Transverse free vibration of simply supported right triangular thin plates: A highly accurate simplified solution

Journal of Sound and Vibration (1990) 139(2), 289-297

TRANSVERSE RIGHT

FREE

VIBRATION

TRIANGULAR ACCURATE

THIN

OF

SIMPLY

PLATES:

SIMPLIFIED

SUPPORTED

A HIGHLY

SOLUTION

H. T. SALIBA Department of Mechanical Engineering, Lakehead University, Thunder Buy, Ontario, Canada P7B 5El (Received 22 May 1989, and in revised form 29 August 1989)

A simplified approach to the solution of the title problem is outlined. A detailed discussion of this solution is provided, making it possible for the reader to generate eigenvalues, as required, for any particular application. Whenever it is possible, the excellent convergence of the solution is demonstrated by comparing results with known exact eigenvalues. The efficiency of the solution is also stressed by showing not only the excellent rate of convergence but also the significant reduction in the number of required building blocks from six, as was reported in references [ 1,2], to only two in the present paper. Consequently, the size of the coefficient matrix involved is shown to be reduced from 94x94 to only 12X 12.

1. INTRODUCTION The introduction

of superposition techniques to the solution of free vibration problems of thin plates has removed many roadblocks that have traditionally presented researchers with insurmountable difficulties. As a result, analytically exact, highly accurate solutions have been obtained for many rectangular thin plate free vibration problems. In an effort to introduce this powerful solution technique to the free vibration analysis of triangular plates, Gorman introduced the six building block arrangement shown in Figure 1 [ 1,2]. This arrangement was later used by the author to introduce yet another new technique permitting the use of superposition in the analysis of thin plates with straight line boundaries in general [3,4]. No serious difficulties were encountered in these analyses. The results were obtained quickly and accurately. However, when the technique was developed beyond the one triangular and one rectangular element combination, serious difficulties prevented the progress of the analysis. This prompted the author to review the solutions for both rectangular and triangular elements. It soon became clear that convergence problems as well as false roots are introduced by the skewed building blocks of the triangular element solution. Meanwhile, in a letter to the editor [5], Gorman indicated that while his solution for the right-angled triangular plate [ 1,2] was highly accurate, it had the serious shortcoming of uncovering false roots which he referred to as “rejection mode” eigenvalues. In an effort to overcome this difficulty, he suggested a new set of building blocks to be used. The modified Got-man method was a step in the right direction in that it virtually eliminated all false roots. However, when more than one triangular element was needed, the coefficient matrix grew very rapidly, which limited its use. Vast amounts of effort have been devoted by the author toward studying this basic problem with a view to eliminating any convergence problem, and thereby reducing the number of required terms in the Fourier series 289 0022-460X/90/110289+08 %03.00/O

0 1990 Academic Press Limited

290

H. T. SALIBA

3

7’ \ Figure

1. Superposition

of building

blocks,

used by German

[ 1,2], in the solution

of right triangular

plates.

expansion. As a result, not only the number of Fourier series terms was reduced considerably with an improved convergence, but the number of required building blocks was also reduced to a minimum, resulting in more reduction in the size of the final coefficient matrix, thereby making the solution useful in the analysis of plates with more complicated geometrical shapes. The object of this paper is to fully discuss this solution, where only two building blocks are needed to solve the title problem.

2. THE

BASIC

BUILDING

BLOCKS

The governing differential equation for the free vibration of rectangular plates in its dimensionless form can be written as

a4W5, 7?)/d714+2+2d4WS,

TJ)/a?72a62+44a4W5,

77)/a54-44A4W(5,

a)=O,

(1)

where symbols are as shown in the Appendix. The basic building blocks used in the present solution are as shown in Figure 2. They consist of two rectangular plates, simply supported along three of their edges. Therefore, their common Levy-type solution is written as

(2) which upon substitution into equation (1) leads to d4Y,(~)/d~4-2~2(n1~)2d2Ym(~)/d~2+~4[(mlr)4-h4]Y,(~)=0,

(3)

E-

(b)

Figure 2. Building blocks used in the present edge displacement; (b) forced edge rotation.

solution

of simply supported

right triangular

plates. (a) Forced

SIMPLY

SUPPORTED

RIGHT

TRIANGULAR

291

PLATES

the solution of which depends on whether A2- (MT)’ is negative or positive. If A2> (mu)’ then (4a)

