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Vibration of rectangular plates with time-dependent boundary conditions
Journal
of Sound and Vibration
VIBRATION
OF
(1979) 62(3), 327-337
RECTANGULAR
DEPENDENT
BOUNDARY
PLATES
WITH
TIME-
CONDITIONS
J. VENKATARAMANA,M. MAITIAND R. K. SRINIVASAN Structural Engineering Division, Vikram Sarabhai Space Centre, Trivandrum-695022, India (Received 16 December
1976, and in revisedform 13 May 1978)
A general method of solution for the vibration of rectangular plates with any type of timedependent boundary conditions is developed by an extension of the method of Mindlin and Goodman[l]. Forillustration, theproblemsofaplatewithdifferenttime-dependent boundary conditions are solved and the closed form solutions for the transverse deflections of the plate are obtained. The non-dimensionalized transverse deflections, (w/a) at the middle of the plate are evaluated numerically for different dimensions of the plate and different forcing functions. These are presented graphically against the non-dimensionalized time, T, for three cases and tabulated for other cases.
1. INTRODUCTION
Many structural problems are associated with time-dependent boundary conditions. These problems are generally solved by the classical method of separation of variables or the Laplace integral transform. Several authors [2-53 have solved vibration problems of rectangular isotropic plates by the energy method and by Levy’s method for time independent boundary conditions. Some workers [6, ?‘I have analyzed the problems of composite and sandwich plates with a time-dependent dynamic pressure under one-dimensional conditions. In this paper, the method developed by Mindlin and Goodman [l] is extended to the vibration of a finite rectangular plate with time-dependent boundary conditions. Following this derivation, three forced vibration problems are solved for a rectangular plate simply supported along two parallel edges with the other two edges supported in any manner. In problems 1 and 2, a simply supported plate, initially at rest, is considered and two of its edges are subjected to an oscillating bending moment and an oscillating transverse displacement, respectively. Problem 3 deals with a plate with one end subjected to an oscillating transverse displacement and the other to an oscillating bending moment. For the above problems, closed form expressions for the transverse, displacement are obtained and evaluated numerically at the centre of the plate.
2. STATEMENT OF THE PROBLEM AND SOLUTION Consider an isotropic rectangular plate of edge lengths (a, b), thickness h, density p and Poisson’s ratio v (a list of nomenclature is given in the Appendix). Let its two parallel sides x = 0, a be simply supported and subjected to a time-dependent displacement or force. With the classical assumptions, the governing equation of the vibration of the plate is DV4w + ph a2w/at2 = 0, where w(x, y, t) is the transverse deflection of the plate and D = Eh3/12(1 - v2). 327
(1)
328
J. VENKATARAMANA
ET AL
2.1. PROBLEM 1
Consider a simply supported plate, initially at rest, two of whose edges, s = 0, a (say) are subjected to an oscillating bending moment of amplitude M(x. Y) and forcing function f’(t). The boundary conditions along the edges x = 0, a and the initial conditions are then, respectively, w = 0,a2w/ax2 w
=
[MAX, y)p]
awlat
=
= 0
for
f’(t)
x
=
0, U,
(4
for t = 0,
(3)
where f(t) is a known function oft. To make the boundary conditions (2) homogeneous, one takes (4)
w(x, Y, r) = Ux, Y,r) + [MO, Y)H,(x) + Ma, Y)HA(X)] f’(r).
where H,(x) and H,(x) are the unknown functions to be determined. Substituting equation (4) into the boundary conditions (2) gives UO, Y,r) + [MO, Y)H,(O) + No, Y)~,KY f(t) =
U@,Y,t) + [MO,Y) H,(a)
+ fm4
I =0 .v=a
Y) fq4-j
+ Mb,
Y)
m
0,
= 0,
d2fJ,(x)dX2
II
f(t)
=
\=o
Jwo,Y) fW D
’
d2H,(x) + M(a, Y) --$g--
Without any loss of generality, one may assume H,(x) = H,(x) = 0 at x = 0, a, d2Ho(x)/dx2 = l/D d*H,(x)/dx*
= 0
and and
d*H,(x)/dx* d2H,(x)/dx2
= 0
at
x = 0,
= l/D
at
x = a.
Expressions for H,(x) and H,(x) satisfying these equations and conditions at x = 0, a then are 1 ax x2 x3 H,(x)
= D
- 7
+ T
1
- SQ ,
H,(x)=&ax+;].
With the above expressions, the boundary conditions and initial conditions of the problem reduce to u(x,
y, t)
=
y, q/ax* = 0 at x = 0,U,
a*cqX,
ux, Y, 0) = f(O) Ax, Y),
i&G Y, 0) = j_(O) Ax, Y),
(5) (6)
where J(x, y) = M(0, y) H,(x) + M(0, y) H,(x), and the superscript dot denotes the differentiation with respect to time. The vibration equation (1) now can be written as V4U(x, Y, t) + (pm
a2ub,
Y, q/at2
= Fb,
Y, t),
(7)
F(x, Y,r) = - [f(t) V4J(x, Y) + M/D) J(x, Y) h,]. Hence, the problem reduces to the solution of equation (7) with the homogeneous boundary and initial conditions(5) and(6). Applying the classical separation of variable technique, one takes
u(x, y, t)
=
sin
f El =
1
c) y
k;,(Y?4
TIME-DEPENDENT BOUNDARY CONDITIONS
329
and equation (7) becomes
Multiplying both sides by sin (mrrx/a), integrating with respect to x from 0 to a and using the orthogonal property gives
(8) WY,
0 = a‘f
F(x, y,t)sine)dx.
