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= Cauchy stress tensor =
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Introduction
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In continuum mechanics, the Cauchy stress tensor \boldsymbol\sigma,
true stress tensor, or simply called the stress tensor is a second
order tensor named after Augustin-Louis Cauchy. The tensor consists
of nine components \sigma_{ij} that completely define the state of
stress at a point inside a material in the deformed state, placement,
or configuration. The tensor relates a unit-length direction vector n
to the stress vector T(n) across an imaginary surface perpendicular to
n:
:\mathbf{T}^{(\mathbf n)}= \mathbf n \cdot\boldsymbol{\sigma}\quad
\text{or} \quad T_j^{(n)}= \sigma_{ij}n_i.
where,
:\boldsymbol{\sigma}=
\left[{\begin{matrix}
\sigma _{11} & \sigma _{12} & \sigma _{13} \\
\sigma _{21} & \sigma _{22} & \sigma _{23} \\
\sigma _{31} & \sigma _{32} & \sigma _{33} \\
\end{matrix}}\right]
\equiv \left[{\begin{matrix}
\sigma _{xx} & \sigma _{xy} & \sigma _{xz} \\
\sigma _{yx} & \sigma _{yy} & \sigma _{yz} \\
\sigma _{zx} & \sigma _{zy} & \sigma _{zz} \\
\end{matrix}}\right]
\equiv \left[{\begin{matrix}
\sigma _x & \tau _{xy} & \tau _{xz} \\
\tau _{yx} & \sigma _y & \tau _{yz} \\
\tau _{zx} & \tau _{zy} & \sigma _z \\
\end{matrix}}\right]
The SI units of both stress tensor and stress vector are N/m2,
corresponding to the stress scalar. The unit vector is dimensionless.
The Cauchy stress tensor obeys the tensor transformation law under a
change in the system of coordinates. A graphical representation of
this transformation law is the Mohr's circle for stress.
The Cauchy stress tensor is used for stress analysis of material
bodies experiencing small deformations: It is a central concept in the
linear theory of elasticity. For large deformations, also called
finite deformations, other measures of stress are required, such as
the Piola-Kirchhoff stress tensor, the Biot stress tensor, and the
Kirchhoff stress tensor.
According to the principle of conservation of linear momentum, if the
continuum body is in static equilibrium it can be demonstrated that
the components of the Cauchy stress tensor in every material point in
the body satisfy the equilibrium equations (Cauchy's equations of
motion for zero acceleration). At the same time, according to the
principle of conservation of angular momentum, equilibrium requires
that the summation of moments with respect to an arbitrary point is
zero, which leads to the conclusion that the stress tensor is
symmetric, thus having only six independent stress components, instead
of the original nine.
There are certain invariants associated with the stress tensor, whose
values do not depend upon the coordinate system chosen, or the area
element upon which the stress tensor operates. These are the three
eigenvalues of the stress tensor, which are called the principal
stresses.
Euler�Cauchy stress principle � stress vector
======================================================================
The Euler-Cauchy stress principle states that 'upon any surface (real
or imaginary) that divides the body, the action of one part of the
body on the other is equivalent (equipollent) to the system of
distributed forces and couples on the surface dividing the body', and
it is represented by a field \mathbf{T}^{(\mathbf{n})}, called the
stress vector, defined on the surface S and assumed to depend
continuously on the surface's unit vector \mathbf n.
To formulate the Euler-Cauchy stress principle, consider an imaginary
surface S passing through an internal material point P dividing the
continuous body into two segments, as seen in Figure 2.1a or 2.1b (one
may use either the cutting plane diagram or the diagram with the
arbitrary volume inside the continuum enclosed by the surface S).
Following the classical dynamics of Newton and Euler, the motion of a
material body is produced by the action of externally applied forces
which are assumed to be of two kinds: surface forces \mathbf F and
body forces \mathbf b. Thus, the total force \mathcal F applied to a
body or to a portion of the body can be expressed as:
:\mathcal F = \mathbf b + \mathbf F
Only surface forces will be discussed in this article as they are
relevant to the Cauchy stress tensor.
When the body is subjected to external surface forces or 'contact
forces' \mathbf F, following Euler's equations of motion, internal
contact forces and moments are transmitted from point to point in the
body, and from one segment to the other through the dividing surface
S, due to the mechanical contact of one portion of the continuum onto
the other (Figure 2.1a and 2.1b). On an element of area \Delta S
containing P, with normal vector \mathbf n, the force distribution is
equipollent to a contact force \Delta \mathbf F exerted at point P and
surface moment \Delta \mathbf M. In particular, the contact force is
given by
:\Delta\mathbf F= \mathbf T^{(\mathbf n)}\,\Delta S
where \mathbf T^{(\mathbf n)} is the 'mean surface traction'.
Cauchy's stress principle asserts that as \Delta S becomes very small
and tends to zero the ratio \Delta \mathbf F/\Delta S becomes d\mathbf
F/dS and the couple stress vector \Delta \mathbf M vanishes. In
specific fields of continuum mechanics the couple stress is assumed
not to vanish; however, classical branches of continuum mechanics
address non-polar materials which do not consider couple stresses and
body moments.
The resultant vector d\mathbf F/dS is defined as the 'surface
traction', also called 'stress vector', 'traction', or 'traction
vector'. given by
\mathbf{T}^{(\mathbf{n})}=T_i^{(\mathbf{n})}\mathbf{e}_i at the point
P associated with a plane with a normal vector \mathbf n:
:T^{(\mathbf{n})}_i= \lim_{\Delta S \to 0} \frac {\Delta F_i}{\Delta
S} = {dF_i \over dS}.
This equation means that the stress vector depends on its location in
the body and the orientation of the plane on which it is acting.
This implies that the balancing action of internal contact forces
generates a 'contact force density' or 'Cauchy traction field'
\mathbf T(\mathbf n, \mathbf x, t) that represents a distribution of
internal contact forces throughout the volume of the body in a
particular configuration of the body at a given time t. It is not a
vector field because it depends not only on the position \mathbf x of
a particular material point, but also on the local orientation of the
surface element as defined by its normal vector \mathbf n.
Depending on the orientation of the plane under consideration, the
stress vector may not necessarily be perpendicular to that plane,
'i.e.' parallel to \mathbf n, and can be resolved into two components
(Figure 2.1c):
* one normal to the plane, called 'normal stress'
:\mathbf{\sigma_\mathrm{n}}= \lim_{\Delta S \to 0} \frac {\Delta
F_\mathrm n}{\Delta S} = \frac{dF_\mathrm n}{dS},
:where dF_\mathrm n is the normal component of the force d\mathbf F to
the differential area dS
* and the other parallel to this plane, called the 'shear stress'
:\mathbf \tau= \lim_{\Delta S \to 0} \frac {\Delta F_\mathrm s}{\Delta
S} = \frac{dF_\mathrm s}{dS},
:where dF_\mathrm s is the tangential component of the force d\mathbf
F to the differential surface area dS. The shear stress can be further
decomposed into two mutually perpendicular vectors.
Cauchy�s postulate
====================
According to the 'Cauchy Postulate', the stress vector
\mathbf{T}^{(\mathbf{n})} remains unchanged for all surfaces passing
through the point P and having the same normal vector \mathbf n at P,
i.e., having a common tangent at P. This means that the stress vector
is a function of the normal vector \mathbf n only, and is not
influenced by the curvature of the internal surfaces.
