Introduction

A tensor is a multi-dimensional construct which allows engineers and physicists to go beyond what mere scalars and 2 or 3 dimensional vectors can represent.  One could argue that scalars and vectors are just the simplest, most basic tensors. But more often, tensors are considered to be beyond vectors because tensors can have many more components than we think of normal vectors as having.  One of the strengths of this concept is that we can map, or transform, between coordinate systems in the tensor realm.  Some operations such as the dot product or cross product can be viewed as a tensor mapping, although tensor mappings may also be much more complicated than that.

Tensors are important because they provide a mathematical framework for formulating and solving problems in areas such as general mechanics (i.e. stress and strain relationships), fluid mechanics, electrodynamics, and general relativity.  We will begin our discussion by discussing the mathematics of tensors, and then we will examine how engineers use them.

Tensor space

 Recall that vectors can be defined in terms of component directions.  In 3 dimensional cartesian coordinate space, we normally use the i, j, and k unit vectors of a cartesian coordinate system, which define the x, y and z directional components of a vector.  Tensors can be simple enough so as to include such common vectors but can be much more complex than that, with many more components than the simple 3 of the i, j, and k just spoken of.  Each of the various components of the tensor are referred to as a basis.  Thus, if our simple cartesian coordinate system vector were considered to be a simple tensor space, i, j and k unit vectors would be the basis functions.   However, a complicated tensor might have many more basis functions than three.  Therefore, rather than use i, j, and k, we tend to write our unit vectors as e₁ in the x₁ direction, e₂ in the x₂ direction, and e₃ in the x₃ direction, where e₁ would correspond to our first direction (i.e. what is usually called the i unit vector or x direction), e₂ to our second direction (usually called the j unit vector or y direction) and e₃, to the third direction (usually called the k unit vector or z direction) but allows us to go to e₄, e₅, and so on to e_n if we have need of further basis functions.

Recall that vectors occur within a vector space or vector field.  For example, the i, j, and k unit vectors, representing the x, y, and z directions occur within the cartesian coordinate system vector space.  It is common to study situations in which a different tensor could occur at each and every point of an object.  The stress within an object might vary from one location to another, and would similarly exist within a much larger tensor space or tensor field.

Consider a cubic element of a material which is experiencing complex stresses.  Let us consider the cubic element as oriented such that one side faces in the positive x₁ direction (designate it side 1) and one side faces in the positive x₂ direction (designate it as side 2) and the third side faces in the positive x₃ direction (which will logically be called side 3).  Each of these side designations would be a basis.  Now consider that on the side we designated 1, there could be a stress in the x₁ direction, a stress in the x₂ direction and a stress in the x₃ direction.  We could designate the first of these as σ ₁₁, meaning it is the stress in direction 1 on face 1, and the second of these as σ₁₂, meaning it is the stress in direction 2 on face 1, and the third of these as σ₁₃, meaning it is the stress in direction 3 on face 1.  Similarly, we could designate the three directional stresses on face 2 as σ₂₁, σ₂₂, and σ₂₃, and likewise for face 3.  The result can be seen in the following figure.

We could also show, on each face, a tensor, designated by T.  Each tensor would be the combination of the individual stress components on each face.

Te

The stress example that was just discussed would have 9 stress terms, each with 2 subscripts (combinations of 1, 2 and 3), and each of which would be a basis of this tensor space.  It is therefore easy to see how this could be assembled into a matrix type of form.  However, in tensor mathematics, we call this an array, rather than a matrix, for reasons which will become apparent in a few moments. 

A tensor may be represented as a potentially multidimensional array.  Think of multiple cubic elements of the type just discussed, combined in a 3 by 3 by 3 arrangement of these elements, as shown.  

