Deformation and Strain

2 Deformation and Strain

2.1 Motion of a continuum
For a single particle 𝑃 , whe can define the path line of the particle as 𝒙(𝑡), which is the trajectory
followed by the particle as a function of time 𝑡. In continuum mechanics, we can identify a
particle by the position 𝑿 it occupies at some reference time 𝑡0 . Then, the path lines o every
particle in the continuum can be described by
𝒙 = 𝒙(𝑿, 𝑡) (2.1)

where 𝒙 = 𝑥1 𝒆1̂ + 𝑥2 𝒆2̂ + 𝑥3 𝒆3̂ is the position vector at time 𝑡 for 𝑃 , which was at 𝑿 = 𝑋1 𝒆1̂ +
𝑋2 𝒆2̂ + 𝑋3 𝒆3̂ at time 𝑡0 . The vector 𝑿 is know as the material coordinates of the particle and
serves to identify each particle in a body. Eq. (2.1) defines the motion for a continuum.

Example 2.1: Considering the motion

𝒙 = 𝑿 + 𝑘𝑡𝑋3 𝒆2̂

where 𝑘 is a constant, draw the configuration at time 𝑡 for a body which, at time 𝑡 = 0, has
the shape of a cube of unit sides as shown in the figure.
𝑥3

𝐶
𝐵
′
𝐶
𝐵′

𝑂
𝐴 𝑥2
𝑥1 ′
𝑂 ′
𝐴

Solution:
Expanding the motion equation we have
𝒙 = 𝑋1 𝒆1̂ + 𝑋2 𝒆2̂ + 𝑋3 𝒆3̂ + 𝑘𝑡𝑋3 𝒆2̂

Analyzing the motion of the point 𝑂 we have

𝑥3

𝑃 (𝑡0 )
𝑃 (𝑡)
𝒆3̂ 𝑿
𝒙
𝒆1̂
𝑂 𝒆̂
2 𝑥2
𝑥1

Figure 2.1: Motion of a point 𝑃 .

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Material and spatial descriptions

𝒙 = 0𝒆1̂ + 0𝒆2̂ + 0𝒆3̂ + 𝑘𝑡0𝒆2̂

thus, the point 𝑂 remains at the origin. For point 𝐴 at time 𝑡 we have (𝑋1 , 𝑋2 , 𝑋3 ) = (0, 1, 0),
thus
𝒙 = 0𝒆1̂ + 1𝒆2̂ + 0𝒆3̂ + 𝑘𝑡0𝒆2̂

which gives (0, 1, 0), that is, point 𝐴 also remains at the same position. For point 𝐵 at time
𝑡 we have (𝑋1 , 𝑋2 , 𝑋3 ) = (0, 1, 1), thus
𝒙 = 0𝒆1̂ + 1𝒆2̂ + 1𝒆3̂ + 𝑘𝑡1𝒆2̂

which gives (0, 1 + 𝑘𝑡, 1), that is, point 𝐵 moves along the 𝑥2 axis. For point 𝐶 at time 𝑡 we
have (𝑋1 , 𝑋2 , 𝑋3 ) = (0, 0, 1), thus
𝒙 = 0𝒆1̂ + 0𝒆2̂ + 1𝒆3̂ + 𝑘𝑡1𝒆2̂

which gives (0, 𝑘𝑡, 1), that is, point 𝐶 also moves along the 𝑥2 axis. The same rationale can be
applied to points 𝑂′ , 𝐴′ , 𝐵′ and 𝐶 ′ . The resultin configuration is shown in the figure below.
𝑥3

𝐶
𝐵
𝑘𝑡 𝐶 ′
𝐵′

𝑂
𝐴 𝑥2
𝑥1 ′
𝑂
𝐴′

2.2 Material and spatial descriptions

For a body in motion, a field 𝑓 associated to the body may vary with time. The field 𝑓 can
be a scalar, vector or tensor field. We can describe change of field 𝑓 according to two different
perspectives:
Following de particles: The field is represented as function of the material coordinates 𝑿
and time 𝑡.
𝑓 = 𝑓(𝑋1 , 𝑋2 , 𝑋3 , 𝑡) = 𝑓(𝑿, 𝑡) (2.2)

This representation is known as material description of the field. It can also be called as the
Lagraingian description or the reference description.
Observing the changes at fixed locations: In this case, the field is represented as a function
of the spatial coordinates 𝒙 and time 𝑡.
𝑓 = 𝑓(𝑥1 , 𝑥2 , 𝑥3 , 𝑡) = 𝑓(𝒙, 𝑡) (2.3)

This representation is known as the spatial or Eulerian description. Note that the spatial
description represents the field at a fixed location as a function of time. So it cannot represent
the field changes of a particle.

Example 2.2: Given the motion of a body

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Material derivative

𝑥1 = 𝑋1 + 𝑡, 𝑥2 = 𝑋2 + 2𝑡, 𝑥3 = 𝑋3

determine the material description of the field 𝑓 = 𝑥21 + 𝑥22 + 𝑥23 . Then, find the rate of change
of 𝑓 with respect to time in a material and a spatial description.
Solution:
The material description of the field is given by substituting the motion equations into the
field equation, that is
𝑓 = 𝑓(𝑋1 , 𝑋2 , 𝑋3 , 𝑡)
= (𝑋1 + 𝑡)2 + (𝑋2 + 2𝑡)2 + 𝑋32

The rate of change of 𝑓 with respect to time in a material description is given by

𝜕𝑓 𝜕
= [(𝑋1 + 𝑡)2 + (𝑋2 + 2𝑡)2 + 𝑋32 ]
𝜕𝑡 𝜕𝑡
= 2(𝑋1 + 𝑡) + 4(𝑋2 + 2𝑡)
= 2𝑋1 + 4𝑋2 + 10𝑡

The rate of change of 𝑓 in a spatial description can be found by substituting

𝑋1 = 𝑥1 − 𝑡, 𝑋2 = 𝑥2 − 2𝑡

into the result expressed in the material description, that is

𝜕𝑓
= 2𝑋1 + 4𝑋2 + 10𝑡
𝜕𝑡
= 2(𝑥1 − 𝑡) + 4(𝑥2 − 2𝑡) + 10𝑡
= 2𝑥1 + 4𝑥2

2.3 Material derivative

The time derivative of a quantity 𝑓 of a material particle is called the material derivative. When
the material description of 𝑓 is used, we have
𝑓 = 𝑓(𝑋1 , 𝑋2 , 𝑋3 , 𝑡) (2.4)

and
𝐷𝑓 𝜕𝑓
= (2.5)
𝐷𝑡 𝜕𝑡
In the other hand, when the spatial description of 𝑓 is used, we have
𝑓 = 𝑓(𝑥1 , 𝑥2 , 𝑥3 , 𝑡) (2.6)

where the spatial coordinates are related to the material coordinates by

𝑥1 = 𝑥1 (𝑋1 , 𝑋2 , 𝑋3 , 𝑡), 𝑥2 = 𝑥2 (𝑋1 , 𝑋2 , 𝑋3 , 𝑡), 𝑥3 = 𝑥3 (𝑋1 , 𝑋2 , 𝑋3 , 𝑡) (2.7)

Then, the material derivative of 𝑓 can be written as

𝐷𝑓 𝜕𝑓 𝜕𝑓 𝜕𝑥1 𝜕𝑓 𝜕𝑥2 𝜕𝑓 𝜕𝑥3
= + + + (2.8)
𝐷𝑡 𝜕𝑡 𝜕𝑥1 𝜕𝑡 𝜕𝑥2 𝜕𝑡 𝜕𝑥3 𝜕𝑡

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Deformation gradient

𝜕𝑥
In Cartesian coordinates 𝜕𝑥 𝜕𝑥2
𝜕𝑡 , 𝜕𝑡 and 𝜕𝑡 are the components of the velocity vector 𝒗 of the
1 3

particle. then the material derivative can be written as

𝐷𝑓 𝜕𝑓 𝜕𝑓 𝜕𝑓 𝜕𝑓
= + 𝑣1 + 𝑣2 + 𝑣3 (2.9)
𝐷𝑡 𝜕𝑡 𝜕𝑥1 𝜕𝑥2 𝜕𝑥3

or in direct notation
𝐷𝑓 𝜕𝑓
= + 𝒗 · ∇𝑓 (2.10)
𝐷𝑡 𝜕𝑡
This equation is only valid for fields described in the spatial coordinates. On the other hand,
this equation can be applied in any coordinate system provided that the corresponding gradient
operator is considered.

