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Some exact solutions for a class of compressible non-linearly elastic materials
International Journal of Non-Linear Mechanics 42 (2007) 321 – 329 www.elsevier.com/locate/nlm
Some exact solutions for a class of compressible non-linearly elastic materials F.J. Rooney a,∗ , M.M. Carroll b a Bishop O’Dowd, 9500 Stearns Ave, Oakland, CA 94618, USA b Mechanical Engineering and Materials Science MS 321, Rice University, P.O. Box 1892, Houston, TX 77251 1892, USA
Received 2 October 2006; received in revised form 18 January 2007; accepted 18 January 2007
Abstract In this paper we consider exact solutions for plane and axisymmetric deformations for a class of compressible elastic materials we call coharmonic. The coharmonic materials are derived from the harmonic materials by using Shield’s inverse deformation theorem. The governing equations for the coharmonic material show the same kind of simplification associated with the harmonic materials. The equations reduce to first-order linear equations depending on an arbitrary harmonic function. They are intractable in general, so various ansätze are investigated. Boundary value problems for the coharmonic materials are compared with the same problems for harmonic materials. For certain boundary value problems, the harmonic materials exhibit well-known problematic behaviour which limits their use as models of material behaviour. The corresponding solutions for the coharmonic materials do not display these non-physical features. 䉷 2007 Elsevier Ltd. All rights reserved. Keywords: Non-linear elasticity; Exact solutions; Coharmonic materials; Shield’s theorem
1. Introduction Shield’s inverse deformation theorem states for any deformation, without body forces, possible in a material with a given strain energy W then the inverse deformation is possible, without body forces, in the material with the strain energy W ∗ , which is derived from the original one by the formula W ∗ (F) = (det F)W (F−1 ). The coharmonic materials are derived from the harmonic materials by using Shield’s transformation [1]. The material we are considering is the Shield transform of the harmonic strain energy function W (i1 , i2 , i3 ) = 2(f (i1 ) − i2 − (1 − )i3 + 1 + 2). John [2] originally introduced the harmonic materials and showed that in plane strain the equilibrium equations reduce to a pair of Cauchy–Riemann equations. Thus any harmonic function furnishes a solution of the equations. The definition of the coharmonic materials as Shield transforms of the ∗ Corresponding author.
E-mail address: [email protected] (F.J. Rooney). 0020-7462/$ - see front matter 䉷 2007 Elsevier Ltd. All rights reserved. doi:10.1016/j.ijnonlinmec.2007.01.007
harmonic materials implies that the equilibrium equations for the coharmonic materials reduce to a pair of Cauchy–Riemann equations also. The only difference is that the roles of the reference and current coordinates are interchanged. The equilibrium equations reduce to a system of linear first-order equations depending on an arbitrary harmonic function or alternatively a linear integral equation depending on the arbitrary function. Ogden and Isherwood [3,4] used the integral equation to solve some problems for circular annuli, Jafari et al. [5] analysed the problem of pressurized tubes, Varley and Cumberbatch [6] investigated the stresses around elliptical holes. In three dimensions Abeyaratne and Horgan [7] solved the pressurized sphere problem and Carroll [8] added several other solutions. For axisymmetric deformations Hill and Arrigo [9] reduced the problem for harmonic materials to a system of linear first-order equations again depending on an arbitrary harmonic function. In a recent paper Murphy [10] utilized the Shield transform to examine cavitation for a number of compressible materials including the coharmonic material considered herein. In this paper we solve problems involving the coharmonic functions by making choices for the arbitrary function and investigating the consequences. The solutions for pressurized spheres and tubes are interesting in the fact that for the stresses they have the same form as for the problem for linear elasticity
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and do not exhibit the problematic behaviour, i.e. infinite hoop stress at finite pressure, that the harmonic materials have.
whence tan =
2. Preliminaries We will be considering the strain energy function given in the form i2 W (i1 , i2 , i3 ) = 2 i3 f − i2 + (1 + 2)i3 − 1 − , i3 (1) where is the infinitesimal shear modulus and is another constitutive constant. The condition that the stresses and strain energy vanish in the undeformed state leads to the conditions f (3) = 0,
f (3) = (1 + ).
(2)
The infinitesimal bulk modulus is given by = − 43 + 2f (3). The Cauchy stress is given by jW jW jW jW −1 −1 T = i3 i2 I + i3−1 + i3 V− V . ji2 ji3 ji1 ji2
where the capital letters are cartesian coordinates in the reference configuration and the lower case letters are the coordinates of the same point in the current configuration. The inverse deformation gradient and rotation tensor can be expressed in the forms X x Xy 0 −1 F = Yx Y y 0 , (7) 0 0 1 and
R=
cos sin 0 − sin cos 0 . 0 0 1
(8)
The inverse stretch tensor is then given by V
−1
= RF =
−1
Substituting this expression into the equation for the Cauchy stress (5) leads to i2 i2 T=g I − g V−1 , (13) i3 i3
(4)
(6)
Xy cos + Yy sin 0 Xx cos + Yx sin −Xx sin + Yx cos −Xy sin + Yy cos 0 , 0 0 1 (9)
(11)
For plane strain we can use the Cayley–Hamilton theorem to express V in the form i2 − 1 I − V−1 . V = i3 (12) i3
Substituting into the equilibrium equation gives i2 div g RT = 0, i3
Assume a displacement field given implicitly by the equations Z = z,
and i2 = tr(V−1 ) = (Xx + Yy )2 + (−Xy + Yx )2 + 1. i3
(3)
3. Plane strain
Y = Y (x, y),
(10)
where i2 i2 i2 g =f + 2 1 + 3 − . i3 i3 i3
Using the expression for the strain energy (1) this leads to 1 i2 i2 + 1 + 2 I − V − f V−1 . (5) T= f 2 i3 i3 i3
X = X(x, y),
−Xy + Yx , Xx + Y y
(14)
(15)
If the components of the equilibrium equation (15) are written explicitly i2 i2 g cos + −g sin = 0, (16) i3 i3 x y i2 i2 g sin + g cos = 0. (17) i3 i3 x y As pointed out by John [2] for the case of harmonic materials, these have the form of Cauchy–Riemann equations for the function i2 i F(z) = g e . (18) i3 Thus any analytic function of the variable z = x + iy leads to a solution of the equilibrium equations. i2 g = |F(z)|, = arg(F(z)). (19) i3 If we denote the function inverse to g by G then i2 = G(|F(z)|). i3
(20)
Recalling the expression for given in (10) cos =
X x + Yy i2 i3
−1
,
sin =
−Xy + Yx i2 i3
−1
,
(21)
leads to the equation jZ F(z) = (G(|F(z)|) − 1). jz |F(z)|
(22)
F.J. Rooney, M.M. Carroll / International Journal of Non-Linear Mechanics 42 (2007) 321 – 329
Let H (x, y) be the Airy stress function. The components of the Cauchy stress can be written in the form Txx = H,yy ,
Txy = −H,xy ,
Tyy = H,xx ,
(23)
and the equilibrium equations are satisfied identically. From these relations we deduce Txx + Tyy = ∇ 2 H .
3.1. F (z) = k Consider first the case where F(z) = k,
jZ = (G(k) − 1), jz
(25)
Z = (G(k) − 1)z + (¯z).