Each of the individual small cubes would be just like the cube that we just examined, which had 9 basis functions.  Our array of basis terms would now need to exist in 3 dimensions.  And the components would now all have 3 subscripts rather than 2, as we would have to designate a new basis index to tell us which of the small cubes we were on.  We would need a 3-dimensional matrix to capture all of the terms.  Therefore, we call it an array, since mathematically, there is no such thing as a 3-dimensional matrix.  Also, in some situations (say quantum mechanics) a fourth dimension of time could become relevant.  Clearly what we are dealing with here goes well beyond simple vector and matrix mathematics. 

 

Tensor notation and transformations

A tensor is defined as having a rank, where the rank is based on how many basis components it has.   In other words, a scalar quantity, since it has no directional component, would be of rank 0.  A single uni-directional tensor is of rank 1, having only one basis component.  A tensor in two-dimensional (cartesian xy coordinate space) is of rank 2, having i (x direction) and j (y direction) basis components.  A tensor in three-dimensional space is of rank 3, having i, j and k (z direction) basis components.  If we had a multidimensional tensor with n basis components, we would call it a tensor of rank n.  If our tensor occurs in m dimensional space (we will most usually be working in three-dimension space, so m will usually be 3) then the total number of basis components will be mⁿ.  At the end of the previous section we were dealing with 3 sets of basis components and a 3-dimensional space, so 3³ = 27 would be the total number of combinations of basis components describing the system would be 27.  This may be explained in a more visual sense by viewing the end of either of the two video links provided in the previous section.

One might ask why any of this is useful to us.  Tensors provide us with a mathematical means of formulating and solving problems in areas of physics such as elastic mechanics of materials, fluid mechanics, electromagnetism, and even general relativity.  We are going to examine the mathematical concepts first, and then proceed on to some physical application for engineering.

A tensor is invariant under a change of coordinate systems.   In other words, if you define a tensor in one coordinate system (say the standard cartesian coordinate system, xyz) and then transform it to a different coordinate system (say a spherical coordinate system or an alternative orthogonal coordinate system, which we might call x’y’z’ ), the tensor itself does not change, merely the components that combine to represent it, since the components will be expressed in whatever coordinate system we are using.  And the components in the first system and the components in the second system will change according to specific mathematical formulae, which we refer to as a transformation.

 

Let us say that we wish to transform from expressing the tensor T in the xyz coordinate system, to an alternative x’y’z’ coordinate system.   Let us say that multiplication of T by Q is the transformation operation that changes the magnitude of T in xyz to the magnitude in x’y’z’.   Then multiplication by Q^-1 (the inverse of Q) is the transformation that changes the basis components from x’y’z’ to xyz.  If we apply the transformation [Q][T] = [T’] we have the same tensor we started with, but written in terms of different basis components.  On other words, T and T’ are the same tensor, however, we are expressing its components differently, but they still combine to create the original tensor.  And the result of applying both of those transformations would bring us back to where we started:   [Q^-1][Q][T] = [T], or the same tensor we started with, written in terms of the same basis components.

At this point it seems that we should discuss how the components of a tensor are notated.  The notation that we will use for a tensor is referred to as Einstein Notation.  It will involve indices written as subscripts and superscripts.  Let us say that we are in 3 dimensional (xyz) coordinate space.  We might write the components in this space as a_ij where i is known as the free index and j as the dummy index.  The free index only occurs once, i.e. if it is i, then it must be i all of the time.  Meanwhile, j, must vary between 1, 2 and 3, if we are in 3-dimensional space.  In other words, we might say a_ij  = a_i1 + a_i2 + a_i3.  This is akin to saying that a vector associated with any face (designated i) is the sum of three component vectors, one in the x direction, one in the y direction, and one in the z direction.   From the standpoint of operations which we can apply, the product of a_ijb_j =  a_i1b₁ + a_i2b₂ + a_i3b₃. 

