This article was downloaded by: [vinod varghese] On: 16 April 2013, At: 00:37 Publisher: Taylor & Francis Informa Ltd Registered in England and Wales Registered Number: 1072954 Registered office: Mortimer House, 37-41 Mortimer Street, London W1T 3JH, UK Journal of Thermal Stresses Publication details, including instructions for authors and subscription information: http://www.tandfonline.com/loi/uths20 Thermoelastic Problems of Thin Circular and Rectangular Plates K. S. Parihar a & Sunita S. Patil a a SSBT's College of Engineering & Technology, Bambhori, Jalgaon, India Version of record first published: 04 Oct 2010. To cite this article: K. S. Parihar & Sunita S. Patil (2010): Thermoelastic Problems of Thin Circular and Rectangular Plates, Journal of Thermal Stresses, 33:10, 907-924 To link to this article: http://dx.doi.org/10.1080/01495739.2010.510689 PLEASE SCROLL DOWN FOR ARTICLE Full terms and conditions of use: http://www.tandfonline.com/page/terms-and-conditions This article may be used for research, teaching, and private study purposes. Any substantial or systematic reproduction, redistribution, reselling, loan, sub-licensing, systematic supply, or distribution in any form to anyone is expressly forbidden. The publisher does not give any warranty express or implied or make any representation that the contents will be complete or accurate or up to date. The accuracy of any instructions, formulae, and drug doses should be independently verified with primary sources. The publisher shall not be liable for any loss, actions, claims, proceedings, demand, or costs or damages whatsoever or howsoever caused arising directly or indirectly in connection with or arising out of the use of this material.
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This article was downloaded by: [vinod varghese]On: 16 April 2013, At: 00:37Publisher: Taylor & FrancisInforma Ltd Registered in England and Wales Registered Number: 1072954 Registered office: Mortimer House,37-41 Mortimer Street, London W1T 3JH, UK
Journal of Thermal StressesPublication details, including instructions for authors and subscription information:http://www.tandfonline.com/loi/uths20
Thermoelastic Problems of Thin Circular andRectangular PlatesK. S. Parihar a & Sunita S. Patil aa SSBT's College of Engineering & Technology, Bambhori, Jalgaon, IndiaVersion of record first published: 04 Oct 2010.
To cite this article: K. S. Parihar & Sunita S. Patil (2010): Thermoelastic Problems of Thin Circular and Rectangular Plates,Journal of Thermal Stresses, 33:10, 907-924
To link to this article: http://dx.doi.org/10.1080/01495739.2010.510689
PLEASE SCROLL DOWN FOR ARTICLE
Full terms and conditions of use: http://www.tandfonline.com/page/terms-and-conditions
This article may be used for research, teaching, and private study purposes. Any substantial or systematicreproduction, redistribution, reselling, loan, sub-licensing, systematic supply, or distribution in any form toanyone is expressly forbidden.
The publisher does not give any warranty express or implied or make any representation that the contentswill be complete or accurate or up to date. The accuracy of any instructions, formulae, and drug doses shouldbe independently verified with primary sources. The publisher shall not be liable for any loss, actions, claims,proceedings, demand, or costs or damages whatsoever or howsoever caused arising directly or indirectly inconnection with or arising out of the use of this material.
THERMOELASTIC PROBLEMS OF THIN CIRCULARAND RECTANGULAR PLATES
K. S. Parihar and Sunita S. PatilSSBT’s College of Engineering & Technology, Bambhori, Jalgaon, India
In this paper an attempt is made to determine the temperature, displacement andstress functions of a thin circular plate by applying finite Hankel transform andLaplace transform techniques. This plate that is assumed to be in the plane state ofstress is subjected to axisymmetric boundary conditions. As a further simplification,special cases of the third kind of boundary condition are used on the two planesurfaces, while zero temperature is maintained on the outer curved surface of the thincircular plate. A particular case of the boundary conditions is discussed in detail, andnumerical results are presented graphically. A mathematically similar problem is thatof determining temperature distribution, displacement and stress functions on an edge ofa thin rectangular plate with the stated boundary conditions. The results are obtainedby applying finite Marchi–Fasulo transform and Laplace transform techniques.
Keywords: Heat conduction; Laplace transform; Plane state of stress
INTRODUCTION
Roy Choudhury [1] determined quasi-static thermal stresses in a thin circularplate subjected to transient temperature applied along the circumference of a circleover the upper face with lower face at zero temperature and the fixed circular edgethermally insulated. Such types of problems (see also [2–4]) are of great importancefor engineering applications. For instance, they make it possible to determine thestate of stress in circular plates constituting foundations of containers for hot gasesor liquid, in foundations for furnaces etc., in other words in all cases in which eithertemperature or heat flux is given on the planes bounding the plate.
