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264 Rheology: fracture and flow
τ
τ s
τ d
t s
time
Figure 8.6. The shear stress along a stick–slip fault as a function of time.
shows that the static friction is larger than the dynamic friction (because the elastic shear
stress G is larger than zero). The shear strain of the fault is
τ s − τ d
= (8.9)
G
which is proportional to the difference between static and dynamic friction. If the sheared
zone has the width h on both sides of the fault the displacement during fault slip is w =
2h . The fault in Figure 8.5 is in a state of simple shear as shown by the example in
Figure 3.2. If the shear strain accumulates linearly in time as =˙ t, at a constant strain
rate ˙ , the time between each fault slip is
τ s − τ d
t s = (8.10)
G ˙
which is inversely proportional to the strain rate. The fault therefore accumulates stress
over a long period of time if the strain rate is very small. Figure 8.6 shows the shear stress
increasing linearly with time, from τ d to τ s , until the fault slips.
Exercise 8.1 What is the displacement w along a fault when τ s − τ d = 100 MPa,
G = 30 GPa and h = 100 m?
8.4 The slider-block model of stick–slip motion
Another simple model for this stick–slip motion of faults is the movement of a block drawn
by a spring over a surface as shown in Figure 8.7. The block has the weight m and the free
tip of the spring moves with a constant velocity v. The block does not begin to move
until the force from the spring exceeds the static friction. The static friction is the force
that opposes the initiation of sliding, while dynamic friction is the force that opposes the
movement of the block during motion. Both the static and the dynamic friction follow
Amontons’ law, and they are proportional to the weight of the block. The coefficient of
static friction and dynamic friction are denoted μ s and μ d , respectively. Newton’s second
law gives the following equation for the movement of the block:
2
d x
m = k (x s + vt − x) − μ d mg (8.11)
dt 2