 

A free index can only occur once in each term and must be the same on both sides of the equal sign.  Dummy indices can occur no more than 3 times in a term.  Thus, in the term a_ijb_j, i would be a free index and j a dummy index.  The same would be true for a_jib_j.  If you had an expression x_i = a_ijb_j then i would be a free index and j a dummy index.  Whereas in the expression a_i = a_kib_kjx_j, i would be a free index because it appears only once and is on both sides of the equation, but k and j would be dummy indices because they both appear twice.

 

How would we determine the free and dummy indices of the multiplicative expression a_ij(b_i + c_j + d_k)?  Consider how the individual terms combine.

• a_ij and b_i : i will occur twice and j once
• a_ij and c_j : j will occur twice and i once
• a_ij and d_k : i, j and k all occur once
• Therefore i and j are dummy indices and k is a free index

Let us consider some operations that are valid in tensor math:

• a_ij(x_j + y_j) = a_ijx_j + a_ijy_j
• a_ijx_iy_j =  a_ijy_jx_i
• a_ijx_ix_i =  a_jix_ix_i
• (a_ij + a_ji)x_ix_j =  2a_ijx_ix_j
• (a_ij + a_ji)x_ix_j =  0

 

However, one must be careful, because the following are NOT valid:

• a_ij(x_i + y_j) not = to  a_ijx_i + a_ijy_j
• a_ijx_iy_j not = to  a_ijx_jy_i
• (a_ij + a_ji)x_iy_j not = to 2a_ijx_iy_j

 

There is one additional operator that needs to be defined, and that is the Kronecker Delta (d).  

• d_ij = 0 when i is not equal to j
• d_ij = 1 when i is equal j

Contravariant and covariant tensors

Consider a tensor, T, in a normal cartesian xyz coordinate system with unit vectors in the i, j and k directions.  That same tensor can also be represented in an alternative orthogonal coordinate system, x’y’z’ with components in the i’, j’, and k’ unit vector directions.  The tensor could be represented in either system:

 

T = ai + bj  + ck            Or              T = di’ + ej’ + fk’

 

Whichever of these we use, T is still T, the same tensor in both.  It is just represented by different amplitudes of a, b, and c in the i, j and k directions, or amplitudes d, e, and f in the i’, j’, and k’ directions.

 

Now we need to introduce two new terms, covariant and contravariant.  Our basis components my change in either a covariant or contravariant fashion.  Keep in mind, that in all cases, the actual tensor, T, does not change.  T is still T.  We are only changing the way we represent it’s components. 

 

Let us begin our discussion of covariant and contravariant by considering only  tensors of rank 1 (i.e. simple vectors). 

 

Let us examine Contravariant first. The indices of a contravariant component are written as superscripts.  So, whereas in everything we have done thus far, we wrote our indices as subscripts (e.g.  a_i) we would now write them as superscripts (e.g. aⁱ).  Contravariant means that the basis vector and its associated amplitude vary contra, or counter, to each other.   As one gets larger, the other gets smaller.   Here is an example.

 

Consider a coordinate system made up of axes x¹ and x², where x¹ and x² are NOT perpendicular to each other the way they are in a standard cartesian xy coordinate system.  And let us say that we have a vector V represented in this coordinate system as shown:

Now let us consider that we wish to represent vector V by its components in these two coordinates, x¹ and x².   We will call the unit vector which is directed along the x¹ axis e₁ and the unit vector directed along the x² axis e₂.  Remember, these unit vectors are of magnitude 1.  That is why we called them unit vectors.

If we wished to describe V in terms of these coordinates, we could say that V was made up of 2 unit vectors e₂ and 3 unit vectors e₁. 

 

Written as a vector equation,  V = 3e₁ + 2e₂.

Now consider if we wish to change our system such that e₂ transforms to twice the magnitude of e₂.  We will call this new one e₂’ (which we would speak as e-two-prime), such that  e₂’ = 2e₂.

Nothing has happened to V.  It is still the same as it ever was.  But we would now write it as the vector equation  V = 3e₁ + 1e₂’.

