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# 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

-=-=-=-=-=-=-=-=-=-=-=-=-=

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.

# Introduction to Tensors

## Introduction to TensorsEdit

We've been working with stress and in particular looking as stress at a point and the impact of rotating the reference frame. Using the tools in the previous sections it is possible to identify the principal stresses and the orientation of the reference frame relative to the principal axis, which allows determination of the stress state in any orientation. This allows determination of the orientation and value of the maximum shear and normal forces, which are critical for engineering design. As you've see, computing this information requires either extensive use of equations or geometry/trigonometry. In this section tensors are going to be introduced, which will allow for a more elegant means of addressing coordination transformations.

Lets begin by thinking about vectors, such as

Here is a rank 2 tensor that relates the two vectors and . Tensors are geometric objects that describe the linear relationship between scalars, vectors, and other tensors. Are the vector's and tensors? Although vectors can be tensors, in this case they are not because and do not act to map linear spaces onto each other. The rank of a tensor is the number of indexes needed to describe it, therefore is a rank two tensor because it requires and , , to describe it.

Tensors are used frequently, to represent the intrinsic physical properties of materials. A good example is the electrical conductivity, , a rank two tensor that expresses the current density in a material induced by the application of an electric field, .

There are many other tensors that represent materials properties including the thermal conductivity, diffusivity, permittivity, dielectric susceptibility, permeability, and magnetic susceptibility to name a few. We will see that stress, and strain also are tensors. Stress relates the surface normal to an arbitrary imaginary surface, , to the stress vector at that point, , as was discussed in the previous section.

### Tensor TransformationsEdit

The vectors that represent material properties also must be able to transform. This is useful to allow coordinate transformation, which essentially are rotations. It also allows the tensors that represent material responses to transform according to crystallographic symmetry; these can involve rotations, mirror operations, and inversions. Because these transformations involve linear one-to-one mapping, the transformation themselves are enabled by transformation tensors.

In the section above we rotated vector to by applying transformation tensor

Consider now that there is a second vector that we'll call is related to by the rank two materials property tensor

This tells us that the transformation of to causes the transformations from to , to , and to , where the vector transformations are given by the above equations, and the tensor transformation is given by and . Note that these solutions are really double sums over and , due to Einstein Notation.

Because the order of the summation is not important, we can writeː

This is a Tensor that relates and . Since it is a double sum, each term in has nine elements and the total tensor mapping the relationship between and must have a total of terms as .

### Tensor SymmetryEdit

The nature of a tensor is determined by it's application. There are subsets of tensors that we can classify according to their symmetry properties.

Symmetric tensors have a structure such asː where

Antisymmetric Tensors have a structure such asː where

Note that the main diagonal of an antisymmetric tensor must be zero and the overall symmetry or antisymmetry depends on the reference frame selected. Any second rank tensor can be expressed as a sum of a symmetric and antisymmetric tensor asː

We will find this useful in the next section dealing with strain. Meanwhile, any symmetric tensor can be transformed by rotations to be aligned along its principal axis, such thatː

The properties of tensors are highly tied to the crystal symmetry of the material they represent. For example, let's say that we have two vector properties and in a crystal which are related by a tensor . If we rotate the reference farm according to a symmetry element of the crystal then .

<FIGURE> "Rotating a Simple Cubic Crystal" (Due to symmetry, the crystal will periodically rotate such that it is practically at the same orientation it started at.)

Let's examine this by looking at a simple cubic crystal. When rotated, this crystal will periodically rotate back on itself, and the properties of the relevant tensors should do the same. By applying this theory to each possible crystal formations, we can develop simplified tensors for each, which represent this symmetry.

Crystal
Formation |
Tensor | Number of
Independent Components |
---|---|---|

Cubic | 1 | |

Tetragonal
Hexagonal Trigonal |
2 | |

Orthorhombic | 3 | |

Monoclinic | 4 | |

Triclinic | 6 |

### Tensor Contractions and Invariant Relations in StressEdit

Much of this discussion has been about property relations, but here our interest is in the stress tensor; a symmetric tensor that can therefore be arranged to be aligned in the principal axis. Let's now rederive the 3D stress relationships using tensors. The stresses normal to an oblique plane are writtenː

Here the is the direction of the normal to the plane and is the original stress state. If the oblique plane is a principal direction, with a normal stress of , then we can write our equation asː

By combining these two equations, we getː

#### Kronecker DeltaEdit

Additionally, there is a handy expression called a Kronecker Delta that has the propertiesː

When applied to a tensor, the Kronecker Delta is said to "contract" the tensor's rank by two. This turns a 4th rank tensor into a 2nd rank, a 3rd rank tensor into a 1st rank, etc... For the purpose of this text, we won't be using this expression often, but in applying this to our previous equation we can replace the scalar with the contraction of the second rank tensor such thatː

The rule for contraction here is to replace with , and remove the Kronecker Delta term.

