12.005 Lecture Notes 5
Quantities in Different Coordinate Systems
How to express quantities in different coordinate systems?
x3'
x3
Direction Cosines
P
Axis
φ11
x1
x2'
φ23
x2
φ12
x1'
x1'
x1
x2
x3
α11
α12
α13
x2'
α21
α22
α23
x3'
α31
α32
α33
Figure 5.1
Figure 5.1
Figure by MIT OCW.
Direction cosine ij is cosine of angle φ ij between primed axis i and unprimed axis j.
If xˆi ' and x̂ j represent unit vectors that are the axes of two coordinate systems with the
same origin, they are related by the equation
3
3
j=1
j=1
xˆ i ' = ∑ α ij xˆ j , x i ' = ∑ α ij x j → α ij x j
where αij is the cosine of the angle between the primed axis xˆi ' and the unprimed axis x̂ j .
For example, α12 is the cosine of the angle between xˆ1 ' and x̂ 2 . αij represents a 9component matrix called the transformation matrix. Unlike the stress tensor, it is not
symmetric (αij ≠ αji).
⎡α 11 α 12 α 13 ⎤
α ij = ⎢⎢α 21 α 22 α 23 ⎥⎥
⎢⎣α 31 α 32 α 33 ⎥⎦
In matix equations, the transformation law is written
⎡ x1 ' ⎤ ⎡α 11 α 12 α 13 ⎤ ⎡ x1 ⎤
⎥⎢ ⎥
⎢ x '⎥ = ⎢α
⎢ 2 ⎥ ⎢ 21 α 22 α 23 ⎥ ⎢ x 2 ⎥
⎢⎣ x3 '⎥⎦ ⎢⎣α 31 α 32 α 33 ⎥⎦ ⎢⎣ x3 ⎥⎦
The inverse transformation law is written
xˆ i = α ji xˆ j '
Consider the following transformation of coordinates:
x2
x1'
x2'
30o
x3, x3'
Figure 5.1a
Figure 5.1a
Figure by MIT OCW.
x1
The transformation matrix is
⎡ 30° 60° 0⎤
α ij = ⎢⎢120° 30° 0⎥⎥
⎢⎣ 0
0 0⎥⎦
The explicit transformation equations are
xˆ1 ' = cos 30° xˆ1 + cos 60° xˆ 2
xˆ 2 ' = cos120° xˆ1 + cos 30° xˆ 2
Since xˆi ' and x̂ j are both unit length, these equations are easy to verify from the picture.
First-order tensors
First-order tensors or vectors have two components in 2D coordinates and three
components in 3D coordinates. They transform according to the same laws as coordinate
axes because coordinate axes are themselves vectors.
If uj is a vector in the x̂ j coordinate system and ui’ is a vector in the xˆi ' coordinate system,
then the following equations describe their transformation:
u i ' = α ij u j
u i = α ji u j '
Note that αij is positive if the angle is measured counterclockwise from xˆi ' to x̂ j . It is
negative if the angle is measured clockwise.
Second-order tensors
The transformation law for second-order tensors like stress and strain is more
complicated than the transformation law for first-order tensors. It may be derived as
follows:
1. Begin with the vector transformation of traction Ti to Tk ' :
Tk ' = α ki Ti
2. Rewrite Tk ' and Ti using Cauchy’s formulas:
Tk ' = σ ' kl nl ' and Ti = σ ij n j
Substitute Cauchy’s formulas into the original transformation equation:
σ ' kl nl ' = α kiσ ij n j
3. Transform the normal vector nj to nl’ and substitute into the previous equation:
n j = α lj nl '
σ ' kl nl ' = α kiσ ij α lj nl '
4. Cancel the nl’ term on each side and group the αs:
σ ' kl = α kiα jl σ ij
Note that changing the position of the last α term changes the order of its
subscripts.
In vector notation, the equation is
σ ' = ασα T
where the double under ~ denote second-rank tensors and the superscript T
denotes the transpose of matrix α.
Mohr’s Circle
We explained before that an object resting on a slope will slide down when the shear
traction on the slope is greater than or equal to the product of the normal traction and the
coefficient of friction.
