13. The Classical Theory of Elastic Deformation [12]

Suppose we have an undeformed, continuous material that is at rest. Each point within the solid can be identified by a vector x extending to the point from the origin of some fixed coordinate system. Figure 72 shows what happens when the material is deformed. The point originally at x moves to a new position x'.

 

Figure 72. A continuous material at rest is deformed.

 

Let's define the displacement as
u(x) = x' - x
Clearly, u depends on the original vector position x.

Consider now two points within an elastic solid. If x and y are two neighboring points in an undeformed elastic solid , as illustrated in Figure 73. Deformation results in the following configuration
 

x' = x + u(x) and y' = y + u(y)

 

Figure 73.

 

These relations are very general and could describe a uniform translation (when u is constant). Under uniform translation, the internal configuration doesn't change; there is only a rigid translational movement of the entire specimen. We are interested in internal deformations, therefore, the vector separation would be of most interest here.

y' -x' = y - x + u(y) - u(x)

If the undeformed separation is small, we can write the following expansion:

(12.1)

Note that the partial derivative is evaluated at the initial x position. The u displacements are generally small; therefore we can use a first-order approximation. If the solid only undergoes a uniform translation, the partial derivative disappears.

The partial derivative in Equation 12.1 can be separated into its even and odd components as

(12.2)

where the two tensors, e_ij and O_ij, are real numbers that depend on the initial position. The odd tensor O_ij represents a local rigid rotation; and therefore doesn't relate to the deformation of the solid that we are interested in. The internal deformation is represented by the even tensor, e_ij, which is defined as the elastic strain tensor.

The elastic strain tensor is zero whenever the material is only subjected to a rigid rotation or a simple translation. We return now to the two nearby points in the solid at x and y. A small elastic deformation results in

(12.3)

Example 1

Figure 74 illustrates what happens when an elastic deformation occurs for an arbitrary vector a that is directed along the e₁ axis. Let's suppose the two local points that we are looking at are located at the starting and end points of vector a, so that y - x = a. Under the elastic deformation, this vector becomes

(12.4)

 

Figure 74. A small elastic deformation of the vector a.

 

There are two changes that occur to a as a result of the deformation:
(1) It's length changes to

(12.5)

where

(12.6)

(2) a' has a different direction than a because of the non-zero elements e₂₁ and e₃₁.

Example 2.

Consider Figure 75 that shows how two vectors that are originally orthogonal are altered by the elastic deformation.

 

Figure 75. The effect of a small elastic deformation on Two Initially Orthogonal vectors, a and b.

 

Under the deformation, the two vectors become

(12.7)

The angle between a' and b' is given by their dot product which is equal to a'b' cos q'₁₂ and approximately equal to

(12.8)

Therefore, under the deformation, the right angle becomes

(12.9)

Example 3.

Under an elastic deformation, a volume V = abc will change to

(12.10)

The trace of the strain tensor gives the fractional change in the volume. Using the definition of the elastic strain tensor provides another way of writing this in terms of the displacement vector u

(12.11)

This may also be viewed as a change in the density of the solid:

(12.12)

The Stress Tensor

When we talked about fluid-fluid and solid-vapor interfaces, we introduced the stress tensor. Let's use the following to define an elastic stress tensor:

(12.13)

where F_i is the i^th component of the force acting on the surface dA which is directed outward and normal to the surface.

Hooke's law describes small deformations of elastic materials. It states that the local stress is proportional to the local strain (elastic deformation). This stress opposes further deformation. In basic physics, we learn Hooke's law as F = -kx as it applies to springs. When the spring is stretched a distance x by a force F, an equal, but opposite force opposes that stretching. In elastic solids, Hooke's law is expressed using tensor notation, where T_ij and e_ij are components of a fourth-rank tensor. Due to symmetry, the maximum number of independent elements is 21. The symmetry of a crystalline, elastic solid further reduces the number of independent elements. Here, we'll just consider the isotropic elastic solid where Hooke's law and the lack of a preferred direction by be used. In this case T_ij is a linear combination of two symmetric tensors

(12.14)

where d_ij is the Kronecker delta function which is zero when i is not equal to j and one when i equals j.

Hooke's law is written:

(12.15)

where l and m are the Lame coefficients, or moduli. Again, the minus sign indicates the restoring nature of these forces. In calculations, it becomes preferable to use a traceless tensor which is written

(12.16)

By using this, Hooke's law is rewritten:

(12.17)

The elastic moduli K, m, and l have dimensions of force per unit length.

Hooke's law as written above gives the stress resulting from a given strain. One can also invert this to give the strain induced by a given stress. By taking the trace of the first equation in Equation 12.17, you get

(12.18)

Then, by eliminating Tr e, the desired form is obtained

(12.19)

Example 4.

A small hydrostatic pressure dp is applied to a uniform medium at rest. The stress tensor is that of an ideal stationary fluid with no convective terms,

(12.20)

Applying Equation 12.19 gives the induced, isotropic strain,

(12.21)

Since there are no off-diagonal elements, the object retains its shape, but its volume will change according to Equation 12.11,

(12.22)

Inspection of this equation reveals that K is simply the bulk modulus.

Example 5.

Consider an elastic material in the shape of a long, slender bar under a uniform axial stress,

(12.23)

 

T_zz = p
and all other elements are zero. To achieve this, the pressure p is applied to the two ends of the bar and the lateral sides are unstressed. In this case, Equation 12.19 shows that the bar undergoes uniform axial compression,

(12.24)

Equation 12.24 is usually written

(12.25)

and E is the Young's modulus. In the transverse direction, there is an expansion that occurs as well that is given by

(12.26)

The ratio of the transverse expansion to the longitudinal compression defines the Poisson's ratio s via

(12.27)

The Poisson's ratio is usually between 0 and 0.5.

Example 6.

Consider Figure 76 in which a long, slender bar is subjected to a uniform shear stress

(12.28)

T_xy = T_yx = -f
and all other elements would be zero.

 

Figure 76. A long, slender bar is deformed under a uniform shear stress. The induced force F_I per unit area is shown for each side.

Equation 12.29 implies

(12.29)

If this is compared with the deformation we looked at in Example 2, we see that the original right angles change by an amount

(12.30)

Here is coefficient m is the shear modulus.

Of the four quantities, E, K, m, and s, only two are independent. Therefore, K and m are eliminated by these relations:

(12.31)

These may be used to rewrite Equation 12.19 in a more convenient form:::

(12.32)

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