** **

** **

** **

**4. Kinetics**

** **

** **

** **

Our next objective is to outline the mathematical formulas that describe internal and external forces acting on a solid. Just as there are many different strain measures, there are several different definitions of internal force. We shall see that internal forces can be described as a second order tensor, which must be symmetric. Thus, internal forces can always be quantified by a set of six numbers, and the various different definitions are all equivalent.

##### 4.1 Surface traction and internal body force

Forces can be applied to a solid body in two ways.

(i) A force can be applied to its boundary: examples include fluid pressure, wind loading, or forces arising from contact with another solid.

(ii) The solid can be subjected to *body forces*, which act on the interior
of the solid. Examples include
gravitational loading, or electromagnetic forces.

The *surface
traction vector***t** at a point
on the surface represents the force acting on the surface per unit area of the
deformed solid.** **

Formally,
let *dA* be an element of area on a
surface. Suppose that *dA* is subjected to a force $dP$. Then

$t=\underset{dA\to 0}{\mathrm{lim}}\frac{dP}{dA}$

The
resultant force acting on any portion *S *of
the surface of the deformed solid is

$P={\displaystyle \underset{S}{\int}t\text{\hspace{0.17em}}dA}$

**Surface traction, like `true stress,’ should be
thought of as acting on the deformed solid.**

**Normal**** and shear tractions**

The traction vector is often resolved into components acting normal and tangential to a surface, as shown in the picture.

The normal component is referred to as the **normal traction**, and the tangential
component is known as the **shear traction**.

Formally, let **n** denote a unit vector normal to the surface. Then

${t}_{n}=\left(t\cdot n\right)n\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{t}_{t}=t-{t}_{n}$

The ** body
force vector **denotes the external force acting on the interior of a
solid, per unit mass.

Formally, let *dV*
denote an infinitesimal volume element within the deformed solid, and let $\rho $ denote the mass density (mass per unit
deformed volume). Suppose that the
element is subjected to a force $dP$. Then

$b=\frac{1}{\rho}\underset{dV\to 0}{\mathrm{lim}}\frac{dP}{dV}$

The resultant body force
acting on any volume *V* within the **deformed** solid is

$P={\displaystyle \underset{V}{\int}\rho \text{\hspace{0.17em}}b\text{\hspace{0.17em}}dV}$

#### 4.2 Traction acting on planes within a solid

Every plane in the interior of a solid is subjected to a distribution of traction. To see this, consider a loaded, solid, body in static equilibrium. Imagine cutting the solid in two. The two parts of the solid must each be in static equilibrium. This is possible only if forces act on the planes that were created by the cut.

The **internal
traction vector T(n)** represents the force per unit area acting on a section
of the deformed body across a plane with outer normal vector **n**.

Formally,
let *dA* be an element of area in the
interior of the solid, with normal **n**. Suppose that the material on the underside of
*dA* is subjected to a force $d{P}^{(n)}$ across the plane *dA*. Then

$T(n)=\underset{dA\to 0}{\mathrm{lim}}\frac{d{P}^{(n)}}{dA}$

Note
that internal traction is the *force per
unit area of the deformed solid*, like `true stress’

**The resultant force acting on any internal volume V with boundary surface A within a deformed solid is**

$P={\displaystyle \underset{A}{\int}T(n)dA+}{\displaystyle \underset{V}{\int}\rho \text{\hspace{0.17em}}b\text{\hspace{0.17em}}dV}$

The
first term is the resultant force acting on the internal surface *A*, the second term is the resultant body
force acting on the interior *V*.

**Newton****’s third law (every action has an equal and
opposite reaction) requires that**

$T(-n)=-T(n)$

To see this, note that the forces acting on planes separating two adjacent volume elements in a solid must be equal and opposite.

** Traction acting on different planes passing
through the same point are related, in order to satisfy **

** **

Let $\left\{{e}_{1},{e}_{2},{e}_{3}\right\}$ be a Cartesian basis. Let ${T}_{i}({e}_{1})$,
${T}_{i}({e}_{2})$,
${T}_{i}({e}_{3})$ denote the components of traction acting on
planes with normal vectors in the ${e}_{1}$,
${e}_{2}$,
and ${e}_{3}$ directions, respectively. Then, the traction components ${T}_{i}(n)$ acting on a surface with normal **n** are given by

