Page 78 - Physical Principles of Sedimentary Basin Analysis

60 Linear elasticity and continuum mechanics

3.12 Thermal stress
We saw in the preceding section that the linear elastic stress–strain relation was obtained
by adding several strain contributions. Isotropic thermal expansion can be added to the list
of strain contributions. The coefficient of thermal expansion β is defined by

1 ∂V
β = (3.110)
V ∂T p
where the expansion is measured at a constant pressure p (or a constant average normal
stress). We have from (3.13) that volume strain is given by ε kk , and it then follows that
V
ε kk ≈ ≈ β T. (3.111)
V
The average strain contribution in each spatial direction is therefore

(4) 1
ε = β T δ ij (3.112)
ij
3
and the stress–strain relationship (3.98) becomes generalized to
1 ν 1
− ε ij = (1 + ν) σ ij − σ kk δ ij − β T δ ij . (3.113)
E E 3
It is often convenient to have the inverse of equation (3.113), in other words, the stress
tensor expressed by the strain tensor, which is
1
− σ ij = 2Gε ij + λε kk δ ij − λβ T (3.114)
ν
in terms of the Lamé coefficients.

3.13 Thermal stress compared with lithostatic stress
The horizontal stress down into the crust is calculated in Exercise 3.13 assuming that the
rock behaves like a linear elastic material. It is shown that the (isotropic) horizontal stress
is proportional to the vertical (lithostatic) stress σ b as
ν
σ h = σ b (3.115)
1 − ν
when the rock is confined in the horizontal direction (ε xx = ε yy = 0), and the horizontal
stress state is isotropic (σ h = σ xx = σ yy ). We can do the same calculation with thermal
stress too. From equation (3.113) the strain in the horizontal plane is
1 ν 1
− ε xx =−ε yy = (1 + ν)σ h − (2σ h + σ b ) − β T. (3.116)
E E 3
Zero horizontal strain ε xx = ε yy = 0 gives the horizontal stress

ν Eβ T
σ h = σ b + . (3.117)
1 − ν 3(1 − ν)