At this stage, in the general case, one introduces the boundary conditions along the edges y = 0 and y = b. As specific illustrations of the general procedure we consider here two types of boundary conditions along y = 0 and y = b. 2.1.1. Case 1 Let the edges y = 0, b be simply supported: i.e., MJ= 0, M, = 0 for y = 0, b. Proceeding as before, the boundary and initial conditions on Y,(y, t) are
y,(O,0 = - ; fW,,
K, =
s s
= [M(O,O)H,(x) + M(a, O)H,(x)] sin 0
ym(b, t) = - ;
fw,,
K, =
’[M(O,@H,(x) + M(a, b)H,(x)]
sin
0
a*Y,(Y, t)
a4.2 y=b
f f WL
H,(x)
y,(Y, 0) =
-
f .f wm,
Yv-,
+ Mb,
+
a*hm Y) ay*
Iy=b
HA(x) 1
YvyJ,
(9) I., = f: H,(x)sin[y)dx,L,
= 1: H,(x)sin[y)dx.
To make the boundary conditions homogeneous at the edges y = 0, b, we assume Y,(Y, 6 =
V,(Y, t) +
f @)CH,,W +
H&Y)~.
As before, substituting equation (10) in the above boundary conditions (9) yields H,,(y)
= d*H,,(y)/dy*
= 0 for y = 0,
(10)
J.
330 H,,(y)
=
ET AL
VENKATARAMANA
d’H,,(y)/dy’
= Ofor y = h,
H,,(y) = -(2/a)K,,
d2H,,(y)/dy2
= -(2/a)K,
for y = 0,
H,,(y) = -(~/UK,,
d2H,,(y)/dy2
= -(2/a)K,
for y = b,
and thence one obtains expressions for H,,(y) and H,,(y) which make the boundary conditions on y”(y, t) homogeneous. Then equation (8) reduces to
where G(Y,t)=G(y,t)-f(t)[~-2(~~~+(~!1N(y)]-~N(Y)j(t), NY) = H,,(Y)
+ H,,(Y).
With I/;,(y, t) = 2 Z,,(y)X,,,(t), this equation becomes fl=l f
[{3
- 2(33
+ [(3
- P’jZ,““lx,n”
n=l
+ ph Z D
= G(y, t).
(11)
>I
If the expression in the curly brackets { } on the left-hand side of equation (11) becomes zero, then equation (11) reduces to = GY, t),
or
d2X,,J) O2 ___
dt2
+
ph
s b
x,,(t) = F&t),
0
Solution of this equation is
s b
C(Y, W,,,,(y)dy/
0
Z,~,,(Y)dy.
f
X,,(t) = A,,(t) cos (4&) + D,, sin (4,t)
+ 41
F&4 sin {q,,(t
- 4) dz,
mns 0
where A, and J3, are constants to be evaluated by applying the initial conditions a
&JO) = - f(O)
{wo,y)L
; {WA
, +
Ma,
Y)&)
Y)L, + Ma, yW,f
+ MY) ] Z,,(Y) dY/ +
The function Z,,,,(y) is obtained from the solution of the equation
with the boundary conditions Z,,(y) = d2Z,(y)/dy2
= 0 for y = 0, b.
(12)
TIME-DEPENDENT BOUNDARY CONDITIONS
331
Hence, Z,,(y)
= sin B,,Y, B,, =
Jp2 - (md42,
where the characteristic equation is sin /?,,,b = 0. Therefore, the deflection of the plate for this problem is w(x, y, t) = jJ sin ( ‘F I”=
1
\
u
\f f
sin (B,,y)XJt)
+ NYU”@) + J(x, Y)f(0 1
/C.=l
IfM(y) = constant = M(say) w(x, y, t) = f m=l
sin ‘F (
5 X,,(t) sin (B,y) -y >G =1
3 (cos mlt - l)f(t)
(--$ >
f@Hx2
+ g
(13)
ax),
where 1 P
X,,(t) = 4
F3(z) sin [q,,(t - z)]dz.
mnJ 0
2.1.2. Case 2 As Case 2, one assumes that the edges y = 0 and y = b are clamped: i.e., w = 0 and LJw/ay = 0 for y = 0, b. Proceeding in the same way as in Case 1, one obtains the deflection of the plate as w(x, y, t) = jJ sin [y){ I”=1
il
_(cosh Cl,,b _ cos /jJ,,
X,,,,,($cosh
a,,,y - sin B,,d
(&“dsinh %“y - %I”sin &“Iy) (8, sinh a,nnb - a,,, sin /I,,,&)
-(2M/a)(a/mn)3(cos
m7t - l)f(t)
I
1
+ (M/2D)f(t)(x2
- ax),
(14)
where the characteristic equation is
[cash&,,,b) ~0s (B,J - II/sin h&,,,b)sin(B,J) = b,f,.- Bf,J/2~,,,&,, . 2.1.3. Simple harmonic time dependence In particular, with f(t) = sin ot, one obtains the following non-dimensional for the transverse displacement for Case I :
expression
sin (mwX) sin (m7cY)
cos m7r - 1) sin (QT) sin (m7rX) + $
(X2 - X) sin (QT),
where C, = sin (QT) - (&k+r2)(1/[m2 + (a/Q2n2]} sin {[m” + (~/W21~/~}, C, = [m2 + (a/b)2n2]2, C, = 1 - (f22a2/n4)/[m2
+ (a/b)2n2]2.
(15)
332 2.1.
J. VENKATARAMANA PROBLEM
E 7 .4 L.