When we doubled the magnitude of the basis, we halved the number that we needed to represent the component.  Thus, the two changed counter, or contra, to each other.  From this comes the term contravariant.

Now we need to examine covariant.  Indices for covariant components have the indices represented as subscripts, in other words, like all of the components we had represented in earlier discussions.  Logically, covariant means that the basis vector and its associated amplitude vary similar to each other.      Here is an example, beginning with the same coordinate system and vector V as before.  But this time, instead of our dimensions being determined by drawing lines parallel to the coordinate axes, we will draw the lines perpendicular to the coordinate axes.

And the vector equation is V = b₁e₁ + b₂e₂.

Recall that the mathematical operation known as a dot product is the projection of a vector onto an axis.  In this case therefore the amplitudes b₁ and b₂ could be written as the dot product of V and the unit vectors e₁ and e₂, as follows:

b₁  =  V dot e₁         and     b₂  =  V dot e₂

Recall that V = b₁e₁ + b₂e₂, and substitute this into the expressions above, and we get

b₁  =  V dot e₁    =   (b₁e₁ + b₂e₂) dot e₁  =  b₁e₁ dot e₁  +  b₂e₂ dot e₁

and    

b₂  =  V dot e₂    =   (b₁e₁ + b₂e₂) dot e₂  =  b₁e₁ dot e₂  +  b₂e₂ dot e₂

What if we transformed e₂ into an e₂’, such that the new component was twice as large, just as in the contravariant example, such that:

e₂’  =  2e₂

Then our vector equation becomes V = b₁e₁ + b₂’e₂’

And

b₁’  =  b₁e₁ dot e₁  +  b₂’e₂’ dot e₁

and    

b₂’  =  b₁e₁ dot e₂‘ +  b₂’e₂’ dot e₂’

Since b₂’ = b₂/2    and     e₂’  =  2e₂     by substitution, these become

b₁’  =  b₁e₁ dot e₁  +  (b₂/2)2e₂ dot e₁   =  b₁e₁ dot e₁  +  b₂e₂ dot e₁     which from above, is simply b₁.   So b₁’  =  b₁

and    

b₂’  =  b₁e₁ dot 2e₂ +  (b₂/2)2e₂ dot 2e₂  =  2 [b₁e₁ dot e₂  + b₂e₂ dot e₂ ]  which from above is simply twice b₂.  So b₂’ = 2b₂.

So when we doubled the basis vector, e₂’ = 2e₂, the magnitude also doubled b₂’ = 2b₂. So, they both changed in the same direction by the same amount.  They are thus referred to as covariant. 

 

Coordinate transformations

Consider a point, (A) in a coordinate system.  In this figure, I have shown the coordinate system as a 2D cartesian system, in which the location of A could be given by A = (a¹, a²) where a¹ is the distance along the horizontal axis and a² is the distance along the vertical axis. However, our coordinate space could be in 3D, or mathematically, it could be in even higher dimensions.   So if we want to be completely general, for an n-dimensional system, we could express our point A, as being A = (a¹, a², a³,…..aⁿ).  Note that we have applied the indices as superscripts, not subscripts, which we have shown may be necessary in contravariant systems.   In order to avoid confusion with a squared, cubed, or higher order term, if something is raised to a power, it will need to be placed in parentheses.  In other words, (a¹)² would be a-super-one squared and (b²)⁴ would be b-super-two to the fourth power. 

 

Now consider another point in our system, B = (b₁,  b₂, b₃, …….b_n).  And the vector between them to be V.

V would be expressed as the vector (written in column matrix form) as shown below.  We might also write this as the vector from A to B, or AB.

Where x¹ is the difference, in the first basis direction, between the position of A and B, or on other words, x¹ = b¹ – a¹, and similarly for the other basis directions.

