Definitions

# Stress (physics)

Stress is a measure of the average amount of force exerted per unit area. It is a measure of the intensity of the total internal forces acting within a body across imaginary internal surfaces, as a reaction to external applied forces and body forces. It was introduced into the theory of elasticity by Cauchy around 1822. Stress is a concept that is based on the concept of continuum. In general, stress is expressed as

$sigma = frac\left\{F\right\}\left\{A\right\} ,$

where

$sigma$ is the average stress, also called engineering or nominal stress, and
$F$ is the force acting over the area $A$.

The SI unit for stress is the pascal (symbol Pa), which is a shorthand name for one newton (Force) per square metre (Unit Area). The unit for stress is the same as that of pressure, which is also a measure of Force per unit area. Engineering quantities are usually measured in megapascals (MPa) or gigapascals (GPa). In Imperial units, stress is expressed in pounds-force per square inch (psi) or kilopounds-force per square inch (ksi).

As with force, stress cannot be measured directly but is usually inferred from measurements of strain and knowledge of elastic properties of the material. Devices capable of measuring stress indirectly in this way are strain gauges and piezoresistors.

## Stress as a tensor

In its full form, linear stress is a rank-two tensor quantity, and may be represented as a 3x3 matrix. A tensor may be seen as a linear vector operator - it takes a given vector and produces another vector as a result. In the case of the stress tensor $sigma_\left\{ij\right\}$, it takes the vector normal to any area element and yields the force (or "traction") acting on that area element. In matrix notation:

$F_i=sum_\left\{j=1\right\}^3 sigma_\left\{ij\right\} A_j$

where $A_j$ are the components of the vector normal to a surface area element with a length equal to the area of the surface element, and $F_i$ are the components of the force vector (or traction vector) acting on that element. Using index notation, we can eliminate the summation sign, since all sums will be the same over repeated indices. Thus:

$F_i=sigma_\left\{ij\right\} A_j ,$

Just as it is the case with a vector (which is actually a rank-one tensor), the matrix components of a tensor depend upon the particular coordinate system chosen. As with a vector, there are certain invariants associated with the stress tensor, whose value does not depend upon the coordinate system chosen (or the area element upon which the stress tensor operates). For a vector, there is only one invariant - the length. For a tensor, there are three - the eigenvalues of the stress tensor, which are called the principal stresses. It is important to note that the only physically significant parameters of the stress tensor are its invariants, since they are not dependent upon the choice of the coordinate system used to describe the tensor.

If we choose a particular surface area element, we may divide the force vector by the area (stress vector) and decompose it into two parts: a normal component acting normal to the stressed surface, and a shear component, acting parallel to the stressed surface. An axial stress is a normal stress produced when a force acts parallel to the major axis of a body, e.g. column. If the forces pull the body producing an elongation, the axial stress is termed tensile stress. If on the other hand the forces push the body reducing its length, the axial stress is termed compressive stress. Bending stresses, e.g. produced on a bent beam, are a combination of tensile and compressive stresses. Torsional stresses, e.g. produced on twisted shafts, are shearing stresses.

In the above description, little distinction is drawn between the "stress" and the "stress vector" since the body which is being stressed provides a particular coordinate system in which to discuss the effects of the stress. The distinction between "normal" and "shear" stresses is slightly different when considered independently of any coordinate system. The stress tensor yields a stress vector for a surface area element at any orientation, and this stress vector may be decomposed into normal and shear components. The normal part of the stress vector averaged over all orientations of the surface element yields an invariant value, and is known as the hydrostatic pressure. Mathematically it is equal to the average value of the principal stresses (or, equivalently, the trace of the stress tensor divided by three). The normal stress tensor is then the product of the hydrostatic pressure and the unit tensor. Subtracting the normal stress tensor from the stress tensor gives what may be called the shear tensor. These two quantities are true tensors with physical significance, and their nature is independent of any coordinate system chosen to describe them. In fact, the extended Hooke's law is basically the statement that each of these two tensors is proportional to its strain tensor counterpart, and the two constants of proportionality (elastic moduli) are independent of each other. Note that In rheology, the normal stress tensor is called extensional stress, and in acoustics is called longitudinal stress.

Solids, liquids and gases have stress fields. Static fluids support normal stress but will flow under shear stress. Moving viscous fluids can support shear stress (dynamic pressure). Solids can support both shear and normal stress, with ductile materials failing under shear and brittle materials failing under normal stress. All materials have temperature dependent variations in stress related properties, and non-newtonian materials have rate-dependent variations.

