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Continuum mechanics
Figure 1.1 Stress in a loaded deformable material body assumed as a continuum.
Figure 1.2 Axial stress in a prismatic bar axially loaded

In continuum mechanics, stress is a measure of the average force per unit area of a surface within a deformable body on which internal forces act. In other words, it is a measure of the intensity of the internal forces acting between particles of a deformable body across imaginary internal surfaces [1]. These internal forces are produced between the particles in the body as a reaction to external forces applied on the body. External forces are either surface forces or body forces. Because the loaded deformable body is assumed as a continuum, these internal forces are distributed continuously within the volume of the material body, i.e., the stress distribution in the body is expressed as a piecewise continuous function of space coordinates and time.

The SI unit for stress is the pascal (symbol Pa), which is equivalent to one newton (force) per square meter (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).

For the simple case of a body axially loaded, e.g., a prismatic bar subjected to tension or compression by a force passing through its centroid the stress $\sigma\,\!$, or intensity of the distribution of internal forces, can be obtained by dividing the total tensile or compressive force $F\,\!$ by the cross-sectional area $A\,\!$ where it is acting upon. In this case the stress $\sigma\,\!$ is represented by a scalar called engineering stress or nominal stress that represents an average stress ($\sigma_\mathrm{avg}\,\!$) over the area, meaning that the stress in the cross section is uniformly distributed. Thus, we have

$\sigma_\mathrm{avg} = \frac{F_\mathrm n}{A}\approx\sigma\,\!$

In general, however, the stress is not uniformly distributed over a cross section of a material body, and consequently the stress at a point on a given area is different from the average stress over the entire area. Therefore, it is necessary to define the stress not over a given area but at a specific point in the body (Figure 1.1). According to Cauchy, the stress at any point in an object, assumed to be a continuum, is completely defined by the nine components $\sigma_{ij}\,\!$ of a second order tensor of type (2,0) known as the Cauchy stress tensor, $\boldsymbol\sigma\,\!$:

$\boldsymbol{\sigma}= \left[{\begin{matrix} \sigma _{11} & \sigma _{12} & \sigma _{13} \ \sigma _{21} & \sigma _{22} & \sigma _{23} \ \sigma _{31} & \sigma _{32} & \sigma _{33} \ \end{matrix}}\right] \equiv \left[{\begin{matrix} \sigma _{xx} & \sigma _{xy} & \sigma _{xz} \ \sigma _{yx} & \sigma _{yy} & \sigma _{yz} \ \sigma _{zx} & \sigma _{zy} & \sigma _{zz} \ \end{matrix}}\right] \equiv \left[{\begin{matrix} \sigma _x & \tau _{xy} & \tau _{xz} \ \tau _{yx} & \sigma _y & \tau _{yz} \ \tau _{zx} & \tau _{zy} & \sigma _z \ \end{matrix}}\right] \,\!$

The 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. For large deformations, also called finite deformations, other measures of stress are required, such as the first and second Piola-Kirchhoff stress tensors, 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.

The determination of the internal distribution of stresses, i.e., stress analysis, is required in engineering, e.g., civil engineering and mechanical engineering, for the study and design of structures, e.g., tunnels, dams, mechanical parts, and structural frames among others, under prescribed or expected loads. To determine the distribution of stress in the structure it is necessary to solve a boundary-value problem by specifying the boundary conditions, i.e. displacements and/or forces on the boundary. Constitutive equations, such as Hooke's Law for linear elastic materials, are used to describe the stress:strain relationship in these calculations. A boundary-value problem based on the theory of elasticity is applied to structures expected to deform elastically, i.e. infinitesimal strains, under design loads. When the loads applied to the structure induce plastic deformations, the theory of plasticity is implemented.

Approximate solutions for boundary-value problems can be obtained through the use of numerical methods such as the Finite Element Method, the Finite Difference Method, and the Boundary Element Method, which are implemented in computer programs. Analytical or close-form solutions can be obtained for simple geometries, constitutive relations, and boundary conditions.

The stress analysis can be simplified in cases where the physical dimensions and the distribution of loads allows the structure to be assumed as one-dimensional or two-dimensional. For a two-dimensional analysis a plane stress or a plane strain condition can be assumed.

Alternatively, experimental determination of stresses can be carried out using the photoelastic method.

In design of structures, calculated stresses are restricted to be less than an specified allowable stress, also known as working or designed stress. Allowable stress is chosen as some fraction of the yield strength or of the ultimate strength of the material of which the structure is made. The ratio of the ultimate stress to the allowable stress is defines as the factor of safety. Laboratory tests are usually performed on material samples in order to determine the yield strength and the ultimate strength that the material can withstand before failure.

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.

## Definition of stress

In the mechanics of a continuum, a material body can be acted upon by external forces that produce motion and which are of two kind: surface forces and body forces.

Surface forces or contact forces, designated by $\ F_i\,\!$ (force per unit area), act on the bounding surface as a result of mechanical contact between bodies, or they may also represent the force which an imaginary surface within the body exerts on the adjacent surface. The intensity of surface forces is related, i.e. is inversely proportional, to the area of contact, as will be seen in this article.

Body forces are external forces originating from sources outside of the body, and they represent actions at a distance from the surrounding.[2] Body forces include gravitational forces, electromagnetic forces, and inertial forces, and they are distributed over the entire volume of a body, i.e. acting on every point in the body. In the case of gravitational and inertial forces, the intensity of the force depends on or is proportional to the mass density $\rho\,\!$ of the material. These two forces are specified in terms of force per unit mass ($b_i\,\!$) or per unit volume ($p_i\,\!$). These two specifications are related through the material density by the equation $\rho b_i = p_i\,\!$. Similarly, the intensity of electromagnetic forces depends upon the strength (electric charge) of the electromagnetic field.

