CHAPTER 15

CHARACTERISTICS, APPLICATIONS, AND PROCESSING OF POLYMERS

PROBLEM SOLUTIONS

Stress-Strain Behavior

15.1 From the stress–strain data for poly(methyl

methacrylate) shown in Figure 15.3, determine the

modulus of elasticity and tensile strength at room temperature [20°C (68°F)] , and compare these values with those

given in Table 15.1.

Solution

From Figure 15.3, the elastic modulus is the slope in the

elastic linear region of the 20°C curve, which is

 

E =D (stress)

D (strain)=

30 MPa -

0 MPa

9 ´

10-

3 -

0=

3.3 GPa

(483,

000 psi)

The value range cited in Table 15.1 is 2.24 to 3.24 GPa (325,000 to 470,000 psi). Thus, the plotted value is a little on

the high side.

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

hich is 52 MPa (7500 psi). This

value lies within the range cited in Table 15.1—

48.3 to 72.4 MPa (7000 to 10,500 psi).

15.2 Compute the elastic moduli for the following polymers, whose stress-strain behaviors may be

observed in the “Tensile Tests” module of Virtual Materials Science and Engineering (VMSE): (a) high-density

polyethylene, (b) nylon, and (c) phenol-formaldehyde (bakelite). How do these values compare with those

presented in Table 15.1 for the same polymers?

Solution

The elastic modu

lus is the slope in the linear elastic region (Equation 6.10) as

 

E =D s

D e=

s

2 - s

1

e

2 - e

1

Since all the stress-strain curves pass through the origin, we make take s

1

= 0 and e

1

= 0. Determinations of s

2 and e

2

are possible by moving the cursor to some arbitrary point in the linear region of the curve and then reading

corresponding values in the “Stress” and “Strain” windows that are located below the plot.

(a) For high-density polyethylene, we selected s

2

= 6.0 MPa with its corresponding e

2

= 0.0152. Therefore,

 

E = s

2 - s

1

e

2 - e

1

=

6.

0 MPa -

0 MPa

0.

0152 -

0 =

395 MPa =

0.

395 GPa

The elastic modulus (average) for high-

density polyethylene given in Table 15.1 is 1.08 GPa, which is

significantly higher than this value.

(b) For nylon, we selected s

2

= 24.3 MPa with its corresponding e

2

= 0.0085. Therefore,

 

E = s

2 - s

1

e

2 - e

1

=

24.

3 MPa -

0 MPa

0.

0085 -

0 =

2860 MPa =

2.

86 GPa

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the VMSE nylon lies within this range.

(c) For phenol-formaldehyde (bakelite), we selected s

2

= 33.0 MPa with its corresponding e

2

= 0.0068.

Therefore,

 

E = s

2 - s

1

e

2 - e

1

=

33.

0 MPa -

0 MPa

0.

0068 -

0 =

4850 MPa =

4.

85 GPa

The elastic modulus range for the phenol-

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therefore, this value for the VMSE phenol-formaldehyde lies just above the maximum value for this range.

15.3 For the nylon polymer, whose stress strain behavior may be observed in the “Tensile Tests” module

of Virtual Materials Science and Engineering (VMSE), determine the following:

(a) the yield strength, and

(b) the approximate ductility, in percent elongation.

How do these values compare with those for the nylon material presented in Table 15.1?

Solution

(a) The yield strength corresponds to the first maximum (just beyond the initial linear-elastic region) of the

stress-strain curve. This reading in the stress window located below the plot as the curser point is dragged along the

stress-

strain curve is 84 MPa.

(b) The approximate percent elongation corresponds to the strain at fracture multiplied by 100 (i.e., 44.5%)

minus the maximum elastic strain (i.e., value of strain at which the linearity of the curve ends multiplied by 100—in

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For nylon 6,6, the range of values of yield strength presented in Table 15.1 is

0 3D WR 0 3D

therefore, the value for the VMSE

nylon lies slightly above the upper value. The ductility range for nylon 6,6 is

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15.4 For the phenol-formaldehyde (Bakelite) polymer, whose stress strain behavior may be observed in

the “Tensile Tests” module of Virtual Materials Science and Engineering (VMSE), determine the following:

(a) the tensile strength, and

(b) the approximate ductility, in percent elongation.

How do these values compare with those for the phenol-

formaldehyde material presented in Table 15.1?

Solution

(a) The tensile strength corresponds to the stress at which fracture occurs. This reading in the stress

window located below the plot as the curser point is dragged along the stress-

strain curve is 52 MPa.

(b) The approximate percent elongation corresponds to the strain at fracture multiplied by 100 (i.e., 1.29%)

minus the maximum elastic strain (i.e., value of strain at which the linearity of the

curve ends multiplied by 100—in

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For phenol-

formaldehyde, the range of values of tensile strength presented in Table 15.1 is 34.5 MPa to 62.1

MPa; therefore, the value for the VMSE

material (52 MPa) lies within this range. The ductility range for phenol-

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Viscoelastic Deformation

15.5 In your own words, briefly describe the phenomenon of viscoelasticity.

The explanation of viscoelasticity is given in Section 15.4.

