Structural Analysis and Synthesis_Solution Manual [Part2].pdf

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Chapter 10 Note to instructors

Problems Problems 10.5 and 10.6 could be posed in terms of overall overall strain strain instead instead of stress if the instructor desires. Answers for both approaches are given below. Answers

Problem 10.1

See solution on p. 42. Problem 10.2

Orientations of principal stress axes will vary slightly among students. See solutions on p. 42: s1 s2 s3

71 toward toward 212  toward 056 ¼ 18 toward  toward 323 : ¼ 7 toward ¼

Problem 10.3

Orientations of principal stress axes will vary slightly among students. See solutions on p. 42: s1 s2 s3

7 toward toward 118  toward 209 ¼ 10 toward  toward 353 . ¼ 78 toward ¼

Problem 10.4

Strain: Strain: Regional east–west shortening after the Mississippian. Regional east–west extension followed shortening. Orientation of faults in part governed by bedding surfaces during shortening. Stress: Stress: Early deformation: s1 Later deformation: s1

¼ ¼

east---west, s 2 ¼ north---south, s 3 ¼ vertical: vertical: vertical, s 2 ¼ north---south, s 3 ¼ east---west: east---west : Problem 10.5

Note that events 2 and 3 below are permissively the same event; it depends on whether granodiorite is considered to be folded. Strain: Strain: North–south shortening in southeast block between Mississippian and Cretaceous (F1 folds). East–west shortening in southeast block between Mississippian and Cretaceous but after north–south shortening (Cretaceous granodiorite could be folded) (F2 folds). East–west shortening through map area, post-Eocene, pre-Miocene (F 3 folds). East–west extension, post-Miocene (normal faults).

ROWLAND ROWLAND / Struct Structural ural Analysis Analysis and and Synthesi Synthesiss xxxxxxxx xxxxxxxxxx_4 xx_4_10-1 _10-17 7 Final Final Proof Proof page 41

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---------------------------------------------------------------------------------------------------------- Structural Analysis Analysis and Synthesis Synthesis ----------------------------------------------------------------------------------------

41

Stress: Stress: D1 : D2 : D3 : D4 :

s1 s1 s1 s1

north---south, s 2 ¼ east---west, s 3 ¼ vertical: vertical: east---west, s 2 ¼ north---south, s 3 ¼ vertical: vertical: ¼ east---west, s 2 ¼ north---south, s 3 ¼ vertical: vertical: ¼ vertical, s 2 ¼ north---south, s 3 ¼ east---west: east---west : ¼ ¼

Problem 10.6

(1) (2) (3) (4)

north---south, s 3 ¼ east---west (s1 ¼ vertical): vertical): north---south, s 3 ¼ east---west (s1 ¼ vertical): vertical): vertical): ¼ north---south, s 3 ¼ east---west (s2 ¼ vertical): The parado paradox x is that that s 1 and s 2 both have the same orientation. Stress ellipsoid orientations can be derived from answers 2 and 3 above. If the magnitudes of s1 and s 2 are very close to one another, minor perturbations of the stress field could cause these stress axes to switch. (Note: Duebendorfer doesn’t believe this. He interprets the strike slip faults to be extensional transfer faults.)

s2 s2 s1

¼ ¼

Problem 10.7

(1) (2) (3) (4)

See solu solutio tions ns on p. 43 43 (a and and b). See soluti solution on on on p. 43 43 (c). (c). s 3 . Clockwise Clockwise rotation rotation of s Vertical principal stress could increase due to loading associated with voluminous magmatism. Vertical principal stress could decrease due to crustal thinning associated with normal faulting. Problem 10.8

(1) (2) (3) (4)

See soluti solution on on on p. 44. Not reproducible reproducible for manual. manual. Not reproducible reproducible for manual. manual. Key points points:: (a) Development Development of structur structures es proceeds sequenti sequentially ally from south south to north. north. (b) s1 oriented oriented due north–south north–south and s3 oriented oriented due east–w east–west est early early in collisional history. (c) About About a 25–30 25–30 clockwise rotation in horizontal stress directions later in history as Southeast Asia is ‘‘extruded’’ from the main part of Asia. (d) In the far north north s 1 and s 2 switch orientations such that s 1 is vertical. 8

Problem 10.9

See solution on p. 44. Problem 10.10

See solution on p. 45.