2
A simply supported plate is initially at rest; its edges x = 0, a (say) are then forced into oscillating displacements, the forcing function being .f’(t). Hence the boundary conditions along the edges x = 0, a are
w = A(x, y)af(t), a2~/dxZ = 0 for x = 0, a.
(16)
Proceeding as before, with the particular values A(x, y) = A and f’(t) = sin ox, one obtains, as the non-dimensional expression for the transverse displacement in Case 1, w(X, Y T) = !$
-$y; 2.3.
PROBLEM
f f m3(1 - cosCmz(l - ‘OS nn)C1 sin (m7cX)sin (nnY) m=l n=l 2 3
(1 - cos m7r)sin (OT) sin (mnX)
+ A
sin (QT).
(17)
3
A simply supported plate is initially at rest: one edge (x = 0) is then subjected to an oscillating displacement and the other one (x = a) to an oscillating bending moment. The boundary conditions along the two edges of the plate then can be written as
w
a2w/ax2 = 0 for
w =
A(x,
=
a2wjax2 = [M(x,
0,
y)af(c),
y)/D]fft)
x
for
= x
0, =
a.
(18)
Proceeding as in Problem I, with the particular values A(x, y) = A, M(x, y) = M and f(t) = sin wt, one obtains, for the non-dimensional transverse displacement in Case I,
W(X, Y, T) =
1 -n~cm’ r)c,c, sin (mnX) sin (nzY) $&f f m3’ m=l
n=l
-L! 7 1 c, 7cLum* m= 1
2
3
sin (mnX) sin (LIT) + AC, sin (QT), ~ ’ ’
(19)
where C, = 1 - cos(mn) + { 1 + (aM/D)/6A) cos(mk) + [(aM/D)/6A][(6/mZn2)
- 11.
c, = 1 - {I + (aM/D)/6A}X + {(aM/D)/64X3.
3. NUMERICAL
RESULTS
AND
CONCLUSIONS
With the assumed values u/h = 100, aM/D = A = 1, X = Y = 0.5, expressions ( 15), (17) and (19) have been evaluated at different time (T) intervals for different sets of values of a/b and 51.These are presented in the Tables I, 2 and 3 respectively. Graphs of the nondimensional transverse displacements for each problem, plotted against T, are shown for some values of a/b and B in Figures 1,2 and 3. For Problem 1, the amplitude of W is minimum for the square plate (u/b = 1). For a < b, the amplitude of W is greater than that for a square plate and the period decreases slightly as T increases. The case a > b gives the maximum amplitude of W and the period decreases further. For Problems 2 and 3, there is a marginal change in the amplitude of W for different a/b
0.0 0.4 0.8 l-2 1.6 2.0 2.4 2.8 3.2 3.6 4.0 4.4 4.8 5.2 5.6 6.0
T
a
OJIOOO 0*0041 OGO87 OGO89 0.0171 - 0.0145 0.0264 0.1278 0.0426 -0.1909 - 0.4842 - 0.6232 - 06090 -0.3411 0.2201 0.9737
6=‘,“=’ O@OOO om40 00086 0.0096 0.0172 -0.0159 0.0284 0.1225 0.0530 -0.1845 - 0.4842 - 0.6702 - 0.6233 - 0.2666 0.3895 1.2018
a jj = o-5, 61 = 1.0
Non-dimensional a
OGOOO 0.0042 OGO87 om86 0.0198 - 0.0231 O-0387 0.1589 O-0106 - 0.3229 - 0.6359 - 0.6642 -0.3151 0.405 1 1.2832 2.1100
T;=2.0,P=
1-o
TABLE 1 transverse displacement
0 WOO OJIO81 o-01 34 OGOS1 0.0109 - 0.0600 owo2 0.2382 - 0.2588 - 0.5465 -0.8517 -0-5950 0.1844 1.1492 1.8864 1.9113
5; = 1.0, P = 2.0
a
-
-
2.2133 2.1071
1.3963
OGIOO om79 0.0133 oGO47 O-0117 0.0622 0.0449 0.2281 0.0034 0.5391 0.9229 0.6756 0.2257
a 6 = 0.5, a = 2.0
W x lo3 (X = Y = 0.5)
OGOOO OGO82 0.0134 oh42 0.0162 - 0.0782 0.068 1 0.2950 -0-1219 -0.8177 - 1.0755 -0.3280 l-1423 2-5361 2.8180 1.6813
a 6 = 2.0, c2 = 2.0
0.0 0.4 0.8 1.2 1.6 2.0 2.4 2.8 3.2 3.6 4.0 4.4 4.8 5.2 5.6 6.0
T
-
-
-
-
____~ OGOOO 0.01260 0.03963 0.08760 0.29624 0.67621 0.2363 2.22007 1.7994 0.3472 3.3143 5.6298 6.73055 6.2158 3.0153 0.2800
a 6=l,Q=l
TABLE 2
=
0.5, Q = 1
OGOOO 0.0I 70 - 0.0354 0.0983 - 0.2981 O-7210 0.1553 -2.1286 - 1.9154 0.2397 3.5301 5.9569 6.9577 5.8601 2.3414 - 1.2943
~~.~~_
a
&