15.6 For some viscoelastic polymers that are subjected to stress relaxation tests, the stress decays with

time according to

 

s(t) = s

(0) exp -t

t

æ

è ç

ö

ø ÷

(15.10)

where σ(t) and σ(0) represent the time-

dependent and initial (i.e., time = 0) stresses, respectively, and t and τ

denote elapsed time and the relaxation time; τ is a time-independent constant characteristic of the material. A

specimen of some viscoelastic polymer the stress relaxation

of which obeys Equation 15.10 was suddenly pulled in

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function of time. Determine Er

(10) for this material if the initial stress level was 2.76 MPa

(400 psi), which dropped

to 1.72 MPa (250 psi) after 60 s.

Solution

This problem asks for a determination of the relaxation modulus of a viscoelastic material, which behavior is

according to Equation 15.10--i.e.,

 

s(t) = s

(0) exp -t

t

æ

è ç

ö

ø ÷

We want to determine s

(10), but it is first necessary to compute t from the data provided in the problem statement.

Thus, solving for t from the above expression,

 

t =- t

lns(t)

s(

0)

é

ë ê

ù

û ú

=-

60 s

ln

1.

72 MPa

2.

76 MPa

é

ë ê ù

û ú

=

127 s

Therefore,

 

s

(10) =

(2.76 MPa) exp -

10 s

127 s

æ

è ç

ö

ø ÷ =

2.55 MPa

Now, using Equation 15.1

 

Er

(10) =s (

10)

e

0

=

2.

55 MPa

0.

6=

4.25 MPa

(616 psi)

15.7

In Figure 15.28, the logarithm of Er(t) versus the logarithm of time is plotted for polyisobutylene at a

variety of temperatures. Make a plot of Er

(10) versus temperature and then estimate its Tg.

Solution

Below is plotted the logarithm of Er

(10) versus temperature.

The glass-transition temperature is that temperature corresponding to the abrupt decrease in log Er

(10), which for

this polyisobutylene material is about -

60°C.

15.8

On the basis of the curves in Figure 15.5, sketch schematic strain–time plots for the following

polystyrene materials at the specified temperatures:

(a)

Amorphous at 120°C

(b)

Crosslinked at 150°C

(c)

Crystalline at 230°C

(d) Crosslinked at

50°C

Solution

(a) Amorphous polystyrene at 120°

C behaves a rubbery material (Figure 15.8, curve C); therefore, the strain-

time behavior would be as Figure 15.5c.

(b) Crosslinked polystyrene at 150°

C behaves as a viscoelastic material (Figure 15.8, curve B); therefore, the

strain-

time behavior will be as Figure 15.5c.

(c) Crystalline polystyrene at 230°

C behaves as a viscous liquid (Figure 15.8, curve A); therefore, the strain-

time behavior will be as Figure 15.5d.

(d) Crosslinked polystyrene at 5

0°

C behaves in a glassy manner (Figure 15.8, curve B); therefore, the strain-

time behavior will be as Figure 15.5b.

15.9 (a) Contrast the manner in which stress relaxation and viscoelastic creep tests are conducted.

(b) For each of these tests, cite the experimental parameter of interest and how it is determined.

Solution

(a) Stress relaxation tests are conducted by rapidly straining the material elastically in tension, 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), respectively. The relaxation modulus is the ratio of

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specif

ic time (Equation 15.2).

15.10 Make two schematic plots of the logarithm of relaxation modulus versus temperature for an

amorphous polymer (curve C in Figure 15.8).

(a) On one of these plots demonstrate how the behavior changes with increasing molecular weight.

(b) On the other plot, indicate the change in behavior with increasing crosslinking.

Solution

(a) This portion of the problem calls for a plot of log Er

(10) versus temperature demonstrating how the

behavior changes with increased molecular weight. 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.

Fracture of Polymers

Miscellaneous Mechanical Considerations

15.11 For thermoplastic polymers, cite five factors that favor brittle fracture.

Solution

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

temperature, (2) an increase in strain rate, (3) the presence of a sharp notch, (4) increased sp

ecimen thickness, and (5)

modifications of the polymer structure.

15.12 (a)

Compare the fatigue limits for polystyrene (Figure 15.11) and the cast iron for which fatigue

data are given in Problem 8.20.

(b)

Compare the fatigue strengths at 10

6 cycles fo

r poly(ethylene terephthalate) (PET, Figure 15.11) and

red brass (Figure 8.34).

Solution

(a) The fatigue limits for polystyrene and the cast iron are 10.5 MPa (1500 psi) and 193 MPa (28,000 psi),

respectively.

(b) At 10

6 cycles, the fatigue strengths

of PET and red brass are 15 MPa (2175 psi) and 115 MPa (16,700 psi

), respectively.

Deformation of Semicrystalline Polymers

15.13 In your own words, describe the mechanisms by which semicrystalline polymers (a) elastically

deform and (b) plastically deform, and (c) by which elastomers elastically deform.

(a) and (b) The mechanisms by which semicrystalline polymers elastically and plastically deform are

described in Section 15.7.

(c) The explanation of the mechanism by which elastomers elastically

deform is provided in Section 15.9.

Factors That Influence the Mechanical Properties of Semicrystalline

Polymers

Deformation of Elastomers

15.14 Briefly explain how each of the following influences the tensile modulus of a semicrystalline

polymer and why:

(a) Molecular weight

(b) Degree of crystallinity

(c) Deformation by drawing

(d) Annealing of an undeformed material

(f) Annealing of a drawn material

Solution

(a) The tensile modulus is not directly influenced by a polymer's molecular weight.

