Page 90 - Physical Principles of Sedimentary Basin Analysis

72 Linear elasticity and continuum mechanics

when it is rewritten in terms of the effective stress (3.128). We see that in order to solve for
effective stress we must know the fluid pressure (or we must solve for both effective stress
and fluid pressure simultaneously).

Exercise 3.28 Solve equation (3.176) for the displacement in the vertical direction caused
by gravity in a 1D crust, assuming linear elasticity and zero horizontal strain. The bulk
density is constant. There is no displacement at the top of the column (z = 0), and the
top of the column has zero (vertical) stress.
Solution: A 1D crust with no lateral strain (ε xx = ε yy = 0) is the starting point. The linear
elastic stress–strain relationship (3.98) is then
du
σ zz =−(2G + λ)ε zz =−(2G + λ) (3.178)
dz
where u is the displacement in the vertical direction. We get
2
dσ zz d u
=−(2G + λ) = g (3.179)
dz dz 2
when the stress–strain relationship (3.178) is inserted into equation (3.176) for the force
balance. Equation (3.179) is integrated twice, and gives the vertical displacement
g 1 2
u(z) =− z + Az + B . (3.180)
(2G + λ) 2
It follows from σ zz (z=0) = 0 that du/dz(z=0) = 0, which implies that A = 0. Zero
displacement at the surface gives that B = 0. The solution for the displacement is therefore
g 2
u(z) =− z (3.181)
2(2G + λ)
or alternatively
z

2
u(z) =−u 0 (3.182)
h
1 2
where the coefficient u 0 = gh /(2G + λ) is the maximum displacement at the base
2
(z = h) of the column. Notice that the displacement is upwards. The vertical strain is
du z
ε zz = =−ε 0 (3.183)
dz h
where the maximum strain is ε 0 = gh/(2G + λ). The stress in the vertical direction is
found from the displacement, equation (3.178), and it is
z
σ zz =−(2G + λ)ε zz = σ 0 (3.184)
h
where the maximum stress σ 0 found at the base of the column is σ 0 = gh.Thisis
precisely the stress caused by the weight of the column. What is the displacement at the
depth h = 1000 m when E = 50 GPa and ν = 0.25?