Course E+P P L.2

Course for undergraduates:
«Elasticity and Plasticity»
Section: Plasticity
Lesson № 2
Chapter contents
Chapter 9. Plasticity Theory.
9.1. Assumptions of Plasticity Theory.
9.2. The Tangent and Plastic Modulus.
9.3. Friction Block Models.
9.4. Stress Analysis for Plasticity. Deviatoric Stress. A State of Pure Shear.
9.5. The Octahedral Stresses.
2
Assumptions of Plasticity Theory [18]
Regarding the above test results then, in formulating a basic plasticity theory with which to begin, the
following assumptions are usually made:
(1) the response is independent of rate effects
(2) the material is incompressible in the plastic range
(3) there is no Bauschinger effect
(4) the yield stress is independent of hydrostatic pressure
(5) the material is isotropic
The first two of these will usually be very good approximations, the other three may or may not be, depending
on the material and circumstances. For example, most metals can be regarded as isotropic. After large plastic
deformation however, for example in rolling, the material will have become anisotropic: there will be distinct
material directions and asymmetries.
3
Together with these, assumptions can be made on the type of hardening and on whether elastic deformations
are significant. For example, consider the hierarchy of models illustrated in Figure 1 below, commonly used
in theoretical analyses.
Figure 1. Simple models of elastic and plastic deformation.
4
In (a) both the elastic and plastic curves are assumed linear. In (b) work-hardening is neglected and the yield
stress is constant after initial yield. Such perfectly-plastic models are particularly appropriate for studying
processes where the metal is worked at a high temperature – such as hot rolling – where work hardening is
small. In many areas of applications the strains involved are large, e.g. in metal working processes such as
extrusion, rolling or drawing, where up to 50% reduction ratios are common. In such cases the elastic strains
can be neglected altogether as in the two models (c) and (d). The rigid/perfectly-plastic model (d) is the
crudest of all – and hence in many ways the most useful. It is widely used in analysing metal forming
processes, in the design of steel and concrete structures and in the analysis of soil and rock stability.
Figure 1. Simple models of elastic and plastic deformation.
5
The Tangent and Plastic Modulus
Stress and strain are related through 𝜎 = 𝐸𝜀 in the elastic region, E being the Young’s modulus, Figure 2.
Figure 2. The tangent modulus.
6
The tangent modulus K is the slope of the stress-strain curve in the plastic region and will in general change
during a deformation. At any instant of strain, the increment in stress dσ is related to the increment in strain dε
through (the symbol ε here represents the true strain (the subscript t has been dropped for clarity); as
mentioned, when the strains are small, it is not necessary to specify which strain is in use since all strain
measures are then equivalent)
(1)
After yield, the strain increment consists of both elastic, 𝜀 𝑒 , and plastic, 𝑑𝜀 𝑝 , strains:
(2)
The stress and plastic strain increments are related by the plastic modulus H:
(3)
and it follows that
(4)
7
Friction Block Models
Some additional insight into the way plastic materials respond can be obtained from friction block models.
The rigid perfectly plastic model can be simulated by a Coulomb friction block, Figure 3.
Figure 3. (a) Friction block model for the rigid perfectly plastic material, (b)
response of the rigid-perfectly plastic model.
8
No strain occurs until σ reaches the yield stress Y. Then there is movement – although the amount of movement
or plastic strain cannot be determined without more information being available. The stress cannot exceed the
yield stress in this model:
(5)
If unloaded, the block stops moving and the stress returns to zero, leaving a permanent strain, Figure 3(b).
The linear elastic perfectly plastic model incorporates a free spring with modulus E in series with a friction
block, Figure 4.
Figure 4. Friction block model for the elastic perfectly plastic material.
9
The spring stretches when loaded and the block also begins to move when the stress reaches Y, at which time
the spring stops stretching, the maximum possible stress again being Y. Upon unloading, the block stops
moving and the spring contracts.
The linear elastic plastic model with linear strain hardening incorporates a second, hardening, spring with
stiffness H, in parallel with the friction block, Figure 5.
Figure 5. Friction block model for a linear elastic-plastic material with linear strain hardening; (a)
stress-free, (b) elastic strain, (c) elastic and plastic strain, (d) unloading.
10
Figure 5. Friction block model for a linear elastic-plastic material with linear strain hardening; (a)
stress-free, (b) elastic strain, (c) elastic and plastic strain, (d) unloading.
Once the yield stress is reached, an ever increasing stress needs to be applied in order to keep the block
moving – and elastic strain continues to occur due to further elongation of the free spring. The stress is then
split into the yield stress, which is carried by the moving block, and an overstress σ ―Y carried by the
hardening spring.
11
Upon unloading, the block “locks” – the stress in the hardening spring remains constant whilst the free spring
contracts. At zero stress, there is a negative stress taken up by the friction block, equal and opposite to the
stress in the hardening spring.
The slope of the elastic loading line is E. For the plastic hardening line,
(6)
It can be seen that H is the plastic modulus.
12
Vectors, Tensors and the Index Notation
The equations governing three dimensional mechanics problems can be quite lengthy. For this reason, it is
essential to use a short-hand notation called the index notation (or indicial or subscript or suffix notation).
Consider first the notation used for vectors.
Vectors
Vectors are used to describe physical quantities which have both a magnitude and a direction associated with
them. Geometrically, a vector is represented by an arrow; the arrow defines the direction of the vector and the
magnitude of the vector is represented by the length of the arrow. Analytically, in what follows, vectors will be
represented by lowercase bold-face Latin letters, e.g. a, b.
The dot product of two vectors a and b is denoted by a ⋅ b and is a scalar defined by
(7)
θ here is the angle between the vectors when their initial points coincide and is restricted to the range 0 ≤ θ ≤ π
.
13
Cartesian Coordinate System
So far the short discussion has been in symbolic notation (or absolute or invariant or direct or vector notation
), that is, no reference to ‘axes’ or ‘components’ or ‘coordinates’ is made, implied or required. Vectors exist
independently of any coordinate system. The symbolic notation is very useful, but there are many circumstances
in which use of the component forms of vectors is more helpful – or essential. To this end, introduce the vectors
𝐞1 , 𝐞2 , 𝐞3 having the properties
(8)
so that they are mutually perpendicular, and
(9)
so that they are unit vectors. Such a set
of orthogonal unit vectors is called an
orthonormal set, Figure 6.
Figure 6. An orthonormal set of base vectors and
Cartesian coordinates.
14
Figure 6. An orthonormal set of base vectors and Cartesian coordinates.
This set of vectors forms a basis, by which is meant that any other vector can be written as a linear
combination of these vectors, i.e. in the form
(10)
15
where 𝑎1 , 𝑎2 𝑎𝑛𝑑 𝑎3 are scalars, called the Cartesian components or coordinates of a along the given three
directions. The unit vectors are called base vectors when used for this purpose. The components
𝑎1 , 𝑎2 𝑎𝑛𝑑 𝑎3 are measured along lines called the 𝑥1 , 𝑥2 𝑎𝑛𝑑 𝑥3 axes, drawn through the base vectors.
Note further that this orthonormal system 𝐞1 , 𝐞2 , 𝐞3 is right-handed, by which is meant
(11)
In the index notation, the expression for the vector a in terms of the components 𝑎1 , 𝑎2 , 𝑎3 and the
corresponding basis vectors 𝐞1 , 𝐞2 , 𝐞3 is written as
(12)
This can be simplified further by using Einstein’s summation convention, whereby the summation sign is
dropped and it is understood that for a repeated index (i in this case) a summation over the range of the index
(3 in this case (2 in the case of a two-dimensional space/analysis )) is implied. Thus one writes 𝐚 = 𝑎𝑖 𝐞𝑖 . This
can be further shortened to, simply, 𝒂𝒊 .
16
The dot product of two vectors u and v, referred to this coordinate system, is
(13)
The dot product of two vectors written in the index notation reads
(14)
17
The repeated index i is called a dummy index, because it can be replaced with any other letter and the sum is
the same; for example, this could equally well be written as
(15)
Introduce next the Kronecker delta symbol 𝜹𝒊𝒋 , defined by
(16)
Note that 𝛿11 = 1 but, using the index notation, 𝛿𝑖𝑖 = 3. The Kronecker delta allows one to write the
expressions defining the orthonormal basis vectors (Eqns. 8, 9) in the compact form
(17)
18
Matrix Notation
The symbolic notation v and index notation 𝒗𝒊 𝐞𝒊 (or simply 𝒗𝒊 ) can be used to denote a vector. Another
notation is the matrix notation: the vector v can be represented by a 3×1 matrix (a column vector):
(18)
Matrices will be denoted by square brackets, so a shorthand notation for this matrix/vector would be [v]. The
elements of the matrix [v] can be written in the index notation 𝒗𝒊 .
Note the distinction between a vector and a 3×1 matrix: the former is a mathematical object independent of
any coordinate system, the latter is a representation of the vector in a particular coordinate system – matrix
notation, as with the index notation, relies on a particular coordinate system.
19
As an example, the dot product can be written in the matrix notation as
(19)
Here, the notation 𝐮T denotes the 1×3 matrix (the row vector). The result is a 1×1 matrix, 𝒖𝒊 𝒗𝒊 .
The matrix notation for the Kronecker delta 𝜹𝒊𝒋 is the identity matrix
(20)
20
Then, for example, in both index and matrix notation:
(21)
Matrix – Matrix Multiplication
When discussing vector transformation equations further below, it will be necessary to multiply various
matrices with each other (of sizes 3×1, 1×3 and 3×3). It will be helpful to write these matrix multiplications in
the short-hand notation.
First, it has been seen that the dot product of two vectors can be represented by 𝐮T 𝐯 or 𝒖𝒊 𝒗𝒊 . Similarly, the
matrix multiplication 𝐮 𝐯 T gives a 3×3 matrix with element form 𝒖𝒊 𝒗𝒋 or, in full,
21
(22)
This operation is called the tensor product of two vectors, written in symbolic notation as u ⊗ v (or simply
uv).