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9.1.2007 3:37pm Compositor Name: PDjeapradaban

---------------------------------------------------------------------------------------- Structural Analysis and Synthesis -------------------------------------------------------------------------------------------------------Solutions

Problem 10.1

Problem 10.2

Problem 10.3



-------------------------------------------------------------------------------------------------------- Structural Analysis and Synthesis -------------------------------------------------------------------------------------

43

Problem 10.7

Explanation for conversion of a stike-slip fault from sinistral to dextral during one tectonic episode: _______________

Clockwise rotation of s 3 ______________________________________________________________________________________________________ Speculations about the geologic factors involved in the structural development of this region: _____________________

Vertical principal stress could increase due to loading associated with volcanism

______________________________________________________________________________________________________

Vertical principal stress could decrease due to crustal thinning due to normal faulting


44


---------------------------------------------------------------------------------------- Structural Analysis and Synthesis -------------------------------------------------------------------------------------------------------Problem 10.8

Orientations of the two horizontal principal stresses will vary slightly because faults are not perfectly straight. It is probably best to tell the students to use a best-fit line to approximate the overall orientation of a given fault. For the Himalayan Frontal thrust, s3 is probably vertical; however, you may wish to accept a WNW oriented s3 instead of the solution shown here as the models of Tapponier et al. (1982) show approximately E-W extension.

Problem 10.9



-------------------------------------------------------------------------------------------------------- Structural Analysis and Synthesis ------------------------------------------------------------------------------------Problem 10.10

Yes, they are compatible (NW-SE shortening axes). The extension axes are different, but they are not incompatible (two directions of extension).

45




Reports will vary. Listed below, from oldest to youngest, are the main geologic events embodied in the Bree Creek Quadrangle. There are a few relationships that are ambiguous with respect to timing. These are intentional and are designed to force the student into presenting evidence for his or her conclusions. 1. Deposition of Paleozoic rocks. 2. Folding of Paleozoic rocks about east–west axes. 3. Folding of Paleozoic rocks about north–south axes (this event is arguably correlative with the folding of Eocene rocks, but we favor separating these events). 4. Intrusion of the Cretaceous granodiorite. 5. Uplift and erosion to expose granodiorite and folded Paleozoic rocks. 6. Deposition of Paleocene to Eocene rocks. 7. Folding of these rocks about north–south axes. This was probably contemporaneous with thrusting along the Mirkwood fault. 8. Uplift and erosion. 9. Deposition of Miocene rocks. Either non-deposition or erosion of Tg unit prior to Tr deposition on northeast block. 10. Normal faulting on all faults (possibly contemporaneous with tilting below). 11. Regional northward tilting. 12. Erosion: complete removal (or original non-deposition) of Miocene rocks in vicinity of Gandalf’s Knob. 13. Deposition of Pliocene unit. 14. Tilting to north.



Chapter 12 Answers and solutions

Problem 12.1

Problem 12.2

Problem 12.3

Higher strain rates in the hinges of folds result in brittle deformation. The limbs of folds, experiencing slower strain rates, deform continuously without fracturing. This problem reinforces the concept that strain rate is a very important aspect of rheology; rocks that fracture at high strain rates will fold under relatively lower strain rates. Problem 12.4

These problems are intended to encourage students and their lab instructor to creatively explore the rheological properties of various materials.



