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Axisymmetric deformation of a short magneto-strictive current-carrying cylinder
PergamonPress. Prinled in Greal Britain Inf.I. Engng Sri..1919, Vol.13, pp.1099-1109.
AXISYMMETRIC DEFORMATION OF A SHORT MAGNETO-STRICTIVE CURRENT-CARRYING CYLINDER A. CHAKRABARTIt and A. AMARNATH Department of Applied Mathematics, Indian Institute of Science, Bangalore-560012,India Abstract-The title-problem has been reduced to that of solving a Fredholm integral equation of the second kind. One end of the cylinder is assumed to be fixed, while the cylinder is deformed by an axial current. The vertical displacement on the upper flat end of the cylinder has been determined from an iterative solution of the Fredholm equation valid for large values of the length. The radial displacement of the curved boundary has also been determined at the middle of the cylinder, by using the iterative solution. INTRODUCTION THE PROBLEMof
the deformation of magneto-strictive materials have become of interest after the corresponding constitutive equations have been derived by Lewis[l] (see [2]). Giri[2] has considered the torsion produced in a short inhomogeneous magneto-strictive cylinder under the influence of an axial current. However, there will be other types of deformation produced in the medium because of the presence of the current (see [l]). The second type of deformation mentioned above is the axisymmetric one, and its determination is a more difIlcult problem than the one considered by Giri. However, the problem of determining the axisymmetric deformation pattern in a homogeneous magneto-strictive current-carrying short cylinder is not very difficult. In this paper we have solved this problem when one end of the cylinder is kept fixed and all the other boundaries are left free. The method followed here is that of representing axisymmetric biharmonic stress-functions (see [3]) by means of suitable integrals involving Bessel functions. The problem under consideration has been solved in three stages. (i) The first stage is that of determining axisymmetric deformation produced in the cylinder when the curved boundary is assumed constrained and the flat-ends fixed together with the Maxwell stress in free space. This will produce certain normal stress on the curved boundary of the cylinder, but we want this boundary to be stress-free. So we consider the following problems: (ii) The second problem is to determine the axisymmetric deformation in the cylinder when we apply a stress equal in magnitude but opposite in direction to the one obtained from problem (i) on the curved boundary, whereas as in problem (i), the flat-ends are assumed fixed. We easily observe that both problem (i) and problem (ii) will give rise to certain normal stress on the upper flat-end of the cylinder along with the existing Maxwell stress component. But we want that end also to be stress-free. Thus we superpose with the solutions of (i) and (ii), the solution of the following problem. (iii) The final stage is to determine the axisymmetric deformation in the cylinder when the curved boundary is assumed stress-free, the lower end fixed and on the upper end is applied an equal and opposite normal stress to that obtained as a result of superposition of this stress in problems (i) and (ii) on that boundary. The solutions of the problems (i) and (ii) are very easy. In fact, the solution to problem (i) can be derived from a similar problem considered by Lewis[l] (see Section 2). The solution of problem (ii) then becomes still easier (see Section 3). The solution of problem (iii) is our main problem, to arrive at the final form of the deformation pattern and for this purpose we have employed axisymmetric biharmonic stress-function in our analysis. The analysis of the whole problem is similar to that of Watanabe and Atsumi [4] in solving a mixed boundary value problem of a row of penny-shaped cracks in a long circular cylinder. We find (see Section 4) that problem (iii) is reduced to a Fredholm integral equation of the second kind similar to that obtained in 141. We have solved this integral equation iteratively for large values of the ratio of the length of the cylinder to its radius. We have also made use of this iterative solution in determining the vertical displacement of the flat-end of the cylinder, and by a similar technique-the radial displacement on the curved boundary at the middle of the cylinder has also been determined. tPresent address: Department of Mathematics, University of Dundee, Dundee DDl 4HN, Scotland.
1100
A.CHAKRAB~~
andA. AM~NATH
It is observed that the free term of the Fredholm integral equation obtained here involves both the magneto-strictive coefficients (see Lewisfl]) in a complicated manner, whose values for particular materials are not available to the authors. It is for this reason that we have made no attempt in solving the Fredholm equation numerically. However, after determining the vertical displacement on the upper flat-end and the radial displacement on the curved boundary of the cylinder we can determine approximate values of the magneto-strictive coefficients for a particular material. 1. BASIC
EQUATIONS ANDRELATIONS
Basic e~~at~o~sgove~i~g the ~rna~lde~o~at~~n of isotropic magneto-~tricfive medium
The basic equations that govern the small deformation of an isotropic magneto-strictive medium are: (1.1)
(1.3) 1 T:j = /.J H,H, -- ~~~HkH~ 2 1
C
(i, j, k = 1,2,3)
(1.4)
where 1;1is the stress tensor, p the mass density, U the particle displacement, l!Jiand Hi are the electric and the magnetic intensities, W the electric displacement, B, the magnetic induction and Ji the current density, Sij is the Kronecker delta and eijkis the alternating tensor, r:j the Maxwell stress tensor. The equations that express stresses in terms of strains and Bi in terms of Hi and strains are: cj = hSk&j •t2G& t m&&& Bs = pHi -
2(mskk&j
-i-
+ 2nHiH,
IZn&)Hj,
0.5) (1.6)
where Sij is the strain-tensor, A and G are elastic constants, I* is the permeability of the medium, m and n are the magneto-stri~tive constants, 2.SOLUTIONOFPROBLEM(i)
In this section we consider the problem of the axisymmetric deformation of a short cylinder carrying a current J, = j. in the axial direction whose flat-ends are fixed and whose curved boundary is constrained. The components of the magnetic field are obtained from (1.2) as H,=O=H,
and
Ho=$.