## Cauchy's stress principle

Cauchy's stress principle asserts that when a continuum body is acted on by forces, i.e. surface forces and body forces, there are internal reactions (forces) throughout the body acting between the material points. Based on this principle, Cauchy demonstrated that the state of stress at a point in a body is completely defined by the nine components $sigma_\left\{ij\right\}$ of a second-order Cartesian tensor called the Cauchy stress tensor, given by

$begin\left\{align\right\}$
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] end{align}

where

$mathbf\left\{T\right\}^\left\{\left(mathbf\left\{e\right\}_1\right)\right\}$, $mathbf\left\{T\right\}^\left\{\left(mathbf\left\{e\right\}_2\right)\right\}$, and $mathbf\left\{T\right\}^\left\{\left(mathbf\left\{e\right\}_3\right)\right\}$ are the stress vectors associated with the planes perpendicular to the coordinate axis,
$sigma_\left\{11\right\}$, $sigma_\left\{22\right\}$, and $sigma_\left\{33\right\}$ are normal stresses, and
$sigma_\left\{12\right\}$, $sigma_\left\{13\right\}$, $sigma_\left\{21\right\}$, $sigma_\left\{23\right\}$, $sigma_\left\{31\right\}$, and $sigma_\left\{32\right\}$ are shear stresses.

The Voigt notation representation of the Cauchy stress tensor takes advantage of the symmetry of the stress tensor to express the stress as a 6-dimensional vector of the form

$begin\left\{align\right\}$
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_{31} & sigma_{12} end{bmatrix}^T end{align} The Voigt notation is used extensively in representing stress-strain relations in solid mechanics and for computational efficiency in numerical structural mechanics software.

Derivation of Cauchy's stress tensor
Considering a body subjected to surface forces $mathbf\left\{F\right\}$ and body forces $mathbf\left\{f\right\}$ per unit of volume, with an imaginary plane dividing the body into two segments (Figure 1). A small area $Delta A ,$ in one of the segments, passing through a point $P ,$, and with a normal vector $mathbf\left\{n\right\} ,$ is acted upon by a force $Delta F ,$ resulting from the action of the material in one side of the area (right segment) onto the other side (left segment). The distribution of force on $Delta A ,$ is, however, not always uniform, as there may be a moment $Delta M ,$ at $P ,$ due to the force $Delta F ,$, as shown in the Figure. As $Delta A ,$ becomes very small and tends to zero the ratio $Delta F / Delta A ,$ becomes $dF/dA$, and the moment $Delta M$ vanishes. The vector $dF/dA ,$ is defined as the stress vector $mathbf\left\{T\right\}^\left\{\left(mathbf\left\{n\right\}\right)\right\} ,$ at point $P ,$ associated with a plane with a normal vector $mathbf\left\{n\right\} ,$:
$mathbf\left\{T\right\}^\left\{\left(mathbf\left\{n\right\}\right)\right\}= lim_\left\{Delta A to 0\right\} frac \left\{Delta F\right\}\left\{Delta A\right\} = \left\{dF over dA\right\} ,$

By Newton's third law, the stress vectors acting upon opposite sides of the same surface are equal in magnitude and of opposite direction. Thus,

$- mathbf\left\{T\right\}^\left\{\left(mathbf\left\{n\right\}\right)\right\}= mathbf\left\{T\right\}^\left\{\left(- mathbf\left\{n\right\}\right)\right\}$

The stress vector, not necessarily being perpendicular to the plane on which it acts, can be resolved into two components: one normal to the plane, called normal stress, and the other parallel to this plane, called the shearing stress. The latter, can be further decomposed into two mutually perpendicular vectors.

The state of stress at a point would be defined by all the stress vectors $mathbf\left\{T\right\}^\left\{\left(mathbf\left\{n\right\}\right)\right\} ,$ associated with all planes (infinite number of planes) that pass through that point. However, by just knowing the stress vectors on three mutually perpendicular planes, the stress vector on any plane passing through that point can be found through coordinate transformation equations. Assuming a material element (Figure 2) with planes perpendicular to the coordinate axes of a cartesian coordinate system, the stress vectors associated with each of the element planes, i.e. $mathbf\left\{T\right\}^\left\{\left(mathbf\left\{e\right\}_1\right)\right\}$, $mathbf\left\{T\right\}^\left\{\left(mathbf\left\{e\right\}_2\right)\right\}$, and $mathbf\left\{T\right\}^\left\{\left(mathbf\left\{e\right\}_3\right)\right\}$ can be decomposed into components in the direction of the three coordinate axes:

