A Brief Review of Elasticity and Viscoelasticity for Solids 1 Introduction

24 H. T. Banks, S. H. Hu and Z. R. Kenz / Adv. Appl. Math. Mech., 3 (2011), pp. 1-51
Note that
(
cos(ωt) = sin ωt + π )
,
2
and thus we can rewrite the above expression as
(
σ(t) = ηϵ 0 ω sin ωt + π )
.
2
The stress amplitude is linear in the strain amplitude: σ 0 = ηϵ 0 ω, which is dependent
on the frequency ω. The stress is out of phase with the strain, and strain lags stress by
a 90 degree phase lag.
• A linear viscoelastic solid.
With the sinusoidal strain (3.10), the stress as a function of time appears complicated
in the first few cycles. But a steady state will eventually be reached in which
the resulting stress is also sinusoidal, having the same angular frequency ω but retarded
in phase by an angle δ. This is true even if the stress rather than the strain is
the controlled variable. The cyclic stress is written as
σ(t) = σ 0 sin(ωt + δ), (3.11)
where the phase shift δ is between 0 and π/2, and the stress amplitude σ 0 depends on
the frequency ω. By an identity of trigonometry, we can rewrite (3.11) as
σ(t) = σ 0 cos(δ) sin(ωt) + σ 0 sin(δ) cos(ωt). (3.12)
Thus the stress is the sum of an in-phase response and out-of-phase response.
We consider the energy loss for a linear viscoelastic material such as described by
the stress (3.11) in response to the strain input (3.10). Let l be the length of a rod with
cross sectional area a. When the solid is strained sinusoidally, according to (3.10), the
solid elongates as
∆l(t) = lε 0 sin(ωt).
By (3.12), the force on the rod is
F(t) = aσ(t) = aσ 0 cos(δ) sin(ωt) + aσ 0 sin(δ) cos(ωt).
During a time interval dt, the solid elongates by d∆l, and the work done on the rod is
In one full cycle the work done is
∫ 2π
ω
W =alε 0 σ 0 ω cos(δ)
0
=πalε 0 σ 0 sin(δ).
F(t)d∆l = F(t) d∆l
dt dt = lε 0ωF(t) cos(ωt)dt.
∫ 2π
ω
sin(ωt) cos(ωt)dt + alε 0 σ 0 ω sin(δ)
0
cos(ωt) cos(ωt)dt