# Stress (physics)

Template:Continuum mechanics In Continuum mechanics, 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

$\ \sigma = \frac{P}{A} \,$

where

$\ \sigma$ is the average stress, also called engineering or nominal stress, and
$\ P$ 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 types

There are two basic stresses: normal stress, $\ \sigma$, acting normal to the surface under consideration, and shearing stress, $\ \tau$, acting parallel to the stressed surface. All other stresses produced on a body by different load conditions are similar or a combination of them. 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, it is termed tensile stress; or it is named compressive stresses if the forces push the body reducing its length. 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. The term normal stress in rheology is called extensional stress, and in acoustics is called longitudinal stress.

Solids, liquids and gases have stress fields. Static fluids support normal stress (hydrostatic pressure) 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.

The von Mises stress is derived from the distortion energy theory and is a simple way to combine stresses in three dimensions to calculate failure criteria of ductile materials. In this way, the strength of material in a 3-D state of stress can be compared to a test sample that was loaded in one dimension.

## Cauchy's stress principle

File:Internal forces in a body.png
Figure 1. Internal forces in a body
File:Stress components in element.png
Figure 2. Components of stress in three dimensions
File:Stress vector on a plane.png
Figure 3. Stress vector acting on a plane with normal vector n

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.

Considering a body subjected to surface forces $\ \mathbf{F}$ and body forces $\ \mathbf{f}$ 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 $\mathbf{n} \,$ 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 $\mathbf{T}^{(\mathbf{n})} \,$ at point $P \,$ associated with a plane with a normal vector $\mathbf{n} \,$:

$\mathbf{T}^{(\mathbf{n})}= \lim_{\Delta A \to 0} \frac {\Delta F}{\Delta A} = {dF \over dA} \,$

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

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

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 $\mathbf{T}^{(\mathbf{n})} \,$ 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. $\ \mathbf{T}^{(\mathbf{e}_1)}$, $\ \mathbf{T}^{(\mathbf{e}_2)}$, and $\ \mathbf{T}^{(\mathbf{e}_3)}$ can be decompose into components in the direction of the three coordinate axes:

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

In index notation this is

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

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

$\ \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

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

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.

The stress vector $\mathbf{T}^{(\mathbf{n})} \,$ at any point associated with a plane of normal vector $\ \mathbf{n}$ can be expressed as a function of the stress vectors on the planes perpendicular to the coordinate axes, i.e. stress tensor $\ \sigma_{ij}$. 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 $\ \mathbf{n}$ (Figure 3). The stress vector on this plane is denoted by $\ \mathbf{T}^{(\mathbf{n})}$. The stress vectors acting on the faces of the tetrahedron are denoted as $\ \mathbf{T}^{(\mathbf{e}_1)}$, $\ \mathbf{T}^{(\mathbf{e}_2)}$, and $\ \mathbf{T}^{(\mathbf{e}_3)}$, and are by definition the components of the stress tensor $\ \sigma_{ij}$. From equilibrium of forces, i.e. Newton's second law, we have

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

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

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

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

\ \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 as a scalar equation

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

## Equilibrium equation and symmetry of the stress tensor

The state of stress as defined by the stress tensor is an equilibrium state if the following conditions are satisfied:

$\frac {\partial {\sigma_{11}}} {\partial {x_{1}}} + \frac {\partial {\sigma_{21}}} {\partial {x_{2}}} + \frac {\partial {\sigma_{31}}} {\partial {x_{3}}} = f_{1}$
$\frac {\partial {\sigma_{12}}} {\partial {x_{1}}} + \frac {\partial {\sigma_{22}}} {\partial {x_{2}}} + \frac {\partial {\sigma_{32}}} {\partial {x_{3}}} = f_{2}$
$\frac {\partial {\sigma_{13}}} {\partial {x_{1}}} + \frac {\partial {\sigma_{23}}} {\partial {x_{2}}} + \frac {\partial {\sigma_{33}}} {\partial {x_{3}}} = f_{3}$

$\sigma_{ij}$ are the components of the tensor, and f 1 , f 2 , and f 3 are the body forces (force per unit volume).

