Solutions Manual for Polymer Science and Technology [3 ed.] 9780137039555, 0133845591, 9780133845594

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Solutions Manual for

Polymer Science and Technology Third Edition

Joel R. Fried

Upper Saddle River, NJ • Boston • Indianapolis • San Francisco New York • Toronto • Montreal • London • Munich • Paris • Madrid Capetown • Sydney • Tokyo • Singapore • Mexico City This text is associated with Fried/Polymer Science and Technology, Third Edition (9780137039555) Copyright 2014, Pearson Education, Inc. Do not redistribute.

The author and publisher have taken care in the preparation of this book, but make no expressed or implied warranty of any kind and assume no responsibility for errors or omissions No liability is assumed for incidental or consequential damages in connection with or arising out of the use of the information or programs contained herein. Visit us on the Web: InformIT.com/ph Copyright © 2015 Pearson Education, Inc. This work is protected by United States copyright laws and is provided solely for the use of instructors in teaching their courses and assessing student learning. Dissemination or sale of any part of this work (including on the World Wide Web) will destroy the integrity of the work and is not permitted. The work and materials from it should never be made available to students except by instructors using the accompanying text in their classes. All recipients of this work are expected to abide by these restrictions and to honor the intended pedagogical purposes and the needs of other instructors who rely on these materials. ISBN-10: 0-13-384559-1 ISBN-13: 978-0-13-384559-4 This text is associated with Fried/Polymer Science and Technology, Third Edition (9780137039555) Copyright 2014, Pearson Education, Inc. Do not redistribute.

SOLUTIONS TO PROBLEMS IN POLYMER SCIENCE AND TECHNOLOGY, 3RD EDITION TABLE OF CONTENTS Chapter 1 Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 7 Chapter 11 Chapter 12 Chapter 13

1 5 14 24 28 36 40 51 52

CHAPTER 1 1-1 A polymer sample combines five different molecular-weight fractions, each of equal weight. The molecular weights of these fractions increase from 20,000 to 100,000 in increments of 20,000. Calculate M n , M w , and M z . Based upon these results, comment on whether this sample has a broad or narrow molecular-weight distribution compared to typical commercial polymer samples. Solution Fraction # 1 2 3 4 5 Σ 5

M n = ∑Wi N = i =1

Mi (×10-3) 20 40 60 80 100 300

Wi 1 1 1 1 1 5

Ni = Wi/Mi (×105) 5.0 2.5 1.67 1.25 1.0 11.42

5 = 43,783 1.142 × 10−4

5

Mw =

∑W M i

i =1

∑W 5

∑W M i =1 5

i

300,000 = 60,000 5

=

4 × 108 + 16 × 108 + 36 × 108 + 64 × 108 + 100 × 108 = 73,333 3 × 105

2 i

∑W M i =1

=

i

i =1

Mz =

i

5

i

i

M z 60,000 = = 1.37 (narrow distribution) M n 43,783

1-2 A 50-gm polymer sample was fractionated into six samples of different weights given in the table below. The viscosity-average molecular weight, M v , of each was determined and is included in the table. Estimate the number-average and weight-average molecular weights of the original sample. For these calculations, assume that the molecular-weight distribution of each fraction is extremely narrow and can 1 This text is associated with Fried/Polymer Science and Technology, Third Edition (9780137039555) Copyright 2014, Pearson Education, Inc. Do not redistribute.

be considered to be monodisperse. Would you classify the molecular weight distribution of the original sample as narrow or broad? Fraction 1 2 3 4 5 6

Solution Let M i ≈ M v

6

M n = ∑Wi N = i =1

Weight (gm) 1.0 5.0 21.0 15.0 6.5 1.5

Fraction

Wi

Mi

1 2 3 4 5 6 Σ

1.0 5.0 21.0 15.0 6.5 1.5 50.0

1,500 35,000 75,000 150,000 400,000 850,000

Mv 1,500 35,000 75,000 150,000 400,000 850,000 Ni = Wi/Mi (×106) 667 143 280 100. 16.3 1.76 1208

WiMi 1500 175.000 627,500 2,250,000 2,600,000 1,275,000 7,929,000

50.0 = 41,322 1.21 × 10−3

6

Mw =

∑W M i

i =1

6

∑W

i

=

7,930,000 = 158,600 50.0

i

i =1

M w 158, 600 = = 3.84 (broad distribution) Mn 41,322

1-3 The Schultz–Zimm [11] molecular-weight-distribution function can be written as

W (M ) =

a b +1 M b exp ( − aM ) Γ ( b + 1)

where a and b are adjustable parameters (b is a positive real number) and Γ is the gamma function (see Appendix E) which is used to normalize the weight fraction. (a) Using this relationship, obtain expressions for M n and M w in terms of a and b and an expression for M max , the molecular weight at the peak of the W(M) curve, in terms of M n . Solution Mn =

∫

∞

0

∞

WdM

∫ (W 0

M ) dM

let t = aM 2 This text is associated with Fried/Polymer Science and Technology, Third Edition (9780137039555) Copyright 2014, Pearson Education, Inc. Do not redistribute.

∫

∞

0

∞ 1 ∞ b 1 a b +1 a b +1 b d t a t exp ( −t ) dt = − = Γ ( b + 1) = 1 exp t a t ( ) ( ) ( ) ∫ Γ ( b + 1) 0 Γ ( b + 1) a b +1 ∫0 Γ ( b + 1)

WdM =

∞

∫0 (W

M ) dM =

∞ a b +1 a b +1 1 b −1 − = d t a t a t exp ( ) ( ) ( ) Γ ( b + 1) ∫0 Γ ( b + 1) a b

∫

∞

0

t b −1 exp ( −t ) dt =

a b +1 1 Γ (b) = Γ ( b + 1) a b

a a Γ (b) = bΓ ( b ) b

Mn =

1 b = ab a

Mw =

∫

∞

0

∫

WMdM

∞

0

∞

= ∫ WMdM = 0

WdM

b +1 ∞ a b +1 a b +1 Γ ( b + 2 ) t exp − = = t d t a a ( ) ( ) ( ) Γ ( b + 1) ∫0 Γ ( b + 1) a b + 2

( b + 1) Γ ( b + 1) = b + 1 aΓ ( b + 1) a (b) Derive an expression for Mmax, the molecular weight at the peak of the W(M) curve, in terms of M n . Solution dW a b +1 = bM b −1 exp ( − aM ) + M b ( − a ) exp ( − aM ) = 0 dM Γ ( b + 1) bM b − a = aM b

b = M a = M n (i.e., the maximum occurs at M n ) a

(c) Show how the value of b affects the molecular weight distribution by graphing W(M) versus M on the same plot for b = 0.1, 1, and 10 given that M n = 10,000 for the three distributions. Solution b a= 10,000 b a W=

0.1 1×10-5

1 1×10-4

10 1×10-3

a b +1 M b exp ( − aM ) dM Γ ( b + 1)

where Γ ( b + 1) = ∫

∞

0

( aM )

b

exp ( − aM ) dM .