These acting external forces (surface and body forces) are then transmitted from point to point within the material body, leading to the generation of internal forces. The transmission of such forces is governed by the conservation laws of linear and angular momenta (Newton's Second Law of motion). For bodies in static equilibrium, these laws are related to the principles of equilibrium of forces and moments, respectively.

The measure of the intensity of this internal forces acting within the material body across imaginary surfaces is called stress. In other words, stress is a measure of the average quantity of force exerted per unit area of the surface on which these internal forces act. For example, if we compare a force applied to a small area and a distributed load of the same resulting magnitude applied to a larger area, we find that the effects or intensities of these two forces are locally different because the stresses are not the same.

Stress is related to deformations in the body. This relationship is expressed through constitutive equations.

### Stress in a prismatic bar

Figure 1.3 Normal stress in a prismatic (straight member of uniform cross-sectional area) bar. The stress or force distribution in the cross section of the bar is not necessarily uniform. However, an average normal stress $\sigma_\mathrm{avg}\,\!$ can be used
Figure 1.4 Shear stress in a prismatic bar. The stress or force distribution in the cross section of the bar is not necessarily uniform. However, an average shear stress $\tau_\mathrm{avg}\,\!$ is a reasonable approximation.[3]

First the simple case of a prismatic bar subjected to an axial force $F_\mathrm n\,\!$ will be examined. These axial forces can be produced either by tension or compression (Figures 1.2 and 1.3). Considering a cross sectional area perpendicular to the axis of the bar, from the equilibrium of forces the resultant normal force $F_\mathrm n\,\!$ can be found. The intensity of internal forces, or stress $\sigma\,\!$, in the cross sectional area can then be obtained by dividing the total normal force $F_\mathrm n\,\!$, e.g. tensile force if acting outward to the plane or compressive force if acting inward to the plane, by the cross-sectional area $A\,\!$ where it is acting upon. In this case the stress $\sigma\,\!$ is a scalar quantity called engineering or nominal stress that represents an average stress ($\sigma_\mathrm {avg}\,\!$) over the area, i.e. the stress in the cross section is uniformly distributed. Thus, we have

$\sigma_\mathrm{avg} = \frac{F_\mathrm n}{A}\approx\sigma\,\!$

A different type of stress is obtained when transverse forces $F_\mathrm\,\!$ are applied to the prismatic bar as show in Figure 1.4. Considering the same cross section as before, from static equilibrium, the internal force has a magnitude equal to $F_\mathrm s\,\!$ and in opposite direction parallel to the cross section. $F_\mathrm s\,\!$ is called the shear force. Dividing the shear force $F_\mathrm s\,\!$ by the area $A\,\!$ of the cross section we obtain the shear stress. In this case the shear stress $\tau\,\!$ is a scalar quantity representing an average shear stress ($\tau_\mathrm{avg}\,\!$) in the section, i.e. the stress in the cross section is uniformly distributed.

$\tau_\mathrm{avg}= \frac{F_\mathrm s}{A}\approx\tau\,\!$

In general, however, the stress is not uniformly distributed over the cross section of a material body, and consequently the stress at a point on a given area is different from the average stress over the entire area. In Figure 1.3, the normal stress is observed in two planes $m-m\,\!$ and $n-n\,\!$ of the axially loaded prismatic bar. The stress on plane $n-n\,\!$, which is closer to the point of application of the load $F\,\!$, varies more across the cross section than that of plane $m-m\,\!$. However, if the cross sectional area of the bar is very small, e.g. a slender bar, the variation of stress across the area is small and the normal stress can be approximated by $\sigma_\mathrm {avg}\,\!$. On the other hand, the variation of shear stress across the section of a prismatic bar cannot be assumed uniform.

Therefore, it is necessary to define the stress at a specific point in the surface.

## Cauchy's stress principle

Figure 2.1 Internal forces in a body

Consider a material body, as seen in Figure 2.1 at right, in equilibrium 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. A small area $\Delta A\,\!$ in one of the segments, passing through a point P, and with a normal unit vector $\mathbf{n}\,\!$ is acted upon by a force $\Delta \mathbf F\,\!$ resulting from the action of the material on one side of the area (left segment) onto the other side (right segment).

The distribution of force on the area $\Delta A\,\!$ is, however, not always uniform, as there may be a moment $\Delta \mathbf M\,\!$ at $P\,\!$ due to the force $\Delta \mathbf F\,\!$, as shown in Figure 2.1. Cauchy's stress principle states that as $\Delta A\,\!$ becomes very small and tends to zero the ratio $\Delta \mathbf F / \Delta A\,\!$ becomes $d \mathbf F/dA\,\!$, and the moment $\Delta \mathbf M\,\!$ vanishes. The resultant vector $d \mathbf F/dA\,\!$ is defined as the stress vector or traction vector $\mathbf{T}^{(\mathbf{n})}=T_i^{(\mathbf{n})}\mathbf{e}_i\,\!$ at point $P\,\!$ associated with a plane with a normal vector $\mathbf{n}\,\!$:

$T^{(\mathbf{n})}_i= \lim_{\Delta A \to 0} \frac {\Delta F_i}{\Delta A} = {dF_i \over dA}\,\!$

This equation means that the stress vector depends on the location in the body and the orientation of the plane on which it is acting.

By Newton's third law of motion, the stress vectors acting on opposite sides of the same surface are equal in magnitude and opposite in direction. This is Cauchy's Fundamental Lemma [4][5][6], and it is expressed as

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

The state of stress at a point in the body is then defined by all the stress vectors $\mathbf{T}^{(\mathbf{n})}\,\!$ associated with all planes (infinite number of planes) that pass through that point [1]. However, according to Cauchy's fundamental theorem [7], by just 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.

