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Introduction
Currently, there is no suitable open source text for materials science that not only includes the basics of the topic, as seen in W.D. Callister's Materials Science and Engineering: An Introduction, but also covers additional concepts such as the derivation of Mohr's circle and an introduction to tensors. Never the less, these concepts are key for student to gain an understanding of more advanced topics in materials science and engineering. This text is built as an open-source companion for the currently available texts. It attempts to address these additional topics and give further detail on their application.
While this text will go in depth on the linear elastic behavior of materials, there are currently no plans to include boundary conditions, which are essential for advanced analysis. The non-linear response often observed in polymeric and biological materials also is neglected here. Thus, this content should only be considered as an introductory overview of the topic.
A basic level of understanding on the subject of materials science, including stress and strain, is required to understand the real-world applications of this text. While the resulting equations and concepts given here can be immediately applied by the reader, a basic understanding of the mathematical foundations presented here is recommended for advanced applications.
SPB -- April 2022
Introducing Stress
Introduction
editWe will begin by developing the constitutive equations that describe the 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 Hooke's Law holds 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. First, there is surface force which can either be point forces or distributed forces that are applied over a surface. Second, there is body force which is 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 over the area 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 be further projected into the two orthogonal directions identified in Figure 1 as 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 is 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 by to specify that the normal stress is applied to the surface with a normal in the direction with a force projected in the direction. The tangential components and specify the surface having a normal with forces projected in the and directions, respectively. Cutting an infinitesimal cuboid, the stresses are defined in all three directions as shown in Figure 2. For comparison, the notation used in some textbooks will write normal stresses , whereas here we will use . These textbooks also use to denote shear stress, such as whereas here we will 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 for 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 Stress
editWe will begin by simplifying the picture we are working with. The plane stress condition is observed for a 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 plane stress condition, our goal is 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:
- [1 & 2]
where the elements and are the projection of the A in the original orientation, shown in Figure 3 (a), and and are the total stresses in the and directions, where . Then, dividing by A yields:
- [3 & 4]
Projecting the total stresses shown in Figure 3 (b) into the normal direction in the coordinate yields
- [5]
In a similar fashion, we project tangential to the plane and yield
- [6]
Resulting in
- [7]
and
- [8]
It is known that and therefore only needs determining. To do so, we define a new area that is rotated by /2 relative to our original plane as shown in Figure 4. In this new orientation,
- [9]
and
- [10]
Projecting the total stress in the normal direction yields
- [11]
Substituting Equations 9 and 10 for and into Equation 11 for yields
- [12]
The well-known trigonometric identities
are applied to Equations 7, 8, and 12 for , , and respectively, resulting in
- [13]
- [14]
and
- [15]
Principal Stress
editThere are numerous immediate results that come from this derivation, from which we can gain greater insights. One result that comes 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 . Within 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 /4, (3) the maximum and minimum normal stresses are shifted from each other by /2, and (4) the maximum and minimum shear stresses are shifted by /4 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 to be 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 Equation 8 for to determine the angle ( ) needed to rotate the tensor into which is principal,
Resulting in
- [16]
It is observed graphically by plotting in Figure 5 that adjacent roots are each separated by /2. 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
which can be combined with the Pythagorean Theorem, and Equation 16,
These can be further combined which yields
- [17]
and
- [18]
These equations tell us for a given stress state, , what rotation is needed to align with the principal axis.
Substituting these equations into Equation 13 for , determines the principal stresses
Resulting in
- [19]
Use Equation 19 in Equation 16 to find for .
To find the maximum shear stress, we take the derivative with respect to theta of our simplified Equation 15 for and set it equal to .
Resulting in an expression for :
- [20]
Notice that Equation 20 and Equation 16 are negative reciprocals which means that and are shifted by /2. This is indicative of
which implies that and are separated by /4. Through substitution of Equation 20 into Equation 15, we arrive at an expression for :
- [21]
Mohr's Circle
editA convenient means of visualizing angular relationships is through Mohr's circle, which we derive here. Rearrange Equation 13 for and Equation 15 for ,
- [22]
- [23]
Square both expressions,
Next, add them together to yield
The resulting expression is the equation for a circle:
- [24]
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 the projection onto the x-axis identifies and . The projection onto the y-axis identifies . Rotating the bisection is equivalent to transforming the stress state by , i.e., a rotation 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 bisection on the diagram by is equivalent to rotating the system by /2, which can be imagined as rotating the cuboid faces until the system is back in registry, i.e., it returns to the original stress state. Further, rotating the bisection on the diagram by /2 is equivalent to rotating by /4, 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 the circle.
