Page 67 - Physical Principles of Sedimentary Basin Analysis

3.7 Principal stress 49

Working with the stress tensor is considerably simplified in the principal system. The stress
vector on any plane oriented with the normal vector n in the principal system is

S = σn = (σ 1 n 1 ,σ 2 n 2 ,σ 3 n 3 ) T (3.47)

and the normal stress on the plane becomes
2 2 2
σ = n · S = σ 1 n + σ 2 n + σ 3 n . (3.48)
1 2 3
The shear stress on the plane is most easily given to the power of two:
2 2
2
2
2
2 2
2
2 2
τ = (σ 1 − σ 3 ) n n + (σ 1 − σ 2 ) n n + (σ 2 − σ 3 ) n n (3.49)
1 3 1 2 3 2
as shown in Note 3.4.
Note 3.2 Real eigenvalues. We have that σn i = σ i n i and we want to show that the eigen-
values σ i are real for a real and symmetric matrix. Some more notation is needed before
∗
we can do that. The complex conjugate of a scalar a and a matrix A are denoted a and A ,
∗
respectively. The combined operation of complex conjugation and transposing is denoted
†
∗ T
A = (A ) . The stress tensor does not change by taking the transpose and the conjugate,
σ † = σ, because it is symmetric and real. The transpose and conjugate of σn i = σ i n i
† ∗ † † ∗ †
is therefore n σ = σ n and right multiplication by n i gives n σn i = σ n · n i .Left
i i i i i i
†
†
†
multiplication of σn i = σ i n i by n gives n σn i = σ i n ·n i . We have now two expressions
i
i
i
† ∗ † ∗
for n σn i and their difference is (σ − σ i )n · n i = 0, which implies that σ = σ i because
i
i
i
i
†
n · n i = 0.
i
Note 3.3 Orthogonal eigenvectors. The orthogonality of the eigenvectors of a real and
symmetric matrix are shown by a similar reasoning as in Note 3.2.Let σn i = σ i n i and
σn j = σ i n j . Taking the transpose of the first equation before right multiplication by n j ,
T T
and then left multiplication of the second equation by n gives n σn j = σ i n i · n j =
i i
σ j n i · n j , and the difference becomes (σ i − σ j )n i · n j = 0. The eigenvectors for different
eigenvalues are therefore orthogonal. In the case that the eigenvalues are equal, σ i = σ j ,
it is not possible to conclude that n i · n j = 0. But any linear combination of eigenvectors
with the same eigenvalue is a new eigenvector, and it is therefore possible to construct two
new eigenvectors that are orthogonal, for instance m i = n i and m j = n j − an i , where
a = (n i · n j )/(n i · n i ).
Note 3.4 Figure 3.9 shows the shear stress, normal stress and the stress vector, and the size
of the shear stress follows from the Pythagorean theorem:
2
2
2
2
2
2
2
2

τ = S − σ = (σ 1 n 1 ) + (σ 2 n 2 ) + (σ 3 n 3 ) − σ 1 n + σ 2 n + σ 3 n 2 2 . (3.50)
1 2 3
We expand the last parentheses and regroup the terms to get
2 2 2 2 2 2 2 2 2 2
τ = σ n (1 − n ) + σ n (1 − n ) + σ n (1 − n )
1
3
1 1
2 2
3 3
2
2 2 2 2 2 2
− 2σ 1 σ 2 n n − 2σ 2 σ 3 n n − 2σ 1 σ 3 n n . (3.51)
2 3
1 3
1 2