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  106 and 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

And
)
+ E f Vf
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Ecl = Em(1  V f
19.7 GPa = Em (1  0.25) + E f (0.25)

Ect =

(1
3.66 GPa =
EmE f
 V f ) E f V f Em
Em E f
(1  0.25)E f  0.25Em
Solving these two expressions simultaneously for Em and Ef leads to

Em = 2.79 GPa (4.04  105 psi)
E f = 70.4 GPa (10.2  106 psi)
Th


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16.D3 It is desired to produce a continuous and oriented carbon fiber-reinforced epoxy having a modulus of
elasticity of at least 83 GPa (12  106 psi) in the direction of fiber alignment. The maximum permissible specific
gravity is 1.40. Given the following data, is such a composite possible? Why or why not? Assume that composite
specific gravity may be determined using a relationship similar to Equation 16.10a.
Modulus of Elasticity
[GPa (psi)]
Carbon fiber
1.80
260 (37 × 106)
Epoxy
1.25
2.4 (3.5 × 105)
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SpecificGravity
Solution
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  V f ) + E f V f
83 GPa = (2.4 GPa)(1  V f ) + (260 GPa)(V f )

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 (or density), , we employ the following modified form of
Th
Equation 16.10b
c = m(1  V f

)
+  f Vf
1.40 = 1.25(1  V f ) + 1.80(V f )
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.
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18.2 A copper wire 100 m long must experience a voltage drop of less than 1.5 V when a current of 2.5 A passes
through it. Using the data in Table 18.1, compute the minimum diameter of the wire.
Solution
For this problem, given that a copper wire 100 m long must experience a voltage drop of less than 1.5 V
when a current of 2.5 A passes through it, we are to compute the minimum diameter of the wire. Combining
Equations 18.3 and 18.4 and solving for the cross-sectional area A leads to
Il
Il
=
V
V
A=
2
then
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d 
-1
From Table 18.1, for copper  = 6.0  107 (-m) . Furthermore, inasmuch as A =    for a cylindrical wire,
2 
d 2
Il
   =
2 
V
or

d=

4 Il
V
When values for the several parameters given in the problem statement are incorporated into this expression, we get

d =
(4)(2.5 A)(100 m)

()(1.5 V) 6.0  10 7 (  m)1

-3
Th
= 1.88  10 m = 1.88 mm

18.10 (a) Calculate the drift velocity of electrons in germanium at room temperature and when the magnitude of the
electric field is 1000 V/m. (b) Under these circumstances, how long does it take an electron to traverse a 25-mm (1in.) length of crystal?
Solution
(a) The drift velocity of electrons in Ge may be determined using Equation 18.7.
Since the room
temperature mobility of electrons is 0.38 m2/V-s (Table 18.3), and the electric field is 1000 V/m (as stipulated in the
problem statement),
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vd = e E
= (0.38 m2/V- s)(1000 V/m) = 380 m/s

(b) The time, t, required to traverse a given length, l (= 25 mm), is just

t =
l
25  103 m
=
= 6.6  10-5 s
vd
380 m/s
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18.11 At room temperature the electrical conductivity and the electron mobility for copper are 6.0  107 (-m)-1

and 0.0030 m2/V-s, respectively. (a) Compute the number of free electrons per cubic meter for copper at room
temperature. (b) What is the number of free electrons per copper atom? Assume a density of 8.9 g/cm 3.
Solution
(a) The number of free electrons per cubic meter for copper at room temperature may be computed using
Equation 18.8 as
n=

 e  e
6.0  10 7 (  m)1
= 
(1.602  1019 C)(0.003 m2 /V- s)
= 1.25  1029 m-3

(b) In order to calculate the number of free electrons per copper atom, we must first determine the number
of copper atoms per cubic meter, NCu. From Equation 4.2 (and using the atomic weight value for Cu found inside
Th
the front cover—viz. 63.55 g/mol)
N Cu =
=
(6.022

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N A 
ACu
 10 23 atoms / mol)(8.9 g/cm3)(10 6 cm 3 / m3)

63.55 g/mol
= 8.43  1028 m-3
(Note: in the above expression, density is represented by ' in order to avoid confusion with resistivity which is
designated by .) And, finally, the number of free electrons per aluminum atom is just n/NCu
n
N Cul
=
1.25  10 29 m3
= 1.48
8.43  10 28 m3
18.15 Determine the electrical conductivity of a Cu-Ni alloy that has a yield strength of 125 MPa (18,000 psi). You

will find Figure 7.16 helpful.
Solution
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We are asked to determine the electrical conductivity of a Cu-Ni alloy that has a yield strength of 125 MPa.
From Figure 7.16b, the composition of an alloy having this tensile strength is about 20 wt% Ni.
For this
composition, the resistivity is about 27  10-8 -m (Figure 18.9). And since the conductivity is the reciprocal of the
resistivity, Equation 18.4, we have
=
1
1
=
= 3.70  10 6 ( - m) -1
 27  108   m
18.29 (a) The room-temperature electrical conductivity of a silicon specimen is 5.93  10–3 (Ω-m)–1. The hole

concentration is known to be 7.0  1017 m–3. Using the electron and hole mobilities for silicon in Table 18.3,
compute the electron concentration. (b) On the basis of the result in part (a), is the specimen intrinsic, n-type
extrinsic, or p-type extrinsic? Why?
Solution
(a) In this problem, for a Si specimen, we are given values for p (7.0  1017 m-3) and [-3 (m)-1], while values for h and e (0.05 and 0.14 m2/V-s, respectively) are found in Table 18.3. In order to solve for
Th
n we must use Equation 18.13, which, after rearrangement, leads to
=
n=
  p| e |  h
| e | e
5.93  103 (  m)1  (7.0  1017 m3)(1.602  1019 C)(0.05 m2 / V - s)
 (1.602  1019 C)(0.14 m2 / V - s)
= 1.44  1016 m-3

(b) This material is p-type extrinsic since p (7.0  1017 m-3) is greater than n (1.44  1016 m-3).
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18.33
At temperatures near room temperature, the temperature dependence of the conductivity for intrinsic
germanium is found to equal
 Eg 
  CT 3/ 2 exp

 2kT 
(18.36)
where C is a temperature-independent constant and T is in Kelvins. Using Equation 18.36, calculate the intrinsic

electrical conductivity of germanium at 150°C.
Solution
It first becomes necessary to solve for C in Equation 18.36 using the room-temperature (298 K)
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conductivity [2.2 (-m)-1] (Table 18.3). This is accomplished by taking natural logarithms of both sides of
Equation 18.36 as
ln  = ln C 
Eg
3
lnT 
2
2 kT
and after rearranging and substitution of values for Eg (0.67 eV, Table 18.3), and the room-temperature

conductivity, we get
Eg
3
ln C = ln  + lnT +
2
2 kT
3
0.67 eV
= ln (2.2) + ln (298) +

2
(2)(8.62  105 eV / K)(298 K)
= 22.38

Th
Now, again using Equation 18.36, we are able to compute the conductivity at 423 K (150C)
ln  = ln C 
Eg
3
ln T 
2
2 kT
3
0.67 eV
= 22.38  ln (423 K) 

2
(2)(8.62  105 eV / K)(423 K)

= 4.12
which leads to
 = e4.12 = 61.6 (-m)-1.
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