MSE 230 Solutions for Assignment 13 Fall 2014 1. 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)Ef + VfEm

3.66 GPa = EmEf

(1 - 0.25)Ef + 0.25Em

Solving these two expressions simultaneously for Em and Ef leads to

Em = 2.79 GPa Ef = 70.4 GPa

2. On the graph below, sketch a qualitatively correct plot of Elong/Etran versus volume fraction of metal for a layered metal/ceramic composite material (E is the modulus of the composite, and "long" and "tran" refer to loading in the longitudinal and transverse directions respectively). Hint: assume that the pure ceramic and pure metal are isotropic. Elong/Etran represents the degree of anisotropy in the Young’s modulus of the composite as a function of the volume fraction of metal and ceramic. For pure ceramic (0) and pure metal (1) the material is isotropic. As one adds layers of metal and increases their thickness, the magnitude of anisotropy increases until there is the same volume fraction of metal and ceramic, and then decreases again with increasing metal volume fraction.

3. This problem asks us to determine whether or not it is possible to produce a continuous and

oriented carbon fiber-reinforced epoxy having a modulus of elasticity of at least 83 GPa in the direction of fiber alignment, and a maximum specific gravity of 1.40. We will first calculate the minimum volume fraction of fibers to give the stipulated elastic modulus, and then the maximum volume fraction of fibers possible to yield the maximum permissible specific gravity; if there is an overlap of these two fiber volume fractions then such a composite is possible.

With regard to the elastic modulus, from Equation (16.10b)

Ecl = Em(1 - Vf) + EfVf

83 GPa = (2.4 GPa)(1 - Vf) + (260 GPa)(Vf)

Solving for Vf yields Vf = 0.31. Therefore, Vf > 0.31 to give the minimum desired

elastic modulus. Now, upon consideration of the specific gravity, ρ, we employ the following

relationship:

ρc = ρm(1 - Vf) + ρfVf

1.40 = 1.25(1 - Vf) + 1.80(Vf)

Longitudinal Tran sverse

Elong

Etran

0 1vol. frac. of metal

And, solving for Vf from this expression gives Vf = 0.27. Therefore, it is necessary for Vf < 0.27 in order to have a composite specific gravity less than

1.40. Hence, such a composite is not possible since there is no overlap of the fiber

volume fractions as computed using the two stipulated criteria. 4. We are asked to determine the electrical conductivity of a Cu-Ni alloy that has a yield strength of 125

MPa. From Figure 7.16(b), the composition of an alloy having this yield strength is about 20 wt%

Ni. For this composition, the resistivity is about 27 x 10-8 Ω-m (Figure 18.9). And since the

conductivity is the reciprocal of the resistivity, Equation (18.4), we have

σ = 1ρ =

127 x 10-8 Ω-m

= 3.70 x 106 (Ω-m)-1

5. The (a) and (b) portions of the problem ask that we make schematic plots on the same graph for the

electrical resistivity versus composition for lead-tin alloys at both room temperature and 150°C;

such a graph is shown below.

(c) Upon consultation of the Pb-Sn phase diagram (Figure 9.8) we note upon extrapolation of the

two solvus lines to at room temperature (e.g., 20°C), that the single phase α phase solid solution

exists between pure lead and a composition of about 2 wt% of Sn-98 wt% Pb. In addition, the

composition range over which the β phase is between approximately 99 wt% Sn-1 wt% Pb and

pure tin. Within both of these composition regions the resistivity increases in accordance with

Equation (18.11); also, in the above plot, the resistivity of pure Pb is represented (schematically)

as being greater than that for pure Sn, per the problem statement.

Furthermore, for compositions between these extremes, both α and β phases coexist, and

alloy resistivity will be a function of the resisitivities the individual phases and their volume

fractions, as described by Equation (18.12). Also, mass fractions of the α and β phases within the

two-phase region of Figure 9.8 change linearly with changing composition (according to the lever

rule). There is a reasonable disparity between the densities of Pb and Sn (11.35 g/cm3 versus 7.3

g/cm3). Thus, phase volume fractions will not exactly equal mass fractions, which means that the

resistivity will not exactly vary linearly with composition. In the above plot, the curve in this region

has been depicted as being linear for the sake of convenience.

At 150°C, the curve has the same general shape, and is shifted to significantly higher

resistivities inasmuch as resistivity increases with rising temperature [Equation (18.10) and Figure

18.8]. In addition, from Figure 9.8, at 150°C the solubility of Sn in Pb increases to approximately

10 wt% Sn--i.e., the α phase field is wider and the increase of resistivity due to the solid solution

effect extends over a greater composition range, which is also noted in the figure. The resistivity-

temperature behavior is similar on the tin-rich side, where, at 150°C, the β phase field extends to

approximately 2 wt% Pb (98 wt% Sn). And, as with the room temperature case, for compositions

within the α + β two-phase region, the plot is approximately linear, extending between resistivity

values found at the maximum solubilities of the two phases.

6. (a) This germanium material to which has been added 5 x 1022 m-3 Sb atoms is n-type since Sb is a

donor in Ge. (Antimony is from group VA of the periodic table--Ge is from group IVA.)

(b) Since this material is n-type extrinsic, Equation (18.16) is valid. Furthermore, each Sb will

donate a single electron, or the electron concentration is equal to the Sb concentration since all of

the Sb atoms are ionized at room temperature; that is n = 5 x 1022 m-3, and,

σ = n⎜e⎜µe

= (5 x 1022 m-3)(1.602 x 10-19 C)(0.1 m2/V-s)

= 800 (Ω-m)-1

7. This question asks that we compare and then explain the difference in the temperature dependence of

the electrical conductivity for metals and intrinsic semiconductors. For a pure metal, this

temperature dependence is just

σ = 1

ρo + aT

[This expression comes from Equations (18.4) and (18.10).] That is, the electrical conductivity

decreases with increasing temperature.

By way of contrast, for an intrinsic semiconductor

ln σ ≅ C - Eg2kT

Or, with rising temperature, the conductivity increases.

The temperature behavior for metals is best explained by consulting Equation (18.8)

σ = n⎜e⎜µe

As the temperature rises, n will remain virtually constant, whereas the mobility (µe) will decrease,

because the thermal scattering of free electrons will become more efficient. Since |e| is

independent of temperature, the net result will be diminishment in the magnitude of σ .

For an intrinsic semiconductor, Equation (18.15) describes the conductivity; i.e.,

σ = n⎜e⎜(µe + µh) = p⎜e⎜(µe + µh)

Both n and p will increase with rising temperature, rather dramatically, since more thermal energy becomes available for valence band-conduction band electron excitations. The magnitudes of µe

and µh will diminish somewhat with increasing temperature (because of the thermal scattering of

electrons and holes), which effect will be overwhelmed by the increase in n and p. The net result

is that σ increases with temperature.

8. Increasing the dopant concentration increases the charge carrier density (n). Therefore, the electron concentration will plateau at a higher level.

9. Please review section 18.15 in Callister.