Week 4

Dynamic Analysis I. Read pages 98-122 in Chapter 3: Dynamic Analysis.

You are expected to read all the sections listed below. Information from the sections in italics will be discussed in class. You are expected to read the other sections and you may be called on in class to answer questions based on that material.

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You should become familiar with the following terms during this weeks lectures and readings:

force differential stress hydrostatic stress lithostatic stress
stress gradient load mass Mohr stress diagram
newton normal stress prinicpal stress directions shear stress
stress stress ellipse/ellipsoid fundamental stress equations stress field
stress tensor triaxial stress uniaxial stress weight

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You should be able to answer the questions below following this week:

  1. What is the difference between weight and mass?
  2. What are the units of: a) mass; b) weight; c) load; d) force?
  3. Determine the mass of a cube of rock (volume 1m3) with a density of 2700 kg/m3.
  4. The vertical stress acting on a point within the earth is 40 Mpa, gravity is 9.8 m/s2, and the density of the overlying rock is 2700 kg/m3. How far is the point below the earth's surface?
  5. The vertical stress acting on a point within the earth is 30 Mpa, the horizontal stress is 50 Mpa. What is the magnitude and orientation of stress acting on a plane oriented 65 degrees (relative to vertical)?
  6. Repeat question 5 for a plane oriented 20 degrees to vertical.
  7. Two rocks were deformed under the stress conditions described below. Both deformed by faulting. in each case the normal stress (sn) acting on the fault was 40 MPa. Use the Mohr circle for stress to determine: a) What were the orientations of the faults (relative to s1) in each case?; b) What was the shear stress acting on each failure plane? Rock A was deformed with s1 = 60 MPa and s3 = 10 MPa; Rock B was deformed with s1 = 60 MPa and s3 = 30 MPa.
  8. A vertical stress of 600 MPa and a horizontal stress of 1,000 MPa act on a plane that makes an angle of 70o to vertical. a) Use a force balance to calculate the magnitude and orientation of the stress vector acting on the plane; b) Use a Mohr stress diagram to determine sn and ss for the plane.
  9. The Texas Gulf coast is an area of concave upward normal faults. Vertical stress and fault angle vary with depth on a fault as shown in the table below. The horizontal stress is uniform throughout the area and is 100 MPa. Determine: a) What is sn on the fault plane at 5 km?; b) ss is greatest on the fault at which depth?

Depth (km)

Vertical Stress (MPa)

Fault angle (from horizontal)
1
200
60
3
690
40
5
1250
15

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Dynamic Analysis I

Deformation depends upon:

Strain is the response of rock to stresses generated by forces

Force = mass x acceleration

Units of Force

Stress = force / area

Units of Stress

 

Lithostatic stress calculation

 

Calculate stress acting on a point 10 km below the earth's surface:

 

Stress on a plane:

 

What is the magnitude and orientation of stresses acting on a plane within the earth?

Setting up a force balance to calculate stress on an inclined plane. Vertical stress = 15,000 Pa, horizontal stress = 5,000 Pa, plane oriented 65 degrees to vertical stress. This problem will make a lot more sense if you have the diagrams from the lecture.

1. visualize the problem & set up the force balance;

2. calculate horizontal (Sx) and vertical (Sz) components of stress acting on the plane;
Fox = Fxo
5,000 (OZ) = Sx (n)
5,000 (n Cos 65) = Sx (n)
2113 Pa = Sx
 
Foz = Fzo
15,000 (OX) = Sz (n)
15,000 (n Sin 65) = Sz (n)
13,594 Pa = Sz
 
3. calculate the magnitude and orientation of stress (s) acting on the plane;
magnitude of stress vector acting on plane = square root of (Sz2 + Sx2)
stress = 13,758 Pa
tan b = Sx / Sz
b = arctan (0.1554)
b = 9 degrees (angle between stress vector and maximum principal stress)
 
4. determine normal and shear stresses acting on the plane
Cos 16 = normal stress/stress vector
13,758 Cos 16 = normal stress
13,213 Pa = normal stress
 
Sin 16 = shear stress/stress vector
13,758 Sin 16 = shear stress
3,830 Pa = shear stress

 

Stress Tensor

 

Hydrostatic Stress

 

Fundamental Stress Equations

sn = {(s1 + s3)/2} - {(s1 - s3)/2}Cos 2q

ss = {(s1 - s3)/2}Sin 2q

 

Mohr Stress Diagram

Procedure for plotting a Mohr stress diagram

  1. axes are calibrated in units of stress, x-axis = normal stress, y-axis = shear stress
  2. locate principal stresses on the x-axis (shear stress is zero at these points)
  3. locate the center of the Mohr circle
  4. measure the angle 2q clockwise from the x-axis for positive angles (counterclockwise for negative angles)
  5. draw a radius with this angle
  6. the point where the radius intersects the circumference of the circle indicates the normal and shear stress acting on a plane with that orientation

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