Normal stress effects induced during circular shear of a compressible non-linear elastic cylinder

Normal stress effects induced during circular shear of a compressible non-linear elastic cylinder

Pergamon hr. J. Non-Linear Vol. 30, No. 3, pp. 323-339, 1995 Copyri&at 0 1995 Elsevier Science Ltd Printed in Great Britain. All rights reserved 002...

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Pergamon

hr. J. Non-Linear

Vol. 30, No. 3, pp. 323-339, 1995 Copyri&at 0 1995 Elsevier Science Ltd Printed in Great Britain. All rights reserved 002%7462/95 $9.50 + 0.00

Mechanics,

0020-7462(94)00043-3

NORMAL STRESS EFFECTS INDUCED DURING CIRCULAR SHEAR OF A COMPRESSIBLE NON-LINEAR ELASTIC CYLINDER A. S. Wineman* and W. K. Waldron Jr+ *Department of Mechanical Engineering and Applied Mechanics, The University of Michigan, Ann Arbor, MI 48109, U.S.A.

’ Polymers Division, National Institute of Standards and Technology, Gaithersburg, MD 20899, U.S.A. (Received 6 November 1993; in revised form 29 August 1994)

Abstract-The response of compressible non-linear isotropic elastic materials is studied using the problem of the circular shearing of a hollow cylinder. The hollow cylinder’s inner surface is tixed and its outer surface is allowed to rotate but not move radially. The present work is concerned with several issues not considered in previous studies, namely, how the load condition and material properties affect the distribution of local volume change and the local shear response. The study is carried out using a generalization of the Blatz-Ko constitutive equation. First, the homogeneous deformation of simple shear superposed on triaxial extension is considered because it can be compared to the local deformation of a material element in a hollow cylinder subjected to circular shear. It is shown that for plane strain without normal tractions, the shear modulus depends on the magnitude of shear K and a loss of monotonicity in the 5i2 vs K curve is possible for certain material parameters. In circular shear, the cylindrical surfaces rotate about the centerline which produce shear strains in the circumferential direction and associated normal stresses in the radial and circumferential directions. The compressibility of the material allows the cylindrical surfaces to undergo radial expansion or contraction. The radial distribution of shear strain, radial and circumferential stretch ratios, and local volume changes are determined for different values of the material parameters. It is shown that the local volume changes vary monotonically with radius. Regions of volume increase and decrease depend on the values of the material parameters. It is also shown that there is a value of the material parameter for which the local shear stress-shear strain response is found to be non-monotonic, which can lead to kinks in the moment-rotation relation.

1. INTRODUCTION

The problem of the circular shearing of a hollow cylinder has been used to study a variety of aspects of the response of non-linear isotropic elastic materials. In this problem, a hollow cylinder is fixed over its inner cylindrical surface, and a uniform distribution of shear tractions is applied to the other surface, acting in the circumferential direction. The cylindrical surfaces rotate about the centerline, producing shear strains in the circumferential direction and associated normal stresses in the radial and circumferential directions. When the material is incompressible, these normal stresses are determined to within a pressure term which arises from the incompressibility constraint. The normal stresses are found because the pressure field can be chosen to satisfy the radial equilibrium equation. The problem of circular shear can then be used to study the interaction of non-linear shear response and the spatial variation of shear deformation. Tao et al. [l] carried out such a study for generalized neo-Hookean materials. They showed that for certain values of the material parameters there can be a boundary layer like distribution of shear deformation near the inner cylindrical boundary. When the material is compressible, the normal stresses produce additional phenomena. The compressibility of the material may allow the cylindrical surfaces to undergo radial expansion or contraction. Thus, a radial distribution of shear strain, radial and circumferential stretch ratios, and local volume changes may develop. Mioduchowski and Haddow [2] considered the case in which the outer boundary does not displace radially, and obtained results for two strain energy functions, those proposed by Levinson and Burgess [3] and by Blatz and Ko [4]. They discussed an approximate Contributed by W. F. Ames.

