gp202mooc hw4 solutions

GP 202 Reservoir Geomechanics

Spring 2014, OpenEdX

Homework 4 – Estimating Limits on Shmax

SOLUTIONS Please direct any questions to the forum on the OpenEdX Courseware page

Question 1: Which Stress States Are Possible?

Which stress states are possible in each of the following cases at depth of 12000 ft?

Case 1: μ = 0.6, Sv = 12000 psi, Shmin = 8000 psi

(d) Normal and/or strike-slip faulting. In this case, Sv > Shmin; thus, the minimum

principal stress cannot be the vertical stress, so only reverse faulting is not

possible.




possible.




possible.


(e) Any faulting regime is possible. In this case, Sv = Shmin; thus, depending on the

value of Shmax, any of the three faulting regimes are possible because if Shmax > Sv

= Shmin, both reverse and strike-slip faulting are possible, and if Shmax = Sv = Shmin,

any of the three faulting regimes could take place due to perturbations of any of

the components of the stress around an equal value.

Question 2: Constraints on Shmax


Since normal and/or strike-slip faulting are possible in this case, we will refer to

equations 4.45 and 4.46 on Lecture 6 pg. 29 to determine the constraints on Shmax

based on frictional faulting theory. In addition, since we know the value of Shmin

and Shmax cannot be less Shmin, we know that the lower bound for Shmax must be

Shmin = 8000 psi.

To calculate the upper bound on Shmax, we must use the relationship for strike-slip

faulting equation (equation 4.46).

(Shmax – Pp) / (Shmin – Pp) = 3.1 (Equation 4.46)

Shmax = 3.1(Shmin – Pp) + Pp = 13712 psi

In the case of strike-slip faulting, Shmax must be greater than Sv = 12000 psi, but

cannot be greater than 3.1(Shmin - Pp) + Pp, thus 8000 psi ≤ Shmax ≤ 13712 psi.






Shmin = 9500 psi.












Shmin = 10000 psi.








Since all faulting regimes are possible in this case, we will refer to all equations

on Lecture 6 pg. 29 to determine the constraints on Shmax based on frictional

faulting theory. In addition, since we know the value of Shmin and Shmax cannot be

less Shmin, we know that the lower bound for Shmax must be Shmin = 12000 psi.

In this case, since reverse faulting is possible and is the most compressive stress

regime, we must use the relationship for reverse faulting to calculate the upper

bound of Shmax.

(Shmax – Pp) / (Sv – Pp) = 3.1 (Equation 4.47)

Shmax = 3.1(Sv – Pp) + Pp = 26112 psi

In the case of reverse faulting, Shmax must be greater than Sv = Shmin = 12000 psi,

but cannot be greater than 3.1(Sv - Pp) + Pp, thus 12000 psi ≤ Shmax ≤ 26112 psi.

gp202mooc hw4 solutions

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