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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 (293 K)], 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

3

(stress) 30MPa 0MPa3.3GPa

(strain) 9 10 0E −

∆ −= = =

∆ × −

The value range cited in Table 15.1 is 2.24 to 3.24 GPa. Thus, the plotted value is a little on the high

side.

The tensile strength corresponds to the stress at which the curve ends, which is 52 MPa. This

value lies within the range cited in Table 15.1—48.3 to 72.4 MPa.

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

2 1

2 1

E

σ σσ∈ ∈ ∈

−∆= =

∆ −

Since all the stress–strain curves pass through the origin, we make take σ1 = 0 and ∈1 = 0.

Determinations of σ2 and ∈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 σ2 = 6.0 MPa with its corresponding ∈2 = 0.0152.

Therefore,

2 1

2 1

6.0 MPa 0 MPa 395 MPa 0.395 GPa 0.0152 0

Eσ σ∈ ∈

− −= = = =

− −

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 σ2 = 24.3 MPa with its corresponding ∈2 = 0.0085. Therefore,

2 1

2 1

24.3 MPa 0 MPa 2860 MPa 2.86 GPa 0.0085 0

Eσ σ∈ ∈

− −= = = =

− −

The elastic modulus range for nylon 6,6 given in Table 15.1 is 1.58 GPa to 3.80 GPa;

therefore, the value for the VMSE nylon lies within this range.

(c) For phenol-formaldehyde (bakelite), we selected σ2 = 33.0 MPa with its corresponding

∈2 = 0.0068. Therefore,

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2 1

2 1

33.0 MPa 0 MPa 4850 MPa 4.85 GPa 0.0068 0

Eσ σ∈ ∈

− −= = = =

− −

The elastic modulus range for the phenol-formaldehyde given in Table 15.1 is 2.76 GPa to

4.83 GPa; therefore, this value for the VMSE phenol-formaldehyde lies just above the maximum value

for this range.

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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 this case about 4%); this gives a value of about 40%EL.

For nylon 6,6, the range of values of yield strength presented in Table 15.1 is 44.8 MPa to

82.8 MPa; therefore, the value for the VMSE nylon lies slightly above the upper value. The ductility

range for nylon 6,6 is 15%EL to 300%EL; therefore, our value (40%EL) lies within this range.

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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 this case about 1%); this gives a value of about 0.3%EL.

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-formaldehyde is 1.5%EL to 2.0%EL; therefore, our value (0.3%EL) lies

below the lower limit.

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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.

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

with time according to

( ) (0) exp t

tσ στ

⎛ ⎞= −⎜ ⎟⎝ ⎠ (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 whose stress

relaxation obeys Equation 15.10 was suddenly pulled in tension to a measured strain of 0.6;

the stress necessary to maintain this constant strain was measured as a function of time.

Determine Er(10) for this material if the initial stress level was 2.8 MPa, which dropped to

1.75 MPa 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.,

( ) (0) exp ttσ στ

⎛ ⎞= −⎜ ⎟⎝ ⎠

We want to determine σ(10), but it is first necessary to compute τ from the data provided in the

problem statement. Thus, solving for τ from the above expression,

60127.7s

( ) 1.75MPaln ln(0) 2.8MPa

tt

τσσ

− −= = =

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

Therefore,

10s(2.8MPa)exp 2.58MPa

127.7s⎛ ⎞

− =⎜ ⎟⎝ ⎠

Now, using Equation 15.1

0

(10) 2.58MPa(10) 4.3MPa0.6rE σ

∈= = =

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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.

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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 125°C

(b) Crosslinked at 160°C

(c) Crystalline at 230°C

(d) Crosslinked at 50°C

Solution

(a) Amorphous polystyrene at 125°C behaves as a rubbery material (Figure 15.8, curve C); therefore,

the strain–time behavior would be as Figure 15.5c.

(b) Crosslinked polystyrene at 160°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 50°C behaves in a glassy manner (Figure 15.8, curve B); therefore, the

strain–time behavior will be as Figure 15.5b.

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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 stress measured after 10 s and strain (Equation 15.1); creep modulus is the ratio

of stress and strain taken at a specific time (Equation 15.2).

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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.

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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

specimen thickness, and (5) modifications of the polymer structure.

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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 106 cycles for 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 and 193 MPa, respectively.

(b) At 106 cycles, the fatigue strengths of PET and red brass are 15 MPa and 115 MPa, respectively.

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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.

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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.

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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.

