Polymer Fracture And Failure

Question 1

What are the causes of polymer failure?

Answer 1

Incorrect material selection

Chemical and environment interactions - leads to rapid crazing, cracking, fracture and product failure

Response to long term loads - creep and fatigue, strength and stiffness of polymers in service conditions usually a lot lower than on a data sheet

Processing errors - incorrect processing leads to high residual stresses, degradation, embrittlement, inhomogenity, introduction of faults, defects or contaminants

Inappropriate design - long term loads, chemicals, high speed loading, fatigue loads - shortens service life

Effect of additives - too much our too little or incorrect additive, non-uniform dispersion, migration to surface, unanticipated secondary effects

Question 2

What are some examples of unintentional additives?

Answer 2

Extraneous lint, dirt etc.

Residual solvent

Water

Ionic impurities from water

Trace metal from extruder barrel

Impurities in intentional additives

Question 3

What are the two natures of loads?

Answer 3

Impact - hammering, accidental drop

Fatigue - vibration, rotation in loaded components

Question 4

What are the types of mechanical failure in polymers?

Answer 4

Excessive deformation

Ductile failure - encountered in materials that can undergo large irreversible plastic deformation before fracturing (tend to design products with a yield criteria in mind as yielding is the onset of failure)

Brittle failure - low strains and negligible permanent deformation, occurs in components with geometrical discontinuities (stress concentrations)

(No failure is entirely brittle or ductile, fraction depends on speed of loading and temperature of sample)

Crazing - occurs at strain level below that required for brittle fracture, isn’t catastrophic but undesirable

Question 5

What is crazing?

Answer 5

Crazes are made up of micro-cavities whose surfaces are joined by highly oriented material

They initiate near structural discontinuities like impurities and are collectively visible at strained surface as they become large enough to reflect light

They are not cracks and can sustain loads but can become cracks if the fibrils break.

Question 6

What is LEFM?

Answer 6

Linear elastic fracture mechanics

Observing linear elastic behaviour in brittle polymers, fibre-reinforced materials, specimens of great thickness or below Tg

Question 7

What is SIF?

Answer 7

Stress intensity Factor (K)

Describes the linear elastic area in front of the crack tip

Critical value of SIF is called fracture or crack toughness Klc (static loading) or Kld (dynamic loading)

Kii or Kiii refer to crack opening modes, mode I is highest importance

Question 8

What are the three crack opening modes?

Answer 8

Mode 1 - opening like splitting

Mode 2 - in-plane shearing (pushing back or forward)

Mode 3 - out of plane shear (splitting sideways)

Question 9

In LEFM, how does stress and strain grow?

Answer 9

They increase linearly together until crack initiation when there is an abrupt drop of load

Question 10

What’s the formula for stress intensity factor K?

Answer 10

K = external stress x diameter of flaw

Question 11

What is EPFM?

Answer 11

Elastic Plastic Fracture Mechanics

Observing toughness for non-negligible elastic-plastic material and extensive plastic area in front of crack tip

Question 12

What is CTOD and what is a measure of it?

Answer 12

Crack tip opening displacement

Based on assumption that for ductile materials, fracture process is not controlled by stress intensity but by plastic deformation in front of the crack tip

Measure of it is widening at crack tip

Question 13

What is chain scission?

Answer 13

Fracture on an atomic level - bonds breaking

Caused by high stress concentrations, non-uniform loads causing some bonds to have a high load

Non-uniformity is more noticeable in amorphous polymers - crystalline tend to distribute stress more evenly

Question 14

What is chain disentanglement?

Answer 14

Molecules separate from each other intact.

Likelihood depends on length of molecules and degree to which they are entangled

Question 15

What are the two types of failure of polymers like type I and type II?

Answer 15

Type I - tend to fail by crazing if craze initiation stress is less than yield stress

Type II - tend to fail by shear yielding if yield stress is less than craze initiation stress

If both stresses are equal then both mechanisms can occur simultaneously

Question 16

What are some properties of Type I and II polymers?

Answer 16

Type I - brittle just below Tg, low initiation and crack propagation energies so low unnotched and notched impact strength

Type II - high crack initiation and propagation energies so high unnotched impact strength but low notched, shows a brittle to ductile transition Tbt

Question 17

What materials are most prone to crazing?

Answer 17

Amorphous, glassy polymers such as PS, SAN, PMMA, PVC

Semi crystalline polymers (PE, PP, PETP)

Epoxy resins

Question 18

What are the 3 stages of craze propagation?