Depending on the orientation of the plane under consideration, the stress vector may not necessarily be perpendicular to that plane, and can be resolved into two components:

• one normal to the plane, called normal stress $\sigma_\mathrm{n} \,\!$
$\mathbf{\sigma_\mathrm{n}}= \lim_{\Delta A \to 0} \frac {\Delta F_\mathrm n}{\Delta A} = \frac{dF_\mathrm n}{dA}\,\!$

where $dF_\mathrm n\,\!$ is the normal component of the force $d \mathbf F\,\!$ to the differential area $dA\,\!$

• and the other parallel to this plane, called the shearing stress $\tau \,\!$.
$\mathbf \tau= \lim_{\Delta A \to 0} \frac {\Delta F_\mathrm s}{\Delta A} = \frac{dF_\mathrm s}{dA}\,\!$

where $dF_\mathrm s\,\!$ is the tangential component of the force $d \mathbf F\,\!$ to the differential surface area $dA\,\!$. The shear stress can be further decomposed into two mutually perpendicular vectors.

Figure 2.2 Components of stress in three dimensions

Assuming a material element (Figure 2.2) 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 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, the normal stress is denoted by $\sigma_{11}\,\!$, and the two shear stresses are denoted as $\sigma_{12}\,\!$ and $\sigma_{13}\,\!$:

$\mathbf{T}^{(\mathbf{e}_1)}= T_1^{(\mathbf{e}_1)}\mathbf{e}_1 + T_2^{(\mathbf{e}_1)} \mathbf{e}_2 + T_3^{(\mathbf{e}_1)} \mathbf{e}_3 = \sigma_{11} \mathbf{e}_1 + \sigma_{12} \mathbf{e}_2 + \sigma_{13} \mathbf{e}_3\,\!$
$\mathbf{T}^{(\mathbf{e}_2)}= T_1^{(\mathbf{e}_2)}\mathbf{e}_1 + T_2^{(\mathbf{e}_2)} \mathbf{e}_2 + T_3^{(\mathbf{e}_2)} \mathbf{e}_3=\sigma_{21} \mathbf{e}_1 + \sigma_{22} \mathbf{e}_2 + \sigma_{23} \mathbf{e}_3\,\!$
$\mathbf{T}^{(\mathbf{e}_3)}= T_1^{(\mathbf{e}_3)}\mathbf{e}_1 + T_2^{(\mathbf{e}_3)} \mathbf{e}_2 + T_3^{(\mathbf{e}_3)} \mathbf{e}_3=\sigma_{31} \mathbf{e}_1 + \sigma_{32} \mathbf{e}_2 + \sigma_{33} \mathbf{e}_3\,\!$

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 $\sigma_{ij}\,\!$ 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

$\boldsymbol{\sigma}= \sigma_{ij} = \left[{\begin{matrix} \mathbf{T}^{(\mathbf{e}_1)} \ \mathbf{T}^{(\mathbf{e}_2)} \ \mathbf{T}^{(\mathbf{e}_3)} \ \end{matrix}}\right] = \left[{\begin{matrix} \sigma _{11} & \sigma _{12} & \sigma _{13} \ \sigma _{21} & \sigma _{22} & \sigma _{23} \ \sigma _{31} & \sigma _{32} & \sigma _{33} \ \end{matrix}}\right] \equiv \left[{\begin{matrix} \sigma _{xx} & \sigma _{xy} & \sigma _{xz} \ \sigma _{yx} & \sigma _{yy} & \sigma _{yz} \ \sigma _{zx} & \sigma _{zy} & \sigma _{zz} \ \end{matrix}}\right] \equiv \left[{\begin{matrix} \sigma _x & \tau _{xy} & \tau _{xz} \ \tau _{yx} & \sigma _y & \tau _{yz} \ \tau _{zx} & \tau _{zy} & \sigma _z \ \end{matrix}}\right] \,\!$

where

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

The first index $i\,\!$ indicates that 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 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{align} \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.

### Relationship stress vector - stress tensor

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. in terms of the components of the stress tensor $\sigma_{ij}\,\!$. In tensor form this is:

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

To prove the expression, 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 2.3). 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 $\sigma_{ij}\,\!$. This tetrahedron is sometimes called the Cauchy tetrahedron. From equilibrium of forces, i.e. Newton's second law, we have

$\mathbf{T}^{(\mathbf{n})}dA - \mathbf{T}^{(\mathbf{e}_1)}dA_1 - \mathbf{T}^{(\mathbf{e}_2)}dA_2 - \mathbf{T}^{(\mathbf{e}_3)}dA_3 = \rho \left( \frac{h}{3}dA \right) \mathbf{a}\,\!$
Figure 2.3. Stress vector acting on a plane with normal vector n.
A note on the sign convention: The tetrahedron is formed by slicing a parallelepiped along an arbitrary plane n. So, the force acting on the plane n is the reaction exerted by the other half of the parallelepiped and has an opposite sign.

where the right hand side of the equation represent the product of the mass enclosed by the tetrahedron and its acceleration: $\rho\,\!$ 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 by projecting $dA\,\!$ into each face (using the dot product):

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

and then can be substituted into the equation to cancel out dA:

$\mathbf{T}^{(\mathbf{n})} - \mathbf{T}^{(\mathbf{e}_1)}n_1 - \mathbf{T}^{(\mathbf{e}_2)}n_2 - \mathbf{T}^{(\mathbf{e}_3)}n_3 = \rho \left( \frac{h}{3} \right) \mathbf{a}\,\!$

To consider the limiting case as the tetrahedron shrinks to a point, h must go to 0 (intuitively, plane $\mathbf{n}$ is translated along $\mathbf{n}$ towards O). As a result, the RHS approaches 0.