324

A. S. Wineman and W. K. Waldron Jr

numerical solution in which the cylinder is divided into a number of coaxial thin-walled tubes of equal undeformed wall thickness. Distributions of the stresses and the ratio of the current thickness to the undeformed wall thickness, that is, radial stretch ratios were obtained. Ertepinar [S] discussed the case for which the outer surface was free of normal tractions and could displace radially. He obtained numerical results for stresses and stretches using the Levinson-Burgess strain energy function, but only for slightly compressible materials, and in a limited range of material parameters. Carroll and Horgan [6] discussed the analytical solution for a Blatz-Ko material while Carroll et al. [7] considered a Varga [S] material. Haughton [9] considered the problem for the boundary condition in which the outer surface was fixed against a radial displacement. A numerical solution was obtained using the Levinson-Burgess strain energy function, and for a full range of values of the model parameter. He also considered necessary and sufficient conditions on the strain energy density function so that there is pure circular shear, that is, no radial displacement. Polignone and Horgan [lo] carried out a similar study. (Similar analyses were carried out for torsion of solid cylinders [l l] and anti-plane shear of circular tubes [12].) Simmonds and Warne [ 131 considered the case in which cylindrical surfaces were fixed against radial displacement, with the goal of determining the relation between the moment applied to a surface and its rotation. Various material conditions were considered, including incompressibility, compressibility, isotropy and orthotropy. They study is analytical and results are obtained for a number of models. The present work is concerned with several issues not considered in previous studies, namely, how the load condition and material properties affect the distribution of local volume change and the local shear response. (Global volume changes have been measured in cylindrical samples subjected to torsional deformations [14]-[16].) The study of the distribution of local volume change is motivated by the possibility that its experimental measurement may be useful in determining the strain energy density function. The local shear stress-shear strain response is also studied because of the possibility of it being non-monotonic. Such a response could lead to kinks in the moment-rotation relation, as discussed by Simmonds and Warne [ 133. The study is carried out using the generalization the Blatz-Ko constitutive equation discussed in the review article by Beatty [17]. This is presented in Section 2. Each material element of the sheared cylinder undergoes a local homogeneous deformation composed of shear superposed on unequal biaxial stretch. The shear stress-shear strain response in such a deformation is explored in Section 3. There is particular emphasis on the influence of the biaxial stretch on the monotonicity of the response. The governing equations of the problem of the circular shear of a cylinder are presented in Section 4. Section 5 contains the reduction to a boundary value problem for the radial distribution of the local volume ratio and circumferential stretch. Details of the numerical method and the results are presented in Section 6. Section 7 contains some concluding remarks. 2. CONSTITUTIVE

EQUATION

The deformation gradient at a material element is denoted by F and J = det F denotes the ratio of its current volume to its volume in the reference configuration. B = FFr denotes the left Cauchy-Green strain tensor and B -’ denotes its inverse. The generalized Blatz-Ko constitutive equation, as discussed by Beatty [17], has the form u = PO(J)1 + YB

-

PC41-f)

J

B- 1 ,

(2.1)

where /I&(J) = &[ -fJ”_’

+ (1 -f)J_““‘],

(2.2)

and q = -2v,/( 1 - v,). Scalars p0 and v, denote, respectively, the shear modulus and Poisson’s ratio for infinitesimal deformations from the undeformed configuration. f is a constant such that 0
325

Normal stress effects 3. SIMPLE

SHEAR

SUPERPOSED

ON TRIAXIAL

EXTENSION

The homogeneous deformation of shear superposed on triaxial extension is described by the mapping x1 = &Xl + K&Xz x2

=

12x2

x3

=

13x3,

(3.1)

and is illustrated in Fig. 1. The deformation gradient is

n,

F=

1

K&

0

L2

0

0

J-3

0

Lo

and the left Cauchy-Green

1

(3.2)

2

strain tensor and its inverse are 1 B=FFT=

2

Kg A2+ 0K22;

A; 0 , 0 20 I Ki2

(3.3)

and

n;2 B-1

=

(F-‘)I$-’

=

-KiL2 0

-KA;'

AT2+ K2AT2 0

0

,

0

(3.4)

Ai2 1

respectively. The invariants are .J1 = trB = Izf + A;(1 + K2)+ Ai,

.I2 = trB-1=1;2(1+ Js=J= det F =

K2)+1y2 +AT~, (3.5)

A1A2A3.

