Page 323 - Physical Principles of Sedimentary Basin Analysis

9.8 Elastic and viscous deformations 305

Subtracting the elastic strain (9.103) from the total strain (t) gives the viscous strain
e −t/t e t

v (t) = (t) − e (t) = ˙ (t ) e t /t e dt . (9.104)
t e 0
These expressions for the elastic and viscous strain can be used to make similar expres-
sions for the elastic and viscous deflections. The strains are related to the deflections
by equations (9.93) and (9.94), which can be inserted into equations (9.103)to(9.104).
Integration with respect to x two times then gives

e −t/t e t

w e (t) = w(t) − ˙ w(t ) e t /t e dt (9.105)
t e 0
and
e −t/t e t

w v (t) = ˙ w(t ) e t /t e dt . (9.106)
t e 0
We see that the elastic and viscous strain and deflections are related to the total strain and
deflections, respectively, by similar relationships.
Note 9.7 Equation (9.102) is multiplied by e t/t e , and we get
1
˙ e e t/t e + v e t/t e =˙ e t/t e . (9.107)
t e
The left-hand side of the equation can be rewritten as follows using differentiation with
respect to time:
∂
e e t/t e =˙ e t/t e (9.108)
∂t
which is then straightforward to integrate. The integral of the right-hand side with respect
to time is done by integration by parts:
t t t
1

˙ (t ) e t /t e dt = (t ) e t /t e − ˙ (t ) e t /t e dt . (9.109)
0 0 t e 0
An important point is that the initial total strain (or deformation) is equal to the initial
elastic strain (or deformation), (t =0) = e (t =0), because it takes time to accumulate
viscous (permanent) strain.
t
Exercise 9.9 Show that w v (t) = w v (0) + (1/t e ) w e (t )dt .

0
Hint: use that ˙ v = e /t e .
Exercise 9.10 (a) Find the deflection of a viscoelastic beam of length l that is clamped at
one end and loaded at the free end with a time dependent force F(t). (See Figure 9.5, and
also Exercise 9.4.)
(b) What are the elastic and viscous parts of the deflection?
(c) At what time is the viscous (permanent) deflection equal to the elastic deflection when
the load is constant F(t) = F 0 ?