4
4
(28)
Using the constitutive equations (5) leads to = F(z)
jZ . j¯z
(37)
j2 H jZ =k . j¯z2 j¯z
(38)
Integrating (38) and using (36) gives jH = 41 k((G(k) − 1)z + (¯z)) + 1 (z). j¯z
(39)
Let H = 41 (2g(G(k)) − kG(k) + k)z¯z + h(z, z¯ ).
(40)
(29)
T · n = t (s) = t1 (s) + it2 (s) on C,
(30)
with s the arc length of the boundary curve and n is the unit normal to C. The boundary conditions on the stresses take the form dy dx dy dx Txx − Txy + i Txy − Tyy = t1 + it2 on C. ds ds ds ds (31) Using the definition of the stresses in terms of the Airy function eventually leads to jH 2 (32) = i (t1 (s) + it2 (s)) ds on C. j¯z In particular if the traction on the boundary is constant (Txx = S1 , Tyy = S2 ), then (32) becomes on C.
j2 h = 0. jzj¯z
(41)
The solution of this is
If tractions are specified on a boundary curve C in the form
jH 1 = ((S1 + S2 )z − (S1 − S2 )¯z)) j¯z 4
j2 H = 2g(G(k)) − kG(k) + k, jzj¯z
Substituting into the first equation leads to
j2 H 4 2 = Tyy − Txx − 2iTxy . j¯z
j¯z2
(36)
and (26)
Using the definition of the Airy stress functions, the other stress components can be expressed in the form
4
(35)
The equations for the Airy stress function (28) and (29) become
since i2 i2 i2 g = g d = |F(z)| d(G(|F(z)|)) i3 i3 i3 = |F(z)|G(|F(z)|) − G(|F(z)|) d|F(z)|. (27)
j2 H
(34)
or
or in terms of the complex representation jH 4 = |F(z)|G(|F(z)|) jzj¯z − 2 G(|F(z)|) d|F(z)| + |F(z)|,
k is a real constant.
In this case Eq. (22) becomes (24)
Hence, in view of the constitutive equation (5), i2 i2 i2 Txx + Tyy = 2g −g −1 i3 i3 i3
323
(33)
h = (z) + ¯ (¯z).
(42)
This is compatible with Eq. (39) if 4 (¯z) = (¯z) k
and
1 1 (z) = (k − kG(k) + g(G(k)))z. 2 (43)
3.1.1. Simply connected region with constant boundary conditions The function (z) is holomorphic inside the region and satisfies (¯z) + 41 (2g(G(k)) − kG(k) + k)z = 41 ((S1 + S2 )z − (S1 − S2 )¯z)),
(44)
on the boundary. The only solution to this is the trivial one (¯z) = − 41 (S1 − S2 )¯z, 2g(G(k)) − kG(k) + k = S1 + S2 .
(45)
The stresses are constant and the displacement is homogeneous Z = (G(k) − 1)z −
S 1 − S2 z¯ , k
where k is obtained from (45)2 .
(46)
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3.1.2. Circular annulus under constant pressure Consider a circular annulus with inner radius r1 and outer radius r2 , with pressure p1 at the inner boundary and p2 at the outer. The boundary conditions are: (¯z) = − 41 (2p1 + )z
on z¯z = r12 ,
(47)
(¯z) = − 41 (2p2 + )z
on z¯z = r22 ,
(48)
where = 2g(G(k)) − kG(k) + k.
(49)
The solution is (¯z) =
The deformation, as predicted by Shield’s theorem, is in a form inverse to (57) r=
C R0 R+ , 2 R
where the constants are given in terms of the applied pressures by (r22 − r12 )(2 − p1 )(2 − p2 ) H (R0 ) = , R0 4(r22 (2 − p1 ) − r12 (2 − p2 ))
1 , z¯
(50)
(r22 r12 )(p1 − p2 )R0 2(r22 (2 − p1 ) − r12 (2 − p2 ))
p1 r12 =2 r22
− p2 r22 . − r12
(51)
To express this in polar coordinates recall the formulae
.
(62)
On the other hand, the Cauchy stresses have a different form Trr = 2
where
(61)
and C=
r12 r22 (p2 − p1 ) 2(r22 − r12 )
(60)
Q1 r 2 − Q2 , Q3 r 2 + Q 2
T = 2
Q1 r 2 + Q2 , Q3 r 2 − Q 2
(63)
with Q1 = p2 r12 (2 − p1 ) − p1 r22 (2 − p2 ), Q2 = 2r22 r12 (p1 − p2 ),
(64) (65)
Trr + T = Txx + Tyy ,
(52)
Q3 = 2(r22 (2 − p1 ) − r12 (2 − p2 )).
T − Trr − 2iTr = e−2i (Tyy − Txx − 2iTxy ).
(53)
For the problem of internal pressure (p2 = 0) only, the expression for the hoop stress (64)2 becomes infinite at the inner surface at a finite value of the pressure given by r12 p∞ = 1 − 2 . (66) r2
Using the solution for , (50), and the equations for the Airy stress function, (37) and (38), gives the Cauchy stresses in the forms r12 r22 1 2 2 Trr = 2 p1 r1 − p2 r2 + (p2 − p1 ) 2 , (54) r r2 − r12 T =
r22
1 − r12
p1 r12 − p2 r22 − (p2 − p1 )
Tr = 0,
r12 r22 r2
,
(55) (56)
which are identical in form to the stresses in the same problem in the infinitesimal theory. The displacements, however, have the forms R = (G(k) − 1)r +
, r
= ,
=
1 r12 r22 (p2 − p1 ) . k r22 − r12
p1 r12 − p2 r22 r22 − r12
(67)
on the boundary. The function
m z=L w+ w
(68)
maps the exterior of the ellipse with semi axes L(1 + m) and L(1 − m) onto the exterior of the unit circle in the w plane (|w| 1). So the problem becomes that of finding a function (w) holomorphic in |w| 1 with d L
m (w) ¯ =− w+ on |w| = 1. (69) d¯z 4 w
(58)
Then
,
3.1.3. Harmonic materials The solution for harmonic materials is due to Jafari et al. [5]. For a material with strain energy function given by W = 2(H (i1 − 1) − i3 ).
d (¯z) = − z d¯z 4
(57)
where A and k are given by the equations 2g(G(k)) − kG(k) + k =
3.1.4. Elliptic hole under constant stress at infinity Consider a stress free elliptic hole with constant stresses at infinity. The problem is then to find a holomorphic function (z) in the region outside the ellipse that satisfies
(59)
d L (w) ¯ =− d¯z 4
1 + mw¯ . w¯
Using the mapping function gives
d¯z m =L 1− 2 . dw¯ w¯
(70)
(71)
F.J. Rooney, M.M. Carroll / International Journal of Non-Linear Mechanics 42 (2007) 321 – 329
Then
j2 H 4w¯ 2 j d j d 4 2 =4 (w) ¯ =− ( w) ¯ . j¯z j¯z d¯z L(w¯ 2 − m) jw¯ d¯z
(72)
In terms of the stresses Tyy − Txx − 2iTxy = −
(mw¯ 2 − 1) . w¯ 2 − m
(73)
As |w| → ∞
(85) (86)
3.3. F (z) = ez Eq. (22) in this case gives
Tyy − Txx − 2iTxy = −m.
(74)
The stresses at infinity are Tyy = (1 − m) ,
Txx = (1 + m),
Txy = 0.
m L Z = (G (k) − 1)L w + − w k
1 + mw¯ . w¯
jZ = 2eiy (G (ex ) − 1). jz
(87)
Let (75)
The displacement is given by
So the stresses have the forms 5 2 Trr = r − G(r) + 2 r 2 G (r) dr, 3 r 2 2 T = rG (r) − r − G(r) − 2 r 2 G (r) dr. 3 r
325
Z = R(x)eiy .