(2.1)
Since we are concerned with the axisymmetric deformation produced in the cylinder whose flat-ends are fixed and the curved boundary is constrained in such a manner that the radial displacement vanishes there, we assume the displacement vector to be of the form u = (uo(r), 0,O).
(2.2)
Equations (l.l), (1.5), (2.1) and (2.2) give for the displacement vector u(r): uO(r)= -(m - n)r
H&) - H$a) 4(A +2G) I’
t2.3)
where “a” is the radius of the cylinder. From (1.5) and (2.3) we deduce that the radial tension
Axisymmetric deformation of a short magneto-strictive current-carrying cylinder
1101
required to keep the cylinder constrained is
T&,
=(m
+n$p+gIXa),
(2.4)
where the presence of Maxwell stress in free space has been taken and the permeability of free space has been taken to be unity. In this problem we also find that the normal stress T,, on the upper flat-end of the cylinder is given by T,, = -h(m -n)
2&(r) - H&Z) + mH2(r)+ H:(r) B 2 2(h + 2G) I
(2.5)
For the curved boundary to be free of stress we apply a surface traction which is equal and opposite to that given in (2.4). We see that apart from the stresses produced in a current carrying cylinder, account must be taken of the Maxwell stress tensor which exists in free space. 3. SOLUTION
OF PROBLEM (ii)
In this section we solve the problem of the axisymmetric deformation in the cylinder when the curved boundary is subjected to a normal stress, which is equal in magnitude to that given by (2.4) but opposite in direction. As before, the flat-ends are assumed to be fixed. It is easily verified that the solution of this problem is given by:
and
(3.1)
where c = -i(mt n t l)%(a). We deduce the following expression for the component 7ZZof the stress AC
7*z = h+G
(3.2)
Thus we see that the application of a normal stress TVon the curved surface of the cylinder gives rise to the normal stress given by (3.2) on the upper flat-end of the cylinder. 4. SOLUTION
OF PROBLEM (iii)
This problem consists of determining the axisymmetric deformation of a short circular cylinder when the lower boundary is fixed and all the other boundaries are free. In Section 2 we saw that an axial current J, = J0 gave rise to stresses T,, on the curved surface and T,, on the upper surface. Since we want the curved surface to be free, we dealt, in Section 3, with the problem of a cylinder whose flat-ends were fixed and the curved surface was subjected to a stress equal in magnitude but opposite in direction to the stress obtained in Section 2. Thus, we came to the conclusion that the application of a normal stress to the curved surface also gives rise to a normal stress on the upper flat-end of the cylinder. In order to see that all the conditions of our problem are satisfied we superpose the solutions of problems (i) and (ii) with the solution of the problem discussed in this section. Here we determine the axisymmetric deformation in the cylinder when the curved boundary is assumed to be stress free, the lower end is fixed and to the upper end there is applied a normal stress equal and opposite to that obtained after superposition of the stress obtained in problems (i) and (ii). We have chosen the origin of cylindrical polar coordinates (r, 0, z) at the centre of the flat-end of the cylinder, which is fixed and the z-axis along the axis of the cylinder. The length of the cylinder has been taken to be “I” and the radius “a”. The boundary conditions are: on
z=O,
U,=O=r=
(4.1)
1102
A. CHAKRABARTI and A. AMARNATH
and on 2 = 1,
+iHXa) =f(r) II
5’. REDUCTION
r=a
(4.2)
0
(4.3)
and 7,z = 0.
(4.4)
TV
on
(say)
TO A FREDHOLM
=
INTEGRAL
EQUATION
In order to solve the problem posed in Section (4) we represent the axisymmetric biharmonic stress function ,y(r, z) in cylindrical polar coordinates (see [5]), by means of an infinite series and an integral involving Bessel functions, as follows: ,y(r, z) = 2
[A&/Lr)
n=* +
+ Cnr/3nl@nr)l sin &z
m a [A (a) cash (YZ+ B(a) sinh az + zC(a) cash cxz+ zD((Y)sinh cYz].h(cwr)da, I0 (5.1)
where A(a), B(a), C(a), D(a) are functions of (Yand A. and C, are constants to be determined from the given boundary conditions, and
On=?, n=1,2 ,.... The corresponding components of stress and displacement can be obtained by using their expressions in terms of x (see Sneddon[S], p. 452) and (5.1). Using these expressions in [5] and the boundary condition (4.1), we get:
A(a)=D(a)=O.
