CHAPTER 6
MECHANICAL PROPERTIES OF METALS
PROBLEM SOLUTIONS
Concepts of Stress and Strain
6.1 (a) Equations 6.4a and 6.4b are expressions for normal ( σ ′ ) and shear ( τ ′ ) stresses, respectively, as
a function of the applied tensile stress (s) and the inclination angle of the plane on which these stresses are taken (θ
of Figure 6.4). Make a plot showing the orientation parameters of these expressions (i.e., cos2 θ and sin θ cos θ)
versus θ.
(b) From this plot, at what angle of inclination is the normal stress a maximum?
(c) At what inclination angle is the shear stress a maximum?
Solution
(a) Below are plotted curves of cos2q (for s¢) and sinq cosq (for t¢) versus q.
(b) The maximum normal stress occurs at an inclination angle of 0°.
(c) The maximum shear stress occurs at an inclination angle of 45°.
Stress–Strain Behavior
6.2 A specimen of copper having a rectangular cross section 15.2 mm × 19.1 mm is pulled in tension with
44,500 N force, producing only elastic deformation. Calculate the resulting strain.
Solution
This problem calls for us to calculate the elastic strain that results for a copper specimen stressed in tension.
The cross-sectional area is just (15.2 mm) ´ (19.1 mm) = 290 mm2; also, the elastic modulus for Cu is given in
Table 6.1 as 110 GPa. Combining Equations 6.1 and 6.5 and solving for the strain yields
F
A
σ
F
ε= = 0 =
E
E A0 E
=
(
44,500 N
= 1.39 × 10−3
)(110 × 109 N/m2 )
2.90 × 10−4 m2
6.3 An aluminum bar 125 mm long and having a square cross section 16.5 mm.) on an edge is pulled in
tension with a load of 66,700 N and experiences an elongation of 0.43 mm. Assuming that the deformation is
entirely elastic, calculate the modulus of elasticity of the aluminum.
Solution
This problem asks us to compute the elastic modulus of aluminum. For a square cross-section, A0 = b02 ,
where b0 is the edge length. Combining Equations 6.1, 6.2, and 6.5 and solving for the modulus of elasticity, E,
leads to
F
A
Fl0
σ
E =
= 0=
Δl b 20 Δl
ε
l0
Substitution into this expression, values of l0, b0 Dl, and F given in the problem statement yields the following
E=
(66,700 N) (125 × 10−3 m)
(16.5 × 10−3 m)2 (0.43 × 10−3 m)
= 71.2 × 109 N/m2 = 71.2 GPa
6.4 Consider a cylindrical nickel wire 2.0 mm in diameter and 3 × 104 mm long. Calculate its elongation
when a load of 300 N is applied. Assume that the deformation is totally elastic.
Solution
In order to compute the elongation of the Ni wire when the 300 N load is applied we must employ
Equations 6.2, 6.5, and 6.1. Combining these equations and solving for ∆l leads to the following
Δl = l0ε = l0
σ l0 F
=
=
E EA0
l0 F
⎛d ⎞
Eπ ⎜ 0 ⎟
⎝ 2⎠
2
=
4l0 F
Eπ d02
since the cross-sectional area A0 of a cylinder of diameter d0 is equal to
⎛d ⎞
A0 = π ⎜ 0 ⎟
⎝ 2⎠
2
Incorporating into this expression values for l0, F, and d0 in the problem statement, and realizing that for Ni, E = 207
GPa (Table 6.1), and that l0 = 3 ´ 104 mm = 30 m, the wire elongation is
Δl =
(4)(30 m)(300 N)
(207 × 109 N/m2 )(π )(2 × 10−3 m)2
= 0.0138 m = 13.8 mm
6.5 A cylindrical rod of steel (E = 207 GPa) having a yield strength of 310 MPa is to be subjected to a
load of 11,100 N. If the length of the rod is 500 mm, what must be the diameter to allow an elongation of 0.38 mm?
Solution
This problem asks us to compute the diameter of a cylindrical specimen of steel in order to allow an
elongation of 0.38 mm. Employing Equations 6.1, 6.2, and 6.5, assuming that deformation is only elastic, we may
write the following expression:
σ =
F
=
A0
F
⎛ d2 ⎞
π⎜ 0⎟
⎝ 4 ⎠
= Eε = E
Δl
l0
And upon simplification
F
⎛ d2 ⎞
π 0
⎜ ⎟
⎝ 4 ⎠
=E
Δl
l0
Solving for the original cross-sectional area, d0, leads to
d0 =
4l0 F
π E Δl
Incorporation of values for l0, F, E, and Dl provided in the problem statement yields the following:
d0 =
(4) (500 × 10−3 m) (11,100 N)
(π ) (207 × 109 N/m2 )(0.38 × 10−3 m)
= 9.5 ´ 10-3 m = 9.5 mm
6.6 Consider a cylindrical specimen of a steel alloy (Figure 6.22) 8.5 mm in diameter and 80 mm long that
is pulled in tension. Determine its elongation when a load of 65,250 N is applied.
Solution
This problem asks that we calculate the elongation ∆l of a specimen of steel the stress-strain behavior of
which is shown in Figure 6.22. First it becomes necessary to compute the stress when a load of 65,250 N is applied
using Equation 6.1 as
σ =
F
=
A0
F
⎛d ⎞
π⎜ 0⎟
⎝ 2⎠
2
=
65,250 N
⎛ 8.5 × 10−3 m ⎞
π⎜
⎟
2
⎝
⎠
2
= 1.15 × 109 N/m2 = 1150 MPa
Referring to Figure 6.22, at this stress level we are in the elastic region (shown in the inset) on the stress-strain
curve, which corresponds to a strain of 0.0054. Now, utilization of Equation 6.2 to compute the value of Dl:
Δl = ε l0 = (0.0054)(80 mm) = 0.43 mm
6.7 In Section 2.6, it was noted that the net bonding energy EN between two isolated positive and negative
ions is a function of interionic distance r as follows:
EN = −
A B
+
r rn
(6.30)
where A, B, and n are constants for the particular ion pair. Equation 6.30 is also valid for the bonding energy
between adjacent ions in solid materials. The modulus of elasticity E is proportional to the slope of the interionic
force–separation curve at the equilibrium interionic separation; that is,
⎛ dF ⎞
E ∝⎜ ⎟
⎝ dr ⎠ r
0
Derive an expression for the dependence of the modulus of elasticity on these A, B, and n parameters (for the twoion system), using the following procedure:
1. Establish a relationship for the force F as a function of r, realizing that
F=
dEN
dr
2. Now take the derivative dF/dr.
3. Develop an expression for r0, the equilibrium separation. Because r0 corresponds to the value of r at the
minimum of the EN-versus-r curve (Figure 2.10b), take the derivative dEN/dr, set it equal to zero, and solve for r,
which corresponds to r0.
