Cauchy stress tensor
Components of stress in three dimensions
Common symbols	σ
SI unit	pascal (Pa)
Other units	Pound per square inch (psi),bar
InSI base units	Pa =kg⋅m−1⋅s−2
Behaviour under coord transformation	tensor
Dimension	${\displaystyle {\mathsf{L}}^{-1}\mathsf{M}{\mathsf{T}}^{-2}}$

Incontinuum mechanics, theCauchy stress tensor (symbol⁠${\displaystyle \mathit{\sigma }}$⁠, named afterAugustin-Louis Cauchy), also calledtrue stress tensor^[1] or simplystress tensor, completely defines the state ofstress at a point inside a material in thedeformed state, placement, or configuration. The second ordertensor consists of nine components${\displaystyle {\sigma }_{ij}}$ and relates a unit-lengthdirection vectore to thetraction vectorT^(e) across a surface perpendicular toe:

${\displaystyle {\mathbf{T}}^{(\mathbf{e})}=\mathbf{e}\cdot \mathit{\sigma }\text{or}{T}_j^{(\mathbf{e})}=\sum _i{\sigma }_{ij}{e}_i.}$^[a]

TheSI unit of both stress tensor and traction vector is thenewton per square metre (N/m²) orpascal (Pa), corresponding to the stress scalar. The unit vector isdimensionless.

The Cauchy stress tensor obeys thetensor transformation law under a change in the system of coordinates. A graphical representation of this transformation law is theMohr's circle for stress.

The Cauchy stress tensor is used forstress analysis of material bodies experiencingsmall deformations: it is a central concept in thelinear theory of elasticity. For large deformations, also calledfinite deformations, other measures of stress are required, such as thePiola–Kirchhoff stress tensor, theBiot stress tensor, and theKirchhoff stress tensor.

According to the principle ofconservation 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 ofconservation of angular momentum, equilibrium requires that the summation ofmoments with respect to an arbitrary point is zero, which leads to the conclusion that thestress tensor is symmetric, thus having only six independent stress components, instead of the original nine. 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 theKnudsen number is close to one,⁠${\displaystyle K_n\to 1}$⁠, or the continuum is anon-Newtonian fluid, which can lead to rotationally non-invariant fluids, such aspolymers.

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 threeeigenvalues of the stress tensor, which are called theprincipal stresses.

TheEuler–Cauchy stress principle states thatupon any surface that divides the body, the action of one part of the body on the other is equivalent (equipollent) to the system of distributed forces and couples on the surface dividing the body,^[2] and it is represented by a field⁠${\displaystyle {\mathbf{T}}^{(\mathbf{n})}}$⁠, called thetraction vector, defined on the surface${\displaystyle S}$ and assumed to depend continuously on the surface's normal unit vector⁠${\displaystyle \mathbf{n}}$⁠.^{[3][4]: p.66–96}

To formulate the Euler–Cauchy stress principle, consider a surface${\displaystyle S}$ passing through an internal material point${\displaystyle P}$ dividing the continuous body into two segments, as seen in Figure 2.1a or 2.1b (one may use either the cutting plane diagram or the diagram with the arbitrary volume inside the continuum enclosed by the surface⁠${\displaystyle S}$⁠).

Following the classical dynamics ofNewton andEuler, the motion of a material body is produced by the action of externally appliedforces which are assumed to be of two kinds:surface forces${\displaystyle \mathbf{F}}$ andbody forces⁠${\displaystyle \mathbf{b}}$⁠.^[5] Thus, the total force${\displaystyle \mathcal{F}}$ applied to a body or to a portion of the body can be expressed as:

${\displaystyle \mathcal{F}=\mathbf{b}+\mathbf{F}}$

Only surface forces will be discussed in this article as they are relevant to the Cauchy stress tensor.

When the body is subjected to external surface forces orcontact forces⁠${\displaystyle \mathbf{F}}$⁠, followingEuler's equations of motion, internal contact forces and moments are transmitted from point to point in the body, and from one segment to the other through the dividing surface⁠${\displaystyle S}$⁠, due to the mechanical contact of one portion of the continuum onto the other (Figure 2.1a and 2.1b). On an element of area${\displaystyle \mathrm{\Delta }S}$ containing${\displaystyle P}$, with normalvector⁠${\displaystyle \mathbf{n}}$⁠, the force distribution is equipollent to a contact force${\displaystyle \mathrm{\Delta }\mathbf{F}}$ exerted at point${\displaystyle P}$ and surface moment${\displaystyle \mathrm{\Delta }\mathbf{M}}$. In particular, thecontact force is given by

${\displaystyle \mathrm{\Delta }\mathbf{F}={\mathbf{T}}^{(\mathbf{n})}\mathrm{\Delta }S,}$

where${\displaystyle {\mathbf{T}}^{(\mathbf{n})}}$ is themean surface traction.

