CHAPTER Stress and Strain Transformation

 

 

CHAPTER22

Stress and Strain
Transformation

2.1 INTRODUCTION

In Chapter 1 we defined stress and strain states at any point within the
solid body as having six distinctive components, i.e. three normal and
three shear components, with respect to an arbitrary coordinate system.
The values of these six components at the given point will change with
the rotation of the original coordinate system. It is therefore important
to understand how to perform stress or strain transformations between
two coordinate systems, and to be able to determine the magnitudes
and orientations of stress or strain components that result. One key
reason for stress or strain transformation is that the strains are normally
measured in the laboratory along particular directions, and they must be
transformed into a new coordinate system before the relevant stresses
can be re-calculated. In this chapter we discuss the stress/strain trans-
formation principles and the key role they play in the stress calculation
of a drilled well at any point of interest; whether vertical, horizontal
or inclined.

2.2 TRANSFORMATION PRINCIPLES

Let’s consider the cube of Figure 1.2, and cut it in an arbitrary way such
that the remaining part will form a tetrahedron. The reason for choosing a
tetrahedron for this analysis is that a shape with four sides has the least
number of planes to enclose a point. Figure 2.1 shows the stresses acting on
the side and cut planes of the tetrahedron. The stress acting on the cut
plane is denoted by S, which can be resolved into three components along
the respective coordinate axes, assuming n defines the directional normal
to the cut plane.

Petroleum Rock Mechanics
DOI 10.1016/B978-0-12-385546-6.00002-4

(cid:1) 2011 Elsevier Ltd.
All rights reserved.

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14

Transformation Principles

Figure 2.1 Normal and shear stresses acting on a tetrahedron.

Assuming the cut plane to have an area of unity, i.e. A = 1, the areas of

the remaining cube sides can be expressed as (Figure 2.2):

A ¼ 1
A1 ¼ cosðn; yÞ
A2 ¼ cosðn; xÞ
A3 ¼ cosðn; zÞ

ð2:1Þ

Figure 2.2 The areas of the side and cut planes.

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Stress and Strain Transformation

15

Since the tetrahedron remains in equilibrium, we use the concept of force
balance to determine the magnitude of the stresses acting on the cut plane.
A force balance in the x direction is given by:

X

Fx ¼ 0

or

SxA (cid:2) sxA2 (cid:2) txyA1 (cid:2) txzA3 ¼ 0

Repeating the force balance for the other two directions, and inserting the
expressions for the areas given in Equation 2.1, the stresses acting on the
cut plane, after some manipulation, can be given by:
2

3

3

2

3

2

4

5 ¼

4

Sx
Sy
Sz

txy
sx
txy sy
txz

txz
tyz
tyz sz

5

4

cosðn; xÞ
cosðn; yÞ
cosðn; zÞ

5

ð2:2Þ

Equation 2.2 is known as Cauchy’s transformation law (principle), which can
also be shortened to:

S½

(cid:3) ¼ s½

(cid:3) n½ (cid:3)

where [S] represents the resulting stress vector acting on area A, assuming
that the initial coordinate system of the cube will remain unchanged, and
[n] is the vector of direction cosines.

By rotating our coordinate system, all stress components may change in
order to maintain the force balance. For simplicity, we will first study a
coordinate transformation and its effect on the stress components in a two-
dimensional domain, before proceeding towards a general three-dimen-
sional analysis.

2.3 TWO-DIMENSIONAL STRESS TRANSFORMATION

Figure 2.3 shows a steel bar under a tensile load F, where the tensile
stress can simply be determined across the plane p-q, normal to the
loading problem,
applied load. Although, this is a one-dimensional
the stress state is two-dimensional where a side load of zero actually
exists. To develop the idea of coordinate transformation, we examine the
stresses acting on plane m-n, which has an arbitrary orientation relative to
the applied load.

16

Two-Dimensional Stress Transformation

Amn

Apq

F

F

θθ

(a)

FS

θ

F

FN

(b)

spq ¼ F
Apq

Figure 2.3 (a) Tensile force applied on a bar, (b) Projected forces parallel and normal
to surface m-n.

The stress acting on plane p-q is simply expressed as:

Applying force balance on plane m-n will project the applied force F into
the normal force FN and shear force FS, i.e.:

FN ¼ Fcos u

;

FS ¼ Fsin u

The resulting normal and shear stresses acting on plane m-n will be:

smn ¼ FN
Amn

¼ Fcos u
Apq
cosu

¼ spqcos2 u

tmn ¼ FS
Amn

¼ Fsin u
Apq
cosu

¼ spqsin u cos u

Introducing the following trigonometric identities:

2sin ucos u ¼ sin2 u
ð
cos2 u ¼ 1

Þ
1 þ cos2 u

sin2 u ¼ 1

ð

Þ
1 (cid:2) cos2 u

2

2

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Stress and Strain Transformation

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σ
pq

σσ
pq

τ
mn

σ

mn

τ
mn

σ
pq

σ

mn

Figure 2.4 Normal and shear stresses acting in the bar.

The above two stress components can therefore be simplified to
(Figure 2.4):

smn ¼ 1

2

spqð1 þ cos2uÞ

tmn ¼ 1

spqsin2u

2

ð2:3Þ

The relation between the normal and shear stresses can most easily be
illustrated using Mohr’s Circle in which normal stress appears on the
horizontal axis, shear stress corresponds to the vertical axis and the
circle diameter extends to spq as shown in Figure 2.5. By rotating the
imaginary plane, any combination of shear and normal stresses can be
found. Mohr’s circle is used to determine the principal stresses, as
well as for implementing failure analysis using Mohr-Coulomb crite-
rion, which is going to be introduced and discussed in detail
in
Chapter 5.