(88)
Substituting into (87) leads to (76)
Rx + R = 2(G (ex ) − 1),
(89)
with solution 3.2. F (z) = z
R(x) = 2e−x G(ex ) − 2.
Since |F (z)| = r it makes more sense to do the calculation in polar coordinates. Thus jZ R 1 1 (77) = Rr + + i R r − R ei( −) , jz 2 r r where the subscripts denote differentiation with respect to the indicated variable. Substituting Eq. (77) into Eq. (22) gives 1 R 1 Rr + + i R r − R ei( −) 2 r r = ei (G (r) − 1).
(78)
Txx + Tyy = ex G (ex ) − 2G(ex ) + ex ,
(91)
and Tyy − Txx − 2iTxy = F (¯z)
jZ , j¯z
(92)
which becomes Tyy − Txx − 2iTxy = ex G (ex ) − 2G(ex ) + ex . Tyy = ex G (ex ) − 2G(ex ) + ex ,
R Rr + = 2(G (r) − 1), r
(93)
Txx = Txy = 0.
(94)
(79) 4. Axisymmetric deformations
with solution 2 2 R(r) = 2 r 2 G (r) dr − r. r 3
(80)
To determine the stresses note 1 jZ R i( +) 1 R 3i = Rr − e Rr − e . = j¯z 2 r 2 r So the equation for the Airy function becomes j2 H jZ R 3i −i 1 4 2 = F (¯z) = re Rr − e . j¯z j¯z 2 r
(81)
(82)
(83)
r 2 G (r) dr.
Assume a displacement field given implicitly by the equations R = R(r, z),
Thus
4 7 T − Trr − 2iTr = rG (r) − r − 2 3 r
This represents the straightening of an annular sector. The stresses are determined from
The stresses are then
This implies that = 2 and
Trr + T = rG (r) − 2G(r) + r,
(90)
(84)
= ,
Z = Z(r, z),
(95)
where the capital letters are cylindrical coordinates in the reference configuration and the lower case letters are the coordinates of the same point in the current configuration. The inverse deformation gradient and rotation tensor can be expressed in the forms Rr 0 Rz −1 R F = 0 (96) 0 , r Zr 0 Zz and R=
cos 0 0 1 − sin 0
sin . 0 cos
(97)
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F.J. Rooney, M.M. Carroll / International Journal of Non-Linear Mechanics 42 (2007) 321 – 329
The inverse stretch tensor is then given by −1 V−1 = RF Rr cos + Zr sin = 0 −Rr sin + Zr cos
0
R r
0
From the formula for the angle of rotation
Rz cos + Zz sin , 0 −Rz sin + Zz cos (98)
cos =
sin =
whence tan =
−Rz + Zr , Rr + Z z
(99)
and i2 = tr(V−1 ) = i3
R (Rr + Zz )2 + (−Rz + Zr )2 + . r
(100)
and 1 i2 i2 j sin + f sin f jr i3 r i3 j i2 − cos = 0. f jz i3 Following Hill and Arrigo [9] we let i2 2 2 i2 cos = p − q , f sin = 2pq. f i3 i3
p = z
where ∇ 2 = 0.
the following equations become i2 i2 cos = 2z − 2r , f sin = 2 r z . f i3 i3 These can be written in the form 2 r z i2 = 2r + 2z , tan = 2 . f i3
z − 2r
.
(110)
(Rr + Zz )2 + (−Rz + Zr )2
Using Eqs. (100), (106), (108)–(110) yields the first-order system of equations
2z − 2r R
2z − 2r F ( 2z + 2r ), =
2z + 2r r
2z + 2r
(111)
and −Rz + Zr +
2 z r 2 z r R = 2 F ( 2z + 2r ). 2 2 r
z + r
z + 2r
(112)
4.1. = kz The system of Eqs. (111) and (112) reduces to R = F (k 2 ), r
R = 13 F (k 2 )r + r , (102)
(103)
(104)
Rz − Zr = 0,
(113)
Z = 13 F (k 2 )z + z ,
(114)
where is an arbitrary harmonic function. 4.1.1. Inflation of a sphere A simple example of the foregoing is spherical expansion or compaction. Consider a sphere with inner radius s1 and outer and p2 on the radius s2 , with pressure p1 at the inner surface √ outer. Assume depends only on s = r 2 + z2 . Since is harmonic it is (s) =
A , s
(115)
so in terms of S = (105)
(106)
1 A S = cs − 2 , 3 s
(107)
(108)
√ R 2 + Z 2 the deformation is
= ,
= ,
(116)
where c = F (k 2 ). For this deformation the stretch and inverse stretch tensor are given by V=
Let F be the function inverse to the function f . Then Eq. (107)1 can be solved for i2 /i3 to give i2 = F ( 2r + 2z ). i3
(109)
with solution
with the solution q = r ,
−Rz + Zr
Rr + Zz +
The equilibrium equations (101) and (102) reduce to 1 qr + q + pz = 0, r
,
(Rr + Zz )2 + (−Rz + Zr )2
Rr + Zz +
Using these equations and the constitutive equations (5) in the equilibrium equations leads to two equations, 1 i2 i2 j cos + f (cos − 1) f jr i3 r i3 j i2 − sin = 0, (101) f jz i3
pr − qz = 0,
R r + Zz
s3 1 3 3 cs
+ V−1 =
+ 2A
(es ⊗ es )
s3 1 3 3 cs
−A
(e ⊗ e + e ⊗ e ),
2A c + 3 (es ⊗ es ) 3 s c A (e ⊗ e + e ⊗ e ). + − 3 s3
(117)
(118)
F.J. Rooney, M.M. Carroll / International Journal of Non-Linear Mechanics 42 (2007) 321 – 329
Substituting into the constitutive equation leads to the expressions A 2 1 Tss = f (c) + 1 + 2 − c − 3 3 s 1 2A (119) − f (c) c+ 3 , 3 s A 2A 1 1 c− 3 c+ 3 T =T =f (c)+1+2− 3 s 3 s 1 A − f (c) c− 3 . 3 s (120)
The stresses have the forms
The boundary conditions
The system of Eqs. (111) and (112) becomes 2 R A Rr + Zz − = −F , r r2
Tss = −p1 Tss = −p2
on s = s1 on s = s2
and (121)
yield two equations to determine the constants c and A. For the particular materials with =0, the boundary conditions become −p1 1 2A (122) = f (c) + 1 − f (c) c+ 3 , 2 3 s1
−p2 1 2A = f (c) + 1 − f (c) c+ 3 . 2 3 s2
(123)
Solving these for the constants leads to the expressions for the stresses Tss =
p1 s13 − p2 s23 s23 − s13
T = T =
s 3 s 3 (p2 − p1 ) 1 + 1 23 , s3 s2 − s13
p1 s13 − p2 s23 s23 − s13
s 3 s 3 (p2 − p1 ) 1 − 1 2 3 , 2(s2 − s13 ) s 3
(124)
(125)
(126)
A=−
1 s13 s23 (p2 − p1 ) . 2f (c) s23 − s13
T = T = 2
,
(130)
AS 6 + BS 3 + 2 , (BS 3 + 1)(BS 2 − 2)
(131)
where A and B are constants depending on the pressures and internal and external radii. As in the case of the cylindrical problem, the hoop stress becomes infinite at a finite (< 2) value of pressure in the case of internal pressure only. 4.2. = A log r
(132)
and −Rz + Zr = 0.
