An Introduction to Continuum Mechanics (Mathematics in Science and Engineering)

4.

In Exercises 6-8, r(x)

= X -

o.

5.

37

THE DIVERGENCE THEOREM. STOKES' THEOREM

6. (a) Show that Vr = I. (b) Let e = r/lrl. Compute (Ve)e. 7. Let a E "f-, SELin, and define qJ: Iff ~

~

by

qJ = a . (r x Sr).

Compute VqJ. 8. Let u be the vector field on Iff - {o} defined by u

= r/lrl 3 •

Show that u is harmonic. Find a scalar field whose gradient is u. 9. Let u be a class C 2 vector field. Show that (a) div{(Vu)u} = Vu' VuT + u· (V div u), (b) Vu' VuT = div{(Vu)u - (div u)u} + (div U)2. 10. Let u and v be smooth. Show that div(u x v) = v· curl u - u· curl v.

5. THE DIVERGENCE THEOREM. STOKES' THEOREM We use the term regular region in the sense of Kellogg [1].1 Roughly speaking, a regular region is a closed region [Jt with piecewise smooth boundary aiYt. It is important to note that .JfI may be bounded or unbounded. In the former case we write vol([Jt) for the volume of [Jt. Divergence Theorem. Let [Jt be a bounded regular region, and let qJ: fYt v: [Jt ~ 1/, and S: rJt ~ Lin be smooth fields. Then

r

J

qJn

dA

=

iJ9I

r VqJ dV,

J

9I

r v - n dA = Jrdiv v dV,

J

iJ9I

9I

r Sn dA Jr div S dV,

J

iJ9I

=

9I

where n is the outward unit normal field on arJt. 1

See also Gurtin [1, §5].

~

~,

II.

38

TENSOR ANALYSIS

Proof The results concerning cp and v are classical! and will not be verified here. To establish the last relation, let a be a vector. Then

a'

f

Sn dA

f

= fa. Sn dA =

Mt

iJ9t

=

(STa) · n dA

iJ9t

f div(STa) dV = f (div S)· a dV = a' f div S dV, 9t

9t

9t

which implies the desired result, since a is arbitrary.

0

Our reason for defining div S the way we did should be clear from the foregoing proof. We will often find it important to deduce local field equations from global formulations of balance laws. This procedure is greatly facilitated by the Localization Theorem. Let be a continuous scalar or vector field on an open set fJIl in S. Then given any X o E fJIl, (xo)

=

lim

a~O

vo

I~O

)

a

i

(I)

dV, no

where Oa (8 > 0) is the closed ball of radius 8 centered at

Xo'

Therefore,

for every closed ball 0 c fJIl, then

Proof

Va

I(xo) - ~va ino dV I,

= vol(Oa)' Then

t, ::;; ~ Va

i

1(xo) - (x)j dVx

no

which tends to zero as 8 1

o.

Let

t, = where

=

::;;

sup I(x o) - (x) I

xen o -+

0, since is continuous.

See, e.g., Kellogg [1, Chapter 4].

0

if

5, THE DIVERGENCE THEOREM. STOKES' THEOREM

39

Note that the localization theorem and the divergence theorem together yield the following interesting interpretation of the divergence: div v(x) div S(x)

= lim

~-o vo

= lim

~-o vo

l~n)

l~n)

~ ~

r v . dA, r So dA. Jilfi

Jilfi

0

6

6

We will make use of Stokes' theorem, but only in the following very special form. Stokes' Theorem.' Let v be a smooth vector field on an open set [j£ in Iff. Further, let n be a disc in [j£ (Fig. 3), let 0 be a unit normal to n, and let the bounding circle c(O"), 0 ~ 0" ~ 1, be oriented so that [c(O) x c(O")] • 0 > 0, Then

In(curl v) .

0

dA =

i

V

0<0"<1.

i

v . dx.

The integral '

dX

represents the circulation of v around c; it sums the tangential component of v around the curve c and can be used to give a physical interpretation of curl v. Indeed, if we let Xo denote the center and b the radius of n, and write n = n~, c = c~, then by Stokes' theorem and an argument similar to that used to derive (1),

J

• 1 ( ) - I' C6 V • dx n cur v X o - im A(A) , ilo"<5

~-o

where A(n~) is the area ofn~. Considering different choices for 0, we conclude that curl v lies perpendicular to the plane in which the circulation per unit n

Figure 3 I

See, e.g., Kellogg [I, Chapter 4, §4].

C

II.

40

TENSOR ANALYSIS

area is greatest, and that the magnitude of curl v is the circulation per unit area in that plane. EXERCISES

1.

Let [lA be a bounded regular region, and let v, w: [lA -4 be smooth. Show that: (a)

(b) (c) (d) 2.

"r and S: [lA -4 Lin

10'" v ® 0 dA = I", v- dV, Ioo'll (So) ® v dA = Io'll [(div S) ® v + SVvT ] dV, 10'" v· So dA = (v - div S + S· Vv) dV, 10'" v(w' 0) dA = Io'll [v div w + (Vv)w] dV.

I",

Let v be a smooth vector field on an open region

[lA. Show

r v'odA = 0

J09 for every regular region

{!/J c [lA

if and only if div v =

SELECTED REFERENCES

Bartle [1, Chapter 7]. Chadwick [1, Chapter 1]. Dieudonne [1, Chapters 8, 10]. Gurtin [1, §§5, 6]. Kellogg [1, Chapter 4]. Nickerson, Spencer, and Steenrod [1, Chapters 6-8, 10]. Truesdell and Toupin [1, Appendix by J. L. Ericksen].

o.

that

CHAPTER

III Kinematics

6. BODIES. DEFORMAnONS. STRAIN Bodies have one distinct physical property: they occupy regions of euclidean space G. Although a given body will occupy different regions at different times, and no one of these regions can be intrinsically associated with the body, we will find it convenient to choose one such region, ffl say, as reference, and to identify points of the body with their positions in ffl. Formally, then, a body ffl is a (possibly unbounded) regular region in G. We will sometimes refer to ffl as the reference configuration. Points p E ffl are called material points; bounded regular subregions of P'A are called parts. Continuum mechanics is, for the most part, a study of deforming bodies. Mathematically, a body is deformed via a mapping f that carries each material point p into a point x

= f(p).

The requirement that the body not penetrate itself is expressed by the assumption that f be one-to-one. As we shall see later, det Vf represents, locally, the volume after deformation per unit original volume; it is therefore reasonable to assume that det Vf # O. Further, a deformation with det Vf < 0 cannot be reached by a continuous process starting in the reference configuration; that is, by a continuous one-parameter family f.,. (0 ~ (J ~ 1) 41

III.

42

KINEMATICS

of deformations with f o the identity, f1 = f, and det Vf a never zero. Indeed, since det Vf a is strictly positive at (T = 0, it must be strictly positive for all (T. We therefore require that det Vf >

o.

(1)

The above discussion should motivate the following definition. By a def ormation of fJI we mean a smooth, one-to-one mapping f which maps fJI onto a closed region in C, and which satisfies (1). The vector u(p) = f(p) - p

(2)

represents the displacement ofp (Fig. 1). When u is a constant, f is a translation; in this case f(p)

= p + u.

F(p)

= Vf(p)

The tensor (3)

is called the deformation gradient and by (I) belongs to Lin ". A deformation with F constant is homogeneous. In view of (4.9), every homogeneous deformation admits the representation f(p)

= f(q) + F(p - q)

(4)

for all p, q E fJI, and conversely, a point field f on fJI that satisfies (4) with F E Lin + is a homogeneous deformation. For any given value of q the right side of (4) is well defined for all p E Iff. Thus any homogeneous deformation of fJI can be extended to form a homogeneous deformation of Iff. We therefore consider homogeneous deformations as defined on all of Iff. For future use we note the following properties of homogeneous deformations: (i) Given a point q and a tensor F E Lin +, there is exactly one homogeneous deformation f with Vf = F and q fixed [i.e., f(q) = q].

Figure 1

6.

BODIES. DEFORMATIONS. STRAIN

43

(ii) If f and g are homogeneous deformations, then so also is fog and

V(f g) = (Vf)(Vg). 0

Moreover, if f and g have q fixed, then so does fog. Proposition. Let f be a homogeneous deformation. Then given any point q we can decompose f as follows:

f=d 1 og=god 2 , where g is a homogeneous deformation with qfixed, while d, and d 2 are translations. Further, each of these decompositions is unique. Proof (Uniqueness) Assume that the first decomposition f = d, 0 g holds. Then Vf = (Vd1)(Vg) and Vd1 = I (because d 1 is a translation), so that Vf = Vg. Therefore, by property (i) above, g is uniquely determined. Moreover, since d 1 = fog-I, this implies that d 1 is uniquely determined. That f = g 0 d 2 is also unique has an analogous proof. (Existence) By hypothesis,

f(p) = f(q)

+ F(p

- q).

Since g must fix q and have Vg = Vf (= F) (cf. the previous paragraph),

g(p) = q

+ F(p

- q).

Define

To complete the proof we must show that d 1 and d 2 are translations. Let 00 =

f(q) - q.

Then, since

we have

+ F(q + F-1(p q + F-1(f(q) + F(p -

+ 00' p + F-1o O ' 0

d1(p) = f(q)

q) - q) = p

d 2(p) =

q) - q) =

The last proposition allows us to concentrate on homogeneous deformations with a point fixed. An important example of this type of deformation is a rotation about q:

f(p) = q + R(p - q) with R a rotation.

III.

44

---

------

--

KINEMATICS

------

•q Figure 2

---

----

A second example is a stretch from q, for which f(p)

=

q + U(p - q)

with U symmetric and positive definite. If, in particular, U

=

I + (A. - l)e ® e

with A. > 0 and Ie I = 1, then f is an extension of amount A. in the direction e. Here the matrix of U relative to a coordinate frame with e = e 1 has the simple form

0 0] 001

A. [U] = 0 1 0, [ and the corresponding displacement, shown in Fig. 2, has components (u 1 , 0, 0) with

Properties (i) and (ii) of homogeneous deformations, when used in conjunction with the polar decomposition theorem, yield the following Proposition. Let f be a homogeneous deformation with q.fixed. Then f admits the decompositions

r=

g

0

SI

= S2

0

g,

where g is a rotation about q, while SI and S2 are stretches from q. Further, each of these decompositions is unique. In fact, if F = RU = VR is the polar decomposition ofF = Vf, then

Vg = R, Thus any homogeneous deformation (with a fixed point) can be decomposed into a stretch followed by a rotation, or into a rotation followed by a stretch. The next theorem yields a further decomposition of either of these stretches into a succession of three mutually orthogonal extensions.

6.

BODIES. DEFORMATIONS. STRAIN

45

Proposition.

Every stretch f from q can be decomposed into a succession of three extensions from q in mutually orthogonal directions. The amounts and directions of the extensions are eigenvalues and eigenvectors of V = Vf, and the extensions may be performed in any order. Proof

Since V

E

Psym, we conclude from the spectral theorem that

V =

L Aiei (8) e, i

with {ei } an orthonormal basis of eigenvectors and Ai > 0 the eigenvalue associated with e, (Ai> 0 since V is positive definite). In view of the identities (1.2h,4' where

Vi = I + (Ai - l)ei (8) e.. Let f i (i = 1, 2, 3) be the extension from q of amount Ai in the direction e., By property (ii) of homogeneous deformations, f 1 f 2 f 3 is a homogeneous deformation with q fixed and deformation gradient V 1 V 2 V 3 = U. But f also has q fixed and Vf = U. Thus by property (i) of homogeneous deformations, 0

f

=

f1 0 f2

0

0

f3 .

As a matter of fact, it is clear that f = f a ( l ) 0 f a (2 ) 0 f a ( 3 )

for any permutation

(J

of {I, 2, 3}. D

In view of the last proposition every stretch can be decomposed into a succession of extensions, the amounts of the extensions being the eigenvalues ..1.1' . 1. 2, . 1. 3 of U. For this reason we refer to the Ai as principal stretches. Note that, since the stretch tensors V and V have the same spectrum, the stretch s, (ofthe proposition on p. 44) has the same principal stretches as the stretch S2. Note also that by (2.10) the principal invariants of V take the form 'l(V)

= ..1.1 + . 1. 2 + . 1. 3,

liV) = . 1. 1..1.2 + . 1. 2..1.3 + . 1. 3..1.1,

'3(V) = . 1. 1..1.2..1.3•

We now turn to a study of general deformations of f!l. To avoid repeated hypotheses we assume for the remainder of the section that f is a deformation of f!l. Since f is one-to-one its inverse f- 1 : f(f!l) -. f!l exists. Moreover, by (1),

46

III.

KINEMATICS

Vf(p) is invertible at each point p of fJI, and we conclude from the smoothinverse theorem (page 22) that f - 1 is a smooth map. Two other important properties of fare

° = f(fJlt, f(fJI) f(ofJI)

=

(5) of(fJI),

where f(fJI)o denotes the interior of f(fJI). We leave the proof of (5) as an exercise. The concept of strain is most easily introduced by expanding the deformation f about an arbitrary point q E fJI: f(p) = f(q) + F(q)(p - q) + o(p - q), where F is the deformation gradient (3). Thus in a neighborhood of a point q and to within an error of o(p - q) a deformation behaves like a homogeneous deformation. This motivates the following terminology: let F = RU = VR be the pointwise polar decomposition of F; then R is the rotation tensor, U the right stretch tensor, and V the left stretch tensor for the deformation f. R(p) measures the local rigid rotation of points near p, while U(p) and V(p) measure local stretching from p. Since U and V involve the square roots of FTF and FFT, their computation is often difficult.For this reason weintroduce the right and left Cauchy-Green strain tensors, C and B, defined by B

= V2 = FFT ;

(6)

in components, C ij

_ ~ ofm ofm -

L...--' m

0Pi OPj

Note that B = RCR T;

(7)

thus, since R is a rotation,

(cf. (2.10) and Exercise 2.3). Recall that the angle () between two nonzero vectors u and v is defined by u·v cos () = - [u/lvl

(0

~

()

~

zr).

6.

47

BODIES. DEFORMATIONS. STRAIN

f(p)

f(.")

C

f (p + ae)

9,

f(p + ad)

Figure 3

Let d and e be unit vectors and let p E

Proposition.

£i. Then

as ex

--+ 0,

If(p + exe) - f(p) I --+ IU( ) I lexl pe, and the angle between f(p + «d) - f(p) and f(p between U(p)d and U(p)e (Fig. 3).

+ exe) -

(8)

f(p) tends to the angle

Proof For convenience, we write F for F(p) and U for U(p). Then given any vector u,

f(p as ex

--+ O.

+ zu) =

f(p)

+ exFu + o(ex)

Let

d", = f(p

+ exd)

+ exe)

- f(p),

e",

=

f(p

d, = exFd + o(ex),

e",

=

zFe

- f(p).

Then

+ o(ex),

and

d", •2 e", ex

--+ Fd

• Fe = RUd . RUe = Ud • Ue,

since the rotation tensor R is orthogonal. Taking d = e leads us to (8). Next, let ()'" designate the angle between d, and e",. Then cos () = d",' e", = d", • e", • ~ • ~ --+ Ud· Ue '" Id",lIe,,1 ex 2 Id",1 [e, I IUdIIUel' which is the cosine of the angle between Ud and Ue. (Note that since U is invertible, Ud, Ue ¥= 0, and since f is one-to-one, d, e, ¥= 0 for ex ¥= O. Hence the last computation makes sense.) Finally, since the cosine has a continuous inverse on [0, n], ()'" tends to the angle between Ud and Ue. 0

III.

48

KINEMATICS

Figure 4

The above proposition shows that the stretch tensor U measures local distance and angle changes under the deformation. In particular, since Iex I is the distance between p and q = p + exe, (8) asserts that to within an error that vanishes as ex approaches zero, IU(p)el is the distance between p and q after the deformation per unit original distance. The next result shows that U also determines the deformed length of curves in Pi. In this regard, note that given any curve c in Pi, f c is the deformed curve f(c(o)), 0 :::; a :::; 1 (Fig. 4). 0

Proposition.

Given any curve c in i!4,

length(f c) = 0

Proof.

f

IU(c(u))c(u)I da.

(9)

By definition, length(f c) = 0

f Id~

f(c(u))\ da.

But by the chain rule, :u f(c(u)) = F(c(u))c(u) = R(c(u))U(c(u))c(u), where R(p) is the rotation tensor. Thus, since R is orthogonal,

Id~

f(C(U))!

=

IU(c(u))c(u)!.

0

A deformation that preserves distance is said to be rigid. More precisely, f is rigid if

If(p) - f(q)1 = Ip - ql

(10)

6.

49

BODIES. DEFORMA nONS. STRAIN

for all p, q E fJl. This condition imposes severe restrictions; indeed, as the next theorem shows, a deformation f is rigid if and only if: (i) f is homogeneous and (ii) Vf is a rotation.

Theorem (Characterization of rigid deformations).

The following are

equivalent:

(a) (b)

f is a rigid deformation. f admits a representation of the form f(p) = f(q) + R(p - q)

for all p, q

E

fJl with R a rotation.

(c) F(p) is a rotation for each p E fJl. (d) U(p) = Ifor each p E fJl. (e) For any curve c in fJl, length(c) = length(f c). 0

Proof

We will show that (a) ~ (b)

(a) ~ (b). differentiate

= (c) = (d) = (e) = (b).

Let f be a rigid deformation. If we use (4.2h and (4.3) to [f(p) - f(q)J . [f(p) - f(q)J = (p - q) . (p - q),

first with respect to q and then with respect to p, we find that

= p - q, Vf(q)TVf(p) = I.

Vf(q)T[f(p) - f(q)J

(11)

Taking q = pin (11h we see that Vf(p) is orthogonal at each p; hence (11h implies that Vf(p) = Vf(q) for all p and q, so that Vf is constant. Finally, since det Vf > 0, Vf is a rotation. Thus f is a homogeneous deformation with R = Vf a rotation. Conversely, assume that (b) holds. Then, since R is orthogonal, [f(p) - f(q)J . [f(p) - f(q)J = R(p - q) . R(p - q) = (p - q) . (p - q) and (a) follows. (b) = (c) = (d). If (b) holds, then Vf = R, so that (c) is satisfied. Assume next that F(p) is a rotation. Then C(p) = F(plF(p) = I. But U(p)2 = C(p), and by the square-root theorem (page 13) the tensor U(p) E Psym which satisfies this equation is unique; hence U(p) = I. (d) = (e). This is an immediate consequence of (9). (e) = (b). This is the most delicate portion of the proof. Assume that (e) holds. It clearly suffices to show that F

is a constant rotation.

(12)

50

Ill.

KINEMATICS

Thus choose Po E rj and let 0 be an open ball centered at Po and sufficiently small that f(O) is contained in an open ball r in f(91) (cf. the discussion at the end of the proof). Let p, q E 0 (p :F q), and let c be the straight line from q to p: c(o) = q + e(p - q),

O::;u::;1.

Then, trivially, Ip - ql

=

length(c),

and since the end points of f care f(q) and f(p), 0

length(f c) 0

~

If(p) - f(q) I;

hence (e) implies that If(p) - f(q) I ::; Ip - q].

(13)

Next, since f(q) and rep) lie in the open ball r c £(91), the straight line b from Ceq) to f(p) lies in f(91). Consider the curve c in 91 that maps into b:

O::;u::;1.

c(u) = £-l(b(u»,

Then the argument used to derive (13) now yields the opposite inequality If(p) - f(q)1

~

Ip - ql·

Thus (10) holds for all p, q E 0, and f restricted to 0 is a rigid deformation. The argument given previously [in the proof of the assertion (a) => (b)] therefore tells us that (12) holds on o. We have shown that (12) holds in some neighborhood of each point of Bi. Thus, in particular, the derivative of F exists and is zero on rj; since 91 is connected, this means that F is constant on 91. Thus (12) holds on 91. D We remark that U in (d) can be replaced by C, V, or B without impairing the validity of the theorem. We now construct the open ball 0 used in the above proof. By (5)1' £(91) contains an open ball r centered at f(po). Moreover, since f is continuous, f- l(n is an open neighborhood of Po and hence contains an open ball 0 centered at Po. Trivially, C(O) c r, so that 0 has the requisite properties. It follows from (b) of the last theorem that every rigid deformation is a translation followed by a rotation or a rotation followed by a translation, and conversely; thus (as a consequence of the first two propositions of this section) every homogeneous deformation can be expressed as a rigid deformation followed by a stretch, or as a stretch followed by a rigid deformation.

6.

51

BODIES. DEFORMAnONS. STRAIN

It is often important to convert integrals over f(glI) to integrals over glI. The next proposition, which we state without proof, gives the corresponding transformation law. 1

Proposition. Let f be a deformation of f!4, and let field on f(f!4). Then given any part glI of f!4,

f

J'(i¥)

f

qJ(x) dVx =

f

qJ

be a continuous scalar

q>(f(p)) det F(p) dVp ,

i¥

qJ(x)m(x) dA x

=

(14)

f

iJf(i¥)

q>(f(p))G(p)n(p) dA p ,

iJi¥

where

G

=

(det F)F-r,

while m and n, respectively, are the outward unit normal fields on of(glI) and ogll.

Given a part glI,

f

vol(f(glI)) =

dV

J'(i¥)

represents the volume of glI after it is deformed under f. In view of (14)1> vol(f(glI)) =

L

det F dV,

(15)

and therefore, by the localization theorem (5.1), (16) where (lei is the closed ball of radius b and center at p. Thus det F gives the volume after deformation per unit original volume. We say that f is isochoric (volume preserving) if given any part glI, vol(f(glI)) = vol(glI). An immediate consequence of this definition is the following

Proposition.

A deformation is isochoric

det F

if and only if

=

1.

(17)

1 For (14)1' cf., e.g., Bartle [I, Theorem 45.9]; for (14h, cf., e.g., Truesdell and Toupin [I, Eq. (20.8)].

52

III.

KINEMATICS

EXERCISES

1. Establish properties (i) and (ii) of homogeneous deformations. 2. A homogeneous deformation of the form Xl

= PI

+ YP2'

is called a pure shear. For this deformation compute: (a) the matrices of F, C, and B; (b) the list § c of principal invariants of C (or B); (c) the principal stretches. 3. Compute C, B, and § c for an extension of amount A in the direction e. 4. Show that a deformation is isochoric if and only if det C = 1. 5. Show that

C = I + Vu + VuT + VuT Vu. 6. Show that a deformation is rigid if and only if § c = (3, 3, 1). 7. Show that the principal invariants of C are given by 11(C) =

lic)

=

Ai + A~ + AL AiA~

+ A~A~ + A~Ai,

13(C) = AiA~A~

with Ai the principal stretches. 8. A deformation of the form Xl

=

X2

= f2(PI, P2),

fl(PI,

P2),

is called a plane strain. Show that for such a deformation the principal stretch A3 (in the P3 direction) is unity. Show further that the deformation is isochoric if and only if the other two principal stretches, A" and Ap , satisfy 1

A" =

T. fJ

6. 9.

BODIES. DEFORMATIONS. STRAIN

53

Let f l and f 2 be deformations of fJl with the same right Cauchy-Green strain tensors. Show that there exists a rigid deformation g such that

f 2 = go fl' 10. Establish the following analogs of (14h:

f f

f f f

v(x) . m(x) dA" =

of(&')

T(x)m(x) dA"

=

of(&')

f

v(f(p))' G(p)n(p) dA p ,

o&'

T(f(p))G(p)n(p) dA p ,

(18)

o.~

(x - 0) x T(x)m(x) dA"

=

of(&')

(f(p) - 0) x T(f(p))G(p)n(p) dA p •

o&'

Here v and T are continuous fields on f(fJl) with v vector valued and T tensor valued. 11. Consider the hypothesis and notation of the proposition on page 47. The number Id", x e",1 IlXd x lXel represents the ratio of the area dA" at x = f(p) spanned by d, = f(p + IXd) - f(p) and e", = f(p + lXe) - f(p) to the area dAp spanned by lXe and IXd (Fig. 5). Define dA" = lim Id", x e",l. dA p ",-+0 IlXd x lXel Use the identity (Sa) x (Sb)

f

(det S)S-T(a x b)

=

•

Figure 5

III.

54

KINEMATICS

to show that m(x) dA; dA

=

G( p)o(p),

p

where m(x) and o(p) are the unit normals

. d", x e", m(x) = Itm I' ",-0

ld'"

x e",

dxe o(p) = [d x e]" [Cf. (14h,J 12. Establish (5). 13. Let f!l be the closed half-space f!l = {piO :;;; PI < co}

and consider the mapping f on f!l defined by Xl =

1

-

--,

PI

+1

X3 = P3'

(a) Verify that f is one-to-one and det Vf > O. (b) Compute f(f!l) and use this result to demonstrate that f is not a deformation. (c) Show that (5)2 is not satisfied. 7. SMALL DEFORMATIONS

We now study the behavior of the various kinematical fields when the displacement gradient Vu is small. Since f(p)

+ u(p),

=

p

=

I + Vu;

it follows that F

(1)

hence the Cauchy-Green strain tensors C and B, defined by (6.6), obey the relations C = I + Vu + VuT + VuT Vu,

B

=

I

+ Vu + VuT + Vu VuT •

(2)

7.

55

SMALL DEFORMA nONS

When the deformation is rigid, C = B = I and

Vu + VuT + VuVuT

=

O.

(3)

Moreover, in this case Vu is constant, because F is. The tensor field (4)

is called the infinitesimal strain; clearly

C = I + 2E + VuT Vu,

Proposition.

(5)

+ 2E + Vu VuT .

B= I

Let f. (0 <s < eo) be a one-parameter family of deformations

with

IVu.1 = e. Then

+ o(e)

2E. = C. - I as s

~

O. Further,

=

B. - I

+ o(e)

(6)

if each f. is rigid, then VUE =

Vui'

-

+ o(e).

(7)

Proof The result (6) is a trivial consequence of (2), while (7) follows from (3). 0 This proposition asserts that to within an error of order o(e) the tensors

2E., C. - I, and B. - I coincide. It asserts, in addition, that to within the same error the displacement gradient corresponding to a rigid deformation is skew. The above discussion should motivate the following definition: An infinitesimal rigid displacement of f!4 is a vector field u on f!4 with Vu constant and skew; or equivalently, a vector field u that admits the representation u(p)

=

u(q)

+ W(p

- q)

(8)

for all p, q E f!4, where W is skew (cf. the proposition on page 36). Of course, using the relation between skew tensors and vectors, we can also write u in the form u(p) = u(q)

+ co x

(p - q)

with cothe axial vector corresponding to W.

III.

56

KINEMATICS

Theorem (Characterization of infinitesimal rigid displacements). a smooth vector field on f!4. Then the following are equivalent:

(a) (b)

u is an infinitesimal rigid displacement. u has the projection property: for all p, q

E

f!4,

(p - q) . [u(p) - u(q)] = (c) (d)

Let u be

o.

Vu(p) is skew at each p E f!4. The infinitesimal strain E(p) = 0 at each p E f!4.

Proof

(a)

=>

(b).

Let u be rigid. Then (8) implies

(p - q) • [u(p) - u(q)J = (p - q) . W(p - q) = 0,

since W is skew. (b) => (a). If we differentiate the expression in (b) with respect to p, we arrive at u(p) - u(q)

+ VU(p)T(p

- q) = 0,

and this result, when differentiated with respect to q, yields - Vu(q) - Vu(p? = O.

(9)

Taking p = q tells us that Vu(p) is skew; hence (9) implies that Vu(p) = Vu(q)

for all p, q E 14, and Vu is constant. Thus (a) holds. (a) ¢> (c). Trivially, (a) implies (c). To prove the converse assertion assume that H(p) = Vu(p) is skew at each p E 14. We must show that H is constant. Let n be an open ball in f!4. Choose p, q E n and let c(a) = q

+ a(p

- q),

0

~ a ~

1,

so that c describes the straight line from q to p. Then u(p) - u(q) =

so that

i

Vu(x) dx

=

f f

(p - q) • [u(p) - u(q)] =

H(c(a»c(a) do =

f

H(c(a»(p - q) do,

(p - q) . H(c(a»(p - q) do = 0,

since H is skew. Thus u has the projection property on n, and the argument given previously [in the proof of the assertion (b) => (a)] tells us that H is constant on n. But n is an arbitrary open ball in f!4; thus H is constant on f!4. (c) ¢> (d). This is a trivial consequence of (4). 0

7.

SMALL DEFORMA nONS

57

EXERCISES

1. Under the hypotheses of the proposition containing (6) show that

E. = U. - I + 0(8) = V. - I + 0(8), det F. - 1 = div U.

+ 0(8).

Give a physical interpretation of det F. - 1 in terms of the volume change in the deformation f e : 2. Let u and v be smooth vector fields on rJI and suppose that u and v correspond to the same infinitesimal strain. Show that u - v is an infinitesimal rigid displacement. For the remaining exercises u is a smooth vector field on rJI and E is the corresponding infinitesimal strain. Also, in 3 and 5, rJI is bounded.

3. Define the mean strain E by vol(rJI)E =

L

E dV.

Show that vol(rJI)E

=

"21 Jr

oal

(u

® n + n ® u) dA,

so that E depends only on the boundary values of u. 4. Let

w

=

t(Vu - VuT ) .

Show that

5. (Korn's inequality)

IEI 2

+ IWl 2 =

IEl 2

-

IWI 2 = Vu' VuT •

Let u be of class C 2 and suppose that

u=O Show that

IVuI2 ,

on

i}rJI.

III.

58

KINEMATICS

6. Consider the deformation defined in cylindrical coordinates Xl

= r cos e,

PI = R cos

X2

= r sin e,

P2

e, = R sin e,

P3

=

Z,

by

e= e

r = R,

+

rLZ,

z

z.

=

A deformation of this type is called a pure torsion; it describes a situation in which a cylinder with generators parallel to the Z-axis is twisted uniformly along its length with cross sections remaining parallel and plane. The constant rL represents the angle of twist per unit length. Show that the corresponding displacement is given by

/3 P2(COS /3 -

uI(p) = PI(COS

uip)

=

/3, 1) + PI sin /3, 1) - P2 sin

/3

= rLP3'

U3(P) = O.

Show further that both Vu and u approach zero as a fact,

-+

0, and that, in

UI(P) = -rLP2P3 + o(rL),

(10)

uip) = rLPIP3 + o(rL)

as

rL -+

O.

8. MOTIONS Let fJd be a body. A motion of fJd is a class C 3 function x:fJd x

~-+g

with x(', t), for each fixed t, a deformation of fJd (Fig. 6). Thus a motion is a smooth one-parameter family of deformations, the time t being the parameter. We refer to ' x = x(p, t) as the place occupied by the material point p at time t, and write fJd t

= x(fJd, t)

1 We carefully distinguish between the motion x and its values x, and between the reference map p and material points p.

8.

59

MOTIONS

• x = X(P. t)

Figure 6

for the region of space occupied by the body at t. It is often more convenient to work with places and times rather than with material points and times, and for this reason we introduce the trajectory

At each t, x(', t) is a one-to-one mapping of flI onto inverse p(', t): flit

flit;

hence it has an

flI

~

such that x(p(x, r), t)

Given (x, t)

E

= x,

p(x(p, r), t) = p.

!Y,

p = p(x, t) is the material point that occupies the place x at time t. The map p: !Y ~ flI so defined is called the reference map of the motion. We call

.

a x(p, t)

x(p, t)

= at

x(p, t)

= :t22

the velocity and x(p, t)

60

III.

KINEMATICS

the acceleration. Using the reference map p we can describe the velocity x(p, r) as a function vex, t) of the place x and time t. Specifically,

v: ,'Y -+

'f~

is defined by

vex, t)

=

x(p(x, t), t)

and is called the spatial description of the velocity. The vector vex, t) is the velocity of the material point which at time t occupies the place x. More generally, any field associated with the motion can be expressed as a function of the material point and time with domain fJ4 x IR, or as a function of the place and time with domain ,'Y. We therefore introduce the following terminology: a material field is a function with domain fJ4 x IR; a spatial field is a function with domain :Y. The field i is material, the field v is spatial. It is a simple matter to transform a material field into a spatial field, and vice versa. We define the spatial description

Q",(p, t)

=

Q(x(p, t), t).

Clearly, (Q"')d = Q.

c:

Smoothness Lemma. field is of class en (n

~

The reference map p is of class Thus a material 3) if and only if its spatial description is of class en,

The proof of this lemma will be given at the end of the section. Given a material field
oto
dl(p, t) =

for the derivative with respect to time t holding the material point p fixed,and

V
=

Vp
for the gradient with respect to p holding t fixed. <1> is called the material time derivative of
F = Vx

8.

61

MOTIONS

is the deformation gradient in the motion x. Since the mapping p ~ x(p, t) is a deformation of 14, (1) det F > O. Similarly, given a spatial field 0 we write

a

O'(x, t) = at O(x, t) for the derivative with respect to t holding the place x fixed, and grad O(x, t) = V"O(x, t) for the gradient with respect to x holding t fixed. 0' is called the spatial time derivative of 0, grad 0 the spatial gradient of O. We define the spatial divergence and the spatial curl, div and curl, to be the divergence and curl operations for spatial fields, so that grad is the underlying gradient. Similarly, Div and Curl designate the material divergence and the material curl computed using the material gradient V. The notation introduced above is summarized in Table 1. It is also convenient to define the material time derivative of a spatial field O. Roughly speaking, n represents the time derivative of 0 holding the material point fixed. Thus to compute n we transform 0 to the material description, take the material time derivative, and then transform back to the spatial description:

n

(2)

Q = «0",)").; that is, Q(x, t)

=

a

at O(x(p, t), t)lp~p(",tl'

The next proposition shows that the material time derivative commutes with both the material and spatial transformations. Table 1

Material field l1> Domain Arguments Gradient with respect to first argument Derivative with respect to second argument (time) Divergence Curl

~x~

Material point p and time t

Spatial field n

.r Place x and time t grad

n


n'

Divl1> Curll1>

divn curl n

III.

62

Proposition.

KINEMATICS

Let et> be a smooth material field, Q a smooth spatial field. Then

(<1»,,= (et>,,)' == <1>",

(3)

(0)", = (Q",)' == 0 ....

Proof If we take the material description of (2), we arrive at (3)2' Also, by definition,

(et>,,). = «et>,,)...)')., = (<1>).,.

Note that, by (3h with

D

= v,

Q

(v)... = (v",), = ii,

so that vis the spatial description of the acceleration. The relation between the material and spatial time derivatives is brought out by the following

.Proposition. Let cp and u be smooth spatial fields with cp scalar valued and u vector valued. Then ciJ

=

cp' + v . grad cp,

U = u'

+ (grad

(4)

u)v.

Thus, in particular,

v = Vi + (grad v)v. Proof

(5)

By the chain rule, ciJ(x, t) = :t cp(x(p, t), t)lp=p(".t) =

[grad cp(x, t)] • x(p(x, r), t) + cp'(x, t)

= v{x, t) • grad cp(x, t) + cp'(x; t),

a

u(x, t) = at u(x(p, t), t) Ip=p(". t) =

[grad u(x, t)]x(p(x, t), t) + u'(x, t)

=

[grad u(x, t)]v(x, t)

+ u'(x, t).

D

A simple application of (4) is expressed in the next result, which gives the material time derivative of the position vector r: tff -+ "f/ defined by r(x) = x - o.

8. Proposition. r(x, t)

63

MOTIONS

Consider the position vector as a spatial field by defining

= r(x)for every (x, t) E .'Y. Then t

=

v.

(6)

Proof Since r' = 0 and grad r = I, (4)2 with U = r yields (6). [This result can also be arrived at directly by noting that r... (p, z) = x(p, t) - 0.] 0

Proposition.

Let u be a smooth spatial vector field. Then (7)

V(u...) = (grad u)",F, where F is the deformation gradient. Proof

By definition, u...(p, t) = u(x(p, t), t);

that is,

u",C t) = u(', r) a X(', t). Thus the chain rule (3.11) tells us that V(u...) is the gradient grad u ofu times the gradient F = Vx of x. 0 The spatial field L=gradv is called the velocity gradient.

Proposition F= L... F,

(8)

F = (grad v)...F. Proof

Since x is by definition C3, . 0 F(p, t) = Vx(p, t) = Vit(p, t) = Vv...(p, t),

ot

and (8)1 follows from (7) with u = v. Similarly, F(p, t) = Vx(p, t) = Vv",(p, t), and taking u =

vin (7) we are led to (8h.

0

Given a material point p, the function s: IR s(t)

--+ S

defined by

= x(p, t)

is called the path line of p. Clearly, s is a solution of the differential equation

S(t) = v(s(t), t),

64

Ill.

KINEMATICS

and conversely every maximal solution of this equation is a path line. (A solution is maximal provided it is not a portion of another solution.) On the other hand, if we freeze the time at t = r and look at solution curves of the vector field v(', r), we get the streamlines of the motion at time r, Thus each streamline is a maximal solution s of the differential equation S(A) = V(S(A), r).

