Solutions to Problems

Chapter 1 1.1 No, because some cubes of this size would contain no biotite at all, while others would contain as much as, say, 20 percent biotite. 1.2 Yes. Such a material might, for example, display a gradual change in density from one point to another. Natural examples would be vertical col• umns of atmosphere or water or granite, which increase in density downwards on account of the weight of the overlying material. No natural material provides a perfect example, however, because no natural material is perfectly continuous on all scales. 1.3 The answer depends on the scale on which homo• geneity is judged. Suppose the discontinuities are cracks along grain boundaries in a granite. On scales smaller than the grain size, the rock is inhomogeneous with respect to crack frequency, and the answer to the question is no. On scales much larger than the grain size, all samples of the rock might contain practically the same number of cracks, so that the rock is approximately homogene• ous with respect to crack frequency, and the answer to the question is yes. 1.4 It depends on whether the fluid is considered to be part of the material or not. In either case, there will be abrupt and essentially discontinuous changes in strength and other material properties across pore boundaries. In this sense a rock with fluid filled pores is as discontinuous as a rock with dry pores. 1.5 In each picture structure is present on the scale of the labeled material constituents, as pointed out in the figure caption. But structure is also shown on a smaller scale within regions occupied by some of the labeled constituents. Thus in Figure 1.2c there is structure within the regions of folded sediment, defined by the configuration of bedding planes; in

275 Stress and Strain

Figure 1.2e there is structure within the mica films, defined by the arrangement of the boundaries of the mica grains; and in Figure 1.2g there is atomic structure within the regions of normal crystal, de• fined by the arrangement of atoms. 1.6 It cannot be homogeneous on every scale, but it can be homogeneous on a scale large compared with the scale of the discontinuities by which the structure is defined. The cracked granite mentioned in Problem 1.3 is an example, if we consider samples of granite that are very large compared with the grain size.

1. 7 If a depositional basin is gradually moved closer to the earth's rotational pole, younger faunas may show increasingly cold-water characteristics.

Chapter 2 2.1 An infinite number. , , 2.2 x Y R 3.4 2.0 S 4.1 1.7 2.3 y' 3

2.4 y'

276 Solutions to problems

~ ~ ~ To sum vectors P and Q, one vector (P) is drawn as in the previous problem, with its tail at the origin. The other vector (c:b is then drawn with the same length and orientation as in the previous problem, but with its tail at the head of the first vector. To ~ ~ -* construct a vector (P + Q) equal to the sum of P and Q a new vector is constructed with its tal! at the origin and its head at the head of vector Q.

2.5 y'

~ ~ . To construct the vector (P - Q), we proce~ III similar fashion except that now we add to P the ~ ~ vector -Q. P is constructed as above. Then a ~ vector -Q is drawn, with its tail at the head of P, -* with length and orien~tion equal to Q but with direction opposite to Q. The vector then con• ~ structed from the origin to the head of -Q is the ~ ~ vector (P - Q). It is identical in length, orientation, -* ~ and direction to (P - Q) in the xy coordinate sys- tem (Figure 2.2); this reflects the fact that the distance between two particles is independent of the choice of coordinate system. 2.6 See drawing at top of page 278. 2.7 Nothing. The velocity field at time t tells us nothing about the velocities at times between t and (t + 1) yr. Without this information, we cannot predict how far each grain will have moved by the end of a year.

Chapter 3 3.1 In Figure 3.3 the scale for the displacement vectors (the relation between the length of an arrow on the paper and the magnitude of the displacement rep• resented) is the same as the map scale shown on the

277 Stress and Strain

Y

\ \ \ \ \ \ \ Va' \ \ Problem 2.6 \ \ \ \ \ \

~------X o 2 cm/yr I I Scale for vectors

coordinate axes. In Figure 3.4, however, the scale for the displacement vectors is not equal to the map scale, so a separate scale for the displacement vectors has to be specified. If drawn at the map scale, the vectors in Figure 3.4 would be invisible.

3.2 The drawing below shows one of the displacement vectors from Figure 3.3. From this drawing it can be seen that the displacement of any particle in• creases its x coordinate by 0.44 and decreases its

Y

3

iYi = -0.87 2 b ~ XfYf

a = 0.44

278 Solutions to problems

Y coordinate by 0.87. Equations 3.2 are therefore

Xf = Xi + 0.44,

Yf = Yi - 0.87. 3.3 The reference frame is fixed to the footwall side of the shear zone. This is indicated by the fact that the displacement vectors on the footwall block (dots) are null vectors (i.e., they have zero magni• tude). If the same relative movement of the blocks is represented by displacement vectors in a refer• ence frame fixed to the hanging wall block, the displacement field would have the appearance of Figure 3.5 turned upside down. 3.4 The thermal state of a rock body is more like a state of stress, because the temperature at any instant is specified by a number that is independent of prior temperatures. If, however, we say that a rock is currently 1000 degrees colder than when it crystallized, this thermal quantity is more akin to a state of strain. 3.5 It is unsound because a state of strain in such cases compares two configurations separated by a long time interval .1 t. The stress quantity responsible for the strain is, in general, neither the state of stress at the beginning of the interval nor at the end, but instead the infinite number of states of stress corresponding to the infinite number of instants in the interval .1 t. In general, the state of stress will be different at different instants, so no single state of stress can be correlated with a state of strain. (The exception to this is the situation where the strain is small and takes place instantane• ously, discussed in Chapter 25).

Chapter 4 4.1 The quantity analogous to the position is the instantaneous vertical force on the quartz grain. At the moment a grain is first deposited, the instantaneous force will be related to the depth of the water. The quantity analogous to displacement is the' change in vertical force between any two instants in the history of burial. The change in

279 Stress and Strain

vertical force will depend mainly on the thickness of sediments laid over the grain during the interval of time considered. Between any particular times, there will be a whole sequence of different vertical forces exerted on the grain, and this could be called the vertical force path, analogous to the path describing the motion of a particle. The force path could be a simple sequence of increasing force, or it could involve intervals of decreasing force if deposition were interrupted by erosion. The full force history would be expressed by what we could call a dated vertical force path (e.g., a graph of ver• tical force plotted against time). 4.2 See drawing below. 4.3 .See drawing below. 4.4 See drawing at top of page 281.

Chapter 5 5.1 See drawing at top of page 28l.

5.2 The constructi

~""" / \

\ ~

280 Solutions to problems

o -+ FBA .------P (b)

o 500 kg I ! I ! ! I (a) Scale for force vectors Problem 4.4 Problem 5.1

(compressive). The sign of Fs is negative. (Fs is the shearing force exerted on the material below plane P by the material above plane P. This force, together with the equal and opposite shearing force exerted on the material above P by the material below P, make a clockwise, and thus negative pair.)

-+~: F" : p ~-+ F,

o 5000 10,000 kg [ ! I I I I I I I I I Scale for force vectors

5.3 An acceleration of 1 cm/yr/0.5 Myr is 1cm 1 yr X 500,000 yr or 1cm 3.15 X 107 sec X 500,000(3.15 X 107 sec) or 2.0 X 10- 21 cm/sec2 . A cubic meter of quartzite (p = 2.6) has mass

m = p V = 2.6 X 1003 = 2.6 X 106 g.

281 Stress and Strain

The inertial force necessary to accelerate the cube is F = rna = 2.6 X 106 g X 2.0 X 10- 21 cm/sec2 = 5.2 X 10-15 dyn. This force is insignificant compared, for example, to the gravitational force of 2.5 X 109 dyn previ• ously calculated for the same cube of quartzite. 5.4 By Equation 5.2 F=pgHA = 2.85 X 980 X (30 X 105 ) X (3 X 1017 ) = 2.5 X 1027 dyn. 5.5 If the density of the whole column is 0.5 percent higher than 2.6, then by Equation 5.2 the force at each depth would be 0.5 percent higher than that shown in Figure 5.2, a negligible difference. If the density increases gradually, from 2.600 at the sur• face to 2.613 at 5 km, then the difference between the new calculated curve and the curve in Figure 5.2 would be even smaller. 5.6 With open joints along the sides of the cube, it can safely be assumed that the vertical surface forces acting across the sides of the cube are zero. The total force on the base -of the cube can then be calculated simply as the sum of the weight of the cube and the weight of the atmosphere above it. If open joints are absent, however, there may also exist vertical shearing forces on the sides of the cube, and these would contribute to the total force on the base of the cube. Calculating this force would be impossible unless the shearing forces are specified. 5.7 Fn = F cos e, Fs = F sin e.

Chapter 6 6.1 The force calculated in Problem 5.4 was 2.5 X 1027 dyn. The area is 3 X 107 km2 or 3 X 1017 cm2 .