Cauchy�s fundamental lemma
============================
A consequence of Cauchy's postulate is 'Cauchy�s Fundamental Lemma',
also called the 'Cauchy reciprocal theorem', which states that the
stress vectors acting on opposite sides of the same surface are equal
in magnitude and opposite in direction. Cauchy's fundamental lemma is
equivalent to Newton's third law of motion of action and reaction, and
is expressed as
:- \mathbf{T}^{(\mathbf{n})}= \mathbf{T}^{(- \mathbf{n})}.
Cauchy�s stress theorem�stress tensor
======================================================================
'The state of stress at a point' in the body is then defined by all
the stress vectors T(n) associated with all planes (infinite in
number) that pass through that point. However, according to 'Cauchy�s
fundamental theorem', also called 'Cauchy�s stress theorem', merely by
knowing the stress vectors on three mutually perpendicular planes, the
stress vector on any other plane passing through that point can be
found through coordinate transformation equations.
Cauchy's stress theorem states that there exists a second-order tensor
field �(x, t), called the Cauchy stress tensor, independent of n, such
that T is a linear function of n:
:\mathbf{T}^{(\mathbf n)}= \mathbf n \cdot\boldsymbol{\sigma}\quad
\text{or} \quad T_j^{(n)}= \sigma_{ij}n_i.
This equation implies that the stress vector T(n) at any point 'P' in
a continuum associated with a plane with normal unit vector n can be
expressed as a function of the stress vectors on the planes
perpendicular to the coordinate axes, 'i.e.' in terms of the
components '�ij' of the stress tensor �.
To prove this expression, consider a tetrahedron with three faces
oriented in the coordinate planes, and with an infinitesimal area d'A'
oriented in an arbitrary direction specified by a normal unit vector n
(Figure 2.2). The tetrahedron is formed by slicing the infinitesimal
element along an arbitrary plane with unit normal n. The stress vector
on this plane is denoted by T(n). The stress vectors acting on the
faces of the tetrahedron are denoted as T(e1), T(e2), and T(e3), and
are by definition the components '�ij' of the stress tensor �. This
tetrahedron is sometimes called the 'Cauchy tetrahedron'. The
equilibrium of forces, 'i.e.' Euler's first law of motion (Newton's
second law of motion), gives:
:\mathbf{T}^{(\mathbf{n})} \, dA - \mathbf{T}^{(\mathbf{e}_1)} \, dA_1
- \mathbf{T}^{(\mathbf{e}_2)} \, dA_2 - \mathbf{T}^{(\mathbf{e}_3)} \,
dA_3 = \rho \left( \frac{h}{3}dA \right) \mathbf{a},
where the right-hand-side represents the product of the mass enclosed
by the tetrahedron and its acceleration: '�' is the density, a is the
acceleration, and 'h' is the height of the tetrahedron, considering
the plane n as the base. The area of the faces of the tetrahedron
perpendicular to the axes can be found by projecting d'A' into each
face (using the dot product):
:dA_1= \left(\mathbf{n} \cdot \mathbf{e}_1 \right)dA = n_1 \; dA,
:dA_2= \left(\mathbf{n} \cdot \mathbf{e}_2 \right)dA = n_2 \; dA,
:dA_3= \left(\mathbf{n} \cdot \mathbf{e}_3 \right)dA = n_3 \; dA,
and then substituting into the equation to cancel out d'A':
:\mathbf{T}^{(\mathbf{n})} - \mathbf{T}^{(\mathbf{e}_1)}n_1 -
\mathbf{T}^{(\mathbf{e}_2)}n_2 - \mathbf{T}^{(\mathbf{e}_3)}n_3 = \rho
\left( \frac{h}{3} \right) \mathbf{a}.
To consider the limiting case as the tetrahedron shrinks to a point,
'h' must go to 0 (intuitively, the plane n is translated along n
toward 'O'). As a result, the right-hand-side of the equation
approaches 0, so
: \mathbf{T}^{(\mathbf{n})} = \mathbf{T}^{(\mathbf{e}_1)} n_1 +
\mathbf{T}^{(\mathbf{e}_2)} n_2 + \mathbf{T}^{(\mathbf{e}_3)} n_3.
Assuming a material element (Figure 2.3) with planes perpendicular to
the coordinate axes of a Cartesian coordinate system, the stress
vectors associated with each of the element planes, 'i.e.' T(e1),
T(e2), and T(e3) can be decomposed into a normal component and two
shear components, 'i.e.' components in the direction of the three
coordinate axes. For the particular case of a surface with normal unit
vector oriented in the direction of the 'x'1-axis, denote the normal
stress by '�'11, and the two shear stresses as '�'12 and '�'13:
:\mathbf{T}^{(\mathbf{e}_1)}= T_1^{(\mathbf{e}_1)}\mathbf{e}_1 +
T_2^{(\mathbf{e}_1)} \mathbf{e}_2 + T_3^{(\mathbf{e}_1)} \mathbf{e}_3
= \sigma_{11} \mathbf{e}_1 + \sigma_{12} \mathbf{e}_2 + \sigma_{13}
\mathbf{e}_3,
:\mathbf{T}^{(\mathbf{e}_2)}= T_1^{(\mathbf{e}_2)}\mathbf{e}_1 +
T_2^{(\mathbf{e}_2)} \mathbf{e}_2 + T_3^{(\mathbf{e}_2)}
\mathbf{e}_3=\sigma_{21} \mathbf{e}_1 + \sigma_{22} \mathbf{e}_2 +
\sigma_{23} \mathbf{e}_3,
:\mathbf{T}^{(\mathbf{e}_3)}= T_1^{(\mathbf{e}_3)}\mathbf{e}_1 +
T_2^{(\mathbf{e}_3)} \mathbf{e}_2 + T_3^{(\mathbf{e}_3)}
\mathbf{e}_3=\sigma_{31} \mathbf{e}_1 + \sigma_{32} \mathbf{e}_2 +
\sigma_{33} \mathbf{e}_3,
In index notation this is
:\mathbf{T}^{(\mathbf{e}_i)}= T_j^{(\mathbf{e}_i)} \mathbf{e}_j =
\sigma_{ij} \mathbf{e}_j.
The nine components '�ij' of the stress vectors are the components of
a second-order Cartesian tensor called the 'Cauchy stress tensor',
which completely defines the state of stress at a point and is given
by
:\boldsymbol{\sigma}= \sigma_{ij} = \left[{\begin{matrix}
\mathbf{T}^{(\mathbf{e}_1)} \\
\mathbf{T}^{(\mathbf{e}_2)} \\
\mathbf{T}^{(\mathbf{e}_3)} \\
\end{matrix}}\right] =
\left[{\begin{matrix}
\sigma _{11} & \sigma _{12} & \sigma _{13} \\
\sigma _{21} & \sigma _{22} & \sigma _{23} \\
\sigma _{31} & \sigma _{32} & \sigma _{33} \\
\end{matrix}}\right] \equiv \left[{\begin{matrix}
\sigma _{xx} & \sigma _{xy} & \sigma _{xz} \\
\sigma _{yx} & \sigma _{yy} & \sigma _{yz} \\
\sigma _{zx} & \sigma _{zy} & \sigma _{zz} \\
\end{matrix}}\right] \equiv \left[{\begin{matrix}
\sigma _x & \tau _{xy} & \tau _{xz} \\
\tau _{yx} & \sigma _y & \tau _{yz} \\
\tau _{zx} & \tau _{zy} & \sigma _z \\
\end{matrix}}\right],
where '�'11, '�'22, and '�'33 are normal stresses, and '�'12, '�'13,
'�'21, '�'23, '�'31, and '�'32 are shear stresses. The first index 'i'
indicates that the stress acts on a plane normal to the 'Xi' -axis,
and the second index 'j' denotes the direction in which the stress
acts (For example, �12 implies that the stress is acting on the plane
that is normal to the 1st axis i.e.;'X'1 and acts along the 2nd axis
i.e.;'X'2). A stress component is positive if it acts in the positive
direction of the coordinate axes, and if the plane where it acts has
an outward normal vector pointing in the positive coordinate
direction.
