Strain Tensor

8 Following the simplified (and restrictive) introduction to strain, we now turn our attention to a rigorous presentation of this important deformation tensor.

9 The approach we will take in this section is as follows:

1. Define Material (fixed, Xj) and Spatial (moving, Xj) coordinate systems.

2. Introduce the notion of a position and of a displacement vector, U, u, (with respect to either coordinate system).

3. Introduce Lagrangian and Eulerian descriptions.

4. Introduce the notion of a material deformation gradient and spatial deformation gradient e yy

Strain Tensor Line Element

5. Introduce the notion of a material displacement gradient and spatial displacement gradient.

6. Define Cauchy's and Green's deformation tensors (in terms of (dX)2 and (dx)2 respectively.

7. Introduce the notion of strain tensor in terms of (dx)2 — (dX)2 as a measure of deformation in terms of either spatial coordinates or in terms of displacements.

4.2.1 Position and Displacement Vectors; (x, X)

10 We consider in Fig. 4.3 the undeformed configuration of a material continuum at time t = 0 together with the deformed configuration at coordinates for each configuration.

11 In the initial configuration P0 has the position vector

which is here expressed in terms of the material coordinates (X1,X2,X3).

12 In the deformed configuration, the particle P0 has now moved to the new position P and has the following position vector x = xiii + X2i2 + X3i3 (4.9)

which is expressed in terms of the spatial coordinates.

13 Note certain similarity with Fig. 4.1, and Eq. 4.2-4.5 where the strains are defined in terms of l and l0 rather than the displacement Al.

14 The relative orientation of the material axes (OX1X2X3) and the spatial axes (ox1x2x3) is specified through the direction cosines a^.

15 The displacement vector u connecting P0 and P is the displacement vector which can be expressed in both the material or spatial coordinates

again Uk and uk are interrelated through the direction cosines ik = aKIK. Substituting above we obtain u = uk (aK Ik ) = Uk Ik = U ^ Uk = aK uk (4.11)

16 The vector b relates the origin o with respect to O. From geometry X + u = b + x, thus u = b + x — X or if the origins are the same (superimposed axis), Fig. 4.4:

M Example 4-1: Displacement Vectors in Material and Spatial Forms

Figure 4.3: Position and Displacement Vectors
Figure 4.4: Position and Displacement Vectors, b = 0

With respect to superposed material axis Xi and spatial axes xi, the displacement field of a continuum body is given by: xi = Xi, x2 = X2 + AX3, and x3 = AX2 + X3 where A is constant.

1. Determine the displacement vector components, u, in both the material and spatial form.

2. Determine the displacements of the edges of a cube with edges along the coordinate axes of length dXi = dX, and sketch the displaced configuration for A =1/2.

3. Determine the displaced location of material particles which originally comprises the plane circular surface X1 = 0, X22 + X| = 1/(1 - A2) if A = 1/2.

Solution:

1. From Eq. 4.12-c the displacement field can be written in material coordinates as

ui

= xi

- Xi = 0

u2

= x2

- X2 = AX3

u3

= x3

- X3 = AX2

2. The position vector can be written in matrix form as

xi

1

0

0

r Xi

x2

=

0

1

A

l X2

x3

0

A

1

Xi X2 X3

that is X1 = x1, X2 = (x2 - Ax3)/(1 - A2), and X3 = (x3 - Ax2)/(1 - A2).

1 - A2 0 0

0

0

r xi

1

-A

l x2

-A

1

{ x3

3. The displacement field can now be written in spatial coordinates as ui = xi - Xi = 0

4. The displacements for the edge of the cube are determined as follows:

= Xi,X2 = X3 = 0, ui = U2 = U3 = 0 0, X2 = X2, ui = U2 = 0, U3 = AX2. = X2 = 0,X3 = X3, Ui = U3 = 0,U2

AX3, thus points along this edge are displaced in the X2 direction proportionally to their distance from the origin.

