Polymer Fracture and Failure Flashcards

Question

What are the 2 fracture/failure mechanics of polymers?

Answer 1

Linear elastic fracture mechanics - brittle Elastic plastic fracture mechanics - ductile

Answer 2

Observes linear elastic material behaviour in brittle polymer materials, filled or fibre-reinforced material systems, or specimens of great thickness or below the glass transition temperature Tg LEFM with small-scale yielding, in the case where plastic zone at the crack tip is taken into consideration.

Answer 3

 As a parameter, the stress intensity factor (SIF) Kdescribing the linear elastic area in front of the crack tip is used.  The critical value of the SIF under conditions of plane strain is called the fracture or crack toughness KIc (static loading), KId (dynamic loading), KII or KIII (I, II, III refer to crack-tip-open modes). In practice, mode I has the highest importance.

Answer 4

Mode I (opening) Mode II (in-plane shear) Mode III (out-of-plan shear)

Answer 5

The tensile stress increases linearly with the strain, followed by an abrupt drop of the load at the instant of crack growth initiation.  When characterising the crack instability by the stress, a solely stress-dependent term, the stress intensity factor K, is used. K = σi√a where σi is the external stress, a is the diameter of the flaw. This is called the critical stress intensity factor Kc in the case of failure.

Answer 6

For evaluating the toughness in the case of non-negligible elastic-plastic material behaviour and extensive plastic area in front of the crack tip, the concepts of elastic-plastic fracture mechanics have to be used.

Answer 7

the crack-tip opening displacement (CTOD) and the J-integral.

Answer 8

The fracture process is not controlled by stress intensity any more, but by plastic deformation in front of the crack tip. A measure of this is the widening at the crack tip, called the crack-tip-opening displacement or crack-opening displacement δ. evaluated against the crack resistance concept.

Answer 9

After the load or the deformation has been determined J (the energy) can be determine to ascertain whether there is unstable crack growth, stable/unstable crack growth or stable crack growth.

Answer 10

Composition(polymer or additives), Molecular weight(weight distribution or cross linking) and intermolecular order(crystallinity, orientation, degree of fusion or thermal transitions)

Answer 11

Natural polymers or synthetic polymers

Answer 12

Found in plants and animals  proteins,  cellulose,  starch,  natural rubber

Answer 13

Synthesises by chemical methods  plastic (polyethylene, polypropylene, PVC, Polystyrene…),  synthetic fibres (nylon 6,6, Kevlar… ) and  synthetic rubbers (polybutadiene)

Answer 14

two or more monomers polymerised together

Answer 15

 random – A and B randomly positioned along chain  alternating – A and B alternate in polymer chain  block – large blocks of A units alternate with large blocks of B units  graft – chains of B units grafted onto A backbone

Answer 16

1.Linear Polymers: These polymers consist of long and straight chains.  high density polythene,  PVC, 2.Branched Polymers: These polymers contain linear chains having some branches,  low density polythene. 3. Cross-linked Polymers: These are usually formed from the monomers with more than two functional groups and contain strong covalent bonds between various linear polymer chains,  vulcanized rubber,  urea-formaldehyde resins,  Epoxy.

Answer 17

 High molecular weight, or long chain length, is the single most important property of polymers that distinguished them from other materials, such as metals and ceramics.  Molecular weight in the range of about 10,000 – 100,000 accounts for the ability of polymers to perform in many applications.

Answer 18

Resistance to stress in tensile, flexural and shear modes, toughness, creep resistance, and environmental stress cracking are some of the product performance criteria that are strongly affected by molecular weight.

Answer 19

When tensile strength increases the molecular weight increases

Answer 20

A polymer normally does not have a single MW. The number of monomer units comprising a polymer chain is not exactly the same for each and every molecule in a particular polymer.  There is actually a distribution of chain length, which is called molecular weight distribution (MWD). MWD = Mw / Mn The broader the distribution, the higher polydispersity index.

Answer 21

Characteristic of material viscoelasticity used to describe stress relaxation of materials with time

Answer 22

As the temperature increases and the relaxation modulus decreases logarithmicly, the polymer moves from a glassy state to a leathery state and then it continues to be rubbers -> rubbery flow -> viscous flow at the lowest modulus and highest temperatures.

Answer 23

Crystalline isostatic - High modulus at low temperature, sharp drop near T_m (melting temperature) as the material softens. Lightly crosslinked atactic- Gradual decrease in modulus with temperature, less sharp than crystalline due to crosslinking, but still softens at high temperatures then plateaus in the rubber region. Amorphous polystyrene -Sharp drop at Tg, showing a glass-to-rubber transition, then a slight plateau in the rubbery state before decreasing further with increasing temperature.

Answer 24

The higher the molecular weight with a increasing temperature, the longer it shall plateau in the rubbery region.

