Response of circular plates to thermal impact

Journal of Sound and Vibration (1984) 95(2), 213-222

RESPONSE

OF CIRCULAR Y.

PLATES TO THERMAL

NAKAJO

AND

IMPACT

K. HAYASHI

Department of Mechanical Engineering, Sophia University, Tokyo, Japan (Received

12 May 1983, and in revised form 29 September 1983)

The thermally induced vibrations of simply supported circular plates are investigated analytically. The solution is composed of two parts; the first is obtained by neglecting the inertia term and the second represents the vibrating component which oscillates about the first, due to the effect of the inertia. One of the reasons for separating the solution in two parts is to obtain the complete solution expediently. But when this is represented in non-dimensional form, the quasi-static part has the particular desirable feature that it is independent of the strength of the thermal shock and thus will give an index for the displacement due to the temperature moment if (and only if) the inertia force has a small effect. The basic equation is governed by only one non-dimensional parameter B which contains the density, the thermal diffusivity of the material and the radius of the plate in its numerator, and the flexural rigidity and thickness of the plate in its denominator, all to various powers. B represents the strength of the thermal shock. As E increases the non-dimensional amplitude of the vibration becomes large and the natural frequency decreases.

1. INTRODUCTION

Supersonic travel, space technology and nuclear engineering are recent areas of engineering in which problems of extreme temperatures are encountered. Moreover, in these cases the materials are subjected not only to high temperatures but also to great differences of temperatures which cause thermal stresses. When the temperatures vary rapidly, the heat conduction will become unsteady, as will the thermal stresses. Unsteady thermal stresses can be regarded as “thermal shock” when the variation is rapid. Thermal shock can induce vibration of considerable magnitude, so that the effect of the inertia on the vibration cannot be regarded as negligible compared to that of the stiffness associated with the quasistatic displacement. The behavior of rectangular plates and beams subjected to sudden heat input on one side were analyzed, with the effect of inertia taken into account, by Boley [l-3] and Mindlin et al. [4]. In this paper a similar problem is analyzed for a simply supported circular plate, by using the method of separation of variables. The solution is described by non-dimensional parameters and variables so that comparisons with measurements may be carried out easily.

2. BASIC

EQUATION

The hypotheses are as follows. (1) A uniform heat flux is applied to one side of a simply supported circular plate of constant thickness. On the other side of the plate, the heat flux is zero (adiabatic). (2) There is no in-plane restraint, and hence no membrane force. (3) The distribution of temperature appears only along the direction of the thickness. (4) The material constants such as the modulus of elasticity, Poisson’s ratio, etc., are independent of the temperature. (5) The deflection is described in the form of the product 213 0022--460X/84/140213+ IO %03.00/O @ 1984 Academic Press Inc. (London) Limited

214

Y. NAKAJO

AND

K. HAYASHI

of the space-dependent function and the time-dependent function. (6) The solution obtained by disregarding the effect of inertia is called the “quasi-static part” and denoted by the subscript st and the rest of solution, depending on the effect of the inertia, is called the “inertial part” and denoted by the subscript in. The entire solution is obtained by adding these two parts. MY

\

+ psdxdyh

/W,+*dy ay

1

\ Qx Figure

I. Forces

In

Figure 1 the equilibrium equation can be obtained,

Ox, 0, : shearing forces

+dQ,dx ax

and moments

acting

of forces

on the infinitesimal

and moments

DV4W +pha*W/at*=

-{l/(1

plate element.

are shown.

From

this the basic

- v)}V2n/l,,

where the symbols have the following meanings: M, moment; moment calculated h, thickness of the plate; MT, temperature

(1)

p, density from

of the material;

h/* cxEBz dz;

MT= I position along thickness; w, deflection of the plate;

z,

(Y, coefficient t, time;

(2)

-h/2

of thermal

expansion;

E, modulus

of elasticity;

V* = a*/ar* +(l/r)d/ar; r, position

along radius;

D, flexural

(3)

rigidity, D= Eh3/12(1

-v*);

(4)

v, Poisson’s ratio; 0, temperature. The temperature moment MT can be calculated if the distribution in the z-direction is obtained. From the hypothesis the basic equation boundary conditions and initial condition are, respectively a*e/az*-(i/K)ae/at ae/az=O

where

K

atz=-h/2, is the thermal

do/az=-Q/k diffusivity,

of the temperature of heat conduction,

=o,

atz=h/2,

k is the thermal

conductivity,

(5) 8=0

att=O, and

(6a-c)

Q is the heat flux.