Chapter 13 Answers

Problem 13.1 sn

¼

77:53 MPa; s s

¼

35:95MPa

Problem 13.2 See solution on facing page

Problem 13.3 sn

¼

29 MPa; s s

¼

12MPa

Problem 13.4 (1) (2) (3)

See solution on p. 50. m ¼ 0:53 u ¼ 59

Problem 13.5 Maximum spacing is 56.45 m (see solution on p. 50)

Problem 13.6 m ¼ 0:90 (since students will measure slightly different u angles, the value of m will vary widely; most students will report values of 0.8 to 1.2 for m )

Problem 13.7 Pore pressure

¼

38 MPa

Problem 13.8

Pore pressure

¼

2.5 MPa



-------------------------------------------------------------------------------------------------------- Structural Analysis and Synthesis ------------------------------------------------------------------------------------Solutions

Problem 13.1 s1 ¼ 100 MPa; s 3 sn ¼ (s1 ¼

20 MPa; u

þ s3 )=2 þ [(s1 

¼

32

s3 )=2]cos2u

(100 þ 20)=2 þ [(100  20)=2]cos64

ss ¼ (s1 ¼

¼

¼

77:53MPa

 s3 )=2 s i n 2u

[(100  20)=2]sin64

¼

35:95 MPa

Problem 13.2 Involves substitution into above equations (Problem 13.1). Plane 1 2 3 4 5

u

sn

32 45 90 135 180

8 8 8 8 8

77.5 60 20 60 100

MPa MPa MPa MPa MPa

ss

36 40 0  40 0

MPa MPa MPa MPa MPa

49


50



Problem 13.4

Problem 13.5

At failure (from Problem 13.4): s1 ¼ 50MPa ¼ 5:1  106 kg=m2

Weight of tuff per m 3 : (2:0 g=cm3 )(106 cm3 =m3 )(kg=1000 g) ¼ 2000kg=m3

Weight supported by pillar: [(pr2 )(20m)(2000 kg=m3 )]=[(2:5 m2 )p] ¼ 5:1  106 kg=m2

r2 ¼ [p(2:5 m2 )(5:1  106 kg=m2 )]=[p(20m)(2000 kg=m3 )] r ¼ 28:23 m d ¼ 2r ¼ 56:46 m maximum spacing



-------------------------------------------------------------------------------------------------------- Structural Analysis and Synthesis -------------------------------------------------------------------------------------Problem 13.7

Problem 13.8

51


Chapter 14 Answers and solutions

Problem 14.1 (1) (2)

0.10 c ¼ 45 ; g

¼

tan c

¼

1:0

Problem 14.2

Problem 14.3





53

Problem 14.4

Results will vary slightly due to errors inherent in measurement. (1) e ¼ [11  10 ]=10 ¼ [1:75  3:5 ]=3:5 ¼ 0:50. (2) e ¼ [11  10 ]=10 ¼ [3:5  3:5 ]=3:5 ¼ 0. (3) 1 þ e1 : 1 þ e2 ¼ 2 : 1. (4) Diagonal dike on left falls into zone 2. Thus this segment of the dike has undergone early shortening followed by elongation, but with net shortening. Diagonal dike on right falls into zone 1b. Thus this segment of the dike has undergone early shortening followed by elongation, with net elongation. See below. 00

00

00

00

00

00

Problem 14.5

Problem 14.6

Constructions will vary. Overall ellipse shape should be similar to that shown. Ratio ¼ 2 : 1; orientation of maximum principal strain ¼ 293 . 8


54





-------------------------------------------------------------------------------------------------------- Structural Analysis and Synthesis -------------------------------------------------------------------------------------Problem 14.8

Note measured values will differ slightly, but overall ratios should be close.

Semi-major axis 10 9 7 7 11 9 7 – x=8.6

Semi-minor axis 7 7 5 4.5 8 7 5 – x=6.2

1 +e2 : 1 +e3 1.4 : 1.0

Semi-major axis 24 19 17 11.5 18 8.5 15 – x=16.1

Semi-minor axis 8 6 5.5 4 6 3 5 – x=5.4

1 +e1 : 1 +e3 3.0 : 1.0

strain ellipsoid plots in field of apparent constriction

55




This chapter is designed to introduce students to the principles and a few fundamental techniques of cross-section balancing. We strongly urge instructors to augment the exercises in this chapter with local examples of unbalanced cross sections. For obvious reasons, we felt that it would be inappropriate to include in this chapter examples of unbalanced cross sections from the literature.