Ishihara et al. [5] studied the heat conduction problem to determine thetemperature distribution and discuss the thermoelastic deformation in thin circularplates subjected to partially distributed and axisymmetric heat supply on theouter curved surface. Navlekar and Khobragade [6] studied the direct steady-stateand transient thermoelastic problem to determine the temperature, displacementand stress functions of a thin circular plate of thickness h occupying the spaceD � 0 ≤ r ≤ a, 0 ≤ z ≤ h with the known boundary conditions. The finite Hankeland Laplace transform techniques are used to find the solution of the problem.
Received 30 June 2009; accepted 20 April 2010.The authors are grateful to the anonymous referee for several helpful suggestions.Address correspondence to K. S. Parihar, SSBT’s College of Engineering & Technology,
A numerical estimate for the temperature distribution is obtained and depictedgraphically.
The present paper is concerned with the determination of temperature,displacement and stress functions of a thin circular plate of thickness h occupyingthe space D � 0 ≤ r ≤ a, 0 ≤ z ≤ h. Special cases of the third kind of boundarycondition [
T�r� z� t�+ �
�zT�r� z� t�
]z=h
= f�r� t� (1)
[T�r� z� t�+ �
�zT�r� z� t�
]z=0
= g�r� t� (2)
are used where T�r� z� t� is the temperature of the plate and f� g are known functionsof rand t. Mathematically, a third kind of boundary condition is the one in whicha linear combination of the temperature and its normal derivative is prescribedat the boundary surface. The physical significance of the third kind of boundarycondition is that the boundary surface under consideration dissipates heat byconvection according to Newton’s law of cooling (i.e., heat transfer is proportionalto temperature difference) to a surrounding temperature, which varies both withtime and position along the boundary surface. Boundary conditions of the type (1)–(2) have been considered in [6–8]. Also, T�r� z� t� satisfied the initial condition
T�r� z� 0� = 0 (3)
while zero temperature is maintained on outer curved surface of the thin circularplate; that is,
T�a� z� t� = 0 (4)
The boundary conditions (1), (2), (4) and the initial condition (3) suggest that thecircular plate is subject to axisymmetric heating.
The differential equation governing the displacement function U�r� z� t� as inNowacki ([4], Chap.1) is
�2U
�r2+ 1
r
�U
�r= �1+ ��atT (5)
where � and at are Poisson’s ratio and the linear coefficient of thermal expansion ofthe material of the plate, respectively, and
U�a� z� t� = 0 (6)
The stress functions �rr and ��� are given by
�rr = −21r
�U
�r ��� = −2
�2U
�r2(7)
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THERMOELASTIC OF THIN PLATES 909
where is the Lame’ constant, while each of the stress functions �rz� �zz� ��z are zerowithin the plate in the plane state of stress. Also, it follows from (5) and (7) that
��� = −2�1+ �� atT − �rr (8)
which obviates the need to calculate ��� separately.Singru and Khobragade [9] studied the direct steady-state thermo-elastic
problem to determine the temperature, displacement and stress functions of athin rectangular plate occupying the space D∗ � 0 ≤ x ≤ a, −b ≤ y ≤ b with thestated boundary conditions. More recently, Ghadle and Khobragade [10] consideredthe problem of determining temperature gradient, temperature distribution,displacement and stress functions on the edge x = a of a thin rectangular plateoccupying the space D∗ with the stated boundary conditions, by applying finiteMarchi–Fasulo transform and Laplace transform techniques. The numerical resultsfor unknown temperature gradient are calculated and depicted graphically. Theproblem addressed in [10] is reconsidered in the present paper in order to highlightsome new features.
The reason for combining the study of the circular and the rectangular platesin a single paper is purely mathematical. More precisely, the expressions for whichthe inverse Laplace transforms are to be found are similar in the two problems. Weuse the theory of residues for calculating the inverse Laplace transforms [11].
THE CIRCULAR PLATE PROBLEM
Consider the thin circular plate of thickness h occupying the space D � 0 ≤r ≤ a, 0 ≤ z ≤ h. The temperature T�r� z� t� satisfies the differential equation
�2T
�r2+ 1
r
�T
�r+ �2T
�z2= 1
k
�T
�t(9)
where k is the thermal diffusivity of the material of the plate. Equations (1)–(7) and(9) constitute the formulation of the problem under consideration.
Let �T denote the finite Hankel transform of T with respect to the variable rand let T ∗ denote the Laplace transform of T with respect to the variable t, then wehave
�T ∗��n� z� s� =∫ �
0e−stdt
∫ a
0rT�r� z� t� J0��nr�dr (10)
Using the two transforms and equations (9), (1)–(4) we get
d2
dz2�T ∗ − q2T ∗ = 0 (11)
q2 = �2n +s
k(12)[
�T ∗��n� z� s�+d
dz�T ∗��n� z� s�
]z=h
= f ∗��n� s� (13)
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910 K. S. PARIHAR AND S. S. PATIL[�T ∗��n� z� s�+
d
dz�T ∗��n� z� s�
]z=0
= g∗��n� s� (14)
where �n are the positive roots of the transcendental equation
J0��na� = 0 (15)
We use the principle of superposition and split the problem into the following twoproblems:
Problem 1. To find the function T�r� z� t� = Tf �r� z� t� satisfying theequations (9), (1)–(4) with g�r� t� ≡ 0.
Problem 2. To find the function T�r� z� t� = Tg�r� z� t� satisfying theequations (9), (1)–(4) with f�r� t� ≡ 0.
Once the solutions of the Problem 1 and Problem 2 are found, the solutionT�r� z� t� satisfying the equations (9), (1)–(4) is given by
T�r� z� t� = Tf �r� z� t�+ Tg�r� z� t� (16)
THE SOLUTION OF PROBLEM 1
The solution of equations (11)–(14) with g�r� t� ≡ 0 is given by
�T ∗f ��n� z� s� = f ∗��n� s�
�sinh�qz�− q cosh�qz� �1− q2� sinh�qh�
(17)
where q is given by (12). We now find the inverse Laplace transform of G∗ given by
Thus, if L−1 denotes the inverse Laplace transform operator we have
L−1[�G∗��n� z� s�
] = sum of residues of estG∗ at the poles
s = −2mn� m = 1� 2� � � � � � � � � � and s = −�2
n (25)
Taking the inverse finite Hankel transform and the inverse Laplace transformin (17) and using the convolution theorem of Laplace transform together withequations (18)–(25) we can write
Tf �r� z� t� = T1�r� z� t�+ T2�r� z� t� (26)
T1�r� z� t� =4ka2h
�∑n=1
J0��nr�
J 21 ��na�
�∑m=1
�−1�m+1Bm�z�Amn�t� (27)
where Bm�z� is given by (23) and we have
Amn�t� =∫ t
0f ��n� t
′�e2mn�t−t′�dt′ (28)
f ��n� t� =∫ a
0rf�r� t�J0��nr�dr (29)
T2�r� z� t� =2ke−z
a2 sinh�h�
�∑n=1
J0��nr�
J 21 ��na�
∫ t
0f ��n� t
′�e�2n�t−t′�dt′ (30)
where mn and �n are given by (20)–(21).The function Amn�t� in (28) may be written also in the form
Amn�t� =12mn
[f ��n� t�− f ��n� 0�e
2mnt −∫ t
0e
2mn�t−t′� �
�t′f��n� t
′�dt′]
(31)
which is obtained from (28) by carrying out an integration by parts. The form (31)is more suitable when we want to differentiate the series in (27) term by term.
The solution of Problem 1 was attempted earlier in [6]. Due to erroneousinverse Laplace transform, the solution in [6] differs from the one given by (26)–(30) on two counts. In [6], T2�r� z� t� ≡ 0 and the factor
[1+ � h
m��2]is replaced by[
� hm��2�1− �2n�
]in T1�r� z� t� given by (27)–(29) together with (23). To ensure that the
solution given above by (26)–(30) is the correct solution, we carry out the followingverifications.
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912 K. S. PARIHAR AND S. S. PATIL
Using (27), (23) and (31) while carrying out term-by-term differentiation of theseries in (27) we can show that
�2T1
�r2+ 1
r
�T1
�r+ �2T1
�z2− 1
k
�T1
�t= − 4
a2h
�∑n=1
f ��n� t�J0��nr�
J 21 ��na�
�∑m=1
�−1�m+1Bm�z�
= 2e−z
a2 sinh�h�
�∑n=1
f ��n� t� J0��nr�
J 21 ��na�
(32)
where we have obtained the last step by using the Fourier series expansion
sinh��z� = 2�h2
sinh��h��∑
m=1
m�−1�m+1 sin�m�zh�[
�2 + �m�h�2] (33)
with � = 1. Also from (30) we have
�2T2
�r2+ 1
r
�T2
�r+ �2T2
�z2− 1
k
�T2
�t= − 2e−z
a2 sinh�h�
�∑n=1
f ��n� t�J0��nr�
J 21 ��na�
(34)
Thus using (26), (32) and (34) we find that T = Tf satisfies the differentialequation (9).
Also, using (27) and (31) we find
T1�r� z� t� + �
�zT1�r� z� t� =
2a2
�∑n=1
J0��nr�f ��n� t� sinh��nz�
J 21 ��na� sinh��nh�
− 4�ka2h2
�∑n=1
J0��nr�
J 21 ��na�
�∑m=1
m�−1�m+1 sin�m�zh�
k[�2n +
(m�h
)2]×
[f ��n� 0�e
2mnt +∫ t
0e−2mn�t−t′� �
�t′f ��n� t
′�dt′]
(35)
where we have made use of (33) with � = �n. From (35) we now have[T1�r� z� t�+
�
�zT1�r� z� t�
]z=h
= 2a2
�∑n=1
J0��nr�f ��n� t�
J 21 ��na�
= f�r� t� (36)[T1�r� z� t�+
�
� zT1�r� z� t�
]z=0
= 0 (37)
Also, from (30) we find that[T2�r� z� t�+
�
� zT2�r� z� t�
]≡ 0 (38)
It follows from (26) and (36)–(38) that T = Tf satisfies the conditions (1)–(2) withg�r� t� ≡ 0. Finally, it is a simple matter to verify that the solution T = Tf satisfiesthe conditions (3) and (4). Thus, it is verified that the solution T = Tf of Problem 1
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THERMOELASTIC OF THIN PLATES 913
given by (26)–(30) satisfies the differential equation (9), the initial condition (3) andthe boundary conditions (1)–(2), and (4) with g�r� t� ≡ 0.
DETERMINATION OF DISPLACEMENT AND STRESS FUNCTIONS
The solutions of Problem 1 given by (26)–(30) is of the form
T�r� z� t� = Tf �r� z� t� =�∑n=1
En�z� t� J0��nr� (39)
Using (5), (6) and (39) we have
U�r� z� t� = −�1+ ��at
�∑n=1
En�z� t�J0��nr�
�2n(40)
Now, using (7) and (40) we find
�rr�r� z� t� = −2�1+ ��at
�∑n=1
En�z� t���nr�−1J1��nr� (41)
����r� z� t� = −2�1+ ��at
�∑n=1
En�z� t�[J0��nr�− ��nr�
−1J1��nr�]
(42)
The expression for ��� given by equation (42) may be obtained also by using (39),(41) and (8).
A SPECIAL CASE OF PROBLEM 1
The function f�r� t� in Problem 1, introduced through the boundary condition(1), can be arbitrarily prescribed. As in [6] we take
For this choice of f�r� t� we can get the corresponding results for the steady-statecase discussed separately in [6]. Substituting f�r� t� from (43) into (27)–(30) we get
T1�r� z� t� =4kHah
�∑n=1
J0��nr�
�2nJ21 ��na�
[∫ a
0J0��nx�dx −
4�n
J1��na�
]
×�∑
m=1
�−1�m+1Bm�z�
�2mn − 1�
[1− e−t + �e−2mnt − 1�/2
mn
](44)
T2�r� z� t� =2kHe−z
a sinh�h�
�∑n=1
J0��nr�
�2nJ21 ��na�
[∫ a
0J0��nx�dx −
4�n
J1��na�
]× ��2
n − 1�−1[1− e−t + �e−�2nt − 1�/�2
n
](45)
where Bm�z�� 2mn and �2
n are given by (23), (20) and (21). Thus the temperatureT =Tf is given by (26), (44) and (45).
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914 K. S. PARIHAR AND S. S. PATIL
Figure 1 Variation of T = −Tf �r� z� t�/� given by Eqs. (26), (44)–(45) and (50) with z for t = 1 andr = 0�5, 1.0, 1.5. The values of constants are a = 2, h = 1, k = 0�86.
In the steady-state case we encounter the series
�∑m=1
�−1�m+1Bm�z�
2mn
= h
2k�1− �2n�
[e−z
sinh�h�+ sinh��nz�− �n cosh��nz�
sinh��nh�
](46)
where we have used partial fractions and the Fourier series expansion (33) for�= 1 and � = �n. For the special case (43), let Tf0�r� z� denote the limiting value ofTf �r� z� t� as t → � then using (26) and (44)–(46) we get
Tf0�r� z� =2Ha
�∑n=1
J0��nr� �sinh��nz�− �n cosh��nz�
�2n�1− �2n�J21 ��na� sinh��nh�
[∫ a
0J0��nx�dx −
4�n
J1��na�
](47)
The expression for the steady-state temperature Tf0�r� z� has been given in [6] as
Tf0�r� z� =2a2
�∑n=1
F��n�J0��nr� �sinh��nz�− �n cosh��nz�
�1− �2n�J21 ��na� sinh��nh�
(48)
F��n� =∫ a
0rF�r�J0��nr�dr � F�r� = �r2 − ra�H (49)
It is easily verified that Tf0�r� z� given by (48)–(49) reduces to that given by (47) onsimplification.
Returning to the transient case set
� = 4� ka�h2H�−1 (50)
Numerical results are obtained for the temperature Tf �r� z� t�/� given by (26), (44),(45) and (50). The variation of −Tf �r� z� t�/� with z is shown in Figure 1 fort = 1� a = 2 and r = 0�5� 1�0 and 1�5.
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THERMOELASTIC OF THIN PLATES 915
While carrying out the numerical work, it is observed that the contributionof T2 dominates over that of T1. It is evident from (39) and (41) that theexpressions for �2�1+ ��at
−1 �rr and �2�1+ ��at −1 ��� may be obtained from the
temperature expression Tf given by (26), (44), and (45) simply by replacing J0��nr�by −��nr�
−1J1��nr� and[��nr�
−1J1��nr�− J0��nr�], respectively. The variations of
�2�1+ ��at −1 �rr and �2�1+ �� at
−1 ��� with z are shown in Figures 2 and 3,respectively.
THE SOLUTION OF PROBLEM 2
The solution of equations (11)–(14) with f�r� t� ≡ 0 is given by
T ∗g ��n� z� s� = −g∗��n� s�
�sinh q�z− h�− q cosh q�z− h�
�1− q2� sinh�qh�(51)
Comparing (51) with (17) we find that the expression for T ∗g can be obtained from
that of T ∗f by replacing z by �z− h� and f ∗��n� s� by −g∗��n� s�. Therefore, we can
write the solution of Problem 2 at once using (26)–(30). Thus we have
Tg�r� z� t� = T3�r� z� t�+ T4�r� z� t� (52)
T3�r� z� t� =4ka2h
�∑n=1
J0��nr�
J 21 ��na�
�∑m=1
�−1�m+1Bm�z− h�Dmn�t�� (53)
Dmn�t� = −∫ t
0g��n� t
′�e2mn�t−t′�dt′ (54)
g��n� t� =∫ a
0rg�r� t�J0��nr�dr (55)
Figure 2 Variation of �rr = �2�1+ ���at −1�rr �r� z� t� with z for t = 1. The expression for �rr is
obtained form T in Figure 1 simply by replacing Jo��nr� by ��nr�−1J1��nr�. The values of r and various
constants are the same as in Figure 1.
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916 K. S. PARIHAR AND S. S. PATIL
Figure 3 Variation of ��� = �2�1+ ���at −1 ����r� z� t� with z for t = 1. It follows from Eq. (8) that
��� = T − �rr � and so no new calculations are required for ���. The values of r and various constantsare the same as in Figures 1–2.
T4�r� z� t� = −2ka2
e−�z−h�
sinh�h�
�∑n=1
J0��nr�
J 21 ��na�
∫ t
0g��n� t
′�e−�2n�t−t′�dt′� (56)
where Bm�z�� mn and �n are given by (23), (20) and (21) in which �n are thepositive roots of the transcendental equation (15). The solution of Problem 2 wasattempted earlier in [7]. Due to erroneous inverse Laplace transform the solution in[7] differs from the one given by (52)–(56) on two counts. In [7], T4�r� z� t� ≡ 0 andthe factor
[1+ � h
m��2]is replaced by
[� hm��2�1− �2n�
]in T3�r� z� t� given by (53)–(55)
together with (23). Since we have deduced the solution of Problem 2 from that ofProblem 1, the solution of Problem 2 given by (52)–(56) is the correct solution andno verification is called for.
