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, B1 (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 dxB1 dx (4.53b)
dX dX

 Figure 4.5: Undeformed and Deformed Configurations of a Continuum
in which the second order tensor
lj dxi dx3
VxXXV 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 CauchyGreen deformation tensor. It relates (dx)2 to (dX)2.
42 Inspection of Eq. 4.54 and Eq. 4.58 yields
■ Example 45: 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 = FcF where F was defined in Eq. 4.24 as

■ 1 
0 
0 

_ 
0 
1 
A 
(4.60b) 

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. 4.57a and 4.51 this difference is expressed as
in which the second order tensor dX dX  Sij) dXidXj = 2EijdXidXj dX(Fc•F  IftdX = 2dXEdX
= \_( 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 StVenant 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 46: 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 4.53b 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. 4.68a 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 4.62b and 4.68b respectively. If the displacement gradients are small, the finite strain tensors in Eq. 4.62b and 4.68b 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 47: 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. 4.12c 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 
(4.79a) 

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. 4.62b 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. 4.57a, 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. 4.53b, 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.

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