Geokon 4300 Series, Инструкция по эксплуатации

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18
For Example:
Three gauges are set in borehole. The first is at 0 ° ( σ
0
), the second at 45 ° ( σ
45
) and the third at
90 ° ( σ
90
), measured counterclockwise from 0. The uniaxial stress changes for each gauge are
determined by the reading change times the calibration factor. Substitute the constants into the
equations to obtain the magnitude of the changes of the two secondary principal stresses σ
1
relative to 0 ° .
Stress Changes:
Gauge 1, σ
0
= 600 psi
Gauge 2, σ
45
= 800 psi
Gauge 3, σ
90
= 300 psi
Calculate the values for constants, a and b:
a = σ
0
+ σ
90/
2 = 600 + 300/2 = 450
b = [( σ
45
– a)
2
+ ( σ
0
– a)
2
]
1/2
= [(800-450)
2
+ (600-450)
2
]
1/2
= 380.79
σ
1
= 3/2a + 3/4b = 3 x 450/2 + 3 x 380.79/4 = 960.59 psi
σ
2
= 3/2a – 3/4b = 3 x 450/2 – 3 x 380.79/4 = 389.41 psi
sin 2 θ = -0.92
θ = 33.40 °
σ
1
Direction: since σ
45
> a and σ
0
> σ
90
, then 135 < θ < 180 ° . Therefore, θ = 180 – 33.40 =
146.6 ° . This is measured clockwise from σ
0
.
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.