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Chapter Outline Review of Atomic Structure - peopleVirginiaEDU

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Description

CHAPTER 15

CHARACTERISTICS,

APPLICATIONS,

AND PROCESSING OF POLYMERS PROBLEM SOLUTIONS

Stress-Strain Behavior 15

the elastic modulus is the slope in the elastic linear region of the 20°C curve,

∆ (stress) 30 MPa − 0 MPa = = 3

The value range cited in Table 15

24 to 3

the plotted value is a little on the high side

The tensile strength corresponds to the stress at which the curve ends,

This value lies within the range cited in Table 15

3 to 72

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Viscoelastic Deformation 15

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which behavior is according to Equation 15

⎛ t⎞ σ(t) = σ(0) exp ⎜− ⎟ ⎝ τ⎠

We want to determine σ(10),

but it is first necessary to compute τ from the data provided in the problem statement

solving for τ from the above expression,

Therefore,

⎛ 10 s'⎞ σ(10) = (3

4 s'⎠

Er (10) =

σ (10) 1

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The glass-transition temperature is that temperature corresponding to the abrupt decrease in log Er(10),

which for this PMMA material is about 115°C

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(a) Crystalline polystyrene at 70°C behaves in a glassy manner (Figure 15

the strain-time behavior would be as Figure 15

(b) Amorphous polystyrene at 180°C behaves as a viscous liquid (Figure 15

the strain-time behavior will be as Figure 15

(c) Crosslinked polystyrene at 180°C behaves as a rubbery material (Figure 15

the strain-time behavior will be as Figure 15

(d) Amorphous polystyrene at 100°C behaves as a leathery material (Figure 15

the straintime behavior will be as Figure 15

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holding the strain level constant,

and then measuring the stress as a function of time

For viscoelastic creep tests,

a stress (usually tensile) is applied instantaneously and maintained constant while strain is measured as a function of time

(b) The experimental parameters of interest from the stress relaxation and viscoelastic creep tests are the relaxation modulus and creep modulus (or creep compliance),

The relaxation modulus is the ratio of stress measured after 10 s'and strain (Equation 15

creep modulus is the ratio of stress and strain taken at a specific time (Equation 15

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Such a plot is given below

Increasing molecular weight increases both glass-transition and melting temperatures

(b) We are now called upon to make a plot of log Er(10) versus temperature demonstrating how the behavior changes with increased crosslinking

Such a plot is given below

Increasing the degree of crosslinking will increase the modulus in both glassy and rubbery regions

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Fracture of Polymers Miscellaneous Mechanical Considerations 15

five factors that favor brittle fracture are as follows: (1) a reduction in temperature,

and (5) modifications of the polymer structure

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the fatigue strengths for nylon 6 and 2014-T6 aluminum are 11 MPa (1600 psi) and 200 MPa (30,000 psi ),

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Deformation of Semicrystalline Polymers 15

(c) The explanation of the mechanism by which elastomers elastically deform is provided in Section 15

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Factors That Influence the Mechanical Properties of Semicrystalline Polymers Deformation of Elastomers 15

(b) Tensile modulus increases with increasing degree of crystallinity for semicrystalline polymers

This is due to enhanced secondary interchain bonding which results from adjacent aligned chain segments as percent crystallinity increases

This enhanced interchain bonding inhibits relative interchain motion

(c) Deformation by drawing also increases the tensile modulus

The reason for this is that drawing produces a highly oriented molecular structure,

and a relatively high degree of interchain secondary bonding

(d) When an undeformed semicrystalline polymer is annealed below its melting temperature,

(e) A drawn semicrystalline polymer that is annealed experiences a decrease in tensile modulus as a result of a reduction in chain-induced crystallinity,

and a reduction in interchain bonding forces

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This effect is explained by increased chain entanglements at higher molecular weights

(b) Increasing the degree of crystallinity of a semicrystalline polymer leads to an enhancement of the tensile strength

this is due to enhanced interchain bonding and forces

in response to applied stresses,

interchain motions are thus inhibited

(c) Deformation by drawing increases the tensile strength of a semicrystalline polymer