\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,

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

Or, in matrix form we have

$\left[{\begin{matrix} T^{(n)}_1 & T^{(n)}_2 & T^{(n)}_3\end{matrix}}\right]=\left[{\begin{matrix} n_1 & n_2 & n_3 \end{matrix}}\right]\cdot \left[{\begin{matrix} \sigma _{11} & \sigma _{12} & \sigma _{13} \ \sigma _{21} & \sigma _{22} & \sigma _{23} \ \sigma _{31} & \sigma _{32} & \sigma _{33} \ \end{matrix}}\right] \,\!$

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

### Transformation rule of the stress tensor

It can be shown that the stress tensor is a second order tensor, which is a statement of how it transforms under a change of the coordinate system. From an $x_i\,\!$ system to an $x^'_i\,\!$ system, the components $\sigma_{ij}\,\!$ in the initial system are transformed into the components $\sigma^'_{ij}\,\!$ 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 $\mathbf A\,\!$ is the rotation matrix with components $a_{ij}\,\!$. In matrix form this is

$\left[{\begin{matrix} \sigma^'_{11} & \sigma^'_{12} & \sigma^'_{13} \ \sigma^'_{21} & \sigma^'_{22} & \sigma^'_{23} \ \sigma^'_{31} & \sigma^'_{32} & \sigma^'_{33} \ \end{matrix}}\right]=\left[{\begin{matrix} a_{11} & a_{12} & a_{13} \ a_{21} & a_{22} & a_{23} \ a_{31} & a_{32} & a_{33} \ \end{matrix}}\right]\left[{\begin{matrix} \sigma_{11} & \sigma_{12} & \sigma_{13} \ \sigma_{21} & \sigma_{22} & \sigma_{23} \ \sigma_{31} & \sigma_{32} & \sigma_{33} \ \end{matrix}}\right]\left[{\begin{matrix} a_{11} & a_{21} & a_{31} \ a_{12} & a_{22} & a_{32} \ a_{13} & a_{23} & a_{33} \ \end{matrix}}\right] \,\!$
Figure 2.4 Transformation of the stress tensor

Expanding the matrix operation, and simplifying some terms by taking advantage of 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}\,\!

A graphical representation of this transformation of stresses, for a two-dimensional (plane stress and plane strain) and a general three-dimensional state of stresses, is the Mohr's circle for stresses

### Normal and shear stresses

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

\begin{align} \sigma_\mathrm{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, $\tau_\mathrm{n}\,\!$, acting in the plane formed by the two vectors $\mathbf{T}^{(\mathbf{n})}\,\!$ and $\mathbf n\,\!$, can then be found using the Pythagorean theorem, thus

\begin{align} \tau_\mathrm{n} &=\sqrt{ \left( T^{(n)} \right)^2-\sigma_\mathrm{n}^2} \ &= \sqrt{T_i^{(n)}T_i^{(n)}-\sigma_\mathrm{n}^2} \ \end{align} \,\!

where $\left( T^{(n)} \right)^2 = T_i^{(n)}T_i^{(n)} = \left( \sigma_{ij}n_j \right) \left( \sigma_{ik}n_k \right)=\sigma_{ij}\sigma_{ik}n_jn_k\,\!$

## Equilibrium equations and symmetry of the stress tensor

Figure 4. Continuum body in equilibrium

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

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

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.

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

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 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.

A stress vector parallel to the normal 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} \ 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}+2\sigma_{12}\sigma_{23}\sigma_{31}-\sigma_{12}^2\sigma_{33}-\sigma_{23}^2\sigma_{11}-\sigma_{13}^2\sigma_{22} \ \end{align} \,\!

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. The principal stresses are unique for a given stress tensor. Therefore, from the characteristic equation it is seen that the coefficients $I_1\,\!$, $I_2\,\!$ and $I_3\,\!$, called the first, second, and third stress invariants, respectively, have always the same value regardless of the orientation of the coordinate system chosen.

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 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 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 may 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, working and thinking in the principal coordinate system is often very useful when considering the state of the elastic medium at a particular point.

## 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

$\tau_\mathrm{max}=\frac{1}{2}\left|\sigma_\mathrm{max}-\sigma_\mathrm{min}\right|\,\!$

Assuming $\sigma_1\ge\sigma_2\ge\sigma_3\,\!$ then

$\tau_\mathrm{max}=\frac{1}{2}\left|\sigma_1-\sigma_3\right|\,\!$

The normal stress component acting on the plane for the maximum shear stress is non-zero and it is equal to

$\sigma_\mathrm{n}=\frac{1}{2}\left(\sigma_1+\sigma_3\right)\,\!$

## Stress deviator tensor

The stress tensor $\sigma_{ij}\,\!$ 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, $p\delta_{ij}\,\!$, 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

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

The deviatoric stress tensor can be obtained by subtracting the hydrostatic stress tensor from the 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} 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 $s_{ij}\,\!$ 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 \end{align} \,\!
\begin{align} 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 $s_{kk}=0\,\!$, 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

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

## Octahedral stresses

Figure 6. Octahedral stress planes

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.

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\mathbf{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{n}^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-\sigma_1)^2\right]^{1/2} = \tfrac{1}{3}\sqrt{2I_1^2-6I_2} = \sqrt{\tfrac{2}{3}J_2} \end{align} \,\!

## Analysis of stress

The analysis of stress within a body implies the determination at each point of the body of the magnitudes of the nine stress components. In other words, it is the determination of the internal distribution of stresses. A stress analysis is required in engineering, e.g., civil engineering and mechanical engineering, for the study and design of structures, e.g., tunnels, dams, mechanical parts, and structural frames among others, under prescribed or expected loads.

To determine the distribution of stress in the structure it is necessary to solve a boundary-value problem by specifying the boundary conditions, i.e. displacements and/or forces on the boundary. Constitutive equations, such as e.g. Hooke's Law for linear elastic materials, are used to describe the stress:strain relationship in these calculations. A boundary-value problem based on the theory of elasticity is applied to structures expected to deform elastically, i.e. infinitesimal strains, under design loads. When the loads applied to the structure induce plastic deformations, the theory of plasticity is implemented.

