Kinematics: Bodies, Deformations, and Strain

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"
=
T(f(p))G(p)n(p) dA p ,
(18)
o.~
of(&')
f
v(f(p))' G(p)n(p) dA p ,
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 <lid ofa material field (p, t) f---+ <II(p, t) by
<II.(x, t) = <II(p(x, t), t),
and the material description Q", of a spatial field (x, t) f---+ Q(x, t) by
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 <II we write
oto <II(p, t)
dl(p, t) =
for the derivative with respect to time t holding the material point p fixed,and
V<II(p, t)
=
Vp <II(p, t)
for the gradient with respect to p holding t fixed. <1> is called the material
time derivative of <II, V<II the material gradient of <II. In particular, the material
field
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
Spatial field n
~x~
.r
Material point p
and time t
Place x and
time t
grad
n
<b
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 <I> be a smooth field on the trajectory of a steady motion. Then <I> is
steady if
<1>'
=
0
(1)
[in which case we consider <I> as a function x ....... <I>(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
= !<L +
U) = !<grad v + grad vT ) ,
W = !<L - U) = !<grad v - grad vT).
Then
L=D+W
and
v(x) = v(y) + W(y)(x - y) + D(y)(x - y) + o(x - y).
Thus in a neighborhood of a given point y and to within an error of o(x - y) a
general velocity field is the sum of a rigid velocity field
X 1--+ v(y)
+ 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: ( ) = e . D(x, t e.
~-o u~
(8)
t
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
J~(t)
e • [v(x
+ lXe, t)
c5~(t)
+ lXe, which is IX, (6)
- vex, t)]
IX
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] =
0
[~y ~ ~]
0
0
0
(relative to the above frame), and a trivial computation shows that
[WD + DW] =
0
[
+ fJ)
Y(IX
-Y(IX + fJ)
O
~
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 <I> be a smooth spatial field, and assume
that <I> is either scalar valued or vector valued. Thenfor any part f!IJ and time t,
:f
: f~t
t
t
~,
<I> dV
<I> dV
f
=f
=
~t
e,
(d>
+ <I> div v) dV,
(5)
<1>' dV
+
f
JO~t
<l>v • n dA.
10. TRANSPORT THEOREMS. VOLUME. ISOCHORIC MOTIONS
Proof
d
dt
79
In view of (2),
i
<D dV
= dd
t
WI,
=
i
<D",det F dV
=
WI
i
(<D...det F)'dV
WI
r (ci> + <D div v)", det F dV = i (ci> + <D div v) dV,
JWI
WI,
which is (5)1' To derive (5h assume that <D is scalar valued. Then (8.4)1 yields
the identity
ci> + <D div v = <D' + div(<Dv),
and (5h follows from (5)1 and the divergence theorem. The proof for <D
vector valued is exactly the same. 0
Note that
i
<D'dV =
WI,
i:
WI,
t
<D(x, t) dv" = [ :
r
i
<D(x, r) dv"J
WI,
.
t=t
Thus (5h asserts that the rate at which the integral of <D over f!lJ t is changing
is equal to the rate computed as if f!lJ t were fixed in its current position plus
the rate at which <D is carried out of this region across its boundary.
EXERCISES
1.
Let Pbe a smooth spatial scalar field with
Show that
o:
+ div(<pv) = O.
ip'
2.
p= 0 and define <p = p'(det F)
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 <D vector valued.
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),
k(a)
(7)
= 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
~ f.
dt
u· dx
=
c,
f. (Ii + LTu) . dx.
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)