RESPONSE

Solving

equation

OF

PLATES

(5) with equations

TO THERMAL

215

IMPACT

(6a) and (6b) gives the distribution

of temperature

as [51 (7) where

7 is non-dimensional

time defined

as

r = tct/h’. Hence

the temperature

moment

is obtained

from equation

T

MT is not dependent

n4k on r, the basic equation

V= r4kw/

non-dimensional

variables

R = r/a,

192Qaa*,

(9)

’

DV4w +pha2w/at2 Now, the following

(2) as follows:

=48(1 -v2)aQD

M

Since

(8)

(1) can be rewritten

as

= 0.

(1)’

and parameter

can be introduced:

mT = -r4kMT/192(1

- v)DQa,

B = pa4K2/ Dh’. Here a is the radius of the circular again in the non-dimensional form

plate.

Then

(104

the basic equation

(1)’ can be rewritten

V4V+BV”=0. Differentiations respectively.

with respect

The

boundary

conditions

OF THE

be denoted

by primes

SOLUTION

(a’v/aR*)(l,

part is to be derived.

7) + mT = 0.

The basic equation

for the quasi-static

v4 v,,= 0. The boundary

conditions

equations

part is

(11)’

Vy,(l)+m,=O.

( 12a, b)’

(1 l)‘, (12a)’ and (12b)‘, V,, is V,, = (mT/2)(

It is next assumed

(12a, b)

are v,,(l) = 0,

From

and dots,

are

V(1,7)“0, First the quasi-static

(11)

to R and T will henceforth

3. DERIVATION

( 1Oa-c)

that the solution

V= V,, + ? P,(R)T&) ?I=, Here the a, are the coefficients and the T, are time-dependent (11) can be written as

1 - R2).

can be obtained = f (a,mr n=,

(13)

in the form

+ TAP,,

03 V,, = C a,P,m,. n=,

(14, 15)

of the Fourier series, the P,, are space-dependent functions, functions. By using equations (1 l)‘, (14) and (15), equation T,V4P,

= - BP,(a,m,

+ T,)“.

216

Y. NAKAJO

Then separating

the variables

AND

K. HAYASHI

gives

(I/Pn)V4Pn=-B(a,m,+T,)‘jT,=q4,. for P,, the basic equation

Hence

(16)

is V4Pn = q4,P,.

In terms of a new variable

A defined

(17)

by A = q,R

equation

(18)

(17) becomes V*,P,-Pn=O,

where V’, = d*/dA * +( I/ h)d/dA.

(17)’ that P, can be separated

It is next assumed

as (19)

P, = P*, +P”Z, with P,,, and Pn2 satisfying,

respectively, v:p,,

Since equations respectively,

(20a)

and

+p,,

(20b)

P,I = GJoO)

+

=o,

are

v:P”,+P*,=o.

Bessel

Pn2

C,Y,(A),

where Jo, Y,,, I,, and & are Bessel functions. P, = The boundary

conditions

From conditions

differential

=

Hence,

GOa, b) equations,

C&(A)

+

GKoO

the solutions

h

(21%

C,Jo(h)+C,Y,(A)+C,I,(A)+C,~(h).

(22)

for P,, are P,(R = 1) = 0,

P,(R = 0)

finite,

(23a, b)

P;(R = 1) = 0,

PE( R = 0)

finite.

(23~~ d)

(23b) and (23d), C2 and C., must be zero. Hence,

Conditions (23a) and (23~) then lead to the equation qn can be determined: Jo(qn) -IJo

P, is (22)’

from which the characteristic

IO(%)

- J2(qn)I

Uo(qn) +Mqn)l

(23a)

equation

(25) into equation P. = G{Jo(qnW

From

equation

values

(24)

= ”

C, = -[Jo(qn)lIo(qnW,. Substituting

b)

P, is

P, = C,J,(A) + C&(A).

From equation

are,

(16), the basic

equation

(25)

(22)’ gives (22)”

- [Jo(qn)/Io(qn)lIo(q,R)}.

for

T, can be constructed:

since

- B(a,m,

+

Tn)‘;! Tn = q:, T, +(qz/B)T, From equations

= -a”mT.

(26)

(9) and (10~) mT=(a)(l

+v)

f j=1,3,5...

(-7r4e-j2”+).

(27)

RESPONSE

From equations

OF PLATES

TO THERMAL

217

IMPACT

(13) and (15) ; a,P,=;(l-R’). n=l

If P, forms an orthogonal case) can have a Fourier if the following equations P,P,,, dA := 0

set on the interval (0, 1), an arbitrary function (f( 1 - R2) in this expansion with respect to P,,. P,, can be said to be orthogonal are satisfied:

(I

P,,P, dA # 0

(n # m),

A

P,,P,,, dA =

P,P,,,2rR

dR >

A

(29a. b) From the results in the Appendix, these equations norm of P, is required to make P, orthonormal: r I.1 IIPJ2 = 1 P,P,, dA = 1 P,P,2nR

are found

to be satisfied.