Thus, using the components of the stress tensor
:\begin{align} \mathbf{T}^{(\mathbf{n})} &=
\mathbf{T}^{(\mathbf{e}_1)}n_1 + \mathbf{T}^{(\mathbf{e}_2)}n_2 +
\mathbf{T}^{(\mathbf{e}_3)}n_3 \\
& = \sum_{i=1}^3 \mathbf{T}^{(\mathbf{e}_i)}n_i \\
&= \left( \sigma_{ij}\mathbf{e}_j \right)n_i \\
&= \sigma_{ij}n_i\mathbf{e}_j
\end{align}
or, equivalently,
:T_j^{(\mathbf n)}= \sigma_{ij}n_i.
Alternatively, in matrix form we have
:\left[{\begin{matrix}
T^{(\mathbf n)}_1 & T^{(\mathbf n)}_2 & T^{(\mathbf
n)}_3\end{matrix}}\right]=\left[{\begin{matrix}
n_1 & n_2 & n_3
\end{matrix}}\right]\cdot
\left[{\begin{matrix}
\sigma _{11} & \sigma _{12} & \sigma _{13} \\
\sigma _{21} & \sigma _{22} & \sigma _{23} \\
\sigma _{31} & \sigma _{32} & \sigma _{33} \\
\end{matrix}}\right].
The Voigt notation representation of the Cauchy stress tensor takes
advantage of the symmetry of the stress tensor to express the stress
as a six-dimensional vector of the form:
:\boldsymbol{\sigma} = \begin{bmatrix}\sigma_1 & \sigma_2 &
\sigma_3 & \sigma_4 & \sigma_5 & \sigma_6 \end{bmatrix}^T
\equiv \begin{bmatrix}\sigma_{11} & \sigma_{22} & \sigma_{33}
& \sigma_{23} & \sigma_{13} & \sigma_{12} \end{bmatrix}^T.
The Voigt notation is used extensively in representing stress-strain
relations in solid mechanics and for computational efficiency in
numerical structural mechanics software.
Transformation rule of the stress tensor
==========================================
It can be shown that the stress tensor is a contravariant second order
tensor, which is a statement of how it transforms under a change of
the coordinate system. From an 'xi'-system to an ' xi' '-system, the
components '�ij' in the initial system are transformed into the
components '�ij' ' in the new system according to the tensor
transformation rule (Figure 2.4):
:\sigma'_{ij}=a_{im}a_{jn}\sigma_{mn} \quad \text{or} \quad
\boldsymbol{\sigma}' = \mathbf A \boldsymbol{\sigma} \mathbf A^{T},
where A is a rotation matrix with components 'aij'. In matrix form
this is
:\left[{\begin{matrix}
\sigma'_{11} & \sigma'_{12} & \sigma'_{13} \\
\sigma'_{21} & \sigma'_{22} & \sigma'_{23} \\
\sigma'_{31} & \sigma'_{32} & \sigma'_{33} \\
\end{matrix}}\right]=\left[{\begin{matrix}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23} \\
a_{31} & a_{32} & a_{33} \\
\end{matrix}}\right]\left[{\begin{matrix}
\sigma_{11} & \sigma_{12} & \sigma_{13} \\
\sigma_{21} & \sigma_{22} & \sigma_{23} \\
\sigma_{31} & \sigma_{32} & \sigma_{33} \\
\end{matrix}}\right]\left[{\begin{matrix}
a_{11} & a_{21} & a_{31} \\
a_{12} & a_{22} & a_{32} \\
a_{13} & a_{23} & a_{33} \\
\end{matrix}}\right].
Expanding the matrix operation, and simplifying terms using the
symmetry of the stress tensor, gives
:\sigma_{11}' =
a_{11}^2\sigma_{11}+a_{12}^2\sigma_{22}+a_{13}^2\sigma_{33}+2a_{11}a_{12}\sigma_
{12}+2a_{11}a_{13}\sigma_{13}+2a_{12}a_{13}\sigma_{23},
:\sigma_{22}' =
a_{21}^2\sigma_{11}+a_{22}^2\sigma_{22}+a_{23}^2\sigma_{33}+2a_{21}a_{22}\sigma_
{12}+2a_{21}a_{23}\sigma_{13}+2a_{22}a_{23}\sigma_{23},
:\sigma_{33}' =
a_{31}^2\sigma_{11}+a_{32}^2\sigma_{22}+a_{33}^2\sigma_{33}+2a_{31}a_{32}\sigma_
{12}+2a_{31}a_{33}\sigma_{13}+2a_{32}a_{33}\sigma_{23},
:\begin{align}
\sigma_{12}' =
&a_{11}a_{21}\sigma_{11}+a_{12}a_{22}\sigma_{22}+a_{13}a_{23}\sigma_{33}\\
&+(a_{11}a_{22}+a_{12}a_{21})\sigma_{12}+(a_{12}a_{23}+a_{13}a_{22})\sigma_{
23}+(a_{11}a_{23}+a_{13}a_{21})\sigma_{13},
\end{align}
:\begin{align}
\sigma_{23}' =
&a_{21}a_{31}\sigma_{11}+a_{22}a_{32}\sigma_{22}+a_{23}a_{33}\sigma_{33}\\
&+(a_{21}a_{32}+a_{22}a_{31})\sigma_{12}+(a_{22}a_{33}+a_{23}a_{32})\sigma_{
23}+(a_{21}a_{33}+a_{23}a_{31})\sigma_{13},\end{align}
:\begin{align}
\sigma_{13}' =
&a_{11}a_{31}\sigma_{11}+a_{12}a_{32}\sigma_{22}+a_{13}a_{33}\sigma_{33}\\
&+(a_{11}a_{32}+a_{12}a_{31})\sigma_{12}+(a_{12}a_{33}+a_{13}a_{32})\sigma_{
23}+(a_{11}a_{33}+a_{13}a_{31})\sigma_{13}.\end{align}
The Mohr circle for stress is a graphical representation of this
transformation of stresses.
Normal and shear stresses
===========================
The magnitude of the normal stress component '�'n of any stress vector
T(n) acting on an arbitrary plane with normal unit vector n at a given
point, in terms of the components '�ij' of the stress tensor �, is the
dot product of the stress vector and the normal unit vector:
:\begin{align}
\sigma_\mathrm{n} &= \mathbf{T}^{(\mathbf{n})}\cdot \mathbf{n} \\
&=T^{(\mathbf n)}_i n_i \\
&=\sigma_{ij}n_i n_j.