(133)
Assume R = R(r)
and
Z = − z.
(134)
The first equation becomes 2 dR A R , − − = −F dr r r2 with solution R = r log r −
r
1 F t
(135)
A2 t2
dt .
(136)
4.3. = A/s In this case the system of Eqs. (111) and (112) becomes z 2 − r 2 A2 z2 − r 2 R R r + Zz + , (137) = F s2 r s2 s4 2zr 2zr R −Rz + Zr + 2 = 2 F s r s
(127)
(129)
A2 s4
.
(138)
Letting R = rG(s)
(128)
4.1.2. Harmonic materials The solution for pressurized harmonic spheres is also due to Abeyaratne and Horgan [7]. The deformation is of the form 1 c2 s = c1 S + 2 . 3 S
(BS 3 + 1)2
and
with the constants determined from the expressions p1 s13 − p2 s23 1 f (c) + 1 − cf (c) = , 3 s23 − s13
AS 6 − 2BS 3 − 1
This represents cylindrical inversion.
which have the exact same form as in the infinitesimal case. The radial deformation is given by 1 A S = cs − 2 , 3 s
Tss = 2
327
and
Z = −zG(s).
The equations reduce to 2 A G(s) 1 G (s) − . = − F s s s4 This has solution s 1 A2 G(s) = −s dt, F t2 t4
(139)
(140)
(141)
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F.J. Rooney, M.M. Carroll / International Journal of Non-Linear Mechanics 42 (2007) 321 – 329
or in terms of the spherical coordinates s 1 A2 2 dt. F S = −s t2 t4 This represents inversion of a hollow sphere. The radial Cauchy stress has the form 2 s 1 4A2 A dt − 2 Tss = √ F 5 2 t4 t A 2 s 2A2 S1 1 A dt , − 3 − F 2 2 s t4 s1 s1 t
(142)
where pˆ = p1 /2 and = s2 /s1 . (143)
and the boundary conditions supply, in principle, two equations to determine the constants S1 and A2 . As an example consider a material with f given by a power law
x N . (144) f (x) = 3 This implies that f must have the form 3 x N+1 f (x) = −1 . N +1 3 We have A2 i2 f = 4, i3 s so in this case 2 2 1/N A A F =3 4 . 4 s s
N −2 p−((N ˆ −2)/(N +1))(4+4/N −1) s13 1 T = + 2 N +1 2(1+4/N −1) s3 2 3 3(N +2N +4) (p−((N ˆ −2)/(N +1))( −1) + N (5N +2) 1−−(1+4/N) 1+1/N s4 × 14 , (151) s
(145)
4.3.1. Harmonic materials For harmonic materials the expression for the radial displacement is S 1 B2 s = −S 2 dt. (152) F t2 t4 In particular, S2 1 B2 2 s1 dt, − F s2 = S2 2 t4 S12 S1 t
(153)
and the radial Cauchy stress has the form B2 1 Tss = 2 2 − 1, 2 s S
(146)
(154)
and the hoop stress is B 2 dS 1 T = 3 − 1. 2 s S ds
(147)
The radial Cauchy stress has the form ⎞ ⎛ (N+1)/N 1 N − 2 ⎝ 6N A2 s13 A2 2A2 S1 ⎠ 1 + Tss = + − N + 4 s14 2 N +1 s3 s12 2 1+1/N N(5N + 2) A + , (148) (N + 1)(N + 4) s 4 and the hoop stress is ⎞ ⎛ (N+1)/N 1 N − 2 ⎝ 3N A2 s13 A2 A 2 S1 ⎠ 1 + 2 T = + − N + 4 s14 2 N +1 s3 s1 1+1/N 3(N 2 + 2N + 4) A2 + . (149) (N + 1)(N + 4) s 4 For purely internal pressure the boundary conditions lead to the expressions 1 N−2 p−((N−2)/(N ˆ +1))(4+4/N −1) s13 Tss = + 2 N +1 1+4/N −1 s3 1+4/N 3 (pˆ − ((N − 2)/(N + 1))( − 1)) + 1+4/N − 1 1+1/N s4 × 14 , (150) s
(155)
Under internal pressure only the boundary conditions yield −
B2 p1 = 2 2 −1 2 s1 S1
and
0=
B2 − 1. s22 S22
(156)
Let sˆ1 =
s1 , S1
sˆ2 =
s2 , S2
=
S2 , S1
pˆ =
p1 . 2
The boundary conditions reduce to 1 − pˆ 2 4 2 sˆ1 . B = sˆ2 S2 and sˆ2 = 2 Equating the two expressions for sˆ2 gives 1 − pˆ 1 ˆ sˆ1 2 (1 − p) dt. sˆ1 = sˆ1 − F 2 t4 2 1 t
(157)
(158)
(159)
Now assume F is given by the power law (147). Then the equation can be solved as 4/N−1 N +4 3 − 1−pˆ 2/N −1 . (160) = sˆ1 3N (1−p) ˆ 1/N (4/N+1 − 1) The radial stretch at the inner boundary is ds = 2sˆ1 −F (sˆ1 2 (1−p)) ˆ = 2sˆ1 −3sˆ1 2/N (1−p) ˆ 1/N , dS
(161)
F.J. Rooney, M.M. Carroll / International Journal of Non-Linear Mechanics 42 (2007) 321 – 329
which vanishes when sˆ1 2/N−1 = or when pˆ = 1 −
2 (1 − p) ˆ 1/N
1 N +4
,
(162)
2N4/N+1 − 2N + (N + 4)4/N+2 4/N−1
2 . (163)
At this pressure it can be seen from (155) that the hoop stress becomes infinite, whereas the corresponding hoop stress for the coharmonic material (151) does not exhibit this type of instability. References [1] M.M. Carroll, F.J. Rooney, Implications of Shield’s inverse deformation theorem for compressible elastic materials, ZAMP 56 (2005) 1048–1060.
329
[2] F. John, Plane strain problems for a perfectly elastic material of harmonic type, Commun. Pure Appl. Math. 13 (1960) 239–296. [3] D.A. Isherwood, R.W. Ogden, Finite plane strain problems for compressible elastic solids: general solution and volume changes, Rheol. Acta 16 (1977) 113–122. [4] R.W. Ogden, D.A. Isherwood, Solution of some finite plane strain problems for compressible elastic solids, Quart. J. Mech. Appl. Math. 21 (1978) 219–249. [5] A.H. Jafari, R. Abeyaratne, C.O. Horgan, The finite deformation of a pressurized circular tube for a class of compressible materials, ZAMP 35 (1984) 227–246. [6] E. Varley, E. Cumberbatch, Finite deformations of elastic materials surrounding cylindrical holes, J. Elasticity 10 (1980) 341–405. [7] R. Abeyaratne, C.O. Horgan, The pressurized hollow sphere problem in finite elastostatics for a class of compressible materials, Int. J. Solids Struct. 20 (1984) 715–723. [8] M.M. Carroll, Finite strain solutions in compressible finite elasticity, J. Elasticity 20 (1988) 65–92. [9] J.M. Hill, D.J. Arrigo, On axially symmetric deformations of perfectly elastic compressible materials, Quart. J. Mech. Appl. Math. 49 (1996) 19–28. [10] J.G. Murphy, Inverse radial deformations and cavitation in finite compressible elasticity, Math. Mech. Solids 8 (2003) 639–650.