Then

the

dR

Jo

JA

l-1

= 27rC:

+2ARJ,,(q,R)Io(q,R)

-

where A = -Jo(qn)/Io(qn) a, =

dR (30)

(see the Appendix).

a, can be determined

from

( Pn;( 1 - R2))

(26) and (27)

+d

T,

=

_

B

One can find a particular which leads to the result

solution

n4(1 +‘)

a,

4

solution

f”?,=

,=,f,

,.,

e-j2r’T.

(32)

9. .

7” by using

the method

undetermined

a,Br4( 1 + V) e I zn2r f ,=I, 3.5,... - 4(j47r4B + q:)

of equation

C5 and C, are arbitrary T,

The initial

conditions

= C5

(32) is

constants

and p’-

qt/B.

Wb) Hence,

the complete

a,Br4( 1 + V) e _I.z&T ,=1,3,5,... 4(j4n4B +qt) .

eipr+ C, emi@ -

which must be satisfied

V’,,(T=O)+

coefficients,

(33a)

T = C, ei@T+ C e-iPT ” 6 9 where

(31)

.

p’,

T,

The homogeneous

+A2RI:(q.R)

Jo

&GIJdqn> - AIdsnM2,

=

From equations

1 RJ;(qnR)

f P,T,(7=0) “=I

solution

is (34)

are

=O,

whence

T,,(~=0)=0,

Wa)

V;,( T = 0) + F P,T,( 7 = 0) = 0, “=I

whence

T,( T = 0) = 0.

(35b)

Equations (22)“, (30) and (31) show that the constant C, can be any value so that one may choose the value of 1 for C,. Hence, the arbitrary constants C5 and C, can be

218

Y.

determined

and

NAKAJO

AND

K. HAYASHI

T,, becomes x

T, = S, cos pr -;

sin PT -

aB7r4( 1 + v) f ,=,.3,5 ,... 4( j4rr4B + 4:) ’

s,= By substituting be determined.

equations

aB7rl( 1 + V)

, ,&,... 4(j4a4B sz =

f

+9:)

_,ZnA7 e

1 + v)

(j2r2)uBr4(

,-1.3,5 ,...

.

4(j4r4B+9i)

(22)“, (27), (3 1) and (36) into equation

(36)

’

(14) the solution

(37) can

4. RESULTS Plots of non-dimensional solutions are shown in Figure 2. (The non-dimensional solutions are plotted against non-dimensional time with B as the parameter.) B =0 corresponds to the quasi-static part, obtained by neglecting the inertia. This part is not affected by B (because the basic equation for this part does not contain B as described in equation (1 l)‘), so that it is a part common to all solutions, for any value of B. B=co

/ /’

021

0

0.2

o-4 Non-dimensional

Figure 2. Non-dimensional

f

B

-0.4

solution

for B = 0 (----),

I

06

00

IO

time T 0.

I (- - -), I (-----),

co(-)

I

.o E

z

-0.3

-

B E s 7ii

-0.2

-

E 5 I E 2 .;i %

0

1 Not?-dimensional

2 parameter

3 B

Figure 3. The amplitude against B.

at R = I).

RESPONSE

OF PLATES

TO THERMAL

Non-dimens~onol

Figure

4. The non-dimensional

porometer

natural

219

IMPACT

B

frequency

against

B

corresponds, inversely, to the effect of inertia being infinite (but this solution is not so useful). B = 1 and B = 0.1 are intermediate values between the previous two extreme ones. The results show that the amplitude and the period of the vibration become large with increase of B. This implies that the role of inertia is very important in the case of sudden heating. B is thus the only parameter governing the thermally induced vibrations of circular plates. Figures 3 and 4 show the maximum amplitude and the natural frequency against B, respectively, in order to make the relation between the vibration characteristics and B more evident. 5. DISCUSSION

AND

CONCLUSION

The merits of this classical type of analysis are that (1) the method is suitable even though the temperature distribution along the thickness may change (however, the coefficients of the Fourier expansion will then also change because the temperature and inertial parts of the solution are obtained moment MT varies), (2) the quasi-static separately so that the ratio of these two parts can be determined easily, (3) the nondimensional basic equation is governed by only one parameter B so that comparisons with measurements can be easily made. According to equation (lOd), B is large for a plate with high density, small thickness, large radius, high thermal diffusivity and low flexural rigidity. With the non-dimensional time T given by equation (8), it is evident that the actual natural frequency of the plate is low for high thermal diffusivity and small thickness even though B has the same value. The natural frequency is high for small B provided that K/h* has the same value.

REFERENCES B. A. BOLEY 1956Journal ofthe Aeronautical

Sciences

23, 179-l 8 I.

