EAS 326-04 Name: ____________________________

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ANSWER KEY

EAS 326-03 MIDTERM EXAMThis exam is closed book and closed notes. It is worth 150 points; the value of eachquestion is shown at the end of each question. At the end of the exam, you will find twopages of potentially useful equations.

1. The following figure shows a triangle in its initial state (left) and its final state after ithas been deformed by simple shear parallel to line a. The lengths of the lines beforeand after are given. Use the figure to answer parts 1 - 3.

lin e c

a'

b'

c '

a' = 208b' = 92c' = 21 6

a = 147b = 147c = 14 7

line a

line b

49

46

85 b = tan5 = 0.0875

b = (92/147)2 = 0.392

a = tan44 = 0.966a = (208/147)

2 = 2.002

c = tan41 = 0.869

c = (216/147)2 = 2.159

LNFELong axis ofstrain ellipse

LNFE

a. Construct the Mohr's Circle for finite strain which describes this deformation. Clearlylabel the axes as well as the position of each line on the Mohr's Circle. (Hint: theperpendicular bisector of the chord of a circle goes through the center of the circle)[20 points]

1 2 3

1

1

3 = 2.58

3 = 1/2.58 = 0.3876

1 = 0.39

1 = 1/0.39 = 2.564 b

a

c 2b = 168

2c = 21

2a = 25

2 max shear = 42

2 LNFE = 64

Ave = 98/150, s.d. = 21

EAS 326-04 Name: ____________________________

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ANSWER KEY

b. Determine the orientations of the principal axes of strain relative to line a. [10points]

Line Shear strain, Quadraticelongation,

=

=1

a 0.966 2.002 0.482 0.499

b 0.0875 0.392 0.223 2.553

c -0.869 2.159 0.403 0.463

As you can see from the Mohrs Circle for strain, the 2 angle for line a is 25 so thelong axis of the strain ellipse will be oriented 12.5 from a.

c. Determine the orientations of the two lines of no finite extension (LNFE). Showthese on the Mohr's circle and on the diagram of the triangle. [10 points]

This is a simple shear deformation so the shear planes, and one of the two lines ofno finite elongation (LNFE) is horizontal. Both lines must be oriented at 2 = 64 sothe two LNFE are 32 on either side of the long axis of the strain ellipse.

d. Determine the orientation of the line of maximum angular shear. [10 points]

The line of maximum angular shear, , and shear strain, , can be found on theMohrs Circle by drawing a line from the origin which is tangent to the circle, asshown above. This line is oriented at 2 = 42 (or 21 from the principal extensionalaxis) and has an angular shear, = 48.

2. Below are three hypothetical stress strain curves. Assume that the deformation in caseI is by Coulomb failure and that the deformation beyond yield in cases II and III isgoverned by crystal plastic mechanisms. In all cases, the confining pressure was 1kbar.

50

(MPa)

2 4 6 8 2 4 6 8 2 4 6 8

Stra in (%)

case I25C

case II

case III600 C

300 C100

150

200

EAS 326-04 Name: ____________________________

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ANSWER KEY

a. Describe the deformation mechanism(s) for each case. Include in your account adescription of what happens at the atomic or molecular level. [30 points]

case I

In case I, the rock begins deforming elastically. Bonds between oppositely chargedions are stretched or shorten, but none are broken. If the stress is removed before itreaches 200 Mpa, the rock sample will return to its initial undeformed state, i.e., thedeformation is non-permanent.

At 200 MPa, the rock fractures. It would be a relatively clean break across all thebonds in the material at once. The rock on either side of the fracture is pretty muchundeformed.

case II

In case II, elastic deformation occurs as before in case I. When the yield stress isreached at about 160 MPa, the rock begins to deform permanently, but withoutrupturing. Because the rock is at 300C the deformation mechanism that isresponsible for the permanent solid-state flow is probably dislocation glide. Once thedislocations begin to move, they are immediately impeded by impurity atoms, theinterference of their self stress field with that of other dislocations, and by jogsproduced by intersecting dislocations in different glide planes and systems. Thus itrequires more stress for the dislocations to keep moving and the deformation is saidto be strain (or work) hardening.

case III

Case III also starts out with some elastic deformation but because the temperaturesare significantly higher the yield stress (about 75 MPa in this example) issubstantially lower. When plastic deformation does start, it is not accompanied bystrain hardening. Because the temperatures are higher, diffusion allows dislocationsto climb over and around obstacles so they can keep moving freely. This isdislocation glide and climb.