(5.2)
Similarly, the first of the conditions (4.2) leads to: &3((u) = -C((Y)[~Y + (~1coth al], and the second condition of (4.2) takes the form: _ _ o a3C(a)[l + H(al)lJdar) da + “z, (-l)“P%A.ld/Lr) I
(5.3)
+ C,{2(2 - v)lo(pnr) + pnrl@nr)}l = f(r),
(0 =Sr < a)
(5.4)
where H(d)
=
al + sinh al cash al _ 1 3
sinh* cul
and v is Poisson’s ratio. Applying the boundary conditions on the curved surface of the cylinder we obtain the following results: T,=O
z, pf,[AJ@,a)
+ C,{&&(Pna)
r=a
on
implies that
+ 2(1- v)l~(Pna))lsin &Z
=
(~~C(a).l~(aa)[l coth al sinh (YZ-z cash WZ]da.
Axisymmetric defo~ation of a short magneto-strictive c~ent-c~yi~
1103
cylinder
This is a Fourier sine series whose Fourier coefficients are given by PZ,IAJI(&U) -t C,{P&(&a)
t 2(1- v)l@,u)}] = (-1)““;
lb” a5$j
;y;n ” Jl(oa) da. (5.5)
Similarly, TV= 0 on r = a implies that nz, B.M.P~a{l@~a) - /3&@,a)}-
&u*Cn{(l - 2v)j3J&?“u) + &&(&a)}]
cos &,.z
m =
f Cl
aaC(a)[a(al
coth cul+ (2~ - 1)}cash LYZ - CY’Z sinh cuz][au&(aa) - J,(aa)] drr.
This is a Fourier cosine series whose Fourier coefficients are given by An&u {&&(/La >- [email protected] 1)+ C&u ‘((1- 2v)PJo(j3,~ ) + p :al,(p,a 1) _
Y-1)“” g3n
“$ **x Iwa3nC(a) 0
[~2n(J,(au)- au.&,(au))- ~{a’+ ~~}~~(a~)] da.
(cu’+B:)
(5.6)
The coefficients A, and C, can be determined by solving (5.5) and (5.6). Let au = w; &a = &, ; UC’,(U)= C(u/u); L = I/u and p = r/a. Then, the expressions for A,, and C,, are found to be A,=
’
M.(&))L
*4.(-l)“. s’,
+j,(u){2(1-
V)*6:- 2v(l-
V)U28”- u’sEy,(s”)uJo(u) x {s:Io(&) + 2(1- v)&I4&))1 du,
x{u*,g~,,(fi,,)-(~
-
v)an(uz+8’.)wn)H du,
(5.7)
(5.8)
where
(5.9) Equation (5.4) in terms of u, I%, Cl, L and p takes the form:
I
a
u’C,(u) sinh uL{l + ~(u~)}~~(~~) du
0
+ 2
n-1
(-l)“~%[AJo(&p) + C,{2(2- v)~o(I%,P)+ &pI~(&p)}] = a3f(r) = F(p)
(say).
(5.10)
Let us assume that (5.11)
u”C,(u) sinh uL =
where +([) is an unknown function to be determined. Substituting (5.11) in (5.10) and interchanging the orders of integration we get
+ 2 (-l)“‘tlf$A&&np) + G{2(2 - V)lo(@,p)+ ~“p~l(~nP)~l = F(p), n-1
(0 5 P < 1).
(5.12)
A. CHAKRABARTI and A. AMARNATH
1104
Hence by the Abel inversion formula we get
(5.13) The following integrals were made use of in obtaining (5.13):
The expressions for A,, and C,, after using (5.11) take the form:
(5.14)
(5.15)
The foi~owing integrals were made use of in obtaining (5.14) and (5.15): 0)
Axisymmetric deformation of a short magneto-strictive current-carrying cylinder
1105
Now, substituting (5.14) and (5.15) in (5.13) and making use of the following substitutions:
we get the following Fredholm integral equation of the second kind for the unknown function d(5): (5.16) where (5.17) and the Kernel K(& 5) is of the form:
go AT2 I(5 + 5)“” -
it5 + 02’+*- (5 - 5)2’+21g:+
(5 - .$)‘“‘]G:.
(5.18) In the derivation of (5.18), we made use of the following expressions: (i) The integral representation of Riemann zeta function (5.19) (ii)
2vl*(nn/L)K,(nV/L) 2M(nn/L)
+ n’~‘/L’lo(nlr/L)Ko(n?rlL) NW/L)
- n2r2/L2 - 3 + 2v
I (5.20)
N(nn/L)
(iii) (5.21)
M(K)
= {K*
+ 2(1-
V)}II(K)KI(K)
+
KILO&,
N(K)={2(1-V)fK2}1:(K)-K*I$K).
6. ITERATIVE
SOLUTION
(5.22) 1
OF THE INTEGRAL
EQUATION
In this section we present an iterative solution of the Fredholm integral equation which we obtained in Section 5. The analysis employed here is similar to that discussed in [41.The solution of (5.16) for values of L * 1 can be written as:
I106
A. CHAKRABARTI and A. AMARNATH
where
- A Iy
+A+G
H&l) +;
H:(a))]. (6.3)
Now, using (2.1), (6.3) and p = r/a in (6.2) we obtain
G(5) = 40(t) = + [ {
An+;?mG+1(A:2G))53_(2An+G(~(:~~:(A+2G)]5]7i(~~~G)
= PI!!? - P&
(6.4)
where jZa’ ‘I= -7r(A +2G) .2
“=
-41r(A
+;(A
An +32mG+; (qq,
5
+2G)
[2An + G(m + 3n) + (A + 2G)].