4. Finally, substitute this expression for r0 into the relationship obtained by taking dF/dr.
Solution
This problem asks that we derive an expression for the dependence of the modulus of elasticity, E, on the
parameters A, B, and n in Equation 6.30. It is first necessary to take dEN/dr in order to obtain an expression for the
force F; this is accomplished as follows:
⎛ A⎞
⎛ B⎞
d⎜− ⎟ d⎜ ⎟
dE
⎝ r⎠
⎝ n⎠
F= N =
+ r
dr
dr
dr
=
A
nB
−
r (1 + 1) r (n + 1)
The second step is to set this dEN/dr expression equal to zero and then solve for r (= r0). The algebra for this
procedure is carried out in Problem 2.10, with the result that
⎛ A⎞
r0 = ⎜ ⎟
⎝ nB ⎠
1/(1 − n)
Next it becomes necessary to take the derivative of the force (dF/dr), which is accomplished as follows:
⎛ A⎞
⎛ nB ⎞
d⎜ ⎟ d⎜−
⎝ r2 ⎠
⎝ r n + 1 ⎟⎠
dF
=
+
dr
dr
dr
= −
2A
r3
+
(n)(n + 1)B
r (n + 2)
Now, substitution of the above expression for r0 into this equation yields
⎛ dF ⎞
2A
(n)(n + 1)B
+
⎜⎝ dr ⎟⎠ = −
3/(1 − n)
(n + 2)/(1 − n)
⎛ A⎞
⎛ A⎞
r0
⎜⎝ nB ⎟⎠
⎜⎝ nB ⎟⎠
which is the expression to which the modulus of elasticity is proportional.
6.8 Using the solution to Problem 6.7, rank the magnitudes of the moduli of elasticity for the following
hypothetical X, Y, and Z materials from the greatest to the least. The appropriate A, B, and n parameters (Equation
6.30) for these three materials are tabulated below; they yield EN in units of electron volts and r in nanometers:
Material
A
B
n
X
1.5
7.0 × 10–6
8
Y
2.0
1.0 × 10–5
9
3.5
–6
7
Z
4.0 × 10
Solution
This problem asks that we rank the magnitudes of the moduli of elasticity of the three hypothetical metals
X, Y, and Z. From Problem 6.7, it was shown for materials in which the bonding energy is dependent on the
interatomic distance r according to Equation 6.30, that the modulus of elasticity E is proportional to
E∝−
2A
3/(1 − n)
⎛ A⎞
⎜⎝ nB ⎟⎠
+
(n)(n + 1)B
⎛ A⎞
⎜⎝ nB ⎟⎠
(n + 2)/(1 − n)
For metal X, A = 1.5, B = 7 ´ 10-6, and n = 8. Therefore,
E ∝−
(2)(1.5)
⎡
1.5
⎢
⎢ (8) 7 × 10−6
⎣
(
)
⎤
⎥
⎥
⎦
3/(1 − 8)
+
(8)(8 + 1) (7 × 10−6 )
⎡
⎤
1.5
⎢
⎥
−6
⎣ (8) (7 × 10 ) ⎦
(8 + 2)/(1 − 8)
= 830
For metal Y, A = 2.0, B = 1 ´ 10-5, and n = 9. Hence
E ∝−
(2)(2.0)
⎡
2.0
⎢
⎢ (9) 1 × 10−5
⎣
(
)
⎤
⎥
⎥
⎦
3/(1 − 9)
+
= 683
(9)(9 + 1) (1 × 10−5 )
⎡
⎤
2.0
⎢
⎥
−5
⎣ (9) (1 × 10 ) ⎦
(9 + 2)/(1 − 9)
And, for metal Z, A = 3.5, B = 4 ´ 10-6, and n = 7. Thus
E ∝−
(2)(3.5)
⎡
3.5
⎢
⎢ (7) 4 × 10−6
⎣
(
)
⎤
⎥
⎥
⎦
3/(1 − 7)
+
(7)(7 + 1) (4 × 10−6 )
⎡
⎤
3.5
⎢
⎥
−6
⎣ (7) (4 × 10 ) ⎦
= 7425
Therefore, metal Z has the highest modulus of elasticity.
(7 + 2)/(1 − 7)
Elastic Properties of Materials
6.9 A cylindrical specimen of steel having a diameter of 15.2 mm and length of 250 mm is deformed
elastically in tension with a force of 48,900 N. Using the data contained in Table 6.1, determine the following:
(a) The amount by which this specimen will elongate in the direction of the applied stress.
(b) The change in diameter of the specimen. Will the diameter increase or decrease?
Solution
(a) We are asked, in this portion of the problem, to determine the elongation of a cylindrical specimen of
steel. To solve this part of the problem requires that we use Equations 6.1, 6.2 and 6.5. Equation 6.5 reads as
follows:
σ = Eε
Substitution the expression for s from Equation 6.1 and the expression for e from Equation 6.2 leads to
F
Δl
=E
l0
⎛ d02 ⎞
π⎜ ⎟
⎝ 4⎠
In this equation d0 is the original cross-sectional diameter. Now, solving for Dl yields
Δl =
4 Fl0
π d02 E
And incorporating values of F, l0, and d0, and realizing that E = 207 GPa (Table 6.1), leads to
Δl =
(4)(48,900 N) (250 × 10−3 m)
= 3.25 × 10−4 m = 0.325 mm
(π ) (15.2 × 10−3 m)2 (207 × 109 N/m2 )
(b) We are now called upon to determine the change in diameter, Dd. Using Equation 6.8 (the definition of
Poisson's ratio)
ν= −
εx
Δd/d0
= −
εz
Δl/l0
From Table 6.1, for steel, the value of Poisson's ratio, n, is 0.30. Now, solving the above expression for ∆d yields
Δd = −
ν Δl d0
(0.30)(0.325 mm)(15.2 mm)
= −
l0
250 mm
= –5.9 ´ 10-3 mm
The diameter will decrease since Dd is negative.
6.10 A cylindrical specimen of a metal alloy 10 mm in diameter is stressed elastically in tension. A force of
15,000 N) produces a reduction in specimen diameter of 7 × 10–3 mm. Compute Poisson’s ratio for this material if
its elastic modulus is 100 GPa.
Solution
This problem asks that we compute Poisson's ratio for the metal alloy. From Equations 6.5 and 6.1
εz =
σ
F
=
=
E A0 E
F
2
⎛d ⎞
π⎜ 0⎟ E
⎝ 2⎠
=
4F
π d02 E
Since the transverse strain ex is equal to
εx =
Δd
d0
and Poisson's ratio is defined by Equation 6.8, then
ν= −
εx
Δd/d0
d Δd π E
= −
= − 0
⎛ 4F ⎞
εz
4F
⎜ 2 ⎟
⎝ π d0 E ⎠
Now, incorporating values of d0, Dd, E and F from the problem statement yields the following value for Poisson's
ratio
ν=−
(10 × 10−3 m)(−7 × 10−6 m) (π )(100 × 109 N/m2 )
= 0.367
(4)(15,000 N)
6.11 Consider a cylindrical specimen of some hypothetical metal alloy that has a diameter of 10.0 mm. A
tensile force of 1500 N produces an elastic reduction in diameter of 6.7 × 10–4 mm. Compute the elastic modulus of
this alloy, given that Poisson’s ratio is 0.35.
Solution
This problem asks that we calculate the modulus of elasticity of a metal that is stressed in tension.
