HW4 F11 key - University of Idahosimkat/course_materials/geol542/HW4_F… ·...
Transcript of HW4 F11 key - University of Idahosimkat/course_materials/geol542/HW4_F… ·...
Geol 542: Advanced Structural Geology Fall 2011
Problem Set #4: Principal Stresses and Mohr Circles 1. At some point in the Earth's crust, the three principal stresses are: σ1 = 100 MPa; σ2 = 80 MPa; σ3 = 65 MPa
(compression is positive). [16]
a) What term can we use to describe this state of stress? (2) Polyaxial or triaxial. (2) b) What is the mean normal stress? (2)
σmean = 1/3(σ1 + σ2 + σ3) = 1/3(100 + 80 + 65) MPa = 81.67 MPa (2)
c) What is the maximum shear stress? (2)
σs (max) = 1/2(σ1 -‐ σ3) = 1/2(100 -‐ 65) MPa = 17.5 MPa (2)
d) What is the differential stress? (2)
σdiff= (σ1 -‐ σ3) = (100 -‐ 65) MPa = 35 MPa (2)
e) What are the three deviatoric stresses? (3)
σ1 (dev) = σ1 – σmean = 100 – 81.67 MPa = 18.33 MPa (1) σ2 (dev) = σ2 – σmean = 80 – 81.67 MPa = -‐1.67 MPa (1) σ3 (dev) = σ3 – σmean = 65 – 81.67 MPa = -‐16.67 MPa (1)
f) Use the calculated deviatoric stress to determine in what directions the rock will contract, and in what
directions the rock will extend. (3)
The rock will contract in the σ1 direction (the deviatoric σ1 is positive) and extend in the σ2 and σ3 directions (the deviatoric σ2 and σ3 are negative). (3)
g) Given the result above, comment on whether or not tensile stresses are needed in the Earth in order to
allow extension (e.g., through normal faulting), or if extension is possible with all-‐round compression. (2)
In this all-‐round compressive stress field, the rock still extends in two directions, so tensile stresses are not required in order for extension to occur. In fact, in ANY stress field, there will be extension in at least one principal stress direction (the σ3 direction, where compression is positive). (2)
2. Use graph paper to construct a Mohr circle, as directed below, then answer the related questions. Remember, a
Mohr diagram can only work if the σn and σs axes have an identical scale. Use an entire graphing page for your Mohr diagram (i.e., in landscape format). [44]
a. Construct a Mohr diagram to represent a state of stress in which the principal stresses are
measured to be: σ1 = 40 MPa, σ2 = 22 MPa, and σ3 = 18 MPa (compression is positive). (6)
See next page. (6)
Geol 542: Advanced Structural Geology Fall 2011
b. If the minimum compressive stress, σ3, is the vertical (lithostatic) component of stress, use your Mohr diagram to read off the approximate normal (σn) and shear (σs) stresses acting on a fault plane dipping at 40° (remember, θ is defined in physical space as the angle between the σ1 direction and the normal vector n to the plane). Assume that only σ1 and σ3 resolve onto the fault plane (i.e., the strike of the fault is parallel to σ2). Include a sketch showing your fault configuration. (5)
In this configuration, the dip of the fault is 40° but the value of θ is 50° (or -‐50°), so 2θ = 100° (or -‐100°). Choosing the negative θ values, the corresponding (σn, σs) values are (27, -‐11) MPa. (5)
c. What would the normal and shear stress be on a fault plane having the same strike and dip as above but dipping in the opposite direction? (2)
A fault dipping opposite to the negative θ values would have positive θ values with the same absolute magnitude, so the corresponding (σn, σs) values are (27, 11) MPa. (2)
Geol 542: Advanced Structural Geology Fall 2011
d. In geologic problems, why would we even care about the sign of σs? (2)
The sign of σs indicates the sense of shearing in the defined coordinate system. This will allow the correct sense of slip to be determined for conjugate fault sets for any stress configuration. (2)
e. Corroborate your result in (b) by using the Mohr equations (of the form shown below) to calculate the precise magnitudes of the shear and normal stresses. (6)
σn = σ1 cos
2 θ + σ3 sin2 θ
σs = (σ1 – σ3) sin θ cos θ
σn = σ1 cos2 θ + σ3 sin
2 θ = 40 cos2 [-‐50°] + 18 sin2 [-‐50°] MPa = 16.53 + 10.56 MPa = 27.09 MPa (3) σs = (σ1 – σ3) sin θ cos θ = (40 – 18) sin [-‐50°] cos [-‐50°] MPa = -‐10.83 MPa (3)
f. In this state of stress, in which directions would the rock contract and in which directions would the rock extend (show your calculations). (6)
σ1 (dev) = σ1 – σmean = 40 – 26.67 MPa = 13.33 MPa – contraction in this direction (2) σ2 (dev) = σ2 – σmean = 22 – 26.67 MPa = -‐4.67 MPa – extension in this direction (2) σ3 (dev) = σ3 – σmean = 18 – 26.67 MPa = -‐8.67 MPa – extension in this direction (2)
g. What would we call this type of tectonic environment based on the relative orientations of the
principal stresses in 3D space? (2)
As σ1 is horizontal and σ3 is vertical, this is a compressional tectonic environment. (2)
h. Now assume the fault exists within saturated rocks having a fluid pressure of 15 MPa. Draw a new Mohr circle representing this effective stress condition on the SAME graph as before. (4)
See graph. The Mohr circle shifts to the left by 15 MPa with no change in diameter. (4)
Geol 542: Advanced Structural Geology Fall 2011
i. How has σn and σs changed for this fault in response to the addition of fluid? Intuitively, what would this mean for the likelihood of sliding along the fault? (2)
By shifting the Mohr circle, σn is smaller by 15 MPa but σs is exactly the same as before. The ratio of shear to normal stress is thus larger, resulting in a larger propensity for sliding along a fault by frictional failure. (2)
j. How did the differential stress and deviatoric stresses change in response to the addition of a fluid pressure? Show your calculations. (7)
In dry rocks: σdiff= (σ1 -‐ σ3) = (40 -‐ 18) MPa = 22 MPa (1) σ1 (dev) = σ1 – σmean = 40 – 26.67 MPa = 13.33 MPa (calculated earlier) σ2 (dev) = σ2 – σmean = 22 – 26.67 MPa = -‐4.67 MPa (calculated earlier) σ3 (dev) = σ3 – σmean = 18 – 26.67 MPa = -‐8.67 MPa (calculated earlier)
In wet rocks: σdiff= (σ1 -‐ σ3) = (25 -‐ 3) MPa = 22 MPa – differential stress has not changed. (1) σ1 (dev) = σ1 – σmean = 25 – 11.67 MPa = 13.33 MPa – deviatoric stress has not changed. (3) σ2 (dev) = σ2 – σmean = 7 – 11.67 MPa = -‐4.67 MPa – deviatoric stress has not changed. (1) σ3 (dev) = σ3 – σmean = 3 – 11.67 MPa = -‐8.67 MPa – deviatoric stress has not changed. (1)
k. Given your answers to i and j above, how does the addition of fluid pressure affect the way rocks in the Earth’s crust deform (in terms of the distortion they experience)? (2)
The addition of a fluid pressure does not change the way the crust distorts (i.e., shape change); however, it does make the distortion more likely to occur by shear failure. It also causes a smaller change in volume (despite the identical change in shape), so affects the deformation in this manner. (2)
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