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**Authors :**Sunil D. Bagde -
**Paper ID :**IJERTV8IS100071 -
**Volume & Issue :**Volume 08, Issue 10 (October 2019) -
**Published (First Online):**16-10-2019 -
**ISSN (Online) :**2278-0181 -
**Publisher Name :**IJERT -
**License:**This work is licensed under a Creative Commons Attribution 4.0 International License

#### Thermal Stresses in a Finite Solid Cylinder

(Magnetothermodynamics Stresess in A Solid Cylinder)

Sunil D. Bagde

Department of Mathematics, Gondwana University, Gadchiroli – 442605 (M/S), India

Abstract This paper is concerned an analytical method for determining the thermal stresses in a finite solid cylinder under a sudden temperature change to a constant temperature with the help of integral transform technique.

Keywords Hankel transform, Laplace transform, magneto thermo stresses, solid cylinder, thermo elasticity.

INTRODUCTION

Kaliski and Nowacki [1] considered the half-space problem of magneto thermo elastic waves produced by a thermal shock in a perfectly conducting medium. The problem of magneto thermo elasticity related to an infinite cylindrical region was solved by Dhaliwal and Singh [2]. Noda et al. [3] give a combined formulation of the two theories of generalized thermo elasticity to discuss the problem of an infinite solid with a cylindrical of spherical hole.

In this paper, an attempt has been made to determine the magneto thermo stresses of finite solid cylinder with help of Hankel Transform and Laplace transform technique

NOMENCLATURE

Non dimensional quantities

L

L

R / a, (tc ) / a, * /(ET( ,t))

z z z

z z z

* /(ET( ,t)), h* h /(T ( ,t)H )

I -STATEMENT OF THE PROBLEM

z

z

Consider a long finite solid cylinder of radius with perfect conductivity placed initially in an axial magnetic field H (0, 0, H ) .

Let this cylinder be subjected to a rapid change in temperature T ( ,t) produced by the absorption of an electromagnetic pulse or ay pulse radiant energy.

Assuming that the magnetic permeability, of the solid

cylinder equals the magnetic permeability of the medium around it and omitting. Displacement maxwell equations for a perfectly conducting elastic body are given by

r r h

r r U r

(1)

T v, t – Temperature charge (absolute temperature minus reference temperature).

J Curl h ,

t Curl e ,

e H

t

t

U

U

~ – Displacement vector

u – Radial displacement

An initial magnetic field vector H (0,0, Hz ) in cylindrical

polar co-ordinate (r, , z) .are

, – Radial stress and circumferential stress.

U u(r,t), 0, 0 ,

e 0, H

u ,0

(2)

, t, a – Density, time and radius of solid cylinder.

– Coefficient of linear thermal expansion.

r

h (0, 0, h )

r h

(0, z 0,)

z t

(3)

, G – Lame constants

z J r

E, v

– Youngs modulus and Poissons ratio

Let magneto elastic dynamic equation of the solid cylinder

1

1

– Magnetic permeability

H – Magnetic intensity vector H

r o, o, 2

c – Perturbation of electric field vector.

becomes

r

r

r ( ) fr

2 v p t 2

, (4)

H – Perturbation of magnetic field vector (o, o, hz)

C1 ( lG) / p – Elastic wave speed

Where fr is defined as

Cz

Ht / p – Magnetic interference wave speed

r r 2 u u

z

z

t t f

( j H ) H 2

(5)

CL

c1 ct – Magneto thermo elastic wave speed.

r z r r r

The radial stress and the circumferential stress of a solid

2u (r, t) 1 2u (r, t) 1

d d

ud (r, t)

cylinder subjected to a thermal shock load are T (r,t), are

r 2

r r r 2

(15)

2C

2C

d

d

L

L

s

s

1 2u (r, t) 2u (r, t)

u

u

r (r,t) ( 2G) r

u

r

E

1 2v

T (r,t)

(6)

t 2 t 2

SOLUTION OF THE PROBLEM Substituting Eqs (5) to (6) into (4) the basic displacement

ud (0,t) 0

2G ud (a, t) 1 u (r, t) 0

(16)

(17)

equations of magneto thermo elastic motion is expressed as

r r d

1 u(r,t) 1 us (r, 0)

ud (r,0) us (r,0) 0

r

r r2 t

1 2u(r,t) E

2T (r,t)

(7)

ud (r, 0) us (r, 0)

u (0,t) u (0,t) V

(18)

C2

t2

(1 2 )B t

, 0 r a,t 0

t

t d s 0

L

z

z

Where B 2 H 2

Omitting the Maxwell tensor on the surface of the solid cylinder the corresponding boundary conditions are

Where (12).

us (r,t)

is the known static solution shown in Eq.

u(0,t) 0

(8)

The solution of the homogeneous formula of Eq. (14) assuming us (r,t) 0 is given by

r (a,t)

( 2G) u

r

u

r

E

(1 2v)

T (r,t) 0

(9)

Ud0 (r,t) f (r)exp(iwt)

(19)

r 0

Where

f (r)

and w are the characteristic function and

The initial conditions are

u(r,0)

(10)

natural frequency respectively.

Substituting Eq. (19) into the Eq. (15) and utilizing Eqs.