Cauchy's stress principle asserts^{[6]: 47–102} that as${\displaystyle \mathrm{\Delta }S}$ tends to zero the ratio${\displaystyle \mathrm{\Delta }\mathbf{F}/\mathrm{\Delta }S}$ becomes${\displaystyle d\mathbf{F}/dS}$ and the couple stress vector${\displaystyle \mathrm{\Delta }\mathbf{M}}$ vanishes. In specific fields of continuum mechanics the couple stress is assumed not to vanish; however, classical branches of continuum mechanics address non-polar materials which do not consider couple stresses and body moments.

The resultant vector${\displaystyle d\mathbf{F}/dS}$ is defined as thesurface traction,^[7] also calledstress vector,^[8]traction,^[4] ortraction vector.^[6] given by${\displaystyle {\mathbf{T}}^{(\mathbf{n})}=T_i^{(\mathbf{n})}{\mathbf{e}}_i}$ at the point${\displaystyle P}$ associated with a plane with a normal vector⁠${\displaystyle \mathbf{n}}$⁠:

${\displaystyle T_i^{(\mathbf{n})}=\underset{\mathrm{\Delta }S\to 0}{lim}\frac{\mathrm{\Delta }{F}_i}{\mathrm{\Delta }S}=\frac{d{F}_i}{dS}.}$

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

This implies that the balancing action of internal contact forces generates acontact force density orCauchy traction field^[5]${\displaystyle \mathbf{T}(\mathbf{n},\mathbf{x},t)}$ that represents a distribution of internal contact forces throughout the volume of the body in a particularconfiguration of the body at a given time⁠${\displaystyle t}$⁠. It is not avector field because it depends not only on the position${\displaystyle \mathbf{x}}$ of a particular material point, but also on the local orientation of the surface element as defined by its normal vector⁠${\displaystyle \mathbf{n}}$⁠.^[9]

Depending on the orientation of the plane under consideration, the stress vector may not necessarily be perpendicular to that plane,i.e. parallel to⁠${\displaystyle \mathbf{n}}$⁠, and can be resolved into two components (Figure 2.1c):

• one normal to the plane, callednormal stress

  ${\displaystyle {\sigma }_{\mathrm{n}}=\underset{\mathrm{\Delta }S\to 0}{lim}\frac{\mathrm{\Delta }{F}_{\mathrm{n}}}{\mathrm{\Delta }S}=\frac{d{F}_{\mathrm{n}}}{dS},}$

where${\displaystyle d{F}_{\mathrm{n}}}$ is the normal component of the force${\displaystyle d\mathbf{F}}$ to the differential area${\displaystyle dS}$

• and the other parallel to this plane, called theshear stress

  ${\displaystyle \tau =\underset{\mathrm{\Delta }S\to 0}{lim}\frac{\mathrm{\Delta }{F}_{\mathrm{s}}}{\mathrm{\Delta }S}=\frac{d{F}_{\mathrm{s}}}{dS},}$

where${\displaystyle d{F}_{\mathrm{s}}}$ is the tangential component of the force${\displaystyle d\mathbf{F}}$ to the differential surface area⁠${\displaystyle dS}$⁠. The shear stress can be further decomposed into two mutually perpendicular vectors.

According to theCauchy Postulate, the stress vector${\displaystyle {\mathbf{T}}^{(\mathbf{n})}}$ remains unchanged for all surfaces passing through the point${\displaystyle P}$ and having the same normal vector${\displaystyle \mathbf{n}}$ at⁠${\displaystyle P}$⁠,^[7][10] i.e., having a commontangent at⁠${\displaystyle P}$⁠. This means that the stress vector is a function of the normal vector${\displaystyle \mathbf{n}}$ only, and is not influenced by the curvature of the internal surfaces.

A consequence of Cauchy's postulate isCauchy's Fundamental Lemma,^[1][7][11] also called theCauchy reciprocal theorem,^{[12]: p.103–130} which states that the stress vectors acting on opposite sides of the same surface are equal in magnitude and opposite in direction. Cauchy's fundamental lemma is equivalent toNewton's third law of motion of action and reaction, and is expressed as

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

The state of stress at a point in the body is then defined by all the stress vectorsT⁽ⁿ⁾ associated with all planes (infinite in number) that pass through that point.^[13] However, according toCauchy's fundamental theorem,^[11] also calledCauchy's stress theorem,^[1] merely by knowing the stress vectors on three mutually perpendicular planes, the stress vector on any other plane passing through that point can be found through coordinate transformation equations.

