CONTINUUM MECHANICS for ENGINEERS

γ ()= t γ ( cos ωt+ isinωt)= γ e
12
o o
(9.6-2)
where i = −1.
It is understood that physically the real part of the resulting
stress corresponds to the real part of the applied strain, and likewise, the
imaginary parts of each are directly related. The stress resulting from the
excitation prescribed by Eq 9.6-2 will have the same frequency, ω, as the
imposed strain. Therefore, expressing the stress as
ω t
12()= t ei ∗
σ σ
(9.6-3)
where σ * is complex, the response will consist of two parts: a steady-state
response which will be a function of the frequency ω, and a transient response
that decays exponentially with time. It is solely with the steady-state
response that we concern ourselves in the remainder of this section.
Substituting Eq 9.6-2 and Eq 9.6-3 into the fundamental viscoelastic constitutive
equation given by Eq 9.3-1, and keeping in mind the form of the
operators {P} and {Q} as listed by Eq 5.12-7 we obtain
N
∑
k=
0
Canceling the common factor e iωt we solve for the ratio
(9.6-4)
(9.6-5)
which we define as the complex modulus, G * (iω), and write it in the form
(9.6-6)
The real part, G′(ω), of this modulus is associated with the amount of energy
stored in the cube during a complete loading cycle and is called the storage
modulus. The imaginary part, G″(ω), relates to the energy dissipated per cycle
and is called the loss modulus.
In terms of the complex modulus, the stress σ 12 as assumed in Eq 9.6-3
may now be written
∑
iωt N
k iωt k iωt o k
k=
0
( ) = ( )
∗
σ p iω e γ q iω e
k
∗
σ
=
γ
o
N
∑
k=
0
N
∑
k=
0
( )
p iω
k
( )
q iω
∗
σ ∗
= G ( iω)= G′( ω)+ iG′′(
ω)
γ
o
12 = ( ) o = [ ′( )+ ′′( ) ]
∗
i t
σ G iω γ e G ω iG ω γ e
k
k
k
ω iωt o
(9.6-7)