In (5.18) if we retain the powers of l/L only up to order l/L”, then, after making use of (5.20) and (5.21) we find that it can be written in the form
+o
($2,>
(6.5)
where g,,=4gA+2G&
gl=8g;+2GI,
gzl=12g:+2G:,
g,, = 4Og; + 12G:, g3l= 12g: + 2G:, g4, = 2og: + 2G:,
Now
Using (6.4) and (6.5) we get
g.,z= 240g: + 56G:,
gx = 112g: + 30G;, g43= 504g: + 140~7: i
(6.6)
Axisymmetric deformation of a short magneto-strictive current-carrying cylinder +$3
[&, (p,
($+~)-P’(~+~)}t&+*
($$)-P*
($+f)] 5’
PI--_ ()
+g43
I
Simii~ly we obtain
7. DETERMINATION
OF THE DISPLACEMENTS
In this section we deduce the expression for the vertical displacement on the upper flat-end of the cylinder and the radial displacement on the curved boundary for large values of the length of the cylinder. For the sake of convenience we determine the latter quantity at the middle of the cylinder. From Sneddon[S] and by (5.1) we obtain: a!??(a) sinh ~4. JO(W) da u”CI(u) sinh uL. J&m) drc.
(7.1)
Using (5.11) in (7.1) we get after interchanging the orders of integration d& 06p
(7.2)
The expression for #(& can be obtained from (6.1) after utilizing the expressions for #&I, &Z(T),4&), 43(t) and 44(S). Thus we find that the final form of tiz]r=l can be written as:
(7.3)
1108
A. CHAKRABARTI and A. AMARNATH
Again the quantity u,(z = l/2, r = a) takes the form
(7.4) (Using [Sl, (5.1), (3.1) and (5.11)). Now we have
~.~(~-y)+o($), ($+$)}]+
-P2
(7.5)
_ J,(u) sin ut du=~~,sinh(~)K,(~).(-l)‘, 0 sinh uL/2
I
mJI(u) . uL. cash UL . sin u[ 4n = = x & (2k + 1) sinh ((2k ;l)‘@)K, sinh uL . sinh uL/2 I0 -:
$, sinh (2_kqK,
(F)
((2k;
W)
(_ljk
““)
(-l)k.
. (--I)~,
m muL . Jl(u) . sin ut du = F c (2k i- 1) sinh ((2k ;1)T~&((2k; cash uL/2 k 1 I0
(7.6)
The values of the integrals in (7.6) were obtained by using the calculus of residues after a proper choice of the function and the contour for each integral. Substituting (7.5) and (7.6) in (7.4), after performing the necessary integrations we arrive at the result
I(
&
)
2(C,+3C2+5C1)+
(
$
)
4Wz+60C3
t120C~(~~~sinh(~)~-;~,(l~k+‘K~((lk~1’~)
’
L (Cl i- cz + Cd (2k + 1)~
+ ((2k &r)
-(((2k+Ll)n)l
(6C,+20C~)+120C,(~2k~l~n)l)cosh((2k;1)li) (c,+3C,+5CS+((2k~l)n)9(6c,+60c,)
+ 120C3(t2k fljTr}
sinh ((2kl
““)] -j’allE
where c,=[P~t~~-~)+~-~+3~LL,(~-~)1,
c*=-
II
p,t--&y
I
,
c,=q. 5?rL
:iy
‘)
Axisymmetric deformation of a short magneto-strictive current-carrying cylinder
1109
CONCLUSION
This paper suggests that we can determine the magneto-strictive coefficients of a material approximately by means of a simple experiment. We have calculated the vertical displacement at the upper flat-end of the cylinder and the radial displacement on the curved surface of the cylinder at half of its length. In order to calculate the magneto-strictive coefficients we can take a particular type of material whose Lame’s constants are known, determine the quantities by passing a current in the axial direction of a moderately long cylinder ZG~~=~,~=I,Z and ~~~~~~ whose one end is fixed. Comparing these experimentally determined quantities with those obtained analytically above we obtain the magneto-strictive coefficients approximately by solving two simultaneous equations. Acknowledgement-One of us (A. Amarnath) is grateful to the Indian Institute of Science, Bangalore, India, for the invaluable opportunities and the C.S.I.R. for financial support. The authors are grateful to the referees for many valuable suggestions. REFERENCES [l] J. A. LEWIS, Quart. appl. Math. 20 (1962). [2] R. R. GIRI, Bull. Cal. Math. Sot. 60 (1968). [3] A. E. H. LOVE, A treatise on the mafhematical theory of elasticity, 4th Edn, pp. 274-276. Dover (1944). [4] K. WATANABE and A. ATSUMI, Int. J. Engng. Sci. 10 (1972). [5] I. N. SNEDDON, Fourier Transforms, p. 452. McGraw-Hill, New York (1951). (Received 5 October 1973)
AXISYMMETRIC DEFORMATION OF A SHORT MAGNETO-STRICTIVE CURRENT-CARRYING CYLINDER A. CHAKRABARTIt and A. AMARNATH Department of Applied Mathematics, Indian Institute of Science, Bangalore-560012,India Abstract-The title-problem has been reduced to that of solving a Fredholm integral equation of the second kind. One end of the cylinder is assumed to be fixed, while the cylinder is deformed by an axial current. The vertical displacement on the upper flat end of the cylinder has been determined from an iterative solution of the Fredholm equation valid for large values of the length. The radial displacement of the curved boundary has also been determined at the middle of the cylinder, by using the iterative solution. INTRODUCTION THE PROBLEMof