Cauchy's stress theorem states that there exists a second-ordertensor fieldσ(x, t), called the Cauchy stress tensor, independent ofn, such thatT is a linear function ofn:

${\displaystyle {\mathbf{T}}^{(\mathbf{n})}=\mathbf{n}\cdot \mathit{\sigma }\text{or}{T}_j^{(\mathbf{n})}={\sigma }_{ij}{n}_i.}$

This equation implies that the stress vectorT⁽ⁿ⁾ at any pointP in a continuum associated with a plane with normal unit vectorn can be expressed as a function of the stress vectors on the planes perpendicular to the coordinate axes,i.e. in terms of the componentsσ_ij of the stress tensorσ.

To prove this expression, consider atetrahedron with three faces oriented in the coordinate planes, and with an infinitesimal area dA oriented in an arbitrary direction specified by a normal unit vectorn (Figure 2.2). The tetrahedron is formed by slicing the infinitesimal element along an arbitrary plane with unit normaln. The stress vector on this plane is denoted byT⁽ⁿ⁾. The stress vectors acting on the faces of the tetrahedron are denoted asT^(e₁),T^(e₂), andT^(e₃), and are by definition the componentsσ_ij of the stress tensorσ. This tetrahedron is sometimes called theCauchy tetrahedron. The equilibrium of forces,i.e.Euler's first law of motion (Newton's second law of motion), gives:

${\displaystyle {\mathbf{T}}^{(\mathbf{n})}dA-{\mathbf{T}}^{({\mathbf{e}}_1)}d{A}_1-{\mathbf{T}}^{({\mathbf{e}}_2)}d{A}_2-{\mathbf{T}}^{({\mathbf{e}}_3)}d{A}_3=\rho \left(\frac{h}{3}dA\right)\mathbf{a},}$

where the right-hand-side represents the product of the mass enclosed by the tetrahedron and its acceleration:ρ is the density,a is the acceleration, andh is the height of the tetrahedron, considering the planen 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):

${\displaystyle d{A}_1=\left(\mathbf{n}\cdot {\mathbf{e}}_1\right)dA=n_{1}dA,}$

${\displaystyle d{A}_2=\left(\mathbf{n}\cdot {\mathbf{e}}_2\right)dA=n_{2}dA,}$

${\displaystyle d{A}_3=\left(\mathbf{n}\cdot {\mathbf{e}}_3\right)dA=n_{3}dA,}$

and then substituting into the equation to cancel out dA:

${\displaystyle {\mathbf{T}}^{(\mathbf{n})}-{\mathbf{T}}^{({\mathbf{e}}_1)}{n}_1-{\mathbf{T}}^{({\mathbf{e}}_2)}{n}_2-{\mathbf{T}}^{({\mathbf{e}}_3)}{n}_3=\rho \left(\frac{h}{3}\right)\mathbf{a}.}$

To consider the limiting case as the tetrahedron shrinks to a point,h must go to 0 (intuitively, the planen is translated alongn towardO). As a result, the right-hand-side of the equation approaches 0, so

${\displaystyle {\mathbf{T}}^{(\mathbf{n})}={\mathbf{T}}^{({\mathbf{e}}_1)}{n}_1+{\mathbf{T}}^{({\mathbf{e}}_2)}{n}_2+{\mathbf{T}}^{({\mathbf{e}}_3)}{n}_3.}$

Assuming a material element (see figure at the top of the page) with planes perpendicular to the coordinate axes of a Cartesian coordinate system, the stress vectors associated with each of the element planes,i.e.T^(e₁),T^(e₂), andT^(e₃) can be decomposed into a normal component and two shear components,i.e. components in the direction of the three coordinate axes. For the particular case of a surface with normalunit vector oriented in the direction of thex₁-axis, denote the normal stress byσ₁₁, and the two shear stresses asσ₁₂ andσ₁₃:

${\displaystyle {\mathbf{T}}^{({\mathbf{e}}_1)}=T_1^{({\mathbf{e}}_1)}{\mathbf{e}}_1+T_2^{({\mathbf{e}}_1)}{\mathbf{e}}_2+T_3^{({\mathbf{e}}_1)}{\mathbf{e}}_3={\sigma }_{11}{\mathbf{e}}_1+{\sigma }_{12}{\mathbf{e}}_2+{\sigma }_{13}{\mathbf{e}}_3,}$

${\displaystyle {\mathbf{T}}^{({\mathbf{e}}_2)}=T_1^{({\mathbf{e}}_2)}{\mathbf{e}}_1+T_2^{({\mathbf{e}}_2)}{\mathbf{e}}_2+T_3^{({\mathbf{e}}_2)}{\mathbf{e}}_3={\sigma }_{21}{\mathbf{e}}_1+{\sigma }_{22}{\mathbf{e}}_2+{\sigma }_{23}{\mathbf{e}}_3,}$

${\displaystyle {\mathbf{T}}^{({\mathbf{e}}_3)}=T_1^{({\mathbf{e}}_3)}{\mathbf{e}}_1+T_2^{({\mathbf{e}}_3)}{\mathbf{e}}_2+T_3^{({\mathbf{e}}_3)}{\mathbf{e}}_3={\sigma }_{31}{\mathbf{e}}_1+{\sigma }_{32}{\mathbf{e}}_2+{\sigma }_{33}{\mathbf{e}}_3,}$

In index notation this is

${\displaystyle {\mathbf{T}}^{({\mathbf{e}}_i)}=T_j^{({\mathbf{e}}_i)}{\mathbf{e}}_j={\sigma }_{ij}{\mathbf{e}}_j.}$

The nine componentsσ_ij of the stress vectors are the components of a second-order Cartesian tensor called theCauchy stress tensor, which can be used to completely define the state of stress at a point and is given by

whereσ₁₁,σ₂₂, andσ₃₃ are normal stresses, andσ₁₂,σ₁₃,σ₂₁,σ₂₃,σ₃₁, andσ₃₂ are shear stresses. The first indexi indicates that the stress acts on a plane normal to theX_i -axis, and the second indexj denotes the direction in which the stress acts (for example,σ₁₂ implies that the stress is acting on the plane that is normal to the 1st axis i.e.,X₁, and acts along the 2nd axis i.e.,X₂). A stress component is positive if it acts in the positive direction of the coordinate axes, and if the plane where it acts has an outward normal vector pointing in the positive coordinate direction.

Thus, using the components of the stress tensor

or, equivalently,

${\displaystyle T_j^{(\mathbf{n})}={\sigma }_{ij}{n}_i.}$

Alternatively, in matrix form we have

TheVoigt notation representation of the Cauchy stress tensor takes advantage of thesymmetry of the stress tensor to express the stress as a six-dimensional vector of the form:

The Voigt notation is used extensively in representing stress–strain relations in solid mechanics and for computational efficiency in numerical structural mechanics software.

It can be shown that the stress tensor is acontravariant second order tensor, which is a statement of how it transforms under a change of the coordinate system. From anx_i-system to an x_i'-system, the componentsσ_ij in the initial system are transformed into the componentsσ_ij′ in the new system according to the tensor transformation rule (Figure 2.4):

${\displaystyle {\sigma }_{ij}^{\prime }=a_{im}{a}_{jn}{\sigma }_{mn}\text{or}{\mathit{\sigma }}^{\prime }=\mathbf{A}\mathit{\sigma }{\mathbf{A}}^{\mathsf{\text{T}}},}$

whereA is arotation matrix with componentsa_ij. In matrix form this is

Expanding thematrix operation, and simplifying terms using thesymmetry of the stress tensor, gives

TheMohr circle for stress is a graphical representation of this transformation of stresses.

The magnitude of thenormal stress componentσ_n of any stress vectorT⁽ⁿ⁾ acting on an arbitrary plane with normal unit vectorn at a given point, in terms of the componentsσ_ij of the stress tensorσ, is thedot product of the stress vector and the normal unit vector:

The magnitude of the shear stress componentτ_n, acting orthogonal to the vectorn, can then be found using thePythagorean theorem:

where

${\displaystyle {\left(T^{(\mathbf{n})}\right)}^2=T_i^{(\mathbf{n})}{T}_i^{(\mathbf{n})}=\left({\sigma }_{ij}{n}_j\right)\left({\sigma }_{ik}{n}_k\right)={\sigma }_{ij}{\sigma }_{ik}{n}_j{n}_k.}$

According to the principle ofconservation 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:

${\displaystyle {\sigma }_{ji,j}+F_i=0,}$

where${\displaystyle {\sigma }_{ji,j}=\sum _j{\mathrm{\partial }}_j{\sigma }_{ji}}$

For example, for ahydrostatic fluid in equilibrium conditions, the stress tensor takes on the form:

${\displaystyle {\sigma }_{ij}=-p{\delta }_{ij},}$

where${\displaystyle p}$ is the hydrostatic pressure, and${\displaystyle {\delta }_{ij}}$ is theKronecker delta.

Derivation of equilibrium equations

Consider a continuum body (see Figure 4) occupying a volume${\displaystyle V}$, having a surface area⁠${\displaystyle S}$⁠, with defined traction or surface forces${\displaystyle T_i^{(n)}}$ per unit area acting on every point of the body surface, and body forces${\displaystyle F_i}$ per unit of volume on every point within the volume${\displaystyle V}$. Thus, if the body is inequilibrium the resultant force acting on the volume is zero, thus: ${\displaystyle {\int }_S{T}_i^{(n)}dS+{\int }_V{F}_{i}dV=0}$ By definition the stress vector is${\displaystyle T_i^{(n)}={\sigma }_{ji}{n}_j}$, then ${\displaystyle {\int }_S{\sigma }_{ji}{n}_{j}dS+{\int }_V{F}_{i}dV=0}$ Using theGauss's divergence theorem to convert a surface integral to a volume integral gives ${\displaystyle {\int }_V{\sigma }_{ji,j}dV+{\int }_V{F}_{i}dV=0}$ ${\displaystyle {\int }_V({\sigma }_{ji,j}+F_i)dV=0}$ For an arbitrary volume the integral vanishes, and we have theequilibrium equations ${\displaystyle {\sigma }_{ji,j}+F_i=0}$

According to the principle ofconservation of angular momentum, equilibrium requires that the summation ofmoments with respect to an arbitrary point is zero, which leads to the conclusion that the stress tensor issymmetric, thus having only six independent stress components, instead of the original nine:

${\displaystyle {\sigma }_{ij}={\sigma }_{ji}}$

Derivation of symmetry of the stress tensor

Summing moments about pointO (Figure 4) the resultant moment is zero as the body is in equilibrium. Thus, where${\displaystyle \mathbf{r}}$ is the position vector and is expressed as ${\displaystyle \mathbf{r}=x_j{\mathbf{e}}_j}$ and${\displaystyle {\epsilon }_{ijk}}$ is theLevi-Civita symbol. Knowing that${\displaystyle T_k^{(n)}={\sigma }_{mk}{n}_m}$ and using Gauss's divergence theorem to change from a surface integral to a volume integral, we have The second integral is zero as it contains the equilibrium equations. This leaves the first integral, where${\displaystyle x_{j,m}={\delta }_{jm}}$, therefore ${\displaystyle {\int }_V({\epsilon }_{ijk}{\sigma }_{jk})dV=0}$ For an arbitrary volume V, we then have ${\displaystyle {\epsilon }_{ijk}{\sigma }_{jk}=0,}$ which is satisfied at every point within the body. Expanding this equation we have ${\displaystyle {\sigma }_{12}={\sigma }_{21}}$,${\displaystyle {\sigma }_{23}={\sigma }_{32}}$, and${\displaystyle {\sigma }_{13}={\sigma }_{31}}$ or in general ${\displaystyle {\sigma }_{ij}={\sigma }_{ji}}$ This proves that the stress tensor is symmetric

However, in the presence of couple-stresses, i.e. moments per unit volume, the stress tensor is non-symmetric. This also is the case when theKnudsen number is close to one,⁠${\displaystyle K_n\to 1}$⁠, or the continuum is a non-Newtonian fluid, which can lead to rotationally non-invariant fluids, such aspolymers.

At every point in a stressed body there are at least three planes, calledprincipal planes, with normal vectors⁠${\displaystyle \mathbf{n}}$⁠, calledprincipal directions, where the corresponding stress vector is perpendicular to the plane, i.e., parallel or in the same direction as the normal vector⁠${\displaystyle \mathbf{n}}$⁠, and where there are no normal shear stresses⁠${\displaystyle {\tau }_{\mathrm{n}}}$⁠. The three stresses normal to these principal planes are calledprincipal stresses.

The components${\displaystyle {\sigma }_{ij}}$ 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 certaininvariants 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 themagnitude of the vector is a physical quantity (a scalar) and is independent of theCartesian coordinate system chosen to represent the vector (so long as it isnormal). 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 oreigenvectors.

A stress vector parallel to the normal unit vector${\displaystyle \mathbf{n}}$ is given by:

${\displaystyle {\mathbf{T}}^{(\mathbf{n})}=\lambda \mathbf{n}={\sigma }_{\mathrm{n}}\mathbf{n},}$

where${\displaystyle \lambda }$ is a constant of proportionality, and in this particular case corresponds to the magnitudes${\displaystyle {\sigma }_{\mathrm{n}}}$ of the normal stress vectors or principal stresses.

