Kinematics of Deformation

I. Material and Spatial Coordinates

Consider the figure below depicting a chunk of material in the reference (material or

undeformed) configuration shown as Ro and the same chunk of material after loads have

been applied in a spatial (deformed) configuration at time t shown as Rt. P is a particle in

the reference configuration and the same particle in the spatial configuration is p. The

particle P has coordinates X and these coordinates may be used as a label to uniquely

identify P as time progresses. In the spatial configuration the particle p has coordinates

x . Note that both X and x are measured in the same vector basis 1 2 3 e ,e ,e . The

coordinates X are referred to as the Lagrangian or material coordinates, the coordinates

x are referred to as the Eulerian or spatial coordinates.

The spatial configuration may be related to the reference configuration through a

functional relationship

x x X ,t (1) or

k k Kx x X ,t (2)

p

P

Ro

1

3

2

Rt

Where x is in Rt and X is in Ro. We assume that the function (1) is continuous and

differentiable and is a one-to-one mapping so that a unique inverse exists, i.e.

X X x,t (3) or

K K kX X x ,t (4)

Note the difference in subscript for the reference and spatial quantities. As a consequence

of the existence of the inverse we must have a non-zero Jacobian where the Jacobian is

given by

k k ,KK

xJ det det x

X

(5)

Note that the values of x for a fixed value of X are the points in space occupied by the

particle P during the motion of the body. The values of X for a fixed value of x tell us

the particles that pass through the position x during the motion of the body.

Any property of the motion can be described in terms of the reference or the spatial

coordinates.

Example 1:

The motion of a body is given by

k kK K kx a X b (6)

where

kK k

t 1 0

a 0 t 1 { b } { t ,1,0 }

0 1 1

(7)

a) Calculate J and show that it is positive

b) Obtain the inverse relationship for the motion

c) A scalar field within the solid is given by the reference description

L 1( X ,t ) X . Find the spatial description.

a)

t 1 0

J det 0 t 1 t( t 1 ) 0 for t 0

0 1 1

(8)

b) from (6) we can see that

1 1

K kK k kK kX a x a b

(9)

or

1 1

2 2

2

3 3

1 t( t 1 )

t( t 1 )X 1 t 1 1 x

1 1X 0 t t x

t t 1 ( t 1 )X 0 t t x

1

( t 1 )

(10)

c) the spatial description is obtained by expressing X in terms of x as follows:

L 1

E E 1 1 2 3

( X ,t ) X

( X( x,t ),t ) ( x,t ) X ( x , x , x )

(11)

Using the results of part b) we obtain

E 1 2 31 t( t 1 )

( x,t ) t 1 x x xt( t 1 ) t( t 1 )

(12)

End of Example 1

II. Eulerian vs. Lagrangian Descriptions and the Material Derivative

Consider a property expressed as a function in the Lagrangian (reference)

description in which each particle is labeled by its original position X .

L IX ,t (13)

Let x be the spatial position of the material point or particle

i i Ix x X ,t (14)

We may calculate velocities and accelerations

i

i

2

i

i 2

xv

t

xa

t

(15)

that are functions of the particle tag,I

X , and time t . In general, this description of motion

is useful for a solid but is too complicated to use for a fluid and we instead use the

Eulerian description. Expressing using an Eulerian description we have

E ix ,t (16) where

i i Ix x X ,t (17)

Now, in order to express our fundamental principles we must have

L I E i i IX ,t x x ( X ,t ),t (18)

and

iL E E

i

x

t t x t

(19)

or

L E Ei

i

vt t x

(20)

or

L EE

vt t

(21)

Local rate

of change

Time derivative following

material particle Convective change

III. The Displacement Vector, the Displacement Gradient Tensor and Two Possible Strain Tensors

Consider the figure shown below. Particle P is shown in the reference configuration with

coordinate X and in the spatial configuration (labeled as p) with coordinate x . The

displacement vector of P may be described in reference coordinates as

U( X ,t ) x( X ,t ) X (22)

note that the capital U is used to remind us that the displacement vector is written in

reference coordinates. We may just as well write u( X ,t ) x( X ,t ) X and the meaning

would be the same. We may also write the displacement vector in spatial coordinates as

u( x,t ) x X( x,t ) (23)