Let equations (3.3) and (3.4) be substituted into the constitutive introduce the non-dimensionalization

equation (2.1) and (3.6)

and (3.7)

N(e2) 1

undeformed

Fig. 1. Simple shear superposed on triaxial extension.

326

The non-dimensional

A. S. Wineman and W. K. Waldron Jr

components of the stress tensor are

623 = 631 = 0.

(3.8)

These stresses satisfy the universal relation (cf. Wineman and Gandhi [18]) a11

c22 = 512

-

2: - (1 - K2)Ii KA;

(3.9)

>*

As seen in Fig. 1, K = tan y represents the amount of shear deformation. Thus a nondimensional shear modulus p= Ci2/K can be introduced. According to the expression for tii2 in equation (3.8), this shear modulus depends on &, L2, J and f: For simple shear, A1 = A2 = & = 1, J = 1 and hence ,u = 1. For equal triaxial extension in which 1i = A2 = A3 = 1 and J = 13, the shear modulus is given by p= f /A+ (1 -f )/A’.It is again a constant. However, p increases when the volume decreases (A < 1, J < 1) and p decreases when volume increases (1 > 1, J > 1). Similar results occur during plane strain equal biaxial extension, that is, R1 = A2 = 1, 1L3= 1. Thus, the shear response can soften or stiffen, depending on the underlying stretch. If this underlying stretch changes with the amount of shear deformation, the shear modulus also changes. In order to illustrate this, consider the response when the normal tractions N(e2) and N(n) shown in Fig. 1 are maintained at zero. The unit normal of the upper plane remains in the e2 direction and N(e2) = 022. The slanted sides of the parallelopiped are described by Xi = (x1 - Kx2)/ll = constant. The unit normal on this surface is grad X1 e, - Ke2 n = )gradX, 1= (1 + K2)l12 ’

(3.10)

Normal traction N(n) is calculated to be N(n) =

qTm =

cl1 + K2a22 - 2Ka12

1+K2



(3.11)

When N(e2) and N(n) vanish, a22 = 0

(3.12)

N(n) = al1 - 2Ka12 = 0.

(3.13)

and by equation (3.1 l),

Substituting the stresses from (3.8) into equations (3.12) and (3.13) results in two equations for 1,) A2 and 3L3in terms of K. A third equation can be obtained by specifying either N(eJ = a33 = 0 or A3 = 1. Setting a22 = 0 in relation (3.9) and combining with equation (3.13), we obtain the universal relation 1: = A; (1 + K2),

(3.14)

(see Wineman and Rajagopal [19]). Relation (3.14) and equation (3.12) are simpler equations for Ii, A2 and & in terms of K than are equations (3.13) and (3.12). Attention will be confined to the case when A3 = 1. c?r2 in equation (3.8) becomes al2

-

=

(f 1

A, +

(1 -f) ;lJ1

1

2

K.

(3.15)

Normal stress effects

327

Forf = 1, it is possible to solve equations (3.12) and (3.14) analytically for A1and AZfor any value of q in equation (2.2). From this solution, expressions for ai2 and J as functions of K are found to be 512

=

(3.16)

&

.I = (1 + zc2p4.

(3.17)

For f= 0, it is possible to solve equations (3.12) and (3.14) analytically for Ai and A2 if q = - 1 (v, = 0.25). This leads to the expressions K cl2

=

1 +

(3.18)

2K2’

(3.19) Note that the shear modulus p now depends on the shear K. Equations (3.12) and (3.14) were solved numerically for A1and & in terms of K for various values offwith q = - 1 (v, = 0.25). The results were used to calculate g12 and J in terms of K. Figure 2 shows the shear stress 5 12 vs K forf= 1,O and l/2. Forf= 1, e12 monotonically approaches the horizontal asymptote Oi 2 = 1. Forf = 0, Cl2 reaches a local maximum of $214 at K = $12 and then decreases asymptotically to zero with increasing K. For f = l/2, Cl2 reaches a local maximum of &3 at K = fi and then decays to a value of l/2 with increasing K. For anyf# 1, it can be shown that cYi2reaches a local maximum and dhen approaches a horizontal asymptote S12 =J: Figure 3 shows the stretch ratios and volume ratio vs K for f = 0. J and rZ1increase significantly with increasing K,while A2increases slightly at first, reaches a local maximum at K = 0.8 and then decreases slowly. For other values of J; J and A1 vary in a similar manner, while A2 decreases monotonically. In general, a material element undergoes neither simple shear nor shear in the absence of normal tractions. However, the results presented here suggest that non-monotonic shear response may be possible, depending on how A1 and A2 or J vary.