τ

τ
mn

o

2θ

0.5σ
pq

σ

mn

σ

Figure 2.5 Mohr’s circle for the stress state of plane p-q.

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18

Stress Transformation in Space

Note 2.1: Stresses are transformed according to a squared trigonometric law.
This is because: (a) The criterion for transformation is force balance and not
stress balance according to Newton’s second law; and (b) Both the force
and the area have to be transformed in space. This results in a squared
transformation law.

2.4 STRESS TRANSFORMATION IN SPACE

We have presented how the tractions are transformed using the same
coordinate system. We will now develop a formulation for the stress
transformation in a three-dimensional domain – from the coordinate
system (x, y, z) to a new system (x’, y’, z’), as shown in Figure 2.6.

The transformation is performed in two stages. Firstly, the x’ axis is
rotated to align with the cut plane normal n, and then the stress compo-
nents are calculated (see Figure 2.6).

The tractions for the element along the old coordinate axes, i.e. x, y
and z can be written using the Cauchy’s transformation law as follows:

3

5 ¼

2

4

2

4

0

0

0

Sx
Sy
Sz

x

y

z

txy
sx
txy sy
txz

txz
tyz
tyz sz

3

2

5

4

3

5

cosðx
cosðy
cosðz

0; xÞ
0; yÞ
0; zÞ

ð2:4Þ

We then transform the tractions into the new coordinate system, i.e. (x’,
y’, z’), which is Sx

! Sx

0 .
y

y

0

0

Figure 2.6 Stress transformation from one coordinate system to another.

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Stress and Strain Transformation

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It can be noted that the Cauchy’s transformation law is similar to the
transformation of the force components. Now the transformation of the
area remains, which is carried out when the stresses related to the new
coordinate system are found.

Considering the equilibrium condition of the tetrahedron, and using

Newton’s second law for the first stress component, we can write:
X

0 ¼ 0

Fx

or

0

Sx

0 ¼ Sx

0

xcosðx

x

0; xÞ þ Sx

0

ycosðx

0; yÞ þ Sx

0

zcosðx

0; zÞ

Assuming s
form as:

0 (cid:4) Sx

0

x

0 , the above equation can be expressed in a matrix
x

(cid:2)
0 ¼ cosðx

0; xÞ

s

x

cosðx

0; yÞ

cosðx

0; zÞ

ð2:5Þ

3

5

2
(cid:3) Sx
4
Sx
Sx

0

0

0

x

y

z

By combining Equations 2.4 and 2.5, s

2

4

cosðx
cosðy
cosðz

0; xÞ
0; yÞ
0; zÞ

3
2
T sx
5
4

x

3

2

0 can be given by:
0; xÞ
0; yÞ
0; zÞ

cosðx
cosðy
cosðz

4

5

txy
txy sy
txz

txz
tyz
tyz sz

3

5

s

0 ¼

x

Equation 2.6 presents a general stress transformation relationship for one
of the stress components. To find the rest, the above method will be
repeated five times. The final three-dimensional stress transformation
equation becomes:

h i
s0

¼ q½ (cid:3) s½

(cid:3) q½ (cid:3)T

where:

h i
s0

¼

2

4

q½ (cid:3) ¼

2

4

0

x
0

s
t
t

x
0

x

0

y

0
z

3

5

0
z
0

t
0
x
s
t

y
0

0

y
0

t
0
x
t
0
y
0 s
z

y

z
0
z

and

cosðx
cosðy
cosðz

0; xÞ
0; xÞ
0; xÞ

cosðx
cosðy
cosðz

0; yÞ
0; yÞ
0; yÞ

cosðx
cosðy
cosðz

0; zÞ
0; zÞ
0; zÞ

3

5

ð2:6Þ

ð2:7Þ

ð2:8Þ

20

Stress Transformation in Space

Note 2.2: The complex derivation of the general stress transformation
equation is the result of two processes: (1) determining traction along a
new plane, and (2) rotation of the coordinate system. This is equivalent to
performing a force balance, and also transforming the area.

It can easily be shown that the direction cosines will satisfy the following
equations:

cos2ðx
cos2ðy
cos2ðz

0; xÞ þ cos2ðx
0; xÞ þ cos2ðy
0; xÞ þ cos2ðz

0; yÞ þ cos2ðx
0; yÞ þ cos2ðy
0; yÞ þ cos2ðz

0; zÞ ¼ 1
0; zÞ ¼ 1
0; zÞ ¼ 1

ð2:9Þ

As an example, we now assume that stresses are known in the coordinate
system (x, y, z), and we would like to find the transformed stresses in the
new coordinate system (x’, y’, z’) where the first coordinate system is
rotated by an angle of u around the z-axis to create the second one.

This is a two-dimensional case, because the z-axis remains unchanged

as shown in Figure 2.7.

Using Equation 2.8 and Figure 2.7, it can be seen that:

¼ u

¼ 90

(cid:5) þ u

(cid:4)

(cid:5)

0; y
x

¼ 90

(cid:5) (cid:2) u

(cid:4)
y

(cid:5)

0; y
(cid:5)

(cid:4)

¼ u

(cid:4)

(cid:5)

0; z
x
(cid:5)
0; z

(cid:4)
y

(cid:4)

(cid:5)

(cid:5)

¼ 90

(cid:5)

¼ 90

(cid:5)

¼ 90

0; y
z

(cid:5)

¼ 90

0; z
z

(cid:5)

¼ 0

y’

σ
y’

y

τ
x’y’

τ
x’y’

σ
x’

x’

o

o

x

θ

x

Figure 2.7 Stress components before and after transformation.

(cid:4)

(cid:4)

(cid:5)
0; x
x
(cid:5)

0; x
y

(cid:4)

(cid:5)
0; x
z

y

σ
y

τ
xy

τ
xy

σσ
x

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