the deformation of magneto-strictive materials have become of interest after the corresponding constitutive equations have been derived by Lewis[l] (see [2]). Giri[2] has considered the torsion produced in a short inhomogeneous magneto-strictive cylinder under the influence of an axial current. However, there will be other types of deformation produced in the medium because of the presence of the current (see [l]). The second type of deformation mentioned above is the axisymmetric one, and its determination is a more difIlcult problem than the one considered by Giri. However, the problem of determining the axisymmetric deformation pattern in a homogeneous magneto-strictive current-carrying short cylinder is not very difficult. In this paper we have solved this problem when one end of the cylinder is kept fixed and all the other boundaries are left free. The method followed here is that of representing axisymmetric biharmonic stress-functions (see [3]) by means of suitable integrals involving Bessel functions. The problem under consideration has been solved in three stages. (i) The first stage is that of determining axisymmetric deformation produced in the cylinder when the curved boundary is assumed constrained and the flat-ends fixed together with the Maxwell stress in free space. This will produce certain normal stress on the curved boundary of the cylinder, but we want this boundary to be stress-free. So we consider the following problems: (ii) The second problem is to determine the axisymmetric deformation in the cylinder when we apply a stress equal in magnitude but opposite in direction to the one obtained from problem (i) on the curved boundary, whereas as in problem (i), the flat-ends are assumed fixed. We easily observe that both problem (i) and problem (ii) will give rise to certain normal stress on the upper flat-end of the cylinder along with the existing Maxwell stress component. But we want that end also to be stress-free. Thus we superpose with the solutions of (i) and (ii), the solution of the following problem. (iii) The final stage is to determine the axisymmetric deformation in the cylinder when the curved boundary is assumed stress-free, the lower end fixed and on the upper end is applied an equal and opposite normal stress to that obtained as a result of superposition of this stress in problems (i) and (ii) on that boundary. The solutions of the problems (i) and (ii) are very easy. In fact, the solution to problem (i) can be derived from a similar problem considered by Lewis[l] (see Section 2). The solution of problem (ii) then becomes still easier (see Section 3). The solution of problem (iii) is our main problem, to arrive at the final form of the deformation pattern and for this purpose we have employed axisymmetric biharmonic stress-function in our analysis. The analysis of the whole problem is similar to that of Watanabe and Atsumi [4] in solving a mixed boundary value problem of a row of penny-shaped cracks in a long circular cylinder. We find (see Section 4) that problem (iii) is reduced to a Fredholm integral equation of the second kind similar to that obtained in 141. We have solved this integral equation iteratively for large values of the ratio of the length of the cylinder to its radius. We have also made use of this iterative solution in determining the vertical displacement of the flat-end of the cylinder, and by a similar technique-the radial displacement on the curved boundary at the middle of the cylinder has also been determined. tPresent address: Department of Mathematics, University of Dundee, Dundee DDl 4HN, Scotland.
1100
A.CHAKRAB~~
andA. AM~NATH
It is observed that the free term of the Fredholm integral equation obtained here involves both the magneto-strictive coefficients (see Lewisfl]) in a complicated manner, whose values for particular materials are not available to the authors. It is for this reason that we have made no attempt in solving the Fredholm equation numerically. However, after determining the vertical displacement on the upper flat-end and the radial displacement on the curved boundary of the cylinder we can determine approximate values of the magneto-strictive coefficients for a particular material. 1. BASIC
EQUATIONS ANDRELATIONS
Basic e~~at~o~sgove~i~g the ~rna~lde~o~at~~n of isotropic magneto-~tricfive medium
The basic equations that govern the small deformation of an isotropic magneto-strictive medium are: (1.1)
(1.3) 1 T:j = /.J H,H, -- ~~~HkH~ 2 1
C
(i, j, k = 1,2,3)
(1.4)
where 1;1is the stress tensor, p the mass density, U the particle displacement, l!Jiand Hi are the electric and the magnetic intensities, W the electric displacement, B, the magnetic induction and Ji the current density, Sij is the Kronecker delta and eijkis the alternating tensor, r:j the Maxwell stress tensor. The equations that express stresses in terms of strains and Bi in terms of Hi and strains are: cj = hSk&j •t2G& t m&&& Bs = pHi -
2(mskk&j
-i-
+ 2nHiH,
IZn&)Hj,
0.5) (1.6)
where Sij is the strain-tensor, A and G are elastic constants, I* is the permeability of the medium, m and n are the magneto-stri~tive constants, 2.SOLUTIONOFPROBLEM(i)
In this section we consider the problem of the axisymmetric deformation of a short cylinder carrying a current J, = j. in the axial direction whose flat-ends are fixed and whose curved boundary is constrained. The components of the magnetic field are obtained from (1.2) as H,=O=H,
and
Ho=$.
