composite Structures 48 (2000) 265±273
natural frequency but also the amplitude of the response. The in¯uence of ®ber orientation of the face layer is much signi®cant than the inner layer. The successful application of the SEM in transient analysis of laminated composite plates with the eect of rotatory inertia shows that the SEM is an eective and accurate numerical method. Appendix A Matrices in Eq. (8): 2
523be 6 3465
6 6 6 6 6 6 6 6 B e6 I0 6 6 6 6 6 6 6 6 4
2
ÿ8b2e 693
131be 6930
ÿ29b2e 13860
2b3e 3465
2b2e 315
ÿb3e 1155
29b2e 13860
ÿb3e 4620
128be 315
0
4be 63
32b3e 3465
8b2e 693 523be 3465
3
7 7 7 7 7 7 2 ÿ2be 7 7 315 7 7; ÿb3e 7 7 1155 7 7 ÿ19b2e 7 7 2310 7 5 2b3e 3465
105be
523be 6 3465
6 6 6 6 6 6 6 6 e B 8 I2 6 6 6 6 6 6 6 6 4
4be 63
sym:
2 ÿ278 6 6 6 6 6 6 6 6 e B 7 I2 6 6 6 6 6 6 6 6 4
19b2e 2310
ÿ13 210
256 105be
ÿ8 21
22 105be
1 70
ÿ2be 45
ÿ8 105
4be 315
ÿ1 70
be 126
ÿ512 105be
0
256 105be
ÿ8 105
ÿ128be 315
8 21
4be 315
ÿ278 105be
13 210
sym:
3 7 7 7 7 7 7 7 7 7; 7 7 7 7 7 7 7 5
ÿ2be 45 19b2e 2310
4be 63
ÿ8b2e 693
131be 6930
ÿ29b2e 13860
2b3e 3465
2b2e 315
ÿb3e 1155
29b2e 13860
ÿb3e 4620
128be 315
0
4be 63
32b3e 3465
8b2e 693
sym:
523be 3465
3
7 7 7 7 7 7 ÿ2b2e 7 7 315 7 7: ÿb3e 7 7 1155 7 7 ÿ19b2e 7 7 2310 7 5 2b3e 3465
273
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