${T}_{i}(n)={T}_{i}({e}_{1}){n}_{1}+{T}_{i}({e}_{2}){n}_{2}+{T}_{i}({e}_{3}){n}_{3}$

where ${n}_{i}$ are the components of **n. **

To see this, consider the forces acting on the infinitessimal
tetrahedron shown in the figure. The
base and sides of the tetrahedron have normals in the $-{e}_{2}$,
$-{e}_{1}$ and $-{e}_{3}$ directions.
The fourth face has normal **n**. Suppose the volume of the tetrahedron is *dV*, and let $d{A}_{1}$, $d{A}_{2}$, $d{A}_{3}$, $d{A}_{n}$ denote the areas of the faces. Assume that the material within the
tetrahedron has mass density $\rho $ and is subjected to a body force **b**. Let **a** denote the acceleration of the center of mass of the tetrahedron.
Then, **F =**

*m*

**a**for the tetrahedron requires that

$T(n)d{A}^{(n)}+T(-{e}_{1})d{A}_{1}+T(-{e}_{2})d{A}_{2}+T(-{e}_{3})d{A}_{3}+\rho bdV=\rho dVa$

Recall that $T(-{e}_{i})=-T({e}_{i})$ and divide through by $d{A}_{n}$:

$T(n)-T({e}_{1})\frac{d{A}_{1}}{d{A}^{(n)}}-T({e}_{2})\frac{d{A}_{2}}{d{A}^{(n)}}-T({e}_{3})\frac{d{A}_{3}}{d{A}^{(n)}}+\rho b\frac{dV}{d{A}^{(n)}}=\rho \frac{dV}{d{A}^{(n)}}a$

Finally, let $d{A}_{n}\to 0$. We can show that

$\frac{d{A}_{1}}{d{A}^{(n)}}={n}_{1}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{d{A}_{2}}{d{A}^{(n)}}={n}_{2}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{d{A}_{3}}{d{A}^{(n)}}={n}_{3}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\underset{d{A}^{(n)}\to 0}{\mathrm{lim}}\frac{dV}{d{A}^{(n)}}=0$

so

$T(n)=T({e}_{1}){n}_{1}-T({e}_{2}){n}_{2}-T({e}_{3}){n}_{3}$

or, using index notation

${T}_{i}(n)={T}_{i}({e}_{1}){n}_{1}+{T}_{i}({e}_{2}){n}_{2}+{T}_{i}({e}_{3}){n}_{3}$

The significance of this result is that the tractions acting on planes with normals in the ${e}_{1}$, ${e}_{2}$, and ${e}_{3}$ directions completely characterize the internal forces that act at a point. Given these tractions, we can deduce the tractions acting on any other plane. This leads directly to the definition of the Cauchy stress tensor in the next section.

#### 4.3 The Cauchy (true) stress tensor

** **

Consider a solid which deforms under external loading. Let $\left\{{e}_{1},{e}_{2},{e}_{3}\right\}$ be a Cartesian basis. Let ${T}_{i}({e}_{1})$, ${T}_{i}({e}_{2})$, ${T}_{i}({e}_{3})$ denote the components of traction acting on planes with normals in the ${e}_{1}$, ${e}_{2}$, and ${e}_{3}$ directions, respectively, as outlined in the preceding section

**Define the components of the Cauchy stress tensor ${\sigma}_{ij}$ by**

$\begin{array}{c}{\sigma}_{ij}={T}_{j}({e}_{i})\\ \equiv \{\begin{array}{l}{\sigma}_{11}={T}_{1}({e}_{1})\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\sigma}_{12}={T}_{2}({e}_{1})\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\sigma}_{13}={T}_{3}({e}_{1})\\ {\sigma}_{21}={T}_{1}({e}_{2})\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\sigma}_{22}={T}_{2}({e}_{2})\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\sigma}_{23}={T}_{3}({e}_{2})\\ {\sigma}_{31}={T}_{1}({e}_{3})\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\sigma}_{32}={T}_{2}({e}_{3})\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\sigma}_{33}={T}_{3}({e}_{3})\end{array}\end{array}$

Then, the traction ${T}_{i}(n)$ acting on any plane with normal **n** follows as

$T(n)=n\cdot \sigma \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{or}{T}_{i}(n)={n}_{j}{\sigma}_{ji}$

To see this, recall the last result from the preceding section

${T}_{i}(n)={T}_{i}({e}_{1}){n}_{1}+{T}_{i}({e}_{2}){n}_{2}+{T}_{i}({e}_{3}){n}_{3}$

and substitute for ${T}_{i}({e}_{j})$ in terms of the components of the Cauchy stress tensor

${T}_{i}(n)={\sigma}_{1i}{n}_{1}+{\sigma}_{2i}{n}_{2}+{\sigma}_{3i}{n}_{3}={n}_{j}{\sigma}_{ji}$

The
Cauchy stress tensor completely characterizes the internal forces acting in a
deformed solid. The physical
significance of the components of the stress tensor is illustrated in the figure:
${\sigma}_{ji}$ represents the *i*th component of traction acting on a plane with normal in the ${e}_{j}$ direction.

