CHAPTER 15
CHARACTERISTICS, APPLICATIONS, AND PROCESSING OF POLYMERS
Page 1/21
CLOSED-BOOK PRACTICE
CHAPTER 15: CHARACTERISTICS, APPLICATIONS, AND PROCESSING OF POLYMERS CONCEPT CHECK
is not necessary to specify 1. When citing the ductility as percent elongation for semicrystalline polymers, it is the specimen gage length, as in the case with metals. Why so? Ans:
The reason that it is not necessary to specify specimen gage length when citing percent elongation for semicrystalline polymers is because, for semicrystalline polymers that experience necking, the neck normally propagates along the entire gage length prior to fracture (thus, called a propagating or running neck); thus, there is no localized (snapping) necking as with metals and the magnitude of the percent elongation is independent of gage length. 2. An amorphous polystyrene, which is is deformed at 120C, will exhibit which of the behaviors shown in Fig. 15.5?
); therefore, the Ans: Amorphous polystyrene at 120C behaves as a rubbery material (Fig. 15.8, curve C ); strain-time behavior would be as Fig. 15.5c 15.5c. 3. For the following pair pair of polymers, do the following: (1) state whether it is possible to decide if one p olymer 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 explain why not. Syndiotactic polystyrene having a number-average molecular weight of 400,000 g/mol Isotactic polystyrene having a number-average molecular weight of 650,000 g/mol Ans: No, it is not possible. possible. Both syndiotactic and isotactic polystyrene have a tendency to crystallize, and, therefore, we assume that they have approximately the same crystallinity. Furthermore, since tensile modulus is virtually independent of molecular weight, we would expect both materials to have approximately the same modulus.
pair of polymers, do the following: 4. For the following pair (1) state whether it is possible to decide if one p olymer 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 explain why not. Syndiotactic polystyrene having a number-average molecular weight of 600,000 g/mol Isotactic polystyrene having a number-average molecular weight of 500,000 g/mol Ans:
possible. The syndiotactic polystyrene has the higher tensile strength. Both syndiotactic and Yes, it is possible. isotactic polymers tend to crystallize. Therefore, we assume that both materials have approximately the same crystallinity. However, tensile modulus increases with increasing molecular weight, and the syndiotactic PS has the higher molecular weight (600,000 g/mol versus 500,000 g/mol for the isotactic material).
ME 46100: ENGINEERING MATERIALS
CLOSED-BOOK PRACTICE
CHAPTER 15
CHARACTERISTICS, APPLICATIONS, AND PROCESSING OF POLYMERS
Page 2/21
5. For the following following pair of polymers, polymers, plot and label schematic stress-strain stress-strain curves on the same graph: graph: 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 85C Hint : Synthetic poly(styrene-butadiene) copolymers are commonly called styrne-butadiene rubbers (SBR). The first version, called Neolite, was developed by Goodyear for used as car tires. Their glass transition temperatures (T (T g ) are 65C for S-SBR (solution) and 50C for E-SBR (emulsion), respectively. Ans:
Shown below are the stress-strain curves for the two poly(styrene-butadiene) random copolymers. The copolymer tested at 20C will display elastomeric behavior (curve C of Fig. 15.1, i.e., elastomeric) 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 85C is below the glass transition temperature of the other poly(styrene-butadiene) copolymer, the stress-strain behavior under these conditions is as curve A curve A of of Fig. 15.1 (i.e., b rittle).
6. In terms of molecular structure, explain why phenol-formaldehyde (Bakelite) will not be an elastomer. The molecular structure for phenol-formaldehyde is presented in Table 14.3 . Ans:
The molecules in an elastomer must be two-dimensional chains that are lightly crosslinked and capable of being twisted and kinked in the unstressed state. Phenol-formaldehyde has a rigid three-dimensional (network or heavily crosslinked) structure, consisting of trifunctional repeat units, which does not meet these criteria for chain conformation and flexibility.
ME 46100: ENGINEERING MATERIALS
CLOSED-BOOK PRACTICE
CHAPTER 15
CHARACTERISTICS, APPLICATIONS, AND PROCESSING OF POLYMERS
Page 3/21
7. For the following pair of polymers, plot and label schematic specific volume-temperature curves on the same graph: 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 Ans:
Shown below are the specific volume-temperature curves for the polypropylene and polystyrene materials. Since both polymers are 25% crystalline, they will exhibit behavior similar to curve B in Fig. 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 h as a higher weight-average molecular weight.