(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, the tensile

modulus increases.

(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.

15.15 Briefly explain how each of the following influences the tensile or yield strength of a semicrystalline

polymer and why:

(a) Molecular weight

(b) Degree of crystallinity

(c) Deformation by drawing

(d) Annealing of an undeformed material

Solution

(a) The tensile strength of a semicrystalline polymer increases with increasing molecular weight. 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. Again, 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.

15.16 Normal butane and isobutane have boiling temperatures of –

0.5 and –

12.3°C (31.1 and 9.9°F),

respectively. Briefly explain this behavio

r on the basis of their molecular structures, as presented in Section 14.2.

Solution

Normal butane has a higher melting temperature as a result of its molecular structure (Section 14.2). 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.

15.17 The tensile strength and number-average molecular weight for two poly(methyl methacrylate)

materials are as follows:

Tensile Strength

(MPa) Number-Average

Molecular Weight (g/mol)

107

40,000

170

60,000

Estimate the tensile strength at a number-

average molecular weight of 30,000 g/mol.

Solution

This problem gives us the tensile strengths and associated number-average molecular weights for two

poly(methyl methacrylate) materials and then asks that we estimate the tensile strength for

 

M n

= 30,000 g/mol.

Equation 15.3 cites the dependence of the tensile strength on

 

M n. Thus, 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:

 

107 MPa = TS¥ -A

40,

000 g /mol

 

170 MPa = TS¥ -A

60,

000 g /mol

Thus, the values of the two constants are: TS

= 296 MPa and A

= 7.56 ´

10

6 MPa-g/mol. Substituting these values

into Equation 15.3 for

 

M n

= 30,000 g/mol leads to

 

TS = TS¥ -A

30,

000 g /mol

 

=

296 MPa -

7.

56 ´

10

6 MPa - g /mol

30,

000 g /mol

= 44 MPa

15.18 The tensile strength and number-average molecular weight for two polyethylene materials are as

follows:

Tensile Strength (MPa)

Number-Average Molecular Weight (g/mol)

85

12,700

150

28,500

Estimate the number-average molecular weight that is required to give

a tensile strength of 195 MPa.

Solution

This problem gives us the tensile strengths and associated number-average molecular weights for two

polyethylene materials and then asks that we estimate the

 

M n that is required for a tensile

strength of 195 MPa.

Equation 15.3 cites the dependence of the tensile strength on

 

M n. Thus, 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:

 

85 MPa = TS¥ -A

12,

700 g /mol

 

150 MPa = TS¥ -A

28,

500 g /mol

Thus, the values of the two constants are: TS

= 202 MPa and A

= 1.489 ´

10

6 MPa-g/mol. Solving for

 

M n in

Equation 15.3 and substituting TS

= 195 MPa as well as the above values for TS

and A leads to

 

M n =A

TS¥ - TS

 

=

1.

489 ´

10

6 MPa - g /mol

202 MPa -

195 MPa=

213,

000 g/mol

15.19

For each of the following pairs of polymers, do the following: (1) state whether or not it is possible

to decide whether one polymer has

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state why.

(a) Random acrylonitrile-butadiene copolymer

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acrylonitrile-

butadiene copolymer with 5% of possible sites crosslinked

(b)

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isotactic polypropylene with a deg

ree of polymerization of 3000

(c) Branched polyethylene with a number-

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isotactic poly(vinyl chloride) with a number-

average molecular weight of 200,000 g/mol

Solution

For each of four pairs of polyme

rs, we are to do the following: (1) determine whether or not it is possible to decide

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(a) No, it is not possible. The random acrylonitrile-butadiene copolymer will tend to a lower degree of

crystallinity than the alternating acrylonitrile-butadiene copolymer inasmuch as random copolymers don’t normally

crystallize. On this basis only, the alternating material would have a higher modulus inasmuch as tensile modulus

increases with degree of crystallinity. On the other hand, the random copolymer has a higher degree of crosslinking

(10% versus 5% for the alternating copolymer), and, on this basis only would have the higher tensile modulus—an

increase in crosslinking leads to an increase in E. Thus, this determination is not possible; with regard to degree of

crystallinity the alternating material has the higher E, whereas the random copolymer would have a higher E value on

the basis of degree of crosslinking.

(b) Yes, it is possible. The linear and isotactic polypropylene will have a greater tensile modulus. Linear

polymers are more likely to crystallize that branched ones. The likelihood of crystallization for both syndiotactic and

isotactic polypropylene is about the same, and, therefore, degree is crystallization is not a factor. Furthermore, tensile

modulus is relatively insensitive to degree of polymerization (i.e., molecular weight)—the fact that branched PP has

the higher molecular weight is not important.

(c) No, it is not possible. Linear polymers have higher degrees of crystallization (and higher tensile moduli)

than branched polymers—on this basis, the PVC material should have the higher value of E. On the other hand, PVC

has a more complex repeat unit structure than does polyethylene, which means that, on this basis, the PE would have

a higher degree of crystallinity and also a greater tensile modulus. Also, tensile modulus is relatively independent of

number-average molecular weight. Therefore, this determination is not possible since it is not possible to determine

which of the two materials has the greater degree of crystallinity.