5. For the circular surface, and by direct substitution of X2 = (x2 — Ax3)/(1 — A2), and X3 = (x3 — Ax2)/(1 — A2) in X2 + X| = 1/(1 — A2), the circular surface becomes the elliptical surface

= (1 - A2) or for A = 1/2 expressed in its principal axes, X* (at n/4), it has the equation x2

When

Vector Deformation And Jacobian

4.2.1.1 Lagrangian and Eulerian Descriptions; x(X, t) , X(x, t)

17 When the continuum undergoes deformation (or flow), the particles in the continuum move along various paths which can be expressed in either the material coordinates or in the spatial coordinates system giving rise to two different formulations:

Lagrangian Formulation: gives the present location Xi of the particle that occupied the point (XiX2X3) at time t = 0, and is a mapping of the initial configuration into the current one.

Eulerian Formulation: provides a tracing of its original position of the particle that now occupies the location (x1,x2,x3) at time t, and is a mapping of the current configuration into the initial one.

and the independent variables are the coordinates xi and t. is (X, t) and (x, t) are the Lagrangian and Eulerian variables respectivly.

19 If X(x, t) is linear, then the deformation is said to be homogeneous and plane sections remain plane.

20 For both formulation to constitute a one-to-one mapping, with continuous partial derivatives, they must be the unique inverses of one another. A necessary and unique condition for the inverse functions to exist is that the determinant of the Jacobian should not vanish

dxi dXi

For example, the Lagrangian description given by

has the inverse Eulerian description given by

■ Example 4-2: Lagrangian and Eulerian Descriptions

The Lagrangian description of a deformation is given by xi = Xi +X3(e2 — 1), x2 = X2 +X3(e2— e-2), and x3 = e2X3 where e is a constant. Show that the jacobian does not vanish and determine the Eulerian equations describing the motion. Solution:

The Jacobian is given by

Inverting the equation

1

0

(e2 - 1)

0

1

(e2 - e-2)

= e2 ^ 0

0

0

e2

1

1

0

(e-2 - 1)

( Xi =

xi

+ (e-2

- 1)x3

=

0

1

(e-4 - 1)

^ { X2 =

x2

+ (e-4

- 1)x3

0

0

e-2

X3 =

e-

2x3

4.2.2 Gradients

21 Partial differentiation of Eq. 4.17 with respect to Xj produces the tensor dxi/dXj which is the material deformation gradient. In symbolic notation dxi/dXj is represented by the dyadic

The matrix form of F is

xi x2

L dXi a dX2

a ax3

axi axi ax2 axi axs axi axi

axi ax ax2

ax3 ax3

22 Similarly, differentiation of Eq. 4.18 with respect to Xj produces the spatial deformation gradient

22 Similarly, differentiation of Eq. 4.18 with respect to Xj produces the spatial deformation gradient dX dX dX dX, H = XVx = ei + e2 + = '

dxi dx2 dx3 dx

The matrix form of H is

Xi X2 X3

dXi

dXi

dXi '

dxi dx2

0X2

1X2

\dXi

Ma

x

8x3

dx-

dxi

dx2

dx3 _

23 The material and spatial deformation tensors are interrelated through the chain rule dx, dX„ dX, dx4

dXj dxk dx- dXk

24 The deformation gradient characterizes the rate of change of deformation with respect to coordinates. It reflects the stretching and rotation of the domain in the infinitesimal neighborhood at point x (or X).

25 The deformation gradient is often called a two point tensor because the basis e» ® Ej has one leg in the spatial (deformed), and the other in the material (undeformed) configuration.

26 F is a tensor of order two which when operating on a unit tangent vector in the undeformed configuration will produce a tangent vector in the deformed configuration. Similarly H is a tensor of order two which when operating on a unit tangent vector in the deformed configuration will produce a tangent vector in the undeformed configuration.

4.2.2.1.1 I Change of Area Due to Deformation 27 In order to facilitate the derivation of the Piola-Kirchoff stress tensor later on, we need to derive an expression for the change in area due to deformation.