Consider next the particle Q located at reference coordinate X dX and spatial

coordinate x dx . We denote the displacement vector at point P as U( X ) and the

displacement vector at point Q asU( X dX ) . Note that we are using the reference

description. We have

U( X dX ) U( X ) dx dX (24)

Or, using indicial notation

q

P Q

p

Ro

1

3

2

Rt

i

i k k i k i k i i

change in dX

U ( X dX ) U ( X ) dU ( X ) dx dX (25)

Assuming the k

dX are small, the left hand side can be expanded in a first order Taylor

series resulting in

i

i

k i k i i

kchange in dX

UdX dU ( X ) dx dX

X

(26)

where

i

k

Udisplacement gradient tensor( referencecoordinates )

X

(27)

Note that some physical meaning for the displacement gradient tensor may be found in

equation (26). The length of the fiber pq in the spatial configuration may be found from

the expression

2i i

ds dx dx (28)

and the length of the fiber PQ in the reference configuration may be found from the

expression

2i i

dS dX dX (29)

The change in the fiber length is thus

2 2i i i i

ds dS dx dx dX dX (30)

Using (25) to eliminate i

dx we have

i i i

dx dU dX (31)

Substituting this into (30) we arrive at

2 2 i i i i i ids dS dU dX dU dX dX dX (32)

using equation (26) we have

2 2 i i

k i j i i i

k j

U Uds dS dX dX dX dX dX dX

X X

(33)

Re-arranging

2 2 i i i ik j k i i j

k j k j

U U U Uds dS dX dX dX dX dX dX

X X X X

(34)

factoring j k

dX dX out of each term (be careful with dummy indices) we arrive at

j2 2 i i k

j k

j k k j

UU U Uds dS dX dX

X X X X

(35)

or

2 2

jk j kds dS 2E dX dX (36)

where

j k i i

jk

k j j k

U U U U1E Green Lagrange Strain Tensor

2 X X X X

(37)

Normally this expression will be written with uppercase indices to remind us that the

Green-Lagrange strain tensor of a function of reference coordinates, i.e.:

J K I IJK

K J J K

U U U U1E

2 X X X X

(38)

Suppose we wish to use the spatial description where the particle p is located at spatial

coordinate x and the particle q is located at spatial coordinate x dx . We denote the

displacement vector at point p as u( x ) and the displacement vector at point q as

u( x dx ) . We have

u( x dx ) u( x ) dx dX (39)

Or, using indicial notation

i

i k k i k i k i i

change in dX

u ( x dx ) u ( x ) du ( x ) dx dX (40)

Assuming the k

dx are small the left hand side can be expanded in a first order Taylor

series resulting in

ik i k i i

k

udx du ( x ) dx dX

x

(41)

where

i

k

udisplacement gradient tensor ( spatial coordinates )

x

(42)

Following the same procedure as before, we form the change in the fiber length (squared)

2 2i i i i

ds dS dx dx dX dX (43)

and use (25) to eliminate i

dX

i i i

dX dx du (44)

thus we arrive at

2 2 i i i i i ids dS dx dx dx du dx du (45)

Next, we use (41) to eliminate i

du

2 2 i ii i i k i j

k j

u uds dS dx dx dx dx dx dx

x x

(46)

re-arranging

2 2 i i i ik j k i j i

k j k j

u u u uds dS dx dx dx dx dx dx

x x x x

(47)

and factoring j k

dx dx out of each term (be careful with dummy indices) we arrive at

j2 2 k i i

j k

k j k j

u u u uds dS dx dx

x x x x

(48)

or

2 2

jk j kds dS 2 e dx dx (49)

where

j k i i

jk

k j k j

u u u u1e Almansi Eulerian Strain Tensor

2 x x x x

(50)

IV. Infinitesimal Strain

Infinitesimal strain may be defined as

i

j

u1

x

(51)

and

i

j

U1

X

(52)

Suppose we have a function expressed in reference coordinates:

iX (53) Next suppose that we need to calculate the gradient of this function with respect to the

spatial coordinates

i

j i j

X

x X x

(54)