Next, the matrix multiplication
(23)
22
is a 3×1 matrix with elements 𝐐 𝐮
which can be expressed as 𝐮T 𝐐T
𝑖
𝑖
≡ 𝑄𝑖𝑗 𝑢𝑗 . The elements of [Q][u] are the same as those of 𝐮T 𝐐T ,
≡ 𝑢𝑗 𝑄𝑖𝑗 .
The expression [u][Q] is meaningless, but 𝐮T 𝐐 is a 1×3 matrix with elements
𝐮T 𝐐
𝑖
≡ 𝑢𝑗 𝑄𝑗𝑖 .
This leads to the following rule:
1.
if a vector pre-multiplies a matrix [Q] → the vector is the transpose 𝐮T
2.
if a matrix [Q] pre-multiplies the vector → the vector is [u]
3.
if summed indices are “beside each other”, as the j in 𝑢𝑗 𝑄𝑗𝑖 or 𝑄𝑖𝑗 𝑢𝑗 → the matrix is [Q]
4.
if summed indices are not beside each other, as the j in 𝑢𝑗 𝑄𝑖𝑗 → the matrix is the transpose, 𝐐T
Finally, consider the multiplication of 3×3 matrices. Again, this follows the “beside each other” rule for the
summed index.
For example, [A][B] gives the 3×3 matrix
𝐀𝐓 𝐁 𝒊𝒋 = 𝐴𝑘𝑖 𝐵𝑘𝑗 .
𝐀 𝐁
𝒊𝒋
= 𝐴𝑖𝑘 𝐵𝑘𝑗 , and the multiplication 𝐀𝐓 𝐁 is written as
23
There is also the important identity
(24)
Note also the following:
(i) if there is no free index, as in 𝑢𝑖 𝑣𝑖 , there is one element
(ii) if there is one free index, as in 𝑢𝑗 𝑄𝑗𝑖 , it is a 3×1 (or 1×3 ) matrix
(iii) if there are two free indices, as in 𝐴𝑘𝑖 𝐵𝑘𝑗 , it is a 3×3 matrix
24
Vector Transformation Rule
Introduce two Cartesian coordinate systems with base vectors 𝐞𝒊 and 𝐞′𝒊 and common origin o, Figure 7.
Figure 7. A vector represented using two different coordinate systems.
The vector u can then be expressed in two ways:
(25)
25
Note that the 𝒙′𝒊 coordinate system is obtained from the 𝒙𝒊 system by a rotation of the base vectors. Figure 7
shows a rotation θ about the 𝒙𝟑 axis (the sign convention for rotations is positive counterclockwise).
Concentrating for the moment on the two dimensions 𝒙𝟏 − 𝒙𝟐 , from trigonometry (refer to Figure 8),
Figure 8. Geometry of the 2D coordinate transformation.
26
(26)
and so
(27)
27
In matrix form, these transformation equations can be written as
(28)
The 2×2 matrix is called the transformation matrix or rotation matrix [Q]. By premultiplying both sides
of these equations by the inverse of [Q], 𝐐−1 , one obtains the transformation equations transforming from
𝒖𝟏 𝒖𝟐 𝑻 to 𝒖′𝟏 𝒖′𝟐 𝑻 :
(29)
It can be seen that the components of [Q] are the directions cosines, i.e. the cosines of the angles between the
coordinate directions:
(30)
28
It is straight forward to show that, in the full three dimensions, Figure 9, the components in the two coordinate
systems are also related through
Figure 9. two different coordinate systems in a 3D space.
(31)
29
Orthogonality of the Transformation Matrix [Q]
From (31), it follows that
(32)
and so
(33)
A matrix such as this for which 𝐐T = 𝐐−1 is called an orthogonal matrix.
30
Tensors
A second-order tensor (to be called simply a tensor in what follows) A may be defined as an operator that
acts on a vector u generating another vector v, so that T(u) = v , or
(34)
The second-order tensor T is a linear operator, by which is meant
(35)
for scalar α. In a Cartesian coordinate system, the tensor T has nine components and can be represented in the
matrix form
(36)
31
The rule (34), which is expressed in symbolic notation, can be expressed in the index and matrix notation
when T is referred to particular axes:
(37)
Again, one should be careful to distinguish between a tensor such as T and particular matrix representations of
that tensor. The relation (34) is a tensor relation, relating vectors and a tensor and is valid in all coordinate
systems; the matrix representation of this tensor relation, Eqn. (37), is to be sure valid in all coordinate
systems, but the entries in the matrices of (37) depend on the coordinate system chosen.
(34)
Note also that the transformation formulae for vectors, Eqn. (31), is not a tensor relation; although (31) looks
similar to the tensor relation (37), the former relates the components of a vector to the components of the same
vector in different coordinate systems, whereas (by definition of a tensor) the relation (37) relates the
components of a vector to those of a different vector in the same coordinate system.
(31)
32
For these reasons, the notation 𝒖𝒊 = 𝑸𝒊𝒋 𝒖′ in Eqn. (31) is more formally called element form, the 𝑸𝒊𝒋 being
elements of a matrix rather than components of a tensor. This distinction between element form and index
notation should be noted, but the term “index notation” is used for both tensor and matrix-specific
manipulations in these notes.
Tensor Transformation Rule
Consider now the tensor definition (34) expressed in two different coordinate systems:
(38)
From the vector transformation rule (31),
(39)
33
Combining 38-39,
(40)
and so
(41)
Note that 𝑸𝒎𝒊 𝑸𝒋𝒊 𝒖𝒋 = 𝜹𝒎𝒋 𝒖𝒋 = 𝒖𝒎 . Comparing with (38), it follows that
(42)
34
Stress Analysis for Plasticity
This section on from the analysis of three dimensional stress. The plastic behaviour of materials is often
independent of a hydrostatic stress and this feature necessitates the study of the deviatoric stress.
!!! A detailed analysis of the basic concepts of the stress state is in Appendix A.
Analysis of Three Dimensional Stress and Strain
Here, the fully three dimensional stress state is examined.
The Traction Vector and Stress Components
Consider a traction vector t acting on a surface element, Figure 10.
Figure 10. Components of
the traction vector.
35
Figure 10. Components of the traction vector.
36
Introduce a Cartesian coordinate system with base vectors 𝐞𝑖 so that one of the base vectors is a normal to the
surface and the origin of the coordinate system is positioned at the point at which the traction acts. For
example, in Figure 6, the 𝐞3 direction is taken to be normal to the plane, and a superscript on t denotes this
normal:
Figure 6. An orthonormal set of base vectors and Cartesian coordinates.
37
(43)
Each of these components 𝒕𝒊 is represented by 𝝈𝒊𝒋 where the first subscript denotes the direction of the normal
and the second denotes the direction of the component to the plane. Thus the three components of the traction
vector shown in Figure 10 are 𝝈𝟑𝟏 , 𝝈𝟑𝟐 , 𝝈𝟑𝟑 :
(44)
The first two stresses, the components acting tangential to the surface, are shear stresses whereas 𝝈𝟑𝟑 , acting
normal to the plane, is a normal stress.
Consider the three traction vectors 𝐭 e1 , 𝐭 e2 , 𝐭 e3 acting on the surface elements whose outward normals are
aligned with the three base vectors 𝐞𝑗 , Figure 11(a). The three (or six) surfaces can be amalgamated into one
diagram as in Figure 11(b).
Figure 11. The three traction vectors acting at a point; (a) on mutually orthogonal planes, (b) the
38
traction vectors illustrated on a box element.
Figure 11. The three traction vectors acting at a point; (a) on mutually
orthogonal planes, (b) the traction vectors illustrated on a box element.
39
In terms of stresses, the traction vectors are
(45)
The components of the three traction vectors, i.e. the stress components, can now be displayed on a box element
as in Figure 12.
Figure 12. The nine stress components with respect to a Cartesian coordinate system.
40
Note that the stress components will vary slightly over the surfaces of an elemental box of finite size.
However, it is assumed that the element in Figure 12 is small enough that the stresses can be treated as
constant, so that they are the stresses acting at the origin.
The nine stresses can be conveniently displayed in 3×3 matrix form:
(46)
It is important to realise that, if one were to take an element at some different orientation to the element in
Figure 12, but at the same material particle, for example aligned with the axes 𝒙′𝟏 , 𝒙′𝟐 , 𝒙′𝟑 shown in Figure 13,
one would then have different tractions acting and the nine stresses would be different also.
41
Figure 13. the stress components with respect to a Cartesian coordinate system
different to that in Figure 12.
The stresses acting in this new orientation can be represented by a new matrix:
(47)
42
Cauchy’s Law
Cauchy’s Law, states that the normal to a surface, 𝐧 = 𝑛𝒊 𝐞𝒊 , is related to the traction vector 𝐭
acting on that surface, according to
𝒏
= 𝑡𝒊 𝐞𝒊
(48)
Writing the traction and normal in vector form and the stress in 3×3 matrix form,
(49)
and Cauchy’s law in matrix notation reads
(50)
43
Note that it is the transpose stress matrix which is used in Cauchy’s law. Since the stress matrix is symmetric,
one can express Cauchy’s law in the form
(51)
Cauchy’s law is illustrated in Figure 14; in this figure, positive stresses 𝜎𝑖𝑗 are shown.
Figure 14. Cauchy’s Law; given the stresses and the normal to a plane, the traction
vector acting on the plane can be determined.
44
Normal and Shear Stress
It is useful to be able to evaluate the normal stress 𝜎𝑁 and shear stress 𝜎𝑆 acting on any plane, Figure 15.
Figure 15. The normal and shear stress acting on an
arbitrary plane through a point.