This effect is due to the highly oriented chain structure that is produced by drawing,

which gives rise to higher interchain secondary bonding forces

(d) Annealing an undeformed semicrystalline polymer produces an increase in its tensile strength

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There is more of an opportunity for van der Waals bonds to form between two molecules in close proximity to one another than for isobutane because of the linear nature of each normal butane molecule

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Equation 15

using the data provided in the problem statement,

we may set up two simultaneous equations from which it is possible to solve for the two constants TS∞ and A

These equations are as follows: 50 MPa = TS∞ −

A 30,000 g / mol

A 50,000 g / mol

the values of the two constants are: TS∞ = 300 MPa and A = 7

Substituting these values into Equation 15

A 40,000 g / mol

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Equation 15

using the data provided in the problem statement,

we may set up two simultaneous equations from which it is possible to solve for the two constants TS∞ and A

These equations are as follows: 90 MPa = TS∞ −

A 20,000 g / mol A 40,000 g / mol

the values of the two constants are: TS∞ = 270 MPa and A = 3

Solving for M n in Equation 15

A TS∞ − TS

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we are to do the following: (1) determine whether or not it is possible to decide which has the higher tensile modulus

note which has the higher tensile modulus and then state the reasons for this choice

and (3) if it is not possible to decide,

The linear and isotactic poly(vinyl chloride) will display a greater tensile modulus

Linear polymers are more likely to crystallize that branched ones

In addition,

polymers having isotactic structures will normally have a higher degree of crystallinity that those having atactic structures

Increasing a polymer's crystallinity leads to an increase in its tensile modulus

In addition,

tensile modulus is independent of molecular weight--the atactic/branched material has the higher molecular weight

The block styrene-butadiene copolymer with 10% of possible sites crosslinked will have the higher modulus

Block copolymers normally have higher degrees of crystallinity than random copolymers of the same material

A higher degree of crystallinity favors larger moduli

In addition,

the block copolymer also has a higher degree of crosslinking

increasing the amount of crosslinking also enhances the tensile modulus

Branched polyethylene will tend to have a low degree of crystallinity since branched polymers don't normally crystallize

The atactic polypropylene probably also has a relatively low degree of crystallinity

atactic structures also don't tend to crystallize,

and polypropylene has a more complex repeat unit structure than does polyethylene

Tensile modulus increases with degree of crystallinity,

and it is not possible to determine which polymer is more crystalline

Furthermore,

tensile modulus is independent of molecular weight

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we are to do the following: (1) determine whether or not it is possible to decide which has the higher tensile strength

then note which has the higher tensile strength and then state the reasons for this choice

and (3) if it is not possible to decide,

The linear and isotactic material will have the higher tensile strength

Both linearity and isotacticity favor a higher degree of crystallinity than do branching and atacticity

and tensile strength increases with increasing degree of crystallinity

Furthermore,

the molecular weight of the linear/isotactic material is higher (100,000 g/mol versus 75,000 g/mol),

and tensile strength increases with increasing molecular weight

Alternating copolymers tend to be more crystalline than graft copolymers,

and tensile strength increases with degree of crystallinity

However,

the graft material has a higher degree of

and tensile strength increases with the percentage of crosslinks

The network polyester will display a greater tensile strength

Relative chain motion is much more restricted than for the lightly branched polytetrafluoroethylene since there are many more of the strong covalent bonds for the network structure

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will be greater than for a polytetrafluoroethylene having the same molecular weight and degree of crystallinity

The replacement of one fluorine atom within the PTFE repeat unit with a chlorine atom leads to a higher interchain attraction,

Furthermore,

poly(vinyl chloride) is stronger than polyethylene (Table 15

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both of which have a molecular weight of 100,000 g/mol

These two materials are elastomers and will have curves similar to curve C in Figure 15

However,

the curve for the material having the greater number of crosslinks (20%) will have a higher elastic modulus at all strains

(b) Shown below are the stress-strain curves for the two polypropylene materials

These materials will most probably display the stress-strain behavior of a normal plastic,

