VISCOSITY
 Plastic deformation occurs by dislocation
motions in crystallic structures.
 For noncrystallic structures plastic
deformations are due to viscous flow.
 The characteristic property for viscous flow,
viscosity, is a measure of a noncrystalline
material’s resistance to deformation.
Plate A
F
L
V
When a tangential force
(F) acts on the plate,
the plate moves with
respect to the bottom.
 The velocity of the liquid particles in each
layer is a function of the distance L. Thus the
rate at which the particles change their
position is the measure of the rate of flow.
velocity
gradient
dV
dL
=
dγ
dt
rate of
flow
dV
 Newton expresses: F = η
A
dL
Since τ = F / A
η : coefficient of viscosity
1
τ = η dV
dL
&
τ=η
dγ
dt
2
Unit of viscosity is Pa.s (Pascal-seconds)
(N.s/m2)
 The liquids that follow equations (1) & (2) are
termed as Newtonian Liquids.

Newtonian
liquids
η
 Viscosity also varies with temperature.
A: Constant
1 = A . e-E/RT
η
E: Energy of activation
R: Gas constant
T: Absolute temperature
 When solid particles are introduced into
Newtonian liquids, viscosity increases.
1) η = η0 (1+2.5 Ø)
η0: Coefficient of
viscosity of the
parent liquid.
2) η = η0 (1+2.5 Ø + 14.1 Ø2) Ø: Volume
concentration of
suspended particles
NON-NEWTONIAN MATERIALS
 In certain materials, τ-dγ/dt does not obey
the linearity described by Newton, i.e.
Viscosity may vary with the rate of shear
strain.
Dilatant:
η
increases
Newtonian
with increasing dγ/dt or
τ (clay)
η
Newtonian: (all liquids)
Pseudoplastic: η
decreases with dγ/dt or
τ (most plastics)
 The relationship between dγ/dt & τ can be
described by the following general equation.
dγ
= τn . 1
η
dt
If n=1 → Newtonian
n > 1 → Pseudoplastic
n < 1 → Dilatant
Fresh cement pastes & mortars, have highly
concentrated solid particles in the liquid
medium. Such a behaviour is described by
Bingham’s equation.
τ = τy + η dγ
dt
dγ
dt
τy
τ
(Upto τy there is no flow)
VISCOELASTICITY & RHEOLOGICAL
CONCEPTS
 Viscoelastic behaviour, as the name implies,
is a combination of elasticity & viscosity.
 Such a behaviour can be described by
Rheological Models consisting of springs (for
elasticity) & dashpots (for viscosity).
Load
(Load)
t0
Strain
Strain
t1
Time
(Elastic)
ε = σ/E
t0
t0
Strain
t1
t1
Time
Time
(Viscous)
dε/dt = σ/γ
(Viscoelastic)
t0
t1
Time
 Models to explain the viscoelastic behavior:
• Maxwell Model
• Kelvin Model
• 4-Element Model (Burger’s Model)
1. Maxwell Model:
σ
 A spring & a dashpot
connected in series.
k=E
β = 1/η
σ
 The stress on each element
is the same:
σspring = σdashpot
 However, the deformations
are not the same:
εspring ≠ εdashpot
 When the force (stress) is applied the spring
responds immediately and shows a
deformation εspring = σ/E
 At the same time the dashpot piston starts to
move at a rate βσ = σ/η
and the displacement of the piston at time t, is
given by:

 dashpot   dt   dt

0
0
t
t
 Therefore the total displacement becomes:


 t   spring   dashpot    dt
E 0
t
P
ε
εdashpot
εspring
t
εdashpot→ viscous permanent def.
 Relaxation: An important mode of behavior for
viscoelastic materials can be observed when a
material is suddenly stroked to ε0 & this strain
is kept constant.
σ
ε
σ0
ε0
t
Instead of this loading pattern strain is kept constant.
t
   spring   dashpot
d d s d d


dt
dt
dt
&
d 1 d 


dt E dt 
If ε is constant →
d
E

dt

d s 1 d
s  

E
dt
E dt
d d 

dt


d
1 d 
0
 0
dt
E dt 
Solving this
differential
equation
   0 .e
E
 t

Where :
σ0 = Eε0
ε
t rel 
ε0

E
Relaxation time is a parameter
of a viscoelastic material.
t
 If the body is left under
σ
σ0
Slope @ t=0
0.37σ0
trel
t
a constant strain, the
stress gradually
disappears (relaxes). This
pehenemenon can be
observed in glasses &
some ceramics.
2. Kelvin Model:
 Consists of a spring & a
σ
1/η
E
σ
dashpot connected in
parallel.
 In this case the
deformations are equal but
the stresses are different.
 εspring = εdashpot
 σspring ≠ σdashpot
 σ = σspring + σdashpot
E. spr
d dash
.
dt
σ
d
  E  
dt
σ0
t
E

t 

0 
  d
 
 
 
