EAS 326-03 Name: ____________________________

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EAS 326-03 FINAL EXAMThis exam is closed book and closed notes. It is worth 150 points; the value of each ques-tion is shown at the end of each question. At the end of the exam, you will find two pages ofpotentially useful equations.

1. a. What are the similarities and differences among mylonites, breccia, and gouge? Inyour answer, be sure to explain the processes by which each form and where in thecrust you would expect to find each one. [15 pts.]

Mylonites, gouge and breccia all form in fault zones.Gouge and breccia are usually unfoliated (although gougemay have a clayey foliation) and are the result of brittlegrinding and milling of the rock particles in the faultzone. Gouge is generally non-cohesive and forms in the upfew kilometers of the crust. Breccia can be more cohesiveand may form somewhat deeper in the upper crust. Mylonitesare always foliated and have had their grain size reducedby crystal plastic mechanisms, brittle grinding, or morelikely a combination of the two. For example, it is commonin some mylonites to see ductile deformation of quartz andbrittle deformation of feldspars. Mylonites generally forma depths below 8-10 km (depending on the heat flow).

!b. Define “S-C fabrics” and state in which of the above rock types you would expect tofind S-C fabrics. [10 pts.]

S-C fabrics are coevally forming foliation (S) and shear(C) planes that are common in mylonites. They are excellentsense of shear indicators. Kinematically analogous featuresare also found in brittle shear zones.

EAS 326-03 Name: ____________________________

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!c. Suppose that, midway between two C planes, the S planes make an angle of 34°with respect to the C planes. What would the angular shear and shear strain be atthat point (i.e., midway between the two C planes)? [15 pts.]

From class, we have the relationship (initially from Ramsay& Graham) that:

†

tan2 ¢ q =2g rearranging:

†

g =2

tan2 ¢ q =

2tan2* 34

= 0.808

The angular shear is:

†

y = tan-1 g = tan-1 0.808 = 38.9°

!d. In a different rock, you found some S planes that were oriented at 67° to the some Cplanes. What would you conclude about the relationship between these two foliationsin this rock and why? [10 pts.]

You would conclude that the two foliations are most likelyunrelated to each other. Probably the S planes formedfirst. We know that the axes of the infinitesimal strainellipse are oriented at 45° to the boundaries of the shearzone and that, with increasing shearing and strain, the an-gles between shear zone boundary and the finite strain el-lipse are less (remember our virtual card deck experiment).Thus there is no way that, if the S and C planes formed atthe same time that the S planes could be at a higher anglethan 45°.

EAS 326-03 Name: ____________________________

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!2. Riedel shears are produced in a right-lateralbrittle shear zone. The pertinent propertiesof the rock and the stresses on it are shownin the table at the right. For all of the fol-lowing questions, assume that the orienta-tions and magnitudes of the stresses areconstant (even if that assumption may notbe very realistic…).

!a. The shear zone is shown below. Drawthe orientations of the principal stressesand both sets of Riedel shears. Be sure to show (and label) their correct angular re-lations with respect to the shear zone boundaries. [10 pts.]

m = 0.466 = tan f

f = 25°

2q = 90 + 25 = 115°

s1

f/2

90 - f/2

! b. Plot the Mohr’s Circle for stress and the failure envelopes for Coulomb fracture and

for slip along pre-existing faults. Be sure to label your diagram! [15 pts.]

Coefficient Symbol Amount

internal friction µ 0.466

sliding friction µs 0.700

cohesion So 50 MPa

max. principal stress s1 258 MPa

confining pressure Pc 42 MPa

EAS 326-03 Name: ____________________________

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!c. How much, and in what direction, will the synthetic (R) and antithetic (R') Riedel

shears rotate before they become inactive? [10 pts.]

Both the R and R' shears will rotate clockwise because theshear zone is right lateral. The R' shears will rotate (115- 89)/2 = 13° before they become inactive (note that the R’shears plot above the X-axis because the way we derivedMohr’s Circle, left lateral shear is positive). The Rshears will rotate (162 - 115)/2 = 23.5° before they becomeinactive.

d. How much shear strain or angular shear will the shear zone have to experience be-fore the synthetic Riedels become inactive? You can solve this problem graphicallyor trigonometrically. [10 pts.]

The figure below shows the graphical construction:

You can also solve the problem trigonometrically:

†

tan ¢ q =y

ytanq

- y tany

rearranging to solve for y:

†

tany =1

tanq-

1tan ¢ q

=1

tan(12.5)-

1tan(12.5 + 23.5)

= 3.1343 \ y = 72.3°

EAS 326-03 Name: ____________________________

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!3. A characteristic feature of thrust belts is the duplex. How would you distinguish be-tween and “in sequence” duplex that follows Dahlstrom’s rules and an out of sequenceduplex that does not? Make a careful sketch of each type to illustrate your answer. [15pts.]

A normal in-sequence duplex is shown below. Notice thatthere is no truncation off beds on the roof thrust. Allfaults obey Dahlstrom’s rules and cut upsection in the di-rection of translation.

