Download Problem Set #3: Pressure and Stress 1. Consider a region of the

Geol 542: Advanced Structural Geology Fall 2011 Problem Set #3: Pressure and Stress -­‐3
1. Consider a region of the Earth’s crust where the rock density is a depth-­‐independent 2400 kg m . a) What is the vertical stress (lithostatic stress or overburden) at a depth of 5 km? (3) The vertical stress is the lithostatic stress: σv = ρrgh = 2400 kg/m3 x 9.8 m/s2 x 5000 m = 1.176 x 108 kg m-­‐1 s-­‐2 = 117.6 MPa (3) b) In the absence of any tectonic stresses, what are the magnitudes of principal stresses in the horizontal plane? (2) Lithostatic stress is isotropic, so σvert = σhoriz = 117.6 MPa (2) c) If the rocks are water saturated, and the water table is at a depth of 500m, what is the magnitude of 3
the hydrostatic pressure at this 5 km depth? Assume the density of water is 1000 kg/m . (4) If the water table is at 500 m depth, then at 5 km depth, there is a 4500 m water column present. Therefore: pf = ρwgd = 1000 kg/m3 x 9.8 m/s2 x 4500 m = 44.1 MPa (4) d) What is the magnitude of the effective stress at 5 km depth given this water pressure? (2) The effective stress: σeff = σv – pf = 117.6 MPa – 44.1 MPa = 73.5 MPa (2) 2. Now consider a region of the Earth’s crust where a borehole is being drilled into a hydrocarbon reservoir. The 3
stratigraphy consists of the following rock types: a 1 km thick shale layer (density 2100 kg/m ), a 1.5 km thick 3
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sandstone layer (density 2400 kg/m ) and a limestone unit (density 2600 kg/m ) of unknown thickness but which extends deeper than the maximum depth of the borehole. a) Calculate the lithostatic stress that exists along the walls of the borehole at depths of (i) 1 km, (ii) 2 km and (iii) 3 km (the bottom of the borehole). (15) The vertical stress is the lithostatic stress: (i) At 1 km depth: σv = ρrgh = 2100 kg/m3 x 9.8 m/s2 x 1000 m = 20.58 MPa (3) Geol 542: Advanced Structural Geology Fall 2011 (ii) At 2 km depth: σv = ρrgh = σv (top 1 km) + σv (next 1 km) = 20.58 MPa + (2400 kg/m3 x 9.8 m/s2 x 1000 m) = 20.58 MPa + 23.52 MPa = 44.1 MPa (6) (iii) At 3 km depth: σv = ρrgh = σv (top 1 km) + σv (next 1.5 km) + σv (bottom 0.5 km) = 20.58 MPa + (2400 kg/m3 x 9.8 m/s2 x 1500 m) + (2600 kg/m3 x 9.8 m/s2 x 500 m) = 20.58 MPa + 35.28 MPa + 12.74 MPa = 68.6 MPa (6) b) Now calculate the effective stress at each of these depths, assuming the rocks are saturated up to the Earth’s surface. (9) (i) At 1 km depth: pf = ρwgh = 9.8 MPa, so σeff = 20.58 MPa – 9.8 MPa = 10.78 MPa (ii) At 2 km depth: pf = ρwgh = 19.6 MPa, so σeff = 44.1 MPa – 19.6 MPa = 24.5 MPa (iii) At 3 km depth: pf = ρwgh = 29.4 MPa, so σeff = 68.6 MPa – 29.4 MPa = 39.2 MPa (3) (3) (3) c)
Assuming the borehole is cased down to the depth of hydrocarbon recovery 3 km down, what density of drilling mud would the borehole operators need to use to ensure that the borehole neither collapses nor explodes? (5) Drilling mud needs to push out against the 39.2 MPa of confining stress present at a depth of 3 km (i.e., λ = 1). So: ρmudgh = 39.2 MPa ρmud = 39.2 MPa / gh = 39.2 x 106 kg m-­‐1 s-­‐2 / 9.8 m s-­‐2 x 3000 m ρmud = 1,333.33 kg/m3 (5) If you mistakenly forgot to add the effect of hydrostatic pressure (i.e., dry rocks), the drilling mud density ends up being 2,333.33 kg/m3. 3. Force is a vector. Traction is a vector. Stress is a tensor. Yet tractions are defined as the limit as area ΔA -­‐> 0 of the ratio of force divided by area: ΔF
ΔA → 0 ΔA
t (n ) = lim
Force divided by area is the definition of stress, but traction vectors are not stress; they are simply one component of the concept we call stress. In fact, an infinite number of tractions describe the state of stress at a point, P. €