Let us assume that we also have another coordinate system, also consisting of n basis directions.   It is possible to move from one system to the other using a Transformation Function, F, where F consists of n functions, each of which would act to convert one basis in the original system to a new basis in the new system.   If each of these functions is Real (in the mathematical sense) and have continuous second partial derivatives, and are invertible, then F is called a Coordinate Transformation.  If the original system, and thus the coordinates being input to F are rectangular coordinates, then we call the output coordinates (the new system) either affine or curvilinear:

• Affine if F is made up of linear functions
• Curvilinear if F is made up of non-linear functions

The most common curvilinear coordinate system is the polar coordinate or spherical system which you should have familiarity within previous studies.  However, this is not the only possible curvilinear system, just the most common one.

Additional resources for tensor calculus

Applications of tensors

 

 

 

 

Traction vector and stress components

The traction vector, T, is simply the stress on a surface and is calculated in the traditional manner, by dividing the force, F, on the surface’s area, by the area, A, such that T = F/A.  So T has units of stress (e.g. MPa in the metric system or psi in the Imperial system). 

Normal stresses are simply the components of the traction vector that are normal (perpendicular) to the area's surface, divided by the area of the surface. We often use the letter n to represent the unit vector normal to the surface.

Shear stresses are simply the components of the traction vector that are parallel to the area's surface, divided by the area of that surface. We often use the letter s for the unit vector parallel to the surface.

These can both be found by the dot product of T and the appropriate unit vector (recall that the dot product projects one vector onto another, so in this case it projects T onto an axis in a particular direction, giving us the component of T in that direction). Thus, σ_n = T dot n and σ_s = T dot s.  Recognize that σ_n and σ_s are each scalar values defining the stress in their respective directions, and their direction is in the direction of the unit vector n or s, respectively.  This is the natural result of the dot product operations (dot products always produce scalar results).

Note that in 3 dimensions, there are in fact an infinite number of s vectors parallel to the surface.  It is fairly common to specify one parallel to the page and a second one perpendicular to it (i.e. coming out of the page).

Consider the following figure and develop an expression for the traction vector, T, by setting the sum of forces in the positive direction equal to the forces in the negative direction, in both x (horizontal) and y (vertical), as follows.

σ_xxAcosθ + τ_xyAsinθ =  T_xA  and τ_xyAcosθ + σ_yyAsinθ =  T_yA 

Cancel A out of all terms, and recognise that multiplying by cosθ is the same as taking the dot product with unit vector n_x, and similarly, multiplying by sinθ is the same as taking the dot product with unit vector n_y.  Thus, our equations simplify to:

σ_xxn_x + τ_xyn_y  =  T_x       and     τ_xyn_x + σ_yyn_y  =  T_y      

Taken together, these can be combined into one vector expression T = σ dot n.

Let us now move to 3 dimensions, but initially by considering only a surface element.  We will choose a horizontal (i.e. like a table top) surface.  The normal (perpendicular) to this surface is upward, which we generally call the z direction in a cartesian system and it is normally the third coordinate.  We will, therefore, choose to call this axis x₃ rather than z, and we will call this surface 3.  The unit vector normal to this surface will be designated e₃.  And the traction vector applied to it would be t^e3.  The stresses on this surface would be σ_ij where the first index would be 3 because it is on surface 3, and the second index would indicate which direction the stress pointed.  By the logic previously discussed in 2 dimensions, we could express this as   t^e3 = σ₃₁e₁  +  σ₃₂e₂ + σ₃₃e₃.  This is represented in the following figure:

If this is true for a horizontal, table-top surface, then let us move on and consider a cubic element, and use the logic from the horizontal surface with traction t^e3 to extend our model to the vertical surface which is parallel to the page (i.e. has normal unit vector e₁ and experiences traction t^e1) and the vertical surface perpendicular to the page (i.e. has normal unit vector e₂ and experiences traction t^e2).  This is shown in the following figures:

These three faces assemble into the cubic element as shown.

