DOI : 10.17577/IJERTV2IS70634
- Open Access
- Total Downloads : 501
- Authors : C. M. Jadhav
- Paper ID : IJERTV2IS70634
- Volume & Issue : Volume 02, Issue 07 (July 2013)
- Published (First Online): 22-07-2013
- ISSN (Online) : 2278-0181
- Publisher Name : IJERT
- License:
This work is licensed under a Creative Commons Attribution 4.0 International License
Thermal Stresses in Semi-Infinite Solid Rectangular Plate with Heat Source
C. M. Jadhav
Department of Mathematics, Dadasaheb Rawal College Dondaicha, NMU Jalgaon University, Jalgaon [M.S.], INDIA
This paper is concerned with three dimensional inverse thermoelastic problem of semi-infinite solid rectangular plate due to internal heat source. To determine unknown temperature distribution, displacement and thermal stresses on edge z=h of the rectangular plate due to internal heat sources with known third kind boundary and initial condition by applying Fourier and Laplace transform technique.
The results are obtained in terms of infinite series and
Consider a solid rectangular plate from fig. 1 occupying the space D: 0 x < , 0 y < , 0 z h.The displacement components ux, uy, uz in the x, y, z direction respectively are in the integral form as in [1] u = 1 2U + 2 U 2U + T dx (2.1)
x 0 E y2 z2 x2
x 0 E y2 z2 x2
y 0 E z2 x2 y2
y 0 E z2 x2 y2
u = 1 2U + 2U 2U + T dy (2.2)
u = h 1 2U + 2U 2U + T dz (2.3)
z 0 E x2 y2 z2
the numerical calculations are carried out by using MATHCAD -7 software and shown graphically.
Keywords: Three dimensional inverse thermoelastic problem, solid rectangular plate, integral transform.
-
Introduction
Many heat conduction problem encountered in engineering application involve time as independent variable. The goal of analysis is to determine the variation in temperature displacement and stresses with known boundary condition. Tanigawa et al. [1] have studied thermal stress analysis of a rectangular plate and its thermal stress intensity factor for compressive
stress field. Khobragade et al. [4] to determine an
where E, and are the Young modulus, the poisson ratio and the linear coefficient of thermal
expansion of the material of the plate respectively, U(x, y, z ,t) is the Airy stress function which satisfies the differential equation.
2 2 2
2 2 2
2 + 2 + 2 2 U x, y, z, t =
x y z
2 2 2
2 2 2
E 2 + 2 + 2 T(x, y, z, t) (2.4)
x y z
Here T(x, y, z, t) denotes the temperature of the thin rectangular plate satisfying the following differential equation.
inverse unsteady-state thermoelastic problem of thick
2T + 2 T + 2T + (x,y,z,t) = 1 T
(2.5)
rectangular plate. Lamba et al. [9] have studied thermoelastic problem of thin rectangular plate due to partially distributed heat supply. Himanshu S.Roy and N.W.Khobragade have discussed transient thermoelastic problem of infinite rectangular slab with second kind boundary condition.
In the present paper, an attempt has been made
x2 y2 z2 k t
Where k is thermal conductivity and is the thermal diffusivity of the material of the plate and (x, y, z, t) is heat generated within the rectangular plate for t>0 subject to initial conditions
T x, y, z, 0 = 0 (2.6)
The boundary conditions
completely the inverse unsteady state thermoelastic
T(x,y,z,t)
= F1 y, z, t (2.7)
problem of semi-infinite solid rectangular plate with
x x=0
internal heat source applied for upper plane surface
T(x,y,z,t)
y y=0
= F2 x, z, t (2.8)
with third kind boundary conditions. To determine the
temperature, displacement and thermal stresses on the
T(x, y, z, t) + c T(x,y,z,t)
z
z=0
= 0 (2.9)
edge z= h of semi- infinite rectangular plate due to internal heat by applying Fourier and Laplace transform
T(x, y, z, t) z=h = g x, y, t Unknown) (2.10) The interior condition
technique.