Since i i i

X x u this becomes

i i

j i j

( x u )x X x

(55)

or

i i

j i j j

x u( )

x X x x

(56)

The first term on the right hand side becomes the Kronecker delta thus

iij

j i j

u( )

x X x

(57)

Now, for infinitesimal strain i

j

u1

x

and the last term is negligible leaving

ij

j i jx X X

(58)

Thus for infinitesimal strain there is no distinction between reference or spatial

coordinates in derivatives, furthermore the strain tensors become

j k i i

jk

k j j k

j k i i

k j j k

0

j k

k j

U U U U2E

X X X X

U U U U

x x x x

U U

x x

(59)

and

j k i i

jk

k j k j

0

j k

k j

u u u u2e

x x x x

u u

x x

(60)

And at k k Kx x X ,t , ju = jU hence the strain tensors are the same.

For infinitesimal strain there is no distinction between the spatial and reference

coordinates and the Green-Lagrange and Almansi-Euler strain tensors are the same.

Example 2:

Consider a rectangular block of dimensions a b c wherec a,b . Suppose that the

block deformation is described by

1 1 2

2 2 1

3 3

x X Xb

x X Xa

x X

(61)

This deformation is shown graphically below:

(i) Determine the Green-Lagrange strain tensor and the Almansi-Euler strain tensor.

(ii) Show that they are identical as 0

(i) In order to calculate the Green-Lagrange strain tensor we must have U( X ) . We begin

by using the motion (given by equation (61) to obtain

1 1 2 3 1 1 2 3 1 2

2 1 2 3 2 1 2 3 2 1

3 1 2 3 3 1 2 3 3

U ( X ,X ,X ) x ( X ,X ,X ) X Xb

U ( X ,X ,X ) x ( X ,X ,X ) X Xa

U ( X ,X ,X ) x ( X ,X ,X ) X 0

(62)

Now

2 2 2 2

31 1 2

11

1 1 1 1

2 2 2 2

32 1 2

22

2 2 2 2

2 2

3 31 2

33

3 3 3 3

UU U U1 1E

X 2 X X X 2 a

UU U U1 1E

X 2 X X X 2 b

U UU U1E

X 2 X X X

2

3 31 2 1 1 2 2

12

2 1 1 2 1 2 1 2

3 3 31 1 1 2 2

13

3 1 1 3 1 3 1 3

32 1 1 2 2

23

3 2 2 3 2

0

U UU U U U U U1 1E

2 X X X X X X X X 2 a b

U U UU U U U U1E 0

2 X X X X X X X X

UU U U U U1E

2 X X X X X X

3 3

3 2 3

U U0

X X

(63)

In order to calculate the Almansi-Euler strain tensor we must have u( x ) . We begin by

inverting the motion (given by equation (61) to obtain

1 1 22 2

2 1 22 2

3 3

ab aX x x

ab ab

b abX x x

ab ab

X x

(64)

and u( x ) is now found as

1 1 2 3 1 1 1 2 3 1 22 2

2 1 2 3 2 2 1 2 3 1 22 2

3 1 2 3 3 3 1 2 3

b abu ( x , x , x ) x X ( x , x , x ) x x

b ab ab

ab au ( x , x , x ) x X ( x , x , x ) x x

a ab ab

u ( x , x , x ) x X ( x , x , x ) 0

(65)

Finally, this may be used to calculate the Almansi-Euler strain tensor

2 2 2 2 22 2

31 1 2

11 2 2 2

1 1 1 1

2 2 2 22

32 1 2

22 2 2

2 2 2 2

uu u u1 1 be

x 2 x x x ab 2 ab ab

uu u u1 1 ae

x 2 x x x ab 2 ab

22

2

33

3

3 32 1 1 1 2 2

12 222

1 2 1 2 1 2 1 2

13

23

ab

e 0

u uu u u u u u1 1 ( a b ) ( a b )e

2 x x x x x x x x 2 ab ab

e 0

e 0

(66)

(ii) Now, as 0 we keep only terms linear in and arrive at the only non-zero strains

12 12

1E e

2 a b

(67)

Note that as expected, the Green-Lagrange and Almansi-Euler strains are identical. We

will use the symbol ij

to refer to the infinitesimal strain. Thus for this example we may

say that the only non-zero infinitesimal strain component is

12

1

2 a b

(68)