(2.1)
Since we are concerned with the axisymmetric deformation produced in the cylinder whose flat-ends are fixed and the curved boundary is constrained in such a manner that the radial displacement vanishes there, we assume the displacement vector to be of the form u = (uo(r), 0,O).
(2.2)
Equations (l.l), (1.5), (2.1) and (2.2) give for the displacement vector u(r): uO(r)= -(m - n)r
H&) - H$a) 4(A +2G) I’
t2.3)
where “a” is the radius of the cylinder. From (1.5) and (2.3) we deduce that the radial tension
Axisymmetric deformation of a short magneto-strictive current-carrying cylinder
1101
required to keep the cylinder constrained is
T&,
=(m
+n$p+gIXa),
(2.4)
where the presence of Maxwell stress in free space has been taken and the permeability of free space has been taken to be unity. In this problem we also find that the normal stress T,, on the upper flat-end of the cylinder is given by T,, = -h(m -n)
2&(r) - H&Z) + mH2(r)+ H:(r) B 2 2(h + 2G) I
(2.5)
For the curved boundary to be free of stress we apply a surface traction which is equal and opposite to that given in (2.4). We see that apart from the stresses produced in a current carrying cylinder, account must be taken of the Maxwell stress tensor which exists in free space. 3. SOLUTION
OF PROBLEM (ii)
In this section we solve the problem of the axisymmetric deformation in the cylinder when the curved boundary is subjected to a normal stress, which is equal in magnitude to that given by (2.4) but opposite in direction. As before, the flat-ends are assumed to be fixed. It is easily verified that the solution of this problem is given by:
and
(3.1)
where c = -i(mt n t l)%(a). We deduce the following expression for the component 7ZZof the stress AC
7*z = h+G
(3.2)
Thus we see that the application of a normal stress TVon the curved surface of the cylinder gives rise to the normal stress given by (3.2) on the upper flat-end of the cylinder. 4. SOLUTION
OF PROBLEM (iii)
This problem consists of determining the axisymmetric deformation of a short circular cylinder when the lower boundary is fixed and all the other boundaries are free. In Section 2 we saw that an axial current J, = J0 gave rise to stresses T,, on the curved surface and T,, on the upper surface. Since we want the curved surface to be free, we dealt, in Section 3, with the problem of a cylinder whose flat-ends were fixed and the curved surface was subjected to a stress equal in magnitude but opposite in direction to the stress obtained in Section 2. Thus, we came to the conclusion that the application of a normal stress to the curved surface also gives rise to a normal stress on the upper flat-end of the cylinder. In order to see that all the conditions of our problem are satisfied we superpose the solutions of problems (i) and (ii) with the solution of the problem discussed in this section. Here we determine the axisymmetric deformation in the cylinder when the curved boundary is assumed to be stress free, the lower end is fixed and to the upper end there is applied a normal stress equal and opposite to that obtained after superposition of the stress obtained in problems (i) and (ii). We have chosen the origin of cylindrical polar coordinates (r, 0, z) at the centre of the flat-end of the cylinder, which is fixed and the z-axis along the axis of the cylinder. The length of the cylinder has been taken to be “I” and the radius “a”. The boundary conditions are: on
z=O,
U,=O=r=
(4.1)
1102
A. CHAKRABARTI and A. AMARNATH
and on 2 = 1,
+iHXa) =f(r) II
5’. REDUCTION
r=a
(4.2)
0
(4.3)
and 7,z = 0.
(4.4)
TV
on
(say)
TO A FREDHOLM
=
INTEGRAL
EQUATION
In order to solve the problem posed in Section (4) we represent the axisymmetric biharmonic stress function ,y(r, z) in cylindrical polar coordinates (see [5]), by means of an infinite series and an integral involving Bessel functions, as follows: ,y(r, z) = 2
[A&/Lr)
n=* +
+ Cnr/3nl@nr)l sin &z
m a [A (a) cash (YZ+ B(a) sinh az + zC(a) cash cxz+ zD((Y)sinh cYz].h(cwr)da, I0 (5.1)
where A(a), B(a), C(a), D(a) are functions of (Yand A. and C, are constants to be determined from the given boundary conditions, and
On=?, n=1,2 ,.... The corresponding components of stress and displacement can be obtained by using their expressions in terms of x (see Sneddon[S], p. 452) and (5.1). Using these expressions in [5] and the boundary condition (4.1), we get:
A(a)=D(a)=O.