Approximate solutions for boundary-value problems can be obtained through the use numerical methods such as the Finite Element Method, the Finite Difference Method, and the Boundary Element Method, which are implemented in computer programs. Analytical or close-form solutions can be obtained for simple geometries, constitutive relations, and boundary conditions.

Alternatively, experimental determination of stresses can be carried out using the photoelastic method.

In design of structures, calculated stresses are restricted to be less than an specified allowable stress, also known as working or designed stress, that is chosen as some fraction of the yield strength or of the ultimate strength of the material which the structure is made of. The ratio of the ultimate stress to the allowable stress is defined as the factor of safety. Laboratory test are usually performed on material samples in order to determine the yield strength and the ultimate strength that the material can withstand before failure.

All real objects occupy a three-dimensional space. The stress analysis can be simplified in cases where the physical dimensions and the loading conditions allows the structure to be assumed as one-dimensional or two-dimensional. For a two-dimensional analysis a plane stress or a plane strain condition can be assumed.

### 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 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

$\sigma_\mathrm{true} = (1 + \varepsilon_\mathrm e)(\sigma_\mathrm e)\,\!$,

where

$\varepsilon_\mathrm e\,\!$ is the nominal (engineering) strain, and
$\sigma_\mathrm e\,\!$ is nominal (engineering) stress.

The relationship between true strain and engineering strain is given by

$\varepsilon_\mathrm{true} = \ln(1 + \varepsilon_\mathrm e)\,\!$.

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

Figure 7.1 Plane stress state in a continuum.

### Plane stress

A state of plane stress exists when one of the three principal $\left(\sigma_1, \sigma_2, \sigma_3 \right)\,\!$, stresses is 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. In this case, 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 other three non-zero components remain constant over the thickness of the plate. The stress tensor can then be approximated by:

$\sigma_{ij} = \begin{bmatrix} \sigma_{11} & \sigma_{12} & 0 \ \sigma_{21} & \sigma_{22} & 0 \ 0 & 0 & 0 \end{bmatrix} \equiv \begin{bmatrix} \sigma_{x} & \tau_{xy} & 0 \ \tau_{yx} & \sigma_{y} & 0 \ 0 & 0 & 0 \end{bmatrix}\,\!$.

The corresponding strain tensor is:

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

in which the non-zero $\varepsilon_{33}\,\!$ 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.

Figure 7.2 Plane strain state in a continuum.

### 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 (Figure 7.2). 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.

## Stress transformation in plane stress and plane strain

Consider a point $P\,\!$ in a continuum under a state of plane stress, or plane strain, with stress components $(\sigma_x, \sigma_y, \tau_{xy})\,\!$ and all other stress components equal to zero (Figure 7.1, Figure 8.1). From static equilibrium of an infinitesimal material element at $P\,\!$ (Figure 8.2), the normal stress $\sigma_\mathrm{n}\,\!$ and the shear stress $\tau_\mathrm{n}\,\!$ on any plane perpendicular to the $x\,\!$-$y\,\!$ plane passing through $P\,\!$ with a unit vector $\mathbf n\,\!$ making an angle of $\theta\,\!$ with the horizontal, i.e. $\cos \theta\,\!$ is the direction cosine in the $x\,\!$ direction, is given by:

$\sigma_\mathrm{n} = \frac{1}{2} ( \sigma_x + \sigma_y ) + \frac{1}{2} ( \sigma_x - \sigma_y )\cos 2\theta + \tau_{xy} \sin 2\theta\,\!$
$\tau_\mathrm{n} = -\frac{1}{2}(\sigma_x - \sigma_y )\sin 2\theta + \tau_{xy}\cos 2\theta\,\!$

These equations indicate that in a plane stress or plane strain condition, one can determine the stress components at a point on all directions, i.e. as a function of $\theta\,\!$, if one knows the stress components $(\sigma_x, \sigma_y, \tau_{xy})\,\!$ on any two perpendicular directions at that point. It is important to remember that we are considering a unit area of the infinitesimal element in the direction parallel to the $y\,\!$-$z\,\!$ plane.

Figure 8.1 - Stress transformation at a point in a continuum under plane stress conditions.
Figure 8.2 - Stress components at a plane passing through a point in a continuum under plane stress conditions.

The principal directions (Figure 8.3), i.e. orientation of the planes where the shear stress components are zero, can be obtained by making the previous equation for the shear stress $\tau_\mathrm{n}\,\!$ equal to zero. Thus we have:

$\tau_\mathrm{n} = -\frac{1}{2}(\sigma_x - \sigma_y )\sin 2\theta + \tau_{xy}\cos 2\theta=0\,\!$

and we obtain

$\tan 2 \theta_\mathrm{p} = \frac{2 \tau_{xy}}{\sigma_x - \sigma_y}\,\!$

This equation defines two values $\theta_\mathrm{p}\,\!$ which are $90^\circ\,\!$ apart (Figure 8.3). The same result can be obtained by finding the angle $\theta\,\!$ which makes the normal stress $\sigma_\mathrm{n}\,\!$ a maximum, i.e. $\frac{d\sigma_\mathrm{n}}{d\theta}=0\,\!$