Example 2.3: Find the material derivative of the temperature field 𝑇 = 𝛼(𝑥1 + 𝑥2 ) considering
the motion
𝑥1 = 𝑋1 + 𝑘𝑡𝑋2 , 𝑥2 = (1 + 𝑘𝑡)𝑋2 , 𝑥3 = 𝑋3

The components of the velocity field is given by

𝜕𝑥1 𝜕𝑥2 𝜕𝑥3
𝑣1 = = 𝑘𝑋2 , 𝑣2 = = 𝑘𝑋2 , 𝑣3 = =0
𝜕𝑡 𝜕𝑡 𝜕𝑡
and the components of the gradient of the temperature field is
𝜕𝑇 𝜕𝑇 𝜕𝑇
= 𝛼, = 𝛼, =0
𝜕𝑥1 𝜕𝑥2 𝜕𝑥3

Then, the material derivative of the temperature field is

𝐷𝑓 𝜕𝑓
= + 𝒗 · ∇𝑓
𝐷𝑡 𝜕𝑡
𝜕𝑓 𝜕𝑓 𝜕𝑓 𝜕𝑓
= + 𝑣1 + 𝑣2 + 𝑣3
𝜕𝑡 𝜕𝑥1 𝜕𝑥2 𝜕𝑥3
= 𝑘𝑋2 𝛼 + 𝑘𝑋2 𝛼
= 2𝑘𝑋2 𝛼
𝑥2
Considering that 𝑋2 = 1+𝑘𝑡 , the final expression for the material derivative of 𝑇 is
𝐷𝑓 𝑥
= 2𝑘𝛼 2
𝐷𝑡 1 + 𝑘𝑡

2.4 Deformation gradient

A spatial coordinate at a material location 𝑿 + d𝑿 can be expressed using a Taylor series as

𝜕𝑥𝑖 1 𝜕 2 𝑋𝑖
𝑥𝑖 (𝑿 + d𝑿) = 𝑋𝑖 + d𝑋𝑗 + d𝑋𝑗 d𝑋𝑘 + ⋯ (2.11)
𝜕𝑋𝑗 2 𝜕𝑋𝑗 𝜕𝑋𝑘

Considering only the first-order tems, we have

𝜕𝑥𝑖
𝑥𝑖 (𝑿 + d𝑿) = 𝑋𝑖 + d𝑋𝑗 (2.12)
𝜕𝑋𝑗

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Deformation gradient

𝑥3

𝑃
𝑃′
d𝑿
𝑃 (𝑡) ′
𝒆3̂ 𝑿 𝑃 (𝑡)
d𝒙
𝒙
𝒆1̂
𝑂 𝒆̂
2 𝑥2
𝑥1

Figure 2.2: Corresponding infinitessimal changes in the reference and the spatial coordinates.

Using direct notation, we can express the spatial coordinates as

𝜕𝒙
𝒙(𝑿 + 𝚫𝑿) = 𝑿 + d𝑿 (2.13)
𝜕𝑿
𝜕𝒙
The second-order tensor 𝑭 = 𝜕𝑿 is called the deformation gradient tensor. It relates the change
in the position of a particle in the material coordinates to the change in the position in the
spatial coordinates. Thus, considering an infinitesimal change d𝑿 in the reference configuration,
Figure 2.2, under a given motion, the change in the position of the particle in the spatial
coordinates is given by
d𝒙 = 𝑭 d𝑿 (2.14)
The deformation gradient 𝑭 can be uniquely decomposed into the product of a rotation tensor
and a symmetric tensor by means of the polar decomposition as
𝑭 = 𝑹𝑼 Right polar decomposition (2.15)

and
𝑭 = 𝑽𝑹 Left polar decomposition (2.16)

where
• 𝑹 is the rotation tensor, an orthogonal tensor which represents the rigid body rotation
part of the deformation;
• 𝑼 is the right stretch tensor, a symmetric positive-definite tensor that represents the pure
stretching in the material in the reference configuration; and
• 𝑽 is the left stretch tensor, a symmetric positive-definite tensor that represents the pure
stretching in the material in the current configuration.
Tensors 𝑼 and 𝑽 can be calculated as
√√
𝑼= 𝑭𝑇𝑭 = 𝑪
√ √ (2.17)
𝑽 = 𝑭𝑭𝑇 = 𝑩
The products 𝑪 = 𝑭 𝑇 𝑭 and 𝑩 = 𝑭 𝑭 𝑇 are called the right and left Cauchy-Green deformation
tensors, respectively.

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Material strain tensor

𝑥3

𝑃
𝒖
𝒆3̂ 𝑃 (𝑡)
𝑿
𝒆1̂ 𝒙
𝑂 𝒆̂
2 𝑥2
𝑥1

Figure 2.3: Displacement vector.

Once 𝑼 or 𝑽 is known, the rotation tensor 𝑹 can be calculated from the polar decomposition.
For the case of rigid body motion, we have 𝑼 = 𝑽 = 𝑰 and 𝑹 = 𝑭 . For the case of translational
motion we have 𝑭 = 𝑰.
Considering the displacement field 𝒖 = 𝒙 − 𝑿, Figure 2.3, the tensor 𝑭 can be written in terms
of the displacement
𝜕𝒙
𝑭 =
𝜕𝑿
𝜕 (2.18)
= (𝒖 + 𝑿)
𝜕𝑿
= 𝛁𝑿 𝒖 + 𝑰

where the operator 𝛁𝑿 represents the gradient operator with respect to the material coordinates
𝑿.
For the case of pure translation, we have 𝒖 constant and its gradient is zero, then 𝑭 = 𝑰.

2.5 Material strain tensor

To evaluate the change in the internal geometry of the body, we can consider the change in the
length of the infinitesimal increments in both configurations. For convenience, we can consider
the squared lengths, i.e., d𝑿 · d𝑿 and d𝒙 · d𝒙. Then,
d𝒙 · d𝒙 − d𝑿 · d𝑿 = (𝑭 d𝑿) · (𝑭 d𝑿) − d𝑿 · d𝑿
= d𝑿 · (𝑭 𝑇 𝑭 d𝑿) − d𝑿 · d𝑿
(2.19)
𝑇
= d𝑿 · (𝑭
⏟⏟⏟ 𝑭 −⏟𝑰)
⏟ d𝑿
2𝑬

From this relationship we can extract the material strain tensor, also known as the Green-
Lagrange strain tensor:
1
𝑬 = (𝑭 𝑇 𝑭 − 𝑰) (2.20)
2
which fully describes the strain state at a point. This tensor contains information about not
only the normal strains (stretching or compression along the principal directions) but also shear
strains (distortion or change in the angle between two originally perpendicular directions).
The diagonal components of 𝑬, such as 𝐸11 , 𝐸22 , and 𝐸33 , represent normal strains (stretching
or compression) along each principal direction 𝑋1 , 𝑋2 , and 𝑋3 , respectively. The off-diagonal