An example of a motion x (of 8) is furnished by the mapping defined in cartesian components by Xl = PIe X2

l

>,

= P2 el ,

The deformation gradient F is given by 12

[F(p, I)]

~ e~ [

while the velocity i has components

o. Thus, since the reference map p is given by

the spatial description of the velocity has components vI(x, t) = 2x l t, V2(X,

t)

=

X2,

and the velocity gradient L has the matrix 2t

[L(x, t)J =

[

~

8.

65

MOTIONS

The streamlines of the motion at time t are solutions of the system Sl(A)

=

2tS 1( A),

S2(A)

=

S2(A),

S3(A) = 0, so that

Sl(A) S2(A)

= y 1 e 2 t\ = Y2 e',

S3(A) = Y3 is the streamline passing through (Yl, Y2, Y3) at A = O. We close this section by giving the

Proof ofthe Smoothness Lemma. It suffices to show that p is of class C 3 , for then the remaining assertion in the lemma follows trivially. Consider the mapping

defined by 'I'(p, t) = (x(p, t), t).

It follows from the properties of x that 'I' is class C 3 and one-to-one; in fact, 'I'-l(X, t)

=

(p(x, r), t).

Thus to complete the proof it suffices to show that the derivative D'I'(p, t): 11 x IR

--+

11 x IR

is invertible at each (p, t), for then the smooth-inverse theorem (page 22) tells us that 'I' - 1 is as smooth as '1', and hence that p is of class C 3 • Since x(p + h, t + r) = x(p, t) + F(p, t)h + x(p, t)r + 0(8) as

8

=

(h 2 + r 2 ) 1!2

--+

0, it follows that

'I'(p + h, t + r) = 'I'(p, t) + (F(p, t)h + x(p, t)r, r) + 0(8). Thus D'I'(p, t) [h, r] = (F(p, t)h for all h

E

11 and r

E

IR.

+ x(p, t)r, r)

(9)

66

III.

KINEMATICS

To show that D'I'(p, t) is invertible, it suffices to show that D'I'(p, t) [h, r]

=0

(10)

implies

= 0,

h

r

=

o.

Thus assume that (10) holds. Then by (9), r

=

F(p, t)h = 0,

0,

0

and, since F(p, t) is invertible, h = 0.

EXERCISES

1. A motion is a simple shear if the velocity field has the form v(x, t) = v1(x2)el in some cartesian frame. Show that for a simple shear div v = 0,

v = v',

(grad v)v = 0,

In the next two exercises D and W, respectively, are the symmetric and skew parts of grad v. 2. Prove that

C=

2F TD",F.

3. Let v be a class C 2 velocity field. Show that div v = (div v): + IDI2

-

/WI 2 •

4. Consider the motion of $ defined by

X3

=

P3'

in some cartesian frame. Compute the spatial velocity field v and determine the streamlines. 5. Consider the motion x defined by x(p, t)

=

Po + U(t)[p - Po],

where 3

U(t)

=

L Q(j(t)ej ® e, i= 1

9.

TYPES OF MOTIONS. SPIN. RATE OF STRETCHING

67

with !Xi> 0 smooth. (Here {ei } is an orthonormal basis.) Compute p, v, and L, and determine the streamlines. 6. Define the spatial gradient and spatial time derivative of a material field and show that X'

7. Consider a surface

[I'

= 0,

grad x = I.

in fJI of the form [I'

= {p E .@lqJ(p) = O},

where .@ is an open subset of fJI and qJ is a smooth scalar field on .@ with VqJ never zero on [1'. Let x be a motion of fJI. Then, at time t, [I' occupies the surface [1', =

{x E .@,Il/J(x, t) = O},

where .@, = x(.@, t) and l/J(x, t)

=

qJ(p(x, t».

Show that: (a) VqJ(p) is normal to [I' at p E [1'; (b) grad l/J(x, t) is normal to [1', at x E [1',; (c) VqJ = F T (grad l/J)..., and hence grad l/J(x, t) never vanishes on Y't; (d) IVqJ 12 = (grad l/J)... • B(grad l/J)_, where B = FFT is the leftCauchyGreen strain tensor; (e) l/J' = - v • grad l/J. 9. TYPES OF MOTIONS. SPIN. RATE OF STRETCHING A motion x is steady if for all time t and v' = 0

everywhere on the trajectory ff. Note that !!I = fJlo x IR, because the body occupies the same region fJlo for all time. Also, since the velocity field v is independent of time, we may consider v as a function xr->v(x) on fJlo . Thus in a steady motion the particles that cross a given place x all cross x with the same velocity v(x). Of course, for a given material point p the velocity will generally change with time, since x(p, t) = v(x(p, t». Consider now a (not necessarily steady) motion x and choose a point p with x(p, T) E ofJI, at some time T. Then (6.5)2 with f = x(', T) implies that

68

III.

KINEMATICS

P E iJ!!J and a second application of (6.5h, this time with f = x(', r), tells us that x(p, t) E iJ!!Jt for all t. Thus a material point once on the boundary lies on the boundary for all time. If the boundary is independent of time (iJ!!J t = iJ!!J o for all r), as is the case in a steady motion, then x(p, t), as a function of t, describes a curve on iJ!!J o , and x(p, t) is tangent to iJ!!J o . Thus we have the following Proposition. In a steady motion the velocity field is tangent to the boundary; i.e., v(x) is tangent to iJ!!J o at each x E iJ!!J o .

In a steady motion path lines and streamlines satisfy the same autonomous differential equation s(t)

= v(s(t».

Thus, as a consequence of the uniqueness theorem for ordinary differential equations, we have the following Proposition. In a steady motion every path line is a streamline and every streamline is a path line.

Let be a smooth field on the trajectory of a steady motion. Then is steady if <1>'

=

0

(1)

[in which case we consider as a function x ....... (x) on !!JoJ. Proposition. Let qJ be a smooth, steady scalar field on the trajectory of a steady motion. Then the following are equivalent: (a)

qJ is constant on streamlines; that is, given any streamline s,

d

dt qJ(s(t» = 0 for all t.

(b) q, = O. (c) v· grad qJ = O. Proof

Note first that, by (8.4)1 and (1),

q, =

v • grad qJ;

thus (b) and (c) are equivalent. Next, for any streamline s, :t qJ(s(t»

=

s(t) • grad qJ(s(t»

=

v(s(t» • grad qJ(s(t»,

(2)

so that (c) implies (a). On the other hand, since every point of !!J o has a streamline passing through it, (a) and (2) imply (c). D

9.

TYPES OF MOTIONS. SPIN. RATE OF STRETCHING

69

x(q, t)

x(a, t)

Figure 7

x(P,

t)

The number bet) = Ix(p, t) - x(q, t) I

(3)

represents the distance at time t between the material points p and q. Similarly, the angle OCt) at time t subtended by the material points a, p, q is the angle between the vectors x(a, t) - x(p, t) and x(q, t) - x(p, t). (See Fig. 7.) A motion x is rigid if

a

at Ix(p, t) - x(q, t)! = 0

(4)

for all materials points p and q and each time t. Thus a motion is rigid if the distance between any two material points remains constant in time. Theorem (Characterization of rigid motions). Let x be a motion, and let v be the corresponding velocity field. Then the following are equivalent:

(a) x is rigid. (b) At each time t, v(', t) has the form ofan infinitesimal rigid displacement of fJI,; that is, v(', t) admits the representation

Vex, t)

=

v(y, t) + W(t)(x - y)

for all x, y E fJI where Wet) is a skew tensor. " (c) The velocity gradient L(x, t) is skew at each (x, t) Proof.

(5) E

.'!I.

If we use (3) to differentiate b(t)2, we find that bet) J(t) = [x(p, t) - x(q, t)] • [x(p, t) - x(q, t)],

or equivalently, letting x and y denote the places occupied by p and q at time t, bet) J(t) = (x - y) • [vex, t) - v(y, t)J.

(6)

By (6), (4), and the fact that bet) 'I- 0 for p 'I- q, x is rigid if and only if v(-, t) has the projection property at each time t. The equivalence of (a), (b), and (c) is therefore a direct consequence of the theorem characterizing infinitesimal rigid displacements (page 56). D

70

III.

KINEMATICS

Let ro(t) be the axial vector corresponding to W(t); then (5) becomes v(x, t) = v(y, t)

+ ro(t)

x (x - y),

which is the classical formula for the velocity field of a rigid motion. The vector function ro is called the angular velocity. Note that curl v

= 2ro,

which gives a physical interpretation of curl v, at least for rigid motions. For convenience, we suppress the argument t and write v(x)

v(y)

=

+ ro x (x - y).

Assume ro ¥- O. Then for fixed y the velocity field XHro x (x - y) vanishes for x on the line {y + O(rolO( E

~}

and represents a rigid rotation about this line. Thus given any fixed y, v is the sum of a uniform velocity field with constant value v(y) and a rigid rotation about the line through y spanned by roo For this reason we calli = sp{ro} the spin axis. For future use, we note that 1= l(t) can also be specified as the set of all vectors e such that We=O. As we have seen, a rigid motion is characterized (at each time) by a velocity gradient which is both constant and skew. We now study the case in which the gradient is still constant, but is symmetric rather than skew. Thus consider a velocity field of the form v(x)

= D(x - y)

with D a symmetric tensor. By the spectral theorem D is the sum of three tensors of the form O(e ® e,

lei = 1,

with corresponding e's mutually orthogonal. It therefore suffices to limit our discussion to the velocity field

= O(e ® e)(x - y). Relative to a coordinate frame with e = e., v has components v(x)

(Vl' 0, 0),

(7)

9.

TYPES OF MOTIONS. SPIN. RATE OF STRETCHING

---

-----

---

---- ----

•y Figure 8

71

--- ---

and is described in Fig. 8. What we have shown is that every velocity field with gradient symmetric and constant is (modulo a constant field) the sum of three velocity fields of the form (7) with" axes" e mutually perpendicular. Now consider a general velocity field v. Since L = grad v, it follows that v(x)

=

v(y) + L(y)(x - y) + o(x - y)

as x --+ y, where y is a given point, and where we have suppressed the argument t. Let D and W, respectively, denote the symmetric and skew parts ofL: D

= !
U) = !
W = !
+ W(y)(x - y)

and a velocity field of the form

x 1--+ D(y)(x - y). For this reason we call W(y, t) and D(y, r), respectively, the spin and the stretching, and we use the term spin axis at (y, t) for the subspace I of ~ consisting of all vectors e for which W(y, t)e

= O.

[Of course, I has dimension one when W(y, t) ¥further motivated by the following

O.J

The term stretching is

III.

72

KINEMATICS

.11,

Figure 9

Proposition. Let x

o

iJlt with t afixed time, let e be a unit vector, and let c5,,.(r) (for IX sufficiently small) denote the distance at time t between the material points that occupy the places x and x + lXe at time t (Fig. 9). Then E

. J~(t) 1im s: ( ~-o

u~

t

)

)

= e . D(x, t e.

(8)

Further, ifd is a unit vector perpendicular to e, and ifO~(t) is the angle at time r subtended by the material points that occupy the places x + ze, x, x + IXd at time t, then

lim e~(t)

= -2d' D(x, t)e.

~-o

Proof Since c5aCt) is the distance between x and x with y = x + lXe implies e • [v(x

J~(t)

+ lXe, t)

+ lXe, which is IX, (6)

- vex, t)]

IX

c5~(t)

But lim! [v(x

e-e OIX

+

lXe, t) - vex, t)]

= L(x,

and e : L(x, t)e = e • D(x, t)e

[since W(x, t) is skew]. Thus (8) holds.

t)e

(9)

9.

TYPES OF MOTIONS. SPIN. RATE OF STRETCHING

73

Next, let p, q, and a denote the material points that occupy the places x, y = x + ze, and z = x + ad, respectively, at time t. Further, let u(r) = x(q, r) - x(p, r), w(.)

=

x(a, r) - x(p, r),

so that

u(t)

= «e,

li(t) = v(y, t) - v(x, t),

w(t)

=

ad,

w(t) = v(z, t) - v(x, t).

Thus 1 - (u w)o(t) = eo [v(z, t) - v(x, t)J + d [v(y, t) - v(x, t)J. a 0

0

Further, u oW

cos Oa. = [ullw!' and, as u and ware orthogonal at time t, (u w)o(t) 0

= Iu(t) IIw(t) I

(cos Oa.)o(t)

On the other hand, since sin Oit) = 1, (cos Oa.)°(t) = - Oa.(t), and the above relations imply that

-aOa.(t) = eo [v(x

+ ad, t)

If we divide by a and let a part for d, that

--+

- v(x, t)J

+ do [v(x + ae, t)

- v(x, t)J.

0, we conclude, with the aid of(9) and its counter-

lim Oit) = -e L(x, t)d - do L(x, t)e 0

= -d [L(x, t) + L(x, t?Je 0

= -2d D(x, t)e. 0

0

Using the spin W we can establish the following important relations for the acceleration v.

Proposition

v = V' + t gradiv') + 2Wv, v = v' + t grad(v 2 ) + (curl v)

(10) x v.

74

III.

KINEMATICS

Since

Proof

t gradtv"),

2Wv = (grad v - grad vT)v = (grad v)v -

(11)

(10)1 follows from (8.5). The result (10)2 follows from (10)1 and the fact that curl v is twice the axial vector corresponding to W. 0

A motion is plane if the velocity field has the form v(x, t) =

V1(X 1, X2'

t)e l +

V2(XI, X 2,

t)e2

in some cartesian frame.

Proposition.

I n a plane motion

WD + DW = (div v)W. Proof

(12)

Clearly, D and W have matrices of the form [D]

=

[~ ~ ~], o

0

[W] =

[~y ~ ~]

0

0

0

0

(relative to the above frame), and a trivial computation shows that

[WD + DW] =

0

[

Y(IX

-Y(IX + fJ) O

+ fJ)

0]

~

~

=

(IX

+ P)[W]·

But div v = tr L = tr D = IX

+ P,

0

and the proof is complete.

EXERCISES

1. It is often convenient to label material points by their positions at a given time r. Suppose that a material point p occupies the place y at r and X at an arbitrary time t (Fig. 10):

y

=

x(p, r),

X

= x(p, t).

Roughly speaking, we want x as a function of y. Thus, since p = p(y, r),

we have x = x(p(y, r], t).

9.

TYPES OF MOTIONS. SPIN. RATE OF STRETCHING

75

-

X,(', I)

Figure 10

We call the function defined by xlY, t) = x(p(y, r), t) the motion relative to time r ; xly, t) is the place occupied at time t by the material point that occupies y at time r, Let Fly, t) = Vyx.(y, t), where Vy is the gradient with respect to y holding t fixed. Also let Ft=RtUt denote the right polar decomposition of F" and define

C, = (U t )2, (a)

Show that

a

v(x, t) = at xlY, t) (b)

provided x = xt(y, t). Use the relation xC t) = x.(-, t)

0

(13)

x(', r) to show that

Fly, t)F(p, r) = F(p, r), where y = x(p, r), and then appeal to the uniqueness of the polar decomposition to prove that Fly, r) = Uly, r) = Rly, r) = I.

76

III.

KINEMATICS

(c) Show that (14)

C(p, t) = F(p, ,yciy, t)F(p, T).

(d) Show that

a

L(y, T) = at FlY, t)lt=t

=

[~ot Diy, t) + ot~ RlY, t)] t=t ,

a

D(y, T) = at Diy, t)lt=t'

(15)

a

W(y, T) = at Riy, Olt=t·

(16)

(e) Show that

an + 2 :l

ut

n+2 Fiy, t)lt=t = grad aInley, T),

where a(nl is the spatial description of the material time derivative of x of order n + 2. 2. Let x be a motion and suppose that for some fixed T, xt(y, t)

= q(t) + Q(t)(y - z),

(17)

with z and q(t) points and Q(t) E Orth +. Show that x is rigid. 3. Let x be a rigid motion. Show that x, has the form (17). Exercises 2 and 3 assert that, given a motion x and a time deformation at each t if and only if x is a rigid motion.

T,

xl, t) is a rigid

4. Let x be a COO motion. The tensors

an

AnCy, T) = at n Ct(y, t)lt=t

(n = 1,2, ...)

(18)

are called the Rivlin-Ericksen tensors. (a) Show that At = 2D. (b) Show, by differentiating (14) with respect to t, that (19)

(c)

where c(nl is the nth material time derivative of C, and where we have omitted the subscript {) from C and F. Verify that

10. 5.

TRANSPORT THEOREMS. VOLUME. ISOCHORIC MOTIONS

77

Show that the acceleration field of a rigid motion has the form

v(x, t)

= V(y, t)

+ m(t)

x (x - y)

+ ro(t)

x [CO(t) x (x - y)]

with co the angular velocity.

10. TRANSPORT THEOREMS. VOLUME. ISOCHORIC MOTIONS Let x be a motion of !!l. Given a part

for the region of space occupied by

f!j>

f!j>,

we write

at time t. Thus

represents the volume of f!j> at time t. Using the deformation gradient F = Vx we can also express vol(f!j>t) as an integral over f!j> itself [cf. (6.15)J:

Thus

:t

vol(f!j>t) =

fa. (det F)· dY.

(1)

Next, (3.14) and (8.8)1 imply that (det F)"

= (det F) tr(FF- 1) =

(det F) tr L",.

But tr L = tr grad v = div v, so that (det F)· = (det F)(div v)",

(2)

and (1) becomes

d vol(f!j>t) = dt

f ~

(div v)",det F dV =

f

~,

div v dY.

III.

78

KINEMATICS

Thus we have the following Theorem

(Transport of volume). :t vol(f!IJ,) =

f~

For any part f!IJ and time t,

(det F)' dV =

f~,

div v dV =

f

Jo~,

v· n dA.

(3)

Thus (det F)' and div v represent rates of change of volume per unit volume: (det F)' is measured per unit volume in the reference configuration; div v is measured per unit volume in the current configuration. We say that x is isochoric if d

(4)

dt vol(f!IJ,) = 0

for every part f!IJ and time t. As a direct consequence of (3) and (4) we have the following Theorem (Characterization of isochoric motions). equivalent:

The following are

(a) . x is isochoric. (b) (det F)' = O. (c) div v = O. (d) For every part f!IJ and time t,

f

Jo~,

v· ndA

= O.

Warning: For a motion to be isochoric the volume of each part must be constant throughout the motion; it is not necessary that the volume of a part during the motion be equal to its volume in the reference configuration. Note that rigid motions are isochoric, a fact which follows from (c) of the above theorem and (c) of the theorem on page 69. The computations leading to (3) are easily generalized; the result is

Reynolds' Transport Theorem. Let be a smooth spatial field, and assume that is either scalar valued or vector valued. Thenfor any part f!IJ and time t,

:f : f~t t

t

~,

dV

dV

f =f =

~t

e,

(d>

+ div v) dV, (5)

<1>' dV

+

f

JO~t

v • n dA.

10. TRANSPORT THEOREMS. VOLUME. ISOCHORIC MOTIONS

Proof d dt

79

In view of (2),

i


= dd

t

WI,

=

i


=

WI

i

(
WI

r (ci> + +
JWI

WI,

which is (5)1' To derive (5h assume that
ci> +
i


WI,

i: WI,

t


i


WI,

. t=t

Thus (5h asserts that the rate at which the integral of
EXERCISES

1.

Let Pbe a smooth spatial scalar field with Show that

o:

+ div(
ip'

2.

p= 0 and define

Prove that (for f!J bounded)

i

(2Wv

+ v div

v) dV =

91,

r [v(v' n) -

Jo9iJ,

tv 2n] dA,

where n is the outward unit normal to of!J1> so that in an isochoric motion with v = 0 on of!Jt ,

i

WvdV = O.

91,

3.

Derive (5)2 for
80

III.

KINEMATICS

11. SPIN. CIRCULATION. VORTICITY As we have seen, a rigid motion is characterized by a skew velocity gradient; that is, the velocity field is determined (up to a spatially uniform vector field) by its spin W

=

-!(grad v - grad vT).

More generally, given any motion the spin W(x, t) describes the local rigid rotation of material points currently near x. As our next theorem shows, the evolution of W with time is govemed by the field

J = -!(grad v - grad

vT) ,

which represents the skew part of the acceleration gradient. The statement and proof of this theorem are greatly facilitated if we introduce the notation (1)

for any spatial tensor field G, where F is the deformation gradient. Theorem

(Transport of spin).

The spin W satisfies the differential equations

(W F) · = J F ,

(2)

W+DW+WD=J, where D is the stretching. Proof

Recall (8.8): L = FF- 1,

F=

(grad t)F,

where for convenience we have omitted the subscript Since 2W = L - LT,

m

from L and grad

and hence 2WF = FTP - PTF = F Tgrad vF - F T grad vTF, which implies (2)1' Next, by (1) and (2)1' FTWF + FTWF + FTWF = F TJF. Thus

W + F-TFTW + WFF- 1 = J, and, since L = FF- 1 and L T = F-TFT, this equation implies W + LTW + WL = J.

v.

II. SPIN. CIRCULATION. VORTICITY

81

Thus as L = D + W, it follows that

W + (D which implies (2h.

- W)W

+ W(D + W) =

J,

0

We now introduce two important (and somewhat related) definitions. A motion is irrotational if

W=o, or equivalently, if curl v =

o.

A spatial vector field g is the gradient of a potential if there exists a spatial scalar field tl. such that g(x, t) = grad z(x, t) for all (x, t) on the trajectory of the motion. For a large class of fluids, in particular, inviscid fluids under a conservative body force, the acceleration vis the gradient of a potential. When this is the case the potential, tl. say, is C 2 in x, because vis C 1 , and thus grad grad tl. is symmetric; therefore

J = t[grad grad

tl. -

(grad grad

tl.)T]

= 0,

and we have the following consequence of (2)1. Lagrange-Cauchy Theorem. A motion with acceleration the gradient of a potential is irrotational if it is irrotational at one time. Proof: Since J = 0, we conclude from (2)1 that (W F ) " = 0,

(3)

so that WF(p, t) is independent of t. But at some time r, W(x, r) = 0 for all x in f!4" and hence WF(p, r) = 0 for all P in f!4. Thus WF(p, t) = 0 for all p and t, and, since F is invertible, (1) implies that W = O. Hence the motion is irrotational. 0 As is clear from (9.12) and (2h, for plane, isochoric motions a result stronger than (3) holds. Proposition. a potential,

For a plane, isochoric motion with acceleration the gradient of

W=o.

82

III.

KINEMATICS

Figure 11

Let x be a motion of !!A. By a material curve we mean a curve c in !!A. Choose an arbitrary material point c(O') on c. At time t, c(O') will occupy the place x(c(O'), t).

Thus the material points of the curve form a curve ct(O')

= x(c(O'), t),

(4)

in !!At (Fig. 11). When c (and hence e.) is closed,

f.

v(x, t) • dx

Ct

gives the circulation around C at time t; this integral sums the tangential component of the velocity around the curve Ct.

Theorem (Transport of circulation). Then :t

f.

v(x, t) • dx =

Ct

Proof

Let

f.

C

be a closed material curve.

v(x, t) • dx.

(5)

C:t

By definition,

f.

_

v(x, t) • dx =

i

l

0

v(ct(O'), t) . : ct(O') da; uO'

our proof will involve differentiating the right side under the integral. We therefore begin by investigating the partial derivatives (%t)v(ct(O'), t) and (02/ot oO')ct(O'). Note that ct(O') has derivatives with respect to 0' and t which are jointly continuous in (0', t). In particular, by (4), :t ct(O') = i(c(O'), t) = v(ct(O'), t).

(6)

11.

SPIN. CIRCULATION. VORTICITY

83

Clearly, this relation has a derivative with respect to a which is jointly continuous in (a, t). Thus we can switch the order of differentiation; that is,

a2

at aa ct(a)

exists and by (6) a2 a2 a at aa ct(a) = aa at ct(a) = aa v(ct(a), t).

Note also that (6) implies

a

= x(c(a), t) = v(ct(a), t).

at v(ct(a), t)

These identities can be used to transform the left side of (5) as follows: d dt

f.

c, v(x,

t) . dx

d

= dt

f. = f. =

fl v(cla), t) . -----a;;act(a) do 0

v(x, t) • dx

+

c,

fl v(ct(a), t) • aaa v(ct(a), t) do 0

v(x, t) • dx + !{v 2(c t (l), t) - v 2(ct(O), t)}.

c,

But c (and hence ct) is closed; thus ct( 1)

=

ct(O) and the last term vanishes.

D

We say that the motion preserves circulation if

: f. t

v(x, t) • dx = 0

c,

v

for every closed material curve c and all time t. When = grad IX the right side of (5) vanishes, as c, is closed [cf. (4.8)], and we have

Kelvin's Theorem. Assume that the acceleration is the gradient ofa potential. Then the motion preserves circulation. A curve h in !!It is a vortex line at time t if the tangent to h at each point x on h lies on the spin axis of the motion at (x, t). Since the spin axis at (x, t) is the set of all vectors e such that W(x, t)e = 0, h is a vortex line if and only if

W(h(a), t) for 0

s

a

~

1.

d:~)

= 0

III.

84

KINEMATICS

Theorem (Transport of vorticity). Assume that the acceleration is the gradient of a potential. Then vortex lines are transported with the motion; that is,for any material curve c, if e, is a vortex line at some time t = r, then c, is a vortex linefor all time t. Proof

Let

C

be a material curve. Then by (4),

d du ct(u) = F(c(a), t)k(a), (7) k(a)

= d~~).

If c, is a vortex line, then

d

0= W(c.(a), r) du ct(a) = W(c.(a), -r)F(c(u), -r)k(a), so that trivially WF(c(u), -r)k(a) = O.

(8)

Moreover, by (3), WF is constant in time, so that (8) is valid for any t, Therefore, multiplying by F(c(a), t)-T and using (1), W(ct(u), t)F(c(u), t)k(u) = 0, and this relation, with (7), implies that c, is a vortex line for all t.

0

We close this section by listing an important property of irrotational, isochoric motions; this result follows from (c) of the theorem on page 78 and the proposition on page 32.

Theorem.

The velocity field ofan irrotational, isochoric motion is harmonic: ~v

= O.

EXERCISES

We use the notation

w = curl v,

u = (det

1. Show that in a plane motion (uW)" = uJ,

so that when it is the gradient of a potential, (uW)"

= O.

Ft.

85

SELECTED REFERENCES

2. Establish the identity (ow): = uLw + u curl

Note that when

v.

vis the gradient of a potential this reduces to (ow): = uLw.

3. Assume that (9), that

(9)

vis the gradient of a potential. Show, as a consequence of w(x, t)

=

[det Fly, t)] - 1 Fly, t)w(y, t),

where x is the place occupied at time t by the material point which occupies y at time r (cf. Exercise 9.1). 4. Let u be a smooth spatial vector field and c a material curve. Show that

~

dt

f. u· dx f. (Ii + LTu) . dx. =

c,

c,

5. Let v be the gradient of a potential q>. Show that

v = grad ( q>' + v;). so that the acceleration is also the gradient of a potential. SELECTED REFERENCES

Chadwick [1, Chapter 2]. Eringen [1, Chapter 2]. Germain [1, Chapters 1, 5]. Serrin [1, §§1l, 17,21-23,25-29]. Truesdell [1, Chapter 2]. Truesdell and Noll [1, §§21-25]. Truesdell and Toupin [1, §§13-149].

(10)

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CHAPTER

IV Mass. Momentum

12. CONSERVATION OF MASS One of the most important properties of bodies is that they possess mass. We here consider bodies whose mass is distributed continuously. No matter how severely such a body is deformed, its mass is the integral of a density field; that is, given any deformation f there is a density field Pr on f( fJi) such that the mass m(&l) of any part &l is given by

f.

m(&l) =

Pr dV.

r(&')

Since the mass of a part cannot be altered by deforming the part, m(&l) is independent of the deformation f. We now make the above ideas precise. A mass distribution for fJi is a family of smooth density fields

Pr: f(fJi)

-+

IR+,

one for each deformation f, such that

f.

r(&')

Pr dV =

f.

PI dV == m(&l)

1(&')

87

(1)

88

IV.

MASS. MOMENTUM

for any part f!/' and all deformations f and g. The number Pr(x) represents the density at the place x E f(~) in the deformation f, and equation (I) expresses conservation ofmass. We denote by Po the density field Pr when f(p) = p for all p E~. Thus Po(p) gives the density at p when the body is in the reference position. Note that, by the localization theorem, m(nel) . Po(p) = lim l(n )' s

el-O VO

where n el is the ball of radius /j centered at p. As our next result shows, the reference density Po determines the density in all deformations. Proposition.

Let f be a deformation of ~ and let F

= Vf. Then

Pr(x) det F(p) = Po(p) provided x Proof

(2)

= f(p). By (1) and the definition of Po,

r

Pr(x) dY.,

=

Jr(9'l

f

Po(p) dVp '

9'

On the other hand, if we change the variable of integration on the left side from x to p, we arrive at Lpr(f(p» det F(p) dv.,. Thus

L

[p,(f(p» det F(p) - Po(p)] dVp = 0

for every part f!/', and (2) follows from the localization theorem.

0

Given a motion x of ~ we will always write p(x, t) for the density at the place x E ~t in the deformation x(., t). Thus p::r

-+

IR+

is defined by p(x, r) = PX(.,t)(x); we will refer to P as the density in the motion x. In view of (1), m(f!/') =

f

9',

p(x, r) dV"

==

f

9',

P dV,

89

12. CONSERVATION OF MASS

and we have the following

Theorem.

F or every part f!J' and time t, d dt

f

=

p dV

O.

(3)

~t

If F = Vx is the deformation gradient in the motion, then (2) takes the form

p(x, t) det F(p, t)

=

Po(p)

(4)

provided x = x(p, t). Thus p is the spatial description of po/det F, and we conclude from the smoothness lemma that p is smooth on !Y.

Theorem (Local conservation of mass)

p + p div v = 0, p' + div(pv) = O. Proof

(5)

By (4),

+ p( det

p( det F):

F). = 0,

which with (10.2) yields (5k Next, by (8.4)1'

P=

p'

+ v· grad p,

and (5)1' when combined with this relation, implies (5h.

D

Since a motion is isochoric if and only if div v = 0, (5) has the following

Corollary.

A motion is isochoric

if and only if p = O.

Equation (3) expresses conservation of mass for a region f!jJt that moves with the body. It is often more convenient, however, to work with afixed region &/ called a control volume. Of course, material will generally flow into and out of fit, a fact which will become clear in the derivation of the resulting conservation law. By a control volume at time t we mean a bounded regular region &/ with &/

!fl t

c

for all r in some neighborhood of t, Thus for J sufficiently small we have the situation shown in Fig. 1. Let n denote the outward unit normal to 8&/. By the divergence theorem,

rdiv(pv) dV Jr pv : n dA.

J9I

=

091

90

IV.

MASS. MOMENTUM

Figure 1

Further, Lp'(X, t) dVx = L :t p(x, t) dv., = :t Lp(X, t) dVx '

where the last step uses the fact that .o/l is independent of t. Conservation of mass (5h, when combined with the relations above, yields the following

Theorem (Conservation of mass for a control volume). volume at time t. Then d dt

f ~

f

p(x, t) dVx = -

Let ~ be a control

p(x, t)v(x, t)· n{'x) dA x •

(6)

o~

Since n is the outward unit normal to a~, pv • n represents the mass flow, per unit area, out of ~ across its boundary. Thus (6) asserts that the rate of increase of mass in ~ is equal to the mass flow into ~ across its boundary. Lemma.

Let
f


9,

f


(7)

9

Proof We simply change the variable of integration from x to p in the left side of (7); the result is

f

9,


f


9

and, in view of (4), this relation implies (7).

D

If we differentiate (7) with respect to t we arrive at the integral of ] the integral of p over 9,. Thus we have the following important

91

12. CONSERVATION OF MASS

Theorem. Let be a smooth spatial field. Then given any part f!J',

:t f

f

p dV =

&',

(8)


&',

In other words, to differentiate

f

p dV

&',

with respect to time we simply differentiate under the integral sign treating the "mass measure" p dV as a constant. Note that if we take == 1 in (8) we recover (3). EXERCISES

Derive (5)1 and (6) using (3) and Reynold's transport theorem (l0.5). 2. Show that a deformation f is isochoric if and only if 1.

Pr(x)

= Po(p)

whenever X = f(p). 3. Let be a smooth spatial field. Show that for any part f!J',

f

=

(x, t)p(x, t) dVx

f

(x.(y, t), t)p(y, T) dYy,

&"

&',

where x., defined in Exercise 9.1, is the motion relative to time 4.

T.

Prove Kelvin's theorem: Of all motions of a body which, at a given time t, (a) correspond to a given steady density p, (b) have;?lt assigned, (c) have v· n prescribed on iJ;?Io one with velocity the gradient of a potential yields the least kinetic energy at time t. More precisely, let f!It be a bounded regular region of space, let p > 0 be a smooth scalar field on f!It, let A. be a scalar field on iJf!It, let d be the set of all smooth vector fields v on f!It that satisfy div(pv) = 0

in

f!It,

v • n = A. on

and define the kinetic energy % on d by

%{v}

=

L~

pdV.

Show that if v E d has the form v

= grad cp,

iJf!It,

92

IV.

MASS. MOMENTUM

then

X{v}

X{g}

~

for all g E d, with equality holding only when v = g.

13. LINEAR AND ANGULAR MOMENTUM. CENTER OF MASS Let x be a motion of fll. Given a part 9 of fll, the linear momentum 1(9, t) and the angular momentum a(9, t) (about the origin 0) of 9 at time tare defined by 1(9, t) =

f f

vp dV,

@,

a(9, t) =

(1)

r x vp dV,

@,

where r: Iff ~

r

is the position vector

=x -

r(x)

Proposition.

(2)

o.

For every part 9 and time t,

=

i(9, t)

f f

vp dV,

@,

a(9, t) =

(3)

r x vp dV.

@,

Proof

The identity (12.8) yields (3)1 and a(9, t) =

f

(r x v)"p dV.

@,

But by (8.6),

(r x v)"

=r x

and, since v x v = 0, (3}z follows.

v+ v x

v,

0

Assume now that fJ8 is bounded, so that m(fll) is finite. Then the center of mass e(r) at time t is the point of space defined by e(r) -

0 =

m(~)

L,r p dV.

(4)

13. LINEAR AND ANGULAR MOMENTUM. CENTER OF MASS

93

(It is not difficult to show that this definition is independent of the choice of origin 0.) If we differentiate (4) with respect to t and use (12.8) and (8.6), we

find that

m(~) LtVP dV,

ci(t) =

so that ci represents the average velocity of the body. This result and (1)1 imply that l(gjI, t) = m(gjI)ci(t).

(5)

Thus the linear momentum of a body gjI is the same as that of a particle of mass m( gjI) attached to the center ofmass of gjI.

EXERCISES

In these exercises :14 is bounded. 1.

Show that IX(t) - z = _1_ m(gjI)

f

(x - z)p(x, t) dV,.,

(6)

ga,

for any point z, so that IX(t) is independent of the choice of origin. 2. Other types of momenta of interest are the angular momentum a.(t) relative to a moving point z(t) and the spin angular momentum aspin(t):

asPin(t)

=

f

(7)

r" x V"P dV.

gat

.

Here rz(x, t) = x - z(t)

(8)

is the position vector from z(t), r,,(x, t) is the position vector from IX(t), and V" =

i" =

V -

ci

is the velocity relative to IX. For convenience, let I(t) = l(gjI, t) and a(t) = a(gjI, t). Show that a z = aspin a

=

aspin

+ (IX

- z) x I,

+ (IX - 0) x i.

94

IV.

MASS. MOMENTUM

The term «x - z) x I is usually referred to as the orbital angular momentum about z; it represents the angular momentum /JI would have if all of its mass were concentrated at the center of mass. 3. Consider a rigid motion. By (9.17),

Xo(y, t)

=

q(t)

+ Q(t)(y

- z)

(9)

for all y E /JI o and all t. (Here X o is x, at r = 0; i.e., X o is the motion taking the configuration at time r = 0 as reference.) (a)

Show that

= q(t) + Q(t)[(X(O) - z]

(X(t)

(10)

and hence [noting that xo(y, t) is well defined for all y E C]

!X(t) = xo((X(O), t). What is the meaning of this result? (b)

Show that the angular velocity co(t) is the axial vector ofQ(t)Q(t)T and that

v.. = co x r... (c)

A vector function k on

~

(11)

rotates with the body if k(t) = Q(t)k(O)

(12)

for all t. [Note that Q(O) = I, since xo(y, 0) = y for all y.] Show that k rotates with the body if and only if

k= (d)

co x k.

