The purpose of this chapter is to summarize the equations that govern the response of solids to mechanical or thermal loading. The following topics will be addressed in turn:
1. The mathematical description of shape changes in a solid;
2. The mathematical description of internal forces in a solid;
3. Equations of motion for deformable solids;
4. Concepts of mechanical work and power for deformable solids; and the important principle of virtual work
In this section, we list the various mathematical formulas that are used to characterize shape changes in solids (and in fluids). The formulas might look scary at first, but they are mostly just definitions. You might find it helpful to refresh your memory on vectors and matrices (Appendix A), and to read the brief discussion of Tensors (Appendix B) and Index Notation (Appendix C) before wading through this section.
As you work through the various definitions, you should bear in mind that shape changes near a point can always be characterized by six numbers. These could be could be the six independent components of the Lagrangian strain, Eulerian strain, the left or right stretch tensors, or your own favorite deformation measure. Given the complete set of six numbers for any one deformation measure, you can always calculate the components of other strain measures. The reason that so many different deformation measures exist is partly that different material models adopt different strain measures, and partly because each measure is useful for describing a particular type of shape change.
The displacement vector u(x,t) describes the motion of each point in the solid. To make this precise, visualize a solid deforming under external loads. Every point in the solid moves as the load is applied: for example, a point at position x in the undeformed solid might move to a new position y at time t. The displacement vector is defined as
We could also express this formula using index notation, which is discussed in detail in Appendix C, as
Here, the subscript i has values 1,2, or 3, and (for example) represents the three Cartesian components of the vector y.
The displacement field completely specifies the change in shape of the solid. The velocity field would describe its motion, as
Examples of some simple deformations
These quantities are defined by
Displacement Gradient Tensor: is a tensor with components
Deformation Gradient Tensor:
where I is the identity tensor, with components described by the Kronekor delta symbol:
and represents the gradient operator. Formally, the gradient of a vector field u(x) is defined so that
(for more details see Appendix B), but in practice the component formula is more useful.
Note also that
The rules of differentiation using index notation are described in more detail in Appendix C.
The concepts of displacement gradient and deformation gradient are introduced to quantify the change in shape of infinitesimal line elements in a solid body. To see this, imagine drawing a straight line on the undeformed configuration of a solid, as shown in the figure. The line would be mapped to a smooth curve on the deformed configuration. However, suppose we focus attention on a line segment dx, much shorter than the radius of curvature of this curve, as shown. The segment would be straight in the undeformed configuration, and would also be (almost) straight in the deformed configuration. Thus, no matter how complex a deformation we impose on a solid, infinitesimal line segments are merely stretched and rotated by a deformation.The infinitesimal line segments dx and dy are related by
Written out in as a matrix equation, we have
To derive this result, consider an infinitesimal line element dx in a deforming solid. When the solid is deformed, this line element is stretched and rotated to a deformed line element dy. If we know the displacement field in the solid, we can compute dy=[x+dx+u(x+dx)]-[x+u(x)] from the position vectors of its two end points
Expand as a
We identify the term in parentheses as the deformation gradient, so
The inverse of the deformation gradient arises in many calculations. It is defined through
Suppose that two successive deformations are applied to a solid, as shown. Let
map infinitesimal line elements from the original configuration to the first deformed shape, and from the first deformed shape to the second, respectively, with
The deformation gradient that maps infinitesimal line elements from the original configuration directly to the second deformed shape then follows as
Thus, the cumulative deformation gradient due to two successive deformations follows by multiplying their individual deformation gradients.
To see this, write the cumulative mapping as and apply the chain rule
The Jacobian is defined as
It is a measure of the volume change produced by a deformation. To see this, consider the infinitessimal volume element shown with sides dx, dy, and dz in the figure above. The original volume of the element is
Here, is the permutation symbol. The element is mapped to a paralellepiped with sides dr, dv, and dw with volume given by
For any physically admissible deformation, the volume of the deformed element must be positive (no matter how much you deform a solid, you can’t make material disappear). Therefore, all physically admissible displacement fields must satisfy J>0
If a material is incompressible, its volume remains constant. This requires J=1.
If the mass density of the material at a point in the undeformed solid is , its mass density in the deformed solid is
Derivatives of J. When working with constitutive equations, it is occasionally necessary to evaluate derivatives of J with respect to the components of F. The following result (which can be proved e.g. by expanding the Jacobian using index notation) is extremely useful
The Lagrange strain tensor is defined as
The components of Lagrange strain can also be expressed in terms of the displacement gradient as
The Lagrange strain tensor quantifies the changes in length of a material fiber, and angles between pairs of fibers in a deformable solid. It is used in calculations where large shape changes are expected.