!!! A detailed analysis of the basic concepts of the stress state is in Appendix A.
45
For this purpose, note that the stress acting normal to a plane is the projection of 𝐭
𝐧
in the direction of n ,
(52)
The magnitude of the shear stress acting on the surface can then be obtained from
(53)
46
Proof of Cauchy’s Law
Cauchy’s law can be proved using force equilibrium of material elements. First, consider a tetrahedral freebody, with vertex at the origin, Figure 16. It is required to determine the traction t in terms of the nine stress
components (which are all shown positive in the diagram).
Figure 16. Proof of Cauchy’s Law.
47
The components of the unit normal, 𝑛𝑖 , are the direction cosines of the normal vector, i.e. the cosines of the
angles between the normal and each of the coordinate directions:
(54)
Let the area of the base of the tetrahedron, with normal n, be ∆S . The area ∆𝑆1 is then ∆S cosα , where α is the
angle between the planes, as shown to the right of Figure 16; this angle is the same as that between the vectors
n and 𝐞1 , so ∆𝑆1 = 𝑛1 ∆𝑆, and similarly for the other surfaces:
(55)
The resultant surface force on the body, acting in the 𝒙𝒊
direction, is then
(56)
For equilibrium, this expression must be zero, and one
arrives at Cauchy’s law.
Figure 16. Proof of Cauchy’s Law.
48
The Stress Tensor
Cauchy’s law (51) is of the same form as (37) and so by definition the stress is a tensor. Denote the stress tensor
in symbolic notation by σ . Cauchy’s law in symbolic form then reads
(57)
Further, the transformation rule for stress follows the general tensor transformation rule (42):
(58)
(51)
(37)
(42)
49
As with the normal and traction vectors, the components and hence matrix representation of the stress changes
with coordinate system, as with the two different matrix representations (46) and (47). However, there is only
one stress tensor σ at a point. Another way of looking at this is to note that an infinite number of planes pass
through a point, and on each of these planes acts a traction vector, and each of these traction vectors has three
(stress) components. All of these traction vectors taken together define the complete state of stress at a point.
(46)
(47)
50
Isotropic State of Stress
Suppose the state of stress in a body is
(59)
One finds that the application of the stress tensor transformation rule yields the very same components no
matter what the new coordinate system. In other words, no shear stresses act, no matter what the orientation of
the plane through the point. This is termed an isotropic state of stress, or a spherical state of stress. One
example of isotropic stress is the stress arising in a fluid at rest, which cannot support shear stress, in which
case
(60)
where the scalar p is the fluid hydrostatic pressure. For this reason, an isotropic state of stress is also referred
to as a hydrostatic state of stress.
51
Isotropy and Anisotropy
Many materials have a strong direction-dependence. The classic example is wood, which has a clear structure
–along the grain, along which fine lines can be seen, and against the grain, Figure 16.2.
Figure 16.2. Wood.
The wood is stiffer and stronger along the grain than against the grain. A material which has this directiondependence of mechanical (and physical) properties is called anisotropic.
52
Figure 16.3 shows stress-strain curves for human ligament tissue; in one test, the ligament is stretched along
its length (the longitudinal direction), in the second, across the width of the ligament (the transverse
direction).
Figure 16.3. Anisotropic response of human ligament.
53
It can be seen that the stiffness is much higher in the longitudinal direction. Another example is bone – it is
much stiffer along the length of the bone than across the width of the bone. In fact, many biological
materials are strongly anisotropic.
A material whose properties are the same in all directions is called isotropic. In particular, the relationship
between stress and strain at any single location in a material is the same in all directions. This implies that if a
specimen is cut from an isotropic material and subjected to a load, it would not matter in which orientation
the specimen is cut, the resulting deformation would be the same – as illustrated in Figure 16.4. Most metals
and ceramics can be considered to be isotropic.
Figure 16.4. Illustration of Isotropy; the relationship between stress and strain is the same no
matter in what “direction” the test specimen is cut from the material.
As usual an anisotropic material can have a Poisson’s ratio greater than 0.5.
54
Homogeneous Materials
The term homogeneous means that the mechanical properties are the same at each point throughout the
material. In other words, the relationship between stress and strain is the same for all material particles. Most
materials can be assumed to be homogeneous.
In engineering applications, it is sometimes beneficial to design materials/components which are specifically
not homogeneous, i.e. inhomogeneous. Such materials whose properties vary gradually throughout are called
Functionally Graded Materials, and have been gaining popularity since the 1980s-90s in advanced
technologies.
Note that a material can be homogeneous and not isotropic, and vice versa – homogeneous refers to different
locations whereas isotropy refers to the same location.
55
Principal Stresses
For certain planes through a material particle, there are traction vectors which act normal to the plane, as in
Figure 17. In this case the traction can be expressed as a scalar multiple of the normal vector, 𝐭 𝐧 = 𝜎𝐧.
Figure 17. A purely normal traction vector.
!!! A detailed analysis of the basic concepts of the stress state is in Appendix A.
56
From Cauchy’s law then, for these planes,
(61)
This is a standard eigenvalue problem from Linear Algebra: given a matrix 𝜎𝑖𝑗 , find the eigenvalues σ and
associated eigenvectors n such that Eqn. (61) holds.
To solve the problem, first re-write the equation in the form
(62)
57
or
(63)
This is a set of three homogeneous equations in three unknowns (if one treats σ as known). From basic linear
algebra, this system has a solution (apart from 𝑛𝑖 = 0) if and only if the determinant of the coefficient matrix
is zero, i.e. if
(64)
Evaluating the determinant, one has the following cubic characteristic equation of the stress tensor σ ,
(65)
58
and the principal scalar invariants of the stress tensor are
(66)
(𝐼3 is the determinant of the stress matrix.) The characteristic equation (65) can now be solved for the
eigenvalues σ and then Eqn. (63) can be used to solve for the eigenvectors n.
Now another theorem of linear algebra states that the eigenvalues of a real (that is, the components are real),
symmetric matrix (such as the stress matrix) are all real and further that the associated eigenvectors are
mutually orthogonal. This means that the three roots of the characteristic equation are real and that the three
associated eigenvectors form a mutually orthogonal system. This is illustrated in Figure 18; the eigenvalues
are called principal stresses and are labelled 𝜎1 , 𝜎2 , 𝜎3 and the three corresponding eigenvectors are called
principal directions, the directions in which the principal stresses act. The planes on which the principal
stresses act (to which the principal directions are normal) are called the principal planes.
(63)
(65)
59
Figure 18. the three principal stresses acting at a point and the three
associated principal directions 1, 2 and 3.
Once the principal stresses are found, as mentioned, the principal directions can be found by solving Eqn.
(63), which can be expressed as
(67)
60
Each principal stress value in this equation gives rise to the three components of the associated principal
direction vector, 𝑛1 , 𝑛2 , 𝑛3 . The solution also requires that the magnitude of the normal be specified: for a unit
vector, n ⋅ n = 1. The directions of the normals are also chosen so that they form a right-handed set.
Invariants
The principal stresses 𝜎1 , 𝜎2 , 𝜎3 are independent of any coordinate system; the 0𝑥1 𝑥2 𝑥3 axes to which the stress
matrix in Eqn. (61) is referred can have any orientation – the same principal stresses will be found from the
eigenvalue analysis. This is expressed by using the symbolic notation for the problem: σn = σ n , which is
independent of any coordinate system. Thus the principal stresses are intrinsic properties of the stress state at
a point. It follows that the functions 𝐼1 , 𝐼2 , 𝐼3 in the characteristic equation Eqn. (65) are also independent of
any coordinate system, and hence the name principal scalar invariants (or simply invariants) of the stress.
The stress invariants can also be written neatly in terms of the principal stresses:
(68)
61
Also, if one chooses a coordinate system to coincide with the principal directions, Figure 19, the stress matrix
takes the simple form
Figure 19. Principal directions.
(69)
62
Note that when two of the principal stresses are equal, one of the principal directions will be unique, but the
other two will be arbitrary – one can choose any two principal directions in the plane perpendicular to the
uniquely determined direction, so that the three form an orthonormal set. This stress state is called axisymmetric. When all three principal stresses are equal, one has an isotropic state of stress, and all directions
are principal directions – the stress matrix has the form (69) no matter what orientation the planes through the
point.
63
Maximum and Minimum Stress Values
Normal Stresses
The three principal stresses include the maximum and minimum normal stress components acting at a point. To
prove this, first let 𝐞1 , 𝐞2 , 𝐞3 be unit vectors in the principal directions. Consider next an arbitrary unit normal
vector 𝐧 = 𝑛𝑖 𝐞𝑖 . From Cauchy’s law (see Figure 20 – the stress matrix in Cauchy’s law is now with respect to
the principal directions 1, 2 and 3), the normal stress acting on the plane with normal n is
(70)
Figure 20. normal stress acting on a plane
defined by the unit normal n.
64
Thus
(71)
Since 𝑛12 + 𝑛22 + 𝑛32 = 1 and, without loss of generality, taking 𝜎1 ≥ 𝜎2 ≥ 𝜎3 , one has
(72)
Similarly,
(73)
Thus the maximum normal stress acting at a point is the maximum principal stress and the minimum normal
stress acting at a point is the minimum principal stress.
