In this Chapter
2.2 Body Forces and Surface Forces
The forces on an element of a solid are of two types: body forces and surface forces. Body forces act throughout the volume of the solid. The magnitude of the body force on an element is thus directly pro- portional to its volume or mass. An example is the downward force of gravity, that is, the weight of an element, which is the product of its mass and the accel- eration of gravity g. Since density ρ is mass per unit volume, the gravitational body force on an element is also the product of ρg and the element’s volume. Thus the downward gravitational body force is g per unit mass and ρg per unit volume.
The densities of some common rocks are listed in Appendix B, Section B.5. The densities of rocks depend on the pressure; the values given are zero-pressure densities. Under the high pressures encountered deep in the mantle, rocks are as much as 50% denser than the zero-pressure values. The varia- tion of density with depth in the Earth is discussed inChapter 4. Typical mantle rocks have zero-pressure densities of 3250 kg m−3. Basalt and gabbro, which are the principal constituents of the oceanic crust, have densities near 2950 kg m−3. Continental igneous rocks such as granite and diorite are significantly lighter with densities of 2650 to 2800 kg m−3. Sedimentary rocks are generally the lightest and have the largest varia- tions in density, in large part because of variations in porosity and water content in the rocks.
Surface forces act on the surface area bounding an element of volume. They arise from interatomic forces exerted by material on one side of the surface onto material on the opposite side. The magnitude of the surface force is directly proportional to the area of the surface on which it acts. It also depends on the
Figure 2.1 Body and surface forces acting on a vertical column of rock.
orientation of the surface. As an example, consider the force that must act at the base of the column of rock at a depth y beneath the surface to support the weight of the column, as illustrated inFigure 2.1. The weight of the column of cross-sectional area δA, is ρgyδA. This weight must be balanced by an upward surface force
σyyδAdistributed on the horizontal surface of area δA at depth y. We are assuming that no vertical forces are acting on the lateral surfaces of the column and that the density ρ is constant; σyy is thus the surface force per unit area acting perpendicular to a horizontal surface, that is, stress. Since the forces on the column of rock must be equal if the column is in equilibrium, we find
σyy= ρgy. (2.1)
The normal force per unit area on horizontal planes increases linearly with depth. The normal stress due to the weight of the overlying rock or overburden is known as the lithostatic stress or pressure.
To find a typical value for stress in the lithosphere, let us determine the lithostatic stress on a horizontal plane at the base of the continental crust. Assume that the crust is 35 km thick and that its mean density is
“9781107006539AR” — 2013/12/3 — 19:53 — page 97 — #111
2.2 Body Forces and Surface Forces 97
Figure 2.2 A continental block “floating” on the fluid mantle.
2750 kg m−3; fromEquation (2.1)we find that
σyy= 2750 kg m−3× 10 m s−2× 3.5 × 104m = 9.625 × 108Pa= 962.5 MPa.
The SI unit for pressure or stress is the pascal (Pa). Pressures and stresses in the Earth are normally given in megapascals (MPa); 1 megapascal= 106pascals.
Think of continents as blocks of wood floating on a sea of mantle rock, as illustrated inFigure 2.2. The mean density of the continent, say ρc= 2750 kg m−3, is less than the mean upper mantle density, say
ρm = 3300 kg m−3, so that the continent “floats.”
Archimedes’ principle applies to continents; they are
buoyed up by a force equal to the weight of mantle rock displaced. At the base of the continent σyy = ρcgh, where ρcis the density of the continent and h is its thickness. At this depth in the mantle, σyy is ρmgb, where ρmis the mantle density and b is the depth in the mantle to which the continent “sinks.” Another statement of Archimedes’ principle, also known as
hydrostatic equilibrium, is that these stresses are equal.
Therefore we find
ρch= ρmb. (2.2)
The height of the continent above the surrounding mantle is h− b = h − ρc ρm h= h 1− ρc ρm . (2.3) Using the values given earlier for the densities and the thickness of the continental crust h = 35 km, we find from Equation (2.3) that h− b = 5.8 km. This analysis is only approximately valid for deter- mining the depth of the oceans relative to the con- tinents, since we have neglected the contribution of the seawater and the oceanic crust. The application
of hydrostatic equilibrium to the continental crust is known as isostasy; it is discussed in more detail in Chapter 5.
Problem 2.1
An average thickness of the oceanic crust is 6 km. Its density is 2900 kg m−3. This is overlain by 5 km of water (ρw = 1000 kg m−3) in a typical ocean basin. Determine the normal force per unit area on a horizontal plane at the base of the oceanic crust due to the weight of the crust and the overlying water.
