12.005 Lecture Notes 19
Stress and strain from a screw dislocation
x3
ce
rfa t
u
S ul
fa
Fault
displacement S
D
o
ati
c
o
sl
s
Di axi
x2
n
S/2
S/2
x1
Figure 19.1
Figure by MIT OCW.
Need to get traction = 0 at surface. First consider ∞ medium.
Assume:
u1 = u3 = 0
u2 =
Sθ
2π
∇ ⋅ u = 0 ⇒ no compression, only shear
Symmetry ⇒ cylindrical coordinates, r, θ , z
with z parallel to x2 axis;
Solution is σ θ z =
ur = uθ = 0
µ ∂uz µ S
=
r ∂θ 2π r
We can get σ ij by coordinate transformation.
How to get traction on surface?
Trick – image dislocation
x3
2D
Image dislocation
Surface
D
Actual dislocation
x1
Figure 19.2
Figure by MIT OCW.
Solution for matched image dislocation is whole space gives σ i3 = 0 on surface of ½
space!
Shear strain at surface:
x3
2D
Surface
D
θ
Figure 19.3
Figure by MIT OCW.
r
x1
r 2 = x12 + x32
S
From each dislocation ε zθ =
2π r
Rotating strain tensor
ε12 = ε zθ cosθ = ε zθ
x3
r
At surface
⎡
⎤
2D − x3
SD
S ⎢ x3
⎥=
ε =
+
2
2
2
2
2
⎥
2π ⎢ x1 + x3
2D − x3 + x12 ⎦ π (D + x1 )
⎣
D
12
(
)
where
x3
is actual dislocation
x + x32
2
1
2D − x3
(2D − x ) + x
2
3
2
1
is image dislocation
Displacement
u2 =
x1
∫ε
D
12
dx1 =
∞
x ⎞
S⎛ 2
⎜1− tan −1 1 ⎟
D⎠
2⎝ π
Aside – slip discontinuity objectionable?
σ 12 → ∞ along x1 = 0 as x3 → 0
(stress singularity at tip of fault)
Alternative model
0
Apply uniform σ 12
Cut 0 ≤ x3 ≤ D, set σ 12 = 0
1/ 2
⎛
⎞
2
x1 ⎟
S ⎜⎛ x1 ⎞
u2 = ⎜⎜1+ 2 ⎟⎟ −
2 ⎜⎝ D ⎠
D⎟
⎝
⎠
Displacement (meters)
2
6
1
3.5
0
0
2
4
6
8
10
Distance from fault (km) NE
Figure 19.4
Figure by MIT OCW.
Virtually indistinguishable!
St. Venant’s principle
Elastostatics – if boundary tractions on a part S1 of the boundary of S are replaced by
statically equivalent traction dist, effects on stress dist are negligible at pts whose
distance from S1 is large compare to size of S.
Usual context – long beam under end load (non-uniform)
See's average end load
Figure 19.5
Figure by MIT OCW.
Apply to loading 1/2 space
Figure 19.6
Pt source approximates in seismology.