282 Solutions to problems

The mean stress a is therefore Total force _ 2.5 X 1027 dyn Total area - 3 X 1017 cm2

= 8.33 X 109 dyn/cm2, (or 8.33 kbar, 833 MPa, or 0.833 GPa). 6.2 Neglecting atmospheric pressure, the stress across the vertical faces of the cube is zero. This is so because the force across these faces is zero (i.e., the kilogram weight pushes only on the top of the cube, not on the sides). The stress is therefore o dyn = 0 dyn cm-2. 1 cm 2 If atmospheric pressure (assumed to be exactly 1 atm in magnitude) is taken into account, then there is a horizontal force on each vertical face equal to the product of the area of the face and the atmospheric pressure: Force = area X pressure = 1 cm2 X 1 atm = 1 atm cm2 or 1.01 X 106 dyn. The stress on each face is found by dividing this force by the area of the face:

Stress = force = 1.01 X 106 dyn area 1 cm2

= 1.01 X 106 dyn cm- 2, or 1 atm. In other words, the stress on each vertical face is exactly equal to the pressure of the atmosphere adjacent to it. A stress of 1 atm can also be written as 1.01 bar, 1.01 X 105 Pa, or 0.101 MPa. 6.3

I d = 650 a ctn = 4.23 kbar = 0.423 GPa 0 5 10 kbar p-.J4---:------'--- I I I I I I I ~ 0.5 P ds = 9.06 kbar = 0.906 GPa 0 1.0 GPa

an = a cos e as = a sin e

283 Stress and Strain

The equations have the same form as those obtained in Problem 5.7 because the stress vector and the force vector are resolved in exactly the same way. 6.4 I

I d --> I an = -0.423 GPa 0 5 10 kbar p--~------~-~ LI LI~~I~I~I~~I~I P d, = -0.906 GPa 0 0.5 1.0 GPa

In Problem 6.3 the sign of the normal component was positive (compression) and the sign of the shear component was also positive (the force exerted by the material above P on the material below P is directed to the left, so that this stress vector, together with its equal and opposite vector (not shown) representing the force per area exerted by material below P on that above, make an anticlock• wise pair). In the sketch above, both components have negative signs, and it is seen that this simply reverses the direction of d. 6.5 According to Figure 5.2, the force acting vertically. on each square meter of a horizontal plane at 5-km

Lithostatic pressure

2 0 E ."'- "S'" :::T .!: 3' a. 0 !!

2

Lithostatic pressure

284 Solutions to problems

depth is 1.29 X 107 kg. The area of a square meter is 104 cm2 . The stress at 5-km depth is therefore

Stress = force _= 1.29 X 107 kg = 1290 kg/cm2 area 104cm2

= 1.26 kbar or 126 MPa. Since force varies linearly with depth in Figure 5.2, stress will also vary linearly with depth, and the stress-versus-depth curve can be constructed simply by drawing a straight line between a stress of 1.26 kbar at 5 km and a stress of 0 kbar at 0 km, as above. Notice that the slope of this line indicates a vertical stress gradient of about 1 kbar/4km or 100 MPa/4km, useful figures to remember for rough calculations of the lithostatic pressure at various depths in the crust.

Chapter 7 7.1

o 1 kbar II" III " II In (b) the principal stresses are equal and the stress ellipse is a circle; all radii are parallel to principal directions. In (c) 02 is zero, and the stress ellipse degenerates into a line. This is the ellipse repre• senting a two-dimensional state of uniaxial compression. 7.2 Using the construction described for Figure 7.3, a vertical stress vector is found to operate upon a plane dipping 19 degrees to the east:

W------i I----o---E

285 Stress and Strain

7.3

W--+-+-----~~----4_4_~-E

o 2 kbar I I I Scale for vectors

Concentric circles of radii 2 kbar and 1.6 kbar are drawn. The fault plane and its normal (pole) are constructed. Points m and n are identified as in Figure 7.6, and a point P is found on the stress ellipse. The total stress vector is then constructed by drawing a line from P to the center of the circles. The magnitudes are about: a = 1.72 kbar, an = 1.71 kbar, as = -0.18 kbar. 7.4 A triaxial stress ellipsoid is a bona fide, three• dimensional ellipsoid like the example shown in Figure 7.7, but the "ellipsoid" for biaxial and uniaxial stress are degenerate special cases. For biaxial stress, the "ellipsoid" is an ellipse, and for uniaxial stress the "ellipsoid" is a straight line. 7.5 (a) No; (b) No; (c) Yes; (d) Yes; (e) No; (f) Yes.

Chapter 8 8.1 In (a) the forces operating across the horizontal and vertical faces of the cube are, respectively, represented by arrows six times and four times as long as the cube edge. These arrows represent force vectors of 6 X 106 dyn and 4 X 106 dyn. To find the stresses, these forces must be divided by 4 cm2 , the area of each cube face. This yields stresses of 1.5 X 106 dyn/cm2 and 1.0 X 106 dyn/cm2 acting across the horizontal and vertical faces of the cube, and these are represented by arrows of appropriate length as in diagram (a). When one considers only the left half of the cube,

286 Solutions to problems

..... F ..... F ..... a ...... ~a a :- ...... • ..... F ..... F a ...... F F (b)

Scale for force vectors' ..... o 1 2 3 4 X 106 dyn F I I I I I o 1 2 3 4 X 106 dyn!cm2 (a) Scale for stress vectors

as in diagram (b) below, the forces across its hori• zontal surfaces are just half as large as those across the horizontal faces of the whole cube. These forces are accordingly represented by arrows half as long as those in (a). Notice, however, that the stress vectors across the horizontal surfaces of the half-cube are exactly the same as the stress vectors across the horizontal faces of (a), because the areas as well as the forces have been reduced by a factor of 2. 8.2 By Equation 5.2 the force acting across the bottom of the cube due to its weight is F=pgHA or F=pgV. For a cube 1 cm on a side, with density of, say, 2.6, this force is (2.6) (980) (1) = 2548 dyn.

The area of the bottom of the cube is 1 cm 2 , so the stress is

Force = 2548 = 2548 dyn/cm2 Area 1 = 2.548 X 10-6 kbar. This extra stress due to body forces is thus only 0.00025 percent of the total stress (1 kbar), and is therefore negligible. 8.3 (a) (1,0); (b) (0, 1); (c) (0.707, 0.707).

287 Stress and Strain

8.4

Plane P

.+_-'---;-____{3'----'---"'r- - --x Acos{3 :8 ~----Am ------'! aD (the normal to plane P in the figure) divides triangle aBC into two similar triangles (OBD and ODC). Angles OCD and OBD are therefore equal to 0' and {3, respectively. It can thus be seen that the area of the vertical side of the triangle is A cosO', or AI. 8.5 The diagram below shows that point p, as determined by the construction of Figure 7.6, must have coordinates Ox = 01 cosO' = 101, and Oy = 02 cos {3 = m 02. If p has these coordinates, then a line drawn from the origin to point p will represent the magnitude and direction of the total stress vector on plane P, as claimed without proof in the last chapter.

Chapter 9 9.1 The stress ellipse can be used to provide more com• plete information about the stress at a point, be• cause it can be drawn oriented in such a way as to record the principal stress directions as well as the magnitudes. Problem 8.5 Problem 9.2

o 1.0 kbar I..u..u..w..u.. Pole to P

as

x, al direction 1.0 kbar 0.5

288 Solutions to problems

9.3 Point P is labeled with a capital letter P in Figure 9.2 to remind us that a point in a Mohr diagram repre• sents a plane in the object under stress.

9.4 /Normal toP /Normal to Q / 0 / "~OP =45 = 600 U1 direction '-.",~Q Plane P Plane Q

Us, MPa

-4

For P: un = 6.35 MPa, Us = 3.35 MPa For Q: un = 4.75 MPa, Us = 2.90 MPa

The principal mistake made in solving this kind of problem is to use the angle between the plane and the al direction as () instead of the angle between the pole to the plane and the al direction. It pays to sketch the lines and planes involved first and then draw the Mohr diagram. 9.5 The an and as axes for a Mohr diagram are con• structed and calibrated in kilo bars. Points P and Q are then plotted having coordinates equal to the given normal and shear stresses. Since planes P and Q are perpendicular to each other, points P and Q representing these planes on the Mohr circle must be at opposite ends of a diameter of the circle. This

Us (kbar)

U2 direction +1 Plane P

Un (kbar) U1 direction ------~---r-----

Plane Q \ -1 \ \ \lQ

289 Stress and Strain

diameter is drawn in, and the circle can then be constructed. The magnitudes of a1 and a2 can then immediately be read off the an axis, at the points where it is intersected by the circle. a1 is about 2.9 kbar and a2 is about 0.5 kbar. The orientations of a1 and a2 in space cannot be determined from the data given, but their orienta• tions relative to planes P and Q can be determined as follows. A line drawn in any direction (right diagram above) is labeled a1. Then the angle 28p is measured in the Mohr diagram, divided in half, and plotted in the same sense (counterclockwise) from the a1 direction (lower diagram). This establishes the orientation of the normal to P and hence of planes P and Q relative to the directions of a1 and a2. In tqis example the a1 direction is inclined at about 57 and 33 degrees to the planes P and Q, respec• tively, in such a way that the shear stress is positive on P and negative on Q. 9.6

as for plane P ~-'-I-....

as for plane Q t----\-'~----=,.,

an for plane P

The shear stress on any two perpendicular planes P and Q is always equal in magnitude but opposite in sense (or sign) because perpendicular planes are always represented by two points at opposite ends of a diameter of the Mohr circle. Such points always have as coordinates of equal absolute magni• tude but opposite signs. The sum of the normal stresses on perpendicular planes is a constant or "invariant" quantity because the points representing the two normal stresses on the an axis are always equidistant from the center of the circle. Thus an for plane P + an for plane Q 2 the mean stress, and therefore

290 Solutions to problems

an for P.+ an for Q = al + a2 for any choice of P and Q so long as they are per• pendicular .

