Solutions of practice problems_A6.docx

8/10/2019 Solutions of practice problems_A6.docx

1/13

14.5 Below, molecular weight data for a polypropylene material are tabulated. Compute (a) the number-average

molecular weight, (b) the weight-average molecular weight, and (c) the degree of polymerization.

Molecular Weight

Range (g/mol ) xi wi

8,00016,000 0.05 0.02

16,00024,000 0.16 0.10

24,00032,000 0.24 0.20

32,00040,000 0.28 0.30

40,00048,000 0.20 0.27

48,00056,000 0.07 0.11

Solution

(a) From the tabulated data, we are asked to compute Mn, the number-average molecular weight. This is

carried out below.

Molecular wt

Range MeanMi xi xiMi

8,000-16,000 12,000 0.05 600

16,000-24,000 20,000 0.16 3200

24,000-32,000 28,000 0.24 6720

32,000-40,000 36,000 0.28 10,080

40,000-48,000 44,000 0.20 8800

48,000-56,000 52,000 0.07 3640

____________________________

Mn = xiMi = 33,040 g/mol

(b) From the tabulated data, we are asked to compute Mw , the weight-average molecular weight.

Molecular wt.

Range MeanMi w

i w

iM

i

8,000-16,000 12,000 0.02 240

16,000-24,000 20,000 0.10 2000

24,000-32,000 28,000 0.20 5600

32,000-40,000 36,000 0.30 10,800


2/13

40,000-48,000 44,000 0.27 11,880

48,000-56,000 52,000 0.11 5720

___________________________

Mw = wiMi = 36,240 g/mol

(c) Now we are asked to compute the degree of polymerization, which is possible using Equation 14.6.

For polypropylene, the repeat unit molecular weight is just

m= 3(AC) + 6(AH)

= (3)(12.01 g/mol) + (6)(1.008 g/mol) = 42.08 g/mol

AndDP=

Mn

m=

33,040 g/mol

42.08 g/mol= 785


3/13

14.7 Is it possible to have a poly(methyl methacrylate) homopolymer with the following molecular weight data and

a of polymerization of 527? Why or why not?

Molecular Weight

Range (g/mol ) wi xi

8,000

20,000 0.02 0.05

20,00032,000 0.08 0.15

32,00044,000 0.17 0.21

44,00056,000 0.29 0.28

56,00068,000 0.23 0.18

68,00080,000 0.16 0.10

80,00092,000 0.05 0.03

Solution

This problem asks if it is possible to have a poly(methyl methacrylate) homopolymer with the given

molecular weight data and a degree of polymerization of 527. The appropriate data are given below along with a

computation of the number-average molecular weight.

Molecular wt.

Range MeanMi xi xiMi

8,000-20,000 14,000 0.05 700

20,000-32,000 26,000 0.15 3900

32,000-44,000 38,000 0.21 7980

44,000-56,000 50,000 0.28 14,000

56,000-68,000 62,000 0.18 11,160

68,000-80,000 74,000 0.10 7400

80,000-92,000 86,000 0.03 2580

_________________________

Mn = xiMi = 47,720 g/mol

For PMMA, from Table 14.3, each repeat unit has five carbons, eight hydrogens, and two oxygens. Thus,

m= 5(AC) + 8(AH) + 2(AO)


4/13

= (5)(12.01 g/mol) + (8)(1.008 g/mol) + (2)(16.00 g/mol) = 100.11 g/mol

Now, we will compute the degree of polymerization using Equation 14.6 as

DP =Mn

m =

47,720 g/mol

100.11 g/mol = 477

Thus, such a homopolymer isnot possiblesince the calculated degree of polymerization is 477 (and not 527).


5/13

14.25 The density and associated percent crystallinity for two polytetrafluoroethylene materials are as follows:

(g/cm3) crystallinity (%)

2.144 51.3

2.215 74.2

(a) Compute the densities of totally crystalline and totally amorphous polytetrafluoroethylene.

(b)Determine the percent crystallinity of a specimen having a density of 2.26 g/cm3.