End of Example 2

The infinitesimal strain tensor (used in linear elasticity) is

j k

jk

k j

u u1

2 x x

(69)

and, as discussed above, no distinction need be made between the reference or spatial

coordinates. It is obvious from inspection of equation (69) that the infinitesimal strain

tensor is symmetric, i.e.

jk kj

(70)

This is also true of the Green-Lagrange and Almansi-Euler strain tensors. A geometrical

interpretation for infinitesimal strains may be found in the following diagram:

The notation in this diagram is that 1 2 3 1 2 3u u, u v, u w, x x, x y, x z . The first

interpretation is that for material fibers aligned with the x-direction

xx

udx

length ux

length dx x

(71)

A similar interpretation may be made for yy

and zz

. Another interpretation may be

arrived at by inspection of the change in angle of the material fibers aligned with the x

and y directions:

xy xy XY

2 (72)

where XY

is the angle between the material fibers aligned with the x and y directions

(originally equal to 2

). Another interpretation may be seen from this diagram by

considering the quantity

yx z

1 v u

2 x y

(73)

undeformed element

deformed element

This quantity looks like a rigid body rotation. If we define

zy x

xz y

1 w v

2 y z

1 u w

2 z x

(74)

Then we may define a skew-symmetric tensor that represents rigid body rotation as

z y

ij ji z x

y x

0

0

0

(75)

or

ji

ij

j i

uu1

2 x x

(76)

and it is easy to see that the displacement gradient may be written as the sum of two

tensors

iij ij

j

u

x

(77)

with the first term representing deformation and the second representing rigid body

rotation. This expression is important because it indicates that in infinitesimal strain the

stress should be a function of the infinitesimal strain (ij

) and not of the displacement

gradient ( i

j

u

x

) because the displacement gradient includes rigid-body motion.

It is instructive to consider the case where the strains are all zero and integrate these

strains to demonstrate the presence of rigid body motion. We start with.

ij

0 0 0

0 0 0

0 0 0

(78)

Using the normal strains we have:

xx

yy

zz

u0 0 u f y,z

x

v0 0 v g x,z

y

w0 0 w h x, y

z

(79)

and using the shear strains we have:

xy

yz

xz

1 u v f g0 0 F( z )

2 y x y x

1 v w g h0 0 H( x )

2 z y z y

1 u w f h0 0 G( y )

2 z x z x

(80)

Equation (80) indicates that

G ( y ) F ( z )

H ( x ) G ( y )

F ( z ) H ( x )

(81)

Equation (81) can only be satisfied trivially, i.e.

F ( z ) 0 F( z ) A

G ( y ) 0 G( y ) B

H ( x ) 0 H( x ) C

(82)

This may be used in equation (80) to obtain

o

o

1

o

1

1

f gA f f ( y ,z ) Ay f ( z )

y x

g g( x ,z ) Ax g ( z )

g hC g g( x ,z ) Cz g ( x )

z y

h h( x , y ) Cy h ( x )

f hB f f ( y ,z ) Bz f ( y )

z x

h h( x , y ) Bx h ( y )

(83)

Comparing the two expressions for f,g, and h we arrive at

o

o

o

u f y,z u Ay Bz

v g x,z v Ax Cz

w h x, y w Bx Cy

(84)

whereo

u , ov , and

ow are rigid body displacements. We may express A, B, and C in

terms of the rotation tensor components as follows

x

y

z

1 w vC

2 y z

1 u wB

2 z x

1 v uA

2 x y

(85)

And finally we may express the displacement field in terms of the rigid body

displacement and rotation as

o z y

o z x

o y x

u f y,z u y z

v g x,z v x z

w h x, y w x y

(86)

This result indicates that pure rigid body motion results in zero strains (as we suspected).

V. Velocity Gradient Tensor, the Rate of Deformation Tensor and the Spin Tensor.

There are some materials where the rate of deformation (rather than the deformation

itself) is of interest in forming constitutive models. One such example is a fluid where the

stress is thought to be a function of the velocity gradient (this is, of course,

experimentally verified for many fluids). We write the velocity vector as (note the

unfortunate duplication of notation for velocity and displacement)

i i 1 2 3 v v e ue ve we (87)

and define the velocity gradient tensor as

j

jk

k

vl

x

(88)

A geometrical interpretation for infinitesimal strains may be found in the following

diagram.