(5.2)
Similarly, the first of the conditions (4.2) leads to: &3((u) = -C((Y)[~Y + (~1coth al], and the second condition of (4.2) takes the form: _ _ o a3C(a)[l + H(al)lJdar) da + “z, (-l)“P%A.ld/Lr) I
(5.3)
+ C,{2(2 - v)lo(pnr) + pnrl@nr)}l = f(r),
(0 =Sr < a)
(5.4)
where H(d)
=
al + sinh al cash al _ 1 3
sinh* cul
and v is Poisson’s ratio. Applying the boundary conditions on the curved surface of the cylinder we obtain the following results: T,=O
z, pf,[AJ@,a)
+ C,{&&(Pna)
r=a
on
implies that
+ 2(1- v)l~(Pna))lsin &Z
=
(~~C(a).l~(aa)[l coth al sinh (YZ-z cash WZ]da.
Axisymmetric defo~ation of a short magneto-strictive c~ent-c~yi~
1103
cylinder
This is a Fourier sine series whose Fourier coefficients are given by PZ,IAJI(&U) -t C,{P&(&a)
t 2(1- v)l@,u)}] = (-1)““;
lb” a5$j
;y;n ” Jl(oa) da. (5.5)
Similarly, TV= 0 on r = a implies that nz, B.M.P~a{l@~a) - /3&@,a)}-
&u*Cn{(l - 2v)j3J&?“u) + &&(&a)}]
cos &,.z
m =
f Cl
aaC(a)[a(al
coth cul+ (2~ - 1)}cash LYZ - CY’Z sinh cuz][au&(aa) - J,(aa)] drr.
This is a Fourier cosine series whose Fourier coefficients are given by An&u {&&(/La >- [email protected] 1)+ C&u ‘((1- 2v)PJo(j3,~ ) + p :al,(p,a 1) _
Y-1)“” g3n
“$ **x Iwa3nC(a) 0
[~2n(J,(au)- au.&,(au))- ~{a’+ ~~}~~(a~)] da.
(cu’+B:)
(5.6)
The coefficients A, and C, can be determined by solving (5.5) and (5.6). Let au = w; &a = &, ; UC’,(U)= C(u/u); L = I/u and p = r/a. Then, the expressions for A,, and C,, are found to be A,=
’
M.(&))L
*4.(-l)“. s’,
+j,(u){2(1-
V)*6:- 2v(l-
V)U28”- u’sEy,(s”)uJo(u) x {s:Io(&) + 2(1- v)&I4&))1 du,
x{u*,g~,,(fi,,)-(~
-
v)an(uz+8’.)wn)H du,
(5.7)
(5.8)
where
(5.9) Equation (5.4) in terms of u, I%, Cl, L and p takes the form:
I
a
u’C,(u) sinh uL{l + ~(u~)}~~(~~) du
0
+ 2
n-1
(-l)“~%[AJo(&p) + C,{2(2- v)~o(I%,P)+ &pI~(&p)}] = a3f(r) = F(p)
(say).
(5.10)
Let us assume that (5.11)
u”C,(u) sinh uL =
where +([) is an unknown function to be determined. Substituting (5.11) in (5.10) and interchanging the orders of integration we get
+ 2 (-l)“‘tlf$A&&np) + G{2(2 - V)lo(@,p)+ ~“p~l(~nP)~l = F(p), n-1
(0 5 P < 1).
(5.12)
A. CHAKRABARTI and A. AMARNATH
1104
Hence by the Abel inversion formula we get
(5.13) The following integrals were made use of in obtaining (5.13):
The expressions for A,, and C,, after using (5.11) take the form:
(5.14)
(5.15)
The foi~owing integrals were made use of in obtaining (5.14) and (5.15): 0)
Axisymmetric deformation of a short magneto-strictive current-carrying cylinder
1105
Now, substituting (5.14) and (5.15) in (5.13) and making use of the following substitutions:
we get the following Fredholm integral equation of the second kind for the unknown function d(5): (5.16) where (5.17) and the Kernel K(& 5) is of the form:
go AT2 I(5 + 5)“” -
it5 + 02’+*- (5 - 5)2’+21g:+
(5 - .$)‘“‘]G:.
(5.18) In the derivation of (5.18), we made use of the following expressions: (i) The integral representation of Riemann zeta function (5.19) (ii)
2vl*(nn/L)K,(nV/L) 2M(nn/L)
+ n’~‘/L’lo(nlr/L)Ko(n?rlL) NW/L)
- n2r2/L2 - 3 + 2v
I (5.20)
N(nn/L)
(iii) (5.21)
M(K)
= {K*
+ 2(1-
V)}II(K)KI(K)
+
KILO&,
N(K)={2(1-V)fK2}1:(K)-K*I$K).
6. ITERATIVE
SOLUTION
(5.22) 1
OF THE INTEGRAL
EQUATION
In this section we present an iterative solution of the Fredholm integral equation which we obtained in Section 5. The analysis employed here is similar to that discussed in [41.The solution of (5.16) for values of L * 1 can be written as:
I106
A. CHAKRABARTI and A. AMARNATH
where
- A Iy
+A+G
H&l) +;
H:(a))]. (6.3)
Now, using (2.1), (6.3) and p = r/a in (6.2) we obtain
G(5) = 40(t) = + [ {
An+;?mG+1(A:2G))53_(2An+G(~(:~~:(A+2G)]5]7i(~~~G)
= PI!!? - P&
(6.4)
where jZa’ ‘I= -7r(A +2G) .2
“=
-41r(A
+;(A
An +32mG+; (qq,
5
+2G)
[2An + G(m + 3n) + (A + 2G)].