Also, the analysis of relationships among the temperature, displacement andstress functions carried out previously for Problem 1 can be extended verbatimto Problem 2. Thus the expressions for �2�1+ ��at
−1 �rr and �2�1+ ��at −1 ���
may be obtained from the temperature expression Tg given by (52)–(56) simply byreplacing J0��nr� by −��nr�
where Ti�i = 1� 2� 3� 4� are given by (27)–(30), (53)–(56) and (57). The special case(57) satisfies the condition f�r� t� ≡ ehg�r� t�. Then, it follows from (30) and (56) that
T2�r� z� t�+ T4�r� z� t� ≡ 0 (59)
Using (58)–(59), (27)–(29), (53)–(55) and (57) we find
T�r� z� t� = 16kah
�∑n=1
J0��nr�
�3nJ1��na�
�∑m=1
[1+ �−1�m+1eh
]Bm�z�
�2mn − 1�
×[1− e−t + �e−2mnt − 1�/2
mn
](60)
where Bm�z� is given by (23).In the limiting case as t → � we need the series
�∑m=1
�−1�m+1Bm�z− h�
2mn
= −�∑
m=1
Bm�z�
2mn
= h
2k�1− �2n�
[e−�z−h�
sinh�h�
+ sinh ��n�z− h� − �n cosh ��n�z− h�
sinh��nh�
](61)
We have deduced the above result from the series (46). If we denote by T0�r� z� thelimiting value of T�r� z� t� as t → � and make use of the relations (60)–(61) and (46)we get
The expression for the steady-state temperature T0�r� z� has been given in [7] as
T0�r� z� =2a2
�∑n=1
F0��n�
�1− �2n�
[sinh��nz�sinh��nh�
− �ncosh��nz�sinh��nh�
]J0��nr�
J 21 ��na�
− 2a2
�∑n=1
G0��n�
�1− �2n�
[sinh ��n�z− h�
sinh��nh�− �n
cosh ��n�z− h�
sinh��nh�
]J0��nr�
J 21 ��na�
(63)
F0��n� =∫ a
0rF0�r�J0��nr�dr G0��n� =
∫ a
0rG0�r�J0��nr�dr� (64)
F0�r� = �a2 − r2�eh G0�r� = �a2 − r2� (65)
It is easily verified that T0�r� z� given by (63)–(65) reduces to that given by (62) onsimplification.
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918 K. S. PARIHAR AND S. S. PATIL
Figure 4 Variation of T ∗ = −T�r� z� t�/� given by Eqs. (60) and (66) with z for t = 1 and r = 0�5, 1.0,1.5. The values of constants are a = 2, h = 1 and k = 0�86�
Returning to the transient case the variation of −T�r� z� t�/� with z is shownin Figure 4 for t = 1� a = 2 and r = 0�5� 1�0 and 1�5, where
� = 16k�/�ah2� (66)
and T is given by (60). The variation of �2�1+ ���at −1 �rr with r is shown
in Figure 5 for t = 1� h = 1 and z = 0�0� 0�5 and 1�0. As pointed out earlier,the expression for �2�1+ ��at
−1 �rr may be obtained from that of T simply byreplacing J0��nr� by −��nr�
−1J1��nr�.While calculating the value of �rr for r = 0 the limit formula
limx→0
[x−1J1�x�
] = 12
(67)
is employed.
Figure 5 Variation of �∗rr = �2�1+ ���at
−1 �rr �r� z� t� with r for t = 1. The expression for �∗rr is
obtained from T ∗ in Figure 4 simply by replacing Jo��nr� by ��nr�−1J1��nr�. While calculating �∗
rr forr = 0 the limit formula (67) is used for z = 0�0� 0�5, 1.0. The values of constants are the same as inFigure 4.
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THERMOELASTIC OF THIN PLATES 919
THE RECTANGULAR PLATE PROBLEM
Consider a thin rectangular plate occupying the space D∗ � 0 ≤ x ≤ a�−b ≤y ≤ b� where x� y are Cartesian coordinates. The displacement components ux anduy in the x and y directions are represented in the integral form as
ux =∫ [
1E
(�2U
�y2− �
�2U
�x2
)+ �T
]dx (68)
uy =∫ [
1E
(�2U
�x2− �
�2U
�y2
)+ �T
]dy (69)
where � and � are the Poisson ratio and the linear coefficient of thermal expansion ofthe material of the plate, respectively, and U�x� y� is the Airy stress function, whichsatisfies the relation (
�2
�x2+ �2
�y2
)2
U = �E
(�2
�x2+ �2
�y2
)T (70)
where E is Young’s modulus of elasticity and T is the temperature of the plate (see[9, 10]). The temperature T satisfies the differential equation
�2T
�x2+ �2T
�y2= 1
k
�T
�t(71)
the initial condition
T�x� y� 0� = 0 (72)
and the boundary conditions[T�x� y� t�+ �
�xT�x� y� t�
]x=0
= h�y� t� (73)[T�x� y� t�+ k1
�
�yT�x� y� t�
]y=b
= 0 (74)[T�x� y� t�+ k2
�
�yT�x� y� t�
]y=−b
= 0 (75)
and the interior condition[T�x� y� t�+ �
�xT�x� y� t�
]x=�
= f�y� t� (76)
where 0 < � < a� k is the thermal diffusivity of the material of the plate and k1� k2are the radiation constants on the two edges y = b�−b of the rectangular plate. Thestress components in terms of U are given by
�xx =�2U
�y2 �yy =
�2U
�x2 �xy = − �2U
�x�y� (77)
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920 K. S. PARIHAR AND S. S. PATIL
Equations (68)–(77) constitute the mathematical formulation of the problem underconsideration.