28 If we consider two material element dX« = dXiei and dX(2) = dX2 e2 emanating from X, the rectangular area formed by them at the reference time t0 is dA0 = dX(1) xdX(2) = dX1 dX2e3 = dA0e3 (4.30)

29 At time t, dX« deforms into dx(1) = FdX(1) and dX(2) into dx(2) = FdX(2), and the new area is dA = FdX(1) x FdX(2) = dX1 dX2 Fe 1 x Fe 2 = dA0 Fe 1 x Fe 2 = dAn

where the orientation of the deformed area is normal to Fe 1 and Fe2 which is denoted by the unit vector n. Thus,

and recalling that a-bxc is equal to the determinant whose rows are components of a, b, and c,

or e3FTn det(F)

and Ft n is in the direction of e3 so that dA

Ft n = —0 det Fe3 ^ dAn = dA0 det(F)(F^1)J e3 d A

which implies that the deformed area has a normal in the direction of (F 1)Te3. A generalization of the preceding equation would yield

4.2.2.1.2 | Change of Volume Due to Deformation 30 If we consider an infinitesimal element it has the following volume in material coordinate system:

in spatial cordiantes: If we define d^0 = (dX1 e1 xdX2 e2 >dX3 e3 = dX1 dX2 dX3

d^ = (dx1e1 xdx2e2 ftdx3e3

F = dxi F = dXj e then the deformed volume will be d^ = (F1 dX1 xF2 dX2 )^F3 dX3 = (F1 xF2 ^F3 )dX1dX2dX3

and J is called the Jacobian and is the determinant of the deformation gradient F

dxi dxi dxi dXi 8X2 SX3

dx2 dx2 dx2

dXi 8X2 SX3

8x3 8x3 8x3

8X1 8X2 8X3

and thus the Jacobian is a measure of deformation.

31 We observe that if a material is incompressible than det F = 1.

■ Example 4-3: Change of Volume and Area

For the following deformation: x\ = X\X\, x2 = —A3X3, and x3 = A2X2, find the deformed volume for a unit cube and the deformed area of the unit square in the Xi — X2 plane. Solution:

" A1 0

0

[F] =

0 0

-A3

0 A2

0

det F =

A1A2A3

4.2 Straii

Ay AAq no AAn

A1A2A3 1

A3 0

0 1

0

0 1

=

to

-1 !

0

32 We now turn our attention to the displacement vector ui as given by Eq. 4.12-c. Partial differentiation of Eq. 4.12-c with respect to Xj produces the material displacement gradient

The matrix form of J is j dui

L dXi 8X2 8X3 J

L dXi 8X2 8X3 J

dui dXi du2 dXi du3 dXi dui

Ou3 8X3

dui dXj

33 Similarly, differentiation of Eq. 4.12-c with respect to Xj produces the spatial displacement gradient _

The matrix form of K is

L dxi dx2 dx3 J

L dxi dx2 dx3 J

dui

dui

dui

dxi

dx2

dx3

du2

du2

du 2

dxi

dx2

dx3

du 3

du 3

du 3

dxi

dx2

4.2.2.3 Examples

■ Example 4-4: Material Deformation and Displacement Gradients

A displacement field is given by u = XiX|ei + XlX2e2 + X|X3e3, determine the material deformation gradient F and the material displacement gradient J, and verify that J = F — I. Solution:

1. Since x = u + X, the displacement field is given by x = Xi(1 + X32) ei + X2(1 + X2) e2 + X3(1 + X22) e3 (4.48)

xi x2 x3

2. Thus

dxi ax!

dx2 dXi dx3 dXi

dx

dX1

ei + d

dxi

8x1

8X2

8X3

8x2

8x2

8X2

8X3

8x3

8x3

8X2

8X3

2 3

dx dX2

dxi dX

2X1X2 0

2X1X3 0

3. The material deformation gradient is: dui dXi

r 8nx1

duxi

duxi

dXi dux2

dux2

dux2

dXi dux3

dX2 8ux3

8ux3

L dXi

8X2

8X3

X32

0

2

2Xi X2

X2

2X2 X3

We observe that the two second order tensors are related by J = F — I.

4.2.3 Deformation Tensors

34 The deofrmation gradients, previously presented, can not be used to determine strains as embedded in them is rigid body motion.

35 Having derived expressions for -jX and ^X we now seek to determine dx2 and dX2 where dX and dx correspond to the distance between points P and Q in the undeformed and deformed cases respectively.