65
Shear Stresses
Next, it will be shown that the maximum shearing stresses at a point act on planes oriented at 45° to the
principal planes and that they have magnitude equal to half the difference between the principal stresses. First,
again, let 𝐞1 , 𝐞2 , 𝐞3 be unit vectors in the principal directions and consider an arbitrary unit normal vector 𝐧 =
𝑛𝑖 𝐞𝑖 . The normal stress is given by Eqn. (71),
(74)
Cauchy’s law gives the components of the traction vector as
(75)
and so the shear stress on the plane is, from Eqn. (53),
(76)
(53)
66
Using the condition 𝑛12 + 𝑛22 + 𝑛32 = 1 to eliminate 𝑛3 leads to
(77)
The stationary points are now obtained by equating the partial derivatives with respect to the two variables 𝑛1
and 𝑛2 to zero:
(78)
One sees immediately that 𝑛1 = 𝑛2 = 0 (so that 𝑛3 = ±1) is a solution; this is the principal direction 𝐞3 and
the shear stress is by definition zero on the plane with this normal. In this calculation, the component 𝑛3 was
eliminated and 𝜎𝑆2 was treated as a function of the variables 𝑛1 , 𝑛2 . Similarly, 𝑛1 can be eliminated with
𝑛2 , 𝑛3 treated as the variables, leading to the solution 𝐧 = 𝐞1 , and 𝑛2 can be eliminated with 𝑛1 , 𝑛3
treated as the variables, leading to the solution 𝐧 = 𝐞2 . Thus these solutions lead to the minimum shear stress
67
value 𝜎𝑆2 = 0.
A second solution to Eqn. (78) can be seen to be 𝑛1 = 0, 𝑛2 = ± 1 2 (so that 𝑛3 = ± 1 2) with
1
corresponding shear stress values 𝜎𝑆2 = 𝜎2 − 𝜎3 2 . Two other solutions can be obtained as described
4
earlier, by eliminating 𝑛1 and by eliminating 𝑛2 . The full solution is listed below, and these are evidently the
maximum (absolute value of the) shear stresses acting at a point:
(79)
Taking 𝜎1 ≥ 𝜎2 ≥ 𝜎3 , the maximum shear stress at a point is
(80)
68
and acts on a plane with normal oriented at 45° to the 1 and 3 principal directions. This is illustrated in Figure
21.
Figure 21. Maximum shear stress at a point.
69
Mohr’s Circles of Stress
For the 3D case one has the conditions
(81)
Solving these equations gives
(82)
70
Taking 𝜎1 ≥ 𝜎2 ≥ 𝜎3 , and noting that the squares of the normal components must be positive, one has that
(83)
and these can be re-written as
(84)
71
If one takes coordinates 𝜎𝑁 , 𝜎𝑆 , the equality signs here represent circles in 𝜎𝑁 , 𝜎𝑆 stress space, Figure 22.
Figure 22. Admissible points in stress space.
72
Each point 𝜎𝑁 , 𝜎𝑆 in this stress space represents the stress on a particular plane through the material particle
in question. Admissible 𝜎𝑁 , 𝜎𝑆 pairs are given by the conditions Eqns. (84); they must lie inside a circle of
1
1
centre
𝜎1 + 𝜎3 , 0 and radius 𝜎1 − 𝜎3 . This is the large circle in Figure 22. The points must lie outside
2
the circle with centre
1
2
1
2
2
𝜎2 + 𝜎3 , 0 and radius
𝜎1 + 𝜎2 , 0 and radius
1
2
1
2
𝜎2 − 𝜎3 and also outside the circle with centre
𝜎1 − 𝜎2 ; these are the two smaller circles in the figure. Thus the admissible
points in stress space lie in the shaded region of Figure 22.
Figure 22. Admissible points in stress space.
73
Three Dimensional Strain
The strain 𝜀𝑖𝑗 , in symbolic form ε , is a tensor and as such it follows the same rules as for the stress tensor. In
particular, it follows the general tensor transformation rule (58); it has principal values ε which satisfy the
characteristic equation (65) and these include the maximum and minimum normal strain at a point. There are
three principal strain invariants given by (66) or (68) and the maximum shear strain occurs on planes oriented
at 45° to the principal directions.
(58)
(65)
(66)
(68)
74
Governing Equations of Three Dimensional Elasticity
Hooke’s Law and Lamé’s Constants
The three-dimensional Hooke’s law for isotropic linear elastic solids can be expressed in index notation as
(85)
where
(86)
are the Lamé constants (µ is the Shear Modulus).
Eqns. (85) can be inverted to obtain
(87)
75
Navier’s Equations
The governing equations of elasticity are Hooke’s law (Eqn. 85), the equations of motion, from Eqn. (87.2)
(87.2)
we have
(88)
and the strain-displacement relations,
(89)
76
Substituting (89) into (85) and then into (88) leads to the 3D Navier’s equations
(90)
These reduce to the 2D plane strain Navier’s equations, by setting 𝑢3 = 0 and
𝜕
𝜕𝑥3
= 0. They do not reduce to
the plane stress equations since the latter are only an approximate solution to the equations of elasticity which
are valid only in the limit as the thickness of the thin plate of plane stress tends to zero.
77
Deviatoric Stress
Any state of stress can be decomposed into a hydrostatic (or mean) stress 𝜎𝑚 𝐈 and a deviatoric stress s,
according to
(91)
where
(92)
and
(93)
78
In index notation,
(94)
In a completely analogous manner to the derivation of the principal stresses and the principal scalar invariants
of the stress matrix, one can determine the principal stresses and principal scalar invariants of the deviatoric
stress matrix. The former are denoted 𝑠1 , 𝑠2 , 𝑠3 and the latter are denoted by 𝐽1 , 𝐽2 , 𝐽3 . The characteristic
equation analogous to Eqn. (65) is
(95)
and the deviatoric invariants are (compare with (66), (68)) (unfortunately, there is a convention (adhered to by
most authors) to write the characteristic equation for stress with a +𝐼2 𝜎 term and that for deviatoric stress
with a −𝐽2 𝑠 term; this means that the formulae for 𝐽2 in Eqn. (95) are the negative of those for 𝐼2 in Eqn. (66))
(65)
79
(66)
(68)
(96)
Since the hydrostatic stress remains unchanged with a change of coordinate system, the principal directions of
stress coincide with the principal directions of the deviatoric stress, and the decomposition can be expressed
with respect to the principal directions as
(97)
80
Note that, from the definition Eqn. (93), the first invariant of the deviatoric stress, the sum of the normal
stresses, is zero:
(98)
The second invariant can also be expressed in the useful forms
(99)
and, in terms of the principal stresses,
(100)
Further, the deviatoric invariants are related to the stress tensor invariants through
(101)
81
A State of Pure Shear
The stress state at a point is one of pure shear if for any one coordinate axes through the point one has only
shear stress acting, i.e. the stress matrix is of the form
(102)
Applying the stress transformation rule (58) to this stress matrix and using the fact that the transformation
′
′
′
matrix Q is orthogonal, i.e. 𝐐𝐐T = 𝐐T 𝐐 = 𝐈, one finds that the first invariant is zero, 𝜎11
+ 𝜎22
+ 𝜎33
= 0.
Hence the deviatoric stress is one of pure shear.
(58)
82
The Octahedral Stresses
Examine now a material element subjected to principal stresses 𝜎1 , 𝜎2 , 𝜎3 as shown in Figure 23. By
definition, no shear stresses act on the planes shown.
Figure 23. Stresses acting on a material element.
83
Consider next the octahedral plane; this is the plane shown shaded in Figure 24, whose normal 𝐧𝑎 makes
equal angles with the principal directions. It is so-called because it cuts a cubic material element (with faces
perpendicular to the principal directions) into a triangular plane and eight of these triangles around the origin
form an octahedron.
Figure 24. The octahedral plane.
84
Next, a new Cartesian coordinate system is constructed with axes parallel and perpendicular to the octahedral
plane, Figure 25. One axis runs along the unit normal 𝐧𝑎 ; this normal has components 1 3 , 1 3 , 1 3
angle 𝜃0 the normal direction makes with the 1 direction can be obtained from 𝐧𝑎 ∙ 𝐞1 = cos 𝜃0 , where 𝐞1 =
1,0,0 is a unit vector in the 1 direction, Figure 25. To complete the new coordinate system, any two
perpendicular unit vectors which lie in (parallel to) the octahedral plane can be chosen.
Choose one which is along the projection of the 1
axis down onto the octahedral plane. The
components of this vector are 𝐧𝑐 =
2 3 , − 1 6 , − 1 6 . The final unit vector
𝐧𝑏 is chosen so that it forms a right hand Cartesian
coordinate system with 𝐧𝑎 and 𝐧𝑐 , i.e. 𝐧𝑎 × 𝐧𝑏 =
𝐧𝑐 .
Figure 25. A new Cartesian coordinate system.
85
In summar
(103)
Figure 25. A new Cartesian coordinate system.
86
To express the stress state in terms of components in the a, directions, construct the b, c stress transformation
matrix:
(104)
and the new stress components are
(105)
87
Now consider the stress components acting on the octahedral plane, 𝜎𝑎𝑎 , 𝜎𝑎𝑏 , 𝜎𝑎𝑐 , Figure 26. Recall from
Cauchy’s law, Eqn. (51), that these are the components of the traction vector 𝐭 𝐧𝑎 acting on the octahedral
plane, with respect to the (a, b, c) axes:
(106)
(51)
Figure 26. The stress vector σ and its components.
88
The magnitudes of the normal and shear stresses acting on the octahedral plane are called the octahedral
normal stress 𝜎𝑜𝑐𝑡 and the octahedral shear stress 𝜏𝑜𝑐𝑡 . Referring to Figure 26, these can be expressed as
(106)
The octahedral normal and shear stresses on all 8 octahedral
planes around the origin are the same.
Figure 26. The stress vector σ and its components.
89
Note that the octahedral normal stress is simply the hydrostatic stress. This implies that the deviatoric stress has
no normal component in the direction 𝐧𝑎 and only contributes to shearing on the octahedral plane. Indeed,
from Eqn. 8.2.15,
(107)
The σ’s on the right here can be replaced with s’s since 𝜎𝑖 − 𝜎𝑗 = 𝑠𝑖 − 𝑠𝑗 .