8. For the following pair of polymers, do the following: (1) state whether 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 explain why not. 3 Isotactic polystyrene that has a density of 1.12 g/cm and a weight-average molecular weight of 150,000 g/mol 3 Syndiotactic polystyrene that has a density of 1.10 g/cm and a weight-average molecular weight of 125,000 g/mol Ans: Yes, it is possible to determine which of the two polystyrenes has the higher T m. 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. 9. During the winter months, the temperature in some parts of Alaska may go as low as 55C. Of the elastomers: natural isoprene, styrene-butadiene, acrylonitrile-butadiene, chloroprene, and polysiloxane, which would be suitable for automobile tires unde r these conditions? Why? Ans:
From Table 15.4, only natural polyisoprene, poly(styrene-butadiene), and polysiloxane have useful temperature ranges that extend to below 55C. At temperatures below the lower useful temperature range limit, the other elastomers listed in this table become brittle, and, therefore, are not suitable for automobile tires. ME 46100: ENGINEERING MATERIALS
CLOSED-BOOK PRACTICE
CHAPTER 15
CHARACTERISTICS, APPLICATIONS, AND PROCESSING OF POLYMERS
Page 4/21
10. Silicone polymers may be prepared to exist as liquids at room temperature. Cite differences in molecular structure between them and the silicone elastomers. Hint : You may want to consult Sections 14.5 and 15.9. Ans:
The liquid silicones will have low molecular weights and very little crosslinking, whereas the molecular weights for the elastomers will be much higher with some crosslinking. 11. State whether the molecular weight of a polymer that is synthesized by addition polymerization is relatively high, medium, or relatively low for the following situations. Ans:
(a) Rapid initiation, slow propagation, and rapid termination The molecular weight will be relatively low. (b) Slow initiation, rapid propagation, and slow termination The molecular weight will be relatively high. (c) Rapid initiation, rapid propagation, and slow termination A medium molecular weight will be achieved. (d ) Slow initiation, slow propagation, and rapid termination The molecular weight will be low or medium. 12. Nylon 6,6 may be formed by means of a condensation polymerization reaction in which hexamethylene diamine [NH -(CH ) -NH ] and adipic 2
2 6
2
acid react with one another with the formation of water as a by-product. Write out this reaction in the manner of Eq. 15.9. Note: The structure for adipic acid is shown at right. Ans:
The following represents the reaction between hexamethylene diamine and adipic acid to produce nylon 6,6 with water as a byproduct.
13. (a) Why must the vapor pressure of a plasticizer be relatively low? (b) How will the crystallinity of a polymer be affected by the addition of a plasticizer? Why? (c) How does the addition of a plasticizer influence the tensile strength of a polymer? Why? Ans:
(a) If the vapor pressure of a plasticizer is not relatively low, the plasticizer may vaporize, which will result in an embrittlement of the polymer. (b) The crystallinity of a polymer, to which a plasticizer has been added, will be diminished, inasmuch as the plasticizer molecules fit in between the pol ymer molecules, causing more misalignment of the latter. (c) The tensile strength of a polymer will be diminished when a plasticizer is added. As the plasticizer molecules force the polymer chain molecules apart, the magnitudes of the secondary interchain bonds are lessened, which weakens the material since strength is a function of the magnitude of these bonds.
ME 46100: ENGINEERING MATERIALS
CLOSED-BOOK PRACTICE
CHAPTER 15
CHARACTERISTICS, APPLICATIONS, AND PROCESSING OF POLYMERS
Page 5/21
14. For a rubber component that is to be vulcanized in its final form, should vulcanization be carried out before or after the forming operation? Why? Hint : You may want to consult Section 15.9. Ans:
Vulcanization of a rubber component should be carried out prior to the forming operation since, once it has been vulcanized, plastic deformation (and thus forming) is not possible since chain crosslinks have been introduced. QUESTIONS & PROBLEMS
Viscoelastic D eformation On the basis of the curves in Fig. 15.5, sketch schematic strain-time plots for polystyrene materials at the following specified temperatures. Ans:
(a) crystalline at 70 C
Crystalline polystyrene at 70C behaves in a glassy manner (Fig. 15.8, curve A); therefore, the strain-time behavior would be as Fig. 15.5b (i.e., elastic ).
(b) amorphous at 180 C
Amorphous polystyrene at 180C behaves as a viscous liquid (Fig. 15.8, curve C ); therefore, the strain-time behavior will be as Fig. 15.5d (that is viscous).
(c) crosslinked at 180 C
Crosslinked polystyrene at 180C behaves as a rubbery material (Fig. 15.8, curve B); therefore, the strain-time behavior will be as Fig. 15.5c (i.e., viscoelastic ).
(d ) amorphous at 100 C
Amorphous polystyrene at 100C behaves as a leathery material (Fig. 15.7); therefore, the strain-time behavior will also be as Fig. 15.5c (that is, viscoelastic ).