15.20

For each of the following pairs of polymers, do the following: (1) state whether or not it is possible

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the higher tensile strength and then

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state why.

(a) Syndiotactic polystyrene having a number-

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polystyrene having a number-

average molecular weight of 500,000 g/mol

(b) Random acrylonitrile-

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acrylonitrile-

butadiene copolymer with 5% of possible sites crosslinked

(c) Network polyester; lightly branched polypropylene

Solution

For each of three pa

irs of polymers, we are to do the following: (1) determine whether or not it is possible to

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then state the reasons for this choice;

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

(a) Yes it is possible. The syndiotactic polystyrene has the higher tensile strength. Syndiotactic polymers

are more likely to crystallize than atactic ones; the greater the crystallinity, the higher the tensile strength.

Furthermore, the syndiotactic also has a higher molecular weight; increasing molecular weight also enhances the

strength.

(b) No it is not possible. The random acrylonitrile-butadiene copolymer has more crosslinking; increased

crosslinking leads to an increase in strength. However, the block copolymeric material will most likely have a higher

degree of crystallinity; and increasing crystallinity improves the strength.

(c) Yes it is possible. The network polyester will display a greater tensile strength. Relative chain motion is

much more restricted than for the lightly branched polypropylene since there are many more of the strong covalent

bonds for the network structure.

15.21 Would you expect the tensile strength of polychlorotrifluoroethylene to be greater than, the same as,

or less than that of a polytetrafluoroethylene specimen having the same molecular weight and degree of

crystallinity? Why?

Solution

The strength of a polychlorotrifluoroethylene having the repeat unit structure

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

at

traction, and, thus, a stronger polymer. Furthermore, poly(vinyl chloride) is stronger than polyethylene (Table 15.1)

for the same reason.

15.22 For each of the following pairs of polymers, plot and label schematic stress–strain curves on the

same graph [i.e., make separate plots for parts (a), (b), and (c)] .

(a) Isotactic and linear polypropylene having a weight-

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atactic and linear polypropylene having a weight-

average molecular weight of 100,000 g/mol

(b) Bra

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poly(vinyl chloride) having a degree of polymerization of 2000

(c) Poly(styrene-butadiene) random copolymer having a number-

average molecular weight of 100,000

g/mol

and 10% of the available sites crosslinked and tested at 20°C; poly(styrene-butadiene) random copolymer

having a number-

average molecular weight of 120,000 g/mol and 15% of the available sites crosslinked and

tested at -

85°C. Hint: poly(styrene-butadiene) copolymers may exhibit elastomeric behavior.

Solution

(a) Shown below are the stress-strain curves for the two polypropylene materials. These materials will

display the stress-strain behavior of a normal plastic, curve B

in Figure 15.1. However, the isotactic/linear will have a

higher degree of crystallinity (since isotactic are more likely to crystallize than atactic/linear), and therefore, will have

a higher tensile modulus and strength. Furthermore, the isotactic/linear also has a higher molecular weight which

also leads to an increase in strength.

(b) Shown below are the stress-strain curves for the two polyvinyl chloride materials. The branched PVC

will probably display the stress-strain behavior of a plastic, curve B

in Figure 15.1. However, the heavily crosslinked

PVC will undoubtedly have a higher tensile modulus, and, also a higher strength, and will most likely fail in a brittle

manner--as curve A

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(c) Shown below are the stress-strain curves for the two poly(styrene-butadiene) random copolymers. The

copolymer tested at 20°C will display elastomeric behavior (curve C

of Figure 15.1) inasmuch as it is a random

copolymer that is lightly crosslinked; furthermore, the temperature of testing is above its glass transition

temperature. On the other hand, since -

85°C is below the glass transition temperature of the poly(styrene-butadiene)

copolymer, the stress-strain behavior under these conditions is as curve A

of Figure 15.1.

15.23 List the two molecular characteristics that are essential for elastomers.

Solution

Two molecular characteristics essential for elastomers are: (1) they must be amorphous, having chains that

are extensively coiled

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15.24 Which of the following would you expect to be elastomers and which thermosetting polymers at

room temperature? Justify each choice.

(a) Epoxy having a network structure

(b) Lightly crosslinked poly(styrene-butadiene) random copolymer that has a glass-transition

temperature of -

50°C

(c) Lightly branched and semicrystalline polytetrafluoroethylene that has a glass-transition temperature

of -

100°C

(d) Heavily crosslinked poly(ethylene-propylene) random copolymer that has a glass-transition

temperature of 0°C

(e) Thermoplastic elastomer that has a glass-

transition temperature of 75°C

Solution

This question asks us to choose from a list of materials those that would be expected to be elastomers and

those that would be thermosetting polymers.

(a) Epoxy having a network structure will be a thermoset polymer since it has a network structure. It would

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

(b) A lightly crosslinked poly(styrene-butadiene) random copolymer that has a glass-transition temperature

of -

50°

C will be an elastomer since it 1) is a random copolymer, 2) is lightly crosslinked, and 3) is being used at a

temperature above its glass transition. All three of these criteria are requisites for an elastomer.

(c) Lightly branched and semicrystalline PTFE would be neither an elastomer nor a thermoset. It is not

crosslinked nor does it have a network structure.

(d) A heavily crosslinked poly(ethylene-propylene) random copolymer would be a thermoset inasmuch as it

is heavily crosslinked.

