In this Chapter
2.7 Basic Ideas about Strain
Equation (2.1)with ρg= 26.5 MPa km−1the vertical component of stress σyyis also given as a function of depth inFigure 2.19. The measured stresses correlate reasonably well with 0.6 σyy.
Another technique used to determine the orienta- tion of crustal stresses is the observation of wellbore
breakouts. Wellbore breakouts are the result of local-
ized failure around a borehole in response to hori- zontal compression. Compression produces spallation zones along the wellbore at the azimuth of minimum principal stress where the circumferential compressive stress is a maximum. The spallation zones can be used to infer the directions of the horizontal principal stresses (Gough and Bell,1982).
Observations of wellbore breakouts can be obtained from borehole televiewer data. The borehole tele- viewer is an ultrasonic well-logging tool which can image the orientation and distribution of fractures as well as the orientation of stress-induced wellbore breakouts.
Problem 2.19
An overcoring stress measurement in a mine at a depth of 1.5 km gives normal stresses of 62 MPa in the N–S direction, 48 MPa in the E–W direc- tion, and 51 MPa in the NE–SW direction. Deter- mine the magnitudes and directions of the principal stresses.
M
Problem 2.20
The measured horizontal principal stresses at a depth of 200 m are given inTable 2.1as a function of distance from the San Andreas fault. What are the
Table 2.1 Stress Measurements at 200 m Depth vs.
Distance from the San Andreas Fault
Distance Maximum Minimum
from Principal Principal
Fault (km) Stress (MPa) Stress (MPa)
2 9 8
4 14 8
22 18 8
34 22 11
values of maximum shear stress at each distance? A MATLAB code for solving this problem is given in Appendix D.
2.7 Basic Ideas about Strain
Stresses cause solids to deform; that is, the stresses pro- duce changes in the distances separating neighboring small elements of the solid. In the discussion that fol- lows we describe the ways in which this deformation can occur. Implicit in our discussion is the assumption that the deformations are small.
Figure 2.20shows a small element of the solid in the shape of a rectangular parallelepiped. Prior to defor- mation it has sides δx, δy, and δz. The element may be deformed by changing the dimensions of its sides while maintaining its shape in the form of a rectangular par- allelepiped. After deformation, the sides of the element are δx−εxxδx, δy−εyyδy, and δz−εzzδz. The quantities εxx, εyy, and εzzare normal components of strain; εxx is the change in length of the side parallel to the x axis
divided by the original length of the side, and εyyand εzzare similar fractional changes in the lengths of the sides originally parallel to the y and z axes, respectively. The normal components of strain εxx, εyy, and εzzare assumed, by convention, to be positive if the deforma- tion shortens the length of a side. This is consistent with the convention that treats compressive stresses as positive.
If the deformation of the element inFigure 2.20 is so small that squares and higher order products of the strain components can be neglected in computing the change in volume of the element, the fractional change in volume (volume change divided by original volume) is εxx+ εyy+ εzz. This quantity is known as the dilata- tion ; it is positive if the volume of the element is
decreased by compression.
Problem 2.21
Uplift and subsidence of large areas are also accom- panied by horizontal or lateral strain because of the curvature of the Earth’s surface. Show that the lat- eral strain ε accompanying an uplift y is given by
ε= y
R , (2.74)
where R is the radius of the Earth.
Problem 2.22
The porosity φ of a rock is defined as its void volume per unit total volume. If all the pore spaces could be closed, for example, by subjecting the rock to a suffi- ciently large pressure, what would be the dilatation? For loose sand φ is about 40%, and for oil sands it is usually in the range of 10 to 20%.Table 2.2gives the porosities of several rocks.