ME 46100: ENGINEERING MATERIALS
CLOSED-BOOK PRACTICE
CHAPTER 15
CHARACTERISTICS, APPLICATIONS, AND PROCESSING OF POLYMERS
Page 6/21
The Maxwell visco-elastic model can be represented by a Newtonian viscous damper and a Hookean elastic spring E connected in series, as shown in the diagram. Derive the differential equation governing the time-varying stress t and time-varying strain
E
(t ) (t )
.
t
Ans:
constitutive laws: spring: spring t E spring t & damper : strain compatibility:
spring t damper t
t
iso-stress condition (or stress transmission):
d t dt
spring t damper t dt E d
d t dt
damper t
d spring t dt
d damper t dt
d damper t dt
spring t damper t
t
d t dt
1
d t
E
dt
t
The model is closer to fluid in nature and is commonly used to explain the relaxation behavior of polymers. The Kelvin-Voigt visco-elastic model can be represented by a Newtonian viscous damper and a Hookean elastic spring E connected in parallel, as shown in the diagram. Derive the differential equation governing the time-varying stress t and time-varying strain
(t ) (t )
.
t
Ans:
spring: constitutive laws: damper: stress equilibrium:
damper t
dt
spring t damper t
t
E t
d damper t
iso-strain condition (or strain compatibility): t
E
E spring t
spring t
E spring t
t
d damper t dt
spring t damper t
t
d t dt
The model is closer to solid in nature and is commonly used to explain the creep behavior of polymers.
ME 46100: ENGINEERING MATERIALS
CLOSED-BOOK PRACTICE
CHAPTER 15
CHARACTERISTICS, APPLICATIONS, AND PROCESSING OF POLYMERS
Page 7/21
The Maxwell visco-elastic model can be represented by a Newtonian viscous damper and a Hookean elastic spring E connected in series, as shown in the diagram. The differential equation governing the time-varying stress t and time-varying strain
is:
t
1
d t
E
dt
t
E
(t )
d t dt
(t )
. In a relaxation test,
0 1 t ,
t
where
1 loading and 1 t is the Heaviside unit step function defined as: 1 t 0
initial stress: lim t 0 . The relaxation stress response is then: t 0
a constant-strain
when
t 0
when
t 0
E 0e
Et
t
0 is
. Assume zero
.
(a) Determine the time constant of the relaxation stress response. (b) Find the relaxation modulus Er t from the relaxation stress response. (c ) Plot the relaxation stress response
and relaxation modulus E t
t
r
the duration: 0 t 4 . (d ) Graphically, will the relaxation stress response exponential decaying?. (e) Obtain the special values:
be
vs. time t schematically within
classified as exponential saturation or
t
lim t , Er ,0 lim Er t and Er , lim Er t . t
t 0
t
Ans: st
(a) By definition, the time constant of this 1 -order model is: st
E
(b) By definition, the relaxation modulus of this 1 -order model is:
Er t
t 0
Ee
Er t
Et
(c ) See the plot on next page. (d ) The relaxation stress response:
E 0e
t
Et
is exponential decaying.
lim t lim E 0e Et 0 t t Et E (e) special values: Er ,0 lim E r t lim Ee t 0 t 0 Et E r t lim Ee 0 Er , lim t t
ME 46100: ENGINEERING MATERIALS
CLOSED-BOOK PRACTICE
CHAPTER 15
CHARACTERISTICS, APPLICATIONS, AND PROCESSING OF POLYMERS
Page 8/21
The Kelvin-Voigt visco-elastic model can be represented by a Newtonian viscous damper and a Hookean elastic spring E connected in parallel, as shown in the diagram. The differential equation governing the time-varying stress strain
t
is:
0 1t , where
t
d t dt
0 is
and
t
E t t .
In
a
(t )
time-varying
(t )
creep test,
E
a constant-stress loading and 1 t is the
1 Heaviside unit step function defined as: 1 t 0
the creep strain response is then:
t
0
when
t 0
when
t 0
. Assume zero initial strain: lim t 0 , t 0
1 e . E Et
(a) Determine the time constant of the creep strain response. (b) Find the creep modulus Ec t from the creep strain response. (c ) Plot the creep strain response
t and
creep modulus
Ec t
vs. time t schematically within the
duration: 0 t 4 . (d ) Graphically, will the creep strain response decaying. (e) Obtain the special values:
ME 46100: ENGINEERING MATERIALS
t be classified as exponential saturation or exponential
lim t , Ec,0 lim Ec t and Ec, lim Ec t . t
t 0
t
CLOSED-BOOK PRACTICE
CHAPTER 15
CHARACTERISTICS, APPLICATIONS, AND PROCESSING OF POLYMERS
Page 9/21
Ans: st
(a) By definition, the time constant of this 1 -order model is: st
(b) By definition, the creep modulus of this 1 -order model is:
E
Ec t
0
t
Ec t
E
1 e Et
(c ) See the plot below. (d ) The creep strain response:
t
0
1 e is exponential saturation. E Et
0 0 Et 1 lim t lim e t t E E E (e) special values: Ec ,0 lim Ec t lim Et t 0 t 0 1 e E E Ec , lim Ec t lim Et t t 1 e
ME 46100: ENGINEERING MATERIALS
CLOSED-BOOK PRACTICE
CHAPTER 15
CHARACTERISTICS, APPLICATIONS, AND PROCESSING OF POLYMERS
Page 10/21
(a) Compare 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. Ans:
(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, respectively. The relaxation modulus is the ratio of stress measured after 10s and strain (Eq. 15.1), that is, E r (10); creep modulus is the ratio of stress and strain taken at a specific time t 1 (Eq. 15.2), i.e., E c(t 1). Make two schematic plots of the logarithm of relaxation modulus vs. temperature for an amorphous polymer (curve C in Fig. 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 Ans:
(a) This portion of the problem calls for a plot of log E r (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 E r (10) versus temperature demonstrating how the behavior changes with increased crosslinking. Such a plot is also given below. Increasing the degree of crosslinking will increase the modulus in both glassy and rubbery regions.