\end{align}
The magnitude of the shear stress component '�'n, acting orthogonal to
the vector n, can then be found using the Pythagorean theorem:
:\begin{align}
\tau_\mathrm{n} &=\sqrt{ \left( T^{(\mathbf n)}
\right)^2-\sigma_\mathrm{n}^2} \\
&= \sqrt{T_i^{(\mathbf n)}T_i^{(\mathbf n)}-\sigma_\mathrm{n}^2},
\end{align}
where
:\left( T^{(\mathbf n)} \right)^2 = T_i^{(\mathbf n)} T_i^{(\mathbf
n)} = \left( \sigma_{ij} n_j \right) \left(\sigma_{ik} n_k \right) =
\sigma_{ij} \sigma_{ik} n_j n_k.
Cauchy's first law of motion
==============================
According to the principle of conservation of linear momentum, if the
continuum body is in static equilibrium it can be demonstrated that
the components of the Cauchy stress tensor in every material point in
the body satisfy the equilibrium equations.
:
\sigma_{ji,j}+ F_i = 0
For example, for a hydrostatic fluid in equilibrium conditions, the
stress tensor takes on the form:
: {\sigma_{ij}} = -p{\delta_{ij}} ,
where p is the hydrostatic pressure, and {\delta_{ij}}\ is the
kronecker delta.
: !Derivation of equilibrium equations
|Consider a continuum body (see Figure 4) occupying a volume V,
having a surface area S, with defined traction or surface forces
T_i^{(n)} per unit area acting on every point of the body surface, and
body forces F_i per unit of volume on every point within the volume V.
Thus, if the body is in equilibrium the resultant force acting on the
volume is zero, thus: :\int_S T_i^{(n)}dS + \int_V F_i dV = 0 By
definition the stress vector is T_i^{(n)} =\sigma_{ji}n_j, then
:\int_S \sigma_{ji}n_j\, dS + \int_V F_i\, dV = 0 Using the Gauss's
divergence theorem to convert a surface integral to a volume integral
gives :\int_V \sigma_{ji,j}\, dV + \int_V F_i\, dV = 0 :\int_V
(\sigma_{ji,j} + F_i\,) dV = 0 For an arbitrary volume the integral
vanishes, and we have the 'equilibrium equations' :\sigma_{ji,j} + F_i
= 0
Cauchy's second law of motion
===============================
According to the principle of conservation of angular momentum,
equilibrium requires that the summation of moments with respect to an
arbitrary point is zero, which leads to the conclusion that the stress
tensor is symmetric, thus having only six independent stress
components, instead of the original nine:
:\sigma_{ij}=\sigma_{ji}
: !Derivation of symmetry of the stress tensor
Summing moments about point 'O' (Figure 4) the resultant moment is
zero as the body is in equilibrium. Thus, :\begin{align} M_O
&=\int_S (\mathbf{r}\times\mathbf{T})dS + \int_V
(\mathbf{r}\times\mathbf{F})dV=0 \\ 0 &=
\int_S\varepsilon_{ijk}x_jT_k^{(n)}dS + \int_V\varepsilon_{ijk}x_jF_k
dV \\ \end{align} where \mathbf{r} is the position vector and is
expressed as :\mathbf{r}=x_j\mathbf{e}_j Knowing that T_k^{(n)}
=\sigma_{mk}n_m and using Gauss's divergence theorem to change from a
surface integral to a volume integral, we have :\begin{align} 0 &=
\int_S \varepsilon_{ijk}x_j\sigma_{mk}n_m\, dS +
\int_V\varepsilon_{ijk}x_jF_k\, dV \\ &= \int_V
(\varepsilon_{ijk}x_j\sigma_{mk})_{,m} dV +
\int_V\varepsilon_{ijk}x_jF_k\, dV \\ &= \int_V
(\varepsilon_{ijk}x_{j,m}\sigma_{mk}+\varepsilon_{ijk}x_j\sigma_{mk,m})
dV + \int_V\varepsilon_{ijk}x_jF_k\, dV \\ &= \int_V
(\varepsilon_{ijk}x_{j,m}\sigma_{mk}) dV+ \int_V
\varepsilon_{ijk}x_j(\sigma_{mk,m}+F_k)dV \\ \end{align} The seco
nd
integral is zero as it contains the equilibrium equations. This leaves
the first integral, where x_{j,m}=\delta_{jm}, therefore :\int_V
(\varepsilon_{ijk}\sigma_{jk}) dV=0 For an arbitrary volume V, we then
have :\varepsilon_{ijk}\sigma_{jk}=0 which is satisfied at every point
within the body. Expanding this equation we have
:\sigma_{12}=\sigma_{21}, \sigma_{23}=\sigma_{32}, and
\sigma_{13}=\sigma_{31} or in general :\sigma_{ij}=\sigma_{ji} This
proves that the stress tensor is symmetric
However, in the presence of couple-stresses, i.e. moments per unit
volume, the stress tensor is non-symmetric. This also is the case when
the Knudsen number is close to one, K_{n}\rightarrow 1, or the
continuum is a non-Newtonian fluid, which can lead to rotationally
non-invariant fluids, such as polymers.
Principal stresses and stress invariants
======================================================================
At every point in a stressed body there are at least three planes,
called 'principal planes', with normal vectors \mathbf{n}, called
'principal directions', where the corresponding stress vector is
perpendicular to the plane, i.e., parallel or in the same direction as
the normal vector \mathbf{n}, and where there are no normal shear
stresses \tau_\mathrm{n}. The three stresses normal to these principal
planes are called 'principal stresses'.
The components \sigma_{ij} of the stress tensor depend on the
orientation of the coordinate system at the point under consideration.
However, the stress tensor itself is a physical quantity and as such,
it is independent of the coordinate system chosen to represent it.
There are certain invariants associated with every tensor which are
also independent of the coordinate system. For example, a vector is a
simple tensor of rank one. In three dimensions, it has three
components. The value of these components will depend on the
coordinate system chosen to represent the vector, but the magnitude of
the vector is a physical quantity (a scalar) and is independent of the
Cartesian coordinate system chosen to represent the vector (so long as
it is normal). Similarly, every second rank tensor (such as the stress
and the strain tensors) has three independent invariant quantities
associated with it. One set of such invariants are the principal
stresses of the stress tensor, which are just the eigenvalues of the
stress tensor. Their direction vectors are the principal directions or
eigenvectors.
A stress vector parallel to the normal unit vector \mathbf{n} is given
by:
:\mathbf{T}^{(\mathbf{n})} = \lambda \mathbf{n}=
\mathbf{\sigma}_\mathrm n \mathbf{n}
where \lambda is a constant of proportionality, and in this particular
case corresponds to the magnitudes \sigma_\mathrm{n} of the normal
stress vectors or principal stresses.