An out of sequence duplex is shown below. It is character-ized by truncation on the roof thrust and “beheaded” anti-clines, resulting in the fault locally cutting down-sectionin the direction of translation.

truncated anticlines

faults cut down section in direction of translation1 = first block to move; 3 = last block to move

123

!4. a. Thrusting in a mountain range has produced 50% shortening (e = –0.5) during aplane strain, constant volume deformation. Assume that the crust prior to deformationwas a sea level and was 35 km thick, and that the crustal density is 2600 kg m-3 andthat of the mantle 3200 kg m-3. How high will the mountain belt be and thick will its rootbe (assuming no erosion)? What is the basic assumption of your calculation? [15 pts.]

50% shortening in plane strain, constant volume means thatthe crustal thickness will double from 35 to 70 km as shownbelow. Thus, we know that ∆hcrust = (70–35) = 35 km.

70 km 35 kmrcrust = 2600 kgm-3

rmantle = 3200 kgm-3

∆E

∆hcrust = (70 - 35) = 35 km

EAS 326-03 Name: ____________________________

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From differential isostasy, we know that:

†

DE = Dhcrust + Dhmantle and rcrustDhcrust + rmantleDhmantle = 0

Thus, we have two equations with two unknowns, and we cansolve them for ∆E:

†

DE =Dhcrust rmantle - rcrust( )

rmantle

= 6.56km

and the root will be 35 km - 6.56 km = 28.44 km. Note thatthe extreme elevation obtained in this problem results pri-marily because I gave you an unrealistically low averagedensity for the crust!

The basic assumption here is Airy Isostasy and that thecrust has no lateral strength (i.e., flexural rigidity).

!b. Assume that the crust beneath this mountain range has a geothermal gradient of30°C/km and has a granitic composition with the following parameters: Co = 101.6

GPa-ns–1; n = 3.4; and the activation energy, Q = 139 kJ mol–1. Furthermore youshould assume a geologically reasonable strain rate. Calculate the maximum differ-ential stress that crust can support at 20 km depth. Does this deformation conformto Anderson’s Law? [15 pts.]

The solution to this problem calls for application of powerlaw creep:

†

˙ e = Co s1 -s 3( )n exp -QRT

Ê

Ë Á

ˆ

¯ ˜ or

†

s1 -s 3( ) =˙ e

Co exp -QRT

Ê

Ë Á

ˆ

¯ ˜

Ê

Ë

Á Á Á Á

ˆ

¯

˜ ˜ ˜ ˜

1n

†

=10-14 s-1

101.6GPa-ns-1 exp -139kJmol-1

8.3144 ¥10-3 kJmol-1K-1( ) 20km ¥ 30°Ckm-1( ) + 273( )KÊ

Ë

Á Á

ˆ

¯

˜ ˜

Ê

Ë

Á Á Á Á Á Á

ˆ

¯

˜ ˜ ˜ ˜ ˜ ˜

13.4

= 7.209MPa

This deformation, of course, has nothing to do with Anderson’sLaw because it is well below the free surface and because thedeformation is not Coulomb fracture.

EAS 326-03 Name: ____________________________

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!5. Define the following terms, mentioning, where appropriate, the processes responsiblefor, or the kinematic significance of, the feature (use sketches freely!):

a. Transposition [10 pts.]—Transposition is the conversion of an original planar fab-ric (e.g., bedding) into a second planar compositional lay-ering which has no stratigraphic relationship to the first.The processes involved include isoclinal folding and shear-ing out of the fold limbs. One can recognize transpositionoften by identifying remnant fold hinges, etc.

!b. crenulation cleavage [10 pts.]—

Crenulation cleavage is a secondary cleavage superimposedon an earlier cleavage. In the drawing below, the crenula-tion cleavage is S2:

S1

S2

The processes involved are primarily microfolding and pres-sure solution.

!c. parallel folding [10 pts.]—

Parallel folds have Class 1b dip isogons. The thickness ofthe layer measured perpendicular to bedding is constant.The shear is parallel to the layers (a process known asflexural slip/flow) and therefore, the bedding surfaces arelines of no finite elongation.

!d. antitaxial, sigmoidal veins [10 pts.]—

antitaxial-> (e.g., quartz vein in Limestone)new material added at the vein walls

sigmoidal->tips grow outward with time.The center of the vein rotates in direction of shear

tips at 45° to shear zone,parallel to infinitesimal

shortening direction

EAS 326-03 Name: ____________________________

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!6. Compare and contrast mode II (sliding) and mode III (tearing) cracks with edge andscrew dislocations in crystals. [20 pts.]

Mode II cracks are similar to edge dislocations because theslip vector is perpendicular to the tip line of the crack,just as in an edge dislocation the Burgers vector is per-pendicular to the tangent vector. Likewise, in a mode IIIcrack the slip vector is parallel to the tip line just asin the case of a screw dislocation the Burgers vector isparallel to the tangent vector.