However,

the syndiotactic polypropylene has a higher molecular weight and will also undoubtedly have a higher degree of crystallinity

it will have a higher strength

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(c) Shown below are the stress-strain curves for the two polyethylene materials

The branched

polyethylene will display the behavior of a normal plastic,

On the other hand,

the heavily crosslinked polyethylene will be stiffer,

and more brittle (curve A of Figure 15

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having chains that are extensively coiled and kinked in the unstressed state

and (2) there must be some crosslinking

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(a) Linear and crystalline polyethylene would be neither an elastomer nor a thermoset since it is a linear polymer

(b) Phenol-formaldehyde having a network structure would be a thermosetting polymer since it has a network structure

It would not be an elastomer since it does not have a crosslinked chain structure

Heavily crosslinked polyisoprene having a glass transition temperature of 50°C would be a

thermosetting polymer because it is heavily crosslinked

It would not be an elastomer since it is heavily crosslinked and room temperature is below its Tg

(d) Lightly crosslinked polyisoprene having a glass transition temperature of –60°C is both an elastomer and a thermoset

It is an elastomer because it is lightly crosslinked and has a Tg below room temperature

It is a thermoset because it is crosslinked

(e) Linear and partially amorphous poly(vinyl chloride) is neither an elastomer nor a thermoset

In order to be either it must have some crosslinking

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This problem asks that we compute the fraction of possible crosslink sites in 15 kg of

Given the butadiene repeat unit in Table 14

we may calculate its molecular weight as follows: A(chloroprene) = 4(AC) + 5(AH) + ACl

53 g / mol

For the vulcanization of polychloroprene,

there are two possible crosslink sites per repeat unit—one for each of the two carbon atoms that are doubly bonded

Furthermore,

each of these crosslinks forms a bridge between two repeat units

Therefore,

we can say that there is the equivalent of one crosslink per repeat unit

Let us now calculate the number of moles of sulfur (nsulfur) that react with the chloroprene,

by taking the mole ratio of sulfur to chloroprene,

and then dividing this ratio by 5

this yields the fraction of possible sites that are crosslinked

Thus nsulfur =

2 mol 32

06 g / mol

And nsulfur

4 mol = 0

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we are asked to compute the weight percent sulfur necessary for complete crosslinking,

four sulfur atoms participate in each crosslink

The acrylonitrile and butadiene repeat units are shown in Table 14

from which it may be noted that there are two possible crosslink sites on each butadiene repeat unit (one site at each of the two carbon atoms that are doubly bonded),

and no possible sites for acrylonitrile

since it is an alternating copolymer,

the ratio of butadiene to acrylonitrile repeat units is 1:1

for each pair of combined butadiene-acrylonitrile repeat units which crosslink,

eight sulfur atoms are required,

the sulfur-to-(acrylonitrilebutadiene) ratio is 4:1

one mole of the combined acrylonitrile-butadiene repeat units

In order for complete crosslinking,

four moles of sulfur are required

for us to convert this composition to weight percent,

it is necessary to convert moles to mass

The acrylonitrile repeat unit consists of three carbon atoms,

the butadiene repeat unit is composed of four carbons and six hydrogens

This gives a molecular weight for the combined repeat unit of m(acrylonitrile-butadiene) = 3(AC) + 3(AH) + AN + 4(AC) + 6(AH)

in one mole of this combined repeat unit,

Furthermore,

0 mol)(32

the concentration of S in weight percent CS (using Equation 4

5 wt% 128

24 g + 107

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five sulfur atoms participate in each crosslink