Knowing that T_i^{(n)}=\sigma_{ij}n_j and n_i=\delta_{ij}n_j, we have
:\begin{align}
T_i^{(n)} &= \lambda n_i \\
\sigma_{ij}n_j &=\lambda n_i \\
\sigma_{ij}n_j -\lambda n_i &=0 \\
\left(\sigma_{ij}- \lambda\delta_{ij} \right)n_j &=0 \\
\end{align}
This is a homogeneous system, i.e. equal to zero, of three linear
equations where n_j are the unknowns. To obtain a nontrivial
(non-zero) solution for n_j, the determinant matrix of the
coefficients must be equal to zero, i.e. the system is singular. Thus,
:\left|\sigma_{ij}- \lambda\delta_{ij} \right|=\begin{vmatrix}
\sigma_{11} - \lambda & \sigma_{12} & \sigma_{13} \\
\sigma_{21} & \sigma_{22} - \lambda & \sigma_{23} \\
\sigma_{31}& \sigma_{32} & \sigma_{33} - \lambda \\
\end{vmatrix}=0
Expanding the determinant leads to the 'characteristic equation'
:\left|\sigma_{ij}- \lambda\delta_{ij} \right| = -\lambda^3 +
I_1\lambda^2 - I_2\lambda + I_3=0
where
:\begin{align}
I_1 &= \sigma_{11}+\sigma_{22}+\sigma_{33} \\
&= \sigma_{kk} = \text{tr}(\boldsymbol{\sigma}) \\
I_2 &= \begin{vmatrix}
\sigma_{22} & \sigma_{23} \\
\sigma_{32} & \sigma_{33} \\
\end{vmatrix}
+ \begin{vmatrix}
\sigma_{11} & \sigma_{13} \\
\sigma_{31} & \sigma_{33} \\
\end{vmatrix}
+
\begin{vmatrix}
\sigma_{11} & \sigma_{12} \\
\sigma_{21} & \sigma_{22} \\
\end{vmatrix} \\
&=
\sigma_{11}\sigma_{22}+\sigma_{22}\sigma_{33}+\sigma_{11}\sigma_{33}-\sigma_{12}
^2-\sigma_{23}^2-\sigma_{31}^2
\\
&=
\frac{1}{2}\left(\sigma_{ii}\sigma_{jj}-\sigma_{ij}\sigma_{ji}\right)
= \frac{1}{2}\left[ \left( \text{tr}(\boldsymbol{\sigma}) \right)^{2}
- \text{tr}(\boldsymbol{\sigma}^{2}) \right] \\
I_3 &= \det(\sigma_{ij}) = \det(\boldsymbol{\sigma})\\
&=
\sigma_{11}\sigma_{22}\sigma_{33}+2\sigma_{12}\sigma_{23}\sigma_{31}-\sigma_{12}
^2\sigma_{33}-\sigma_{23}^2\sigma_{11}-\sigma_{31}^2\sigma_{22}
\\
\end{align}
The characteristic equation has three real roots \lambda_i, i.e. not
imaginary due to the symmetry of the stress tensor. The \sigma_1 =
\max \left( \lambda_1,\lambda_2,\lambda_3 \right), \sigma_3 = \min
\left( \lambda_1,\lambda_2,\lambda_3 \right) and
\sigma_2=I_1-\sigma_1-\sigma_3, are the principal stresses, functions
of the eigenvalues \lambda_i. The eigenvalues are the roots of the
characteristic polynomial. The principal stresses are unique for a
given stress tensor. Therefore, from the characteristic equation, the
coefficients I_1, I_2 and I_3, called the first, second, and third
'stress invariants', respectively, always have the same value
regardless of the coordinate system's orientation.
For each eigenvalue, there is a non-trivial solution for n_j in the
equation \left(\sigma_{ij}- \lambda\delta_{ij} \right)n_j =0. These
solutions are the principal directions or eigenvectors defining the
plane where the principal stresses act. The principal stresses and
principal directions characterize the stress at a point and are
independent of the orientation.
A coordinate system with axes oriented to the principal directions
implies that the normal stresses are the principal stresses and the
stress tensor is represented by a diagonal matrix:
:\sigma_{ij}=
\begin{bmatrix}
\sigma_1 & 0 & 0\\
0 & \sigma_2 & 0\\
0 & 0 & \sigma_3
\end{bmatrix}
The principal stresses can be combined to form the stress invariants,
I_1, I_2, and I_3. The first and third invariant are the trace and
determinant respectively, of the stress tensor. Thus,
:\begin{align}
I_1 &= \sigma_{1}+\sigma_{2}+\sigma_{3} \\
I_2 &=
\sigma_{1}\sigma_{2}+\sigma_{2}\sigma_{3}+\sigma_{3}\sigma_{1} \\
I_3 &= \sigma_{1}\sigma_{2}\sigma_{3} \\
\end{align}
Because of its simplicity, the principal coordinate system is often
useful when considering the state of the elastic medium at a
particular point. Principal stresses are often expressed in the
following equation for evaluating stresses in the x and y directions
or axial and bending stresses on a part. The principal normal stresses
can then be used to calculate the von Mises stress and ultimately the
safety factor and margin of safety.
:\sigma_{1},\sigma_{2}= \frac{\sigma_{x} + \sigma_{y}}{2} \pm
\sqrt{\left (\frac{\sigma_{x} - \sigma_{y}}{2}\right)^2 + \tau_{xy}^2}
Using just the part of the equation under the square root is equal to
the maximum and minimum shear stress for plus and minus. This is shown
as:
:\tau_{max},\tau_{min}= \pm \sqrt{\left (\frac{\sigma_{x} -
\sigma_{y}}{2}\right)^2 + \tau_{xy}^2}
Maximum and minimum shear stresses
======================================================================
The maximum shear stress or maximum principal shear stress is equal to
one-half the difference between the largest and smallest principal
stresses, and acts on the plane that bisects the angle between the
directions of the largest and smallest principal stresses, i.e. the
plane of the maximum shear stress is oriented 45^\circ from the
principal stress planes. The maximum shear stress is expressed as
:\tau_\max=\frac{1}{2}\left|\sigma_\max-\sigma_\min\right|
Assuming \sigma_1\ge\sigma_2\ge\sigma_3 then
:\tau_\max=\frac{1}{2}\left|\sigma_1-\sigma_3\right|
When the stress tensor is non zero the normal stress component acting
on the plane for the maximum shear stress is non-zero and it is equal
to
\sigma_\mathrm{n}=\frac{1}{2}\left(\sigma_1+\sigma_3\right)