Some exact solutions for a class of compressible non-linearly elastic materials F.J. Rooney a,∗ , M.M. Carroll b a Bishop O’Dowd, 9500 Stearns Ave, Oakland, CA 94618, USA b Mechanical Engineering and Materials Science MS 321, Rice University, P.O. Box 1892, Houston, TX 77251 1892, USA
Received 2 October 2006; received in revised form 18 January 2007; accepted 18 January 2007
Abstract In this paper we consider exact solutions for plane and axisymmetric deformations for a class of compressible elastic materials we call coharmonic. The coharmonic materials are derived from the harmonic materials by using Shield’s inverse deformation theorem. The governing equations for the coharmonic material show the same kind of simplification associated with the harmonic materials. The equations reduce to first-order linear equations depending on an arbitrary harmonic function. They are intractable in general, so various ansätze are investigated. Boundary value problems for the coharmonic materials are compared with the same problems for harmonic materials. For certain boundary value problems, the harmonic materials exhibit well-known problematic behaviour which limits their use as models of material behaviour. The corresponding solutions for the coharmonic materials do not display these non-physical features. 䉷 2007 Elsevier Ltd. All rights reserved. Keywords: Non-linear elasticity; Exact solutions; Coharmonic materials; Shield’s theorem
1. Introduction Shield’s inverse deformation theorem states for any deformation, without body forces, possible in a material with a given strain energy W then the inverse deformation is possible, without body forces, in the material with the strain energy W ∗ , which is derived from the original one by the formula W ∗ (F) = (det F)W (F−1 ). The coharmonic materials are derived from the harmonic materials by using Shield’s transformation [1]. The material we are considering is the Shield transform of the harmonic strain energy function W (i1 , i2 , i3 ) = 2(f (i1 ) − i2 − (1 − )i3 + 1 + 2). John [2] originally introduced the harmonic materials and showed that in plane strain the equilibrium equations reduce to a pair of Cauchy–Riemann equations. Thus any harmonic function furnishes a solution of the equations. The definition of the coharmonic materials as Shield transforms of the ∗ Corresponding author.
E-mail address: [email protected] (F.J. Rooney). 0020-7462/$ - see front matter 䉷 2007 Elsevier Ltd. All rights reserved. doi:10.1016/j.ijnonlinmec.2007.01.007
harmonic materials implies that the equilibrium equations for the coharmonic materials reduce to a pair of Cauchy–Riemann equations also. The only difference is that the roles of the reference and current coordinates are interchanged. The equilibrium equations reduce to a system of linear first-order equations depending on an arbitrary harmonic function or alternatively a linear integral equation depending on the arbitrary function. Ogden and Isherwood [3,4] used the integral equation to solve some problems for circular annuli, Jafari et al. [5] analysed the problem of pressurized tubes, Varley and Cumberbatch [6] investigated the stresses around elliptical holes. In three dimensions Abeyaratne and Horgan [7] solved the pressurized sphere problem and Carroll [8] added several other solutions. For axisymmetric deformations Hill and Arrigo [9] reduced the problem for harmonic materials to a system of linear first-order equations again depending on an arbitrary harmonic function. In a recent paper Murphy [10] utilized the Shield transform to examine cavitation for a number of compressible materials including the coharmonic material considered herein. In this paper we solve problems involving the coharmonic functions by making choices for the arbitrary function and investigating the consequences. The solutions for pressurized spheres and tubes are interesting in the fact that for the stresses they have the same form as for the problem for linear elasticity
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and do not exhibit the problematic behaviour, i.e. infinite hoop stress at finite pressure, that the harmonic materials have.
whence tan =
2. Preliminaries We will be considering the strain energy function given in the form i2 W (i1 , i2 , i3 ) = 2 i3 f − i2 + (1 + 2)i3 − 1 − , i3 (1) where is the infinitesimal shear modulus and is another constitutive constant. The condition that the stresses and strain energy vanish in the undeformed state leads to the conditions f (3) = 0,
f (3) = (1 + ).
(2)
The infinitesimal bulk modulus is given by = − 43 + 2f (3). The Cauchy stress is given by jW jW jW jW −1 −1 T = i3 i2 I + i3−1 + i3 V− V . ji2 ji3 ji1 ji2
where the capital letters are cartesian coordinates in the reference configuration and the lower case letters are the coordinates of the same point in the current configuration. The inverse deformation gradient and rotation tensor can be expressed in the forms X x Xy 0 −1 F = Yx Y y 0 , (7) 0 0 1 and
R=
cos sin 0 − sin cos 0 . 0 0 1
(8)
The inverse stretch tensor is then given by V
−1
= RF =
−1
Substituting this expression into the equation for the Cauchy stress (5) leads to i2 i2 T=g I − g V−1 , (13) i3 i3
(4)
(6)
Xy cos + Yy sin 0 Xx cos + Yx sin −Xx sin + Yx cos −Xy sin + Yy cos 0 , 0 0 1 (9)
(11)
For plane strain we can use the Cayley–Hamilton theorem to express V in the form i2 − 1 I − V−1 . V = i3 (12) i3
Substituting into the equilibrium equation gives i2 div g RT = 0, i3
Assume a displacement field given implicitly by the equations Z = z,
and i2 = tr(V−1 ) = (Xx + Yy )2 + (−Xy + Yx )2 + 1. i3
(3)
3. Plane strain
Y = Y (x, y),
(10)
where i2 i2 i2 g =f + 2 1 + 3 − . i3 i3 i3
Using the expression for the strain energy (1) this leads to 1 i2 i2 + 1 + 2 I − V − f V−1 . (5) T= f 2 i3 i3 i3
X = X(x, y),
−Xy + Yx , Xx + Y y
(14)
(15)
If the components of the equilibrium equation (15) are written explicitly i2 i2 g cos + −g sin = 0, (16) i3 i3 x y i2 i2 g sin + g cos = 0. (17) i3 i3 x y As pointed out by John [2] for the case of harmonic materials, these have the form of Cauchy–Riemann equations for the function i2 i F(z) = g e . (18) i3 Thus any analytic function of the variable z = x + iy leads to a solution of the equilibrium equations. i2 g = |F(z)|, = arg(F(z)). (19) i3 If we denote the function inverse to g by G then i2 = G(|F(z)|). i3
(20)
Recalling the expression for given in (10) cos =
X x + Yy i2 i3
−1
,
sin =
−Xy + Yx i2 i3
−1
,
(21)
leads to the equation jZ F(z) = (G(|F(z)|) − 1). jz |F(z)|
(22)
F.J. Rooney, M.M. Carroll / International Journal of Non-Linear Mechanics 42 (2007) 321 – 329
Let H (x, y) be the Airy stress function. The components of the Cauchy stress can be written in the form Txx = H,yy ,
Txy = −H,xy ,
Tyy = H,xx ,
(23)
and the equilibrium equations are satisfied identically. From these relations we deduce Txx + Tyy = ∇ 2 H .
3.1. F (z) = k Consider first the case where F(z) = k,
jZ = (G(k) − 1), jz
(25)
Z = (G(k) − 1)z + (¯z).
