MECHANICAL PROPERTIES OF MATERIALS
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Transcript MECHANICAL PROPERTIES OF MATERIALS
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