0.6

0.2

Fig. 2. Shear without normal tractions withf=

1, 0 and l/2: shear stress vs shear.

A. S. Wineman and W. K. Waldron Jr

328

A s--

3.0

2.5

-

v.=O.25

I

I

0.0

1.0

2.0

A J

0.5

3.0

K Fig. 3. Shear without normal tractions withf=

4. FORMULATION

0: stretch ratios and volume ratio vs shear.

OF CIRCULAR

SHEAR

Let (R, 0, Z) denote the reference coordinates of a particle and (I, 8, z) the coordinates of the particle in the deformed state. Circular shear in compressible elastic materials can be described by the mapping I = r(R) e=o+g(R) z = z.

(4.1)

r(R) and g(R) are assumed to have the smoothness required for all necessary mathematical

operations. It is assumed that the inner surface of the cylinder is fixed and the outer surface can rotate. Let Ri and R, denote the inner and outer radii, respectively, of the undeformed cylinder. Then r(R) and g(R) must satisfy the conditions g(Ri) = 0,

(4.2)

r(Ri) = Rip

(4.3)

r(R,) = R,.

(4.4)

The deformation gradient is calculated as

F=[{’

rR

;]

=

[ ijr

i.

i].

(4.5)

where I’ = dr/dR and g’ = dg/dR. The local shear measure, radial and circumferential stretch ratios are (4.6)

respectively.

AI = r’,

(4.7)

Apf,

(4.8)

Normal stress effects

329

The local deformation of a material element of the homogeneous deformation of simple shear superposed Section 3. The radial direction corresponds to the X, direction corresponds to the X, direction. I, correlates The local shear angle y is given by

cylinder can be compared to the on triaxial extension discussed in direction and the circumferential with & and A0correlates with AI.

K = tany. The left Cauchy-Green

tensor and its inverse are

and

A;‘+ K2rZz9 B-1

= (FT)-‘F-1

=

-

0

-K&'

KIi2

I,2

0

0

1

0 ,

1

respectively. The invariants are calculated to be

(4.11)

J1 = trB = A,”+ A: (1 + K2) + 1, J2 = trB_’ = Ai2(1 + K2) + I;’

+ 1,

(4.12)

J,=J=detF=A,&.

Using equations (4.10) and (4.1 l), the non-dimensional stress components from the constitutive equation (2.1) with respect to cylindrical coordinates are found to be

a,,= &J(J)+ ; (1;) - (1 -f) J

oee= Be(J)+ ; (A,”+

*zz=

&(J)

+ f _ J

cr,gz=

czr= 0.

-2

(A, + K21i2),

K2A.,Z)-

(l -f) ?--’

(4.13)

Note that by equation (4.12), A, = J/& and the non-zero stresses in (4.13) can be expressed in terms of the material element volume ratio J, shear K and stretch ratio A0as follows:

~~.=~~(J)+f(;)-(l-I)(~+~),

(4.14)