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Spatial strain tensor

components, such as 𝐸12 , 𝐸13 , and 𝐸23 , represent shear strains. These terms quantify the
distortion in the material by measuring how much the original right angle between two directions
has changed due to deformation.
Then, the Green-Lagrange tensor is written in terms of the displacement as
1 𝑇
𝑬 = ((𝛁𝑿 𝒖 + 𝑰) (𝛁𝑿 𝒖 + 𝑰) − 𝑰)
2
(2.21)
1
= (𝛁𝑿 𝒖 + 𝛁𝑿 𝒖𝑇 + 𝛁𝑿 𝒖𝛁𝑿 𝒖𝑇 )
2

2.6 Spatial strain tensor

Analogously to the derivation of the material strain tensor,

d𝑿 = 𝑭 −1 d𝒙 (2.22)
The difference between the squared differential lengths in the spatial and material configurations
is expressed as:

d𝒙 · d𝒙 − d𝑿 · d𝑿 = d𝒙 · d𝒙 − (𝑭 −1 d𝒙) · (𝑭 −1 d𝒙)
= d𝒙 · d𝒙 − d𝒙 · (𝑭 −𝑇 𝑭 −1 d𝒙)
(2.23)
= d𝒙 · 𝑰
⏟⏟ (𝑭 −𝑇 𝑭⏟⏟
−⏟⏟⏟ −1
) d𝒙
2𝒆

where 𝒆 is the spatial deformation tensor, commonly known as the Eulerian or Almansi strain
tensor:
1 −1
𝒆 = (𝑰 − (𝑭 𝑭 𝑇 ) ) (2.24)
2
This tensor characterizes the strain at a specific point within the spatial configuration, offering
a local measure of strain in the deformed state. Since the Almansi tensor uses the deformed
configuration as the reference, it provides a direct measure of the deformation that the material
undergoes, which can be interpreted as the “true” strain at that point in the deformed body. For
this reason, the Almansi strain tensor is often used in the context of large nonlinear deformations.

2.7 Infinitesimal deformation

The Green-Lagrange strain tensor was defined in terms of the displacement field 𝒖 as
1
𝑬 = (𝛁𝑿 𝒖 + 𝛁𝑿 𝒖𝑇 + 𝛁𝑿 𝒖𝛁𝑿 𝒖𝑇 ) (2.25)
2
In the case of small deformations, the displacement gradients are small (i.e. the components of
𝛁𝑿 𝒖 are close to zero). Consequently, the quadratic term becomes negligible and the Green-
Lagrange tensor simplifies to the infinitely small strain tensor
1
𝜺 = (𝛁𝑿 𝒖 + 𝛁𝑿 𝒖𝑇 ) (2.26)
2
This tensor represnts the symmetric part of the displacement gradient tensor. The components
of 𝜺 are

1 𝜕𝑢 𝜕𝑢𝑗
𝜀𝑖𝑗 = ( 𝑖 + ) (2.27)
2 𝜕𝑋𝑗 𝜕𝑋𝑖

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Infinitesimal deformation

𝜀33
𝑥3
𝜀32
𝑥1 𝜀31 𝜀23
𝑥2 𝜀13 𝜀22
𝜀21
𝜀12
𝜀11

Figure 2.4: Components of the infinitely small strain tensor.

The tensor 𝜺 can be written in full form as

𝜕𝑢 1 𝜕𝑢1 𝜕𝑢2 1 𝜕𝑢1 𝜕𝑢3

( 2 ( 𝜕𝑋2 + 𝜕𝑋1 ) 2 ( 𝜕𝑋3 + 𝜕𝑋1 ))
1
(
(
𝜕𝑋1
))
( 1 𝜕𝑢2 𝜕𝑢1 𝜕𝑢2 1 𝜕𝑢2 𝜕𝑢3 )
𝜺=(
( 2 ( 𝜕𝑋1 + 𝜕𝑋2 ) 𝜕𝑋2 2 ( 𝜕𝑋3 + )
𝜕𝑋2 )) (2.28)
(
( ))
( 1 𝜕𝑢3 𝜕𝑢1 1 𝜕𝑢3 𝜕𝑢2 𝜕𝑢3 )
2 ( 𝜕𝑋1 + 𝜕𝑋3 ) 2 ( 𝜕𝑋2 + 𝜕𝑋3 ) 𝜕𝑋3
( )
Figure 2.4 shows a representation of the components of the strain tensor 𝜺.

Note 2.1: For the case of small deformations, we have 𝜺 ≈ 𝑬 and 𝜺 ≈ 𝑭 − 𝑰.

The non-symmetric part of the displacement gradient tensor is called the infinitesimal rotation
tensor and is given by
1
𝝎 = (𝛁𝑿 𝒖 − 𝛁𝑿 𝒖𝑇 ) (2.29)
2
which can be written in full form as

1 𝜕𝑢1 𝜕𝑢2 1 𝜕𝑢1 𝜕𝑢3

( 0 2 ( 𝜕𝑋2 − 𝜕𝑋1 ) 2 ( 𝜕𝑋3 − 𝜕𝑋1 ))
(
( ))
( 𝜕𝑢 𝜕𝑢1 1 𝜕𝑢2 𝜕𝑢3 )
𝝎=(
(
1
2 ( 𝜕𝑋1 −
2
𝜕𝑋2 ) 0 2 ( 𝜕𝑋3 − )
𝜕𝑋2 )) (2.30)
(
( ))
( 1 𝜕𝑢3 𝜕𝑢1 1 𝜕𝑢3 𝜕𝑢2 )
2 ( 𝜕𝑋1 − 𝜕𝑋3 ) 2 ( 𝜕𝑋2 − 𝜕𝑋3 ) 0
( )
Tensor 𝝎 captures the local rotation of a material due to deformation. It is related to the
rigid-body-like rotation that occurs within the material as it deforms. This rotation does not
contribute to the strain energy, as it does not change the relative distances between neighboring
points, but it does describe the orientation change.

Note 2.2: For small deformations we have 𝜺 ≈ 𝑬 ≈ 𝒆.

Example 2.4 : Given the displacement field shown for the body in the figure, determine the
following at the material point 𝑿 = (1, 1, 0): the displacement field 𝒖, the Green-Lagrange
tensor 𝑬, the infinitesimal strain tensor 𝜺, and the infinitesimal rotation tensor 𝝎. The body
has a square shape with side length 1 in the reference configuration. Assume that there is no
displacement in the 𝑥3 direction,

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Infinitesimal deformation

0.006
𝑥2
0.004
𝐷 𝐶

𝑥1 0.004
𝐴 𝐵
0.005

Solution::
Assuming a bilinear variation of the displacement field, we define
𝑢1 (𝑋1 , 𝑋2 ) = 𝑐1 + 𝑐2 𝑋1 + 𝑐3 𝑋2 + 𝑐4 𝑋1 𝑋2
𝑢2 (𝑋1 , 𝑋2 ) = 𝑐5 + 𝑐6 𝑋1 + 𝑐7 𝑋2 + 𝑐8 𝑋1 𝑋2

To determine the constants 𝑐𝑖 , we use the boundary conditions at points 𝐴, 𝐵, 𝐶, and 𝐷

𝑢1 (𝐴) = 𝑐1 = 0
𝑢1 (𝐵) = 𝑐1 + 𝑐2 = 0.005
𝑢1 (𝐶) = 𝑐1 + 𝑐2 + 𝑐3 + 𝑐4 = 0.006
𝑢1 (𝐷) = 𝑐1 + 𝑐3 = 0