$mathbf\left\{T\right\}^\left\{\left(mathbf\left\{e\right\}_1\right)\right\}= sigma_\left\{11\right\} mathbf\left\{e\right\}_1 + sigma_\left\{12\right\} mathbf\left\{e\right\}_2 + sigma_\left\{13\right\} mathbf\left\{e\right\}_3 ,$
$mathbf\left\{T\right\}^\left\{\left(mathbf\left\{e\right\}_2\right)\right\}= sigma_\left\{21\right\} mathbf\left\{e\right\}_1 + sigma_\left\{22\right\} mathbf\left\{e\right\}_2 + sigma_\left\{23\right\} mathbf\left\{e\right\}_3 ,$
$mathbf\left\{T\right\}^\left\{\left(mathbf\left\{e\right\}_3\right)\right\}= sigma_\left\{31\right\} mathbf\left\{e\right\}_1 + sigma_\left\{32\right\} mathbf\left\{e\right\}_2 + sigma_\left\{33\right\} mathbf\left\{e\right\}_3 ,$

In index notation this is

$mathbf\left\{T\right\}^\left\{\left(mathbf\left\{e\right\}_i\right)\right\}= sigma_\left\{ij\right\} mathbf\left\{e\right\}_j$

The nine components $sigma_\left\{ij\right\}$ of the stress vectors are the components of a second-order Cartesian tensor called the Cauchy stress tensor, which completely defines the state of stresses at a point and it is given by

$sigma_\left\{ij\right\}= left\left[\left\{begin\left\{matrix\right\} mathbf\left\{T\right\}^\left\{\left(mathbf\left\{e\right\}_1\right)\right\}$
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

$sigma_\left\{11\right\}$, $sigma_\left\{22\right\}$, and $sigma_\left\{33\right\}$ are normal stresses, and
$sigma_\left\{12\right\}$, $sigma_\left\{13\right\}$, $sigma_\left\{21\right\}$, $sigma_\left\{23\right\}$, $sigma_\left\{31\right\}$, and $sigma_\left\{32\right\}$ are shear stresses.

The first index $i ,$ indicates the stress acts on a plane normal to the $x_i ,$ axis, and the second index $j ,$ denotes the direction in which the stress acts. 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.

### Relationship stress vector - stress tensor

The stress vector $mathbf\left\{T\right\}^\left\{\left(mathbf\left\{n\right\}\right)\right\} ,$ at any point associated with a plane of normal vector $mathbf\left\{n\right\}$ 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 of the stress tensor $sigma_\left\{ij\right\}$. In tensor form this is:

$T_j^\left\{\left(n\right)\right\}= sigma_\left\{ij\right\}n_i$

Derivation of the stress vector as a function of the stress tensor
For this, we consider a tetrahedron with three faces oriented in the coordinate planes, and with an infinitesimal area $dA$ oriented in an arbitrary direction specified by a normal vector $mathbf\left\{n\right\}$ (Figure 3). The stress vector on this plane is denoted by $mathbf\left\{T\right\}^\left\{\left(mathbf\left\{n\right\}\right)\right\}$. The stress vectors acting on the faces of the tetrahedron are denoted as $mathbf\left\{T\right\}^\left\{\left(mathbf\left\{e\right\}_1\right)\right\}$, $mathbf\left\{T\right\}^\left\{\left(mathbf\left\{e\right\}_2\right)\right\}$, and $mathbf\left\{T\right\}^\left\{\left(mathbf\left\{e\right\}_3\right)\right\}$, and are by definition the components of the stress tensor $sigma_\left\{ij\right\}$. From equilibrium of forces, i.e. Newton's second law, we have

$mathbf\left\{T\right\}^\left\{\left(mathbf\left\{n\right\}\right)\right\}dA - mathbf\left\{T\right\}^\left\{\left(mathbf\left\{e\right\}_1\right)\right\}dA_1 - mathbf\left\{T\right\}^\left\{\left(mathbf\left\{e\right\}_2\right)\right\}dA_2 - mathbf\left\{T\right\}^\left\{\left(mathbf\left\{e\right\}_3\right)\right\}dA_3 = rho left\left(frac\left\{h\right\}\left\{3\right\}ds right\right) mathbf\left\{a\right\}$

where the right hand side of the equation represent the body forces acting on the tetrahedron: $rho$ is the density, $mathbf\left\{a\right\} ,$ is the acceleration, and $h ,$ is the height of the tetrahedron, considering the plane $mathbf\left\{n\right\}$ as the base. The area of the faces of the tetrahedron perpendicular to the axes can be found by projecting $dA$ into each face (dot product):