Answer 18

Craze initiation/nucleation - usually on surface grooves or imperfections in bulk of polymer

Craze propagation

Craze breakdown

Question 19

What indicates that crazing is time dependent?

Answer 19

There can be significant lag time between load application and first visible craze.

Question 20

Why do crazes tend to form at the surface of polymers?

Answer 20

Defects are usually found at surface

Injection moulding and machining usually results in a different surface structure to bulk material

Environmental attack like diffusion, degradation etc usually starts and penetrates from surface

Question 21

Explain the process of craze nucleation

Answer 21

Plastic deformation starts at a local stress concentration

Yield is non-linear and polymer glasses soften with strain so there will be localised deformation as the strain increases

Material further is undeformed so some lateral stresses develop

EITHER: Strain hardening response of material will stabilise the strain localisation process OR micro shear band will spread out

If lateral stresses become high enough, material in deformation zone will cavitation and craze nucleation occurs

Question 22

Explain the process of craze propagation

Answer 22

Craze grows by:

Craze tip displaces into increased material by forming voids ahead of the craze tip or by coalescence with a separate, smaller craze

OR

Craze widens while extending craze matter eg. Molecular strands may rupture or disentangle giving rise to a crack

Question 23

What can lead to craze slowing?

Answer 23

If a craze hits an obstacle such as a zone of higher orientation or end of specimen

If a craze has grown to such an extent that the initiating crack is blunted and rendered uncritical

If a craze interacts with another craze or the environment which can change the conditions that have led to its growth or initiation (eg. Stress relaxation or penetrating into a lower stress zone adjacent to another craze)

Question 24

What causes a craze to turn into a crack?

Answer 24

Molecular strand degrades and eventually fails under the tensile load

Question 25

What causes craze breakdown?

Answer 25

Slippage and disentanglement of molecular coils leading to fibril rupture

Breakage of entangled chains leading to fibril rupture

Fibril-matrix separation

Question 26

What criteria in terms of molecular weight states that a material will be brittle?

Answer 26

If molecular weight is lower than double the molecular weight for entanglement

Question 27

How does molecular weight effect craze behaviour and an example?

Answer 27

Fibrillated crazing tends to occur in only high molecular weight polymers

In lower molecular weight polymers, cracks without fibrillation occurs because of the missing intermolecular connections and strength

Eg. In a PS blend of 50-50 of over 2Me and under 2Me, the craze happens only in the higher molecular weight regions and crack in the lower ones

Question 28

What is rubber toughening?

Answer 28

On loading, elastomeric phase concentrates and makes the stress distribution more complex in surrounding matrix

Yield stress is reduced and toughening occurs by the resultant plastic deformation either by crazing or formation of shear bands

Question 29

How does crazing cause rubber toughening?

Answer 29

Dispersed rubber particles act as craze initiators.

The rubber particles must be:
- small enough to prevent a crack propagation
- large enough not to be engulfed by a craze
- in the order of microns

Question 30

How does shear yielding mechanism cause rubber toughening?

Answer 30

Dispersed rubber particles act as initiations for the shear bands.

Need to be less than a few tenths of a micron

Shear bands are barriers to craze or crack propagation so delays failure

Question 31

How can a rubbery second-phase increase toughness?

Answer 31

Makes craze/shear yielding initiation easier by providing sites for nucleation, lowering the stress required for craze and shear yielding formation

Question 32

What factors denote the contribution of either crazing or shear yielding?

Answer 32

Rubber particle size, dispersion and concentration
Matrix
Temperature

Question 33

How could you tell the difference between crazing and shear yielding?

Answer 33

Dilatometry - crazes increase volume, shear yielding doesn’t change it

Visibly - crazes show a stress whitening caused by the light scattering due to the refractive index difference in crazed and uncrazed areas

Question 34

What are some selection criteria for rubber toughened plastics?

Answer 34

• rubber should exist as a discrete second phase, usually as spheroids
• rubber phase should be dispersed effectively
• adhesion between rubber and matrix polymer should be optimised
• optimum particle size and distribution may vary according to yield characteristics of matrix
• elastomer Tg should be well below minimum expect service temperature of the toughened plastics

Question 35

What is the microstructure of HIPS and how does rubber toughening benefit the material?

Answer 35

Continuous polystyrene (PS) matrix with lots of dispersed salami-like domains which each contain many spherical PS sub domains separated by a polybutadiene phase

The polystyrene component provides the rigidity while the polybutadiene phase brings five times higher toughness than pure PS

The hierarchical salami morphology can accommodate displacements to avoid formation of flaws.

Question 36

What is the common trade off with rubber content in toughening?