Answer 25

Glass transition - amorphous materials Melt transition- crystalline materials the transition is a change of state relating to changing temperature or pressure semi crystalline can show both.

Answer 26

Both transition types originate from (inter) molecular forces - thus typically Tg~0.5=0.65Tm The glass transition is not a phase transition and arises from the decoupling of the vibrational and translational motions ie at and above the glass transition temperature the flow of entire chains is possible (for polymers). It is the temperature at which the polymer transforms from rubber to glass in a LONG RANGE SEGMENTAL MOTION. Lowering temperature reduces free volume.

Answer 27

 At temperatures above Tg, 10 to 50 repeat units of the polymer backbone are relatively free to move in cooperative thermal motion to provide conformational rearrangement of the backbone.  Below Tg, the motion of these individual chains segments becomes frozen with only small scale molecular motion remaining, involving individual or small groups of atoms.  Below Tg polymers are hard and glassy  Above Tg polymers are soft and leathery

Answer 28

The actual volume of the molecules stays the same through Tg, but the free volume (the volume through which they can move) increases.  Free volume is the space in a solid or liquid sample which is not occupied by polymer molecules, i.e. the “empty-space” between molecules.

Answer 29

 Both Tm and Tg increase with increasing chain stiffness  Chain stiffness increased by presence of  Bulky side-groups  Polar groups or sidegroups  Chain double bonds and aromatic chain groups  Regularity of repeat unit arrangements – affects Tm only

Answer 30

Polymers exhibit rate-dependent viscoelastic deformation, which is a direct result of their molecular structure. Two neighbouring molecules, or different segments of a single molecule that is folded back upon itself, experience weak attractive force called Van der Waals bonds. These secondary bonds resist any external force that attempts to pull the molecules apart.

Answer 31

Deforming a polymer requires cooperative motion among molecules. The material is relatively compliant if the imposed strain rate is sufficiently low to provide molecules sufficient time to move.  At faster strain rate, however, the forced molecular motion produces friction, and a higher stress is required to deform the material.  If the load is removed, the material attempts to return to its original shape, but molecular entanglements prevent instantaneous elastic recovery.

Answer 32

1. Crystalline materials have their molecules arranged in repeating patterns. As such, they all tend to have highly ordered and regular structures. 2. Amorphous materials have their molecules arranged randomly and in long chains which twist and curve around one-another, making large regions of highly structured morphology unlikely. 3. The morphology of most polymers is semi-crystalline: a combination with the tangled and disordered regions surrounding the crystalline areas.

Answer 33

 For the most part, crystallinity is intentional and can improve the strength, toughness and thermal stability of the plastics.  However, crystallinity can produce shrinkage, unexpected internal stress, and environmental stress-cracking as well.  Polymers have variab

Answer 34

% crystallinity = [ρc(ρs – ρa) / ρs (ρc – ρa )] x 100 where ρs is the density of a specimen for which the percentage crystallinity is to be determined, ρa is the density of the totally amorphous polymer, and ρc is the density of perfectly crystalline polymer.

Answer 35

Intermolecular order - Crystallinity Processing requires heating above the melting point. For high melting polymers, like nylon, processing at high temperature possesses the danger of degradation, especially if traces of water are present.  For an amorphous polymer, the polymer is in the rubbery or “soft” state above Tg. This limits its service temperature range.  For a crystalline polymer the crystalline form retains dimensional stability almost up to the melting point. Thus, crystalline polymers can be used at higher service temperatures.

Answer 36

The polymer chains are chemically bonded to each other via crosslinks.  Crosslinking raises the Tg and increase melt viscosity above Tg. The higher of the degree of crosslinks, the higher of the modulus (the more brittle) of the polymers.

Answer 37

Many moulded products have at least some orientation resulting from processing.  In some cases, orientation is intentional and is required for the product to perform such as fibres and heat shrink tubing. Much of the time, however, orientation is not wanted and processing is carried out so as to minimise it. Frozen-in stress in a plastic product is the result of such orientation. In their normal state most polymer molecules are random coils with no particular shape, physically intertwined with each other.

Answer 38

Processing at high speed and high shear rate extends these random coils into an orientated configuration.  Rapid cooling during processing does not allow enough time for complete relaxation, leaving a substantial amount of orientation and resultant frozen-in stress in the part.  The higher the molecular weight, the longer it takes for orientated molecules to relax at any given temperature. With time or temperature, this stress can be relieved, allowing the molecules to become more like their original relaxed, coiled configuration.

Answer 39

The frozen-in stress can be removed by heating above the material’s softening temperature.  The greater the shrinkage and distortion on heating, the greater the frozen-in stress. Generally, the smaller the change on heating, the less likely is the part to fail.