The strain components of a small element of solid can be related to the displacement of the element. In order to simplify the derivation of this relationship, we consider the two-dimensional example inFigure 2.21. Prior to deformation, the rectangular element occu- pies the position pqrs. After deformation, the element is in the position pqrs. It is assumed to retain a rect- angular shape. The coordinates of the corner p before
Table 2.2 Rock Porosities
Rock Porosity (%) Hasmark dolomite 3.5 Marianna limestone 13.0 Berea sandstone 18.2 Muddy shale 4.7 Repetto siltstone 5.6
Figure 2.21 Distortion of the rectangular elementpqrs into the rectangular elementpqrs.
strain are x and y; after strain the corner is displaced to the location denoted by pwith coordinates x, y. The displacement of the corner p as a result of the strain or deformation is
wx(x, y)= x − x (2.75) in the x direction and
wy(x, y)= y − y (2.76) in the y direction. Displacements in the negative x and y directions are considered positive to agree with the sign convention in which positive strains imply a contraction. Corner q at x+ δx, y is displaced to position q with coordinates x+ δx, yas a result of the deformation. Its displacement in the x direction is
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2.7 Basic Ideas about Strain 113
Similarly, the displacement of corner s in the y direc- tion wy(x, y+ δy) is given by the difference in the y coordinates of sand s
wy(x, y+ δy) = y + δy − (y+ δy). (2.78) In writing Equations (2.77) and (2.78), we have assumed that the strains (δx− δx)/δxand (δy− δy)/ δyare small.
Since δx and δy are infinitesimal, we can expand
wx(x+ δx, y) and wy(x, y+ δy) as wx(x+ δx, y) = wx(x, y)+ ∂wx ∂x δx (2.79) wy(x, y+ δy) = wy(x, y)+ ∂wy ∂y δy. (2.80)
Substitution ofEquation (2.79) intoEquation (2.77) and subtraction ofEquation (2.75)yield
δx= δx+∂wx
∂x δx. (2.81)
Similarly, substitution of Equation (2.80)intoEqua- tion (2.78)and subtraction ofEquation (2.76)yield
δy= δy+∂wy
∂yδy. (2.82)
From the definitions of the strain components and Equations (2.81)and(2.82), we find
εxx≡ δx− δx δx = ∂wx ∂x (2.83) εyy≡ δy− δy δy = ∂wy ∂y. (2.84)
In three-dimensional strain, the third strain component
εzzis clearly given by εzz= δz− δz δz = ∂wz ∂z . (2.85)
The components of strain in the x, y, and z direc- tions are proportional to the derivatives of the asso- ciated displacements in the respective directions. The dilatation is given by = ∂wx ∂x + ∂wy ∂y + ∂wz ∂z . (2.86)
We have so far considered strains or deformations that do not alter the right angles between line elements
Figure 2.22 Distortion of a rectangle into a parallelogram by a strain field involving shear.
that are mutually perpendicular in the unstrained state. Shear strains, however, can distort the shapes of small elements. For example, Figure 2.22shows a rectangular element in two dimensions that has been distorted into a parallelogram. As illustrated in this figure, the shear strain εxyis defined to be one-half of the decrease in the angle SPQ
εxy≡ −12(φ1+ φ2), (2.87) where φ1and φ2are the angles through which the sides
of the original rectangular element are rotated. The sign convention adopted here makes εxynegative if the original right angle is altered to an acute angle. As in the case of stress, the shear strain is symmetric so that
εyx= εxy.Figure 2.22shows that the angles φ1and φ2
are related to the displacements by tan φ1=−wy (x+ δx, y) δx = φ1 (2.88) tan φ2=−wx (x, y+ δy) δy = φ2. (2.89)
In Equations (2.88) and (2.89), we assume that the rotations are infinitesimal so that the tangents of the angles are very nearly equal to the angles themselves.