(a)
(b)
F racture of Polymers /Mi scellaneous Mechanical Characteri stics For thermoplastic polymers, cite five factors that favor brittle fracture. Ans:
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 to raise the glass transition temperature T g . ME 46100: ENGINEERING MATERIALS
CLOSED-BOOK PRACTICE
CHAPTER 15
CHARACTERISTICS, APPLICATIONS, AND PROCESSING OF POLYMERS
Page 11/21
F actors That I nfluence the Mechanical Properties of Semicrystalline Polymers Explain how each of the following factors influences the tensile modulus of a semicrystalline polymer and why. Ans:
(a) Molecular weight: The tensile modulus is not directly influenced by a polymer’s molecular weight. (b) Degree of crystallinity: 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: 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 ) Annealing of an undeformed material: When an undeformed semicrystalline polymer is annealed below its melting temperature, the tensile modulus increases. (e) Annealing of a drawn material: A drawn semicrystalline polymer that is annealed experiences a decrease in tensile modulus as a result of a reduction in chain-induced crystallinity, thus a reduction in interchain bonding forces. Explain how each of the following factors influences the tensile or yield strength of a semicrystalline polymer and why. Ans:
(a) Molecular weight: 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) Degree of crystallinity: 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: 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 of an undeformed material: Annealing an undeformed semicrystalline polymer produces an increase in its tensile strength. (e) Annealing of a drawn material: On the other hand, annealing a drawn semicrystalline polymer produces a reduction in its tensile strength, but an increase in ductility. The figures at right show the basic mer structures of polychlorotrifluoroethylene (PCTFE) and polytetrafluoroethylene (PTFE). If all other conditions are the same (e.g., molecular weight and degree of crystallinity), would you expect the tensile strength of PCTFE to be (a) greater than, (b) about the same as, or (c ) less than that of PTFE? Explain why.
PCTFE
PTFE
Ans:
If all other conditions are the same (e.g., molecular weight and degree of crystallinity), the tensile strength of a polychlorotrifluoroethylene (PCTFE) having the repeat unit structure will be (a) greater than that of a polytetrafluoroethylene (PTFE). This effect is explained by increased chain entanglements at higher ME 46100: ENGINEERING MATERIALS
CLOSED-BOOK PRACTICE
CHAPTER 15
CHARACTERISTICS, APPLICATIONS, AND PROCESSING OF POLYMERS
Page 12/21
molecular weights. The replacement of a fluorine (F) atom within the PTFE repeat unit with a heavier chlorine (Cl) atom leads to a higher interchain attraction, and thus, a stronger polymer. Note: As shown in Table 15.1, polyvinyl chloride (PVC) is stronger than polyethylene (PE) for the same reason. For each of the following pairs of polymers, do the following: (1) state whether it is possible to decide whether o ne polymer has a higher tensile modulus than the other; (2) if this is possible, note which has the hi gher tensile modulus and cite the reason(s) for your choice; and (3) if it is not possible to decide, state why. Ans:
(a) Branched and atactic poly(vinyl chloride) with a weight-average molecular weight of 100,000 g/mol Linear and isotactic poly(vinyl chloride) having a weight-average molecular weight of 75,000 g/mol Yes, it is possible. The linear and isotactic poly(vinyl chloride) will display a greater tensile modulus. Linear polymers are more likely to crystallize that branched ones. In addition, polymers having isotactic structures will normally have a higher degree of crystallinity than those having atactic structures. Increasing a polymer ’s crystallinity leads to an increase in its tensile modulus. In addition, tensile modulus is independent of molecular weight. Hence the atactic/branched material has the higher molecular weight. (b) Random styrene-butadiene copolymer with 5% of possible sites crosslinked Block styrene-butadiene copolymer with 10% of possible sites crosslinked Yes, it is possible. The block styrene-butadiene copolymer with 10% of possible sites crosslinked will have the higher modulus. Block copolymers normally have higher degrees of crystallinity than random copolymers of the same material. A higher degree of crystallinity favors larger moduli. In addition, the block copolymer also has a higher degree of crosslinking; increasing the amount of crosslinking also enhances the tensile modulus. (c) Branched polyethylene with a number-average molecular weight of 100,000 g/mol Atactic polypropylene with a number-average molecular weight of 150,000 g/mol No, it is not possible. Branched polyethylene will tend to have a low degree of crystallinity since branched polymers don’t normally crystallize. The atactic polypropylene probably also has a relatively low degree of crystallinity; atactic structures also don’t tend to crystallize, and polypropylene has a more complex repeat unit structure than does polyethylene. Tensile modulus increases with degree of crystallinity, and it is not possible to determine which polymer is more crystalline. Furthermore, tensile modulus is independent of molecular weight. For each of the following pairs of pol ymers, do the following: (1) state whether it is possible to decide whether o ne polymer has a higher tensile strength than the other; (2) if this is possible, note which has the hi gher tensile strength and cite the reason(s) for your choice ; and (3) if it is not possible to decide, state why. Ans:
(a) Linear and isotactic poly(vinyl chloride) with a weight-average molecular weight of 100,0 00 g/mol Branched and atactic poly(vinyl chloride) having a weight-average molecular weight of 75,000 g/mol Yes, it is possible. The linear and isotactic material will have the higher tensile strength. Both linearity and isotacticity favor a higher degree of crystallinity than do branching and atacticity; and tensile strength increases with increasing degree of crystallinity. Furthermore, the molecular weight of the linear/isotactic material is higher (100,000 g/mol vs. 75,000 g/mol), and tensile strength increases with increasing molecular weight. ME 46100: ENGINEERING MATERIALS
CLOSED-BOOK PRACTICE
CHAPTER 15
CHARACTERISTICS, APPLICATIONS, AND PROCESSING OF POLYMERS
Page 13/21
(b) Graft acrylonitrile-butadiene copolymer with 10% of possible sites crosslinked Alternating acrylonitrile-butadiene copolymer with 5% of possible sites crosslinked No, it is not possible. Alternating copolymers tend to be more crystalline than graft copolymers, and tensile strength increases with degree of crystallinity. However, the graft material has a higher degree of crosslinking, and tensile strength increases with the percentage of crosslinks. (c) Network polyester Lightly branched polytetrafluoroethylene 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 polytetrafluoroethylene since there are many more of the strong covalent bonds for the network structure. For each of the following pairs of polymers, plot and label schematic stress – strain curves on the same graph. Ans:
(a) Polyisoprene having a number-average molecular weight of 100,000 g/mol and 10% of available sites crosslinked Polyisoprene having a number-average molecular weight of 100,000 g/mol and 20% of available sites crosslinked Shown at right are the schematic stress-strain curves for the two polyisoprene materials, both of which have a molecular weight of 100,000 g/mol. These two materials are elastomers and will have curves similar to curve C in Fig. 15.1. However, the curve for the material having the greater number of crosslinks (20%) will have a higher elastic modulus at all strains. (b) Syndiotactic polypropylene having a weight-average molecular weight of 100,000 g/mol Atactic polypropylene having a weight-average molecular weight of 75,000 g/mol Shown at right are the stress-strain curves for the two polypropylene materials. These materials will most probably display the stress-strain behavior of a normal plastic, curve B in Fig. 15.1. However, the syndiotactic polypropylene has a higher molecular weight and will also undoubtedly have a higher degree of crystallinity; therefore, it will have a higher strength. (c) Branched polyethylene having a number-average molecular weight of 90,000 g/mol Heavily crosslinked polyethylene having a number-average molecular weight of 90,000 g/mol Shown at right are the stress-strain curves for the two polyethylene materials. The branched polyethylene will display the behavior of a normal plastic, curve B in Fig. 15.1. On the other hand, the heavily crosslinked polyethylene will be stiffer, stronger, and more brittle (curve A of Fig. 15.1). ME 46100: ENGINEERING MATERIALS
CLOSED-BOOK PRACTICE
CHAPTER 15
CHARACTERISTICS, APPLICATIONS, AND PROCESSING OF POLYMERS
Page 14/21
Deformation of E lastomers List the two molecular characteristics that are essential for elastomers. Ans:
Two molecular characteristics essential for elastomers are: (1) They must be amorphous, having chains that are extensively coiled and kinked in the unstressed state. (2) There must be some crosslinking. Which of the following would you expect to be elastomers and which thermosetting polymers at room temperature? Justify each choice. Ans:
(a) Linear and highly crystalline polyethylene would be neither an elastomer nor a thermoset since it is a linear polymer. (b) Phenol-formaldehyde having a network structure would be a thermosetting polymer since it has a network structure. It would not be an elastomer since it does not have a crosslinked chain structure. (c) Heavily crosslinked polyisoprene having a glass transition temperature of 50C would be a thermosetting polymer because it is heavily crosslinked. It would not be an elastomer since it is heavily crosslinked and room temperature is below its T g . (d ) Lightly crosslinked polyisoprene having a glass transition temperature of 60C is both an elastomer and a thermoset. It is an elastomer because it is lightly crosslinked and has a T g below room temperature. It is a thermoset because it is crosslinked. (e) Linear and partially amorphous poly(vinyl chloride) is neither an elastomer nor a thermoset. In order to be either it must have some crosslinking.
