Lesson 10 Web

Lesson 10: Continuum Mechanics

Introduction

In the electrodynamics lessons you treated matter largely as a continuum that responds to fields through polarization, magnetization, conductivity, and permittivity. Those macroscopic descriptions already assumed that the discrete atomic world could be smoothed into continuous fields of density, stress, and deformation. Continuum mechanics makes that assumption precise and develops it into a complete, self-contained theory of the motion and deformation of solids and fluids.

Deformations

We now consider how to model deformations mathematically. What is a deformation? Begin with a region within , called .

Figure 10.1 The region representing our deformable body before deformation.

We can define a point within as defined by the position vector .

Figure 10.2 Establishing a position vector to a point in the original configuration.

At some later time the system occupies a new region R.

Figure 10.3 A region representing a new configuration of the object in question.

We can now shift our point from to P in the new region, described by the new position vector .

Figure 10.4 Establishing a new position vector to a point in the new configuration.

This shift of a point from one region (and configuration) to another, is what we call a deformation.

Figure 9.5 The deformation of an original configuration to a new configuration.

The act of the point described by transforming into the point described by is given by the set of equations

(10.1)

Mathematically, this is how we characterize a deformation. It is important to realize that it is the physical object that is deforming and not the space it is in. The regions and R are configurations of the body in and not the body itself.

If (10.1) is a set of continuous transformations, then we can write the inverse transformations

(10.2)

We call the undeformed region. We call the point with coordinates a tag and we use this to identify the material that will be displaced by the point P having coordinates in the deformed region R.

Following the trajectories of individual points starting with is what we call a Lagrangian approach. This approach was introduced by Leonhard Euler in 1752. Watching the grid of points defined by and measuring various properties at these points is what we call an Eulerian approach. This approach was introduced by Jean le Rond d’Alembert in 1749.

Deformation and Displacement

Say we examine any curve in the configuration space , we can label this .

Figure 10.6 A curve in .

We denote a point on this curve as .

Figure 10.7 A point on the curve in .

We allow this curve to be transformed into a new curve C in the region R. Allow the point to transform into the point P.

Figure 10.8 Transforming the curve in into the curve C in R. Note the corresponding transformation of the point into P.

As above the coordinates of are and those of P is .

If we write the displacement from to P as

Figure 10.9 Establishing the displacement from to P.

We can specify the position vector to the point and for the point P.

Figure 10.10 Establishing the position vectors for and P.

If we shift along by a line element we have

Figure 10.11 Moving the point by the line element .

Now we find the corresponding point along C, this will be .

Figure 10.12 Moving the point P by the line element .

This produces the new displacement .

Figure 10.13 Establishing a new displacement.

This situation allows us to write

(10.3)

and

(10.4)

(10.5)

This represents the local change in length and angle.

If we can write

(10.6)

then we can write the Nabla operators for

(10.7)

and for R,

(10.8)

This gives us the results

(10.9)

and

(10.10)

So we can write

(10.11)

Exercise 10.1: Explain these steps.

Here we call a displacement gradient.

From this and (10.5) we can derive the statement

(10.12)

If we define a tensor Ψ

(10.13)

then we can write

(10.14)

So (consider referring to (10.13))

(10.15)

The tensor is called the deformation gradient.

Exercise 10.2: Explain the steps in the derivation of (10.15).

Exercise 10.3: Show that .

The Transformation of a Volume Element

When a body is deformed, we can think of it as a change in the volume element according to (10.1).

Figure 10.13 A deformation as a change in volume elements.

The volume element associated with is ,

(10.16)

From this, we can see that,

(10.17)

How do we characterize these component vectors?

(10.18)

We can then rewrite (10.17)

(10.19)

If we define the Jacobian, J

(10.20)

then we can rewrite (10.17) again,

(10.21)

this is the ratio of the volume element to its material volume.

Exercise 10.4: Write out the determinant of the Jacobian.

Exercise 10.5: If a deformation represents a change in coordinates, then what is the volume element for spherical coordinates.

Stretch

Let’s say we have an object whose initial length is given as . After a time the length becomes l. The ratio of the new length from the initial length is what we call the stretch of the object, .

(10.22)

We can then write a quantity

(10.23)

We can call this quantity the extension of our object.

If we switch from a generic object to a length element, then we rewrite (10.22)

(10.24)

This is very hard to calculate. Instead we can write,

(10.25)

This is a characteristic difficulty of the kinematics of deformation, squares of lengths are easily calculated, while the lengths are not.

By applying (10.13), (10.14), and the result of Exercise 10.3

(10.26)

Conjugate Tensors

If we define a tensor according to its tensors product,

(10.27)

then we can write its conjugate

(10.28)

We state a theorem without proof.