We can express wy(x+ δx, y) and wx(x, y+ δy) in terms of the spatial derivatives of the displacements
according to wy(x+ δx, y) = ∂wy ∂x δx (2.90) wx(x, y+ δy) = ∂wx ∂y δy, (2.91)
where, for simplicity, we assume wx(x, y) = 0 and wy(x, y) = 0. Substitution of Equations (2.90) and
(2.91) into Equations (2.88) and (2.89) and further substitution of the resulting expressions for φ1and φ2
intoEquation (2.87)yield
εxy= 1 2 ∂wy ∂x + ∂wx ∂y (2.92) as the relation between shear strain and the spatial derivatives of displacements. In the engineering litera- ture, γxy= 2εxyis often used. Care should be exercised in dealing with these quantities.
Shear strain can also lead to a solid-body rotation of the element if φ1 = φ2. The solid-body rotation ωzis defined by the relation
ωz= −1
2(φ1− φ2). (2.93) Substitution ofEquations (2.88)and(2.89)intoEqua- tion (2.93)gives ωz=1 2 ∂wy ∂x − ∂wx ∂y . (2.94)
The rotation of any element can be resolved in terms of the shear strain and the solid-body rota- tion. FromEquations (2.87) and(2.93), the angle φ1
through which a line element parallel to the x axis is rotated is
φ1= −(εxy+ ωz), (2.95) and the angle φ2through which a line element in the y
direction is rotated is
φ2= ωz− εxy. (2.96) Thus, in the absence of solid-body rotation, εxyis the clockwise angle through which a line element in the
x direction is rotated. It is also the counterclockwise
angle through which a line element in the y direction is rotated.
Figure 2.23 Sketch of(a) pure shear strain that involves no solid-body rotation of elements and(b) simple shear strain that includes such rotation.
If the amount of solid-body rotation is zero, the dis- tortion is known as pure shear. In this case, illustrated inFigure 2.23a, φ1= φ2 (2.97) ∂wy ∂x = ∂wx ∂y (2.98)
and the shear strain is
εxy= ∂wx
∂y = ∂wy
∂x . (2.99)
The case of simple shear, shown inFigure 2.23b, com-
bines solid-body rotation and shear in such a manner that
φ1=
∂wy
∂x = 0. (2.100)
From Equation (2.94), the amount of solid-body rotation is ωz= − 1 2 ∂wx ∂y , (2.101)
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2.7 Basic Ideas about Strain 115
and the shear strain is
εxy= 1
2
∂wx
∂y . (2.102)
Simple shear is often associated with strike–slip faulting.
The equations given for two-dimensional strains and solid-body rotation about one axis can be readily generalized to three dimensions. A pure shear strain in the xz plane has an associated shear strain component given by εxz= εzx= 1 2 ∂wz ∂x + ∂wx ∂z (2.103) and a pure shear strain in the yz plane corresponds to
εyz= εzy= 1 2 ∂wz ∂y + ∂wy ∂z . (2.104)
A solid-body rotation about the x axis ωxis related to displacement derivatives by ωx= 1 2 ∂wz ∂y − ∂wy ∂z . (2.105)
Similarly, a solid-body rotation about the y axis is
ωy=1 2 ∂wx ∂z − ∂wz ∂x . (2.106)
The strain components εxx, εyy, εzz, εxy, εxz, and εyz are sufficient to describe the general infinitesimal deformation of solid elements subjected to stresses. The solid-body rotations ωx, ωy, and ωzdo not alter distances between neighboring elements of a solid and, therefore, do not involve stresses. Accordingly, the strain components and their associated stresses are of primary concern to us in subsequent chapters.
Just as it was important to know the stresses on area elements whose normals make arbitrary angles with respect to x, y axes, so it is essential to know the fractional changes in length and the rotation angles of arbitrarily inclined line elements. For simplicity we consider the two-dimensional case. We wish to deter- mine the strains in the x, ycoordinate system, which is inclined at an angle θ with respect to the x, y coor- dinate system, as shown inFigure 2.24a. As a result of
the strain field εxx, εyy, εxyand the solid-body rotation ωz, the line elements PR and PQ experience changes in
length and rotations. Line element PR is parallel to the
xaxis, and PQ is parallel to the yaxis. The extension in length of PR divided by the original length δxis the strain component −εxx; the counterclockwise angle of rotation of PR is the angle φ1 = −εxy − ωz. This is illustrated in Figure 2.24b. The extension in length
of PQ divided by the original length δy is the strain component−εyy; the clockwise rotation of PQ is the angle φ2= ωz − εxy. This is shown inFigure 2.24c.