Solving the system of equations we have

𝑐1 = 0, 𝑐2 = 0.005, 𝑐3 = 0, 𝑐4 = 0.001

Symilarly, for the displacement field 𝑢2 , we have

𝑢2 (𝐴) = 𝑐5 = 0
𝑢2 (𝐵) = 𝑐5 + 𝑐6 = 0.004
𝑢2 (𝐶) = 𝑐5 + 𝑐6 + 𝑐7 + 𝑐8 = 0.004
𝑢2 (𝐷) = 𝑐5 + 𝑐7 = 0

Solving the system yields

𝑐5 = 0, 𝑐6 = 0.004, 𝑐7 = 0, 𝑐8 = 0

Thus, the displacement field becomes

𝑢 = (0.005𝑋1 + 0.001𝑋1 𝑋2 )𝒆1̂ + 0.004𝑋1 𝒆2̂

Next, we compute the displacement gradient tensor

0.005 + 0.001𝑋2 0.001𝑋1 0

( )
𝛁𝑿 𝒖 = (
( 0.004 0 0))
( 0 0 0)

Evaluating at point 𝑿 = (1, 1, 0) we have

9/22
Compatibility equations

0.006 0.001 0
( )
𝛁𝑿 𝒖 = (
(0.004 0 0) )
( 0 0 0)
The Green-Lagrange tensor, defined as 𝐸 = 12 (𝛁𝑿 𝒖 + 𝛁𝑿 𝒖𝑇 + 𝛁𝑿 𝒖𝛁𝑿 𝒖𝑇 ) is then calculated
as

0.0060185 0.000502 0.0

( )
𝑬=(
( 0.000502 0.004008 0.0))
( 0.0 0.0 0.0)
For the infinitesimal strain tensor 𝜺 = 12 (𝛁𝑿 𝒖 + 𝛁𝑿 𝒖𝑇 ), we have

0.006 0.0005 0.0

( )
𝜺=(
(0.0005 0.004 0.0))
( 0.0 0.0 0.0)
Finally, the infinitesimal rotation tensor 𝝎 = 12 (𝛁𝑿 𝒖 − 𝛁𝑿 𝒖𝑇 ) is given by

0 0.0005 0
(
(−0.0005 )
)
𝝎=( 0 0)
( 0 0 0)

Exercices 2.1:
1. A body is subjected to a deformation given by the displacement field

𝒖 = (𝑋12 + 𝑋22 )𝒆1̂ + 𝑋1 𝑋2 𝒆2̂

Determine the Green-Lagrange tensor 𝑬, the infinitesimal strain tensor 𝜺 and the infinitesimal
rotation tensor 𝝎.

2.8 Compatibility equations

For a continuous displacement field 𝒖, it is possible to find the corresponding strain field by
derivation of the displacement components. For example, for the case of the infinitesimal strain
tensor

1 𝜕𝑢 𝜕𝑢𝑗
𝜀𝑖𝑗 = ( 𝑖 + ) (2.31)
2 𝜕𝑋𝑗 𝜕𝑋𝑖

On the other hand, given a compatible strain field, it is possible to find the corresponding
displacement field 𝒖 by integration of the strain components. A deformation field (or displace-
ment field) is called compatible if it represents a physically realizable deformation, meaning it
can be achieved without introducing discontinuities or gaps in the material.

Example 2.5: Given a vector field 𝒗, it is said that 𝒗 is a potential field if there is a scalar
field 𝜑 such that
𝒗 = 𝛁𝜑

Find the compatibility equations that the vector field 𝒗 must satisfy to be a potential field.

10/22
Compatibility equations

Considering the components of 𝒗, we can write:

𝜕𝜑 𝜕𝜑 𝜕𝜑
𝑣1 − = 0, 𝑣2 − = 0, 𝑣3 − =0
𝜕𝑋1 𝜕𝑋2 𝜕𝑋3

For an arbitrary vector field, these equations are not sufficient to guarantee that 𝒗 is a
potential field. Thus, the vector field 𝒗 should satisfy addtional conditions called compatibility
equations. These conditions can be obtained by deriving the previous equations with respect
to 𝑋2 and 𝑋3 , that is

𝜕𝑣1 𝜕2𝜑 𝜕𝑣2 𝜕 2 𝑣2 𝜕𝑣2 𝜕 2 𝑣2

− = 0, − = 0, − =0
𝜕𝑋1 𝜕𝑋12 𝜕𝑋2 𝜕𝑋1 𝜕𝑋2 𝜕𝑋3 𝜕𝑋1 𝜕𝑋3
𝜕𝑣1 𝜕 2 𝑣1 𝜕𝑣2 𝜕2𝜑 𝜕𝑣2 𝜕 2 𝑣2
− = 0, − = 0, − =0
𝜕𝑋2 𝜕𝑋2 𝜕𝑋1 𝜕𝑋2 𝜕𝑋22 𝜕𝑋3 𝜕𝑋2 𝜕𝑋3
𝜕𝑣1 𝜕 2 𝑣1 𝜕𝑣2 𝜕 2 𝑣2 𝜕𝑣3 𝜕2𝜑
− = 0, − = 0, − =0
𝜕𝑋3 𝜕𝑋3 𝜕𝑋1 𝜕𝑋3 𝜕𝑋3 𝜕𝑋2 𝜕𝑋3 𝜕𝑋32

At this point we can consider the Schwarz theorem (equality of mixed derivatives). Thus, from
the above equations we have

𝜕 2 𝑣2 𝜕 2 𝑣2 𝜕 2 𝑣2 𝜕 2 𝑣2 𝜕 2 𝑣2 𝜕 2 𝑣2
= , = , and =
𝜕𝑋1 𝜕𝑋2 𝜕𝑋2 𝜕𝑋1 𝜕𝑋1 𝜕𝑋3 𝜕𝑋3 𝜕𝑋1 𝜕𝑋2 𝜕𝑋3 𝜕𝑋3 𝜕𝑋2

Then, the following equations can be derived from the previous ones
𝜕𝑣1 𝜕𝑣2 𝜕𝑣1 𝜕𝑣3 𝜕𝑣2 𝜕𝑣3
− = 0, − = 0, − =0
𝜕𝑋2 𝜕𝑋1 𝜕𝑋3 𝜕𝑋1 𝜕𝑋3 𝜕𝑋2

As can be seen, these equations represent certain relations between the components of the
vector field 𝒗 that must be satisfied for 𝒗 to be a potential field. Then, they are called
compatibility equations.

Considering the infinitesimal strain tensor

𝜕𝑢 1 𝜕𝑢1 𝜕𝑢2 1 𝜕𝑢1 𝜕𝑢3

( 2 ( 𝜕𝑋2 + 𝜕𝑋1 ) 2 ( 𝜕𝑋3 + 𝜕𝑋1 ))
1
(
(
𝜕𝑋1
))
( 1 𝜕𝑢2 𝜕𝑢1 𝜕𝑢2 1 𝜕𝑢2 𝜕𝑢3 )
𝜺=(
( 2 ( 𝜕𝑋1 + 𝜕𝑋2 ) 𝜕𝑋2 2 ( 𝜕𝑋3 + )
𝜕𝑋2 )) (2.32)
(
( ))
( 1 𝜕𝑢3 𝜕𝑢1 1 𝜕𝑢3 𝜕𝑢2 𝜕𝑢3 )
2 ( 𝜕𝑋1 + 𝜕𝑋3 ) 2 ( 𝜕𝑋2 + 𝜕𝑋3 ) 𝜕𝑋3
( )
due to symmetry, only six components are independent. Thus, we can write the following set of
equations
𝜕𝑢1 1 𝜕𝑢 𝜕𝑢2
𝜀11 − = 0, 𝜀12 − ( 1 + ) = 0,
𝜕𝑋1 2 𝜕𝑋2 𝜕𝑋1
𝜕𝑢2 1 𝜕𝑢 𝜕𝑢3
𝜀22 − = 0, 𝜀13 − ( 1 + ) = 0, (2.33)
𝜕𝑋2 2 𝜕𝑋3 𝜕𝑋1
𝜕𝑢3 1 𝜕𝑢 𝜕𝑢3
𝜀33 − = 0, 𝜀23 − ( 2 + )=0
𝜕𝑋3 2 𝜕𝑋3 𝜕𝑋2