(e) A thermoplastic elastomer that has a glass-

transition temperature of 75°C is neither an elastomer nor a

thermoset. Since it is a thermoplastic it is not a thermoset. Furthermore, room temperature is below its glass-

transition temperature, and, therefore, it will not display elastomeric behavior.

15.25 Ten kilogram of pol

ybutadiene is vulcanized with 4.8 kg sulfur. What fraction of the possible

crosslink sites is bonded to sulfur crosslinks, assuming that, on the average, 4.5 sulfur atoms participate in each

crosslink?

Solution

This problem asks that we compute the fr

action of possible crosslink sites in 10 kg of polybutadiene when

4.8 kg of S is added, assuming that, on the average, 4.5 sulfur atoms participate in each crosslink bond. Given the

butadiene repeat unit in Table 14.5, we may calculate its molecular weight as follows:

A

(butadiene) = 4(AC

) + 6(AH

)

= (4)(12.01 g/mol) + 6(1.008 g/mol) = 54.09 g/mol

Which means that in 10 kg of butadiene there are

 

10,

000 g

54.

09 g/mol =

184.9 mol = nbuta.

For the vulcanization polybutadiene, 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 butadiene, by taking the mole ratio of sulfur to butadiene,

DQG WKHQ GLYLGLQJ WKLV UDWLR E\ DWRP V SHU FURVVOLQN WKLV \ LHOGV WKH IUDFWLRQ RI SRVVLEOH VLWHV WKDW DUH FURVVOLQNHG

Thus

 

nsulfur =

4800 g

32.

06 g /mol=

149.7 mol

And

 

fraction sites crosslinked =

nsulfur

nbuta

4.5 =

149.

7 mol

184.

9 mol

4.

5=

0.180

15.26 Compute the weight percent sulfur that must be added to completely crosslink an alternating

chloroprene-acrylonitrile copolymer, assuming that five sulfur atoms participate in each crosslink.

Solution

For an alternating chloroprene-acrylonitrile copolymer, we are asked to compute the weight percent sulfur

necessary for complete crosslinking, assuming that, on the average, five sulfur atoms participate in each crosslink.

The chloroprene and acryloni

trile repeat units are shown in Table 14.5, from which it may be noted that there are two

possible crosslink sites on each chloroprene repeat unit (one site at each of the two carbon atoms that are doubly

bonded), and no possible sites for acrylonitrile; also, since it is an alternating copolymer, the ratio of chloroprene to

acrylonitrile repeat units is 1:1. Thus, for each pair of combined chloroprene-acrylonitrile repeat units that crosslink,

ten sulfur atoms are required, or, for complete crosslinking, the sulfur-to-(chloroprene-

acrylonitrile) ratio is 5:1.

Now, let us consider as our basis, one mole of the combined chloroprene-acrylonitrile repeat unit. In order

for complete crosslinking, five moles of sulfur are required. Thus, 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, three

hydrogen atoms, and one nitrogen atom; the chloroprene repeat unit is composed of four carbons, five hydrogens,

and one chlorine. This gives a molecular weight for the combined repeat unit of

m(chloroprene-

acrylonitrile) = 3(AC

) + 3(AH

) + AN

+ 4(AC

) + 5(AH

) + ACl

= 7(12.01 g/mol) + 8(1.008 g/mol) + 14.007 g/mol + 35.45 g/mol = 141.59 g/mol

Or, in one mole of

this combined repeat unit, there are 141.59 g. Furthermore, for complete crosslinking 5.0 mol of

sulfur is required, which amounts to (5.0 mol)(32.06 g/mol) = 160.3 g. Thus, the concentration of S in weight percent

CS is just

 

CS =

160.

3 g

160.

3 g +

141.

59 g´

100 =

53.1 wt%

1

5.27

The vulcanization of polyisoprene is accomplished with sulfur atoms according to Equation 15.4. If

57 wt% sulfur is combined with polyisoprene, how many crosslinks will be associated with each isoprene repeat

unit if it is assumed that, on the average, six sulfur atoms participate in each crosslink?

Solution

This problem asks for us to determine how many crosslinks form per isoprene repeat unit when 57 wt%

sulfur is added. If we arbitrarily consider 100 g of the vulcanized material, 57 g will be

sulfur and 43 g will be

polyisoprene. Next, let us find how many moles of sulfur and isoprene correspond to these masses. The atomic

weight of sulfur is 32.06 g/mol, and thus,

 

# moles S =

57 g

32.

06 g /mol=

1.78 mol

Now, in each isoprene repeat unit there are five carbon atoms and eight hydrogen atoms. Thus, the molecular weight

of a mole of isoprene units is

(5)(12.01 g/mol) + (8)(1.008 g/mol) = 68.11 g/mol

Or, in 43 g of polyisoprene, the number of moles is equal to

 

# moles isoprene =

43 g

68.

11 g /mol=

0.631 mol

Therefore, the ratio of moles of S to the number of moles of polyisoprene is

 

1.

78 mol

0.

631 mol:

1 =

2.82 :

1

When all possible sites are crosslinked, the ratio of the number of moles of sulfur to the number of moles of isoprene

LV WKLV LV EHFDXVH WKHUH DUH WZR FURVVOLQN VLWHV per repeat unit and each crosslink is shared between repeat units

on adjacent chains, and there are 6 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. Or,

 

fraction of repeat unit sites crosslinked =

2.