-------------------------------------------------------------------------------------------------------- Structural Analysis and Synthesis ------------------------------------------------------------------------------------- Answers and solutions

Problem 15.1

Panel

Hanging wall

Footwall

A

Ramp

Flat

B

Flat

Flat

C

Flat

Ramp

D

Flat

Flat

E

Ramp

Flat

F

Flat

Flat

G

Flat

Ramp

H

Flat

Flat

I

Flat

Flat

J

Ramp

Flat

K

Flat

Flat

L

Ramp

Flat

M

Flat

Flat

N

Flat

Ramp

O

Flat

Flat

P

Flat

Ramp

Q

Flat

Flat

Problem 15.2

57


58



Note to instructor.

The problems present in this cross section can be stated in several different ways; i.e., if there is no hanging wall flat corresponding to a certain footwall flat, the reverse is also true. You will have to use your judgment in determining whether or not the students understand the concepts embodied in this exercise.

Line lengths are not equal. (Note: scale may have changed during final printing .) Top of Jurassic ¼ 210 mm Base of Jurassic ¼ 211 mm Top of Triassic ¼ 245 mm Base of Triassic ¼ 247 mm Top of Permian ¼ 185 mm Base of Permian ¼ 188 mm A – no hanging wall flat corresponding to this footwall flat. B – no hanging wall ramp corresponding to this footwall ramp. C – no hanging wall flat corresponding to this footwall flat. D – hanging wall flat at the base of the Triassic unit is far too long. E – no hanging wall ramp corresponding to this footwall ramp. F – no footwall ramp that cuts off both the Jurassic and Triassic. Bottom line: There are three footwall ramps but only two hanging wall ramps. This discrepancy leads to problems with different units cut off in hanging wall and footwall ramps, number of hanging wall vs. footwall flats, and length of footwall vs. hanging wall flats. Section does not meet the template constraint. Problem 15.4



-------------------------------------------------------------------------------------------------------- Structural Analysis and Synthesis ------------------------------------------------------------------------------------Problem 15.5

Fault-bend fold. About 36% shortening.

Problem 15.6

(1) Fault-propagation fold. (2) About 37%. (3) See map. Would hit unit Te at crest of anticline.

59




We encourage the instructor to augment the exercises in this chapter with personal hand specimens and thin sections of deformed rocks. Answers and solutions

Problem 16.1

Quartz deformed by crystal plastic processes as evidenced by highly elongate grain shapes (ribbons). Feldspar deformed by microcracking as evidenced by conspicuous loss of cohesion. Temperature of deformation was probably between 300 and 450 C. Quartz grains are largely unrecrystallized and show only minor evidence for recovery. 8

Problem 16.2

Pressure solution (diffusive mass transfer).

IP ¼ interpenetrating grain boundaries SG ¼ straight grain boundaries



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61

Problem 16.3

SG ¼ subgrains NG ¼ new grains Problem 16.4

Ultramylonite. Porphyroclasts make up less than 10% of rock. If the protolith was a coarse-grained granite, one would expect to see feldspar porphyroclasts. The absence of feldspar porphyroclasts indicates that the feldspars have undergone grain-size reduction. Problem 16.5

Mylonite. Porphyroclasts make up between 10 and 50% of rock.


62



(1) The feldspar exhibits a discrete microfracture and therefore was deformed by brittle deformation processes (cataclasis). The quartz forms highly elongate ribbons and therefore was deformed by intracrystalline plastic deformation processes (dislocation glide and probably limited dislocation climb). (2) Temperatures high enough for quartz to deform plastically but not high enough for feldspar to deform plastically. Deformation therefore occurred between approximately 300 and 450 . (3) Dextral shear sense. 8

AF = antithetic fracture SP = sigma porphyroclast DP = possible delta porphyroblast (weak)

Problem 16.7

(1) Delta porphyroclast. (2) Shear sense ¼ reverse-sense motion; southeast side up. Problem 16.8

Shear sense ¼ normal-sense motion; southwest side down. Problem 16.9

Pre-135 Ma southeast-over-northwest reverse movement (thrusting) on shear zone C. Southeast-side down normal movement on shear zone B between 135 and 65 Ma. Post-65 Ma right-slip faulting on shear zone A.