Combining Equations 6.5 and 6.1 leads to
F
A
σ
F
E=
= 0 =
=
εz
ε z A0ε z
F
⎛d ⎞
ε zπ ⎜ 0 ⎟
⎝ 2⎠
2
=
4F
ε zπ d02
From the definition of Poisson's ratio, (Equation 6.8) and realizing that for the transverse strain, ex=
Δd
leads to
d0
ε
Δd/d
Δd
εz = − x = −
= −
ν
ν
d0ν
Therefore, substitution of this expression for ez into the above equation for E yields
E=
4F
4F
4Fν
=
= −
2
⎛
⎞
π
d0 Δd
ε zπ d0
Δd
2
⎜ − d ν ⎟ π d0
⎝ 0 ⎠
Incorporation of values for F, n, d0, and Dd given in the problem statement (and realizing that Dd is negative) allows
us to calculate the modulus of elasticity as follows:
E= −
(4)(1500 N)(0.35)
π (10 × 10−3 m)(−6.7 × 10−7 m)
=1011 Pa = 100 GPa
6.12 A cylindrical metal specimen 15.0 mm in diameter and 150 mm long is to be subjected to a tensile
stress of 50 MPa; at this stress level, the resulting deformation will be totally elastic.
(a) If the elongation must be less than 0.072 mm, which of the metals in Table 6.1 are suitable candidates?
Why?
(b) If, in addition, the maximum permissible diameter decrease is 2.3 × 10–3 mm when the tensile stress of
50 MPa is applied, which of the metals that satisfy the criterion in part (a) are suitable candidates? Why?
Solution
(a) This part of the problem asks that we ascertain which of the metals in Table 6.1 experience an
elongation of less than 0.072 mm when subjected to a tensile stress of 50 MPa. The maximum strain that may be
sustained, (using Equation 6.2) is just
ε=
Δl 0.072 mm
=
= 4.8 × 10−4
l0
150 mm
Since the stress level is given (50 MPa), using Equation 6.5 it is possible to compute the minimum modulus of
elasticity that is required to yield this minimum strain. Hence
E=
σ
50 MPa
=
= 104.2 GPa
ε 4.8 × 10−4
Which means that those metals with moduli of elasticity greater than this value are acceptable candidates—namely,
Cu, Ni, steel, Ti and W.
(b) This portion of the problem further stipulates that the maximum permissible diameter decrease is 2.3 ´
10-3 mm when the tensile stress of 50 MPa is applied. This translates into a maximum lateral strain ex(max) as
ε x (max) =
Δd
−2.3 × 10−3 mm
=
= − 1.53 × 10−4
d0
15.0 mm
But, since the specimen contracts in this lateral direction, and we are concerned that this strain be less than 1.53 ´
10-4, then the criterion for this part of the problem may be stipulated as −
Now, Poisson’s ratio is defined by Equation 6.8 as
ε
ν=− x
εz
Δd
< 1.53 × 10−4 .
d0
For each of the metal alloys let us consider a possible lateral strain, ε x =
Δd
. Furthermore, since the deformation is
d0
elastic, then, from Equation 6.5, the longitudinal strain, ez is equal to
εz =
σ
E
Substituting these expressions for ex and ez into the definition of Poisson’s ratio we have
Δd
εx
d
ν=− = − 0
σ
εz
E
which leads to the following:
−
Δd ν σ
=
d0
E
Using values for n and E found in Table 6.1 for the six metal alloys that satisfy the criterion for part (a), and for s =
50 MPa, we are able to compute a −
−
−
Δd
for each of Cu, Ni, steel, Ti and W as follows:
d0
Δd
(0.34)(50 × 106 N/m2 )
(copper) =
= 1.55 × 10−4
d0
110 × 109 N/m2
Δd
(0.34)(50 × 106 N/m2 )
(titanium) =
= 1.59 × 10−4
d0
107 × 109 N/m2
−
Δd
(0.31)(50 × 106 N/m2 )
(nickel) =
= 7.49 × 10−5
d0
207 × 109 N/m2
−
Δd
(0.30)(50 × 106 N/m2 )
(steel) =
= 7.25 × 10−5
d0
207 × 109 N/m2
−
Δd
(0.28)(50 × 106 N/m2 )
(tungsten) =
= 3.44 × 10−5
d0
407 × 109 N/m2
Thus, copper and titanium alloys will experience a negative transverse strain greater than 1.53 ´ 10-4. This means
that the following alloys satisfy the criteria for both parts (a) and (b) of this problem: nickel, steel, and tungsten.
6.13 Consider the brass alloy for which the stress–strain behavior is shown in Figure 6.12. A cylindrical
specimen of this material 10.0 mm in diameter and 101.6 mm long is pulled in tension with a force of 10,000 N. If it
is known that this alloy has a value for Poisson’s ratio of 0.35, compute (a) the specimen elongation and (b) the
reduction in specimen diameter.
Solution
(a) This portion of the problem asks that we compute the elongation of the brass specimen. The first
calculation necessary is that of the applied stress using Equation 6.1, as
σ =
F
=
A0
F
⎛d ⎞
π⎜ 0⎟
⎝ 2⎠
2
=
10,000 N
⎛ 10 × 10−3 m ⎞
π⎜
⎟
2
⎝
⎠
2
= 127 × 106 N/m2 = 127 MPa
From the stress-strain plot in Figure 6.12, this stress corresponds to a strain of about 1.5 ´ 10-3. From the definition
of strain, Equation 6.2
Δl = ε l0 = (1.5 × 10−3 ) (101.6 mm) = 0.15 mm
(b) In order to determine the reduction in diameter ∆d, it is necessary to use Equation 6.8 and the definition
of lateral strain (i.e., ex = ∆d/d0) as follows
Δd = d0ε x = − d0ν ε z = − (10 mm)(0.35) (1.5 × 10−3 )
= –5.25 ´ 10-3 mm
6.14 A cylindrical rod 500 mm long and having a diameter of 12.7 mm is to be subjected to a tensile load.
If the rod is to experience neither plastic deformation nor an elongation of more than 1.3 mm when the applied load
is 29,000 N, which of the four metals or alloys listed in the following table are possible candidates? Justify your
choice(s).
Material
Modulus of Elasticity (GPa)
Yield Strength (MPa)
Tensile Strength (MPa)
Aluminum alloy
70
255
420
Brass alloy
100
345
420
Copper
110
210
275
Steel alloy
207
450
550
Solution
This problem asks that we ascertain which of four metal alloys will not (1) experience plastic deformation,
and (2) elongate more than 1.3 mm when a tensile load of 29,000 N is applied. It is first necessary to compute the
stress using Equation 6.1; a material to be used for this application must necessarily have a yield strength greater
than this value. Thus,
σ =
F
=
A0
F
⎛d ⎞
π⎜ 0⎟
⎝ 2⎠
2
=
29,000 N
⎛ 12.7 × 10−3 m ⎞
π⎜
⎟
2
⎝
⎠
2
= 230 MPa
Of the metal alloys listed, aluminum, brass and steel have yield strengths greater than this stress.
Next, we must compute the elongation produced in each of aluminum, brass, and steel by combining
Equations 6.2 and 6.5 in order to determine whether or not this elongation is less than 1.3 mm. For aluminum (E =
70 GPa = 70 ´ 103 MPa)
Δl = ε l0 =
σ l0 (230 MPa)(500 mm)
=
= 1.64 mm
E
70 × 103 MPa
Thus, aluminum is not a candidate because its Dl is greater than 1.3 mm.
For brass (E = 100 GPa = 100 ´ 103 MPa)
Δl =
σ l0 (230 MPa)(500 mm)
=
= 1.15 mm
E
100 × 103 MPa
Thus, brass is a candidate. And, for steel (E = 207 GPa = 207 ´ 103 MPa)
Δl =
σ l0 (230 MPa)(500 mm)
=
= 0.56 mm
E
207 × 103 MPa
Therefore, of these four alloys, only brass and steel satisfy the stipulated criteria.
