Solution Exercise 2
Parameter values for use in this exercise are listed below. At the end of this solution, a section titled Supplemental information for Exercise 2 – Elastic Properties Explanation describes elastic properties.
|
σv |
= |
vertical stress |
|
ρ |
= |
2,500 kg/m3 (average rock density) |
|
g |
= |
10 m/s2 (gravity acceleration) |
|
h |
= |
3,000 m (depth) |
|
mdol |
= |
5 (dolomite Poisson number) |
|
mmud |
= |
4 (mudstone Poisson number) |
a) Vertical stress for the dolostone and mudstone
σv = ρ g h
σv = 2500 kg 10 m 3000 𝑚/m3s2 = [75,000,000 kg] 1 Pa 𝑀𝑃𝑎 = 75 Mpa
m3 s2 m3 s2 1 Kg / 𝑚 s2 1x106
σv = 75 Mpa
b) Horizontal stress for the dolostone
σhdol = σv/(mdol−1) = 75 MPa / (5 − 1) = 19 MPa
Horizontal stress for the mudstone
σh mud = σv/(mmud−1) = 75 MPa / (4 − 1) = 25 MPa
c) σv and σh are the larger and smaller stresses, respectively, so they correspond to the principal stresses σ1 and σ3, respectively. With the values of σv and σh known for each rock, the Mohr circles can be plotted in the Mohr diagram as shown below. Using the given equations, the mudstone and dolostone failure envelopes are plotted by calculating the shear stress τ at two points, for example, when σn is zero and 10 MPa (blue and brown dots, respectively). With these two points the failure envelopes can be drawn on the Mohr diagram.
At the end of Section Exercise 2, a subsection titled Supplemental information for Exercise 2 – Stress State Explanation provides information that is useful to the developing the following answers.
d) Under the represented stress state, neither rock will undergo fracturing because neither Mohr circle intersects the shear failure envelope.
e) The cohesion (C) is determined where the failure envelope intersects the shear stress axis, so it is 75 MPa for the dolostone and 15 MPa for the mudstone.
f) C is much smaller for the mudstone, so it will undergo shear fracturing under smaller tectonic stresses than the dolostone.
Click to return to where text linked to Exercise 2