9.7 o. (lb/in2) 3

2

-+-+'---t==---F---+'-~--On (lb/in2)

9.8 0,

IP

The angle e shown in the Mohr diagram above is the angle between the normal to plane P and the al direction. But the angle e is also the angle between the stress vector (/ on plane P and the normal to P, as explained in the text of this chapter. Therefore (/ is always parallel to the al direction. This is not true for more general states of ,stress.

9.9 4 3 3 2 2 1 234 1 1

-1 -1 °mean -2 -2 -3 -3 On (kbar) -4 + t t Total stress Deviatoric stress + Mean stress

291 Stress and Strain

Chapter 10 10.1 z

y

x

10.2 z

y

x

From the figure it can be seen that ax = (cosa)a, ay = (cosJ3)a, az = (cos'Y)a, .or ax = la, ay = rna, az = na. 10.3 axxCD axyCD axzCD ayxEB ayy 8 ayz 8 azxCD azy 8 azz CD 10.4 2.5 kbar ~.5kb~ 1.5 krnrrJ 2.5 kbar -3.0 kbar -1.0 kbar 1.5 kbar -1.0 kbar 6.0 kbar

292 Solutions to problems

10.5. Conventions used earlier in book Tensor conventions Compressive stresses Compressive stresses positive negative Anticlockwise shear Shear stresses positive stresses positive where shear com• ponent and outward• directed normal to plane are both posi• tive, or both negative. Shear stresses on perpen• Shear stresses on perpen• dicular planes have dicular planes have opposite signs same signs 10.6. See drawing below. 10.7. If the cube is infinitesimally small, then the stress vectors acting on the left face of the cube will be equal and opposite to those acting on the right face of the cube, as shown belqw. As components of the stress tensor, however, these new vectors have exactly the same magnitudes and signs as the vectors on the right face of the cube. The tensor

-080 kbar 0 0 ] [ -6 kbar 1 kbar _ 1 kbar -4 kbar

The x2 X3 plane is a principal plane of stress.

Problem 10.6 X1

~ -5 kbar -2 kbar 1 kbar] J!l -2 kbar -8 kbar 3kbar _""" t1 kbar 3 kbar -6 kbar 4 2

Scale for stress vectors? 2 4 (both diagrams)

X1

293 Stress and Strain

-+ Oyy

-y

x

<,!omponent ayy , for example, thus represents both the force on the right-hand y face pushing the cube to the left and the equal force on the left-hand y face pushing the cube to the right.

10.8 e12 e13

ejj represents e21 e22 e23 r'e31 e32 e33 e12 e13 ek I likewise, e23 represents e21 e22 r'e31 e32 e33

elk bk represents ell bl + el2b2 + el3 b3 which, in turn expands to: e11 bl + e12 b2 + e13 b3 e21 bl + e22 b2 + e23 b3 e31 bl + e32 b2 + e33 b3·

Chapter 11 11.1 Because vectors PI,~ ~all, ~a21, and ~a31 are stress vectors. To obtain the corresponding forces, which do balance, one has to multiply each stress by the

294 Solutions to problems

area it acts upon, as per Equation 11.1. The dia• gram is constructed to scale for the situation where the area of P is unity and the direction cosines of the normal to P (Ii) are 0.707,0.500, and 0.500. 11.2 Since ex, {3, and 'Yare 45, 60, and 60 degrees, the direction cosines Ii are 0.707, 0.500, and 0.500, respectively. Equations 11.2 can then be written out, PI = (-0.4) (0.707) + (-0.4) (0.5) + (-0.35) (0.5), P2 = (-0.4) (0.707) + (0.45) (0.5) + (-0.5) (0.5), P3 = (-0.35) (0.707) + (-0.5) (0.5) + (-0.2) (0.5), and the Pi are evaluated as follows: PI = -0.66, P2 = -0.31, P3 = -0.60. This is in fact the case shown in Figure 11.3. The negative signs of the Pi magnitudes can be inter• preted as corresponding to the fact that each vector Pi points in the direction of a negative coordinate axis. Thus in Figure 11.3 the component PI points in the negative Xl direction. P2 and P3 are shown but not labeled. (The magnitudes of the vectors that are shown in Figure 11.3 can be com• pared by measuring their lengths with a ruler because the diagram is an isometric projection, with equal distortion of lengths parallel to all three coordinate axes.) 11.3 Since the components of pare known from the last problem, the absolute magnitude of the total stress, Ipl, can be found by Equation 10.2: Ipi = (P1 2 + P22 + P32 )1/2 = [(-0.66)2 + (-0.31)2 + (-0.60)2] 1/2 = 0.94 kbar.

295 Stress and Strain

11.4 (a) Vi' = lij Vj. Writing this for i = 3, we get

V3' = 13j Vj,

and expanding for j = 1, 2, 3, and summing,

V3 ' = 131 V1 + 132 V2 + 133 V3·

This is the formula for the new V3 ' component as a function of the old components Vi and the direc• tion cosines 1ij. (b) Where Vi = 1,1,0

and lij = [:::

131 the formula gives

V3 ' = (1) (1) + (0) (1) + (0) (0) = 1. (c) The old coordinate axes are drawn in solid lines, in the standard orientation used throughout this book (but any orientation would do), and the new axes are then drawn, in dashed lines, after examining the values of the lij. Thus for example, III = 0, 112 = 0, and 113 = -1. This means that the positive end of the new Xl' axis makes an angle of cos- I (0) or 90 degrees with the positive end of the old Xl axis, an angle of cos- 1 (0) or 90 degrees with the positive end of the old x2 axis, and an angle of cos-I(-I) or 180 degrees with the posi• tive end of the old x3 axis. From this information the correct position of the new Xl' axis can be drawn in, as shown in th.e diagram below. The new X2' and X3' axes are located using a similar pro• cedure, and it turns out that the particular lij speci• fied here describe a simple relationship between the old and new axes. The new axes and the old ones are related by a 90-degree rotation around the x2 = X2' axis. ~ The vector V is drawn in, and it can be seen that if its components Vi relative to the old axes are (1,1,0), its components V/ relative to the new axes must be (0, 1, 1). In particular, its component

296 Solutions to problems

~ __-=-- __...... (1.1.0) = 1.1 (0.1.1)= Vi I I I I +x, I I I I I I I+x,

V3 = 0 has been transformed with change of axes into a component V3' = 1, as obtained in (b) above using formula 11.8. Notice that the vector V is itself unchanged by the change of axes, but its components have changed from (1, 1,0) to (0, 1, 1).

11.5 UU' = hpljqupq. Writing this for i = 2 and j = 3, we get

a23' = 12p13 q a pq . Then expanding for p = 1, 2, 3 and summing, we get

U23' = 121 13q U1q + 12213qu2q + 123 13q u 3q. Finally, expanding each of these terms for q = 1, 2, 3 and summing, we get the fully expanded formula:

U23' = 121131 U11 + 121 132U12 + 121 133U13

+ 122131 U21 + 122132U22 + 122133 u23

+ 123131 u31 + 123132u 32 + 123 133 u33·

Chapter 12 12.1 It would be a circle, because Uyy and u zz are exactly equal, as shown in Figure 12.5.

297 Stress and Strain

12.2 The "level of isotropic points" would rise from 565 m below the surface to about 283 m below the surface. This can be read off Figure 12.5 by constructing a new line for Oyy that is parallel to the old line but passes through the point for 50 bar at O-km depth. This new Oyy line crosses the ozz line at about 283m. 12.3 The arrangement of principal stress trajectories and the axial ratios of the stress ellipses should look exactly like those in Figure 12.3. The North• South vertical plane will be the plane of 01 and 03, with 0, everywhere vertical and 03 everywhere horizontal.

A sketch of the 01 trajectories after faulting can be prepared, as shown by drawing a short 01 line at each intersection of the grid formed by planes of maximum shear stress in Figure 12.8. To be parallel to the 01 direction locally, each line must bisect the right angle between the planes of maximum shear stress and be oriented such as to give the correct sense of shear on the shear planes.

Chapter 13 13.1 In Figures 13.1b, c and 13.3 the x direction is fixed to lines of particles, and the y direction is simply defined as perpendicular to x at all times. The x direction corresponds throughout the deformation to a particular "material line," but the y direction does not. Where a rock body is being deformed, as in Fig• ures 13.1b, c and 13.3, it is generally only possible to fix one reference direction parallel to a material 298 Solutions to problems

line, because material lines in initially perpendicular directions are generally not perpendicular after deformation. For example, the material line per• pendicular to x at time 1 in Figure 13.3 is likely to rotate away from this position as folding pro• ceeds and to lie at a low angle to x by time 4. 13.2 Yes. This is probably a common situation at shallower levels in the crust, or wherever brittle processes contribute to the deformation on a microscopic scale. Average stresses over large areas may change smoothly, while local stresses on a granular or subgranular scale are fluctuating violently as microfracturing occurs. 13.3

I

2 3 4 Time The simplest way to determine the magnitude of uyx for each time is to draw a circle around the stress ellipse (as in Figure 7.3), observe where the y axis intersects this circle (point p' in Figure 7.3), find from this the outer end of the total stress vector (point P in Figure 7.3), and finally estimate the magnitude of the shear component of this vector. The resulting graph should look something like the one shown.