In (5.18) if we retain the powers of l/L only up to order l/L”, then, after making use of (5.20) and (5.21) we find that it can be written in the form
+o
($2,>
(6.5)
where g,,=4gA+2G&
gl=8g;+2GI,
gzl=12g:+2G:,
g,, = 4Og; + 12G:, g3l= 12g: + 2G:, g4, = 2og: + 2G:,
Now
Using (6.4) and (6.5) we get
g.,z= 240g: + 56G:,
gx = 112g: + 30G;, g43= 504g: + 140~7: i
(6.6)
Axisymmetric deformation of a short magneto-strictive current-carrying cylinder +$3
[&, (p,
($+~)-P’(~+~)}t&+*
($$)-P*
($+f)] 5’
PI--_ ()
+g43
I
Simii~ly we obtain
7. DETERMINATION
OF THE DISPLACEMENTS
In this section we deduce the expression for the vertical displacement on the upper flat-end of the cylinder and the radial displacement on the curved boundary for large values of the length of the cylinder. For the sake of convenience we determine the latter quantity at the middle of the cylinder. From Sneddon[S] and by (5.1) we obtain: a!??(a) sinh ~4. JO(W) da u”CI(u) sinh uL. J&m) drc.
(7.1)
Using (5.11) in (7.1) we get after interchanging the orders of integration d& 06p
(7.2)
The expression for #(& can be obtained from (6.1) after utilizing the expressions for #&I, &Z(T),4&), 43(t) and 44(S). Thus we find that the final form of tiz]r=l can be written as:
(7.3)
1108
A. CHAKRABARTI and A. AMARNATH
Again the quantity u,(z = l/2, r = a) takes the form
(7.4) (Using [Sl, (5.1), (3.1) and (5.11)). Now we have
~.~(~-y)+o($), ($+$)}]+
-P2
(7.5)
_ J,(u) sin ut du=~~,sinh(~)K,(~).(-l)‘, 0 sinh uL/2
I
mJI(u) . uL. cash UL . sin u[ 4n = = x & (2k + 1) sinh ((2k ;l)‘@)K, sinh uL . sinh uL/2 I0 -:
$, sinh (2_kqK,
(F)
((2k;
W)
(_ljk
““)
(-l)k.
. (--I)~,
m muL . Jl(u) . sin ut du = F c (2k i- 1) sinh ((2k ;1)T~&((2k; cash uL/2 k 1 I0
(7.6)
The values of the integrals in (7.6) were obtained by using the calculus of residues after a proper choice of the function and the contour for each integral. Substituting (7.5) and (7.6) in (7.4), after performing the necessary integrations we arrive at the result
I(
&
)
2(C,+3C2+5C1)+
(
$
)
4Wz+60C3
t120C~(~~~sinh(~)~-;~,(l~k+‘K~((lk~1’~)
’
L (Cl i- cz + Cd (2k + 1)~
+ ((2k &r)
-(((2k+Ll)n)l
(6C,+20C~)+120C,(~2k~l~n)l)cosh((2k;1)li) (c,+3C,+5CS+((2k~l)n)9(6c,+60c,)
+ 120C3(t2k fljTr}
sinh ((2kl
““)] -j’allE
where c,=[P~t~~-~)+~-~+3~LL,(~-~)1,
c*=-
II
p,t--&y
I
,
c,=q. 5?rL
:iy
‘)
Axisymmetric deformation of a short magneto-strictive current-carrying cylinder
1109
CONCLUSION
This paper suggests that we can determine the magneto-strictive coefficients of a material approximately by means of a simple experiment. We have calculated the vertical displacement at the upper flat-end of the cylinder and the radial displacement on the curved surface of the cylinder at half of its length. In order to calculate the magneto-strictive coefficients we can take a particular type of material whose Lame’s constants are known, determine the quantities by passing a current in the axial direction of a moderately long cylinder ZG~~=~,~=I,Z and ~~~~~~ whose one end is fixed. Comparing these experimentally determined quantities with those obtained analytically above we obtain the magneto-strictive coefficients approximately by solving two simultaneous equations. Acknowledgement-One of us (A. Amarnath) is grateful to the Indian Institute of Science, Bangalore, India, for the invaluable opportunities and the C.S.I.R. for financial support. The authors are grateful to the referees for many valuable suggestions. REFERENCES [l] J. A. LEWIS, Quart. appl. Math. 20 (1962). [2] R. R. GIRI, Bull. Cal. Math. Sot. 60 (1968). [3] A. E. H. LOVE, A treatise on the mafhematical theory of elasticity, 4th Edn, pp. 274-276. Dover (1944). [4] K. WATANABE and A. ATSUMI, Int. J. Engng. Sci. 10 (1972). [5] I. N. SNEDDON, Fourier Transforms, p. 452. McGraw-Hill, New York (1951). (Received 5 October 1973)