To visualize the physical significance of E, suppose we mark out an imaginary tensile specimen with (very short) length on our deforming solid, as shown in the picture. The orientation of the specimen is arbitrary, and is specified by a unit vector m, with components . Upon deformation, the specimen increases in length to . Define the strain of the specimen as
Note that this definition of strain is similar to the definition you are familiar with, but contains an additional term. The additional term is negligible for small . Given the Lagrange strain components , the strain of the specimen may be computed from
We proceed to derive this result. Note that
is an infinitesimal vector with length and orientation of our undeformed specimen. From the preceding section, this vector is stretched and rotated to
The length of the deformed specimen is equal to the length of dy, so we see that
Hence, the strain for our line element is
giving the results stated.
The Eulerian strain tensor is defined as
Its physical significance is similar to the Lagrange strain tensor, except that it enables you to compute the strain of an infinitesimal line element from its orientation after deformation.
Specifically, suppose that n denotes a unit vector parallel to the deformed material fiber, as shown in the picture. Then
The proof is left as an exercise.
The infinitesimal strain tensor is defined as
where u is the displacement vector. Written out in full
The infinitesimal strain tensor is an approximate deformation measure, which is only valid for small shape changes. It is more convenient than the Lagrange or Eulerian strain, because it is linear.
Specifically, suppose the deformation gradients are small, so that all . Then the Lagrange strain tensor is
so the infinitesimal strain approximates the Lagrange strain. You can show that it also approximates the Eulerian strain with the same accuracy.
Properties of the infinitesimal strain tensor
For small strains, the engineering strain of an infinitesimal fiber aligned with a unit vector m can be estimated as
(see below for more details)
The infinitesimal strain tensor is closely related to the strain matrix introduced in elementary strength of materials courses. For example, the physical significance of the (2 dimensional) strain matrix
is illustrated in the figure.
To relate this to the infintesimal strain tensor, let be a Cartesian basis, with parallel to x and parallel to y as shown. Let denote the components of the infinitesimal strain tensor in this basis. Then
For a general strain tensor (which could be any of , or , among others), the diagonal strain components are known as `direct’ strains, while the off diagonal terms are known as ‘shear strains’
The shear strains are sometimes reported as ‘Engineering Shear Strains’ which are related to the formal definition by a factor of 2 i.e.
This factor of 2 is an endless source of confusion. Whenever someone reports shear strain to you, be sure to check which definition they are using. In particular, many commercial finite element codes output engineering shear strains.
The volumetric infinitesimal strain is defined as
The deviatoric infinitesimal strain is defined as
The volumetric strain is a measure of volume changes, and for small strains is related to the Jacobian of the deformation gradient by . To see this, recall that
The deviatioric strain is a measure of shear deformation (shear deformation involves no volume change).
The infinitesimal rotation tensor is defined as
Written out as a matrix, the components of are
Observe that is skew symmetric: .
A skew tensor represents a rotation through a small angle. Specifically, the operation rotates the infinitesimal line element through a small angle about an axis parallel to the unit vector . (A skew tensor also sometimes represents an angular velocity).
To visualize the significance of , consider the behavior of an imaginary, infinitesimal, tensile specimen embedded in a deforming solid. The specimen is stretched, and then rotated through an angle about some axis q. If the displacement gradients are small, then .
The rotation of the specimen depends on its original orientation, represented by the unit vector m. One can show (although one would rather not do all the algebra) that represents the average rotation, over all possible orientations of m, of material fibers passing through a point.
As a final remark, we note that a general deformation can always be decomposed into an infinitesimal strain and rotation
Physically, this sum of and can be regarded as representing two successive deformations a small strain, followed by a rotation, in the sense that
first stretches the infinitesimal line element, then rotates it.
The three principal values and directions of the infinitesimal strain tensor satisfy
Clearly, and are the eigenvalues and eigenvectors of . There are three principal strains and three principal directions, which are always mutually perpendicular.
Their significance can be visualized as follows.
1. Note that the decomposition can be visualized as a small strain, followed by a small rigid rotation, as shown in the picture.
