Lesson1-Stress

## Introduction

Consider a body subjected to external forces (Figure 1). The external forces tend to pull the body apart, crush it over, or in general deform it and produce rupture. The body resists against deformation and rupture because internal forces (i.e. actions and reactions between various particles) are developed within the body. The major concern of Mechanics of Materials is determination of these internal forces that balance the effect of the externally applied forces.

Assume the body is in static equilibrium. Let an arbitrary section be cut through the body, and completely separates it into two parts. Because the body is in equilibrium, each part of it must also be in equilibrium and hence the external forces on one side of the arbitrary cut must be balanced by the internal forces at the cut. We can use statics to find the resultant force and the resultant couple moment at any position in the section (Figure 2a), but we wish to determine the distribution of internal forces transmitted from one part of the body to the other through this plane (Figure 2b), and infinitely many force distributions yields the same resultant force and the resultant couple moment.

In some special cases, the internal forces are the same at every single point of the cross section. For example, if a prismatic bar is under tension or compression by external forces uniformly distributed over the ends, the internal forces are also uniformly distributed over any cross section (Figure 3). However, in general, the internal forces vary in magnitude and direction from point to point (Figure 2b). In Mechanics of Materials, it is essential to determine the intensity of the internal forces (= the magnitude of force per area) at each point of the section because resistance to deformation depends on these intensities. The intensity of the internal forces is called stress.

## Stress Components

Because infinitely many cross-sections can be considered, the question is: how can we determine the stress on any arbitrary section?

Let the section be perpendicular to the x-axis. Consider an infinitesimally small area $\Delta A_x$ surrounding point $Q$. We use the subscript $x$ to indicate the direction of the normal of this plane. 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:

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

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.

• 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,

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. 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

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 $\sigma_z$).

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 $\tau_{zy}$. These 9 components specify the stress state at a point.

What does that mean?

Obviously we can choose any three mutually perpendicular axes as our coordinate system. If we know the stress state (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 6, we will discuss how we can convert the components of stress in a given coordinate system to a new coordinate system.

### Stress Tensor

We can arrange nine stress components in a matrix form

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

but to identify the components of stress we need two indices. Stress is a new type of quantities called second-rank tensor.

## Average Stress

Consider a prismatic bar under uniaxial loading, as in Figure 1. Consider a cross-section perpendicular to the line of action of forces. Let

$P$ = magnitude of the load

$A$ = cross-sectional area of the member.

Then, the average normal stress on the cross-section is given by

The average stress is a good approximation of the normal (component of) stress at any point on the cross-section $\sigma_{\rm avg}\approx \sigma_x$, unless:

1. The cross-section is very close to the ends where the loads are applied (Saint-Venant's Principle)
2. The loading is eccentric. That is, the line of action of the two equal forces does not pass through the center of such cross-sections

In other words, except the forces are eccentric or the section is near the ends, the distribution of stress is almost uniform.