Page 197 - Mechanics of Asphalt Microstructure and Micromechanics

Fundamentals of Phenomenological Models 189

This actually represents that there is no coupling between volumetric strain and
deviatory stresses:
s = 2 Gd
ij ij (6-117)
σ = 3 Kε
ii ii
G is shear modulus and K is the bulk modulus:

G = 1 Q 2 , K = 1 Q 3 , E = 9 KG , ν = 3 K − 2 G
+
2 P 3 P 3 KG 6 K + 2 G
2 3
6.3.7 Complex Modulus and Compliance
It would be interesting to see the responses of a viscoelastic material to a cyclic loading,
σ σ
for example the material modeled with the Maxwell model ε = + , subjected a
R η
stress σ = σ cos ωt. By solving the governing equation, one can obtain the following
0
δ
strain response: ε = ε cos( ω − ) :
t
0
Where tanδ = R and ε = σ 0 . (6-118)
η 0 1 + 1
2
R 2 ωη 2
For a more general case, when the load is σ = σ e ω it
0
ω
t isin
Where e it = cos ω + ω t
The strain response is:
ε = ε cos( ω − )
δ
t
0
Or ε = ε e i( ωδ )
−
t
0
ε
*
Or ε = ( e − δi )e ω i t = ε e ω i t
0
Denoting ε = ε e −i δ = ε (cos − i sin δ) and placing both the s and e in the general
δ
*
0 0
governing equation, one can have the following equation:
σ
ω
iω
[p + i p +ω ( ) p +ω 2 ...] e it = [q + i q +ω ( ) q + ...]ε e ω (6-119)
* it
2
i
0 1 2 0 0 1 2 2
or σ σ
δ
iδ
E = 0 e = 0 [cos + isin ] = E + iE = E e iδ (6-120)
δ
*
*
1
2
ε 0 ε 0
6.3.8 Energy Dissipation
It would be interesting to see how energy is dissipated in one loading cycle (T is the
period):
T dε
∫
ΔW = σ dt (6-121)
0 dt
ΔW = ∫ 0 T εσ ω sin ω tcos( ω t −δ) dt
0
0
ΔW = πσ ε sin δ
00
ΔW = πε 2 E
0 2