$dA_1= left\left(mathbf\left\{n\right\} cdot mathbf\left\{e\right\}_1 right\right)dA = n_1dA ,$
$dA_2= left\left(mathbf\left\{n\right\} cdot mathbf\left\{e\right\}_2 right\right)dA = n_2dA ,$
$dA_3= left\left(mathbf\left\{n\right\} cdot mathbf\left\{e\right\}_3 right\right)dA = n_3dA ,$

Thus, taking the limit when $h to 0 ,$ and replacing the previous equations, we have

$begin\left\{align\right\} mathbf\left\{T\right\}^\left\{\left(mathbf\left\{n\right\}\right)\right\} &= mathbf\left\{T\right\}^\left\{\left(mathbf\left\{e\right\}_1\right)\right\}n_1 + mathbf\left\{T\right\}^\left\{\left(mathbf\left\{e\right\}_2\right)\right\}n_2 + mathbf\left\{T\right\}^\left\{\left(mathbf\left\{e\right\}_3\right)\right\}n_3$
& = sum_{i=1}^3 mathbf{T}^{(mathbf{e}_i)}n_i &= left(sigma_{ij}mathbf{e}_j right)n_i &= sigma_{ij}n_imathbf{e}_j end{align}

or, equivalently,

$T_j^\left\{\left(n\right)\right\}= sigma_\left\{ij\right\}n_i$

In matrix form we have

$left\left[\left\{begin\left\{matrix\right\} T^\left\{\left(n\right)\right\}_1 & T^\left\{\left(n\right)\right\}_2 & T^\left\{\left(n\right)\right\}_3end\left\{matrix\right\}\right\}right\right]=left\left[\left\{begin\left\{matrix\right\} n_1 & n_2 & n_3 end\left\{matrix\right\}\right\}right\right]cdot left\left[\left\{begin\left\{matrix\right\} sigma _\left\{11\right\} & sigma _\left\{12\right\} & sigma _\left\{13\right\} sigma _\left\{21\right\} & sigma _\left\{22\right\} & sigma _\left\{23\right\} sigma _\left\{31\right\} & sigma _\left\{32\right\} & sigma _\left\{33\right\} end\left\{matrix\right\}\right\}right\right]$

This equation expresses the components of the stress vector acting on an arbitrary plane with normal vector $mathbf\left\{n\right\}$ at a given point in terms of the components of the stress tensor, $sigma_\left\{ij\right\}$, at that point.

### Transformation rule of the stress tensor

It can be shown that the stress tensor is a second order tensor; this is, under a change of the coordinate system, from an $x_i$ system to an $x^\text{'}_i$ system, the components $sigma_\left\{ij\right\}$ in the initial system are transformed into the components $sigma^\text{'}_\left\{ij\right\}$ in the new system according to the tensor transformation rule:

$sigma^\text{'}_\left\{ij\right\}=a_\left\{im\right\}a_\left\{jn\right\}sigma_\left\{mn\right\}$

where $a_\left\{ij\right\} ,$ is a rotation matrix. In matrix form this is

$left\left[\left\{begin\left\{matrix\right\}$
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]

An easy visualization of this transformation for 2D and 3D stresses for simple rotations is Mohr's circle

### Normal and shear stresses

The magnitude of the normal stress component, $sigma_n$, of any stress vector $mathbf\left\{T\right\}^\left\{\left(mathbf\left\{n\right\}\right)\right\}$ acting on an arbitrary plane with normal vector $mathbf\left\{n\right\}$ at a given point in terms of the component of the stress tensor $sigma_\left\{ij\right\}$ is the dot product of the stress vector and the normal vector, thus

$begin\left\{align\right\}$
sigma_n &= mathbf{T}^{(mathbf{n})}cdot mathbf{n} &=T^{(n)}_in_i &=sigma_{ij}n_in_j end{align}

The magnitude of the shear stress component, $tau_n$, can then be found using the Pythagorean theorem, thus

$begin\left\{align\right\}$
tau_n &=sqrt{(T^{(n)})^2-sigma_n^2} &= sqrt{T^{(n)}T^{(n)}-sigma_n^2}

end{align}

## Equilibrium equations and symmetry of the stress tensor

When a body is in equilibrium the components of the stress tensor in every point of the body satisfy the equilibrium equations,

$$
sigma_{ji,j}+ F_i = 0

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 $T_i^\left\{\left(n\right)\right\}$ 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:
$int_S T_i^\left\{\left(n\right)\right\}dS + int_V F_i dV = 0$
By definition the stress vector is $T_i^\left\{\left(n\right)\right\} =sigma_\left\{ji\right\}n_j$, then
$int_S sigma_\left\{ji\right\}n_j , dS + int_V F_i , dV = 0$
Using the Gauss's divergence theorem to convert a surface integral to a volume integral gives
$int_V sigma_\left\{ji,j\right\} , dV + int_V F_i , dV = 0$
$int_V sigma_\left\{ji,j\right\} + F_i , dV = 0$
For an arbitrary volume the integrand vanishes, and we have the equilibrium equations
$sigma_\left\{ji,j\right\} + F_i = 0$

At the same time, 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, i.e.