Tensile Properties
6.15 Figure 6.22 shows the tensile engineering stress–strain behavior for a steel alloy.
(a) What is the modulus of elasticity?
(b) What is the proportional limit?
(c) What is the yield strength at a strain offset of 0.002?
(d) What is the tensile strength?
Solution
(a) Shown below is the inset of Figure 6.22.
The elastic modulus is just the slope of the initial linear portion of the curve; or, from Equation 6.10
E=
σ 2 − σ1
ε 2 − ε1
Inasmuch as the linear segment passes through the origin, let us take both s1 and e1 to be zero. If we arbitrarily take
e2 = 0.005, as noted in the above plot, s2 = 1050 MPa. Using these stress and strain values we calculate the elastic
modulus as follows:
E=
σ 2 − σ 1 (1050 − 0) MPa
=
= 210 × 103 MPa = 210 GPa
ε 2 − ε1
(0.005 − 0)
The value given in Table 6.1 is 207 GPa.
(b) The proportional limit is the stress level at which linearity of the stress-strain curve ends, which is
approximately 1370 MPa.
(c) As noted in the plot below, the 0.002 strain offset line intersects the stress-strain curve at approximately
1600 MPa.
(d) The tensile strength (the maximum on the curve) is approximately 1950 MPa, as noted in the following
plot.
6.16 A load of 140,000 N is applied to a cylindrical specimen of a steel alloy (displaying the stress–strain
behavior shown in Figure 6.22) that has a cross-sectional diameter of 10 mm.
(a) Will the specimen experience elastic and/or plastic deformation? Why?
(b) If the original specimen length is 500 mm, how much will it increase in length when this load is
applied?
Solution
This problem asks us to determine the deformation characteristics of a steel specimen, the stress-strain
behavior for which is shown in Figure 6.22.
(a) In order to ascertain whether the deformation is elastic or plastic, we must first compute the stress, then
locate it on the stress-strain curve, and, finally, note whether this point is on the elastic or elastic + plastic region.
Thus, from Equation 6.1
σ =
F
F
140,000 N
=
=
= 1782 MPa
2
2
A0
⎛ d0 ⎞
⎛ 10 × 10−3 m ⎞
π⎜ ⎟
π⎜
⎟
⎝ 2⎠
2
⎝
⎠
The 1782 MPa point is beyond the linear portion of the curve, and, therefore, the deformation will be both elastic
and plastic.
(b) This portion of the problem asks us to compute the increase in specimen length. From the stress-strain
curve, the strain at 1782 MPa is approximately 0.017. Thus, from Equation 6.2
Δl = ε l0 = (0.017)(500 mm) = 8.5 mm
6.17 A cylindrical specimen of stainless steel having a diameter of 12.8 mm and a gauge length of 50.800
mm is pulled in tension. Use the load–elongation characteristics shown in the following table to complete parts (a)
through (f).
Load (N)
Length (mm)
0
50.800
12,700
50.825
25,400
50.851
38,100
50.876
50,800
50.902
76,200
50.952
89,100
51.003
92,700
51.054
102,500
51.181
107,800
51.308
119,400
51.562
128,300
51.816
149,700
52.832
159,000
53.848
160,400
54.356
159,500
54.864
151,500
55.880
124,700
56.642
Fracture
(a) Plot the data as engineering stress versus engineering strain.
(b) Compute the modulus of elasticity.
(c) Determine the yield strength at a strain offset of 0.002.
(d) Determine the tensile strength of this alloy.
(e) What is the approximate ductility, in percent elongation?
(f) Compute the modulus of resilience.
Solution
This problem calls for us to make a stress-strain plot for stainless steel, given its tensile load-length data,
and then to determine some of its mechanical characteristics.
(a) The data are plotted below on two plots: the first corresponds to the entire stress-strain curve, while for
the second, the curve extends to just beyond the elastic region of deformation.
(b) The elastic modulus is the slope in the linear elastic region (Equation 6.10)—i.e.,
E=
σ 2 − σ1
ε 2 − ε1
Because the stress-strain curve passes through the origin, to simplify the computation let us take both s1 and e1 to be
zero. If we select s2 = 400 MPa, its corresponding strain on the plot e2 is about 0.002. Thus, the elastic modulus is
equal to
E=
400 MPa − 0 MPa
= 200 × 103 MPa = 200 GPa
0.002 − 0
(c) For the yield strength, the 0.002 strain offset line is drawn dashed in the lower plot. It intersects the
stress-strain curve at approximately 750 MPa.
(d) The tensile strength is approximately 1250 MPa, corresponding to the maximum stress on the complete
stress-strain plot.
(e) Ductility, in percent elongation, is just the plastic strain at fracture, multiplied by one-hundred. The
total fracture strain at fracture is 0.115; subtracting out the elastic strain (which is about 0.003) leaves a plastic strain
of 0.112. Thus, the ductility is about 11.2%EL.
(f) From Equation 6.14, the modulus of resilience is just
Ur =
σ 2y
2E
which, using values of sy and E computed above, (750 MPa = 750 ´ 106 N/m2 and 200 GPa = 200 ´ 109 N/m2,
respectively) give a modulus of resilience of
Ur =
(750 × 106 N/m2 )2
(2) (200 × 109 N/m2 )
= 1.40 × 106 N/m2 = 1.40 × 106 J/m3
6.18 A cylindrical metal specimen 15.00 mm in diameter and 120 mm long is to be subjected to a tensile
force of 15,000 N.
(a) If this metal must not experience any plastic deformation, which of aluminum, copper, brass, nickel,
steel, and titanium (Table 6.2) are suitable candidates? Why?
(b) If, in addition, the specimen must elongate no more than 0.070 mm, which of the metals that satisfy the
criterion in part (a) are suitable candidates? Why? Base your choices on data found in Table 6.1.
Solution
(a) In order to determine which of the metals in Table 6.2 do not experience plastic deformation we need to
compute the applied engineering stress; any metal in the table that has a yield strength greater than this value will
not plastically deform.
The applied engineering stress is computed using Equation 6.1 as follows:
σ=
F
A0
which, for a cylindrical specimen of diameter d0 takes the form:
σ=
F
⎛d ⎞
π⎜ 0⎟
⎝ 2⎠
2
For F = 15,000 N and d0 = 15.00 mm (15 ´ 10-3 m) the stress is equal to
σ=
15,000 N
⎛ 15 × 10−3 m ⎞
π⎜
⎟
2
⎝
⎠
2
= 84.9 × 106 N/m2 = 84.9 MPa
Of those 6 alloys in Table 6.2, only nickel, steel, and titanium have yield strengths greater than 84.9 MPa.
(b) Because none of these three metals has experienced plastic deformation it is possible to compute the
elongation Dl by combining Equations 6.2 and 6.5. From Equation 6.5 it is the case that
ε=
σ
E
in which E is the modulus of elasticity. If we incorporate the definition of the strain (Equation 6.2) into this
expression, the following results:
Δl σ
=
l0 E
And solving for the elongation Dl leads to
lσ
Δl = 0
E
In order to calculate the elongation for each of these metals we incorporate the original length 120 mm (120 ´ 10-3
m), the stress computed above (84.9 MPa = 84.9 ´ 106 N/m2), and, for each of the three metals, its modulus of
elasticity found in Table 6.1.
For Ni (E = 207 GPa = 207 ´ 109 N/m2), therefore:
Δl =
(120 × 10−3 m)(84.9 × 106 N/m2 )
207 × 109 N/m2
= 4.9 × 10−5 m = 0.049 mm
Thus, Ni satisfies this criterion since its elongation is less than 0.070 mm.