0.0000 0.0104 - 0*0400 0.0970 - 0.3838 0.8914 0.0807 - 2.8200 - 1.7078 1.6080 5.4821 6.8894 6.1848 2.4141 -2.3212 - 6.8981
6 = 2.0, a = 1 a __._~_ O.oooO O-00143 -0.1131 0.1610 - 0.5903 I.4980 0.1143 - 4.9881 -2.3152 4.5467 10.3251 9.5980 2.4255 - 7.2948 - 15.8307 - 14.0685
g=l,Q=2 a .~
OGOOO 0.0101 -0.1084 0.1718 -0.61419 15648 - 0.0140 -4.81 19 - 2.5726 4.3880 10.9304 10.2623 2.43567 - 8.7956 - 17.5931 - 15-200
7; a = 03, Q = 2
Non-dimensional transverse displacement W y. 15 (X = Y = 03) u
0~0000 - 0.0027 --0.1 117 O-l867 - 0.7608 1.9663 -0.2310 - 6.3630 _ I.5844 7.7416 13.9134 8.6004 - 3.9215 _ 18.0563 _ ‘2.3704 - 13.521’
s= 2.0, !2 =
2
0.0 0.4 0.8 I.2 1.6 2.0 2.4 2.8 3.2 3-6 4.0 4.4 4.8 5.2 5.6 6.0
T
a
0aooO + 0@062 -0.0199 0.0436 -0.1481 0.3379 0.1029 - 1.1094 - 0.8990 0.1730 1.6547 2.8109 3.3608 3.1049 1.5081 -0.1344
6=‘,“=1
OGO84 -0-0177 0.04900 -0.1491 0.3603 0.0773 - l-0637 - 0.9569 o-1193 1.7627 2.9740 3.4741 2.9275 I.1 724 - 0.6398
oaooo
a & = 0.5, a = 1
Non-dimensional
TABLE 3
04000 om5 1 - 0~0200 0.0483 -0.1919 04455 0@4001 - 1.4488 - 0.8531 O-8026 2.7371 34400 3.0895 1.2090 - 1.1525 - 3.4346
a 6 = 2.0, f2 = I
transverse displacement a
OGOOO OGOO6 - 0.0566 0.0802 - 0.2949 0.7488 0.0568 - 2.4924 - 1.1565 0.2707 5.1565 4.7930 1.2115 -3.6410 - 7.9022 - 7.0189
+,f2=2
-
-
-
-
06000 om49 0.0542 0.0856 0.3068 0.0783 om73 2.4044 1.285 2.1914 5.4587 5.1257 1.2168 4.3897 8.7549 7.5826
a g = 0.5, 51 = 2
W x 15 (X = Y = 0.5)
2,0, sz = 2 0~0ooo -0-0015 - 0.0559 0.0931 -0.3801 0.9830 -0-1068 -3.1793 -0.7913 3.8662 6.9485 4.2951 - 1.9556 -9.0111 -11.1636 - 6.7447
;=
0.8 -
“0 f
0.60.4-
0.2-
-0.2
-
-0.4
-
-0.6
-
-0.8
I I.0
-1.0
I 2.0
I 3.0
I 4.0
I 6.0
I
5.0
I 7.0
r
Figure 1. Transverse
displacement
vs. time (Problem I
I
1). a/b:
----,
I
I
0.5; -----,
I
I
1.0; ~ -I
7.06.05.04.03.02.0l.O-
-2.o-3.o-4.O\
-5*o-6.O-
\, -7.0
1 I.0
I 2.0
I 3.0
I 4.0
I 5.0
6.0
I 7.0
T
Figure
2. Transverse 60
I
displacement I
vs. time (Problem I
I
2). Key as Figure I
I
1
I
5.04.03.02.0I.0Lo s -1.0-2.0 -3.o-
\ \
-4.o-
\
-5-o-6.0
Figure
I I.0
3. Transverse
I 2.0
displacement
I 3.0
I 4.0
OS. time (Problem
I 5.0
1 6.0
I 7.0
3). Key as Flgure
I
‘-,
2.
TIME-DJL’SNDENT
BOUNDARY
CONDITIONS
337
ratios but the periods for u/b 5 1 are less than those for a square plate (a/b = 1) as T increases. With decrease ofa, the period of oscillations for all problems increases and a reduction of amplitude of W is observed as T increases.
REFERENCES
1. R. D. MINDLIN and L. E. GOOUMAN1950 Transactions of the American Society of Mechanical Engineers, Journal of Applied Mechanics 17,371-380. Beam vibrations with time-dependent boundary conditions. 2. D. YOUNG 1950 Transactions of the American Society of Mechanical Engineers, Journal of Applied Mechanics 17,448450. Vibration of rectangular plates by the Ritz method. 3. M. V. BARTON1951 Transactions of the American Society of Mechanical Engineers, Journal of Applied Mechanics 73, 129-131. Vibrations of rectangular and skew cantilever plates. 4. Y. C. DAS 1961 Journal of Aeronautical Society of India 13, 111-125. On the transverse vibration of rectangular isotropic plates. 5. M. LEVY1899 Comptes Rendus 129, 535-537. Sur l’equilibre elastique dune plaque. 6. C. T. SUN and J. M. WHITNEY 1974 Journal of the Acoustical Society of America 55, 1003-1008. Forced vibrations of laminated composite plates in cylindrical bending. 7. Y. Y. Yu 1960 Transactions of the American Society of Mechanical Engineers, Journal of Applied Mechanics 27, 535-540. Forced flexural vibrations of sandwich plates in plane strain.