THE USE OF INTEGRAL TRANSFORMS
We make use of the Laplace transform and the Marchi–Fasulo integraltransform as discussed in [9, 10]. Let
Pn�y� = Qn cos any −Wn sin any (78)
Qn = �k1 + k2�an cos anb Wn = 2 cos anb + �k2 − k1�an sin anb (79)
The function Pn�y� satisfies the differential equation
then it is easily verified that Pn�y�, n = 1� 2� � � � are orthogonal on �−b� b� and wecan define the Marchi–Fasulo transform of f�y� as
f �an� =∫ b
−bf�y�Pn�y�dy (83)
with the inversion formula
f�y� =�∑n=1
f �an�
An
Pn�y� (84)
where
An =∫ b
−b�Pn�y�
2dy = b�Q2n +W 2
n �+ �Q2n −W 2
n �sin�2anb�
2an
(85)
Now consider∫ b
−b
�2T
�y2Pn�y�dy =
[�T
�yPn�y�− TP ′
n�y�
]b
−b
+∫ b
−bTP ′′
n �y�dy
= −P ′n�b�
[k1
�T
�y+ T
]y=b
+ P ′n�−b�
[k2
�T
�y+ T
]y=−b
− a2n
∫ b
−bTPn�y�dy (86)
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THERMOELASTIC OF THIN PLATES 921
where we have made use of the integration by parts and the relations (80)–(81).Making use of the conditions (74)–(75), equation (86) reduces to
∫ b
−b
�2T
�y2Pn�y�dy = −a2
n
∫ b
−bTPn�y�dy (87)
Taking Marchi–Fasulo transform with respect to the variable y and theLaplace transform with respect to the variable t in equations (71), (73), (76) andusing (72), (87) we get
d2
dx2T ∗ − q2T ∗ = 0 q2 = a2
n +s
k(88)
[T ∗�x� an� s�+
d
dxT ∗�x� an� s�
]x=0
= h∗�an� s�� (89)[T ∗�x� an� s�+
d
dxT ∗�x� an� s�
]x=�
= f ∗�an� s� (90)
The solution of the differential equation (88) satisfying the conditions (89)–(90) maybe written
T ∗�x� an� s� = f ∗�an� s��sinh qx − q cosh qx
�1− q2� sinh q�
− h∗�an� s��sinh q�x − ��− q cosh q�x − ��
�1− q2� sinh q�(91)
This form of solution is similar to the one given by (17) and (51). Therefore thesolution of the problem stated by (71)–(76) may at once be written as
The constants an are the positive zeros of the transcendental equation (82).The solution of the rectangular plate problem was attempted earlier in [10].
Due to erroneous inverse Laplace transform, the solution in [10] differs from theone given by (92)–(99) on four counts.
In [10], T6�x� y� t� ≡ 0, T8�x� y� t� ≡ 0 and Dm�x� in T5�x� y� t� and T7�x� y� t� isreplaced by Lm�x� given by
Lm�x� =m�
��1−m2�
[sin
(m�x
�
)−m cos
(m�x
�
)](100)
Substituting T�x� y� t� from (92)–(96) into (70) we can find the Airy stress functionU�x� y� t�. Indeed, we have
With this choice we can get the results for the steady-state case studied separatelyin [9]. The expressions in (102) satisfy the condition f�y� t� ≡ e�h�y� t�. When thiscondition is used in (94) and (96) together with (99) we get
T6�x� y� t�+ T8�x� y� t� ≡ 0 (103)
Then using (92)–(93), (95) and (102) we find
T�x� y� t� = 2k�
�∑n=1
MnPn�y�
An
�∑n=1
[1+ �−1�m+1e�
]��2
mn − 1�Dm�x�
×[1− e−t + �e�
2mnt − 1�/�2
mn
](104)
Mn = 16�k1 + k2� cos�anb�
a4n
[{3− �anb�
2}sin�anb�− 3�anb� cos�anb�
](105)
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THERMOELASTIC OF THIN PLATES 923
where An� �n and �mn are given by (85), (79) and (98). The function Pn�y� is givenby (78) where an are the positive zeros of the transcendental equation (82).