Answer 36

More rubber content = higher impact strength but lower tensile and bending strength which is a common conflict

By controlling size of rubber domains to produce a bimodal size distribution, surface of larger particles are more likely to initiate crazing while smaller particles can reduce average ligament thickness between particles which increases toughness

Question 37

What is the structure of ABS and how does rubber toughening benefit it?

Answer 37

Continuous polystyrene-co-acrylonitrile glassy phase with droplets of dispersed polybutadiene latex

Emulsion polymerisation is used industrially as it yields a fine graft control and good distribution which toughens more.

Question 38

What factors influence the mechanical performance of ABS polymers?

Answer 38

Matrix composition and molecular weight
Type of rubber used
Rubber phase volume fraction
Rubber particle size and internal structure
Rubber cross linking density

Question 39

How does bimodal size distribution help the toughness of ABS?

Answer 39

Big particles increase toughness, small particles increase surface finish so a mix of a range of sizes is good (ideal for ABS is 0.3-0.3um)

Question 40

How does the rubber phase increase the toughness of ABS?

Answer 40

Stress concentrations in equatorial regions of rubber particles cause deformation processes

With increasing deformation, stresses are restricted and more mechanical energy can be dissipated before fracture causing multiple crazing

Only if temp is over 70C does shear deformation start to dominate

Question 41

What is the effect of temperature and strain rate on deformation of polymers?

Answer 41

Temperature:
Tg - glass transition - below is typically brittle so have limited deformation and above, polymer chains are more mobile so there is increased ductility and deformation.
Tm - melting temp - above this, chain mobility is increased significantly so viscosity decreases a lot and allows substantial deformation

Strain rate:
Viscoelastic - polymers are viscous and elastic.
Low strain rates - more elastic behaviour, allowing for significant deformation and recovery
High strain rates - more viscous behaviour, permanent deformation

Question 42

What are the causes of premature polymer product failure?

Answer 42

Incorrect material selection - temp limits, mech properties, service life, durability

Environmental factors - temperature, humidity, UV, chemicals etc can cause degradation, promote oxidation, thermal expansion or physical damage

Manufacturing defects - poor processing or quality control can leave improper moulding, voids, inclusions, uneven additives which affects structural integrity

Mechanical stresses - excessive mechanical stresses beyond material capability

Response to long term loads - creep or fatigue, strength in service lower than data sheet

Inappropriate design of all of these

Question 43

Describe the yield process for a thermoplastic

Answer 43

Stress is applied, polymer chains slide past each other - plastic deformation

Yield point is when enough stress is applied that it makes the polymer chains move causing permanent deformation

May exhibit strain hardening or softening dependent on the material

Question 44

What is the mechanism for craze formation?

Answer 44

Occurs in high tensile stress regions, localised not bulk

Polymer chains are aligned and stretched in the direction of applied stress

Microvoids are formed and connect, causing a network of fibrillation structures called crazes

Question 45

What’s the difference between a craze and a crack?

Answer 45

Crazes are fine, interconnected micro cracks or fibrils — Cracks are larger, more distinct fractures
Crazing is reversible deformation, shape can be recovered — Cracks are irreversible, shape can’t be recovered
Crazes propagate in a network pattern — Cracks propagate linearly and extend throughout material
Crazing can enhance toughness and strength — Cracking decreases toughness and strength

Question 46

How can you distinguish crazes and cracks?

Answer 46

Optical Microscopy - observe deformation pattern in material (craze will be fine, interconnected fibrils and cracks will be larger, more distinct features)

Scanning Electron Microscopy - provides higher magnification of surface morphology

Question 47

What is the effect of temperature, strain rate and microstructure on a rubber toughened polymer?

Answer 47

Temperature - Tg affects brittle or ductile
Strain rate - better energy dissipation and relaxation at low strain so better toughness
Molecular structure - long, flexible chains have higher toughness as they can absorb energy, prescience of

Question 48

What is the rubber toughening mechanism in ABS?

Answer 48

Soft rubbery phase is introduced into a brittle matrix material which improves the impact toughness

Due to energy dissipative and deformation process initiated at the particles - as there are stress concentrations at the particles

Question 49

How do we achieve the optimised rubber toughening effect?

Answer 49

Consider:

• rubber phase selection to ensure a good compatibility with matrix (should have good plastic deformation)
• rubber particle size and distribution (smaller will improve toughness by increasing crack deflection)(well distributed means better consistency in toughening)
• compatibility and adhesion using surface modification or coupling agents (crucial for efficient stress transfer and energy dissipation)
• rubber content (higher will be better toughness)