Knowing that${\displaystyle T_i^{(n)}={\sigma }_{ij}{n}_j}$ and⁠${\displaystyle 1}$⁠, we have

This is ahomogeneous system, i.e. equal to zero, of three linear equations where${\displaystyle n_j}$ are the unknowns. To obtain a nontrivial (non-zero) solution for⁠${\displaystyle n_j}$⁠, the matrix determinant of the coefficients must be equal to zero, i.e. the system is singular. Thus,

Expanding the determinant leads to thecharacteristic equation

${\displaystyle \left|{\sigma }_{ij}-\lambda {\delta }_{ij}\right|=-{\lambda }^3+I_1{\lambda }^2-I_2\lambda +I_3=0}$

where

The characteristic equation has three real roots${\displaystyle {\lambda }_i}$ (i.e. with a zeroimaginary component) due to the stress tensor being symmetric. The${\displaystyle {\sigma }_1=max\left({\lambda }_1,{\lambda }_2,{\lambda }_3\right)}$,${\displaystyle {\sigma }_3=min\left({\lambda }_1,{\lambda }_2,{\lambda }_3\right)}$ and⁠${\displaystyle {\sigma }_2=I_1-{\sigma }_1-{\sigma }_3}$⁠, are the principal stresses, functions of the eigenvalues⁠${\displaystyle {\lambda }_i}$⁠. The eigenvalues are the roots of thecharacteristic polynomial. The principal stresses are unique for a given stress tensor. Therefore, from the characteristic equation, the coefficients${\displaystyle I_1}$,${\displaystyle I_2}$ and${\displaystyle I_3}$, called the first, second, and thirdstress invariants, respectively, always have the same value regardless of the coordinate system's orientation.

For each eigenvalue, there is a non-trivial solution for${\displaystyle n_j}$ in the equation⁠${\displaystyle \left({\sigma }_{ij}-\lambda {\delta }_{ij}\right)n_j=0}$⁠. These solutions are the principal directions oreigenvectors defining the plane where the principal stresses act. The principal stresses and principal directions characterize the stress at a point and are independent of the orientation.

A coordinate system with axes oriented to the principal directions implies that the normal stresses are the principal stresses and the stress tensor is represented by a diagonal matrix:

The principal stresses can be combined to form the stress invariants,⁠${\displaystyle I_1}$⁠,⁠${\displaystyle I_2}$⁠, and⁠${\displaystyle I_3}$⁠. The first and third invariant are the trace and determinant respectively, of the stress tensor. Thus,

Because of its simplicity, the principal coordinate system is often useful when considering the state of the elastic medium at a particular point. Principal stresses are often expressed in the following equation for evaluating stresses in the x and y directions or axial and bending stresses on a part.^{[14]: p.58–59} The principal normal stresses can then be used to calculate thevon Mises stress and ultimately the safety factor and margin of safety.

${\displaystyle {\sigma }_1,{\sigma }_2=\frac{{\sigma }_x+{\sigma }_y}{2}\pm \sqrt{{\left(\frac{{\sigma }_x-{\sigma }_y}{2}\right)}^2+{\tau }_{xy}^2}.}$

Using just the part of the equation under thesquare root is equal to the maximum and minimum shear stress for plus and minus. This is shown as:

${\displaystyle {\tau }_{max},{\tau }_{min}=\pm \sqrt{{\left(\frac{{\sigma }_x-{\sigma }_y}{2}\right)}^2+{\tau }_{xy}^2}.}$

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${\displaystyle {45}^{\circ }}$ from the principal stress planes. The maximum shear stress is expressed as

${\displaystyle {\tau }_{max}=\frac{1}{2}\left|{\sigma }_{max}-{\sigma }_{min}\right|.}$

Assuming${\displaystyle {\sigma }_1\ge {\sigma }_2\ge {\sigma }_3}$ then

${\displaystyle {\tau }_{max}=\frac{1}{2}\left|{\sigma }_1-{\sigma }_3\right|}$

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

${\displaystyle {\sigma }_{\text{n}}=\frac{1}{2}\left({\sigma }_1+{\sigma }_3\right).}$

Derivation of the maximum and minimum shear stresses[8]: p.45–78 [11]: p.1–46 [13][15]: p.111–157 [16]: p.9–41 [17]: p.33–66 [18]: p.43–61