Generalizing from 2D to 3D
editGeneralizing 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 intercepts 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.
Define , , and to be the direction cosine between , , and and the normal to the stress. Using the unit vectors , , and parallel to , , and , we have
The projection of stress along , , and direction give the total stresses , , and :
In the biaxial derivation, the area is projected into three directions, producing the triangles in Figure 7 which have areas , and . We can now equate the forces in the two reference frames:
So,
- [25]
By a similar process, the and components yield
- [26]
- [27]
These equations rearrange to
- [28]
- [29]
- [30]
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
to zero and solving for the eigenvalues and subsequent eigenvectors.
Upon rearranging, we get
- [31]
The three roots of this cubic equation give the principal stresses, , , and . The principle stresses, once determined, are substituted back into the secular Equations 28-30 to determine the eigenvectors corresponding to , also recognizing that .
Solving the cubic equation is not the focus of this text, but Equation 31 is important because the coefficients in front of the principal stress must be invariant, i.e., the same principal coordinates must exist no matter the orientation of the coordinate system. From the cubic equation, the three invariants are
- [32]
- [33]
- [34]
This is useful because these invariant relations determine the relationship between stresses in different orientations, i.e. given , you can now directly determine , , and .
Now, we will generalize our solution to include not only the principal stresses. Just as we did earlier, we can write out the total forces:
- [35]
- [36]
- [37]
Which gives the total stress:
- [38]
From this, the projection onto the normal component is:
- [39]
Substituting Equations 34-36 into Equation 38 gives us:
Which simplifies to,
- [40]
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 the 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 Equation 39 for projection is rewritten as:
- [41]
Squaring this equation gives us:
- [42]
We can then use the principal components and substitute Equations 34-36 into Equation 37 to get:
- [43]
After much algebra and putting Equations 41 & 42 into Equation 40, we get:
- [44]
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 ). Here are the values of , and for these three planes:
By convention, , and therefore our maximum shear stress is:
Note that we know there are two planes of maximum shear stress, rotated /2 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 as seen in Figure 8.
Introduction to Tensors
Introduction to Tensors
editWe have been working with stress and in particular looking at 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 for the determination of the stress state in any orientation. This also allows for the determination of the orientation and value of the maximum shear and normal forces, which are critical for engineering design. As you have seen, computing this information requires either extensive use of equations or geometry/trigonometry. In this section, tensors are introduced, which allows for a more elegant means of addressing coordination transformations.
We will begin by thinking about vectors, such as
This vector is represented using the coordinates, but it just as well could have been expressed relative to a different set, which we will call . The direction cosines, the cosine between all the axis of the two coordinate systems, allows us to rewrite the vector
where is the cosine between and and can be rewritten as which allows forː
- [1]
- [2]
- [3]
recognizing that for the cosine of an angle, . Equations 1-3 can be written in a compact form, known as Einstein notation:
- [4]
In Einstein notation, if a subscript is seen two or more times on a side of an equation, a summation is performed. In the example above, the shows up twice on the right side of the equation, but the only once. This means that this equation becomes:
In this equation, is a dummy variable; substituting the value 1, 2, or 3 in for returns the equations above.
Here, is a rank two tensor that relates the two vectors and . Tensors are geometric objects that describe the linear relationship between scalars, vectors, and other tensors. The rank of a tensor is the number of indexes, or directions needed to describe it which directly translates to the dimensionality of the respective array. Therefore, is a rank two tensor because it requires and ( ) to describe it, and as such, is defined by a two dimensional array. Other texts may refer to rank as dimensionality or order as the terms can be used interchangeably. The table below may be helpful in understanding the concept of rank:
Name Rank/dimensionality/order of tensor Example Scalar Zero Vector One or Matrix Two
Are the vectors 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.