The tractions can be represented by their components as follows:

t^e1 = σ₁₁e₁  +  σ₁₂e₂  +  σ₁₃e₃

t^e2 = σ₂₁e₁  +  σ₂₂e₂  +  σ₂₃e₃

t^e3 = σ₃₁e₁  +  σ₃₂e₂  +  σ₃₃e₃

Or, stated in simplest form, t^ei = σ_ije_j  where i and j both range from 1 to 3.

The nine stresses can be written in matrix form, as follows (this is often referred to as the stress tensor for this point in the material.)

We must now recognise that at this precise point in the material, we could have selected a different element at a different orientation, using an alternative x’y’z’ orthogonal coordinate system, rather than our normal xyz cartesian coordinate system.  Our system would now be represented by axes x₁’x₂’x₃’.

But the material is actually the same, and the loads are actually the same, we have merely expressed the stresses in a different coordinate system.  Therefore, by the logic we discussed in our tensor discussion previously, there must be a transformation between the two which does not change the overall state.  We will get to that transformation operation in a few minutes.

Cauchy’s Law

Cauchy’s Law states that there is a relationship between the traction vector acting on a surface, the normal to that surface, and the transpose of the stress tensor at the point.  This is expressed as t_i = σ_jin_j.  Note that the stress tensor σ_ji is the transpose of the stress tensor σ_ij, as defined in the previous section.  Recall that the transpose is created by exchanging the rows and columns, so the transpose of the stress matrix in the previous section would be

If we wish to find the traction on any plane, which cuts through our element at the point in question, then we can use Cauchy to find it.  The figure below shows a slice taken through the cubic element that we have been previously working with.  And this slice is identified by the normal vector n and it experiences the traction t⁽ⁿ⁾ as shown.

Once we have found the traction on this plane, we can find the normal stress (stress perpendicular to the slice) the same way we did in the previous section, using the relationship σ_n = n dot t⁽ⁿ⁾

To find the net shear stress acting on this slice, we recognise that the normal component (perpendicular to the slice) and the shear component (parallel to the slice) have to add to form the traction.  And since they are perpendicular to each other, Pythagoras will help us, in that σ_shear²  +  σ_normal² = (t⁽ⁿ⁾)².

 

Example 1

The stress at a point in a material is given by the following stress matrix, in units of MPa.

$\sigma =\left(\begin{array}{ccc}1& \mathrm{-1}& 0\\ -1& 5& 0\\ 0& 0& 4\end{array}\right)MPa$

Find the traction, normal stress, and shear stress components on a plane whose normal vector is in the direction of

$d=\left(\begin{array}{c}1\\ 1\\ 1\end{array}\right)$

First, we must recognise that d is not a unit vector, and therefore we must find the unit vector that represents its direction. We will call this n, since we want the vector to be normal to the plane in question.  We find the unit vector by dividing by the total magnitude of d, which by Pythagoras is the square root of the sum of the components or (1 + 1 + 1)^1/2 = 3^1/2.  So

$n=\frac{1}{3^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\left(\begin{array}{c}1\\ 1\\ 1\end{array}\right)$

Also, note that the transpose of the stress matrix is the same as the stress matrix.  (i.e. σij = σji in this particular case).  Therefore, applying Cauchy, we can find the traction on the surface

$t_n=\sigma n=\frac{1}{3^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\left(\begin{array}{ccc}1& \mathrm{-1}& 0\\ -1& -5& 0\\ 0& 0& 4\end{array}\right)\left(\begin{array}{c}1\\ 1\\ 1\end{array}\right)=\frac{1}{3^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\left(\begin{array}{c}0\\ -6\\ 4\end{array}\right)$

This is the traction on that face.  The normal stress on the surface requested could then be found by the dot product of t_n and the normal to that surface ( unit vector n).