W%(f+(J-j+

(4.15)

~~~=A(J)+~~+~)-(l-f)(~).

(4.16)

czz = &(J) - (1 - 2f) f .

(4.17)

0

Define k? = R/RI, f = r/Rip and Z = Z/Ri as non-dimensional variables. Note that the non-dimensional stresses in equations (4.14)-(4.17) can be regarded as functions of r only. The equilibrium equations with respect to cylindrical coordinates reduce to (4.18)

330

A. S. Wineman and W. K. Waldron Jr

d& 25, x+7=0.

(4.19)

Since the functions which describe the deformation in mapping (4.1) depend on I?, it is convenient to express the equilibrium equations in terms of the radial position R in the reference configuration. Using equations (4.7), (4.8) and 1, = J/&, the equilibrium equations become (4.20) (4.21)

5. THE BOUNDARY

VALUE

PROBLEM

Equilibrium equation (4.19) can be solved for the circumferential shear stress distribution

A -rtl c7=7 where n;i is the non-dimensional by

(5.1)

moment per unit length applied to the outer shell defined

(5.2) and M is the dimensional moment per unit length applied to the outer shell. Using equation (4.8), the shear stress can be expressed, in terms of the radius R, as (5.3) An explicit expression for the local shear measure can found by substituting die from equation (4.15) into equation (5.3) and rearranging, that is

K=M_

(5.4)

R’Jf’

where (1 -j-)/J’.

s^=j+

(5.5)

Substitute from equ_ation: (5.4) and (5.5) into the expression for dII in equation (4.14) and denote the result by I?,~= & (ii, J, A,). An expression for dc?,,/dR in equation (4.20) can be obtained by first writing a;,, dJ d%r _aR+--+--. & dR aJ dR

a:,,, d& a& dR

(5.6)

This can be rearranged to give (5.7) By equations (4.14), (4.16), (5.4) and (5.5), (5.8) Substituting into the equilibrium equation (4.20) and rearranging, we obtain

d5r dR -

n;i2

_

~4fy

~2

R”n;^

_

2;)

.

(5.9)

Normal stress effects

331

From equations (4.14), (5.4) and (5.5), (5.10) (5.11) and (5.12) By equation (2.2): B. =

z=

-f(q

- 1) Jq-2 - (1 -f)(q

+ 1) J-(q+2),

(5.13)

Differentiating equation (4.8) and using I, = J/A, gives (5.14) Substitution of equations (5.9)-(5.14) into equation (5.7) gives an equation of the form

$ =F(J, &, R). Equation (5.15), together with equation (5.14), form a system of non-linear ordinary differential equations for J(R) and A,(R). Boundary conditions are obtained from equations (4.3), (4.4) and (4.8) which imply &(Ri) = [email protected](.&)= 1,

(5.16)

where Ri = 1 and R, = R,/Ri. Equations (5.14), (5.15) and (5.16) form a boundary value problem for J(R) and A,@) for each specified value of a. The shear deformation measure K(R) can be obtained directly from equation (5.4). By equations (4.2), (4.6) and 1, = J/[email protected], the rotation g(R) can be obtained from g(R) =

;sdR.

(5.17)

I Haughton [9] and Polignone and Horgan [lo] derived the necessary and sufficient conditions on the strain energy function for circular shear without radial displacement or local volume change. Their results indicate that this was possible for the constitutive equation in equations (2.1) and (2.2) whenf= 3/4. It is instructive to derive this result in the present context. Thus, let r = 8, Rr I R I &. By equations (4.7), (4.8) and (4.12), I, = & = J = 1. Equations (4.14) and (4.16) reduce to $I, = Ml)

+f-

(1 -f)

(1 + K2)

(5.18)

and a&r = /Y&(l)+f(l

+ K2) - (1 -f).

(5.19)

By equations (5.4) and (5.5), -

K=F.

(5.20)

Substituting equations (5.18) and (5.19) into equilibrium equation (4.18) or (4.20), and making use of equation (5.20), we obtain $

C--4(1 -f)

+ l] = 0.

This is satisfied if f = 3/4 for any value of the applied moment M.

(5.21)

A. S. Wineman

332

6. NUMERICAL

and W. K. Waldron

SOLUTION

AND

Jr RESULTS

