Page 58 - Physical Principles of Sedimentary Basin Analysis

40 Linear elasticity and continuum mechanics

F
A
dx
θ
h

Figure 3.2. A rectangular block is subjected to a tangential force F over its top surface with area A.
The shear force has moved the upper surface a distance dx relative to the lower fixed surface. The
height of the block is h.

Exercise 3.3 Show that the work needed to shear the block in Figure 3.2 by an angle θ is
1
2
W = Gθ V , where V = Ah is the volume of the block.
2

3.4 Strain
We have seen that the strain of a stretched rod is ε = dl/l 0 , when the rod’s initial length
is l 0 and it is stretched a distance dl. A point x along the initial rod becomes displaced a
distance u(x) = xdl/l 0 by the stretching. Such a displacement function makes it possible
to obtain the strain locally by measuring the difference in displacement of two nearby
points. The strain at a position x along the rod is the difference in displacement between
two points x and x + dx relative to the distance dx separating the points, ε = (u(x +
dx) − u(x))/dx ≈ ∂u/∂x (assuming that dx is small). Notice that it is differences in
the displacement field that produce strain. For example, a constant displacement does not
result in any strain, only a pure translation. The use of a (scalar) displacement function to
obtain the strain can be generalized to three dimensions by use of the displacement vector
u i = u i (x 1 , x 2 , x 3 ), i = 1, 2, 3. The difference in displacement between two nearby points
separated by (dx 1 , dx 2 , dx 3 ) can now be written

∂u i
du i = dx j
∂x j

1 ∂u i ∂u j 1 ∂u i ∂u j
= + + − dx j
2 ∂x j ∂x i 2 ∂x j ∂x i

= ε ij + R ij dx j . (3.8)

(Notice that the summation convention is used, where there is summation over all pairs
of equal indices.) The partial derivative ∂u i /∂x j has been rewritten as a sum of a
symmetric part
1 ∂u i ∂u j
ε ij = + (3.9)
2 ∂x j ∂x i