The first interpretation is that for material fibers aligned with the x-direction

xx

udx

rateof length uxllength dx x

(89)

A similar interpretation may be made for yyl and

zzl . Another interpretation may be

arrived at by inspection of the change in angle of the material fibers aligned with the x

and y directions:

xy XY

2l rate of (90)

where XY

is the angle between the material fibers aligned with the x and y directions

(originally equal to 2

).

Another interpretation may be seen from this diagram by considering the quantity

yx z

1 v uW

2 x y

(91)

This quantity looks like a rigid body angular velocity. If we define

zy x

xz y

1 w vW

2 y z

1 u wW

2 z x

(92)

Then we may define a skew-symmetric tensor that represents rigid body rotation as

z y

ij ji z x

y x

0

W W 0

0

(93)

undeformed element

deformed element

or

ji

ij

j i

uu1W

2 x x

(94)

This tensor is known as the spin tensor or the vorticity tensor and it is the anti-symmetric

part of the velocity gradient. We also define the rate of deformation tensor as the

symmetric part of the velocity gradient, i.e.

ji

ij

j i

uu1D

2 x x

(95)

It is easy to see that the velocity gradient may be written as the sum of two tensors

iij ij

j

uD W

x

(96)

This expression is important because it indicates that the stress should be a function of

the rate of deformation (ij

D ) and not of the velocity gradient ( i

j

u

x

) because the velocity

gradient includes rigid-body motion.

VI. The Deformation Gradient Tensor

Consider again the functional relationship

k k Kx x X ,t (97)

We may form the total differential

kk K

K

xdx dX

X

(98)

or

k kK K

dx F dX (99)

where

kkK

K

xF deformation gradient tensor

X

(100)

Equation (99) provides the interpretation that the deformation gradient tensor evaluated at

a material point describes the transformation of infinitesimal line elements in the

neighborhood of that material point.

Now, if kKJ det F is non-zero then kKF is non-singular and an inverse exists and

KK k Kk k

k

XdX dx G dx

x

(101)

where

1G F (102)

The deformation gradient may be related to the displacement gradient by writing the

displacement vector in reference coordinates

u( X ,t ) x( X ,t ) X (103)

Then the gradient with respect to the reference coordinates is formed

i i i

K K K

u x X

X X X

(104)

or

iiK iK

K

uF

X

(105)

This is sometime written as

ou F (106)

where the subscript o indicates that the differentiation is with respect to the reference

coordinates. An alternative, but common, expression is to define H as the displacement

gradient tensor

H F (107)

It is also possible to relate the deformation gradient to the Green-Lagrange strain tensor

as follows:

T1

E F F2

(108)

Example 3:

The displacement vector has components given by:

21 2 2 2 3 3 2 3 1u X u 2X X , u X X X (109)

Where 1 2 3

X ,X ,X are reference coordinates.

(i) Determine the exact change in length of the fiber with endpoints

P 1 2 3X X ,X ,X 1,1,1 and Q 1 2 3X X ,X ,X 1,1,1.01 .

(ii) Compare this calculation with the same calculation using the linear strain tensor

(iii) Compare this calculation with the same calculation using the deformation

gradient tensor

(i) By inspection k k

dS dX dX 0.01 (note Q P

dX X X ).

At P

X

1 2 3 1 2 3 pu ,u ,u 1,2,2 x , x , x ( 2,3,3 ) x (110)

At Q

X

1 2 3 1 2 3 qu ,u ,u 1,2.02,2.01 x , x , x ( 2,3.02,3.02 ) x (111)

The fiber length in the spatial frame is thus

q p

dx x x (0.0,0.02,0.02 ) (112)

or

2 2

k kds dx dx .02 .02 0.0283 (113)

Finally, the exact change in length of the fiber is

ds dS .0183 (114)

(ii) The pertinent component of the linear strain tensor is

333 2

3

uX

X

(115)

Recalling the definition given by equation (89) we have

33 2

ds dSX

dS

(116)

