report
haojin
viscoelasticity.pdf
Viscoelasticity
Rate
Strain rate
Units [time]-1
Similar definitions for every
x
v
dt
dx
x
1
ij
Viscoelastic behavior
σ
t
Reversible BUT after some time
ε0 = elastic instantaneous
τ = characteristic time to reach ε(t0)/e
(e=2.718)
Time dependent (creep) modulus
at constant stress
Time dependent (relaxation)
modulus at constant strain
Moduli are functions of time
E(t), G(t), K(t)
and the corresponding
compliances
C(t), J(t), D(t)
e.g.,
Relaxation shear modulus:
G(t)=σ(t)/0
Creep compliance:
J(t)= ε(t)/s0
Notation gets messy
Basic assumptions (many times not valid ☺)
1. Linear viscoelasticity
2. Boltzman’s superposition principle
3. Correspondence principle
(elasticity-viscoelasticity)
4. Time-temperature equivalance
Linear viscoelasticity
)()()()(
)()(
)()(
2121
22
11
tttt
THEN
tt
tt
IF
ss
s
s
Boltzman’s superposition principle
Based on linear-viscoelasticity assumption
It is the basis of the beautiful math
behind viscoelasticity
transformFourierindicates
Gi
dtGGt
t
*
)(*)(*)(*)()(*
)()()0()0()( 0
s
s
G
Viscoelastic Characterization
Based on frequency:
1) Low: Creep or relaxation test ( < 1Hz )
2) Medium: DMA (0.1 – 100 Hz)
3) High: Wave propagation (Split Hopkinson Bar) (100 Hz and higher)
• Dynamic Mechanical Analysis (DMA) ▫ Applied strain: = o cos(t) ▫ Response with lag: σ = σo cos(t + )
• Complex Modulus ▫ Storage and Loss components
E*() = E() + iE()
• Dissipation (Loss) Factor tan = E / E
Stress Relaxation Creep
DMA result for viscoelastic material
A practical approximation
• If the viscoelastic moduli (and compliances) change slowly with time then:
)(
1 )(
tJ tG
• In general moduli and compliances are inverse to each other only for their
Fourier transforms
Correspondence principle
)( 3)(3
)( 33
tC I
FL
ItE
FL t
EI
FL
3
3
Elastic solution for deflection
Approximate viscoelastic solution
for deflection
)( 3)(3
)(* 33
C Ε I
FL
I
FL
Viscoelastic solution
for deflection
Same but with
Fourier transforms
Time temperature equivalance
Typical thermal activation based
temperature dependence
Assumes a single activation – i.e. single relaxation mechanism (NOT TRUE)
More practical is to use WLF
Williams-Landel-Ferry
equation
Shift factor
Plot αΤ vs ΔΤ
and fit to get
C1 and C2
Time – temperature
superposition
For amorphous plastics
From experimental data to models
Oops … not linear
Isochronous
stress-strain curve
FIG 21
Intermittent stresses
Viscoelastic solutions
• It is OK to use raw data for creep strain for our
problem (pipe under pressure)
• But this is NOT possible if we want a solution to a
more general problem i.e.
• If pressure varies with time (what about our case?)
• If stress varies with location (what about a bending
problem?)
• In those cases we need a full model for the moduli
and compliances
From experimental data to models
Example: Voigt model
τ = retardation time
Simple Voigt model can not model relaxation
Voigt model
Generalized Voigt
τ = retardation time
Can not model relaxation
The Relaxation Spectrum
If n is sufficiently large the summation can be approximated by the
integral over a continuous distribution of relaxation times.This is called
the “Relaxation Spectrum”. It provides a continuous function of the
relaxation time rather than a discrete set.
0
d)]/t)[exp((F)t(G
Using a logarithmic scale, and a continuous spectrum H(), where
H=F and H dln()=Fd, the relaxation modulus can be related to
H():
)(lnd)]/t)[exp((H)t(G
where H() is the RELAXATION SPECTRUM
Generalized Voigt-Kelvin Model
Similar arguments can be applied for a generalized Voigt-Kelvin model.
n
1i
/t
i
o
n
1i
/t
io
/t
ioi
)e1(J )t(
)t(J
)e1(J)t(
)e1(J)t(
i
i
i
Effect of soil above the pipe
D
q
3
223.0
t
D
E
q
q
3
03.0
t
D
E
q
D+d
gravel duncompacte
and clay,sandy compacted
clay,sandy for kN/m 20 and9, 2,
pressure soil horizontal
pressure soil vertical
091.0
15.0
2
3
H
V
H
V
P
P
d
DP E
EDtE
DP d
modulus elastic
lengthunit per load
Ε
q
C
When a pipe is loaded by two opposite forces F, the
points A, B, C and D are subjected to the maximum
moments which are all equal to:
The stresses due to this loading are:
1) If every point in the cross section is elastically loaded
(stress distribution is linear)
2) If every point in the cross section has plastically
yielded (all points are subjected to the yield stress)
FRM
2 0
lengthpipe, 12
where
2 ,
2
3
0 maxmax
L Lt
I
I
tM
t
y sss
F
F
A
B
D
R
t
2
max
0 4
Lt
M Y s
t M0
smax
t M0
sY
When F is large the pipe
deforms by yielding
at points A, B, C and D
Yielding