Melting & Glass Transition Temperatures The table at right shows the glass transition temperatures T g and melting temperatures T m of several polymers. (a) Which polymer(s) would be suitable for the fabrication of cups to contain hot coffee, which has a temperature slightly lower than 95C (203F)? Why? (b) Which polymer(s) would be best suited for use as ice cube trays? Why? Ans:
(a) At its glass transition temperature T g , an amorphous polymer begins to soften. The maximum temperature of hot coffee is probably slightly below 100C (212F). Of the polymers listed, only polystyrene (PE) and polycarbonate (PC) have glass transition temperatures of 100C or above and would be suitable for this application. (b) In order for a polymer to be suited for use as an ice cube tray it must have a glass-transition temperature T g below 0C (32F). Of those polymers listed in the table, only low-density and high-density polyethylene (LDPE & HDPE), polytetrafluoroethylene (PTFE), and polypropylene (PP) satisfy this criterion. ME 46100: ENGINEERING MATERIALS
CLOSED-BOOK PRACTICE
CHAPTER 15
CHARACTERISTICS, APPLICATIONS, AND PROCESSING OF POLYMERS
Page 15/21
F actors That I nfluence Melting and Glass Transition Temperatures For each of the following pairs of polymers, plot and label schematic specific volume-versus-temperature curves on the same graph. Ans:
(a) Linear polyethylene with a weight-average molecular weight of 75,000 g/mol Branched polyethylene with a weight-average molecular weight of 50,000 g/mol Shown at right are specific volume vs. temperature curves for the two polyethylene materials. The linear polyethylene will be highly crystalline, and, therefore, will exhibit behavior similar to curve C in Fig. 15.18. The branched polyethylene will be semicrystalline, and, therefore its curve will appear as curve B in this same figure. Furthermore, since the linear polyethylene has the greater molecular weight, it will also have the higher melting tempe rature. (b) Spherulitic poly(vinyl chloride) of 50% crystallinity and having a degree of polymerization of 5,000 Spherulitic polypropylene of 50% crystallinity and degree of polymerization of 10,000 Shown at right are specific volume vs. temperature curves for the poly(vinyl chloride) and polypropylene materials. Since both are 50% crystalline, they will exhibit behavior similar to curve B in Fig. 15.18. However, since the polypropylene has the greater molecular weight it will have the higher melting temperature. Furthermore, polypropylene will also have the higher glass-transition temperature inasmuch as its CH3 side group is bulkier than the Cl for PVC. (c) Totally amorphous polystyrene having a degree of polymerization of 7,000 Totally amorphous polypropylene having a degree of polymerization of 7,000 Shown at right are specific volume-versus-temperature curves for the polystyrene and polypropylene materials. Since both are totally amorphous, they will exhibit the behavior similar to curve A in Fig. 15.18. However, since the polystyrene repeat unit has a bulkier side group than polypropylene (Table 14.3), its chain flexibility will be lower, and, thus, its glass-transition temperature will be higher.