$sigma_\left\{ij\right\}=sigma_\left\{ji\right\}$

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,
$begin\left\{align\right\}$
M_O &=int_S (mathbf{r}timesmathbf{T})dS + int_V (mathbf{r}timesmathbf{F})dV=0 0 &= int_Svarepsilon_{ijk}x_jT_k^{(n)}dS + int_Vvarepsilon_{ijk}x_jF_k dV end{align} where $mathbf\left\{r\right\}$ is the position vector and is expressed as

$mathbf\left\{r\right\}=x_jmathbf\left\{e\right\}_j$

Knowing that $T_k^\left\{\left(n\right)\right\} =sigma_\left\{mk\right\}n_m$ and using Gauss's divergence theorem to change from a surface integral to a volume integral, we have

$begin\left\{align\right\}$
0 &= int_S varepsilon_{ijk}x_jsigma_{mk}n_m , dS + int_Vvarepsilon_{ijk}x_jF_k , dV &= int_V (varepsilon_{ijk}x_jsigma_{mk})_{,m} dV + int_Vvarepsilon_{ijk}x_jF_k , dV &= int_V (varepsilon_{ijk}x_{j,m}sigma_{mk}+varepsilon_{ijk}x_jsigma_{mk,m}) dV + int_Vvarepsilon_{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 second integral is zero as it contains the equilibrium equations. This leaves the first integral, where $x_\left\{j,m\right\}=delta_\left\{jm\right\}$, therefore
$int_V \left(varepsilon_\left\{ijk\right\}sigma_\left\{jk\right\}\right) dV=0$

For an arbitrary volume V, we then have

$varepsilon_\left\{ijk\right\}sigma_\left\{jk\right\}=0$

which is satisfied at every point within the body. Expanding this equation we have

$sigma_\left\{12\right\}=sigma_\left\{21\right\}$, $sigma_\left\{23\right\}=sigma_\left\{32\right\}$, and $sigma_\left\{13\right\}=sigma_\left\{31\right\}$

or in general

$sigma_\left\{ij\right\}=sigma_\left\{ji\right\}$

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, $K_\left\{n\right\}rightarrow 1$, e.g. Non-Newtonian fluid, which can lead to rotationally non-invariant fluids, such as polymers.

## Principal stresses and stress invariants

The components $sigma_\left\{ij\right\}$ 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 length of the vector is a physical quantity (a scalar) and is independent of the coordinate system chosen to represent the vector. 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. When the coordinate system is chosen to coincide with the eigenvectors of the stress tensor, the stress tensor is represented by a diagonal matrix:

$sigma_\left\{ij\right\}=$
begin{bmatrix} sigma_1 & 0 & 0 0 & sigma_2 & 0 0 & 0 & sigma_3 end{bmatrix}

where $sigma_1$, $sigma_2$, and $sigma_3$, are the principal stresses. These principal stresses may be combined to form three other commonly used invariants, $I_1$, $I_2$, and $I_3$ , which are the first, second and third stress invariants, respectively. The first and third invariant are the trace and determinant respectively, of the stress tensor. Thus, we have

$begin\left\{align\right\}$
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, working and thinking in the principal coordinate system is often very useful when considering the state of the elastic medium at a particular point.

Derivation of principal stresses and stress invariants
At every point in a stressed body there are at least three planes, called principal planes, with normal vectors $mathbf\left\{n\right\}$, called principal directions, where the corresponding stress vector is parallel or in the same direction as the normal vector $sigma_n$ and where there are no normal shear stresses $tau_n$. Thus,
$mathbf\left\{T\right\}^\left\{\left(mathbf\left\{n\right\}\right)\right\} = lambda mathbf\left\{n\right\}= mathbf\left\{sigma\right\}_n mathbf\left\{n\right\}$

where $lambda$ is a constant of proportionality, and in this particular case corresponds to the magnitudes $sigma_n$ of the normal stress vectors or principal stresses.