2. The formula indicates that a vector n is mapped to another, parallel vector by the strain.
3. Thus, if you draw a small cube with its faces perpendicular to on the undeformed solid, this cube will be stretched perpendicular to each face, with a fractional increase in length . The faces remain perpendicular to after deformation.
4. Finally, w rotates the small cube through a small angle onto its configuration in the deformed solid.
There are two Cauchy-Green deformation tensors defined through
The Right Cauchy Green Deformation Tensor
The Left Cauchy Green Deformation Tensor
They are called `left’ and `right’ tensors because of their relation to the `left’ and ‘right’ stretch tensors defined below. They can be regarded as quantifying the squared length of infinitesimal fibers in the deformed configuration, by noting that if a material fiber in the undeformed solid is stretched and rotated to in the deformed solid, then
The definitions of these quantities are
The Right Stretch Tensor
The Left Stretch Tensor
The Rotation Tensor
To calculate these quantities you need to remember how to calculate the square root of a matrix. For example, to calculate the square root of C, you must
1. Calculate the eigenvalues of C we will call these , with n=1,2,3. Since C and B are both symmetric and positive definite, the eigenvalues are all positive real numbers, and therefore their square roots are also positive real numbers.
2. Calculate the eigenvectors of C and normalize them so they have unit magnitude. We will denote the eigenvectors by . They must be normalized to satisfy
3. Finally, calculate , where denotes a dyadic product (See Appendix B). In components, this can be written
4. As an additional bonus, you can quickly compute the inverse square root (which is needed to find R) as
To see the physical significance of these tensors, observe that
1. The definition of the rotation tensor shows that
2. The multiplicative decomposition of a constant tensor can be regarded as a sequence of two homogeneous deformations U, followed by R. Similarly, is R followed by V.
3. R is proper orthogonal (it satisfies and det(R)=1), and therefore represents a rotation. To see this, note that U is symmetric, and therefore satisfies , so that
where are the three (mutually perpendicular) eigenvectors of U. (By construction, these are identical to the eigenvectors of C). If we interpret as basis vectors, we see that U is diagonal in this basis, and so corresponds to stretching parallel to each basis vector, as shown in the figure below.
are known as the right and left polar decomposition of F. (The right and left refer to the positions of U and V). They show that every homogeneous deformation can be decomposed into a stretch followed by a rigid rotation, or equivalently into a rigid rotation followed by a stretch. The decomposition is discussed in more detail in the next section.
The principal stretches can be calculated from any one of the following (they all give the same answer)
The principal stretches are also related to the eigenvalues of the Lagrange and Eulerian strains. The details are left as an exercise.
There are two sets of principal stretch directions, associated with the undeformed and deformed solids.
To visualize the physical significance of principal stretches and their directions, note that a deformation can be decomposed as into a sequence of a stretch followed by a rotation.
Note also that
The decomposition can be visualized in much the same way. In this case, the directions are first rotated to coincide with . The cube is then stretched parallel to each to produce the same shape change.
We could compare the undeformed and deformed cubes by placing them side by side, with the vectors and parallel, as shown in the figure.
The polar decompositions and provide a way to define additional strain measures. Let denote the principal stretches, and let and denote the normalized eigenvectors of U and V. Then one could define strain tensors through
The correspoinding Eulerian strain measures are
Another strain measure can be defined as
This can be computed directly from the deformation gradient as
and is very similar to the Lagrangean strain tensor, except that its principal directions are rotated through the rigid rotation R.
We now list several measures of the rate of deformation. The velocity gradient is the basic measure of deformation rate, and is defined as
It quantifies the relative velocities of two material particles at positions y and y+dy in the deformed solid, in the sense that
The velocity gradient can be expressed in terms of the deformation gradient and its time derivative as
To see this, note that
and recall that , so that
The stretch rate tensor is defined as
The spin tensor is defined as
A general velocity gradient can be decomposed into the sum of stretch rate and spin, as
The stretch rate quantifies the rate of stretching of material fibers in the deformed solid, in the sense that
is the rate of stretching of a material fiber with length l and orientation n in the deformed solid. To see this, let , so that
Finally, take the dot product of both sides with n, note that since n is a unit vector must be perpendicular to n and therefore . Note also that , since W is skew-symmetric. It is easiest to show this using index notation: . Therefore
The spin tensor W can be shown to provide a measure of the average angular velocity of all material fibers passing through a material point.