(105)
90
Example 1:
The state of stress at a point with respect to a Cartesian coordinates system 0𝑥1 𝑥2 𝑥3 is given by:
Determine:
(a)
the traction vector acting on a plane through the point whose unit normal is
(b)
the component of this traction acting perpendicular to the plane
(c)
the shear component of traction on the plane
Solution:
(a) From Cauchy’s law,
91
so that
(b) The component normal to the plane is
(c) The shearing component of traction is
92
Example 2:
The stress at a point is given with respect to the axes 0𝑥1 𝑥2 𝑥3 by the values
Determine (a) the principal values, (b) the principal directions (and sketch them).
Solution:
(a)
The principal values are the solution to the characteristic equation
which yields the three principal values
93
(b)
The eigenvectors are now obtained from Eqn. (E.1). First, for
(E.1)
and using also the equation
leads to
94
Similarly, for
one has, respectively,
and
which yield
95
The principal directions are sketched in Figure E.1.
Figure E.1. Principal directions.
Note that the three components of each principal direction, 𝑛1 , 𝑛2 , 𝑛3 , are the direction cosines: the cosines
of the angles between that principal direction and the three coordinate axes. For example, for 𝜎1 with 𝑛1 =
0, 𝑛2 = − 3 5 , 𝑛3 = 4 5, the angles made with the coordinate axes 𝑥1 , 𝑥2 , 𝑥3 are, respectively, 90, 126.87°
and 36.87°.
96
APPENDIX A
Stress State [10]
It should thus be clear that the stress state is very important—we must understand how the material
element responds to the deformation caused by the stress components. On the other hand, in the case of
inhomogeneous stress distribution on cut planes, which is most common in reality, it is required to
analyze the stress state of a deforming body from one point inside the body to the other. Generally, the
stress condition of a point inside a deforming body is often defined by a cubic element. Further
understanding stress components in relation to any complex stress state would be essential to fully grasp
the stress and stress tensor concepts.
97
Stresses on an Arbitrary Inclined Plane
Let’s investigate the necessary condition by analyzing the stress state at a point inside the deformable
body. Suppose that the point to be analyzed is O. Usually, three mutually perpendicular planes XOY ,
YOZ, and ZOX (see Figure 1) are set up to analyze its stress state.
Figure 1. Stress components on three mutually perpendicular planes
98
Stress components on each plane are divided into one normal stress, symbolized by 𝜎, vertical to the
plane and two shear stresses by 𝜏, parallel to the coordinate axes. In order to identify what plane the
stress components act on, one subscript is used for the normal stress and two subscripts for each of the
shear stresses. The subscript for the normal stress denotes its acting direction.
The first of the two subscripts of the shear stress denotes the normal direction of the acting plane of the
shear stress, and the second the acting direction of the shear stress (see Figure 1). The value of all stress
components is not arbitrary but is determined by the equilibrium between stress components on the cut
plane and the associated external force (see the example in Figure 2).
Figure 2. Relationship between forces and stresses on a
plane cut out of a loaded body under uniaxial tension
99
Customarily, positive normal stresses are supposed to be tensile ones and negative normal stresses are
compressive. Normally, it doesn’t matter whether a shear stress is positive or negative, because shear
stresses always exist in pairs. Nevertheless, because some materials show anisotropic yield behavior
and/or strength differential in tension and compression, the change in loading direction or stress state
might change the yielding behavior in value of the shear stresses. Therefore, for some materials, when
the stress direction changes (e.g., from tension to compression or vice versa), we must still define
whether the shear stresses are positive or negative, based on the action direction. Namely, when the
second subscript of shear stress implies the positive direction of the axis, this shear stress is positive,
and vice versa. Thus, stress components on three planes in a xyz coordinate system can be expressed in
the matrix form (see Figure 1).
From Figure 1, we know that the nine stress components on the three mutually perpendicular planes
share a common feature. That is, shear stresses exist in pairs with the same value and the subscripts
composed of two identical English letters in opposite order. Namely,
(1)
100
Equation (1) means that there are only six independent stress components in a symmetric form (Figure
1), which can be represented by a matrix as follows:
(2)
In the case that, on the three mutually perpendicular planes, there are only three normal stresses, called
the principal stresses: 𝜎1 , 𝜎2 and 𝜎3 , without any shear stress, Equation (2) becomes
(3)
which represents the principal coordinate system.
101
Obviously, once the six stress components at the point O are given with respect to the x, y, and z
coordinate axis, the stress state at the point O is fixed. Any change in the value of one of the stress
components, as long as it is not due to the conversion of coordinate system, would mean a change in the
stress state at the point O. In other words, the stress state at a point must be described with six stress
components or three principal stresses.
102
Stress Components on an Oblique Plane
It has been proved that given six stress components or three principal stresses at a point, the normal and
shear stresses on any oblique plane relative to the x, y, and z coordinate axes can be determined.
On the oblique plane represented by a triangle
ABC (see Figure 3), the normal N (ON) is
denoted by directional cosines (l, m, n) with
three angles 𝛼𝑥 , 𝛼𝑦 and 𝛼𝑧 formed between ON
and separately OX, OY , and OZ.
Figure 3. Stress components on an oblique plane
103
Let ΔA denote the area of the triangle ABC, ∆𝐴𝑥 , ∆𝐴𝑦 and ∆𝐴𝑧 the areas formed by projecting ΔA,
respectively, on the three coordinate surfaces:
(4)
Let the resultant stress on the triangle ABC be denoted by S, which has a direct stress component 𝜎𝑁
normal to the plane ABC and a shear stress component 𝜏 on it. Thus, the equilibrium of the forces on the
tetrahedron OABC in the direction OX, OY , and OZ, respectively, can be described by
(5)
where 𝑆𝑥 , 𝑆𝑦 , and 𝑆𝑧 are the components of the resultant stress S in parallel with OX, OY , and
OZ, respectively.
104
Simplification of Equation (5) gives
(6)
According to the principle of parallel hexahedron, the resultant stress S on the oblique plane is
(7)
The normal stress 𝜎𝑁 on the oblique plane can be expressed with the components 𝑆𝑥 , 𝑆𝑦 , and 𝑆𝑧 as
follows:
(8)
Substitution of the components 𝑆𝑥 , 𝑆𝑦 , and 𝑆𝑧 in the forms of Equation (6) into Equation (8) gives
(9)
105
Further, the shear stress 𝜏 on the oblique plane is
(10)
The stress components in Equation (7) to Equation (10) are defined based on the arbitrary coordinate
system. It means that if stress components on three mutually perpendicular planes are given, the
resultant stress S, the normal stress 𝜎𝑁 and the shear stress 𝜏 on a plane in any direction can all be
determined.
As regards the principal coordinate system, there are only three normal stresses 𝜎1 , 𝜎2 , and 𝜎3 without
any shear stress on three mutually perpendicular planes (see Figure 4).
Figure 4. Three principal stresses on an
oblique plane
106
Therefore, the resultant stresses 𝑆1 , 𝑆2 , and 𝑆3 parallel to the three principal axes are
(11)
The resultant stress S on the oblique plane can then be obtained as follows:
(12)
The normal stress on the oblique plane is
(13)
The shear stress on the oblique plane is
(14)
107
From these equations, it is clear that the sufficient and necessary condition to describe the stress state at
a point inside material is that the six stress components or the three principal stresses on three mutually
perpendicular planes relative to an x, y, z coordinate system must be known.
108
Special Stresses
There are several special stresses often involved in the analysis of material deformation, and their
physical meanings need to be understood clearly.
Principal stresses
When defining three mutually perpendicular planes at a point inside a deformed material body, we can
always find a case in which there are stresses normal to the planes without any shear stresses on them.
In this situation, the three mutually perpendicular planes are called the principal planes while the three
stresses normal to them are principal stresses, which are commonly expressed by 𝜎1 , 𝜎2 , and 𝜎3
arranged in a descending order of 𝜎1 > 𝜎2 > 𝜎3 .
Principal shear stresses
In the principal coordinate system, there are three particular planes 𝜎1 𝑂𝜎2 , 𝜎2 𝑂𝜎3 , and 𝜎1 𝑂𝜎3 . Based on
them, three maximum shear stresses exist in turn, equaling 𝜏12 = 𝜎1 − 𝜎2 2 , 𝜏23 = 𝜎2 − 𝜎3 2 and
𝜏13 = 𝜎1 − 𝜎3 2. Such planes are commonly termed the principal shear planes, and there are only
three principal shear stresses without any normal stress on them.
109
Maximum shear stress
Although the value of a shear stress at a point inside a deforming material body varies with the direction
of its acting plane, the maximum shear stress can always be found and is known as the maximum shear
stress 𝜏𝑚𝑎𝑥 . The normal direction of its acting plane is vertical to the principal stress 𝜎2 and forms 45°
between the principal stresses 𝜎1 and 𝜎3 (see Figure 5). 𝜏𝑚𝑎𝑥 can be expressed by
(15)
where 𝜎1 , 𝜎3 are the maximum and the minimum
principal stresses, respectively.
Figure 5. Maximum shear stress and relationship among principal stresses
110
Octahedral stresses
In a coordinate system, whose axes extend in the principal directions, there are eight octahedral planes,
of which the normal vectors form equal angles against the coordinate axes. These planes form an
octahedron (see Figure 6). The direction cosine of the octahedral plane is expressed by
(16)
Stresses on the octahedral plane are called
octahedral stresses including normal and shear
stresses.
Figure 6. Octahedral stress planes
111
By substituting Equation (16) into Equation (13), we obtain the octahedral normal stress as follows
(17)
where 𝜎𝑚 is the mean stress. This equation means that the octahedral normal stress is equal to the mean
stress. From Equation (6) to Equation (14), we achieve the octahedral shear stress as follows:
(18)
Expressed by means of six stress components, Equation (18) becomes
(19)
The resultant stress on the octahedral plane (see Figure 6) is expressed by
(20)
112
Common Stress States
Because each of the three principal stresses has the probability of being positive or negative, the stress
state can be divided into nine types: a state with three tension stresses, with two tension stresses, with
two tensions and one compression stress, with one tension stress, with one tension and one compression
stress, with one tension and two compression stresses, with one compression stress, with two
compression stresses and with three compression stresses (see Figure 7).