However, cracks are a permanent rupture across the mate-rial, whereas the movement of a dislocation across a crys-tal leaves a perfect crystal in its wake (i.e., there is noplane across which bonds are permanently broken).

EAS 326-03 Name: ____________________________

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Potentially Useful Equations

Note that not all of these equations are needed forthe exam and that some of them have not, or willnot, be covered in class at all.

†

s DT =aEDT1-u

†

s ij =

s11 s12 s13

s 21 s 22 s 23

s 31 s 32 s 33

È

Î

Í Í Í

˘

˚

˙ ˙ ˙

†

Vi =kij

hdPP

dx j

Ê

Ë Á Á

ˆ

¯ ˜ ˜

†

s n* =

s1* + s 3

*

2Ê

Ë Á

ˆ

¯ ˜ +

s1* -s 3

*

2Ê

Ë Á

ˆ

¯ ˜ cos2q

†

s s =s1

* -s 3*

2Ê

Ë Á

ˆ

¯ ˜ sin2q

†

¢ l =¢ l 3 + ¢ l 12

Ê

Ë Á

ˆ

¯ ˜ -

¢ l 3 - ¢ l 12

Ê

Ë Á

ˆ

¯ ˜ cos2 ¢ q

†

¢ g =¢ l 3 - ¢ l 12

Ê

Ë Á

ˆ

¯ ˜ sin2 ¢ q

†

tan ¢ q = tanql3

l1

= tanqS3

S1

†

s s = So + s n*m

†

˙ e = Co s1 -s 3( )n exp -QRT

Ê

Ë Á

ˆ

¯ ˜

†

˙ e = Co T( )D s1 -s 3( )

dn

†

U = -C1

r+

C2

r12

†

Plith = rgdz0

zÚ

†

f =Vf

Vf + Vs

†

f = fo exp -az( )

†

D v =Vfinal -Vinitial

Vinitial

†

e =lf - li

li

†

e =sin f + q( )

sinf-1

†

S =lf

li

= l

†

l = S2

†

¢ l =1l

†

sin2q = 2sinq cosq

†

cos2 q =1+ cos2q

2Ê

Ë Á

ˆ

¯ ˜

†

sin2 q =1- cos2q

2Ê

Ë Á

ˆ

¯ ˜

†

Ui = Uoi+ Eijdx j

†

U1

U2

U3

È

Î

Í Í Í

˘

˚

˙ ˙ ˙

=

Uo1

Uo2

Uo3

È

Î

Í Í Í

˘

˚

˙ ˙ ˙ +

E11 E12 E13

E21 E22 E23

E31 E32 E33

È

Î

Í Í Í

˘

˚

˙ ˙ ˙

dx1

dx2

dx3

È

Î

Í Í Í

˘

˚

˙ ˙ ˙

†

s m* =

s1 + s 2 + s 3 - 3Pf( )3

†

s1* = Co + Ks 3

*

†

K =1+ sinf1- sinf

; Co = 2So K

EAS 326-03 Name: ____________________________

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†

st = 0.85s n*

†

st = 50 MPa + 0.6s n*

†

a + b( ) =1- l f( )m f + b

1- l( )k +1

R = 8.3144 x 10–3 kJ/mol °K = 1.9872 x 10–3 kcal/mol °K

°K = °C + 273.16°

1 MPa = 106 kg/m s2 = 10 bars

g = 9.8 m/s2 = 980 cm/s2

†

cosa = cos(trend)cos(plunge)cosb = sin(trend)cos(plunge)cosg = sin(plunge)

†

cosa = sin(strike)sin(dip)cosb = -cos(strike)sin(dip)cosg = cos(dip)

†

tan2 ¢ q =2g

†

pi = s ij l j

†

p1 = s11l1 + s12l2 + s13l3

p2 = s 21l1 + s 22l2 + s 23l3

p3 = s 31l1 + s 32l2 + s 33l3

†

Ld = 2pT E6Eo

3

†

Ld = 2pTh S -1( )6ho 2S2( )

3

†

C ≡1r

†

CG = CmaxCmin

†

b = q - f + 180° - 2g( )

†

f = tan-1 -sin g -q( ) sin 2g -q( ) - sinq[ ]cos g -q( ) sin 2g -q( ) - sinq[ ] - sing

Ï Ì Ô

Ó Ô

¸ ˝ Ô

˛ Ô

†

f = q = tan-1 sin 2g( )2cos2 g( ) -1

Ï Ì Ó

¸ ˝ ˛

†

g = tany = 2tan d2

Ê

Ë Á

ˆ

¯ ˜

†

g = tany @ 0.0175 d( )

†

s = 2h tan d2

Ê

Ë Á

ˆ

¯ ˜

†

s @ 0.0175h d( )

S DM = 0 = DMw + DMs + DMc +DMm + DMa

0 =D(rwhw) + D ( r shs) + D(rchc) +D(rmhm) + D(raha)

DE = Dhw + Dhs + Dhc + Dhm + Dha