We first determine the strain component−εxx. The displacement of R to R in Figure 2.24b is the net
result of the combined elongations and rotations of δx and δy. The x component of the displacement of R relative to R arises from the elongation of δx in the
x direction, −εxxδx, and the rotation of δy through the clockwise angle φ2. The latter contribution to
the displacement is φ2δy, which, according toEqua- tion (2.96), is (ωz−εxy)δy. Thus the total x component of the displacement of Rwith respect to R is
−εxxδx+ (ωz− εxy)δy.
The y component of the displacement of R with respect to R is the sum of the elongation of δy,−εyyδy, and the contribution from the rotation of δx, which, withEquation (2.95), is φ1δx = −(εxy+ ωz)δx. Thus the total y component of displacement of R with respect to R is
−εyyδy− (εxy+ ωz)δx.
For small strains, the change in length of PR is the sum of the x component of RRprojected on the line PR,
[−εxxδx+ (ωz− εxy)δy] cos θ,
and the y component of RRprojected on the line PR, [−εyyδy− (εxy+ ωz)δx] sin θ.
The strain component εxxis thus −εxx = [−εxx δx+ (ωz− εxy)δy] cos θ δx + [−εyyδy− (εxy+ ωz)δx] sin θ δx . (2.107) Since δx δx = cos θ δy δx = sin θ (2.108)
Figure 2.24 (a) The transformation of coordinates x, y through an angle θ to x,y.(b) The transformation of the strain components onto thexaxis.(c) The transformation of the strain components onto the yaxis.
Equation (2.107)can be rewritten as
εxx = εxxcos2θ+ εyysin2θ+ 2εxysin θ cos θ. (2.109) Using Equation (2.36), we can further rewrite Equation (2.109)as
εxx = εxxcos2θ+ εyysin2θ+ εxysin 2θ. (2.110) This has the same form as the transformation of the normal stress given inEquation (2.37).
We next determine the strain component−εyy. As can be seen inFigure 2.24c, the component of the dis-
placement of Q with respect to Q in the negative x direction is the sum of the elongation of δx,−εxxδx, and the contribution from the rotation of δy,−φ2δy=
−(ωz− εxy)δy, that is,
−εxxδx− (ωz− εxy)δy.
The y component of the displacement of Q with respect to Q is the sum of the elongation of δy,−εyyδy,
and the contribution due to the rotation of δx, −φ1δx= (εxy+ ωz)δx, that is,
−εyyδy+ (ωz+ εxy)δx.
After projection of these displacements onto the line
PQ, the strain component εyycan be written as −εyy = −[εxxδx+ (ωz− εxy)δy] sin θ δy + [−εyyδy+ (ωz+ εxy)δx] cos θ δy . (2.111) Since δx δy = sin θ δy δy = cos θ, (2.112)
Equation (2.111)can be put in the form
εyy = εxxsin2θ+ εyycos2θ− 2εxysin θ cos θ. (2.113)
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2.7 Basic Ideas about Strain 117
Figure 2.25 Geometrical determination of(a) φ1and(b) φ2. By substitutingEquation (2.36)intoEquation (2.113),
we get
εyy = εxxsin2θ+ εyycos2θ− εxysin 2θ. (2.114)
Problem 2.23
Derive Equation (2.114)from Equation (2.110)by using the substitution θ= θ + π/2. Why can this be done?
We now turn to the determination of the shear strain, εxy, and the solid-body rotation ωz in the new coordinate system. We first determine the angle
φ1 = −εxy − ωz from the geometrical relationships shown inFigure 2.25a. For sufficiently small strain, φ1
is given by
φ1 = −εxy− ωz =R
V
δx . (2.115)
FromFigure 2.25a, we can see that
RV= RU− VU = RU− TS, (2.116) and
RU= [−εyyδy− (εxy+ ωz)δx] cos θ (2.117) TS= [−εxxδx+ (ωz− εxy)δy] sin θ. (2.118) By combining Equations (2.108) and (2.115) with (2.118), we obtain
εxy+ ωz = (εyy− εxx)sin θ cos θ
+ εxy(cos2θ− sin2θ )+ ωz. (2.119)
The angle φ2 can be found from the geometrical relationships shown inFigure 2.25b; it is given by
φ2= ωz− εxy = US
δy . (2.120)
FromFigure 2.25b, it is seen that
US= UT+ TS, (2.121) and
UT= −[−εxxδx− (ωz− εxy)δy] cos θ (2.122) TS= [−εyyδy+ (ωz+ εxy)δx] sin θ. (2.123) By combining Equations (2.112) and (2.120) with (2.123), we obtain
ωz− εxy = (εxx− εyy)sin θ cos θ
− εxy(cos2θ− sin2θ )+ ωz. (2.124) By adding and subtracting Equations (2.119) and (2.124), we can find separate equations for ωz and εxy:
ωz = ωz (2.125)
εxy = (εyy− εxx)sin θ cos θ+ εxy(cos2θ− sin2θ ). (2.126) The solid-body rotation is invariant to the coordi- nate transformation, as expected, because it represents a rotation of an element without deformation. By introducing Equations (2.36) and (2.39) into Equa- tion (2.126), we obtain
This has the same form as the transformation of the shear stress given inEquation (2.40).
Just as there are principal axes of stress in a solid, there are principal axes of strain. In the principal strain axis coordinate system, shear strain components are zero. Setting εxy = 0 in Equation (2.127) gives the direction of one of the principal axes of strain as
tan 2θ = 2εxy
εxx− εyy
. (2.128)
We have already shown, in connection with principal stress axes, that if θ is a principal axis direction, so is θ + π/2. The fractional changes in length along the directions of the principal strain axes are the
principal strains. With θ given by Equation (2.128), Equation (2.110)determines the principal strain ε1 =
εxx. The principal strain ε2 is identified with εyy. By a procedure analogous to the one used in deriving Equation (2.51)we find ε1,2= 12(εxx+ εyy)± εxy2 +14(εxx− εyy)2 1/2 . (2.129) It is convenient to have formulas for the normal and shear strains at an angle θ with respect to the ε1princi-
pal strain axis. Taking εxy= 0, εxx= ε1, and εyy= ε2
inEquations (2.109)and(2.127), we obtain
εxx= ε1cos2θ+ ε2sin2θ (2.130)
εxy= −12(ε1− ε2)sin 2θ. (2.131)
Problem 2.24
Show that the principal strains are the minimum and the maximum fractional changes in length.
Problem 2.25
Show that the maximum shear strain is given by
1
2(ε1− ε2). What is the direction in which the shear
strain is maximum?
Principal axes of strain can also be found for arbi- trary three-dimensional strain fields. With respect to these axes all shear strain components are zero. The normal strains along these axes are the principal strains ε1, ε2, and ε3. One can introduce the concept
of deviatoric strain in analogy to deviatoric stress by
referring the strain components to a state of isotropic
strain equal to the average normal strain e. In three
dimensions
e≡ 13(εxx+ εyy+ εzz)=13 . (2.132) The average normal strain and the dilatation are invariant to the choice of coordinate axes. The devi- atoric strain components, denoted by primes, are
εxx= εxx− e εyy= εyy− e εzz = εzz− e εxy= εxy εxz = εxz εyz= εyz. (2.133)