Chapter 14 14.1 At this scale, the whole shell would be represented by two points on the map, one in its depositional site and one in the mountain belt. The displace• ment field for the shell would condense into a single arrow connecting these two points. 14.2 Drawing this displacement field reminds one that displacement vectors must be drawn as straight 299 Stress and Strain

lines connecting the initial and final positions of particles. The displacement vectors thus pass through the interior of the earth and are not the same as the curved, tangential paths that the particles actually followed. 14.3

This can be shown by considering a picture like Figure 14.4d. Suppose that this figure were redrawn as shown, with different elongations for each of the initially horizontal lines. It is clear that this would introduce nonparallelism of some sets of initially parallel lines (e.g., the initially vertical set in Fig- ure 14.4). Hence any deformation in which lines of the same orientation suffer different elongations is an inhomogeneous deformation. 14.4

Line m was initially perpendicular to line h. After deformation m is at 60 degrees to h. m has thus

300 Solutions to problems

been rotated through 30-degrees from its initial position relative to h, and this is the angular shear I/J of line h. The shear strain of h is "I = tan I/J = tan 300 = 0.577. The sign of the shear strain is positive in this case, because the sense of shear of lines parallel to h is anticlockwise. (The sign of the shear strain for line m, on the other hand, is negative.)

14.5 8

0.8 -0.2 0.64 -0.223

14.6 € 8 = 1 + € A =82 "I = tan I/J

For Figure 14.4c

Originally horizontal 0.21 12.1 1.46 0.19 31 0 0.60 grid lines Originally vertical -0.02 0.98 0.96 -0.20 -310 -0.60 grid lines

For Figure 14.4d

Originally horizontal -0.39 0.61 0.37 -0.49 o grid lines Originally vertical 0.70 1.70 2.89 0.53 o grid lines

14.7 € s € "I 0.0 1.0 1.0 0.0 -1.1 To determine I/J and "I, it is necessary to find the position in the deformed state (Figure 14.5c) of

301 Stress and Strain

some line of particles perpendicular to AB in the undeformed state (Figure 14.5a). A convenient line is the line (CD in the drawing above) which would be perpendicular to AB and half-way down the specimen in Figure 14.5a. 14.8 Line C

\{t = 45° / tP 7 (a) V /

\{t = 45°lP / (b) (a) The angle I/J is 45 degrees and the shear strain of line cis -1.0, as shown in the figures. (b) In this case the final angle I/J is again 45 degrees and the shear strain of line c is thus again -1.0. The displacements, displacement history, and final configuration of the deck are different from those in (a), but the shear strain of line c is identical to that in (a). 14.9 See drawing at top of page 303.

Chapter 15 15.1 The principal strains can be determined by measur• ing the principal diameters of the strain ellipsoid in Figure 15.1 and dividing each of these numbers 302 Solutions to problems

P I I ~B

~A\ Problem 14.9 o OP lOA DB 1 00 :·1 'ITA 1= 1 'lT81 and 11'A 1= 11'8 1

by the diameter of the original circle. Principal strains obtained this way are 1.20 and 0.83, ex• pressed as stretches, and we can designate them 8 1 and 83; resepectively, assuming that the intermediate principal strain direction is perpendicular to the plane of the paper. It is not possible to find the traces of the circular sections from the information given because the principal strain perpendicular to the page is un• known. For each possible value of this principal strain the circular section will make a different angle with the 8 1 direction. If the strain is assumed to be plane strain (82 = 1), then the circular sec• tions will make angles of about 40 degrees with the 8 1 direction.

15.2 1'e = 3 1'e = 2 1'e = 4 1'e = 1

Line C

Lines in the undeformed deck which will become parallel to the S, direction for various shears of the cards (1'e) The card deck exercise will show that different lines of particles lie along the 8 1 direction for different shears of the deck. The position of some of these lines of particles in the undeformed deck are shown in the figure. These lines are drawn assuming that the top card will be moved to the right in performing the deformation-i.e., that the 303 Stress and Strain

ends of the deck will be rotated clockwise. Notice that while the ends of the deck are rotating clock• wise with respect to the plane of the cards, the instantaneous 8 1 direction is rotating anticlockwise with respect to its material position the moment before. This phenomenon is discussed further in Chapter 24.

15.3 S1 S1 S1

(a) .. (b) (c)

15.4 Strain + Translation + Rotation Rotation + Translation + Strain (a -> b) (b -> c) (c -> d) (a->b) (b->c) (c->d)

b c @Jf o~

Translation + Rotation + Strain (a -> b) (b -> c) (c -> d)

15.5 Diagrams (a) show a two-dimensional view of a simple shear deformation of a cube, as portrayed in three dimensions in Figure 15.5 of this chapter. Diagrams (b) show that the same net result-i.e., the same deformation-as in (a) can be obtained by a pure shear, giving the center picture, followed by a rigid-body rotation. Thus simple shear is

304 Solutions to problems

y

(a)

v v

'------'---x """"'------x (b) equivalent to pure shear plus rotation, and the same principle applies to any rotational deformation: any rotational deformation can be resolved into a pure strain and a pure rotation, plus in most cases a pure translation. (In the figure the translational component has been suppressed or removed by choosing reference axes with an origin fixed to a particle of the body.)

15.6

(c) In order to view the deformation of Figure 15.2 as a pure strain, as shown in diagrams (a); it is neces• sary to choose reference axes that are parallel to the principal directions of strain in the deformed and undeformed states (this removes the rotational com-

305 Stress and Strain

ponent of the deformation), and to let the origin of these axes travel with a particle of the body• for example, the center particle (this removes the translational component of the deformation). To view the deformation as a pure strain plus a rotation, as in diagrams (b), one chooses reference axes that are parallel to the principal directions of strain in the undeformed state but that are not parallel to the principal directions of strain in the deformed state, with an origin that is again fixed to a particle of the body. To view the deformation as a pure strain plus a translation, as in diagrams (c), one chooses axes that are parallel to the principal directions of strain in both states, but that have an origin at different particles of the body in the two states. In diagrams (c) this latter is done by placing the origin at the center particle in the undeformed body and placing it somewhere else in the deformed body.

Chapter 16 16.1 There are an infinite number of different shear strains associated with a line L in three dimensions because there are an infinite number of other lines normal to L in the undeformed state. L will have undergone a different shear strain with respect to each of these different reference lines in the de• formed state. One way to get around this difficulty is to arbitrarily select the reference line for which the shear strain of L is maximum.

16.2 (a) The lines of maximum shear strain are taken as bisecting the right angles between €1 and €2 direc• tions, as in all cases of infinitesimal strain. (b) The magnitudes of the maximum shear strains are±(€l -€2)or±0.002.

16.3 From the Mohr diagram opposite the coordinates of points P and Q can be read off to give the normal and half-shear strains of lines P and Q at 30 degrees to the €1 direction. The coordinates are P (0.0005, 0.00087), Q (0.0005, -0.00087). The normal and

306 Solutions to problems

"( pP 2 E, direction .001 ~00 Q

-001 .001

-.001 shear strains are therefore

LineP 0.0005 0.0017 Line Q 0.0005 -0.0017

"(/2

Calibrated orthogonal axes are laid out, and points Rand S are located. A line is drawn from R to S, and its point of intersection with the € axis is found. A circle is drawn with this point as center, and the intersections of the circle with the € axis gives the magnitudes of €1 and €2 as about 0.00090 and -0.00158, respectively. The angle 8R is then read off as 23 degrees measured clockwise from the € axis toward point R on the Mohr diagram. This means that line R itself should be plotted 23 degrees clockwise from the €1 direction. It will be seen that this gives the right (negative) sense of shear on R. Line S is (essentially) perpendicular to Rand has the opposite sense of shear.

307 Stress and Strain

Chapter 17 17.1 No part of a Mohr circle for finite strain can plot on the left side (-A') of the ordinate, because values of A and hence A', can never be negative.

17.2 'Yo , Slope = ~ = 'Yo Slope=~ , X", ,'Y 'Y' 'Yo

X' X2'

'YP' Slope = V = 'YP X2 ' +X1' 2 (a) (b) (c)

(a) For any two lines (P and Q) that are perpendicu• lar to each other in the deformed state, the values of 'Y' will be equal as shown in diagram (a). However, the values of A' for the same two lines will, in gen• eral, be different, and this means that the magnitudes of the shear strains 'Y will also be different, since for each line the shear strain , 'Y=-'Y . A' Statement (a) is therefore not true for finite strain. (b) The two lines at 45 degrees to the Al direction are represented by points P and Q in diagram (b). The magnitudes of the shear strains for these lines are 'Y' lA', or the slopes of lines OP and OQ drawn from the origin through points P and Q. It is evi• dent from diagram (b) that lines P and Q are not lines of maximum shear strain since there are other points (to the left of P and Q on the circle) for which the ratio 'Y'/A' would be higher. Statement (b) is therefore not true for finite strain. (c) In diagram (c), two lines are drawn from the origin that are tangent to the Mohr circle and intersect it at points Rand S. These lines have the maximum slopes possible, and the points Rand S therefore represent the two lines of maximum shear strain. They lie at equal angles e'Y max' to the

308 Solutions to problems

X' direction, where

O,mu' ~ ~ co,-1 t~:: F:].