The principal stresses $\sigma_1\,\!$ and $\sigma_2\,\!$, or minimum and maximum normal stresses $\sigma_\mathrm{max}\,\!$ and $\sigma_\mathrm{min}\,\!$, respectively, can then be obtained by replacing both values of $\theta_\mathrm{p}\,\!$ into the previous equation for $\sigma_\mathrm{n}\,\!$. This can be achieved by rearranging the equations for $\sigma_\mathrm{n}\,\!$ and $\tau_\mathrm{n}\,\!$, first transposing the first term in the first equation and squaring both sides of each of the equations then adding them. Thus we have

\begin{align} \left[ \sigma_\mathrm{n} - \tfrac{1}{2} ( \sigma_x + \sigma_y )\right]^2 + \tau_\mathrm{n}^2 &= \left[\tfrac{1}{2}(\sigma_x - \sigma_y)\right]^2 + \tau_{xy}^2 \ (\sigma_\mathrm{n} - \sigma_\mathrm{avg})^2 + \tau_\mathrm{n}^2 &= R^2 \end{align}\,\!

where

$R = \sqrt{\left[\tfrac{1}{2}(\sigma_x - \sigma_y)\right]^2 + \tau_{xy}^2} \quad \text{and} \quad \sigma_\mathrm{avg} = \tfrac{1}{2} ( \sigma_x + \sigma_y )\,\!$

which is the equation of a circle of radius $R\,\!$ centered at a point with coordinates $[\sigma_\mathrm{avg}, 0]\,\!$, called Mohr's circle. But knowing that for the principal stresses the shear stress $\tau_\mathrm{n} = 0\,\!$, then we obtain from this equation:

$\sigma_1 =\sigma_\mathrm{max} = \tfrac{1}{2}(\sigma_x + \sigma_y) + \sqrt{\left[\tfrac{1}{2}(\sigma_x - \sigma_y)\right]^2 + \tau_{xy}^2}\,\!$
$\sigma_2 =\sigma_\mathrm{min} = \tfrac{1}{2}(\sigma_x + \sigma_y) - \sqrt{\left[\tfrac{1}{2}(\sigma_x - \sigma_y)\right]^2 + \tau_{xy}^2}\,\!$
Figure 8.3 - Transformation of stresses in two dimensions, showing the planes of action of principal stresses, and maximum and minimum shear stresses.

When $\tau_{xy}=0\,\!$ the infinitesimal element is oriented in the direction of the principal planes, thus the stresses acting on the rectangular element are principal stresses: $\sigma_x = \sigma_1\,\!$ and $\sigma_y = \sigma_2\,\!$. Then the normal stress $\sigma_\mathrm{n}\,\!$ and shear stress $\tau_\mathrm{n}\,\!$ as a function of the principal stresses can be determined by making $\tau_{xy}=0\,\!$. Thus we have

$\sigma_\mathrm{n} = \frac{1}{2} ( \sigma_1 + \sigma_2 ) + \frac{1}{2} ( \sigma_1 - \sigma_2 )\cos 2\theta\,\!$
$\tau_\mathrm{n} = -\frac{1}{2}(\sigma_1 - \sigma_2 )\sin 2\theta\,\!$

Then the maximum shear stress $\tau_\mathrm{max}\,\!$ occurs when $\sin 2\theta = 1\,\!$, i.e. $\theta = 45^\circ\,\!$ (Figure 8.3):

$\tau_\mathrm{max} = \frac{1}{2}(\sigma_1 - \sigma_2 )\,\!$

Then the minimum shear stress $\tau_\mathrm{min}\,\!$ occurs when $\sin 2\theta = -1\,\!$, i.e. $\theta = 135^\circ\,\!$ (Figure 8.3):

$\tau_\mathrm{min} = -\frac{1}{2}(\sigma_1 - \sigma_2 )\,\!$

## Mohr's circle for stress

The Mohr's circle, named after Christian Otto Mohr, is a two-dimensional graphical representation of the state of stress at a point. The abscissa, $\sigma_\mathrm{n}\,\!$, and ordinate, $\tau_\mathrm{n}\,\!$, of each point on the circle are the normal stress and shear stress components, respectively, acting on a particular cut plane with a unit vector $\mathbf n\,\!$ with components $\left(n_1, n_2, n_3 \right)\,\!$. In other words, the circumference of the circle is the locus of points that represent state of stress on individual planes at all their orientations.

Karl Culmann was the first to conceive a graphical representation for stresses while considering longitudinal and vertical stresses in horizontal beams during bending. Mohr's contribution extended the use of this representation for both two- and three-dimensional stresses and developed a failure criterion based on the stress circle.

### Mohr's circle for plane stress or plane strain

Figure 9.1. Mohr's circle for plane stress and plane strain conditions (double angle approach).
Figure 9.2. Mohr's circle for plane stress and plane strain conditions (Pole approach). Any straight line drawn from the pole will intersect the Mohr circle at a point that represents the state of stress on a plane inclined at the same orientation (parallel) in space as that line.

Given known stress components $\sigma_x\,\!$. $\sigma_y\,\!$, and $\tau_{xy}\,\!$ at a point $P\,\!$ for any two perpendicular planes in a continuum body under plane stress, or plane strain (Figures 8.1 and 8.2) the Mohr circle of stress may be constructed. Once the Mohr circle is drawn it can be used to find the stress state on any other plane passing through that point in the body.

According to the sign convention for engineering mechanics, in disciplines such as mechanical engineering and structural engineering, which is the one used in this article, for the construction of the Mohr circle the normal stresses are positive if they are outward to the plane of action (tension), and shear stresses are positive if they rotate clockwise about the point in consideration. In geomechanics, i.e. soil mechanics and rock mechanics, however, normal stresses are considered positive when they are inward to the plane of action (compression), and shear stresses are positive if they rotate counterclockwise about the point in consideration. [15][16][17][18]

To construct the Mohr circle of stress for a state of plane stress, or plane strain, first we plot two points in the $\sigma_\mathrm{n}:\tau_\mathrm{n}\,\!$ space corresponding to the known stress components on both perpendicular planes, i.e. $A(\sigma_y, \tau_{xy})\,\!$ and $B(\sigma_x, -\tau_{xy})\,\!$ (Figure 9.1 and 9.2). We then connect points $A\,\!$ and $B\,\!$ by a straight line and find the midpoint $O\,\!$ which corresponds to the intersection of this line with the $\sigma_\mathrm{n}\,\!$ axis. Finally, we draw a circle with diameter $\overline{AB}\,\!$ and centre at $O\,\!$.