-
Statement of the problem
T(x, y, z, t) + c T(x,y,z,t)
z
z=
= f x, y, t (Known)
(2.11)
The components in term of U(x, y, z, t) are given by
= 2U + 2U (2.12)
where p = l
xx y2
z2
= 2U + 2U (2.13)
g x, y, t =
K , x K , y 2k
yy z2
x2
,=0 2
= 2U + 2U (2.14)
1 l+1
l sin ph cp cos ph
zz x2
y2
l=1 1c2p2
the equations (2.1) to (2.14) constitute the
t f , , t PI
mathematical formulation of the problem under consideration.
-
Solution of the problem
The Fourier integral transform of f(x), 0 x < is
0
c dPI
dz
z=
z=
e 2+2 +p2 tt dt dd
K , x K , y 2k
defined to be
,=0 2
F = K , x F(x)dx
(3.1)
1 l+1
l sin p h cp cos p h
x=0
then inverse formula
l=1 1c2p2
F x = K , x F d (3.2) ×
=0
t PI
c dPI
e 2+2+p2 tt dt
Where K , x = 2 cos x
(3.3)
0
+
+
dd
z=0
dz z=0
By applying the Fourier transform two times and
,=0
K , x K , y L1 PI
Laplace transform to equation (2.5) and their inverses, we obtain
2
2
d2 T p2 T = + (3.4)
dz k
Where p2 = 2 + 2 + s , = +
t e 2 +2 +p2 tt dt dd (3.10)
0
0
where
1 z = sin pz cp cos pz
1 2
t =
1 = K(, x) x=0 F1 , 2 = K(, y) y=0 F2 , upper limit vanishes
dT ,,z,t
1
0
0
t f , , t PI z=
T , , z, t + c
dz
z=0
= 0 (3.5)
c dPI
dz
z=
e 2+2 +p2 tt dt
T , , z, t z=h = g , , z, t (3.6)
dT ,,z,t
2 z = sin p z cp cos p z
T , , z, t + c
dz
z=
= f , , t
2 t =
t PI
c dPI
e 2+2+p2 tt dt
(3.7)
The equation (3.4) is a second order differential
0 z=0
dz z=0
equation whose solution is in form
t = t
e 2 +2 +p2 tt dt
0
0
3
T = Aepz + Bepz + PI (3.8)
(Q )
2 2
2 2
Where PI = k
D p
, D d
dz
and A, B are
-
Determination of the airy stress function
Substituting the value of T(x, y, z, t) from equation
constant. Using equation (3.5), (3.7) in (3.8) we obtain the values of A and B substituting these
(3.9) in the equation (2.4), one obtains
values (3.8) and then apply inverse of Laplace transform and Fourier integral transform. We
U x, y, z, t = E
K , x K , y 2k
2
2
obtain
,=0
,=0
T x, y, z, t =
K , x K , y 2k
2
2
0
0
1 l+1
,=0
,=0
l=1 p2 1c2p2
l sin pz cp cos pz
1
1
l+1
1 1c2p2 l sin pz cp cos pz
t f , , t PI z=
l=
t f , , t PI
c dPI
+E
+E
dz
z=
e 2+2 +p2 tt dt dd
0
c dPI
z=
e 2+2+p2 tt dt dd
,=0
K , x K , y 2k
2
2
dz z=
1 l+1
l sin p z cp cos p z
,=0
K , x K , y 2k
2
2
l=1 p2 1c2p2
×
1 l+1
l=1 1c2p2 l sin p z cp cos p z
t PI
c PI
e 2+2+p2 tt dt
0
× dd
z=0
dz z=0
t PI
c dPI
e 2+2+p2 tt dt
E
K , x K , y L1 PI
0 z=0
dz z=0
=0
,=0
,=0
dd + K , x K , y L1 PI
t e 2 +2+p2 tt dt dd (3.9)
t e 2 +2+p2 tt dt dd (4.1)
0
0
U x, y, z, t =
E
E
0
,=0
K , x K , y 2k
2
2
1 l+1
2k
1 l+1
l z t z t