Hence

33 2

dS( ) dS( X ) ds dS (117)

Evaluating this at point P we have

2

ds dS dS( X ) .01(1.0 ) 0.01 (118)

(iii) note that Q PdX X X 0,0,0.01 thus the formula k kK Kdx F dX reduces to

k k3 3

dx F dX (119)

And we need only the 13 23 33

F ,F ,F components of the deformation gradient. The

displacement field ( equation(109)) yields the spatial coordinates in terms of the reference

coordinates

21 1 2 2 2 2 3 3 3 2 3 1x X X x X 2X X , x X X X X (120)

And the components of the deformation gradient we need are

31 213 23 2 33 2

3 3 3

xx xF 0 F 2X F 1 X

X X X

(121)

Evaluating these at point P we have

13 23 33

F 0 F 2 F 2 (122)

the k

dx can now be calculated

3Q

3 P

3Q

3P

1 13 3

X

2 23 3

X

X

3 33 3

X

dx F dX 0

dx F dX ( 2 )(.01 ) .02

dx F dX ( 2 )(.01 ) .02

(123)

and finally

2 2

k kds dx dx .02 .02 0.0283 (124)

Thus the change of length of the fiber is

ds dS .0183 (125)

As expected, this result is in better agreement with the exact answer than is the solution

based upon the infinitesimal strain tensor.

End of Example 3

VII. The Polar Decomposition of the Deformation Gradient

Tensor and the right Cauchy-Green Strain Tensor

Recall that for infinitesimal strain we decomposed the displacement gradient into

a rigid body rotation and a deformation. The purpose of this decomposition was to isolate

the deformation part of the displacement gradient for use in constitutive equations (stress

vs. strain relationships). The deformation gradient tensor may similarly be decomposed

using the polar decomposition theorem. That theorem states that the deformation gradient

may be uniquely decomposed into the form

F RU (126)

Where U is a positive-definite symmetric tensor (orthogonal eigenvectors and real

eigenvalues) containing the deformation information and R is a proper orthogonal

rotation tensor ( T 1R R and R 1 ). Since U is positive-definite symmetric it has three

real positive eigenvalues and three mutually orthogonal eigenvectors. We denote the

eigenvalues by 1 2 3, , and these are termed the principal stretches. The eigenvectors

are denoted by 1 2 3u ,u ,u and these are termed the principal directions.

Equation (99) may be written as

k kK K kL LK K

dx F dX R U dX (127)

If we define

L LK K

dy U dX (128)

then

k kL L

dx R dy (129)

Now equations (128) and (129) describe successive transformations of the fiber dX and

the interpretation of these successive transformations may be described as a stretching

followed by a rotation. Specifically, if our basis vectors are aligned with the principal

directions of U then we have

1 1 1 2 2 2 3 3 3

dy dX , dy dX , dy dX (130)

These expressions indicate that the components of the intermediate fiber dy are obtained

from stretching i 1 or compressing i 1 the original fibersdX .Note that in

general dy will not be in the same direction as dX . dx is then obtained from dy through

a rigid body rotation as expressed in equation (129).

Calculation of U given F is accomplished as follows. First, equation (126) is multiplied

by TF

T T

T T

T

F F ( RU ) ( RU )

U R RU

U U

(131)

But TU U so

2 TU F F (132)

Now the difficulty is in finding U given 2U . If 2U is expressed using the principal

directions then 2U is diagonal and U is found trivially, otherwise some effort must be

expended. Assuming U has been obtained R may be found by manipulating equation

(126) to obtain

1R FU (133)

An alternative tensor that avoids the difficulties associated with computing U is the right

Cauchy-Green tensor defined as

2 TC U F F (134)

A parallel development begins by invoking an alternative form of the polar

decomposition theorem

F VR (135)

Where V is a symmetric positive definite tensor and following the same procedure as

above we arrive at the definition for the left Cauchy-Green tensor

2 Tb V FF (136)

The notation used above for C and b is meant to imply what frame the strain tensor

resides in. To see this we write the deformation gradient in dyadic form:

ii J

J

x F e EX

(137)

