Page 76 - Physical Principles of Sedimentary Basin Analysis

58 Linear elasticity and continuum mechanics

The stress–strain relationship (3.98) is the Lamé equation, and the two parameters G and
λ are called the Lamé coefficients. Notice that both Lamé coefficients are expressed by
Young’s modulus and Poisson’s ratio.

Exercise 3.13
(a) The vertical stress into the crust is σ zz = b gz (where b is the rock bulk density and
g is the constant of gravity), and assume that the rock behaves as a linear elastic material.
Show that the isotropic horizontal stress σ h = σ xx = σ yy then becomes

ν
σ h = b g (3.100)
1 − ν
when there is zero strain in the horizontal direction (ε xx = ε yy = 0). Hint: start with
equations (3.89) and (3.90).
(b) Is the horizontal stress less than or larger than the vertical (lithostatic) stress?
Exercise 3.14 Show that the Lamé equation (3.98) is the inverse of equation (3.96). Hint:
1
first, show that σ kk =− (1 − 2ν)ε kk .
E
Exercise 3.15 Show that the bulk modulus can be expressed as follows by Young’s
modulus (E) and Poisson’s ratio (ν)
E
K = . (3.101)
3(1 − 2ν)
Hint: use definition (3.13) for the volume strain and let the pressure be defined as the
1
average normal stress, p = σ kk .
3
Exercise 3.16 Show that the bulk modulus becomes
2
K = G + λ (3.102)
3
when expressed with the Lamé coefficients.

Exercise 3.17 Show that the Lamé coefficients (3.99) and (3.95) can be inverted to the
following expressions for Poisson’s ratio and Young’s modulus:

λ G(2G + 3λ)
ν = and E = . (3.103)
2(G + λ) G + λ
Hint: notice that λ = 2Gν/(1 − 2ν).

Exercise 3.18
(a) Let σ 1 =−σ 2 in Figure 3.14, and show that σ xy =−σ 1 = σ 2 and that the normal stress
is zero.
(b) Show that
1 1
1 =− (1 + ν) σ 1 and 2 = (1 + ν) σ 1 . (3.104)
E E