: !Derivation of the maximum and minimum shear stresses
|The normal stress can be written in terms of principal stresses
(\sigma_1\ge\sigma_2\ge\sigma_3) as :\begin{align} \sigma_\mathrm{n}
&= \sigma_{ij}n_in_j \\ &=\sigma_1n_1^2 + \sigma_2n_2^2 +
\sigma_3n_3^2\\ \end{align} Knowing that \left( T^{(n)}
\right)^2=\sigma_{ij}\sigma_{ik}n_jn_k, the shear stress in terms of
principal stresses components is expressed as :\begin{align}
\tau_\mathrm{n}^2 &= \left( T^{(n)} \right)^2-\sigma_\mathrm{n}^2
\\
&=\sigma_1^2n_1^2+\sigma_2^2n_2^2+\sigma_3^2n_3^2-\left(\sigma_1n_1^2+\sigma
_2n_2^2+\sigma_3n_3^2\right)^2
\\
&=(\sigma_1^2-\sigma_2^2)n_1^2+(\sigma_2^2-\sigma_3^2)n_2^2+\sigma_3^2-\left
[\left(\sigma_1-\sigma_3\right)n_1^2+\left(\sigma_2-\sigma_2\right)n_2^2+\sigma_
3\right]^2
\\ &=
(\sigma_1-\sigma_2)^2n_1^2n_2^2+(\sigma_2-\sigma_3)^2n_2^2n_3^2+(\sigma_1-\sigma
_3)^2n_1^2n_3^2
\\ \end{align} The maximum shear stress at a point in a continu
um
body is determined by maximizing \tau_\mathrm{n}^2 subject to the
condition that : n_in_i=n_1^2+n_2^2+n_3^2=1 This is a constrained
maximization problem, which can be solved using the Lagrangian
multiplier technique to convert the problem into an unconstrained
optimization problem. Thus, the stationary values (maximum and minimum
values)of \tau_\mathrm{n}^2 occur where the gradient of
\tau_\mathrm{n}^2 is parallel to the gradient of F. The Lagrangian
function for this problem can be written as :\begin{align}
F\left(n_1,n_2,n_3,\lambda\right) &= \tau^2 + \lambda
\left(g\left(n_1,n_2,n_3\right) - 1 \right) \\ &=
\sigma_1^2n_1^2+\sigma_2^2n_2^2+\sigma_3^2n_3^2-\left(\sigma_1n_1^2+\sigma_2n_2^
2+\sigma_3n_3^2\right)^2+\lambda\left(n_1^2+n_2^2+n_3^2-1\right)\\
\end{align} where \lambda is the Lagrangian multiplier (which is
different from the \lambda use to denote eigenvalues). The extreme
values of these functions are : \frac{\partial F}{\partial n_1}=0
\qquad \frac{\partial F}{\partial n_2}=0 \qquad \frac{\partial
F}{\partial n_3}=0 thence :\frac{\partial F}{\partial n_1} =
n_1\sigma_1^2-2n_1\sigma_1\left(\sigma_1 n_1^2+\sigma_2 n_2^2+\sigma_3
n_3^2\right)+\lambda n_1 = 0 :\frac{\partial F}{\partial n_2} =
n_2\sigma_2^2-2n_2\sigma_2\left(\sigma_1 n_1^2+\sigma_2 n_2^2+\sigma_3
n_3^2\right)+\lambda n_2 = 0 :\frac{\partial F}{\partial n_3} =
n_3\sigma_3^2-2n_3\sigma_3\left(\sigma_1 n_1^2+\sigma_2 n_2^2+\sigma_3
n_3^2\right)+\lambda n_3 = 0 These three equations together with the
condition n_in_i=1 may be solved for \lambda, n_1, n_2, and n_3 By
multiplying the first three equations by n_1,\,n_2, and n_3,
respectively, and knowing that \sigma_\mathrm{n} =
\sigma_{ij}n_in_j=\sigma_1n_1^2 + \sigma_2n_2^2 +\sigma_3n_3^2 we
obtain :n_1^2\sigma_1^2-2\sigma_1n_1^2\sigma_\mathrm{n}+n_1^2\lambda=0
:n_2^2\sigma_2^2-2\sigma_2n_2^2\sigma_\mathrm{n}+n_2^2\lambda=0
:n_3^2\sigma_3^2-2\sigma_1n_3^2\sigma_\mathrm{n}+n_3^2\lambda=0 Adding
these three equations we get :\begin{align}
\left[n_1^2\sigma_1^2+n_2^2\sigma_2^2+n_3^2\sigma_3^2\right]-2\left(\sigma_1n_1^
2+\sigma_2n_2^2+\sigma_3n_3^2\right)\sigma_\mathrm{n}+\lambda\left(n_1^2+n_2^2+n
_3^2\right)&=0
\\
\left[\tau_\mathrm{n}^2+\left(\sigma_1n_1^2+\sigma_2n_2^2+\sigma_3n_3^2\right)^2
\right]-2\sigma_\mathrm{n}^2+\lambda&=0
\\
\left[\tau_\mathrm{n}^2+\sigma_\mathrm{n}^2\right]-2\sigma_\mathrm{n}^2+\lambda
&=0 \\ \lambda &= \sigma_\mathrm{n}^2-\tau_\mathrm{n}^2
\end{align} this result can be substituted into each of the first
three equations to obtain :\begin{align} \frac{\partial F}{\partial
n_1} = n_1\sigma_1^2-2n_1\sigma_1\left(\sigma_1 n_1^2+\sigma_2
n_2^2+\sigma_3
n_3^2\right)+\left(\sigma_\mathrm{n}^2-\tau_\mathrm{n}^2\right) n_1
&= 0 \\
n_1\sigma_1^2-2n_1\sigma_1\sigma_\mathrm{n}+\left(\sigma_\mathrm{n}^2-\tau_\math
rm{n}^2\right)
n_1 &= 0 \\
\left(\sigma_1^2-2\sigma_1\sigma_\mathrm{n}+\sigma_\mathrm{n}^2-\tau_\mathrm{n}^
2\right)
n_1 &= 0 \\ \end{align} Doing the same for the other two equations
we have :\frac{\partial F}{\partial
n_2}=\left(\sigma_2^2-2\sigma_2\sigma_\mathrm{n}+\sigma_\mathrm{n}^2-\tau_\mathr
m{n}^2\right)
n_2 = 0 :\frac{\partial F}{\partial n_3}=
\left(\sigma_3^2-2\sigma_3\sigma_\mathrm{n}+\sigma_\mathrm{n}^2-\tau_\mathrm{n}^
2\right)
n_3= 0 A first approach to solve these last three equations is to
consider the trivial solution n_i=0. However, this option does not
fulfill the constraint n_in_i=1. Considering the solution where
n_1=n_2=0 and n_3 \neq 0, it is determine from the condition n_in_i=1
that n_3=\pm1, then from the original equation for \tau_\mathrm{n}^2
it is seen that \tau_\mathrm{n}=0. The other two possible values for
\tau_\mathrm{n} can be obtained similarly by assuming :n_1=n_3=0 and
n_2 \neq 0 :n_2=n_3=0 and n_1 \neq 0 Thus, one set of solutions for
these four equations is: :\begin{align}
n_1&=0,\,\,n_2&=0,\,\,n_3&=\pm1,\,\,\tau_\mathrm{n}&=0
\\
n_1&=0,\,\,n_2&=\pm1,\,\,n_3&=0,\,\,\tau_\mathrm{n}&=0
\\
n_1&=\pm1,\,\,n_2&=0,\,\,n_3&=0,\,\,\tau_\mathrm{n}&=0
\end{align} These correspond to minimum values for \tau_\mathrm{n} and
verifies that there are no shear stresses on planes normal to the
principal directions of stress, as shown previously. A second set of
solutions is obtained by assuming n_1=0,\, n_2\neq0 and n_3 \neq 0.
