On Strain and Straining

MICHAEL HAYES

Communicated by K. R. RAJAGOPAL

For M~IRIN and SIOBH~N

1. Introduction

This paper is about strain and straining, both finite and infinitesimal. In developing the theory of strain and straining, CAUCHV introduced the material and spatial strain-ellipsoids and the quadric of stretching. In particular, for the defor- mation x = x(X) which takes the particle X to the place x, he showed that any three mutually orthogonal directions at X go into three conjugate directions of the material strain ellipsoid at x. Also the principal axes of the spatial strain ellip- soid at X are deformed into the principal axes of the material strain ellipsoid at x. Further, any pair of orthogonal directions at X such that one direction is along a principal axis of the spatial strain ellipsoid at X and the other is in a principal plane of this ellipsoid suffer no shear. There are results corresponding to this in the theory of straining-pairs of orthogonal directions instantaneously directed along the principal axes of the quadric of stretching at x suffer no shearing. Also suffering no shearing are those pairs of orthogonal directions at x which are such that one direction is in a principal plane ~r (say) of the stretching quadric and the second direction is along a principal axis of the quadric and orthogonal to the plane Jr.

Here (Section 3) it is shown that at every point x there is at least one circle of directions such that any pair of directions, whether orthogonal or not, instantan- eously directed along the radii of this circle suffers no shearing. For any motion there is either an infinity of such circles, two such circles, or just one circle. Further it is shown that the circle(s) found are indeed the only plane circle(s) of directions for which there is no shearing. We call the corresponding planes "special".

For strain (Section 4) it is shown that the planes of the central circular sections of the spatial strain ellipsoid at X are deformed into the planes of the central circular sections of the material strain ellipsoid at x. Further, pairs of material elements lying in the central circular section(s) of the spatial strain ellipsoid at X undergo no shear. The only special planes at X are those containing those central circular sections of the spatial strain ellipsoid.

Finally infinitesimal strain is considered (Section 5). There is at least one

266 M. HAYES

circle of directions at a point such that directions along the radii of this circle under- go no shear in the deformation.

If the special planes are determined experimentally they give some insight into the local deformation. If there are only two special planes and these are known, then one may conclude that the principal stretchings (or stretches) are all unequal and the intermediate principal axis of stretching (or stretch) is deter- mined since it is common to both planes. Also, the other principal axes are along the divisors (internal and external) of the angle between the unit normals to the special planes.

If there is only one special plane at the particle then two of the principal stretchings (or stretches) are equa l - the third principal stretching (or stretch) being attained in a direction normal to this special plane.

2. Equations and Definitions

The notation and treatment follows TRUESDELL & TOUPIN [l ]. Let the motion be given by

x = x ( X , t ) , x i = x i (XA, t ) , (2.1)

where x is the place occupied by the particle X at time t. Then

dx = F d X , dxi : FiA dxA, F,'A = c~xi/C~XA, (2.2)

where F is the deformation gradient.

Strain

The left and right Cauchy-Green strain tensors B and C are given b y

n = F F T, C = F T F , B i j = Xi,AXj,A, CAB = Xi,AXi, B . (2.3)

The principal stretches at X are denoted by 2a, 22, 23. They are given by the positive square roots of the eigenvalues of C.

The infinitesimal sphere at X:

dXadXA 1, (2.4)

is deformed into the material strain ellipsoid at x:

~ j ' dx~ dxj = ~ 2 . (2 .5)

The spatial strain ellipsoid at X:

CAB dXA d X s = f12, tfll "~ 1, (2.6)

is deformed into the infinitesimal sphere at x:

dxi dx~ = / 5 z . (2.7)

From (2.2)

On Strain and Straining

Strain ing

267

d x i = Ui, A d X A : vi, j dx), (2.8)

where a dot (-) is used to denote the material time derivative, and vi are the components of particle velocity. The stretching tensor is d given by

2dij : Vi, j -~ l)j,i. (2.9)

It may be shown [1, Section 82] that if n is a unit vector instantaneously along a material element dx : d x n , at x, then

dx : dijninj dx . (2.10)

The quadric of stretching [1, Section 82] at x is

dij x i x j = constant. (2.11)

This may be open or closed.