(13)

Use the identity

f x (d x f)

=

(f 21 - f ® f)d

to show that

as p in

=

Jco,

where J(t) =

(e)

fBI,(r;1 -

r.. ® r..)p dV

(14)

is the inertia tensor of /JI t relative to the center of mass. Show that J(t) = Q(t)J(O)Q(t)T,

(15)

and use this fact to prove that the matrix [J(t)] of J(t)-relative to any orthonormal basis {e;(t)} that rotates with the body-is independent of t.

95

SELECTED REFERENCE

(f)

Construct an orthonormal basis {ei(t)} that rotates with the body and has J1

[J] = [

0

o

J2

0

0] O.

0

J3

In this case {ei(t)} is called a principal basis and the corresponding numbers J; are called moments of inertia. Let w;(t) denote the components of ro(t) with respect to {e;(t)}. Show that the components of aspin(t) relative to this basis are given by

+ (J 3 J 2 W2 + (J 1 -

(aspin)! = J l w!

J 2 )W 2 W 3 ,

(a.Pin) 2 =

J 3)WI W3,

(a.Pinh

= J 3 W3 +

(J 2

-

J l )W I W 2 '

4. Define the kinetic energy S(t) and the relative kinetic energy S ,,(t) by

S ,,(t)

=

t Jr ,v; DI

(a)

Prove Konig's theorem: S = SOl

(b)

+ tm(81)ci2.

Consider a rigid motion. Use the identity (d x £)2

= d· (£21 - f

(8) £)d

to prove that

5. Show that

J = (tr M)I - M, where M(t) =

JrDI, f" (8) f"P dV

is called the Euler tensor.

SELECTED REFERENCE

Truesdell and Toupin [1, §§155-171].

p dV.

This page intentionally left blank

CHAPTER

v Force

14. FORCE. STRESS. BALANCE OF MOMENTUM During a motion mechanical interactions between parts of a body or between a body and its environment are described by forces. Here we shall be concerned with three types offorces: (i) contact forces between separate parts of a body; (ii) contact forces exerted on the boundary of a body by its environment; (iii) body forces exerted on the interior points of a body by the environment. One of the most important and far reaching axioms in continuum mechanics is Cauchy's hypothesis concerning the form of the contact forces. Cauchy assumed 1 the existence of a surface force density s(n, x, t) defined for each unit vector n and every (x, t) in the trajectory ff of the motion (Fig. 1). This field has the following property: Let!/' be an oriented surface in !!4t with positive unit normal n at x. Then s(n, x, t) is the force, per unit area, exerted across!/' upon the material on the negative side of!/' by the material on the positive side. Cauchy's hypothesis is quite strong; indeed, if ~ is an oriented surface tangent to !/' at x and having the same positive unit normal there, then the force per unit area at x is the same on ~ as on !/' (Fig. 2). 1 Noll [3) has shown that Cauchy's hypothesis actually follows from balance of linear momentum under very general assumptions concerning the form of the surface force s. (Cf. also Truesdell [I, p. 136); Gurtin and Williams [1).)

97

98

V.

FORCE

s(n, x, I)

+

Figure 2

Figure I

To determine the contact force between two separate parts rY' and!!} at time t (Fig. 3) one simply integrates s over the surface of contact

!/t = rY't n !!}t; thus

f

s(o",

X,

t) dA"

==

Sf',

f

s(o) dA

Sf',

gives the force exerted on rY' by!!} at time t. Here 0" is the outward unit normal to orY't at x. For points on the boundary of !!It, s(o, X, t)-with n the outward unit normal to o!!lt at x-gives the surface force, per unit area, applied to the body by the environment. This force is usually referred to as the surface traction. In any case, given a part rY',

i

s(o) dA

iJ~,

represents! the total contact force exerted on rY' at time t (Fig. 4). The environment can also exert forces on interior points of 14, a classical example being the force field due to gravity. Such forces are determined by a vector field b on .r; b(x, t) gives the force, per unit volume, exerted by the environment on x. Thus for any part rY' the integral

f

J~,

b(x, t) dV" ==

i

~,

b dV

gives that part of the environmental force on rY' not due to contact. 1 Here it is tacit that, given any part [7; and time t, 5(n., x, r) is an integrable function ofx on of!l'" This assumption of integrability actually follows from the momentum balance laws (I) and our hypotheses (i) and (ii) concerning force systems, Indeed, (i) trivially implies integrability on surfaces of polyhedra, and this is all that is needed to establish the existence of the stress T; (9) and (11) then yield integrability on of!l', for any part [7;,

14.

FORCE. STRESS. 8ALANCE OF MOMENTUM

Figure 3

99

Figure 4

The above discussion should motivate the following definitions. Let AI be the set of all unit vectors. By a system of forces for 9B during a motion (with trajectory ff) we mean a pair (s, b) of functions

b: ff

s: AI x ff ~ "f/,

with (i) (ii)

"//,

~

s(n, x, t), for each n E .K and t, a smooth function of x on 9B 1 ; b(x, t), for each t, a continuous function of x on 9B t •

We call s the surface force and b the body force; and we define the force f(&>, t) and moment m(&>, t) (about 0) on a part &> at time t by f(&>, t) =

i

s(n) dA +

i

r x s(n) dA +

ofJ',

m(&>, t) =

f

b dV,

fJ',

o!¥.,

f

r x b dV.

fJ',

Here n is the outward unit normal to iJ&>t and r is the position vector (13.2). The basic axioms connecting motion and force are the momentum balance laws. These laws assert that for every part &>and time t, f(&>, t)

=

l(&>, t),

(1)

m(&>, t) = a(&>, t); they express, respectively, balance of linear momentum and balance of angular momentum [cf. (13.1)J. Tacit in the statement of axiom (1) is the existence of an observer (frame of reference) relative to which the motion and forces are measured. The existence of such an observer is nontrivial, since the momentum balance laws are generally not invariant under changes in observer (cf. Exercise 21.3). Observers relative to which (1) hold are called inertial; in practice the fixed stars are often used to define the class of inertial observers.

v.

100

FORCE

An obvious consequence of (1)1 and (13.5) is that f(PA, t) = m(PA)iX(t),

provided PA is bounded. Thus the totalforce on afinite body is equal to the mass of the body times the acceleration of its mass center. By virtue of (13.3) the laws of momentum balance can be written as follows:

r

Jo&'t

sen) dA

f

+

fP,

f =f

b dV =

r r x sen) dA + ffP,r x b dV J0&',

vp dV,

e,

fP,

(2)

r x vp dV.

If we introduce the total body force

b* = b - pv,

(3)

which includes the inertial body force - pv, and define

f*(&" t) = m*(&', t) =

r

J ofP,

r

JofP,

sen) dA

+

f

r x sen) dA

fP,

b* dV,

+

f

fP,

(4)

r x b* dV,

then (1) takes the simple form

f*(&" t) = 0,

(5)

Our next result gives a far less trivial characterization of the momentum balance laws. Recall that an infinltesimal rigid displacement (ofS) is a mapping w: S -+ 1/ of the form w(x) = wo

+ W(x

- 0)

(6)

with Wo a vector and Wa skew tensor [cf. (7.8)]. Theorem of Virtual Work. Let (s, b) be a system offorces for PA during a motion. Then a necessary and sufficient condition that the momentum balance laws be satisfied is that given any part &' and time t,

r

s(n)· w dA +

J OfPt

f

fP,

b*· w dV = 0

(7)

for every infinitesimal rigid displacement w. Proof

Note first that by (13.2) we can write (6) in the form w

=

Wo

+ ro x r,

(8)

14.

101

FORCE. STRESS. BALANCE OF MOMENTUM

with co the axial vector corresponding to W. Let cp(wo, co) denote the left side of (7) with w given by (8). Then cp(wo, co) involves integrals of the fields s • w = s· W o

+ s . (co

x r),

where we have written s for s(n), and since

k·(co x r)

=

co·(r x k),

it follows that

= h, . w = s· w

Wo • S + m- (r x s), Wo • b; + co· (r x b*).

Thus by (4), cp(wo, co) =

t) + co· m*(&', t),

W o • f*(&"

and cp( Wo, co) = 0 for all vectors

Wo

and co if and only if (5) hold.

0

The next theorem is one of the central results of continuum mechanics. Its main assertion is that s(n) is linear in n. Cauchy's Theorem (Existence of stress). Let (s, b) be a system offorcesfor f1l during a motion. Then a necessary and sufficient condition that the momentum balance laws be satisfied is that there exist a spatial tensor field T (called the Cauchy stress) such that (a) for each unit vector n,

s(n) = Tn; (b) (c)

(9)

T is symmetric; T satisfies the equation of motion

+b=

div T

Proof

pt.

(10)

In stating condition (c) we do not assume explicitly that T(x, t) is smooth in x,

(11)

so that, a priori, it is not clear whether or not (c) makes sense. We therefore begin our proof by showing that (11) is a direct consequence of (9). Indeed, by (9), for an orthonormal basis {eJ,

L s(ei) e e, = L (Tei) ® ei; i

i

thus, in view of Exercise 1.6c, T(x, t) =

L s(e;, x, r) ® e.,

and (11) follows from (12) and property (i) of force systems.

(12)

v.

102

FORCE

We are now in a position to establish the necessity and sufficiency of (a)-(c). N ecessit y. Assume that (1) are satisfied. For convenience, we fix the time t and suppress it as an argument in most of what follows. The proof will proceed in a number of steps. Assertion I. Given any x unit vector k with

E [fBI>

any orthonormal basis {e.], and any (i = 1,2,3),

(13)

if follows that

s(k, x) = -

L (k· ei)s( -ei' x).

(14)

i

Proof First let x belong to the interior of [fBr' Let b > 0 and consider the tetrahedron '§s with the following properties: the faces of '§~ are 9' ~, 9' 1s» 9' 2~, and 9' 3~' where k and - e, are the outward unit normals to (J'§s on 9's and 9'i~' respectively; the vertex opposite to 9'~ is x; the distance from x to 9'~ is b (Fig. 5). Clearly, '§~ is contained in the interior of [fBI for all sufficiently small say 0 s 00 . Next, b*(x, t), defined by (3), is continuous in x, since b(x, r), p(x, t), and v(x, t) have this property; hence b*(·, t) is bounded on '§~o (for t fixed). Thus if we apply (5) 1 to the part f!jJ which occupies the region '§s at time t, and use (4) 1, we conclude that

s,

IL,,,s(n) dA I :s; "vol(,§~) for all 0 :s; 00 , where" is independent of

o.

~----~~.e2

Figure 5

(15)

14.

FORCE. STRESS. BALANCE OF MOMENTUM

103

Let A(15) denote the area of [1'6' Since A(15) is a constant times 15 2 , while vol(~6) is a constant times 15 3 , we may conclude from (15) that

A~15) Lf~(O) as 15

-+

dA

-+

0

s(k) dA +

L:

f

O. But

f

JiJ~6

f9'6

s(o) dA =

i

s( -ei) dA,

9"6

and, since s(o, x) is continuous in x for each fixed

0,

A~15) f9'6S(k) dA -+ s(k, x), A~15)

L'6

S(

-e;) dA

-+

(k· ei)s( -ei> x),

where we have used the fact that the area of [l'i6 is A(15)k • ei' Combining the above relations we conclude that (14) holds as long as x lies in the interior of fIl t • But x 1-+ s(o, x) is continuous on f!4 t • Thus, by continuity, (14) must hold everywhere on f!4 t • Assertion 2 (Newton's law of action and reaction). map 01-+ s(o, x) is continuous on .AI and satisfies

=

s(o, x)

For each x E f!4 t the

-s( -0, x),

(16)

Proof Since the right side of(14) is a continuous function ofk, s(k, x) is continuous on the set of all unit vectors k consistent with (13). Thus, since this is true for every choice of orthonormal basis {e.},s(k, x) must be a continuous function ofk everywhere on JV: Therefore, ifin (14) we let k -+ e 1, we arrive at s(e 1, x) = -s( -e 1 , x), and again, since the basis rei} is arbitrary, (16) must hold. Assertion 3.

Given any x E f!4 t and any orthonormal basis {eJ, s(k, x)

=

L: (k· ei)s(ei, x)

(17)

i

for all unit vectors k. Proof Choose an orthonormal basis {e.} and a unit vector k that does not lie in a coordinate plane (that is, in a plane spanned by two e.), Then there is no i such that k· e, = 0, and we can define a new orthonormal basis rei} by ei

=

[sgn(k· ei)]ei'

104

V.

Then k ei > 0 for i the basis {eJ yields 0

s(k, x) = -

I

=

FORCE

1,2,3, and Assertion 1 together with (16) applied to

I

(k i\)s( -ei, x) =. 0

i

(k ei)s(ei' x) = 0

i

I

(k e;)s(ei , x). 0

i

Thus (17) holds as long as k does not lie in a coordinate plane. But the map Ol--+s(o, x) is continuous on A: Thus (17) holds for all unit vectors k, and Assertion 3 is proved. Choose an orthonormal basis {ei} and let T(x, t) be defined by (12). Then T(x, t) is smooth in x, and, in view of (17), (9) holds. By (9), balance of linear momentum (2)1 takes the form

i

To dA +

a~

f

=

b dV

~

f vp

dV,

~

or equivalently, using the divergence theorem,

f

(divT

+b-

pv)dV = O.

~,

By the localization theorem, this relation can hold for every part & and time t only if the equation of motion (10) is satisfied. To complete the proof of necessity we have only to establish the symmetry ofT. Let w be any smooth vector field on !!Jr' We then conclude, with the aid of the divergence theorem and (4.2)5' that

i

ToowdA =

iJ~t

i f

f

(TTw)oodA =

iJ~t

=

div(TTw)dV

~t

(w div T 0

+ To grad w) dV.

~t

Thus, by (3), (9), and (10),

i

s(o) w dA 0

iJ~t

+

f

b* w dV 0

~t

=

f

To grad w dV.

(18)

e,

In particular, if we take w equal to the infinitesimal rigid displacement (6), then grad w = Wand (7) implies

f

ToW dV = 0

~t

for every part &, so that ToW

= O.

Since this result must hold for every skew tensor W, (f) of the proposition on page 6 yields the symmetry of T.

14.

FORCE. STRESS. BALANCE OF MOMENTUM

105

Tn

Figure 6

Sufficiency. Assume that there exists a symmetric spatial tensor field T consistent with (9) and (10). Clearly, (18) holds in the present circumstances. Since T is symmetric, and since the gradient of an infinitesimal rigid displacement is skew, when w is such a field the right side of(18) vanishes. We therefore conclude from the theorem of virtual work that the momentum balance laws hold. 0

Actually, one can show that (a) and (c) are equivalent to balance oflinear momentum (1)1; and that granted (1)1' the symmetry of T is equivalent to balance of angular momentum (1)2. Let T = T(x, t) be the stress at a particular place and time. If Tn

=

[n] = 1,

ern,

then a is a principal stress and n is a principal direction, so that principal stresses and principal directions are eigenvalues and eigenvectors ofT. Since T is symmetric, there exist three mutually perpendicular principal directions and three corresponding principal stresses. Consider an arbitrary oriented plane surface with positive unit normal n at x (Fig. 6). Then the surface force Tn can be decomposed into the sum of a normal force (0 • Tn)n

=

(n ® o)Tn

and a shearing force (I - n ® n)Tn, and it follows that n is a principal direction if and only if the corresponding shearing force vanishes. A fluid at rest is incapable of exerting shearing forces. In this instance Tn is parallel to n for each unit vector n, and every such vector is an eigenvector ofT. In view of the discussion given in the paragraph following the spectral theorem, T has only one characteristic space, 1/ itself, and (c) of the spectral theorem implies that

T =-nI

v.

106

FORCE

10I

t

I

I I

Pressure

Pure tension

t

t

-----Pure shear

t

Figure 7

with n a scalar. n is called the pressure of the fluid. Note that in this case the force (per unit area) on any surface in the fluid is -no. Two other important states of stress are: (a) Pure tension (or compression) with tensile stress a in the direction e, where lei = 1: T

=

rr(e ® e).

(b) Pure shear with shear stress, relative to the direction pair (k, 0), where k and 0 are orthogonal unit vectors: T = -r(k ®

0

+ 0 ® k),

The surface force fields corresponding to the above examples (with T constant) are shown in Fig. 7. EXERCISES

In Exercises 1, 4, and 8, fll is bounded. 1. The moment mit) about a moving point z(t) is defined by mz(t) =

r r, x s(o) dA + f r, x b dV,

JiJ9I,

91,

where r z is the position vector from z [cf. (13.8)]. (a) Let f(t) = f(fll, t). Show that, for y: ~ -+ $,

m, = Illy + (y - z) x f. (b) Let l(t) = l(fll, t). Show that [cf. (13.7)]

2. Prove that the momentum balance laws (1) hold for every choice of origin (constant in time) if they hold for one such choice.

14. FORCE. STRESS. BALANCE OF MOMENTUM

107

3. Show that the Cauchy stress is uniquely determined by the force system. 4. Consider a rigid motion with angular velocity roo Let {ej(t)} denote a principal basis of inertia and let J, denote the corresponding moments of inertia (relative to e) (cf. Exercise 13.3f). Further, let wlt) and mj(t) denote the components of ro(t) and m..(t) relative to {ei(t)}. Derive Euler's equations

m 1 = J 1Wl mz

=

m3 =

+ (J 3 -

JZ)WZ W3'

Jzwz + (J 1 - J 3)WIW3,

+

J 3 W3

(Jz -

Jd W I W Z '

These relations supplemented by f

m(91)li

=

constitute the basic equations of rigid body mechanics. When f and m.. are known they provide a system of nonlinear ordinary differential equations for ro and ~. 5. Let T be the stress at a particular place and time. Suppose that the corresponding surface force on a given plane !/ is perpendicular to g, while the surface force on any plane perpendicular to g vanishes. Show that T is a pure tension. 6. Let T be a pure shear. Compute the corresponding principal stresses and directions. 7. Prove directly that s(o, x) = -s( -0, x) by applying (2)1 to the part [JJ that occupies the region fYt o at time t. Here fYt o is the rectangular region which is centered at x, which has dimensions (j x (j X (jz, and which has n normal to the (j x (j faces (Fig. 8).

Figure 8

Ii

8. Prove Da Silva's theorem: At a given time the total moment on a body can be made to vanish by subjecting the surface and body forces to a suitable rotation; that is, there exists an orthogonal tensor Q such that

i

O~t

r x Qs(o) dA

+

f

~t

r x Qb dV = O.

V.

108

FORCE

Show in addition that

r

(QTr) x s(n) dA +

JoflB•

i

(QTr) x b dV = O.

flB,

What is the meaning of this relation? 9.

Suppose that at time t the surface traction vanishes on the boundary ofJB t • Show that at any x E afJBt the stress vector on any plane perpendicular to afJB t is tangent to the boundary.

15. CONSEQUENCES OF MOMENTUM BALANCE By Cauchy's theorem (and Exercise 14.3), to each force system consistent with the momentum balance laws there corresponds exactly one symmetric tensor field T consistent with (14.9) and (14.10). Conversely, the force system (s, b) is completely determined by the stress T and the motion x. Indeed, the surface force s is given by (14.9), while the body force b is easily computed using the equation of motion (14.10) [with p computed using (12.4)]: s(n) = Tn,

b=

pv -

div T.

This observation motivates the following definition. By a dynamical process we mean a pair (x, T) with . (a) (b) (c)

x a motion, T a symmetric tensor field on the trajectory ff of x, and T(x, t) a smooth function of x on fJB t •

Further, ifv and p are the velocity field and density corresponding to x, then the list (v, p, T) is called a flow. In view of the above discussion, to each force system consistent with the momentum balance laws there corresponds exactly one dynamical process (or equivalently, exactly one flow), and conversely. For the remainder ofthis section (v, p, T) is a flow and (s, b) is the associated force system. Theorem (Balance of momentum for a control volume). volume at time t. Then at that time

Let 9i be a control

r s(n) dA + Jutrb dV = dd Jutrvp dV + Jrout(pv)v' n dA,

Jout

t

(1)

r r x s(n) dA + Jutrr x b dV = ddt Jutrr x vp dV + Jfoutr x (pv)v' n dA.

Jout

109

15. CONSEQUENCES OF MOMENTUM BALANCE

Figure 9

We will prove only (1)\. First of all, by (4.2)4' (8.5), and (12.5h,

Proof

pv

pv' + p(grad v)v = (pv)' - p'v + (grad v)(pv) = (pv)' + v div(pv) + (grad v)(pv) = (pV)' + div(v ® pv).

=

Thus, since

L

div T dV = 191Tn dA = 191 s(n) dA,

L(pv)'

f

91

div(v ® pv) dV =

dV =

:t Lpv

dV,

i~

..

!::f; -,

r (v ® pv)n dA = JriJ91(pv)v· n dA,

JiJ91

if we integrate the equation of motion (14.10) over f7l we arrive at (1k

D

Equation (1)\ asserts that the totalforce on the control volume f7l is equal to the rate at which the linear momentum ofthe material in f7l is increasing plus the rate ofoutflow ofmomentum across of7l. Equation (1h has an analogous interpretation. A flow (v, p, T) is steady if fJl1 = fJlo for all t and Vi

= 0,

o' = 0,

T' = O.

In this case we call fJI0 the flowregion. Note that our definitions are consistent: in a steady flow the underlying motion is a steady motion (cf. Section 9). As an example of the last theorem consider the steady flow of a fluid through a curved pipe (Fig. 9), and let f7l be the control volume bounded by the pipe walls and the cross sections marking the ends ofthe pipe. Assume that the stress is a pressure and that the velocity, density, and pressure at the entrance and exit have the constant values and respectively, with e\ perpendicular to the entrance cross section and ez perpendicular to the exit cross section. Since the flow is steady,

~dt

f

91

vpdV = O.

v.

110

FORCE

Further, v- n must vanish at the pipe walls; therefore

i

(pv)v-ndA =

P2v~A2e2

- PlviA1e l,

iJEJt

where AI and A 2 are the areas of the entrance and exit cross sections, respectively. Let f denote the total force exerted by the fluid on the pipe walls. Then the total force on the control volume 91 is

and (1)1 yields

= (1t 1 + Plvi)A1e1

f

- (1t2

+ P2v~)A2e2.

A similar analysis, based on conservation of mass (12.6), tells us that P 1v1A 1 = P2V2A2 == em,

and hence f = (1tlAI

+ emv 1)e1

(1t2A2 + mv 2)e2.

-

Thus we have a simple expression for the force on the pipe walls in terms of conditions at the entrance and exit. Although the assumptions underlying this example are restrictive, they are, in fact, good approximations for a large class of physical situations.

Theorem of Power Expended. For every part

i

sen) - v dA

o<

+

f

b - v dV =

&It

f

(!J

and time t,

T - D dV + :

&It

t

f~

P dV,

(2)

&It

where

D

=

!(grad v + grad vT)

is the stretching. Proof

Since T is symmetric, (a) of the proposition on page 6 implies T - grad v = T - D.

Also, by (12.8),

so that (14.3) yields

f

&It

b. - v dV

=

f

&It

b - v dV -

~ dt

f

&It

2

v P dV.

2

15.

CONSEQUENCES OF MOMENTUM BALANCE

Equation (14.18) (with w identity. D

111

v) and the above relations yield the desired

=

The terms

f

~

p

dV

fT. D dV

and

~.

~.

are called, respectively, the kineticenergy and the stress power of fJ/ at time t. The theorem of power expended asserts that the power expended on fJ/ by the surface and body forces is equal to the stress power plus the rate of change of kinetic energy. A flow is potential if the velocity is the gradient of a potential: v = grad (fJ. Note that (fJ is a spatial field of class C 2 (because v is C 2 ) ; hence curl v = curl grad tp = 0 and potential flows are irrotational. Conversely, by the potential theorem (page 35), if a flow is irrotational, and if {fit is simply connected at some (and hence every) time t, then the flow is potential. For a body force b, the field b/p represents the body force per unit mass. We say that the body force is conservative with potential [3 if

b/p = -grad

[3.

(3)

If the flow is also steady the equation of motion (14.10) implies that b' = 0; in this case we will require that

[3' =

O.

Bernoulli's Theorem. Consider afiow (v, p, T) with stress a pressure -nI and body force conservative with potential [3. (a)

Ifthefiow is potential,

grad ((fJ'

(b)

grad n = O.

(4)

Ifthefiow is steady,

v. (c)

+ ~2 + [3) + ~

grad(~

+ [3) + ~

v . grad n

=

O.

(5)

Ifthefiow is steady and irrotational,

grad(~

+

[3) + ~

grad n = O.

(6)

112

V. FORCE

By hypothesis,

Proof

T = -nI, so that div T = -grad n; thus the equation of motion (14.10) takes the form - grad n + b = pv,

(7)

or equivalently, by (3),

v = - .!. grad n -

grad p.

p

(8)

On the other hand, by (9.10)1' for a potential (and consequently irrotational) flow

v = v' + ! gradtv")

=

grad ( q/ +

~2);

(9)

for a steady flow v•v= v•

grad(~)

(10)

(where we have used the fact that v· Wv = 0, since W is skew); for a steady, irrotationa1 flow

v=

grad(~}

(11)

The relations (9)-(11), when combined with (8), yield the desired results (4)-(6).

0 EXERCISES

1. Consider a statical situation in which a (bounded) body occupies the region fJlo for all time. Let b: fJlo -+ 1/ and T: fJlo -+ Sym with T smooth

satisfy divT + b = O. Define the mean stress T through vol(fJloyr =

r T dV.

J£to

113

SELECTED REFERENCES

(a)

(Signorini's theorem) Show that T is completely determined by the surface traction Tn and the body force b as follows:

vol(86'o)T (b)

=

io~o

(Tn ® r) dA +

r b ® r dV.

J~o

Assume that b = 0 and that iJ86'o consists of two closed surfaces Yo and Y 1 with Y 1 enclosing Yo (Fig. 10). Assume further that Yo and Y 1 are acted on by uniform pressures no and n1' so that sen) = -non

on

Yo,

sen) = -n1n

on

Y

1,

Figure 10

with no and n 1 constants. Show that T is a pressure of amount

where 00 and 01 are, respectively, the volumes enclosed by Yo and Y 1• 2. Prove (l)z. 3. Derive the following counterpart of (2) for a control volume 9l:

i

o~

s(n)· v dA

=

I

~

+

r b· v dV

J~

d T . D dV + -d

I

2

-v P dV t ~ 2

+

I

SELECTED REFERENCES

Gurtin and Martins [1]. Noll [3, 4]. Truesdell [1, Chapter 3]. Truesdell and Toupin [1, §§195-238].

o~

2

-pv

2

(v . n) dA.

This page intentionally left blank

CHAPTER

VI Constitutive Assumptions. Inviscid Fluids

16. CONSTITUTIVE ASSUMPTIONS The axioms laid down for force-namely the laws ofmomentum balanceare common to most bodies in nature. These laws, however, are insufficient to fully characterize the behavior of bodies because they do not distinguish between different types of materials. Physical experience has shown that two bodies of the same size and shape subjected to the same motion will generally not have the same resulting force distribution. For example, two thin wires of the same length and diameter, one of steel and one of copper, will require different forces to produce the same elongation. We therefore introduce additional hypotheses, called constitutive assumptions, which serve to distinguish different types of material behavior. Here we will consider three types of constitutive assumptions. (i) Constraints on the possible deformations the body may undergo. The simplest constraint is that only rigid motions be possible, a constraint which forms the basis of rigid body mechanics. Another example is the assumption of incompressibility, under which only isochoric deformations are permissible. This assumption is realistic for liquids such as water under normal flow 115

116

VI.

CONSTITUTIVE ASSUMPTIONS. IN VISCID FLUIDS

conditions and even describes the behavior of gases like air provided the velocities are not too large. (ii) Assumptions on the form of the stress tensor. The most widely used assumption of this type is that the stress be a pressure, an assumption valid for most fluids when viscous effects are negligible. (iii) Constitutive equations relating the stress to the motion. Here a classical example is the equation of state of a gas giving the pressure as a function of the density. We now make these ideas precise. A material body is a body f!J together with a mass distribution and a family C(J of dynamical processes. C(J is called the constitutive class of the body; it consists of those dynamical processes consistent with the constitutive assumptions of the body. A dynamical process (x, T) is isochoric if x(', t)

is an isochoric deformation

(1)

at each t; a material body is incompressible if each (x, T) E C(J is isochoric. By (1), given any part f!JJ,

vol(f!JJt ) = vol(f!JJ) for all t, which implies (10.4). Thus every motion of an incompressible body is isochoric. The restriction (1), however, implies more: not only does each motion preserve volume, but the volume of any part throughout the motion must be the same as its volume in the reference configuration. Further, in view of (6.17) and (c) of the theorem characterizing isochoric motions (page 78), every flow (v, p, T) of an incompressible body must have det F

= 1,

div v =

o.

(2)

Also, by (2)1 and (12.4), p(x, t) = Po(p) for x = x(p, t). Thus, in particular, when Po is constant,

in this case we refer to Po as the density of the body. A dynamical process (x, T) is Eulerian if the stress T is a pressure; that is, if

T = -1tI with n a scalar field on the trajectory of x.

16. CONSTITUTIVE ASSUMPTIONS

117

Figure 1

An ideal fluid is a material body consistent with the following two constitutive assumptions: (a) The constitutive class is the set of all isochoric, Eulerian dynamical processes. (b) The density Po is constant. Thus an ideal fluid is an incompressible material body for which the stress is a pressure in every flow. Note that for an ideal fluid the pressure is not determined uniquely by the motion: an infinite number of pressure fields correspond to each motion. This degree of flexibility in the pressure is common to all incompressible materials. That such a property is physically reasonable can be inferred from the following example. Consider a ball composed of an ideal fluid under a time-independent uniform pressure 11: (Fig. 0, and assume, for the moment, that all body forces, including gravity, are absent. Then the ball should remain in the same configuration for all time. Moreover, since the ball is incompressible, an increase or decrease in pressure should not result in a deformation. Thus the same "motion" should correspond to all uniform pressure fields. This argument is easily extended to nonuniform pressure fields and arbitrary motions, but for consistency the requirement of null body forces must be dropped to insure satisfaction of the equation of motion. We next introduce a material body, called an elastic fluid, in which compressibility effects are not ignored, and for which the pressure is completely specified by the motion. Here the constitutive class is defined by a smooth response function

giving the pressure when the density is known: 11:

= ft(p).

(3)

That is, the constitutive class is the set of all Eulerian dynamical processes (x, - nI) which obey the constitutive equation (3). We will consistently write (v, p, 11:) in place of (v, p, - nI)

118

VI.

CONSTITUTIVE ASSUMPTIONS. INVISCID FLUIDS

for the flow of an ideal fluid or of an elastic fluid. (Of course, P = Po in the former case.) An inviscid fluid is often characterized by the requirement that it be incapable of exerting shearing forces; in this sense both ideal and elastic fluids are inviscid.

EXERCISE

1. Consider a material body f:!4 with constitutive class C(j. A simple constraint for f:!4 is a function y: Lin" -+ ~

such that each dynamical process (x, T) E C(j satisfies y(F) = O.

(4)

For such a material one generally lays down the following constraint axiom: the stress is determined by the motion only to within a stress N that does no work in any motion consistent with the constraint. [The rate at which a stress N does work is given by the stress power, per unit volume, N'D (cf. page 111), where D is the stretching.] We now make this idea precise. Let PlJ be the set of all possible stretching tensors; that is, PlJ is the set of all tensors D with the following property: for some C 2 function F: ~ -+ Lin + consistent with (4), D is the symmetric part of L = F(t)F(t)-l

at some fixed time t. Let fJt = PlJ.1;

that is, fJt

= {NESymIN'D = 0 forall DEPlJ}.

Then the constraint axiom can be stated as follows: If (x, T) belongs to C(j, then so also does every dynamical process of the form (x, T + N) with N(x, t) E fJt for all (x, t) in the trajectory of x. We call fJt the reaction space.

17.

119

IDEAL FLUIDS

An incompressible material can be defined by the simple constraint y(F) = det F - 1.

(5)

Show that this constraint satisfies the constraint axiom if and only if the corresponding reaction space is the set of all tensors of the form - nI, n E IR; i.e., if and only if the stress is determined by the motion at most to within an arbitrary pressure field. Show further that the constitutive assumptions of an ideal fluid are consistent with the constraint axiom.

17. IDEAL FLUIDS Consider an ideal fluid with density Po. The basic equations are the equation of motion (15.7) and the constraint equation (16.2)z: -grad n

+b=

Pov,

(1)

div v = O. When the body force is conservative with potential

v = -grad(~

+

p, (15.8) implies

p),

and the acceleration is the gradient of a potential. We therefore have the following corollary of the results established in Section 11.

Theorem (Properties of ideal fluid motion).

The flow of an ideal fluid under a conservative body force has the following properties: (a) (b) (c)

If the flow is irrotational at one time, it is irrotational at all times. The flow preserves circulation. Vortex lines are transported with the fluid.

As we shall see, the equations of motion of a fluid are greatly simplified when the flow is assumed to be irrotational. The above theorem implies that the motion of an ideal fluid under a conservative body force is irrotational if the flow starts from a state of rest. This furnishes the usual justification for the assumption ofirrotationality. For a plane flow in an infinite region a much stronger assertion can be made. Indeed, if the flow is steady and in a uniform state at infinity [i.e., if grad vex) ~ 0 as x ~ 00], and if each streamline begins or ends at infinity, then, since W is constant on streamlines (cf. the proposition on page 81), the flow must necessarily be irrotational.

VI.

120

CONSTITUTIVE ASSUMPTIONS. INVISCID FLUIDS

Bernoulli's Theorem for IdealFluids, Let (v, Po, n) be a fiow ofan idea/fluid under a conservative body force with potential 13. (a)

If the flow is potential (v = grad cp),

+ ~ + ~ + fJ)

grad ( cp' (b)

=

o.

If the flow is steady, 2

+ ~ + 13)' =

(v

2

Po

0,

(2)

so that (v 212) + (nl Po) + fJ is constant on streamlines. (c) If the flow is steady and irrotational,

v2

n

2

Po

- + - + 13 = const

(3)

everywhere. Proof. Conclusion (a) follows from (15.4) (with P = Po). To establish (b) and (c) let

v2 2

n Po

1]' =

0,

v' grad

I'[ =

= - + - + 13·

I'[

For a steady flow

and by (15.5),

0;

hence ~ = 1]'

+ v ' grad I]

= 0,

which is (2). Finally, for a steady, irrotational flow (15.6) yields grad which, with

1]' =

0, implies (3).

I] =

0,

D

By Bernoulli's theorem, for a steady, irrotational flow under a conservative body force equations (1) reduce to

div v = 0, 2

v

-2

n

curl v = 0,

+ - + 13 = Po

(4)

const.

17.

121

IDEAL FLUIDS

These equations are also sufficient to characterize this type of flow. Indeed, suppose that (v, n) is a steady solution of (4) on a region of space flI o . Then, since W = 0, (9.10)1 implies

v=

v'

+ grad(~)

grad(~)

=

=

-grad(~

+ p),

and the basic equations (1) (with b/po = -grad p) are satisfied. In a steady motion the velocity is tangent to the boundary (cf. the proposition on page 68); thus (4) should be supplemented by the boundary condition

v'n=O

on

ofll o .

(5)

When the flow is nonsteady we are left with the system (1) to solve. The basic nonlinearity of these equations is obscured by the material time derivative in (1)1. Indeed, in terms of spatial operators these equations take the form v'

+ (grad v)v = div v

=

- grad n

+ b,

(6)

0,

where, for convenience, we have written n for n/po and b for b/po. Equations (6) are usually referred to as Euler's equations. EXERCISES

1.

Show that in the flow of an ideal fluid the stress power is zero (cf. page 111 ).

2. Consider the flow of a bounded ideal fluid under a conservative body force, and suppose that for each t and each x E ofllt> v(x, t) is tangent to ofll r • Show that

d -d t

f

v2 dV

= 0,

111,

so that the kinetic energy is constant. 3. Consider a homogeneous motion of the form

x(p, t)

= Po +

F(t) [p - Po]'

Show that if x represents a motion of an ideal fluid under conservative body forces, then F(t)F(t) -1 must be symmetric at each t. 4. Consider a steady, irrotational flow of an ideal fluid over an obstacle fJl (Fig. 2); that is, ~ is a bounded regular region whose interior lies outside the flow region flI o and whose boundary is a subset of ofll o . Assume that

122

VI.