Statement (c) is therefore true for finite strain.

17.3 NI S2 r--t-'/LineL w Sl = 2.5 S2 = 0.5 Al = 6.25 A2 = 0.25 1?A;Kkt 1-1 At' = 0.16 A2 = 4.0 30 L ___ -' sl "I'

AL = 1.18 "IL = -1.70 At = 0.85 "IL = -1.445 SL = 0.92

€L =-0.08

Elongation = -8 percent (shortening) Shear strain = - 1.5 (clockwise)

17.4

Lineh 0.65 0.577 1.53 0.89 S2 = 0.66 Linem 1.73 -0.577 0.58 -0.33

h Sl = 1.39

Al = 1.92 A2 = 0.49 ~,,; = 10°

309 Stress and Strain

17.5 The two lines for which the elongation is zero will be two lines for which A and A' are 1.0. These lines can be found from the Mohr circle of the last prob• lem, as in the figure below.

Point representing line for which e = 0

-+----+---~-----+_---=~-A'

Point representing line for which e = 0

Chapter 18 18.1 ac __~b~ __~e ____ +d_ abed T T T T x TO. • f I x o 1 2 3 in. o 1 2 in.

Undeformed state Deformed state (a) (b)

x-x I 3 in.

Displacement field (c)

Displacement field, Displacement field, undeformed state deformed state -0.75 -2.25 0 -0.75 -1.5 -2.25 u 0\-1.5/ u f f f f T • I 0 2 3 in. X o in. x

(a) (b)

Coordinate transformation field, Coordinate transformation field, undeformed state deformed state

0 0.25 0.50 0.75 x 023 x T T f T , . . . 0 2 3 in. X o in. x

(c) (d)

310 Solutions to problems

1 18.2 (a) X=-X4 X=4x u = -0.75X u =-3x ax 1 (b) ax =4 ax 4 ax au ~=-3 ax= -0.75 ax

18.3 z (0. 0.1)

o v C 2

Displacement field. deformed state

z 2

Coordinate transformation field. undeformed state

z (0. O. 1) 2 • A B (0.1.1)

(0. o. 0) (0.1.0) =*"------,,---,-V 2

Coordinate transformation field. deformed state

311 Stress and Strain

Z=z €3, S3, i\3 direction u = OX+ OY + OZ Undeformed state v = OX+ OY + OZ w=OX+OY-1Z Displacement Y=y equations u =Ox+Oy+Oz v =Ox+Oy+Oz w = Ox + Oy-1z

Deformed state x = 1X+ OY+OZ Y =OX+ 1Y+OZ 1 z = OX+ OY+"2Z Coordinate transformations X = 1x + Oy + Oz Y = Ox + 1y + Oz x =x axis Z=Ox+Oy+2z

au ilu au 0 0 0 ax ax ax 0 0 ax ay az ax aY az av av av 0 0 0 k k 11-'. 0 0 ax aY az ax ay az aw aw aw az az az 1 0 0 _1 0 0 ax ay az 2 ax ay az "2

au au au ax ax ax 0 0 0 0 0 ax ay az ax av a;- av av ay ay av 0 0 0 E 0 0 ax Tv az ax av az aw aw aw az az az 0 0 -1 0 0 2 ax ay az ax av az

Problem 18.4

18.5 x = IX + OY + OZ

y = OX + 1 Y + ~ Z

z = OX + OY + lZ

X = Ix + Oy + Oz

1 Y = Ox + ly -zz Z = Ox + Oy + lz

312 Solutions to problems

u = OX+ OY+ OZ V=OX+OY+~Z

w = OX+ OY+ OZ

u = Ox + Oy + Oz

1 V = Ox + Oy +"2Z

w = Ox + Oy + Oz

Chapter 19 19.1 z

u = -O.OOlx + Oy + Oz } v = OK - 0.002y + Oz Displacement 0.003 z = OK + Oy + 0.003z equations

au au 1.0 ax ay o ::] [-0.001 [ av av ax av ~az = 0 -0.002 aw aw aw 0 ax av az o Displacement gradient matrix x

19.2 z

u=OK+Oy+Oz } v = OK + Oy + 0.003z Displacement w = OK + Oy - O.OOlz equations

au au au 1.0 ax av az 0 0 av av av ax av az 0 0.003 aw aw aw ~[: 0 -0.001 ax Tv" az Displacement gradient matrix x

313 Stress and Strain

19.3 ell = exx = -0.001 ell =exx = 0 e22 = e yy = 0 en = eyy = -0.002 e33 = ezz 0.003 e33 = ezz =-0.001 e12 = exy 0 e12 = exy 0 =e e2l = eyx 0 e2l yx 0 e13 = exz 0 e13 = exz 0 = e31 = ezx 0 e3l ezx 0 e e23 = eyz 0 e23 = yz 0.0015 = e 11]2 = ezy 0 e32 zy 0.0015

0 [-~001 -0.002 ~ ~.0015l 0 ~.003J [~ 0.0015 -0.001 'J Tensor components of strain Tensor components of strain for deformation of for the deformation Problem 19.1 of Problem 19.2

Chapter 20 20.1 z

/' €2 /' /'

...... -+-...L-----'._-,--,-_ v. €2

IP

314 Solutions to problems

Direction cosines: For P (cos30, cos300, cosO) = (0.866,0.5,0) = (/1' ml, nd For lP (cos 300, cos210, cosO) = (0.5, -0.866,0) = (/2, m2, n2)·

Components of strain: o -0.001 o

By Formula 20.1: Ep = (0.866)2(0.001) + (0.5)2(-0.001) = 0.0005. By Formula 20.2: 'Yp = 2[(0.866) (0.5) (0.001) + (0.5) (-0.866) (-0.001] = 0.0017.

Chapter 21 21.1 Rewriting (21.6) with i and j set equal to 2 and 3, respectively, we have

Then, since

---aU! ------au! aU2 aU2~ aX2 aX3 aX2 aX3 aU3 aU3 ----- aX2 aX3 .

315 Stress and Strain

21.2 Rewriting (21.8) with i, j = 2,2, we have c ---aXk aXk 22 - aX2 aX2' and this expands to ax )2 C22= (~

Likewise, aXk aXk aX1 aX1 aX2 aX2 c23=-- =--+-- aX2 aX3 ax 2 aX3 aX2 ax 3 +--aX3 aX3 aX2 aX3

1 1 21.3 Ell ="2[Cll -8ll ] ="2[Cll -1]

1 1 E31 ="2 [C31 -<')31] = "2C31

E32 = ~ [C32 -<')32] = ~ C32

Chapter 22 22.1 Figure U·I

22.1 (0,0,0) 22.2 (1.45,1.1,0.5) = (h, t2, t3)

316 Solutions to problems

22.3 (0, -1, 1) 22.4 (0,0,0)

Problem 22.2

X3 =l"/~/~",,/X2=X2,€2 Using new axes, the strain in Figure 22.5 is represented '- /' by diagram (ai, or more simply I '- I by diagram (bl I I ~

(a) (bl u, = 1X, + OX2 + OX3 x, = 2X, + OX2 + OX3 U2 = OX, + OX2 + OX3 X2 = OX, + 1X2 + OX3 U3 = OX, + OX2 - 0.5X3 X3 = ox, + OX2 + 0.5X3

u, = 0.5x, + OX2 + OX3 x, = 0.5x, + OX2 + OX3 U2 = Ox, + OX2 + OX3 X2 = OX, + 1X2 + OX3 U3 = Ox, + OX2 - 1X3 X3 = Ox, + OX2 + 2x3

Displacement equations Coordinate transformations

0 0 [au, ] 0 [ax] ax; a~ = [2~ [~ 0 J,J 0 ~.J

0 0 [0.5 [ax;] =[0.5 tax;aXj J = 0 0 ax. 0 0 0 -n J 0 0 ~J Displacement gradients Deformation gradients

0 0 [E;a 0 [c;a= 1 =[r 0 J.375J [~ 0 LJ 0 0 [e;j] 375 0 [C;j] = [ ~.25 =n· 0 -tJ 0 n Finite strain tensors Deformation tensors

317 Stress and Strain

22.3 Til = I ipljq Tpq. Setting i,j = 1, 1, we get Tu' = 11pl1qTpq. Then letting p successively take the values 1, 2, and 3, and summing, we get Tu' = IU /1qT1q + 112/1qT2q + h3hqT3q. Finally, we expand each of the three terms above, letting q successively take the values 1, 2, and 3, and this yields the fully ex• panded formula:

Tu' = lu/uTu + IU /12T12 + IU /13 T 13

+ 112/U T 21 + 112/12T22 + 112/13 T 23

+ 113 /U T 31 + 113 /12 T 32 + 113 /13 T 33' 22.4 Putting appropriate direction cosines and Cij into the above formula we obtain: Cu ' = 0 + o + o

+ 0 + (0.707)2(2.13) + (0.707)2(1.88)

+ 0 + (0.707)2(1.88) + (0.707)2(2.13), or Cn ' = 4. 22.5 For this problem i andj are 2 and 3, respectively, and Formula 22.3 becomes 2(E23 ) cos () = ------'---==--'------[1 + 2(E22 )]-1/2 (1 + 2E33 )1/2 2(0.94) [1 + 2(0.56)]-1/2 [1 + 2(0.56)]1/2 = 0.887 and () = 27.5 degrees. The correctness of this answer can be confirmed by drawing the front face of the deformed cube in Figure 22.5 to scale.