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Compatibility equations

Deriving these equations twice with respect to 𝑋1 , 𝑋2 and 𝑋3 we get a new set of equations
where the Schwarz theorem can be used to obtain the following compatibility equations

𝜕 2 𝜀11 𝜕 2 𝜀22 𝜕 2 𝜀12

2
+ 2
−2 =0
𝜕𝑋2 𝜕𝑋1 𝜕𝑋1 𝜕𝑋2
𝜕 2 𝜀11 𝜕 2 𝜀33 𝜕 2 𝜀13
2
+ 2
−2 =0
𝜕𝑋3 𝜕𝑋1 𝜕𝑋1 𝜕𝑋3
𝜕 2 𝜀22 𝜕 2 𝜀33 𝜕 2 𝜀23
+ − 2 =0
𝜕𝑋32 𝜕𝑋22 𝜕𝑋2 𝜕𝑋3
(2.34)
𝜕 2 𝜀11 𝜕 𝜕𝜀 𝜕𝜀 𝜕𝜀
− (− 23 + 13 + 12 ) = 0
𝜕𝑋2 𝜕𝑋3 𝜕𝑋1 𝜕𝑋1 𝜕𝑋2 𝜕𝑋3
𝜕 2 𝜀22 𝜕 𝜕𝜀 𝜕𝜀 𝜕𝜀
− ( 23 − 13 + 12 ) = 0
𝜕𝑋1 𝜕𝑋3 𝜕𝑋2 𝜕𝑋1 𝜕𝑋2 𝜕𝑋3
𝜕 2 𝜀33 𝜕 𝜕𝜀 𝜕𝜀 𝜕𝜀
− ( 23 + 13 − 12 ) = 0
𝜕𝑋1 𝜕𝑋2 𝜕𝑋3 𝜕𝑋1 𝜕𝑋2 𝜕𝑋3

For 2D space, only the following compatibility equation is required

𝜕 2 𝜀11 𝜕 2 𝜀22 𝜕 2 𝜀12

2
+ 2
−2 =0 (2.35)
𝜕𝑋2 𝜕𝑋1 𝜕𝑋1 𝜕𝑋2

Example 2.6: Given the infinitessimal strain field,

0 𝑘𝑋1 𝑋2 0
( )
𝜺=(
(𝑘𝑋1 𝑋2 0 0))
( 0 0 0 )
verify if the compatibility equations are satisfied. Then, by attempting integration, show that
there is no continuous displacement field that corresponds to this strain tensor field.
Solution:
Using the first compatibility equation, we have:

𝜕 2 𝜀11 𝜕 2 𝜀22 𝜕 2 𝜀12

𝐶= 2
+ 2
−2
𝜕𝑋2 𝜕𝑋1 𝜕𝑋1 𝜕𝑋2

where 𝐶 should equal zero. Performing the calculations

𝜕20 𝜕20 𝜕2
𝐶= 2
+ 2
−2 (𝑘𝑋1 𝑋2 )
𝜕𝑋2 𝜕𝑋1 𝜕𝑋1 𝜕𝑋2
𝐶 = −2𝑘
Since 𝐶 ≠ 0, the compatibility equations are not satisfied.
Next, let’s attempt to find a displacement field that corresponds to the given strain tensor
field. Considering the strain components
𝜕𝑢1 𝜕𝑢2
𝜀11 = =0 and 𝜀22 = =0
𝜕𝑋1 𝜕𝑋2

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Typical strain states

Integrating with respect to 𝑋1 and 𝑋2 , respectively, we obtain

𝑢1 = 𝑓(𝑋2 , 𝑋3 ) and 𝑢2 = 𝑔(𝑋1 , 𝑋3 )

where 𝑓 and 𝑔 are arbitrary functions.

Next, for the components 𝜀12 , we have
1 𝜕𝑢 𝜕𝑢2
𝜀12 = ( 1 + ) = 𝑘𝑋1 𝑋2
2 𝜕𝑋2 𝜕𝑋1

Applying the derivatives we get

1 𝜕 𝜕
( 𝑓(𝑋2 , 𝑋3 ) + 𝑔(𝑋1 , 𝑋3 )) = 𝑘𝑋1 𝑋2
2 𝜕𝑋2 𝜕𝑋1

which can be rewritten as

̄
𝑓(𝑋 2 , 𝑋3 ) + 𝑔(𝑋
̄ 1 , 𝑋3 ) = 𝑘𝑋1 𝑋2

where 𝑓 ̄ and 𝑔 ̄ are new arbitrary functions. Note that the left-hand side of the equation
cannot have a term 𝑋1 𝑋2 , which indicates that there is no continuous displacement field that
corresponds to the given strain tensor field.

Exercices 2.2:
1. Check if the following tensor field qualifies as a compatible infinitesimal strain field,

( 𝑘𝑋22 𝑘𝑋1 0 )
(𝑘𝑋 𝑘𝑋
𝜺=( 0 ) )
( 1 2 )
2
( 0 0 𝑘𝑋 2 )
2. Show that the tensor field below is a compatible,

( 2𝑘𝑋1 32 𝑘𝑋2 0 )
(
𝜺=( 3 )
( 2 𝑘𝑋2 𝑘𝑋1 0 ) )
( 0 0 𝑘𝑋 3 )

2.9 Typical strain states

In addition to the general strain tensor, there are specific strain states frequently encountered
in materials, each describing a unique deformation pattern. Examples include:
Uniaxial strain: A state of strain in which the material is stretched or compressed in one
direction while the other two directions remain unchanged. E.g.,

𝜀 0 0
( 11 )
𝜺=(
( 0 0 0)
) (2.36)
( 0 0 0)
Pure shear: Also known as simple shear, this strain state occurs when the material is sheared
with no change in volume. For example

13/22
Volumetric strain

0 𝜀12 0 𝜀 0 0
( ) ( )
𝜺=(
(𝜀21 0 0) ) or 𝜺′ = (
(0 −𝜀 0)
) (2.37)
( 0 0 0 ) ( 0 0 0)
where 𝜀 = 𝜀′11 = −𝜀′22 .
Plane strain: A state of strain in which the material is deformed in a plane, with no change
in thickness in the direction perpendicular to the plane. E.g.,

𝜀 𝜀 0
( 11 12 )
𝜺=(
(𝜀21 𝜀22 0)) (2.38)
( 0 0 0 )
Volumetric strain: A state of strain in which the material is compressed or expanded in all
directions. E.g.,

𝜀 0 0
( )
𝜺=(
(0 𝜀 0) ) (2.39)
(0 0 𝜀 )
where 𝜀 = 𝜀11 = 𝜀22 = 𝜀33 .

