Question

A major fault in a reservoir has strike N35W and dip 60° from horizontal. Shmin in this area acts along 020 and in situ stresses are Shmin = 70 MPa, SHmax=80 MPa, S. = 65 MPa. Assume the pore pressure Pp = 30 MPa. If you have the knowledge of a fault transecting the reservoir. Determine if the fault likely to slip if the drilling operation will cause the respective frictional coefficient to be j = 0.7?

Answer 1

To determine if a fault is likely to slip during a drilling operation, we need to calculate the resolved shear stress and compare it to the product of the frictional coefficient and the effective stress normal to the fault plane. In this case, the fault is not likely to slip.

Step-by-step explanation:

A major fault in a reservoir with a strike N35W and a dip of 60° from horizontal can be analyzed to determine if it is likely to slip during a drilling operation. To determine if the fault will slip, we need to calculate the resolved stress on the fault plane. The resolved shear stress (τr) can be calculated as:

τr = (Shmax - Shmin) / 2

Substituting the given values, we get:

τr = (80 MPa - 70 MPa) / 2 = 5 MPa

The resolved shear stress should be greater than or equal to the product of the frictional coefficient (μ) and the effective stress normal to the fault plane. The effective stress can be calculated as:

(Shmin - Pp)

Substituting the given values, we get:

(Shmin - Pp) = (70 MPa - 30 MPa) = 40 MPa

Since the resolved shear stress (5 MPa) is less than the product of the frictional coefficient (0.7) and the effective stress (40 MPa), the fault is not likely to slip during the drilling operation.

Answer 2

The fault described is a reverse fault, and the drilling operation is not likely to cause it to slip.

Step-by-step explanation:

The fault described, with a strike of N35W and a dip of 60° from horizontal, is a reverse fault, as the hanging wall moves upwards. Reverse faults occur during compression and shortening of the Earth's crust. In this case, the fault is likely to slip if the drilling operation causes the frictional coefficient, j, to be less than or equal to the ratio of the maximum shear stress to the normal stress on the fault plane.

First, we need to calculate the shear stress on the fault plane. Using the formula:

Shear stress = (Shmax - Shmin)/2 = (80 MPa - 70 MPa)/2 = 5 MPa

Next, we calculate the normal stress on the fault plane:

Normal stress = Shmin - Pp = 70 MPa - 30 MPa = 40 MPa

Finally, we can compare the frictional coefficient (j) to the ratio of shear stress to normal stress:

j = Shear stress / Normal stress = 5 MPa / 40 MPa = 0.125

Since the frictional coefficient (j) is greater than the ratio of shear stress to normal stress, the fault is not likely to slip during the drilling operation.