Plot W(M) versus M Hint:

∫

∞

0

x n exp ( − ax ) dx = Γ ( n + 1) a n +1 = n ! a n +1 (if n is a positive interger).

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1-4 (a) Calculate the z-average molecular weight, M z , of the discrete molecular weight distribution described in Example Problem 1.1. Solution 3

Mz =

∑W M i =1 3

i

2 i

∑W M i =1

i

1(10,000 ) + 2 ( 50,000 ) + 2 (100,000 ) 2

= i

2

1(10,000 ) + 2 ( 50,000 ) + 2 (100,000 )

2

= 80,968

(b) Calculate the z-average molecular weight, M z , of the continuous molecular weight distribution shown in Example 1.2. Solution 105

Mz

∫ = ∫

3

M 2 dM

10 105 103

MdM

( M 3) = ( M 2) 3

105

2

103 105

= 66,673

103

(c) Obtain an expression for the z-average degree of polymerization, X z , for the Flory distribution described in Example 1.3.

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Solution ∞

Xz =

∑ X 2W ( X )

∞

∑X

3

p x −1

∑ XW ( X ) ∑ X

2

p x −1

1 ∞ 1

=

1 ∞ 1

Let ∞

A = ∑ Xp x −1 = 1 + 2 p + 3 p 2 +

=

1

1 1− p

(geometric series)

∞

B = ∑ X 2 p x −1 = 1 + 22 p + 32 p 2 + 1

∞

C = ∑ X 3 p x −1 = 1 + 23 p + 32 p 2 + 1

Can show that B (1 − p ) = A (1 + p ) Therefore B =

1+ p

(1 − p )

3

∞

∞

∞

x =1

x =1

x =1

Write C (1 − p ) = ∑ 3 X 2 p x −1 − ∑ 3 Xp x −1 + ∑ p x −1 = 3B − 3 A2 +

Therefore C =

1 1 + 4 p + p2 = 3 1− p (1 − p )

1 + 4 p + p2

(1 − p ) ∞

and finally X z =

4

∑X

3

p x −1

∑X

2

p x −1

1 ∞

2 1 + 4 p + p2 1 + 4 p + p2 C 1 + 4 p + p (1 − p ) = = = = 4 1 − p2 B (1 − p )(1 + p ) (1 − p ) (1 + p ) 3

1

Mz = Mo X z

CHAPTER 2 2.1 If the half-life time, t1/2, of the initiator AIBN in an unknown solvent is 22.6 h at 60°C, calculate its dissociation rate constant, kd, in units of reciprocal seconds. Solution [ I] = [ I]o exp ( −kd t )

[ I] [ I]o

=

1 = exp ( −kd t ) 2

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− kd t = ln (1 2 ) = −0.693

kd =

0.693 0.693 h = = 8.52 × 10−5 s -1 t 22.6 h 3600 s

2.2 Styrene is polymerized by free-radical mechanism in solution. The initial monomer and initiator concentrations are 1 M (molar) and 0.001 M, respectively. At the polymerization temperature of 60°C, the initiator efficiency is 0.30. The rate constants at the polymerization temperature are as follows: kd = 1.2 × 10-5 s-1 kp = 176 M-1 s-1 kt = 7.2 × 107 M-1 s-1 Given this information, determine the following: (a) Rate of initiation at 1 min and at 16.6 h Solution Ri = 2 fkd [ I ] = 2 ( 0.30 ) (1.2 × 10−5 ) [ I ] = 7.2 × 10−6 [ I ]

[ I] = [ Io ] exp ( −kd t ) at 1 min: [ I] = 0.001( 0.9993) = 0.0009993 M

Ri = ( 7.2 × 10−6 ) ( 0.0009993) = 7.19 × 10−9 M s -1

at 16.6 h: [ I] = 0.001( 0.488) = 0.000488 M

Ri = ( 7.2 × 10−6 ) ( 0.000448 ) = 3.51 × 10−9 M s -1

(b) Steady-state free-radical concentration at 1 min Solution 12

fk [ IM x ⋅] = d kt at 1 min:

[ I]

12

( 0.30 ) (1.2 × 10−5 )

[ IM x ] =

7.2 × 107

12

( 0.0009993)

12

= 7.08 × 10−9 M

(c) Rate of polymerization at 1 min Solution Ro = kp [ IM x ⋅][ M ]

[ M ] = [ M ]o exp ( −kp [ IM x ⋅] t ) = (1) exp −176 ( 7.08 × 10−9 ) 60 Ro = 176 ( 7.08 × 10−9 ) ( 0.9999 ) = 1.24 × 10−6 M s −1

= 0.9999 M

(d) Average free-radical lifetime, τ, at 1 min, where τ is defined as the radical concentration divided by the rate of termination Solution

τ=

[ IM x ⋅] 2 2k t [ IM x ⋅]

=

1 1 = = 0.981 s 7 2k t [ IM x ⋅] 2 ( 7.2 × 10 )( 7.08 × 10−9 )

6 This text is associated with Fried/Polymer Science and Technology, Third Edition (9780137039555) Copyright 2014, Pearson Education, Inc. Do not redistribute.

(e) Number-average degree of polymerization at 1 min Solution Rp 1.24 × 10−6 1.24 × 10−6 1.24 × 10−6 Xn = = = == = 172 Rt 2k t [ IM x ⋅]2 2 ( 7.2 × 107 )( 7.08 × 10−9 )2 7.22 × 10−9 2.3 It has been reported that the rate of a batch photopolymerization of an aqueous acrylamide solution using a light-sensitive dye is proportional to the square of the monomer concentration, [M]2, and the square root of the absorbed light-intensity, I1/2. Note that, although this polymerization is free radical, the apparent kinetics appear not to be typical of usual free-radical polymerization for which the rate of polymerization is proportional to the first power of monomer concentration and to the square root of the initiator concentration (eq. (2.25)). The following polymerization mechanism has been proposed to explain the observed kinetics: Initiation M+D

k1 ,hν

R +M

k2

→R

→ RM1

Propagation RM1 + M . . . . . . . . .

k3

RM n + M

k3

→ RM 2

→ RM n +1

Termination RM n + RM n

R

k5

k4

→P

→S

where M, monomer D, dye P, terminated polymer S, deactivated initiator Show that this mechanism appears to be correct by deriving an equation for the rate of propagation in terms of [M], I, and the appropriate rate constants. The following assumptions may be made: 1. Equal reactivity in the propagation steps 2. Steady-state concentration of R• and RMn• 3. k2 > k2 [ M ] , we have Ro = k3 1 2 [ M ] I1 2 or Ro ∝ [ M ] I1 2 k 4 k5 See G.K. Oster, G. Oster, & G. Prati, JACS 79, 595 (1957).