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15.16 Normal butane and isobutane have boiling temperatures of –0.5 and –12.3°C (272.5 and

260.7 K) respectively. Briefly explain this behavior 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.

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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)

108 40,000

175 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

nM = 30,000 g/mol. Equation 15.3 cites the dependence of the tensile strength on nM . 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:

108MPa40,000g/mol

ATS∞= −

175MPa60,000g/mol

ATS∞= −

Thus, the values of the two constants are: TS∞ = 309 MPa and A = 8.04 × 106 MPa-g/mol. Substituting

these values into Equation 15.3 for nM = 30,000 g/mol leads to

TS = TS∞ −

A30,000 g /mol

68.04 10 MPa-g/mol309 MPa

30,000g/mol×

= −

= 41 MPa

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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)

90 12,700

160 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 nM that is required for a tensile

strength of 195 MPa. Equation 15.3 cites the dependence of the tensile strength on nM . 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:

90 MPa12,700g/mol

ATS∞= −

160MPa28,500g/mol

ATS∞= −

Thus, the values of the two constants are: TS∞ = 216 MPa and A = 1.604 × 106 MPa-g/mol. Solving for

nM in Equation 15.3 and substituting TS = 195 MPa as well as the above values for TS∞ and A leads to

=nAM

TS TS∞ −

61.604 10 MPa-g/mol 76,380g/mol

216 MPa 195MPa×

= =−

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15.19 For each of the following pairs of polymers, do the following: (1) state whether it is possible

to decide whether one polymer has a higher tensile modulus than the other; (2) if this is

possible, note which has the higher tensile modulus and then cite the reason(s) for your

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

(a) Random acrylonitrile-butadiene copolymer with 10% of possible sites crosslinked;

alternating acrylonitrile-butadiene copolymer with 5% of possible sites crosslinked

(b) Branched and syndiotactic polypropylene with a degree of polymerization of 5000; linear

and isotactic polypropylene with a degree of polymerization of 3000

(c) Branched polyethylene with a number-average molecular weight of 250,000 g/mol; linear

and isotactic poly(vinyl chloride) with a number-average molecular weight of 200,000 g/mol

Solution

For each of four pairs of polymers, we are to do the following: (1) determine whether or not it is

possible to decide which has the higher tensile modulus; (2) if so, note which has the higher tensile

modulus and then state the reasons for this choice; and (3) if it is not possible to decide, then state why.

(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.

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15.20 For each of the following pairs of polymers, do the following: (1) state whether it is possible

to decide whether one polymer has a higher tensile strength than the other; (2) if this is

possible, note which has the higher tensile strength and then cite the reason(s) for your choice;

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

(a) Syndiotactic polystyrene having a number-average molecular weight of 600,000 g/mol;

atactic polystyrene having a number-average molecular weight of 500,000 g/mol

(b) Random acrylonitrile-butadiene copolymer with 10% of possible sites crosslinked; block

acrylonitrile-butadiene copolymer with 5% of possible sites crosslinked

(c) Network polyester; lightly branched polypropylene

Solution

For each of three pairs of polymers, we are to do the following: (1) determine whether or not

it is possible to decide which has the higher tensile strength; (2) if it is possible, 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, 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.

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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 attraction, and, thus, a stronger polymer. Furthermore, poly(vinyl chloride)

is stronger than polyethylene (Table 15.1) for the same reason.

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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-average molecular weight of 120,000

g/mol; atactic and linear polypropylene having a weight-average molecular weight of 100,000

g/mol

(b) Branched poly(vinyl chloride) having a degree of polymerization of 3000; heavily

crosslinked poly(vinyl chloride) having a degree of polymerization of 3000.

(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, Figure 15.1; these are the typical characteristics of a heavily crosslinked polymer.

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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.

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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 and kinked in the unstressed state; and (2) there must be some

crosslinking.

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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.

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15.25 Ten kilogram of polybutadiene is vulcanized with 5.0 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 fraction of possible crosslink sites in 10 kg of

polybutadiene when 5.0 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 g54.09 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, and then dividing this ratio by 4.5

atoms per crosslink; this yields the fraction of possible sites that are crosslinked. Thus

sulfur5000g 155.9 mol

32.06g/moln = =

And

sulfur

buta

155.9 mol184.9 molfraction sites crosslinked 0.187

4.5 4.5

nn

= = =

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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 acrylonitrile 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%

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15.27 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 g32.06 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 g68.11 g /mol

= 0.631 mol

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

1.78 mol0.631 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 is 6:1; 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 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 /16 /1

= 0.470

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

added to ensure that 10% of possible sites will be crosslinked; assume that, on the average,

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 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 sulfur atoms; thus, there must be 3 × 10 or 30 sulfur atoms added for every

100 isoprene repeat units. In terms of moles, it is necessary to add 30 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) + 10(AH)

= (5)(12.01 g/mol) + (10)(1.008 g/mol) = 70.13 g/mol

The mass of sulfur added (mS) is

mS = (30 mol)(32.06 g/mol) = 961.8 g

While for isoprene

mip = (100 mol)(70.13 g/mol) = 7013 g

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

SS

S ip

961.8g100 100 12.1wt%

961.8g 7013gm

Cm m

= × = × =+ +

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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:

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Crystallization

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

polypropylene (Figure 15.17) at 140°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 exp ( )ny kt− = −