As demonstrated in the previous section, the radius $R\,\!$ of the circle is $R = \sqrt{\left[\tfrac{1}{2}(\sigma_x - \sigma_y)\right]^2 + \tau_{xy}^2}\,\!$, and the coordinates of its centre are $\left[\tfrac{1}{2}(\sigma_x + \sigma_y), 0\right]\,\!$.

The principal stresses are then the abscissa of the points of intersection of the circle with the $\sigma_\mathrm{n}\,\!$ axis (note that the shear stresses are zero for the principal stresses).

Using the Mohr circle one can find the stress components $(\sigma_\mathrm{n}, \tau_\mathrm{n})\,\!$ on any other plane with a different orientation $\theta\,\!$ that passes through point $P\,\!$. For this, two approaches can be used:

The first approach relies on the fact that the angle $\theta\,\!$ between two planes passing through $P\,\!$ is half the angle between the lines joining their corresponding stress points $(\sigma_\mathrm{n}, \tau_\mathrm{n})\,\!$ on the Mohr circle and the centre of the circle (Figure 9.1). In other words, the stresses $(\sigma_\mathrm{n}, \tau_\mathrm{n})\,\!$ acting on a plane at an angle $\theta\,\!$ counterclockwise to the plane on which $\sigma_x\,\!$ acts is determined by traveling counterclockwise around the circle from the known stress point $(\sigma_x, \tau_{xy})\,\!$ a distance subtending an angle $2\theta\,\!$ at the centre of the circle (Figure 9.1).

The second approach involves the determination of a point on the Mohr circle called the pole or the origin of planes. Any straight line drawn from the pole will intersect the Mohr circle at a point that represents the state of stress on a plane inclined at the same orientation (parallel) in space as that line. Therefore, knowing the stress components $\sigma\,\!$ and $\tau\,\!$ on any particular plane, one can draw a line parallel to that plane through the particular coordinates $\sigma_\mathrm{n}\,\!$ and $\tau_\mathrm{n}\,\!$ on the Mohr circle and find the pole as the intersection of such line with the Mohr circle. As an example, let's assume we have a state of stress with stress components $\sigma_x,\!$, $\sigma_y,\!$, and $\tau_{xy},\!$, as shown on Figure 9.2. First, we can draw a line from point $B\,\!$ parallel to the plane of action of $\sigma_x\,\!$, or, if we choose otherwise, a line from point $A\,\!$ parallel to the plane of action of $\sigma_y\,\!$. The intersection of any of these two lines with the Mohr circle is the pole. Once the pole has been determined, to find the state of stress on a plane making an angle $\theta\,\!$ with the vertical, or in other words a plane having its normal vector forming an angle $\theta\,\!$ with the horizontal plane, then we can draw a line from the pole parallel to that plane (See Figure 9.2). The normal and shear stresses on that plane are then the coordinates of the point of intersection between the line and the Mohr circle.

### Mohr's circle for a general three-dimensional state of stresses

Figure 7. Mohr's circle for a three-dimensional state of stress

To construct the Mohr's circle for a general three-dimensional case of stresses at a point, the values of the principal stresses $\left(\sigma_1, \sigma_2, \sigma_3 \right)\,\!$ and their principal directions $\left(n_1, n_2, n_3 \right)\,\!$ must be first evaluated, as explained previously.

Considering the principal axes as the coordinate system, instead of the general $x_1\,\!$, $x_2\,\!$, $x_3\,\!$ coordinate system, and assuming that $\sigma_1 > \sigma_2 > \sigma_3\,\!$, then the normal and shear components of the stress vector $\mathbf T^{(\mathbf n)}\,\!$, for a given plane with unit vector $\mathbf n\,\!$, satisfy the following equations

\begin{align} \left( T^{(n)} \right)^2 &= \sigma_{ij}\sigma_{ik}n_jn_k \ \sigma_\mathrm{n}^2 + \tau_\mathrm{n}^2 &= \sigma_1^2 n_1^2 + \sigma_2^2 n_2^2 + \sigma_3^2 n_3^2 \end{align}\,\!
$\sigma_\mathrm{n} = \sigma_1 n_1^2 + \sigma_2 n_2^2 + \sigma_3 n_3^2\,\!$

Knowing that $n_i n_i = n_1^2+n_2^2+n_3^2 = 1\,\!$, we can solve for $n_1^2\,\!$, $n_2^2\,\!$, $n_3^2\,\!$, using the Gauss elimination method which yields

\begin{align} n_1^2 &= \frac{\tau_\mathrm{n}^2+(\sigma_\mathrm{n} - \sigma_2)(\sigma_\mathrm{n} - \sigma_3)}{(\sigma_1 - \sigma_2)(\sigma_1 - \sigma_3)} \ge 0\ n_2^2 &= \frac{\tau_\mathrm{n}^2+(\sigma_\mathrm{n} - \sigma_3)(\sigma_\mathrm{n} - \sigma_1)}{(\sigma_2 - \sigma_3)(\sigma_2 - \sigma_1)} \ge 0\ n_3^2 &= \frac{\tau_\mathrm{n}^2+(\sigma_\mathrm{n} - \sigma_1)(\sigma_\mathrm{n} - \sigma_2)}{(\sigma_3 - \sigma_1)(\sigma_3 - \sigma_2)} \ge 0 \end{align}\,\!