Figure 7. Different stress
types
113
Uniaxial stress state denotes the one that has only one principal stress; biaxial stress state has two
principal stresses and, by analogy, triaxial stress state has three principal stresses. In analyzing sheetmetal stamping and thin-walled tube forming, stress state can be regarded approximately as a plane
stress state viz. biaxial stress state, while in forging, mainly a triaxial stress state.
If a stress state includes both tension and compression stresses, i.e., a state with stresses having opposite
signs, lower forming forces are required. The reverse is true, i.e., a state with stresses all in compression
or tension requires a higher forming force.
114
Stress Tensors and Deviatoric Stress Tensors
When a coordinate system is made to rotate around the origin, changes in all stress components must
take place in order to meet the requirements of the tensor. In an x, y, and z Cartesian coordinate system,
the stress state at a point inside a deforming body can be represented by a matrix of the second order as
follows:
(21)
where 𝝈 is the stress tensor variable at each point of the deformed body, and 𝜎𝑖𝑗 is the stress
components defined in the Cartesian coordinate system.
This stress tensor is symmetric. Letting 𝜎 denote the principal stress, we can obtain the three principal
stresses from the following equation:
(22)
115
Equation (22) can be expanded into
(23)
where 𝐽1 , 𝐽2 , and 𝐽3 are invariants of the stress tensor independent of the coordinate system.
Equation (23) can be expressed by
(24)
According to continuum mechanics and plasticity theory, the stress tensor can be decomposed into two
parts, the spherical or hydrostatic stress tensor and the deviatoric stress tensor as follows:
(25)
116
The hydrostatic stress tensor denotes the isotropic tension or compression state, which cannot cause any
change in the shape of the stressed body, but can change the volume during elastic deformation.
The deviatoric stress tensor means that the hydrostatic stress 𝜎𝑚 is removed from the normal stresses of
the stress tensor:
(26)
The deviatoric stress tensor is also a symmetric one, which determines the changes in the shape of the
stressed body.
117
Figure 8 shows three cases—simple tension, drawing, and extrusion—which, different in stress
components though, have the same deviatoric stress tensors.
Figure 8. Analyses of stress states in connection with three tensors of stress components:
(a) simple tension, (b) drawing, and (c) extrusion
118
Figure 8. Analyses of stress states in connection with three tensors of stress components:
(a) simple tension, (b) drawing, and (c) extrusion
It means that the three forming processes result in similar changes in the shape just like in the processes
of axial elongation and lateral shrink.
119
Similar to the stress tensor, the deviatoric stress tensor also has three invariants:
(27)
where 𝜎1′ , 𝜎2′ , and 𝜎3′ are three principal deviatoric stresses.
120
The invariant 𝐽2′ can be used to determine whether the plastic deformation in a stressed body happens
while the invariant 𝐽3′ shows the deformation type. For example, simple tension, pure torsion, and
simple compression, respectively, correspond to 𝐽3′ > 0, 𝐽3′ = 0 and 𝐽3′ < 0.
121
Nominal Strain and True Strain
The plastic deformation of a stressed body is always measured by nominal strains, also known as
engineering strains, or true strains, and logarithmic strains. The former is the ratio of the factual
increased dimension of the body to its initial dimension (see Figure 9):
(28)
where 𝑙0 is the original length, and 𝑙1 is the final increased size.
As for homogeneous stretch deformation, according to the volume
conservation theory, the reduction percentage of area 𝜓 = 𝐹0 − 𝐹1 𝐹0 (𝐹0
and 𝐹1 are section areas before and after deformation) is equivalent to the
nominal strain e, which also belongs to the nominal strain category.
Figure 9. Change in length of a stretched bar
122
The main disadvantage of using the nominal strain is the invariability of base length, which does not
reflect the fact that the base length keeps changing in forming processes. As a result, the total strain does
not equal the sum of strains in the whole process. For example, in Figure 10, a bar with the base length
of 50 cm is stretched to 90 cm.
The total strain is e = (90 - 50)/50 = 80%. However, the entire
stretch process consists of two stages, in which the stock is
firstly stretched from 50 cm to 80 cm and then to 90 cm. In this
case, the nominal strain of the first stage is 𝑒1 =
80 − 50 50 = 60% and the second 𝑒2 = 90 − 80 80 =
12.5%. The sum of them, viz. total strain, therefore, should be
𝑒1 + 𝑒2 = 60 + 12.5 % = 72.5%, not 80%.
Figure 10. Changes in length at different stretch stages
123
Called logarithmic strain as well, the true strain is the ratio of factual increased dimension of the stressed
body to its initial dimension in terms of natural logarithm, viz. 𝜀 = 𝑙𝑛 𝑙1 𝑙0 . Why it is considered true is
due to its ability to present the ratio of an infinite small instant increment of a stressed element to its
instantaneous size. It means that the total strain in any forming process can be achieved through
integration:
(29)
Of course, the integration should be carried out on the condition that the principal direction of the plastic
strain remains basically unchanged.
The logarithmic strain has additivity, which makes it possible to describe an accumulative process of
deformation for a stressed element. For instance,
(30)
80 𝑑𝑙
90 𝑑𝑙
90
Figure 10 shows the case where 𝜀 = 50 𝑙 + 80 𝑙 = 𝑙𝑛 50 = 0.59. It means that the sum of the true
strains at various deforming stages equals the total true strain.
124
Table 1 lists some calculated results to compare nominal strains with true strains, where the positive
values mean elongation and negative compression.
Table 1. Comparisons between nominal strains and true strains
125
From it, we discover that the difference between the two kinds of strains increases with an increase in
their absolute values. It should be pointed out that the two sorts of strains do very little to differentiate
from one another when the absolute values of both are small enough. Equation (30) can be written in a
series form as follows:
(31)
When 𝜀 < 1, this series is convergent. Omitting the terms of nominal strains that are higher than the
second order, the difference between the true strain and the nominal strain can then be expressed by the
following:
(32)
If 𝜀 < 0.1, the absolute difference between the true strain and the nominal strain is smaller than 0.005,
viz. 0.5%, which appears so tiny that 𝜀 ≈ e becomes acceptable.
126
Equation (32) can also be presented in the form of Figure 11, in which the curve shows the nominal
strain e always larger than the true strain 𝜀, and 𝜀 = e is tenable only on the condition that there is no
deformation at all, viz. 𝑙 𝑙0 = 1.
Figure 11. Comparison between the true strain curve and the nominal strain
The deviation of the nominal strain e from the true strain 𝜀 increases with the increase in deformation. In
the case of superplastic deformation, where the elongation reaches and even exceeds 2000%, the
127
deviation seems much higher.
Strain Components as Functions of Infinitesimal Displacements
Based on practical measurements, the above-discussed strains face certain limitation in their application.
In order to facilitate quantitative analysis in common cases, it is needed to have a general representation
of strain components for either linear deformation or angular distortion. Obviously, strain is not linked to
the rigid movement of a stressed element, but to the displacement of the points inside the stressed body.
In other words, strains can be expressed with functions of displacement, viz. geometric equations. This
method begins with analyzing infinitesimal deformation of the stressed element. Actually, the results
therefrom will not be limited to small deformation but will be suitable for large plastic deformation
because a large plastic deformation process can be considered to be an aggregate of numerous steps of
tiny deformation.
128
Figure 12 shows a small deforming body, whose deformation is supposed to be homogeneous, with all
lines remaining straight before and after deformation and all planes continuous without any warpage
caused by deformation.
Figure 12. Displacements of a point inside a deforming body
129
A and B are two neighboring points on the diagonal of the body before deformation and dx, dy and dz are
the lengths of the body. As the result of the body’s deformation, the diagonal AB moves to A′B′. (In
Figure 12, the deforming body, whose shape and size can all be imagined to have changed, is not shown
for the purpose of visual clarity.) Let (x, y, z) and (x + dx, y + dy, z + dz) be coordinates of the points A
and B, (x + u, y + v, z + w) the coordinates of the point A′, and u, v and w the coordinates of displacement
of A, viz. AA′ projected on x, y and z axes, respectively. Infinitesimal u, v, w are functions of x, y, and z.
Similarly, (x + dx) + (u + du), (y + dy) + (v + dv), and (z + dz) + (w + dw) are the coordinates of the point
B′. Here, du, dv, and dw denote the displacements of the point B relative to the point A projected on x, y,
and z axes.
Since u is assumed to be a continuous function of coordinates x, y, and z, viz. u = f (x, y, z), (u + du) is
also that of coordinates x + dx, y + dy, and z + dz, viz. (u + du) = f {(x + dx), (y + dy), (z + dz)}, which,
further by using Taylor’s expansion, can then be represented as
(33)
Because u is an infinitesimal amount, the last higher-order terms can be dropped. Then Equation (33)
becomes
(34)
130
Similarly, from Figure 12, we have
(35)
(36)
where, (𝜕u/𝜕x)dx is the change in the original length dx or AC. Thus, the direct strain in the direction OX
can be represented as 𝜀𝑥 , i.e.
(37)
Similarly, we have direct strains 𝜀𝑦 = 𝜕𝑣 𝜕𝑦 and 𝜀𝑧 = 𝜕𝑤 𝜕𝑧 in the directions OY and OZ,
respectively.
131
The angular strains in XOZ plane are equal to ∠JA′C′ and ∠F′A′M (see Figure 13).