If we arbitrarily consider 100 g of the vulcanized material,

let us find how many moles of sulfur and isoprene correspond to these masses

The atomic weight of sulfur is 32

06 g/mol,

3 g = 1

41 mol 32

06 g / mol

in each isoprene repeat unit there are five carbon atoms and eight hydrogen atoms

the molecular weight of a mole of isoprene units is (5)(12

the number of moles is equal to

7 g = 0

793 mol 68

11 g / mol

Therefore,

the ratio of moles of S to the number of moles of polyisoprene is 1

78 : 1 0

793 mol

When all possible sites are crosslinked,

the ratio of the number of moles of sulfur to the number of moles of isoprene is 5:1

this is because there are two crosslink sites per repeat unit and each crosslink is shared between repeat units on adjacent chains,

and there are 5 sulfur atoms per crosslink

Finally,

to determine the fraction of sites that are crosslinked,

we just divide the actual crosslinked sulfur/isoprene ratio by the completely crosslinked ratio

fraction of repeat unit sites crosslinked =

78 /1 = 0

356 5 /1

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Table 14

For each of these units there are two possible crosslink sites

one site is associated with each of the two carbon atoms that are involved in the chain double bond

Since 10% of the possible sites are crosslinked,

for each 100 isoprene repeat units 10 of them are crosslinked

actually there are two crosslink sites per repeat unit,

but each crosslink is shared by two chains

Furthermore,

on the average we assume that each crosslink is composed of 3

In terms of moles,

it is necessary to add 35 moles of sulfur to 100 moles of isoprene

The atomic weight of sulfur is 32

06 g/mol,

while the molecular weight of isoprene is A(isoprene) = 5(AC) + 8(AH)

While for isoprene mip = (100 mol)(68

the concentration of sulfur in weight percent (Equation 4

3) is just

mS mS + mip

× 100 =

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Crystallization 15

One way to solve this problem is to take two values of percent recrystallization (which is just 100y,

Equation 10

then set up two simultaneous equations,

from which n and k may be determined

In order to expedite this process,

we will rearrange and do some algebraic manipulation of Equation 10

First of all,

Now taking natural logarithms ln (1 − y) = − kt n

Or − ln (1 − y) = kt n

which may also be expressed as ⎛ 1 ln ⎜ ⎝1 −

⎞ n ⎟ = kt y⎠

Now taking natural logarithms again,

leads to ⎡ ⎛ 1 ⎞⎤ ln ⎢ ln ⎜ ⎟⎥ = ln k + n ln t ⎣ ⎝ 1 − y ⎠⎦

which is the form of the equation that we will now use

From the 150°C curve of Figure 15

let us arbitrarily choose two percent crystallized values of 20% and 80% (i

The corresponding time values are t1 = 220 min and t2 = 460 min (realizing that the time axis is scaled logarithmically)

our two simultaneous equations become ⎡ ⎛ 1 ⎞⎤ ln ⎢ ln ⎜ ⎟⎥ = ln k + n ln (220) ⎣ ⎝ 1 − 0

20 ⎠⎦

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⎡ ⎛ 1 ⎞⎤ ln ⎢ ln ⎜ ⎟⎥ = ln k + n ln (460) ⎣ ⎝ 1 − 0

80 ⎠⎦

from which we obtain the values n = 2

2 x 10-7

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Melting and Glass Transition Temperatures 15

would be suitable for the fabrication of cups to contain hot coffee

At its glass transition temperature,

an amorphous polymer begins to soften

The maximum temperature of hot coffee is probably slightly below 100°C (212°F)

Of the polymers listed,

only polystyrene and polycarbonate have glass transition temperatures of 100°C or above (Table 15

and would be suitable for this application

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Of those polymers listed in Table 15

and polypropylene satisfy this criterion

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Factors That Influence Melting and Glass Transition Temperatures 15

The linear polyethylene will be highly crystalline,

will exhibit behavior similar to curve C in Figure 15

The branched polyethylene will be semicrystalline,

therefore its curve will appear as curve B in this same figure

Furthermore,

since the linear polyethylene has the greater molecular weight,

it will also have the higher melting temperature

Shown below are specific volume-versus-temperature curves for the poly(vinyl chloride) and

Since both are 50% crystalline,

they will exhibit behavior similar to curve B in Figure 15

However,

since the polypropylene has the greater molecular weight it will have the higher melting

Furthermore,

polypropylene will also have the higher glass-transition temperature inasmuch as its CH3 side group is bulkier than the Cl for PVC

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(c) Shown below are specific volume-versus-temperature curves for the polystyrene and polypropylene materials

Since both are totally amorphous,

they will exhibit the behavior similar to curve A in Figure 15

However,

since the polystyrene repeat unit has a bulkier side group than polypropylene (Table 14

its chain flexibility will be lower,

its glass-transition temperature will be higher

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it is possible to determine which polymer has the higher melting temperature

The linear polyethylene will most likely have a higher percent crystallinity,

a higher melting temperature than the branched polyethylene

The molecular weights of both materials are the same and,

molecular weight is not a consideration

it is possible to determine which polymer has the higher melting temperature

Of these two polytetrafluoroethylene polymers,

the PTFE with the higher density (2

a higher melting temperature than the lower density PTFE

The molecular weights of both materials are the same and,

molecular weight is not a consideration

it is possible to determine which polymer has the higher melting temperature

The linear polyethylene will have the greater melting temperature inasmuch as it will have a higher degree of crystallinity

polymers having a syndiotactic structure do not crystallize as easily as those polymers having identical single-atom side groups

With regard to molecular weight,

it is about the same for both materials (8000),

it is not possible to determine which of the two polymers has the higher melting temperature

The syndiotactic polypropylene will have a higher degree of crystallinity than the atactic material

On the basis of this effect alone,

the syndiotactic PP should have the greater Tm,

since melting temperature increases with degree of crystallinity

However,

the molecular weight for the syndiotactic polypropylene (500,000 g/mol) is less than for the atactic material (750,000 g/mol)

and this factor leads to a lowering of the melting temperature

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the elastic modulus may be enhanced by increasing the number of crosslinks (while maintaining the molecular weight constant)

this will also enhance the glass transition

the modulus-glass transition temperature behavior would appear as

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Elastomers Fibers Miscellaneous Applications 15

For the silicone polymers,

this backbone chain is composed of silicon and oxygen atoms that alternate positions

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and (2) they must have chain configurations/structures that will allow for a high degrees of crystallinity

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(3) high tensile and tear strengths

(4) resistance to moisture/chemical attack

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Polymerization 15

the reactant species have the same chemical composition as the monomer species in the molecular chain

This is not the case for condensation polymerization,

wherein there is a chemical reaction between two or more monomer species,

There is often a low molecular weight by-product for condensation polymerization

such is not found for addition polymerization

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Since the chemical formulas are provided in this equation we may calculate the molecular weights of each of these materials as follows: MW(ethylene glycol) = 2( AC ) + 6( AH ) + 2( AO ) = 2 (12

07 g/mol

MW(terephthalic acid) = 8( AC ) + 6( AH ) + 4( AO ) = 8 (12

13 g/mol

13 g / mol

Equation 15

each mole of terephthalic acid used requires one mole of ethylene glycol,

39 mol)(62

(b) Now we are asked for the mass of the resulting polymer

Inasmuch as one mole of water is given off for every repeat unit produced,

39 mol)(18

02 g/mol

The mass of poly(ethylene terephthalate) is just the sum of the masses of the two reactant materials [as computed in part (a)] minus the mass of water released,

or mass [poly(ethylene terephthalate)] = 20

0 kg + 7

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The chemical equation for this reaction is the answer to Concept Check 15

From this equation we may calculate the molecular weights of these molecules

MW(adipic) = 6( AC ) + 10( AH ) + 4( AO )

14 g/mol

MW(hexamethylene) = 6( AC ) + 16( AH ) + 2( AN )

21 g/mol

MW(nylon) = 12( AC ) + 22( AH ) + 2( AN ) + 2( AO )

00 g/mol)

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37 mol 226

32 g / mol

according to the chemical equation given above,

each mole of nylon 6,6 that is produced requires one mole each of adipic acid and hexamethylene diamine,

with two moles of water as the by-product

The masses

914 g = 12

269 g = 10

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Polymer Additives 15

whereas a pigment does not dissolve,

but remains as a separate phase

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Forming Techniques for Plastics 15

and (4) the geometry and size of the finished product

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For compression molding,

both heat and pressure are applied after the polymer and necessary additives are situated between the mold members

For transfer molding,

the solid materials (normally thermosetting in nature) are first melted in the transfer chamber prior to being forced into the die

for injection molding (normally used for thermoplastic materials),

the raw materials are impelled by a ram through a heating chamber,

and finally into the die cavity

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Fabrication of Fibers and Films 15

they must be capable of forming a viscous liquid when heated,

which is not possible for thermosets

mechanical elongation must be possible

inasmuch as thermosetting materials are,

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the one that was formed by extrusion and then rolled would have the higher strength

Both blown and extruded materials would have roughly comparable strengths

however the rolling operation would further serve to enhance the strength of the extruded material

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DESIGN QUESTIONS 15

D1 (a) Several advantages of using transparent polymeric materials for eyeglass lenses are: they have relatively low densities,

they are relatively easy to grind to have the desired contours

they are less likely to shatter than are glass lenses

wraparound lenses for protection during sports activities are possible

and they filter out more ultraviolet radiation than do glass lenses

The principal disadvantage of these types of lenses is that some are relatively soft and are easily scratched (although antiscratch coatings may be applied)

Plastic lenses are not as mechanically stable as glass,

(b) Some of the properties that are important for polymer lens materials are: they should be relatively hard in order to resist scratching

they should be shatter resistant

they must have a relatively high index of refraction such that thin lenses may be ground for very nearsighted people

and they should absorb significant proportions of all types of ultraviolet radiation,

which radiation can do damage to the eye tissues

(c) Of those polymers discussed in this chapter and Chapter 4,

likely lens candidates are polystyrene,

these three materials are not easily crystallized,

Upon consultation of their fracture toughnesses (Table B

polycarbonate is the most superior of the three

Commercially,

the two plastic lens materials of choice are polycarbonate and allyl diglycol carbonate (having the trade name CR-39)

Polycarbonate is very impact resistant,

Furthermore,

PC comes in both normal and high refractive-index grades

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D2 There are three primary requirements for polymeric materials that are utilized in the packaging of food products and drinks

these are: (1) sufficient strength,

(2) barrier protection--that is,

being resistant to permeation by oxygen,

and (3) being nonreactive with the food/drink contents--such reactions can compromise the integrity of the packaging material,

or they can produce toxic by-products

With regard to strength,

poly(ethylene terephthalate) (PET or PETE) and oriented polypropylene (OPP) have high tensile strengths,

linear low-density polyethylene (LLDPE) and low-density polyethylene (LDPE) have high tear strengths,

while those polymers having the best impact strengths are PET and poly(vinyl chloride) (PVC)

Relative to barrier characteristics,

ethylene vinyl alcohol (EVOH) and poly(vinylidene chloride) (PVDC) copolymers are relatively impermeable to oxygen and carbon dioxide,

whereas high-density polyethylene (HDPE),

and LDPE are impervious to water vapor

Most common polymers are relatively nonreactive with food products,

exceptions are acrylonitrile and plasticizers used in PVC materials

The aesthetics of packaging polymers are also important in the marketing of food and drink products

Some will be colored,

many are adorned with printing,

others need to be transparent and clear,

and many need to be resistant to scuffing

On the basis of the preceding discussion,

examples of polymers that are used for specific applications are as follows: PET(E) for soda pop containers

PVC for beer containers

LDPE and HDPE films for packaging bread and bakery products

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D3 The primary reasons that the automotive industry has replaced metallic automobile components with polymer and composite materials are: polymers/composites (1) have lower densities,

and afford higher fuel efficiencies

(2) may be produced at lower costs but with comparable mechanical characteristics

(3) are in many environments more corrosion resistant

and (5) are thermally insulating and thus reduce the transference of heat

These replacements are many and varied

Several are as follows: Bumper fascia are molded from an elastomer-modified polypropylene

Overhead consoles are made of poly(phenylene oxide) and recycled polycarbonate

Rocker arm covers are injection molded of a glass- and mineral-reinforced nylon 6,6 composite

Torque converter reactors,

are made from phenolic thermoset composites that are reinforced with glass fibers

Air intake manifolds are made of a glass-reinforced nylon 6,6

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