Chapter 23 23.1 It appears to be homogeneous on the scale of regions big enough to be visible in Figure 23.1. Thus originally straight and parallel cleavage traces

318 Solutions to problems

remain straight and parallel within the deformation bands. There are an infinite number of ways to divide a region into two kinds of homogeneous strain fields. Two of these are shown in (b) above.

23.2

I I I I I I +H I I I I I I .I..+.L I I I I I I I I I I I I ./ III I I I t++ ~ TTt -t-t-T• I I I I I I I I I I I I -t-t-T- I I I -r-rt --1---1--.1-- I I I I I I I I I I I TTT (a) I I I ~(b) I I I

Chapter 24 24.1

I Isf I Up I I I I I N T + f-t-j~ I I I I I I --+-5Down I --li- I I

(a)

f I Up

N----$-1-I ~ - $11I "- --+- I s / Down I

(b)

319 Stress and Strain

1 I lSI N--- i-t-t~ --IJt ---- Down

24.2 Yes. An example is shown in Figure 24.5, where after fifteen or sixteen increments of deformation the total elongation of a line is zero.

24.3 0.14 = 2.2 X 10- 16 sec-I. 20 X 3.15 X 1013 sec This strain rate corresponds to the slope of a line drawn on Figure 24.5 from the left end-point to the the right end-point of the curve. Actual strain• rates during the progressive deformation will correspond to the slopes of the curve at various points along it. Some of these strain rates will be greater than 2.2 X 10-16 sec-I, and others will be less. The maximum strain rate occurs at the end of the progressive deformation, when line 4 is ex• tending at a rate of about 1.2 X 10- 15 sec-I. The minimum strain rate occurs at 7 or 8 Myr, and equals zero. 24.4 It would be impossible to know whether the pro• gressive deformation was coaxial or noncoaxial, because a Flinn diagram tells nothing about the relative orientations of principal strain directions and material lines.

24.5.

90°L---+---+-__~ __-r __ -+ __ ~ U5 1.0 1.5 2.0 2.5 3~ Shear of cards

320 Solutions to problems

Chapter 25 25.1 1.0 I ~Granite (E = 500 kbarl

/ Sandstone II bedding Stress / (E = 100 kbarl (kbarl 0.5 I / / I Sandstone 1 bedding / (E = 70 kbarl /

o -0.005 Strain, €

25.2 e13 = 51311 011 + 51312 012 + 51313 013 + 51321 021 + 51322022 + 51323023 + 813310 31 + 81332032 + 51333033, e5 = S51 01 + 55202 + 553 0 3 + 554 0 4 + 555 0 5 +556°6· 25.3 The volume of the cube in Figure 25.3 is 1.0 before deformation. After deformation, its volume is (1 + €3)(1 + €1)(1 + €2), where €2 = €1· If no volume change has occurred, then this product must still equal 1.0, so that (1 + €3) (1 + €1)2 = 1 or (1 + €3) (1 + 2€1 + €1 2) = 1 or 1 + 2fI + €1 2 + €3 + 2€1 €3 + €3 €1 2 = 1. Dropping the terms that are squares or products of the strains (on the ground that such terms are neg• ligible for very small strains) we have 1 + 2€1 + €3 = 1 or €1 1 --=v=- €3 2 25.4 According to Equation 25.11, the strain of a hori• zontalline is

321 Stress and Strain

Setting a33 equal to pgh, and a11 equal to a22, we obtain 1 e11 = E [a11 - v(a11 + pgh)].

If the strain e11 is assumed to be zero this becomes 1 0= E [a11 - v(a11 + pgh)].

Since the product of these two terms is zero, one of them must by itself be zero. It is clearly not the 1 IE term, so that [a11 - v(a11 + pgh)] = 0, or a11 (1 - v) - v pgh = 0, and v a11 = 1 ' v pgh.

Chapter 26 26.1 Young's modulus and the rigidity modulus for solids relate stresses to strains. This is impossible for fluids because for a given stress there are an infinite number of strains, each corresponding to a different time interval. 26.2 (a) o o o (In this simple example the components of the strain-rate tensor are equal to the respective velocity gradients. )

(b) a11 - P = 2(1022 P) (10- 14 sec- 1 )

= 2(1022 dyn sec) (10-14 sec- 1 ) cm2 dyn = 2 X 108 --2 = 0.2 kbar = 200 bar cm a33 - P = 2(1022 P) (-0.5 X 10-14 sec 1) = -100 bar. 322 Solutions to problems

All other components of the viscous stress tensor equal zero: 00 bar o [ [aij-Pij] = 0 o o o (c) al - a3 = au - a33 = (au - P ) - (a33 - p)

= 200 bar - (-100 bar) = 300 bar. (d) Since each viscous stress component is propor• tional to the viscosity, then lowering the viscosity by two orders of magnitude would lower each vis• cous stress component by two orders of magnitude and the difference (au - p) - (a33 - p) would equal only 3 bar. 26.3 Equations 26.7 can be written with only one mater• ial constant because of the assumption of incompres• sibility. If a Hookean solid is assumed incompressible, Poisson's ratio becomes 0.5 at small strains, and Equations 25.11 and 25.12 can also be written with only one material constant (E).

Chapter 27 27.1 Kinetic energy.

27.2

27.3

e3 Strain

By Equation 27.7, Wo = ; Ee32. By Equation 27.4,

a3 = Ee3, or e3 = a3/E. Substituting this for one of the e3 's in Equation 27.7, we get 323 Stress and Strain

or 1 Wo = 2 e303 , the area under the stress-strain curve. 27.4 (a) No. It depends only on the final state of strain. (b) No. The net work done depends only on the final state of strain. By "net work done" we mean the work done on the Hookean material minus the work done by it on the surrounding material. Thus if we shorten a Hookean sample by 0.002 of its original length and then allow it to expand back to a final strain of 0.001, the net work done will be the work done on the sample to shorten it to 0.002/0 minus the work done by the sample on the surroundings as it expands back to 0.001/0. The net work done will be exactly the same as if it were shortened directly to 0.001 of its initial length. (c) Yes. If two Newtonian samples are taken to the same state of strain by different paths in the same time, the one deformed at higher rates of strain will generate more heat.

324 References

References

Bayly, M. B. (1969) Anisotropy of viscosity in suspensions of parallel flakes. Jour. Composite Materials, 3: pp.705- 708. Bayly, M. B. (1970) Viscosity and anisotropy estimates from measurements on chevron folds. Tectonophysics, 9: 459-474. Birch, F. (1966) Compressibility; elastic constants. Hand· book of Physical Constants (S. P. Clark, ed.). Geol. Soc. Amer. Mem., 97. pp.97-173. Bombolakis, E. G. (1968a) Analysis of stress at a point. N. S. F. Advanced Science Seminar in Rock Mechanics for College Teachers of Structural Geology (R. E. Riecker ed.). Air Force Cambridge Research Laboratory, Bed• ford, Massachusetts. pp. 31-58. Bombolakis, E. G. (1968b) Mohr circle analysis of infinitesi• mal strain. N. S. F. Advanced Science Seminar in Rock Mechanics for College Teachers of Structural Geology (R. E. Riecker, ed.), Air Force Cambridge Research Laboratory. Bedford, Massachusetts. pp. 119-134. Bombolakis, E. G. (1968c).Some elastic relations between stress and strain. N. S. F. Advanced Science Seminar in Rock Mechanics for College Teachers of Structural Geology (R. E. Riecker, ed.). Air Force Cambridge Research Laboratory, Bedford, Massachusetts. pp.135- 150. Brace, W. F. (1961) Mohr construction in the analysis of large geologic strain. Geol. Soc. Amer. Bull., 72: 1059-80. Brace, W. F. (1968) Review of Coulomb-Navier fracture criterion. N. S. F. Advanced Science Seminar in Rock Mechanics for College Teachers of Structural Geology (R. E. Riecker, ed.), Air Force Cambridge Research Laboratory, Bedford, Massachusetts. pp. 59-96. Carter, N. L. and Raleigh, C. B. (1969) Principal stress directions from plastic flow in crystals. Geol. Soc. Amer. Bull., 80: 1231-1264. Chapple, W. M. (1968) Finite homogeneous strain and re• lated topics. N. S. F. Advanced Science Seminar in Rock Mechanics for College Teachers of Structural Geology (R. E. Riecker, ed.). Air Force Cambridge Research Laboratory, Bedford, Massachusetts. 317-352. Clark, S. P. (1966) Handbook of Physical Constants. Geol. Soc. Amer. Mem., 97. 325 Stress and Strain