CONSTITUTIVE ASSUMPTIONS. IN VISCID FLUIDS

Figure 2

the body force is zero. Show that the total force exerted on fJi by the fluid is equal to

18. STEADY, PLANE, IRROTATIONAL FLOW OF AN IDEAL FLUID For a steady, plane flow the velocity field has the form vex) =

Vt(Xt,

+ V2(Xt, x2)e2,

x2)e t

(1)

and the flow region can be identified with a region fJi in ~2. For convenience, we identify v and x with vectors in ~2 and write, in place of (1), vex) = (Vt(x),

vix»,

Then the basic equations (17.4) reduce to oV t oX t

+

oV2 _ OX2 -

0 ,

(2) 1t

!(vf + vD + - = const Po

with 1t also a field on fJi. Here, for convenience, we have assumed that the body force is zero. For the moment let {J

=

-V2'

18. STEADY, PLANE, IRROTATIONAL FLOW OF IDEAL FLUID

123

Then the first two relations in (2) are the Cauchy-Riemann equations:

orx Op OX2 = - OX 1; these are necessary and sufficient conditions that

g(z) = rx(X1' X2)

+ ip(X1' x 2)

be an analytic function of the complex variable

on r7l (considered as a region in the complex plane C). Thus we have the following

Consider a steady, plane, irrotational flow of an ideal fluid. Let C be defined by

Theorem.

g: ~

--+

(3)

Then g is an analytic function. Conversely, any analyticfunction g generates, in the sense of(3), a solution of(2)1.2' The function g is called the complex velocity. Let c = (C1' C2) be a curve in~. Then c can also be considered as a curve

in C. Given any complex function h on

i

h(z) dz

=

f

~

we write

h(e(u»c(u) do ;

in particular, we define r(c) = ig(Z) dz.

(4)

To interpret r(c) note first that k(u) = (C2(U),

-c 1(u»

is normal to c at c(u). Therefore when c is the boundary curve of quirement that v· n = 0 on o~ [cf. (17.5)] becomes

(5) ~

the re(6)

Thus in this instance

ig(Z)dZ = f(V 1 - iV2)(C1 + iC2)du = f(V 1C1 + v2c2)du,

124

VI.

CONSTITUTIVE ASSUMPTIONS. INVISCID FLUIDS

-ll1I

Figure 3

and

r(c) =

i

V(X) • dx

is the circulation around c. Next,

Ikl =

Jci + c~,

so that Ik(O") I do is the element of arc length along c. The vector f(c) = -

f

n(c(O"»k(O") da

(7)

therefore represents the integral along c of the surface force - no (0 = k/ Ik I) with respect to arc length. In fact, for a simple closed curve oriented counterclockwise as shown in Fig. 3, f(c) gives the total force (per unit length in the X3direction) on the material inside c. Blasius-Kutta-Joukowski Theorem. Consider a steady, plane, irrotational flow of an ideal fluid in a region qj whose boundary is a simple closed curve c. Let g be the complex velocity ofthe flow. Then

f

ipo "g(z) 2 dz. fl(C) - ific) = 2

(8)

If in addition qj is the region exterior to c and the velocity is uniform at infinity in the sense that g(z)

-+

a

as

z-+oo

(9)

with a real, then (10)

fl(c) = 0,

Proof

Since c is closed,

fkdO" = 0,

18. STEADY, PLANE, IRROTATJONAL FLOW OF IDEAL FLUID

125

and we conclude from (7) and the Bernoulli equation (2h that

= P°

f(e)

2

f

v2 k da.

Thus 1 fl(e) - if2(e) = Po I V2(C2 + ic 1) do = ipo I\2(C 1 - ic 2) do, 2 0 2 0

On the other hand, 192 dz =

f

(Vl - iV2)2(l\

+ ic2) da,

and a simple calculation using (6) yields (vi + V~)(Cl

- ic 2) = (Vl - iV2)2(C1 + ic 2)·

Thus (8) holds. Assume now that fJl is exterior to e and that (9) holds. Assume further that the origin 0 lies inside e. Then, since g is analytic and (9) holds, the Laurent expansion of g about the origin has the form g(z) =

U

+ -IXZ1 + -1XZ22 + ...

for any z E fJl. Hence (4) and Cauchy's theorem of residues imply F(c) = 2ni1X 1.

Next, since g is analytic, we can compute g2 by termwise multiplication. Thus 2 2 2UIX1 ( ) =u +--+ gz z

lXi + 2UIX2 Z2

+"',

and a second application of Cauchy's theorem tells us that 192 dz

= 4niuIX 1 = 2ur(e). 0

Therefore, in view of (8), (10) holds.

If fJl is simply connected, then g can be written as the derivative of an analytic function w: g

=

dw dz

126

VI.

CONSTITUTIVE ASSUMPTIONS. INVISCID FLUIDS

(This result extends to multiply connected regions provided one is willing to admit multivalued functions.) The function w: ~ -+ C is called the complex potential; the real and imaginary parts of w, denoted qJ and t/J, respectively, generate the velocity v through

Let s

=

(SI' S2) be a streamline so that

s(t)

= v(s(t».

(11)

Then

and t/J is constant on streamlines. For this reason t/J is called the stream function. Let c be a curve. Then c is essentially a segment of a streamline s if there is a smooth one-to-one mapping r of [0, IJ onto a closed interval of IR such that c(a) = s(-r(a»,

O~a~1.

(13)

Proposition. Consider a flow with a complex potential. Let c be a curve in the flow region and assume that v is nowhere zero on c. Then the following are equivalent: (a) (b) (c) (d)

v l(c(a»cia) = vic(a»c l(a)for 0 ~ a ~ 1. v(c(a» is parallel to c(a)for 0 ~ a ~ 1. c is essentially a segment of a streamline. t/J is constant on c; that is,for 0 ~ a ~ 1,

d da t/J(c(a» = O. Proof

We will show that (d) - (a) - (b) - (c).

(d) - (a).

This follows from the identity [cf. (12)J dd t/J(c(a» = -vic(a»c l(a) a

+ v l(c(a»cia).

(a) - (b). Here we use the fact that (a) and (b) are each equivalent to the assertion that k(a)· v(c(a» = 0 for 0 ~ a ~ 1, where k, defined by (5), is normal to the curve c.

18. STEADY, PLANE, IRROTATIONAL FLOW OF IDEAL FLUID

127

(c) => (b). If c is essentially a segment of a streamline s, then (11), (13), and the chain rule yield c(o) = s(r(u»t(u) = v(s("C(u»)t(u) = i(u)v(c(u»,

(14)

and (b) is satisfied. (b) => (c). Assume that (b) holds. Then, since v(c(u» and c(u) never vanish, c(u)

= tX(u)v(c(u»,

o s o s 1,

with tX a continuous function on [0, 1] which never vanishes. Let "C(u) =

J:

tX().) d)',

let s be the streamline which passes through c(O) at time t = 0, so that s satisfies (11) and s(O) = c(O); and let d be the curve in d(u)

[R2 defined

=

by

s("C(u»,

0 :s; a :s; 1.

(15)

Then d(O) = c(O) and, using the argument (14), ci(u) = tX(u)v(d(u»,

O:S;u:S;1.

Thus c and d satisfy the same differential equation and the same initial condition. We therefore conclude from the uniqueness theorem for ordinary differential equations that c = d, so that, by (15),c is essentially a segment of a streamline. 0 For problems involving flows about a stationary region one usually assumes, as the boundary condition, that the boundary of the region be a streamline. In view ofthe last proposition, this insures that the velocity field be tangent to the boundary, or equivalently, that the boundary be impenetrable to the fluid (cf. the proposition on page 68). As an example consider the region exterior to the unit circle and assume that the velocity is uniform and in the xrdirection at infinity, so that (9) is satisfied. Consider the flow generated by the complex potential

128

VI.

CONSTITUTIVE ASSUMPTIONS. INVISCID FLUIDS

Figure 4

In terms of cylindrical coordinates z = re" the stream function tf; has the form

tf;(r, e) = a(r - ~)

sin

e,

so that tf; has the constant value zero on the unit circle. Further, the complex velocity

satisfies (9). The streamlines are shown in Fig. 4. For this flow F(unit circle) = 0, and by the Blasius-Kutta-Joukowski theorem there is no net force on the boundary. Another flow with the desired properties is generated by w(z)

= a(z + ~ + iy log z).

This potential has a stream function

tf;(r, e) = a[(r - ~)

sin e

+ y log rJ'

which again vanishes on the circle r = 1. Further, w generates a complex velocity g(z)

= a

1 ( 1 - Z2

+ ~iY) ,

which satisfies (9). A point at which g(z) = 0 is called a stagnation point. Points z on the unit circle are given by the equation z = e". A necessary and sufficient condition that there exist a stagnation point at z = ei9 is that .

II

sm o = -

Y

2'

and hence that - 2 ~ Y ~ 2. Assume Y ~ O. Then: (a) for 0 ~ Y < 2 there are two stagnation points on the cylinder; (b) for Y = 2 there is one stagnation

18.

STEADY, PLANE, IRROTATIONAL FLOW OF IDEAL FLUID

129

(a)

Figure 5

point on the cylinder; (c) for y > 2 there are no stagnation points on the cylinder. These three cases are shown in Fig. 5 with the stagnation points denoted by S. For these flows r(unit circle) = - 2n«y provided the unit circle has a counter-clockwise orientation; thus there is no drag (force in the xI-direction), but there is a lift (force in the x2-direction) equal to 2npo«2 y. The problem of flow about an airfoil is far more complicated than the simple flow presented in Fig. 5,but the results are qualitatively the same. There the circulation is adjusted as shown in Fig. 6 to insure two stagnation points with the aft stagnation point coincident with the (sharp) trailing edge (KuttaJoukowski condition).

Figure 6

130

VI.

CONSTITUTIVE ASSUMPTIONS. INVISCID FLUIDS EXERCISES

1.

Show that the complex potential w(z) = _Ce-inpzP+l

represents the flow past a wedge (Fig. 7) of angle 21X, where C and {3 ({3 > -1) are real constants and {3n

IX

= 1 + {3'

Determine the stagnation points (if any).

Figure 7

2. Sketch the streamlines of the complex potential w(z)

=

Cz 2 ,

where C is a real constant, and show that the absolute value of the velocity is proportional to the distance from the origin. 3. Consider a steady, plane, irrotational flow of an ideal fluid. Let c be a curve in the flow region with

Let no be the pressure at a point at which the velocity is Vo' Show that the force (8) on c is given by .

fl(C) - if2(C) = -

Po (no + -2-o2) [(YA V

. YB) + I(XA - x B)]

IPO . +2

f.

cg

2

dz.

19. ELASTIC FLUIDS The basic equations for the flow of an elastic fluid are the equation of motion -grad n

+b=

pt,

(1)

ELASTIC FLUIDS

131

p' + div(pv) = 0,

(2)

19.

conservation of mass and the constitutive equation n

= ft(p).

(3)

We assume that ft has a strictly positive derivative, and we define functions " > 0 and e on ~ + by 2( ) _ dft(p) p - ----;[P'

tc

e(p)

=

r

,,2(0

p.

(4)

e:

(

where p* is an arbitrarily chosen value of the density. The function ,,(p) is called the sound speed; the reason for this terminology will become apparent when we discuss the acoustic equations. In a flow e(p) may be interpreted as a spatial field. We may therefore conclude from the chain rule that ,,2(p) 1 dft(p) 1 grad e(p) = - - grad p = - -d- grad p = - grad n. p

p

p

P

(5)

Thus when the body force is conservative with potential 13, (1) takes the form

v = - grad(e(p) + 13), and the acceleration is the gradient of a potential. We therefore have the following corollary of the results established in Section 11.

Theorem (Properties of elastic fluid motion).

The flow of an elastic fluid under a conservative body force has the following properties: (a) (b) (c)

If the flow is irrotational at one time, it is irrotational at all times. The flow preserves circulation. Vortex lines are transported with the fluid.

Thus, in particular, the flow of an elastic fluid under a conservative body force is irrotational provided the flow starts from rest.

Bernoulli's Theorem for Elastic Fluids. Let (v, p, n) be a flow of an elastic fluid under a conservative body force with potential 13. Let e be defined by (4). (a)

If the flow is potential,

grad(llJ'

+ ~ + e(p) + 13) = o.

132

VI. (b)

CONSTITUTIVE ASSUMPTIONS. INVISCID FLUIDS

If the flow is steady,

(~

+ e(p) +

p)" = 0,

(6)

so that (v 2 j2) + e(p) + p is constant on streamlines. (c) If the flow is steady and irrotational,

v2

2 + e(p) + p =

const

(7)

everywhere. Proof

Conclusion (a) follows from (15.4) and (5). Let '1 =

v2

2 + e(p) + p.

For a steady flow '1' = 0,

v • grad '1 = 0,

where we have used (15.5) and (5); hence ~ =

'1' + v • grad '1 = 0,

which is (6). Finally, for a steady, irrotational flow, (5) and (15.6) yield grad '1 = 0, which with '1' = implies (7). 0

°

If we define

then (1)-(3) take the form v'

+ (grad v)v + rt.(p) grad p = p' + div(pv) =

bjp,

(8)

0.

Equations (8) constitute a nonlinear system for p and v and are generally quite difficult to solve. Further, for many problems of interest the solution will not be smooth because of the appearance of shock waves (surfaces across which the velocity suffers a jump discontinuity), and the notion of a weak solution must be introduced. A careful discussion ofthese matters, however, is beyond the scope of this book. We now consider flows which are close to a given rest state with constant density Po. We therefore assume that Ip - PoI,Igrad p I, lvi, and Igrad vi are small.' Let [) be an upper bound for these fields. Then, since rt.(p) grad p = rt.(Po) grad p

+ [rt.(p)

- rt.(Po)] grad p,

I Since our interest lies only in deriving the asymptotic form of the field equations in the limit as (j ..... 0, it is not necessary to work with dimensionless quantities.

133

19. ELASTIC FLUIDS

and since rx. is continuous, we have, formally, rx.(p) grad p = rx.(Po) grad p

as <>

~

+ 0(<»

O. Similarly, div(pv) = Po div v + 0(<», (grad v)v = 0(15).

Thus if we assume b = 0 and neglect terms of 0(<» in (8), we arrive at the linear system v'

+ rx.(Po) grad

p = 0,

p' + Po div v

=

O.

These equations are usually called the acoustic equations; they constitute an approximate system of field equations appropriate to small departures from a state of rest. If we take the divergence of the first equation and the spatial time derivative of the second, and then eliminate the velocity, we obtain the classical wave equation p"

= K 2 (po)

!:ip,

where A = div grad is the spatial laplacian. Thus within the framework of the linearized acoustic equations disturbances about a rest state with density Po propagate with speed K(Po). This should motivate our use of the term "sound speed" for K(p). We now return to the general nonlinear system (8). The ratio v m=--

K(p)

of fluid speed v = Ivj

to sound speed is called the Mach number, and a flow is subsonic, sonic, or supersonic at (x, t) according as m(x, t) is < 1, = 1, or > 1. Consider a steady flow with b = O. Since p' = 0,

p = v· grad

p,

and (8)1 implies that K

V•

2

(p)

K

2

(p) .

V = - - - v •grad p = - - - p. p p

Thus v • (pv)' = v· (pv + pv) = p(v • v)(1 -

m

2

).

134

VI.

CONSTITUTIVE ASSUMPTIONS. INVISCID FLUIDS

Further, if we differentiate v 2 = v· v, we find that

vv = v· v. Similarly, v(pvr

= v· (pv)".

The last three identities yield the following Proposition.

In a steady flow of an elastic fluid under vanishing body forces (9)

whenever v #- O. Equation (9) demonstrates one of the chief qualitative differences between subsonic and supersonic flow. Consider an arbitrary streamline 0 (Fig. 8) and a point x on a. Then the quantity p(x)v(x) represents the mass flow, per unit area, across the plane through x perpendicular to a. For subsonic flow an increase in v along 0 is associated with an increase in mass flow; for supersonic flow an increase in v corresponds to a decrease in mass flow. Thus if v increases along 0 from subsonic to supersonic values, then the mass flow increases in the subsonic range to a maximum at m = 1, and subsequently decreases along the remainder of 0, which is supersonic. This explains, at least qualitatively, the difference between subsonic and supersonic nozzles. Consider the nozzles shown in Fig. 9. Assume that the density and velocity are constant over each cross section, so that the total mass flow across each such section is pvA, with A the corresponding area. (Although this assumption can be satisfied exactly only when A is constant, it is reasonable when A varies slowly with length.) By conservation of mass (cf. the anlysis on page 110),pvA is independent of position along the nozzle, and hence pv is inversely proportional to A. In the subsonic nozzle the flow is always in the subsonic range, and the decreasing area from entrance to exit results in an increasing mass flow and a concomitant increase in velocity. A nozzle which increases the velocity from subsonic to supersonic values cannot have this design; indeed, in the supersonic region a decreasing area would result in a decreasing velocity. A nozzle with the desired properties is the de Laval nozzle. There the velocity is subsonic at the entrance and increases with decreasing area to a value of Subsonic, 1''' increases

m= I

Supersonic. p" decreases

Figure 8. Streamline with v increasing.

19.

135

ELASTIC FLUIDS

~l

Y",<"c""U" U '('M

1

""

C''',"

I

I

I

Entrance

Exit

I

I

•

I

I

I

"""""""","'"

~>""""

Subsonic nozzle

~ I

I

Entrance

Exit

i

I

r';~

de Laval nozzle (subsonic-supersonic)

Figure 9

'1

= 1 at the throat (point of minimum area); the subsequent increasing area then accelerates the fluid into the supersonic range. By Bernoulli's theorem, when the flow is steady and irrotational, and when the body force is conservative,

'in

div(pv)

=

0,

curl v = 0, (10)

2

v

2 + e(p) + p = C =

const.

Of course, as for an ideal fluid, (10) should be supplemented by the boundary condition v•n = 0 on 0 fJI 0 • (11) Equations (10) also suffice to characterize this type of flow. Indeed, (10)1 implies (2), and if 1t is defined by (3), then (10h and (9.10)1 imply

v=

grad(~)

= -grad(e(p)

+ P),

which, by (5), implies (1). For a steady potential flow the basic equations reduce to a single nonlinear second-order equation for the velocity potential cpo To derive this equation note first that, by (10)1' v· grad p

+ p div v =

0,

(12)

and if we take the inner product of (8)1 with v and use (12) to eliminate the term involving grad p, we arrive at v • (grad v)v =

K

2

(p) div V.

136

VI.

CONSTITUTIVE ASSUMPTIONS. INVISCID FLUIDS

Here, for convenience, we have assumed that b = O. Thus, if (13)

v = grad cp, then cp satisfies the partial differential equation grad cp' (grad/ cp) grad cp

=

,,2(p) Acp,

(14)

where grad? = grad grad, or equivalently, in components,

L ocp ocp

02cp

=

,,2(p)

i.] OXi oX j OXi oX j

L 02~. i

OXi

Further, by (11) and (13), the corresponding boundary condition is

ocp = 0

on

00

:lf1]J

U"""o,

where

Ocp

- = 00

0°

grad cp

is the normal derivative on o[JIo' The sound speed K(p) in (14) can also be expressed as a function of grad cp by solving the Bernoulli equation

e(p)

C _ Igrad cpl2 2

=

[cf. (10h with P = 0] for p as a function of grad cp as follows. Since K > 0, (4h implies that dejdp > 0, so that e(p) is an invertible function of p ; we may therefore write p = e- 1 ( C

_Igra~

C(12)

and use this relation to express K(p) as a function

A(gradtp) K

=

K

(-l(C s -

Igrad2 CPI2))

(15)

of grad cpo Of Course C and hence R will generally vary from flow to flow. EXERCISES

1.

An ideal gas is an elastic fluid defined by a constitutive equation of the form (16)

137

SELECTED REFERENCES

where x > 0 and y > 1 are strictly positive constants. [Note that an ideal gas is not an ideal fluid. Actually, (16) represents the isentropic behavior of an ideal gas with constant specific heats.] (a)

Show jhat the sound speed K = K(p) is given by K

(b)

2

= yn/p.

Show that the relation (15) is given by

y- 1 j(2(grad qJ) = -2- (v 2 with v a constant. 2. Consider the ideal gas (16) in equilibrium (v body force

-

[grad qJ1 2 )

=

0) under the gravitational

b = -pge3' where g is the gravitational constant. Assume that n(x) = no = constant at

X3

= O.

Determine the pressure distribution as a function of height x 3 • 3. Show that for an elastic fluid the stress power of a part &> at time t is

f

T'DdV

=

~,

-f

ttbp di/

~,

(cr. page 111), where 1

v=P

is the specific volume (i.e., the volume per unit mass). SELECTED REFERENCES

Courant and Friedrichs [1]. Hughes and Marsden [1, §§9-11, 13, 14]. Lamb [1, Chapters 1-10]. Meyer [1, §6]. Serrin [1, §§15-18, 35,45,46].

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CHAPTER

VII Change in Observer. Invariance of Material Response

20. CHANGE IN OBSERVER Two observers viewing a moving body will generally record different motions for the body; in fact, the two recorded motions will differ by the rigid motion which represents the movement of one observer relative to the other. We now make this idea precise. Let x and x* be motions of fJI. Then x and x* are related by a change in observer if x*(p, t) = q(t) + Q(t) [x(p, r) - 0]

(1)

for every material point p and time t, where q(t) is a point of space and Q(t) is a rotation. That is, writing f(x, t)

= q(t) + Q(t)(x -

0),

(2)

then at each time t the deformation x*(., t) is simply the deformation x(', t) followed by the rigid deformation f~·, t): x*(', t) = f(·, t) x(', t). 0

139

(3)

140

VII.

OBSERVER CHANGE. INVARIANCE OF RESPONSE

o

X(o,'>j

f(o, I)

,,*(,/)

• Figure I

For convenience, we write !!Ai for the region of space occupied by !!A at time t in the motion x*: !!Ai = x*(!!A, t);

then the relation between x and x* is illustrated by Fig. 1. We now determine the manner.in which the various kinematical quantities transform under a change in observer. Letting

F

=

Vx,

F* = \lx*

and differentiating (1) with respect to p, we arrive at F*(p, t) = Q(t)F(p, r),

(4)

which is the transformation law for the deformation gradient. Note that, since det Q = 1, (4) implies det F*

=

det F.

(5)

Next, let F = RU = VR,

F* = R*U* = V*R*

be the polar decompositions ofF and F*. Then (4) implies F* = R*U* = QRU, and, since QR is a rotation, we conclude from the uniqueness of the polar decomposition that R* = QR, Also,

U* =

u.

141

20. CHANGE IN OBSERVER ~

91,

:/~I

Figure 2

and therefore, since Y

•

fJI~

f( -, I)

= RUR T , = QYQT.

y*

From these relations it follows that the Cauchy-Green strain tensors C = U 2, B = v-, C* = U*2, B* = y*2 transform according to C*

=

C,

Intuitively it is clear that if x E !!AI and x* E x*

=

!!A~

are related through

f(x, t),

(6)

then x and x* must correspond to the same material point. To verify that this is indeed the case, note that, by (3), x(-, t)

= f(·, t) - 1 x*(-, t), 0

which is easily inverted to give pC t)

=

p*C t) f(·, t), 0

where p and p* are the reference maps corresponding to the motions x and x*, respectively (Fig. 2). Thus p(x, t)

= p*(f(x, t), t),

or equivalently, by (6), p(x, t) = p*(x*, t),

(7)

so that x and x* correspond to the same material point. As a consequence of (1), x*(p, t) = q(t)

+ Q(t)x(p, t) + Q(t) [x(p, t)

- o].

If we take p = p*(x*, t) in the left side and p = p(x, t) in the right side of this relation-a substitution justified by (7) provided x* and x are related by (6)-we conclude that

v*(x*, t) = cj'
+ Q(t)v(x, t) + Q(t)(x

- 0),

(8)

142

VII.

OBSERVER CHANGE. INVARIANCE OF RESPONSE

where

are the spatial descriptions of the velocity in the two motions. Equation (8) is the transformation law for the velocity. Now let L = grad v,

L* = grad v*.

Ifwe take x* = f(x, t) in (8) and differentiate with respect to x using the chain rule and the fact that grad f = Q, we arrive at L*(x*, t)Q(t) = Q(t)L(x, t)

+ Q(t),

so that the velocity gradient transforms according to L*(x*, t)

=

Q(t)L(x, t)Q(t)T

+ Q(t)Q(t)T.

(9)

Since QQT = I, it follows that QQT

+ QQT

=

0,

or equivalently that QQT = _(QQT)T,

and the tensor QQT is skew. Thus the symmetric part of (9) is the following transformation law for the stretching: (10)

D*(x*, t) = Q(t)D(x, t)Q(t?, where

EXERCISES

1.

Show that the spin W transforms according to W*(x*, t)

2.

=

Q(t)W(x, t)Q(t?

+ Q(t)Q(t)T.

A spatial tensor field A is indifferent if, during any change in observer, A transforms according to A*(x*, t) = Q(t)A(x, t)Q(t)T. Assume that A is smooth and indifferent. (a)

Show that and t.

Ais indifferent if and only if A(x,

t) =

P(x, t)1 for all x

21. INVARIANCE UNDER A CHANGE IN OBSERVER

(b)

143

Show that the tensor fields

A= A-

+ AW, A= A + LTA + AL WA

o

(c)

6.

are indifferent. In the case of the stress T [cf. (21.1)], T and Tare called, respectively, the corotational and convected stress rates. Show that o

•

6.

•

W=W, W=W+DW+WD.

(11)

3. Use (9.14) to show that

C:(y*, t)

= Q(T)C.(y, t)Q(r?,

and use this result to show that the Rivlin-Ericksen tensors (9.18) are indifferent.

21. INVARIANCE UNDER A CHANGE IN OBSERVER One of the main axioms of mechanics is the requirement that material response be independent ofthe observer. This axiom is never stated explicitly in elementary books on mechanics, probably because it seems so obvious; nonetheless it is instructive to look at a simple example from that subject. A thin elastic string weighted at the end is spun with constant angular velocity and suffers an elongation of amount D. The axiom above asserts that the force in the string which produces this extension is the same as the force required to produce the elongation {) when the string is held in one place. Indeed, the two motions of the string differ only by a change in observer; that is, an observer spinning with the string sees the string undergoing the same motion as a "still observer" sees when he extends the string by hand. We now give a precise statement of this axiom within our general framework. To do this we must first deduce the manner in which we would expect the stress to transform under a change in observer. Thus let (x, T) and (x*, T*) be dynamical processes with x and x* related through (20.1). IfT and T* are also related through this change in observer, then the corresponding surface force fields should transform as shown in Fig. 3. Thus if

Q

n* = Qn,

=

we would expect that s*(n*)

=

Qs(n).

Q(t)

144

VII.

OBSERVER CHANGE. INVARIANCE OF RESPONSE

Figure 3

But s(n)

=

Tn,

s*(n*) = T*n*,

and hence T*n*

=

QTQT n*.

Since this relation should hold for every unit vector n*, we would expect the following transformation law for the stress tensor: T*(x*, t) = Q(t)T(x, t)Q(t)T.

(1)

Of course, in (1) it is understood that x* = f(x, t) with f given by (20.2). This discussion should motivate the following definition: (x, T) and (x*, T*) are related by a change in observer if there exist C3 functions q:IR-.S,

Q: IR -. Orth '

such that (a) the transformation law (20.1) holds for all p E fJI and t E IR; (b) the transformation law (1) holds for all (x, t) in the trajectory of x. We say that the response of a material body is independent of the observer provided its constitutive class ce has the following property: if a process (x, T) belongs to ce, then so does every dynamical process related to (x, T) by a change in observer. The content of this definition is, of course, the axiom discussed previously. Theorem.

The response ofideal fluids and elastic fluids are independent ofthe

observer.

The proof of this result makes use of the following lemma, which is a direct consequence of (16.2) and (20.5).

145

SELECTED REFERENCES

Lemma. A dynamical process related to an isochoric dynamical process by a change in observer is itself isochoric. Proofofthe Theorem. Let ~ be the constitutive class of an ideal fluid, let (x, T) E~, and let (x*, T*) be related to (x, T) by a change in observer. To show that (x*, T*)E~ we must show that (x*, T*) is isochoric and Eulerian. The former assertion follows from the lemma; to verify the latter note that, since T = - nI, (l) yields

T* = QTQT = _nQQT = -ni. Therefore ideal fluids are independent of the observer. The proof that elastic fluids also have this property proceeds in the same manner and is left as an exercise. D EXERCISES

1. Prove that the response of an elastic fluid is independent of the observer. In the next two exercises we consider the change in observer defined by (20.1). 2. Show that

div.; T*(x*, t) = Q(t) div, T(x, t). 3. Show that the acceleration transforms according to v*(x*, t)

=

Q(t)v(x, t)

+ q(t) + 2Q(t)v(x,

t)

+ Q(t)(x

- 0).

Show further that if the body force transforms according to b*(x*, t)

=

Q(t)b(x, t),

then div T*

+ b* + k

=

p*v*,

where k(x*, t) = p(x, t)[q(t)

+ 2Q(t)v(x, t) + Q(t)(x -

0)].

Because of the additional term k, the equation of motion is not invariant under all changes in observer. The observers for whom this equation transforms "properly" are exactly those with Q and q constant. Such observers are called Galilean and have the property of being accelerationless with respect to the underlying inertial observer. SELECTED REFERENCES

Noll [1, 21Truesdell and Noll [1, §§17-19J.

This page intentionally left blank

CHAPTER

VIII Newtonian Fluids. The Navier-Stokes Equations

22. NEWTONIAN FLUIDS Friction in fluids generally manifests itself through shearing forces which retard the relative motion of fluid particles. The fluids discussed thus far never exhibit shearing stress, and for this reason are incapable of describing frictional forces of this type. A measure of the relative motion of fluid particles is furnished by the velocity gradient L

= grad v,

a fact which motivates our considering constitutive equations of the form

T = -nI

+ CELl

(1)

Materials defined by relations of this type with C linear are called Newtonian fluids; they furnish the simplest-and most useful-model of viscous fluid behavior. Here we will use the term Newtonian fluid to mean incompressible Newtonian fluid; thus, since tr L = div v, we limit our discussion to fields L with tr L = O. 147

148

VIII.

NEWTONIAN FLUIDS. NAVIER-STOKES EQUATIONS

Since C is linear, C[O]

=

0; thus

T = -ni when L = 0, and a Newtonian fluid at rest behaves like an ideal fluid. Also, as for an ideal fluid, the pressure tt is arbitrary (i.e., not determined by the motion). This flexibility in n leads to a certain ambiguity regarding the response function C; given any linear, scalar-valued function [3(L) of L, we can rewrite (1) in the form T = - {n

+ [3(L)}I + Cp[L],

where Cp[L] = C[L]

+ [3(L)I.

Since n is arbitrary, n can absorb the term [3(L). Thus our constitutive equation (1) is essentially unaltered by replacing the term C[L] by Cp[L]. To remove this ambiguity, we normalize C by demanding that tr C[L] = O.

(2)

(This is equivalent to replacing C in (1) by C p with [3(L) = -~ tr C[L].) Then C has domain Lin, = {L E Linltr L = O} and codomain Sym, = {T E Sym Itr T = OJ.

In view of the restriction (2), the pressure n is uniquely determined by the stress. In fact, if we take the trace of (1) and use (2), we conclude that

n

=

-1trT.

We define the extra stress To by To = T + nI = T - 1(trT)I, so that To is the traceless part of the stress tensor. Our constitutive equation (1) then takes the simple form To

=

C[L].

(3)

The above discussion should motivate our next definition. A Newtonian fluid is an incompressible material body consistent with the following constitutive assumptions: (a)

There exists a linear response function

C: Lin,

--+

Sym.,

22.

NEWTONIAN FLUIDS

149

such that the constitutive class is the set of all isochoric dynamical processes (x, T) which obey the constitutive equation (3). (b) The density Po is constant. The dimensions of the spaces Lin., and Sym., are 8 and 5, respectively, so that the matrix of C relative to a coordinate frame has 40 entries. Thus at first sight it would appear that it .akes 40 independent constants to specify a Newtonian fluid. The next theorem shows that, in actuality, the response is determined by a single constant. Theorem. A necessary and sufficient condition that the response of a N ewtonian fluid be independent of the observer is that its response function C have the form (4)

for every L

E

Lin., , where

D

l(L + LT ) .

=

T he scalar constant Jl is called the viscosity of the fluid. Proof (Sufficiency) Assume that (4) holds. Let (x, T) belong to the constitutive class ~ of the fluid. Then (x, T) is isochoric and

To

=

2JlD.

Let (x*, T*) be related to (x, T) by a change in observer. Then (x*, T*) is isochoric (cf. the lemma on page 145). Moreover, by (20.10) and (21.1),

and (1.4h yields tr T* = tr(QTQT) = tr(TQTQ) = tr T. Therefore T6 = T* - ~tr =

Thus (x*, T*)

T*)I = QTQT - !{tr T)QQT = QToQT

Q(2JlD)QT = 2JlD*.

E ~

(5)

and the response is independent of the observer.

The proof of necessity is facilitated by the following Lemma.

Let L o E Ling. Then there exists a motion x with velocity gradient L= L o

and with x(', t) isochoric at each time t.

(6)

150 Proof

VIII.

NEWTONIAN FLUIDS. NAVIER-STOKES EQUATIONS

Take F(t)

=

eL o t

so that F is the unique solution of F(O) = I.

(7)

By (36.2), det F =

e(lrLo)t

= 1.

Thus

= q + F(t)[p -

x(p, t)

q]

defines a motion with deformation gradient F and with x(·, t) isochoric at each t. Further, (6) follows from (7)1 and the identity (8.8)1. We now return to the proof of the theorem. To establish the necessity of (4) we assume that the response is independent of the observer.

(8)

Let L o E Lin, be arbitrary, let x be the motion constructed in the lemma, and let T = To be the constant field defined by (3). Then, clearly, (x, T) E CC. Let (x*, T*) be related to (x, T) by a change in observer. Then by (8), (x*, T*) E CC and T~

= C[L*].

(9)

But T~ = QToQT,

L* = QLQT

+ QQT

[cf. (20.9), (21.1), and (5)]; hence (9) yields QToQT

C[QLQT

=

+ QQT],

and we conclude from (3) and (6) that QC[LO]QT

= C[QLoQT + QQT].

(10)

Clearly, this relation holds for every L o E Lin, (the domain of C) and every C 3 function Q: ~ --+ Orth +. Fix L o and take Q(t)

=

e- Wo t ,

where Wo

= 1(L o -

L~).

Then Q(t) is a rotation, since W 0 is skew, and Q(O) = I,

Q(O) = -Wo

22. NEWTONIAN FLUIDS

151

(cf. the discussion in Section 36). Using this function Q in (10) at t = 0 yields C[Lo] = C[Lo - W 0] = C[D o],

(11)

where Do

=

t(L o + L~).

Thus C is completely determined by its restriction to Symo. Next, let Q be a constant function with value in Orth +. Then (10) with L o = Do (Do E Syrnj) implies that (12)

Since this relation must hold for every Do E Sym., and every Q E Orth +, the restriction of C to Sym., is isotropic; we therefore conclude from the representation (37.26) that for all Do E Sym.,;

D

Note that the results (11) and (12) are valid also for nonlinear functions C; we appealed to the linearity of C only in the last step of the argument. By (4) the constitutive equation (1) takes the form

T = -7tI + 2.uD.

(13)

This equation must be supplemented by the equation of motion Po[v'

+ (grad v)v] = div T + b

and the incompressibility condition div v = O. In view of (4.5), 2 div D

= div(grad v + grad vT ) = dV + grad div v = dV,

where d = div grad is the spatial laplacian. Thus the above equations reduce to Po[v' + (grad v)v] = .u dv - grad n + b, (14) div v = O. These relations are the Navier-Stokes equations; given .u, Po, and b they constitute a nonlinear system of partial differential equations for the velocity v and the pressure n. If we define the kinematical viscosity v by

v = .u/Po and write no = n/po,

b o = b/po,

152

VIII.

NEWTONIAN FLUIDS. NAVIER-STOKES EQUATIONS

we can rewrite the Navier-Stokes equations in the form v' + (grad v)v = v ~v - grad 1to + b o,

(15)

div v = O.

If the flow is steady, and if we neglect the nonlinear term (grad v)v, then (15) reduce to v ~v = grad 1to - b o, div v = O. Solutions ofthis equation are called Stokes flows and are presumed to describe slow or creeping flows of Newtonian fluids. We now return to the general Navier-Stokes equations (15). One of the basic differences between flows described by these equations and flows of ideal or elastic fluids is the way circulation and spin are transported.

Theorem (Transport of spin and circulation).

Consider the flow of a Newtonian fluid under a conservative body force. Then

W + DW + WD

= v~W,

(16)

M(x, t) • dx.

(17)

and given any closed material curve c,

:t i t) .

dx

v(x,

=

v

Ct

i

Ct

If in addition the flow is plane,

w = v~W. Proof

(18)

Since the left side of (15)1 represents V, and since bo

v=

v ~v - grad(1to

= -

grad

/3,

+ /3).