Tensors are used frequently to represent the intrinsic physical properties of materials. A good example is electrical conductivity, , which is a rank two tensor that expresses the current density in a material, , induced by the application of an electric field, .
Both and are vectors since they have both a magnitude and direction. Interestingly, the off-axis terms in implies cross interactions between the vectors, e.g., the current response in the direction is influenced by the electric field in the and directions, which is indeed true.
There are many other tensors that represent material 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 Transformations
editThe vectors that represent material properties also must be able to transform. This is useful for coordinate transformation, which are essentially 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 transformations themselves are enabled by transformation tensors.
In Equation 4 we rotated vector to by applying transformation tensor
What if we want to reverse this? We can simply reverse the equation:
- [5]
Note that there are implications here regarding the inversion of . Since
we have , which means that transposing yields the inverse of , written as .
Consider now that there is a second vector, that we will call , which is related to by the rank two material property tensor :
- [6]
In a transformed coordinate system, we can express this as
- [7]
So we can now write
This tells us that the transformation of to causes the transformations from to , to , and to , where the vector transformations are given by Equations 4 & 5, and the tensor transformation is given by
- [8]
and
- [9]
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
- [10]
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 Symmetry
editThe nature of a tensor is determined by its 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
- [11]
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 related to the crystal symmetry of the material they represent. For example, say that we have two vector properties and in a crystal which are related by a tensor . If we rotate the reference frame according to a symmetry element of the crystal, then
We will examine this by looking at a simple cubic crystal such as the one in Figure 1. 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 Tensors 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 Stress
editMuch 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. We will now rederive the 3D stress relationships using tensors. The stresses normal to an oblique plane are writtenː
- [12]
Here, 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
- [13]
By combining Equations 12 & 13, we getː
- [14]
Kronecker Delta
editAdditionally, there is a handy expression called the 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 will not be using this expression often, but in applying this to Equation 14 , 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 Equation 14, we replace the with our Kronecker Delta expansion to getː
- [15]
This equation can be entirely summed over , and because is normal to the plane, this makes equal to and our equation evolves toː
- [16]
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
Which yields the same result as returned before.
The Three Invariants
editWe 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 Strain
editIn this section, we are going to revisit strain to consider it in the infinitesimal limit and to investigate its relationship to displacement using a tensor notation. When we began our discussion, we examined average strain, engineering strain in terms of linear elongation (Figure 1a) and shear deformations (Figure 1b).
Infinitesimal Strain
editWhile average strain generally looks at the strain of a volume, we will now consider how a point on an elastic body moves and how points near it do also.
We will begin in 2D. Say there is a point ( ) on an elastic body that is located at coordinate such as in Figure 2 (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 . We call the displacement vector.
Looking at Figure 2, 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:
- [3]
- [4]
- [5-8]
This allows us to write our infinitesimal displacements using Einstein Notation:
- [9]
Displacement Tensors
editWhat is the physical significance of this? This is easier to see looking in special directions. Considering the points , where and where as seen in Figure 3. Then after the deformation:
How do we interpret this? In the case of , we have . Thus, based on Equations 3 & 5 and Equations 4 & 7 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 Equation 3 with Equation 6 and Equation 4 with Equation 8 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 Tensor
editLet's return to , displacing to . What is the relationship between , and ?
- [10]
From Equations 3, 5-8:
- [11]
Integrating, we have
Which can be rewritten as
- [12]
In a similar fashion, we can also prove that
- [13]
Looking at our tensor, we can see that our deformations also have translations and rotations. We are 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.
- [14]
Here, is the strain tensor and is the rotation tensor. This can be seen schematically in Figure 4. 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:
- [15]
If a deformation is irrotational, or in other words, the directions of the principal axes of strain do not change as a result of displacement, then and
- [16]
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:
- [17]
- [18]
Average Engineering Strain
editNote 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 Equations 17 & 18 do, due to its asymmetry. Textbooks generally like using this "average engineering strain" , but we will not be using this here unless absolutely necessary.
Generalizing 2D to 3D
editThe 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. Equations 3 & 4 are
- [19]
Which in 3D expresses 3 equations , and expands to:
The displacement tensor is:
- [20]
The strain tensor is:
- [21]
The rotation tensor is:
- [22]
Which gives us the displacement:
- [23]
And the new displaced coordinates are: