Refer to Figure 4.9. On arithmetic paper plot seven points showing the peak stre
ID: 1713496 • Letter: R
Question
Refer to Figure 4.9. On arithmetic paper plot seven points showing the peak stress (this occurs approximately at 3% strain in each case) as ordinate versus the confining pressure as the abscissa. Use values of N/m2. The ordinate can be labeled maximum principal stress and the abscissa least principal stress
a. Assume a linear relationship (typical of strong rocks). What is the maximum unconfined strength?
b. Assume this rock is subject to lithostatic loads (maximum principal stress) in a 100 m-deep mine. What confining stress should be exerted on the walls of the mine to keep the rock from crushing?
-1800 Bars 4000 3000 2000 1000 300 Bars 800 Bars 700 Bars 600 Bars 400 Bars 2) -200 Bars Figure 4.9. Stress-strain curves for the Crown Point Limestone showing varying confining pressures (after Donath, 1968). (Note: 1 bar=1 atmos- phere 14.7 psi 105 N/m2) 0 4 8 2 6 20 Strain, (Percent)
Explanation / Answer
I understand that Differential stress is nothing but sigma1- sigma3 and least principle stress is nothing but confining pressure sigma 3. Confining pressure values mentioned here are 200,400,600 etc.
Calculate sigma 1 = (Differential stress) + Confining pressure
sigma 1 = 2000+ (approx) + 200 , sigma 3 = 200....................These are normal stress values
Plot sigma1 and sigma 3 on abscissa and then draw circle passing through these points.
Similarly draw circles for remaining curves. Once this is done, plot approximately tangent line to all these circles.
Intercept of this tangent line on ordinate axis will give you maximum unconfined compressive strength.
This value will be approximately around 1000 bars.
Limestone density - 2.30 gm/cm3, unit weight = 23 kN/m3
Mine depth = 100 m..............Lithostatic pressure = 23 X 100 = 2300 kN/m2 = 23 bar
1000 bar >> 23 bar
No confining stress is required on the walls of mine.
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