τ = f sσ n
On a shallow slope, σn is large and the object will not slide. On a steep slow, τ is large
and the object will slide. For any plane with normal n̂ , we can calculate if the plane will
fail if the stress tensor σij at the interface between the object and the slope is known.
Calculate σn and , τ as follows:
Vector and Tensor Notation
T = σ nˆ
Summation Notation
Ti = σ ijn j
σ n = T ⋅ nˆ
σ n = Ti n i
τ = T − σ n nˆ
τ = Ti − Tk n k n i
This method is straightforward but cumbersome. A different approach involves rotating
the coordinate system such that x1’ is along n̂ . In this case σn and τ are much easier to
derive:
σ n = σ '11
τ = σ '12
Mohr’s circle may be derived in two or three dimensions. This lecture explains the
derivation in two dimensions because it is more straightforward and the results are easier
to graph and understand. The derivation assumes that x1, x2, and x3 are principle
directions.
Consider the following figure in which the xi coordinate system is rotated clockwise
about the x3 axis to xi’:
x1
θ
x1'
φ11
Figure 5.2
x2'
x2
φ22
φ21
Figure 5.2
Figure by MIT OCW.
The rotation matrix is:
⎡α 11 α 12 α 13 ⎤ ⎡cos θ
α ij = ⎢⎢α 21 α 22 α 23 ⎥⎥ = ⎢⎢ sin θ
⎢⎣α 31 α 32 α 33 ⎥⎦ ⎢⎣ 0
− sin θ
cos θ
0
0⎤
0⎥⎥
1⎥⎦
The stress tensor σ in the xi coordinate system is transformed to σ ' in the xi’ coordinate
system by the following equation:
σ ' = ασα T
⎡cosθ −sin θ
⎢
σ ' = ⎢ sin θ cosθ
⎢⎣ 0
0
0⎤⎡σ 11 0
0 ⎤⎡ cosθ
⎥⎢
⎥⎢
0⎥⎢ 0 σ 22 0 ⎥⎢−sin θ
1⎥⎦⎢⎣ 0
0 σ 33 ⎥⎦⎢⎣ 0
⎡σ 11 cosθ −σ 22 sin θ 0 ⎤⎡ cosθ
⎢
⎥⎢
0 ⎥⎢−sin θ
= ⎢ σ 11 sin θ σ 22 cosθ
⎢⎣ 0
0
σ 33 ⎥⎦⎢⎣ 0
sin θ
cosθ
0
sin θ
cosθ
0
0⎤
⎥
0⎥
1⎥⎦
0⎤
⎥
0⎥
1⎥⎦
⎡σ 11 cos2 θ + σ 22 sin 2 θ (σ 11 − σ 22 )sin θ cosθ
0 ⎤
⎢
⎥
2
2
= ⎢ (σ 11 − σ 22 )sin θ cosθ σ 11 sin θ + σ 22 cos θ 0 ⎥
⎢
0
0
σ 33 ⎥⎦
⎣
Use the double-angle identities for sine and cosine to simplify the expressions for the
normal stress σ11’ and the shear stress σ12’ become in the new coordinate system:
sin 2θ = 2 sin θ cos θ
cos 2θ = 2 cos 2 θ − 1 = 1 − 2 sin 2 θ
σ 11 ' = σ n =
σ 11 + σ 22
+
σ 11 − σ 22
2
2
σ − σ 22
σ 12 ' = τ = 11
sin 2θ
2
cos 2θ
The expression for the normal stress and shear stress can be shown graphically in shear
space:
τ
Allowable tractions
2θ
σ2
σ1
σn
τ
Figure 5.3
Figure by MIT OCW.
This figure is called Mohr’s circle.
Mohr’s circle plots in stress space. Admonton’s law may also be plotted in stress space
as a line with slope fs. When this line and Mohr’s circle intersect, the criterion for failure
across a plane is met.
A note on signs.
As derived, assumed σ 1 > σ 2 . Simplest case: consider θ = 45 o
σ1
n
σ2
x2'
45 o
x1
θ
=
P
v
x1
x2
x2
Figure 5.4
Figure 5.4
Figure by MIT OCW.
