Interaction with the Hydraulic Fracture Tip

According to Linear Elastic Fracture Mechanics (LEFM), the stress components close to the fracture tip in opening mode for plane elasticity in a homogenous and isotropic material can be approximated in polar coordinates as (Sun and Jin 2012):

37 𝜎_𝑦𝑦 = − 𝐾I √2𝜋𝑟cos 𝜃 2(1 + sin 𝜃 2 sin 3𝜃 2 ) (12) 𝜏_𝑥𝑦= − 𝐾I √2𝜋𝑟sin 𝜃 2 cos 𝜃 2 cos 3𝜃 2 (13)

Where KI is the stress intensity factor and r and θ are the polar coordinates with

origin at the fracture tip.

Considering a hydraulic fracture approaching a natural fracture with angle β in an infinite rectangular medium whose far field stresses are σH and σh with σH > σh and

with σH in the direction of the hydraulic fracture propagation (Fig. 13), and using the

principle of superposition, the total stress can be expressed as: 𝜎_𝑥𝑥′ _{= 𝜎} 𝐻+ 𝜎𝑥𝑥 ₍₁₄₎ 𝜎_𝑦𝑦′ _{= 𝜎} ℎ+ 𝜎𝑦𝑦 ₍₁₅₎ 𝜏_𝑥𝑦′ _{= 𝜏} 𝑥𝑦 ₍₁₆₎

38

Fig. 13 - Hydraulic fracture approaching a natural fracture. Modified after H.Gu and X.Weng (2010).

It can be noticed from Eq. (11) to Eq. (13) that all stress components present a 1 √𝑟⁄ factor and have an infinite value at the fracture tip (r = 0). However, no material can bear an infinite stress and it will deform plastically at a finite stress value. Then, it can be assumed that there is a distance rc from the fracture tip beyond which the

aforementioned equations are valid (Renshaw and Pollard 1995).

For the natural fracture to slip, the Coulomb criterion must be satisfied:

|𝜏_𝑠| ≥ 𝑆_𝑜+ 𝜇 𝜎_𝑛 (17)

Where τs and σn are the shear and normal stress on the natural fracture plane

respectively, So is the cohesion of the fracture plane and µ is the coefficient of friction.

Stresses on the natural fracture plane are calculated by rotating the total stress field by the angle between the natural and hydraulic fracture (β) as follows:

τs θ=β r =rc σn x y σH σh Hydraulic fracture Natural fracture

39 𝜏_𝑠 = − 1₂(𝜎_𝑥𝑥′ _{− 𝜎}

𝑦𝑦′ ) sin 2𝛽 + 𝜏𝑥𝑦′ cos 2𝛽 (18)

𝜎_𝑛 = 1₂(𝜎_𝑥𝑥′ _{+ 𝜎}

𝑦𝑦′ ) − 1₂(𝜎𝑥𝑥′ + 𝜎𝑦𝑦′ ) cos 2𝛽 − 𝜏𝑥𝑦′ sin 2𝛽 (19)

The total stress field (𝜎_𝑥𝑥′ _{, 𝜎}

𝑦𝑦′ , 𝜏𝑥𝑦′ ) is calculated using Eq. (14) to Eq. (16) with θ

= β. KI and rc values still need to be calculated. KI and rc are lamped in a new variable K

following the treatment of H.Gu and X.Weng (2010): 𝐾 = 𝐾I

√2𝜋𝑟𝑐 cos

𝜃

2 (20)

After replacing K in Eq. (11) to Eq. (13), an additional equation is set up considering that for the hydraulic fracture to reinitiate on the opposite side of the natural fracture, the maximum principal stress σ1 must reach the rock tensile strength To:

𝜎₁ = 𝑇_𝑜 (21)

Where σ1 is calculated as follows:

𝜎1 = 1₂(𝜎𝑥𝑥′ + 𝜎𝑦𝑦′ ) + √(𝜎𝑥𝑥 ′ _+𝜎 𝑦𝑦′ 2 ) 2 − 𝜏𝑥𝑦′ 2 (22)

Finally, Eq.(21) becomes a quadratic equation in K and the solution giving σ1 = To

is picked.