82 /

1

6 /

1=

0.470

15.28 For the vulcanization of polyisoprene, compute the weight percent of sulfur that must be added to

HQVXUH WKDW RI SRVVLEOH VLWHV ZLOO EH FURVVOLQNHG DVVXP H WKDW RQ WKH DYHUDge, three sulfur atoms are associated

with each crosslink.

Solution

We are asked what weight percent of sulfur must be added to polyisoprene in order to ensure that 8% of

possible sites are crosslinked, assuming that, on the average, three sulfur atoms are associated with each crosslink.

Table 14.5 shows the repeat unit for cis-isoprene. 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 8% of the

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crosslink sites per repeat unit, but each crosslink is shared by two chains. Furthermore, on the average we assume

that

HDFK FURVVOLQN LV FRP SRVHG RI VXOIXU DWRP V WKXV WKHUH P XVW EH ´

8 or 24 sulfur atoms added for every 100

isoprene repeat units. In terms of moles, it is necessary to add 24 moles of sulfur to 100 moles of isoprene. The

atomic weight of sulfur is 3

2.06 g/mol, while the molecular weight of isoprene is

A

(isoprene) = 5(AC

) + 8(AH

)

= (5)(12.01 g/mol) + (8)(1.008 g/mol) = 68.11 g/mol

The mass of sulfur added (mS) is

mS

= (24 mol)(32.06 g/mol) = 769.4 g

While for isoprene

mip

= (100 mol)(68.11 g/mol

) = 6811 g

Or, the concentration of sulfur in weight percent (Equation 4.3) is just

 

CS =mS

mS + mip

´

100 =

769 .

4 g

769 .

4 g +

6811 g´

100 =

10.1 wt%

15.29

Demonstrate, in a manner similar to Equation 15.4, how vulcanization may occur in a butadiene

rubber.

Solution

The reaction by which a butadiene rubber may become vulcanized is as follows:

Crystallization

15.30

Determine values for the constants n and k (Equation 10.17) for the crystallization of

polypropylene (Figure 15.17) at 160°C.

Solution

In this problem we are asked to determine the values of the constants n and k

(Equation 10.17) for the

crystallization of polypropylene at 160°

C (Figure 15.17). One way to solve this problem is to take two values of

percent recrystallization (which is just 100y

, Equation 10.17) and their corresponding time values, 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 10.17. First of all, we rearrange as follows:

 

1 - y = exp - ktn( )

Now taking natural logarithms

 

ln (

1 - y) = - ktn

Or

 

- ln (

1 - y) = ktn

which may also be expressed as

 

ln

1

1 - y

æ

è ç

ö

ø ÷ = ktn

Now taking natural logarithms again, leads to

 

ln ln

1

1 - y

æ

è ç

ö

ø ÷

é

ë ê

ù

û ú = ln k + n ln t

which is the form of the equation that we will now use. From the 160°

C curve of Figure 15.17, let us arbitrarily choose

two percent crystallized values of 20% and 80% (i.e., y

1

= 0.20 and y

2

= 0.80). The corresponding time values are t

1 =

3400 min and t

2

= 6700 min (realizing that the time axis is scaled logarithmically). Thus, our two simultaneous

equations become

 

ln ln

1

1 -

0.

20

æ

è ç

ö

ø ÷

é

ë ê

ù

û ú = ln k + n ln (

3400 )

 

ln ln

1

1 -

0.

80

æ

è ç

ö

ø ÷

é

ë ê

ù

û ú = ln k + n ln (

6700 )

from which we obtain the values n

= 2.91 and k

= 1.182 ´

10-

11.

Melting and Glass Transition Temperatures

15.31 Name the following polymer(s) that would be suitable for the fabrication of cups to contain hot

coffee: polyethylene, polypropylene, poly(vinyl chloride), PET polyester, and polycarbonate. Why?

Solution

This question asks us to name which, of several polymers, 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 hav

e glass transition temperatures of 100°

C or above (Table 15.2), and would be suitable for this

application.

15.32

Of those polymers listed in Table 15.2, which polymer(s) would be best suited for use as ice cube

trays? Why?

Solution

In order for a polymer to be suited for use as an ice cube tray it must have a glass-transition temperature

below 0°

C. Of those polymers listed in Table 15.2 only low-density and high-density polyethylene, PTFE, and

polypropylene satisfy this criterion.

Factors That Influence Melting and Glass Transition Temperatures

15.33 For each of the following pairs of polymers, plot and label schematic specific volume versus

temperature curves on the same graph [i.e., make separate plots for parts (a), (b), and (c)] .

(a) Spheru

litic polypropylene, of 25% crystallinity, and having a weight-average molecular weight of

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100,000 g/mol

(b) Graft poly(styrene-butadiene) copolym

HU ZLWK RI DYDLODEOH VLWHV FURVVOLQNHG UDQGRP

poly(styrene-

butadiene) copolymer with 15% of available sites crosslinked

(c)

Polyethylene having a density of 0.985 g/cm

3

DQG D GHJ UHH RI SRO\ P HUL]DWLRQ RI SRO\ HWK\ OHQH

having a density of 0.915 g/cm

3

and a degree of polymerization of 2000

Solution

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

materials. Since both polymers are 25% crystalline, they will exhibit behavior similar to curve B

in Figure 15.18.

However, polystyrene will have higher melting and glass transition temperatures due to the bulkier side group in its

repeat unit structure, and since it has a higher weight-average molecular weight.

(b) Shown below are the specific volume-versus-temperature curves for the graft and random poly(styrene-

butadiene) copolymers. Since these materials are graft and random copolymers, both will be highly noncrystalline,

and, thus, will display the behavior similar to curve A

in Figure 15.1

8. However, since the random has the greater

degree of crosslinking, it will also have the higher glass transition temperature.

(c) Shown below are the specific volume-versus-temperature curves for the two polyethylene materials. The

polyethylene ha

ving a density of 0.985 g/cm

3 will be highly crystalline, and, thus, will exhibit a behavior similar to

curve C

in Figure 15.18. On the other hand, the other material, of lower density will have some branching and also be

semicrystalline; thus, its behavior will be similar to curve B

of Figure 15.18. In addition, the melting temperature of

the higher density material will be greater since it has less branching and a higher degree of polymerization.

15.34 For each of the following pairs of polymers,

do the following: (1) state whether or not it is possible

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has the higher melting temperature and then cite reason(s) for your choice; and (

3) if it is not possible to decide,

then state why.

(a)

Isotactic polystyrene that has a density of 1.12 g/cm

3 and a weight-average molecular weight of

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3 and a weight-average molecular weight of

125,000 g/mol

(b)

/ LQHDU SRO\ HWK\ OHQH WKDW KDV D GHJ UHH RI SRO\ P HUL]DWLRQ RI OLQHDU DQG LVRWDFWLF SRO\ SURS\ OHQH

that has a degree of polymerization of 6,500

(c)

Branched and isotactic polystyrene that has a degree of polymerization of 4

OLQHDU DQG LVRWDFWLF

polypropylene that has a degree of polymerization of 7,500

Solution

(a) Yes, it is possible to determine which of the two polystyrenes has the higher Tm . The isotactic

polystyrene will have the higher melting temperature because it has a higher density (i.e., less branching) and also

the greater weight-average molecular weight.

(b) Yes, it is possible to determine which polymer has the higher melting temperature. The polypropylene

will have the higher Tm because it has a bulky phenyl side group in its repeat unit structure, which is absent in the

polyethylene. Furthermore, the polypropylene has a higher degree of polymerization.

(c) No, it is not possible to determine which of the two polymers has the higher melting temperature. The

polystyrene has a bulkier side group than the polypropylene; on the basis of this effect alone, the polystyrene

should have the greater Tm . However, the polystyrene has more branching and a lower degree of polymerization;

both of these factors lead to a lowering of the melting temperature.

15.35 Make a schematic plot showing how the modulus of elasticity of an amorphous polymer depends on

the glass transition temperature. Assume that molecular weight is held constant.

Solution

For an amorphous polymer, 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 temperature. Thus, the

modulus-glass transition temperature behavior would appear as

Elastomers

Fibers

Miscellaneous Applications

15.36 Briefly explain the difference in molecular chemistry between silicone polymers and other polymeric

materials.

Solution

The backbone chain of most polymers consists of carbon atoms that are linked together. For the silicone

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

15.37 List two important characteristics for polymers that are to be used in fiber applications.

Solution

Two important characteristics for polymers that are to be used in fiber applications are: (1) they must have

high molecular weights, and (2) they must have chain configurations/structures that will allow for high degrees of

crystallinity.

15.38 Cite five important characteristics for polymers that are to be used in thin-film applications.

Solution

Five important characteristics for polymers that are to be used in thin-

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KLJ K IOH[LELOLW\ KLJ K W

HQVLOH DQG WHDU VWUHQJ WKV UHVLVWDQFH WR P RLVWXUH FKHP LFDO DWWDFN DQG ORZ J DV

permeability.

Polymerization

15.39 Cite the primary differences between addition and condensation polymerization techniques.

Solution

For addition polymerization, 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, producing the repeating unit. There is often a low molecular weight

by-product for condensation polymerization; such is not found for addition polymerization.

15.40 (

a) How much ethylene glycol must be added to 47.3 kg of terephthalic acid to produce a linear

chain struc

ture of poly(ethylene terephthalate) according to Equation 15.9?

(b) What is the mass of the resulting polymer?

Solution

(a) This problem asks that we determine how much ethylene glycol must be added to 47.3 kg of terephthalic

acid to produce a linea

r chain structure of poly(ethylene terephthalate) according to Equation 15.9. 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

(12.01 g/mol) +

6

(1.008 g/mol) +

2

(16.00 g/mol) =

62.07 g/mol

 

MW(terephthalic acid) =

8(AC) +

6(AH ) +

4(AO)

 

=

8

(12.01 g/mol) +

6

(1.008 g/mol) +

4

(16.00 g/mol) =

166.13 g/mol

The 47.3 kg mass of terephthalic acid equals 47,300 g or

 

47,

300 g

166.

13 g /mol

= 284.72 mol. Since, according to Equation

15.9, each mole of terephthalic acid used requires one mole of ethylen

e glycol, which is equivalent to (284.72

mol)(62.07 g/mol) = 17,673 g = 17.673 kg.

