Once again consider an arbitrary body in equilibrium and pass a section through the body as shown in Figure 1. Let the section be perpendicular to the x-axis.

 (a) (b) Internal forces on the cross section

Figure 1(a,b)

Consider an infinitesimally small area $$\Delta A_x$$ surrounding point $$Q$$ on the section as in Figure 2. We use the subscript $$x$$ for $$\Delta A_x$$ to indicate the $$x$$ axis is perpendicular to this area. A finite small force vector $$\Delta {\bf F}$$ arising from the action of the other part of the body, acts on $$\Delta A_x$$. Let’s resolve $$\Delta{\bf F}$$ into its components along the $$x$$, $$y$$, and $$z$$ axes: $\Delta {\bf F}=\Delta F_x {\bf i}+\Delta F_y {\bf j}+\Delta F_z {\bf k}$

To find the intensity of the internal force at $$Q$$ (or particularly the stress components), we divide the components of $$\Delta{\bf F}$$ by the area and let $$\Delta A_x$$ approach zero \begin{align} \tau_{xx}=\lim_{\Delta A_x\to 0}\frac{\Delta F_x}{\Delta A_x}\\ \tau_{xy}=\lim_{\Delta A_x\to 0}\frac{\Delta F_y}{\Delta A_x}\\ \tau_{xz}=\lim_{\Delta A_x\to 0}\frac{\Delta F_z}{\Delta A_x}\\ \end{align}

The first subscript signifies that the plane is perpendicular to the x-axis and the second subscript indicates the direction of the component of the force.

 Figure 2

#### Collapse the explanation

• Note that $$\Delta A_x\to 0$$ is in contradiction with the fact that materials are composed of atoms and molecules, but keep in mind that

• We assumed the material is continuous and there is no empty space between particles.
• The above definition is very abstract and is never used in practice.

• Because $$\Delta F_x{\bf i}$$ is parallel to the normal to $$\Delta A_x$$, $$\tau_{xx}=\Delta F_x/\Delta A_x$$ shows the intensity of the force that is normal to $$\Delta A_x$$ and is called the normal stress at Q.

• In this course, we denote the normal stress by $$\sigma$$; we can use only a single subscript for $$\sigma$$ which indicates the normal of the plane and the direction of the stress component at the same time. That is, $\sigma_x=\tau_{xx}$

• If the normal stress (or the normal component of $$\Delta{\bf F}$$) causes tension on the surface of the section, it is called tensile stress and if it pushes on the surface, it is referred to as compressive stress.

• Because $$\Delta F_y{\bf j}$$ and $$\Delta F_z{\bf k}$$ act parallel to the surface of the section, $$\tau_{xy}$$ and $$\tau_{xz}$$ which are the intensities of these components are called shear stresses or shearing stresses.

We can repeat what we did to define $$\tau_{xx}$$ and $$\tau_{xy}$$ , to define 6 other components of stress at Q. That is, we consider a plane that crosses Q and is perpendicular to the y axes and an incremental area $$\Delta A_y$$ around Q on this plane (Figure 3). Then we find the incremental force $$\Delta {\bf F’}$$ that acts on $$\Delta A_y$$ , we can define three more components of stress at Q

\begin{align} \tau_{yx}=\lim_{\Delta A_y\to 0}\frac{\Delta F’_x}{\Delta A_y}\\ \tau_{yy}=\lim_{\Delta A_y\to 0}\frac{\Delta F’_y}{\Delta A_y}\\ \tau_{yz}=\lim_{\Delta A_y\to 0}\frac{\Delta F’_z}{\Delta A_y}\\ \end{align} where $$\Delta{\bf F’}=\Delta F’_x{\bf i}+\Delta F’_y {\bf j}+\Delta F’_z{\bf k}$$. Again $$\tau_{yy}$$ is the normal stress because $$\Delta F’_y {\bf j}$$ is normal to $$\Delta A_y$$ and hence we can denote $$\tau_{yy}$$ by $$\sigma_y$$. The other two components $$\tau_{yx}$$ and $$\tau_{yz}$$ are shear stresses. In a similar manner, we can define, $$\tau_{zx}, \tau_{zy}$$, and $$\tau_{zz}$$ (also denoted by $$\sigma_z$$).

 Figure 3

Therefore, stress at a point has 9 components: 3 normal components $$\sigma_x, \sigma_y$$ and $$\sigma_z$$ and 6 shear components $$\tau_{xy},\tau_{xz},\tau_{yx},\tau_{yz},\tau_{zx}$$ and $$\tau_{zy}$$ (Figure 4). These 9 components specify the state of stress acting around a point in a body.

 Figure 4

#### Read more on State of Stress

Obviously we can choose any three mutually perpendicular axes as our coordinate system. If we know the state of stress (the 9 components of stress) at a point in one coordinate system, then we can find the components of stress at that point on any arbitrary plane or in general, its components in any other coordinate system. In chapter 8, we will discuss how we can convert the components of stress in a given coordinate system to a new coordinate system.

### Matrix Representation

We can arrange the nine stress components in a matrix form

$\begin{bmatrix}{\tau_{x x}} & {\tau_{x y}} & {\tau_{x z}} \\{\tau _{y x}} & {\tau _{y y}} & {\tau _{y z}} \\{\tau _{z x}} & {\tau _{z y}} & {\tau _{z z}}\end{bmatrix}\equiv\begin{bmatrix}{\sigma_{x}} & {\tau_{x y}} & {\tau_{x z}} \\{\tau_{y x}} & {\sigma_{y}} & {\tau_{y z}} \\{\tau_{z x}} & {\tau_{z y}} & {\sigma_{z}}\end{bmatrix}$

The above arrangement is called the matrix representation of the stress state at a point (or stress tensor) .

#### Read more on stress tensor

In elementary courses, you have faced two types of quantities: vectors and scalars. Stress is neither a scalar nor a vector. To denote the components of a vector we need only one index; for example, ${\bf F}=F_x{\bf i}+F_y {\bf j}+F_z {\bf k}=\begin{bmatrix} F_x\\ F_y\\ F_z \end{bmatrix}$

but to identify the components of stress we need two indices. Stress is a new type of quantities called second-rank tensor. A second-rank tensor is a quantity whose components transform in a certain way discussed later under a rotation of axes.

### Units of stress

In the International System or SI:

• Force measured in newtons (N)
• Area measured in square meters (m2)
• Because stress is force divided by area, stress is measured in N/m2 . This unit is called a pascal

1 Pa = 1 N/m2

• Because one pascal is quite small, stress is often measured in
• kPa = kilo (103) pascal (kPa),
• MPa = mega (106 ) pascal, or
• GPa = giga (109) pascal.

#### Stress units in the U.S. customary system

In the U.S. customary system:

• Force measured in pounds (lb)
• Area measured in square inches (in2)
• Stress is expressed in pounds per square inch (psi) or kilopounds per square inch (ksi)

1 ksi = 103 psi = 103 lb/in2