Durelli, A. J., Phillips, E. A., and Tsao, C. H. (1958) Introduction to the Theoretical and Experimental Analysis of Stress and Strain. McGraw-Hill, New York. Durney, D. W., and Ramsay, J. G. (1973) Incremental strains measured by syntectonic crystal growths. Gravity and Tectonics (K. A. DeJong and R. Scholten, eds.). Wiley, New York. pp. 67-96. Elliott, D. (1972) Deformation paths in structural geology. Geol. Soc. Amer. Bull., 83: 2621-2638. Eringen, A. C. (1967) Mechanics of Continua. Wiley, New York. Flinn, D. (1962) On folding during three-dimensional pro• gressive deformation. Geol. Soc. London, Quart. Jour., 118: 385-433. Friedman, M. (1964) Petro fabric techniques for the determination of principal stress directions in rocks. State of Stress in the Earth's Crust (W. R. Judd, ed.). American Elsevier, New York. pp. 451-552. Fung, Y. C. (1969) A First Course in Continuum Mechanics. Prentice-Hall, Englewood Cliffs, N. J. Gay, N. C. (1975) In-situ stress measurements in Southern Africa. Tectonophysics, 29: 447-459. Hafner, W. (1951) Stress distributions and faulting. Geol. Soc. Amer. Bull., 62: 373-398. Handin, J. (1966) Strength and ductility. Handbook of Physical Constants (S. P. Clark, ed.). Geol. Soc. Amer. Mem. 97. pp. 223-289. Hawkins, G. A. (1963) Multilinear Analysis for Students of Engineering and Science. Wiley, New York. Hobbs, B. E. (1971) The analysis of strain in folded layers. Tectonophysics, 11: 329-375. Hobbs, B. E., Means, W. D., and Williams, P. F. (1976) An Outline of Structural Geology. John Wiley, New York. Hodge, P. G. (1970) Continuum Mechanics, an Introductory Text for Engineers. McGraw-Hill, New York. Housner, G. W., and Vreeland, T. Jr. (1966) The Analysis of Stress and Deformation. MacMillan, New York. Howard, J. H. (1968) The use of transformation constants in finite homogeneous strain analysis. Am. J. Sci., 266: 497-506. Hubbert, M. K. (1972) Structural Geology. Hafner, New York. Jaeger, J. C. (1969) Elasticity, Fracture and Flow. Third edition. Methuen, London. Jaeger, J. C., and Cook, N. G. W. (1969) Fundamentals of Rock Mechanics. Methuen, London. Johnson, A. M. (1970) Physical Processes in Geology. Freeman-Cooper, San Francisco.

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Lambe, C. G. (1964) Dynamics. English Universities Press, London. Leeman, E. R. (1964) The measurement of stress in rock: I. The principles of rock stress measurement, II. Bore• hole rock stress-measuring instruments; III. The results of some rock stress investigations. J. S. Afr. Inst. Min. Metall., 65: 45-114, 254-284. Long, R. R. (1961) Mechanics of Solids and Fluids. Prentice-Hall, Englewood Cliffs, N. J. Love, A. E. H. (1944) A Treatise on the Mathematical Theory of Elasticity. Dover, New York. Malvern, L. E. (1969) Introduction to the Mechanics of a Continuous Medium. Prentice-Hall, Englewood Cliffs, N.J. Mase, G. E. (1970) Theory and Problems of Continuum Mechanics. Schaum's Outline Series, McGraw-Hill, New York. Nadai, A. (1950) Theory of Flow and Fracture of Solids. McGraw-Hill, New York. Nye, J. F. (1964) Physical Properties of Crystals. Oxford, London. Obert, L., and Duvall, W. I. (1967) Rock Mechanics and the Design of Structures in Rock. Wiley, New York. Paterson, M. S., and Weiss, L. E. (1961) Symmetry concepts in the structural analysis of deformed rocks. Geol. Soc. Amer. Bull., 72: 841-882. Price, N. J. (1966) Fault and Joint Development in Brittle and Semi-brittle Rock. Pergamon, Oxford. Ragan, D. M. (1973) Structural Geology, an Introduction to Geometrical Techniques, Second edition. Wiley, New York. Ramsay, J. G. (1967) Folding and Fracturing of Rocks. McGraw-Hill, New York. Ramsay, J. G., and Graham, R. H. (1970) Strain variation in shear belts. Can. J. Earth Sci., 7: 786-813. Riecker, R. E., ed. (1968) N. S. F. Advanced Science Semi• nar in Rock Mechanics for College Teachers of Struc• tural Geology. Air Force Cambridge Research Laboratories, Bedford, Massachusetts. Spiegel, M. R. (1959) Theory and Problems of Vector Analysis and an Introduction to Tensor Arfalysis. Schaum's Outline Series. McGraw-Hill, New York. Stanley, R. C. (1972) Mechanical Properties of Solids and Fluids. Butterworths, London. Stippes, M., Wempner, G., Stern, M., and Beckett, R. (1961) An Introduction to the Mechanics of Deformable Bodies. Merrill, Columbus, Ohio. Timoshenko, S., and Goodier, J. N. (1951) Theory of

327 Stress and Strain

Elasticity, Second edition. McGraw-Hill, New York. Truesdell, C., and Toupin R. (1960) The classical field theories. Encyclopedia of Physics, 3, (1) (S. Fiugge, ed.). Springer-Verlag, Berlin. pp.226-793. Turner, F. J., and Weiss, L. E. (1963) Structural Analysis of Metamorphic Tectonites. McGraw-Hill, New York. Weiss, L. E., (1972) The Minor Structures of Deformed Rocks. Springer-Verlag, Heidelberg. Wickham, J. S. (1973) An estimate of strain increments in a naturally deformed carbonate rock. Am. J. Sci., 273: 23-47.

328 Conversion table for length units

Example: 1 meter = 3.281 feet.

Centimeters Inches Feet Meters Kilometers Miles

Centi meters 1.0 0.3937 0.0328 0.01 10- 5 6.215 X 10- 6

Inches 2.540 1.0 0.0833 0.0254 2. 54 X 10---5 1.578X 10- 5

Feet 30.48 12.0 1.0 0.3048 3.048 X 10- 4 1.894 X 10- 4

Meters 100.0 39.37 3.281 1.0 10- 3 6.215 X 10- 4

Kilometers 105 3.937 X 104 3281 103 1.0 0.6215

X 105 5280 1609 1.609 1.0 ~ Miles 1.609 63360 ::s ------<\ ("I) til o· ::s S- C!:) O"' t-:l ~ [ c,." ~ c,." o ;jl ~ I» ::s Q.. 00 -~ s·

Conversion table for mass units

Example: 1 kilogram = 2.205 pounds

Grams Kilograms Pounds Tons (short)

Grams 1.0 10- 3 2.205 X 10- 3 1.103 X 10- 6

Kilograms 103 1.0 2.205 1.102 X 10- 3

Pounds 453.5 0.4536 1.0 5.0 X 10- 4

Tons (short) 9.070 X 105 907.2 2000 1.0 Conversion table for energy units

Example: 1 kilogram meter = 2.342 calories.

Kilograms Ergs Joules meters Foot pounds Calories Kilocalories

Ergs 1.0 10- 7 1.02 X 10- 8 7.37 X 10- 8 2.388 X 10- 8 2.388 X 10- 11

Joules 107 1.0 0.102 0.737 0.2388 2.388 X 10- 4

Kilogram meters 9.807 X 107 9.807 1.0 7.231 2.342 2.342 X 10- 3

Foot pounds 1.356 X 107 1.356 0.1383 1.0 0.3238 3.238 X 10- 4 o Calories 4.187 X 107 4.187 0.427 3.088 1.0 10- 3 g

~ Kilocalories 4.187 X 10 10 4187 426.9 3088 103 1.0 o· ::s S- CA:I a" CI:I ;- ...... '" '" 00 '"t-:) i ~ ::s 0- Conversion table for stress units ~ 1 bar nds per square inch. e. Example: = 14.503 pou ::s

Dynes per Kilograms per Kilobars square centimeter Atmospheres square centimeter Bars (kbar) (dyn/cm 2) (atm) (kg/cm 2)

Bars 1.0 10- 3 106 0.9869 1.0197

Kilobars 103 1.0 109 0.9869 X 103 1.0197 X 103

Dynes per square centimeter 10- 6 10- 9 1.0 0.9869 X 10- 6 1.0197 X 10- 6

Atmospheres 1.0133 1.0133 X 10- 3 1.0133X 106 1.0 1.0333

Kilograms per square centimeter 0.9807 0.9807 X 10- 3 0.9807 X 106 0.9678 1.0

Pounds per square inch 6.895 X 10- 2 6.895 X 10- 5 6.895 X 104 6.805 X 10- 2 7.031 X 10- 2

Pascals 10- 5 10- 8 10 0.9869 X 10- 5 1.0197 X 10- 5

Megapascals 10 10- 2 107 9.869 10.197

G i gapasca Is 104 10 1010 0.9869 X 104 1.0197X 104 . ------_. --_._------_._- -- Conversion table for stress units (cont.)