Theorem 10.1: Given a product of a tensor A and a vector

(10.29)

Problem 10.1: Prove this theorem.

Stretch, Revisited

We can apply Theorem 10.1 to rewrite (10.26)

(10.30)

We can then rewrite(10.25)

(10.31)

If we define

(10.32)

We call this Green’s deformation tensor. If we define the unit tangent vector to the line element

(10.33)

then we can write the

(10.34)

Exercise 10.6: Write (10.34) in terms of components.

Exercise 10.7: If we write (10.33) for some normal vector, , then rewrite (10.34).

Principal Stretches

What direction, defined by a unit vector, has the largest ? This becomes equivalent to finding the eigenvector of the largest eigenvalue of the tensor C.

If we write

(10.35)

then we can choose our principal axes so that

(10.36)

We call , , and the principal stretches.

Strain

If we assume that the extension (10.23) is small, then this is the engineering definition of strain. Since we are dealing with , then it seems reasonable to deal with

(10.37)

In fact we can use this to define what is called the Green strain,

(10.38)

We can then construct this quantity,

(10.39)

We can rewrite this using (10.34)

(10.40)

We can then define the Lagrangian strain tensor

(10.41)

Exercise 10.8: Derive these results.

For small extensions the diagonal terms of the tensor are called the normal strains as represented in (10.37). The off-diagonal components are termed shearing strains. It is important to realize that there is no good physical interpretation of the components for large extensions, even though we still call them normal and shearing strains.

Exercise 10.9: Find the principal strains the same way we found the principal stretches of (10.36). How are these principal strains related to the principal stretches?

Exercise 10.10: Show that the Eulerian strain tensor is .

Exercise 10.11: Write both the Lagrangian and Eulerian strain tensors in component form.

Note that the 1/2 factor exists for two related reasons:

Without the 1/2 factor the Eulerian Strain Tensor would differ from the Green Deformation tensor by a factor of 2.

If we examine a uniaxial stretch, the factor of 1/2 allows the Eulerian strain to be equivalent to the Green strain.

Rotation

If we look at the tensor we called the deformation gradient (10.15)

we can see that it represents a change in the length of a line element while measures that change in length. This tensor also represents rotations of a deformable body. We now turn to those.

Before we proceed we present an important theorem.

Theorem 10.2: Given a tensor A and the product whose eigenvectors are . Then we will have a set of mutually orthogonal unit vectors , a new set of unit vectors and coefficients f, g, h such that

(10.42)

We can also write

(10.43)

and

(10.44)

Problem 10.2: Prove (10.42), (10.434), and (10.44).

So, we can write the deformation gradient

(10.45)

If we look at (10.32) we can write

(10.46)

Exercise 10.12: Show this to be true.

Then, by (10.43) we have

(10.47)

The set of eigenvectors take on the principal axes of strain and the scalars are the principal strains, so we write

(10.48)

We now state another theorem.

Theorem 10.3: This is also called the Polar Decomposition Theorem, any second-order tensor may be decomposed into the product of an orthogonal tensor and a symmetric tensor.

Problem 10.3: Prove this theorem.

By this theorem we can rewrite (10.48)

(10.49)

Since and are orthogonal we can make an appeal to our discussion of rigid bodies. There must be some rotation tensor such that

(10.50)

(10.51)

This allows us to write

(10.52)

This leads us to Cauchy’s theorem.

Theorem 10.4: This is also called Cauchy’s Theorem, The deformation at any point may be considered as resulting from a translation, a rigid rotation of the principal axes of strain, and stretches along those axes.

Exercise 10.13: What are the magnitude and axis of rotation?

Small Strain

In practice it is desirable to simplify our analysis of strain by assuming that we are dealing with “small strains.” What does that mean? It means that we adopt two conditions, where the displacement gradients are small so that, for the Lagrangian approach,

(10.53)

and for the Eulerian approach,

(10.54)

Here (10.54) leaves us with the impression that the Lagrangian (or material) gradients and the Eulerian (or spatial) gradients are approximately the same,

(10.55)

Exercise 10.14: Use (10.55) to show that in this approximation the Lagrangian and Eulerian strain tensors are equal.

We can use the result from Exercise 10.14 to arrive at our small strain tensor

(10.56)

(10.57)

Exercise 10.15: If we write , then write the matrix representation of the small strain tensor for Cartesian coordinates. Identify the normal strains and the shearing strains.

Problem 10.4: Derive an expression for the approximate volume change under a small strain.

We can alter our rotation representation using the small strain approximation and arrive at the small rotation tensor

(10.58)

Problem 10.5: Derive the small rotation tensor.