In f in i tes imal S tra in

The displacement u : x - X is assumed to be such that the product terms and higher in 8 u / S X may be neglected in comparison with first degree terms. Then C may be written

Can : O,,IB + 2EaB, (2.12)

where EA, is the infinitesimal strain tensor defined by

2EaB : (~ua/SXn) + (OUB/~XA). (2.13)

The change in length per unit initial length of a material element along the direc- tion N at X is EaBNaNB.

3. Stretching and Shearing

If the angle between two material elements dx ~ and dx (2) at x be denoted by fl, and if dx ~ = dx ~1) n , dx ~2) = dx ~2) m , where n, m are unit vectors, then

dx~ l) dx~ 2) = dx O) dx (2) cos ft. (3.1)

Now taking the material time derivative, and using the equation (2.8) it may be

shown [1, Section 82] that the shearing,/~, of the pair of material elements, in- stantaneously directed along n and m at x, is given by

- (sin t3) ~ : 2dijnimj - dij(niny ~- mimj) cos ft. (3.2)

268 M. HAYE

Clearly, if the pair o f elements is at right angles, cos fl = 0, and

- f l = 2dijnimj, n . m = 0. (3.3)

I t then follows tha t /~ = 0 if n is in a principal plane o f the quadr ic o f stretching (2.11) and m is a long the principal axis no rma l to this plane. This is the well known result.

Here we seek the special plane(s) at x such tha t there is no shearing for all mater ial elements instantaneously in this plane. Thus we seek the plane o f n and m such tha t

2dijnimj = dij(ninj -]- mimj) cos fl, nimi = cos fl, (3.4)

for all n and m , other than n parallel to m. First take fl = ~/2. Then

donim j ----- 0 u m : n. m = 0. (3.5)

Let a be a unit vector at right angles to n in the plane of n and m. Then

mi = cos flni + sin flai, (3.6)

and equat ion (3.4) gives, on using equat ion (3.5),

dij(ninj - aiay) cos fl sin 2 fl = 0. (3.7)

This has to hold for all fl @ 0, :L Hence, the unit vectors n and a must sat isfy

dqninj = d, Ta/a j, d/iain j = O, ajnj = 0. (3.8)

Since d is real and symmetr ic it possesses three eigenvalues d,. (say). We assume tha t these principal stretchings are ordered: dl > d2 > d3, and that the corre- sponding unit eigenvectors of(d/j) are el , e2, e3, respectively. Let n = hiei, a = hie/. Then equat ions (3.8) lead to

(dl - d2) ( h 2 - - t~ 2) = ( d 2 - d3) (h 2 - t)3z), (3.9)

( d ~ - d2) h l h l = (d2 - d 3 ) h 3 h 3 ,

and hence (d~ - d2) (h~ + ih~) 2 = (d2 - d3) (h3 -[- ih3) 2.

Thus

and so

where

( d l - d2 ) 1/2 h l = + ( d 2 - d3) ~/2 h3,

(d~ - d2) I/2 h 1 = -q--(d2 - d3) 1/2 ha,

m + . n = 0 , m + - a = 0 ,

m - . n : O , m - . a = O ,

m • - - { (d , - a~) "2 e~ • (d~ - a~) '/2 e~) (d , - a ~ ) - ~/2.

Hence there are only two special planes zr ~: (say) of possible n and a. sponding unit normals to them are m -+-.

(3 .10)

(3.11)

(3.12)

(3.13)

The corre-

On Strain and Straining 269

R e m a r k

The stretching of the material element directed instantanenously along n at x is dijn~nj.

It may be seen that the stretchings of all material elements in the planes ~• are all equal to d2. We consider z~ +. All unit vectors q (say) in the plane of n§ may be written

where

Now

But

q = h sin ~p + e 2 cos ~, 0 ~ ~p =< 2z~, (3.14)

h = {(d2 - da)'/2e~ - (d~ - d2) ' /2ea}{d l - d3} -1/2. (3.15)

doqiq j = d~jh~h i sin 2 *p + d~je2~e2j cos 2 ~' q- 2d~jh~e2j sin ~p cos ~p. (3.16)

d, jh ,hj = [dl(d2 - d3) + d3(dl - d2)]/(dl - d~) = de ,

dije2ie2j : d2 , (3.17)

diyhie2j : O. Thus

dijqiqj = d2. (3.18)

Suppose now that d~ : dE =~ da. Then (3.11) gives

h3 = a3 = 0 ,

and there is only one possible special plane z~ (say) whose normal is e3. The stretch- ing of any material element in z~ is again d2.

