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)

c·

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)

c·

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

c·

„= , (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:

c·

„= , (5.68)

c

Eq. (5.67) may also be written as

F 1

= µ + µ.

™ (5.69)

„

c·

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)