If we substitute this equation into (11.5), we recover (17), since c is closed. Further, grad

v=

v ~ grad v - grad grad(1to

+ /3),

because ~ grad = grad ~. (Here and in what follows we need the additional assumption that, as functions of position, v is of class C 3 , while 1to and /3 are of class C 2 . ) If we take the skew part of this equation and use the fact that grad grad(1to + /3) is symmetric, we arrive at !
v-

grad

vT )

=

v ~W.

This equation and (11.2h imply (16). Finally, since div v = 0, (16) and (9.12) yield (18). 0

22.

153

NEWTONIAN FLUIDS

Both ideal and elastic fluids preserve circulation. In contrast, (17) tells us that circulation in a Newtonianfluid generally varies with time. Equation (18) asserts that in plane flow the spin satisfies a diffusion equation. Further, if in (16) we neglect terms of second order in grad v, we again arrive at (18), but here for flows which are not necessarily plane. This motivates the standard assertion that spin diffuses in a viscous fluid. Finally, note that by (20.11h we can rewrite (16) as 6.

W

=

V~W.

Suppose that the flow takes place in a region fJI. For the inviscid fluids considered previously the boundary condition was that v . 0 = 0 on afJI, which is simply the requirement that the boundary be impenetrable to the fluid. For a Newtonian fluid we add the restriction that the fluid adhere, without slipping, to the boundary. For a stationary boundary this means that v = 0 on afJI. If the boundary moves, then at each point on the boundary the fluid velocity must coincide with the velocity of the boundary. The theorem of power expended has an interesting consequence for Newtonian fluids. Indeed, by (13), T· D = -nl' D

+ 2J.lIDj2

= 2J.lIDI

2,

since I . D = tr D = div v = 0, and (15.2) yields the following

Theorem (Balance of energy for a viscous fluid). Newtonianfluid. Then

r s(o)· v dA + f b· v dV = :t f

Jo

@,

@,

@,

Consider the flow of a

v; Po dV + 2J.l

f

e,

2

IDI dV (19)

for every part 9.

The term

represents the rate at which the fluid in 9 dissipates energy. The energy equation (19) asserts that the total power expended on 9 must equal the rate of change of kinetic energy plus the rate of energy dissipation. Note that for a finite body, if v vanishes on afllt for all time, and if b == 0, then (assuming J.l > 0)

154

VIII. NEWTONIAN FLUIDS. NAVIER-STOKES EQUATIONS

so that the kinetic energy decreases with time. This result represents a type of stability inherent in viscous fluids; in Section 24 we will establish a stronger form of stability: we will show that under these conditions the kinetic energy actually decreases exponentially with time. It is instructive to write the Navier-Stokes equations in dimensionless form. Consider a solution of (15) with

b

O.

=

Let I and v be numbers with I a typical length (such as the diameter of a body) and v a typical velocity (such as a mainstream velocity). Further, identify points x with their position vectors x - 0 from a given origin 0, and define the dimensionless position vector

the dimensionless time tv

l

=1'

the dimensionless velocity

1 v(x, l) = - v(x, t), v

and the dimensionless pressure

Then grad

v=

I

-

v

[where grad v(x, l)

grad v,

I

Y'=2 V' , v

grad ito

=

I

2 v

.grad no

= V;;y(x, l), etc.], so that (15) reduces to v'

+ (grad v)V =

1 Re L\v - grad ito, (20)

div v = 0, where

Iv Re=v

is a dimensionless quantity called the Reynolds number of the flow.

155

22. NEWTONIAN FLUIDS

Equations (20) show that a solution of the Navier-Stokes equations with a given Reynolds number can be used to generate solutions which have different length and velocity scales, but the same Reynolds number. This fact allows one to model a given flow situation in the laboratory by adjusting the length and velocity scales and the viscosity to give an experimentally tractable problem with the same Reynolds number. Some typical values for the kinematical viscosity are! :

water: v = 1.004 air: v

=

15.05

X

10- 2

cm 2jsec,

X

10- 2

cm 2jsec.

EXERCISES

1. A Reiner-Rivlin jluid 2 is defined by the constitutive assumptions of the Newtonian fluid with the assumption of linearity removed. Use (37.15) to show that the response of a Reiner-Rivlin fluid is independent of the observer if and only if the constitutive equation has the form T = -nI

+ (l(o(.J"o)D + (l(l(.J"o)D 2

with (l(o(.J"o) and (l(l(.J"O) scalar functions of the list.J"o = (0, liD), of principal invariants of D.

2. Consider a Newtonian fluid in a fixed, bounded region assume that

v=o

on

§I

13 (D ))

of space and

G.ry(

for all time. (a)

Show that the rate at which the fluid dissipates energy, i.e.,

can be written in the alternative forms

indicating that all of the e,nergy dissipation is due to spin. At 20°C and atmospheric pressure. Cf., e.g., Bird, Stewart, and Lightfoot [I, p. 8]. There are better models of non-Newtonian fluids (cf. Truesdell and Noll [I, Chapter E] and Coleman, Markovitz, and Noll [I]). 1

2

156 (b)

VIII.

NEWTONIAN FLUIDS. NAVIER-STOKES EQUATIONS

The surface force exerted on expression

of7t by the fluid is given

s = no - 2JlWo = no

by the simple

+ JlO x curl v,

where 0 is the outward unit normal to for a plane portion of the boundary.

of7t.

Establish this relation

23. SOME SIMPLE SOLUTIONS FOR PLANE STEADY FLOW For steady flow in the absence ofbody forces the Navier-Stokes equations (22.15) reduce to po(grad v)v = Jl .1v - grad n, div v

= o.

(I)

Consider the plane velocity field v(x)

= Vl(X l, X 2)el'

(2)

Then (1)2 implies that

so that V l = grad v is

V 1(X2)

and v is a simple shear (see Fig. I). Further, the matrix of

(3)

and div v = 0,

(grad v)v =

o.

Therefore (I) reduces to

02V 1

on

ox~

OXl'

Jl-=-

(4)

23. SOME SIMPLE SOLUTIONS FOR PLANE STEADY FLOW

157

X2

Figure 1

We now consider two specific problems consistent with the above flow. Problem 1 (Flow between two plates). Consider the flow between two infinite flat plates, one at x 2 = 0 and one at Xz = h, with the bottom plate held stationary and the top plate moving in the x 1 direction with velocity v (Fig. 2). Then the boundary conditions are

v 1 (0) = 0,

v 1 (h) = v.

(5)

We assume in addition that there is no pressure drop in the xcdirection, so that 11: == const. Then (4)1 yields Vi

=

(X

+ 1h z ,

and this relation satisfies the boundary conditions (5) if and only if (X = 0 and

P = v/h. Thus the solution of our problem is Vi

= ox-fh.

(6)

In view of (22.13), (3), and (6), the stress T is the constant field [T]

=

-1 :[~ o ~ ~] 0

+ fl;'

1

[~ ~ ~];

0 0 0

v

Figure 2

158

VIII.

NEWTONIAN FLUIDS. NAVIER-STOKES EQUATIONS

that is,

TI 2

1l'U

= T2 1 = h'

Thus the force per unit area exerted by the fluid on the top plate has a normal component tt and a tangential (shearing) component - 1l'U/h. This fact furnishes a method of determining the viscosity of the fluid. Problem 2 (Flow between two fixed plates under a pressure gradient). Again we consider the above configuration, but now we hold the two plates fixed in space (Fig. 3). Then the relevant boundary conditions are (7) If the pressure were constant, then the solution of the first problem would tell us that the velocity is identically zero. We therefore allow a pressure drop; in particular, since v = V(X 2), (4)1 implies that

-<5

~= oX l

with b (the pressure drop per unit length) constant. For this case (4)1 has the solution bx~

VI

= - -

211

+

<X

+ f3X2'

which satisfies (7) if and only if <X = 0 and f3 = <5h/21l. Thus VI

bX 2

= -

211

(h -

x 2)

and the velocity distribution is parabolic. Further, we conclude from (22.13) that the nonzero components of the stress are T ll = T2 2 = T3 3 = -n,

TI 2 = T2 1 =

<5(~ -

X 2}

Thus the shearing stress is a maximum at the walls and vanishes at the center.

Figure 3

24.

159

UNIQUENESS AND STABILITY

The volume discharge Q per unit time through a cross section one unit deep is obtained by integrating V1(X l) from Xl = 0 to Xl = h; the result is Q = h3 b/12/l.

Since the discharge Q is easily measured, as is the pressure drop b, this formula yields a convenient method of determining the viscosity.

EXERCISES

1. Consider the flow of a Reiner-Rivlin fluid (cf. Exercise 1 of Section 22) between two flat plates. Show that the linear velocity profile (2), (6) remains a solution of the underlying equations with no constant. Show further that, in contrast to the linear theory, the normal stresses are no longer equal.

2. Consider the plane, steady flow of a Newtonian fluid of depth h down a flat surface inclined at an angle IX to the horizontal (Fig. 4). Thus the fluid flows under the influence of the gravitational force b = pog(e 1 sin

IX -

e l cos

IX),

and the boundary conditions are v = 0 at

Xl

= 0,

Tel = -

Cel

at

Xl

= h,

where C is the (constant) atmospheric pressure. Assume that the flow has the form (2). Determine the velocity and stress fields in the fluid.

Figure 4

24. UNIQUENESS AND STABILITY In this section we establish theorems of uniqueness and stability for the Navier-Stokes equations. Aside from their intrinsic interest, the proofs of these theorems utilize techniques which are important in their own right.

160

VIII.

NEWTONIAN FLUIDS. NAVIER-STOKES EQUATIONS

The classical viscous flow problem for the Navier-Stokes equations can be posed as follows:

Given: a bounded regular region ~, a kinematic viscosity v > 0, a body force field bon fJl x [0, 00), an initial velocity distribution Vo on ~, and a boundary velocity distribution von iJfJl x [0, 00). Find: a class C Z velocity field v and a smooth pressure field tt on ~ x [0, 00) that satisfy the Navier-Stokes equations Vi

+ (grad v)v =

v M - grad tt

+ b,

div v = 0,

(1)

the initial condition v(x,O)

for every x

E

= vo(x)

fJl, and the boundary condition v

=

v

iJfJl x [0, 00).

on

A pair (v, rr) with these properties will be called a solution. (Here, for convenience, we have written nand b for what we earlier called no and bo .) Uniqueness Theorem. Let viscous flow problem. Then

(VI'

n l ) and (vz, n z) be solutions of the (same)

with IY. spatially constant: grad

IY.

= O.

The proof of this theorem is based on the following Lemma. Let w be a smooth vector field on fJl, let {J be a smooth scalar field on fJ1/, and assume that div w =

°

and that at each point ofiJfJl either w = 0 or {J = 0. Then

J

w· grad {J d V

= 0.

."f

Proof

The proof follows from the identity div({Jw) = {J div w + w . grad {J,

the divergence theorem, and the hypotheses.

0

24.

Let

Proof of Theorem.

Then, since

(VI' 1[d

u(X,O)

and (vz, 1[z) are solutions,

= 0,

u

=

Further, subtracting (1)1 with we obtain the relation u'

161

UNIQUENESS AND STABILITY

+ (grad

o;]lt

0 on V

VI)V I -

= Vz , 1[ =

div u = 0.

x [0, (0),

1[z from (1)1 with V

=

VI' 1[

(2)

=

1[1'

(grad vz)vz = v ~u - grad a,

and, since

we have u' + (grad u)vI + (grad vz)u = v ~u - grad a.

(3)

U we take the inner product of (3) with u, and use the identities u·

~u

= div[(grad U)TU]

u· (grad u)vI =

VI •

(grad

U)TU

=

-

Igrad ulZ, VI •

grad(~z).

(4)

u . (grad vz)u = u • Du, where D = !(grad Vz + grad V1), we conclude that

!(u z), + VI

•

grad(~)

+ u· Du =

v div[(grad u)Tu] - vl grad ul z - u· grad

fl..

U we integrate this relation over ;]It, use the lemma twice (first with w = VI' f3 = U Z/2, then with w = u, f3 = a), and appeal to the divergence theorem, we conclude, with the aid of (2), that 1d 2dt [u]?

r u· DudV s 0,

+J

9f

where

(5)

162

VIII.

NEWTONIAN FLUIDS. NAVIER-STOKES EQUATIONS

Since div v = 0, tr D = 0; thus the lowest proper value of the symmetric tensor D(x, t) is sO. Let -y(x, t) denote this proper value, so that y ~ and

°

Now choose r >

°and let A. = 2 sup y(x, t). XE91

0" r s

(Since

V2

t

is smooth, D is continuous and

A. 2 u·Du> - --u 2

°s A. <

00.)

Thus

9i x [0, r]

on

and (5) becomes on

[0, r].

Therefore on

[0, r],

and hence

But by (2)1'

thus

°

IluI1 2(r) = and u(x, r)

=0

for every x E 9i. Since r was arbitrarily chosen, u == 0 and v1 = (3) implies that grad (X = O. 0

~2'

Finally,

StabilityTheorem. Consider the viscous flow problem with vanishing boundary data and conservative body forces:

v=

0

on

89i x [0, (0),

b = -grad

p.

(6)

Let (v, n) be the solution (if it exists) of this problem. Then there exists a constant A. > such that

°

(7)

163

24. UNIQUENESS AND STABILITY

Proof

In view of (6), (v, n) satisfies the system v'

+ (grad v)v = v dv - grad IX, div v = 0,

on

v = 0

v(x, 0)

(8)

i}fJl x [0, (0),

= vo(x),

where IX = n + {3. If we take the inner product of (8)1 with v and use the identities (4) 1, 2, we arrive at the relation t(v 2 )'

+ v· gradt« + tv 2 ) =

v div{(grad

V)T V} -

v Igrad v1 2 .

If we integrate this relation over fJl, we conclude, with the aid of the divergence theorem, (8h, 3, and the lemma, that

d

dt IIvl1

2

+ 2vllgrad vl1 2

where [grad vl1

2

=

= 0,

L

Igrad vl 2 dV.

By the Poincare inequality (Exercise 1) there exists a constant that

..1. 0

> 0 such

Thus

where A.

= VA. o , and hence

which integrates to give or equivalently, Ilvll(t) s Ilvll(O)e- l ' . In view of (8)4' this inequality clearly implies (7).

0

Remark. It should be emphasized that the preceding two theorems require that the solution be smooth for all time and therefore may be misleading, since there is no global regularity theorem for the Navier-Stokes equations. 1 1

cr. Ladyzhenskaya [I, §6]; Temam [I, Chapter III].

VIII.

164

NEWTONIAN FLUIDS. NAVIER~STOKES

EQUATIONS

EXERCISES

l.

Supply the details of the following proof of the Poincare inequality. Let be a smooth scalar field on fJf with cp = 0 on a~. Since fJf is bounded, cp can be extended to a box ~ (with sides parallel to coordinate planes) by defining cp = 0 on ~ - fJf. Suppose that for x E ~,x\ varies between o: and fJ. Define cp\ = acpjax\. Then tp

cp2(X)

= 2 flcp(e,X2,X3)cp\(e,X2,X3)de,

and hence

r cp2 dV = 2 J~ J",('cp(e, x 2, x 3)cp\(e, x 2, x 3) de dV

x

J!R

~

2

X

s

(L f (L f

cp2(e, x 2, x 3) de dVx ) \/2 CPT(e, x 2, X3) de dVx ) 1/2

2(fJ - rx)(L cp2 dV yl2 (L [grad cp

2 1

dV yl2

which implies the Poincare inequality

L cp2 dV

~

4(fJ - rx)2 L1grad cpl2 dV.

Show that this implies an analogous result for a smooth vector field v on fJf with v = 0 on afJf. 2. Establish (4). SELECTED REFERENCES

Bird, Stewart, and Lightfoot [l, Part I]. Hughes and Marsden [I, §§ 16, 17]. Ladyzhenskaya [I]. Lamb [I, Chapter I]. Noll [I]. Serrin [l, Chapter G]. Temam [I].

CHAPTER

IX Finite Elasticity

25. ELASTIC BODIES

In classical mechanics the force on an elastic spring depends only on the change in length of the spring; this force is independent of the past history of the length as well as the rate at which the length is changing with time. For a body the deformation gradient F measures local distance changes, and it therefore seems natural to define an elastic body as one whose constitutive equation gives the stress T(x, t) at x = x(p, r) when the deformation gradient F(p, t) is known: T(x, t)

= t(F(p, r), p).

(1)

Formally, then, an elastic body is a material body whose constitutive class C(j is defined by a smooth response function

t: Lin" x

fJ4 --+ Sym

as follows: C(j is the set of all dynamical processes (x, T) consistent with (1). Note that for an elastic body the function t is completely determined by the response of the material to time-independent homogeneous motions of the form x(p, r)

= q + F(p - Po);

that is, to homogeneous deformations. 165

(2)

IX.

166

FINITE ELASTICITY

When convenient, and when there is no danger of confusion, we will write T(F) in place of T(F, p). Proposition. A necessary and sufficient condition that the response of an elastic body be independent of the observer is that the response function t satisfy

QT(F)QT = T(QF)

(3)

for every F E Lin + and every Q E Orth + • Proof Choose FE Lin + arbitrarily, let x be the motion (2), and let T be defined by (1), so that (x, T) E ~, the constitutive class of the body. Assume that the response is independent of the observer. Then every (x*, T*) related to (x, T) by a change in observer must also belong to ~. By (20.4) and (21.1), this can happen only if, given any rotation Q,

QTQT

=

T(QF).

Since T = T(F), this clearly implies (3). Conversely, if (3) is satisfied, then, in view of (20.4) and (21.1), the response is independent of the observer. D We assume henceforth that the response is independent of the observer, so that (3) holds.

The importance of the strain tensors U and C is brought out by the next result, which gives alternative forms for the constitutive equation (1). Corollary (Reduced constitutive equations). The response function T is completely determined by its restriction to Psym; infact,

T(F) = RT(U)R T for every F E Lin +, where R E Orth + is the rotation tensor and U right stretch tensor corresponding to F; that is, F = RU is the decomposition ofF. Further, there exist smooth response functions from Psym into Sym such that

(4)

E Psym the right polar * T, - and T T,

T(F) = Ft(U)FT, T(F) = RT(C)R T ,

(5)

T(F) = FT(C)FT , with C

U2

FTF the right Cauchy-Green strain tensor corresponding to F. Proof To derive (4) we simply choose Q in (3) equal to R T. Next, since F = RU, (4) can be written in the form =

=

T(F) = FU- 1T(U)U- 1FT.

25. ELASTIC BODIES

167

T(U) = U- 1T(U)U- 1,

(6)

Thus if we define we are led to (5h. On the other hand, if we let T(C) = T(C 1 / 2 ) , T(C)

= T(C 1 / 2 ) ,

(7)

for C E Psym, then (4) and (5)1 imply (5h,3' Finally, by (6),.T is smooth, while (7) and the smoothness ofthe square root (page 23) yield the smoothness ofT and T. 0 The converse is also true: each of the constitutive equations in (4) and (5) is independent of the observer. We leave the proof of this assertion as an exercise. * T, - and -T depend also It is important to note that the response functions T, on the material point p under consideration. Suppose that we rotate a specimen of material and then perform an experiment upon it. If the outcome is the same as ifthe specimen had not been rotated, then the rotation is called a symmetry transformation. To fix ideas consider the two-dimensional example shown in Fig. 1; there an elastic ring is connected by identical mutually perpendicular elastic springs which meet at the center ofthe ring. The forces required to produce any given deformation are the same for any prerotation of the ring by a multiple of 90°, so that such rotations are symmetry transformations. Moreover, they are the only symmetry transformations, since any other prerotation can be detected by some subsequent deformation. We now apply these ideas to the general elastic body 114. Choose a point p of 114 and let fr denote the homogeneous deformation from p with deformation gradient F: f,(q) = p

+ F(q

- p)

for every q E 114. Consider two experiments: (1) Deform 114 with the homogeneous deformation f r . (2) Rotate 114 with the rotation fQ (Q E Orth +) about p and then deform the rotated body with the homogeneous deformation fF (Fig. 2). In this case the total deformation is fr ° fQ and the deformation gradient is FQ.

Figure 1

IX.

168

FINITE ELASTICITY

Figure 2

The stress at p is

T(F, p) in Experiment 1 and

T(FQ, p) in Experiment 2. If for Q fixed these two stress tensors coincide for each F, then Q is called a symmetry transformation; Q has the property that the response of the material at p is the same before and after the rotation f Q • Thus, more succinctly, a symmetry transformation at p is a tensor Q E Orth + such that

T(F, p)

=

T(FQ, p)

(8)

for every F E Lin ", We write <§p for the set of all symmetry transformations at p and call <§p the symmetry group at p. The next proposition shows that this terminology is consistent.

Proposition. Proof.

<§p

is a subgroup of Orth + .

Let us agree to write in place of

<§

<§p

whenever there is no danger of confusion. [Recall our previous agreement to write T(F) for T(F, p).] Clearly, <§ is a subset of Orth + and I E <§. To show that <§ is a subgroup of Orth + it therefore suffices to show that <§ is closed under both inversion and multiplication: Q E <§ => Q-l E <§,

Q, H Choose Q yields

E

<§ and

E <§ =>

QH

E <§.

FE Lin + arbitrarily. Then (8) with F replaced by FQ- 1

25.

so that Q-1

E '§.

169

ELASTIC BODIES

Next, choose F

E

Lin + and Q, H

E '§.

Then

T(F) = T(FQ) = T«FQ)H) = T(FQH) so that QH

D

E '§.

Our next result shows that (3) (which expresses invariance under observer changes) and (8) together imply the invariance ofT under '§ (cf. Section 37). "'" - * The response functions T, T, T, and T are invariant under '§. Thus, in particular,

Proposition.

QT(F)QT = T(QFQT), QT(C)QT for every Q Proof

E '§,

F

E

=

T(QCQT),

(9)

Lin +, and C E Psym.

Choose Q

E '§.

Then QT E

'§

and (8) implies

We therefore conclude, with the aid of (3), that (9)1 holds. To prove (9h choose C E Psym arbitrarily and let U be the unique square root ofC:

By (6) and (7h,

Thus, since (QUQT)2

=

QU 2QT = QCQT,

(QUQT)-1 = QU- 1QT, we conclude from (9)1 that QT(C)QT = QU- 1QTQT(U)QTQU- 1QT

= (QUQT)-1T(QUQT)(QUQT)-1 = T«QUQT)2) = T(QCQT). We leave as an exercise the proof of the assertions concerning t and We say that the material at p is isotropic if '§p =

Orth"

(so that every rotation is a symmetry transformation), anisotropic if '§p

=I- Orth + .

T. D

170

IX.

FINITE ELASTICITY

This definition, the last proposition, and the proposition on page 230 imply the following

Proposition. Assume that the material at p is isotropic. Then each of the response functions t, t, 1', and f (at p) is an isotropicfunction. The stress

T R = 1'(1) is called the residual stress at p; T R is the stress at p when the body is undeformed. Since C = U = R = I when F = I, (5) implies that

* = T(I) T R = T(I) = T(I). Proposition. Proof

(10)

If the material at p is isotropic, then T R is a pressure. Let Q

E

Orth. Then, since

QTRQT

l' is isotropic,

= QT(I)QT = T(QIQT) = 1'(1) = T R ,

or, equivalently, QT R = T R Q, so that T R commutes with every orthogonal tensor. We therefore conclude from the corollary on page 13 that T R = -ni with 1t a scalar. D By (6.7) the right and left stretch tensors U and V and the right and left Cauchy-Green strain tensors C and B are related by V = RURT ,

B = RCR T ,

where R E Orth + is the corresponding rotation tensor. Therefore when the material is isotropic, (4), (5)z, and the proposition containing (9) yield T(F) = RT(U)R T = T(RURT ) = T(V), T(F) = RT(C)RT

= T(RCRT ) =

T(B),

and thus the constitutive relation T = T(F) can be written in the alternative forms T = T(V), (11)

T = feB),

with l' and T isotropic functions. In view of(37.21), these comments have the following consequence:

Theorem (Constitutive equation for an isotropic material). Assume that the material at p is isotropic. Then the constitutive relation can be written in the form

+ PI(.FB)B + Pi.FB)B- I,

(12)

where B = FF is the left Cauchy-Green strain tensor and scalar functions ofthe list .F B of principal invariants ofB.

Po, PI' and pz are

T T

=

Po(.fB)1

25.

ELASTIC BODIES

171

Further, as is clear from (11) and (37.15), the constitutive equation (12) for an isotropic material point can also be expressed in the forms T = lXo(JB)I + IXt(fB)B + 1X2(fB)B2, T

=

Ko(fv)I

+ Kt(fv)V + K2(f v)V- t,

T = oo(fv)I + 0t(fv)V + 02(fv)V 2, where V = B t / 2 is the left stretch tensor. We now return to the general theory and list the complete system of field equations; this consists of the constitutive equation

T= C=

J?lr(<:))?T,

(13)

)?T)?,

the equation of motion divT

+b=

pt,

(14)

and balance of mass p det F = Po,

(15)

where Po is the density in the reference configuration. When the body is isotropic (13) may be replaced by T = Po(fB)I + Pt(fB)B + P2(fB)B- t, B = )?)?T. We say that the body is homogeneous provided both Po(p) and T(J?, p) are independent ofthe material point p. In this case each of the response functions, as well as the symmetry group I'§p, is independent of p. Consider the homogeneous deformation (2). For a homogeneous body the corresponding stress T = t(J?) is constant, because F is constant. Therefore T satisfies the equation of equilibrium div T = 0 and (x, T) is a solution of (13)-(15) with b = O. Thus a homogeneous body can be deformed homogeneously without body force. EXERCISES

1. Show that each of the constitutive equations in (5) is independent of the

observer. 2. Show that the response functions

t

and

f are invariant under I'§.

IX.

172

FINITE ELASTICITY

3.

Show that an elastic fluid is an isotropic elastic body.

4.

Define the extended symmetry group .Yep to be the set of all tensors A E Lin + such that

T(F, p) for all F (a) (b) (c)

E

= T(FA, p)

(16)

Lin +.

Show that .Yep is a subgroup of Lin +. Give a physical interpretation of .Yep. Let Unim denote the proper unimodular group; that is, Unim = {A E Lin" [det A

= I}.

Show that if.Yep is not contained in Unim, then there exist sequences {Fn} and {G n} with F n, G n E Lin+ such that (i) (ii)

det F, --. 00 but T(F n , p) is the same for all n; det G n --. 0 but T(G n , p) is the same for all n.

Why is this physically unreasonable?

For the remainder of this section we assume that Unim.

(17)

.Yep = Unim

(18)

.Yep (d)

c

Show that an elastic fluid has

for every p E fJl. (e)

Show that, conversely, an elastic body whose response function obeys (18) is an elastic fluid. [For bodies exhibiting more general types of behavior, such as viscoelasticity, the condition (18) furnishes a useful definition of a fluid.]

5. It is often convenient to use a deformed configuration as reference (see Fig. 3). With this in mind, let g be a deformation of fJl, let fbe a deformation of g(fJl), and let

G = Vg(p),

F = Vf(q),

q

= g(p).

Then (under fog) the material point p experiences the stress

T = T(V(f g)(p), p) 0

Let

=

T(FG, p).

Ta be defined by Ta(F, p)

= T(FG, p)

25. ELASTIC BODIES

173

Figure 3

for every FE Lin + ; f l(-, p) is the response function for p when the deformed configuration (under the deformation g) is used as reference. (a) Show that G E Unim belongs to Jrpifandonlyif,foreveryF E Lin +, fl(F, p) = f(F, p) for every deformation g of ~ with Vg(p) = G. (b) Define Jr ig) to be the extended symmetry group taking g as reference; that is, Jr ig) is the set of all HELin + such that fl(F, p) = fl(FH, p)

(l9)

for all F E Lin +. Show that Jrp(g) = GJrpG- I ,

(20)

where G = Vg(p). (Here GJrpG- I is the set of all tensors of the form GHG- I , H E Jr p.) (c) Show that Jr p(g) c Unim.

Show further that if Jrp(g) = Unim

(21)

for some deformation g, then (21) holds for every deformation g.

174

IX. (d)

FINITE ELASTICITY

A material is a solid if there exists a configuration from which any change in shape (i.e., any nonrigid deformation) is detectable by some subsequent experiment. More precisely, the material at p is solid if there exists a deformation g such that Jt'peg) c Orth +,

in which case g is undistorting at p. Assume that p is solid and that the identity map on f!J is undistorting at p (so that Jt'p c Orth "), Let g be any other deformation which is undistorting at p. Show that Jt'peg) = RJt'pR - 1,

where R is the rotation in the polar decomposition of G = Vg(p). Show further that if, in addition, the material at p is isotropic, then Jt'peg)

=

Jt'p(= Orth +),

and G is a similarity transformation:

G = A.Q 6.

with A. > 0 and Q E Orth ". An incompressible elastic body is an incompressible material body defined by a constitutive equation of the form T

=

-nI

+ t(F),

(22)

or, more precisely, for x = x(p, t), T(x, t)

= -1t(p, t)I +

t(F(p, t), p)

with

t: Unim x

f!J -+

Sym

smooth. (As in the case of an incompressible fluid the pressure 1t is not uniquely determined by the motion.) For such a body: (a) Determine necessary and sufficient conditions that the response be independent of the observer. (b) Define the notions of material symmetry and isotropy. (c) Show that in the case of isotropy the constitutive equation can be written in the form (23) with fa = (ll(D), 12 (D), 1). [The quantities 1tappearing in (22) and (23) are not necessarily the same.]

26. SIMPLE SHEAR

175

Two important examples of (23) are the Mooney-Rivlin material for which T = -nI + PoB + PIB- I with

Po and PI constants, and the neo-Hookean material for which T

with

= -nI + PoB

Po constant. 26. SIMPLE SHEAR OF A HOMOGENEOUS AND ISOTROPIC ELASTIC BODY

Let f1l be a homogeneous, isotropic body in the shape of a cube. Consider the deformation x = x(p) defined (in cartesian components) by Xl

= PI + ypz,

where

Y = tan 8 is the shearing strain (Fig. 4). Since this deformation is homogeneous, the corresponding stress T is constant. Thus (x, T) represents a solution of the basic equations with body force b = o. (Cf. the discussion given at the end of the last section.) The matrix corresponding to the deformation gradient F is given by I

Y 0]

[F]=OIO [ 001

x

• e2

e2

0

0

e,

Figure 4

x(~)

e.

IX.

176

FINITE ELASTICITY

and the matrix for the left Cauchy-Green strain tensor B = FFT is

IH

[B]~[l~y' A simple calculation shows that

[Br' det{B - wI)

=

~ [~Y ~YY' H 1

_w 3

+ {3 +

y2)W 2 -

{3 +

y2)W

+ 1,

and hence the list of principal invariants of B is §B

=

(3

+ yl,3 + y2,

1).

We therefore conclude from (25.12) that T = Po{y2)I

+ Pl{y2)B + pz< y2)B - l ,

Hence Tl 3 = T23 = 0 and (l)

where (2) is the generalized shear modulus. The result (1) asserts that the shear stress Tl2 is an odd function of the shear strain y. In linear elasticity theory (cf. page 202) the normal stresses Tl l , T2 2 , and T3 3 in simple shear are zero. Here Tll

= y2 Pl + r

T22 = y2P2 + r

(3)

26. SIMPLE SHEAR

177

Thus for the normal stresses to vanish both /31 and and this, in turn, would imply that J1 = 0. Thus if for

/32 would have to be zero,

y #- 0,

(4)

which is a reasonable assumption, then it is impossible to produce a simple shear by applying shear stresses alone. Further (2)-(4) imply that, for y #- 0,

which is the Poynting effect. In fact, (1)-(3) yield the important result

This relation is independent of the material properties of the body; it is satisfied by every isotropic elastic body in simple shear.

EXERCISES

1.

Show that the components of the unit normals on the slanted faces of the deformed cube are given by nl

= ±(I + y2)-1/2

n2

= +: y(I + y2)-1/2,

n3

=

0,

and use this fact to show that T 22

_

yT12 and 1 + y2

TI 2 1 + y2

represent, respectively, normal and tangential components of the surface force Tn on these faces. 2. Consider the uniform extension defined by

x3 =

WP3'

Show that the corresponding stress T is a pure tension in the direction e.; i.e.,

IX.

178

FINITE ELASTICITY

if and only if (J

=

Po + Pl;.2 + P2;.-2,

0= Po + PlQi + P2 W - 2, where the scalar functions Po, Pl' and (through the invariants J B) .

P2 are each functions of;.2 and w 2

27. THE PIOLA-KIRCHHOFF STRESS The Cauchy stress T measures the contact force per unit area in the deformed configuration. In many problems of interest-especially those involving solids-it is not convenient to work with T, since the deformed configuration is not known in advance. For this reason we introduce a stress tensor which gives the force measured per unit area in the reference configuration. Let (x, T) be a dynamical process. Then given a part 9, we can use (6.18h to write the total surface force on 9 at time t (see Fig. 5) as an integral over 09; the result is

r

Jot?',

Tm dA

=

r

Jot?'

(det F)T",F-Tn dA,

where m and n, respectively, are the outward unit normal fields on 09, and 09, while T", is the material description of T. Thus if we let (1)

then

r

Jot?',

TmdA

=

S: PA x

~

r SndA.

Jot?'

(2)

We call the field -. Lin

defined by (1) the Piola-Kirchhoff stress.' By (2), Sn is the surface force measured per unit area in the reference configuration. Similarly, if b is the body force corresponding to (x, T), then

r b dV = Jt?'r b",(det F) dV = Jt?'r bo dV,

Jt?'t 1

In the literature S is often referred to as thefirst Piola-Kirchhoffstress.

(3)

27.

THE PIOLA-KIRCHHOFF STRESS

179

~-+.D(P)

Figure S

where bo = (det F)b....

(4)

We call bo the reference body force; bo gives the body force measured per unit volume in the reference configuration.

Proposition.

The Piola-Kirchhoffstress S satisfies the balance equations

f

J8~

f

J8~

Sn dA

(x - 0) X Sn dA

+

+ f (x J~

f

bo dV =

J~

f ~

- 0) X bo dV

xPo dV,

=

f ~

(5) (x - 0) X xpo dV

for every part 9. Proof.

By (12.7),

f

tp dV

=

~,

f

xPo dV.

~

This relation, (2), and (3), when combined with balance of linear momentum (14.2)1' imply (5)1. Similarly, (6.18h, (12.7), (14.2h, and an analogous argumentyields (5)2. 0

Proposition. S satisfies the field equations

= Po X, SF T = FS T •

Div S + bo

Proof.

(6)

By (5)1 and the divergence theorem,

L

(Div S

+ bo -

Po x) dV

=

0,

180

IX.

FINITE ELASTICITY

and, since this relation must be satisfied for every part &>, (6)1 must hold. Further, (1) implies T", = (det F)-ISFT ,

0

and (6)z follows from the symmetry of T.

It is important to note that, by (6)z, S generally is not symmetric.

Next, by the symmetry ofT, (a) of the proposition on page 6, (8.8)1' and (1.5)z,

and therefore

f

T . D dV

~.

=

f

(det F)(T",F- T )

•

F dV =

f.s- F

dV.

Thus the stress power of &> at time t is given by

LF S·

dV.

Also, (1), (4), (6.18)1' and (6.14)1 imply

f Tm. v dA = r (Tv)· m dA = f Sn· idA, lB., lB., lB.

f.,

b· v dV

=

f.

(7)

bo • i dV,

while (12.7) yields

We therefore have the following alternative version of (15.2). Theorem of Power Expended.

Given any part &>,

r Sn· i dA + f• bo • it dV = f•S • F dV + ddt .I.r i2 Po dV. lB. 2

(8)

The results established thus far are consequences of balance of momentum and are independent of the particular constitution of the body. Assume now that the body is elastic, so that by (1) and (25.1),S is given by a constitutive equation of the form

S = S(F)

(9)

27. THE PIOLA-l(IRCHHOFF STRESS

181

S(F) = (det F)T(F)F- T.

(10)

with (As before, we do not mention the explicit dependence of S on the material point p.) Choose Q E Orth ". Then det(QF) = det F, (QF)-T = QF- T, so that (25.3) and (10) imply S(QF)

=

det(QF)T(QF)(QF)-T

=

(det F)QT(F)QTQF- T = QS(F).

Thus S(QF) = QS(F)

(11)

for every F E Lin + and Q E Orth ". This relation expresses the requirement that the response be independent of the observer; it can be used to deduce constitutive equations for S analogous to (25.5). It is easier, however, to proceed directly from (25.5h. Indeed, (25.5h, (9), and (10) imply that

S

=

(det F)FT(C),

(12)

and if we define seC) = Jdet C fCc),

(13)

S = FS(C).

(14)

then, since

(12) reduces to We leave it as an exercise to show that this constitutive equation is independent of the observer. Note that, by (13) and the smoothness off (cf. the corollary on page 166), S is a smooth mapping of Psym into Sym. In view of (o), and (14), we can rewrite the basic system offield equations (25.13)-(25.15) in the form

S = FS(C), C = FTF,

Div S

F

+ bo =

=

Vx

(15)

Po X.