2.4 Reactivity ratios for styrene and 4-chlorostyrene are given in Table 2-6. (a) Using these values, plot the instantaneous copolymer composition of poly(styrene-co-4-chlorostyrene) as a function of comonomer concentration in the copolymerization mixture. Solution Styrene, 1; 4-chlorostyrene, 2 Q 1.00 r1 = 1 exp −e1 ( e1 − e2 ) = exp 0.8 ( −0.8 + 0.33) = 0.667 Q2 1.03 r2 =

1.03 exp 0.33 ( −0.33 + 0.8 ) = 1.203 1.00

(b) Comment on the expected monomer sequence distribution in the resulting copolymer. Solution F1 =

r1 f12 + f1 f 2 r1 f12 + 2 f1 f 2 + r2 f 2 2

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1 0.9

r1 = 0.667; r2 = 1.203

0.8 ideal

0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

f1

(c)

Near ideal with copolymer enrichment in the more reactive monomer, 4ClS.

2.5 If the number-average degree of polymerization for polystyrene obtained by the bulk polymerization of styrene at 60°C is 1000, what would be the number-average degree of polymerization if the polymerization were conducted in a 10% solution in toluene (900 g of toluene per 100 g of styrene) under otherwise identical conditions? The molecular weights of styrene and toluene are 104.12 and 92.15, respectively. State any assumptions that are needed. Solution [SH ] 1 1 = +C Xn ( Xn ) [M] o C = 1.25 × 10−5 (Table 2.4) ( X n )o = 1000

[M] =

100 = 0.9604 104.12

[SH ] =

900 = 9.767 92.15

1 1 9.767 = + 0.125 × 10−4 = 1.127 × 10−3 or X n = 887 X n 1000 0.9604

2.6 Assume that a polyesterification is conducted in the absence of solvent or catalyst and that the monomers are present in stoichiometric ratios. Calculate the time (min) required to obtain a numberaverage degree of polymerization of 50 given that the initial dicarboxylic acid concentration is 3 mol L-1 and that the polymerization rate constant is 10-2 L mol-1 s-1.

9 This text is associated with Fried/Polymer Science and Technology, Third Edition (9780137039555) Copyright 2014, Pearson Education, Inc. Do not redistribute.

Solution X n = [ A − A ]o kt + 1 t=

X n −1 50 − 1 = = 1633 s or 27.2 min [ A − A ]o k 3 × 10−2

2.7 Show how the assumption of steady-state free radical concentration, M1 or M2 , can be used to obtain the instantaneous copolymerization equation in the form of eq. (2.45) starting with eq. (2.39). Solution d [ M1 ] [ M1 ] k11 [ ~M1 ⋅] + k21 [ ~M 2 ⋅] [ M1 ] k11 [ ~M1 ⋅] [ ~M 2 ⋅] + k21 = = d [ M 2 ] [ M 2 ] k12 [ ~ M1 ⋅] + k22 [ ~M 2 ⋅] [ M 2 ] k12 [ ~M1 ⋅] [ ~M 2 ⋅] + k22 Steady state in ~ M1 ⋅ gives

[ ~ M1 ⋅] = k21 [ M1 ] [ ~ M 2 ⋅] k12 [ M 2 ]

Substitution in the equation given above, rearranging, and introducing the definitions of the reactivity ratios gives d [ M1 ] [ M1 ] r1 [ M1 ] + [ M 2 ] = d [ M 2 ] [ M 2 ] [ M1 ] + r2 [ M 2 ] 2.8 Show that the ceiling temperature in a free-radical polymerization can be obtained as Tc =

−∆H p

Rln ( Ap [ M ] Adp )

.

Solution Rp = Rdp kp [ M ][ M x ⋅] = kdp [ M x ⋅]

AP exp −

Ep RTc

[ M ] = Adp exp

ln { Ap [ M ] Adp } =

Ep − Edp RTc

=

−

Edp RTc

−∆H p RTc

and Tc =

{

−∆H p

}

R ln ( Ap [ M ] Adp )

2.9 Find the azeotropic composition for the free-radical copolymerization of styrene and acrylonitrile. Solution Substituting f1 = F1 and rearranging eq. 2.45 (using f 2 = 1 − f1 ) gives

( r1 − 2 + r2 ) f12 + ( 2 − 2r2 − r1 + 1) f1 + r2 − 1 = 0 Using reactivity ratios given in Table 2-6 (r1 = 0.290 and r2 = 0.020) and solving the quadratic equation, gives F1 = f1 = 0.580 . 2.10 Describe the copolymer composition that would be expected in the free-radical copolymerization of styrene and vinyl acetate. 10 This text is associated with Fried/Polymer Science and Technology, Third Edition (9780137039555) Copyright 2014, Pearson Education, Inc. Do not redistribute.

Solution Since the reactivity ratio for vinyl acetate is not given in Table 2-6, try Q-e scheme using values given in Table 2-7 1.00 exp 0.80 ( −0.80 + 0.88 ) = 41.0 (styrene) 0.026 0.026 r2 = exp 0.88 ( −0.88 + 0.80 ) = 0.0242 (vinyl acetate) 1.0 Values of 55 and 0.01 are sometimes reported; experimental values range from 18.8 to 60 for r1 and –0.04 to 0.16 for r2 (Polymer Handbook, 4th ed) k k Since r1 = 11 >> r2 = 22 , there is a tendency for consecutive homopolymerization since M1 (styrene) will k12 k21 polymerize until it is completely consumed and then M2 (vinyl acetate) will polymerize. r1 =

2.11 Explain why high pressure favors the propagation step in a free-radical polymerization. How would the rate of termination be affected by pressure? Answer It would be expected that pressure would increase the rate of propagation but decrease the rate of termination. This is supported by studies of styrene polymerization (Ogo, Macromol. Sci.-Rev. Macromol. Chem. Phys. C24, 1 (1984)). One argument that can be made is that pressure increases viscosity and, therefore, the diffusion of long-chain radicals is reduced (i.e., the rate of termination decreases). Kiran & Saraf (J. Supercritical Fluids 3, 198 (1990) discuss volume production arguments. Net volume decreases during propagation and is, therefore, favored at high pressure. On the other hand, net volume increase during termination and unfavored at high pressure. 2.12 From data available in Section 2.2.1, calculate the activation energy for propagation for the freeradical polymerization of styrene. Do you expect the activation energy to be dependent upon solvent in a solution polymerization? Solution Using data for styrene bulk polymerization in Table 2-3 and kp = Ap exp ( − Ep RT ) R = 8.3144 J mol-1 K-1 Plot gives slope = –3.925×103; using this value gives Ep = 32.6 kcal mol-1 Polymer Handbook, 4th ed., cites a value of 31.5 kcal mol-1

11 This text is associated with Fried/Polymer Science and Technology, Third Edition (9780137039555) Copyright 2014, Pearson Education, Inc. Do not redistribute.

6 5 4 3 2 1 0 0.003

0.0031

0.0032

0.0033

0.0034

-1

1/T (K )

2.13 Draw the chemical structures of the two ends of a terminated polystyrene chain obtained by the atom transfer radical polymerization of styrene using 1-phenylethyl chloride (1-PECl) as the initiator, CuCl as the catalyst, and 2,2’-bipyridine as the complexing agent. Solution 1-phenyl chloride initiator Cl CH3

CH2

CH

See Wang & Matyjaszewski, Macromolecules 28, 7901 (1995) or Coessens et al., Progr. Polym. Sci. 26, 337 (2001). 2.14 Show that the rate of polymerization in atom transfer radical polymerization is proportional to the equilibrium constant defined in eq. (2.50). Solution Rp = kp [ Pn ⋅][ M ] = K e

[ Pn − X ][Cu(I)X ] M [ ] [Cu(II)X 2 ]

2.15 Show that azeotropic copolymerization occurs when the feed composition is given as f1 =

Solution At azeotrope:

d [ M1 ]

1 − r1 . 2 − r1 − r2

[ M1 ] d [M2 ] [M2 ] =

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From eq. 2.42: r1 [ M1 ] + [ M 2 ] = [ M1 ] + r2 [ M 2 ] Dividing both side by [M2] gives: 1− r [ M1 ] = 2 [ M 2 ] 1 − r1 Dividing by [M1]+[M2] gives 1 − r2 f1 = (1 − f1 ) 1 − r1 Rearrangement gives 1 − r2 2 − r1 − r2 Substituting values of r1 and r2 from problem 2.9 gives the same result, f1 = 0.580 at the azeotrope. f1 =

2.16 Methyl methacrylate is copolymerized with 2-methylbenzyl methacrylate (M1) in 1,4-dioxane at 60°C using AIBN as the free-radical initiator. (a) Draw the repeating unit of poly(2-methylbenzyl methacrylate). Solution CH3 CH2

C C

O

O CH2 H3 C

(b) From the data given in the table below, estimate the reactivity ratios of both monomers. f1

F1*

0.10 0.25 0.50 0.75 0.90

0.14 0.33 0.52 0.70 0.87

* From 1H-NMR measurements

Solution Data can be fitted to Eq. 2.45 using nonlinear regression analysis. Alternately (and less preferred) is the traditional linearization of the instantaneous copolymerization equation in the form (see Flory, Principle of Polymer Chemistry, pp. 185–189) G = r1 H − r2 where f F G = 1 1− 2 f2 F1 and 2

f F2 H= 1 f 2 F1 A plot of G versus H gives r1 from the slope and r2 from the intercept.

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Solutions reported in I. Erol, C. Soykan, J. Macromol. Sci.: Part A – Pure & Applied Chemistry A39, 953 (2002) give average values of r1 = 1.03 and r2 = 0.77. A nonlinear regression (Matlab) gives r1 = 0.6311 and r2 = 0.5328. CHAPTER 3 3.1 Polyisobutylene (PIB) is equilibrated in propane vapor at 35°C. At this temperature, the saturated vapor pressure (p1o) of propane is 9050 mm Hg and its density is 0.490 g cm-3. Polyisobutylene has a molecular weight of approximately one million and a density of 0.915 g cm-3. The concentration of propane, c, sorbed by PIB at different partial pressures of propane (p1) is given in the following table. Using this information, determine an average value of the Flory interaction-parameter, χ12, for the PIB– propane system. p1 (mm Hg) 496 941 1446 1452

c (g propane/g PIB) 0.0061 0.0116 0.0185 0.0183

Solution CH3 CH2

C CH3

n

M o = 56.11 ; r =

1 106 = 1.78 × 104 and 1 − = 0.9999 ≈ 1 r 56.1

w1 ; w1 = ρ1V1 ; w2 = ρ 2V2 w2 ρV V ρ c = 1 1 or 2 = 1 ρ 2V2 V1 ρ 2 c ρ V1 V 1 =1+ 2 =1+ 1 φ1 = or φ1 ρ2c V1 + V2 V1 p p a1 = 1o = 1 p1 9050 c 0.0061 0.0116 0.0183 0.0185

φ1 0.01126 0.02120 0.03304 0.03339

p1

496 941 1452 1446

lna1

-2.9039 -2.2636 -1.8298 -1.8340 ave.

χ

c=

12

0.6075 0.6381 0.6559 0.6410 0.64

See S. Prager, E. Bagley, and F. A. Long, JACS 75, 2742 (1953).

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3.2 The following osmotic pressure data are available for a polymer in solution: c (g dL-1) 0.32 0.66 1.00 1.40 1.90

h (cm of solvent) 0.70 1.82 3.10 5.44 9.30

Given this information and assuming that the temperature is 25°C and that the solvent density is 0.85 g cm-3, provide the following: (a) A plot of Π/RTc versus concentration, c Solution Π = ρ gh = 0.85 ( 980.665 ) h h (cm)

( RTc ) ×106

0.32 0.66 1.00 1.40 1.90

0.70 1.82 3.10 5.44 9.30

7.310 9.216 10.360 12.985 16.358

Π

c (g dL-1)

18 16 14 12 10 8 6 4 2 0 0

0.2

0.4

0.6

0.8 c (g

1

1.2

1.4

1.6

1.8

2

dL-1 )

(b) The molecular weight of the polymer and the second virial coefficient, A2, for the polymer solution Solution

Π 1 ≈ + 2 A2 c RTc M n

Least squares fit of data ( R 2 = 0.9884 ) gives slope = 5.654 × 10-6 ( M n = 176,866 ) and intercept = 5.276

(A

2

= 2.64 × 10−6 ) .