Returning to our earlier equation, we replace the with our Kronecker Delta expansion to getː

This equation can be entirely summed over , and because is normal to the plane, this makes equal to and our equation evolves toː

This gives us a set of three equations where . By substituting the direction cosines into the left term and using , , and when , we can solve for the non-trivial (non-zero) solution by taking the determinant ofː

#### The Three InvariantsEdit

We also identify the invariant relations. It should be noted that these also can come from the stress tensor. First, let's apply a contraction to ː

This is our first invariant. The second invariant comes from the minors of , which can be used to expand the determinant.

The third invariant is the determinant of where .

# Introducing Strain

This section introduces strain and show tensor symmetry of strain tensor. We also will discuss special subsets of stress and strain including dilatation and deviatoric stresses and strains.

# Average StrainEdit

This is where we talk about average strain

# Infinitesimal StrainEdit

While average strain generally looks at the strain of a volume, we will not consider how a point on an elastic body moves and how points near it do also.

<FIGURE> "2D Strain at a Point" (Description)

Let's begin in 2D. Say there is a point () on an elastic body that is located at coordinate . (This notation will be more convenient when we want to work with tensors.) If we deform the body, then is displaced to which has the coordinates . <FIGURE> We call the displacement vector.

Now let's say that there is a point infinitesimally close to , called , with coordinates . When is displaced to by the deformation, is similarly displaced to with coordinates . Thinking critically, the displacement experienced on a body depends on the position on the body. Therefore . This allows us to use the chain rule to express infinitesimal displacements.

Now define the following terms:

### Displacement TensorsEdit

<FIGURE> "Title" (Description)

What is the physical significance of this? This is easier to see looking in special directions. Considering the points , where and where . <FIGURE> Then after the deformation:

How do we interpret this? In the case of , we have . Thus, based on our initial equations expressing the infinitesimal displacements, we can infer that , and . This tells us that is an expression of uniaxial extension in the direction and is a rotation of around the point . Similarly, in the case of , we can again combine equations which yields , and . Thus is a uniaxial extension in the direction, and is a rotation of around point . These are our displacement tensors.

### The Strain TensorEdit

Let's return to , displacing to . What is the relationship between , and ?

In a similar fashion we can also prove that .

Looking at our picture <FIGURE>, we can see that our deformations also have translations and rotations. We're not interested in these because they do not tell us about material response such as dilatation (change in volume) or distortion (change in shape). Translations and rotations are a part of the field of mechanics called dynamics. Here we are interested in small scale elastic deformations. Our , but we know that our stress tensor is symmetric as . We can therefore rewrite our displacement tensor as a combination of a symmetric and antisymmetric tensor.

<FIGURE> "Visual representation of the displacement tensor ." (Here the displacement tensor has been broken up into the strain tensor , and the rotation tensor .)

Here is the strain tensor and is the rotation tensor. Schematically this looks like <FIGURE>. In the scope of this text, we are only interested in , but it is generally still worth remembering that displacement includes both shear and rotation components:

The strain tensor maps the irrotational displacement at a point to an imaginary plane, with normal in any direction that cuts through the point. Because the strain tensor is a tensor it must transform in the same manner as the stress tensor did in earlier sections of this text. As a reminder:

#### Average Engineering StrainEdit

Note that when we first started to look at this subject we defined shear strain as , which is asymmetric. In terms of our strain tensor, this would be . (It must be rotated back so each side had an angle of .) You may frequently see a matrix written as:

It is sometimes useful to write it this way. However, it is not a tensor, because it does not transform the same as the above equations do, due to its asymmetry. Textbooks generally like this "average engineering strain" , but we will not be using this here unless absolutely necessary.

### Generalizing 2D to 3DEdit

The results we found for our 2D strain tensor can easily be generalized to 3D by writing them in Einstein Notation and using "3" in the place of "2" in the implicit sums.