The relation (6h follows automatically from the fact that S has symmetric values. Also, since the density enters (15) only through its reference value Po, which is assumed known a priori, balance of mass (25.15) need not be included in the list of field equations.

IX.

182

FINITE ELASTICITY

All of the fields in (15) have fJI x IR as their domain, and the operator Div is with respect to the material point p in fJI. In contrast, some of the fields in (25.13)-(25.15) are defined on the trajectory ff, and, more importantly, the operator div is with respect to the place x in the current configuration. For this reason the formulation (15) is more convenient than (25.13)-(25.15) in problems for which the trajectory is not known in advance. The boundary-value problems of finite elasticity are obtained by adjoining to (15) suitable initial and boundary conditions. As initial conditions one usually specifies the initial motion and velocity: x(p, 0)

=

x(p, 0)

xo(p),

=

(16)

vo(p)

with Xo and Vo prescribed functions on fJI. To specify the boundary conditions one considers complementary regular 1 subsets [/1 and [/2 of ofJI (so that

o

where [/IX is the relative interior of [/IX) and then prescribes the motion on [/1' the surface traction on [/2: x =

x

on

[/1

x [0, (0),

on

[/2

x [0, (0)

(17)

with x and s prescribed vector fields on [/1 X [0, (0) and [/2 X [0, (0), respectively. Another form of boundary condition arises when one specifies the surface traction Tm on the deformed surface X([/2' t). A simple example of this type of condition arises when one considers the effects of a uniform pressure no. Here T(x, t)m(x)

=

-nom(x)

for all x E X([/2' t) and all t ~ 0. We can also write this condition as a restriction on the Piela-Kirchhoff stress S; the result is (cf. Exercise 6) on

[/2

x [0, (0).

(18)

Clearly, this is not a special case of (17) because F(p, t) is not known in advance. We can, of course, generalize (17) to include this case by allowing s to be a function s(F, p, t) ofF, p, and t, rather than a function ofp and t only. [Actually, one can show that the combination (det F)F-Tn depends only on the tangential gradient of x on [/2'] 1 Cf. Kellogg [I]; Gurtin [I, p. 14]. Roughly speaking, each' 5/'. is a relatively closed, piecewise smooth surface with boundary a piecewise smooth closed curve.

183

27. THE PIOLA-KIRCHHOFF STRESS

In the statical theory all of the fields are independent of time and the underlying boundary-value problem consists in finding a deformation f that satisfies the field equations S

= FS(C),

C = FTF,

(19)

F = Vf,

Div S + b o = 0, and the boundary conditions on

So

9'1'

=

s

on

9'2

(20)

with f and s prescribed functions on 9'1 and 9'2' respectively. When tractions are prescribed over the entire boundary, that is, when (20) has the form So

=

s

on

(21)

i}[fI,

(5)1 implies that

Jr s dA + iJl1I

i

111

bo dV = 0.

This relation involves only the data and furnishes a necessary condition for the existence of a solution. On the other hand, (5)z yields

r (f -

J

iJl1I

0) x

s dA

+

i

111

(f - 0) x b o dV = 0,

and, because of the presence of the deformation f, is not a restriction on the data, but rather a compatibility condition which will automatically be satisfied by any solution of the boundary-value problem (19), (21). This difference between force and moment balance is illustrated in Fig. 6. As long

-

----~ Figure 6

184

IX.

FINITE ELASTICITY

as the forces acting on f!4 obey balance of forces, we expect a solution, even though moments are not balanced in the reference configuration; the body will simply deform to insure balance of moments in the deformed configuration. EXERCISES

1.

Give the details of the proof of (5h.

2.

Show that

S(FQ)

3.

S(F)Q

=

(22)

for every Q in the symmetry group f§ and every F that Sand S are invariant under f§. Use (11) to show that

S(Q)·

Q=

E

Lin +. Show further

0

(23)

for any smooth function (of time) Q with values in Orth + . 4.

Let f!4 be bounded. Consider a dynamical process and define qJ =

~

L

2

u Po dV,

where u(p, t) = x(p, t) - P is the displacement. Show that

iP =

f ~

iJ2 po dV -

f

~

S· Vu dV +

r

Jo~

u· So dA.

5. Show that the constitutive equation (14) is independent of the observer. 6.

Equation (6.18h holds for any sufficiently regular subsurface Y' of of!):

i

T(x)m(x) dA; =

f(9')

f

T(f(p»G(p)o(p) dA p •

9'

Use this fact to establish (18).

28. HVPERELASTIC BODIES Thus far the only general restriction we have placed on constitutive equations is the requirement that material response be independent of the observer. Thermodynamics also serves to constrain constitutive behavior, and a standard thermodynamic axiom (consistent with a purely mechanical theory) is the requirement of nonnegative work in closed processes. We now study the implications of this axiom for elastic bodies.

28.

185

HYPERELASTIC BODIES

Let (x, T) be a dynamical process, and let b be the corresponding body force. Then given any part f!I', the work on f!I' during a time interval [to, t 1] is given by

f' {r to

Tn' v dA

J{j~t

+

f

+

Lbo'

b· v dV} dt,

~t

or equivalently, in view of (27.7), by

I' {fa~

Sn • i dA

(1)

i dV} dt

with S the Piola-Kirchhoff stress and bo the reference body force. Let us agree to call (x, T) closed during [to, t 1 ] provided (2)

x(p, to) = x(p, t 1),

for all p E 911. If we integrate (27.8) between to and t 1 and use (2h, we see that for processes of this type (1) reduces to

ff tl

to

S . F dV dt.

~

We consider now an elastic body and restrict our attention to dynamical processes belonging to the constitutive class
ff t'

to

S· F dV dt

~

0

(3)

~

in any process which is closed during [to, tIl

Proposition. any p

E

The work is nonnegative in closed processes

if and

only

if given

911 and any time interval [to, t 1], t'

f

S(p, t) • F(p, t) dt ~ 0

(4)

to

in any process which is closed during [to, tIl Proof Let A(p) denote the left side of (4). Clearly, A(p) ~ 0 for all p implies (3). Conversely, assume that (3) holds for every part f!I'. Then, by interchanging the integrations in (3), we may conclude that

LA

dV

~

0

for every part f!I'. In view of the localization theorem (5.1), this implies that A(p) ~ 0 for all p. 0

IX.

186

FINITE ELASTICITY

Our next step will be to describe the class of elastic materials for which the work is nonnegative in closed processes. Anticipating the result, we introduce the following definition: An elastic body is hyperelastic if the Piela-Kirchhoff stress S(F, p) is the derivative of a scalar function a(F, p); that is, S(F, p)

=

Du(F, p),

(5)

where the derivative is with respect to F holding p fixed. The scalar function

u: Lin+

x fJi --+ ~

is called the strain-energy density. (Note that S determines u only up to an arbitrary function of palone.) For (5) to make sense we must interpret the derivative Du as a tensor. By definition Du(F, p) is a linear mapping of Lin into ~,and hence Du(F, p) [A] can be written as the inner product! of a tensor K with A. We here identify Du(F, p) with K and write Du(F, p) . A

in place of Du(F, p) [A].

With this interpretation Du(F, p) is a tensor and (5) has meaning. The components of this tensor are

o

Du(F, P)ij = of.. u(F, p). '}

Theorem. An elastic body is hyperelastic if and only if the work is nonnegative in closed processes. Proof. Assume that fJi is hyperelastic. Consider a closed process during [to, t 1]. Then

d

dt u(F(p, t), p)

.

~

.

= Du(F(p, t), p) • F(p, t) = ;:,(F(p, r), p) • F(p, t),

(6)

and, since

[cf. (2)1]'

f

it

to

S(p, t) • F(p, t) dt =

ft. dd u(F(p, r), p) dt to

t

= u(F(p, t 1 ) , p) - u(F(p, to), p) = O.

(7)

Thus the work is nonnegative (in fact, zero) in closed processes. Conversely, assume that the work is nonnegative in closed processes. 1 The representation theorem for linear forms (page 1) is valid on any finite-dimensional inner-product space.

28. HYPERELASTIC BODIES

Assertion 1.

187

Let F: IR _ Lin + be a C 2 function with (8)

Then ll

f

S(F)·

F dt

(9)

0

=

10

(Here and in what follows we suppress the dependence of S on p.) Let!F = !F{to, t 1) denote the set of all C 2 functions F: IR - Lin + which satisfy (8). As our first step we observe that given any F E !F, Proof

l l

f

S(F) . F dt ;::: 0,

(10)

10

for if x is the motion x(p, t)

=

Po + F(t)(p - Po),

then the corresponding process is closed during [to, t 1 ] and (4) implies (10). Choose F E !F and let F*: IR - Lin + be defined by F*(t) = F{to

+ t1

-

t),

so that F* represents the reversal (in time) of F. Then, by (8), (11)

F*{to) = F(t 1 ) = F{to) = F*(t 1 ) ,

and, since . d F*(t) = dt F(t o + t 1

-

. t) = -F(t o + t 1

-

t),

(12)

it follows that (13)

F*{to) = -F(t 1 ) = -F(t o) = F*{t1)'

By (11) and (13), F* E!F; thus we may use (10) with F replaced by F* and (12) as follows: Os fIIS(F*) • 11'* dt

= - fIIS(F(to + t 1

to

-

t» • F{to

+ t1

-

t) dt

to

=

-

fIIS(F(r» • F(r) dt: 10

This inequality is clearly compatible with (10) only if (9) holds. Our next step will be to show that we can drop condition (8h without affecting the validity of (9). We do this by applying (9) to a family of functions

188

IX.

FINITE ELASTICITY

which satisfy both of (8), but whose limit satisfies only (8)1. This family is constructed in Assertion 2. Let F be a piecewise smooth,' closed curve in Lin + ; that is, let F: [0, 1] --+ Lin + be piecewise smooth and satisfy F(O) = F(I)

== A.

Then there exists a one-parameter family F s (0 < fJ < fJ o) of COO functions --+ Lin + such that:

F ~: IR

(a) F~ = A on (-00, -fJ] u [1 + fJ, (0); (b) IF ~ I and IF~ I are bounded on IR with bounds independent of fJ; (c) as fJ --+ 0, F~ --+ F everywhere on [0, 1], while F~ --+ F at points of continuity of F on (0, 1). Proof We use the Friedrichs-Sobolev method ofmollifiers to construct the family F~. For each fJ > 0 let p~: IR --+ IR be a Coo function with .the following properties:

0;

(i)

p~

(ii)

p~(t)

= 0 whenever It I ~ fJ;

(iii)

f~oo

pit) dt = l.

~

An example of such a family of functions is furnished by

{

pit) =

, ~ exp ( -

fJ2 1 -

t

2)'

It I <

s

0,

with , ~ chosen to insure satisfaction of (iii). Using P~ we define a Coo family F~ as

F~(t) for all t

E

=

f~oo p~(t

- -r)F(r) dt

(14)

IR, where we have extended the domain of F to IR by defining F = A

By (ii) and (iii), for t

~

on

- fJ or t

5

~

tH

Fit) =

(- 00,0) u (1, (0). 1

p~(t

+ fJ, - -r)A dt

= A,

t-~

and (a) follows. 1 F is continuous on [0, I], while Fexists and is continuous at all but a finite number ofpoints where it suffers, at most, jump discontinuities.

28.

189

HYPERELASTIC BODIES

Next, (14) and an integration by parts using (ii) yield Fit) = f:ooPtl(t - T)F(T)dT =

=

f:oo [- :TPtl(t 5:00 pit -

T)]F(T)dT

T)F(T) dt:

(15)

In view of (i) and (iii), we conclude from (14) and (15) that

lFit)1 s

sup

IFI

[0,1]

IFit) I s sup

foo00 pit -

T) dt = sup

-

IFI,

[0,1]

IFI,

(O,l]

and these bounds yield (b). Finally, by (14), (ii), and (iii), Ftl(t) - F(t) =

f:oo Ptl(t -

= J~:tl pit -

T)F(T) dt - F(t)

f:oo Ptl(t -

T) dt

r)[F(r) - F(t)] dt;

hence (ii) and (iii) imply that IFtl(t) - F(t) I ~

sup IF(T) - F(t)I, t€(t-tl,t+tl)

Similarly, using (15), IFtl(t) - F(t) I ~

sup

IF(T) - F(t)l,

t€(t-tl,t+tl)

and the last two inequalities imply (c). Moreover, since F is uniformly continuous on IR (F is continuous on IR and constant outside a compact interval), F tl ~ F uniformly on IR; thus, since det F > 0, there must exist a 150 > such that det Ftl > 0 on IR for 0 < 15 < bo . For this range of 15, F, can be considered as a mapping of IR into Lin +, This completes the proof of Assertion 2. Our last step is

°

Assertion 3. The body is hyperelastic. Proof Let F be an arbitrary piecewise smooth, closed curve in Lin + , and let F tl (0 < 15 < 15 0 ) be the family established in Assertion 2. By (a) of Assertion 2,

190

IX.

FINITE ELASTICITY

as long as C> < 1. Let C>I = min(1, C>o). Then for 0 < C> < C>I, each F" satisfies the hypotheses of Assertion 1 (with to = -1 and t l = 2); hence

r

I

({J,,(t) dt

(16)

0,

=

where (17)

By (a) of Assertion 2 the integral from - 1to we can therefore rewrite (16) as

II

({J" dt

o

+ fO

({J" dt

+

C> and

fl+"

-"

from 1 +

({J" dt

=

o.

C>

to 2 vanishes;

(18)

I

By (b) and (c) of Assertion 2 in conjunction with Lebesgue's dominated convergence theorem I (recall that F has at most a finite number of discontinuities),

f

({J" dt -.

f

S(F) • F dt.

On the other hand, (b) of Assertion 2 and (17) imply that the remaining integrals in (18) converge to zero as C> -. O. Thus

f

S(F) • F dt

= O.

What we have shown is that the integral of S over any piecewise smooth, closed curve in Lin + vanishes. Since Lin + is an open, connected subset of (the vector space) Lin, a standard theorem- in vector analysis tells us that S is the derivative of a smooth scalar function b on Lin +. Thus (5) holds and the proof is complete. 0 Hyperelastic materials have several interesting properties; an example is the following direct consequence of (7).

Proposition.

For a hyperelastic material the work is zero in closed processes.

If we write IT(p, t)

=

b(F(p, r), t)

for the values of b in a process, then (6) implies 0I

2

= S· F,

Cf., e.g., Natanson [I, p. 161]. cr., e.g., Nickerson, Spencer, and Steenrod [I, Theorem 8.4].

28.

191

HYPERELASTIC BODIES

and the theorem of power expended (27.8) has the following important corollary. Theorem (Balance of energy for hyperelastic materials). Each dynamical process for a hyperelastic body satisfies the energy equation

L~

SO • xdA

+

Lbo· x

dV

= :t

L

(a

+ Po ~)

dV

for each part &>. Here S is the Piola-Kirchhoffstress, Po is the reference density, and bo is the reference body force (27.4).

The term Ladv

represents the strain energy of &>. The energy equation asserts that the power expended on &> must equal the rate at which the total energy of &> is changing. As a direct consequence of the above theorem we have the following important Corollary (Conservation of energy). Assume that the body is finite and hyperelastic. Consider a dynamical process for the body corresponding to body force b = 0, and suppose that So·

x=

0

afJI

on

for all time. Then the total energy is constant:

L(a

+ Po ~) dV

=

const.

EXERCISES

1. Consider a hyperelastic body with strain-energy density fJ. (For con-

venience, we suppress dependence on the material point p.) (a) Use (27.23) and (5) to show that fJ(Q) = fJ(I)

(19)

for every Q E Orth + . (Here you may use the connectivity of Orth + to insure the existence of a smooth curve R: [0, 1] -+ Orth + with R(O) = I and R(l) = Q.) (b) Use (27.11) and (5) to show that fJ( QF) = fJ(F)

for every Q E Orth + and F

E

Lin + .

IX.

192 (c)

FINITE ELASTICITY

Use (27.22) and (5) to show that &(FQ)

(d)

= &(F)

for every Q in the symmetry group and every .F E Lin + . Show that &(F) = &(U) and that there exists a function iJ such that &(F) = iJ(C).

(e)

Here U is the right stretch tensor, C the right Cauchy-Green strain tensor. Show that

S=

2DiJ,

where DiJ(C) E Sym with components

o~ .. iJ(C) 1)

(f)

has an interpretation analogous to D&(F). Assume that the material at p is isotropic. Show that iJ(C)

= iJ(B)

with B the left Cauchy-Green strain tensor, and that

[cf. (37.4)J. Let

Show that the Piela-Kirchhoff and Cauchy stress tensors are given by S =2{iJ 1F

+ iJ2 [ (tr B)I -

BJF + (det B)iJ3F- T } ,

T = 2(det F)-1{(det B)iJ3 1 + [iJ1

+ (tr B)iJ2JB -

iJ2B 2 } ,

where we have omitted the arguments of iJk' Extend the results of this section to incompressible elastic bodies (cf. Exercise 25.6). 3. Consider a Newtonian fluid. Show that the work is nonnegative in closed processes if and only if the viscosity u is nonnegative.

2.

28. 4.

193

HYPERELASTIC BODIES

Consider the statical (time-independent) behavior of a homogeneous hyperelastic body without body forces. Show that Div[a(F)I

-

FTS] = 0

and hence that Ia9[6(F)n

5.

-

FTSn]d A

0

=

for every part 9of 9.(This result is of importance in fracture mechanics.) (Principle of stationary potential energy) For a hyperelastic body the (statical) mixed problem discussed in Section 27 can be stated as follows: find a C2 deformation f such that S

=

F

S(F) = Db(F), Div S

+ b,

=

Vf,

=0

on 9 and

f=P

on

Y,,

Sn=B

on Y,.

(20)

Let us agree to call a C2 function f : 9 4 8 with det Vf > 0 kinematically admissible i f f satisfies the boundary condition (20),. Assume that 9 is bounded. Define the potential energy @ on the set of kinematically admissible functions by

-

@{f} = Ia8(Vf) dV - IBb, u dV -

L-

I u dA,

where u(p) = f(p) - p, We say that the variation of @ is zero at f, and write

6@{f} = 0, if

d -

da

@{f

+ ag} la=,

=

0

for every g with f + ag in the domain of @ for all sufficiently small a (that is, for every C 2 function g : 9d -+ Y with g = 0 on Yl). Show that

b@{f) = 0 provided f is a solution of the mixed problem.

IX.

194

FINITE ELASTICITY

29. THE ELASTICITY TENSOR The behavior of the constitutive equation S = S(F) [cr. (27.9)] near F = I is governed by the linear transformation C: Lin

--+ Lin

defined by C

=

(1)

DS(I).

(As before we have suppressed mention of the dependence on the material point p.) C is called the elasticity tensor for the material point p; TOughly speaking, C is the derivative of the Piela-Kirchhoff stress with respect to F at F = I. The importance of this tensor will become apparent in the next section, where we deduce the linearized theory appropriate to small deformations from the reference configuration. For convenience, we assume throughout this section that the residual stress vanishes:

S(I)

= t(I) = O.

(2)

Our first result shows that, because of (2), C could also have been defined as the derivative of r, the response function for the Cauchy stress. Proposition C Proof

=

(3)

Dt(I).

By (27.10), S(F)FT = qJ(F)t(F),

where qJ(F) = det F. If we differentiate this relation with respect to F, using the product rule, we find that S(F)HT

+ DS(F)[H]FT =

qJ(F) Dt(F) [H]

+ DqJ(F) [H]t(F)

for every HELin. Evaluating this expression at F = I, we conclude, with the aid of (2), that DS(I) [H] = Dt(I) [H], which implies (3). 0

29. Proposition

195

THE ELASTICITY TENSOR

(Properties of the elasticity tensor)

(a) C[H] E Symfor every HELin; (b) C[W] = Ofor every W E Skw. Proof.

Assertion (a) follows from the relation C[H]

=

d drx 1'(1 + rxHH.=o

and the fact that l' has symmetric values. To prove (b) choose W E Skw and take Q(t)

= eW t ,

so that Q(t) E Orth + (cf. Section 36). Then (25.3) with F = I and (2) imply t(Q(t» = 0,

and this relation, when differentiated with respect to t, yields Dt(Q(t»

toon = O.

Since Q(O) = I and Q(O) = W, if we evaluate this expression at t = 0 and use (3), we are led to (b). 0 Let

denote the symmetric part of HELin. Then H = E + W, with W skew, and (b) and the linearity of C imply that C[H]

=

C[E];

(4)

hence C is completely determined by its restriction to Sym. Our next step is to establish the invariance properties of C. Proposition.

C is invariant under the symmetry group

'§ for the material at

p.

Proof. In view of (25.9)1' the last theorem in Section 37, and (3), for HELin and Q E e,

QC[H]QT = QDt(l) [H]QT = Dt(QIQT) [QHQT] = Dt(l) [QHQ~] = C[QHQT], and C is invariant under '§.

0

IX.

196

FINITE ELASTICITY

Since C has values in Sym [cf. (a) above], this proposition and (37.22) have the following obvious but important consequence. Theorem. Assume that the material at p is isotropic. Then there exist scalars Jl and A such that

C[E] = 2JlE + A(tr E)I

(5)

for every symmetric tensor E. The scalars Jl = Jl(p) and A = A(p) are called the Lame moduli at p.

We say that C is symmetric if H'C[G] = G' C[H] for all tensors Hand G. Since C[W] = 0 for all skew W, C can never be positive definite in the usual sense. There are, however, important situations in which C restricted to Sym has this property. Thus let us agree to call C positive definite if

E' C[E] > 0 for all symmetric tensors E #- O. Proposition. Assume that the material at p is isotropic. Then C is symmetric. Moreover, C is positive definite if and only if the Lame moduli obey the inequalities: 2Jl + 3A > O.

u > 0,

(6)

Proof Let Hand G be tensors with symmetric parts Us and G s ' respectively. Then, by (a) and (b) of the proposition on page 195, and since I· H, = tr H, (5) implies

H • C[G] = H s ' C[Gs] = 2JlHs ' G s

+ A(tr H)(tr G) =

G . C[H],

so that C is symmetric. Next, choose a symmetric tensor H and let. a=!trH,

H o = H - aI

so that

H

=

H,

+ aI,

tr H o

= O.

Then, since I • H, = 0, H • C[H] = 2Jl(1H o 12

+ 3ct 2 ) + 9A.a 2

= 2Jll H, 12

+ 3a2(2ti + 3A).

Trivially, (6) implies that C is positive definite. Conversely, if C is positive definite, then by choosing H = aI we conclude that 2Jl + 3A > 0, and by choosing H with tr H = 0 we see that Jl > O. 0

197

29. THE ELASTICITY TENSOR

Assume that the material at p is hyperelastic. Then C is sym-

Proposition. metric.

Proof

Since

[ef. (28.5)], the proposition follows from the symmetry of the second derivative. To see this note that

o

of3 ~(I + txH + f3G)

=

+ txH + f3G) . G,

SCI

and hence

02 Otx of3 ~(I

+ txH + f3 G) !a=P=o = DS(I)[H]· G

=

C[H] . G;

by switching the order of differentiation we see that this expression equals

02

of3 ctx ~(I + txH + f3 G ) la=P=o = and the symmetry of C follows.

C[G]' H,

D EXERCISES

1. Show that, as a consequence of (27.11), DS(F) [WF]

for all F

E

Lin + and W

E

= WS(F)

(7)

Skw.

2. Relate C = DS(I) and L = Dt(I) for the case in which (2) is not satisfied. 3. C is strongly elliptic if

A· C[A] > 0 whenever A has the form A = a ® e, a (a) (b)

=1=

0, C

=1=

O.

Show that if C is positive definite, then C is strongly elliptic. Show that for an isotropic material C is strongly elliptic if and only if

2Jl.

+ A. >

O.

198

IX.

FINITE ELASTICITY

SELECTED ~ERENCES

Coleman and Noll [1]. Green and Zerna [1]. Sternberg and Knowles [1]. Truesdell and Noll [1, Chapter OJ. Wang and Truesdell [1].

CHAPTER

x Linear Elasticity

30. DERIVATION OF THE LINEAR THEORY We now deduce the linearized theory appropriate to situations in which the displacement gradient Vu is small. The crucial step is the linearization of the general constitutive equation (1)

for the Piela-Kirchhoff stress [cf. (27.9)]. In order to discuss the behavior of this equation as 0= Vu tends to zero, we consider S(F) as a function of H using the relation

F

=

I

+0

[cf. (7.1)]. Theorem (Asymptotic form of the constitutive relation). residual stress vanishes. Then

S(F) = C[E] as H

-+

+ 0(0)

Assume that the (2)

0, where C is the elasticity tensor (29.1) and

E = 1(H + H T ) is the infinitesimal strain (7.4). 199

(3)

x.

200

LINEAR ELASTICITY

Proof Since the residual stress vanishes, we may conclude from (29.1), (29.2), and (29.4) that

S(F) = S(I

+ H)

+ DS(I) [HJ + o(H) = C[HJ + o(H) = C[EJ + o(H). D = S(I)

Using (2) we can write the asymptotic form of the constitutive equation (1) as

S = C[EJ

+ o(Vu).

(4)

Therefore, if the residual stress in the reference configuration vanishes, then to within terms ofo(Vu) as Vu -. 0 the stress S is a linearfunction ofthe infinitesimal strain E. Also, since C has symmetric values [cf. (a) of the proposition on page 195J, to within the same error S is symmetric. The linear theory of elasticity is based on the stress-strain law (4) with the terms of order o(Vu) neglected, the strain-displacement relation (3), and the equation of motion (27.6)1: S = C[EJ, E

=

t(Vu

Div S

+ VuT) ,

+ bo =

(5)

Poii.

Note that these equations are expressed in terms of the displacement

u(p, t)

=

x(p, t) - p,

rather than the motion X. It is important to emphasize that the formal derivation of the linearized constitutive equation (5)1 was based on the following two assumptions: (a) (b)

The residual stress in the reference configuration vanishes. The displacement gradient is small.

Note that, by (5)1 and the theorem on page 56, E = S = 0 in an infinitesimal rigid displacement. This is an important property of the linearized theory. Given C, Po, and bo , (5) is a linear system of partial differential equations for the fields u, E, and S. By (29.5), when the body is isotropic (5)1 may be replaced by S = 2J-lE + A.(tr E)I. Moreover, when the body is homogeneous, Po, p; and A. are constants.

(6)

201

31. SOME SIMPLE SOLUTIONS

Assume now that f!l is homogeneous and isotropic. Then, since Div(Vu

+ VU T) = L\u + V Div u

and tr E = Div u, the equations (5)2.3 and (6) are easily combined to give the displacement equation of motion Ji. L\u

+ (A + Ji.)V Div u + bo = Poii.

In the statical theory ii = 0 and we have the displacement equation of equilibrium Ji. L\u

+ (A + Ji.)V Div u + bo =

O.

(8)

EXERCISES

1.

Let u be a C 4 solution of (8) with bo = O. Show that Div u and Curl u are harmonic functions. Show further that u is biharmonic; that is,

= O. with b o =

L\L\u 4

Let u be a C solution of (7) O. Show that Div u and Curl u satisfy wave equations. 3. (Boussinesq-Papkovitch-Neuber solution) Let qJ and g be harmonic fields on f!l (with qJ scalar valued and g vector valued) and define

2.

u = g - exV(r . g

+ qJ),

Ji.+A ex = 2(2Ji. + A)'

where r(p) = p - o. Show that u is a solution of (8) with bo = O. 4. Consider the case in which the residual stress S(I) =F O. Show that S(F)

=

8(1)

+

W8(1)

+ C[E] + o(H).

31. SOME SIMPLE SOLUTIONS Assume now that the body is homogeneous and isotropic. Then any statical (i.e., time-independent) displacement field u with E constant generates a solution of the field equations (30.5h. 3 and (30.6) with S constant and

x.

202

LINEAR ELASTICITY

L

Figure I

I

bo = O. We now discuss some particular solutions of this type, using cartesian coordinates where convenient. (a)

Pure shear.

Let u(p)

= YP2 el

(Fig. I), so that the matrices of E and S are

[E]

=

0] 21[0Y 0Y 0,

~ ~], o

000

0

with 't

= p.y.

Thus p. determines the response of the body in shear, at least within the linear theory, and for this reason is called the shearmodulus. Note that, in contrast to the general nonlinear theory (cf. Section 26), the normal stresses are all zero; the only stress present is the shear stress 'to (b) Uniform compression or expansion. Here u(p) = B(p - 0).

Then E

= BI,

S = -nl,

n = -3KB, where

is the modulus of compression. For our third solution it is simpler to work with the stress-strain law (30.6) inverted to give E as a function of S. This inversion is easily accomplished upon noting that tr S

=

(2p.

+ 3A.) tr E,

31. SOME SIMPLE SOLUTIONS

203

1 E = 2 [S A. A. (tr S)IJ. J1. 2J1. + 3

(1)

and hence

(c)

Pure tension.

Here we want the stress to have the form (1

0 0]

[S] = 0 0 O. [ 000 The corresponding strain tensor is then given by

0]

oI

0

o

I

1=

-VG,

with 1

G

= E (1,

and E

=

J1.(2J1. + 3..1.) J1. + A. '

Note that the strain E corresponds to a displacement field of the form u(p) = GP1e 1 + lPzez + lp3e3.

The modulus E is obtained by dividing the tensile stress (1 by the longitudinal strain G produced by it. It is known as Young's modulus. The modulus v is the ratio of the lateral contraction to the longitudinal strain of a bar under pure tension. It is known as Poisson's ratio. If we write Eo and So for the traceless parts of E and S, that is, Eo = E - i{tr E)I,

So = S - :ktr S)I,

(2)

then the isotropic constitutive relation (30.6) is equivalent to the following pair of relations: So = 2J1.Eo , tr S = 3K tr E. Another important form for this stress-strain relation is the one taken by the inverted relation (1) when Young's modulus and Poisson's ratio are used: E =

1

E [(1 + v)S -

v(tr S)I].

(3)

X.

204

LINEAR ELASTICITY

Since an elastic solid should increase its length when pulled, should decrease its volume when acted on by a pure pressure, and should respond to a positive shearing strain by a positive shearing stress, we would expect that E > 0,

Il > 0.

K> 0,

Also, a pure tensile stress should produce a contraction in the direction perpendicular to it; thus v> O. Even though these inequalities are physically well motivated, we will not assume that they hold, for in many circumstances other (somewhat weaker) assumptions are more natural. In particular, we will usually assume that C is positive definite. A simple computation, based on (29.6), shows that this restriction is equivalent to either of the following two sets of inequalities: (i) (ii)

Il > 0, E > 0,

> 0; - 1< v<

K

1-

Some typical values for Young's modulus E and Poisson's ratio v are ' E = 2.1 X lOll N/m 2, E = lOll N/m 2 , E = 0.55 X lOll N/m 2 ,

carbon steel: copper: glass:

v = 0.29, v = 0.33, v = 0.25.

EXERCISES

1. Assume that C is symmetric and define ~(E)

=

tE· C[E]

for every E E Sym. (a)

Show that the stress-strain law S = C[E] can be written as

S (b)

m(E).

Show that, for an isotropic material,

where

~

A. IlIEI 2 + "2 (tr E)2,

~(E)

=

~(E)

= IIIEo 12 + "2 (tr E)2,

~IEI2

I

=

K

s

~(E)

is the smaller and

Cf., e.g., Sokolnikoff [I, p. 70].

~ PIEI 2,

Pthe larger of the numbers Il and 3K/2.

32.

LINEAR ELASTOSTATICS

205

2. Show that C is positive definite if and only if either of the two sets of inequalities (i) or (ii) hold.

32. LINEAR ELASTOSTATICS

The system offield equations for the statical behavior of an elastic bodywithin the framework ofthe linear theory-consists ofthe strain-displacement relation

E = 1
+ VoT ) ,

(1)

the stress-strain relation S

= C[E],

(2)

and the equation of equilibrium Div S + b = 0

(3)

(where we have written b for b o). The elasticity tensor C, which is a linear mapping of tensors into symmetric tensors, will generally depend on position p in 84; writing Cp to emphasize this dependence, we assume henceforth that Cp is a smooth function of p on 84. A list [0, E, S] of fields which are smooth on 84 and satisfy (1)-(3) for a given body force b will be called an elastic state corresponding to b. Note that by (1), (2), and the properties of C, the fields E and S are symmetric. We assume throughout this section that 84 is bounded. Theorem of Work and Energy. to the body force b. Then

r se-

JiJ~

0

Let [u, E, SJ be an elastic state corresponding

dA +

f

b' u dV = 2OU{E},

(4)

IJI

where OU{E}

=~

L

E' C[E] dV

(5)

is the strain energy.

The proof of this theorem is an immediate consequence of the following Lemma. Let S be a smooth symmetric tensor field on 84, let ii be a smooth vector field on 84, and let div S

+ b = 0,

206

X.

LINEAR ELASTICITY

Then

r SO' ii dA + f~ b· ii dV = f~ S • E dV.

Jo~

(6)

Proof By the symmetry ofS, the divergence theorem, (4.2)5' and (a) of the proposition on page 6,

r So' ii dA = Jo~r (Sii)'

Jo~

0

dA

=

f~

Div(Sii) dV

S· Vii = S· {!
+ ViiT )}

=

f~

(ii • Div S + S . Vii) dV,

= S· E.

These relations clearly imply the desired result (6).

0

We can interpret the left side of(4) as the work done by the external forces; (4) asserts that this work is equal to twice the strain energy. Note that when C is positive definite, dIJ{E} ~ 0, and the work is nonnegative. The next theorem is one of the major results of elastostatics; in essence, it expresses the fact that the underlying system offield equations is self-adjoint when C is symmetric. Betti's Reciprocal Theorem. Assume that C is symmetric. Let [u, E, S] and [ii, E, S] be elastic states corresponding to body force fields band b, respectively. Then

r So' ii dA + f~ b . ii dV = Jo~r So' u dA + f~ b' u dV

Jo~

= LS'EdV Proof that

= LS'EdV.

(7)

Since C is symmetric, we conclude from the stress-strain relation

S'E = C[E]'E = C[E]'E = S'E, and (7) follows from (6) and the analogous relation with the roles of the two states reversed. 0 Betti's theorem asserts that given two elastic states, the work done by the external forces ofthe first over the displacement ofthe second equals the work done by the second over the displacement of the first. Let!/' 1 and!/'2 denote complementary regular subsets ofthe boundary of ffl, so that

32.

LINEAR ELASTOSTATICS

207

o

with f/ Il the relative interior of f/ Il' Then the mixed problem of elastostatics can be stated as follows: Given: fJI, f/ l' f/ 2' an elasticity tensor C on fJI, a body force field b on fJI, surface displacements Uon f/ 1> surface tractions s on f/ 2 • Find: An elastic state [u, E, S] that corresponds to b and satisfies the boundary conditions

(8) An elastic state with these properties will be called a solution. Uniqueness Theorem. Assume that C is positive definite. Let [u l , E l , SI] and [u 2, E 2, S2] be solutions of the (same) mixed problem. Then

where w is an infinitesimal rigid displacement of fJI. Proof

Let

Then [w, E, S] is an elastic state that corresponds to null body forces and satisfies the boundary conditions

w=o

f/l'

on

Thus

Sn·w

=

0

on

iJfJI,

and we conclude from (4) that

L

E • C[E] dV

= O.

Since C is positive definite, this relation can hold only if E = 0; this in turn implies that S = 0 and that w is an infinitesimal rigid displacement of fJI (cf. the theorem on page 56). 0 When f/ 1 = iJfJI (f/ 2 =

0) the boundary condition (8) takes the form u=U

and the mixed problem is referred to as the displacement problem. On the other hand, when f/ 2 = iJfJI (f/ 1 = 0), so that

Sn

=

s

on

iJfJI,

x.

208

LINEAR ELASTICITY

we have the traction problem. This last definition, (3), and Cauchy's theorem (page 101) have the following immediate consequence: Proposition. that

A necessary condition that the traction problem have a solution is

is

dA +

i

oBit

where r(p) = p -

0

f b dV = 0, Bit

(JlJI

f

r x s dA +

r x b dV

(9) = 0,

Bit

is the position vector.

Equations (9) insure equilibrium of the external forces applied to fJ4. It is interesting to compare these conditions with the analogous restrictions ofthe general nonlinear theory discussed at the end of Section 27. There the applied forces need not obey balance of moments in the reference configuration [cf. the discussion following (27.21)], as in the nonlinear theory the body is allowed to undergo the possibly large deformation needed to insure moment balance in the deformed configuration. We now show that the solution of the mixed problem, if it exists, can be characterized as the minimum value of a certain functional. By a kinematically admissible state we mean a list" = [0, E, S] with u, E, and S smooth fields on fJ4 that satisfy the field equations

E

=

t
S = C[E],

and the boundary condition

u=o

on

[/'1.