For steel (E = 207 GPa = 207 ´ 109 N/m2) the elongation will also be 0.049 mm since its elastic modulus is
the same as Ni—i.e., steel meets the criterion).
For titanium (E = 107 GPa = 107 ´ 109 N/m2)
Δl =
(120 × 10−3 m)(84.9 × 106 N/m2 )
9
107 × 10 N/m
2
= 9.5 × 10−5 m = 0.095 mm
And, titanium does not satisfy the criterion because its elongation (0.095 mm) is greater than 0.070 mm.
6.19 A cylindrical metal specimen having an original diameter of 12.8 mm and gauge length of 50.80 mm
is pulled in tension until fracture occurs. The diameter at the point of fracture is 8.13 mm, and the fractured gauge
length is 74.17 mm. Calculate the ductility in terms of percent reduction in area and percent elongation.
Solution
This problem calls for the computation of ductility in both percent reduction in area and percent elongation.
Percent reduction in area, for a cylindrical specimen, is computed using Equation 6.12 as
⎛ A0 − Af ⎞
%RA = ⎜
⎟ × 100 =
⎝ A0 ⎠
2
⎛ df ⎞
⎛ d0 ⎞
π⎜ ⎟ − π⎜ ⎟
⎝ 2⎠
⎝ 2⎠
⎛d ⎞
π⎜ 0⎟
⎝ 2⎠
2
2
× 100
in which d0 and df are, respectively, the original and fracture cross-sectional areas. Thus,
2
%RA =
⎛ 12.8 mm ⎞
⎛ 8.13 mm ⎞
π⎜
− π⎜
⎟
⎟⎠
2
2
⎝
⎠
⎝
⎛ 12.8 mm ⎞
π⎜
⎟⎠
2
⎝
2
2
× 100 = 60%
While, for percent elongation, we use Equation 6.11 as
⎛ lf − l0 ⎞
%EL = ⎜
⎟ × 100
⎝ l0 ⎠
=
74.17 mm − 50.80 mm
× 100 = 46%
50.80 mm
6.20 Calculate the moduli of resilience for the materials having the stress–strain behaviors shown in
Figures 6.12 and 6.22.
Solution
This problem asks us to calculate the moduli of resilience for the materials having the stress-strain
behaviors shown in Figures 6.12 and 6.22. According to Equation 6.14, the modulus of resilience Ur is a function of
the yield strength and the modulus of elasticity as
Ur =
σ 2y
2E
The values for sy and E for the brass in Figure 6.12 are determined in Example Problem 6.3 as 250 MPa and 93.8
GPa, respectively. Thus
Ur =
(250 MPa)2
(250 × 106 N/m2 )2
=
(2) (93.8 GPa) (2)(93.8 × 109 N/m2 )
= 3.32 × 105 N/m2 = 3.32 × 105 J/m3
Values of the corresponding parameters for the steel alloy (Figure 6.22) are determined in Problem 6.15 as
1600 MPa and 210 GPa, respectively, and therefore
Ur =
(1600 MPa)2
(1600 × 106 N/m2 )2
=
(2) (210 GPa) (2)(210 × 109 N/m2 )
= 6.10 × 106 N/m2 = 6.10 × 106 J/m3
6.21 A steel alloy to be used for a spring application must have a modulus of resilience of at least 2.07
MPa. What must be its minimum yield strength?
Solution
The modulus of resilience, yield strength, and elastic modulus of elasticity are related to one another
through Equation 6.14, that is
Ur =
σ 2y
2E
Solving for sy from this expression yields
σy =
2U r E
The value of E for steel given in Table 6.1 is 207 GPa. Using this elastic modulus value leads to a yield strength of
σy =
2U r E =
(2)(2.07 MPa) (207 × 103 MPa)
= 926 MPa
True Stress and Strain
6.22 Show that Equations 6.18a and 6.18b are valid when there is no volume change during deformation.
Solution
To show that Equation 6.18a is valid, we must first rearrange Equation 6.17 as
Ai =
A0 l0
li
Substituting this expression into Equation 6.15 yields
σT =
⎛l ⎞
F
F
F ⎛ li ⎞
=
=
=σ⎜ i ⎟
⎜
⎟
A0l0 A0 ⎝ l0 ⎠
Ai
⎝ l0 ⎠
li
From Equation 6.2
l −l
l
ε=i 0= i − 1
l0
l0
Or
li
l0
=1 + ε
Substitution of this expression into the equation above leads to
⎛l ⎞
σ T = σ ⎜ i ⎟ = σ (1 + ε )
⎝ l0 ⎠
which is Equation 6.18a.
For Equation 6.18b, true strain is defined in Equation 6.16 as
⎛l ⎞
ε T = ln ⎜ i ⎟
⎝ l0 ⎠
Substitution the of the li- l0 ratio of Equation 6.2b above leads to Equation 6.18b—viz.
ε T = ln (1 + ε )
(6.2b)
6.23 A tensile test is performed on a metal specimen, and it is found that a true plastic strain of 0.16 is
produced when a true stress of 500 MPa is applied; for the same metal, the value of K in Equation 6.19 is 825 MPa.
Calculate the true strain that results from the application of a true stress of 600 MPa.
Solution
We are asked to compute the true strain that results from the application of a true stress of 600 MPa; other
true stress-strain data are also given. It first becomes necessary to solve for n in Equation 6.19. Taking logarithms
of this expression leads to
log σ T = log K + n log ε T
Next, we rearrange this equation such that n is the dependent variable:
n=
log σ T − log K
log ε T
We now solve for n using the following data given in the problem statement:
sT = 500 MPa
eT = 0.16
K = 825 MPa
Thus
n=
log (500 MPa) − log (825 MPa)
= 0.273
log (0.16)
We now rearrange Equation 6.19 such that eT is the dependent variable; we first divide both sides of the Equation
6.19 by K, which leads to the following expression:
ε Tn =
σT
K
eT becomes the dependent variable by taking the 1/n root of both sides of this expression, as
⎛σ ⎞
εT = ⎜ T ⎟
⎝ K⎠
1/n
Finally, using values of K and n, we solve for the true strain at a true stress of 600 MPa:
⎛σ ⎞
εT = ⎜ T ⎟
⎝ K ⎠
1/n
⎛ 600 MPa ⎞
=⎜
⎝ 825 MPa ⎟⎠
1/0.273
= 0.311
6.24 For a brass alloy, the following engineering stresses produce the corresponding plastic engineering
strains prior to necking:
Engineering Stress (MPa)
Engineering Strain
315
0.105
340
0.220
On the basis of this information, compute the engineering stress necessary to produce an engineering strain of 0.28.