Knowing that $T_i^\left\{\left(n\right)\right\}=sigma_\left\{ij\right\}n_j$ and $n_i=delta_\left\{ij\right\}n_j$, we have

$begin\left\{align\right\}$
T_i^{(n)} &= lambda n_i sigma_{ij}n_j &=lambda n_i sigma_{ij}n_j -lambda n_i &=0 left(sigma_{ij}- lambdadelta_{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,

$left|sigma_\left\{ij\right\}- lambdadelta_\left\{ij\right\} right|=begin\left\{vmatrix\right\}$
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

$left|sigma_\left\{ij\right\}- lambdadelta_\left\{ij\right\} right| = -lambda^3 + I_1lambda^2 - I_2lambda + I_3=0$

where

$begin\left\{align\right\}$
I_1 &= sigma_{11}+sigma_{22}+sigma_{33} &= sigma_{kk} 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_{13}^2 &= frac{1}{2}left(sigma_{ii}sigma_{jj}-sigma_{ij}sigma_{ji}right) I_3 &= det(sigma_{ij}) &= sigma_{11}sigma_{22}sigma_{33}+2sigma_{12}sigma_{23}sigma_{31}-sigma_{12}^2sigma_{33}-sigma_{23}^2sigma_{11}-sigma_{13}^2sigma_{22} end{align}

$I_1$, $I_2$ and $I_3$ are the first, second, and third stress invariants, respectively. Their values are the same (invariant) regardless of the orientation of the coordinate system chosen.

The characteristic equation has three real roots $lambda$, i.e. not imaginary due to the symmetry of the stress tensor. The three roots $lambda_1=sigma_1$, $lambda_2=sigma_2$, and $lambda_3=sigma_3$ are the eigenvalues or principal stresses, and they are the roots of the Cayley–Hamilton theorem. For each eigenvalue, there is a non-trivial solution for $n_j$ in the equation $left\left(sigma_\left\{ij\right\}- lambdadelta_\left\{ij\right\} right\right)n_j =0$. 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 on the orientation of the coordinate system.

If we choose a coordinate system with axes oriented to the principal directions, then the normal stresses will be the principal stresses. Thus, we have

$begin\left\{align\right\}$
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}

## Stress deviator tensor

The stress tensor $sigma_\left\{ij\right\}$ can be expressed as the sum of two other stress tensors:

1. a mean hydrostatic stress tensor or volumetric stress tensor or mean normal stress tensor, $pdelta_\left\{ij\right\}$, which tends to change the volume of the stressed body; and
2. a deviatoric component called the stress deviator tensor, $s_\left\{ij\right\}$, which tends to distort it.

$sigma_\left\{ij\right\}= s_\left\{ij\right\} + pdelta_\left\{ij\right\}$

where $p$ is the mean stress given by

$p=frac\left\{sigma_\left\{kk\right\}\right\}\left\{3\right\}=frac\left\{sigma_\left\{11\right\}+sigma_\left\{22\right\}+sigma_\left\{33\right\}\right\}\left\{3\right\}=tfrac\left\{1\right\}\left\{3\right\}I_1$

The deviatoric stress tensor can be obtained by subtracting the hydrostatic stress tensor from the stress tensor:

$begin\left\{align\right\}$
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}
`  p & 0 & 0 `
`  0 & p & 0 `
`  0 & 0 & p `
end{matrix}}right] &=left[{begin{matrix} sigma_{11}-p & sigma_{12} & sigma_{13} sigma_{21} & sigma_{22}-p & sigma_{23} sigma_{31} & sigma_{32} & sigma_{33}-p 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 $s_\left\{ij\right\}$ are the same as the principal directions of the stress tensor $sigma_\left\{ij\right\}$. Thus, the characteristic equation is

$left| s_\left\{ij\right\}- lambdadelta_\left\{ij\right\} right| = lambda^3-J_1lambda^2-J_2lambda-J_3=0$

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 $s_\left\{ij\right\}$ or its principal values $s_1$, $s_2$, and $s_3$, or alternatively, as a function of $sigma_\left\{ij\right\}$ or its principal values $sigma_1 ,$, $sigma_2 ,$, and $sigma_3 ,$ . Thus,

$begin\left\{align\right\}$
J_1 &= s_{kk}=0 end{align}
$begin\left\{align\right\}$
J_2 &= textstyle{frac{1}{2}}s_{ij}s_{ji} &= -s_1s_2 - s_2s_3 - s_3s_1 &= 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 J_3 &= det(s_{ij}) &= tfrac{1}{3}s_{ij}s_{jk}s_{ki} &= s_1s_2s_3 &= tfrac{2}{27}I_1^3 - tfrac{1}{3}I_1 I_2 + I_3 end{align} Because $s_\left\{kk\right\}=0 ,$, 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