2.10 Volumetric strain

Volumetric strain (also called dilatation or dilatational strain) refers to the relative change in
volume of a material element due to deformation. It is a measure of how much a material
expands or contracts uniformly in all directions. It is defined as
Δ(d𝑉 )
𝜀𝑣 = (2.40)
d𝑉
where d𝑉 is a infinitesimal volume element in the reference configuration. Assuming d𝑉 as a
parallelepiped whose sides have been elongated from the initial lengths d𝑙1 , d𝑙2 , and d𝑙3 , to
d𝑙1 (1 + 𝜀1 ), d𝑙2 (1 + 𝜀2 ), and d𝑙3 (1 + 𝜀3 ), respectively, the dilatation is given by
d𝑙1 (1 + 𝜀1 ) d𝑙2 (1 + 𝜀2 ) d𝑙3 (1 + 𝜀3 ) − d𝑙1 d𝑙2 d𝑙3
𝜀𝑣 =
d𝑙1 d𝑙2 d𝑙3
(2.41)
d𝑙1 d𝑙2 d𝑙3 (𝜀1 + 𝜀2 + 𝜀3 ) − d𝑙1 d𝑙2 d𝑙3 + ⋯
=
d𝑙1 d𝑙2 d𝑙3

For small deformations, the dilatation can be approximated as

𝜀𝑣 = 𝜀1 + 𝜀2 + 𝜀3 (2.42)

Which is also expressed as

𝜕𝑢𝑖
𝜀𝑣 = 𝜀𝑖𝑖 = =𝛁·𝒖 (2.43)
𝜕𝑋𝑖

Since 𝐼1 = trace(𝜺) = 𝜀𝑣 , the first invariant determines if the material is undergoing expansion
(𝐼1 > 0) or contraction (𝐼1 < 0).

Exercices 2.3:
1. Given the motion field

14/22
Engineering strains

𝒖 = 𝑘(𝑋12 + 𝑋22 )𝑡 𝒆1̂ + 𝑘𝑋1 𝑋2 𝒆2̂ + 𝑘𝑋1 𝑋3 𝒆3̂

where 𝑘 = 10−4 , determine the volumetric strain at point 𝑿 = (2, 1, 1) for 𝑡 = 2.

2.11 Engineering strains

Engineering strains are measures that quantify both the relative change in length along a specific
directions (normal strains) and angular changes between initially perpendicural lines (shear
strains). The normal components in a given direction, say 𝑥, is defined as:
Δ𝐿𝑥
𝜀𝑥 = = 𝜀11 (2.44)
𝐿𝑥

where Δ𝐿𝑥 is the change in length and 𝐿𝑥 is the original length in the 𝑥 direction.
The shear strain, for example in the 𝑥𝑦 plane, is defined as
𝛾𝑥𝑦 = tan(Δ𝜃) = 2𝜀12 (2.45)

where Δ𝜃 is the angular change between two initially perpendicular lines. Note that the shear
engineering strain is twice the size of the corresponding component in the infinitesimal strain
tensor. This factor of 2 arises because the component 𝜀12 in the infinitesimal strain tensor
represents the average angular distortion in both the 𝑥 and 𝑦 directions.
Thus, the engineering strains are related to the components of the infinitesimal strain tensor as
follows
1 1
𝜀11 𝜀12 𝜀13 ( 𝜀𝑥 2 𝛾𝑥𝑦 2 𝛾𝑥𝑧 )
(
( )
) (
( )
1
𝜺 = (𝜀21 𝜀22 𝜀23 ) = ( 2 𝛾𝑦𝑥 𝜀𝑦 12 𝛾𝑦𝑧 )
) (2.46)
(1 )
(𝜀31 𝜀32 𝜀33 ) ( 2 𝛾𝑧𝑥
1
𝛾
2 𝑧𝑦 𝜀 𝑧
)
Due to the symmetry of the strain tensor, it is common to use only 6 independent components.
Then, the engineering strain tensor can be represented by a 6-component vector as

( 𝜀𝑥 )
(
( 𝜀𝑦 ) )
(
( )
)
𝜀
𝜺eng =(
(
𝑧 )
(2.47)
( 𝛾𝑦𝑧 ))
(
( )
( 𝑧𝑥 )
𝛾 )
𝛾
( 𝑥𝑦 )

Note 2.3 : Note that depending on the source, the shear components of the engineering strain
tensor can be listed in a different order.

2.12 Strain transformation

The components of the strain tensor 𝜺 can be expressed in a different coordinate system by
applying a transformation tensor 𝑹. This transformation is given by

𝜺′ = 𝑹 𝑇 𝜺𝑹 (2.48)
Here, the components of 𝑹 are the direction cosines of the new coordinate system with respect
to the original coordinate system. Thus,

15/22
Strain transformation

𝑅11 𝑅12 𝑅13 | | |

(
( )
) ( ′)
( 21 22 23 ) ( 1̂ 𝒆2̂ 𝒆3̂ )
(
𝑅 𝑅 𝑅 = 𝒆 ′ ′ (2.49)
)
(𝑅31 𝑅32 𝑅33 ) ( | | | )
where 𝒆′1̂ , 𝒆′2̂ , and 𝒆′3̂ are the unit vectors of the new coordinate system.
When the strain tensor is given in the form of a vector in engineering notation, the transfor-
mation can be performed using a 6 × 6 transformation matrix, as follows
̄
𝜺′ = 𝑹𝜺 (2.50)
where 𝑹̄ is constructed from the components of the transformation tensor 𝑹. For convenience,
let’s assume that the second-order transformation tensor is given by

𝑅 𝑅 𝑅 | | |
( 11 12 13 ) ( )
𝑹=(
(𝑅21 𝑅22 𝑅23 ) )=(
(𝒍 𝒎 𝒏) ) (2.51)
( 𝑅 31 𝑅32 𝑅 33 ) ( | | | )
where 𝒍, 𝒎, and 𝒏 are the unit vectors of the new coordinate system, the transformation matrix
𝑹,̄ used to transform a strain vector in engineering notation, is then given by

( 𝑙21 𝑙22 𝑙23 𝑙2 𝑙 3 𝑙3 𝑙1 𝑙1 𝑙2 )

(
( 2 2 2 )
)
( 𝑚 1 𝑚 2 𝑚 3 𝑚 2 𝑚 3 𝑚 3 𝑚 1 𝑚 1 𝑚 2 )
(
( 𝑛 2
𝑛 2
𝑛 2
𝑛 𝑛 𝑛 𝑛 𝑛 𝑛 )
)
𝑹̄ = (
(
1 2 3 2 3 3 1 1 2 ) (2.52)
(2𝑚1 𝑛1 2𝑚2 𝑛2 2𝑚3 𝑛3 𝑚2 𝑛3 + 𝑚3 𝑛2 𝑚3 𝑛1 + 𝑚1 𝑛3 𝑚1 𝑛2 + 𝑚2 𝑛1 ) )
(
( )
2𝑙
( 1 1 𝑛 2𝑙 𝑛
2 2 2𝑙 𝑛
3 3 𝑙 𝑛
3 2 + 𝑛 𝑙
3 2 𝑙 𝑛
1 3 + 𝑛 𝑙
1 3 𝑙 𝑛
2 1 + 𝑛 2 1 )
𝑙 )
2𝑙1 𝑚1 2𝑙2 𝑚2 2𝑙3 𝑚3 𝑙2 𝑚3 + 𝑙3 𝑚2 𝑙3 𝑚1 + 𝑙1 𝑚3 𝑙1 𝑚2 + 𝑙2 𝑚1
( )

Note 2.4 : The transformation matrix 𝑹̄ presented here is only valid if the shear components
are listed in the same order as shown in Eq. (2.47).