Figure 13. Strains and displacements of the plane XOZ
132
Thus, we have
(38)
Likewise,
(39)
The engineering shear strains in XOZ plane are
(40)
(41)
(42)
133
where 𝛾𝑥𝑧 , 𝛾𝑦𝑧 , and 𝛾𝑥𝑦 are shear strains composed of two sides of the deformed element, with 𝜀𝑥𝑧 =
𝜀𝑧𝑥 , 𝜀𝑦𝑧 = 𝜀𝑧𝑦 , and 𝜀𝑥𝑦 = 𝜀𝑦𝑥 .
The subscripts of the strain components are consistent with those of the stress components.
Then the strain components can also be represented as a strain tensor as follows:
(43)
As a second-order symmetric one, the above-presented strain tensor has the same characteristics as the
stress tensor and fully defines the deformation of a point inside a stressed body. Similar to the stress
tensor, the strain tensor also has three invariants and can be categorized into two: the spherical strain
tensor and the deviatoric strain tensor. The former describes the volume change and the latter the shape
change in a stressed element. In regard to the plastic deformation, the stressed body is often supposed to
be of incompressibility, which is described by 𝜀𝑥 + 𝜀𝑦 + 𝜀𝑧 = 0, meaning the mere existence of deviatoric
strain tensors without any spherical strain tensors.
134
Thus, the relationship between strains and displacements can be represented as
(44)
(45)
135
The Maximum Shear Strains and the Octahedral Strains
From the strain analysis, we have three principal shear strains, viz.
(46)
Assuming 𝜀1 ≥ 𝜀2 ≥ 𝜀3 , the maximum shear strain can then be obtained as follows:
(47)
The normal vector of the octahedral plane forms the same angles of 55°44′ against three principal axes
and with a direction cosine 1 3. The corresponding normal strain on the octahedral plane is
(48)
where 𝐼1 is the first invariant of the strain tensor.
136
Equation (48) implies that the octahedral normal strain is equal to the mean strain. The shear strain on
the octahedral plane can be expressed by
(49)
Expressed in terms of the invariants, Equation (49) turns into
(50)
where 𝐼2 is the second invariant of the strain tensor.
137
Strain Rates and Strain Rate Tensors
In the process of plastic deformation, the distance between two points inside a deforming body never
stops changing. The amount of distance variation determines the value of the strain and its varying
speed decides that of the strain rate. Expressed by 𝑢, the displacement speed between deforming points
and displacement itself are a continuous function of time. Thus, the displacement speeds in three
directions can be expressed by
(51)
If strains are small enough, Equation (51) can be reduced to
(52)
138
As regards two adjacent points inside a deforming body, the strain rate in a direction can be represented
as the ratio of speed difference to the distance between them. In fact, if the speed difference and the
distance approach zero, the strain rate can be expressed with a differential equation as follows:
(53)
Substitution of Equation (52) into Equation (53) gives
(54)
(55)
139
With other strain rates obtained in the same way, all strain rates can be summarized as follows:
(56)
140
These equations imply that the rates of strain components are equal to the derivatives of the
displacement speeds in relation to the corresponding coordinate axes or of the strain 𝜀𝑖𝑗 with respect to
time t. Similar to the strain component, the strain rates can also be defined in the form of tensor.
Following the concept of the principal axes, there are only direct strain rates without shear strain rates.
The maximum shear strain rate 𝛾𝑚𝑎𝑥 can be obtained as well. Mohr circles can find application in
dealing with strain rates, too.
In axisymmetric cases, we can use the spherical coordinates (see Figure 14) to describe the relationships
between the strain rate and the displacement as follows:
Figure 14. Spherical coordinate system
141
(57)
Finally, it should be pointed out that the strain rate and the displacement speed differentiate from each
other in dimension, meaning that they have different units.
142
Incompressibility and Chief Deformation Types
Strictly speaking, some changes in the volume will happen in plastic deformation. Such volume change
in a stressed material element includes two parts: elastic and plastic. Let the volume strain be 𝜃 and a
small hexahedron be cut out of the deforming body with three side lengths dx, dy, and dz in the principal
directions (see Figure 15). Apparently, the volume of this hexahedron is dV = dxdydz.
Figure 15. A small hexahedron cut from a
deforming body along the principal planes
143
Each side length of this small hexahedron will change with the volume changing during plastic
deformation. After deformation, the volume of the deformed hexahedron becomes
(58)
Omitting high order infinitesimals, Equation (58) becomes
(59)
The relative change of the small volume is just the volume strain:
(60)
Even if the small hexahedron is not cut out along the principal axes, which means shear stresses will
exist in the coordinate directions, Equation (60) is still tenable because the volume changes caused by
shear stresses are so small that they can be neglected. Thus the general expression of the volume strain
is
(61)
where the coordinate axes x, y, and z do not specifically denote the principal directions, but are directed
at all directions.
144
Equal to the sum of three direct strains, the volume strain can be presented with a function of
displacement as follows:
(62)
According to continuum mechanics, the volume change is mainly determined by the mean stress 𝜎𝑚 ,
thus, it can be represented as:
(63)
where 𝜈 is Poisson’s ratio, and E is Young’s modulus.
Bridgman et al. have proved that the volume compression of dense solids and liquids belongs to elastic
deformation with an approximately linear relationship between the relative volume change and the
hydrostatic stress. It means that the change in density of a stressed body is caused mainly by elastic
deformation.
145
When the imposed pressure reaches 980 MPa, the decrease in volume is 0.6% for steel, and 1.3% for
copper. In common practices, the amount of plastic deformation in metal forming processes usually
exceeds 10% and even 50%. By contrast, the elastic deformation is too small to be worth considering.
Out of it comes the assumption that deforming body is incompressible. It considerably simplifies the
analysis in practice. The condition that an incompressible body must satisfy is
(64)
Taking account of elastic volume change, Equation (64) satisfies the part of plastic deformation. The
sum of direct plastic strains is zero. This equation includes three types of possible plastic deformation
(see Figure 16).
Figure 16. Three deformation types of a deforming body: (a) tension type, (b) plane state type and
146
(c) compression type
The first is tension in the direction of the maximum principal strain 𝜀1 > 0 and compression in other
two principal directions 𝜀2 < 0 and 𝜀3 < 0. The second is of plane state (pure shear) deformation with
one zero principal strain (𝜀2 = 0) and two other principal strains, which have the same absolute value
but opposite signs (𝜀1 = 𝜀3 ). The third type is composed of compression in the direction of the
principal strain maximal in absolute value and extension in other two principal directions, i.e., 𝜀3 <
0, 𝜀1 > 0 and 𝜀2 > 0. It is notable that the assumption of incompressibility is in contradiction to that of
uncompacted material, but in conformity with the law of mass conservation, viz.
(65)
Letting 𝜌0 and 𝑉0 be the original density and the volume of powder or castings, respectively, they
become 𝜌 and V after compression as follows:
(66)
Equation (66) can be written in the logarithmic form as follows:
(67)
or
(68)
147
where 𝜀𝜌 = 𝑙𝑛 𝜌 𝜌0 is true density, and 𝜀𝑉 = 𝑙𝑛 𝑉 𝑉0 is true volume strain.
The true volume strain can also be expressed in terms of direct strains below
(69)
where 𝜀ℎ , 𝜀𝑟 and 𝜀𝜃 are direct strains in height, radius, and circumference, respectively.
148
Effective Stress, Effective Strain, and Stress Type
Effective Stress
In practices, there may be various stress states. Usually, we acquire knowledge on the deformability of
as-received materials through some simple tests. To establish the relationship between a complex stress
state and a simple experimental stress state, we need a special stress called effective or equivalent stress.
Commonly, it is the flow stress determined in the uniaxial tension test. Of course, it can also be a flow
stress determined through other simple tests (e.g., uniaxial compression or pure torsion). In fact, a
proper yield criterion just provides a possibility to build up an equivalent relationship between a
complex stress state and a simple experimental stress state, i.e., 𝑓 = 𝜎, where f is a function expressing
a complex stress state, and 𝜎 is flow stress decided through a simple test. It is worth noting that the
different yield criterion gives the different relationships between f and 𝜎. For example, if we use the
Mises yield criterion to describe the relationship between a simple test state and a complex stress state,
it becomes
(70)
where 𝜎 is the effective stress, and 𝜎𝑡 is flow stress acquired through the uniaxial tension test.
149
But, if the effective stress is found from the pure torsion test, the above-stated relation becomes
(71)
where 𝜏𝑝 is the flow stress decided through the pure torsion test, which is capable of reflecting a
complex stress state, too.
It is notable that the effective stress can be used in two fields—one in strength theory and the other in
plasticity. When used in strength theory, it is tied up with the state of 𝑓 = 𝜎 ≤ 𝜎𝑡0 . In this case, it is also
termed strength stress. The strength stress corresponds to the equivalent stress of a material element in
the range of elastic strain under any complex stress state. When used in plasticity, it is linked to the state
of 𝑓 = 𝜎 ≥ 𝜎𝑡0 . That is, the effective stress constitutes the sound base for ascertaining the behavior of a
stressed material element under elastic or plastic deformation state.
150
Effective Strain
Used in strength theory and plasticity, though, the effective strain has completely different concepts. In
the former case, the relationship between the simple experimental strain and the strain components
under a complex stress state is established totally on the basis of the concept of equivalent elastic
deformation strain energy under any stress state, i.e.,
(72)
where the stress 𝜎𝑡 is decided through uniaxial tension while 𝜎𝑖𝑗 under complex stress state.
From Equation (73) representing the elastic deformation strain energy, we have
(73)
(74)
(75)
151
Substitution of Equation (74) and Equation (75) into Equation (73) gives
(76)
The strain components in Equation (76) are all of total elastic deformation. Commonly, the
representation of the effective strain used in strength theory has nothing to do with making use of
whichever yield criterion.