APPENDIX : NOMENCLATURE
a, b lengths of sides of rectangular plates x9 Y position co-ordinates A constant, amplitude of transverse displacement M constant, amplitude of bending moment h plate thickness W transverse displacement t time radian frequency w, ; (w, x)/a
Y
E P
ylb
a
modulus of elasticity mass density Poisson’s ratio oa Jp(I - v’)/E
D
Eh3/12C1 - v’)
a: R
h2/12a2 aM/D
V
t/q/d1 - v2YE
of Sound and Vibration
VIBRATION
OF
(1979) 62(3), 327-337
RECTANGULAR
DEPENDENT
BOUNDARY
PLATES
WITH
TIME-
CONDITIONS
J. VENKATARAMANA,M. MAITIAND R. K. SRINIVASAN Structural Engineering Division, Vikram Sarabhai Space Centre, Trivandrum-695022, India (Received 16 December
1976, and in revisedform 13 May 1978)
A general method of solution for the vibration of rectangular plates with any type of timedependent boundary conditions is developed by an extension of the method of Mindlin and Goodman[l]. Forillustration, theproblemsofaplatewithdifferenttime-dependent boundary conditions are solved and the closed form solutions for the transverse deflections of the plate are obtained. The non-dimensionalized transverse deflections, (w/a) at the middle of the plate are evaluated numerically for different dimensions of the plate and different forcing functions. These are presented graphically against the non-dimensionalized time, T, for three cases and tabulated for other cases.
1. INTRODUCTION
Many structural problems are associated with time-dependent boundary conditions. These problems are generally solved by the classical method of separation of variables or the Laplace integral transform. Several authors [2-53 have solved vibration problems of rectangular isotropic plates by the energy method and by Levy’s method for time independent boundary conditions. Some workers [6, ?‘I have analyzed the problems of composite and sandwich plates with a time-dependent dynamic pressure under one-dimensional conditions. In this paper, the method developed by Mindlin and Goodman [l] is extended to the vibration of a finite rectangular plate with time-dependent boundary conditions. Following this derivation, three forced vibration problems are solved for a rectangular plate simply supported along two parallel edges with the other two edges supported in any manner. In problems 1 and 2, a simply supported plate, initially at rest, is considered and two of its edges are subjected to an oscillating bending moment and an oscillating transverse displacement, respectively. Problem 3 deals with a plate with one end subjected to an oscillating transverse displacement and the other to an oscillating bending moment. For the above problems, closed form expressions for the transverse, displacement are obtained and evaluated numerically at the centre of the plate.
2. STATEMENT OF THE PROBLEM AND SOLUTION Consider an isotropic rectangular plate of edge lengths (a, b), thickness h, density p and Poisson’s ratio v (a list of nomenclature is given in the Appendix). Let its two parallel sides x = 0, a be simply supported and subjected to a time-dependent displacement or force. With the classical assumptions, the governing equation of the vibration of the plate is DV4w + ph a2w/at2 = 0, where w(x, y, t) is the transverse deflection of the plate and D = Eh3/12(1 - v2). 327
(1)
328
J. VENKATARAMANA
ET AL
2.1. PROBLEM 1
Consider a simply supported plate, initially at rest, two of whose edges, s = 0, a (say) are subjected to an oscillating bending moment of amplitude M(x. Y) and forcing function f’(t). The boundary conditions along the edges x = 0, a and the initial conditions are then, respectively, w = 0,a2w/ax2 w
=
[MAX, y)p]
awlat
=
= 0
for
f’(t)
x
=
0, U,
(4
for t = 0,
(3)
where f(t) is a known function oft. To make the boundary conditions (2) homogeneous, one takes (4)
w(x, Y, r) = Ux, Y,r) + [MO, Y)H,(x) + Ma, Y)HA(X)] f’(r).
where H,(x) and H,(x) are the unknown functions to be determined. Substituting equation (4) into the boundary conditions (2) gives UO, Y,r) + [MO, Y)H,(O) + No, Y)~,KY f(t) =
U@,Y,t) + [MO,Y) H,(a)
+ fm4
I =0 .v=a
Y) fq4-j
+ Mb,
Y)
m
0,
= 0,
d2fJ,(x)dX2
II
f(t)
=
\=o
Jwo,Y) fW D
’
d2H,(x) + M(a, Y) --$g--
Without any loss of generality, one may assume H,(x) = H,(x) = 0 at x = 0, a, d2Ho(x)/dx2 = l/D d*H,(x)/dx*
= 0
and and
d*H,(x)/dx* d2H,(x)/dx2
= 0
at
x = 0,
= l/D
at
x = a.
Expressions for H,(x) and H,(x) satisfying these equations and conditions at x = 0, a then are 1 ax x2 x3 H,(x)
= D
- 7
+ T
1
- SQ ,
H,(x)=&ax+;].
With the above expressions, the boundary conditions and initial conditions of the problem reduce to u(x,
y, t)
=
y, q/ax* = 0 at x = 0,U,
a*cqX,
ux, Y, 0) = f(O) Ax, Y),
i&G Y, 0) = j_(O) Ax, Y),
(5) (6)
where J(x, y) = M(0, y) H,(x) + M(0, y) H,(x), and the superscript dot denotes the differentiation with respect to time. The vibration equation (1) now can be written as V4U(x, Y, t) + (pm
a2ub,
Y, q/at2
= Fb,
Y, t),
(7)
F(x, Y,r) = - [f(t) V4J(x, Y) + M/D) J(x, Y) h,]. Hence, the problem reduces to the solution of equation (7) with the homogeneous boundary and initial conditions(5) and(6). Applying the classical separation of variable technique, one takes
u(x, y, t)
=
sin
f El =
1
c) y
k;,(Y?4
TIME-DEPENDENT BOUNDARY CONDITIONS
329
and equation (7) becomes
Multiplying both sides by sin (mrrx/a), integrating with respect to x from 0 to a and using the orthogonal property gives
(8) WY,
0 = a‘f
F(x, y,t)sine)dx.