Note the Cauchy stress represents force per unit area of the *deformed *solid. In elementary strength of materials courses
it is called `true stress,’ for this reason.

**HEALTH WARNING: **Some texts define stress as the *transpose* of the definition used here,
so that $T(n)=\sigma \cdot n\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{or}{T}_{i}(n)={\sigma}_{ij}{n}_{j}$. In this case the first index for each stress
component denotes the direction of traction, while the second denotes the
normal to the plane. We will see later
that Cauchy stress is always symmetric, so there is no confusion if you use the
wrong definition. But some stress
measures are *not* symmetric (see
below) and in this case you need to be careful to check which convention the
author has chosen.

**4.4** **Other stress measures $\u2013$ Kirchhoff, Nominal and Material stress tensors**

** **

Cauchy stress ${\sigma}_{ij}$ (the actual force per unit area acting on an actual, deformed solid) is the most physical measure of internal force. Other definitions of stress often appear in constitutive equations, however.

** **

The
other stress measures regard forces as acting on the *undeformed* solid.
Consequently, to define them we must know not only what the deformed
solid looks like, but also what it looked like before deformation. The deformation is described by a
displacement vector $u(x)$ and the associated deformation gradient

$F=I+\nabla u\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{F}_{ij}={\delta}_{ij}+\frac{\partial {u}_{i}}{\partial {x}_{j}}$

as outlined in Section 2.1. In addition, let $J=\mathrm{det}(F)$

** **

We then define the following stress measures

**Kirchhoff stress $\tau =J\sigma \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\tau}_{ij}=J{\sigma}_{ij}$**

**Nominal (First Piola-Kirchhoff) stress ** $S=J{F}^{-1}\cdot \sigma \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{S}_{ij}=J{F}_{ik}^{-1}{\sigma}_{kj}$

**Material
(Second Piola-Kirchhoff) stress ** $\Sigma =J{F}^{-1}\cdot \sigma \cdot {F}^{-T}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\Sigma}_{ij}=J{F}_{ik}^{-1}{\sigma}_{kl}{F}_{jl}^{-1}$

** **

The inverse relations are also useful $\u2013$ the one for Kirchhoff stress is obvious $\u2013$ the others are

$\sigma =\frac{1}{J}F\cdot S\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\sigma}_{ij}=\text{\hspace{0.17em}}\frac{1}{J}{F}_{ik}{S}_{kj}$ $\sigma =\frac{1}{J}F\cdot \Sigma \cdot {F}^{T}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\sigma}_{ij}=\frac{1}{J}{F}_{ik}^{}{\Sigma}_{kl}{F}_{jl}^{}$

** **

The **Kirchoff stress** has no obvious physical significance.

** **

The
**nominal stress tensor** can be
regarded as the internal force per unit undeformed area acting within a solid,
as follows

1. Visualize an element of area *dA *in the deformed solid, with normal **n**, which is subjected to a force $d{P}^{(n)}$ by the internal traction in the solid;

2. Suppose that the element of area d*A* has started out as an element of area $d{A}_{0}$ with normal ${n}_{0}$ in the undeformed solid, as shown in the
figure;

3. Then, the force $d{P}^{(n)}$ is related to the nominal stress by $d{P}_{j}^{(n)}=d{A}_{0}{n}_{i}^{0}{S}_{ij}$

To see this, note that one can show that

$dAn=J{F}^{-T}\cdot d{A}_{0}{n}_{0}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}dA{n}_{i}^{}=J{F}_{ki}^{-1}{n}_{k}^{0}d{A}_{0}$

Recall that the Cauchy stress is defined so that

$d{P}_{i}^{(n)}=dA{n}_{j}{\sigma}_{ji}$

Substituting for $dA{n}_{j}$ and rearranging shows that

$d{P}_{i}^{(n)}=Jd{A}_{0}{n}_{k}^{0}\left({F}_{kj}^{-1}{\sigma}_{ji}\right)=d{A}_{0}{n}_{k}^{0}{S}_{ki}$

The
**material stress tensor **can also be visualized
as force per unit undeformed area, except that the forces are regarded as
acting within the undeformed solid, rather than on the deformed solid. Specifically

1. The infinitesimal force $d{P}^{(n)}$ is assumed to behave like an infinitesimal
material fiber in the solid, in the sense that it is stretched and rotated just
like an small vector d**x** in the solid

2. This means that we can define a (fictitious) force in the reference configuration $d{P}^{(n0)}$ that is related to $d{P}^{(n)}$ by $F\cdot d{P}^{(n0)}=d{P}^{(n)}$ or ${F}_{ij}d{P}_{j}{}^{(n0)}=d{P}_{i}{}^{(n)}$.