The non-linear ordinary differential equations (5.14) and (5.15) were integrated numerically using the fourth order Runge-Kutta method. First, for a prescribed moment I\;i, the volume ratio at the inside radius J(Ri) was estimated. Using this value and the boundary condition &(I?,) = 1, equations (5.14) and (5.15) were integrated for i? E [Bi, I?,]. The value of the circumferential stretch ratio was checked against the boundary condition &(I?& = 1 at the outside radius. Next, iterations based on the secant method were used to adjust the estimate for J(Ri) until the boundary condition &(I?,) = 1 was satisfied within a tolerance of 10-s. In the numerical results presented here, Poisson’s ratio v, = 0.25 and all calculations were done with R, = R,/Ri = 2. Consider first, results for the casef= 1. Figure 4 shows plots of K vsR.For each value of I\;i, shear K is maximum at the inner support and decreases monotonically with increasing i?. For each fixed radius R, K increases with n;i. Plots of .I vs R with f= 1 are shown in Fig. 5. For each moment a, the material element volume is decreased (J < 1) at the inner support and increases monotonically with increasing 1. Near the outer shell material element volume is increased (J > 1) but to a lesser degree. At approximately R = 1.42, material element volume is unchanged. As the moment increases, the volume continues to decrease near the inner boundary and increase near the outer boundary. For n;i = 0.800, the volume ratio J is reduced at the inner support to approximately 0.96. Figure 6 shows plots of the circumferential stretch ratio & vs R. Since & c 1, the cylindrical surfaces move inward which is consistent with the volume change shown in Fig. 5. Note that these results represent a possible deformation for a material whose response can be described using the Blatz-Ko strain energy function. However, this deformation is not necessarily possible for every compressible isotropic elastic material without body forces (see Ericksen [20]). Plots of the stresses c?,~,& and see vs i? are omitted for the purpose of brevity. For each moment n;i, the greatest stress occurs at the inside radius Ri. c?,, is compressive near the inner support where volume decreases, tensile near the outer shell where volume increases. The circumferential stress tee is tensile for all i?. Details of the stress distributions can be found in Waldron [21]. The numerical solution of the boundary value problem was obtained without numerical difficulty for any magnitude of d. As the moment increases, ele remains monotonic in K for all R. Figure 7 is a plot of .I vs i? for moments A = 1.0,2.0,3.0,4.0,5.0 and 6.0. For n;i = 6.0, the volume ratio becomes extremely small at the inner support and J(Bi) z 0.3.

R=0.133 zizz --M=O.400 --__-

A \

a=0533 y=b;gj na=OBOO

f=l u,=O.25

K

Fig. 4. Circumferential

shear with/=

1: shear vs radius for moments 0.667 and 0.800.

of 0.133, 0.267, 0.400, 0.533,

Normal stress effects s=o.133 TzEE T-A&O400 -._‘.-

1.02

333

!=0.533 M=0.667 ...~~‘~____..’ A&O600 ---A--

f=l

1 1.0

1.2

1.4

1.6

1.6

2.0

K Fig. 5. Circumferential shear withf=

1: volume ratio vs radius for moments of 0.133,0.267,0.400, 0.533, 0.667 and 0.800.

M=o.133 p=o.2g7 T-M=0.400 -____

1 .ooo

a=o.533 &!?:W a=0 600 ---A--

A, 0.998

0.997

-

\ \ \ \ ‘\

I

0.996 1.0

1.2

f=l

/ ,I’

.,’

1

u,=o.25

,

,

1.4

I 1.6

,

, 1.8

, 2.0

R Fig. 6. Circumferential shear withf= 1: circumferential stretch ratio vs radius for moments of 0.133, 0.267, 0.400, 0.533, 0.667 and 0.800.

Figure 8 shows J(Ri), y(Ri), and u vs g(&,) for this higher level of applied moments. Note that the shear angle r(Ri) at the inside radius goes to an asymptote of n/2 since the material cannot fold over itself. Now, consider results for f= 0. Figure 9 shows plots of K vs R. K is maximum at the inner support and is larger than whenf= 1. It is also steeper at R = 1 when A? = 0.8. Note that relative to thef= 1 case, the distributions of J and & in Figs 10 and 11 are inverted about the iF axis. This means that each cylindrical surface moves radially outward (f 2 R) and volume increases near the inner support and decreases near the outer shell. NLM 30/3-J

A. S. Wineman and W. K. Waldron Jr

334

M=l.O

M=4.0 -_..Q=5 .. . . . . ..0. . . . . a=s 0 ---L--

f=Fo_-EL_

2.0

_

f=l l/,=0.25

1.5

/’

/’

1.0

J

/,;

y?y)’

J=;&:;;/ _,/ /’ .*/ ,/ /’ ,‘,/

0.5

0.0 1.0

1.2

1.4

1.6

1.8

2.0

K Fig. 7. Circumferential shear withf=

1: volume ratio vs radius for moments of 1.0, 2.0, 3.0, 4.0, 5.0 and 6.0.