4
4
(28)
Using the constitutive equations (5) leads to = F(z)
jZ . j¯z
(37)
j2 H jZ =k . j¯z2 j¯z
(38)
Integrating (38) and using (36) gives jH = 41 k((G(k) − 1)z + (¯z)) + 1 (z). j¯z
(39)
Let H = 41 (2g(G(k)) − kG(k) + k)z¯z + h(z, z¯ ).
(40)
(29)
T · n = t (s) = t1 (s) + it2 (s) on C,
(30)
with s the arc length of the boundary curve and n is the unit normal to C. The boundary conditions on the stresses take the form dy dx dy dx Txx − Txy + i Txy − Tyy = t1 + it2 on C. ds ds ds ds (31) Using the definition of the stresses in terms of the Airy function eventually leads to jH 2 (32) = i (t1 (s) + it2 (s)) ds on C. j¯z In particular if the traction on the boundary is constant (Txx = S1 , Tyy = S2 ), then (32) becomes on C.
j2 h = 0. jzj¯z
(41)
The solution of this is
If tractions are specified on a boundary curve C in the form
jH 1 = ((S1 + S2 )z − (S1 − S2 )¯z)) j¯z 4
j2 H = 2g(G(k)) − kG(k) + k, jzj¯z
Substituting into the first equation leads to
j2 H 4 2 = Tyy − Txx − 2iTxy . j¯z
j¯z2
(36)
and (26)
Using the definition of the Airy stress functions, the other stress components can be expressed in the form
4
(35)
The equations for the Airy stress function (28) and (29) become
since i2 i2 i2 g = g d = |F(z)| d(G(|F(z)|)) i3 i3 i3 = |F(z)|G(|F(z)|) − G(|F(z)|) d|F(z)|. (27)
j2 H
(34)
or
or in terms of the complex representation jH 4 = |F(z)|G(|F(z)|) jzj¯z − 2 G(|F(z)|) d|F(z)| + |F(z)|,
k is a real constant.
In this case Eq. (22) becomes (24)
Hence, in view of the constitutive equation (5), i2 i2 i2 Txx + Tyy = 2g −g −1 i3 i3 i3
323
(33)
h = (z) + ¯ (¯z).
(42)
This is compatible with Eq. (39) if 4 (¯z) = (¯z) k
and
1 1 (z) = (k − kG(k) + g(G(k)))z. 2 (43)
3.1.1. Simply connected region with constant boundary conditions The function (z) is holomorphic inside the region and satisfies (¯z) + 41 (2g(G(k)) − kG(k) + k)z = 41 ((S1 + S2 )z − (S1 − S2 )¯z)),
(44)
on the boundary. The only solution to this is the trivial one (¯z) = − 41 (S1 − S2 )¯z, 2g(G(k)) − kG(k) + k = S1 + S2 .
(45)
The stresses are constant and the displacement is homogeneous Z = (G(k) − 1)z −
S 1 − S2 z¯ , k
where k is obtained from (45)2 .
(46)
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3.1.2. Circular annulus under constant pressure Consider a circular annulus with inner radius r1 and outer radius r2 , with pressure p1 at the inner boundary and p2 at the outer. The boundary conditions are: (¯z) = − 41 (2p1 + )z
on z¯z = r12 ,
(47)
(¯z) = − 41 (2p2 + )z
on z¯z = r22 ,
(48)
where = 2g(G(k)) − kG(k) + k.
(49)
The solution is (¯z) =
The deformation, as predicted by Shield’s theorem, is in a form inverse to (57) r=
C R0 R+ , 2 R
where the constants are given in terms of the applied pressures by (r22 − r12 )(2 − p1 )(2 − p2 ) H (R0 ) = , R0 4(r22 (2 − p1 ) − r12 (2 − p2 ))
1 , z¯
(50)
(r22 r12 )(p1 − p2 )R0 2(r22 (2 − p1 ) − r12 (2 − p2 ))
p1 r12 =2 r22
− p2 r22 . − r12
(51)
To express this in polar coordinates recall the formulae
.
(62)
On the other hand, the Cauchy stresses have a different form Trr = 2
where
(61)
and C=
r12 r22 (p2 − p1 ) 2(r22 − r12 )
(60)
Q1 r 2 − Q2 , Q3 r 2 + Q 2
T = 2
Q1 r 2 + Q2 , Q3 r 2 − Q 2
(63)
with Q1 = p2 r12 (2 − p1 ) − p1 r22 (2 − p2 ), Q2 = 2r22 r12 (p1 − p2 ),
(64) (65)
Trr + T = Txx + Tyy ,
(52)
Q3 = 2(r22 (2 − p1 ) − r12 (2 − p2 )).
T − Trr − 2iTr = e−2i (Tyy − Txx − 2iTxy ).
(53)
For the problem of internal pressure (p2 = 0) only, the expression for the hoop stress (64)2 becomes infinite at the inner surface at a finite value of the pressure given by r12 p∞ = 1 − 2 . (66) r2
Using the solution for , (50), and the equations for the Airy stress function, (37) and (38), gives the Cauchy stresses in the forms r12 r22 1 2 2 Trr = 2 p1 r1 − p2 r2 + (p2 − p1 ) 2 , (54) r r2 − r12 T =
r22
1 − r12
p1 r12 − p2 r22 − (p2 − p1 )
Tr = 0,
r12 r22 r2
,
(55) (56)
which are identical in form to the stresses in the same problem in the infinitesimal theory. The displacements, however, have the forms R = (G(k) − 1)r +
, r
= ,
=
1 r12 r22 (p2 − p1 ) . k r22 − r12
p1 r12 − p2 r22 r22 − r12
(67)
on the boundary. The function
m z=L w+ w
(68)
maps the exterior of the ellipse with semi axes L(1 + m) and L(1 − m) onto the exterior of the unit circle in the w plane (|w| 1). So the problem becomes that of finding a function (w) holomorphic in |w| 1 with d L
m (w) ¯ =− w+ on |w| = 1. (69) d¯z 4 w
(58)
Then
,
3.1.3. Harmonic materials The solution for harmonic materials is due to Jafari et al. [5]. For a material with strain energy function given by W = 2(H (i1 − 1) − i3 ).
d (¯z) = − z d¯z 4
(57)
where A and k are given by the equations 2g(G(k)) − kG(k) + k =
3.1.4. Elliptic hole under constant stress at infinity Consider a stress free elliptic hole with constant stresses at infinity. The problem is then to find a holomorphic function (z) in the region outside the ellipse that satisfies
(59)
d L (w) ¯ =− d¯z 4
1 + mw¯ . w¯
Using the mapping function gives
d¯z m =L 1− 2 . dw¯ w¯
(70)
(71)
F.J. Rooney, M.M. Carroll / International Journal of Non-Linear Mechanics 42 (2007) 321 – 329
Then
j2 H 4w¯ 2 j d j d 4 2 =4 (w) ¯ =− ( w) ¯ . j¯z j¯z d¯z L(w¯ 2 − m) jw¯ d¯z
(72)
In terms of the stresses Tyy − Txx − 2iTxy = −
(mw¯ 2 − 1) . w¯ 2 − m
(73)
As |w| → ∞
(85) (86)
3.3. F (z) = ez Eq. (22) in this case gives
Tyy − Txx − 2iTxy = −m.
(74)
The stresses at infinity are Tyy = (1 − m) ,
Txx = (1 + m),
Txy = 0.
m L Z = (G (k) − 1)L w + − w k
1 + mw¯ . w¯
jZ = 2eiy (G (ex ) − 1). jz
(87)
Let (75)
The displacement is given by
So the stresses have the forms 5 2 Trr = r − G(r) + 2 r 2 G (r) dr, 3 r 2 2 T = rG (r) − r − G(r) − 2 r 2 G (r) dr. 3 r
325
Z = R(x)eiy .