3. This fictitious force is related to material stress by $d{P}_{i}^{(n0)}=d{A}_{0}{n}_{j}^{0}{\Sigma}_{ji}$

To see this, substitute into the expression relating $d{P}^{(n)}$ to nominal stress to see that

${F}_{ik}d{P}_{k}^{(n0)}=d{A}_{0}{n}_{j}^{0}{S}_{ji}$

Finally multiply through by ${F}_{li}^{-1}$, note ${F}_{li}^{-1}{F}_{ik}={\delta}_{lk}$, and rearrange to see that

$d{P}_{l}^{(n0)}=d{A}_{0}{n}_{j}^{0}{S}_{ji}{F}_{li}^{-1}=d{A}_{0}{n}_{j}^{0}{\Sigma}_{jl}$

where we have noted that ${\Sigma}_{jl}={S}_{ji}{F}_{li}^{-1}$

In
practice, it is best not to try to attach too much physical significance to
these stress measures. Cauchy stress is
the best physical measure of internal force $\u2013$ it is the force per unit area acting inside
the deformed solid. The other stress
measures are best regarded as *generalized
forces* (in the sense of Lagrangian mechanics), which are work-conjugate to
particular strain measures. This means
that the stress measure multiplied by the time derivative of the strain measure
tells you the rate of work done by the forces.
When setting up any mechanics problem, we always work with conjugate
measures of motion and forces.

Specifically, we shall show later that the rate of work $\dot{W}$ done by stresses acting on a small material element with volume $d{V}_{0}$ in the undeformed solid (and volume $dV$ in the deformed solid) can be computed as

$\dot{W}={D}_{ij}{\sigma}_{ij}dV={D}_{ij}{\tau}_{ji}d{V}_{0}={\dot{F}}_{ij}{S}_{ji}d{V}_{0}={\dot{E}}_{ij}{\Sigma}_{ji}d{V}_{0}$

where ${D}_{ij}$ is the stretch rate tensor, ${\dot{F}}_{ij}$ is the rate of change of deformation gradient, and ${\dot{E}}_{ij}$ is the rate of change of Lagrange strain tensor. Note that Cauchy stress (and also Kirchhoff stress) is not conjugate to any convenient strain measure $\u2013$ this is the main reason that nominal and material stresses need to be defined. The nominal stress is conjugate to the deformation gradient, while the material stress is conjugate to the Lagrange strain tensor.

** **

** **

** **

** **

**4.5 Stress measures for
infinitesimal deformations**

For
a problem involving *infinitesimal
deformation *(where shape changes are characterized by the infinitesimal
strain tensor and rotation tensor) all the stress measures defined in the
preceding section are approximately equal.

${\sigma}_{ij}\approx {\tau}_{ij}\approx {S}_{ij}\approx {\Sigma}_{ij}$

To
see this, write the deformation gradient as ${F}_{ij}={\delta}_{ij}+\partial {u}_{i}/\partial {x}_{j}$;
recall that *$J=\mathrm{det}(F)\approx 1+\partial {u}_{k}/\partial {x}_{k}$*,
and finally assume that for infinitesimal motions $\partial {u}_{i}/\partial {x}_{j}<<1$. Substituting into the formulas relating
Cauchy stress, Nominal stress and Material stress, we see that

${\sigma}_{ij}=\frac{1}{J}{F}_{ik}{S}_{kj}\approx \frac{1}{1+\partial {u}_{p}/\partial {x}_{p}}\left({\delta}_{ip}+\frac{\partial {u}_{i}}{\partial {x}_{p}}\right){S}_{pj}={S}_{pj}+\mathrm{...}\approx {S}_{pj}$

#### The same procedure will show that material stress and Cauchy stress are approximately equal, to within a term of order $\partial {u}_{i}/\partial {x}_{j}<<1$

#### 4.6 Principal Stresses and directions

For any stress measure, the *principal stresses ${\sigma}_{i}$ and their directions ${n}^{(i)}$*,
with *i*=1..3 are defined such that

${n}^{(i)}\cdot \sigma ={\sigma}_{i}{n}^{(i)}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{or}{n}_{j}^{(i)}{\sigma}_{jk}={\sigma}_{i}{n}_{k}^{(i)}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}(\text{nosumon}i)$

Clearly,

- The
**principal stresses**are the (left) eigenvalues of the stress tensor - The
**principal stress directions**are the (left) eigenvectors of the stress tensor

The term `left’ eigenvector and eigenvalue indicates that the vector multiplies the tensor on the left. We will see later that Cauchy stress and material stress are both symmetric. For a symmetric tensor the left and right eigenvalues and vectors are the same.