1 e

E E dt
E 

ε
(Delayed elasticity)
t
 At each strain increment the spring will extend
by σ/E so that a part of the load is taken over
and the force on the piston decreases. Thus a
final displacement is reached asymtotically and
when the load is removed, there will be an
asymtotic recovery until σ=0.
 When a viscoelastic Kelvin Body is subjected
to a constant stress, σ0, the response could be
obtained by solving the differential equation.

 0 
E 
1 e
E
 t





 0 is reached at

σ
E time t=∞
t ret 
t
ε
0

E
retardation
time
When stressed the elastic
deformation in the spring is
retarded by the viscous
deformation of the dashpot.
ε0
0.63ε0
Retarded elastic strain
(delayed elastic strain)
E
tret
3. Burger’s Model:
 The actual viscoelastic behavior of materials is
very complex. The simplest models, Maxwell &
Kelvin Models, explain the basic characteristics
of viscoelastic behavior.
 The Maxwell Model, for example, has a viscous
character and explains the relaxation behavior of
viscoelastic materials
 The Kelvin Model on the other hand has a solid
character and explains the retarded elasticity
behavior.
 However, none of the mentioned models
completely explain the real behavior of
viscoelastic materials.
 There are other models with different E and η
constants but they are rather complex.
 One such model is given by BURGER, which
consists of a Maxwell Model and Kelvin Model
connected in series.
σ
σ
σ0
E1
t
η1
η2
E2
σ
εvis+εret
ε
ε1
ε1
t
εvis
E2

t

0
0
0 
 

t
1  e 2
E1
1
E2 

Spring Dashpot
(elastic) (viscous)




Kelvin (retarded
elasticity)
Most engineering materials show certain deviations
from the behavior described by the 4-Element Model.
Therefore the deformation equation is usually
approximated as:

 

E


 k 1  e
Instantaneous
elastic
Retarded
elastic
 qt
  
n
t
Viscous
Where “k, β & γ” are material constants & “α,
n” are constants accounting for nonlinearity.
1
k
E

q
1
t ret

E

1

Example 1: For a certain oil, the experimentally
determined shear stress, rate of flow data
provided the following plot. Determine the
viscosity of the oil.
d 1

dγ/dt (1/sec)
   
d
dt 
0.9
dt
0.6
20
N

 33.3 2 sec .
0.6
m
0.3
10
20
30
τ (Pa)
Example 2: When a concrete specimen of 75 cm
in length is subjected to a 150 kgf/cm2 of
constant compressive stress, the following
data were obtained.

t (month)
ε
 B. .t where B is constant.
Assume  
0
0.0006
1
0.0007
E
What will be the total deformation under 150 kgf/cm2
after 6 months?
@t  0   

E
 0.0006
1  10 4
@ t  1    0.0007  0.0006  B.150.1  B 
150
1  10 4
@ t  6months    0.0006 
 150  6  0.0012
150
Example 3: A glass rod of 2.5 cm in diameter &
2.5 m in length, is subjected to a tansile load
of 10 kgf @ 450°C.
a) Calculate the deformation of the rod after
100 hrs.
b) Determine trel (relaxation time)
c) What is the time during which the stress in
the material would decay to 5% of its initial
volume?
η=2x1012 poise @ 450°C & E=1.55x105 kgf/cm2
Assume that the behavior of glass at this
temperature can be approximated by a
Maxwell Model.
a) For Normal Stresses & Strains the viscous
behavior is described by dε/dt=σ/λ where λ is
called “the Coefficient of Viscous Traction” &
equals to “3η”.

10kgf
  2.5 4
2
 2.04 kgf
cm 2
η=2x1012 poise (1 poise = 1 dyne.sec/cm2) & (1 kgf = 106 dyne)
d 
2.04  10 dyne
7 1



3
.
4

10
sec
dt 3 3.2  1012  poise 
6
After 100 hrs the total strain is 3.4x10-7x100x60x60 =
0.1244
Δ = 0.1244x250 = 30.6 cm