At this stage, in the general case, one introduces the boundary conditions along the edges y = 0 and y = b. As specific illustrations of the general procedure we consider here two types of boundary conditions along y = 0 and y = b. 2.1.1. Case 1 Let the edges y = 0, b be simply supported: i.e., MJ= 0, M, = 0 for y = 0, b. Proceeding as before, the boundary and initial conditions on Y,(y, t) are
y,(O,0 = - ; fW,,
K, =
s s
= [M(O,O)H,(x) + M(a, O)H,(x)] sin 0
ym(b, t) = - ;
fw,,
K, =
’[M(O,@H,(x) + M(a, b)H,(x)]
sin
0
a*Y,(Y, t)
a4.2 y=b
f f WL
H,(x)
y,(Y, 0) =
-
f .f wm,
Yv-,
+ Mb,
+
a*hm Y) ay*
Iy=b
HA(x) 1
YvyJ,
(9) I., = f: H,(x)sin[y)dx,L,
= 1: H,(x)sin[y)dx.
To make the boundary conditions homogeneous at the edges y = 0, b, we assume Y,(Y, 6 =
V,(Y, t) +
f @)CH,,W +
H&Y)~.
As before, substituting equation (10) in the above boundary conditions (9) yields H,,(y)
= d*H,,(y)/dy*
= 0 for y = 0,
(10)
J.
330 H,,(y)
=
ET AL
VENKATARAMANA
d’H,,(y)/dy’
= Ofor y = h,
H,,(y) = -(2/a)K,,
d2H,,(y)/dy2
= -(2/a)K,
for y = 0,
H,,(y) = -(~/UK,,
d2H,,(y)/dy2
= -(2/a)K,
for y = b,
and thence one obtains expressions for H,,(y) and H,,(y) which make the boundary conditions on y”(y, t) homogeneous. Then equation (8) reduces to
where G(Y,t)=G(y,t)-f(t)[~-2(~~~+(~!1N(y)]-~N(Y)j(t), NY) = H,,(Y)
+ H,,(Y).
With I/;,(y, t) = 2 Z,,(y)X,,,(t), this equation becomes fl=l f
[{3
- 2(33
+ [(3
- P’jZ,““lx,n”
n=l
+ ph Z D
= G(y, t).
(11)
>I
If the expression in the curly brackets { } on the left-hand side of equation (11) becomes zero, then equation (11) reduces to = GY, t),
or
d2X,,J) O2 ___
dt2
+
ph
s b
x,,(t) = F&t),
0
Solution of this equation is
s b
C(Y, W,,,,(y)dy/
0
Z,~,,(Y)dy.
f
X,,(t) = A,,(t) cos (4&) + D,, sin (4,t)
+ 41
F&4 sin {q,,(t
- 4) dz,
mns 0
where A, and J3, are constants to be evaluated by applying the initial conditions a
&JO) = - f(O)
{wo,y)L
; {WA
, +
Ma,
Y)&)
Y)L, + Ma, yW,f
+ MY) ] Z,,(Y) dY/ +
The function Z,,,,(y) is obtained from the solution of the equation
with the boundary conditions Z,,(y) = d2Z,(y)/dy2
= 0 for y = 0, b.
(12)
TIME-DEPENDENT BOUNDARY CONDITIONS
331
Hence, Z,,(y)
= sin B,,Y, B,, =
Jp2 - (md42,
where the characteristic equation is sin /?,,,b = 0. Therefore, the deflection of the plate for this problem is w(x, y, t) = jJ sin ( ‘F I”=
1
\
u
\f f
sin (B,,y)XJt)
+ NYU”@) + J(x, Y)f(0 1
/C.=l
IfM(y) = constant = M(say) w(x, y, t) = f m=l
sin ‘F (
5 X,,(t) sin (B,y) -y >G =1
3 (cos mlt - l)f(t)
(--$ >
f@Hx2
+ g
(13)
ax),
where 1 P
X,,(t) = 4
F3(z) sin [q,,(t - z)]dz.
mnJ 0
2.1.2. Case 2 As Case 2, one assumes that the edges y = 0 and y = b are clamped: i.e., w = 0 and LJw/ay = 0 for y = 0, b. Proceeding in the same way as in Case 1, one obtains the deflection of the plate as w(x, y, t) = jJ sin [y){ I”=1
il
_(cosh Cl,,b _ cos /jJ,,
X,,,,,($cosh
a,,,y - sin B,,d
(&“dsinh %“y - %I”sin &“Iy) (8, sinh a,nnb - a,,, sin /I,,,&)
-(2M/a)(a/mn)3(cos
m7t - l)f(t)
I
1
+ (M/2D)f(t)(x2
- ax),
(14)
where the characteristic equation is
[cash&,,,b) ~0s (B,J - II/sin h&,,,b)sin(B,J) = b,f,.- Bf,J/2~,,,&,, . 2.1.3. Simple harmonic time dependence In particular, with f(t) = sin ot, one obtains the following non-dimensional for the transverse displacement for Case I :
expression
sin (mwX) sin (m7cY)
cos m7r - 1) sin (QT) sin (m7rX) + $
(X2 - X) sin (QT),
where C, = sin (QT) - (&k+r2)(1/[m2 + (a/Q2n2]} sin {[m” + (~/W21~/~}, C, = [m2 + (a/b)2n2]2, C, = 1 - (f22a2/n4)/[m2
+ (a/b)2n2]2.
(15)
332 2.1.
J. VENKATARAMANA PROBLEM
E 7 .4 L.