6.0

5.0

4.0

[email protected],) r(R)

t/,=0.25

3 .o

M

,/’

2.0

/’ ,I’ //--

___-.

1.0 0.0 0.0

0.5

1.0

1.5

2.0

2.5

8 fJL) Fig. 8. Circumferential shear withf= 1 and a higher level of applied moments: volume ratio and shear angle at the inner support and the moment vs rotation of the outer shell.

Plots of the stresses c&, ti,, and eee vs E are omitted for the purpose of brevity. Results show that the distributions for 8, and &, are interchanged and reversed from the case when f = 1. That is, the radial stress is compressive for all i?, the circumferential stress Geeis tensile near the inner support where volume increases and compressive near the outer shell where volume decreases (for details, see Waldron [21]). This interchange and reversal of response can be associated with the manner in which the stress tensor e depends on the strain tensor B and its inverse B-‘. When f = 0, B depends only on B- ‘. By equations (4.14)-(4.17), the only normal stress which depends on the shear K is a,, and its contribution is negative. When f = 1, e depends on B. Only normal stress rFeedepends on the shear K, and its contribution is positive.

Normalstress effects

1.0

0.8

bo.133 -67 Y-M=0.400 -___-

335

g-o.533 _-..Y=O.667 .._.......... iFae0.800

\\\ I

f=O v.=o.25

0.8 K

1.0

1.2

1.4

1.6

1.8

2.0

R Fig. 9. Circumferential shear withf=

1.25

0: shear vs radius for moments of 0.133, 0.267, 0.400, 0.533, 0.667 and 0.800.

M=o.133 &+=0.267 8=0 400 -..‘.-

M10.533 _-M=O.66? ._._.._.._.___. a=0 800 ---A-,

I

f=O 1.20 I

1.15 J

I I \ t \

u,=O.25

1.10

1.05 1.00

0.95 1.0

1.2

1.4

1.6

1.8

2.0

K Fig. 10. Circumferential shear withf=

0: volume ratio vs radius for moments of 0.133,0.267,0.400, 0.533, 0.667 and 0.800.

The greatest stress and deformation again occur at the inside radius Ri.Figure 12 shows J(Rr), r(Ri) and R vs g(R,,). Note that J(Ri) + 1.25, r(rT,) + 0.7, and B becomes slightly non-linear as g(R,) + 0.3. Numerical difficulties arose as A? approached 0.835. This appeared to be related to an approaching loss of monotonicity in the 5 i2 vs K curve, as shown in Fig. 2. In the circular shear deformation, the most severe deformation occurs at the inner surface where r = 3 = Ri = 1 and by equation (5.1) Cro = fi. Therefore, the moment-shear curve can approach a loss of monotonicity. This is indicated in Fig. 13, which shows a plot of M vs K.

A. S. Wineman and W. K. Waldron Jr

336

1.015

B=O.lSS 8=0.267 --M=O 400 -..‘.-

I

/ _

/--\

/

:

1=0.533 ..M=O .. . . . . . .667 . . .. . 8-O 000 ---2-s

\

\

\

f=O

\\ \ \

:

u,=O.25 \

1 .OlO

A, 1.005

1 .ooo 1.0

1.2

1.6

1.4

1.8

2.0

K Fig. 11. Circumferential shear with f = 0: circumferential stretch ratio vs radius for moments of 0.133, 0.267, 0.400, 0.533, 0.667 and 0.800.

J

a ____-

Fig. 12. Circumferential shear with_/‘= 0: volume ratio and shear angle at the inner support and the moment vs rotation of the outer shell.

As n? approaches 0.835, the plot for i? = I?i = 1 appears to approach a horizontal asymptote which suggests that a practical limit on the applied moment exists. In order to investigate this numerical difficulty, plots were constructed of the circumferential stretch ratio at the outer surface vs the volume ratio at the inside surface for different values of d. Figure 14 shows plots of &(R,) vs J(&) for values of A above and below 0.835. Note that the boundary condition le = 1 at the outer radius cannot be satisfied for ti > 0.835. Simmonds and Wame [13] showed that the problem has a smooth solution, that is, no kinks, if in? is below some critical value. The results in their Fig. 6 indicate that this critical value is very close to 0.835. Thus, the numerical difficulties encountered here can

Normal stress effects

0.9

,

R=l.O KG----R=1.4 -____

Fig. 13. Circumferential shear withf=

;=1.6 ..R=l8 . . . . . . . . .. . . . R=2 0 -_-A--

0: moment vs shear at radii d = 1.0, 1.2, 1.4,1.6, 1.8 and 2.0.

li?=o..500

#=O.SOO -_go.835 ‘..z”..‘.-~“’ M=0.900 ------

XGZO --M=O 700 -..:-

1.10

337

r

1.05

A,(R.)

1 00

