Page 57 - Physical Principles of Sedimentary Basin Analysis

3.3 Shear modulus 39

dr
r
F
A
l 0 dl

Figure 3.1. A rod of length l and cross-section area A is stretched by a force F. The rod is dl longer
after being stretched and its radius is reduced by dr.

the relative volume change is the strain, ε = dV/V 0 , where V 0 is the initial volume of the
body. The stress and the strain are linearly related for linear elastic materials:
stress pV 0
K = =− (3.3)
strain dV
where the constant K is the bulk modulus. The minus sign is added because dV is negative
for a positive (compressive) pressure p. The definition of the bulk modulus is related to the
definition of compressibility
1 dV
C =− (3.4)
V dp
which shows that the bulk modulus (3.3) is an inverse compressibility, K = 1/C.

Exercise 3.2 Show that the compressibility can be expressed as
1 ∂
C = (3.5)
∂p
in terms of the density.

3.3 Shear modulus

A third type of strain is produced by tangential surface forces, as shown in Figure 3.2.The
force F acts parallel to the top surface with area A of the rectangular block, and the top
surface becomes sheared relative to the lower surface. The strain is in this case measured
as the ratio of horizontal displacement to the height of the block, ε = dx/h = tan θ.
The shear stress is the tangential force divided by the surface area, and the stress–strain
relation is
stress F h F F
G = = = ≈ . (3.6)
strain A dx A tan θ A θ
For small deformations, when tan θ ≈ θ, we have that the shear modulus G is the shear
stress over the deformation angle. Young’s modulus, Poisson’s ratio and the shear modulus
are not independent. It turns out that there are only two independent moduli for linear
elastic materials. The shear modulus can be expressed by Young’s modulus and Poisson’s
ratio as
E
G = (3.7)
2(1 + ν)
which is shown in Exercise 3.18.