Table 2 shows the parameters used for the stress analysis close to the propagating

40 σh 2700 psi (18.6 MPa) σH 1.3 σh To - 150 psi (- 1.0 MPa) So 0 psi (0 MPa) µ 0.6

Table 2 - Values for stress analysis induced on the natural fracture by the hydraulic fracture tip.

Fig. 14 and Fig. 15 show the plots for normal and shear stress components on the

natural fracture plane vs σH / σh ratio calculated using Eqs. (18) and (19) for the two

different fracture sets: one intersecting the natural fracture at β = 10 deg. and the other one intersecting the natural fracture at β = 55 deg. The red horizontal dashed line marks the assumed σH / σh ratio equal to 1.3 (Farrington 1955).

According to Fig. 14, for the lower angle of interserction β = 10 deg, the normal stress component is lower than the shear stress on the fracture plane for every σH / σh ratio considered in the plot. For the assumed σH / σh ratio at the reservoir, the actual values are:

σn = 10.6 psi (0.1 MPa), τs = 375.9 psi (2.6 MPa) which accommodates for a high probability for shear slip.

On the other hand, for the higher angle of intersection at β = 55 deg (Fig. 15), the shear stress component resolved on the fracture plane is always greater than the normal stress component. The actual values of tensile and shear components are: σn = 1708.2 psi

(11.8 MPa), τs = 1179.8 psi (8.13 MPa) which decreases the possibility of shear

41

Fig. 14 - Normal and shear stress on the natural fracture plane vs σH / σh ratio for a natural fracture

intersecting at 10 deg. the hydraulic fracture. The red dashed line is the assumed stress ratio at the reservoir.

Fig. 15 - Normal and shear stress on the natural fracture plane vs σH / σh ratio for a natural fracture

intersecting at 10 deg. the hydraulic fracture. The red dashed line is the assumed stress ratio at the reservoir.

42

For both angles of interaction, the normal stress on the fracture planes is never negative. Thus, the type of rupture expected is compressional shear, however normal stress values close to zero for β = 10 deg. suggest that pure shear rupture mode can also be expected.

For the slip criterion, a slight modification of Eq. (17) is used as follows: 𝜇 ≤(|𝜏𝑠|− 𝑆𝑜)

𝜎𝑛 (23)

Furthermore, shear and normal stresses (τs and σn) on the fracture plane are

expressed in terms of σH / σh ratio.

Fig. 16 and Fig. 17 show the criterion curves for a hydraulic fracture crossing a

natural fracture at β = 10 deg. Crossing occurs in the zone below the curves. The red dashed line marks the assumed σH / σh ratio at the reservoir (1.3), the vertical black dashed is the assumed natural fracture friction coefficient (0.6). The intersection of both lines determines the assumed state at the reservoir.

Fig. 16 shows the criterion for two for two different values of fracture cohesion (So). Notice that the reservoir state at the assumed values of σH / σh ratio and natural

fracture friction coefficient (0.6) is above both curves meaning that the natural fracture will slip at this condition. However, notice the effect of the natural fracture cohesion (So).

The higher this value is, the closer it gets to crossing at the assumed σH / σh and friction coefficient at the reservoir.

43

Fig. 16 - Criterion for a hydraulic fracture crossing a natural fracture at an angle of 10 deg. and for two different values of fracture cohesion (So). Crossing occurs in the zone below the curves. The red dashed line marks the assumed σH / σh ratio at the reservoir (1.3), the vertical black dashed is the

assumed natural fracture friction coefficient (0.6). The intersection of both lines determines the assumed state at the reservoir. This point is above both curves meaning that the natural fracture will slip at this condition.