Let ell be the functional defined on the set of kinematically admissible states by

ell{,,} = Olt{E} -

f

b'udV -

Bit

f

s·odA.

(10)

.'72

Principle of Minimum Potential Energy. Assume that C is symmetric and positive definite. Let" = [0, E, S] be a solution of the mixed problem. Then

ell{o} s ell{a}

for every kinematically admissible state a = [ii, E, S], and equality holds only ii = 0 + w with w an infinitesimal rigid displacement of fJ4. Proof

Let

w=ii-o,

E=E-E.

if

32.

209

LINEAR ELASTOSTATICS

Then, since <1 is a solution and <3 is kinematically admissible,

E = -t(Vw + VW T), W =

0

on

(11)

!/'t.

Further, since C is symmetric and S = C[E], E . C[E] = E' C[E] + E' C[E] + E • C[E] + E . C[E] = E' C[E] + E' C[E] + 2S • E; hence o/!{E} - o/!{E}

o/!{E} +

=

Ls'

E dV.

Because <1 is a solution, we conclude from (11h and (6) with ii and by wand E that

rS· E dV = Jr So' w dA + Jrb· w dV = f

JBI

OBI

s· w dA +

f/,

BI

Ereplaced

rb· w dV.

JBI

In view of (10) and the last two relations, cI>{<3} - cI>{<1} = o/!{E}.

Thus, since C is positive definite, cI>{<1}

S;

cI>{<3}

and only when E = 0;

cI>{<1} = cI>{<3}

that is, only when w = ii - u is an infinitesimal rigid displacement.

D

In words, the principle of minimum potential energy asserts that the difference between the strain energy and the work done by the body force and prescribed surface traction assumes a smaller value for the solution of the mixed problem than for any other kinematically admissible state. A standard technique for obtaining approximate solutions is based on the principle of minimum potential energy. We now demonstrate this method, but for convenience limit our attention to the traction problem. Consider an approximate solution of the form N

u(p)

=

L: IXng,.(p),

(12)

n=l

where gl' gz, ... , gN are given vector fields on fJI, and where the scalar constants 1X 1, IX z, .•. , IXN are chosen to make cI>(1X1' IXz, ... , IXN) a minimum. Here

x.

210
LINEAR ELASTICITY

is the function obtained by substituting (12) into the right

side of (10):

where K mn =

L

G n = -!(Vg"

G m • C[Gn] dV,

qJn =

f

b· g" dV

+

fJI

+ VgJ),

r s· g" dA. JofJI

Let K be the matrix with entries K mn • If C is symmetric and positive definite, K will be symmetric and positive semidefinite; thus
KO[ =
EXERCISES

1. Show that if the traction problem has a solution [u, E, S], then [u + w, E, S] is a solution for every infinitesimal rigid displacement w. Does the analogous result hold for the mixed problem in general? 2. Show that for a homogeneous, isotropic elastic body the "infinitesimal volume change"

r u-

JO fJI

D

dA =

r div u dV

J

fJI

32. LINEAR ELASTOSTATICS

211

in an elastic state is given by

where r(p) = p - o. For the remaining exercises C is symmetric. 3.

Assume that C is positive definite. Then C restricted to symmetric tensors is invertible. Let K denote the corresponding inverse, so that the stress-strain relation S = C[E] can be inverted to give E

=

K[S].

(a) Show that K is symmetric. (b) Show that the strain energy can be written in the form O/tK{S} =

(c)

~

Ls.

K[S] dV.

(Principle of minimum complementary energy) A statically admissible stress field S is a smooth symmetric tensor field that

satisfies the equilibrium equation Div S

+b=

0

and the boundary condition on

[/2'

Let 'II be the functional defined on the set of statically admissible stress fields by 'P{S}

=

CfI K {S } -

f

So· fj dA.

9',

Show that if [u, E, S] is a solution of the mixed problem, then 'P{S} ~ 'P{S}

for every statically admissible stress field S. (d) Show that if o = [u, E, S] is a solution of the mixed problem, then {{j}

+

'P{S}

=

O.

212 4.

X.

LINEAR ELASTICITY

(Hu-Washizu principle) An admissible state is a list a = [u, E, S] of smooth fields on 84 with E and S symmetric. Let A be the functional defined on the space of admissible states by

A{a}

=

IJlt{E} -

+

f

f

S·EdV -

~

f

(DivS

+ b)·udV +

f

So·(kdA

Y,

~

(So - s) . u dA.

Y2

Show that

(jA{a} = 0, provided a is a solution of the mixed problem (cf. Exercise 28.5). 5. (Hellinger-Prange-Reissner principle) Assume that C restricted to symmetric tensors is invertible with inverse K. Let d be the set of all admissible states that satisfy the strain-displacement relation (1) and define e on d by era}

= IJltI({S} -

f

S·EdV

~

+

f

+

f

b·udV

+

f

So·(u - u)dA

Y,

~

s·u dA.

Y2

Show that (je{a} =

°

provided a is a solution of the mixed problem. 6.

Show that for a kinematically admissible,

A{a} =
Assuming that [U., E., SvJ tends to a limit [u, E, S] as v -+ 1, show that tr E = 0, S = -nI + 2flE, where t: = -t tr S. Give an argument to support the claim that v = 1corresponds to incompressibility. (Note that, as in the case of ideal and Newtonian fluids, the "pressure" 1t is not uniquely determined by the deformation; that is, E does not uniquely determine n.)

32.

213

LINEAR ELASTOSTATICS

(b) Assuming that the limit state [u, E, S] is approached in a manner which is sufficiently regular to justify interchanges of limits and derivatives, show that

= 0, Div u = O. - Vn

{t ~u

(Cf. the equations on page 152 describing Stokes flows.) 8. An elastic state [u, E, S] is a state of plane strain if u has the form

in some cartesian frame. Consider such a state, and assume that the body is isotropic and that b = O. (a) Show that E and S are functions of (PI' P2), and that E"p = 1(u",p + up,,,), S"p = 2jJ.E"p

+ A.c5"p(E 11 + E 2 2 ) ,

(13)

2

L S"p,p=O,

P=l

while E 13 = E 2 3 = E 3 3 = 0, S33 = A.(E 11

Here (X

(X

and

p have the range

= p and zero otherwise, and

+ E 22 )

= V(Sl1

+ S22)'

of the integers (1, 2), c5"p = 1 when

ou"

u",p = opp' etc.

(b) Show that E (when C 2 ) satisfies the compatibility equation 2E 12 , 1 2 = E 11 , 2 2

+ E22,l1'

(14)

For the remaining exercises fJt is an open simply connected region in 1R 2 • 9. Let E"p (= E p,,) be a C 2 solution of (14) on fJt. Show that there exist displacements u" such that (13)1 hold. 10. Let E"p and S"p be C 2 functions on fJt consistent with (13h, 3' Show that (14) is satisfied if and only if (15)

214

X.

LINEAR ELASTICITY

[Thus for a simply connected region plane elastic states are completely characterized by (13h and (15). Indeed, when these equations are satisfied, Ea.(J can be defined by (13h (assuming invertibility) and, since (14) holds, there exist Ua. that satisfy (13)1'] 11. Let qJ be a scalar field of class C4 on fYt. Define (16) Show that Sa.(J satisfies (13h. Show further that (15) holds if and only if qJ is biharmonic:

The field

qJ is

called an Airy stress function. 12. Let Sa.(J (= S(Ja.) be a C 2 solution of (13)3 and (15) on fYt. Show that there exists a biharmonic function qJ on fYt such that (16) holds.

33. BENDING AND TORSION

A.

BENDING OF A BAR

Consider a (homogeneous, isotropic) cylindrical bar with generators parallel to the P3-axis (Fig. 2). Let the end faces [/0 and [/, be located at P3 = 0 and P3 = I, respectively, with the origin at the centroid of [/0 and with the PIand Pl-axes principal axes of inertia:

f

!/' 0

PI dA

=

f

P2 dA =

!/' 0

f

(1)

PIP2 dA = O.

!/' 0

We assume that the bar is loaded only on the end faces, and by opposing couples about the P2-axis. More precisely, we assume that (a) (b)

body forces are zero; the lateral surface 2 is traction-free; that is,

So

=

0

on

2,

or equivalently, since n3 = 0 on 2, on

2; (2)

33.

215

BENDING AND TORISON

P3

P, Figure 2

(c) the total force on go vanishes and the total moment on go is equipollent to a moment of magnitude m about the negative P2-axis; that is,

f

SodA = 0,

9'0

f

r x So dA = - me2'

9'0

or equivalently, since 0 = -e 3 and r(p) = p -

f

S13 dA

=

9'0

S S23 dA f

S33 dA = 0,

=

9'0

f

9'0

f

P2S33 dA =

f

PlS33 dA = -m.

9'0

0,

(Pl S23 - P2 S13) dA = 0,

(3)

9'0

9'0

We do not specify the loading on g" since balance of forces and moments require that (Exercise 1) (d) the total force on g, vanish and the total moment on gl equal me2; in fact, (3) hold with go replaced by gl. There are many sets of boundary conditions consistent with (c) and (d). We take the simplest possible and assume that SI3(P) = S23(P) = 0,

S33(P) =

"0 + "IPl + "2P2

with "0' "1' and "2 constants. Then (1) and (3) imply that

"1 = where I =

f pi 9'0

dA

-mil,

at

P3 = 0, l,

X.

216

LINEAR ELASTICITY

is the moment of inertia of [/0 about the P2-axis. Our boundary conditions therefore consist of (2) and at

P3 = 0, l.

(4)

A stress field compatible with (2) and (4) is obtained by taking

Sl1(P)

= S22(P) = S12(P) = S13(P) = S23(P) = 0, S33(P) =

mpl

- -/-

for all P in the body. This field clearly satisfies Div S

=

o.

Thus to show that S is actually a solution of our problem we need only construct the corresponding displacement field. Using the stress-strain law in the form (31.3), we see that E 12 (p)

= E 13(P) = E 23(p) = 0,

and the strain-displacement relation (32.1) is easily integrated to give (Exercise 2)

(5) u 3(p)

= -

m E/ P1P3

+ W3(P)

with w(p) an arbitrary infinitesimal rigid displacement. To compare our results with classical beam theory we "fix" [/0 at P1 = P2 = by requiring that

°

u(o) = Vu(o) = O.

Then w(p) == 0 and the displacement of the centroidal axis (i.e., the P3-axis) takes the form mp~ Ul(O, 0, P3) = 2EI'

U2(0, 0, P3) = U3(0, 0, P3) = 0.

33. BENDING AND TORISON

217

[Actually, to insure that w(p) == 0 it suffices to assume that u(o) = 0 and Vu(o) = VU(O)T.] In this solution the maximum stress occurs at points of If for which IpII attains its largest value, e say, and this maximal stress has magnitude mcfl,

Moreover, the maximum deflection ofthe centroidal axis occurs at P3 = I and is given by

ml2 2EI

B.

TORSION OF A CIRCULAR CYLINDER

Consider a circular cylinder of length I and radius a (Fig. 3). As before, the axis of the cylinder coincides with the P3-axis, while f/ 0 and f/, correspond to P3 = 0 and P3 = I. We assume that the end face f/ 0 is held fixed, while the end face f/, is rigidly rotated about the P3-axis through an angle {3. Thus, assuming that the lateral surface If is traction-free, our boundary conditions consist of (2) and the requirement that n(p) = 0

uI(p) = -rxIP2'

at

P3 = 0,

uip) = rxlpl'

U3(P) = 0

at

P3 = I,

with rx = {3/1 the angle of twist per unit length. (The boundary condition at P3 = I represents an infinitesimal rigid rotation of f/, about the P3-axis [cf. (7.10) with P3 = I]). It seems reasonable to expect that under this type of loading the cylinder will twist uniformly along its length; thus, in view of (7.10), we consider a displacement field of the form ul(p) = -rxP2P3,

U2(P) = rxPIP3, U3(P) =

(6)

o. P2

- t.......

Figure 3

PI

_p,

X.

218

LINEAR ELASTICITY

The corresponding stress field, computed using the strain-displacement relation (32.1) and the stress-strain relation (30.6), is given by S11(P)

=

S2ip)

=

S33(P)

S23(P) = /JlXPl,

=

S12(P)

=

0,

(7)

S13(P) = -/JlXP2,

and clearly satisfies Div S = O. Further, S satisfies the boundary condition (2); indeed (2)1.2 are satisfied trivially, and, since o(p) on !fl is proportional to (PI, P2' 0), it follows that on

!fl,

(8)

which is (2h. Thus (6) and (7) comprise the solution of our problem. Next, by (1) and (7),

f

S13 dA

=

~o

f

S23

dA =

~o

f

S33

dA = 0,

~o

so that

f

(9)

SodA = 0,

~o

and the total force on each end face vanishes. In addition,

where] 0 = na 4/2 is the polar moment ofinertia of the cross section. Therefore

f

r x So dA

= -

me3

(10)

~o

with m = /JlX]o'

Further, by balance of moments,

f

r x So dA = me3'

~I

Thus the cylinder is twisted by equal and opposite couples about the P3-axis.

33.

219

BENDING AND TORSION EXERCISES

1. Establish (d). 2. Derive (5). 3. Show that w(p) == 0 in (5) provided u(o) = 0 and Vu(o) = VU(O)T.

In the next two exercises we consider the torsion of an arbitrary cylindrical bar (cf. Fig. 2). 4. Show that (7) is compatible with (8) only if the cross section is circular. 5. Consider the displacement field u1(p) = -r.xP2P3,

U2(P) = r.xPIP3, u 3 (p)

= r.xCP(Pl' P2);

cp is called the warpingfunction. (a) Compute the corresponding stress fieldS and show that Div S = 0 is equivalent to Acp = O. Here etc. (b) Show that (2) is equivalent to

for all p E !fl, where (11)

(c)

is the normal derivative of cp on !fl. Show that (9) and (10) hold, where

m = Kr.x, K=f.l K

f (Pi+P~+PICP'2-P2CP'1)dA;

J.9'o

is called the torsional rigidity of the cross section.

x.

220

LINEAR ELASTICITY

34. LINEAR ELASTODYNAMICS In this section we will establish some of the basic results of linear elastodynamics. Here the basic equations, as derived previously, are E

t(V'u

=

+ V'u T ) , (1)

S = C[E], Div S

+b=

pii

(where we have written p for Po and b for bo). We assume that f!4 is bounded and that p is continuous. Let [u, E, S] be a list offields on f!4 x [0, 00) with u ofclass C 2 and E and S smooth, and suppose that (1) holds with b a given body force field on f!4 x [0, 00). Then [u, E, S] is called an elastic process corresponding to b. Since E is time-dependent, the strain energy Ol/{E} = t

L

E' C[E] dV

depends on time. In fact, when C is symmetric, t(E' C[E]r =

teE' C[E] + E • C[E]) =

E· C[E] = S . E.

Thus (Ol/{E})' =

LS'EdV,

(2)

so that the rate of change of strain energy is equal to the stress power. Theorem of Power and Energy. Assume that C is symmetric. Let [u, E, S] be an elastic process corresponding to the body force b. Then

f

Sn·aidA

+

oBI

f

b'udV = (Ol/{E} + f{ti})·,

(3)

BI

where

is the kinetic energy. Proof b - po,

In view of the lemma on page 205 with ii replaced by 0 and b by

f

oBI

Sn- 0 dA

+

f b·o BI

dV =

f

S . E dV +

BI

f

pii . 0 dV.

BI

34.

221

LINEAR ELASTODYNAMICS

This relation, (2), and the identity (f{o}Y = Lpn'o dV

imply (3).

D

This theorem asserts that the power expended by the external forces equals the rate at which the total energy is changing.

Corollary. Assume that C is symmetric. Let [u, E, SJ be an elastic process corresponding to body force b = 0, and suppose that Sn'o=O

on iJf!4.

(4)

const.

(5)

Then the total energy is constant: OlI{E}

+ f{o}

=

Another interesting corollary may be deduced as follows. Consider an elastic process which starts from an unstrained rest state. Then E(·,O)

=

0,

0(·,0) = 0,

and the total energy is initially zero. Assume that C is positive definite and p > 0, so that the total energy is always nonnegative. Then, if we integrate (3) with respect to time from 0 to r, we arrive at

ItJr o

Sn'o dA dt

+

It Jr 0

iJBI

b·o dV dt

~

O.

(6)

BI

The left side of (6) represents the work done by the external forces in the time interval [0, 't]; (6) asserts that for an elastic process starting from an unstrained rest state this work is always nonnegative. The mixed problem of elastodynamics can be stated as follows: Given: f!4, complementary regular subsets.Z, and !7 2 of iJf!4, an elasticity tensor C on f!4, a density field p on f!4, a body force field b on f!4 x [0, co), surface displacements uon !7 I X [0, co), surface tractions § on !7 2 X [0, co), an initial displacement field U o on f!4, an initial velocity field Vo on f!4. Find: An elastic process [u, E, SJ that corresponds to b, satisfies the initial conditions

u(p,O) = uo(p),

o(p,O) = vo(p)

for every p E f!4, and satisfies the boundary conditions u=

u

on

!7 I

X

[0, co),

Sn = § on

!7 2

X

[0, oo).

An elastic process with these properties will be called a solution.

222

X.

LINEAR ELASTICITY

Uniqueness Theorem. The mixed problem has at most one solution provided C is symmetric and positive definite and p > O.

Proof Let [u, E, S] denote the difference between two solutions. Then [u, E, S] is an elastic process corresponding to b = 0 and satisfies u(-, 0) = u(', 0) = 0, u = 0

on

II'1

[0, (0),

X

Sn = 0

on

II' 2

X

[0, (0).

(7)

Thus (4) holds and we conclude from (5) that

O//{E} + Jr'{u}

=0

(8)

for all time. Here we have used the fact that, by (7)1,2' the total energy is initially zero. But since C is positive definite and p > 0, both the strain energy and kinetic energy are nonnegative; hence (8) implies that Jr' {u} = 0, and this in turn yields u = 0 on ~ x [0, (0). This fact and (7)1 imply that u = 0

on

x [0, (0),

~

0

and the proof is complete.

EXERCISES

Throughout these exercises C is symmetric and

bounded.

~

1. Consider an elastic process and define

e = !E· C[E] q

=

+ !pu 2 ,

-SU.

Show that

e= 2.

- Div q

+ b· U.

Consider an elastic process with null initial data, i.e.,

(Brun's theorem) with

u(p, 0) = 0, for all p E~.

u(p, 0)

=

(9)

0

Show that

O//{E} - Jr'{u} = cI>, where cI>(t) = -1

2

q>(rx,

with s

/3) =

= Sn.

it

[q>(t + 1', t - 1') - q>(t - 1', t + 1')] di,

0

r s(p, rx) • u(p, /3) dA

Jo~

(10)

p

+

r b(p, c) • u(p, /3) dV

J~

p,

223

35. PROGRESSIVE WAVES

3. Use (8) and (10) to establish uniqueness for the mixed problem when C is symmetric (but not necessarily positive definite) and p > O. 4. Show that under the null initial data (9) the equation of motion Div S + b = pii can be written in the form

). * (Div S + where ).(t) = t and

b)

= pu,

* denotes convolution; i.e.,if A and 'P are functions on

14 x [0, (0),

(A * 'P)(p, t) = EA(P, t - r)'P(p, r) dt: 5.

(11)

(Graffi's reciprocal theorem) Let us agree to use the notation (11) when A and 'P are vector fields, but here the product in the integrand is the inner product A(p, t - r)· 'P(p, r), Show that if [u, E, S]_ and [ii, E, S] are

elastic processes corresponding to body forces band b, respectively, and to null initial data, then

r s * ii dA + f~ b * ii dV = Jo~r s * u dA + f~ b* u dV,

Jo~

where s = Sn and

s=

(12)

Sn.

35. PROGRESSIVE WAVES Sinusoidal progressive waves form an important class of solutions to the equations of linear e1astodynamics. We will study these waves under the assumptions of homogeneity and isotropy, and in the absence of body forces. The underlying field equation is then the displacement equation of motion Jl .6u

+ (A. +

(1)

Jl)V Div u = pii.

A vector field u of the form u(p, t) = a sin(r . m - ct),

r = p-

0

(2)

with Im I = 1 is called a sinusoidal progressive wave with amplitude a, direction m, and velocity c. We say that u is longitudinal if a and m are parallel, transverse if a and m are perpendicular. We now determine conditions which are necessary and sufficient that (2) solve (1). By the chain rule, Vu

= (a ® m) cos tp,

x.

224

LINEAR ELASTICITY

where cp(p, t) = r . m - ct,

and hence Div u = (a . m) cos cp, Curl u = (m x a) cos tp, Thus the wave is longitudinal ¢;> Curl u = 0, transverse ¢;> Div u =

o.

Next, L1u V Div u ii

=

-a sin cp,

= -(a' m)m sin cp, = - c 2a sin cp,

and u satisfies (1) if and only if p,a

+ (A. + p,)(a . m)m =

pc2a.

(3)

We call the tensor A(m)

1 p

= - [p,I + (A. + p,)m ® m]

the acoustic tensor; with this definition (3) takes the simple form A(m)a

=

c 2a.

(4)

Thus a necessary and sufficient condition that u satisfy (1) is that c 2 be an eigenvalue and a a corresponding eigenvector of the acoustic tensor A(m). A simple computation shows that A(m)

=

(A.:

2P,)m ® m

+~

(I - m ® m).

But this is simply the spectral decomposition of A(m), and we conclude from (b) of the spectral theorem (page 11) that and

p,jp

are the eigenvalues of A(m), while theline spanned by m and the plane (through 0) perpendicular to m are the corresponding characteristic spaces. Thus we have the following

35. PROGRESSIVE WAVES

225

Theorem. A sinusoidal progressive wave with velocity c will be a solution of( 1) if and only if either

(a) (b)

c 2 = (2J1

+ )..)/p and the

wave is longitudinal, or

c2 = J1/P and the wave is transverse.

This theorem asserts that for a homogeneous and isotropic material only two types of sinusoidal progressive waves are possible: longitudinal and transverse. The two corresponding speeds J(2J1

+ )..)jp

and

are called, respectively, the longitudinal and transverse sound speeds of the material. (Note that these speeds are real when the elasticity tensor is positive definite.) For an anisotropic body the situation is far more complicated. A propagation condition of the form (4) still holds (Exercise 1), but the waves will generally be neither longitudinal nor transverse, and they will generally propagate with different speeds in different directions.

EXERCISES

1. For an anisotropic but homogeneous body the displacement equation of

motion has the form Div C[Vu] = pii

(5)

with C and p independent of position. (Here we have assumed zero body forces.) (a)

Show that the sinusoidal progressive wave (2) solves (5) if and only if the propagation condition (4) holds, where now A(m) is the tensor defined by A(m)k

1

= - C[k ® m]m p

for every vector k. (b) Show that A(m) is positive definite for each direction m if and only if C is strongly elliptic (cf. Exercise 29.3). (c) Show that QA(m)QT

= A(Qm)

for any symmetry transformation Q.

226

X.

LINEAR ELASTICITY

SELECTED REFERENCES

Fichera [1]. Gurtin [1]. Love [1]. Sokolnikoff [1]. Villaggio [1].

Appendix

36. THE EXPONENTIAL FUNCTION The exponential of a tensor can be defined either in terms of its series representation 1 or in terms of a solution to an ordinary differential equation. For our purposes the latter is more convenient. Thus let A be a tensor and consider the initial-value problem

X(t)

= AX(t),

t > 0,

X(O) = I,

(1)

for a tensor function X(t), 0 ~ t < 00. The existence theorem for linear differential equations tells us that this problem has exactly one solution X: [0, 00) ~ Lin, which we write in the form X(t) = e'".

Proposition. Proof

1

For each t ~ 0, eAt belongs to Lin + and det(eAt ) = e1trA)t.

Let X(t)

= e", Since X is continuous and det X(O) = 1,

See, e.g., Hirsch and Smale (I, Chapter 5, §3].

227

(2)

(3)

228

APPENDIX

det X(t) > 0 in some nonempty interval [0, r). Let r be as large as possible; that is, let r = sup{Aldet X(t) > 0 for

0

~

t ~ A}.

To show that X(t) E Lin + for all t ~ 0 we must show that r = the contrary, that r is finite. Then, since X is continuous,

00.

Suppose, to

det X(r) = O.

(4)

We will show that this leads to a contradiction. Since X(t) is invertible for all t E [0, r), (1)1 implies A = X(t)X(t)-1

on (0, r), and we conclude from (3.14) that (det X)" = (tr A) det X on (0, r). This equation with the initial condition (3) has the unique solution det X(t)

=

(5)

e(trAlt

for 0 ~ t < r. In view of the continuity of X, this result clearly implies that det X(r) > 0, which contradicts (4). Thus r = 00 and (5) implies (2). 0 Proposition.

Proof

Let W be a skew tensor. Then eW t is a rotation for each t ~ O.

Let X(t) = eW t for t ~ O. Then, by definition, X=WX,

X(O)

=

I.

Let

Then, since W is skew, Z = XX T + XX T = WXX T + XXTW T = WXX T - XXTW,

and Z satisfies Z=WZ-ZW,

Z(O) = I.

This initial-value problem has the unique solution Z(t) = I for all t

~

O. Thus

X(t)X(t)T = I

and X(t) is orthogonal. But from the preceding proposition X(t) X(t) is a rotation. 0

E

Lin +. Thus

229

37. ISOTROPIC FUNCTIONS EXERCISE

1. Let A E Lin and Vo E "Y. Show that the function

=

eA1v o

Av(t),

t

vet) satisfies the initial-value problem vet)

=

v(O)

= Vo '

> 0,

SELECTED REFERENCE

Hirsch and Smale [1].

37. ISOTROPIC FUNCTIONS Let ~ c Orth. A set d c Lin is invariant under ~ if QAQTEd whenever AEdandQE~.

Proposition.

Thefollowing sets are invariant under Orth: Lin,

Lin +,

Orth,

Orth +,

Sym,

Skw,

Psym.

Proof We will give the proof only for Lin +. Choose A E Lin and Q E Orth. Then det(QAQT) = (det A)(det Q)2 = det A, since [det QI = 1. Thus A E Lin" implies QAQT E Lin+.

(1)

0

Let d c Lin. A scalar function qJ: d --+ IR is invariant under ~ if d is invariant under ~ and qJ(A)

=

qJ(QAQT)

(2)

for every A E d and Q E ~. Similarly, a tensor function G: d --+ Lin is invariant under ~ if d is invariant under ~ and

for every A E d and Q E ~. An isotropic function is a function invariant under Orth.

230

APPENDIX

Proposition. Let <1l be a scalar or tensor function with domain in Lin. Then <1l is isotropic if <1l is invariant under Orth +. Proof

The proof follows from the identity (-Q)A( _Q)T = QAQT

and the fact that for Q E Orth either Q or - Q belongs to Orth + .

0

Theorem (a) det and tr, considered as functions on Lin, are isotropic. (b) The list (2.9) of principal invariants is isotropic; that is, J A

=

(3)

JQAQT

for every A E Lin and Q E Orth, Proof

That det is isotropic follows from (1). Next, for Q E Orth, tr(QAQT) = tr(AQTQ) = tr A,

so that tr is isotropic. By (2.8) and (2.9), since A f-+ (tr A)2 is isotropic, to establish (b) we have only to show that A f-+ tr(A 2) is isotropic. Choose A E Lin and Q E Orth. Then (QAQT)2 = QAQTQAQT = QA 2QT, and hence

For convenience, we write

for the set of all possible lists J

A

as A ranges through the set d. Of course,

J(d) c lR 3 •

Next we establish several important representation theorems for functions with domain .91 in Sym. We assume for the remainder of this section that .91 is invariant under Orth. Representation Theorem for Isotropic Scalar Functions.

A function

(.91 c Sym) is isotropic

if and only if there exists a funcionii:

J(d)

-+

lR such that (4)

for every A E d.

37.

231

ISOTROPIC FUNCTIONS

Proof Assume that qJ is isotropic. To establish the representation (4) it suffices to show that qJ(A) = qJ(B)

(5)

.FA = .Fa.

(6)

whenever

Thus let A, BE Sym and assume that (6) holds. By (6) and the proposition on page 16, A and B have the same spectrum. Thus by the spectral theorem there exist orthonormal bases {ei} and {fi} such that A=

L (JJiei <8> e;,

B

=

L (JJifi <8> f

i·

i

i

Let Q be the orthogonal tensor carrying the basis {fil into the basis {e.}:

Qfi = ei · Then, since

it follows that QBQT = A.

But since qJ is isotropic, qJ(QBQT) = qJ(B); thus (5) holds and qJ admits the representation (4). The converse assertion, that (4) defines an isotropic function, is a trivial consequence of (3). D Transfer Theorem.

Let

G: d

-+

Lin

(d c Sym)

be isotropic. Then every eigenvector of A E d is an eigenvector of G(A). Proof Let e be an eigenvector of A E .91, and let Q E Orth be the reflection across the plane perpendicular to e:

Qe = -e,

Qf= f

if f· e = O.

(7)

Then by the spectral theorem, Q leaves invariant the characteristic spaces of A; hence we may conclude from the commutation theorem that QAQT = A.

Thus, since G is isotropic,

232

APPENDIX

so that Q commutes with G(A). Therefore QG(A)e = G(A)Qe = - G(A)e, and Q transforms G(A)e into its negative. But by (7) this can happen only if G(A)e is parallel to e: G(A)e = we.

D

Thus e is an eigenvector of G(A).

The following lemma, whose statement is based on the spectral theorem, will be extremely useful in what follows. Wang's Lemma. Let A E Sym. (a)

Consider the spectral decomposition (8)

A = LWiei®ei, i

and assume that the eigenvalues Wi are distinct. Then the set {I, A, A 2 } is linearly independent and (9) (b)

Assume that A has exactly two distinct eigenvalues, so that

A

=

w1e®e

+ w2(I -

e®e),

lei =

1.

(10)

Then {I, A} is linearly independent and

sp{I, A}

=

sp{e ® e, I - e ® e}.

(11)

Proof We prove only (a). To establish the linear independence of {I, A, A 2} we must show that

exA 2

+ f3A + yI =

0

(12)

implies

ex = f3 = y = O.

(13)

Assume (12) holds. Acting with (12) on the eigenvector e, leads to the relation

(exwf

+ f3Wi + y)ei

= 0,

so that

exwf

+ f3Wi + Y = 0,

and the Wi are roots of a quadratic equation; since the co, are distinct, this is possible only if (13) hold. Thus {I, A, A2} is linearly independent.

37. ISOTROPIC FUNCTIONS

233

Next, the subspace

.Ye = sple, ® e l , e 2 ® e 2 , e 3 ® e 3 } of Lin has dimension 3. On the other hand, since A2 =

L ro;e

i

® ei ,

(14)

i

as is clear from (1.2h and (8), we may conclude from (1.2)4 and (8) that I, A, A 2 E .Ye. But 3 linearly independent vectors in a vector space ofdimension 3 must necessarily span the space. Thus.Ye = sp{I, A, A 2} and the proof of (a) is complete. We leave the remainder of the proof as an exercise. 0 We are now in a position to establish several important representation theorems for isotropic tensor functions. First Representation Theorem for Isotropic Tensor Functions.

(.91 c Sym)

G: .91 ---. Sym

is isotropic that

A function

if and only if there exist scalar functions ({Jo, ({Jl' ({J2: j(d) ---.!R such (15)

for every A E d. Proof Assume first that G admits the representation (15). Choose A E .91 and Q E Orth. Then by (3), G(QAQT)

= =

({Jo(..¢"QAQT)I + ({Jl(..¢"QAQT)QAQT + ({J2(..¢"QAQT)QAQTQAQT ({JO(..¢"A)QQT + ({Jl(..¢"A)QAQT + ({J2(..¢"A)QA 2QT = QG(A)QT,

so that G is isotropic. To prove the converse assertion assume that G is isotropic. Choose AEd.

Case 1. A has exactly three distinct eigenvalues. Let (8) be the spectral decomposition of A. By the transfer theorem G(A)

=

L Piei ® e., i

and we conclude from (9) that there exist scalars IXo(A), IXl (A), and IXiA) such that (16)

234

APPENDIX

Case 2. A has exactly two distinct eigenvalues. Then A admits the representation (10), and the characteristic spaces for A are sp{e} and {e}.L. By the transfer theorem each of these two subspaces must be contained in a characteristic space for G(A). This is possible only if either: (i) G(A) has as characteristic spaces sp{e} and {e}', or (ii) -r:is the only characteristic space for G(A). By (b) and (c) of the spectral theorem,

(17) in either case [withfi, =F p2incase(i),p1 = p2incase(ii)].Inviewof(1l),(17) can be written in the form (16) with (18) Case 3. A has exactly one distinct eigenvalue. Then by (c) of the spectral theorem in conjunction with the transfer theorem, f is the characteristic space for A and G(A), so that G(A) = PI, and G(A) again admits the representation (16) with oco(A) = Pand oct(A) = oc 2(A) = O.

Since these three cases are exhaustive, we have shown that G, when isotropic, admits the representation (16). In view of the representation theorem for isotropic scalar functions, to complete the proof we have only to show that OCo, oc 1, and OC2 are isotropic: (k

= 0, 1,2)

(19)

for every A E.rd and Q E Orth. Thus choose A E.rd, Q E Orth. Since G is isotropic,

or equivalently,

and, since

it follows that [oco(A) - oco(QAQT)]I

+ [oc 1(A) - OCt(QAQT)]A + [ociA) - oc 2(QAQT)]A2 =

O. (20)

We consider again the three cases studied previously. Case 1. By (a) of Wang's lemma, {I, A, A2} is linearly independent, so that (20) implies (19).

235

37. ISOTROPIC FUNCTIONS

Case 2. By (3) and the proposition on page 16, A and QAQT have the same spectrum. Thus A and QAQT each have exactly two distinct eigenvalues and (18) yields C\:iA) = C\:2(QAQT) = O. Moreover, by (b) of Wang's lemma, {I, A} is linearly independent, and (20) again yields (19). Case 3. Here A

= A.I, so that A =

QAQT and (19) holds trivially.

D

By the Cayley-Hamilton theorem (2.12), A 2 = 11(A)A - liA)I

+ 13(A)A- l ,

provided, of course, A is invertible. Thus, by (2.9), the previous theorem has the following corollary: Second Representation Theorem for Isotropic Tensor Functions. a subset ofthe set ofall invertible, symmetric tensors. Then

Let ,?i be

G: d -+Sym is isotropic if and only if there exist scalar functions that

Po, Pi' P2: J(d) -+ IR such (21)

for every A E d.

For linear functions there is a far simpler result. Representation Theorem for Isotropic Linear Tensor Functions. function

A linear

G: Sym -+ Sym is isotropic

if and only if there exist scalars

f:l and A such that

G(A) = 2f:lA + A.(tr A)I

(22)

for every A E Sym. Proof Let % be the set of all unit vectors. For e E % the tensor e ® e has spectrum {O, 0, I} and characteristic spaces sp{e} and {e}1. • Thus the same argument used to arrive at (17) now leads to the conclusion that there exist functions f:l, A: % -+ IR such that G(e

® e) = 2f:l(e)e ® e + A(e)I

(23)

236

APPENDIX

for every e E.iV. Choose e, f E.iV, and let Q be any orthogonal tensor such that Qe = f. (Clearly, at least one such Q exists.) Since Q(e ® e)QT

= f®

f,

and since G is isotropic,

o=

=

QG(e ® e)QT - G(f ® f)

2[/l(e) - tl(f)]f ® f

+ [A(e) -

A(f)]I.

But {I, f ® f} is linearly independent; thus A(e) = A(f).

/l( e) = J.l(f),

Therefore /l and A must be scalar constants, and we conclude from (23) that

=

G(e ® e)

2/le ® e

+ AI.

(24)

Next, choose A E Sym arbitrarily. By the spectral theorem A admits the representation

A=

L

OJi

e, ® e,

i

with {e.} orthonormal; therefore, in view of (24) and the linearity of G, G(A) =

L OJiG(e

i

® e.)

= 2/lA

+ A.(OJ l + OJ2 + OJ 3 )1.

(25)

i

Since tr A

=

OJ 1

+ OJ2 + OJ3,

(25) implies the desired result (22). The converse assertion, that (22) delivers an isotropic function, is left as an exercise. D Corollary.

Let

Sym, = {Ae Symltr A = O}, and let G: Sym.,

be linear. Then G is isotropic

-+

Sym

if and only if there

exists a scalar J1 such that (26)

for every A E Symo ' Proof Clearly (26) defines an isotropic function. To prove the converse assertion assume that G is isotropic. For A E Sym, A - j{tr A)I belongs to Symo . Thus we can extend G from Sym., to Sym by defining C(A)

=

G(A - j{tr A)I)

37.

