## Strain Decomposition

72 In this section we first seek to express the relative displacement vector as the sum of the linear (Lagrangian or Eulerian) strain tensor and the linear (Lagrangian or Eulerian) rotation tensor. This is restricted to small strains.

73 For finite strains, the former additive decomposition is no longer valid, instead we shall consider the strain tensor as a product of a rotation tensor and a stretch tensor.

### 4.3.1 jLinear Strain and Rotation Tensors

74 Strain components are quantitative measures of certain type of relative displacement between neighboring parts of the material. A solid material will resist such relative displacement giving rise to internal stresses.

75 Not all kinds of relative motion give rise to strain (and stresses). If a body moves as a rigid body, the rotational part of its motion produces relative displacement. Thus the general problem is to express the strain in terms of the displacements by separating off that part of the displacement distribution which does not contribute to the strain.

4.3.1.1 Small Strains

76 From Fig. 4.7 the displacements of two neighboring particles are represented by the vectors uPo and uQo and the vector du, = uQ0 — up0 or du = uQo — uPo (4.100)

is called the relative displacement vector of the particle originally at Q0 with respect to the one originally at P0.

4.3.1.1.1 Lagrangian Formulation

77 Neglecting higher order terms, and through a Taylor expansion du, = ^d—^ — or du =(uVx)p0 dX (4.101)

78 We also define a unit relative displacement vector du,/d— where d— is the magnitude of the differential distance d—i, or d—i = £,d—, then du,- du, d—i du, „ du __„ „ .

d— = d— — = d— ** or d— = uVX'^ = (4.102)

Figure 4.7: Relative Displacement du of Q relative to P

79 The material displacement gradient dX can be decomposed uniquely into a symmetric and an anti-symetric part, we rewrite the previous equation as duj =

dui dXi dui 1 du 2 dX2 + dXi dui i du 3 dX3 ' dXi dX

du2 dX2 I du2 i du3 I dX3^ dX2

We thus introduce the linear lagrangian rotation tensor in matrix form:

du2 dX2 I du2 i du3 I dX3^ dX2

du2 dXi

 0 1 I dui du2 2 [ dX2 dXi 1 I dui du3 2 \ dX3 - dXi

dui dX2

du2 dXi

2 V dX3

80 In a displacement for which Ej is zero in the vicinity of a point P0, the relative displacement at that point will be an infinitesimal rigid body rotation. It can be shown that this rotation is given by the or

linear Lagrangian rotation vector w = -W23ei - W3ie2 - ^i2e3

4.3.1.1.2 Eulerian Formulation si The derivation in an Eulerian formulation parallels the one for Lagrangian formulation. Hence, dui dui = ——dX, or du = K-dx dx,

The unit relative displacement vector will be dui dx, dui du dui = ä--j = nj or — = uVx'^ = Kß

S3 The decomposition of the Eulerian displacement gradient du results in dui =

n dx

dus + dus dxs dxs dus + dxs dus dxs dxi dus dxs

We thus introduced the linear Eulerian rotation tensor

2 I dxs dxi

in matrix form:

2 V dxs dxi dui

. dxs dui dxs dus dxi du3 dxi

dus dxs du 3 dxs and the linear Eulerian rotation vector will be

2 I dxs dxi dus dx3

du-3 dxs

ss or

Example 4-8: Relative Displacement along a specified direction

A displacement field is specified by u = X2X2ei + (X2 — )e2 + X|X3e3. Determine the relative displacement vector du in the direction of the —X2 axis at P(1, 2, —1). Determine the relative displacements uQi — uP for Q1(1,1, —1), Q2(1, 3/2, —1), Q3(1, 7/4, —1) and Q4(1,15/8, —1) and compute their directions with the direction of du. Solution:

2XiX2 X2

thus from Eq. 4.101 du = (uVX)P dX in the direction of —X2 or

 4 1 0 " f 0 1 f -1 {du} = 0 1 2 \ -1 = \ -1 0 -4 4 1 0 J 4

By direct calculation from u we have up = 2ei + e2 - 4e3 uQl = ei - e3

;(-ei - e2 + 3.5e3) -(-ei - e2 + 3.75e3) -(-ei - e2 + 3.875e3)

and it is clear that as Qi approaches P, the direction of the relative displacements of the two particles approaches the limiting direction of du. ■