$sigma_e = sqrt\left\{3~J_2\right\} = sqrt\left\{tfrac\left\{1\right\}\left\{2\right\}~left\left[\left(sigma_1-sigma_2\right)^2 + \left(sigma_2-sigma_3\right)^2 + \left(sigma_3-sigma_1\right)^2 right\right]\right\}$

## Octahedral stresses

Considering the principal directions as the coordinate axes, a plane which normal vector makes equal angles with each of the principal axes, i.e. having direction cosines equal to $|1/sqrt\left\{3\right\}|$, 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 $sigma_\left\{oct\right\}$ and octahedral shear stress $tau_\left\{oct\right\}$, respectively.

Knowing that the stress tensor of point O (Figure 6) in the principal axes is

$sigma_\left\{ij\right\}=$
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:

$begin\left\{align\right\}$
mathbf{T}_{oct}^{(mathbf{n})}&= sigma_{ij}n_imathbf{e}_j &=sigma_1n_1mathbf{e}_1+sigma_2n_2mathbf{e}_2+sigma_3n_3mathbf{e}_3 &=tfrac{1}{sqrt{3}}(sigma_1mathbf{e}_1+sigma_2mathbf{e}_2+sigma_3mathbf{e}_3) end{align}

The normal component of the stress vector at point O associated with the octahedral plane is

$begin\left\{align\right\}$
sigma_{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) 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

$begin\left\{align\right\}$
tau_{oct} &=sqrt{T^{(n)}T^{(n)}-sigma_n^2} &=left[tfrac{1}{3}(sigma_1^2+sigma_2^2+sigma_3^2)-tfrac{1}{9}(sigma_1+sigma_2+sigma_3)right]^{1/2} &=tfrac{1}{3}left[(sigma_1-sigma_2)^2+(sigma_2-sigma_3)^2+(sigma_3-sigma_1)^2right]^{1/2} end{align}

## Analysis of stress

All real objects occupy a three-dimensional space. However, depending on the loading condition and viewpoint of the observer the same physical object can alternatively be assumed as one-dimensional or two-dimensional, thus simplifying the mathematical modelling of the object.

### Uniaxial stress

If two of the dimensions of the object are very large or very small compared to the others, the object may be modelled as one-dimensional. In this case the stress tensor has only one component and is indistinguishable from a scalar. One-dimensional objects include a piece of wire loaded at the ends and viewed from the side, and a metal sheet loaded on the face and viewed up close and through the cross section.

When a structural element is elongated or compressed, its cross-sectional area changes by an amount that depends on the Poisson's ratio of the material. In engineering applications, structural members experience small deformations and the reduction in cross-sectional area is very small and can be neglected, i.e., the cross-sectional area is assumed constant during deformation. For this case, the stress is called engineering stress or nominal stress. In some other cases, e.g., elastomers and plastic materials, the change in cross-sectional area is significant, and the stress must be calculated assuming the current cross-sectional area instead of the initial cross-sectional area. This is termed true stress and is expressed as

$sigma_mathrm\left\{true\right\} = \left(1 + varepsilon_e\right)\left(sigma_e\right) ,$,

where

$varepsilon_e$ is the nominal (engineering) strain, and
$sigma_e$ is nominal (engineering) stress.

The relationship between true strain and engineering strain is given by

$varepsilon_mathrm\left\{true\right\} = ln\left(1 + varepsilon_e\right) ,$.

In uniaxial tension, true stress is then greater than nominal stress. The converse holds in compression.

### Plane stress

A state of plane stress exist when one of the principal stresses is zero, stresses with respect to the thin surface are zero. This usually occurs in structural elements where one dimension is very small compared to the other two, i.e. the element is flat or thin, and the stresses are negligible with respect to the smaller dimension as they are not able to develop within the material and are small compared to the in-plane stresses. Therefore, the face of the element is not acted by loads and the structural element can be analyzed as two-dimensional, e.g. thin-walled structures such as plates subject to in-plane loading or thin cylinders subject to pressure loading. The stress tensor can then be approximated by:

$sigma_\left\{ij\right\} = begin\left\{bmatrix\right\}$
sigma_{11} & sigma_{12} & 0 sigma_{21} & sigma_{22} & 0 0 & 0 & 0end{bmatrix}.