ME 46100: ENGINEERING MATERIALS
CLOSED-BOOK PRACTICE
CHAPTER 15
CHARACTERISTICS, APPLICATIONS, AND PROCESSING OF POLYMERS
Page 16/21
For each of the following pairs of pol ymers, 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. Ans:
(a) Branched polyethylene having a number-average molecular weight of 850,000 g/mol Linear polyethylene having a number-average molecular weight of 850,000 g/mol Yes, it is possible to determine which polymer has the higher melting temperature. The linear polyethylene will most likely have a higher percent crystallinity, and, therefore, a higher melting temperature than the branched polyethylene. The molecular weights of both materials are the same and thus, molecular weight is not a consideration. (b) PTFE having a density of 2.20 g/cm3 and a weight-average molecular weight of 600,000 g/mol Polytetrafluoroethylene having a density of 2.14 g/cm3 and a weight-average molecular weight of 600,000 g/mol Yes, it is possible to determine which polymer has the higher melting temperature. Of these two 3 polytetrafluoroethylene polymers, the PTFE with the higher density (2.20 g/cm ) will have the higher percent crystallinity, and, therefore, a higher melting temperature than the lower density PTFE. The molecular weights of both materials are the same and, thus, molecular weight is not a consideration. (c) Linear polyethylene having a number-average molecular weight of 225,000 g/mol Linear and syndiotactic poly(vinyl chloride) having a number-average molecular weight of 500,000 g/mol Yes, it is possible to determine which polymer has the higher melting temperature. The linear polyethylene will have the greater melting temperature inasmuch as it will have a higher degree of crystallinity; polymers having a syndiotactic structure do not crystallize as easily as those polymers having identical single-atom side groups. With regard to molecular weight, or rather, degree of polymerization, it is about the same for both materials (8000), and therefore, is not a consideration. (d ) Linear and syndiotactic polypropylene having a weight-average molecular weight of 500,000 g/mol Linear and atactic polypropylene having a weight-average molecular weight of 750,000 g/mol No, it is not possible to determine which of the two polymers has the higher melting temperature. The syndiotactic polypropylene will have a higher degree of crystallinity than the atactic material. On the basis of this effect alone, the syndiotactic PP should have the greater T m, since melting temperature increases with degree of crystallinity. However, the molecular weight for the syndiotactic polypropylene (500,000 g/mol) is less than for the atactic material (750,000 g/mol); a lowering of molecular weight generally results in a reduction of melting temperature.
ME 46100: ENGINEERING MATERIALS
CLOSED-BOOK PRACTICE
CHAPTER 15
CHARACTERISTICS, APPLICATIONS, AND PROCESSING OF POLYMERS
Page 17/21
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. Ans:
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 elastic modulus-glass transition temperature behavior would appear something like the figure at right.
E lastomers, F ibers &Miscellaneous Applications Explain the difference in molecular chemistry between silicone pol ymers and other polymeric materials. Ans:
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. List two important characteristics for polymers that are to be used in fiber applications. Ans:
Two important characteristics for polymers that are to be used in fiber applications are: (1) high molecular weights, and (2) chain configurations/structures that allow for high degrees of crystallinity. Cite five important characteristics for polymers that are to be used in thin-film applications. Ans:
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.
Polymerization Cite the primary differences between addition and condensation polymerization techniques. Ans:
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.
ME 46100: ENGINEERING MATERIALS
CLOSED-BOOK PRACTICE
CHAPTER 15
CHARACTERISTICS, APPLICATIONS, AND PROCESSING OF POLYMERS
Page 18/21
Polymer Additives What is the distinction between dye and pigment colorants? Ans:
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.
F orming Techniques for Plastics Cite four factors that determine what fabrication technique is used to form polymeric materials. Ans:
Four factors that determine what fabrication technique is u sed 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. Compare compression, injection, and transfer molding techniques that are used to form plastic materials. Ans:
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.
F abrication of F ibers and F ilms Long polymeric fibers are usually manufactured by melting, spinning and then drawing. Should the polymer material be (a) thermosetting, (b) thermoplastic or (c ) either one? Cite two reasons. Ans:
Fiber materials that are melt spun must be (b) 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. A polyethylene (PE) thin film was formed by blowing whereas another PE thin film was formed by extrusion and then rolled. Determine which PE thin film would have higher mechanical strength. Why? (1) Those formed by blowing. (2) Those formed by extrusion and then rolled. Ans:
Of the two PE’s 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.
ME 46100: ENGINEERING MATERIALS
CLOSED-BOOK PRACTICE
CHAPTER 15
CHARACTERISTICS, APPLICATIONS, AND PROCESSING OF POLYMERS
Page 19/21
FUNDAMENTALS OF ENGINEERING
Amorphous thermoplastics are formed above their (A) glass transition temperatures (B) softening points (C) melting temperatures (D) none of the above Ans: A. Amorphous thermoplastics are formed above their glass transition temperatures.
Which of the following correctly represents, for an amorphous polymer, the sequential change in mechanical state with increasing temperature? (A) Viscous liquid; rubbery solid; glass (B) Viscous liquid; glass; rubbery solid (C) Glass; viscous liquid; rubbery solid (D) Glass; rubbery solid; viscous liquid (E) Rubbery solid; glass; viscous liquid (F) Rubbery solid; viscous liquid; glass Ans: D. As temperature increases, the correct sequential change in mechanical state for an amorphous polymer is as follows: glass, rubbery solid , and viscous liquid .
How does deformation by drawing of a semicrystalline polymer affect its tensile strength? (A) Decreases (B) Increases (C) About the same Ans: B. 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.
During a tensile test on a semicrystalline polymer, once a neck forms in the gauge section of the specimen, continued specimen elongation proceeds by (A) further neck constriction. (B) neck propagation along the specimen gage length. Ans: B. During a tensile test on a semicrystalline polymer, once a neck forms in the gage section of the specimen, continued specimen elongation occurs by the propagation of this neck region along the gauge length. The phenomenon is called propagating or running neck.