(88)
Substituting into (87) leads to (76)
Rx + R = 2(G (ex ) − 1),
(89)
with solution 3.2. F (z) = z
R(x) = 2e−x G(ex ) − 2.
Since |F (z)| = r it makes more sense to do the calculation in polar coordinates. Thus jZ R 1 1 (77) = Rr + + i R r − R ei( −) , jz 2 r r where the subscripts denote differentiation with respect to the indicated variable. Substituting Eq. (77) into Eq. (22) gives 1 R 1 Rr + + i R r − R ei( −) 2 r r = ei (G (r) − 1).
(78)
Txx + Tyy = ex G (ex ) − 2G(ex ) + ex ,
(91)
and Tyy − Txx − 2iTxy = F (¯z)
jZ , j¯z
(92)
which becomes Tyy − Txx − 2iTxy = ex G (ex ) − 2G(ex ) + ex . Tyy = ex G (ex ) − 2G(ex ) + ex ,
R Rr + = 2(G (r) − 1), r
(93)
Txx = Txy = 0.
(94)
(79) 4. Axisymmetric deformations
with solution 2 2 R(r) = 2 r 2 G (r) dr − r. r 3
(80)
To determine the stresses note 1 jZ R i( +) 1 R 3i = Rr − e Rr − e . = j¯z 2 r 2 r So the equation for the Airy function becomes j2 H jZ R 3i −i 1 4 2 = F (¯z) = re Rr − e . j¯z j¯z 2 r
(81)
(82)
(83)
r 2 G (r) dr.
Assume a displacement field given implicitly by the equations R = R(r, z),
Thus
4 7 T − Trr − 2iTr = rG (r) − r − 2 3 r
This represents the straightening of an annular sector. The stresses are determined from
The stresses are then
This implies that = 2 and
Trr + T = rG (r) − 2G(r) + r,
(90)
(84)
= ,
Z = Z(r, z),
(95)
where the capital letters are cylindrical coordinates in the reference configuration and the lower case letters are the coordinates of the same point in the current configuration. The inverse deformation gradient and rotation tensor can be expressed in the forms Rr 0 Rz −1 R F = 0 (96) 0 , r Zr 0 Zz and R=
cos 0 0 1 − sin 0
sin . 0 cos
(97)
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F.J. Rooney, M.M. Carroll / International Journal of Non-Linear Mechanics 42 (2007) 321 – 329
The inverse stretch tensor is then given by −1 V−1 = RF Rr cos + Zr sin = 0 −Rr sin + Zr cos
0
R r
0
From the formula for the angle of rotation
Rz cos + Zz sin , 0 −Rz sin + Zz cos (98)
cos =
sin =
whence tan =
−Rz + Zr , Rr + Z z
(99)
and i2 = tr(V−1 ) = i3
R (Rr + Zz )2 + (−Rz + Zr )2 + . r
(100)
and 1 i2 i2 j sin + f sin f jr i3 r i3 j i2 − cos = 0. f jz i3 Following Hill and Arrigo [9] we let i2 2 2 i2 cos = p − q , f sin = 2pq. f i3 i3
p = z
where ∇ 2 = 0.
the following equations become i2 i2 cos = 2z − 2r , f sin = 2 r z . f i3 i3 These can be written in the form 2 r z i2 = 2r + 2z , tan = 2 . f i3
z − 2r
.
(110)
(Rr + Zz )2 + (−Rz + Zr )2
Using Eqs. (100), (106), (108)–(110) yields the first-order system of equations
2z − 2r R
2z − 2r F ( 2z + 2r ), =
2z + 2r r
2z + 2r
(111)
and −Rz + Zr +
2 z r 2 z r R = 2 F ( 2z + 2r ). 2 2 r
z + r
z + 2r
(112)
4.1. = kz The system of Eqs. (111) and (112) reduces to R = F (k 2 ), r
R = 13 F (k 2 )r + r , (102)
(103)
(104)
Rz − Zr = 0,
(113)
Z = 13 F (k 2 )z + z ,
(114)
where is an arbitrary harmonic function. 4.1.1. Inflation of a sphere A simple example of the foregoing is spherical expansion or compaction. Consider a sphere with inner radius s1 and outer and p2 on the radius s2 , with pressure p1 at the inner surface √ outer. Assume depends only on s = r 2 + z2 . Since is harmonic it is (s) =
A , s
(115)
so in terms of S = (105)
(106)
1 A S = cs − 2 , 3 s
(107)
(108)
√ R 2 + Z 2 the deformation is
= ,
= ,
(116)
where c = F (k 2 ). For this deformation the stretch and inverse stretch tensor are given by V=
Let F be the function inverse to the function f . Then Eq. (107)1 can be solved for i2 /i3 to give i2 = F ( 2r + 2z ). i3
(109)
with solution
with the solution q = r ,
−Rz + Zr
Rr + Zz +
The equilibrium equations (101) and (102) reduce to 1 qr + q + pz = 0, r
,
(Rr + Zz )2 + (−Rz + Zr )2
Rr + Zz +
Using these equations and the constitutive equations (5) in the equilibrium equations leads to two equations, 1 i2 i2 j cos + f (cos − 1) f jr i3 r i3 j i2 − sin = 0, (101) f jz i3
pr − qz = 0,
R r + Zz
s3 1 3 3 cs
+ V−1 =
+ 2A
(es ⊗ es )
s3 1 3 3 cs
−A
(e ⊗ e + e ⊗ e ),
2A c + 3 (es ⊗ es ) 3 s c A (e ⊗ e + e ⊗ e ). + − 3 s3
(117)
(118)
F.J. Rooney, M.M. Carroll / International Journal of Non-Linear Mechanics 42 (2007) 321 – 329
Substituting into the constitutive equation leads to the expressions A 2 1 Tss = f (c) + 1 + 2 − c − 3 3 s 1 2A (119) − f (c) c+ 3 , 3 s A 2A 1 1 c− 3 c+ 3 T =T =f (c)+1+2− 3 s 3 s 1 A − f (c) c− 3 . 3 s (120)
The stresses have the forms
The boundary conditions
The system of Eqs. (111) and (112) becomes 2 R A Rr + Zz − = −F , r r2
Tss = −p1 Tss = −p2
on s = s1 on s = s2
and (121)
yield two equations to determine the constants c and A. For the particular materials with =0, the boundary conditions become −p1 1 2A (122) = f (c) + 1 − f (c) c+ 3 , 2 3 s1
−p2 1 2A = f (c) + 1 − f (c) c+ 3 . 2 3 s2
(123)
Solving these for the constants leads to the expressions for the stresses Tss =
p1 s13 − p2 s23 s23 − s13
T = T =
s 3 s 3 (p2 − p1 ) 1 + 1 23 , s3 s2 − s13
p1 s13 − p2 s23 s23 − s13
s 3 s 3 (p2 − p1 ) 1 − 1 2 3 , 2(s2 − s13 ) s 3
(124)
(125)
(126)
A=−
1 s13 s23 (p2 − p1 ) . 2f (c) s23 − s13
T = T = 2
,
(130)
AS 6 + BS 3 + 2 , (BS 3 + 1)(BS 2 − 2)
(131)
where A and B are constants depending on the pressures and internal and external radii. As in the case of the cylindrical problem, the hoop stress becomes infinite at a finite (< 2) value of pressure in the case of internal pressure only. 4.2. = A log r
(132)
and −Rz + Zr = 0.
