A Brief Review of Elasticity and Viscoelasticity for Solids 1 Introduction

28 H. T. Banks, S. H. Hu and Z. R. Kenz / Adv. Appl. Math. Mech., 3 (2011), pp. 1-51
where H(t) is the Heaviside step function (also called the unit step function in the literature),
t 0 ≥ 0. The solution σ(t) to (3.16) is the relaxation function. With this strain
function, (3.16) can be written as
where δ is the Dirac delta function. Let
σ
η + 1 dσ
κ dt = ε 0δ(t − t 0 ),
ˆσ(s) = L {σ(t)}(s),
where L denotes the Laplace transform. Then taking the Laplace transform of both
sides of the above differential equation we obtain
(
ˆσ(s) = ε 0 e −t 1
0s
η + 1 ) −1 (
κ s = κε0 e −t 0s
s + κ ) −1.
η
Taking the inverse Laplace transform one finds that
[
σ(t) = κ exp − κ ]
η (t − t 0) ε 0 H(t − t 0 ).
The stress relaxation function for the Maxwell model (3.16) is illustrated in Fig. 9 (compare
with Fig. 4).
The creep function corresponds to the creep that occurs under the imposition of a
constant stress given by
σ(t) = σ 0 H(t − t 0 ), and ε(0) = 0,
the solution ε(t) to (3.16) is the creep function. With this stress function, (3.16) can be
written as
dε
dt = σ 0
η H(t − t 0) + σ 0
κ δ(t − t 0).
Then taking the Laplace transform of both sides of the above differential equation we
have that
ˆε(s) = σ 0
η
e −t 0s
s 2 + σ 0
κ
Upon taking the inverse Laplace transform we find
ε(t) =
e −t 0s
[ 1
κ + 1 η (t − t 0)]
σ 0 H(t − t 0 ).
The creep function of Maxwell model (3.16) is illustrated in Fig. 10 (again, compare
with Fig. 5).
s
.