Appendix C.Biaxial Stress Changes

The relationship between the radial deformation of a borehole, U, and two principle stresses in the plane of a borehole has been given by Hast (1958) and Merrill and Peterson (1961). The equation for Plane Stress is:

U = d/Er [(σ₁ + σ₂) + 2 (σ₁ – σ₂) cos 2 θ]

equation 5: Plane Stress

Where;

σ₁ and σ₂ are the principle stresses in the plane of the borehole.

θ is the angle measured counterclockwise from the direction of σ1.

d is the diameter of the borehole.

Er is the Young’s Modulus of the rock

 

If it is assumed that the stress measured across the stressmeter is proportional to the radial deformation that would have occurred in this direction if the stressmeter had not been there, then the term d/Er can be replaced by one reflecting the relationship between the rock modulus and the gauge modulus. Hast (1958) has shown this to be applicable for a uniaxial stressmeter.

For the measurement of stress (σ_R) in any direction (θ) the following equation applies:

Note: θ is measured counterclockwise from σ₁

σ_R = 1/3 (σ₁ + σ₂) + 2/3 (σ₁ – σ₂) cos 2 θ

equation 6: Stress in any Direction

Using this relationship and three uniaxial stress change measurements at 45° to each other, the secondary principle stresses σ₁ and σ₂ and the angle (θ) are given by:

σ₁ = 3/2 a + ¾ b
σ₂ = 3/2 a – ¾ b
θ = 1/2 sin⁻¹ ((a – σ₄₅)/b)

equation 7: Secondary Principle Stresses and Angle

Where;

a = σ₀ + σ₉₀ /2

b = [(σ₄₅ – a)² +(σ₀ – a)²]^½

To determine the θ angles, it must be determined what quadrant the angle lies in. The inequalities to do this are as follows:

If σ45 ≤ a and σ₀ ≥ 90, then 0 ≤ θ ≥ 45°

If σ45 ≤ a and σ₀ ≤ 90, then 45° ≤ θ ≤ 90°

If σ45 ≥ a and σ₀ ≤ 90, then 90° ≤ θ ≤ 135°

If σ45 ≥ a and σ₀ ≥ 90, then 135° ≤ θ ≤ 180°

Note: θ is measured clockwise for σ₀ (this is same as counterclockwise for σ₁).

For Example:

Three gauges are set in a borehole. The first is at 0° (σ₀), the second at 45° (σ₄₅) and the third at 90° (σ₉₀), measured counterclockwise from 0. Determine the uniaxial stress changes for each gauge by the reading change times the calibration factor. Substitute the constants into the equations to obtain the change magnitude of the two secondary principal stresses σ₁ relative to 0°.

Stress Changes:

Gauge 1, σ₀ = 600 psi

Gauge 2, σ₄₅ = 800 psi

Gauge 3, σ₉₀ = 300 psi

Calculate the values for these constants: a and b

a = σ₀ + σ₉₀/2 = 600 + 300/2 = 450

b = [(σ₄₅ – a)² + (σ₀ – a)²]^½ = [(800–450)² + (600–450)²]^½ = 380.79

σ1 = 3/2a + 3/4b = 3 x 450/2 + 3 x 380.79/4 = 960.59 psi

σ2 = 3/2a – 3/4b = 3 x 450/2 – 3 x 380.79/4 = 389.41 psi

sin 2θ = –0.92

θ = 33.40°

σ₁ direction: since σ₄₅ > a and σ₀ > σ₉₀, then 135 < θ < 180°. Therefore, θ = 180 – 33.40 = 146.6°. This is measured clockwise from σ₀.

 

References:

Hast, N.; THE MEASUREMENT OF ROCK PRESSURE IN MINES;

Sveriges Geologiska Undersokning, Arsbok 52, Series C, 3. 1958.

Merrill, R.H. and Peterson, J.R.; DEFORMATION OF A BORE HOLE IN ROCK;

U.S. Bureau of Mines, RI 5881.