Thus we have :\frac{\partial F}{\partial n_2}=
\sigma_2^2-2\sigma_2\sigma_\mathrm{n}+\sigma_\mathrm{n}^2-\tau_\mathrm{n}^2
= 0 :\frac{\partial F}{\partial
n_3}=\sigma_3^2-2\sigma_3\sigma_\mathrm{n}+\sigma_\mathrm{n}^2-\tau_\mathrm{n}^2
= 0 To find the values for n_2 and n_3 we first add these two
equations :\begin{align}
\sigma_2^2-\sigma_3^2-2\sigma_2\sigma_n+2\sigma_3\sigma_n&=0 \\
\sigma_2^2-\sigma_3^2-2\sigma_n\left(\sigma_2-\sigma_3\right)&=0
\\ \sigma_2+\sigma_3&=2\sigma_n \end{align} Knowing that for
n_1=0
:\sigma_\mathrm{n} = \sigma_1n_1^2+\sigma_2n_2^2 +
\sigma_3n_3^2=\sigma_2n_2^2 + \sigma_3n_3^2 and
:n_1^2+n_2^2+n_3^2=n_2^2+n_3^2=1 we have :\begin{align}
\sigma_2+\sigma_3&=2\sigma_n \\
\sigma_2+\sigma_3&=2\left(\sigma_2n_2^2 + \sigma_3n_3^2\right) \\
\sigma_2+\sigma_3&=2\left(\sigma_2n_2^2 +
\sigma_3\left(1-n_2^2\right)\right)&=0 \end{align} and solving for
n_2 we have :n_2=\pm\frac{1}{\sqrt 2} Then solving for n_3 we have
:n_3=\sqrt{1-n_2^2}=\pm\frac{1}{\sqrt 2} and :\begin{align}
\tau_\mathrm{n}^2&=(\sigma_2-\sigma_3)^2n_2^2n_3^2 \\
\tau_\mathrm{n}&=\frac{\sigma_2-\sigma_3}{2}\end{align} The other
two possible values for \tau_\mathrm{n} can be obtained similarly by
assuming :n_2=0,\, n_1\neq0 and n_3 \neq 0 :n_3=0,\, n_1\neq0 and n
_2
\neq 0 Therefore, the second set of solutions for \frac{\partial
F}{\partial n_1}=0, representing a maximum for \tau_\mathrm{n} is
:n_1=0,\,\,n_2=\pm\frac{1}{\sqrt 2},\,\,n_3=\pm\frac{1}{\sqrt
2},\,\,\tau_\mathrm{n}=\pm\frac{\sigma_2-\sigma_3}{2}
:n_1=\pm\frac{1}{\sqrt 2},\,\,n_2=0,\,\,n_3=\pm\frac{1}{\sqrt
2},\,\,\tau_\mathrm{n}=\pm\frac{\sigma_1-\sigma_3}{2}
:n_1=\pm\frac{1}{\sqrt 2},\,\,n_2=\pm\frac{1}{\sqrt
2},\,\,n_3=0,\,\,\tau_\mathrm{n}=\pm\frac{\sigma_1-\sigma_2}{2}
Therefore, assuming \sigma_1\ge\sigma_2\ge\sigma_3, the maximum shear
stress is expressed by
:\tau_\mathrm{max}=\frac{1}{2}\left|\sigma_1-\sigma_3\right|=\frac{1}{2}\left|\s
igma_\mathrm{max}-\sigma_\mathrm{min}\right|
and it can be stated as being equal to one-half the difference between
the largest and smallest principal stresses, acting on the plane that
bisects the angle between the directions of the largest and smallest
principal stresses.
Stress deviator tensor
======================================================================
The stress tensor \sigma_{ij} can be expressed as the sum of two other
stress tensors:
# a 'mean hydrostatic stress tensor' or 'volumetric stress tensor' or
'mean normal stress tensor', \pi\delta_{ij}, which tends to change the
volume of the stressed body; and
# a deviatoric component called the 'stress deviator tensor', s_{ij},
which tends to distort it.
So:
:\sigma_{ij}= s_{ij} + \pi\delta_{ij},\,
where \pi is the mean stress given by
:\pi=\frac{\sigma_{kk}}{3}=\frac{\sigma_{11}+\sigma_{22}+\sigma_{33}}{3}=\tfrac{
1}{3}I_1.\,
Pressure (p) is generally defined as negative one-third the trace of
the stress tensor minus any stress the divergence of the velocity
contributes with, i.e.
:p = \lambda\, \nabla\cdot\vec{u} -\pi = \lambda\,\frac{\partial
u_k}{\partial x_k} -\pi = \sum_k\lambda\,\frac{\partial u_k}{\partial
x_k} -\pi,
where \lambda is a proportionality constant, \nabla is the divergence
operator, x_k is the 'k':th Cartesian coordinate, \vec{u} is the
velocity and u_k is the 'k':th Cartesian component of \vec{u}.
The deviatoric stress tensor can be obtained by subtracting the
hydrostatic stress tensor from the Cauchy stress tensor:
:\begin{align}
\ s_{ij} &= \sigma_{ij} - \frac{\sigma_{kk}}{3}\delta_{ij},\,\\
\left[{\begin{matrix}
s_{11} & s_{12} & s_{13} \\
s_{21} & s_{22} & s_{23} \\
s_{31} & s_{32} & s_{33}
\end{matrix}}\right]
&=\left[{\begin{matrix}
\sigma_{11} & \sigma_{12} & \sigma_{13} \\
\sigma_{21} & \sigma_{22} & \sigma_{23} \\
\sigma_{31} & \sigma_{32} & \sigma_{33}
\end{matrix}}\right]-\left[{\begin{matrix}
\pi & 0 & 0 \\
0 & \pi & 0 \\
0 & 0 & \pi
\end{matrix}}\right] \\
&=\left[{\begin{matrix}
\sigma_{11}-\pi & \sigma_{12} & \sigma_{13} \\
\sigma_{21} & \sigma_{22}-\pi & \sigma_{23} \\
\sigma_{31} & \sigma_{32} & \sigma_{33}-\pi
\end{matrix}}\right].
\end{align}
Invariants of the stress deviator tensor
==========================================
As it is a second order tensor, the stress deviator tensor also has a
set of invariants, which can be obtained using the same procedure used
to calculate the invariants of the stress tensor. It can be shown that
the principal directions of the stress deviator tensor s_{ij} are the
same as the principal directions of the stress tensor \sigma_{ij}.
Thus, the characteristic equation is
:\left| s_{ij}- \lambda\delta_{ij} \right| =
-\lambda^3+J_1\lambda^2-J_2\lambda+J_3=0,
where J_1, J_2 and J_3 are the first, second, and third 'deviatoric
stress invariants', respectively. Their values are the same
(invariant) regardless of the orientation of the coordinate system
chosen. These deviatoric stress invariants can be expressed as a
function of the components of s_{ij} or its principal values s_1, s_2,
and s_3, or alternatively, as a function of \sigma_{ij} or its
principal values \sigma_1, \sigma_2, and \sigma_3. Thus,
:\begin{align}
J_1 &= s_{kk}=0,\, \\
J_2 &= \textstyle{\frac{1}{2}}s_{ij}s_{ji} =
\tfrac{1}{2}\text{tr}(\boldsymbol{s}^2)\\
&= \tfrac{1}{2}(s_1^2 + s_2^2 + s_3^2) \\
&= \tfrac{1}{6}\left[(\sigma_{11} - \sigma_{22})^2 + (\sigma_{22}
- \sigma_{33})^2 + (\sigma_{33} - \sigma_{11})^2 \right ] +
\sigma_{12}^2 + \sigma_{23}^2 + \sigma_{31}^2 \\
&= \tfrac{1}{6}\left[(\sigma_1 - \sigma_2)^2 + (\sigma_2 -
\sigma_3)^2 + (\sigma_3 - \sigma_1)^2 \right ] \\
&= \tfrac{1}{3}I_1^2-I_2 =
\frac{1}{2}\left[\text{tr}(\boldsymbol{\sigma}^2) -
\frac{1}{3}\text{tr}(\boldsymbol{\sigma})^2\right],\,\\
J_3 &= \det(s_{ij}) \\
&= \tfrac{1}{3}s_{ij}s_{jk}s_{ki} = \tfrac{1}{3}
\text{tr}(\boldsymbol{s}^3)\\
&= \tfrac{1}{3}(s_1^3 + s_2^3 + s_3^3) \\
&= s_1s_2s_3 \\
&= \tfrac{2}{27}I_1^3 - \tfrac{1}{3}I_1 I_2 + I_3 =
\tfrac{1}{3}\left[\text{tr}(\boldsymbol{\sigma}^3) -
\text{tr}(\boldsymbol{\sigma}^2)\text{tr}(\boldsymbol{\sigma})
+\tfrac{2}{9}\text{tr}(\boldsymbol{\sigma})^3\right].\,
\end{align}
Because s_{kk}=0, the stress deviator tensor is in a state of pure
shear.
