<<

. 17
( 67 .)



>>

™ (5.38)

In the Maxwell model according to the set-up of Fig. 5.10(a) the strain µ is addi-
tionally composed of the strain in the spring (µs ) and the strain in the damper (µd ):

µ = µs + µd , (5.39)

implying that

µ = µs + µd .
™™ ™ (5.40)


F F
F
F


(a) Spring-dashpot in series (b) Spring-dashpot in parallel

Figure 5.10
A Maxwell (a) and Kelvin-Voigt (b) arrangement of the spring and dashpot.
79 5.3 Linear visco-elastic behaviour

The forces in both the spring and the damper must be the same, therefore, based
on Eq. (5.38), the strain rates in the spring and dashpot satisfy
1

µs =
™ F (5.41)
c
and
F
µd =
™ . (5.42)

Eq. (5.40) reveals:
1 F

µ=
™ F+ . (5.43)
c c·
This is rewritten (by multiplication with c) as
c

F + F = c™ , µ (5.44)

and with introduction of the so-called relaxation time „ :

„= , (5.45)
c
the ¬nal expression is obtained:
1

F+ F = c™ .
µ (5.46)

This differential equation is subject to the condition that for t < 0 the force F and
the strain rate µ vanish. To ¬nd a solution of this differential equation, the force F

is split into a solution Fh of the homogeneous equation:
1

Fh + Fh = 0 (5.47)

and one particular solution Fp of the inhomogeneous Eq. (5.46):

F = F h + Fp . (5.48)

The homogeneous solution is of the form:

Fh = c1 ec2 t , (5.49)

with c1 and c2 constants. Substitution into Eq. (5.47) yields
1 1
Fh = c1 e’t/„ .
c1 c2 ec2 t + c1 ec2 t = 0 ’’ c2 = ’ ’’ (5.50)
„ „
The solution Fp is found by selecting

Fp = C( t) e’t/„ , (5.51)
Fibres: time-dependent behaviour
80

with C( t) to be determined. Substitution into Eq. (5.46) yields
dC ’t/„ C ’t/„ C ’t/„ dC
’e +e = c™
µ ’’ = c et/„ µ,

e (5.52)
„ „
dt dt
1
dFp
„ Fp
dt

hence

c eξ/„ µ( ξ ) dξ .
C= ™ (5.53)

Because the strain rate µ = 0 for all t < 0, it follows that

t
c eξ/„ µ ( ξ ) dξ e’t/„ .
Fp = ™ (5.54)
0
Combining Eqs. (5.49) and (5.54), the solution F is given by
t
’t/„
ceξ/„ µ( ξ ) dξ e’t/„ .
F = c1 e + ™ (5.55)
0
Requiring that for all t < 0 the force satis¬es F = 0 leads to

F( t = 0) = c1 , c1 = 0. (5.56)

Consequently, the solution of the ¬rst-order differential equation Eq. (5.46), is
given by
t
e’(t’ξ )/„ µ ( ξ ) dξ .

F( t) = c (5.57)
0
Apparently, the integral equation as introduced in the previous section, Eq. (5.26),
can be considered as a general solution of a differential equation. In the present
case the relaxation spectrum, as de¬ned in Eq. (5.37) is built up by just one single
Maxwell element and in this case: G( t) = e’t/„ .
To understand the implications of this model, consider a strain history as spec-
i¬ed in Fig. 5.11, addressing a spring-dashpot system in which one end point is
¬xed while the other end point has a prescribed displacement history. The force
response is given in Fig. 5.12 in case t— = 5„ . Notice that in this ¬gure the time
has been scaled with the relaxation time „ , while the force has been scaled with
c· r, with r the strain rate, see Fig. 5.11. Two regimes may be distinguished.
(i) For t < t— the strain proceeds linearly in time leading to a constant strain rate r. In
this case the force response is given by (recall that c· = „ c)
t
F = c· r( 1 ’ e’ „ ) . (5.58)

„ it holds that
For t
t
t
e’ „ ≈ 1 ’ , (5.59)

81 5.3 Linear visco-elastic behaviour


µ µ—
µ ™ r= t—

µ— r




t—
t—
t t
(a) (b)

Figure 5.11
Strain(a) and strain rate (b) as a function of time.




1
c·r
F




0


0 5 10 15
t /„

Figure 5.12
Force response of the Maxwell model.



such that the force in that case is given by
t
F = c· r = c t r. (5.60)

So, for relatively small times t, and constant strain rate, the response is dominantly
elastic. This is consistent with the spring-dashpot con¬guration. For small t only
the spring is extended while the dashpot is hardly active. Furthermore, at t = 0:


F = cr, (5.61)

which implies that the line tangent to the force versus time curve should have a
slope of cr. For larger values of t, but still smaller than t— we have
t
e’ „ ’ 0, (5.62)
Fibres: time-dependent behaviour
82

such that the force is given by

F = c· r, (5.63)

which is a purely viscous response. In this case the spring has a constant extension,
and the force response is dominated by the dashpot. This explains why, at a constant
strain rate, the force curve tends towards an asymptote in Fig. 5.12 for suf¬ciently
large t.
(ii) For times t > t— the strain rate is zero. In that case, the force decreases exponen-
tially in time. This is called relaxation. If F — denotes the force t = t— , the force for
t > t— is given by
— )/„
F = F — e’(t’t . (5.64)

The rate of force relaxation is determined by „ , which explains why „ is called a
relaxation time. At t = t— , the slope of the tangent to the force curve equals
F—
™ =’ .
F (5.65)
t=t— „



5.3.3 Visco-elastic models based on springs and dashpots: Kelvin“Voigt model

A second example of combined viscous and elastic behaviour is obtained for the
set-up of Fig. 5.10(b). In this case the total force F equals the sum of the forces
due to the elastic spring and the viscous damper:

F = c µ + c· µ ,
™ (5.66)

or, alternatively after dividing by c· :
F c
= µ + µ.
™ (5.67)
c· c·
Introducing the retardation time:

„= , (5.68)
c
Eq. (5.67) may also be written as
F 1
= µ + µ.
™ (5.69)


The set-up according to Fig. 5.10(b) is known as the Kelvin“Voigt model. In anal-
ogy with the Maxwell model, the solution of this differential equation is given by
t
1
e’(t’ξ )/„ F( ξ ) dξ .
µ( t) = (5.70)
c· 0
83 5.4 Harmonic excitation of visco-elastic materials

In the case of a constant force F, the strain response is given by
F
( 1 ’ e’t/„ ) .
µ( t) = (5.71)
c
This phenomenon of an increasing strain with a constant force (up to a maximum
„:
of F/c) is called creep. For t
Ft
µ≈ , (5.72)

<<

. 17
( 67 .)



>>