Now taking natural logarithms

ln (1 − y) = − kt n

Or

− ln (1 − y) = kt n

which may also be expressed as

ln 1

1 − y

⎛

⎝ ⎜

⎞

⎠ ⎟ = kt n

Now taking natural logarithms again, leads to

ln ln 1

1 − y

⎛

⎝ ⎜

⎞

⎠ ⎟

⎡

⎣ ⎢

⎤

⎦ ⎥ = ln k + n ln t

which is the form of the equation that we will now use. From the 140°C curve of Figure 15.17, let us

arbitrarily choose two percent crystallized values of 20% and 80% (i.e., y1 = 0.20 and y2 = 0.80). The

corresponding time values are t1 = 30 min and t2 = 70 min (realizing that the time axis is scaled

logarithmically). Thus, our two simultaneous equations become

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1ln ln ln ln 301 0.20

k n⎡ ⎤⎛ ⎞ = +⎜ ⎟⎢ ⎥⎝ ⎠−⎣ ⎦

1ln ln ln ln 70

1 0.80k n⎡ ⎤⎛ ⎞ = +⎜ ⎟⎢ ⎥⎝ ⎠−⎣ ⎦

from which we obtain the values n = 2.22 and k = 4.725 × 10−5.

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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 (373 K).

Of the polymers listed, only polystyrene and polycarbonate have glass transition temperatures of

100°C or above (Table 15.2), and would be suitable for this application.

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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.

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

weight of 75,000 g/mol; spherulitic polystyrene, of 25% crystallinity, and having a weight-

average molecular weight of 100,000 g/mol

(b) Graft poly(styrene-butadiene) copolymer with 10% of available sites crosslinked; random

poly(styrene-butadiene) copolymer with 15% of available sites crosslinked

(c) Polyethylene having a density of 0.980 g/cm3 and a degree of polymerization of 2500;

polyethylene having a density of 0.920 g/cm3 and a degree of polymerization of 2200.

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 15.18.

However, since the random has the greater degree of crosslinking, it will also have the higher glass

transition temperature.

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

materials. The polyethylene having a density of 0.980 g/cm3 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.

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15.34 For each of the following pairs of polymers, do the following: (1) state whether it is possible

to determine whether one polymer has a higher melting temperature than the other; (2) if it is

possible, note which 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 1.15 g/cm3 and a weight-average molecular

weight of 150,000 g/mol; syndiotactic polystyrene that has a density of 1.12 g/cm3 and a

weight-average molecular weight of 125,000 g/mol

(b) Linear polyethylene that has a degree of polymerization of 6000; linear and isotactic

polypropylene that has a degree of polymerization of 7000.

(c) Branched and isotactic polystyrene that has a degree of polymerization of 5000; linear and

isotactic polypropylene that has a degree of polymerization of 8000.

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.

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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

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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.

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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.

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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-film applications are:

(1) low density; (2) high flexibility; (3) high tensile and tear strengths; (4) resistance to

moisture/chemical attack; and (5) low gas permeability.

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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.

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15.40 (a) How much ethylene glycol must be added to 47.5 kg of terephthalic acid to produce a

linear chain structure 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 47.5 kg of

terephthalic acid to produce a linear 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 47.5 kg mass of terephthalic acid equals 47,500 g or 47,500g166.13g/mol

= 285.92 mol. Since,

according to Equation 15.9, each mole of terephthalic acid used requires one mole of ethylene glycol,

which is equivalent to (285.92 mol)(62.07 g/mol) = 17,747 g = 17.747 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 285.92 moles or (285.92 mol)(18.02 g/mol) =

5152 g or 5.152 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)] 47.500kg 17.747 kg 5.152kg 60.095kg= + − =

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

hexamethylene diamine [NH2—(CH2)6—NH2] and adipic acid react with one another with the

formation of water as a byproduct. What masses 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.

C H O(adipic) 6( ) 10( ) 4( )MW A A A= + +

= 6(12.01 g/mol) + 10(1.008 g/mol) + 4(16.00 g/mol) = 146.14 g/mol

C H N(hexamethylene) 6( ) 16( ) 2( )MW A A A= + +

= 6(12.01 g/mol) + 16(1.008 g/mol) + 2(14.01 g/mol) = 116.21 g/mol

C H N O(nylon) 12( ) 22( ) 2( ) 2( )MW A A A A= + + +

= 12(12.01 g/mol) + 22(1.008 g/mol) + 2(14.01 g/mol) + 2(16.00 g/mol)

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= 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

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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.

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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 the polymer is thermoplastic or thermosetting; (2) if thermoplastic, the softening

temperature; (3) atmospheric stability; and (4) the geometry and size of the finished product.

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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 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. For

transfer molding, the solid materials (normally thermosetting in nature) are first melted in the transfer

chamber prior to being forced into the die.

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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 possible 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.

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15.46 Which of the following polyethylene thin films would 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.

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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 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 14, 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.

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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 products and drinks; these are: (1) sufficient strength, to include tensile, tear, and

impact strengths; (2) barrier protection—that is, being resistant to permeation by oxygen, water vapor,

and carbon dioxide; 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), 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.

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

composite materials. Address the following issues: (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 efficiencies; (2) may be produced at lower costs but with comparable

mechanical characteristics; (3) are in many environments more corrosion resistant; (4) reduce noise,

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, 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.