In order for a polymer to behave as an elastomer, which of the following are necessary? (A) It must not crystallize easily; chain bond rotations must be relatively free; the polymer must be above its glass transition temperature. (B) It must not crystallize easily; chain bond rotations must not be relatively free; the polymer must be below its glass transition temperature. (C) It must not crystallize easily; chain bond rotations must be relatively free; the polymer must be below its glass transition temperature. (D) It must not crystallize easily; chain bond rotations must not be relatively free; the polymer must be above its glass transition temperature. Ans: A. In order for a polymer to behave as an elastomer, it must not crystallize easily, chain bond rotations must be relatively free, and the polymer must be above its glass transition temperature. ME 46100: ENGINEERING MATERIALS
CLOSED-BOOK PRACTICE
CHAPTER 15
CHARACTERISTICS, APPLICATIONS, AND PROCESSING OF POLYMERS
Page 20/21
For the two polymers listed below, do the following: indicate if it is not possible to determine if one polymer has a higher tensile strength than the other. On the other hand, if it is possible to make such a determination then note which of the two has the higher tensile strength. Branched polyethylene with a number-average molecular weight of 100,000 g/mol Linear polyethylene with a number-average molecular weight of 80,000 g/mol (A) Not possible (B) Branched polyethylene (C) Linear polyethylene Ans: A. The tensile strength of a polymer increases with both increasing degree of crystallinity and increasing molecular weight. The linear polyethylene will have the higher degree of crystallinity; linear polymers are more crystalline than branched ones. However, the branched polyethylene has the greater molecular weight. Therefore, it is not possible to determine which polymer has the higher tensile strength.
Would you expect a heavily crosslinked polyethylene that has a glass-transition temperature of 0°C to be an elastomer , a thermoset (nonelastomer ), or a thermoplastic polymer at room temperature? (A) Elastomer (B) Thermoset (nonelastomer) (C) Thermoplastic Ans: B. The polyethylene that has a glass-transition temperature of 0C is a thermoset (nonelastomer) since it is heavily crosslinked.
At right are shown and labeled (by letter) three schematic stress-strain curves that may be displayed by polymeric materials at room temperature. Indicate, by letter, the stress-strain behavior expected for a heavily crosslinked polyisoprene. Why? Ans: A. Heavily crosslinked polyisoprene is represented by curve A; it will be quite stiff and brittle due to its high d egree of crosslinking.
Below are shown and labeled three schematic specific volume vs. temperature curves that may be displayed by various types of materials. Indicate, by letter, the specific volume-versustemperature behavior expected for polypropylene that has a spherulitic structure. Why? Ans: B. Polypropylene that has a spherulitic structure is represented by curve B since the spherulitic structure is semicrystalline.
The presence of double-chain bonds, triple-chain bonds, and/or aromatic groups causes the melting point of a polymer to (A) increase (B) decrease Ans: A. The presence of double-chain bonds, triple-chain bonds, and/or aromatic groups, which lower chain flexibility, causes the melting point of a polymer to increase.
ME 46100: ENGINEERING MATERIALS
CLOSED-BOOK PRACTICE
CHAPTER 15
CHARACTERISTICS, APPLICATIONS, AND PROCESSING OF POLYMERS
Page 21/21
For polymeric materials, the addition of polar side atoms or groups causes their glass transition temperatures to (A) increase (B) decrease Ans: A. The addition of polar side atoms or groups causes the glass transition temperatures of polymers to increase.
Elastomers must have some crosslinking (A) True (B) False Ans: True. Elastomers must have some crosslinking.
For fiber applications, a polymer should be (A) highly crystalline (B) semicrystalline (C) noncrystalline Ans: A. For fiber applications, a polymer should be highly crystalline.
For addition polymerization, the reacting monomer species are (A) single and bifunctional (B) multiple monomer Ans: A. For addition polymerization the reacting monomer species are single and bifunctional .
Which polymerization technique is generally used to produce crosslinked and network polymers? (A) Addition polymerization (B) Condensation polymerization Ans: B. Condensation polymerization is generally used to produce crosslinked and network polymers.
Why are plasticizers added to polymers? (A) To improve tensile strength (B) To improve compressive strength (C) To improve flexibility (D) To improve thermal stability (E) To improve toughness (F) Both C and E (G) Both B and D Ans: F. Plasticizers are added to polymers to improve flexibility and toughness.
Injection molding may be used for the fabrication of which type(s) of polymers? (A) Thermosetting (B) Thermoplastic (C) Both thermosetting and thermoplastic Ans: C. Injection molding may be used for the fabrication of both thermosetting and thermoplastic polymers.
ME 46100: ENGINEERING MATERIALS
CLOSED-BOOK PRACTICE