However, unlike used in strength theory, the concept of the effective strain used in plasticity is not a total
elastic but a plastic strain increment. Determination of the relationship between an effective strain
obtained from a simple experiment and strain components involving a complex stress state wholly relies
on an equivalent definition of a subsequent hardening increment under any stress state. Such definition
of equivalent hardening state would fit all the plastic potential functions, the yield functions and the
hardening models. Suppose that a constitutive relation adopts an isotropic hardening model and an
associated flow rule f = g; then the equivalent quantity of a subsequent hardening increment under any
stress state would be the second order of the plastic strain work increment 𝑑2 𝑤 𝑝 (see Equation (77)), i.e.,
(77)
(78)
152
Thus, the effective plastic strain increment can be expressed by
(79)
As a special case, if the yield function is of the Mises and all loading stresses increase proportionally
(see Equation (80)), Equation (79) finally becomes
(80)
(81)
Equation (81) is similar to Equation (76) but with components of plastic strain increment. This effective
plastic strain increment is nothing but a special form of Equation (79) with the plastic potential (the
Mises yield criterion) and is tenable on the condition of proportionally loading, although it is not a
general representation of the effective plastic strain increment.
A plastic constitutive relation involving different hardening models reflects dissimilar subsequent
hardening states (different shapes of subsequent yield surface). Therefore, an equivalent plastic strain
increment would have a different representation to befit such an equivalent hardening state decided by
the associated hardening model. It means that an effective relationship between the plastic strain
increment through a simple test and the plastic strain increments under any stress state would be
determined completely based on the definition of an equivalent subsequent hardening state.
153
Stress Type
In the section of “Stress Tensor and Deviatoric Stress Tensor,” we made reference to the concepts of
compression strain type and elongation strain type. Actually, different stress types can be expressed with
the Lode’s parameter put forward in 1926 as follows:
When 𝜇𝜎 = −1 and 𝜎2 = 𝜎3 , the corresponding stress type is under simple tension state, which is the
same as a uniaxial tension with 𝜎1 = 𝜎𝑡 and 𝜎2 = 𝜎3 = 0, and the equibiaxial compression also belongs
to this stress type. 𝜇𝜎 = 0 and 𝜎3 = 𝜎1 + 𝜎2 2 is under the state of pure shear. Actually, all plane strain
states pertain to this stress type. When 𝜇𝜎 = 1, it denotes the state of simple compression (e.g., the case
of uniaxial compression). Thus, the Lode’s parameter lies in the range of −1 ≤ 𝜇𝜎 ≤ 1.
The expression of Mises yield criterion with stress order 𝜎1 ≥ 𝜎2 ≥ 𝜎3 in terms of the Lode parameter
should be
Compared to Tresca yield criterion, the maximum deviation between these two criteria happens under the
plane strain state, i.e., 𝜇𝜎 = 0.
154
In building up a yield criterion, the Lode’s parameter may be a variable for considering the effects the
stress type brings about on the material yield. But, since it is hard to define the Lode’s parameter by
means of common stress components.
155
Mohr Stress Circles
With Mohr stress circles, the variation of normal and shear stresses can be visually explained when the
direction of the cut plane changes. This method is intrinsically bound up with a circle. There are two
kinds of Mohr circles: one for the two-dimensional stress system and the other for the three-dimensional
system.
156
Mohr Circles for a Two-Dimensional Stress System
The plane stress state at a point of a stressed body can be represented by a Mohr stress circle. Suppose
that the stresses at a point inside a deforming body are at a plane stress state (see Figure 17).
Figure 17. Plane stress state at a point inside a stressed body
157
As in a metal-forming process, the stressed elements of the deforming body mostly show their normal
stresses under compression states, let’s suppose the normal stresses 𝜎𝑥 and 𝜎𝑦 are all of compression and
𝜎𝑥 < 𝜎𝑦 in terms of algebraic values.
There is an arbitrary plane AC with an angle 𝛼 against the plane BC (anticlockwise). Again, there are
normal stress 𝜎𝑥 and shear stress 𝜏𝑥𝑦 acting on the plane BC. Suppose that 𝜎 and 𝜏 are normal and shear
stresses acting on the plane AC. Figure 17 shows the directions of these stresses.
Figure 17. Plane stress state at a point inside a stressed body
158
Let’s consider the equilibrium of a small element ABC cut out of the body, on which the resultant force
in the normal direction of the plane AC can be represented by
(82)
or
(83)
Thus,
(84)
159
which further becomes
(85)
The resultant force parallel to AC can be represented by
(86)
or
(87)
Hence,
(88)
In order to remove the parameter 𝛼, by adding together the squares of Equation (85) and Equation (88)
and simplifying, we obtain
(89)
160
Equation (89) can be rewritten into
(90)
Equation (90) denotes a circle in 𝜎 − 𝜏 coordinate system
with a radius 𝑅 =
𝜎𝑥 + 𝜎𝑦 2,0 .
𝜎𝑥 − 𝜎𝑦
2
2
+
2
𝜏𝑥𝑦
1 2
and the center
The coordinate axes with abscissas and ordinates in a 𝜎 −
𝜏 coordinate system of Mohr circle determine the normal
and shear stresses respectively (see Figure 18).
This circle, often called Mohr stress circle, presents the
variation rule of normal and shear stresses, viz. 𝜎 and 𝜏,
because any point on it implies the stress state acting on
correspondent plane. Drawn by means of a simple
geometric method, the circle is widely used to study the
problem of plane stresses and plane strains.
Figure 18. Mohr circle for a plane stress state
161
Mohr Circles for a Three-Dimensional Stress System
In the study of three-dimensional stress states, Mohr circle (see Figure 19) also finds broad application
because the values of the normal and the shear stresses on an oblique plane can be indicated by a point
on it.
From three groups of data, viz. 𝜎1 − 𝜎2 , 𝜎2 −
𝜎3 , and 𝜎3 − 𝜎1 , generate three circles, on
which points denote the variation range of
the normal and the shear stresses on the
planes vertical to the stresses of 𝜎3 , 𝜎1 , and
𝜎2 . The shaded area between three circles
denotes the normal and the shear stresses on
an oblique plane, of which the normal
direction forms angles of 𝛼1 , 𝛼2 , and 𝛼3
respectively, against stresses 𝜎1 , 𝜎2 , and 𝜎3 ,
associated cosines l, m, and n.
Figure 19. Mohr Circle for the three-dimensional stress state
162
Figure 20 illustrates how three different stress circles on different planes could finally be transferred
onto one plane through rotation.
Such representations are very useful
in helping us to understand their
intrinsic relationships.
Figure 20. Generation of Mohr circle for the three-dimensional
stress state
163
Equations of Force Equilibrium
In a deforming body, stresses at each point inside it vary continuously. In other words, stress is a
continuous function of the coordinate system.
From the body, let’s take a unit cube with
lengths of each side equaling dx, dy, and dz,
of which each plane is parallel to the
coordinate plane (see Figure 21).
Force equilibrium of this unit can then be
established in the following way.
Figure 21. Force equilibrium at a point inside a deforming body
164
Suppose that the stress tensor at the point a is
(91)
The stresses at the other point a′ differentiate from those at the point a by an infinite small amount
because of changes in coordinates. Omitting the higher-order terms, each stress increment can be
represented by partial differentials of coordinate variables, which denote the direction of motion of the
acting plane.
For instance, 𝜕𝜏𝑦𝑧 𝜕𝑦 𝑑𝑦 means an infinite small amount of a shear stress 𝜏𝑦𝑧 that has moved a
distance dy. The shear stress 𝜏𝑦𝑧 is on the plane perpendicular to the y-axis and oriented the same with
z-axis. Thus, taking account of the coordinate differences of dx, dy, and dz of the point a to the point a′,
the stress tensor at the point a′ can be expressed by
(92)
165
Since the hexahedron remains in an equilibrium state, the condition for equilibrium in the direction OX
is
(93)
As regards the directions OY and OZ, we have two similar equations. Finally, three equations can be
simplified into
(94)
Equation (94) can also be rewritten in the form of tensor as follows:
(95)
166
In the case of plane stress or plane strain state, the stress components are independent of a certain axis.
Suppose that the stress acting plane is coincident with 𝑥 − 𝑧 plane. It means the shear stresses directing
y-axis equal zero, i.e., 𝜏𝑥𝑦 = 𝜏𝑧𝑦 = 0. Otherwise a skew would take place causing some change of the
plane. Such a stress condition is represented by the following equilibrium equations
(96)
Likewise, making use of the cylindrical coordinate system (r, 𝜃, z) enables us to obtain equilibrium
conditions of the axial symmetry as follows:
(97)
It should, however, be pointed out that the above-described equilibrium equations, represented by means
of stress components, belong to the force equilibrium, not the stress equilibrium. They have nothing to
do with the factors of mass force inclusive of gravity and inertia. If they did, it would be necessary to
167
include related terms into the equations.
Literature Sources, Recomended
[1]
Norman E. Dowling
Mechanical Behavior of Materials
Pearson Education Limited
[2]
Richard W. Hertzberg, Richard P.
Vinci, Jason L. Hertzberg
Deformation and Fracture Mechanics
of Engineering Materials
John Wiley & Sons, Inc.
[3]
Thomas H. Courtney
Mechanical Behavior of Materials
Waveland Press, Inc.
168
[4]
[5]
Dominique Francois, Andre Pineau, Andre
Zaom
Mechanical Behavior of Materials
Volume 1: Elasticity and Plasticity
Springer Science+Business Media, B.V.
L.S. Srinath
Advanced Mechanics of Solids
Tata McGraw-Hill Publishing
Company Limited
[6]
R.C. Hibbeler
Mechanics of materials
Pearson Education Limited
169
[7]
[8]
W. Johnson, P.B. Mellor
Engineering Plasticity
Ellis Horwood Limited
Mumtaz Kassir
Applied Elasticity and Plasticity
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R.A.C. Slater
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L.M. Kachanov
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174
The End
175