Y,(77)=A,cosh(Pmrl)+Bmsinh(P,rl)+C,sin(Y,77)+Dmcos(Ym~), and if A*< (m.lr)* then Y,(11)=A,cosh(P,rl)+B,sinh(P,rl)+C,,sinh(~,?7)+D,cosh(~,?7),

(4b)

where /3,,,= +m, and y,,, = 4w or y,,, = c$-, whichever is real. A,,,, B,, C, and D,,, are constants to be determined by means of prescribed boundary conditions. It is clearly seen from equation (2) that the simple support conditions along the two opposite edges 5 = 0 and .$= 1 of each building block are automatically satisfied. Satisfying the zero displacement and zero bending moment conditions along the third simply supported edge of each of the two building blocks, 77= 0, leads to A,,, = D, = 0. The fourth edge of the first building block (BBl), n = 1, has a prescribed harmonic lateral displacement given by

and a forbidden bending moment. In the light of these boundary conditions, the complete solution of the first building block is obtained as W1(5, rl) =

F

E,,@ll,[sinh

(&n)+

4, sin

m=1,2,..

+

&,,h,Jsinh

f

(y,,,rl)l sin (me7

(P,,,T)+ 02, sinh (rmrl)l

m=k*+l

sin(m5),

(6)

where the first summation pertains to values of A*> (mr)*, and where f3,, = [/I’, - v+‘(mn)‘]

sinh (&)/[

y’m+ u42(mn)2]

sin (ym),

0,, = -[/3’, - z$‘( mr)‘] sinh (&,)/[ y’m- v4’( mr)‘]

ellm= l/[sinh ML,)+e,, sin (Y~)I,

sinh (y,,,),

e22m = l/IX-h WJ+ e2, sinh (Y,,,)I.

The second building block, (BB2), has a forbidden lateral displacement, scribed harmonic bending moment along its fourth edge, n = 1, given by M,,b2

$

lim -=

k-m

aD

E2,,, sin

(mT().

and a pre-

(7)

m=1,2,..

In view of these boundary conditions, the Levy-type solution for the second building block, W2(5, n), is obtained from equation (6) by replacing E,, by E2,,,, and where elm = -sinh (&,)/sin

(~~1,

e,, = -sinh (&)/sinh

(y,,,),

8Ilm= -l/U%, sinh (Pm)- e,,yf,, sin (y,,,)l, e

22m

=

-l/[Pfi

sinh

(Pm)+

%,r’,

sinh

(y,,,)l.

With the Levy-type solutions of the basic building blocks in place, one may now proceed with the superposition of these building blocks as discussed in the following section.

292

H. 3. SUPERPOSITION

T. SALIBA

OF THE

BASIC BUILDING

BLOCKS

The superposition of the basic building blocks is achieved by first determining the contributions of each individual building block to the relevant boundary conditions and, second, due to the linear nature of the individual building block problems, their total contribution to a given boundary condition is found by adding together their individual contributions. In the forthcoming analysis, reference is made to Figure 3, in which the two basic building blocks of Figure 2 are assembled on top of each other, and where the triangular region of interest is clearly shown to be bounded by two adjacent simply supported edges, and the diagonal (d) which is to have simple support conditions forced on it by adjusting the Fourier coefficients E,, and Ezm appearing in the Levy-type solutions, developed in the previous section, as follows herein.

3.1.

CONTRIBUTIONS

OF

BBI

AND

BB2

TO

DISPLACEMENT

ALONG

(d)

Along this diagonal line, n may be written in terms of .$ as r] = 1 - 5, and therefore W,(&1-5)=

E,,e,,,[sinhp,(l-~)+8,,siny,(1-5)1sin(m~5)

5 m=1.2,...

+

E,,&,,[sinh&,(l-~)+02,,,sinhy,,,(l-,$)]sin(m~&).

f

(8)

m=k*+l

W,([, 1 - 5) is obtained from equation (8) by replacing appropriate values for 8, ,,,, O,, , 0, ,m and 02*,,,.

3.2.