(133)
Assume R = R(r)
and
Z = − z.
(134)
The first equation becomes 2 dR A R , − − = −F dr r r2 with solution R = r log r −
r
1 F t
(135)
A2 t2
dt .
(136)
4.3. = A/s In this case the system of Eqs. (111) and (112) becomes z 2 − r 2 A2 z2 − r 2 R R r + Zz + , (137) = F s2 r s2 s4 2zr 2zr R −Rz + Zr + 2 = 2 F s r s
(127)
(129)
A2 s4
.
(138)
Letting R = rG(s)
(128)
4.1.2. Harmonic materials The solution for pressurized harmonic spheres is also due to Abeyaratne and Horgan [7]. The deformation is of the form 1 c2 s = c1 S + 2 . 3 S
(BS 3 + 1)2
and
with the constants determined from the expressions p1 s13 − p2 s23 1 f (c) + 1 − cf (c) = , 3 s23 − s13
AS 6 − 2BS 3 − 1
This represents cylindrical inversion.
which have the exact same form as in the infinitesimal case. The radial deformation is given by 1 A S = cs − 2 , 3 s
Tss = 2
327
and
Z = −zG(s).
The equations reduce to 2 A G(s) 1 G (s) − . = − F s s s4 This has solution s 1 A2 G(s) = −s dt, F t2 t4
(139)
(140)
(141)
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or in terms of the spherical coordinates s 1 A2 2 dt. F S = −s t2 t4 This represents inversion of a hollow sphere. The radial Cauchy stress has the form 2 s 1 4A2 A dt − 2 Tss = √ F 5 2 t4 t A 2 s 2A2 S1 1 A dt , − 3 − F 2 2 s t4 s1 s1 t
(142)
where pˆ = p1 /2 and = s2 /s1 . (143)
and the boundary conditions supply, in principle, two equations to determine the constants S1 and A2 . As an example consider a material with f given by a power law
x N . (144) f (x) = 3 This implies that f must have the form 3 x N+1 f (x) = −1 . N +1 3 We have A2 i2 f = 4, i3 s so in this case 2 2 1/N A A F =3 4 . 4 s s
N −2 p−((N ˆ −2)/(N +1))(4+4/N −1) s13 1 T = + 2 N +1 2(1+4/N −1) s3 2 3 3(N +2N +4) (p−((N ˆ −2)/(N +1))( −1) + N (5N +2) 1−−(1+4/N) 1+1/N s4 × 14 , (151) s
(145)
4.3.1. Harmonic materials For harmonic materials the expression for the radial displacement is S 1 B2 s = −S 2 dt. (152) F t2 t4 In particular, S2 1 B2 2 s1 dt, − F s2 = S2 2 t4 S12 S1 t
(153)
and the radial Cauchy stress has the form B2 1 Tss = 2 2 − 1, 2 s S
(146)
(154)
and the hoop stress is B 2 dS 1 T = 3 − 1. 2 s S ds
(147)
The radial Cauchy stress has the form ⎞ ⎛ (N+1)/N 1 N − 2 ⎝ 6N A2 s13 A2 2A2 S1 ⎠ 1 + Tss = + − N + 4 s14 2 N +1 s3 s12 2 1+1/N N(5N + 2) A + , (148) (N + 1)(N + 4) s 4 and the hoop stress is ⎞ ⎛ (N+1)/N 1 N − 2 ⎝ 3N A2 s13 A2 A 2 S1 ⎠ 1 + 2 T = + − N + 4 s14 2 N +1 s3 s1 1+1/N 3(N 2 + 2N + 4) A2 + . (149) (N + 1)(N + 4) s 4 For purely internal pressure the boundary conditions lead to the expressions 1 N−2 p−((N−2)/(N ˆ +1))(4+4/N −1) s13 Tss = + 2 N +1 1+4/N −1 s3 1+4/N 3 (pˆ − ((N − 2)/(N + 1))( − 1)) + 1+4/N − 1 1+1/N s4 × 14 , (150) s
(155)
Under internal pressure only the boundary conditions yield −
B2 p1 = 2 2 −1 2 s1 S1
and
0=
B2 − 1. s22 S22
(156)
Let sˆ1 =
s1 , S1
sˆ2 =
s2 , S2
=
S2 , S1
pˆ =
p1 . 2
The boundary conditions reduce to 1 − pˆ 2 4 2 sˆ1 . B = sˆ2 S2 and sˆ2 = 2 Equating the two expressions for sˆ2 gives 1 − pˆ 1 ˆ sˆ1 2 (1 − p) dt. sˆ1 = sˆ1 − F 2 t4 2 1 t
(157)
(158)
(159)
Now assume F is given by the power law (147). Then the equation can be solved as 4/N−1 N +4 3 − 1−pˆ 2/N −1 . (160) = sˆ1 3N (1−p) ˆ 1/N (4/N+1 − 1) The radial stretch at the inner boundary is ds = 2sˆ1 −F (sˆ1 2 (1−p)) ˆ = 2sˆ1 −3sˆ1 2/N (1−p) ˆ 1/N , dS
(161)
F.J. Rooney, M.M. Carroll / International Journal of Non-Linear Mechanics 42 (2007) 321 – 329
which vanishes when sˆ1 2/N−1 = or when pˆ = 1 −
2 (1 − p) ˆ 1/N
1 N +4
,
(162)
2N4/N+1 − 2N + (N + 4)4/N+2 4/N−1
2 . (163)
At this pressure it can be seen from (155) that the hoop stress becomes infinite, whereas the corresponding hoop stress for the coharmonic material (151) does not exhibit this type of instability. References [1] M.M. Carroll, F.J. Rooney, Implications of Shield’s inverse deformation theorem for compressible elastic materials, ZAMP 56 (2005) 1048–1060.
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[2] F. John, Plane strain problems for a perfectly elastic material of harmonic type, Commun. Pure Appl. Math. 13 (1960) 239–296. [3] D.A. Isherwood, R.W. Ogden, Finite plane strain problems for compressible elastic solids: general solution and volume changes, Rheol. Acta 16 (1977) 113–122. [4] R.W. Ogden, D.A. Isherwood, Solution of some finite plane strain problems for compressible elastic solids, Quart. J. Mech. Appl. Math. 21 (1978) 219–249. [5] A.H. Jafari, R. Abeyaratne, C.O. Horgan, The finite deformation of a pressurized circular tube for a class of compressible materials, ZAMP 35 (1984) 227–246. [6] E. Varley, E. Cumberbatch, Finite deformations of elastic materials surrounding cylindrical holes, J. Elasticity 10 (1980) 341–405. [7] R. Abeyaratne, C.O. Horgan, The pressurized hollow sphere problem in finite elastostatics for a class of compressible materials, Int. J. Solids Struct. 20 (1984) 715–723. [8] M.M. Carroll, Finite strain solutions in compressible finite elasticity, J. Elasticity 20 (1988) 65–92. [9] J.M. Hill, D.J. Arrigo, On axially symmetric deformations of perfectly elastic compressible materials, Quart. J. Mech. Appl. Math. 49 (1996) 19–28. [10] J.G. Murphy, Inverse radial deformations and cavitation in finite compressible elasticity, Math. Mech. Solids 8 (2003) 639–650.