(16) and (17)

u(r,0) 0, t 0

We have

d 2 f (r) 1 df (r) 1

Assume that general solution to the basic equation (7) to

(10) may be expressed in the form

dr 2 r 2

k 2

dr

f (r) 0;

r 2

0 r a

(20)

u(r,t) us (r,t) ud (r,t)

(11)

f (0) 0

(21)

Where

u (r,t)

and

u (r,t)

are the static solution and

( 2G) df (a) f (a) 0

(22)

s d dr a

dynamic solutions to Eq. (7) to (10). Solving Eq. (11) we have,

The generalized solution of Eq. (24) is given by

E r B

(12)

f (r) A J (kr) A Y (kr)

(23)

us (r, t) r.T (r, t)dr B1 2

1 1 2 1

(1 2v)r 0

Where unknown constants determined by

r

B1 and

B2 in Eq. (12) may be

Substituting Eq. (23) into Eqs. (21) the corresponding characteristic function Eq. (23) reduces to.

Ba

Ba

E a

B1 2 rT (r, t)dr

0

(1 v)E B(1 v)(1 2v)

B(1 2v)

(13)

f (r) An Jn (knr)

This should satisfy the characteristic equation

k J1 (k a) J (k a) 0

(24)

(25)

B2 0

(14)

n 1 n a( 2 ) 1 n

From equations (7),(10) and (12) and using the boundary condition (8),(9) we get,

Where Jm (knr) is an mth order Bes(s1e4l) function of the first kind, kn (n 1, 2,3……) are the positive roots of eigen equation (25).

And

wn kncL

(26)

By means of the normalization property of eigen function, the

u (r,t) Ud (kn,t)JL (k , r)

(35)

1 n

1 n

constant An in Eq. (24) is determined as

a

kn F(kn )

rfn (r) An J1 (knr)dr

rJ1 (k r)dr

rJ1 (k r)dr

a

a

An 0

2

n

d

d

(27)

Substituting Eq. (16) and (39) into Eq. (12) the general solution of the basic equation (8) becomes

0 u(r,t) E

r,T (r,t)dr B r J (knr)

1

1

r ud (knt)

r ud (knt

Define a finite Hankel transform of

f (r) as

B(1 2v) 0 1

kn F (kn )

(36)

a

f (kn ) Hankel F (r) rfn (r) An J1 (knr)dr

0

Then the inverse of Eqn (28) is given by

f (r) f (kn )J (k r)

(28)

(29)

Equations (35) and (36), are the corresponding magneto thermodynamic stress.

By using Eq. (36) and the basic solution for magneto thermo elastic motion equation (11) reduces to

ust(k )

f (k ) 1 n

u(r,t) (1 v)T (1 v)T n J (k r) cos( w st)

0 0

1 n

0 0

1 n

kn n

Where

a 2

2

kn F (kn )

n

(37)

F (k ) rJ 2 (r)(k r)dr a J 1 (k a)2 1 1 J (k a)2

Where

n 1 n 2 1

n k a 1 n 2

0

n

u (k ) Hankel[r] a

J (k a) 2a J (k a)

k

k

(38)

k

k

(30) st n

0 n 2 1 n n n

Using Eq. (28) and applying a finite Hankel transform to

Eq. (15) we have

From Eqs. (6) and (7) and utilizing the following properties of the Bessel functions.

aJ1 (kna) ( 2 )u (a) u (a) k 2u (k , t)

k J 1 (k r) J1 (knr ) k J (k

) 1 J (k r)

(39)

( 2 )

d a d

n d n

(31)

n 1 n

r n 0 nr

r 1 n

1 ud (kn , t) us (kn , t)

1 k k

C 2 dt 2 dt 2 J (k r) n [J (k r) n J (k r)]

(40)

Where

L

r 1 n

2 0 n

2 2 n

us (kn ,t) Hankel us (r,t)

We get the magneto thermo stresses,

* (r,t) kn J (k r) kn J (k r) ust (kn ) cos(w t)

(41)

r 2(2v 1) 0 n 2 2 n F (k ) n

The first term on the left hand side of Eq. (31) should be the homogeneous boundary condition (17) simplifies to

kn n

* (r,t) kn

J (k r) kn J (k r) ust (kn ) cos(w t)

0

n

0

n

(42)

1 u (k , t)

u (k , t)

kn 2(2v 1)

2 2 n

F (k ) n

L

L

k 2u (k , t)

d n s n

(32) n

n d n

C 2

dt2

dt2

Where * (r,t) (r,t) /(T E) is normalized.

Applying Laplace transforms to Eq. (32), one obtains

i i 0

k 2C 2

2

2

u * (k , p) u *(k , p) n L U

p U

1 V 2

Eq. (41) and (42) are the magneto thermo stress is only dependent on the dynamic term in the basic solution (7) .

d n s n

k 2C 2 p s k 2C 2 p 0 k 2C 2 p 0

Using the proportion of Bessel function;

n L n L n L

(33)

Where p is the Laplace transform parameter. Taking the inverse Laplace transform of Eq. (33) we have

t

ud (kn , t) us (kn , t) knCL us sin knCL (t )dt

J0 (0) 1 and J2 (0) 0

The magneto thermo stress response at the center (R = 0) of solid cylinder reduces to;

V * (0,t) k

V * (0,t) k

0

n

ust (kn ) cos(w t)

(43)

u cos(k C t)

0 sin(k C t)

n L

n L

r 2(2v 1) F (k ) n

0 n L

knCL

(34)

kn n

* (0,t) kn ust (kn ) cos(w t)

(44)

kn n

kn n

Using Eqs (29) and (30) and applying finite inverse Hankel

transform to Eq. (34), the elasto dynamic solution ud (r,t)

2(2v 1) F (k ) n

of Eq. (15) to (18) may be expressed as

Equations (43) to (44) are the magneto thermo stress response at the center (R = 0) of a finite solid cylinder.

CONCLUSION

In this paper, we have investigated the magneto thermo stresses in a finite solid cylinder with the help of the finite Hankel transform and Laplace transform techniques. The expressions that are obtained can be applied to the design of useful structures or machines in engineering application .

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