Question 15.16

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

Tensile Strength MPa	Num. Av. Mol. Wt.(gm/mol)
85	12,700
150	28,500

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

Answer 15.16

This requires Equation 15.3,

TS = TS_infinity - A/M_N

We use the given data to find TS_infinity and A. Then we put TS = 195 MPa and solve for M_N.

There is one tricky bit to this question: the final step requires you to solve

M_N = A/(TS_infinity-TS)

and TS_infinity is quite close in value to TS. Thus an error of 1% in TS_infinity results in an error of 20% in M_N. So your figure for TS_infinity should be accurate to the fourth significant figure.

Question 15.22

Which of the following would you expect to be elastomers and which thermosetting polymers at room temperature? Justify each choice.

1. Linear and crystalline polyethylene
2. Phenol-formaldehyde
3. Heavily cross-linked polyisoprene having a glass-transition temperature of 50 C.
4. Lightly crosslinked poly isoprene having a glass-transition temperature of - 60 C.
5. Linear and partially amorphous polyvinyl chloride.

Answer 15.22

This is a potentially confusing question. For a start, each of the options listed is a single substance -- `linear and crystalline polyethylene' is one material, for example.

Secondly, `elastomers' and `thermosetting polymers' [`thermosets'] are not mutually exclusive options; in fact, most elastomers are thermosets.

Polymers can be divided into two mutually exclusive sets, thermoplasts and thermosets, according to whether they melt or decompose when heated. Thermoplasts tend to have straight or branched chains, possibly with van de Waals cross-links, while thermosets tend to have networked structures or primary-bond cross-links.

Being an elastomer is a property of a material at a given temperature; a single material may be an elastomer at one temperature and not at another. Elastomers always have an amorphous rather than crystalline structure, and their molecules must be cross-linked.

1. Linear crystalline polyethylene is not an elastomer, since it's crystalline, and is not a thermoset, since it lacks cross-links.
2. Phenol-formaldehyde (`bakelite') is a thermoset (p. 505) because of its network structure (p. 458). As we can see from Table 15.1, it is not stretchy, and is therefore not an elastomer.
3. This is a thermoset, (heavily vulcanised rubber) because of the extensive cross-linking. It is not an elastomer, since it is below the glass transition temperature.
4. This is again a thermoset (natural rubber); it is also an elastomer, since what could be more elastic than natural rubber?
5. PVC has no cross-links, and is therefore neither a thermoset nor an elastomer.

Question 15.35

During the winter months, the temperature in some parts of the Yukon may go as low as - 55 C. Of the elastomers natural polyisoprene, styrene-butadiene, acrylonitrile-butadiene, chloroprene and polysiloxane, which would be suitable for automobile tyres under these conditions? Why?

Answer 15.35

If we just look at the temperature ranges, natural polyisoprene, styrene-butadiene and polysiloxane all maintain their properties down to -55 C. But polysiloxane is not a strong material, and would probably not be suitable for tires. Styrene-butadiene is better, but might not have the weather-resistance.

Question 15.44

If you want to produce nylon 6, 6 by condensation polymerisation using hexamethylene diamine and adipic acid, what masses of these two components will be needed to yield 20 kg of completely linear nylon 6, 6?

Answer 15.44

We write down the molecular weights of the three substances and calculate their masses in the ratio of their molecular weights:

MW(adipic acid) = 146

MW(hexamethylene diamine) = 116.21

MW Hence mass of aidpic acid needed = 20 kg * 146/226 = 12.9 kg

and mass of hexamethylene diamine needed = 20 * 116/226 = 10.26 kg.

The most common thing to go wrong was calculating the MW of nylon; it's best to do this from the structural diagram on p. 773. When one molecule of each of the two reactants reacts to make one molecule of nylon, two molecules of water condense out.

Question 16.11

A continuous and aligned fiber-reinforced composite is to be produced consisting of 30 vol% aramid fibres and 70vol% of a polycarbonate matrix. Mechanical characteristics of these two materials are as follows:

	Modulus of Elasticity (GPa)	Tensile Strength (MPa)
Aramid Fibre	131	3,600
Polycarbonate	2.4	65

Also, the stress on the polycarbonate matrix when the aramid fibres fail is 45 MPa. For this composite, compute (a) the longitudinal tensile strength and (b) the longitudinal modulus of elasticity.