■ Example 4-9: Linear strain tensor, linear rotation tensor, rotation vector

Under the restriction of small deformation theory E = E*, a displacement field is given by u = (x1 — x3)2e1 + (x2 + x3)2e2 — x:[x2e3. Determine the linear strain tensor, the linear rotation tensor and the rotation vector at point P(0, 2, —1). Solution:

the matrix form of the displacement gradient is

dui dx,

0

-x2

2

0

-2

0

2

2

2

0

Decomposing this matrix into symmetric and antisymmetric components give:

4.3.2 Finite Strain; Polar Decomposition

85 When the displacement gradients are finite, then we no longer can decompose dX (Eq. 4.101) or du (Eq. 4.109) into a unique sum of symmetric and skew parts (pure strain and pure rotation).

86 Thus in this case, rather than having an additive decomposition, we will have a multiplicative decomposition.

87 we call this a polar decomposition and it should decompose the deformation gradient in the product of two tensors, one of which represents a rigid-body rotation, while the other is a symmetric positive-definite tensor.

88 We apply this decomposition to the deformation gradient F:

where R is the orthogonal rotation tensor, and U and V are positive symmetric tensors known as the right stretch tensor and the left stretch tensor respectively.

89 The interpretation of the above equation is obtained by inserting the above equation into dxi = dxj dXj dxi = RikUkjdXj = VikRkjdXj or dx = R-U-dX = V-R-dX (4.124)

and we observe that in the first form the deformation consists of a sequential stretching (by U) and rotation (R) to be followed by a rigid body displacement to x. In the second case, the orders are reversed, we have first a rigid body translation to x, followed by a rotation (R) and finally a stretching (by V).

To determine the stretch tensor from the deformation gradient

Recalling that R is an orthonormal matrix, and thus RT = R 1 then we can compute the various tensors from

 U= vf7 f (4.126) R = FU-i (4.127) V= FR7 (4.128)

9i It can be shown that

Example 4-10: Polar Decomposition I

4.3 Straii

Given xi = Xi, x2 = — 3X3, x3 = 2X2, find the deformation gradient F, the right stretch tensor U, the rotation tensor R, and the left stretch tensor V. Solution:

From Eq. 4.126

thus

From Eq. 4.127

Finally, from Eq. 4.128

dxi dXi dx2 dXi dx3 dXi dx\

dxi 0x2

1

0

0

1

U2 = ftF =

0

0

2

0

0

-3

0

0

1

0

U

=

0

2

0

0

1

0

0

1

R = FU-1 =

0

0

-3

0

0

2

0

 1 0 0 0 4 0 0 0 9

1

0

0

1

0

0

1

0

0

V = frt =

0

0

-3

0

0

1

=

0

3

0

0

2

0

0

-1

0

0

0

■ Example 4-11: Polar Decomposition II

For the following deformation: xi = A1X1, x2 = —A3X3, and x3 = X2X2, find the rotation tensor. Solution:

 " A1 0 0 [R] = [F][U]-1 = 0 0 -A3 0 A2 0
 A1 0 0 r A1 0 0 0 0 -A3 = 0 A2 0 0 A2 0 0 0 A2 0 Thus we note that R corresponds to a 90° rotation about the e1 axis. Example 4-12: Polar Decomposition III Determine U and U-1 with respect to the ebasis i 001. y ^ matrix Ui obtain th U__e = N[vnormal ized . Ueigeb . vnormalized, 0] i 0.707 0.707 0. y 0.707 e.ie 0. 0. 0. 1., vnormalized=GramSchmidt[{v3, -v2, v1<] F 0.388683 0.92388 0y CSTeigen = Chop[N[vnormalized . CST . vnormalized, 4]] jer ""j® 882Iwi9.lh7,i^sp|ejct to ln[12]:= ln[2]:= the ei basis Ueigen=N[Sqrt[CSTeigen], 4] N[Eigenvalues[CST]] Ueigenminusl= Inverse[Ueigen]
0 0