# Overview of Elasticity of Materials/Introducing Stress

## IntroductionEdit

We will begin by developing the constitutive equations that describe relationship between stress, , and strain, . This is the subset of continuum mechanics that focuses on the purely elastic regime, and in particular, will focus on linear elasticity where Hook's Law hold's true.

The concepts of stress and strain originate by considering the forces applied to a body and its displacement. Beginning with forces, there are two types of forces that can be applied: surface forces that are either point forces or distributed forces that are applied over a surface or body force that are applied to every element of a body not just a surface, i.e., gravity, electric fields, etc.

The body of interest has numerous forces acting on it and these are transmitted through the material. At any point inside the body you can imagine slicing it to observe the forces present on the imagined cut surface, as pictured in **Figure 1**. These forces are the interactions between the material on either side of the imagined cut. We define the stress at a point in the body as the forces acting on the surface of such an imagined cut.

As you recall the stress is defined as the force divided by the area over which it is applied. The force, , is a vector quantity, allowing the components to be projected into the normal and tangential directions. As shown in **Figure 1** the normal component is defined according to the angle yielding a normal stress . The tangential component of the force, , can further be projected into the two orthogonal directions identified in **Figure 1** and and , yielding two orthogonal shear stresses. This is performed according to the angle giving and .

Note here that we've defined the coordinate system such that the direction normal to the cut surface. It is convenient to use instead of because it allows us to pass the indexes to the stress and strain quantities. In this example the normal stress is given to specify that the normal stress is applied to the surface with a normal in the with a force projected in the direction. The tangential components and specify the surface having a normal with forces projected in the and directions. Cutting an infinitesimal cuboid the stresses are defined in all three directions as shown in **Figure Y**. For comparison, the notation used in some textbooks will write normal stresses whereas here we'll use . These textbooks also use to denote shear stress, such as whereas here we'll use . This allows the stress state to be succinctly written in matrix (tensor) form

The imaginary slice taken through point in the body in **Figure 1** could have been any plane, but the force would remain the same. This would result in a new definition of the surface normal, and potentially a new expression the stress. The physical presence of the stress does not change, but the description does, i.e., the coordinate system is modified. The remainder of this section is devoted to expressing the coordinate transformation and analysis of the stresses.

## Plane StressEdit

Let's begin by simplifying the picture we're working with. The plane stress condition is observed for thin 2D object, e.g., a piece of paper, which has no stress out of the plane. This allows us to write . Further there is no shear in the direction such that . For an object in the plain stress condition our goal to determine the state of stress at some point for *any* orientation of the axis.

For this object, the direction with zero force is coming out of the page and the non-zero stress state in the and directions have components , , and .

Imagine a new area defined on a plane rotated about such that the normal, defined is related to by as shown in **Figure 3**.

The components of force on the area is determined by the application of the original stresses to the projection of the new area

**Figure 3 (a)**, and are the total stresses in the and directions, and . Then dividing by A yields

Projecting the total stresses shown in **Figure 3 (b)** into the normal direction in the coordinate we get

It is known that therefore only yet needs determining. To do so we define a new area that is rotated by relative to our original plane as shown in **Figure 4**. In this new orientation

Projecting the total stress in the normal direction yields

Substituting the above equations for and into the equation for yields

The well-known trigonometric identities

### Principal StressEdit

There are numerous immediate results that come from this derivation, from which we can gain greater insights. One results from the equations for is , for all . This means that the trace of the stress tensor is invariant.

A second result is that the maximum normal stresses and shear stresses vary as a sine wave with period . With in this oscillation the normal and shear stresses are shifted by a phase factor that results in (1) the maximum and minimum normal stresses occur when the shear is zero, (2) the maximum and minimum shear stresses are shifted from each other by , (3) the maximum and minimum normal stresses are shifted from each other by , and (4) the maximum and minimum shear stresses are shifted by from the minimum and maximum normal stresses.

Any stress state can be rotated to yield . This diagonalizes the stress tensor and gives normal stresses that are extreme. In this orientation the planes are called the principal planes and the normal stresses are called the principal stresses. The directions that give these principal stresses are called the principal axis. As a matter of convention we define the first principal stress to be the largest and the sequentially smaller principal stresses and , although here we have limited ourselves to 2D plane stress and only enumerate and .

We know in the principal orientation, which means we can use our equation for to determine the angle ( ) needed to rotate the tensor into which is principal,

It is observed graphically by plotting in **Figure 5** that adjacent roots are each separated by . Furthermore, we can now utilize the Pythagorean Theorem to solve for our principal stresses.