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3.3 (a) What is the osmotic pressure (units of atm) of a 0.5 wt % solution of poly(methyl methacrylate) ( M n = 100,000) in acetonitrile (density, 0.7857 g cm-3) at 45°C for which [η] = 4.8×10-3 M0.5? Solution The form of the Mark-Houwink-Sakurata equation shown above (see eq 3.101) indicates that for PMMA in acetonitrile a = 0.5 and, therefore, A2 = 0 (i.e., θ solvent) and from eq. 3.85 we can write RTc atm cm3 0.5 g mol Π= = 82.057 = 1.305 × 10−3 atm 318 K 3 5 Mn mol K 100 cm 10 g (b) What is the osmotic head in units of cm? Solution From eq 3.88, we have h=

1 kg 1000 g m 2 Π 1.305 × 10−3 atm cm3 s 2 = = 1.716 cm ρg 0.7857 g 9.80665 m 9.869 × 10−6 atm m s 2 kg 104 cm 2

(c) Estimate the Flory interaction parameter for polysulfone in methylene chloride. Solution 2 χ12 ∝ (δ1 − δ 2 ) Table 3-3: δ PSF = 9.92 ( cal cm -3 )

12

and δ CH Cl = 9.92 ( cal cm -3 )

12

3

and therefore χ12 ≈ 0

(d) Based upon your answer above, would you expect methylene chloride to be a good or poor solvent for polysulfone? Answer Good solvent (solubility parameters match). 3.4 The osmotic pressure of two samples, A and B, of poly(vinyl pyridinium chloride) CH2

CH N

n Cl

were measured in different solvents. The following data were obtained: Osmotic Pressure Data in Distilled Water Sample c (g mL-1) Π (atm × 103) A 0.002 29 A 0.005 50 B 0.002 31 B 0.005 52 Osmotic Pressure Data in 0.01 N Aqueous NaCl Samplee c (g mL-1) Π (atm × 103) A 0.002 5 A 0.005 13 B 0.002 2 B 0.005 5.5 16 This text is associated with Fried/Polymer Science and Technology, Third Edition (9780137039555) Copyright 2014, Pearson Education, Inc. Do not redistribute.

Discuss these results and account for any features that you consider anomalous. Solution Π 1 = RT + A2 c + c Mn

Sample A A B B

c ( g mL-1) 0.002 0.005 0.002 0.005

Π (atm × 103) 29 50 31 52

Π

Distilled Water 14.5 10.0 15.5 10.4

Sample A A B B

c ( g mL-1) 0.002 0.005 0.002 0.005

Π (atm × 103) 5 13 2 5.5

Π

0.01 N aq NaCl 2.5 2.6 1.0 1.1

As shown by the data for the aqueous solution, Π c increases with dilution which is opposite to expected behavior of a polymer in solution. This may be attributed to dissociation of the chlorine substituent and, thereby, an increase in the effective number of osmotic units at high dilution. When chloride anions are added, dissociation is inhibited and, as shown by the data given in the second table, Π c is essentially independent of concentration ( A2 ≈ 0 ). See, for example, the discussion of osmotic pressure of polyelectrolytes given in Flory, Principles of Polymer Chemistry, pp. 633-635. 3.5

The following viscosity data were obtained for solutions of polystyrene (PS) in toluene at 30°C:

c (g dL-1) 0 0.54 1.08 1.62 2.16

t (s) 65.8 101.2 144.3 194.6 257.0

Using this information, please do the following: (a) Plot the reduced viscosity as a function of concentration Solution

ηi c

= [η ] + kH [η ] c where

ηi =

2

η − η s t − ts ≈ ηs ts

ts = 65.8 s

17 This text is associated with Fried/Polymer Science and Technology, Third Edition (9780137039555) Copyright 2014, Pearson Education, Inc. Do not redistribute.

i

0.538 1.193 1.957 2.906

η

η

c (g dL-1) 0.54 1.08 1.62 2.16

i

0.996 1.105 1.208 1.345

(b) Determine the intrinsic viscosity of this PS sample and the value of the Huggins constant, kH Solution From plot (R2 = 0.9958); we can obtain the following: From eq 3.103 ηi 2 = [η ] + k H [η ] c c

the intercept gives [η] = 0.8760 dL g-1 and the slope (kH[η]) is 0.2130 which gives kH = 0.278 (c) Calculate the molecular weight of PS using Mark–Houwink parameters of a = 0.725 and K = 1.1 × 10dL g-

4

Solution [η ] = KM v a Substituting values, we have 0.876 = (1.1 × 10−4 ) M v 0.725 or M v = 240,350 3.6 Given that the molecular weight of a polystyrene (PS) repeating unit is 104 and that the carboncarbon bond distance is 1.54 Å, calculate the following: (a) The mean-square end-to-end distance for a PS molecule of 1 million molecular weight assuming that the molecule behaves as a freely-rotating, freely-jointed, and volumeless chain. Assume that each link is equivalent to a single repeating unit of PS. Solution 18 This text is associated with Fried/Polymer Science and Technology, Third Edition (9780137039555) Copyright 2014, Pearson Education, Inc. Do not redistribute.

106 = 9.62 × 103 104 l = 2 (1.54 Å ) = 3.08 Å n=

r 2 = nl 2 = 9.13 × 104 Å 2

(b) The unperturbed root-mean-square end-to-end distance, 〈r2〉o1/2, given the relationship for intrinsic viscosity, [η], of PS in a θ solvent at 35°C as

[η ] = 8 × 10−4 M 0.5 where [η] is in units of dL g-1. Solution r2

r2

o

M [η ]θ

=

12 o

23

=

Φ

106 ( 8 × 10−4 )(106 )

0.5

2/3

2.1 × 1021

= 5.26 × 105 Å 2

= 725 Å

(c) The characteristic ratio, CN, for PS. Solution CN =

r2 nl 2

=

5.26 × 105 = 5.76 9.13 × 104

3.7 The use of universal calibration curves in GPC is based upon the principle that the product [η] M, the hydrodynamic volume, is the same for all polymers at equal elution volumes. If the retention volume for a monodisperse polystyrene (PS) sample of 50,000 molecular weight is 100 mL in toluene at 25°C, what is the molecular weight of a fraction of poly(methyl methacrylate) (PMMA) at the same elution volume in toluene at 25°C? The Mark–Houwink parameters, K and a, for PS are given as 7.54 × 10-3 mL g-1 and 0.783, respectively; the corresponding values for PMMA are 8.12 × 10-3 mL g-1 and 0.71. Solution η M=KM v1+ a 7.54 × 10−3 ( 5 × 104 )

1.783

= 8.12 × 10−3 M1.71

or M V = 75,988 3.8 Show that the most probable end-to-end distance of a freely-jointed polymer chain is given as

( 2n

2

3) . 12

Solution

ω (r ) =

b π1 2

3

exp ( −b 2 r 2 ) 4πr 2

The maximum value occurs at dω ( r ) dr

= 8πr

b π1 2

3

dω ( r ) dr

=0

(1 − b r ) exp ( −b r ) = 0 2 2

2 2

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or b 2 r 2 = 1 and 1 2nl r= = b 3

12

3.9 The (reduced or excess) Rayleigh ratio ( Rθ ) of cellulose acetate (CA) in dioxane was determined as a function of concentration by low-angle laser light-scattering measurements. Data are given in the following table. If the refractive index ( no ) of dioxane is 1.4199, the refractive-index increment ( dn dc ) for CA in dioxane is 6.297 × 10-2 cm3 g-1, and the wavelength (λ) of the light is 6328 Å, calculate the weight-average molecular weight of CA and the second virial coefficient (A2) R(θ) × 105 (cm-1) 0.239 0.440 0.606 0.790 0.902

c × 103 (g mL-1) 0.5034 1.0068 1.5102 2.0136 2.517 Solution 2π 2 no2 dn K= N A λ 4 dc

c × 103 0.5034 1.0068 1.5102 2.0136 2.517

2π 2 (1.4199 ) ( 6.297 × 10−2 ) 2

2

=

R ( θ ) ×105

0.239 0.440 0.606 0.790 0.902

6.023 × 10

23

( 6328)

4

(10 )

−8 4

2

= 1.63 × 10−8

Kc R ( θ ) ×106

3.43 3.73 4.06 4.15 4.55

Least squares (R2 = 0.9758) fit of data gives (from the intercept) 20 This text is associated with Fried/Polymer Science and Technology, Third Edition (9780137039555) Copyright 2014, Pearson Education, Inc. Do not redistribute.

Mn =

1 = 313,873 3.186 × 10−6

and (from the slope) 1 A2 = 5.284 × 10−4 = 2.64 × 10−4 mL-mol g -2 2

(

)

3.10 Chromosorb P was coated with a dilute solution of polystyrene in chloroform, thoroughly dried, and packed into a GC column. The column was then heated in a GC oven and maintained at different temperatures over a range from 200°C to 270°C under a helium purge. At each temperature, a small amount of toluene was injected and the time for the solute to elute the column was recorded and compared to that for air. From this information, the specific retention volume was calculated as given in the table below. Using this data, plot the apparent Flory interaction parameter as a function of temperature. T °C 200 210 220 230 240 250 260 270

Vg mL/g-coating 6.55 5.58 4.66 4.07 3.38 2.87 2.88 2.38

Solution 0.46 0.44 0.42 0.4 0.38 0.36 0.34 0.32 0.3 0.0018

0.0019

0.002

0.0021

0.0022

1/T (K-1 )

See J. R. Fried and A. C. Su, "Poly(2,6-dimethyl-1,4-phenylene oxide) Blends Studied by Inverse Gas Chromatography," Adv. Chem. Ser. 211, 59–66 (1986).

21 This text is associated with Fried/Polymer Science and Technology, Third Edition (9780137039555) Copyright 2014, Pearson Education, Inc. Do not redistribute.

3.11 (a) Derive eq (3.72) and (b) develop an expression that can be used to obtain the exchange interaction parameter, X 12 , appearing in the Flory equation of state from inverse gas chromatography measurements. Solution (a) The F-H equation, eq 3.37 ln a1 = ln (1 − φ2 ) + φ2 + χ12φ22

Rearrangement of eq. 3.37 gives a1

ln

= ln a1 − ln φ1 = ln a1 − ln (1 − φ2 ) = 1 − φ1 + χ12 (1 − φ1 )

φ1

2

and it follows that lim ln

φ1 → 0

a1

= 1 + χ12

φ1

Substituting eq. 3.70 (below) into the LHS of the equation above

a1

ln γ = ln ∞ 1

∞

= lim

φ1

φ1 → 0

a1

φ1

= ln

p1o ( B11 − V1 ) 273.16 Rv2 − Vg p1o RT

finally gives the required result eq. 3.72

p1o ( B11 − V1 ) 273.16 Rv2 − −1 Vg p1oV1 RT

χ12 = ln

(b) For the Flory EOS, write eq. 3.61 as

ln a1 =

v* v1 3 − 1 ∆µ1 1 θ 22 v1* X 12 1 1 = ln φ1 + 1 − 1* φ2 + + v1* p1* 3T1 ln 11 3 + − RT v2 RT v v − 1 v1 v

and then rearrangement gives

ln

a1

φ1

= 1−

v1* v11 3 − 1 1 θ 22 v1* X 12 1 1 * * φ + + + − v p 3 T ln 1 2 1 1 * 13 v2 RT v v − 1 v1 v

Noting that φ2 = 1; φ1 = 1; θ 2 =1; θ1 =1; v = v2 ; T = T2 ; α =α 2 and using eq. 3.70 (part a), we have

lim ln

φ1 → 0

a1

φ1

X 12 = RT

p1o ( B11 − V1 ) v1* v11 3 − 1 273.16 Rv2 1 v1* X 12 v1* p1* 1 1 T ln + + + − = ln − = − 1 3 1 o * 13 Vg p1 RT v2 RT v2 RT v2 − 1 v1 v2 v2 v1*

ln

p1o ( B11 − V1 ) v1* v1* p1* v11 3 − 1 273.16 Rv2 1 1 T − − − − + − 1 3 ln 1 o * 13 Vg p1 RT v2 RT v2 − 1 v1 v2

22 This text is associated with Fried/Polymer Science and Technology, Third Edition (9780137039555) Copyright 2014, Pearson Education, Inc. Do not redistribute.

and finally X 12 = RT

273.16 Rv2 v2 v v* − p1o ( B11 − V1 ) 2* − RT 1 − 1* ln * o v1 Vg p1 v1 v2

v2 v − v1* p1* 2* v1* v1

3T1 ln

v11 3 − 1 1 1 + − v21 3 − 1 v1 v2

3.12 Derive the expression for osmotic pressure given by eq. (3.83). Mν 2 Π RT 1+ = c M V1

1 1 Mν 3 2 c + − χ12 c + 2 3 V1

Solution From eq. 3.80, we can write

Π=−

RT ln a1 V1

Substituting eq. 3.36 for lna1 in the Flory-Huggins model gives Π=−

RT 1 ln (1 − φ2 ) + 1 − φ2 + χ12φ2 2 V1 r

Substituting the Taylor series (Appendix E) in the form ln (1 − φ2 ) ≈ −φ2 −

φ2 2 2

−

φ23 3

+

gives Π≈

φ2 φ3 RT 1 φ2 + 2 + 2 − 1 − φ2 − χ12φ2 2 + V1 2 3 r

Since r =

=

φ3 RT φ2 1 + − χ12 φ2 2 + 2 + V1 r 2 3

vM and φ2 = cv (eq. 3.82) V1

we can write

φ2 cv 1 = V1 = cV1 r vM M and finally, we obtain eq. 3.83 as 1 1 c 2 v3 Π = RT + − χ12 cv 2 + + M c 2 3

=

RT Mv 2 1+ M V1

1 1 Mv 3 2 c + − χ12 c + 2 3 V1

3.13 Show how eq. eq. 3.78 for the relationship between the Flory-Huggins interaction parameter and the solubility parameters of polymer and solvent was derived. Eq. 3.78 V 2 χ12 ≅ 1 (δ1 − δ 2 ) RT 23 This text is associated with Fried/Polymer Science and Technology, Third Edition (9780137039555) Copyright 2014, Pearson Education, Inc. Do not redistribute.