(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, this corresponds to 284.72 moles or

(284.72 mol)(18.02 g/mol) = 5130 g or 5.130 kg since

the molecular weight of water is 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)] =

47.300 kg +

17.673 kg -

5.130 kg =

59.843 kg

15.41

Nylon 6,6 may be formed by means of a condensation polymerization reaction in which

hexamethylene diamine [NH

2—(CH

2)

6—NH

2] and adipic acid react with one another with the formation of water

as a byproduct. What m

asses of hexamethylene diamine and adipic acid are necessary to yield 37.5 kg of

completely linear nylon 6,6? (Note: the chemical equation for this reaction is the answer to Concept Check

15.12.)

Solution

This problem asks for us to calculate the masses of hexamethylene diamine and adipic acid necessary to

yield 37.5 kg of completely linear nylon 6,6. The chemical equation for this reaction is the answer to Concept Check

15.12, which is as follows:

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

 

MW(adipic) =

6(AC) +

10(AH ) +

4(AO)

 

=

6

(12.01 g/mol) +

10

(1.008 g/mol) +

4

(16.00 g/mol) =

146.14 g/mol

 

MW(hexamethylene) =

6(AC) +

16(AH ) +

2(AN )

 

=

6

(12.01 g/mol) +

16

(1.008 g/mol) +

2

(14.01 g/mol) =

116.21 g/mol

 

MW(nylon) =

12(AC) +

22(AH ) +

2(AN ) +

2(AO)

 

=

12

(12.01 g/mol) +

22

(1.008 g/mol) +

2

(14.01 g/mol) +

2

(16.00 g/mol)

= 226.32 g/mol

The mass of 37.5 kg of nylon 6,6 equals 37,500 g or

 

m(nylon) =

37,

500 g

226.

32 g /mol=

165.7 mol

Since, 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

corresponding to 165.7 moles of adipic acid and hexamethylene diamine are as follows:

 

m(adipic) =

(165.7 mol)(146.14 g/mol) =

24,

215 g =

24.215 kg

 

m(hexamethylene) =

(165.7 mol)(116.21 g/mol) =

19,

256 g =

19.256 kg

Polymer Additives

15.42 What is the distinction between dye and pigment colorants?

Solution

The distinction between dye and pigment colorants is that a dye dissolves within and becomes a part of the

polymer structure, whereas a pigment does not dissolve, but remains as a separate phase.

Forming Techniques for Plastics

15.43 Cite four factors that determine what fabrication technique is used to form polymeric materials.

Solution

Four factors that determine what fabrication technique is used to form polymeric materials are: (1) whether

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VWDELOLW\ DQG WKH J HRP HWU\ and size of the finished product.

15.44 Contrast compression, injection, and transfer molding techniques that are used to form plastic

materials.

Solution

This question requests that we compare polymer molding techniques. 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. And, 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.

Fabrication of Fibers and Films

15.45 Why must fiber materials that are melt spun and then drawn be thermoplastic? Cite two reasons.

Solution

Fiber materials that are melt spun must be thermoplastic because: (1) In order to be melt spun, they must be

capable of forming a viscous liquid when heated, which is not p

ossible for thermosets. (2) During drawing,

mechanical elongation must be possible; inasmuch as thermosetting materials are, in general, hard and relatively

brittle, they are not easily elongated.

15.46 Which of the following polyethylene thin films w

ould have the better mechanical characteristics: (1)

formed by blowing, or (2) formed by extrusion and then rolled? Why?

Solution

Of the two polymers cited, 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.

DESIGN QUESTIONS

15.D1 (a) List several advantages and disadvantages of using transparent polymeric materials for

eyeglass lenses.

(b) Cite four properties (in addition to being transparent) that are important for this application.

(c) Note three polymers that may be candidates for eyeglass lenses, and then tabulate values of the

properties noted in part (b) for these three materials.

Solution

(a) Several advantages of using transparent polymeric materials for eyeglass lenses are: they have

relatively low densities, and, therefore, are light in weight; 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, and, therefore,

are not as precise optically.

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

in order to resist scratching; they must be impact resistant; 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 1

4, likely lens candidates are polystyrene,

poly(methyl methacrylate), and polycarbonate; these three materials are not easily crystallized, and, therefore, are

normally transparent. Upon consultation of their fracture toughnesses (Table B.5 in Appendix 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, but not as hard as CR-

39. Furthermore, PC comes in

both normal and high refractive-index grades.

15.D2 Write an essay on polymeric materials that are used in the packaging of food products and drinks.

Include a list of the general requisite characteristics of materials that are used for these applications. Now cite a

specific material that is utilized for each of three different container types and the rationale for each choice.

Solution

There are three primary requirements for polymeric materials that are utilized in the packaging of food

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protection--

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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), PVDC, polypropylene, and

LDPE are impervious to water vapor.

Most common polymers are relatively nonreactive with food products, and are considered safe; 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.

15.D3 Write an essay on the replacement of metallic automobile components by polymers and composite

materials. Address the following issu

es: (1) Which automotive components (e.g., crankshaft) now use polymers

and/or composites? (2) Specifically what materials (e.g., high-

density polyethylene) are now being used? (3) What

are the reasons for these replacements?

Solution

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

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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, water outlets, pulleys, and brake pistons, are made from phenolic thermoset

composites that are reinforced with glass fibers.

Air intake manifolds are made of a glass-

reinforced nylon 6,6.