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THERMOELASTIC EFFECT

and

RETARDED ELASTICITY

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In a perfect elastic material the relationship

between stress and strain is linear andindependent of time. However, there arealways some deviations from this perfectelastic behavior.

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When a metal bar is rapidly stretched (line OA), itincreases in volume and its temperature decreases.(Rapid! No time for heat exchange with the environment)

If the specimen is allowed to remain under the loadfor a sufficiently long time it warms up to roomtemperature and expands (line AB)

Mechanical

def.

C

BA

oThermal

def.

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If unloaded at the original rate it will contract (line BC)and its temperature increases. When allowed to cool itcools to room tamperature (line CO).

This is called anadiabatic process. There is no heatexchange of the material with the environment. Inother words the mechanical and thermal deformations

can be identified.

Mechanical

def.

C

BA

oThermal

def.

Adiabatic

line

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If a specimen is stretched at such a rate thatits temperature remains constant (because of

the heat exchange with the environment),youll obtain theisothermal behaviourasrepresented by line OB.

Mechanical

def.

C

BA

oThermal

def.

Isothermal

line

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As seen from the figure Eadiabatic > EisothermalIn real life, the changes are not adiabatic,

there will always be some heat exchange.Thus the - will assume the following shape.

This loop is called asHYSTERESISLOOPand represents the amountof 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.

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Heating and cooling ofa material as a result of

its deformation is calledTHERMOELASTIC EFFECT,this effect is due to a

phenemenon calledRETARDED ELASTICITY.

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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 toelasticity.

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 isanelasticity.

Thermoelastic effect and retarded elasticity are the

aspects ofanelasticity.

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Example 1:The steel prismatic member is

subjected to the following

load combination.P1=900kN P2=-900kN

P3=900kN =0.26, E=200GPa

a) Find the change in volume.b) What must be the

magnitude of thecompressive load if there isto be no change involume?

P2 P2

P3

P3

P1

P1

10cm

5cm

50cm

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a) 1 =50x100

=900*103

A1

P1= 180 MPa = 0.18 GN/m2

P1

A1

(N)

(mm2)

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500x50=

-900*103

A2

-P2= - 0.036 GN/m22 =

P2

A2

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500x100=

900*103

A3

P3= 0.018 GN/m23 =

P3

A3

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= 0.054 GN/m23

=0.18-0.036+0.018

3

1+2+3avg =

V0 = 0.05 x 0.10 x 0.5 = 2.5x10-3 m2

3(1-2*0.26)=

200

3(1-2)

E = 138.96 GN/m2K =

0.054

V/2.5x10-3V = 9.7x10-7 m3138.96 =

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Since = 0.26 then avg = 0 should be satisfied.

b) ForV = 0 = 0.5

or avg = 0

avg = 1 + 2 + 3 = 0.18 + 2 + 0.018 = 0

2 = -0.198 GN/m2

P2 = -0.198 * 500 * 50 = -4950 kN

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Example 2: An aluminum alloy rod 3 cm in diameterand 75 cm in length is subjected to a tensile loadof 2000 kgf. Calculate:

a) Longitudinal strain, lb) Change in length,l

c) Change in diameter,d

Material properties are:

E = 7x105 kgf/cm2 & = 0.33

3 cm75cm

2000 kgf

2000 kgf

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a) E =l

l =

E=

2000/(*32/4)

7x105

l = 4.042x10-4 cm/cm

b) l =L

L L= 4.042x10-4 * 75 =0.0303 cm

c) =longlat lat = . l

d= . l . d

= 0.33 * 4.042x10-4 * 3 = 0.0004 cm

(Tensile)Elongation

Shortening

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d) K =avg

V/V0=

E

3(1-2)

VV0

=avg*3*(1-2)

E

V =V0avg*3*(1-2)

E

= *

= 0.073 cm3

Volume expansion

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Example 3: A steel bar having a diameter of 1cm shows a unit elongation of 0.0007 when a

uniaxial tensile load is applied. Determine theload P. E = 2.1x106 kgf/cm2

= E

& = PA = PA.E

P = . A . E

P = 0.0007 * * 12/4 * 2.1x106 = 1155 kgf