$\sigma n=t_{\mathrm{n }}\text{dot }n=\frac{1}{3^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\left(\begin{array}{c}0\\ -6\\ 4\end{array}\right)\text{dot}\frac{1}{3^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\left(\begin{array}{c}1\\ 1\\ 1\end{array}\right)=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\left(\begin{array}{c}0\\ -6\\ 4\end{array}\right)\text{dot}\left(\begin{array}{c}1\\ 1\\ 1\end{array}\right)=\frac{-6+4}{3}=\raisebox{1ex}{$-2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.=-0.67MPa$

Whatever stress is not normal stress, must be shear stress, so if we subtract the normal stress (which we just found) from the total stress on this surface, then we are left with the shear stress on this surface.  Since we have kept them as vectors, we can simply make this a vector subtraction.

$t_n-{\sigma }_{n}n=\frac{1}{3^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\left(\begin{array}{c}0\\ -6\\ 4\end{array}\right)-\frac{-2}{3}\frac{1}{3^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\left(\begin{array}{c}1\\ 1\\ 1\end{array}\right)=\left(\begin{array}{c}0\\ \mathrm{-3.46}\\ 2.31\end{array}\right)+\left(\begin{array}{c}\mathrm{0.38}\\ 0.38\\ 0.38\end{array}\right)=\left(\begin{array}{c}0.38\\ 3.08\\ 2.69\end{array}\right)MPa$

Recognise that this just gave us three orthogonal components of shear stress, and we can use Pythagoras to combine them into the resultant shear stress term.

t_n  = (0.38² + 3.08² + 2.69²)^1/2  =  4.10 MPa

And now we have the normal and shear stresses on the indicated plane, so we have achieved what was requested. 

 

Principal stresses

We just discussed how at a single point in the material, elements oriented differently would experience different stress components, and how different planes through that point would have different normal and shear stresses.  Which raises the question: is there an orientation of particular interest to us?  If you think back to study of statics or strength of materials, the concept of the principle stresses was introduced.  The maximum principal stress is the highest normal stress in the material and therefore is of particular interest to us.  If we knew the orientation in which these principle stresses occur, then the work we just did would give us these principle stresses.

 

As it turns out, the three principal stresses (σ₁, σ ₂, and σ ₃) are the eigenvalues of the stress tensor and their directions are the associated eigenvectors.    The Von Mises equivalent stress, σ_eq, which is often used in failure mode assessments, is found from (σ₁ – σ₂)² + (σ₂ – σ ₃)² + (σ₃ – σ₁)² = 2σ_eq²

 

Example 2

Let us continue with the material and stress matrix from the previous example.  But now let us find the principal stresses (and their directions) and also the Von Mises equivalent stress.  The stress matrix was:

$\sigma =\left(\begin{array}{ccc}1& \mathrm{-1}& 0\\ -1& 5& 0\\ 0& 0& 4\end{array}\right)MPa$

Let us put this into eigen-form, and set its determinant equal to zero

$\text{det}\left(\begin{array}{ccc}1-\mathrm{\lambda }& -1& 0\\ -1& 5-\mathrm{\lambda }& 0\\ 0& 0& 4-\mathrm{\lambda }\end{array}\right)=0$

If we were to take this determinate, the result would be what is called the characteristic equation.  And the roots of that characteristic equation would be the eigenvalues, each of which would correspond to the three principal stresses (σ₁, σ ₂, and σ ₃).

In practice, the easiest way to do this is to build the stress matrix in MATLAB using the statement

>> A =  [1 -1 0; -1 5 0; 0 0 4]

And then use the Eigen solution command

>> lambda = eig(A)

In this case, the output should be the three solutions -5.16, 1.16 and 4, which would be the principal stresses in MPa. 

