Materials 2 - Section B Flashcards

1
Q

What is the difference between saturated and unsaturated hydrocarbons?

A

Saturated hydrocarbons: each carbon atom singly bonded to 4 other atoms. They are extremely stable.

Unsaturated hydrocarbons: Carbon atoms double and triple bonds. Double’s are ethene’s, triple’s are ethyne’s.

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2
Q

What is primary bonding, how does it deform, and what is it governed by?

A
Primary bonding = covalent bonding:
Sharing of valence electrons
Significant charge density between atoms
Directional nature
Think socialist

Covalent bonds deform elastically and are governed by Hooke’s Law, i.e. a linear relationship between stress and strain (think elastic=spring).

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3
Q

What is secondary bonding, how does it deform, and what is it governed by?

A

Secondary bonding are entanglements:
Many chains are intertwined, think Spaghetti.
Secondary Forces act between the chains, think blocks of cheese. These forces restrict movement and glue the chains together.
These forces can be dipoles, van der Waal forces, or hydrogen bonds.
Much weaker than primary bonds.

They deform in a viscous fashion and are governed by Newton’s Law, i.e. rate of change of strain, which is time dependent.

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4
Q

What is meant by crystalline?

A

Chain folded regions where all chains are aligned.

Very difficult to get all chains aligned, hence polymers are rarely 100% crystalline.

The molecules are uniform and closely packed, hence high density, hard, and stiff.

Application of heat causes crystalline regions to grow.

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5
Q

What is meant by amorphous?

A

Molecules positions are randomly arranged, with chains being twisted and un-aligned. Think of a ball of cables or spaghetti.

They have a low density, and are soft and compliant.

Examples include: natural rubber, latex, elastic bands.

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6
Q

Label the following diagram?

A
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7
Q

Label the following stress-strain graph?

A

Brittle Polymer = Crosslinked and Network

Plastics = Semi-Crystalline

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8
Q

Explain the stress-strain behaviour of “brittle” crosslinked and network polymers?

A

Linear stress-strain behaviour with sudden and catastrophic failure.

They act like glass/ceramic materials.

Max stress of roughly 60 MPa, with a max strain of just above 0, i.e. just above original length.

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9
Q

Explain the stress-strain behaviour of semi-crystalline polymers?

A

They behave similarly to metals in the sense that they have a defined elastic region, a yielding point (peak), and then a periodic of non-elastic behaviour.

Max stress of roughly 30 MPa, with a max strain of roughly 4. I.e. material is 4x longer than original when failure.

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10
Q

Explain the stress-strain behaviour of elastomers?

A

Elastomers are amorphous polymers.

They have enormous strains to failure and the deformation is recoverable across the entire curve. I.e. you can go from the top all the way to 0.

Max stress of roughly 20 MPa, with a max strain of roughly 7-8.

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11
Q

Describe the mechanisms of deformation of crosslinked and network polymers?

A

Think stretching an elastic band.

  1. Initially, the chains are somewhat aligned with lots of crosslinks between the chains.
  2. As stress increases and we move up the stress-strain curve. The polymer chains become more straight and aligned with the loading direction. No bonds are breaking, and hence elastic behaviour is taking place, i.e. reduce the load and the structure will go back to its preferred and initial form/confirmation.
  3. Finally, we get brittle, catastrophic, elastic failure.

Network polymers act similarly to crosslinked, however its the orientation of the primary bonds that change. They become aligned with the loading direction. Everything else is identical, i.e. elastic behaviour etc.

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12
Q

Describe the mechanisms of deformation of semi-crystalline polymers?

A

We have 2 regions, amorphous (not well ordered) and crystalline (well ordered).

Elastic Deformation:

  1. Bonds in the Amorphous regions elongate and become as ordered as they can.
  2. Crystalline regions then align in the direction of loading. Up until this point we still get elastic behaviour as no bonds broken.

Plastic Deformation:

  1. Crystalline region becomes fully aligned.
  2. Blocks in the crystalline regions separate, and slide past one another. This causes bonds to break, which means we get plastic behaviour.
  3. Block separating and sliding mean that deformation can occur at lower stresses (hence the drop on the graph).
  4. Just before failure we see a “fibrillar” structure, so fibre-like, with Amorphous regions very aligned.
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13
Q

Describe the mechanics of the deformation of elastomers and the driving force behind it?