If the strain tensors are expressed in Mandel notation, then the corresponding transformation
matrix is given by
√ √ √
( 𝑙21 𝑙22 𝑙23 2𝑙2 𝑙3 2𝑙3 𝑙1 2𝑙 𝑙 )
( √ √ √ 12 )
(
( 𝑚1 2
𝑚2 2
𝑚3 2
2𝑚2 𝑚3 2𝑚3 𝑚1 2𝑚1 𝑚2 ) )
( √ √ √ )
(
( 𝑛 2
𝑛 2
𝑛 2
2𝑛 𝑛 2𝑛 𝑛 2𝑛 𝑛 )
)
𝑹̄ 𝑀 =(
(
√ 1 √ 2 √ 3 2 3 3 1 1 2
)(2.53)
( 2𝑚1 𝑛1 2𝑚2 𝑛2 2𝑚3 𝑛3 𝑚2 𝑛3 + 𝑚3 𝑛2 𝑚3 𝑛1 + 𝑚1 𝑛3 𝑚1 𝑛2 + 𝑚2 𝑛1 ) )
( √ √ √ )
(
( 2𝑙 𝑛 2𝑙 𝑛 2𝑙 𝑛 𝑙 𝑛 + 𝑛 𝑙 𝑙 𝑛 + 𝑛 𝑙 𝑙 𝑛 + 𝑛 𝑙 )
)
(√ 1 1 √ 2 2 √ 3 3 3 2 3 2 1 3 1 3 2 1 2 1 )
2𝑙1 𝑚1 2𝑙2 𝑚2 2𝑙3 𝑚3 𝑙2 𝑚3 + 𝑙3 𝑚2 𝑙3 𝑚1 + 𝑙1 𝑚3 𝑙1 𝑚2 + 𝑙2 𝑚1
( )
For a 2D case, considering the new coordinate system rotated by an angle 𝜃 counterclockwise
around the 𝑥3 axis (out of the plane), the strain components in the new system can be calculated
in a straightforward manner as
𝜀11 + 𝜀22 𝜀11 − 𝜀22
𝜀′1;2 = ± cos(2𝜃) ± 𝜀12 sin(2𝜃)
2 2
𝜀 − 𝜀22 (2.54)
𝜀′12 = − 11 sin(2𝜃) + 𝜀12 cos(2𝜃)
2

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Principal strains and directions

Example 2.7: Given the following strain tensor, determine its components in a coordinate
system rotated by 30° around the 𝑥1 axis in the original coordinate system.

0.01 0.02 0
( )
𝜺=(
(0.02 0.04 0.05))
( 0 0.05 0.06)
Solution:
The unit vector in the 𝑥′1 direction is the same as in the original coordinate system, thus 𝒆′1̂ =
𝒆1̂ . For the 𝑥′2 and 𝑥′3 directions, we have
𝒆′2̂ = cos(30°)𝒆2̂ + sin(30°)𝒆3̂ = 0.866𝒆2̂ + 0.5𝒆3̂
𝒆′3̂ = − sin(30°)𝒆2̂ + cos(30°)𝒆3̂ = −0.5𝒆2̂ + 0.866𝒆3̂

Then the transformation tensor is

1 0 0
( )
𝑹=(
( 0 0.866 −0.5 )
)
( 0 0.5 0.866 )
Finally, the components of the strain tensor in the new coordinate system are

0.01 0.0173 −0.01

( )
𝜺′ = 𝑹 𝑇 𝜺𝑹 = (
( 0.0173 0.0883 0.0337)
)
( −0.01 0.0337 0.0117)
If we consider the strain tensor as a vector in engineering notation, the 6 × 6 transformation
matrix is given by

( 1 0 0 0 0 0 )
(
( 0 0.75 0.25 0.866 0 0 ) )
(
( )
)
0 0.25 0.75 −0.866 0 0
𝑹̄ = (
( )
( 0 −0.433 0.433 0.5 0 0 ) )
(
( )
)
( 0 0 0 0 0.866 −0.5 )
0 0 0 0 0.5 0.866
( )
Then, the components of the strain vector in the new coordinate system is

( 0.01 )
(
( 0.0883))
(
( )
)
̄ eng 0.0117
𝜺′eng = 𝑹𝜺 =(
( )
( 0.0673))
(
( )
( −0.02 ))
0.0346
( )
Note that the shear components in the engineering strain vector are twice the size of the
corresponding components in the strain tensor, as expected.

2.13 Principal strains and directions

The principal strains are the eigenvalues of the infinitesimal strain tensor 𝜺. These eigenvalues
represent the normal strains along the principal directions, the directions in which the material

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Principal strains and directions

𝜀33
𝑥3 𝑥3′ 𝜀3
𝜀32
𝜀2
𝑥1 𝜀31 𝜀23 𝑥1′ 𝑥2′
𝑥2 𝜀13 𝜀22
𝜀21
𝜀12 𝜀1
𝜀11

Figure 2.5: Strain components in the original and principal strain direction systems

experiences pure stretching or compression without shear, Figure 2.5. The principal strains are
denoted by 𝜀1 , 𝜀2 , and 𝜀3 and are defined such that
𝜺𝒗𝑖 = 𝜀𝑖 𝒗𝑖 (no summation on 𝑖) (2.55)

where 𝒗𝑖 is the eigenvector associated with the eigenvalue 𝜀𝑖 .

The application of the strain tensor 𝜺 to 𝒗𝑖 scales 𝒗𝑖 by 𝜀𝑖 without changing its direction.
Since 𝒗𝑖 is unchanged in direction, the strain tensor 𝜺 has no component that distorts 𝒗𝑖 into
a different direction, as would occur if shear were present. Each value of 𝜀𝑖 is a measure of how
much the material stretches or compresses in the direction 𝒗𝑖 .
The principal strains provide the maximum and minimum normal strains at a point, offering a
way to identify the directions in which the material is most stretched or compressed.
The principal directions are the eigenvectors of the infinitesimal strain tensor 𝜺 and are denoted
by 𝒗1 , 𝒗2 , and 𝒗3 . These directions represent the orientations along which the material experi-
ences pure stretching or contraction with no accompanying shear.
In mechanics and material science, it is common practice to list the eigenvalues of the strain
tensor in descending order, from largest to smallest, such that 𝜀1 ≥ 𝜀2 ≥ 𝜀3 . The corresponding
eigenvectors are also listed in this order, pairing each eigenvalue with its associated principal
direction.
The first invariant of tensor 𝜺 is given by
𝐼1 = trace(𝜺) = 𝜀1 + 𝜀2 + 𝜀3 (2.56)

This invariant is related to the overall strain in the material. and is a measure of the volumetric
change. The second and third invariants of the infinitesimal strain tensor provide information
about shear and combined deformation but are less directly interpreted in plysical terms.
For 2D space, the principal strains can be calculated in a straightforward manner as

1 𝜀 − 𝜀22 2
𝜀1;2 = (𝜀11 + 𝜀22 ) ± √( 11 ) + 𝜀212 (2.57)
2 2

Example 2.8: Find the principal strains and directions for the strain tensor of a solid body
under simples shear as shown in the figure. The body has unitary dimensions in the 𝑥1 and
𝑥2 directions. Consider no strain in the 𝑥3 direction.

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Principal strains and directions

𝑥2
0.004 0.004

𝐷 𝐶

𝑥1
𝐴 𝐵

Solution:
From the figure we can see that the displacement fields in the 𝑥1 and 𝑥2 directions are
𝑢1 = 0.004𝑋2 , 𝑢2 = 0 (2.58)

Then, the strain components are

𝜕𝑢1 𝜕𝑢2 1 𝜕𝑢 𝜕𝑢2
𝜀11 = = 0, 𝜀22 = = 0, 𝜀12 = ( 1 + ) = 0.002 (2.59)
𝜕𝑋1 𝜕𝑋2 2 𝜕𝑋2 𝜕𝑋1

and the strain tensor is given by

0 0.002 0
( )
𝜺=(
(0.002 0 0) ) (2.60)
( 0 0 0 )
The eigenvalues, listed from greater to smaller, and corresponding eigenvectors of this strain
tensor are
𝜀1 = 0.002, 𝜀2 = −0.002, 𝜀3 = 0 (2.61)

and

0.707 0.707 0
( ) ( ) ( )
𝒗1 = (
(0.707)), 𝒗2 = (
(−0.707)), 𝒗3 = (
(0) ) (2.62)
( 0 ) ( 0 ) 1
( )
respectively. Note that the coordinate system of the principal directions is rotated by 45° with
respect to the original coordinate system.
Since this is a 2D problem, we can also calculate the principal strains using the formula

1 𝜀 − 𝜀22 2
𝜀1;2 = (𝜀11 + 𝜀22 ) ± √( 11 ) + 𝜀212 = ±0.002 (2.63)
2 2
which provides the same results as the eigenvalues.