These equations can be compactly written using Einstein notation as:

$\ \sigma_{ji,j}=f_i$

The equilibrium conditions may be derived from the condition that the net force on an infinitesimal volume element must be zero. Consider an infinitesimal cube aligned with the $x_1$, $x_2$, and $x_3$ axes, with one corner at $x_i$ and the opposite corner at $x_i+dx_i$ and having each face of area $dA$. Consider just the faces of the cube which are perpendicular to the $x_1$ axis. The area vector for the near face is $[-dA,0,0]$ and for the far face it is $[dA,0,0]$. The net stress force on these two opposite faces is

$dF_i=\sigma_{1i}([x_1+dx_1,x_2,x_3])\,dA-\sigma_{1i}([x_1,x_2,x_3])\,dA \approx \partial_1\sigma_{1i}\,dV$

A similar calculation can be carried out for the other pairs of faces. The sum of all the stress forces on the infinitesimal cube will then be

$dF_i=\partial_j\sigma_{ji}\,dV$

Since the net force on the cube must be zero, it follows that this stress force must be balanced by the force per unit volume $f_i$ on the cube (e.g., due to gravitation, electromagnetic forces, etc.) which yields the equilibrium conditions written above.

Equilibrium also requires that the resultant moment on the cube of material must be zero. Taking the moment of the forces above about any suitable point, it follows that, for equilibrium in the absence of body moments

$\sigma_{ij}= \sigma_{ji}\,$.

The stress tensor is then symmetric and the subscripts can be written in either order.

### Another approach to find the symmetry of the stress tensor

The fact that the stress is a symmetric tensor follows from some simple considerations. The force on a small volume element will be the sum of all the stress forces over the surface area of that element. Suppose we have a volume element in the form of a long bar with a triangular cross section, where the triangle is a right triangle. We can neglect the forces on the ends of the bar, because they are small compared to the faces of the bar. Let $\vec{A}$ be the vector area of one face of the bar, $\vec{B}$ be the area of the other, and $\vec{C}$ be the area of the "hypotenuse face" of the bar. It can be seen that

$\vec{C}=-\vec{A}-\vec{B}$

Let's say $\sigma(\vec{A})$ is the force on area $\vec{A}$ and likewise for the other faces. Since the stress is by definition the force per unit area, it is clear that

$\sigma(k\vec{A})=k\sigma(\vec{A})$

The total force on the volume element will be:

$\vec{F}=\sigma(\vec{A})+\sigma(\vec{B})-\sigma(\vec{A}+\vec{B})$

Suppose that the volume element contains mass, at a constant density. The important point is that if we make the volume smaller, say by halving all lengths, the area will decrease by a factor of four, while the volume will decrease by a factor of eight. As the size of the volume element goes to zero, the ratio of area to volume will become infinite. The total stress force on the element is proportional to its area, and so as the volume of the element goes to zero, the force/mass (i.e. acceleration) will also become infinite, unless the total force is zero. In other words:

$\sigma(\vec{A}+\vec{B})=\sigma(\vec{A})+\sigma(\vec{B})$

This, along with the second equation above, proves that the $\sigma$ function is a linear vector operator (i.e. a tensor). By an entirely analogous argument, we can show that the total torque on the volume element (due to stress forces) must be zero, and that it follows from this restriction that the stress tensor must be symmetric.

However, there are two fundamental ways in which this mode of thinking can be misleading. First, when applying this argument in tandem with the underlying assumption from continuum mechanics that the Knudsen number is strictly less than one, then in the limit $K_{n}\rightarrow 1$, the symmetry assumptions in the stress tensor may break down. This is the case of Non-Newtonian fluid, and can lead to rotationally non-invariant fluids, such as polymers. The other case is when the system is operating on a purely finite scale, such as is the case in mechanics where Finite deformation tensors are used.

## Principal stresses and stress invariants

The components $\ \sigma_{ij}$ of the stress tensor depend on the orientation of the plane that passes through the point under consideration. This would lead to the incorrect conclusion that the state of stress at a point in a body depends on the viewpoint of the observer, i.e. orientation of the coordinate system. However, every tensor, including stress, has invariants that do not depend on the choice of viewpoint. This means that the stress components seen by one observer are related, via the tensor transformation relations, to those seen by any other observer. The length of a first-order tensor, i.e a vector, is a simple example.