Answer 16.11

This is a simple application of Equation 16.17:

sigma_cl=sigma_m(1-V_f)+sigma_fV_f

= 45 * 0.7 + 3600*0.3 = 1111.5 MPa

And then of Equation 16.10a:

E_cl=E_fV_f+E_m(1-V_f)

=131*0.3+2.4*0.7

=41 MPa.

Question 16.18

A continuous and aligned fibrous reinforced composite having a cross-sectional area of 970 mm² is subjected to an external tensile load. If the stresses sustained by the fibre and matrix phases are 215 MPa and 5.38 MPa respectively, the force sustained by the fibre phase is 76,800 N and the total longitudinal composite strain is 1.56 * 10^-3, then determine:

1. the force sustained by the matrix phase
2. the modulus of elasticity of the composite material in the longitudinal direction
3. the moduli of elasticity for fibre and matrix phases

Answer 16.11

This question is noticeably longer than the others, so it's out of 4.

1. Knowing the force and the stress on the fibres, we can work out the cross-sectional area of the fibres:

   A_f=76.8*10³/(215*10⁶)=357 mm²

   So V_f=357/970=0.368

   So area of matrix phase = 970 - 215 = 615 mm²

   So force sustained by matrix = 5.38 * 615 N = 3,297 N

2. E for the composite is stress/strain, and stress is force/area:

   Stress = (76.8+3.297)/970 kN/mm²=82.6 MPa

   So E = 82.6/(1.56*10^-3)=52.9 GPa

3. E for fibre phase = 215/(1.56*10^-3)=137.8 GPa

   E for matrix phase =5.38/(1.56*10^-3)=3.449 GPa

Question 16.20

Suppose you want to produce an aligned carbon fibre-epoxy matrix composite having a longitudinal tensile strength of 500 MPa. Calculate the volume fraction of fibres necessary if

1. the average fibre diameter and length are 0.015 mm and 2.0 mm respectively; and
2. the fibre fracture strength is 4.0 GPa and
3. the fibre-matrix bond strength is 25 MPa and
4. the matrix stress at composite failure is 7.0 MPa.

Answer 16.20

For some reason the figures on the website aren't the same as those in the book. So there are two possible solutions:

In each case, we need to know whether the fibres are longer than the critical length , l_c. We find this from Equation 16.3:

l_c=sigma_fd/(2 tau_c)

which evaluates to 1.2 mm using the website figures, 0.8 mm using Callister's figures.

For the website, l = 2 mm, so l > l_c and we calculate the volume fraction via Equation 16.18:

sigma_cd=sigma_fV_f(1-l_c/2l)+sigma_m(1-V_f)

So 1500 = 4000V_f(1-1.2/(2*2))+7*(1-V_f)

From which V_f=0.1765

For Callister, l = 0.5mm, so l < l_c and we calculate the volume fraction via Equation 16.19:

sigma_cd=(l tau_c/d)V_f+sigma_m(1-V_f)

=25*0.5*0.001/(0.01*0.001)V_f+7(1-V_f)

From which V_f=0.3966.

Question 16.22

1. From the moduli of elasticity data in Table 16.2 for glass fibre-reinforced polycarbonate composites, determine the value of the fibre efficiency parameters for each of 20, 30 and 40 vol% fibres
2. Estimate the modulus of elasticity for 50 vol% glass fibres.

Answer 16.22

This involves solving Equation 16.20 for K:

E_cd=KE_fV_f+E_mV_m

Hence K = (E_cd-E_mV_m)/E_fV_f

Plugging in numbers, this gives us that

K(20%) = 0.277

K(30%) = 0.319

K(40%)=0.350

The second part of the question asked you to find E_cd for V_f = 50%.

Assuming K varies linearly with V_f is not the way to solve this; the results for the first part show that it doesn't vary linearly. You could either observe that E_cd does vary almost linearly with V_f to conclude that E_cd(50%) = 14.8 MPa, or you could curve-fit for K vs. V_f, using the above three points plus the fact that K = 1 when V_f=1.