36 We consider next the initial (undeformed) and final (deformed) configuration of a continuum in which the material OX1,X2,X3 and spatial coordinates ox1 x2x3 are superimposed. Neighboring particles P0 and Qo in the initial configurations moved to P and Q respectively in the final one, Fig. 4.5.

4.2.3.1 Cauchy's Deformation Tensor; (dX)2

37 The Cauchy deformation tensor, introduced by Cauchy in 1827, B-1 (alternatively denoted as c) gives the initial square length (dX)2 of an element dx in the deformed configuration.

38 This tensor is the inverse of the tensor B which will not be introduced until Sect. 4.3.2.

39 The square of the differential element connecting Po and Q0 is

however from Eq. 4.18 the distance differential dXi is dX

thus the squared length (dX)2 in Eq. 4.51 may be rewritten as dX dx dx-B-1 -dx (4.53-b)

dX dX

Figure 4.5: Undeformed and Deformed Configurations of a Continuum

in which the second order tensor

lj dxi dx3

VxX-XV x

is Cauchy's deformation tensor. It relates (dX)2 to (dx)2. 4.2.3.2 Green's Deformation Tensor; (dx)2

40 The Green deformation tensor, introduced by Green in 1841, C (alternatively denoted as B_1), referred to in the undeformed configuration, gives the new square length (dx)2 of the element dX is deformed.

41 The square of the differential element connecting Po and Q0 is now evaluated in terms of the spatial coordinates

however from Eq. 4.17 the distance differential dxj is ox'

dXj thus the squared length (dx)2 in Eq. 4.55 may be rewritten as

in which the second order tensor

is Green's deformation tensor also known as metric tensor, or deformation tensor or right Cauchy-Green deformation tensor. It relates (dx)2 to (dX)2.

42 Inspection of Eq. 4.54 and Eq. 4.58 yields

■ Example 4-5: Green's Deformation Tensor

A continuum body undergoes the deformation x1 = X1, x2 = X2 + AX3, and x3 = X3 + AX2 where A is a constant. Determine the deformation tensor C. Solution:

From Eq. 4.58 C = Fc-F where F was defined in Eq. 4.24 as

■ 1

0

0

_

0

1

A

(4.60-b)

0

A

1

and thus

C = FCF

" 1 O O

=

O 1 A

O A 1

Strains; (dx)2

- (dX)2

43 With (dx)2 and (dX)2 defined we can now finally introduce the concept of strain through (dx)2 — (dX )2.

4.2.4.1 Finite Strain Tensors

44 We start with the most general case of finite strains where no constraints are imposed on the deformation (small).

4.2.4.1.1 Lagrangian/Green's Strain Tensor

45 The difference (dx)2 — (dX)2 for two neighboring particles in a continuum is used as the measure of deformation. Using Eqs. 4.57-a and 4.51 this difference is expressed as

in which the second order tensor dX dX - Sij) dXidXj = 2EijdXidXj dX-(Fc•F - IftdX = 2dX-E-dX

= \_( dxk dxk_ Eij = 2 [dX. dX. Sij or E =-(Vxx*Vx -I)

is called the Lagrangian (or Green's) finite strain tensor which was introduced by Green in 1841 and St-Venant in 1844.

46 The Lagrangian stress tensor is one half the difference between the Green deformation tensor and I.

47 Note similarity with Eq. 4.4 where the Lagrangian strain (in 1D) was defined as the difference between the square of the deformed length and the square of the original length divided by twice the square of the original length (E = 2

2dXidXj

which gives a clearer physical meaning to the Lagrangian Tensor.

48 To express the Lagrangian tensor in terms of the displacements, we substitute Eq. 4.44 in the preceding equation, and after some simple algebraic manipulations, the Lagrangian finite strain tensor can be rewritten as

dX1 2

f dui du

VdX1 du2

dX2 dX

du; dX1

■ Example 4-6: Lagrangian Tensor

Determine the Lagrangian finite strain tensor E for the deformation of example 4.2.3.2. Solution:

Note that the matrix is symmetric.