The normal stress can be written in terms of principal stresses${\displaystyle ({\sigma }_1\ge {\sigma }_2\ge {\sigma }_3)}$ as Knowing that⁠${\displaystyle {\left(T^{(n)}\right)}^2={\sigma }_{ij}{\sigma }_{ik}{n}_j{n}_k}$⁠, the shear stress in terms of principal stresses components is expressed as The maximum shear stress at a point in a continuum body is determined by maximizing${\displaystyle {\tau }_{\mathrm{n}}^2}$ subject to the condition that ${\displaystyle n_i{n}_i=n_1^2+n_2^2+n_3^2=1.}$ This is a constrained maximization problem, which can be solved using theLagrangian multiplier technique to convert the problem into an unconstrained optimization problem. Thus, the stationary values (maximum and minimum values)of${\displaystyle {\tau }_{\text{n}}^2}$ occur where the gradient of${\displaystyle {\tau }_{\text{n}}^2}$ is parallel to the gradient of⁠${\displaystyle F}$⁠. The Lagrangian function for this problem can be written as where${\displaystyle \lambda }$ is the Lagrangian multiplier (which is different from the${\displaystyle \lambda }$ use to denote eigenvalues). The extreme values of these functions are ${\displaystyle \frac{\mathrm{\partial }F}{\mathrm{\partial }{n}_1}=0\frac{\mathrm{\partial }F}{\mathrm{\partial }{n}_2}=0\frac{\mathrm{\partial }F}{\mathrm{\partial }{n}_3}=0}$ thence ${\displaystyle \frac{\mathrm{\partial }F}{\mathrm{\partial }{n}_1}=n_1{\sigma }_1^2-2n_1{\sigma }_1\left({\sigma }_1{n}_1^2+{\sigma }_2{n}_2^2+{\sigma }_3{n}_3^2\right)+\lambda n_1=0}$ ${\displaystyle \frac{\mathrm{\partial }F}{\mathrm{\partial }{n}_2}=n_2{\sigma }_2^2-2n_2{\sigma }_2\left({\sigma }_1{n}_1^2+{\sigma }_2{n}_2^2+{\sigma }_3{n}_3^2\right)+\lambda n_2=0}$ ${\displaystyle \frac{\mathrm{\partial }F}{\mathrm{\partial }{n}_3}=n_3{\sigma }_3^2-2n_3{\sigma }_3\left({\sigma }_1{n}_1^2+{\sigma }_2{n}_2^2+{\sigma }_3{n}_3^2\right)+\lambda n_3=0}$ These three equations together with the condition${\displaystyle n_i{n}_i=1}$ may be solved for⁠${\displaystyle \lambda }$⁠,⁠${\displaystyle n_1}$⁠,⁠${\displaystyle n_2}$⁠, and⁠${\displaystyle n_3}$⁠. By multiplying the first three equations by⁠${\displaystyle n_1}$⁠,⁠${\displaystyle n_2}$⁠, and⁠${\displaystyle n_3}$⁠, respectively, and knowing that${\displaystyle {\sigma }_{\text{n}}={\sigma }_{ij}{n}_i{n}_j={\sigma }_1{n}_1^2+{\sigma }_2{n}_2^2+{\sigma }_3{n}_3^2}$ we obtain ${\displaystyle n_1^2{\sigma }_1^2-2{\sigma }_1{n}_1^2{\sigma }_{\text{n}}+n_1^2\lambda =0}$ ${\displaystyle n_2^2{\sigma }_2^2-2{\sigma }_2{n}_2^2{\sigma }_{\text{n}}+n_2^2\lambda =0}$ ${\displaystyle n_3^2{\sigma }_3^2-2{\sigma }_1{n}_3^2{\sigma }_{\text{n}}+n_3^2\lambda =0}$ Adding these three equations we get This result can be substituted into each of the first three equations to obtain Doing the same for the other two equations we have ${\displaystyle \frac{\mathrm{\partial }F}{\mathrm{\partial }{n}_2}=\left({\sigma }_2^2-2{\sigma }_2{\sigma }_{\text{n}}+{\sigma }_{\text{n}}^2-{\tau }_{\text{n}}^2\right)n_2=0}$ ${\displaystyle \frac{\mathrm{\partial }F}{\mathrm{\partial }{n}_3}=\left({\sigma }_3^2-2{\sigma }_3{\sigma }_{\text{n}}+{\sigma }_{\text{n}}^2-{\tau }_{\text{n}}^2\right)n_3=0}$ A first approach to solve these last three equations is to consider the trivial solution⁠${\displaystyle