Which in 3D express 3 equation. , and expands to:

The displacement tensor is:

The strain tensor is:

The rotation tensor is:

Which gives us the displacement:

The new displaced coordinates:

Now that we have a symmetric strain tensor with properties analogous to stress, we can examine the other properties these other properties using similar methods as when we analyzed stress. For small strains where , we define the mean stress as:

The total strain tensor can be broken into dilatation and deviatoric components.

In a similar fashion we also have deviatoric stresses and hydrostatic stresses which are analogous to the deviatoric and hydrostatic stresses which are analogous to the deviatoric and dilatation strains. The hydrostatic, or mean, stress is:

Therefore, the deviatoric stress can be reduced because:

The principal components of break down to:

And we know that these are just the maximum shear stresses:

Keep in mind that we took this from:

Where:

<FIGURE> "Title" (Strain Gauges only measures uniaxial strain.)

Strain Gauges only measures uniaxial strain, no stress and no shear.

# Isotropic Response

This section covers linear isotropic response.

We now understand tensor relations and have established two tensors that we need to elasticity; and . In this chapter we will learn how to relate these to each other. In this particular lecture we will begin studying elasticity in the case of isotropic elasticity where we assume that all directions have the same material response.

#### Modulus of ElasticityEdit

In earlier chapters we learned that Hook's Law can relate uniaxial loading to uniaxial strain:

Here, is the modulus of elasticity. Note that we picked the direction arbitrarily. It could have easily been or , or any other direction in this crystal, and found the same material response, . This is what comes from working with an isotropic solid.

#### Poisson's RatioEdit

When we load stress in the direction, there is also a natural contraction in the transverse directions, and . This is expressed as Poisson's Ratio, .

For linear elastic, isotropic materials, normal stresses will not induce shear strain and shear stress will not cause normal stress. In linear elasticity the various contributions to strain can be superimposed. We can use the two above equations for the Modulus of Elasticity and Poisson's Ratio to determine that if a triaxial normal stress is applied, the strain will be:

Again, due to the isotropic nature this relation holds in all the other directions as well. These three constants of proportionality are sufficient to describe isotropic linear response. They take typical values (for common engineering metals)...

There are many other useful relations that we will summarize here. The Block Modulus, , is the ratio of the hydrostatic pressure to the dilatation it produces. , where is the hydrostatic pressure and is the compressibility.

Adding together our earlier equations for the triaxial normal strain, we get:

Applying this to our equations for the Block Modulus results in:

These next solutions are given without proof, because several advanced topics are used to show the relationships.

<TABLE> Moduli Relationships

The stress-strain relationships can be expressed in compact tensor notation.

Looking specifically at a normal strain, for example:

Similarly, for a shear strain, , gets us:

From , we find that:

For a given stress state:

We can only write the strain tensor as:

So we have an expression for strain in terms of the applied load, the stress. This can certainly be useful for predicting the deformation to anticipate when loading in application. There is another application we are interested in. Given we deform a part, what stress does it feel? This answers questions such as, "How much can I bend this part before I reach the yield strength?" Basically we want to take our existing solution and invert it.

To do this, we can take our first triaxial strain equation and rearrange it.

We can additionally take our earlier equation for , and rearrange it to get:

Then combining the two above equations we get:

Thinking about shear stress is simpler:

Taking the above two equations gives us the tensor expression for stress in terms of strain, which is:

Here, is the Lamé Constant. We can analyze these results and extract useful equations. Let's begin by extracting the deviatoric and hydrostatic contributions to stress.

We can show this is true by considering shear and normal components.

Shear:

Normal:

The relationship between the hydrostatic stress and the mean strain is:

Note that we've seen this expression several times before!

## Simplified CasesEdit

Plane stress .

Once again looking at our earlier triaxial strain equations, let's start by adding the first two together.

Plane stress encountered either as loaded sheet, or more likely a pressure vessel.

<Double Check... Much Greater than or Less than???>

Another simplification is Plane Strain, where , typically when one dimension is much greater than the other two so and are much greater than , such as a long rod where strain along the length of the rod is constrained. Here, our third triaxial strain equation rearranges to:

Note that does not equal zero just because equals zero. Substituting this equation into the triaxial strain equations yields us:

# Anisotropic Response

This section covers linear anisotropic response.

# Appendix A: Coordinate Notation

This is a discussion of the notational choices used in the book including the coordinate choices.

# Appendix B: List of Variables

This section lists the variables used throughout the text.