A
  1. An elastomer is an amorphous polymer. In an unstressed state, the elastomer is in its preferred confirmation (lowest energy state), with molecular chains that are twisted and coiled, and cross-linked. Entropy is high.
  2. When a load is applied, the elastic deformation is simply the untwisting and uncoiling of the chains. The primary bonds and chains are elongated. They become straighter and aligned with the loading direction, hence the stiffness increases. Entropy decreases.
  3. Once the stress is released, the chains spring back to their original, twisted form.

The driving force behind elastic deformation is entropy (i.e. the degree of disorder within a system). When the elastomer is stretched and chains untangled, the entropy is low vs the elastomer in its original from, entropy high.

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14
Q

What is drawing?

What does it cause?

A

Think drawing an elastic band.

Drawing stretches the polymer prior to use. It aligns the chains in the direction of stretching, creating a highly orientated molecular structure.

It causes:
A high degree of interchain secondary bonding.
An increase in elastic modulus E in the direction of stretching.
An increase in tensile strength TS in the direction of stretching.
A decrease in ductility (%EL).

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15
Q

What happens when you anneal after drawing?

A

Decreases in chain alignment.

Has the opposite effect of drawing. I.e. decreases E and TS, but increases %EL.

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16
Q

What are the differences between Thermoplastics and Thermosets? Name some examples?

A
Thermoplastics
Little crosslinking
Ductile
Soften with heating
E.g. polyethylene, polycarbonate, polypropylene, polystyrene

Thermosets
Significant crosslinking (10-50% of repeat units)
Hard and brittle
Don’t soften with heating
E.g. vulcanized rubber, epoxies, polyester resin

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17
Q

What influence does temperature and strain rate have on Thermoplastics?

A

Decreasing temperature:
Increases elastic modulus E
Increases tensile strength
Decreases ductility

Increasing strain rate:
<=> decreasing temperature

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18
Q

In general, increasing the temperature in materials and polymers does what?

A

Increases the KE of the molecules.

As energy increases:
Atoms vibrate -> side-groups rotate/vibrate -> whole branches rotate/vibrate.

In condensed polymers, at specific temperatures there’s enough energy to destroy secondary bonds.

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19
Q

What is free volume and what influences it?

A

Free volume is the difference between the total volume and the volume occupied by the polymer chains.

It normally equates for 2-3% of the total volume.

Lower secondary forces = Less cohesion between molecules so more free volume.

Higher temperature = Molecules with more energy, hence more free volume (think Gas vs Solid).

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20
Q

What are transition temperatures?

A

Where at certain temperatures, particular secondary bonds fail completely. All transitions are reversible, with those destroyed secondary bonds able to return.

Obviously less secondary bonds means the polymer is easier to deform.

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21
Q

What is the glass transition temperature?
How does a material behave above and below it?
What influences it?
How does it effect free volume?

A

The glass transition temperature is where crankshaft-like motion of the main chain occurs up to 10 repeat units within the polymer chain. So imagine a crank handle rotating around, that’s molecules in the chain.

Below the glass transition temperature, the material behaves as glass. Above, the material behaves as rubber.

Tg is influenced by chain stiffness. A stiff chain is harder to “crank” due to:

  1. Bulky side groups
  2. Polar side groups
    c. Chain double bonds

Temperatures above the Tg cause the destruction of secondary bonds. This increases free volume as there’s less cohesion between molecules.

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22
Q

How is free volume influenced by the glass transition temperature?

A
  1. Free volume is influenced by secondary bonds.
  2. Above Tg, secondary bonds break.
  3. This increases free volume as less cohesion between molecules.
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23
Q

What happens to the following above the transition temperature?

A
24
Q

What are the characteristics and use-cases for fibres?

And why are they stiff and strong?

A

Very anisotropic (properties that differ based on the direction of measurement)
Fibre length > 100 x diameter
High tensile strength.
Very crystalline.

They can be used for textiles and composites.

Since they’re very crystalline, we know that the chains are very aligned. This means any load applied is taken along the polymer chains and dealt with by primary bonds. Primary bonds > secondary bonds and hence, fibres act very differently to most polymers.

25
Q

What are engineering polymers and their properties?

A

They are semi-crystalline polymers.
The crystalline regions behave like very strong entanglements. This increases the strength and stiffness of the polymer.
They are insensitive to temperatures below the melting point of the crystalline regions.