p2 1c2p2
dd
2
l=1 p2 1c2p2 1 1 2 2
l=1 l 1 z 1 t 2 z 2 t
+ y=0 =0 K" , y d dy
K , x d L1 PI 3 t
E
K , x K , y L1 PI t dd
=0
=0
3 + K , y d dy K , x d
y=0
=0
=0
(4.2)
2k
1 l+1
l z t z t
-
Determination of the displacement
2
l=1 p2 1c2 p2 1 1 2 2
components
+
+
y=0 =0
K , y d dy
Substituting the value of U(x, y, z, t) from equation
K , x d L1 PI 3 t (5.2)
(4.1) in the equation (2.1), (2.2), (2.3) one obtains
=0
2k
ux =
K , x d dx
K" , y d
uz = K , y d K" , x d
x=0 =0
=0
=0
=0 2
2k
1 l+1
l z t z t
1 l+1
2
l=1 p2 1c2 p2 1 1 2 2
p2 1c2 p2
1 h h
K , x d dx
l= l 1 z dz 1 t 2 z dz 2 t
x=0 =0 0 0
=0
K" , y d L1 PI 3 t
x=0
=0
K , x d dx
K , y d
=0
=0
=0
K , y d
K" , x d
2k
=0
=0
1 l+1
l " z t " z t
h L1 PI dz 3 t
2
l=1 p2 1c2p2 1 1 2 2
0
2k
2 L1 PI
=0 K , x d =0 K" , y d 2
x=0 =0 K , x d dx K n , y 3 t
z2
1 l+1
p2 1c2 p2
+
K" , x d dx
l=1 l h z dz t h z dz t
x=0
=0
0 1 1 0 2 2
=0
K , y d 2k
2
2
1 l+1
l z t z t
K , x d
K" , y d h L1 PI dz 3 t
l=1 p2 1c2 p2 1 1 2 2
=0
=0 0
+
K" , x d dx
+
K , x d K , y d 2k
x=0
=0
=0
=0 2
=0 K , y d L1 PI 3 t
1 l+1
+ K , x d dx
K , y d
p2 1c2 p2
x=0
=0
=0
l=1 l h "1 z dz 1 t h "2 z dz 2 t
2k
1 l+1
l z t z t 0 0
2
l=1 p2 1c2 p2 1 1 2 2
+
K , x d
K , y d
=0
=0
+ x=0 =0 K , x d dx =0 K , y d
h L1 PI dz 3 t
0
0
L1 PI 3 t (5.1)
2k
uy =
K , y d dy
K , x d
+ K , x d K , y d 2
2k
y=0
=0
1 l+1
p2 1c2p2
=0
=0
=0
1 l+1
p2 1c2p2
2 l=1 l "1 z 1 t "2 z 2 t
l=1 l h z dz t h z dz t
K , y d dy K , x d
0 1 1 0 2 2
y=0
2 L1 PI
z2
=0
3 t
=0
+ K , x K , y h L1
m,n=1 m n 0
m,n=1 m n 0
(5.3)
y=0
=0
K , y d dy
K" , x d
-
Determination of stress functions
Using (4.1) in (3.14), (3.15) and (3.16) the stress
2k
=0
=0
1 l+1
l z t z t
functions are obtained as
2
l=1 p2 1c2p2 1 1 2 2
xx = E K , x d K" , y d
K , y d dy
K" , x d
=0
1 l+1
=0
y=0 =0
=0
=0
L1
=0
2k
p2 1c2p2
PI
3 t
2 l=1 l 1 z 1 t 2 z 2 t
y=0
y=0
+
=0
K" , y d dy
K , x d
E
K , x d
K" , y d
e p2 +A t
1
=0
=0
2
z0 x y
L PI 3 t
1 e e
erf erf
4 4
=0
=0
=0
=0
E K , x d
K , y d 2k
2
2
×
×
1 l+1
8
1 l+1
l sin p z cp cos p z
l=
l=
1 p2 1c2p2 l "1 z 1 t "2 z 2 t
2 l=1 1c2 p2
=0
=0
E
K , x d
K , y d
e p2+B t
2 L1 PI
t (6.1) 1
x y 2
=0
=0
z2 3
+ 2 e(zz0)erf erf e p +1 t
2k
1p
4
2
4
(7.6)
yy = E K , x d K , y d
g x, y, t = 1 e ez0 erf x
erf y
1 l+1
=0
=0 2
8 1 l+1
4
4
l=
l=
1 p2 1c2p2 l "1 z 1 t "2 z 2 t
=0
=0
=0
=0
E K , x d K , y d
× 2 1 c2p2 l sin ph cp cos ph
l=1
e p2 +A t
2 L1 PI
z2
3 t
1 e 2ez0 erf x
4
erf y
4
=0
=0
E
K" , x d
K , y d 2k
2
2
× 8
=0
=0
1 l+1
1c2p2
1 l+1
l z t z t
2
l=1 l sin p h cp cos p h