In the limiting case as t → � we encounter the series
�∑m=1
�−1�m+1Dm�x�
�2mn
= �
2k�1− a2n�
[e−x
sinh �+ sinh anx − an cosh anx
sinh an�
](106)
where Dm�x� is given by (97). Like the series in (46), the above series has beenevaluated using partial fractions and the Fourier series expansion (33). Also,replacing x by �x − �� we can get a result similar to that of (61). Using these seriesand taking the limit as t → � in (104) we get
T0�x� y� �� =�∑n=1
[�1− a2
n�An sinh an�]−1
MnPn�y�
× [e��sinh anx − an cosh anx�− �sinh an�x − ��− an cosh an�x − ���
]�
(107)
where T0�x� y �� denotes the limiting value of T�x� y� t� as t → �. The steady-statecase has been studied in [9] with � = a. It is shown in [9] that
where Pn�y� is given by (78). It is easily verified that
F0�an� = eaMn G0�an� = Mn (111)
Thus T0�x� y a� given by (108) and (111) is the same as the one given by (107) with� = a.
Returning to the transient case, we set
k1 = 1� k2 = 0� b = 1� an =12xn (112)
Then (82) reduces to
2+ xn cot xn = 0 (113)
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924 K. S. PARIHAR AND S. S. PATIL
Figure 6 Variation of �T = T�a� y� t�/� with y for t = 1� 10. The expressions for T�a� y� t� and � aregiven by Eqs. (104)–(105) and (114). The values of constants are k1 = 1� k2 = 0� b = 1� a = 2� � = 1�5and k = 0�86.
Taking a = 2� � = 1�5� k = 0�86 and
g�y� t� = T�a� y� t� � = 8�k1 + k2��k�/�� (114)
the variation of g�y� t�/� with y is shown in Figure 6 for t = 1� 10�
REFERENCES
1. S. K. Roy Choudhury, A Note on the Quasi-Static Stresses in a Thin Circular PlateDue to Transient Temperature Applied Along the Circumference of a Circle Over theUpper Face, Bull. Sci. Acad. Poln. Ser. Tech., vol. 20, pp. 21–24, 1972.
2. N. Noda, R. B. Hetnarski, and Y. Tanigawa, Thermal Stresses, 2nd ed., Taylor andFrancis, New York, 2003.
3. R. B. Hetnarski and M. R. Eslami, Thermal Stresses – Advanced Theory and Applications,Springer, New York, 2009.
4. W. Nowacki, Thermoelasticity, Addison-Wesley, New York, 1962.5. M. Ishihara, Y. Tanigawa, R. Kawamura, and N. Noda, Theoretical Analysis of
Thermoelastic Deformation of a Circular Plate Due to a Partially Distributed HeatSupply, J. Therm. Stresses, vol. 20, pp. 203–225, 1997.
6. A. A. Navlekar and N. W. Khobragade, Thermoelastic Problem on Upper PlaneSurface of a Thin Circular Plate, Bull. Cal. Math. Soc., vol. 99, pp. 27–36, 2007.
7. N. W. Khobragade and M. S. Warbhe, Thermoelastic Problem of a Thin Circular Plate,Bull. Cal. Math. Soc., vol. 98, pp. 329–338, 2006.
8. L. C. Andrews and B. K. Shivamoggi, Integral Transforms for Engineers, Prentice-Hallof India Pvt. Ltd., New Delhi, 2003.
9. S. S. Singru and N. W. Khobragade, Thermoelastic Problem of a Thin RectangularPlate in Marchi–Fasulo Transform Domain-I, Bull. Cal. Math. Soc., vol. 99.pp. 223–230, 2007.
10. K. P. Ghadle and N. W. Khobragade, Study of an Inverse Transient ThermostaticProblem of a Thin Rectangular Plate, Bull. Cal. Math. Soc., vol. 100, pp. 1–10, 2008.
11. R. V. Churchil and J. W. Brown, Complex Variables and Applications, 5th Ed., Mc GrawHill, New York, 1990.