M
Problem 2.2
A mountain range has an elevation of 5 km. Assum- ing that ρm = 3300 kg m−3, ρc = 2800 kg m−3, and that the reference or normal continental crust has a thickness of 35 km, determine the thickness of the continental crust beneath the mountain range. Assume that hydrostatic equilibrium is applicable. A MATLAB code for solving this problem is given inAppendix D.
Problem 2.3
There is observational evidence from the continents that the sea level in the Cretaceous was 200 m higher than today. After a few thousand years, however, the seawater is in isostatic equilibrium with the ocean basins. What was the corresponding increase in the depth of the ocean basins? Take ρw= 1000 kg m−3 and the density of the displaced mantle to be ρm = 3300 kg m−3.
A more realistic model for the depth of the ocean basins is illustrated in Figure 2.3. The continental
Figure 2.3 Isostasy of the continental crust relative to an ocean basin.
crust has a thickness hcc and a density ρcc; its upper surface is at sea level. The oceanic crust is covered with water of depth hw and density ρw. The oceanic crust has a thickness hocand density ρoc. The mantle den- sity is ρm. Application of the principle of isostasy to the base of the continental crust gives
ρcchcc= ρwhw+ ρochoc+ ρm(hcc− hw− hoc). (2.4) The depth of the ocean basin relative to the continent is given by hw= (ρm− ρcc) (ρm− ρw) hcc− (ρm− ρoc) (ρm− ρw) hoc. (2.5) Taking hcc= 35 km, hoc= 6 km, ρm= 3300 kg m−3, ρw= 1000 kg m−3, ρcc= 2800 kg m−3, and ρoc= 2900 kg m−3, we find hw= 6.6 km.
Subsidence of the surface of the continental crust often results in the formation of a sedimentary basin. Assume that the surface of the continental crust is initially at sea level and, as it subsides, sediments are deposited so that the surface of the sediments remains at sea level. One cause of the subsidence is the thin- ning of the continental crust. As the crust is thinned, isostasy requires that the surface subside. A simple model for this subsidence applicable to some sedimen- tary basins is the crustal stretching model (McKenzie, 1978). This two-dimensional model is illustrated in Figure 2.4. A section of continental crust with an ini- tial width w0 is stretched to a final width wb. The stretching factor α is defined by
α=wb w0
. (2.6)
In order to conserve the volume of the stretched con- tinental crust we assume a constant crustal density ρcc and require that
Figure 2.4 Illustration of the crustal stretching model for the formation of a sedimentary basin. A section of continental crust of initial widthw0, illustrated in (a), is stretched by a stretching
factor α= 4 to a final width wb to form the sedimentary basin illustrated in (b ).
wbhcb= w0hcc, (2.7) where hcc is the initial thickness of the continental crust and hcb is the final thickness of the stretched crust. The combination of Equations (2.6)and (2.7) gives
hcb= hcc
α . (2.8)
The surface of this stretched continental crust sub- sides and is assumed to be covered with sediments of density ρs(ρs< ρcc)to sea level. The sediments have a thickness hsband the lower boundary of the sediments is referred to as basement. Application of the princi- ple of isostasy to the base of the reference continental crust gives
“9781107006539AR” — 2013/12/3 — 19:53 — page 99 — #113
2.2 Body Forces and Surface Forces 99
Figure 2.5 Thickness of a sedimentary basinhsbas a function of the crustal stretching factor α.
The combination of Equations (2.8) and (2.9) gives the thickness of the sedimentary basin in terms of the stretching factor as hsb= hcc ρm− ρcc ρm− ρs 1− 1 α . (2.10)
The thickness of the sedimentary basin is given as a function of the stretching factor in Figure 2.5 for
hcc = 35 km, ρm= 3300 kg m−3, ρcc = 2800 kg m−3, and ρs = 2500 kg m−3. The maximum thickness of the sedimentary basin for an infinite stretching factor is hsb= 22 km.
M
Problem 2.4
A sedimentary basin has a thickness of 4 km. Assuming that the crustal stretching model is appli- cable and that hcc = 35 km, ρm = 3300 kg m−3, ρcc = 2750 kg m−3, and ρs = 2550 kg m−3, deter- mine the stretching factor. A MATLAB code for solving this problem is given inAppendix D.
M
Problem 2.5
A sedimentary basin has a thickness of 7 km. Assuming that the crustal stretching model is appli- cable and that hcc = 35 km, ρm = 3300 kg m−3, ρcc= 2700 kg m−3, and ρs = 2450 kg m3, determine the stretching factor. A MATLAB code for solving this problem is given inAppendix D.