2
A simply supported plate is initially at rest; its edges x = 0, a (say) are then forced into oscillating displacements, the forcing function being .f’(t). Hence the boundary conditions along the edges x = 0, a are
w = A(x, y)af(t), a2~/dxZ = 0 for x = 0, a.
(16)
Proceeding as before, with the particular values A(x, y) = A and f’(t) = sin ox, one obtains, as the non-dimensional expression for the transverse displacement in Case 1, w(X, Y T) = !$
-$y; 2.3.
PROBLEM
f f m3(1 - cosCmz(l - ‘OS nn)C1 sin (m7cX)sin (nnY) m=l n=l 2 3
(1 - cos m7r)sin (OT) sin (mnX)
+ A
sin (QT).
(17)
3
A simply supported plate is initially at rest: one edge (x = 0) is then subjected to an oscillating displacement and the other one (x = a) to an oscillating bending moment. The boundary conditions along the two edges of the plate then can be written as
w
a2w/ax2 = 0 for
w =
A(x,
=
a2wjax2 = [M(x,
0,
y)af(c),
y)/D]fft)
x
for
= x
0, =
a.
(18)
Proceeding as in Problem I, with the particular values A(x, y) = A, M(x, y) = M and f(t) = sin wt, one obtains, for the non-dimensional transverse displacement in Case I,
W(X, Y, T) =
1 -n~cm’ r)c,c, sin (mnX) sin (nzY) $&f f m3’ m=l
n=l
-L! 7 1 c, 7cLum* m= 1
2
3
sin (mnX) sin (LIT) + AC, sin (QT), ~ ’ ’
(19)
where C, = 1 - cos(mn) + { 1 + (aM/D)/6A) cos(mk) + [(aM/D)/6A][(6/mZn2)
- 11.
c, = 1 - {I + (aM/D)/6A}X + {(aM/D)/64X3.
3. NUMERICAL
RESULTS
AND
CONCLUSIONS
With the assumed values u/h = 100, aM/D = A = 1, X = Y = 0.5, expressions ( 15), (17) and (19) have been evaluated at different time (T) intervals for different sets of values of a/b and 51.These are presented in the Tables I, 2 and 3 respectively. Graphs of the nondimensional transverse displacements for each problem, plotted against T, are shown for some values of a/b and B in Figures 1,2 and 3. For Problem 1, the amplitude of W is minimum for the square plate (u/b = 1). For a < b, the amplitude of W is greater than that for a square plate and the period decreases slightly as T increases. The case a > b gives the maximum amplitude of W and the period decreases further. For Problems 2 and 3, there is a marginal change in the amplitude of W for different a/b
0.0 0.4 0.8 l-2 1.6 2.0 2.4 2.8 3.2 3.6 4.0 4.4 4.8 5.2 5.6 6.0
T
a
OJIOOO 0*0041 OGO87 OGO89 0.0171 - 0.0145 0.0264 0.1278 0.0426 -0.1909 - 0.4842 - 0.6232 - 06090 -0.3411 0.2201 0.9737
6=‘,“=’ O@OOO om40 00086 0.0096 0.0172 -0.0159 0.0284 0.1225 0.0530 -0.1845 - 0.4842 - 0.6702 - 0.6233 - 0.2666 0.3895 1.2018
a jj = o-5, 61 = 1.0
Non-dimensional a
OGOOO 0.0042 OGO87 om86 0.0198 - 0.0231 O-0387 0.1589 O-0106 - 0.3229 - 0.6359 - 0.6642 -0.3151 0.405 1 1.2832 2.1100
T;=2.0,P=
1-o
TABLE 1 transverse displacement
0 WOO OJIO81 o-01 34 OGOS1 0.0109 - 0.0600 owo2 0.2382 - 0.2588 - 0.5465 -0.8517 -0-5950 0.1844 1.1492 1.8864 1.9113
5; = 1.0, P = 2.0
a
-
-
2.2133 2.1071
1.3963
OGIOO om79 0.0133 oGO47 O-0117 0.0622 0.0449 0.2281 0.0034 0.5391 0.9229 0.6756 0.2257
a 6 = 0.5, a = 2.0
W x lo3 (X = Y = 0.5)
OGOOO OGO82 0.0134 oh42 0.0162 - 0.0782 0.068 1 0.2950 -0-1219 -0.8177 - 1.0755 -0.3280 l-1423 2-5361 2.8180 1.6813
a 6 = 2.0, c2 = 2.0
0.0 0.4 0.8 1.2 1.6 2.0 2.4 2.8 3.2 3.6 4.0 4.4 4.8 5.2 5.6 6.0
T
-
-
-
-
____~ OGOOO 0.01260 0.03963 0.08760 0.29624 0.67621 0.2363 2.22007 1.7994 0.3472 3.3143 5.6298 6.73055 6.2158 3.0153 0.2800
a 6=l,Q=l
TABLE 2
=
0.5, Q = 1
OGOOO 0.0I 70 - 0.0354 0.0983 - 0.2981 O-7210 0.1553 -2.1286 - 1.9154 0.2397 3.5301 5.9569 6.9577 5.8601 2.3414 - 1.2943
~~.~~_
a
&
0.0000 0.0104 - 0*0400 0.0970 - 0.3838 0.8914 0.0807 - 2.8200 - 1.7078 1.6080 5.4821 6.8894 6.1848 2.4141 -2.3212 - 6.8981
6 = 2.0, a = 1 a __._~_ O.oooO O-00143 -0.1131 0.1610 - 0.5903 I.4980 0.1143 - 4.9881 -2.3152 4.5467 10.3251 9.5980 2.4255 - 7.2948 - 15.8307 - 14.0685
