1. THERMOELASTIC EFFECT and RETARDED ELASTICITY

2. 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. ε σ

3. σ ε
Mechanical def. C B A o
Thermal def.
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)

4. σ
Adiabatic line A B C o ε
Mechanical def.
Thermal def.
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.

5. 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 ε
Mechanical def. C B A o
Thermal def.
Isothermal line

6. 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. ε σ
This loop is called as “HYSTERESIS LOOP” and represents the amount of heat dissipated (energy loss) during loading and unloading.
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.

7. σ 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.

8. 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”.

9. Example 1:
The steel prismatic member is subjected to the following load combination.
P1=900kN P2=-900kN
P3=900kN ν=0.26, E=200GPa
Find the change in volume.
What must be the magnitude of the compressive load if there is to be no change in volume? P2 P3 P1
10cm
5cm
50cm

10. σ1 =
50x100 =
900*103 A1 P1
= 180 MPa = 0.18 GN/m2
(N)
(mm2) P1 A1

11. 500x50 =
-900*103 A2
-P2
= GN/m2
σ2 = A2 P2

12. 500x100 =
900*103 A3 P3
= GN/m2
σ3 = A3 P3

13. = GN/m2 3 =
σ1+σ2+σ3
σavg =
V0 = 0.05 x 0.10 x 0.5 = 2.5x10-3 m2
3(1-2*0.26) =
200
3(1-2ν) E
= GN/m2
K =
0.054
ΔV/2.5x10-3
ΔV = 9.7x10-7 m3 =

14. ν = 0.5
For ΔV = 0
or σavg = 0
Since ν = 0.26 then σavg = 0 should be satisfied.
σavg = σ1 + σ2 + σ3 = σ = 0
σ2 = GN/m2
P2 = * 500 * 50 = kN

15. Longitudinal strain, εl Change in length, Δl Change in diameter, Δd
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:
Longitudinal strain, εl
Change in length, Δl
Change in diameter, Δd
Material properties are:
E = 7x105 kgf/cm2 & ν = 0.33
3 cm
75 cm
2000 kgf

16. Shortening E = σ εl εl = E = 2000/(π*32/4) 7x105 εl = 4.042x10-4 cm/cm
(Tensile) Elongation
εl = ΔL L
ΔL = 4.042x10-4 * 75 = cm
ν =
εlong
εlat
εlat = ν . εl
Δd = ν . εl . d
= 0.33 * 4.042x10-4 * 3 = cm
Shortening

17. Volume expansion σavg E K = = ΔV/V0 3(1-2ν) ΔV σavg*3*(1-2ν) = V0 E
= cm3
Volume expansion

18. Example 3: A steel bar having a diameter of 1 cm shows a unit elongation of when a uniaxial tensile load is applied. Determine the load P. E = 2.1x106 kgf/cm2 σ
σ = P A P
ε =
&
ε = E
A.E
P = ε . A . E
P = * π * 12/4 * 2.1x106 = 1155 kgf