Eq. 3.34 ∆H m = kT χ12 n1φ2 Eq. 3.77

∆H m = V (δ1 − δ 2 ) φ1φ2 Equating equations 3.34 and 3.77 and noting that k = R/NA and φ1 = ν 1 V V v V V 2 N 2 N 2 χ12 = (δ1 − δ 2 ) A φ1 = (δ1 − δ 2 ) A 1 = 1 (δ1 − δ 2 ) RT n1 RT n1 V RT 2

CHAPTER 4 4.1 Show that σ = σ (1 + ε ) for an incompressible material. T

Solution F F F Lo + L − Lo F L ∆L 1+ σ T = = σ (1 + ε ) = = = A Ao Lo Ao Lo Ao Lo For an incompressible material, AL = Ao Lo or Ao = A ( L Lo ) Substituting this expression for Ao gives

σT =

F L F Lo = Ao Lo A L

L F = Lo A

4.2 A tensile strip of polystyrene that is 10 cm in length, 5 cm in width, and 2 cm in thickness is stretched to a length of 10.5 cm. Assuming that the sample is isotropic and deforms uniformly, calculate the resulting width and the % volume change after deformation. Solution Vo = 10 × 5 × 2 = 100 cm3 eq. 4.45, v = − ε T ε L

ε L = ln ( L Lo ) = ln (10.5 10 ) = 0.04879 Table 4.13, v = 0.35 for PS Assuming isotropic,

ε T = ln (W Wo ) = ln (T To ) = −vε L = −0.01708 W = 5 ( 0.9831) = 4.92 cm T = 2 ( 0.9831) = 1.97 cm

24 This text is associated with Fried/Polymer Science and Technology, Third Edition (9780137039555) Copyright 2014, Pearson Education, Inc. Do not redistribute.

V = 10.5 × 4.92 × 1.97 = 101.77 cm3 ∆V = 101.77 − 100 = 1.77 cm3 or 1.77%

4.3 A polymer has a crystalline growth parameter (n) of 2 and a rate constant (k) of 10-2 s-2 at 100°C. The polymer is melted and then quenched to 100°C and allowed to crystallize isothermally. After 10 s, what is the percent crystallinity of the sample? Solution φ = 1 − exp ( −kt n )

φ = 1 − exp −102 (10 )

2

= 1 − exp ( −1) = 1 − 0.3679 = 0.6321 or 63.2% crystalline

4.4 What is the % volume change that is expected at 100% elongation of natural rubber, assuming that no crystallization occurs during deformation? Solution ε = ln ( L Lo ) = ln 2 = 0.6931 ∆v = (1 − 2v ) ε Vo where v = 0.49 (Table 4.13)

∆V = (1 − 0.98 )( 0.6931) = 0.0139 or 1.39% Vo

4.5 Give your best estimate for the weight fraction of plasticizer required to lower the Tg of poly(vinyl chloride) (PVC) to 30°C. Assume that the Tg of PVC is 356 K and that of the plasticizer is 188 K. No other information is available. Solution Eq. 4.33 (1 − W1 ) ln ( 356 188) 303 ln = 188 W1 ( 356 188 ) + 1 − W1 solve for W1 gives W1 = 0.151 or 15.1 wt% 4.6 Show that the inverse rule of mixtures given by eq.4.34 can be obtained from the generalized relationship given by eq.4.32 when Tg,1 ≈ Tg,2 Solution Substituting ∆Cp,1 = constant Tg ,1 and ∆Cp,2 = constant Tg ,2 into eq. 4.32 and multiplying numerator and denominator by Tg,2 gives eq. 4.33. Next, let ln (1 + x ) ≈ x where x = (Tg Tg,1 ) − 1 Substitute into eq. 4.33 gives

25 This text is associated with Fried/Polymer Science and Technology, Third Edition (9780137039555) Copyright 2014, Pearson Education, Inc. Do not redistribute.

Tg Tg,1 Tg =

W2 (Tg,2 − Tg,1 ) + W1Tg,2 + W2Tg,1 W1 (Tg,2 Tg,1 ) + W2

=

W2 −1 =

Tg,2 Tg,1

−1

W1 (Tg,2 Tg,1 ) + W2

W2Tg,2 + W1Tg,2

W1 (Tg,2 Tg,1 ) + W2

=

(W1 + W2 ) Tg,2

W1 (Tg,2 Tg,1 ) + W2

=

Tg,2

W1 (Tg,2 Tg,1 ) + W2

1 W1 W2 = + Tg Tg,1 Tg,2

4.7 Polytetrafluoroethylene has been reported to exhibit a negative Poisson ratio. Explain why this polymer exhibits this unusual behavior. Teflon thickens upon elongation due to a rotation of crystals perpendicular to the draw direction. Solution See B. W. Ludwig and M. W. Urban, Polymer 35, 5130 (1994). 4.8 A sample of poly(ethylene terephthalate) is reported to be 20% crystalline. (a) What is the expected density of this sample? Solution Using eq. 4.6 and densities give in Table 4-5

ρ = φ ( ρc − ρ a ) + ρ a = 0.2 (1.396 − 1.280 ) + 1.280 = 1.303 g cm -3 (b) What is the expected specific heat increment of this semicrystalline sample? Solution From eq. 4.26 ∆Cp = (1 − φ ) ( ∆Cp ) = 0.8 ( ∆Cp ) am am Simha–Boyer rule: ( ∆Cp )

am

≈ 115 J g -1 Tg

From Table 4-3, Tg = 342 K = 0.336 J g -1 K -1 ( ∆Cp )am = 115 342 ∆Cp = ( 0.8 ) 0.336 = 0.269 J g -1 K -1 = 0.269 ( 0.2387 ) = 0.06416 cal g -1 K -1 (d) What is the expected heat of fusion of this sample? Solution From eq. 4.25 and Table 4-4 ∆Q = φ∆H f = 0.2 ( 6.431) kcal mol-1 = 1.286 kcal mol-1 =1.286 ( 0.2387 ) =0.307 kJ mol-1 4.9 Twenty wt % of a styrene oligomer having a number-average degree of polymerization of 7 is mixed with a commercial polystyrene sample having a number-average molecular weight of 100,000.

26 This text is associated with Fried/Polymer Science and Technology, Third Edition (9780137039555) Copyright 2014, Pearson Education, Inc. Do not redistribute.

(a) What is the Tg (K) of the styrene oligomer? Solution Use of the Fox-Flory relation, eq 4.27, and parameters given in Table 4-11 M n = 7 × 104 = 728

1.2 × 105 = 373 − 165 = 208 K 728

Tg,1 = 373 −

(b) What is the Tg (K) of the polystyrene mixture. Solution 1.2 × 105 = 372 K 105 Use eq. 4.34 as best approximation since Tgs are far apart Tg,2 = 373 −

ln

Tg

=

165

0.8ln ( 372 165 )

0.2 ( 372 165 ) + 0.8

= 0.5199

Tg = 1.6819 (165 ) = 277.5 K 4.10

The 1% secant modulus of a polystyrene sample is 3 GPa.

