Page 77 - Physical Principles of Sedimentary Basin Analysis

3.11 Linear stress–strain relations 59

y –σ 2 x

σ yx σ xy σ
σ 1 σ 1 σ 1 xy
σ xy σ yx

–σ
–σ 2 2
(a) (b)
Figure 3.14. (a) The square is compressed in the horizontal direction and stretched in the vertical
direction. (b) The inner (rotated) square has shear stress σ xy =−σ 1 = σ 2 and zero normal stress,
when σ 1 =−σ 2 . This stress condition is called pure shear as opposed to the stress condition in
Figure 3.2, which is called simple shear.

(c) Show that

xy = 1 (3.105)
which gives that the shear modulus is

E
G = . (3.106)
2(1 + ν)
Solution: (a) The force balance on the triangle in Figure 3.14b, in the direction of the shear
stress, is
√ 1 1
2σ xy =−√ σ 1 + √ σ 2 (3.107)
2 2
which gives that σ xy =−σ 1 = σ 2 , when σ 1 =−σ 2 . The force balance in the x-direction
shows that the normal stress is zero.
(b) The strain in the x- and z-directions is

1 ν 1
− 1 = σ 1 − σ 2 = (1 + ν)σ 1 (3.108)
E E E
ν 1 1
− 2 =− σ 1 + σ 2 =− (1 + ν)σ 1 . (3.109)
E E E
(Recall the sign convention saying that compressive stresses are positive. The sign reversal
is also applied to the shear stress–strain relationship, see preceding section.)
(c) Strain is rotated as a matrix in the same way as stress. The normal strain is therefore
zero ( xx = 0) and the shear strain is xy =− 1 = 2 . Shear strain and shear stress
1 1
are then related as xy =− 1 = (1 + ν)σ 1 =− (1 + ν)σ xy , which gives the shear
E E
modulus (3.106), when compared with the relation σ xy =−2G xy .