n_i=0}$⁠. However, this option does not fulfill the constraint⁠${\displaystyle n_i{n}_i=1}$⁠. Considering the solution where${\displaystyle n_1=n_2=0}$ and⁠${\displaystyle n_3\ne 0}$⁠, it is determine from the condition${\displaystyle n_i{n}_i=1}$ that⁠${\displaystyle n_3=\pm 1}$⁠, then from the original equation for${\displaystyle {\tau }_{\text{n}}^2}$ it is seen that⁠${\displaystyle {\tau }_{\text{n}}=0}$⁠.The other two possible values for${\displaystyle {\tau }_{\text{n}}}$ can be obtained similarly by assuming ${\displaystyle n_1=n_3=0}$ and${\displaystyle n_2\ne 0}$ ${\displaystyle n_2=n_3=0}$ and${\displaystyle n_1\ne 0}$ Thus, one set of solutions for these four equations is: These correspond to minimum values for${\displaystyle {\tau }_{\mathrm{n}}}$ and verifies that there are no shear stresses on planes normal to the principal directions of stress, as shown previously. A second set of solutions is obtained by assuming⁠${\displaystyle n_1=0}$⁠,⁠${\displaystyle n_2\ne 0}$⁠ and⁠${\displaystyle n_3\ne 0}$⁠. Thus we have ${\displaystyle \frac{\mathrm{\partial }F}{\mathrm{\partial }{n}_2}={\sigma }_2^2-2{\sigma }_2{\sigma }_{\text{n}}+{\sigma }_{\text{n}}^2-{\tau }_{\text{n}}^2=0}$ ${\displaystyle \frac{\mathrm{\partial }F}{\mathrm{\partial }{n}_3}={\sigma }_3^2-2{\sigma }_3{\sigma }_{\text{n}}+{\sigma }_{\text{n}}^2-{\tau }_{\text{n}}^2=0}$ To find the values for${\displaystyle n_2}$ and${\displaystyle n_3}$ we first add these two equations Knowing that for${\displaystyle n_1=0}$ ${\displaystyle {\sigma }_{\text{n}}={\sigma }_1{n}_1^2+{\sigma }_2{n}_2^2+{\sigma }_3{n}_3^2={\sigma }_2{n}_2^2+{\sigma }_3{n}_3^2}$ and ${\displaystyle n_1^2+n_2^2+n_3^2=n_2^2+n_3^2=1}$ we have and solving for${\displaystyle n_2}$ we have ${\displaystyle n_2=\pm \frac{1}{\sqrt{2}}}$ Then solving for${\displaystyle n_3}$ we have ${\displaystyle n_3=\sqrt{1-n_2^2}=\pm \frac{1}{\sqrt{2}}}$ and The other two possible values for${\displaystyle {\tau }_{\text{n}}}$ can be obtained similarly by assuming ⁠${\displaystyle n_2=0}$⁠,⁠${\displaystyle n_1\ne 0}$⁠ and${\displaystyle n_3\ne 0}$ ⁠${\displaystyle n_3=0}$⁠,⁠${\displaystyle n_1\ne 0}$⁠ and${\displaystyle n_2\ne 0}$ Therefore, the second set of solutions for⁠${\displaystyle \frac{\mathrm{\partial }F}{\mathrm{\partial }{n}_1}=0}$⁠, representing a maximum for${\displaystyle {\tau }_{\text{n}}}$ is ${\displaystyle n_1=0,n_2=\pm \frac{1}{\sqrt{2}},n_3=\pm \frac{1}{\sqrt{2}},{\tau }_{\text{n}}=\pm \frac{{\sigma }_2-{\sigma }_3}{2}}$ ${\displaystyle n_1=\pm \frac{1}{\sqrt{2}},n_2=0,n_3=\pm \frac{1}{\sqrt{2}},{\tau }_{\text{n}}=\pm \frac{{\sigma }_1-{\sigma }_3}{2}}$ ${\displaystyle n_1=\pm \frac{1}{\sqrt{2}},n_2=\pm \frac{1}{\sqrt{2}},n_3=0,{\tau }_{\text{n}}=\pm \frac{{\sigma }_1-{\sigma }_2}{2}}$ Therefore, assuming⁠${\displaystyle {\sigma }_1\ge {\sigma }_2\ge {\sigma }_3}$⁠, the maximum shear stress is expressed by ${\displaystyle {\tau }_{max}=\frac{1}{2}\left|{\sigma }_1-{\sigma }_3\right|=\frac{1}{2}\left|{\sigma }_{max}-{\sigma }_{min}\right|}$ and it can be stated as being equal to one-half the difference between the largest and smallest principal stresses, acting on the plane that bisects the angle between the directions of the largest and smallest principal stresses.