4.2.4.1.2 Eulerian/Almansi's Tensor

49 Alternatively, the difference (dx)2 - (dX)2 for the two neighboring particles in the continuum can be expressed in terms of Eqs. 4.55 and 4.53-b this same difference is now equal to

in which the second order tensor

E* =1 (s„_ X X Eij 2 V j dxi dxj or E* = -(I - VxX^XVx

is called the Eulerian (or Almansi) finite strain tensor.

50 The Eulerian strain tensor is one half the difference between I and the Cauchy deformation tensor.

51 Note similarity with Eq. 4.5 where the Eulerian strain (in 1D) was defined as the difference between the square of the deformed length and the square of the original length divided by twice the square of the deformed length (E* = i I l2 0 I). Eq. 4.68-a can be rewritten as

52 For infinitesimal strain it was introduced by Cauchy in 1827, and for finite strain by Almansi in 1911.

52 For infinitesimal strain it was introduced by Cauchy in 1827, and for finite strain by Almansi in 1911.

53 To express the Eulerian tensor in terms of the displacements, we substitute 4.46 in the preceding equation, and after some simple algebraic manipulations, the Eulerian finite strain tensor can be rewritten as _

54 Expanding

dui dx1

dui dxi

dui du2 2 V dx2 dxi

du2 xi

4.2.4.2 Infinitesimal Strain Tensors; Small Deformation Theory

55 The small deformation theory of continuum mechanics has as basic condition the requirement that the displacement gradients be small compared to unity. The fundamental measure of deformation is the difference (dx)2 — (dX)2, which may be expressed in terms of the displacement gradients by inserting Eq. 4.65 and 4.71 into 4.62-b and 4.68-b respectively. If the displacement gradients are small, the finite strain tensors in Eq. 4.62-b and 4.68-b reduce to infinitesimal strain tensors and the resulting equations represent small deformations.

56 For instance, if we were to evaluate e + e2, for e = 10~3 and 10_1, then we would obtain 0.001001 « 0.001 and 0.11 respectively. In the first case e2 is "negligible" compared to e, in the other it is not.

4.2.4.2.1 Lagrangian Infinitesimal Strain Tensor

57 In Eq. 4.65 if the displacement gradient components dX are each small compared to unity, then the third term are negligible and may be dropped. The resulting tensor is the Lagrangian infinitesimal strain tensor denoted by

Note the similarity with Eq. 4.7.

dXi fi

1 ui u2

4.2.4.2.2 Eulerian Infinitesimal Strain Tensor

58 Similarly, inn Eq. 4.71 if the displacement gradient components du are each small compared to unity, then the third term are negligible and may be dropped. The resulting tensor is the Eulerian infinitesimal strain tensor denoted by

= 1 Í dui + duj ij 2 {dxj + dxi or E* = ^(uVx + Vxu) 2 S '

K+Kc

59 Expanding

E

*

dui

ii =

dxi

*

1 / dui

i2 =

du^ dx1

4.2.4.3 Examples

■ Example 4-7: Lagrangian and Eulerian Linear Strain Tensors

A displacement field is given by x1 = Xi + AX2,x2 = X2 + AX3, x3 = X3 + AX\ where A is constant. Calculate the Lagrangian and the Eulerian linear strain tensors, and compare them for the case where A is very small. Solution:

The displacements are obtained from Eq. 4.12-c uk = xk — Xk or

U2 U3

xi - X1 = X1 + AX2 - X = AX2 X2 - X2 = X2 + AX3 - X2 = AX3 x3 - X3 = X3 + AXi - X3 = AXi then from Eq. 4.44

0

A

0

0

0

A

2E = (J + Jc ) =

0

0

A

+

A

0

0

(4.79-a)

A

0

0

0

A

0

0

A

0

J = uVx =

0

0

A

(4.78)

A

0

0

To determine the Eulerian tensor, we need the displacement u in terms of x, thus inverting the displacement field given above:

1+A3

1 + A3 A(—Ax1 + A2x2 + x3) 1 + A A(x1 — Ax2 + A2x3)

1 +A3

A2

1

—A

—A

A2

1

1

A

A2

A2

1

—A '

+ 1 + A3

" A2

—A

1

—A

A2

1

1

A2

—A

1

—A

A2

—A

1

A2

2A2 1 — A 1 — A 1 — A 2A2 1 — A 1 - A 1 - A 2A2

as A is very small, A2 and higher power may be neglected with the results, then E* ^ E.