(a) What is the nominal stress (MPa) of this sample at a nominal strain of 0.01? Solution

σ = Eε = 0.01(3 GPa) == 0.03 GPa=30 MPa (b) What is the true stress (MPa) of this sample at a nominal strain of 0.01? Solution

σ T = σ (1 + ε ) = 30(1.01) = 30.3 MPa (c) What is the percent change in volume of this sample at the nominal strain of 0.01? Solution From eq. 4.44 and Table 4-13,

∆V = (1 − 2v ) ε T = (1 − 0.70 ) ε T Vo

From eq. 4.43 ε T = ln (1.01 1) = 0.00995

∆V = 0.3 ( 0.009950 ) = 0.002985 or 0.299% Vo

If the Young’s modulus of a sample of polystyrene is determined to be 3 GPa at room temperature, calculate its shear modulus. Solution From eq. 4.56: G = E 3 = 1 GPa

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4.12 Isotactic poly(methyl methacrylate) has a much lower Tg than the corresponding syndiotactic polymer. How can isotactic and syndioactic PMMA be polymerized? Explain why the isotactic polymer has the lower Tg Solution Syndiotactic PMMA can be obtained by free-radical polymerization at low temperatures (T. G. Fox et al., JACS 80, 1768 (1958); F. A. Bovey, JPS 46, 69 (1960)). Both syndiotactic and isotactic polymers can be obtained from anionic polymerizations. Recent molecular modeling (A. Soldera, Polymer 43, 4269 (2002)) indicate that differences in both intermolecular interactions and bending angle energy along the backbone can be correlated with the higher Tg of the syndiotactic polymer. CHAPTER 5 5.1 Show that E * = σ o ε o and D * = 1 E * . Solution

( E ')

E* = E'=

2

+ ( E ")

2

σo σo ; = E " sin δ cos δ εo εo

Therefore E * =

σo εo

Similarly D * ×D* = D *

2

12

( sin

2

δ + cos 2 δ )

=

σo εo

=

εo 1 = o σ E*

2

where D* = D '− iD " and D* = D '+ iD " D' =

εo εo and cos D " sin δ = δ σo σo

therefore D * =

εo σo

2

12

( sin

2

δ + cos δ ) 2

5.2 Show that the work per cycle per unit volume performed during dynamic tensile oscillation of a o o viscoelastic solid may be given as πσ ε sin δ (eq. (5.30)). Solution 2π

W = ∫ σ *d ε * 0

σ * = σ o sin (ωt + δ )

ε * = ε o sin ωt and d ε * = ε o cos ωtd (ωt )

28 This text is associated with Fried/Polymer Science and Technology, Third Edition (9780137039555) Copyright 2014, Pearson Education, Inc. Do not redistribute.

W = σ oε o ∫

2π

ωt = 0

sin (ωt + δ ) cos ωtd (ωt )

Since sin (ωt + δ ) = sin ωt cos δ + cos ωt sin δ , write W

σ ε o

o

2π

2π

0

0

= cos δ ∫ sin ωt cos ωtd (ωt ) + sin δ ∫ cos 2ωtd (ωt )

Note ∫ sin ω x cos ω xdx =

∫

Therefore

2π

0

1 1 sin 2 ω x and sin 2 θ = − cos 2θ 2 2 2ω

sin ωt cos ωtd (ωt ) = ω

sin 2 ωt 2ω

2π

= 0

1 1 1 1 1 − ( cos 4π ) − + ( cos 0 ) = 0 2 2 2 2 2

This is the elastic response (no work expanded). Next,

W

σ ε o

2π

o

= sin δ ∫ cos 2ωtd (ωt ) 0

Note ∫ cos 2udu = Then

W

σ ε o

o

u sin 2u + 2 4

ωt

= sin δ

2

+

sin 2ωt 4

2π

= sin δ π +

sin ( 4π )

0

4

−0−

sin 0 = π sin δ and 4

W = πσ oε o sin δ

5.3 Given the expression G (t ) = Go exp ( −t τ ) + G1 , show that the compliance function, J(t), can be written in the form J (t ) = A − B exp ( −C t τ ) , where A, B, and C are constants. Solution L J (t ) L G (t ) =

1 p2

L G (t ) =

Go G + 1 p +1 τ p

L J (t ) =

1 p +1 τ = p L G (t ) p ( Go + G1 ) p + ( G1 τ ) 2

Rearrange in form

p+a bp ( p + c )

where

29 This text is associated with Fried/Polymer Science and Technology, Third Edition (9780137039555) Copyright 2014, Pearson Education, Inc. Do not redistribute.

a=

1

τ

; b = Go + G1 ; c = G1 a b = G1 τ ( Go + G1 )

Then J ( t ) = L−1

p+a bp ( p + c )

Use partial fractions, to find inverse p ( A + bB ) + Ac p+a A B p+a = + = = bp ( p + c ) bp p + c bp ( p + c ) bp ( p + c )

Then A + bB = 1 a = Ac or A = a c =

B=

1− A c − a = b cb

J ( t ) = L−1

B=

Go + G1 G1

G1 1 B A t A + L−1 = + B exp ( −ct ) = + B exp − bp p+c b G1 Go + G1 τ

−Go 1− A = b G1 + ( Go + G1 )

J (t ) =

Go 1 − G1 G1 ( Go + G1 )

5.4 For a Maxwell model, show the following: (a) The equation for complex modulus E* (eq. (5.57)) can be obtained from the Fourier transform of the stress-relaxation modulus, Er (eq. (5.49)). (b) A maximum in the loss modulus plotted as a function of frequency occurs at ω = 1 / τ . Solutions (a) eq. 5.49: Er = E exp −

t

τ

∞

∞

0

0

E ′ = ω ∫ sin (ω s )E ( s ) ds = ω E ∫ exp ( − s τ ) sin (ω s ) ds

Fourier transform f ( x ) = exp ( −bx ) ; FS =

α α + b2 2

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(ωτ ) E′ = E 2 (ωτ ) + 1 2

Therefore

∞

Similarly E ′′ = ω E ∫ exp ( − s τ ) cos (ω s )ds 0

Fourier transform f ( x ) = exp ( −bx ) ; Fc (α ) =

b α + b2 2

Get eq. 5.59: E ′′ = ω E

1τ

ω + (1 τ ) 2

2

=E

ωτ 2 (ωτ ) + 1

2 2 2 dE ′′ (ω τ + 1) Eτ − Eτω 2ω wτ = =0 2 dω (ω 2τ 2 + 1)

(b)

(ω τ

2 2

+ 1) Eτ − Eτω 2ω wτ 2 = 0

or ω 2τ 2 = 1 and ω = 1 τ at maximum 5.5 Given the four-element model illustrated below, derive an analytical solution for the strain behavior and sketch ε ( t ) versus time under the following stress conditions: t