Solution

(a) We are asked to compute the densities of totally crystalline and totally amorphous

polytetrafluoroethylene (cand afrom Equation 14.8). From Equation 14.8 let C=% crystallinity

100, such that

C =c (s a)

s

(c

a)

Rearrangement of this expression leads to

c (Cs s) + ca Cs a = 0

in which cand aare the variables for which solutions are to be found. Since two values of sand Care specified

in the problem statement, two equations may be constructed as follows:

c (C1s1 s1) + ca C1s1a = 0

c (C2 s2 s2) + ca C2 s2 a = 0

In which s1= 2.144 g/cm3, s2= 2.215 g/cm

3,C1= 0.513, and C2= 0.742. Solving the above two equations for

aand cleads to

a =s1s2 (C1 C2)

C1s1 C2 s2

=(2.144 g/cm3)(2.215 g/cm3)(0.513 0.742)

(0.513)(2.144 g/cm3) (0.742)(2.215 g/cm3) = 2.000 g/cm3


6/13

And

c =

s1

s2

(C2 C1)

s2

(C2 1) s1 (C1 1)

=(2.144 g/cm3)(2.215 g/cm3)(0.742 0.513)

(2.215 g/cm3)(0.742 1) (2.144 g/cm3)(0.513 1) = 2.301 g/cm3

(b) Now we are to determine the %crystallinity for s= 2.26 g/cm3. Again, using Equation 14.8

% crystallinity =c (s a)s (c a)

100

= (2.301 g/cm3

)(2.260 g/cm3 2.000 g/cm3

)(2.260 g/cm3)(2.301 g/cm3 2.000 g/cm3) 100

= 87.9%


7/13

14.28 Argon diffuses through a high density polyethylene (HDPE) sheet 40 mm thick at a rate of 4.0 107 (cm3

STP)/cm2-s at 325 K. The pressures of argon at the two faces are 5000 kPa and 1500 kPa, which are maintained

constant. Assuming conditions of steady state, what is the permeability coefficient at 325 K?

Solution

This problem asks us to compute the permeability coefficient for argon through high density polyethylene

at 325 K given a steady-state permeability situation. It is necessary for us to Equation 14.9 in order to solve this

problem. Rearranging this expression and solving for the permeability coefficient gives

PMJxP

Jx

P2P1

TakingP1= 1500 kPa (1,500,000 Pa) and P2= 5000 kPa (5,000,000 Pa), the permeability coefficient of Ar through

HDPE is equal to

PM

4.0 10-7(cm3STP)

cm2 - s

(4 cm)

(5,000,000 Pa - 1,500,000 Pa)

4.57 10-13(cm3STP)(cm)

cm2 - s - P a


8/13

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

as follows:

Tensile Strength

(MPa)

Number-Average

Molecular Weight (g/mol)

107 40,000

170 60,000

Estimate the tensile strength at a number-average molecular weight of 30,000 g/mol.

Solution

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

poly(methyl methacrylate) materials and then asks that we estimate the tensile strength for Mn = 30,000 g/mol.

Equation 15.3 cites the dependence of the tensile strength on Mn. 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

andA. These equations are as follows:

107 MPa = TSA

40,000 g/mol

170 MPa = TSA

60,000 g/mol

Thus, the values of the two constants are: TS

= 296 MPa and A = 7.56 106MPa-g/mol. Substituting these

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

TS= TSA

30,000 g/mol

= 296 MPa 7.56 106 MPa- g/ mol

30,000 g/mol

= 44 MPa


9/13

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

follows:

Tensile Strength

(MPa)

Number-Average

Molecular Weight (g/mol)85 12,700

150 28,500

Estimate the number-average molecular weight that is required to give a tensile strength of 195 MPa.

Solution

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

polyethylene materials and then asks that we estimate the Mn that is required for a tensile strength of 195 MPa.