For a simple right triangle with hypotenuse and sides and we know

These can be further combined by yield

These equations tell us for a given stress state what rotation is needed align with the principal axis.

Substituting these equations into , determines the principal stresses

To find the maximum shear stress we take the derivative with respect to theta of our simplified equation for .

Notice that this is the negative reciprocal of our earlier equation when the principal orientation was derived. This is indicative of

### Mohr's CircleEdit

A convenient means of visualizing angular relationships is through Mohr's circle, which we derive here. Rearrange the expression for ,

From this expression Mohr's circle is drawn in **Figure 6**. For a given stress state the center of the circle is and the radius
. A bisecting line intercepts the circle such that that projection onto the x-axis identified and . The projection onto the y-axis identifies . Rotating the bisection is equivalent to transforming the stress state by , i.e., a rotation of by on the diagram is equivalent to rotating by in our equations. This allows the new stress state to be read from the diagram. When the bisector is horizontal, the principal orientation is identified. Rotating the on the diagram by is equivalent to rotating the system by , which can be imagined as rotating the cuboid faces until the system is back in registry, i.e., it comes returns to the original stress state. Further, rotating on the diagram by is equivalent to rotating by , which is known to be the orientation with maximum shear stress.

Thus, from a given initial stress state, , all stress states that can be achieved through rotation are visualized on on the circle.

## Generalizing from 2D to 3DEdit

Generalizing from 2D to 3D we move from a biaxial, plane stress, system to a triaxial system. Determining the principal axis and angular relations is similar to the case of 2D and will be shown below. Note as a matter of convention, when two of the three principal stresses are equal, we call the system "cylindrical", and if all three principal stresses are equal we call the system "hydrostatic" or "spherical".

As in the case of the biaxial system we begin by defining a plane with area that passes through our , , and coordinate system, as shown in **Figure 7**. The plane intercept the axis at ( , , and ) as demonstrated in the figure. To simplify the problem and allow us to make progress toward our derivation, we will say that the plane is one of the principal planes so that the shear stress components are zero. Thus, we only need to consider our principal stress that is normal to the plane.

<FIGURE> Figure 7: "Coordinate Plane"

Define , , and to be the direction cosine between , , and and the normal to stress. Using the unit vectors \hat{i}, \hat{j}, and \hat{k} parallel to , , and we have

The projection of stress along , , and direction give the total stresses , , and

As in the biaxial derivation the area is projected into the three directions giving the triangles in **Figure 7** , and . We can now equate the forces in the two reference frames

By a similar process, the and components yield

These equations rearrange to

This set of equations can be solved for for a particular value of . This set of secular equations can be solved for eigenvalues and eigenvectors . The non-trivial solutions, when and are non-zero, involves setting the determinant

Upon rearranging we get

The three roots of this cubic equation give the principal stresses, \sigma_{p1}, \sigma_{p2}, and \sigma_{p3}. The eigenvalues, once determined are substituted back into the secular equations to determine the eigenvectors corresponding , also recognizing that .

Solving the cubic equation is not the focus of this text, but the equation is important because the coefficients in front of the principal stress must be invariant, i.e., the same principal coordinates must exists no matter the orientation of the coordinate system. From the cubic equation the three invariants are

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Now, let's generalize our solution to include not only the principal stresses. Just as we did earlier we can write out the total forces:

Which gives the total stress:

From this, the projection onto the normal component is:

Further substituting some earlier equations gives us:

The magnitude of the shear component can be determined utilizing , but we cannot easily decompose our shear stress into its constituent elements. Fortunately, we are primarily interested in the maximum shear stress. We know that the plane containing the maximum shear stress is located midway between planes of principal normal stresses. Starting by setting our known stress state as the principal axis such that , , and , our direction cosine is between the principal axis and the normal of the plane with the maximum shear stress. This means that our earlier projection equation is rewritten as:

Taking the square of this equation gives us:

We can then combine this with the principal components to get:

With this solution, we now have three possible planes. One plane bisects , and , another plane bisects , and , and the final plane bisects , and . (Bisecting means , and )

By convention, , and therefore our maximum shear stress is:

Note that we know there are two planes of maximum shear stress, rotated from each other. Thus, the direction cosine above are actually .

Because these axial rotations are decoupled, we can represent 3D stress states using Mohr's Circles.