Finally, if dx ---- d2 : da, then (3.8) is satisfied identically and every plane is a special plane. The stretching of any element in any plane is equal to d2.

4. Finite Strain

Before considering the circular central sections of the material and spatial strain ellipsoids we first recall briefly how central circular sections of ellipsoids are obtained.

Central Circular Sect ions

I f a 2 ~ b 2 ~ c 2, t h e ellipsoid

a2x 2 + bEy 2 + c2z 2 : 1 (4.1)

has only two central circular sections each of radius 1/[ b I. The circles lie on the planes

a2x 2 + b2y2 q_ c2z 2 : b2(x 2 q_ y2 + z2), (4.2)

270 M. HAYES

o r

(a 2 - b2)l/2 x = 4-(b 2 - c2) 1/2 z , (4.3)

and pass through the intermediate axis. Generally, if (aij) is a real positive definite symmetric matrix with eigenvalues

p2 > q2 > r 2, then the two central circular sections of the ellipsoid

aijxixj : 1, (4.4)

are each of radius 1/Iq[ and lie on the planes

aijxixj : q2xixi. (4.5)

Of course if the ellipsoid is a spheroid it has only one central circular section whereas if it is a sphere, every central section is circular.

We assume that the principal stretches are ordered ;t~ > 22 > 2a. We note the identity

(CAB - - 22 OAB) d X A d X , : 22(22 -2 ~Sij - Bi] 1) dxi dxj. (4.6)

Thus particles on those planes which determine the central circular sections of the spatial strain ellipsoid at X, namely

CABdXA dXB : 22 dXA dXa, (4.7)

are deformed into those planes that determine the central circular sections of the material strain ellipsoid at x, namely

BiT 1 dxi dxj = 2 2 2 d x i d x i . (4.8)

In other words, central sections of the spatial strain ellipsoid at X go into central sections of the material strain ellipsoid at x.

It is easily seen that if 21 = 22 @ 2a, each ellipsoid is a spheroid and hence has just one central circular section. The central circular section at X is deformed into the central circular section at x.

Of course, if 2~ = 22 = 23, the ellipsoids are spheres. All central circular sections at X are deformed into central circular sections at x.

Shear

If N and M is a pair of unit vectors along material line elements at X, and 0 is the angle between them, then it may be shown that 7, the increase in the angle between N and M, as a result of the deformation, is given by [1, Section 26]

CAsNAMB cos (0 + ~,) = (CRsNRNs)I/E (CeQMeMo)I/2 , cos 0 = NAMA. (4.9)

We wish to find the special planes, containing N and M at X, such that ~, = 0. Thus

CAsNAMB = (CnsNnNs) 1/2 (CI, o M e M o ) I/2 cos 0. (4.10)

On Strain and Straining 271

This is to be valid for all 0 other than 0 = 0, ~. First take 0 = ~]2. Then we have the condition

C~BNAMB--_O, VN, M : N ' M =O. (4.11)

Next let Q be a unit vector at right angles to N in the plane of N and M. Then

M = cos ON + sin OQ, (4.12)

and on using (4.11)

CAnMAMB = CABNaNB cos 2 0 + CABQAQB sin2 0,

CaBNAMB = CABNANB cos 0,

so that (4.10) becomes

CABNaNB COS 0 ~- (CRsNRNs) 112 (C,4BNANB cos2 0 + CABQAQB sin2 0) 1/2 cos 0.

(4.13)

Thus

CABN,~NB : CABNANB COS 2 0 + CAsQaQn sin2 0. (4.14)

Hence, finally,

CAnNANB : CABQAQn, CABNAQn = O, NAQA = 0. (4.15)

This is precisely the problem solved in Section 3, the only difference being that (C) is positive definite whereas (d) in equation (3.8) is not necessarily so. That the planes of N and Q are those of the central circular sections of the spatial strain ellipsoid at X may be seen as follows.

Equation (4.15)1 expresses the fact that radius vectors to the spatial strain ellipsoid along N and Q are equal in length. Equation (4.15)2 expresses the fact that the normal to the ellipsoid "a t " Q is at right angles to N, so that Q and N are conjugate directions Equation (4.15)a says that Q and N are at right angles. The only way in which all these conditions may be satisfied is that N and Q lie in a central circular section of the spatial strain ellipsoid at X.