Since $\sigma_1 > \sigma_2 > \sigma_3\,\!$, and $(n_i)^2\,\!$ is non-negative, the numerators from the these equations satisfy

$\tau_\mathrm{n}^2+(\sigma_\mathrm{n} - \sigma_2)(\sigma_\mathrm{n} - \sigma_3) \ge 0\,\!$ as the denominator $\sigma_1 - \sigma_2 > 0\,\!$ and $\sigma_1 - \sigma_3 > 0\,\!$
$\tau_\mathrm{n}^2+(\sigma_\mathrm{n} - \sigma_3)(\sigma_\mathrm{n} - \sigma_1) \le 0\,\!$ as the denominator $\sigma_2 - \sigma_3 > 0\,\!$ and $\sigma_2 - \sigma_1 < 0\,\!$
$\tau_\mathrm{n}^2+(\sigma_\mathrm{n} - \sigma_1)(\sigma_\mathrm{n} - \sigma_2) \ge 0\,\!$ as the denominator $\sigma_3 - \sigma_1 < 0\,\!$ and $\sigma_3 - \sigma_2 < 0\,\!$

These expressions can be rewritten as

\begin{align} \tau_\mathrm{n}^2 + \left[ \sigma_\mathrm{n}- \tfrac{1}{2} (\sigma_2 + \sigma_3) \right]^2 \ge \left( \tfrac{1}{2}(\sigma_2 - \sigma_3) \right)^2 \ \tau_\mathrm{n}^2 + \left[ \sigma_\mathrm{n}- \tfrac{1}{2} (\sigma_1 + \sigma_3) \right]^2 \le \left( \tfrac{1}{2}(\sigma_1 - \sigma_3) \right)^2 \ \tau_\mathrm{n}^2 + \left[ \sigma_\mathrm{n}- \tfrac{1}{2} (\sigma_1 + \sigma_2) \right]^2 \ge \left( \tfrac{1}{2}(\sigma_1 - \sigma_2) \right)^2 \ \end{align}\,\!

which are the equations of the three Mohr's circles for stress $C_1\,\!$, $C_2\,\!$ , and $C_3\,\!$, with radii $R_1=\tfrac{1}{2}(\sigma_2 - \sigma_3)\,\!$, $R_2=\tfrac{1}{2}(\sigma_1 - \sigma_3)\,\!$ , and $R_3=\tfrac{1}{2}(\sigma_1 - \sigma_2)\,\!$, and their centres with coordinates $\left[\tfrac{1}{2}(\sigma_2 + \sigma_3), 0\right]\,\!$, $\left[\tfrac{1}{2}(\sigma_1 + \sigma_3), 0\right]\,\!$, $\left[\tfrac{1}{2}(\sigma_1 + \sigma_2), 0\right]\,\!$, respectively.

These equations for the Mohr's circles show that all admissible stress points $(\sigma_\mathrm{n}, \tau_\mathrm{n})\,\!$ lie on these circles or within the shaded area enclosed by them (see Figure 7). Stress points $(\sigma_\mathrm{n}, \tau_\mathrm{n})\,\!$ satisfying the equation for circle $C_1\,\!$ lie on, or outside circle $C_1\,\!$. Stress points $(\sigma_\mathrm{n}, \tau_\mathrm{n})\,\!$ satisfying the equation for circle $C_2\,\!$ lie on, or inside circle $C_2\,\!$. And finally, stress points $(\sigma_\mathrm{n}, \tau_\mathrm{n})\,\!$ satisfying the equation for circle $C_3\,\!$ lie on, or outside circle $C_3\,\!$.

## Alternative measures of stress

The Cauchy stress tensor 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-Kirchhoff tensors are identical. These tensors take their names from Gabrio Piola and Gustav Kirchhoff.

#### 1st Piola-Kirchhoff stress tensor

Whereas the Cauchy stress tensor, $\boldsymbol{\sigma}$, relates forces in the present configuration to areas in the present configuration, the 1st Piola-Kirchhoff stress tensor, $\boldsymbol{P}$ relates forces in the present configuration with areas in the reference ("material") configuration.

$\boldsymbol{P} = J~\boldsymbol{\sigma}\cdot\boldsymbol{F}^{-T}$

where $\boldsymbol{F}$ is the deformation gradient and $J= \det\boldsymbol{F}$ is the Jacobian determinant.

In terms of components with respect to an orthonormal basis, the first Piola-Kirchhoff stress is given by

$P_{iL} = J~\sigma_{ik}~F^{-1}_{Lk} = J~\sigma_{ik}~\cfrac{\partial X_L}{\partial x_k}~\,\!$

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 $\boldsymbol{S}$ 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.

$\boldsymbol{S} = J~\boldsymbol{F}^{-1}\cdot\boldsymbol{\sigma}\cdot\boldsymbol{F}^{-T} ~.$

In index notation with respect to an orthonormal basis,

$S_{IL}=J~F^{-1}_{Ik}~F^{-1}_{Lm}~\sigma_{km} = J~\cfrac{\partial X_I}{\partial x_k}~\cfrac{\partial X_L}{\partial x_m}~\sigma_{km} \!\,\!$

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.

## References

1. ^ a b Chen 2007 p.46
2. ^ Irgens p.43
3. ^ Walter D. Pilkey, Orrin H. Pilkey (1974). Mechanics of solids. p. 292.
4. ^ Atanackovic p.4
5. ^ Liu pp.44-45
6. ^ Irgens p.47
7. ^ Atanackovic 2000 p.9
8. ^ Chatterjee pp. 141-146
9. ^ Atanackovic pp.30-32
10. ^ Jaeger pp. 31-32
11. ^ Ameen pp. 59-60
12. ^ Chen pp. 55-57
13. ^ Wu pp. 60-62
14. ^ Prager pp.48-49
15. ^ Jumikis p.20
16. ^ Holtz p. 434
17. ^ Parry p.8