0.95

I

f=O

I

1

v,=O.25 0.90

I 1.0

I

I

1.1

I

I

1.2

I

I

1.3

I

I 1.4

J(K) Fig. 14. Circumferential shear withf= 0: circumferential stretch ratio at the outer surface vs volume ratio at the inner surface for moments of 0.506, 0.666, 0.766, 0.800, 0.835 and 0.900.

be associated with the loss of monotonicity in the relation between shear stress and shear strain at the inner support and the onset of a non-smooth solution. These results for the case whenf= 0 are undoubtedly due to the loss of ellipticity, which Knowles and Sternberg [22] have shown occurs at sufficiently large deformations when 0
338

A. S. Wineman and W. K. Waldron Jr 7. CONCLUSION

The response of compressible non-linear isotropic elastic materials was studied using the problem of the circular shearing of a hollow cylinder. The hollow cylinder’s inner surface was fixed and its outer surface was allowed to rotate but not move radially. In this work, the effects of load condition and material properties on the distribution of local volume change and shear response were studied using the generalization of the Blatz-Ko constitutive equation discussed in the review article by Beatty [17]. First, the homogeneous deformation of simple shear superposed on triaxial extension was considered because it can be compared to the local deformation of a material element in a hollow cylinder subjected to circular shear. It was shown that for plane strain without normal tractions, the shear modulus depends on the magnitude of shear K and a loss of monotonicity occurs in the 6i2 vs K curve forf# 1. In circular shear, the cylindrical surfaces rotate about the centerline producing shear strains in the circumferential direction and associated normal stresses in the radial and circumferential directions. The compressibility of the material allows the cylindrical surfaces to undergo radial expansion or contraction. For the case when the material parameter f= 1, the shear K is maximum at the inner support and decreases monotonically with increasing i?. For each moment A?, the material element volume is decreased (J < 1) at the inner support and increases monotonically with increasing R. Near the outer shell material element volume is increased (J > 1) but to a lesser degree. The cylindrical surfaces move inward for all R, that is, r I R, which is consistent with the volume change distribution. The greatest stress occurs at the inside radius Ri. a,, is compressive near the inner support where volume decreases and tensile near the outer shell where volume increases. The circumferential stress r& is tensile for all R. For f= 0, K is also maximum at the inner support but is larger than when f = 1. The distributions of J and & are inverted about the i? axis, that is, each cylindrical surface moves radially outward (f 2 R) and volume increases near the inner support and decreases near the outer shell. Further, the distributions for C,, and &, are interchanged and reversed from the case whenf= 1. That is, the radial stress is compressive for all R, the circumferential stress &, is tensile near the inner support where volume increases and compressive near the outer shell where volume decreases. Numerical difficulties arose as h? approached 0.835 which appeared to be related to the loss of monotonicity in the c?i2 vs K relation. The study of the distribution of local volume change and shear response was motivated by the possibility that their experimental measurement may be useful in determining the strain energy density function. The results demonstrated some interesting aspects of the response of compressible non-linear isotropic elastic materials. In particular, the local shear stress-shear strain response was found to be non-monotonic whenf # 1, which can lead to kinks in the moment-rotation relation, as discussed by Simmonds and Warne [13]. Acknowledgement-This work was supported in part by a grant from the National Science Foundation DMR-8708405). The authors are grateful for the useful comments of C. 0. Horgan and G. B. McKenna.

(No.

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