### Note that the eigenvectors of a symmetric tensor are orthogonal. Consequently, the principal Cauchy or material stresses can be visualized as tractions acting normal to the faces of a cube. The principal directions specify the orientation of this special cube.

One can also show that if **${\sigma}_{1}>{\sigma}_{2}>{\sigma}_{3}$,
**then ${\sigma}_{1}$ is the largest normal traction acting on any
plane passing through the point of interest, while ${\sigma}_{3}$ is the lowest.
This is helpful in defining damage criteria for brittle materials, which
fail when the stress acting normal to a material plane reaches a critical
magnitude.

In
the same vein, the largest shear stress can be shown to act on the plane with
unit normal vector ${m}_{\text{shear}}=-({m}_{1}+{m}_{3})/\sqrt{2}$ (at 45^{o} to the ${m}_{1}$ and ${m}_{3}$ axes), and its magnitude is ${\tau}_{\text{max}}=\frac{1}{2}({\sigma}_{1}-{\sigma}_{3})$. This observation is useful for defining yield
criteria for metal polycrystals, which begin to deform plastically when the
shear stress acting on a material plane reaches a critical value.

**4.7 Hydrostatic and Deviatoric Stress; von Mises
effective stress**

** **

Given the Cauchy stress tensor
**$\text{\sigma}$**,
the following may be defined:

The **Hydrostatic stress** is defined as ${\sigma}_{h}=\text{trace(}\sigma )/3\equiv {\sigma}_{kk}/3$

The **Deviatoric stress** **tensor** is defined as ${{\sigma}^{\prime}}_{ij}={\sigma}_{ij}-{\sigma}_{h}{\delta}_{ij}$

The **Von-Mises effective stress **is defined
as ${\sigma}_{e}=\sqrt{\frac{3}{2}{{\sigma}^{\prime}}_{ij}{{\sigma}^{\prime}}_{ij}}$

The
*hydrostatic stress* is a measure of
the pressure exerted by a state of stress.
Pressure acts so as to change the volume of a material element.

The
*deviatoric stress *is a measure of the
shearing exerted by a state of stress. Shear stress tends to distort a solid,
without changing its volume.

The
*Von-Mises effective stress* can be
regarded as a uniaxial equivalent of a multi-axial stress state. It is used in many failure or yield
criteria. Thus, if a material is known
to fail in a uniaxial tensile test (with ${\sigma}_{11}$ the only nonzero stress component) when ${\sigma}_{11}={\sigma}_{crit}$,
it will fail when ${\sigma}_{e}={\sigma}_{crit}$ under multi-axial loading (with several ${\sigma}_{ij}\ne 0$ )

#### The hydrostatic stress and von Mises stress can also be expressed in terms of principal stresses as

$\begin{array}{l}{\sigma}_{h}=\left({\sigma}_{1}+{\sigma}_{2}+{\sigma}_{3}\right)/3\\ {\sigma}_{e}=\sqrt{\frac{1}{2}\left\{{\left({\sigma}_{1}-{\sigma}_{2}\right)}^{2}+{\left({\sigma}_{1}-{\sigma}_{3}\right)}^{2}+{\left({\sigma}_{2}-{\sigma}_{3}\right)}^{2}\right\}}\end{array}$

The hydrostatic and von
Mises stresses are *invariants* of the
stress tensor $\u2013$ they have the same value regardless of the
basis chosen to define the stress components.

**4.8 Stresses near an external surface or edge $\u2013$ boundary conditions on stresses**

Note
that *at an external surface at which
tractions are prescribed, some components of stress are known*. Specifically, let **n** denote a unit vector normal to the surface, and let **t** denote the traction (force per unit
area) acting on the surface. Then the
Cauchy stress at the surface must satisfy

${n}_{i}{\sigma}_{ij}={t}_{j}$

For
example, suppose that a surface with normal in the ${e}_{2}$ direction is subjected to *no* loading. Then (noting
that ${n}_{i}={\delta}_{i2}$ ) it follows that ${\sigma}_{2i}=0$.
In addition, two of the principal stress directions must be parallel to the
surface; the third (with zero stress) must be perpendicular to the surface.

The stress state at an edge is even simpler. Suppose that surfaces with normals in the ${e}_{2}$ and ${e}_{1}$ are traction free. Then ${\sigma}_{1i}={\sigma}_{2i}=0$, so that 6 stress components are known to be zero.