There is a second MATLAB eigen command which yields both the eigen values and the eigen vectors:

>> [V,D] = eig(A)

The eigenvalues (and thus the principle stresses) should be the same.  However, with this command, you will additionally get vectors associated with them as indicated here:

These define the unit vectors of each principle axis (i.e. directions of the principle stresses).  Note that we can confirm that they are unit vectors because the results of each is unity, or one, as proven by:  (0.16² + 0.987² + 0²)^1/2  = 1 

The Von Mises equivalent stress would be

(σ₁ – σ₂)² + (σ₂ – σ ₃)² + (σ₃ – σ₁)² = 2σ_eq²

(-5.16 – 1.16)² + (1.16 – 4)² + (4 + 5.16)² =  4² + 2.84²  +  9.16²  = 2σ_eq²

σ_eq  =  7.35 MPa

Transformation matrix

The transformation matrix is the matrix that transforms from one coordinate system to another.  For example, consider how we would transform from the x₁x₂x₃ coordinate system shown in the figure to the x₁’x₂’x₃’ coordinate system, where we rotate the x₁ and x₃ axes 45 degrees while keeping the x₂ coordinate unchanged. 

 

Recall that cos(45^o) = sin(45^o) = 0.707.  Our transformation matrix, Q, would be:

$Q=\left(\begin{array}{ccc}0.707& 0& 0.707\\ 0& 1& 0\\ -0.707& 0& 0.707\end{array}\right)$

Such that

$\left(\begin{array}{c}{x}_1\prime \\ x_2\prime \\ x_3\prime \end{array}\right)=\left(\begin{array}{ccc}0.707& 0& 0.707\\ 0& 1& 0\\ -0.707& 0& 0.707\end{array}\right)\left(\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right)$

The state of stress at a particular point in the first coordinate system (this is given by the stress tensor σ_ij)  can be transformed to the second coordinate system (call this new stress tensor σ_ij’) using the following matrix operation:   [σ_ij’]  =  [Q]^T[σ_ij][Q] where [Q]^T is the transpose of [Q].

Example 3

The state of stress at a point in a material is given, in an x₁x₂x₃ coordinate system by

$\sigma =\left(\begin{array}{ccc}2& 1& 0\\ 1& 3& -2\\ 0& -2& 1\end{array}\right)$

Assume we want to transform to the x₁’x₂’x₃’ coordinate system where in x₁ and x₃ are rotated 45 degrees while x₂ is held constant (as in the previous discussion).

$Q=\left(\begin{array}{ccc}0.707& 0& 0.707\\ 0& 1& 0\\ -0.707& 0& 0.707\end{array}\right)$

Transform to the new coordinate system.

$\left[{\sigma }_{\mathrm{ij}},\prime \right]={\left[Q\right]}^T\left[{\sigma }_{\mathrm{ij}}\right]\left[Q\right]=\left(\begin{array}{ccc}0.707& 0& -0.707\\ 0& 1& 0\\ 0.707& 0& 0.707\end{array}\right)\left(\begin{array}{ccc}2& 1& 0\\ 1& 3& -2\\ 0& -2& 1\end{array}\right)\left(\begin{array}{ccc}0.707& 0& -0.707\\ 0& 1& 0\\ 0.707& 0& 0.707\end{array}\right)$

$=\left(\begin{array}{ccc}0.707& 0& -0.707\\ 0& 1& 0\\ \begin{array}{c}0.707\end{array}& 0& \begin{array}{c}0.707\end{array}\end{array}\right)\left(\begin{array}{ccc}1.414& 1& \begin{array}{c}1.414\end{array}\\ 2.121& 3& \begin{array}{c}-0.707\end{array}\\ \begin{array}{c}-0.707\end{array}& -2& \begin{array}{c}-0.707\end{array}\end{array}\right)=\left(\begin{array}{ccc}1.500& \begin{array}{c}2.121\end{array}& \begin{array}{c}0.500\end{array}\\ 2.121& 3.000& \begin{array}{c}-0.707\end{array}\\ 0.500& \begin{array}{c}-0.707\end{array}& \begin{array}{c}1.500\end{array}\end{array}\right)$