Figure 2.6 Illustration of the crustal compression model for a mountain belt. A section of continental crust of widthw0, shown in (a), is compressed by compression factor β= 2 to form a mountain belt as shown in (b ).
Problem 2.6
A simple model for a continental mountain belt is the crustal compression model illustrated in Figure 2.6. A section of the continental crust of width w0 is compressed to a width wmb. The com- pression factor β is defined by
β = w0 wmb
. (2.11)
Show that the height of the mountain belt h is given by h= hcc (ρm− ρcc) ρm (β− 1). (2.12) Assuming β = 2, hcc = 35 km, ρm = 3300 kg m−3, and ρcc = 2800 kg m−3, determine the height of the mountain belt h and the thickness of the crustal root b.
Just as there are normal surface forces per unit area on horizontal planes in the Earth, there are also nor- mal surface forces per unit area on vertical planes, as sketched in Figure 2.7. The horizontal normal stress components σxxand σzzcan include large-scale
Figure 2.7 Horizontal surface forces acting on vertical planes.
tectonic forces, in which case σxx = σzz = σyy. On the other hand, there are many instances in which rock was heated to sufficiently high temperatures or was sufficiently weak initially so that the three stresses
σxx, σzz, and σyy are equal to the weight of the over- burden; that is,
pL≡ σxx= σzz= σyy= ρgy. (2.13) When the three normal stresses are equal, they are defined to be the pressure. The balance between pres- sure and the weight of the overburden is known as a
lithostatic state of stress. It is completely equivalent to
the hydrostatic state of stress in a motionless body of fluid wherein pressure forces are exerted equally in all directions and pressure increases proportionately with depth.
We will now show that the continental block illus- trated inFigure 2.2cannot simply be in a lithostatic state of stress. The force balance on the continental block is illustrated in Figure 2.8. A horizontal force is acting on the edge of the block Fm. We assume that this force is due to the lithostatic pressure in the man- tle rock of density ρm. The vertical distribution of this pressure is given inFigure 2.9. The horizontal force Fm is obtained by integrating the lithostatic pressure:
Fm= b 0 pLdy= ρmg b 0 y dy=1 2ρmgb 2. (2.14)
This force is per unit width of the block so that it has dimensions of force per unit length. The total force per unit width is proportional to the area under the stress distribution given inFigure 2.9.
Figure 2.8 Force balance on a section of continental block.
Figure 2.9 The area under the stress versus depth profile is proportional to the total horizontal force on a vertical plane.
We next determine the horizontal force per unit width acting at a typical cross section in the continen- tal block Fc. We assume that the horizontal normal stress acting in the continent σxx is made up of two parts, the lithostatic contribution ρcgy and a constant tectonic contribution σxx,
σxx= ρcgy+ σxx. (2.15) The tectonic contribution is also known as the devi-
atoric stress. The horizontal force Fc is obtained by integrating the horizontal normal stress
Fc= h 0 σxxdy= h 0 (ρcgy+ σxx) dy = 1 2ρcgh 2+ σ xxh. (2.16)
In order to maintain a static balance, the two forces Fc and Fmmust be equal. UsingEquations (2.2),(2.14), and(2.16), we obtain σxx= 1 2 ρmgb2 h − 1 2ρcgh= − 1 2ρcgh 1− ρc ρm . (2.17) A horizontal tensile stress is required to maintain the integrity of the continental block. The horizontal
“9781107006539AR” — 2013/12/3 — 19:53 — page 101 — #115
2.2 Body Forces and Surface Forces 101
tensile stress is a force per unit area acting on vertical planes and tending to pull on such planes. A com-
pressive stress is a normal force per unit area tending
to push on a plane. We consider compressive stresses positive and tensile stresses negative, a convention generally adopted in the geological literature. This is opposite to the sign convention used in most elasticity textbooks in which positive stress is tensional. Taking
h= 35 km, ρm= 3300 kg m−3, and ρc= 2750 kg m−3, we find fromEquation (2.17)that σxx= −80.2 MPa. Typical values for deviatoric stresses in the continents are of the order of 10 to 100 MPa.
M
Problem 2.7
Consider a continental block to have a thickness of 70 km corresponding to a major mountain range. If the continent has a density of 2800 kg m−3 and the mantle a density of 3300 kg m−3, deter- mine the tensional stress in the continental block. A MATLAB code for solving this problem is given in Appendix D.