Pounds per square inch Pascals Megapascals G ipapascals (lb./in. 2 ) (Pa) (MPa) (GPa)

Bars 14.503 105 10- 1 10- 4

Kilobars 14.503 X 103 108 102 10- 1

Dynes per square centimeter 14.503 X 10- 6 10- 1 10- 7 10- 10

Atmospheres 14.695 1.0133 X 105 0.1013 1.0133 X 10- 4

Kilograms per square centimeter 14.223 0.9807 X 105 0.9807 X 10- 1 9.807 X 10- 5

Pounds per square inch 1.0 6.895 X 103 6.895 X 10- 3 6.895 X 10- 6 ~ ::s 5 10- 6 10- 9 Pascals 14.503 X 10- 1.0 ~ !i! Megapascals 145.03 106 1.0 10- 3 g" .... CI:I c,., G igapascals 14.503 X 104 109 103 1.0 ~ c,., ;- . ------'" Index

A

Angular shear, 135

B

Bulk modulus, 248 c Cauchy's deformation tensor, 198-199 Cauchy's formula, 100-103, 111 Coaxial progressive deformation, 225, 237 and fabric, 237 distinguished from rotational deformation, 237 Compliances, 243 Compressibility, 248 Constitutive equations, 239 Configuration, 8, 23 Constriction, 150 Continuity of material, 2 Convention for magnitude of vector, 45-46 for plotting in Mohr diagram, 163-164 for principal stresses, 59, 264 for shearing forces, 39-40 for signs of infinitesimal strain components, 193-194 for sign of stress, 83-84 for signs of stress components, 98, 293 for single suffix notation, 245-246 for stresses on Mohr diagrams, 75-76 for vectors, 12 summation, 97 Conversion tables, 329-333 Coordinate axes, 7, 92, 168 Coordinate system, 7, 8 right-handed, 85 Coordinate transformation field, 171 Coordinate transformations, 170, 173-175

D Deformation coaxial progressive, 225 definitions of, 130, 133, 144-145, 150 distinguished from strain, 144-145, 150 gradient, 172-173, 180 homogeneous, 132,215

335 Index

Deformation (cont.] for infinitesimal elongation, 195 irrotational, 145-146 for infinitesimal normal strain, 154 matrix of gradients, 178-179 for infinitesimal shear strain, 154, 195 noncoaxial progressive, 229-237 for infinitesimal strain tensor, 185-187 one-dimensional, 168-169 for lengths of lines, 213 rotational, 145-146 for normal stress in two dimensions, 72 tensor, 196 for principal strains, 214 three components of, 145, 150 for quadratic elongation of line parallel Deformation tensors, 196-199 to coordinate axis, 212 examples of, 204-211 for stress at depth, 253, 321-322 Dilatation, 138 for shear stress in two dimensions, 72 Dilation, 135, 139, 193 for stress on horizontal planes, 113 Direction cosines, 62-63, 105 for stress on vertical planes, 114, 321- Displacement 322 definition of, 13 for stress vector in three dimensions, distinguished from path, 21 103 equations, 170, 175-177 for transforming tensor components, field, 14, 22 109 gradient, 171-172, 180 for transforming vector cOIIlponents, matrix of gradients, 177-178 109 for weight of a column, 37 Full history indicators, 24, 30 E G Elastic constants, 246-248, 250, 253 Elasticity, 240 Green's deformation tensor, 198-199 Elongation, 133, 139 Energy, 263-273 Engineering shear strain, 182 H ~quations of equilibrium, 65 Equivalent viscosity, 257 Homogeneity Eulerian finite strain tensor, 197-198 of deformation, 132, 215 Extension, 139 of displacement field, 14 of force distribution, 40 of material, 2 F of stress field, 11, 112 of strain field, 215 Finite strain, 226, 236 of velocity field, 11 Finite strain tensors Homogeneous deformation, 215 examples, 204-211 Homogeneous strain, 215 formulae for, 198-199 Hookeian behavior, 240-254 Flattening, 150 Hooke's law Flinn Diagram, 228, 237 for general stress, 243 Fluid, distinction from solid 254 for uniaxial stress, 241 Fluidity, 255 Force body, 36 compressive, 38 inertial, 41 Incremental strain, 224 normal, 39 Indicial notation, 97-98 shear, 39 Infinitesimal strain tensor, 185-186 surface, 36 Invariants tensile, 38 of strain, 193, 195 Formula of stress, 263 for deformation tensors, 199 Irrotational deformation, 145-146 for finite strain tensors, 199 Isotropic points, 116-117 for finite strains in two dimensions, 159 336 Index

K 236 noncoaxial, 229-237 Kronecker delta, 202 Pure rotation, 205 Pure shear, 147, 150 progressive, 226-229 Pure strain, 145, 150, 208 L Pure translation, 205 Lagrangian finite strain tensor, 198-199 Lithostatic pressure, 38 variation with depth, 284-285 Q Quadratic elongation, 135 M

Mohr circle R for infinitesimal strain, 151-159 for finite strain, 159-167 Reciprocal strain ellipsoid, 139-140 for stress, 72-75 Reference frames, 7,8,122,127,148 distinguished from coordinate systems, 7-8,11 Rigid-body rotation, 145 N Rigid-body translation, 145 Rigidity modulus, 247-248 Natural strain, 136-137, 139 Rotation, 145, 205 Newtonian behavior, 254-263 of principal strain directions, 145-146, Newton's laws of motion, 42 232 Normal strain, 152 pure, 145 rigid-body, 145 Rotational deformation, 145-146 p

Path s . configuration, 23 dated, 21 Shear strain, 135 distinguished from spatial changes, S. L Units, 13, 329-333 27-30 Simple shear, 146 indicated by fibrous crystals, 26 progressive, 229-235 particle, 21 Single-state indicators, 24 Path indicators, 24, 25-30 Solid, distinction from fluid, 254 Poise, 255 Stiffnesses, 243 Poisson's ratio, 247 Strain Position vector, 8 biaxial, 141, 149, 210 Pressure, 254 chronological relations to stress, 19, Principal directions 243, 279 of strain, 139-141 classes of, 141-143 of strain-rate, 259 components of infinitesimal, 152 of stress, 56-58 definition of, 17, 23, 133-137 Principal planes distinguished from deformation, 144- of strain, 140 145, 150 of stress, 56-59 distinguished from strain history, 224 Principal strains, 140 distinguished from stress, 17 Principal strain-rates, 259 domain boundaries, 221 Principal stresses, 56, 58-59 ellipsoid, 139-140 Progressive deformation, 224-237 energy, 268-270 coaxial, 237 field, 215-223 distinguished from deformation, 224, finite, 226, 236 337 Index

Strain [conto ] hydrostatic, 59 finite strain tensors, 196, 199 independent components of, 95-96 history, 224-236 maximum shearing, 75-76 homogeneous, 215 mean, 80-82, 263 incremental, 224, 225, 231 Mohr circle for, 70-83 normal, 45 irrotational. Irrotational See orientation relative to strain, 252, 253- deformation 254 measures of, 133-137 orientation relative to strain-rate, 260- natural, 136-137 261 normal, 152 principal, 56, 58-59 path, 224 principal directions of, 56-58 principal directions of, 139-141 principal planes of, 56-59 principal planes of, 140 relation of tensor components to pure, 145, 150, 208 ellipsoid, 92-93 rate, 227 shear, 45 rotational. See Rotational deformation sign convention for tensor components, state of, 18 93-95 state of at a point, 181 superposition of, 116 tensor shear strain, 182 tensor components of, 85-96 total, 224, 226, 236 trajectories, 114-115 triaxial, 142, 149 triaxial, 59 uniaxial, 141, 149, 208 two meanings of, 44-45 Strain ellipsoid, 139-144 uniaxial, 59 equation of, 144 units, 50, 332-333 circular sections of, 143-144 vector, 44-48 Strain field Stress-strain curve, 242, 253 homogeneous, 215 Stretch, 134 periodic, 220 Strain history, 224-237 Structure, definition of, 3 distinguished from strain, 224 Summation convention, 97 Strain-rate, 255 orientation relative to stress, 260-261 Strain-rate tensor, 257-259 T Stress addition of tensors, 116 Tensor notation, 97-98 addition of vectors, 47-48 Tensor quantities axial, 59 relation to vector quantities, 103-104 biaxial, 59 transformation formula for, 104-109 chronological relations to strain, 19, Tensor shear strain, 182 243, 279 Tensor transformation formula, 109 classes of, 59, 79-80 Tensors components of vector, 45-47, 85-90 deformation, 196, 199 concentration, 49, 117 finite strain, 196, 199 constructions relating planes and stress first-order, 109-110, 212 vectors, 54-58, 77-79 fourth-order, 245 conventions for signs, 59, 83-84, 98, second-order, 109,211

0 264,293 Total strain, 224, 226, 236 deficiency, 49 Transformation formula definitions of, 42-45 for tensors, 100, 104-109,.111 deviator, 80-82, 260 for vectors, 109, 111 direct measurement of, 121 Translation, 14, 145, 205 distinguished from strain, 17 pure, 145 ellipse, 53-58, 66-68 rigid-body, 145 ellipsoid, 58-60, 69 Triaxial compression test, 253 equations for normal and shear Two-state indicators, 24-25 components, 72 field, 111-112 history, 122-127

338 Index v Vector addition, 277 components of, 86-87 convention for magnitude, 45-46 displacement, 13 position, 8 stress, 10, 44-48 subtraction, 277 velocity, 9 Velocity gradient, 258 Viscosity, 255 Viscosity coefficients. 259 Viscous stress tensor, 260 w Work to deform rock, 266-267 to raise rock in gravity field, 264-266 y

Young's modulus, 241

339