For small strains the rate of deformation tensor can be approximated by the infinitesimal strain rate, while the spin can be approximated by the time derivative of the infinitesimal rotation tensor
Similarly, you can show that
The rate of deformation tensor can be related to time derivatives of other strain measures. For example the time derivative of the Lagrange strain tensor can be shown to be
Other useful results are
For a pure rotation , or equivalently . To see this, recall that and evaluate the time derivative.
If the deformation gradient is decomposed into a stretch followed by a rotation as then and
. The trace of D is therefore a measure of rate of change of volume.
For small strains the rate of change of Lagrangian strain E is approximately equal to the rate of change of infinitesimal strain :
It is sometimes necessary to invert the relations between strain and displacement that is to say, given the strain field, to compute the displacements. In this section, we outline how this is done, for the special case of infinitesimal deformations.
For infinitesimal motions the relation between strain and displacement is
Given the six strain componets (six, since ) we wish to determine the three displacement components . First, note that you can never completely recover the displacement field that gives rise to a particular strain field. Any rigid motion produces no strain, so the displacements can only be completely determined if there is some additional information (besides the strain) that will tell you how much the solid has rotated and translated. However, integrating the strain field can tell you the displacement field to within an arbitrary rigid motion.
Second, we need to be sure that the strain-displacement relations can be integrated at all. The strain is a symmetric second order tensor field, but not all symmetric second order tensor fields can be strain fields. The strain-displacement relations amount to a system of six scalar differential equations for the three displacement components ui.
To be integrable, the strains must satisfy the compatibility conditions, which may be expressed as
Or, once more equivalently
It is easy to show that all strain fields must satisfy these conditions - you simply need to substitute for the strains in terms of displacements and show that the appropriate equation is satisfied. For example,
and similarly for the other expressions.
Not that for planar problems for which all of these compatibilty equations are satisfied trivially, with the exception of the first:
It can be shown that
(i) If the strains do not satisfy the equations of compatibility, then a displacement vector can not be integrated from the strains.
(ii) If the strains satisfy the compatibility equations, and the solid simply connected (i.e. it contains no holes that go all the way through its thickness), then a displacement vector can be integrated from the strains.
(iii) If the solid is not simply connected, a displacement vector can be calculated, but it may not be single valued i.e. you may get different solutions depending on how the path of integration encircles the holes.
Now, let us return to the question posed at the beginning of this section. Given the strains, how do we compute the displacements?
2D strain fields
For 2D (plane stress or plane strain) the procedure is quite simple and is best illustrated by working through a specific case
As a representative example, we will use the strain field in a 2D (plane stress) cantilever beam with Young’s modulus E and Poisson’s ratio loaded at one end by a force P. The beam has a rectangular cross-section with height 2a and out-of-plane width b. We will show later (Sect 5.2.4) that the strain field in the beam is
We first check that the strain is compatible. For 2D problems this requires
which is clearly satisfied in this case.
For a 2D problem we only need to determine and such that
The first two of these give
We can integrate the first equation with respect to and the second equation with respect to to get
where and are two functions of and , respectively, which are yet to be determined. We can find these functions by substituting the formulas for and into the expression for shear strain
We can re-write this as
The two terms in parentheses are functions of and , respectively. Since the left hand side must vanish for all values of and , this means that
where is an arbitrary constant. We can now integrate these expressions to see that
where c and d are two more arbitrary constants. Finally, the displacement field follows as
The three arbitrary constants , c and d can be seen to represent a small rigid rotation through angle about the axis, together with a displacement (c,d) parallel to axes, respectively.
3D strain fields
For a general, three dimensional field a more formal procedure is required. Since the strains are the derivatives of the displacement field, so you might guess that we compute the displacements by integrating the strains. This is more or less correct. The general procedure is outlined below.
We first pick a point in the solid, and arbitrarily say that the displacement at is zero, and also take the rotation of the solid at to be zero. Then, we can compute the displacements at any other point x in the solid, by integrating the strains along any convenient path. In a simply connected solid, it doesn’t matter what path you pick.
Actually, you don’t exactly integrate the strains instead, you must evaluate the following integral
Here, are the components of the position vector at the point where we are computing the displacements, and are the components of the position vector of a point somewhere along the path of integration. The fact that the integral is path-independent (in a simply connected solid) is guaranteed by the compatibility condition. Evaluating this integral in practice can be quite painful, but fortunately almost all cases where we need to integrate strains to get displacement turn out to be two-dimensional.
(c) A.F. Bower, 2008