Thermally induced vibrations

of beams. B. A. BOLEY and

A. D. BARBER 1957Journal of Applied Mechanics 24, 413-416. Dynamic response of beams and plates to rapid heating. B. A. BOLEY and J. WEINER 1960Theoryof Thermal Stresses. New York: John Wiley & Sons, Inc. See pp. 195-207, 339-345, 406-409. R. D. MINDLIN and L. E. GOODMAN 1950Journal vibratpons with time-dependent boundary conditions.

of Applied

Mechanics

17, 377-380.

Beam

220

Y. NAKAJO

5. H. S. CARSLAW second 6.

and J.

C.

JAEGER

1959

AND

K. HAYASHI

Conduction

of Heat in Solids.

Oxford:

Clarendon

Press,

edition.

D. J. JOHNS 48-83.

1965

Thermal Stress Analyses.

Oxford:

Pergamon

see pp.

Press Ltd, first edition,

APPENDIX

Al.

THE

ORTHOGONALITY

According

to equation

OF P

(17) V4Pn = q”nP,,

V4Pm = q”mPm ;

(Al, A21

hence P,V”P,

- P*V4Pm = (q”n- qL)P,P,.

Therefore

=

P,[R{(1IR)(RP~)‘}‘l’-Pn[RW/R)(Wn>‘~‘l’

=(P,[R{(1IR)(RP~>‘}‘I>‘-(P,[R{(lIR)(RP’,)’}’l)’ - Pi,,[RWIR)(WJ’~‘l +P:,[R~(1IR)(RP’m)‘~‘l = (P,[R~(~IR)(RP~)‘~‘l)‘-(P,[R~(~IR)(RP’,)’~’l)’ -[(P6R){(1IR)(RP~)‘}I’+[(P’,R){(~IR)(RP’,)’}I’ +(P~R)‘{(lIR)(RP’,)‘}-(P’,R)‘{(lIR)(RP’,)’}. The last two terms in equation -iP:,+P;+RP: From the boundary equation (22)”

(A3) cancel

)

conditions

-P,

Jdqn) I

0 ” (q

>

Hence I - 44m)

- P;RP;

)

- P;RP::

P,,,( 1) = P,,( 1) = PL( 1) = P:(l)

cd-J,(qnR))--

244:

each other so that it becomes

-+P&+P:+RP;

(

(A31

RP,P,,, dR

+ P;RP;

dR I

qnIt(qnR)

7

lim 5= R-O R

+P:,RP;

I

.

= 0, and from the result

lim R-O

Pz = P:(O).

(A4) of

RESPONSE

P’,RP;

-

OF

PLATES

TO THERMAL

221

IMPACT

+P;RP;

whence lim (-(l/R)PA

+Pz)=O.

+Px)=lim(-Pz R-+0

R+O

Therefore equations (29a) and (29b) must be satisfied. A2.

THE

NORM

OF

P I

I IIW=

1

P,P, dA=

P,P,2rR

I0

I0

A

RP,P, dR

dR = 23~

I =

24

RJ;(qnR)

+2ARJo(qnR)Io(qnR)

+A2RI:(qnR)

dR,

(A3

I 0

where A= -Jo(qn)/Io(qn).

Then

RJ;(q.R)

dR =

I’0

4

J%qnW + J:CqnRl

dR =

$Jo(q.R)l,(q.R)

[

I 1

o= iiJ&n)

(‘46)

+ J:(qA I

I

RJo(qnR)Io(q,R)

- JdqnWdqnW

1

I 0 =

;tJo(qn)Io(qn)

RI;(qnR)

dR =

0

[

4

0

(A7)

- JdqnYdqn)),

I I

I

U;(qnR) - I:(q,R)I

1

’=#,(q.)

(‘48)

- I:(qn)).

0

Substituting equations (A6), (A7) and (A8) into equation (A5) gives lIPnIl=J~C,{J,(qn)+[Jo(qn)/Io(qn)lI,(q”)}. z.43. a,

COEFFICIENTS

OF

can be determined

THE

FOURIER

EXPANSION

by the equation a, = [P&l -

The numerator

a,

R2MIPnII*.

(A9)

of equation (A9) is I

$0 - R2VXJo(q,R)

+AIo(qnW

A

(R - R’){Jo(q,R)

dA = ~‘2,

+AIo(qnR))

dR;

0 (A101 I

1

R3Jo(q,,R) dR =

I

R2RJo(,R,dR=[~J,,,,R,I:-[~~J2~q~R~]~

0

=+J,(q,)--$J&.). n ”

(Al 1)

222

Y. NAKAJO

AND

K. HAYASHI

Similarly I

JR310(qnW

dR =;

0

I,(q.)--$ Mq,).

(Al21

n

It is thus evident that I

1

J

RJo(qnR)

(A13, A14)

(A12), (A13) and (A14) into equation

(AlO) gives

n

0

Substituting equations equation (3 1).

-$Idqd.

dR = j- J,(qn),

(All),

J0

RIo(qnR)dR =