P
x1'
Consider several cases:
τ
1
0
2
σ2
σ1
σn
4
3
Figure 5.5
Figure by MIT OCW.
0) 2θ = 60o
1) 2θ = 90o
2) 2θ = 180o
3) 2θ = 270o
4) 2θ = 300o
Back to the landslide:
Consider a common experiment in soil mechanics or rock mechanics in which scientists
apply a uniaxial stress σ2 to a cylindrical sample confined by a uniform stress σ1.
x2
x1
x3
σ2
Figure 5.6
σ1
σ1
Figure 5.6
Figure by MIT OCW.
Continue increasing σ2 until failure. n̂ of failure plane typically at θ ; 30o from σ1.
Seal
Rubber jacket
Specimen
Figure 5.7
Confining pressure
Figure 5.7. Triaxial test deformation apparatus
Figure 5.7
Figure by MIT OCW.
55
σ3 = 5000 p.s.i.
50
35
30
25
25
Stress: k.p.s.i.
30
20
15
10
5
0
0
1
40
2
3 4
Millistrains
5
35
Stress: k.p.s.i.
35
Stress: k.p.s.i.
40
σ3 = 0
45
σ3 = 1000 p.s.i.
20
30
25
20
15
15
10
10
5
5
0
0
1
2
3 4
Millistrains
5
0
0
1
2
3 4
5
Millistrains
6
7
Figure 5.8. Stress-strain curves for Rand quartzite at various confining pressures.
Figure 5.8
1) Shear fracture.
2) Extension fracture.
3) Multiple shear fractures.
Figure 5.9
4) Extension fracture produced
by line loads.
5) Longitudinal splitting in
uniaxial compression.
Figure 5.9
Figures by MIT OCW.
9
8
1650
7
Strain (%)
6
685
5
500
4
3
3260
845
235
2
Figure 5.10
1
0
2490
0
0
1000
2000
3000
4000
5000
σ1- σ3 (bars)
Figure 5.10. Stress-strain curves for Carrara marble at various confining
pressures. The numbers on the curves are confining pressures in bars.
Figure 5.10
15
Strain (%)
Figure 5.11
800 oC
10
500 oC
300 oC
5
25 oC
0
0
5
10
15
20
σ1- σ3 (k.bars)
Figure 5.11. Stress-strain curves for granite at a confining pressure of 5 kilobars
and various temperatures.
Figure 5.11
Figures by MIT OCW.
Analyze this using a Mohr circle diagram:
τ
| τ| = µ| σn| + | σ0|
| σ0|
σ2
σ2
σn
σ1
Figure 5.12
Figure 5.12
Figure by MIT OCW.
Mohr circle tangent to failure curve ⇒ sample breaks.
τ
τ
φ = tan-1µ
2θ
σ2
φ
σn
Figure 5.13
σ1
Compressive stress negative
Compressive stress positive
Figure 5.13
Figure by MIT OCW.
σn
µ is the coefficient of friction.
φ = tan −1 µ ; 2θ = 90 o − φ
σ0
Figure 5.14
2θ = 90o
Figure 5.14
Figure by MIT OCW.
µ
0.0
0.6
1.0
lim
µ →∞
θ
45°
30°
23°
0°
Since most rocks have a coefficient of friction of about 0.6, the normal vector to the
failure plane is typically 30°from the direction of the least compressive stress. Another
way of saying this is that the failure plane is 30°from the direction of the most
compressive stress.
Styles of Faulting
Faults are large-scale failure planes. Since the normals to failure planes are in the plane
containing the least compressive and the most compressive stresses and are typically at
30o from the direction of the least compressive stress, different styles of faulting can be
used to infer the directions of σ1, σ2, and σ3.
The following diagram shows the deviatoric stresses associated with thrust faults, normal
faults, and strike-slip faults.
Total Stress
Figure 5.15
Deviatoric
x
β
y
β
Or
Deviatoric
Total
β
Or
Or
"Left Lateral"
Figure 5.15
Figure by MIT OCW.
"Right Lateral"