CONTRIBUTIONS

OF

BBl

AND

882

TO

BENDING

by Ez,,, , and using the

E,,

MOMENT

ABOUT

(d)

The bending moment equation about any line in the plane of a plate was given by Timoshenko [6]. It is reproduced here in its dimensionless form as

Mnb2/aD

a2 W/at2+

= -[e,+’

e2 a2 W/aT2f

e,+ a2 W/at

aq],

(9)

where 8, = cos2 (Y+ v sin2 LY,

e2 = sin’ (Y+

v cos' a,

f$ =

(1 -

v) sin 20,

and where (Yis the angle between the normal to the line and the e-axis. It is clear here that one needs the following derivatives for each term of the summations involved: a2W,/a~2=-E,,e,,,(m~)2[sinhP,(1-~)+e,msin~m(l-~)]sin(m~~), or

= -E,,822,(m~)2[sinh

&(l--&)+

02, sinh y,,,(l-&)I

sin (mn&),

a2W,/a.r72=E,,BI,,[P2,sinhP,(1-5)-8,,y2,siny,(l-5)]sin(m~~), or

= El,&&3?,,

sinh Pm(l -5)+

a2w,/aga?7=E,,e,,,(m~)[p, or

=El,~22m(m~)[P,

&,r’,

coW%J1-5)+4,rm coshP,(1-5)+&,r,

sin (me?, ~0sY~U-~)ICOS(~~ coshy,(l-5)1cos(mrr5).

sinh .ym(1-5)l

b

Figure

3. Superposition

of the two building

blocks of Figure

2.

SIMPLY

SUPPORTED

RIGHT

TRIANGULAR

PLATES

293

Here n has been replaced by its equivalent term (1 - 5). The reader should appreciate the fact that by replacing E,, by Ez, in the above expressions, and by using the appropriate values for 8,, , O,,,,, O1,,,, and &,,, , the derivatives pertaining to W, are obtained. Substituting the above expressions into equation (9) will lead to the contributions of BBl and BB2 to bending moment about the diagonal (d) of Figure 3. The contributions of individual basic building blocks to relevant boundary conditions are now readily available. The next step is to group all of the individual contributions to each of the boundary conditions together and adjust the Fourier coefficients El,,, and E,, to satisfy simultaneously the prescribed boundary conditions which, in this case, are those of simple support along the hypotenuse (d) of the triangular region of interest. However, in order to add these contributions as mentioned above, it is necessary to expand relevant contributions in an appropriate Fourier series. For the problem at hand, this is achieved as follows. Let 1

I=2

I0

sinh P,,, (1 - 5) sin (rnrt) sin (nnf) de.

(10)

Then, I = EXl + EX2, where EXl =p,[coshp,

-cos (m7r-nv)]/[P~+(m7r-nv)*],

EX2=~,[cos(m~+n7r)-cosh~,]/[~f,,+(m~r+n7r)~]. Now let 1 II = 2

Then II = EX3 + EX4,

I0

sin Y,,,(1 - 5) sin (mr[)

d[.

(11)

where

EX3= y,[cos(m~--nr)-cos

y,,,]/[yf,,-(mr-nr)‘],

or if 7: = (mr - nr)2, then EX3 = sin yJ2, EX4=y,,,[cos or if yi = (mr+

sin (~6)

and

y,-cos(mr+n~)]/[-y~-(rn~+n~)~].

nr)2, then EX4= -sin y,,,/2. Also, let 1 III = 2 sinh y,,,( 1 - 5) sin (rnrl) sin (nr[)

I0

de.

(12)

Then 111 = EX5 + EX6, where EX5 and EX6 are obtained from the expressions for EXl and EX2, respectively, by replacing /I,,, by y,,,. To satisfy the zero lateral displacement condition along the diagonal, one requires the sum of all the contributions to this displacement to be zero. It is seen here that the only way to satisfy this requirement is to set the sum of the coefficients of like trigonometric terms to zero, which leads to the following k equations, where k is the number of terms in the Fourier expansions: ; an,m+ m=,

an,m+k

=

O,

n=l,2

,...,

k.

(13)

Here a,,,,=

E,,,,~,,,[EX~+EX~+L~,,,,(EX~+EX~)],

or if A2<(m7r)*, a n.m =E,,t$,,[EXl+EX2+8,,(EX5+EX6)],

8,,,, , t?,,,,, fill ,,,, and &,, are as defined in the solution of BBl, and an,m+kis obtained from the expression for a,,,, by replacing El, by Ez,, and using the appropriate values for 81~19e2m9 e,,,,, and ez2,,,.

294

H.