Next we form the right Cauchy-Green strain tensor

T m i

N m i J

N J

m i

N m i J

N J

m i

N J m i

N J

i i

N J

N J

x x C F F E e e EX X

x x E e e EX X

x x E E e eX X

x x E EX X

(138)

Now, we do the same for the left Cauchy-Green strain tensor

T i m

i J N m

J N

i m

i J N m

J N

i m

i m J N

J N

i m

i m

J J

x x b FF e E E eX X

x x e E E eX X

x x e e E EX X

x x e e

X X

(139)

Inspection of equations (138) and (139) indicates that C is a reference frame based tensor

while b is a spatial frame based tensor. Some authors indicate reference frame based

tensors with capital symbols and spatial frame based tensors with lower case symbols.

Note that the deformation gradient tensor exists in both frames and is often referred to as

a two-point tensor.

The right and left Cauchy Green strain tensors are related to the Green-Lagrange and Almansi strain tensors as

1

E C I2

(140)

(shown below) and

11

e I b2

(141)

Example:

Given the deformation

1 1

2 2 3

3 3 2

x X

x X X

x X X

(142)

Find F, C, U, and R

By inspection

1 0 0

F 0 1

0 1

(143)

and

T 2

2

1 0 0 1 0 0 1 0 0

C F F 0 1 0 1 0 1 0

0 1 0 1 0 0 1

(144)

Now, since 2C U the diagonal form indicates that 2U given by equation (144) is

expressed in principal coordinates thus U is easily obtained as

2

2

1 0 0

U 0 1 0

0 0 1

(145)

Finally, using equation (133)

11 2 2

12 2

1 0 01 0 0

R FU 0 1 0 1 0

0 10 0 1

(146)

or

1 12 22 2

1 12 22 2

1 0 0

R 0 1 1

0 1 1

(147)

To see that R represents a rigid body rotation define

tan (148)

R then becomes a rigid body rotation about the x-direction:

1 0 0

R 0 cos sin

0 sin cos

(149)

End of Example

Recall that the right Cauchy-Green strain tensor may be written as

TC F F (150)

Or, in component form

LK k ,L k ,K k ,L k ,L k ,K k ,KC x x X u X u (151)

or

k k

LK k ,L k ,K

L K

kL k ,L kK k ,K

KL L,K K ,L k ,L k ,K

X XC u u

X X

u u

u u u u

(152)

Recall the definition of the Green-Lagrange strain tensor

i iL KLK

K L L K

u uu u1E

2 X X X X

(153)

Comparing equations (152) and (153) we see that the right Cauchy-Green and Green-

Lagrange strain tensors are related by

LK LK LK

C 2E (154)

These strain tensors will be important when we discuss constitutive equations for the

stress as they both contain information about deformation without rigid body motion. An

additional relationship may found by noticing that

T T T TC F F (H ) ( H ) H H H H (155)

and

T T1 1

E (C ) ( H H H H )2 2

(156)

The last term in equation (156) may be neglected for small strain and the resulting

infinitesimal strain tensor is

T1

E ( H H )2

(157)

Note again that there is no distinction between reference and spatial coordinates when

using the linearized strain tensor.

VIII. Deformation of Surface and Volume Elements

Consider an infinitesimal parallelepiped in the reference configuration with edges ( 1 )

dX , ( 2 )

dX , and ( 3 )

dX . The volume of this parallelepiped is

(1) (2) (3)

(1) (2) (3)

IJK J K

(1) (2) (3)

IJK I J K

dV dX (dX dX )

dX ( e dX dX )

e dX dX dX

(158)

Similarly, consider an infinitesimal parallelepiped in the spatial configuration with edges ( 1 )

dx , ( 2 )dx , and ( 3 )dx . The volume of this parallelepiped is

(1) (2) (3)

ijk i j kdv e dx dx dx (159)

Now, ( k ) ( k )

i i ,I Idx x dX and substituting this into equation (159) yields

( 1 ) (2) (3)

ijk i ,I I j ,J J k ,K K

( 1 ) (2) (3)

ijk i ,I j ,J k ,K I J K

dv e x dX x dX x dX

e x x x dX dX dX

(160)

A useful identity is that if a is a second order tensor

ijk iI jJ kK IJK ije a a a e det( a ) (161)