Problem 2.8
Determine the deviatoric stress in the continent for the oceanic–continental structure inFigure 2.3 by proceeding as follows. Show that the pressure as a function of depth in the continental crust
pcis
pc= ρccgy, (2.18) and that the pressures in the water, in the oceanic crust, and in the mantle beneath the oceanic crust are
p0= ρwgy 0≤ y ≤ hw
= ρwghw+ ρocg(y− hw) hw≤ y ≤ hw+ hoc = ρwghw+ ρocghoc+ ρmg(y− hw− hoc)
hw+ hoc≤ y ≤ hcc. (2.19) Find the net difference in the hydrostatic pres- sure force between the continental and the oceanic crusts F by integrating the pressures over a depth equal to the thickness of the continental crust. The result is F = g[hwhcc(ρm− ρw)+ hochcc(ρm− ρoc) − hwhoc(ρm− ρoc)− 1 2h 2 w(ρm− ρw) − 1 2h 2 oc(ρm− ρoc)− 1 2h 2 cc(ρm− ρcc)]. (2.20) Calculate F for hw= 5 km, ρw= 1000 kg m−3, hoc= 7 km, ρoc = 2900 kg m−3, ρcc = 2800 kg m−3, and ρm= 3300 kg m−3. Find hccfromEquation (2.5). If the elastic stresses required to balance this force are distributed over a depth equal to hcc, determine the stress. If the stresses are exerted in the continental crust, are they tensional or compressional? If they act in the oceanic lithosphere, are they tensional or compressional?
Surface forces can act parallel as well as perpen- dicular to a surface. An example is provided by the forces acting on the area element δA lying in the plane of a strike–slip fault, as illustrated in Figure 2.10. The normal compressive force σxxδA acting on the fault face is the consequence of the weight of the overburden and the tectonic forces tending to press the two sides of the fault together. The tangential or shear force on the element σxzδAopposes the tectonic forces driving the left-lateral motion on the fault. This shear force is the result of the frictional resistance to motion on the fault. The quantity σxzis the tangential surface force per unit area or the shear stress. The first
Figure 2.10 Normal and tangential surface forces on an area element in the fault plane of a strike–slip fault.
Figure 2.11 Normal and tangential forces acting on a rock mass displaced horizontally to the right in a low-angle overthrust fault.
subscript refers to the direction normal to the surface element and the second subscript to the direction of the shear force.
Another example of the resistive force due to a shear stress is the emplacement of a thrust sheet. In zones of continental collision a thin sheet of crystalline rock is often overthrust upon adjacent continental rocks on a low-angle thrust fault. This process is illustrated in Figure 2.11, where the thrust sheet has been emplaced from the left as a consequence of horizontal tectonic forces. Neglecting the influence of gravity, which is considered inSection 8.4, we can write the total hor- izontal tectonic force FT due to a horizontal tectonic stress σxxas
FT = σxxh, (2.21) where h is the thickness of the thrust sheet and FTis a force per unit width of the sheet. This tectonic driving force is resisted by the shear stress σyx acting on the base of the thrust sheet. The total resisting shear force per unit width FRis
FR= σyxL, (2.22)
where L is the length of the thrust sheet.
In many cases it is appropriate to relate the shear stress resisting the sliding of one surface over another to the normal force pressing the sur- faces together. Empirically we often observe that these stresses are proportional to one another so that
σyx= f σyy, (2.23) where σyy is the vertical normal stress acting on the base of the thrust sheet and f , the constant of pro- portionality, is known as the coefficient of friction. Assuming that σyyhas the lithostatic value
σyy= ρcgh, (2.24) and equating the driving tectonic force FT to the resisting shear force, we find that
σxx= f ρcgL. (2.25) This is the tectonic stress required to emplace a thrust sheet of length L. Taking a typical value for the tec- tonic stress to be σxx = 100 MPa and assuming a thrust sheet length L= 100 km and ρc= 2750 kg m−3, we find that the required coefficient of friction is f = 0.036. The existence of long thrust sheets implies low values for the coefficient of friction.
Problem 2.9
Assume that the friction law given inEquation (2.23) is applicable to the strike–slip fault illustrated in Figure 2.10with f = 0.3. Also assume that the nor- mal stress σxxis lithostatic with ρc = 2750 kg m−3. If the fault is 10 km deep, what is the force (per unit length of fault) resisting motion on the fault? What is the mean tectonic shear stress over this depth ¯σzx required to overcome this frictional resistance?
“9781107006539AR” — 2013/12/3 — 19:53 — page 103 — #117