Similar expansions performed following k equations: i

T. SALIBA

for contributions

to the bending moment, lead to the

n=l,2,...,k.

an+k,,m+&+k,,m+k=0~

Here a,,+k,m= -Elm[~l~2Aln+

&AZ,+

(14)

&#&1 and

A,,=-&,(m~)*[EXlfEX2+~,,(EX3+EX4)], A2,,=e*I,[~;(EX1+EX2)-e1,yZ,(Ex3+EX4)], A,, = e,,,(mq)[p,(Yl+

Y2)+%IY,(Y3+

Y4)1,

Y2)+@,,Y,(Y5+

WI.

or if A’> (TWIT)*,then

A3n= e,,,(mr)k(Yl+

an+k,m+k is obtained from the expression for an+k,,, by replacing El,,, by E2,,,, and by using the appropriate values of 01,,,, f+,,, , 01,,,, and 02*,,,pertaining to the solution of BB2. Also, Yl =

(mm + nr)[cosh

Pm - cos (mm + nr)]/[Pi

+ (mr + nr)*],

Y2 = (mr - nn)[cos (mv - nr) -cash &l/[/3:

+ (mr - w-)*1,

Y3 = (m7r + nr)[cos yrn- cos (m7r + n7r)]/[( m7r + nr)’ - yk], or if -yL = (mr + n+r)*,then Y3 = sin -ym/2, and Y4=(m~-n7r)[cos

‘ym-c0s(m7r-na)]/[y~-(m7r~n77)*],

or if y,,, = (mr - nr), then Y4 = -sin y,,,/2, and if y,,, = -( rnr - nr), then Y4 = sin y,,,/2, and where Y5 and Y6 are obtained from the expressions for Yl and Y2 respectively, simply by replacing /3,,, by y,,,. The problem is now reduced to solving the above set of 2k homogeneous algebraic equations. For non-trivial solutions, the determinant of the coefficient matrix is set to zero and values of A2 that satisfy this requirement are the eigenvalues sought. Therefore, the characteristic equation in its matrix form, for the right triangular plate under investigation, may be written as (15) where A,j=O,,,[EXl+EX2+O,,(EX3+EX4)],

Al+k.j=-[e,~2A,n+e2A2n+e3~A3n],

or if A*> mr*, then

and where Ai,j+k and Ai+kj+k are obtained from the expressions for A,j and Ai+kj respectively by using the appropriate values for O,, , O,,,,, 8, ,,,, and t&, pertaining to the solution of BB2.

SIMPLY

SUPPORTED

RIGHT

TRIANGULAR

295

PLATES

Once the eigenvalues have been identified, one of the unknowns, say E,, is set equal to unity and the above set of equations is solved for the remaining 2k - 1 unknowns, which in turn are substituted into the shape function W = W, + W,, for the generation

of the required eigenvectors and consequently 4. NUMERICAL

the mode shapes.

RESULTS AND

CONCLUDING

REMARKS

Numerical results as defined by the eigenvalue parameter A’ are listed in Table 1 for the first four free vibration modes of right triangular plates with aspect ratios ranging from 1.0 to 3.0 as shown, and the corresponding nodal lines are as shown in Figure 4. Although the numerical results provided here are limited to a few modes and plate aspect ratios, if required the reader can easily write a small computer program, using the information provided in this paper, to generate eigenvalues as required for any particular application, However, the reader should keep in mind that in the development of this solution, the following assumptions pertaining to the thin plate small deflection theory were made: plate thickness is small compared to its lateral dimensions; for higher vibration modes, plate thickness is small compared to the distance between nodal lines; lateral displacement W is small compared to the thickness of the plate; negligible rotatory inertia effects; no significant in-plane forces. Fortunately, most plate vibration problems satisfy the above assumptions well enough for most practical purposes. The second precautionary note here is that because of the nature of the hyperbolic functions involved, problems in overflow and underflow, in connection with computation, may be experienced with some higher order arguments. TABLE

I

Eigenvalues for the first four free vibration modes of the simply supported right triangular plate of Figure 2

A* = wa’( p/D)“’ bla

b

Mode

1.0

1.5

2.0

2.5

3.0

1 2 3 4

49.348 98.696 128.305 167.783

34.279 65.615 91.801 107.336

27.759 49.882 74.696 81.310

24.144 41.127 60.541 71.868

21.844 35.613 51.071 66.260

+=1

1.5

2

+=1

1.5

2

2.5

3

2.5

3

(bl

+=I

1.5

2

2-5

3

(c)

Figure

4. Nodal

lines for (a) second,

(b) third and (c) fourth

vibration

mode. Plate aspect

ratios as shown.

