Page 65 - Physical Principles of Sedimentary Basin Analysis

3.7 Principal stress 47

1 ∂u ∂u
j
i
= +
ij
2 ∂x ∂x
j i
1 ∂u k ∂u k
= R ik + R jk
2 ∂x ∂x
j i

1 ∂u k ∂x l ∂u k ∂x l
= R ik + R jk
2 ∂x l ∂x ∂x l ∂x
j i
1 ∂u k −1 ∂u k −1
= R ik R lj + R jk R li . (3.34)
2 ∂x l ∂x l
The last term above can be rewritten using that R −1 = R il
li
∂u k −1 −1 ∂u k ∂u k −1 ∂u l −1
R jk R li = R li R jk = R il R kj = R ik R lj (3.35)
∂x l ∂x l ∂x l ∂x k
which finally gives that = R ik kl R −1 . (The steps are, firstly, swapping sides for R jk
ij lj
−1 −1
and R , using that R = R il and then swapping the indices k and l.)
li li
Exercise 3.5
(a) Show that the 2D stress state represented by the stress tensor

σ xx σ xz
σ = (3.36)
σ xz σ zz
becomes rotated into a stress tensor with components
2
2

σ xx = σ xx cos θ + σ zz sin θ + 2σ xz sin θ cos θ (3.37)
2
2

σ zz = σ xx sin θ + σ zz cos θ − 2σ xz sin θ cos θ (3.38)
2
2

σ xz = (−σ xx + σ zz ) sin θ cos θ + σ xz (cos θ − sin θ) (3.39)
by use of the rotation matrix (2.53). There are only three independent components of the
stress tensor σ in 2D, because σ xz = σ zx .
(b) Show that σ + σ = σ xx + σ zz .
xx zz
(c) Show that shear stress vanishes when the rotation angle is such that
2σ xz
tan θ = . (3.40)
σ xx − σ zz
1 2 2
(Use that sin θ cos θ = sin 2θ and that cos 2θ = cos θ − sin θ.)
2
3.7 Principal stress
It is often an advantage to work with normal stress, because a plane with normal stress
does not have shear stress. The normal stress has to be proportional to the normal vector,
and we have