Find M and N for the norite.
Applications and Investigations in Earth Science (9th Edition)
9th Edition
ISBN:9780134746241
Author:Edward J. Tarbuck, Frederick K. Lutgens, Dennis G. Tasa
Publisher:Edward J. Tarbuck, Frederick K. Lutgens, Dennis G. Tasa
Chapter1: The Study Of Minerals
Section: Chapter Questions
Problem 1LR
Related questions
Question
Please answer part f
![Stress- 10 psi
50
40
30
20
10
0
Stress - 10³ psi
0₁ -P
100
80
60
40
20
1000 lb/in.²
Constant ratio o, :03 15:1
Uniaxial
8
10
Constant ratio 0₁:0₂ 8:1
Axial strain-103 in./in.
(a)
12
-1000 psi
3000 psi
-2000 psi
Uniaxial
12
16
20
Axial strain - 103 in./in.
(b)
4000 lb/in,²
5000 lb/in.³
14
24
16
Lateral pressure
28
18
32
Figure 3.10 Stress difference versus axial strain curves as a function of con-
fining pressure in triaxial compression experiments on sandstone (a) and
norite (b). [From Bieniawski (1972).]
stresses are compressive. In practice, empirical curve fitting is the best proce-
dure for producing a criterion of failure tailored to any given rock type. See, for
example, Herget and Unrug (1976). A satisfactory formula for many purposes
will be afforded by the union of a tension cutoff, 3 = To, and a power law
(Bieniawski, 1974):
(3.15)
0₁.p= 1 + N
qu
The constants N and M will be determined by fitting a curve to the family of
points](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8745b554-e61b-42b2-bb7a-6e5716b8c815%2F2a7c5f0f-fd79-4218-b073-4584c4990e69%2Fg8y9ig_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Stress- 10 psi
50
40
30
20
10
0
Stress - 10³ psi
0₁ -P
100
80
60
40
20
1000 lb/in.²
Constant ratio o, :03 15:1
Uniaxial
8
10
Constant ratio 0₁:0₂ 8:1
Axial strain-103 in./in.
(a)
12
-1000 psi
3000 psi
-2000 psi
Uniaxial
12
16
20
Axial strain - 103 in./in.
(b)
4000 lb/in,²
5000 lb/in.³
14
24
16
Lateral pressure
28
18
32
Figure 3.10 Stress difference versus axial strain curves as a function of con-
fining pressure in triaxial compression experiments on sandstone (a) and
norite (b). [From Bieniawski (1972).]
stresses are compressive. In practice, empirical curve fitting is the best proce-
dure for producing a criterion of failure tailored to any given rock type. See, for
example, Herget and Unrug (1976). A satisfactory formula for many purposes
will be afforded by the union of a tension cutoff, 3 = To, and a power law
(Bieniawski, 1974):
(3.15)
0₁.p= 1 + N
qu
The constants N and M will be determined by fitting a curve to the family of
points
Transcribed Image Text:8. (a) For the sandstone of Figure 3.10, determine the value of op and Sip that
best fits the peak strengths given. (The subscript p denotes "peak.")
(b) Find best fit values of o, and Sir describing the residual strength of the
sandstone.
(c) Find values of M and N to express the peak strength of the sandstone in
Figure 3.10 according to Equation 3.15.
(d) Find op and Sip for the norite of Figure 3.10.
(e) Find o, and Sir for the norite.
(f) Find M and N for the norite.
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