296

H. T. SALIBA

However, this is easily remedied by properly simplifying the appropriate equations. It must be mentioned here that the generation of these eigenvalues is quick and simple. Because of the elimination of false roots, and the reduction of the number of building blocks required for the solution, from six to only two, along with the very high convergence rate of the method, no intricate analysis is required, on the part of the user, in obtaining the required numerical results. The eigenvalues of Table 1 were compared with previously available reliable data where excellent agreements were noted. The first mode eigenvalue for the 30”-60”-90” (b, = 1.7320508) simply supported triangular plate has been deduced as being equal to 30.704 by the method of images [7] and was reported in reference [l] to be 30.7053. Using the present solution, a value of 30.7054 is found. Whenever it was possible, the solution was checked for convergence to known exact eigenvalues. As an example, in the case of a simply supported isosceles right triangle, it is easily shown that the first eigenvalue is 57r*, the second being 10r2, the third 137r2, and the fourth eigenvalue is 177r*. This was confirmed exactly, as shown in Table 1, with only six terms in the Fourier series expansion, which resulted in a coefficient matrix of 12 x 12. Compared to a coefficient matrix of 84 x 84 for the first and second mode eigenvalues, and a coefficient matrix of 96 x 96 for the third and fourth modes, as reported by reference [ 11, this results in significant savings in processing time as well as in storage requirements. An average PC can handle matrices many times the size of this reduced matrix with no difficulties. This makes the use of this solution very attractive for plates with more complicated shapes. However, one must stress the fact that in order to achieve the required convergence, the prescribed harmonic lateral displacement of BBl and the prescribed harmonic moment of BB2, as discussed in the solution of these building blocks, must act along the longest of their edges: i.e., in the solution as presented herein, 4 must be less than or equal to unity. Fortunately, this represents no obstacles in the case of the title problem. A right triangle of aspect ratio two is the same as that of aspect ratio l/2 = 0.5. Therefore, the side length b, in the aspect ratio parameter 4, can always be chosen such that 4 < 1, and the eigenvalue parameter adjusted accordingly. However, precautions must be taken when this method is used in connection with plates of more complicated geometries, such as general triangles, trapezoidal and parallelogram plates.

ACKNOWLEDGMENT The author wishes to acknowledge the financial support of the Natural Science and Engineering Research Council of Canada.

REFERENCES 1983 Journal ofSound and Vibrarion 89, 107-118. A highly accurate analytical solution for free vibration analysis of simply supported right triangular plates. D. J. GORMAN 1986 Journal of Sound and Vibration 106, 419-431. Free vibration analysis of right triangular plates with combinations of clamped-simply supported boundary conditions. H. T. SALIBA 1986 Journal ofSound and Vibration 110,87-97. Free vibration analysis of simply supported symmetrical trapezoidal plates. H. T. SALIBA 1988 Journal of Sound and Vibration 126, 237-247. Transverse free vibration of fully clamped symmetrical trapezoidal plates. D. J. GORMAN 1987 Journal of Sound and Vibration 112, 173-176. A modified superposition method for the free vibration analysis of right triangular plates. S. TIMOSHENKO and S. WOINOWSKY-KRIEGER 1959 7Ireory of Hates and Shells (second edition). New York: McGraw-Hill. A, W. LEISSA 1969 NASA SP-160. Vibration of plates.

1. D. J. GORMAN

2. 3. 4. 5. 6. 7.

SIMPLY

SUPPORTED

APPENDIX:

RIGHT

TRIANGULAR

NOMENCLATURE

plate dimension in x direction plate dimension in y direction D flexural rigidity of plate, = Eh3/ 12( 1 - v*) E Young’s modulus of plate material h plate thickness k number of terms used in solution k* upper subscript limit for first summations of solutions W plate lateral displacement divided by side length a 77 distance along plate y axis divided by side length b, = y/ b A’ eigenvalue, =wa2JJD Poisson ratio of plate material distance along plate x axis divided by side length a, =x/a ; mass of plate per unit area P 4J plate aspect ratio, = b/a circular frequency of plate vibration w

%

PLATES

297