Using this in equation (160) we arrive at

( 1 ) (2) (3)IJK I J K

dv e JdX dX dX (162)

Where we have used equation (5), k ,KJ det x det F . Comparing equations (162) and (158) it is clear that

dv JdV (163)

If J = 1 the motion is said to be isochoric. Another identity may be derived if we consider

a surface patch dS formed by ( 1 )

dX and ( 2 )

dX with unit normal N . The volume dV

may also be written as

(3) (3)

I IdV dSN dX dS dX (164)

where I I

dS N dS . A similar expression may be written in the spatial configuration

where we have a surface patch formed by ( 1 )dx and ( 2 )dx with unit normal n .

(3)i i

dv ds dx (165)

wherei i

ds n ds . Now using equation (163)

( 3 ) (3) ( 3 )

i i I I I I ,i idv ds dx JdV JdS dX JdS X dx (166)

This results in

( 3 ) ( 3 )

i i I I ,i ids dx JdS X dx (167)

or

i I ,i I

ds JX dS (168)

And finally we have (Nansons formula)

i I ,i In ds JX N dS (169)

IX. Equations of Motion for Static, Large Deformation Analysis

We consider next large deformation cases where the spatial and reference coordinates

must be distinguished. We start with the linear momentum equation on Rt:

t t t t

( n )

nk n k

S R S R

T dS gdV n dS g dV 0 (170)

This leads in the usual way to Cauchys equations of motion

nkk

n

g 0x

(171)

In this expression nx are spatial coordinates and this equation is valid for large

deformation but the difficulty is that the spatial coordinates are unknown. This motivates

the following approach where the reference coordinates are used. We begin by

transforming equation (170) to the reference configuration using Nansons formula and the relationship between the two volumes developed in the previous section.

t t o o

ji j i ji I , j I 0 i o

S R S R

n dS g dV JX N dS g JdV 0 (172)

Applying the divergence theorem to the surface integral we have

o o

ji I , j o i o,I

R R

JX dV g JdV 0 (173)

Conservation of mass requires that

o t

o

o o

R R

o

R

dV dV

JdV

(174)

from this we can see that

o

J (175)

Substituting equation (175) into equation (173) we arrive at

o

ji I , j

o i o

IR

JXg dV 0

X

(176)

Following the usual argument that this must be true for arbitrary o

dV we have

ji I , j

o i

I

JXg 0

X

(177)

This expression looks more complicated but it has the advantage that the independent

variables (theI

X ) are known. This expression motivates the definition of the first Piola-

Kirchoff stress

Ii I , j jiP JX (178)

Using the first Piola-Kirchoff stress tensor the linear momentum equation becomes

Iio i

I

Pg 0

X

(179)

One can see the motivation for the use of this stress from this expression. Another

motivation appears if we consider the force on a spatial surface element

j ij j ij I ,i I

Ij I

dF n ds JX N dS

P N dS

(180)

From this expression we see that surface tractions applied to the boundary of the spatial

configuration can be prescribed by specifying Ij IP N on the boundary of the reference

configuration.

IjP is not symmetric. There is another Piola stress, the second Piola-Kirchoff stress that

is symmetric and this stress is given by the formula

IJ J ,i Ii

S X P (181)

or

T 1 TS PG P(F ) (182)

and from this

TP SF (183)

or

Ii IJ iJP S F (184)

Using this second order tensor the linear momentum equation becomes

IJ i ,J

o i

I

S xg 0

X

(185)

The second Piola-Kirchoff stress tensor also has the property that is energetically

conjugate to the Green-Lagrange strain tensor as well as the left Cauchy-Green strain

tensor. This will result in a convenient formulation of hyperelasticity where the strain

energy and second Piola-Kirchoff stress tensor are simply related. One can easily show

that the second Piola-Kirchoff stress tensor is related to the Cauchy stress tensor through

the relationship

1 TJ FSF (186)

The Piola stresses may be viewed as convenient mathematical expressions related to the

Cauchy stress. They do not have the physical significance as does the Cauchy stress in

that they do not directly describe the tractions on the surface of a solid. If one is using the

Piola stresses in an analysis the Cauchy stresses must be calculated to reach any

conclusion involving the physics of the stresses and loading of the material.