Geol 542: Advanced Structural Geology Fall 2011 (a) Explain the fundamental differences between force, traction, and stress in terms of what aspect of a body we assume they are acting upon. See Fig. 6.1 in the textbook to assist you with your answer. (6) Forces are vectors that represent mass times acceleration. They act on particles and indicate the direction of acceleration of the particle in response to that force. Tractions are vectors that represent the limit (as area tends to zero around a point of interest) of the ratio of force over area, and thus act on a surface element having a certain area. This traction vector represents the cumulative result of all forces acting on the body containing the point of interest (whether these forces are external or internal), resolved onto a surface. For the case of equilibrium, an equal but opposite vector on the other side of the surface counterbalances this traction vector (Newton's 3rd law). The combination of these two traction vectors is sometimes called the surface stress. The orientation of these tractions is dependent on the orientation of the surface (whether the surface is real or hypothetical) containing the surface element. So traction is expressed as t(n), where n is the normal vector representing the orientation of the plane. Stress considers the traction vectors that would be acting upon all possible configuration of surfaces in 3D and thus can be considered to be acting upon a volume element. A knowledge of stress allows us to fully characterize how a body will deform in three dimensions in response to the forces acting upon it. (b) Why are there technically an infinite number of tractions that together define the state of stress at a point? [Note: as we will learn in coming lectures, it turns out we can select a mere six tractions out of the population of infinity to fully define the state of stress in three dimensions. Whew!] (4) If you were to consider all possible plane orientations passing through a point in 3D (i.e., an infinite number of planes), there would be a corresponding infinite number of traction vectors for any configuration or combination of forces acting on the body containing the point. Hence, the state of stress around a point is represented by a family of tractions (an infinite number) that form the shape of an ellipsoid around that point when the traction vectors are scaled to their magnitudes. This is the stress ellipsoid and, being a simple mathematical construct, can be fully defined by its 3 principal axes. These axes are defined by 6 traction vectors (or 3 pairs of equal but opposite tractions). These 6 tractions will always be acting on hypothetical planes passing through the point in such a way that the tractions are perpendicular to those planes. For that reason, they are planes of zero shear stress (principal planes) and the 3 pairs of tractions acting upon them are normal stresses ("normal" here meaning "perpendicular to"). They are, in fact, the three principal stresses that we typically use to represent stress. This is why each stress component must be represented by 2 arrows, equal and opposite. They are the two traction vectors acting orthogonal to the principal planes. (4) [Footnote] Stress can also be represented as a tensor of 9 components, which will be addressed later in the semester. In essence, the stress tensor allows us to represent the state of stress acting on an infinitesimally small cube (i.e., a volume element) in some pre-­‐definined coordinate system (typically Cartesian). Each face of the cube will have one normal traction and two shear tractions acting upon it (in the 3 coordinate axis directions). This means the cube will have 18 traction components; however, again invoking Newton's 3rd law, opposing faces of the cube will have equal but opposite traction vectors (for the case of equilibrium), giving us 9 components of stress. There are thus 3 normal components and 6 shear components of stress in the stress tensor. This method of representing stress allows us to define the state of stress in terms of our pre-­‐definined coordinate system, without concern about which directions the principal stresses are acting relative to this coordinate system. [50]