Solution
For this problem we first need to convert engineering stresses and strains to true stresses and strains so that
the constants K and n in Equation 6.19 may be determined. Since sT = s(1 + e) (Equation 6.18a), we convert the
two values of engineering stress into true stresses as follows:
σ T 1 = (315 MPa)(1 + 0.105) = 348 MPa
σ T 2 = (340 MPa)(1 + 0.220) = 415 MPa
Similarly, for strains—we convert engineering strains to true strains using Equation 6.18b [i.e., eT = ln(1 + e)] as
follows:
ε T 1 = ln (1 + 0.105) = 0.09985
ε T 2 = ln (1 + 0.220) = 0.19885
Taking logarithms of both sides of Equation 6.19 leads to
log σ T = log K + n log ε T
which allows us to set up two simultaneous equations for the above pairs of true stresses and true strains, with K and
n as unknowns. Thus
log (348) = log K + n log (0.09985)
log (415) = log K + n log (0.19885)
Solving for n from these two expressions yields
n=
log (348) − log (415)
= 0.256
log (0.09985) − log (0.19885)
We solve for K by substitution of this value of n into the first simultaneous equation as
log K = log σ T − n log ε T
= log(348) − (0.256)[log(0.09985)] = 2.7977
Thus, the value of K is equal to
K = 102.7977 = 628 MPa
We now, converting e = 0.28 to true strain using Equation 6.18b as follows:
ε T = ln (1 + 0.28) = 0.247
The corresponding sT to give this value of eT (using Equation 6.19) is just
σ T = K ε Tn = (628 MPa)(0.247)0.256 = 439 MPa
Now converting this value of sT to an engineering stress using Equation 6.18a gives
σ =
σT
439 MPa
=
= 343 MPa
1 + ε 1 + 0.28
6.25 Find the toughness (or energy to cause fracture) for a metal that experiences both elastic and plastic
deformation. Assume Equation 6.5 for elastic deformation, that the modulus of elasticity is 103 GPa, and that elastic
deformation terminates at a strain of 0.007. For plastic deformation, assume that the relationship between stress
and strain is described by Equation 6.19, in which the values for K and n are 1520 MPa and 0.15, respectively.
Furthermore, plastic deformation occurs between strain values of 0.007 and 0.60, at which point fracture occurs.
Solution
This problem calls for us to compute the toughness (or energy to cause fracture). This toughness is equal to
the area beneath the entire stress-strain curve. The easiest way to make this computation is to perform integrations
in both elastic and plastic regions using data given in the problem statement, and then add these values together.
Thus, we may define toughness using the following equation:
Toughness = ∫ σ d ε
Toughness in the elastic region is determined by integration of Equation 6.5, integrating between strains of 0 and
0.007 (per the problem statement). Thus
0.007
∫ Eε dε
Toughness (elastic) =
0
Eε 2
=
2
=
0.007
0
103 × 109 N/m2 ⎡
2
2⎤
⎣(0.007) − (0) ⎦
2
= 2.52 × 106 N/m2 = 2.52 × 106 J/m3
Now for the plastic region of the stress-strain curve, we integrate Equation 6.19 between the strain limits of
0.007 and 0.60, as follows:
0.60
Toughness (plastic) =
∫ Kε n dε
0.007
K
=
ε (n+1)
(n + 1)
=
0.60
0.007
1520 × 106 N/ m2 ⎡
(0.60)1.15 − (0.007)1.15 ⎤⎦
(1.0 + 0.15) ⎣
= 730 × 106 N/m2 = 7.30 × 106 J/m3
And, finally, the total toughness is the sum of elastic and plastic values, which is equal to
Toughness = Toughness(elastic) + Toughness(plastic)
= 2.52 × 106 J/m3 + 730 × 106 J/m3
= 733 × 106 J/m3 = 7.33 × 108 J/m3
6.26 Taking the logarithm of both sides of Equation 6.19 yields
log σT = log K + n log eT
(6.31)
Thus, a plot of log σT versus log eT in the plastic region to the point of necking should yield a straight line having a
slope of n and an intercept (at log σT = 0) of log K.
Using the appropriate data tabulated in Problem 6.17, make a plot of log σT versus log eT and determine
the values of n and K. It will be necessary to convert engineering stresses and strains to true stresses and strains
using Equations 6.18a and 6.18b.
Solution
This problem calls for us to utilize the appropriate data from Problem 6.17 in order to determine the values
of n and K for this material. From Equation 6.31 the slope of a log sT versus log eT plot will yield n. However,
Equation 6.19 is only valid in the region of plastic deformation to the point of necking; thus, only the 8th, 9th, 10th,
11th, 12th, and 13th data points may be utilized. These data are tabulated below:
s (MPa)
sT(MPa)
log sT
e
eT
797
838
928
997
1163
1236
803
846
942
1017
1210
1310
2.905
2.927
2.974
3.007
3.083
3.117
0.0075
0.010
0.015
0.020
0.040
0.060
7.47 ´ 10-3
9.95 ´ 10-3
1.488 ´ 10-2
1.980 ´ 10-2
3.92 ´ 10-2
5.83 ´ 10-2
The log-log plot with these data points is given below.
log eT
-2.127
-2.002
-1.827
-1.703
-1.407
-1.234
From Equation 6.31, the value of n is equal to the slope of the line that has been drawn through the data
points; that is
n=
Δ log σ T
Δ log ε T
From this line, values of log sT corresponding to log eT (1) = -2.20 and log eT (2) = -1.20 are, respectively, log
sT(1) = 2.886 and log sT (2) = 3.131. Therefore, we determine the value of n as follows:
n=
Δ log σ T log σ T (1) − log σ T (2)
=
Δ log ε T
log ε T (1) − log ε T (2)
=
2.886 − 3.131
= 0.245
−2.20 − (−1.20)
In order to determine the value of K we rearrange Equation 6.31 to read as follows:
log K = log σ T − n log ε T
It is possible to compute log K by inserting into this equation the value for a specific log sT and its corresponding
log eT. (Of course, we have already calculated the value of n.) If we use log sT (1) and log eT (1) values from
above, then K is determined as follows:
log K = log σ T (1) − n log ε T (1)
= 2.886 − (0.245)(−2.20) = 3.425
Which means that
K = 103.425 = 2660 MPa
Elastic Recovery after Plastic Deformation
6.27 A steel alloy specimen having a rectangular cross section of dimensions 19 mm × 3.2 mm has the
stress–strain behavior shown in Figure 6.22. This specimen is subjected to a tensile force of 110,000 N.
(a) Determine the elastic and plastic strain values.
(b) If its original length is 610 mm, what will be its final length after the load in part (a) is applied and then
released?
Solution
(a) We are asked to determine both the elastic and plastic strain values when a tensile force of 110,000 N is
applied to the steel specimen and then released. First it becomes necessary to determine the applied stress using
Equation 6.1; thus
F
F
=
A0 b0d0
σ =
where b0 and d0 are cross-sectional width and depth [19 mm (19 ´ 10-3 m) and 3.2 mm (3.2 ´ 10-3 m),
respectively). Thus
σ =
110,000 N
= 1.810 × 109 N/m2 = 1810 MPa
−3
(19 × 10 m)(3.2 × 10 m)
−3
From Figure 6.22, this point is in the plastic region so the specimen will be both elastic and plastic strains. The total
strain at this point, et, is about 0.020. From this point on the stress-strain curve we are able to estimate the amount
of strain recovery ee from Hooke's law, Equation 6.5 as
εe =
σ
E
And, since E = 207 GPa for steel (Table 6.1)
εe =
1810 MPa
= 0.0087
207 × 103 MPa
The value of the plastic strain, ep is just the difference between the total and elastic strains; that is
ep = et – ee = 0.020 – 0.0087 = 0.0113
(b) If the initial length is 610 mm then the final specimen length li may be determined from a rearranged
form of Equation 6.2 using the plastic strain value as
li = l0(1 + ep) = (610 mm)(1 + 0.0113) = 616.9 mm
Hardness
6.28 (a) A 10-mm-diameter Brinell hardness indenter produced an indentation 2.50 mm in diameter in a
steel alloy when a load of 1000 kg was used. Compute the HB of this material.
