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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<w is sh~ opposite. The absolute magnitudes of Fn and Fs are about 3400 kg and 9400 kg, respectively. The sign of F n is positive Problem 4.2 Problem 4.3 ~""" / \ \ ~ 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.
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