Equation 15.3 cites the dependence of the tensile strength on Mn. 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

andA. These equations are as follows:

85 MPa = TSA

12,700 g/mol

150 MPa = TSA

28,500 g/mol

Thus, the values of the two constants are: TS

= 202 MPa and A= 1.489 106MPa-g/mol. Solving for Mnin

Equation 15.3 and substituting TS= 195 MPa as well as the above values for TS

andAleads to

Mn =A

TSTS

=1.489 106 MPa- g/ mol

202 MPa 195 MPa= 213,000 g/mol


10/13

16.10 For a continuous and oriented fiber-reinforced composite, the moduli of elasticity in the longitudinal and

transverse directions are 19.7 and 3.66 GPa (2.8 106and 5.3 105 psi), respectively. If the volume fraction of

fibers is 0.25, determine the moduli of elasticity of fiber and matrix phases.

Solution

This problem asks for us to compute the elastic moduli of fiber and matrix phases for a continuous and

oriented fiber-reinforced composite. We can write expressions for the longitudinal and transverse elastic moduli

using Equations 16.10b and 16.16, as

Ecl = Em(1 Vf) + EfVf

19.7 GPa = Em(1 0.25) +Ef(0.25)

And

Ect =EmEf

(1 Vf)EfVfEm

3.66 GPa =EmEf

(1 0.25)Ef 0.25Em

Solving these two expressions simultaneously forEmandEfleads to

Em = 2.79 GPa (4.04 105 psi)

Ef = 70.4 GPa (10.2 106 psi)


11/13

16.14 A continuous and aligned fiber-reinforced composite having a cross-sectional area of 1130 mm2(1.75 in.2) is

subjected to an external tensile load. If the stresses sustained by the fiber and matrix phases are 156 MPa (22,600

psi) and 2.75 MPa (400 psi), respectively, the force sustained by the fiber phase is 74,000 N (16,600 lbf) and the

total longitudinal strain is 1.25 10-3, determine

(a) the force sustained by the matrix phase

(b) the modulus of elasticity of the composite material in the longitudinal direction, and

(c) the moduli of elasticity for fiber and matrix phases.

Solution

(a) For this portion of the problem we are asked to calculate the force sustained by the matrix phase. It is

first necessary to compute the volume fraction of the matrix phase, Vm. This may be accomplished by first

determining Vfand then Vmfrom Vm= 1 Vf. The value of Vfmay be calculated since, from the definition of stress

(Equation 6.1), and realizing Vf=Af/Acas

f =Ff

Af=

Ff

VfAc

Or, solving for Vf

Vf =Ff

fAc=

74,000 N

(156 106 N /m2)(1130 mm2)(1 m/1000 mm)2= 0.420

Also

Vm = 1 Vf = 1 0.420 = 0.580

And, an expression for manalogous to the one for fabove is

m =

Fm

Am

=Fm

VmAc

From which

Fm = VmmAc = (0.580)(2.75 106 N/m2)(1.13 10-3m2) = 1802 N (406 lbf)


12/13


13/13

16.15 Compute the longitudinal strength of an aligned carbon fiber-epoxy matrix composite having a 0.25

volume fraction of fibers, assuming the following: (1) an average fiber diameter of 10 10-3mm (3.94 10 -4 in.),

(2) an average fiber length of 5 mm (0.20 in.), (3) a fiber fracture strength of 2.5 GPa (3.625 105psi), (4) a fiber-

matrix bond strength of 80 MPa (11,600 psi), (5) a matrix stress at fiber failure of 10.0 MPa (1450 psi), and (6) a

matrix tensile strength of 75 MPa (11,000 psi).

Solution

It is first necessary to compute the value of the critical fiber length using Equation 16.3. If the fiber length

is much greater than lc, then we may determine the longitudinal strength using Equation 16.17, otherwise, use of

either Equation 16.18 or Equation 16.19 is necessary. Thus, from Equation 16.3

lc =f

d

2 c

=(2.5 103 MPa)(10 103 mm)

2 (80 MPa)

= 0.16 mm

Inasmuch as l>> lc(5.0 mm >> 0.16 mm), then use of Equation 16.17 is appropriate. Therefore,

cl = m

' (1 Vf) +fVf

= (10 MPa)(1 0.25) + (2.5 103MPa)(0.25)

= 633 MPa (91,700 psi)

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Solutions of practice problems_A6.docx