(b) What will be the diameter of an indentation to yield a hardness of 300 HB when a 500-kg load is used?
Solution
(a) We are asked to compute the Brinell hardness for the given indentation. It is necessary to use the
equation in Table 6.5 for HB, where P = 1000 kg, d = 2.50 mm, and D = 10 mm. Thus, the Brinell hardness is
computed as
HB =
=
2P
π D ⎡⎢ D −
⎣
D2 − d 2 ⎤
⎥⎦
(2)(1000 kg)
= 200.5
⎡
(π )(10 mm) 10 mm − (10 mm)2 − (2.50 mm)2 ⎤
⎣⎢
⎦⎥
(b) This part of the problem calls for us to determine the indentation diameter d that will yield a 300 HB
when P = 500 kg. Solving for d from the equation in Table 6.5 gives
d=
⎡
2P ⎤
D2 − ⎢ D −
(HB)
π D ⎥⎦
⎣
2
2
=
⎡
(2)(500 kg) ⎤
(10 mm)2 − ⎢10 mm −
= 1.45 mm
(300)(
π )(10 mm) ⎥⎦
⎣
6.29 (a) What is the indentation diagonal length when a load of 0.60 kg produces a Vickers HV of 400?
(b) Calculate the Vickers hardness when a 700-g load yields an indentation diagonal length of 0.050 mm.
Solution
(a) The equation given in Table 6.5 for Vickers microhardness is as follows:
HV =
1.854P
d12
Here P is the applied load (in kg) and d1 is the diagonal length (in mm). This portion of the problem asks that we
compute the indentation diagonal length. Rearranging the above equation to make d1 the dependent variable gives
d1 =
1.854P
HV
Using values for P and HV given in the problem statement yields the following value for d1:
d1 =
(1.854)(0.60 kg)
= 0.0527 mm
400
(b) For P = 700 g (0.700 kg) and d1 = 0.050 mm, the Vickers hardness value is
HV =
=
1.854P
d12
(1.854)(0.700 kg)
= 519
(0.050 mm)2
6.30 Estimate the Brinell and Rockwell hardnesses for the following:
(a) The naval brass for which the stress–strain behavior is shown in Figure 6.12.
(b) The steel alloy for which the stress–strain behavior is shown in Figure 6.22.
Solution
This problem calls for estimations of Brinell and Rockwell hardnesses.
(a) For the brass specimen, the stress-strain behavior for which is shown in Figure 6.12, the tensile strength
is 450 MPa. From Figure 6.19, the hardness for brass corresponding to this tensile strength is about 125 HB or 85
HRB.
(b) The steel alloy (Figure 6.22) has a tensile strength of about 1950 MPa [Problem 6.15(d)]. This
corresponds to a hardness of about 560 HB or ~55 HRC from the line (extended) for steels in Figure 6.19.
Variability of Material Properties
6.31 Cite five factors that lead to scatter in measured material properties.
Solution
The five factors that lead to scatter in measured material properties are the following: (1) test method; (2)
variation in specimen fabrication procedure; (3) operator bias; (4) apparatus calibration; and (5) material
inhomogeneities and/or compositional differences.
6.32 The following table gives a number of Rockwell G hardness values that were measured on a single
steel specimen. Compute average and standard deviation hardness values.
47.3
48.7
47.1
52.1
50.0
50.4
45.6
46.2
45.9
49.9
48.3
46.4
47.6
51.1
48.5
50.4
46.7
49.7
Solution
The average of the given hardness values is calculated using Equation 6.21 as
18
∑ HRG
i
HRG = i=1
=
18
47.3 + 52.1 + 45.6 . . . . + 49.7
= 48.4
18
And we compute the standard deviation using Equation 6.22 as follows:
18
∑ ( HRG − HRG)
2
i
s=
i=1
18 − 1
1/2
⎡ (47.3 − 48.4)2 + (52.1 − 48.4)2 + . . . . + (49.7 − 48.4)2 ⎤
=⎢
⎥
17
⎢⎣
⎥⎦
=
64.95
= 1.95
17
Design/Safety Factors
6.33 Upon what three criteria are factors of safety based?
Solution
The criteria upon which factors of safety are based are (1) consequences of failure, (2) previous experience,
(3) accuracy of measurement of mechanical forces and/or material properties, and (4) economics.
DESIGN PROBLEMS
6.D1 (a) Consider a thin-walled cylindrical tube having a radius of 65 mm is to be used to transport
pressurized gas. If inside and outside tube pressures are 10.13 and 0.2026 MPa (100 and 2.0 atm), respectively,
compute the minimum required thickness for each of the following metal alloys. Assume a factor of safety of 3.5.
(b) A tube constructed of which of the alloys will cost the least amount?
Yield Strength, sy
Density, r
Unit Mass Cost, c
(MPa)
(g/cm3)
($US/kg)
Steel (plain)
375
7.8
1.65
Steel (alloy)
1000
7.8
4.00
Cast iron
225
7.1
2.50
Aluminum
275
2.7
7.50
Magnesium
175
1.80
15.00
Alloy
Solution
(a) To solve this problem, we use a procedure similar to that used for Design Example 6.2. Alloy yield
strength, tube radius and thickness, and the inside-outside pressure difference are related according to Equation 6.26;
furthermore, labels for the tube dimensions are represented in the sketch of Figure 6.21. For this problem all
parameters in this equation are provided except for the tube thickness, t. Rearranging Equation 6.26 such that t is
the dependent variable leads to
t=
NrΔp
σy
(6.26b)
Tube wall thickness will vary from alloy to alloy since each alloy has a different yield strength, whereas N, r, and Dp
are constant. Their values are provided in the problem statement as follows:
N = 3.5
r = 65 mm = 65 ´ 10-3 m
Dp = 10.13 MPa - 0.2026 MPa = 9.927 MPa
Using Equation 6.26b, we compute the tube wall thickness for these five alloys as follows:
t (plain steel) =
(3.5)(65 × 10−3 m)(9.927 MPa)
= 0.00602 m = 6.02 mm
375 MPa
t (alloy steel) =
(3.5)(65 × 10−3 m)(9.927 MPa)
= 0.00226 m = 2.26 mm
1000 MPa
t (cast iron) =
(3.5)(65 × 10−3 m)(9.927 MPa)
= 0.0100 m = 10.0 mm
225 MPa
t (aluminum) =
(3.5)(65 × 10−3 m)(9.927 MPa)
= 0.00821 m = 8.21 mm
275 MPa
t (magnesium) =
(3.5)(65 × 10−3 m)(9.927 MPa)
= 0.0129 m = 12.9 mm
175 MPa
(b) This portion of the problem asks that we determine at tube constructed of which of the alloys will cost
the least amount. We begin by considering Equation 6.28, which gives tube volume as a function of t. Furthermore,
when the expression for t, from Equation 6.26b, is incorporated into Equation 6.28, the following equation for tube
volume V results:
V = π (2rt + t 2 )L
2⎤
⎡ ⎛
NrΔp ⎞ ⎛ NrΔp ⎞ ⎥
⎢
= π 2r ⎜
⎟ +⎜
⎟ L
⎢
σy ⎠ ⎝ σy ⎠ ⎥
⎝
⎢⎣
⎥⎦
Now tube cost per unit length is dependent on volume, alloy density, and cost per unit mass values according to
Equation 6.29. Substitution of the above expression for V into Equation 6.29 yields
⎛ Vρ ⎞
Cost = ⎜
(c )
⎝ 1000 ⎟⎠
⎧ ⎡
2⎤ ⎫
⎪⎪ ⎢ ⎛ NrΔp ⎞ ⎛ NrΔp ⎞ ⎥ ⎪⎪
⎟ +⎜
⎟ ⎥ Lρ ⎬ c
⎨π ⎢2r ⎜
⎪ ⎢ ⎜⎝ σ y ⎟⎠ ⎜⎝ σ y ⎟⎠ ⎥ ⎪
⎥⎦ ⎪⎭
⎪ ⎢⎣
=⎩
1000
Using this equation, let us now determine the cost per unit length for the plain steel.