26
Q

Increasing Crystallinity of a material …?

A

Increases the elastic modulus.

27
Q

What is meant by polymer viscoelasticity?

A

Viscoelastic behaviour is a time dependent mechanical property/response.

It’s time dependent because amorphous regions which are determined by secondary bonds, need time to find their ideal, lowest energy configuration.

This means the polymer will become more disordered and amorphous over time, hence causing its properties to change.

28
Q

What do graphs a and b show, and what are they known as?

A

Constant stress with a time dependent strain = Creep

Constant strain with a time dependent stress = Stress Relaxation

29
Q

What is the Maxwell Model?

A

The Maxwell model is a spring and dashpot in series. It is used to find the Stress Relaxation at a constant strain.

Constant strain with a time dependent stress = Stress Relaxation

30
Q

What is the Kelvin/Voigt model?

A

The Kelvin/Voigt model is a spring and dashpot in parallel. It is used to find the Creep at a constant stress.

Constant stress with a time dependent strain = Creep

31
Q

What is the Boltzmann Superposition Principle?

What is the equation and how do we apply it?

A

The Boltzmann Superposition Principle describes how to handle creep in materials in service. We know that in service, materials will experience loads of differing size and frequency (how often). We also know that viscoelastic behaviour exists and will effect the material overtime. Hence, it is extremely useful.

32
Q

What is crazing?

A

Crazing is localised plastic deformation. They are essentially small cracks that usually start on the surface and are orientated perpendicular to the tensile axis. They occur whilst in glassy states.

33
Q

List and describe the 3 ways crazes can grow?

A

Crave Nucleation:

  1. Local stresses at the tip of the surface are higher than overall tensile stress applied to the specimen.
  2. Yielding takes place at these stress concentrations.

Fibril Formation:

  1. The stress effectively re-orders the polymer chains into fibrils. This is cold-drawing.
  2. Fibrils remain intact because they contain orientated molecules along the loading direction. Hence, they are stronger (high modulus).

Craze Widening:

  1. The craze boundaries move down into the material, since the stress concentration is at the crack tip.
  2. Craze forms as very thin lamellae in a plane perpendicular to the tensile axis because the surrounding material is at a lower stress level.
34
Q

What is rubber toughening?

A

Glassy polymers are prone to brittle fracture. Rubbery are not.
So we add a second phase of a rubbery particles via blending, or co-polymerization.

The rubber particles initiate plastic deformation around them. Plastic deformation in polymers is crazing, which absorbs energy.

35
Q

What are the 2 parts of a composite?

A

The matrix, a continuous phase that contains…

The reinforcement, a discrete second phase made up of fibres, particles, and laminates.

36
Q

What are the 3 composite classes and their aspect ratios?

A

Monofilaments
Whiskers/Staple Fibres
Particulates

37
Q

What is the relationship between modulus and volume fraction of fibres for axial and transverse loading?

A
38
Q

What does Anisotropic and Isotropic mean?

A

Anisotropic = materials which change their properties with direction, e.g. wood is anisotropic. Load applied transverse is not = load applied axially.

Isotropic = The opposite. So materials which have the same properties in all direction, e.g. metal or glass.

39
Q

Describe the stress-strain relationship of composites, fibres and “un-reinforced” matrices?

A
  1. The composite (matrix and fibres) elongates elastically with a linear stress-strain relationship as depicted by region 1.
  2. As the load increases to a point (εym, σym), matrix yield strain and stress. The matrix can no longer sustain the load elastically, and hence starts deforming plastically (region 2).
  3. Depending on the strain, both the matrix and fibre could deform plastically but rarely occurs.
  4. Eventually, the composite fails, either due to the matrix or the fibre failing (end of region 2). Fibre failing first is more likely.
40
Q

What is off-axis loading

A

Since composites are anisotropic. If we apply a load at an angle to the fibres…

41
Q

Compare continuous and discontinuous fibre reinforcement discussing the following,

Strength?
Toughness?
Anisotropy?

A

Continuous fibres are stronger and tougher if the load is in the axial direction (so parallel to fibres). Hence, they are very anisotropic.

Discontinuous fibres are stronger when loading transversely, but weaker when loading axially. Hence, they are less anisotropic.

Note: Discontinuous fibres are the better option you absolutely know its axial loading only. The delta between axial is far less than the delta between transverse.