l=1 p2 1c2p2 1 1 2 2
=0
=0
E
L1 PI t
K" , x d
K , y d
e p2+B t
+ 1 e(hz0)erf x erf y e p2 +1 t
=0
=0
3 (6.2)
1p2
4
4
2k
(7.7)
zz = E K" , x d K , y d 2
8. Numerical results
=0 =0
1 l+1
l z t z t
For Aluminum metal
l=1 p2 1c2p2 1 1 2 2
Modulus elasticity E = 6.9 × 1011 Poisson
=0
=0
E
L1 PI 3 t
E
K" , x d
=0
=0
K , x d
K , y d
K" , y d 2k
ratio = 0.35, = 12.84 × 106
Thermal Expansion coefficient t = 25.5 × 106
Thermal Diffusivity k=117, c=1
Thermal Conductivity = 3.33 , h=1, = 0.5
=0
0
0
=0 2 l
1 l+1
l z t z t
A=3, B=2, p= , x=2, y=3, z = 0.2
l=1 p2 1c2p2 1 1 2 2
=0
=0
E
L1 PI t
K , x d
K" , y d
T x, y, z, t =
1 e0.5 2e0.50.2erf 2 erf 3
=0
=0
3 (6.3)
4
4
7. Special case
1 l+1
l sin l z l cos l z
Setting
× 8
l 2
f x, y, t = x y (1 e )2et (7.1)
2
l=1 1
x, y, z, t = x y e(zz0)et
l 2+3 t
> 0
(7.2)
e
f = 2
cos x 2
cos y (1 e )2et (7.3)
1 e0.5 2e0.2erf 2
4
erf 3
4
1 l+1
= 2 cos x 2 cos y e(zz0 ) 1
(7.4)
2
2
1 l
s+
× 8
2
l=1
l sin l z l cos p z
1 2 2
(zz )
t
=1p2 cos x cosy e 0 e (7.5)
l 2
Substitute the values in the equation (3.9),(3.10), (4.1),(5.1),(5.2)(5.3),(6.1),(6.2)and(6.3)one obtains
+2 t
e
l 2+1 t
T x, y, z, t =
1 e 2ez0 erf x
erf y
+ 1
1 l 2
e(z0.2)erf 2
4
erf 3 e
4
(7.8)
4
8 1 l+1
4
g x, y, t =
× 2 1 c2p2 l sin pz cp cos pz
1 e0.5 2e0.50.2erf 2 erf 3
l=1
4
4
1 l+1
l sin l l cos l
× 8
l 2
2
l=1 1
l 2+3 t
e
1 e0.5 2e0.2erf 2
4
erf 3
4
×
8
1 l+1
1 l 2
2
l=1
l sin l h l cp cos l h
l 2+2 t
e
+ 1 e h0.2 erf 2
erf 3
1 l 2
4
4
l 2+1 t
e
(7.9)
U x, y, t
= E 1 e0.5 2e0.50.2erf 2
4
erf 3
4
1 l+1
l 2 1 l 2
× 8
2
l=1
l sin l l cos l
l 2+3 t
e
E 1 e0.5 2e0.2erf 2
4
erf 3
4
×
1 l+1
l 2 1 l 2
8
2
l=1
l sin l h l cp cos l h
l 2+2 t
e
+E 1
e h0.2 erf 2
erf 3
1 l 2
4
4
l 2+1 t
e
(7.10)
-
Conclusion
In this paper, we discussed completely the inverse unsteady state thermoelastic problem of semi- infinite solid rectangular plate with internal heat source applied for upper plane surface with known third kind boundary and initial condition. The Fourier and Laplace integral transforms are used to obtain the numerical results. The temperature, Displacement and thermal stresses that are obtained can be applied to the design of useful structure or machines in engineering application. Any particular case of special interest can be derived by assigning suitable value of the parameters and function in the expression.
Acknowledgements
The author expresses their sincere thanks to Dr. B.R.Ahirrao for their expert comments.
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