MECHANICAL DEGRADATION

The polymer chain is ruptured by mechanical
means. The effect is to reduce the polymer
molecular mass.
 This is sometimes done deliberately.
 Example: MASTICATION of natural rubber but if it
occurs in service the polymer properties will be
adversely affected.
 Usually the chain break is made permanent by
oxygen attack and mechanical deg. Is more
accurately described as MECHANOCHEMICAL
DEGREDATION
MECHANICAL DEGRADATION
1)
Shear of chemical bonds
Y
Solid state
X
i.e. break made permanent Results perhaps in
relaxation
e.g. by chemical reaction
most likely with O2
Correctly call this type of degradation as
mechanochemical degradation
Limiting RMM:- under a given set of conditions
related to reduction in viscosity and ability of
chains to more rather than be broken
MECHANICAL DEGRADATION IN SOLUTION

1) Ultrasonics
 cavitation
 bubble forms to collapses
 chain subjected to effectively a shock
wave which shears the chain.
Air bubble
Intensive shear
force

2) Turbulant flow
PolymersSolution
Capillary
Sprayed out high
shear
Mechanical
degradation

3) Violent shaking or stirring
 It will be the highest molecular masses which will
be affected., and hence whilst the average
molecular mass will fall the molecular mass
distribution will change.

a) average value falls
 b) distribution narrows
 In making up solutions for molecular mass
characterization the addition of antioxidants may
not be permitted. I.e. the value of Mn would be
lowered.
Mechanical Degradation in the Solid State

Example
 Mastication: Deliberate break down of chains to reduce
viscosity to aid dispersion of compounding
ingredients.

Example
 In use of N.R. i.e. RMM up to 5X106
 With synthetic polymer RMM controlled in
polymerization reaction in order to avoid reed to
masticate polymer.
EFFICIENCY OF MASTICATION

For a given set of shearing conditions efficiency
of mastication depends on:-

1) Shearing force.

2) Molecular mass of original polymer.

3) Temperature.

4)
Environment.
SHEARING FORCE

Greater the shearing force the greater the
reduction in molecular mass.
MOLECULAR MASS






The greater the molecular mass the greater the
reduction in molecular mass.
Always falls to a limiting value.
At low temperature
Polymer molecules have high viscosity and therefore,
chains are subjected to maximum shearing force (high
efficiency).
As the temperature increases viscosity falls and the
mobility of chains increases and avoid the shearing force
(efficiency decreases).
At high temperature
Degradation is caused by thermal oxidative breakdown
Mastication continually exposing new surfaces for attack
(efficiency increases).
At low temperature
TEMPERATURE
 Polymer molecules have high
viscosity and therefore, chains
are subjected to maximum
shearing force (high
efficiency).
 As the temperature increases
Temperature
viscosity falls and the mobility
of chains increases and avoid
the shearing force (efficiency
decreases).
 At high temperature
 Degradation is caused by thermal oxidative breakdown
 Mastication continually exposing new surfaces for
attack (efficiency increases).
Efficiency


The shearing forces
ruptures the chain.
Efficiency
ENVIRONMENT
Under nitrogen (N2), chain
recombination can occur
(low efficiency)
In air
Under N2 +
benzoquinone
Under N2
Mastication time
In
air (O2 present), the radicls formed will immediately
react with O2, followed by rearrangement reaction which
make the break permanent (high effiiciency)
Adding
benzoquinone under N2 results in increasing the
efficiency because benzoquinone acts as inhibitor and
stabilises the radicals formed (prevents chain
recombination).

Efficiency
In hot mastication, thermal oxidative
degradation is occuring.
 Binzoquinone functions as an antioxidant
(efficiency decreases as benzoquinone
concentration increased).
In air at 140 oC (hot)
Under N2
Benzoquinone concentration
BASIC MECHANISM

1- Shearing force breaks chain
 Example: Natural rubber
CH3
CH3
--------CH2 – C = CH – CH2 - CH2 – C = CH – CH2--------CH3
--------CH2 – C = CH – CH2
i.e.
R
CH3
+
+
CH2 – C = CH – CH2---------
R
2- Under nitrogen, low temperature
R
+
R
R–R
i.e. recombination
3- In presence of oxygen.
R
+
O2
R–O–O
Stable product
Permanent break
4- In presence of radical acceptor
e.g. benzoquinone
R
+O
O
RO
O
Thank You
See You Next Lecture