42
Q

Explain how molecular weight influences the tensile modulus of a semi-crystalline polymer?

A

The Molecular Weight does not influence the tensile modulus.

43
Q

Explain how the degree of crystallinity influences the tensile modulus of a semi-crystalline polymer?

A

As crystallinity increases, so does the tensile modulus. This is because a higher % of crystallinity means more aligned chain segments. This in-turn leads to enhanced secondary interchain bonding which prevents relative interchain motion.

44
Q

Explain how deformation by drawing influences the tensile modulus of a semi-crystalline polymer?

A

Drawing increases the tensile modulus. It aligns chains in the direction of stretching and creates a highly orientated molecular structure. It also creates a high degree of interchain secondary bonding.

45
Q

Explain how annealing of an undeformed material influences the tensile modulus of a semi-crystalline polymer?

A

Annealing an undeformed semi-crystalline polymer below its melting temperature results in an increased tensile modulus.

46
Q

Explain how the annealing of a drawn material influences the tensile modulus of a semi-crystalline polymer?

A

Annealing after drawing decreases chain alignment and it also reverses the effects of drawing. Hence, it will decrease the tensile modulus.

47
Q

Discuss the advantages and disadvantages of composites over polymers?

A

+: Stiffer, stronger and tougher than normal polymers.

-: Less ductile, very anisotropic, and much more difficult and costly to produce.

48
Q

What is the critical fibre length?
What are the criteria?
What influences it?
What is the formula to find it?

A

The critical fibre length is the shortest fibre length for which the fibre fracture strength can be attained in tensile loading.

Used to determine if the composite behaves like a long-fibre or short-fibre. If l > lcrit, we assume long-fibre. If l < lcrit, we assume short-fibre.

Think of the formula: fibre diameter, fibre fracture strength, and fibre-matrix bond strength.

49
Q

What would be the steps to prove, Ec = VfEf + VmEm?

A
  1. Total P = Pf + Pm.
  2. P = σA. So, Pf = σfAf etc.
  3. Vf = Af/Ac, where c = entire composite.
  4. Solve to find σc = Vfσf + Vmσm.
  5. E = σ/ε (stress/strain). So, σf=Ef*ε.
  6. Solve to find Ec.
50
Q

What would be the steps to prove, lcrit formula?

A
  1. Draw triangle. Quote that dσ/dx = 2τ/r = 4τ/d.
  2. Hence, σmax = 4τ/d * 0.5l.
  3. Area’s equal. 0.5lσmax = σmean * l.
  4. Hence, σmean = τl/d
  5. Let l=lcrit, d=df, σmean = σf etc.
  6. Rearrange to form lcrit = (df*σf)/2τ
51
Q

What is the formula for longitudinal/parallel elastic modulus?

What is the formula for transverse/perpendicular elastic modulus?

(Rule of Mixtures)

A

Parallel: Ec = VfEf + VmEm

Perpendicular: Ec = (EfEm) / (VfEf + Vm*Em)
so baso multiplied/parallel

52
Q

What can you do in composite calculations if the composite is aligned?

A

If aligned, εf = εm.

53
Q

What is the formula for Vmin?

A
54
Q

How do we find the “load fraction”?

A

We know, P = σ*A, and load = pressure.
So Lf = Pf/Pc.
Can simplify, cancel out Af/Ac = Vf.

55
Q

With the aid of suitable diagrams, describe the typical nature of stress-strain behaviour for covalently bonded networks and elastomers?

Explain how the differences in the internal structure of these materials lead to the differences in stress-strain behaviour?

A

Linear stress-strain behaviour with a steep gradient. Has sudden and catastrophic failure.

Elastomers are amorphous polymers, with enormous strains to failure and the deformation is recoverable across the entire curve. I.e. you can go from the top all the way to 0.

Crystalline vs amorphous. Crystalline very aligned chains with lots of primary bonding. Hence, can take a lot of stress, but fail suddenly and unexpectedly. Amorphous twisted chains, can be straightened and aligned with far less stress.

56
Q

What would the graphs look like for a polymer under constant stress, below and above Tg?

(no stress applied until time 0)

A

Under constant stress:

The brittle phase would have a constant strain as it behaves elastically. Hence, const stress = const strain.

The ductile phase would have a strain constantly increasing as it behaves plastically and is viscoelastic.