237

ISOTROPIC FUNCTIONS

for every A E Sym. Since G is isotropic, for every A E Sym and Q E Orth, QG(A)QT = G(QAQT - !
t

where we have used the fact that tr is isotropic. Thus G is isotropic, and we conclude from (22) and the fact that G and G coincide on Sym., that G(A) = G(A) = 2JlA + A(tr A)I = 2JlA for every A E Syrnj , 0 Our next step is to discuss the invariance properties of the derivative of a tensor function. Thus let d be an open subset of a subspace 0/1 of Lin, and let G: d -+ Lin be smooth. Theorem

Let G be invariant under

(Invariance of the derivative).

I'§.

Then

(27) for every A

E

d, U

E

0/1, and Q

E I'§.

Proof For (27) to make sense we must show that both d and 0/1 are invariant under I'§. Since d is the domain of G and G is invariant, d is (by definition) invariant under I'§. To see that 0/1 has this invariance, choose U E 0/1, A E d, and Q E I'§. Then for all sufficiently small a, (A + !XU) E d, so that Q(A + !XU)QT E d and hence (QAQT + CXQUQT) E d c 0/1. Thus, since 0/1 is a subspace, QUQT E 0/1, and 0/1 is invariant under I'§. We have only to show that (27) is satisfied. For A E d, U E 0/1, and Q E I'§,

G(Q(A + U)QT) = G(QAQT + QUQT) = G(QAQT) as U

-+

O. But since G is invariant under

+ DG(QAQT)[QUQT] + o(U)

I'§,

G(Q(A + U)QT) = QG(A + U)QT = QG(A)QT + QDG(A)[U]QT = G(QAQT) + QDG(A)[U]QT as U

-+

O. Thus by the uniqueness of the derivative,

which is (27). 0

+ o(U), + o(U)

238

APPENDIX EXERCISES

1. Show that the sets Orth, Orth + , Sym, Skw, and Psym are invariant under

Orth. 2. A scalar function

qJ:

"Y -

is isotropic if

~

qJ(v)

= qJ(Qv)

for all v E "Y and Q E Orth. Show that qJ is isotropic if and only if there exists a function tp: [0, (0) - ~ such that qJ(v)

= tp(lvl)

for all v E "Y. 3. A vector function q: "Y - "Y is isotropic if Qq(v) = q(Qv)

for all v E "Y and Q E Orth. Show that q is isotropic if and only if there exists a function qJ: [0, (0) - ~ such that q(v) =

qJ( IvJ)v

for all v E "Y. 4. Let G: Lin - Lin be defined by G(A)

= A"

(n ~ 1

an integer).

Show that G is isotropic. 5. Show that the mapping

from Lin + into Lin + is isotropic. 6. Show that the function G: Sym - Sym defined by (22) is isotropic. 7. Establish (b) of Wang's lemma.

SELECTED REFERENCES

Gurtin [2]. Martins and Podio-Guidugli [2]. Serrin [1, §59]. Truesdell and Noll (1, §§10-13].

38.

239

GENERAL SCHEME OF NOTATION

38. GENERAL SCHEME OF NOTATION INDEX OF FREQUENTLly USED SYMBOLS Symbol

a .i4 fJI, b bo B C c

C curl div Div D

D

det E

E

8 f

F g C§

H 5s I %{Ii}

L ~

Lin

Lin" m(&') m

n

o(u) 0

Orth Orth " p p &' Psym

Q R

Name

angular momentum body region occupied by body at time t body force reference body force left Cauchy-Green strain tensor right Cauchy-Green strain tensor curve elasticity tensor curl, spatial curl divergence, spatial divergence material divergence derivative stretching determinant Young's modulus infinitesimal strain euclidean point space deformation deformation gradient complex velocity symmetry group displacement gradient list of principal invariants of a tensor S identity tensor kinetic energy velocity gradient linear momentum set of all tensors set of all tensors S with det S > 0 mass of &' subscript indicating material description, also mach number outward unit normal to boundary symbol for "small order u" origin set of all orthogonal tensors set of all rotations (proper orthogonal tensors) material point reference map part of :J4 set of all symmetric, positive definite tensors orthogonal tensor rotation tensor position vector

Place of definition or first occurrence

92 41 58 98 179 46 46 34 194 32,61 30,61 61 21 71 6 203 55 I 42 42 123 168 199 15 3 220 63 92 7 7 87 60,133 37 19 2 7 7 41 59 41 7 7 46 92

240

Symhol IR IR+

S s Sym Skw ,;

sp tr t T

or u 'f/{E} U

V v v

r vol W w

x

x or.

p

cp

{,;} liS) K(p) Ak ). tt

Po P 8(F) Jl. v

ro V ~

Q9

x

ep cp'

APPENDIX

Name

reals strictly-positive reals Piola-Kirchhoff stress stress vector set of all symmetric tensors set of all skew tensors subscript indicating spatial description, also kinematically admissible state span trace time Cauchy stress trajectory displacement strain energy right stretch tensor left stretch tensor spatial description of the velocity speed vector space associated with r! volume spin complex potential place motion center of mass potential of body force potential of flow potential energy principal invariants of a tensor S sound speed principal stretches Lame modulus pressure reference density density in motion strain-energy density viscosity, also shear modulus kinematic viscosity, also Poisson's ratio angular velocity gradient, material gradient laplacian tensor product cross product material time derivative of cp spatial time derivative of cp

Place of definition or first occurrence I I

178 97 7 7 60,208 2 5 58 101 59 42 205 46 46 60 133 I

37 71 126 58 58 92 III III

208 15 131 45 196 106 88 88 186 149,196 151.203 70 29,60 32 4 7 61 61

38.

241

GENERAL SCHEME OF NOTATION GENERAL NOTATION

Expression

~ agp ~

gp U ,F gp n .17 gpc:F XEgp

j: gp .......'1" x ....... j(x)

jog

Meaning

interior of gp boundary of gp closure of gp union of gp and .F intersection of gp and .F gp is a subset of .F x is an element of the set gp j maps the set gp into the set .'1"; gp is the domain, :J7 the codomain the mapping that carries x into j(x); e.g., x ....... Xl is the mapping that carries every real number x into its square composition of the mappingsj andg; that is,(f 0 g) (x) =

{xIR(x) holds}

j(g(x»

the set of all x such that R(x) holds; e.g.,{x10 ~ x is the interval [0, 1]

~

I}

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References

R. G. Bartle [I] The Elements of Real Analysis, 2nd ed. New York: Wiley (1976). R. B. Bird, W. E. Stewart, and E. N. Lightfoot [I] Transport Phenomena. New York: Wiley (1960). R. M. Bowen and C. C. Wang [1] Introduction to Vectors and Tensors, Vol. 1. New York: Plenum (1976). H. W. Brinkmann and E. A. Klotz [I] Linear Alqebra and Analytic Geometry. Reading: Addison-Wesley (1971). P. Chadwick [1] Continuum Mechanics. New York: Wiley (1976). B. D. Coleman, H. Markovitz, and W. Noll [I] Viscometric flows of non-Newtonian fluids. Springer Tracts in Natural Philosophy 5. New York: Springer-Verlag (1966). B. D. Coleman and W. Noll [I] Material symmetry and thermostatic inequalities in finite elastic deformations. Arch. Rational Mech. Anal. 15,87-111 (1964). R. Courant and K. O. Friedrichs [1] Supersonic Flow and Shock Waves. New York: Wiley (Interscience) (1948). J. Dieudonne [1] Foundations of Modern Analysis. New York: Academic Press (1960). A. C. Eringen [I] Nonlinear Theory of Continuous Media. New York: McGraw-Hili (1962). G. Fichera [1] Existence theorems in elasticity. Handbuch der Physik 692' Berlin: Springer-Verlag (1972). 243

244

REFERENCES

W. Fleming [I] Functions of Several Variables. Reading: Addison-Wesley (1965). P. Germain [I] Cours de Mecanique des Milieux Continus, Vol. 1. Paris: Masson (1973). A. E. Green and W. Zerna [I] Theoretical Continuum Mechanics. London and New York: Oxford Univ. Press (1954). M. E. Gurtin [I] The linear theory ofelasticity. Handbuch der Physik 6a 2 . Berlin: Springer-Verlag (1972). [2] A short proof of the representation theorem for isotropic, linear stress-strain relations. J. Elasticity 4, 243-245 (1974). M. E. Gurtin and L. C. Martins [I] Cauchy's theorem in classical physics. Arch. Rational Mech. Anal. 60,305-324 (1976). M. E. Gurtin and W. O. Williams [I] An axiomatic foundation for continuum thermodynamics. Arch. Rational Mech. Anal. 26, 83-117 (1967). P. R. Halmos [I] Finite-Dimensional Vector Spaces, 2nd ed. New York: Van Nostrand-Reinhold (1958). M. W. Hirsch and S. Smale [I] Differential Equations, Dynamical Systems and Linear Algebra. New York: Academic Press (1974). T. J. R. Hughes and J. E. Marsden [I] A Short Course in Fluid Mechanics. Boston: Publish or Perish (1976). O. D. Kellogg [I] Foundations of Potential Theory. Berlin and New York: Springer-Verlag (1929). Reprinted New York: Dover (1953). O. A. Ladyzhenskaya [I] The Mathematical Theory of Viscous Incompressible Flow. New York: Gordon & Breach (1963). H. Lamb [I] Hydrodynamics, 6th ed. London and New York: Cambridge Univ. Press (1932). Reprinted New York: Dover (1945). A. E. H. Love [I] A Treatise on the Mathematical Theory of Elasticity, 4th ed. London and New York: Cambridge Univ, Press (1927). Reprinted New York: Dover (1944). L. C. Martins and P. Podio-Guidugli [I] A variational approach to the polar decomposition theorem. Rend. Accad. Naz. Lincei (Classe Sci. Fis., Mat. Natur.) 66(6), 487-493 (1979). [2] A new proof of the representation theorem for isotropic, linear constitutive relations. J. Elasticity 8, 319-322 (1978). R. E. Meyer [I] Introduction to Mathematical Fluid Dynamics. New York: Wiley (lnterscience) (1971). I. P. Natanson [I] Theory of Functions ofa Real Variable. New York: Ungar (1955). W. Noll [I] On the continuity of the solid and fluid states. J. Rational Mech. Anal. 4,3-81 (1955). [2] A mathematical theory of the mechanical behavior of continuous media. Arch. Rational Mech. Anal. 2, 197-226 (1958). Reprinted in W. Noll, The Foundations of Mechanics and Thermodynamics. Berlin: Springer-Verlag (1974). [3] The foundations of classical mechanics in the light of recent advances in continuum mechanics. The Axiomatic Method, with Special Reference to Geometry and Physics.

REFERENCES

245

Amsterdam: North-Holland Pub!. (1959). Reprinted in W. Noll, The Foundations of Mechanics and Thermodynamics. Berlin: Springer-Verlag (1974). [4] La rnecanique classique, basee sur un axiome d'objectivite. La Methode Axiomatique dans les Mecaniques Classiques el Nouvelles. Paris: Gauthier-Villars (1963). Reprinted in W. Noll, The Foundations of Mechanics and Thermodynamics. Berlin: Springer-Verlag (1974). H. K. Nickerson, D. C. Spencer, and N. E. Steenrod [I] Advanced Calculus. New York: V<'11 Nostrand-Reinhold (1959). J. Serrin [I] Mathematical principles of classical fluid mechanics. Handbuch der Physik 8 Berlin: " Springer-Verlag (1959). I. S. Sokolnikoff [I] Mathematical Theory ofElasticity, 2nd ed. New York: McGraw-Hill (1956). R. Stephenson [I] On the uniqueness of the square-root ofa symmetric positive-definite tensor. J. Elasticity 10,213-214 (1980). E. Sternberg and J. K. Knowles [I] On the existence of an elastic potential for a simple material without memory. Arch. Rational Mech. Anal. 70, 19-30 (1979). F. M. Stewart [I] Introduction 10 Linear Algebra. New York: Van Nostrand-Reinhold (1963). R. Temam [I] Naoier-Stokes Equations, Theory and Numerical Analysis. Amsterdam: North-Holland Pub!. (1977). C. Truesdell [I] A First Course in Rational Continuum Mechanics, Vo!' I. New York: Academic Press (1977). [2] Rational mechanics of deformation and flow (Bingham Medal Address). Proc, 4th lntl. Conqr, Rheol., 1963, 2, 3-50 (1965). C. Truesdell and W. Noll [I] The non-linear field theories of mechanics. Handbuch der Physik 3 3 , Berlin: SpringerVerlag (1965). C. Truesdell and R. A. Toupin [I] The classical field theories. Handbuch der Physik 3,. Berlin: Springer-Verlag (1960). P. Villaggio [I] Qualitatiie Methods in Elasticity. Leyden: Noordhoff(l977). C. C. Wang and C. Truesdell [I] Introduction 10 Rational Elasticity. Leyden: Noordhoff(1973).

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Hints for Selected Exercises

SECTION 1

L Viejo

1.

Consider o/ev) with v =

2.

Take

3.

Show that F = S + T is linear:

8 =

L o/(e;)ei' F(u + v) = Fu + Fv, F(ocu) = ocFu

4.

6.

for all vectors U, v and scalars oc. Fix v and consider Su . v as a function of u. Show that this function is linear and hence can be written as the inner product of a vector a, (which depends on v) and u:

Show that a, is a linear function of v and hence can be written as a, = Av with A a tensor. Define ST = A. Apply each side of the identity to an arbitrary vector v. 247

248 7.

HINTS FOR SELECTED EXERCISES

To verify (1)1 prove that u . (S

14.

+ T)v =

+ TT)U . v,

(ST

which establishes ST + TT as the transpose of S + T. To verify (2)z apply each side of the identity to an arbitrary vector v. Use the identity

to show that

SECTION 2 2. For part (a) use (b) of the first proposition. To derive (2) and (3) use (1) and (1.2)3' 3. Show that QDQT E Sym and f D = fQDQT. 4b. Prove that each eigenvalue of P is either or 1 and then (with the aid of the spectral theorem) consider separately the following four possibilities for the spectrum of P: (0,0,0), (1, 1, 1), (0, 1, 1), (0, 0, 1).

°

5. (only if) Let S commute with every WE Skw. Choose a vector wand let W be the skew tensor whose axial vector is w. Show that W(Sw) = SWw

= 0,

so that Sw belongs to the axis of W and is hence parallel to w. Thus Sw = aw. (Of course, a may depend on the choice of w.) For S E Sym use the spectral theorem and the fact that w is arbitrary to establish (14). For S E Skw,

°

=

w . Sw = aw

2

=> a =

°

and (14) holds trivially. We therefore have the desired result for S E Sym and S E Skw. Complete the proof by showing that if S commutes with every WE Skw, then so also do the symmetric and skew parts of S. 7. Choose Q E Orth +, Q #- R. Use the identity IF - Q 12

= IF 12

-

2F • Q

+3

SECTION

5

249

to show that F - Q 12

1

where F

IF - R 12

-

=

2U . (I - Qo),

= RU is the right polar decomposition of F and Qo = RTQ =F I.

Next, establish the identity 2U • (I - Qo) = U· (Qo - I)(Qo - I)T = tr{(Qo - I)TU(Qo - I)}, and show that U •(I - Qo) > 0 by showing that (Qo - I?U(Qo - I) E Psym. SECTION 3 2. Differentiate AA - 1 = I. 3. Use the chain rule. 6. Differentiate QQT = I. 7. Take Q(t) = eW t• 8c. Show that Dcp(A) [U]

= U·

f

{S(A o

+ IXA) + IX DS(A o + IXA) [A]} da,

and show further that the integrand is equal to d dlX [IXS(A o

+ IXA)].

SECTION 4 2. Show that div(STa) is linear in a and use the representation theorem for linear forms. SECTION 5 l a.

Use the fact that, given any vector a,

(1

o~

v ® n dA)a =

1

(v

o~

® n)a dA =

1

(v

o~

® a)n dA.

250

HINTS FOR SELECTED EXERCISES

SECTION 6 6. 9.

Use (2.7) and (2.10) to show that .fc = (3, 3, 1) if and only if the spectrum of Cis (1, 1, 1). Let and be right polar decompositions of F 1 = Vf 1 and F 2 = Vf2 , and note that U 1(p) = U 2 (p) for all P E .sf. Define g = f 2 fi 1. Choose q E f 1(.sf) arbitrarily and let P = fi 1(q). Show that 0

Vg(q)

=

R 2(p)R 1(p?

and appeal to the characterization theorem for rigid deformations. 10.

To establish (18)1 let v = ~>iei'

r

write

L: ei' r

v . m dA =

J

J

i

ol(go)

Vim

dA,

ol(go)

and use (14.2). For (18)z let a be a vector, write a'

f

Tm dA =

J

ol(go)

r

(TTa)· m dA,

J

ol(go)

and use (18)1' For (18h use the identity a . (r x Tm) = (Tm) . (a x r) = m . TT(a x r)

[r(x) = x - 0]

to write a'

r

J

r x Tm dA

=

ol(go)

r

J

m . TT(a x r) dA

iJf(go)

and appeal to (18)1 again. 12. To verify (5)1 use the continuity of f and f- 1 together with the fact that for a continuous function the inverse image of an open set is open. To prove (5)2 note that, since fJI is closed, f(fJI) = f(~)

u f(J.sf),

and since f(fJI) is closed, f(fJI) = f(fJlt u Jf(.sf).

13b.

Use these identities with (5)1 to verify (5)z. Show that f(fJI) is not closed.

SECTION

251

8

SECTION 7 I.

Consider C and U as functions of H = Vu: C

=

U

t(H),

Use the chain rule and the identity Dt(O) [S]

t

= O(H).

=

0 2 to conclude that

= 2 DO(O) [S]

for all tensors S, and then conclude from (5)1 that O(H) = I + E + o(H).

5. Use Exercise 4 in conjunction with Exercise 4.9b. 6. Use the trigonometric identities cos(E> sin(E>

+ {3) = + {3) =

cos E> cos {3 - sin E> sin {3, sin E> cos {3

+ cos E> sin {3,

together with the estimates

= 1 + 0({3), sin {3 = {3 + 0({3).

cos {3

[It is a simple matter to verify that Vu -+ 0 as a -+ 0 without explicitly computing the gradient. Indeed, one simply notes that u, as a function of a and p, is smooth for (a, p) E ~ X 8; one then obtains the above limit by first evaluating u at a = 0 and then taking its gradient.]

SECTION 8 3. Compute div v using (5) and Exercise 4.9a. 7a. Choose PEg and let c(o), 0::;; c(O) = p. Use the relation

0" ::;;

1, be a smooth curve in g with

d dO" cp(c(O"» = 0 to show that Vcp(p) is normal to c(O).

252

HINTS FOR SELECTED EXERCISES

SECTION 9 Id,

Show that the two terms in the right side of (15) are symmetric and skew, respectively, and use the uniqueness of the expansion L = D + W of L into symmetric and skew parts.

3. Show that IxiY, t) - xiz, t)1 = Iy - z],

so that 4c.

xl,

t) is a rigid deformation.

By (19), (n + I)

(!!1 •

T

T

•

C = F An + 1 F = (C) = (F An F) .

SECTION 10 2.

Use (4.2)4 and (9.11).

SECTION 11 1. Use (9.12), (10.2), and (2h.

2. Take the curl of (9.lOh and use the identity curl(w x v) = (grad w)v - (grad v)w + (div v)w - (div w)v and (10.2). 3.

Fix Yand r, and let g(t)

=

v(xt(y, t), t)w(xiY, t), r).

It suffices to show that

g(t)

=

Fly, t)g('t).

By (9), g satisfies a certain ordinary differential equation. Show that f(t) = Fly, t)g('t) satisfies the same differential equation and f('t) = g('t). Appeal to the uniqueness theorem for ordinary differential equations. 5.

Use (9.10).

SECTION

13

253

SECTION 12 3. Show that p(y, r) = p(x, t) det Fly, t)

and then proceed as in the proof of (7). 4. Let f = g - v and use the divergence theorem to prove that f{v} + f{f}.

:f( {g}

=

SECTION 13 3a. Use (9) and (12.7) to convert (4) to an integral over PA o . 3b. Use (10) to eliminate q(t) from (9) and derive xo(y, t)

= a(t) +

Q(t)[y - a(O)].

By (9.13), v(x o(y, t), t) = xo(y, t) = !i(t)

+ Q(t)[y

- a(O)].

Choose x E PAt arbitrarily, and let y be such that x = xo(y, t), i.e., y - a(O) = Q(t)T[X - a(t)].

Then v(x, t) = !i(t)

+ Q(t)Q(t?[x - a(t)].

Compare the gradients (with respect to x) of the above equation and (9.7).

3c. (If) Define g(t) = Q(t)k(O) and show that g and k satisfy the same differential equation and g(O) = k(O). Appeal to the uniqueness theorem for ordinary differential equations. 3e. To establish (15) transform (14) to an integral over PA o using the same steps as in (3a). Then use (12) (with k = e) and (15) to show that ei(t) • J(t)eit) = ej(O) • J(O)ej(O).

3f.

Choose {ei(O)} to be an orthonormal basis of eigenvectors. Since a.pin(t) =

L Jijwit)ej(t), i.]

where Jij are the components of J(t) relative to {ej(t)}, we may conclude, with the aid of (13), that

a. pin = L [JijWjei + JijWj(co i.j

x ej)].

254

HINTS FOR SELECTED EXERCISES

Show that

4a.

i

v,,-pdV

=0

iJIJ,

and then compute Use (11).

4b.

.Y(

with v = v"- + lX.

SECTION 14 8. Prove that it suffices to find a tensor Q such that G(Q) =

r

Jo

r ® Qs(n) dA

+

i

r ® Qb dV

iJIJ,

@,

is symmetric. Show that G(Q) = G(I)QT, take the left polar decomposition of G(I), and choose QT to make G(I)QT symmetric. Here you will need the following extended version of the polar decomposition theorem 1: given F E Lin there exist unique positive semidefinite, symmetric tensors D, V and a (not necessarily unique) orthogonal tensor R such that F = RD = VR. 9. We must show that n· T(x, t)k = 0, whenever 0 is normal to-and k tangent to-ofJl t at x. Use the symmetry of T and the fact that ofJlt is traction-free.

SECTION 15 la. Use Exercise 5.lb. lb. Use Exercise 5.la. 2. Use components. I

cr., e.g.,

Halmos [I, §83].

SECTION

22

255

SECTION 16 1. Show that q) =

Given any tensor N

E

{D

E

Syml tr D = O}.

Sym, write N

=

-1tI

+ No,

where 1t = - -! tr N, so that tr No = O. Show that N . D = 0 for every DE q) if and only if No = O. SECTION 17

2. Use the divergence theorem to express the term involving b in (15.2) as an integral over iJtJ4 t • 3. Compute grad v. 4. Use Bernoulli's theorem. SECTION 20

2a. Use Exercise 2.5. SECTION 21

2. Use (20.1), (6.18)2' and (1) to show that

f

T*n* dA = Q(t)

iJ&t

f

Tn dA.

iJ&,

Apply the divergence theorem to both sides and then use (6.14)1 to convert the integral over fY'~ to an integral over fY'. SECTION 22

2a. Use Exercises 4.9b and 7.4. 2b. Use coordinates with, say, e3 = n together with the relation div v = 0 to show that (grad V)T n = O.

256

HINTS FOR SELECTED EXERCISES

SECTION 25 For HE Yep with det H # 1 take F, = H" and Gn = H- n (or vice versa). 4e. Choose FE Lin + arbitrarily and take H = (det F)1/3F- 1 to reduce T to a function of the density. Show next that T is isotropic and use this isotropy to reduce the stress to a pressure (cf. the corollary on page 13). 5b. Choose H E Yep. By (16) and (19), Tg(FG- 1) = T(F) = T(FH) = Tg(FHG- 1 ) 4c.

for every F

E

Lin +, so that

fg(F) = Tg(FGHG- 1) for every F E Lin ". Thus H E Yep implies GHG - 1E Ye p(g) and GYep G - 1 C Ye p(g). Similarly, Ye p(g) c GYe p G -I. Use (17) and (20). 5d. Let G = RU be a right polar decomposition of G. Then SC.

Yep(g)

= RUYep U-1R- 1•

Choose Q E Yep. Then Q

= RUQU- 1R- 1

belongs to Ye p(g). Thus, since Q, (QR)U

QE Orth +

(why?),

= (RQ)(QTUQ)

(0() RQRT.

represent polar decompositions ofthe same tensor; hence Q = Thus Q E Yep implies RQR T E Yep(g), and RYepR T c Yep(g). Similarly, Yep(g) c RYepRT. Equation (0() also implies that U = QTUQ. When p is isotropic this result and the corollary on page 13 tell us that U = AI, so that G = AR. 6a. Without loss in generality add the constraint for all F 6c. Define

E

tr T(F) = 0 Unim (cf. the remarks on page 148).

f(F) = T«det F) - 1/3 F)

({3)

for all FE Lin +. Then f(F) = T(F) for all FE Unim, so that f provides an extension of T to Lin +. Show that f(F) must reduce to an isotropic function of B = FFT and use (37.21).

SECTION

29

257

SECTION 27 6. Transform

f

(T +

1t o I)m

dA

.(9',t)

to an integral over!/' ( c !/'2) and use the fact that!/' is arbitrary.

SECTION 28 1a. Integrate u(R)o along a path from R = I to R = Q. lb. Substitute (5) into (27.11) and use the chain rule to verify that DF[u(QF) - u(F)] = O.

Ie,

Integrate this from F = I [using (19)]. Use the chain rule to differentiate u(F) = u(FTF)

with respect to F and appeal to (5) and (27.14). 2. Use the incompressibility of the body to show that (nI)· D

= 0,

so that the term involving the pressure 11: does not enter the expression for the work. Then use the extension (p) to reduce the problem to the one studied in this section. 3. Show that the work done on a part

I

t

to

{!IJ

during [to, t 1] is given by

1

fT. D dV dt iP,

provided the process is closed during this time interval. 4. Use components.

SECTION 29 1. Extend the proof of (b) of the theorem on page 195.

258

HINTS FOR SELECTED EXERCISES

SECTION 30 4.

Use (29.7) with F = I.

SECTION 32 2. Use Exercise 15.1. 9. The following theorem is needed: if w, are class C" (N ~ 1) fields on []It with

then there exists a class C" + 1 field

t/J

on

[]It

such that

The compatibility equation in the form (E ll , 2

-

E 12 , l ) ' 2 = (E 12 , 2

establishes the existence of a field E ll , 2 = (E 12

and hence fields

Ua

t/J

-

E 22 , l ) ' l

such that

+ t/J),l,

such that

Ell

E2 2

= U 1,l' = U2,2'

E 12

+ t/J = U 1,2'

E 12

-

t/J

= U 2,l'

11. Equation (13h, written in the form

implies the existence of fields

and there exists a

qJ such

that

t/J

and () such that

SECTION

34

259

SECTION 33 4. Assume that the boundary of the cross section is parametrized by PI(a) and pz«(1) with (1 the arc lerigth. Then the vector with components

is normal to this boundary at each that

(1.

Use this fact, (7), and (8) to show

5c. Show that

and use the divergence theorem on verify that

g>0

(as a region in

f

S13dA

= O.

f

SZ3dA

= O.

[RZ) and

(11) to

Yo

Similarly,

Yo

SECTION 34 2. Show that

L{

{It(t - T)' C[E(t

+ T)J - E(t

L

{P{Ii(t - T)' ii(t

+ T) -

+ T) . C[E(t -

o(t

+ T) • o(t

T)]} dt dV = 2'¥t{E}(t),

- r)} dt dV = -2.Jt'"{u}(t)

(where we have suppressed the argument p), and that the left sides sum toct>(t). To derive these identities write the integrands on the left sides as derivatives with respect to r.

260

HINTS FOR SELECTED EXERCISES

5. Show that the convolution of the left side of (12) with A(t) = t is equal to

A*

LS * E dV + LpU * ii dV,

where the convolution of two tensor fields is defined using the inner product of tensors in the integrand in (11). Show further, using the commutativity of the convolution operation and the symmetry of C, that

s*E=s*E. In view of the above remarks we have the identity obtained by taking the convolution of (12) with I, and if we differentiate this relation twice with respect to time, we arrive at (12).

SECTION 37

2. Choose U and v with IU I = Iv I, let Q be the orthogonal tensor carrying u into v, and show that qJ(u) = qJ(v). 3. Choose v and let Q be any rotation about v. Show that for this choice of Q, Qq(v) = q(v), and use this fact to prove that q(v) is parallel to v. Thus q(v) = qJ(v)v. Show that qJ is isotropic and use Exercise 1. 4. Use induction.

Index

A

Acceleration, 60 Acoustic equations, 133 Acoustic tensor, 224 Airy stress function, 214 Angle of twist, 58 Angular velocity, 70 Anisotropic material point, 169 Axial vector, 9 B

Balance of energy for hypere1astic material, 191 for linear elastic material, 220 for viscous fluid, 153 Bending of bar, 214-217 Bernoulli's theorem, 111 for elastic fluid, 131 for ideal fluid, 120 Betti's reciprocal theorem, 206 Blasius-Kutta-Joukowski theorem, 124 Body, 41 Body force, 99 Boundary, 35 Boussinesq-Papkovich-Neuber solution, 201

C Cartesian coordinate frame, 2 Cauchy stress, see Stress, Cauchy Cauchy's theorem, 101 Cayley-Hamilton theorem, 16 Center of mass, 92 Chain rule, 26 Characteristic space, 11 Circulation, 82 Closed region, 35 0',21 Commutation of tensors, 3 Commutation theorem, 12 Compatibility equation, 213 Complex potential, 126 Complex velocity, 123 Connected set, 35 Conservation of mass, 88 for control volume, 90 local, 89 Conservative body force, 111 Constitutive assumption, 115-119 Control volume, 89 Cross product, 7 Curl, 32 material, 61 spatial, 61 261

262

INDEX

Curve, 34 c1osed,34 length of, 34

G

D

Da Silva's theorem, 107 Deformation, 42 Deformation gradient, 42 Density, 88 Derivative, 20-21 invariance of, 237 Determinant, 6 derivative of, 23 Differentiable, 21 Displacement, 42 Displacement equation of equilibrium, 201 Displacement equation of motion, 201 Displacement problem of elastostatics, 207 Divergence material, 61 spatial, 61 of tensor field, 30 of vector field, 30 Divergence theorem, 37 Dynamical process, 108 E

Eigenvalue, 11 Eigenvector, 11 Elastic body, 165 incompressible, 174 Elastic fluid, 117, 130-137 Elasticity tensor, 194-197 Elastic process, 220 Elastic state, 205 Equation of motion, 101 Euclidean point space, 1 Eulerian process, 116 Euler's equations for ideal fluids, 121 Euler tensor, 95 Exponential function, 227-229 Extended symmetry group, 172 Extension, 44 Extra stress, 148 F

Finite elasticity, 165-198 Flow, 108 Flow region, 109

Gradient material, 60 of point field, 30 of scalar field, 29 spatial, 61 of vector field, 30 Graffi's reciprocal theorem, 223 H

Harmonic field, 32 Hellinger-Prange-Reissner principle, 212 Homogeneous body, 171 Homogeneous deformation, 42 Hu-Washizu principle, 212 Hyperelastic body, 186 I

Ideal fluid, 117, 119-130 Incompressible body, 116 Inertia tensor, 94 Infinitesimal rigid displacement, 55 characterization of, 56 Infinitesimal strain, 55 Inner product of tensors, 5 of vectors, 1 Integral of vector field around curve, 34 Interior of set, 35 Invariance of function, 229 Invariance of set, 229 Invariance under change in observer, 143145 for elastic body, 166 for Newtonian fluid, 149 Inverse of tensor, 6 Irrotational motion, 81 Isochoric deformation, 51 Isochoric dynamical process, 116 Isochoric motion, 78 Isotropic function definition, 229 representation theorem for linear tensor functions, 235, 236 representation theorem for scalar functions,230 representation theorem for tensor functions, 233, 235 Isotropic material point, 169

263

INDEX

K

Kelvin's theorem, 83 Kinematically admissible state, 208 Kinematical viscosity, 151 Kinetic energy, lIl, 220 Konig's theorem, 95 Kom's inequality, 57 L

Lagrange-Cauchy theorem, 81 Lame moduli, 196 Laplacian, 32 Left Cauchy-Green strain tensor, 46 Left stretch tensor, 46 Lin, 7 Lint, 7 Linear elasticity, 199-226 Linear form, representation theorem, 1 Localization theorem, 38 Longitudinal sound speed, 225 M

Mach number, 133 Mass, 87 Mass distribution, 87 Material body, 116 Material curve, 82 Material description, 60 Material field, 60 Material point, 41 Material time derivative, 60, 61 Mean strain, 57 Minimum complementary energy, principle of,211 Minimum potential energy, principle of, 193,208 Mixed problem of elastodynamics, 221 Mixed problem of elastostatics, 207 Modulus of compression, 202 Momentum angular, 92 balance for control volume, 108 balance of, 99 linear, 92 spin angular, 93 Motion, 58 Motion relative to time T, 75 Multiplicity of eigenvalue, 11

N

Navier-Stokes equations, 151 Newtonian fluid, 148 Normal force, 105

o Observer change, 139 Open region, 35 Origin, 2 Orth,7 Orth", 7 Orthogonal tensor, 7 p

Part of body, 41 Path line, 63 Piola-Kirchhoff stress, see Stress, PiolaKirchhoff Place, 58 Plane motion, 74 poincare inequality ,I 163, 164 Point, 1 Poisson's ratio, 203 Polar decomposition of tensor, 14 Polar decomposition theorem, 14 Position vector, 62 Positive definite tensor, 7 Potential flow, 111 Potential theorem, 35 Power expended, theorem of, 110, 180 Pressure, 106 Principal invariants, 15 list of, 15 Principal stress, see Stress, principal Principal stretch, 45 Progressive wave, 223-226 Psym, 7 Pure torsion, 58 R

Reduced constitutive equation, 166 Reference body force, 179 Reference configuration, 41 Reference density, 88 Regular region, 37 Reiner-Rivlin fluid, 155 Reynolds number, 154 Reynolds' transport theorem, 78

264

INDEX

Right Cauchy-Green strain tensor, 46 Right stretch tensor, 46 Rigid deformation, 48 characterization of, 49 Rigid motion, 69 Rivlin-Ericksen tensors, 76 Rotation, 7 about point, 43 tensor, 46

s Shearing force, 105 Shear modulus, 202 Signorini's theorem, 113 Simple shear of elastic body, 175-178 of Newtonian fluid, 156-159 Simply connected set, 35 Skew tensor, 3 Skw, 7 Small deformation, 54-58 Smooth, 20 Smooth-inverse theorem, 22 Smoothness lemma, 60 Sonic flow, 133 Sound speed, 131 Span, 2 Spatial description, 60 Spatial description of velocity, 60 Spatial field, 60 Spatial time derivative, 61 Spectral decomposition, 12 Spectral theorem, 11 Spectrum, 11 Spin, 71 axis, 71 transport of, 80 Square root of tensor, 13 derivative of, 23 Square-root theorem, 13 Stability theorem for Navier-Stokes equations, 162 Stagnation point, 128 Stationary potential energy, principle of, 193 Steady flow, 109 Steady motion, 67 Stiffness matrix, 210 Stokes flow, 152 Stokes' theorem, 39 Strain energy, 205

Strain-energy density, 186 Stream function, 126 Streamline, 64 Stress Cauchy, 101 Piola-Kirchhoff, 178 principal, 105 shear, 106 tensile, 106 Stress power, III Stretch, 44 Stretching, 71 Subsonic flow, 133 Supersonic flow, 133 Surface force, 99 Surface traction, 98 Sym, 7 Symmetric tensor, 3 Symmetry group, 168 Symmetry transformation, 168 T Tensor, 2 skew part of, 3 symmetric part of, 3 Tensor product, 4 Time, 58 Torsion of circular cylinder, 217-219 Torsional rigidity, 219 Total body force, 100 Trace, 5 Traction problem of elastostatics, 208 Trajectory of motion, 59 Transfer theorem, 231 Transformation law for Cauchy-Green strain tensors, 141 for Cauchy stress, 144 for deformation gradient, 140 for rotation tensor, 140 for spin, 142 for stretch tensors, 140, 141 for stretching, 142 for velocity gradient, 142 Transverse sound speed, 225

u Unimodular group, 172 Uniqueness for linear elastodynamics, 222

265

INDEX

for linear elastostatics, 207 for Navier-Stokes equations, 160

v Vector, 1 Velocity gradient, 63 Virtual work, theorem of, 100 Viscosity, 149 Viscous flow problem, 160 Volume of set, 37

Vortex line, 83 W

Wang's lemma, 232 Wave equation, 133 Work in closed processes, 185 Y

Young's modulus, 203

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