Alternatively' one may proceed as in Section 3 and show that the unit normal to the plane of N and Q, namely the vector product, NaQ, is given by

NAQ = {(2 ] - ).2)1/2 1 ~ (2~ - ).]),12 K } {).~ - 2 ] } - ' / 2 (4 .16)

assuming 21 > ).2 > ).a- Here I, J, K, are the unit eigenvectors of C correspond- ing to 22, 2 2 ).2, ).3 respectively.

The stretches of material elements in the planes of the central circular sections of the spatial strain ellipsoid at X are all equal to ).2- For the stretch in the direction M at X is (CABMAMB) 1/2. Now all such M along radii of the central circular sections may be written

M ~ cos ~p G + sin ~p J, 0 ~ ~p ~< 2~, (4.17)

where (~ : (ZF() .2 _ ).2)1/2 .I -.]- ().2 _ ).2)1/2 K } ().21 - ,/1, 2) 112 (4 .18)

272 H. HavEs

For these

CABMAMB = CABGAGB COS 2 ~) @ 2CABGAJB cos 7~ sin ~o + CABJAJB s in2 7 ~

--- 2 2 COS 2 ~ -~ 2 2 sin 2 ~, • 22. (4.19)

If 21 = 22 @ 23, there is only one central circular section of the spatial strain ellipsoid (now a spheroid) at X. The normal to the plane of the section is K. Again all stretches in this special plane are equal to 22.

Finally if 21 = 2 2 = 2 3 all central sections are circular and the stretches are equal for all directions at X.

5. Infinitesimal Strain

As noted in Section 2, the right Cauchy-Green strain tensor may be approximat- ed by

CAB = 3A~ + 2EAs. (5.1) Then

CaBNAMs = cos 0 + 2EABNAMs,

C~BN~N, = 1 + 2EABN,~NB, (5.2)

(CAsNANB)I/2=- 1 + EABN~N~,

assuming that 0 is the angle between the pair of material elements along N and M at X. Hence the increase in angle, ~,, given by (4.9), may be written

- (sin 0) 7 = 2EABNaMB -- E?Q(NeNQ + MpMQ) cos 0. (5.3)

This is identical in form to (3.2) with E, N, M, 0 corresponding respectively to d, n , m , ft.

Let the principal infinitesimal strains be denoted by E / an d let them be ordered El > / ? 2 > E3. Let the corresponding unit eigenvectors of (E) be i, j , k respec- tively. Then at X the material elements in the planes L ' i (say) whose normals are

{(El - E2) '/2 i ~ (E2 - E3) '/z k) {El - E3) -~/2, (5.4)

undergo no shear. The change in length per unit initial length of a material ele- ment along the unit vector M at X is EABMAM B. For all material elements at X in the planes X ~ the change in length per unit initial length may be shown to be equal to E2.

Of course if E1 = Ez =~ E3 there is only one plane of directions at X which undergoes no shear. The corresponding unit normal is k.

Finally if Et --~ E2 = E3 no pair of elements suffers shear at X.

6. Concluding Remarks

In the case of straining the special planes have unit normals

m ~ = {(dl - d 2 ) l / 2 e l -~- ( d 2 - d3) l /2e3} (dl -- d 3 ) - 1 / 2

On Strain and Straining 273

where we assumed dt > d2 > d3. We note that it is the difference of the prin- cipal stretchings that determines the special planes, so that if the dt are replaced by di + ~ the special planes remain unaltered.

If it can be determined experimentally that there are more than two special planes then one may conclude that d l : d2 -~- d3. Of course, if it can be determin- ed experimentally that there is only one special plane then one may conclude that two of the principal stretchings are equal. If there are only two special planes then all three principal stretchings are unequal. In this case knowledge of the normals to the special planes enables us to determine the ratios

(dl - d2) 1/2 : (d2 - d3) 1/2 : (d3 - d l ) 1/2,

and the direction of the intermediate axis of stretching. The two special planes intersect along this intermediate axis. Also, the other principal axes are along the diagonals of the rhombus formed by the two unit vectors which are along the nor- mals to the special planes.

Similar considerations apply in the cases of finite and infinitesimal strain.

Reference

1. TRUESDELL, C., • R. TOUVIN, Handbuch der Physik, III/1, Springer-Verlag, Berlin Heidelberg New York 1960.

University College Dublin

(Received May 18, 1987)