EAS 326-04 Name: ____________________________

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ANSWER KEY

b. The experiment in case I was done at a confining pressure of 50 Mpa and conjugatefractures were observed to form with an angle of 72 with respect to each other.Calculate the coefficient of internal friction (), the normal and shear stresses on thefracture plane at the instant of failure (n and s, respectively), and the cohesion(So). Plot the Mohrs circle for stress and the Coulomb failure envelop on the graphpaper provided below. [20 points]

2 = 108

= 18So= 58 MPa

3 =Pc= 50 MPa1= 250 MPa

n

s

n = 119 MPa

s= 97 MPa

The differential stress at failure is = (1 3) = 200 Mpa.The confining pressure, Pc = 3 = 50 Mpa, so 1 = 250 Mpa. Asshown in the diagram at the right, 1 bisects the acute anglebetween the conjugate fractures, so = 54. We know thatthe angle of internal friction, = 2 90 = 108 90 = 18.The coefficient of internal friction, = tan = tan 18 = 0.325.The normal and shear stress at failure, as well as the cohesioncan be read directly off of the Mohrs Circle for stress above.You can also calculate them from the equations for MohrsCircle and for the Coulomb failure envelope.

72

36

= 54 1

EAS 326-04 Name: ____________________________

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ANSWER KEY

3. Define the following terms and describe their importance in a geological context [10 pts,each]:

a. Spherical stress

This is a special state of stress encountered in fluids which cannot support shearstress so every plane is perpendicular to a principal stress, and all principal stresses areequal. The mean stress is equal to any of the three principal stresses which is just thepressure in the fluid.

b. First Fresnel zone

The first Fresnel zone defines the minimum horizontal dimension resolvable withacoustic waves used in standard seismic reflection surveys. At horizontal dimensions lessthan the first Fresnel zone reflecting interfaces appear as point sources and producediffractions rather than reflections. Diffractions can be useful to the structural geologistinterpreting seismic reflection profiles because they can be use to identify truncations atfault planes that are too steep to be imaged directly with the reflection technique.

c. Plane strain

Where there is no change in shape or dimensions in the third dimension the strain issaid to be plane strain. The intermediate principal axis of the finite strain ellipse has astretch = 1 and an extension = 0

d. Stylolite

An irregular, sawtooth surface across which soluble rock material (usually theminerals calcite or quartz) has been dissolved, leaving an insoluble residue (usuallyclays). Stylolites are produced by pressure solution: preferential dissolution in a directedstress field. Material is dissolved from points of high concentration and eitherredeposited at points of low stress concentration or if enough water is available flushedout of the system. The former produces a volume constant deformation whereas thelatter results in an overall volume loss on planes approximately perpendicular to themaximum principal stress.

e. Cauchys law

Cauchys law: pi = ijlj , shows that stress, ij , is a second order tensor that relatestwo vectors: pi , the traction (or stress vector) on the plane and lj , the pole to the plane (indirection cosines).

EAS 326-04 Name: ____________________________

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ANSWER KEY

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.

T =ET1

ij =

11 12 13 21 22 23 31 32 33

Vi =kij

dPPdx j

n* =

1* + 3

*

2

+

1* 3

*

2

cos2

s =1* 3

*

2

sin2

= 3 + 12

3 12

cos2

= 3 12

sin2

tan = tan 31

= tan S3S1

s = So + n*

= Co 1 3( )n exp Q

RT

= Co T( )D 1 3( )

dn

U = C1r

+C2r12

Plith = gdz0z

=Vf

Vf +Vs

= o exp az( )

v =Vfinal Vinitial

Vinitial

e =lf lili

e =sin + ( )sin

1

S =lfli

=

= S2

=1

sin2 = 2sin cos

cos2 = 1+ cos22

sin2 = 1 cos22

Ui =Uoi + Eijdx j

U1U2U3

=

Uo1Uo2Uo3

+

E11 E12 E13E21 E22 E23E31 E32 E33

dx1dx2dx3

m* =

1 + 2 + 3 3Pf( )3

1* = Co + K 3

*

K = 1+ sin1 sin

; Co = 2So K

EAS 326-04 Name: ____________________________

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ANSWER KEY

= 0.85 n*

= 50 MPa + 0.6 n*

+ ( ) =1 f( ) f + 1 ( )k +1

R = 8.3144 x 103 kJ/mol K = 1.9872 x 103 kcal/mol K

K = C + 273.16

1 MPa = 106 kg/m s2 = 10 bars

g = 9.8 m/s2 = 980 cm/s2

cos = cos(trend)cos(plunge)cos = sin(trend)cos(plunge)cos = sin(plunge)

cos = sin(strike)sin(dip)cos = cos(strike)sin(dip)cos = cos(dip)

tan2 = 2

pi = ij l j

p1 =11l1 +12l2 +13l3p2 = 21l1 + 22l2 + 23l3p3 = 31l1 + 32l2 + 33l3

Ld = 2TE6Eo

3

Ld = 2T S 1( )6o 2S

2( )3

C 1r

CG = CmaxCmin

= + 180 2( )

= tan1sin ( ) sin 2 ( ) sin[ ]

cos ( ) sin 2 ( ) sin[ ] sin

= = tan1 sin 2( )2cos2 ( ) 1

= tan = 2tan 2

= tan 0.0175 ( )

s = 2h tan 2

s 0.0175h ( )

M = 0 = Mw + Ms + Mc +Mm + Ma

0 =(whw) + ( shs) + (chc) +(mhm) + (aha)

E = hw + hs + hc + hm + ha