We can establish the compatibility equations

(10.59)

If this tensor, called the incompatibility tensor, vanishes then the system is compatible. This means there are no pathologies in it. It is another way of saying that system is integrable. It is important to note that there are integrable theories where this does not vanish, for example in the dislocation of crystal lattices.

The Kinematics of Deformation

Motion in an Eulerian Frame

If we want to study the motion of a nonrigid body we are effectively studying a continuous deformation. Our set of transformations now looks like this,

(10.60)

One could state that any motion governed by this set of transformations is an ordered sequence of transformation where the evolution is governed by an order parameter, t. This is a good example of the fiber bundle approach to the space-time of classical mechanics. We can invert the transformation and get

(10.61)

The transformations of the (10.60)-type are Lagrangian, or material. Those of the (10.61) type are Eulerian, or spatial.

The traditional definition of velocity,

(10.62)

this effectively tags a particle and follows it through its trajectory, thus it is a material derivative.

If we take an Eulerian frame and try the same trick, tagging a point, what happens?

(10.63)

We can find the velocity of the element of material occupying a point,

(10.64)

Once again, we have material derivatives.

If we have a function , then we can write the material derivative

(10.65)

and this allows us to abstract a material derivative operator,

(10.66)

and we can isolate its parts,

Path Lines, Stream Lines and Streak Lines

Instead of simply using equations, we might want to see how a deformable body is moving. We can begin by tagging a particle and following its motion. The resulting visualization is called a path line. We change (10.62) to read,

(10.67)

and we integrate this. A large set of such path lines will allow us to visualize the displacement field over time.

If you want to find the velocity field, then we are finding arcs that are parallel to

(10.68)

These are called stream lines.

If we want to know the locus at some time t all of the particles that have occupied, or will occupy, a point, this is a streak line. We begin by rewriting (10.61) and replace t with some parameter t,

(10.69)

Then we can rewrite (10.61)

(10.70)

Rate of Change of a Volume Element

We have already calculated the change in a volume element in terms of a Jacobian. The determinant of the Jacobian is,

(10.71)

The time derivative of such a determinant is,

(10.72)

Problem 10.6: Derive this result.

We can also write,

(10.73)

Exercise 10.16: Explain each step.

The first of the determinants can then be written

(10.74)

Problem 10.7: Prove this result.

We can write the other determinants in similar ways

(10.75)

(10.76)

So, (10.72) becomes

(10.77)

We can rewrite this,

(10.78)

This result is due to Euler from a work in 1757.

Reynolds Transport Theorem

Say we have some tensor function defined at a point, . It turns out that

(10.79)

is also a function of time, and the limits of integration may be different at different times. We would like to find the time derivative of this, but the variable limits makes this tricky. We can write,

(10.80)

This is a kinematic relation that is valid for any function A. If we apply our definition of the material derivative operator (10.66) and apply the divergence theorem, we may rewrite (10.80)

(10.81)

If we move to a stationary Eulerian frame, then we have the volume , where the volume is bounded by the surface , so we write this,

(10.82)

Here is the velocity of the element passing through the bounding surface. This is the Reynolds Transport Theorem. Note that has a name that we encountered in thermodynamics, we call it the control volume.

Rate of Deformation

Let’s think about the gradient of the velocity field, ; this forms a second rank tensor. We can decompose this tensor into symmetric and antisymmetric parts,

(10.83)

where

(10.84)

(10.85)

This is called the rate of deformation tensor, sometimes it is called the stretch tensor.

And

(10.86)

or,

(10.87)

This tensor is called the spin tensor. Associated with the spin tensor is the vector is called the vorticity vector.

The Dynamics of Deformation

Continuity Equation

If we define density as

(10.88)

We require the density of an element of a body to maintain its mass. Thus we require

(10.89)

So the material, or Lagrangian equation of continuity is,

(10.90)

We can take the material derivative of this,

(10.91)

We can rewrite this,

(10.92)

We can then use (10.79) to get,

(10.93)

This is called the spatial or Eulerian continuity equation.

The Stress Tensor

If we apply a force along a surface , by now it is clear that there must be some sort of tensor that inputs the surface and gives out the force. We call such a tensor the stress tensor and denote it as T or , so that

(10.94)

(10.95)

One way to interpret the components of the stress tensor is to realize that we are locating the i component of force per area along a surface perpendicular to .

The stress tensor is symmetric.

Another way to think about the components of the stress tensor is to think in terms of momentum. Here we are considering the i component of momentum that crosses a unit area perpendicular to per unit time.

Conservation of Momentum

The take-away so far is that the stress tensor is the flux of momentum through a surface. We will write the momentum density in a volume as Π, so the total momentum in a volume is . As momentum flows into and out of the volume the rate of this flow is the integral of the stress tensor for the surface bounding the volume.