The corresponding strain tensor is:

$varepsilon_\left\{ij\right\} = begin\left\{bmatrix\right\}$
varepsilon_{11} & varepsilon_{12} & 0 varepsilon_{21} & varepsilon_{22} & 0 0 & 0 & varepsilon_{33}end{bmatrix}

in which the non-zero $varepsilon_\left\{33\right\}$ term arises from the Poisson's effect. This strain term can be temporarily removed from the stress analysis to leave only the in-plane terms, effectively reducing the analysis to two dimensions.

### Plane strain

If one dimension is very large compared to the others, the principal strain in the direction of the longest dimension is constrained and can be assumed as zero, yielding a plane strain condition. In this case, though all principal stresses are non-zero, the principal stress in the direction of the longest dimension can be disregarded for calculations. Thus, allowing a two dimensional analysis of stresses, e.g. a dam analyzed at a cross section loaded by the reservoir.

## Mohr's circle for stresses

Mohr's circle is a graphical representation of any 2-D stress state and was named for Christian Otto Mohr. Mohr's circle may also be applied to three-dimensional stress. In this case, the diagram has three circles, two within a third.

Mohr's circle is used to find the principal stresses, maximum shear stresses, and principal planes. For example, if the material is brittle, the engineer might use Mohr's circle to find the maximum component of normal stress (tension or compression); and for ductile materials, the engineer might look for the maximum shear stress.

## Alternative measures of stress

The Cauchy stress is not the only measure of stress that is used in practice. Other measures of stress include the first and second Piola-Kirchhoff stress tensors, the Biot stress tensor, and the Kirchhoff stress tensor.

### Piola-Kirchhoff stress tensor

In the case of finite deformations, the Piola-Kirchhoff stress tensors are used to express the stress relative to the reference configuration. This is in contrast to the Cauchy stress tensor which expresses the stress relative to the present configuration. For infinitesimal deformations or rotations, the Cauchy and Piola-Kirchoff tensors are identical. These tensors take their names from Gabrio Piola and Gustav Kirchhoff.

#### 1st Piola-Kirchhoff stress tensor

Whereas the Cauchy stress tensor, $sigma_\left\{ij\right\}$, relates forces in the present configuration to areas in the present configuration, the 1st Piola-Kirchhoff stress tensor, $K_\left\{Lj\right\}$ relates forces in the present configuration with areas in the reference ("material") configuration. $K_\left\{Lj\right\}$ is given by

$K_\left\{Lj\right\}=J X_\left\{L,i\right\} sigma_\left\{ij\right\} !$

where $J$ is the Jacobian, and $X_\left\{L,i\right\}$ is the inverse of the deformation gradient.

Because it relates different coordinate systems, the 1st Piola-Kirchhoff stress is a two-point tensor. In general, it is not symmetric. The 1st Piola-Kirchhoff stress is the 3D generalization of the 1D concept of engineering stress.

If the material rotates without a change in stress state (rigid rotation), the components of the 1st Piola-Kirchhoff stress tensor will vary with material orientation.

The 1st Piola-Kirchhoff stress is energy conjugate to the deformation gradient.

#### 2nd Piola-Kirchhoff stress tensor

Whereas the 1st Piola-Kirchhoff stress relates forces in the current configuration to areas in the reference configuration, the 2nd Piola-Kirchhoff stress tensor $S_\left\{IJ\right\}$ relates forces in the reference configuration to areas in the reference configuration. The force in the reference configuration is obtained via a mapping that preserves the relative relationship between the force direction and the area normal in the current configuration.

$S_\left\{IJ\right\}=J X_\left\{I,k\right\} X_\left\{J,l\right\} sigma_\left\{kl\right\} !$

This tensor is symmetric.

If the material rotates without a change in stress state (rigid rotation), the components of the 2nd Piola-Kirchhoff stress tensor will remain constant, irrespective of material orientation.

The 2nd Piola-Kirchhoff stress tensor is energy conjugate to the Green-Lagrange finite strain tensor.

## Books

• Dieter, G. E. (3 ed.). (1989). Mechanical Metallurgy. New York: McGraw-Hill. ISBN 0-07-100406-8.
• Love, A. E. H. (4 ed.). (1944). Treatise on the Mathematical Theory of Elasticity. New York: Dover Publications. ISBN 0-486-60174-9.
• Marsden, J. E., & Hughes, T. J. R. (1994). Mathematical Foundations of Elasticity. New York: Dover Publications. ISBN 0-486-67865-2.
• L.D.Landau and E.M.Lifshitz. (1959). Theory of Elasticity.
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