g=l,Q=2 a .~
OGOOO 0.0101 -0.1084 0.1718 -0.61419 15648 - 0.0140 -4.81 19 - 2.5726 4.3880 10.9304 10.2623 2.43567 - 8.7956 - 17.5931 - 15-200
7; a = 03, Q = 2
Non-dimensional transverse displacement W y. 15 (X = Y = 03) u
0~0000 - 0.0027 --0.1 117 O-l867 - 0.7608 1.9663 -0.2310 - 6.3630 _ I.5844 7.7416 13.9134 8.6004 - 3.9215 _ 18.0563 _ ‘2.3704 - 13.521’
s= 2.0, !2 =
2
0.0 0.4 0.8 I.2 1.6 2.0 2.4 2.8 3.2 3-6 4.0 4.4 4.8 5.2 5.6 6.0
T
a
0aooO + 0@062 -0.0199 0.0436 -0.1481 0.3379 0.1029 - 1.1094 - 0.8990 0.1730 1.6547 2.8109 3.3608 3.1049 1.5081 -0.1344
6=‘,“=1
OGO84 -0-0177 0.04900 -0.1491 0.3603 0.0773 - l-0637 - 0.9569 o-1193 1.7627 2.9740 3.4741 2.9275 I.1 724 - 0.6398
oaooo
a & = 0.5, a = 1
Non-dimensional
TABLE 3
04000 om5 1 - 0~0200 0.0483 -0.1919 04455 0@4001 - 1.4488 - 0.8531 O-8026 2.7371 34400 3.0895 1.2090 - 1.1525 - 3.4346
a 6 = 2.0, f2 = I
transverse displacement a
OGOOO OGOO6 - 0.0566 0.0802 - 0.2949 0.7488 0.0568 - 2.4924 - 1.1565 0.2707 5.1565 4.7930 1.2115 -3.6410 - 7.9022 - 7.0189
+,f2=2
-
-
-
-
06000 om49 0.0542 0.0856 0.3068 0.0783 om73 2.4044 1.285 2.1914 5.4587 5.1257 1.2168 4.3897 8.7549 7.5826
a g = 0.5, 51 = 2
W x 15 (X = Y = 0.5)
2,0, sz = 2 0~0ooo -0-0015 - 0.0559 0.0931 -0.3801 0.9830 -0-1068 -3.1793 -0.7913 3.8662 6.9485 4.2951 - 1.9556 -9.0111 -11.1636 - 6.7447
;=
0.8 -
“0 f
0.60.4-
0.2-
-0.2
-
-0.4
-
-0.6
-
-0.8
I I.0
-1.0
I 2.0
I 3.0
I 4.0
I 6.0
I
5.0
I 7.0
r
Figure 1. Transverse
displacement
vs. time (Problem I
I
1). a/b:
----,
I
I
0.5; -----,
I
I
1.0; ~ -I
7.06.05.04.03.02.0l.O-
-2.o-3.o-4.O\
-5*o-6.O-
\, -7.0
1 I.0
I 2.0
I 3.0
I 4.0
I 5.0
6.0
I 7.0
T
Figure
2. Transverse 60
I
displacement I
vs. time (Problem I
I
2). Key as Figure I
I
1
I
5.04.03.02.0I.0Lo s -1.0-2.0 -3.o-
\ \
-4.o-
\
-5-o-6.0
Figure
I I.0
3. Transverse
I 2.0
displacement
I 3.0
I 4.0
OS. time (Problem
I 5.0
1 6.0
I 7.0
3). Key as Flgure
I
‘-,
2.
TIME-DJL’SNDENT
BOUNDARY
CONDITIONS
337
ratios but the periods for u/b 5 1 are less than those for a square plate (a/b = 1) as T increases. With decrease ofa, the period of oscillations for all problems increases and a reduction of amplitude of W is observed as T increases.
REFERENCES
1. R. D. MINDLIN and L. E. GOOUMAN1950 Transactions of the American Society of Mechanical Engineers, Journal of Applied Mechanics 17,371-380. Beam vibrations with time-dependent boundary conditions. 2. D. YOUNG 1950 Transactions of the American Society of Mechanical Engineers, Journal of Applied Mechanics 17,448450. Vibration of rectangular plates by the Ritz method. 3. M. V. BARTON1951 Transactions of the American Society of Mechanical Engineers, Journal of Applied Mechanics 73, 129-131. Vibrations of rectangular and skew cantilever plates. 4. Y. C. DAS 1961 Journal of Aeronautical Society of India 13, 111-125. On the transverse vibration of rectangular isotropic plates. 5. M. LEVY1899 Comptes Rendus 129, 535-537. Sur l’equilibre elastique dune plaque. 6. C. T. SUN and J. M. WHITNEY 1974 Journal of the Acoustical Society of America 55, 1003-1008. Forced vibrations of laminated composite plates in cylindrical bending. 7. Y. Y. Yu 1960 Transactions of the American Society of Mechanical Engineers, Journal of Applied Mechanics 27, 535-540. Forced flexural vibrations of sandwich plates in plane strain.
APPENDIX : NOMENCLATURE
a, b lengths of sides of rectangular plates x9 Y position co-ordinates A constant, amplitude of transverse displacement M constant, amplitude of bending moment h plate thickness W transverse displacement t time radian frequency w, ; (w, x)/a
Y
E P
ylb
a
modulus of elasticity mass density Poisson’s ratio oa Jp(I - v’)/E
D
Eh3/12C1 - v’)
a: R
h2/12a2 aM/D
V
t/q/d1 - v2YE