Fig. 17 - Criterion for a hydraulic fracture crossing a natural fracture at an angle of 10 deg. and for two different σh .Crossing occurs in the zone below the curves. The red dashed line marks the assumed

σH / σh ratio at the reservoir (1.3), the vertical black dashed is the assumed natural fracture friction

coefficient (0.6). The intersection of both lines determines the assumed state at the reservoir. This point is above both curves meaning that the natural fracture will slip at this condition.

44

Fig. 17 shows the criterion for two for two different values of σh. At reservoir

conditions, marked by the intersection of the vertical and horizontal dashed lines, the natural fracture will slip since the reservoir state is at above both curves. The effect of increasing σh is similar to the effect of increasing So, in the sense that the criterion curve

gets closer to the reservoir state but in this case the variation is less stronger.

Fig. 18 and Fig. 19 shows the criterion curves for a hydraulic fracture crossing a

natural fracture at β = 55 deg. Crossing occurs in the zone to the right of the curves. The red dashed line marks the assumed σH / σh ratio at the reservoir (1.3), the vertical black dashed is the assumed natural fracture friction coefficient (0.6). The intersection of both lines determines the assumed state at the reservoir.

Fig. 18 shows the criterion for two for two different values of fracture cohesion

(So). For this case, notice that for the assumed σH / σh ratio at the reservoir (1.3), and the

natural fracture friction coefficient (µ = 0.6), natural fracture slippage will occur for the case of So =0 but hydraulic fracture crossing will occur for So =300 psi.

Fig. 19 shows the criterion fro two different values of the reservoir rock tensile

strenght (To). For both curves So =300 psi. For the assumed σH / σh ratio at the reservoir

(1.3) and the natural fracture friction coefficient (µ = 0.6), natural fracture slippage will occur for To = -750 psi but hydraulic fracture crossing will occur for To =-150 psi. Notice

that decreasing the value of the rock tensile strength move the criterion curve to the right increasing the probability of natural fracture slippage.

45

Fig. 18 - Criterion for a hydraulic fracture crossing a natural fracture at an angle of 55 deg. and for two different values of fracture cohesion (So) .Crossing occurs in the zone to the right of the curves. The red dashed line marks the assumed σH / σh ratio at the reservoir (1.3), the vertical black dashed is

the assumed natural fracture friction coefficient (0.6). The intersection of both lines determines the assumed state at the reservoir. This point is to the left of the curve when So =0 and to the right when So = 300 psi. Slip of the natural fracture plane will occur in the first case and crossing of the hydraulic fracture in the second case.

Fig. 19 - Criterion for a hydraulic fracture crossing a natural fracture at an angle of 55 deg. and for two different values of To .Crossing occurs in the zone to the right of the curves. The red dashed line marks the assumed σH/ σh ratio at the reservoir (1.3), the vertical black dashed is the assumed natural

fracture friction coefficient (0.6). The intersection of both lines determines the assumed state at the reservoir. This point is to the right of the curve when To = -150 psi and to the left when To = -750 psi. Crossing of the hydraulic fracture will occur in the first case and slip of the natural fracture in the second case

46 From these analyses we conclude that:

- The lower the angle of interaction (β), the greater the opportunity of natural fracture slippage.

- The greater the fracture cohesion (So,), the lower the opportunity of slippage on the

natural fracture.

- The greater the differential stress (σh), the lower the opportunity of slippage on the natural

fracture, but this effect is less pronounced compared to the effect of increasing fracture cohesion.

- The lower the tensile strength of the reservoir rock (To), the higher the opportunity of

slippage on the natural fracture.

Finally, for the two natural fracture sets considered, and for the parameters used as in Table 2, both fractures will slip in shear. However if a certain value of cohesion exists on the natural fractures such as So = 300 psi, only the set with a lower angle (β = 10 deg.)

will slip, unless the reservoir rock presents a lower value of tensile strenght. For a rock tensile stress equal to -750 psi, both sets will slip in shear.