Page 68 - Physical Principles of Sedimentary Basin Analysis

50 Linear elasticity and continuum mechanics

σ s

τ

Figure 3.9. The stress vector, normal stress and the shear stress on a plane.

2
2
2
The next step is to use that the normal vector has unit length, n + n + n = 1, and to do
1 2 3
2
2
2
2
2
2
2
2
2
the replacements 1 − n = n + n ,1 − n = n + n and 1 − n = n + n . By collecting
1 2 3 2 1 3 3 1 2
terms we get

2 2
2
2 2
2
2 2
2
2
τ = σ − 2σ 1 σ 2 + σ 2 n n + σ − 2σ 2 σ 3 + σ 2 n n + σ − 2σ 3 σ 1 + σ 2 n n
1 2 1 2 2 3 2 3 3 1 3 1
(3.52)
which is (3.49) for the shear stress.
Exercise 3.6 Show that the invariants (3.44) expressed with principal stresses are
I 1 = σ 1 + σ 2 + σ 3 (3.53)
1 2 2 2 2
I 2 = (σ + σ + σ − I ) (3.54)
2
1
1
3
2
I 3 = σ 1 σ 2 σ 3 . (3.55)
Exercise 3.7 Show that the invariants (3.44) for the general stress tensor can be written
(3.56)
I 1 = σ ii
1 2
I 2 = (σ ij σ ij − I ) (3.57)
1
2
I 3 =|σ|. (3.58)
Exercise 3.8 Show that the invariants (3.44) really are invariant under rotation of the stress

tensor. In other words, if σ = RσR −1 is a rotated stress tensor, show that the invariants of

σ and σ are the same.
Solution:
This solution is based on the representation of the identity matrix known as the Kronecker
delta δ ij , where δ ij is one whenever i = j and otherwise zero. Since the Kronecker delta
δ ij is the identity matrix we have that R −1 R kj = δ ij . We will also use that the transpose of
ik
−1 T
the rotation matrix is the inverse, R = R = R ji .
ij ij

(1) The first invariant of σ is

I = σ = R ij σ jk R −1 = R −1 R ij σ jk = δ kj σ jk = σ kk = I 1 . (3.59)
1 ii ki ki
(2) Exercise 3.7 shows that is sufficient to show that J = σ ij σ ij is invariant under rotation

of the stress tensor. The invariant of σ is