Transcript MECHANICAL PROPERTIES OF MATERIALS

THERMOELASTIC EFFECT
and
RETARDED ELASTICITY
 In a perfect elastic material the relationship
between stress and strain is linear and
independent of time. However, there are
always some deviations from this perfect
elastic behavior.
σ
ε
σ
A
o
B
C
Mechanical Thermal
def.
def.
ε
 When a metal bar is rapidly stretched (line OA), it
increases in volume and its temperature decreases.
(Rapid! No time for heat exchange with the environment)
 If the specimen is allowed to remain under the load
for a sufficiently long time it warms up to room
temperature and expands (line AB)
σ
Adiabatic
line
A
o
B
C
Mechanical Thermal
def.
def.
ε
 If unloaded at the original rate it will contract (line BC)
and its temperature increases. When allowed to cool it
cools to room tamperature (line CO).
 This is called an “adiabatic process”. There is no heat
exchange of the material with the environment. In
other words the mechanical and thermal deformations
can be identified.
 If a specimen is stretched at such a rate that
its temperature remains constant (because of
the heat exchange with the environment),
you’ll obtain the “isothermal behaviour” as
represented by line OB.
σ
Isothermal
line
A
o
C
B
Mechanical Thermal
def.
def.
ε
 As seen from the figure Eadiabatic > Eisothermal
 In real life, the changes are not adiabatic,
there will always be some heat exchange.
 Thus the σ-ε will assume the following shape.
This loop is called as “HYSTERESIS
LOOP” and represents the amount
of heat dissipated (energy loss)
during loading and unloading.
σ
The area of a hystresis loop is small especially for metals.
From an engineering point of view, the energy loss leads to
heating, damping of vibrations and also contributes to
friction, particularly in materials like rubber.
ε
 Heating and cooling of
a material as a result of
its deformation is called
THERMOELASTIC EFFECT,
this effect is due to a
phenemenon called
RETARDED ELASTICITY.
σ
ε
In the elastic analysis of metals it is assumed that
elastic strain is a function of stress only. This is
strictly not true since there is time dependence to
elasticity.
 In metals the effect is very small and usually
neglected.
 In polymers the effect is much more significant.
 The general name for this time dependence is
“anelasticity”.

 Thermoelastic effect and retarded elasticity are the
aspects of “anelasticity”.
Example 1:
P1
5cm
10cm
P3
P2
P2
50cm
P3
P1
The steel prismatic member is
subjected to the following
load combination.
P1=900kN P2=-900kN
P3=900kN ν=0.26, E=200GPa
a)
Find the change in volume.
b)
What must be the
magnitude of the
compressive load if there is
to be no change in
volume?
a) σ1 =
P1
A1
900*103(N)
=
= 2180 MPa = 0.18 GN/m2
50x100(mm )
P1
A1
σ2 =
-P2
A2
=
-900*103
500x50
A2
P2
= - 0.036 GN/m2
σ3 =
A3
P3
P3
A3
=
900*103
500x100
= 0.018 GN/m2
σavg =
σ1+σ2+σ3
3
0.18-0.036+0.018
=
= 0.054 GN/m2
3
V0 = 0.05 x 0.10 x 0.5 = 2.5x10-3 m2
200
E
2
=
138.96
GN/m
=
K=
3(1-2ν)
3(1-2*0.26)
138.96 =
0.054
ΔV/2.5x10-3
ΔV = 9.7x10-7 m3
b) For ΔV = 0
ν = 0.5
or σavg = 0
Since ν = 0.26 then σavg = 0 should be satisfied.
σavg = σ1 + σ2 + σ3 = 0.18 + σ2 + 0.018 = 0
σ2 = -0.198 GN/m2
P2 = -0.198 * 500 * 50 = -4950 kN
Example 2: An aluminum alloy rod 3 cm in diameter
and 75 cm in length is subjected to a tensile load
of 2000 kgf. Calculate:
a) Longitudinal strain, εl
b) Change in length, Δl
c) Change in diameter, Δd
Material properties are:
E = 7x105 kgf/cm2 & ν = 0.33
75
cm
2000 kgf
3
cm
2000 kgf
a) E =
σ
εl
σ
εl =
E
2000/(π*32/4)
=
7x105
εl = 4.042x10-4 cm/cm
b) εl =
ΔL
L
εlat
c) ν =
εlong
(Tensile)
Elongation
ΔL = 4.042x10-4 * 75 =0.0303 cm
εlat = ν . εl
Δd = ν . εl . d
Shortening
= 0.33 * 4.042x10-4 * 3 = 0.0004 cm
d) K =
σavg
=
ΔV/V0
E
3(1-2ν)
σavg*3*(1-2ν)
ΔV
=
V0
E
ΔV = V0
σavg*3*(1-2ν)
E
=
= 0.073 cm3
*
Volume expansion
Example 3: A steel bar having a diameter of 1
cm shows a unit elongation of 0.0007 when a
uniaxial tensile load is applied. Determine the
load P. E = 2.1x106 kgf/cm2
ε=
σ
E
&
σ=
P
A
ε=
P
A.E
P=ε.A.E
P = 0.0007 * π * 12/4 * 2.1x106 = 1155 kgf