Exercices 2.4:
1. Given the tensor field

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Infinitesimal strain in cylindrical coordinates

4𝑘𝑋1 3𝑘𝑋2 0
( )
𝜺=(
(3𝑘𝑋2 2𝑘𝑋1 0 ) )
( 0 0 2𝑘𝑋3 )

where 𝑘 = 10−4 , determine the principal strains and directions at point 𝑿 = (1, 1, 1).
Compare the results with the ones obtained using the 2D formula.
2. Given the motion field

𝒖 = 𝑘(𝑋12 + 𝑋22 )𝑡 𝒆1̂ + 𝑘𝑋1 𝑋2 𝒆2̂ + 𝑘𝑋1 𝑋3 𝒆3̂

where 𝑘 = 10−4 , determine the principal strains and directions at point 𝑿 = (2, 1, 1) for
𝑡 = 1 and 𝑡 = 2.

2.14 Infinitesimal strain in cylindrical coordinates

In cylindrical coordinates, the displacement field is given by
𝒖 = 𝑢𝑟 𝒆𝑟̂ + 𝑢𝜃 𝒆𝜃̂ + 𝑢𝑧 𝒆𝑧̂ (2.64)

in the radial, circumferential (angular), and axial directions, respectively. The infinitesimal strain
tensor in cylindrical coordinates can be derived from the displacement gradients, resulting in
the following components:

𝜀 𝜀 𝜀
( 𝑟𝑟 𝑟𝜃 𝑟𝑧 )
𝜺=(
( 𝜃𝑟 𝜀𝜃𝜃 𝜀𝜃𝑧 )
𝜀 ) (2.65)
(𝜀𝑧𝑟 𝜀𝑧𝜃 𝜀𝑧𝑧 )
where each component is given by
𝜕𝑢𝑟
𝜀𝑟𝑟 =
𝜕𝑟
1 𝜕𝑢𝜃 𝑢𝑟
𝜀𝜃𝜃 = +
𝑟 𝜕𝜃 𝑟
𝜕𝑢𝑧
𝜀𝑧𝑧 =
𝜕𝑧
1 1 𝜕𝑢𝑟 𝜕𝑢𝜃 𝑢𝜃 (2.66)
𝜀𝑟𝜃 = ( + − )
2 𝑟 𝜕𝜃 𝜕𝑟 𝑟
1 𝜕𝑢 𝜕𝑢
𝜀𝑟𝑧 = ( 𝑟 + 𝑟 )
2 𝜕𝑧 𝜕𝑧
1 𝜕𝑢 1 𝜕𝑢𝑧
𝜀𝜃𝑧 = ( 𝜃 + )
2 𝜕𝑧 𝑟 𝜕𝜃

Figure 2.6 shows the strain components in an infinitesimal volume.

Example 2.9: The tensor 𝜺 below was computed at the point 𝑿 = (4, 3, 2). Convert this tensor
𝜺 to cylindrical coordinates

0.0136 0.0048 −0.008

( )
𝜺=(
( 0.0048 0.0164 0.006 )) (2.67)
(−0.008 0.006 0.005 )

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Infinitesimal strain in cylindrical coordinates

𝑥3
𝜀𝑧𝑧

𝜀𝑧𝜃
𝜀𝑧𝑟 𝜀𝑟𝑧
𝜀𝜃𝑟 d𝑧
𝜀𝑟𝜃 𝜀𝑟𝑟
𝜀𝜃𝜃 𝜀𝜃𝑧
d𝑟 𝑟 d𝜃
𝑥2
𝑥1

Figure 2.6: Components of the infinitesimal strain tensor in cylindrical coordinates.

Solution:
This transformation can be performed under the assumption of small displacements. For small
displacements, we can use a linear transformation to convert the strain tensor from Cartesian
to cylindrical coordinates. The unit vectors in the cylindrical coordinate system are
𝒆𝑟̂ = cos(𝜃)𝒆1̂ + sin(𝜃)𝒆2̂ , 𝒆𝜃̂ = − sin(𝜃)𝒆1̂ + cos(𝜃)𝒆2̂ , 𝒆𝑧̂ = 𝒆3̂ (2.68)
𝑥 𝑦
where cos(𝜃) = 𝑟 and sin(𝜃) = 𝑟 and 𝑟 = √𝑥2 + 𝑦2 .
Then, the transformation tensor is

cos(𝜃) sin(𝜃) 0
( )
𝑹=(
(− sin(𝜃) cos(𝜃) 0)) (2.69)
( 0 0 1 )
4 3
At the point (4, 3, 2) we have 𝑟 = 5, cos(𝜃) = 5 and sin(𝜃) = 5 we have

0.8 0.6 0
( )
𝑹=(
(−0.6 0.8 0)
) (2.70)
( 0 0 1)
Then, the tensor in cylindrical coordinates is

0.01 0 −0.01
( )
𝜺′ = 𝑹 𝑇 𝜺𝑹 = (
( 0 0.02 0 ) ) (2.71)
( −0.01 0 0.005 )

Exercices 2.5:
1. The tensor 𝜺′ , in cylindrical coordinates, was calculated at (𝑟, 𝜃, 𝑧) = (3, 𝜋6 , 0). Assuming
small displacements, convert this tensor to cartesian coordinates.

0.002 −0.001 0.002

( )
𝜺′ = (
(−0.001 0.001 0 ) ) (2.72)
( 0.002 0 0.003 )

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Infinitesimal strain in spherical coordinates

2.15 Infinitesimal strain in spherical coordinates

In spherical coordinates, the displacement field is given by
𝒖 = 𝑢𝑟 𝒆𝜌̂ + 𝑢𝜃 𝒆𝜃̂ + 𝑢𝜑 𝒆𝜑̂ (2.73)

in the radial, polar, and azimuthal directions, respectively. The infinitesimal strain tensor in
spherical coordinates is given by

𝜀𝜌𝜌 𝜀𝜌𝜃 𝜀𝜌𝜑

(
( )
)
( 𝜀𝜃𝜌 𝜀𝜃𝜃 𝜀𝜃𝜑 )
𝜺=( ) (2.74)
𝜀 𝜀 𝜀
( 𝜑𝜌 𝜑𝜃 𝜑𝜑 )
where each component is given by
𝜕𝑢𝜌
𝜀𝜌𝜌 =
𝜕𝜌
1 𝜕𝑢𝜃 𝑢𝜌
𝜀𝜃𝜃 = +
𝜌 𝜕𝜃 𝜌
1 𝜕𝑢𝜑 𝑢𝜌 𝑢𝜃
𝜀𝜑𝜑 = + + cot(𝜑)
𝜌 sin(𝜃) 𝜕𝜑 𝜌 𝑟
1 1 𝜕𝑢𝜌 𝜕𝑢𝜃 𝑢𝜃 (2.75)
𝜀𝜌𝜃 = ( + − )
2 𝜌 𝜕𝜃 𝜕𝜌 𝜌

1 1 𝜕𝑢𝜌 𝜕𝑢𝜑 𝑢𝜑
𝜀𝜌𝜑 = ( + − )
2 𝜌 sin(𝜃) 𝜕𝜑 𝜕𝜌 𝜌

1 1 𝜕𝑢𝜃 1 𝜕𝑢𝜑 𝑢𝜑
𝜀𝜃𝜑 = ( + − cot(𝜑))
2 𝜌 sin(𝜃) 𝜕𝜑 𝜌 𝜕𝜃 𝜌

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