4.2.5 |Physical Interpretation of the Strain Tensor 4.2.5.1 Small Strain

60 We finally show that the linear lagrangian tensor in small deformation Eij is nothing else than the strain as was defined earlier in Eq.4.7.

61 We rewrite Eq. 4.62-b as

but since dx « dX under current assumption of small deformation, then the previous equation can be rewritten as d«

dXi dXj

Eij dX dX

62 We recognize that the left hand side is nothing else than the change in length per unit original length, d x and is called the normal strain for the line element having direction cosines ;djxr.

63 With reference to Fig. 4.6 we consider two cases: normal and shear strain.

Normal Strain: When Eq. 4.85 is applied to the differential element P0Q0 which lies along the X2

axis, the result will be the normal strain because since "jX1- = "jX?

Eq. 4.85 becomes (with ui xi Xi dx

Figure 4.6: Physical Interpretation of the Strain Tensor

Figure 4.6: Physical Interpretation of the Strain Tensor

Likewise for the other 2 directions. Hence the diagonal terms of the linear strain tensor represent normal strains in the coordinate system.

Shear Strain: For the diagonal terms Eij we consider the two line elements originally located along the X2 and the X3 axes before deformation. After deformation, the original right angle between

' Po gives the unit vector at P in the direction of Q, and M as:

the lines becomes the angle 6. From Eq. 4.101 (dui = (jX^ dXj) a first order approximation n2

dui dX

dX2 du 2

and from the definition of the dot product:

dX2 dX3 dX3 dX2

or neglecting the higher order term

64 Finally taking the change in right angle between the elements as 723 = n/2 — 6, and recalling that for small strain theory y23 is very small it follows that

Therefore the off diagonal terms of the linear strain tensor represent one half of the angle change between two line elements originally at right angles to one another. These components are called the shear strains.

64 The Engineering shear strain is defined as one half the tensorial shear strain, and the resulting tensor is written as

65 We note that a similar development paralleling the one just presented can be made for the linear Eulerian strain tensor (where the straight lines and right angle will be in the deformed state).

4.2.5.2 Finite Strain; Stretch Ratio

66 The simplest and most useful measure of the extensional strain of an infinitesimal element is the stretch or stretch ratio as which may be defined at point P0 in the undeformed configuration or at P in the deformed one (Refer to the original definition given by Eq, 4.1).

67 Hence, from Eq. 4.57-a, and Eq. 4.63 the squared stretch at P0 for the line element along the unit vector m = ^X is given by

Thus for an element originally along X2, Fig. 4.6, m = e2 and therefore dX\/dX = dX3/dX = 0 and dX2/dX = 1, thus Eq. 4.92 (with Eq. ??) yields

and similar results can be obtained for A^ and A^3.

68 Similarly from Eq. 4.53-b, the reciprocal of the squared stretch for the line element at P along the unit vector n = is given by

Again for an element originally along X2, Fig. 4.6, we obtain

69 we note that in general Ae2 = Ae2 since the element originally along the X2 axis will not be along the x2 after deformation. Furthermore Eq. 4.92 and 4.94 show that in the matrices of rectangular cartesian components the diagonal elements of both C and must be positive, while the elements of E must be greater than — 1 and those of E2 must be greater than + 2.

70 The unit extension of the element is dx dX dx dX dX

and for the element P0Q0 along the X2 axis, the unit extension is dx — dX

for small deformation theory E22 << 1, and

-d—- = E(2) = (1 + 2E22) 1 — 1 — 1 + l2E22 — 1

which is identical to Eq. 4.86.

71 For the two differential line elements of Fig. 4.6, the change in angle y23 both Ae2 and Ae3 by

2E23 2E23

Again, when deformations are small, this equation reduces to Eq. 4.90.

0 0

Responses

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    How to calculate the deformation gradient and green strain tensor?
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    Which of the following vectors are linear combinations of X1(4,23), X2(2,1,2) and X3(2,1,0)?
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