You may wish to devote two lab periods to this chapter. Answers and solutions

Problem 17.1

Problem 17.2

Problem 17.3

(a) Reverse-slip movement (with minor sinistral component) along east-striking, 70 N-dipping fault, or reverse-slip movement (with minor dextral component) along N62 W-striking, 30 SSW-dipping fault. (b) Dominantly right-slip movement along N10 W-striking, 80 W-dipping fault, or dominantly left-slip movement along N78 E-striking, 85 S-dipping fault. (c) Dextral-normal slip along N50 E-striking, 50 NW-dipping fault, or sinistralnormal slip along N73 W-striking, 50 SW-dipping fault. 8

8

8

8

8

8

8

8

8

8

8


64



Plots will vary slightly from student to student, particularly where oblique movements are involved. Plots can be generated visually or plotted precisely on stereonets recalling that the pole to the auxiliary plane is the slip direction. Students should assume reasonable dips for boundaries; in the plots below, we assume 90 dip for transform boundaries and 45 dips for convergent and divergent boundary faults. 8

8

Problem 17.5

Event A occurred along the N85 W-striking transform. Right-lateral movement. Event B occurred along the N12 W-striking, ENE-dipping subduction zone. Movement was reverse-slip with a dextral component. 8

8




65

Problem 17.6

(a) I ¼ 36:0. (b) I ¼ 63:4. (c) I ¼ 83:9. (d) I ¼ 19:4. Problem 17.7

Rate ¼ distance/time ¼ 810 km/18 Ma ¼ 45 km/Ma or 45 mm/yr for half-spreading rate. Full spreading rate ¼ 90 mm/yr. Problem 17.8

Age 60 Ma 38 Ma 18 Ma

*

Inclination

Latitude

N-S distance

46 33 25

64.2 52.4 43.0

1312 1045

Ratey

59.6 mm/yr 52.3 mm/yr

*

Equal to degrees change in latitude multiplied by 111 km/degree. Calculated rates are minimum values because paleomagnetic determinations can only detect the north–south component of movement (changes in latitude). y

Problem 17.9

The ‘‘plate game’’: this problem was inspired by a similar exercise devised by the late Dr. Peter Coney of the University of Arizona more than 30 years ago. In the exercise, North America is fast forwarded by about 100 Ma. Western California and Baja have been translated northwest and have collided with southern Alaska; the Basin and Range and Colorado Plateau (along the Rio Grande rift) have separated from the rest of North America. Listed on the next page are calculated latitudes for paleomagnetic data presented. Also shown are the preferred positions of ocean ridge segments and associated transforms based upon correlation


66


---------------------------------------------------------------------------------------- Structural Analysis and Synthesis --------------------------------------------------------------------------------------------------------

of magnetic anomalies. Specific geologic features are named. Students could be asked to determine the ‘‘exact’’ time of rifting and rates of movement of tectonic blocks. We welcome suggestions, additions, deletions, and corrections to this exercise.

Locality P1 P2 P3 P4 P5 P6

Radiometric date (Ma)

Inclination

Latitude (calculated)

120 50 15 2 15 2

51.4 68.0 55.5 58.7 55.5 52.0

32.1 51.1 36.0 39.4 36.0 32.6

1. See above 2. See above 3. Answers will vary. Check for consistency with map. For example, if Colorado Plateau is considered ‘‘fixed’’, the North American Midcontinent is moving relatively southeast, and the North American Cordillera is moving relatively northwest. 4. Information item only (no answer). 5. Basically, this map tracks the separation and northwestern motion of the block west of the San Andreas fault (western California and Baja California) from the rest of North America until its arrival at the Aleutian trench. Because this block is composed of buoyant continental material, it cannot be subducted and therefore collides with and is accreted to Alaska. Basin and Range extension ultimately causes separation of the North American Cordillera from the Midcontinent and the Colorado Plateau. Lesser extension along the Rift Grande rift results in separation of the Colorado Plateau from the Midcontinent.

Structural Analysis and Synthesis_Solution Manual [Part2].pdf

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