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 parallel or in the same direction as the normal vector $\ \sigma_n$ and where there are no normal shear stresses $\ \tau_n$. Thus,

$\ \mathbf{T}^{(\mathbf{n})} = \lambda \mathbf{n}= \mathbf{\sigma}_n \mathbf{n}$

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 $\ T_i^{(n)}=\sigma_{ij}n_j$ 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 homogenous 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} \\  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_{ij}\sigma_{ji}-\sigma_{ii}\sigma_{jj}\right) \\ I_3 &= \det(\sigma_{ij}) \\ &= \sigma_{11}\sigma_{22}\sigma_{33}+2\sigma_{12}\sigma_{23}\sigma_{31}-\sigma_{12}^2\sigma_{33}-\sigma_{23}^2\sigma_{11}-\sigma_{13}^2\sigma_{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 $\ \lambda_3=\sigma_1$ 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 $\ \left(\sigma_{ij}- \lambda\delta_{ij} \right)n_j =0$. 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 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{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}

## Stress deviator tensor

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

1. a mean hydrostatic stress tensor or mean normal stress tensor, $\ p\delta_{ij}$, which tends to change the volume of the stressed body; and
2. a deviatoric component called the stress deviator tensor, $\ s_{ij}$, which tends to distort it.
$\ \sigma_{ij}= s_{ij} + p\delta_{ij}$

where $\ p$ is the mean stress given by

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

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

$\ s_{ij} = \sigma_{ij} - \frac{\sigma_{kk}}{3}\delta_{ij}=\left[{\begin{matrix}  s_{11} & s_{12} & s_{13} \\ s_{21} & s_{22} & s_{23} \\ s_{31} & s_{32} & s_{33} \\ \end{matrix}}\right]=\left[{\begin{matrix} \sigma_{11} & \sigma_{12} & \sigma_{13} \\ \sigma_{21} & \sigma_{22} & \sigma_{23} \\ \sigma_{31} & \sigma_{32} & \sigma_{33} \\ \end{matrix}}\right]-\left[{\begin{matrix} p & 0 & 0 \\ 0 & p & 0 \\ 0 & 0 & p \\ \end{matrix}}\right]=\left[{\begin{matrix} \sigma_{11}-p & \sigma_{12} & \sigma_{13} \\ \sigma_{21} & \sigma_{22}-p & \sigma_{23} \\ \sigma_{31} & \sigma_{32} & \sigma_{33}-p \\ \end{matrix}}\right]$


As it is a second order tensor, the stress deviator tensor also has a set of invariants, which can be obtain using the same procedure used to calculate the stress invariants. 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. This deviatoric stress invariantes can be expressed as a function of the components of $\ s_{ij}$ or its principal values $\ s_1$, $\ s_2$, and $\ s_3$, or alternatevily, 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 \end{align}

\begin{align} J_2 &= \textstyle{\frac{1}{2}}s_{ij}s_{ji} \\ &= -s_1s_2 - s_2s_3 - s_3s_1 \\ &= \textstyle{\frac{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 \\ &= \textstyle{\frac{1}{6}}\left[ \sigma_1 - \sigma_2)^2 + (\sigma_2 - \sigma_3)^2 + (\sigma_3 - \sigma_1)^2 \right ] \\ J_3 &= \det(s_{ij}) \\ &= \textstyle{\frac{1}{3}}s_{ij}s_{jk}s_{ki} \\ &= s_1s_2s_3 \\ \end{align} Because $s_{kk}=0 \,$, the stress deviator tensor is in a state of pure shear.

## Stress conditions

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

$\sigma_{true} = (1 + \varepsilon_e)(\sigma_e) \,$,

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

$\varepsilon_{true} = ln(1 + \varepsilon_e) \,$.

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:

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


The corresponding strain tensor is:

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


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

### 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. It 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.

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

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

where $\ J$ is the Jacobian, and $\ X_{L,i}$ 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 $\ S_{IJ}$ 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.

$S_{IJ}=J X_{I,j} \sigma_{ij} X_{J,i} \!$

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 strain.