A quantity called the equivalent stress or von Mises stress is
commonly used in solid mechanics. The equivalent stress is defined as
:\sigma_\mathrm {vm} = \sqrt{3~J_2} =
\sqrt{\tfrac{1}{2}~\left[(\sigma_1-\sigma_2)^2 + (\sigma_2-\sigma_3)^2
+ (\sigma_3-\sigma_1)^2 \right]}
\,.
Octahedral stresses
======================================================================
Considering the principal directions as the coordinate axes, a plane
whose normal vector makes equal angles with each of the principal axes
(i.e. having direction cosines equal to |1/\sqrt{3}|) is called an
'octahedral plane'. There are a total of eight octahedral planes
(Figure 6). The normal and shear components of the stress tensor on
these planes are called 'octahedral normal stress' \sigma_\mathrm{oct}
and 'octahedral shear stress' \tau_\mathrm{oct}, respectively.
Octahedral plane passing through the origin is known as the '�-plane'
('�' not to be confused with 'mean stress' denoted by '�' in above
section) '.' On the '�-plane', s_{ij} = I/3.
Knowing that the stress tensor of point O (Figure 6) in the principal
axes is
:\sigma_{ij}=
\begin{bmatrix}
\sigma_1 & 0 & 0\\
0 & \sigma_2 & 0\\
0 & 0 & \sigma_3
\end{bmatrix}
the stress vector on an octahedral plane is then given by:
:\begin{align}
\mathbf{T}_\mathrm{oct}^{(\mathbf{n})}&=
\sigma_{ij}n_i\mathbf{e}_j \\
&=\sigma_1n_1\mathbf{e}_1+\sigma_2n_2\mathbf{e}_2+\sigma_3n_3\mathbf{e}_3\\
&=\tfrac{1}{\sqrt{3}}(\sigma_1\mathbf{e}_1+\sigma_2\mathbf{e}_2+\sigma_3\mat
hbf{e}_3)
\end{align}
The normal component of the stress vector at point O associated with
the octahedral plane is
:\begin{align}
\sigma_\mathrm{oct} &= T^{(n)}_in_i \\
&=\sigma_{ij}n_in_j \\
&=\sigma_1n_1n_1+\sigma_2n_2n_2+\sigma_3n_3n_3 \\
&=\tfrac{1}{3}(\sigma_1+\sigma_2+\sigma_3)=\tfrac{1}{3}I_1
\end{align}
which is the mean normal stress or hydrostatic stress. This value is
the same in all eight octahedral planes.
The shear stress on the octahedral plane is then
:\begin{align}
\tau_\mathrm{oct}
&=\sqrt{T_i^{(n)}T_i^{(n)}-\sigma_\mathrm{oct}^2} \\
&=\left[\tfrac{1}{3}(\sigma_1^2+\sigma_2^2+\sigma_3^2)-\tfrac{1}{9}(\sigma_1
+\sigma_2+\sigma_3)^2\right]^{1/2}
\\
&=\tfrac{1}{3}\left[(\sigma_1-\sigma_2)^2+(\sigma_2-\sigma_3)^2+(\sigma_3-\s
igma_1)^2\right]^{1/2}
= \tfrac{1}{3}\sqrt{2I_1^2-6I_2} = \sqrt{\tfrac{2}{3}J_2}
\end{align}
See also
======================================================================
* Critical plane analysis
References
======================================================================
Keith D. Hjelmstad (2005),
[
https://books.google.ca/books?id=LVTYjmcdvPwC&pg=PA103
"Fundamentals of Structural Mechanics"] (2nd edition). Prentice-Hall.
Teodor M. Atanackovic and Ardéshir Guran (2000),
[
https://books.google.ca/books?id=uQrBWdcGmjUC&pg=PA1 "Theory of
Elasticity for Scientists and Engineers"]. Springer.
G. Thomas Mase and George E. Mase (1999),
[
https://books.google.ca/books?id=uI1ll0A8B_UC&pg=PA47 "Continuum
Mechanics for Engineers"] (2nd edition). CRC Press.
Peter Chadwick (1999),
[
https://books.google.ca/books?id=QSXIHQsus6UC&pg=PA95 "Continuum
Mechanics: Concise Theory and Problems"]. Dover Publications, series
"Books on Physics". . pages
Yuan-cheng Fung and Pin Tong (2001)
[
https://books.google.ca/books?id=hmyiIiiv4FUC&pg=PA66 "Classical
and Computational Solid Mechanics"]. World Scientific.
I-Shih Liu (2002),
[
https://books.google.com/books?id=-gWqM4uMV6wC&pg=PA43 "Continuum
Mechanics"]. Springer
Fridtjov Irgens (2008),
[
https://books.google.com/books?id=q5dB7Gf4bIoC&pg=PA46 "Continuum
Mechanics"]. Springer.
Wai-Fah Chen and Da-Jian Han (2007),
[
https://books.google.com/books?id=E8jptvNgADYC&pg=PA46
"Plasticity for Structural Engineers"]. J. Ross Publishing
Bernard Hamrock (2005),
[
https://books.google.com/books?id=jT1XPwAACAAJ "Fundamentals of
Machine Elements"]. McGraw-Hill.
Han-Chin Wu (2005),
[
https://books.google.com/books?id=OS4mICsHG3sC&pg=PA45 "Continuum
Mechanics and Plasticity"]. CRC Press.
Rabindranath Chatterjee (1999),
[
https://books.google.com/books?id=v2F84PwH0BkC&pg=PA111
"Mathematical Theory of Continuum Mechanics"]. Alpha Science.
John Conrad Jaeger, N. G. W. Cook, and R. W. Zimmerman (2007),
[
https://books.google.com/books?id=FqADDkunVNAC&pg=PA10
"Fundamentals of Rock Mechanics"] (4th edition). Wiley-Blackwell.
Mohammed Ameen (2005),
[
https://books.google.ca/books?id=Gl9cFyLrdrcC&pg=PA33
"Computational Elasticity: Theory of Elasticity and Finite and
Boundary Element Methods"] (book). Alpha Science,
William Prager (2004),
[
https://books.google.ca/books?id=Feer6-hn9zsC&pg=PA43
"Introduction to Mechanics of Continua"]. Dover Publications.
Teodor M. Atanackovic and Ardéshir Guran (2000),
[
https://books.google.ca/books?id=uQrBWdcGmjUC&pg=PA1 "Theory of
Elasticity for Scientists and Engineers"]. Springer.
Wai-Fah Chen and Da-Jian Han (2007),
[
https://books.google.com/books?id=E8jptvNgADYC&pg=PA46
"Plasticity for Structural Engineers"]. J. Ross Publishing
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Original Article:
http://en.wikipedia.org/wiki/Cauchy_stress_tensor