Cost (plain steel) =
2
⎧ ⎡
⎫
⎛ (3.5)(65 × 10−3 m)(9.927 MPa) ⎞ ⎛ (3.5)(65 × 10−3 m)(9.927 MPa) ⎞ ⎤ 6
⎪
⎪
−3
⎥ (10 cm 3 /m 3 )(1 m)(7.8 g/cm 3 ) ⎬ (1.65 $US/kg)
+⎜
⎨π ⎢(2)(65 × 10 m) ⎜
⎟
⎟
375
MPa
375
MPa
⎝
⎠
⎝
⎠
⎥⎦
⎪⎩ ⎢⎣
⎪⎭
1000 g/kg
= $33.10
We will not present cost computations for the other alloys. However, the following table lists costs for all five
alloys (along with wall thickness values calculated above.
Alloy
t
(mm)
Cost
($US)
Steel (plain)
6.02
33.10
Steel (alloy)
2.26
29.30
Cast iron
10.0
78.40
Aluminum
8.21
72.20
Magnesium
12.9
156.50
Hence, the alloy steel is the least expensive, the plain steel is next; the most expensive is the magnesium alloy.
6.D2 (a) Gaseous hydrogen at a constant pressure of 0.658 MPa (5 atm) is to flow within the inside of a
thin-walled cylindrical tube of nickel that has a radius of 0.125 m. The temperature of the tube is to be 350°C and
the pressure of hydrogen outside of the tube will be maintained at 0.0127 MPa (0.125 atm). Calculate the minimum
wall thickness if the diffusion flux is to be no greater than 1.25 × 10–7 mol/m2.s. The concentration of hydrogen in
the nickel, CH (in moles hydrogen per cubic meter of Ni), is a function of hydrogen pressure, PH2 (in MPa), and
absolute temperature T according to
⎛ 12,300J/mol ⎞
CH = 30.8 pH exp ⎜ −
⎟⎠
2
RT
⎝
(6.32)
Furthermore, the diffusion coefficient for the diffusion of H in Ni depends on temperature as
⎛ 39,560 J/mol ⎞
DH (m2 /s) = (4.76 × 10−7 )exp ⎜ −
⎟⎠
RT
⎝
(6.33)
(b) For thin-walled cylindrical tubes that are pressurized, the circumferential stress is a function of the
pressure difference across the wall (Δp), cylinder radius (r), and tube thickness (Dx) according to Equation 6.25—
that is
σ=
rΔp
Δx
(6.25a)
Compute the circumferential stress to which the walls of this pressurized cylinder are exposed.
(Note: the symbol t is used for cylinder wall thickness in Equation 6.25 found in Design Example 6.2; in this version
of Equation 6.25 (i.e., 6.25a) we denote wall thickness by Dx.)
(c) The room-temperature yield strength of Ni is 100 MPa, and σy diminishes about 5 MPa for every 50°C
rise in temperature. Would you expect the wall thickness computed in part (b) to be suitable for this Ni cylinder at
350°C? Why or why not?
(d) If this thickness is found to be suitable, compute the minimum thickness that could be used without any
deformation of the tube walls. How much would the diffusion flux increase with this reduction in thickness?
However, if the thickness determined in part (c) is found to be unsuitable, then specify a minimum thickness that you
would use. In this case, how much of a decrease in diffusion flux would result?
Solution
(a) This portion of the problem asks for us to compute the wall thickness of a thin-walled cylindrical Ni
tube at 350°C through which hydrogen gas diffuses. The inside and outside pressures are, respectively, 0.658 and
0.0127 MPa, and the diffusion flux is to be no greater than 1.25 ´ 10-7 mol/m2-s. This is a steady-state diffusion
problem, which necessitates that we employ Equation 5.2. The concentrations at the inside and outside wall faces
may be determined using Equation 6.32, and, furthermore, the diffusion coefficient is computed using Equation
6.33. Solving for Dx (using Equation 5.2)
Δx = −
= −
D ΔC
J
1
1.25 × 10
−7
mol/m2 -s
×
⎛
⎞
39,560 J/mol
×
⎝ (8.31 J/mol-K)(350 + 273 K) ⎟⎠
(4.76 × 10−7 ) exp ⎜ −
⎛
⎞
12,300 J/mol
(30.8)exp ⎜ −
⎟⎠
(8.31
J/mol-K)(350
+
273
K)
⎝
( 0.0127 MPa − 0.658 MPa )
= 0.00366 m = 3.66 mm
(b) Now we are asked to determine the circumferential stress:
σ =
=
r Δp
Δx
(0.125 m)(0.658 MPa − 0.0127 MPa)
0.00366 m
= 22.0 MPa
(c) Now we are to compare this value of stress to the yield strength of Ni at 350°C, from which it is
possible to determine whether or not the 3.66 mm wall thickness is suitable. From the information given in the
problem, we may write an equation for the dependence of yield strength (sy) on temperature (T) as follows:
σ y = 100 MPa −
5 MPa
T − Tr
50°C
(
)
where Tr is room temperature and for temperature in degrees Celsius. Thus, at 350°C
⎛ 5 MPa ⎞
σ y = 100 MPa − ⎜
(350°C − 20°C) = 67 MPa
⎝ 50°C ⎟⎠
Inasmuch as the circumferential stress (22.0 MPa) is much less than the yield strength (67 MPa), this thickness is
entirely suitable.
(d) And, finally, this part of the problem asks that we specify how much this thickness may be reduced and
still retain a safe design. Let us use a working stress by dividing the yield stress by a factor of safety, according to
Equation 6.24. On the basis of our experience, let us use a value of 2.0 for N. Thus
σw =
σy
N
=
67 MPa
= 33.5 MPa
2
Using this value for sw and Equation 6.25a, we now compute the tube thickness as
Δx =
=
r Δp
σw
(0.125 m)(0.658 MPa − 0.0127 MPa)
(33.5 MPa)
= 0.00241 m = 2.41 mm
Substitution of this value into Fick's first law we calculate the diffusion flux as follows:
J =−D
ΔC
Δx
⎡
⎤
39,560 J/mol
= − (4.76 × 10−7 ) exp ⎢ −
⎥ ×
(8.31
J/mol-K)(350
+
273
K)
⎣
⎦
⎡
⎤
12,300 J/mol
(30.8)exp ⎢ −
⎥
(8.31
J/mol-K)(350
+
273
K)
⎣
⎦
0.00241 m
( 0.0127 MPa − 0.658 MPa )
= 1.90 × 10−7 mol/m2 -s
Thus, the flux increases by approximately a factor of 1.5, from 1.25 ´ 10-7 to 1.90 ´ 10-7 mol/m2-s with this
reduction in thickness.