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.
CHAPTER 7
DISLOCATIONS AND STRENGTHENING MECHANISMS
PROBLEM SOLUTIONS
Basic Concepts of Dislocations
Characteristics of Dislocations
7.1 To provide some perspective on the dimensions of atomic defects, consider a metal specimen with a
dislocation density of 105 mm–2. Suppose that all the dislocations in 1000 mm3 (1 cm3) were somehow removed and
linked end to end. How far (in miles) would this chain extend? Now suppose that the density is increased to 109
mm–2 by cold working. What would be the chain length of dislocations in 1000 mm3 of material?
Solution
The dislocation density is just the total dislocation length per unit volume of material (in this case per cubic
millimeters). Thus, the total length in 1000 mm3 of material having a density of 105 mm-2 is just
(105 mm−2 )(1000 mm3 ) = 108 mm = 105 m = 62 mi
Similarly, for a dislocation density of 109 mm-2, the total length is
(109 mm−2 )(1000 mm3 ) = 1012 mm = 109 m = 6.2 × 105 mi
7.2 Consider two edge dislocations of opposite sign and having slip planes that are separated by several
atomic distances, as indicated in the following diagram. Briefly describe the defect that results when these two
dislocations become aligned with each other.
Answer
When the two edge dislocations become aligned, a planar region of vacancies will exist between the
dislocations as:
Slip Systems
7.3 Compare planar densities (Problem 3.35) for the (100), (110), and (111) planes for BCC.
Solution
For the BCC crystal structure, the planar densities of the (100) and (110) planes were determined in
Homework Problem 3.35, which are as follows:
PD100 (BCC) =
PD110 (BCC) =
3
16R2
3
8R2
2
=
0.19
=
R2
0.27
R2
Below is a BCC unit cell, within which is shown a (111) plane.
(a)
The centers of the three corner atoms, denoted by A, B, and C lie on this plane. Furthermore, the (111) plane does
not pass through the center of atom D, which is located at the unit cell center. The atomic packing of this plane is
presented in the following figure; the corresponding atom positions from the Figure (a) are also noted.
(b)
Inasmuch as this plane does not pass through the center of atom D, it is not included in the atom count. One sixth of
each of the three atoms labeled A, B, and C is associated with this plane, which gives an equivalence of one-half
atom.
In Figure (b) the triangle with A, B, and C at its corners is an equilateral triangle. And, from Figure (b), the
xy
area of this triangle is
. The triangle edge length, x, is equal to the length of a face diagonal, as indicated in
2
Figure (a). And its length is related to the unit cell edge length, a, as
x 2 = a2 + a2 = 2a2
or
x=a 2
For BCC, a =
4R
3
(Equation 3.4), and, therefore,
x=
4R 2
3
Also, from Figure (b), with respect to the length y we may write
2
⎛ x⎞
y2 + ⎜ ⎟ = x2
⎝ 2⎠
which leads to y =
x 3
. And, substitution for the above expression for x yields
2
y=
x 3 ⎛ 4R 2⎞ ⎛ 3⎞ 4R 2
=⎜
⎟⎜
⎟=
2
2
3 ⎠⎝ 2 ⎠
⎝
Thus, the area of this triangle is equal to
AREA =
⎛ 1 ⎞ ⎛ 4 R 2 ⎞ ⎛ 4 R 2 ⎞ 8 R2
1
xy = ⎜ ⎟⎜
⎟⎜
⎟=
2
⎝ 2⎠ ⎝
3 ⎠⎝ 2 ⎠
3
And, finally, the planar density for this (111) plane is
PD111 (BCC) =
3
0.5 atom
0.11
=
=
2
2
8R
16 R
R2
3
7.4 One slip system for the BCC crystal structure is {110} 111 . In a manner similar to Figure 7.6b,
sketch a {110}-type plane for the BCC structure, representing atom positions with circles. Now, using arrows,
indicate two different 111 slip directions within this plane.
Solution
Below is shown the atomic packing for a BCC {110}-type plane. The arrows indicate two different 〈111〉
type directions.
7.5 Equations 7.1a and 7.1b, expressions for Burgers vectors for FCC and BCC crystal structures, are of
the form
b=
a
uvw
2
where a is the unit cell edge length. The magnitudes of these Burgers vectors may be determined from the following
equation:
b=
(
)
1/2
a 2 2
u + v + w2
2
(7.11)
determine the values of |b| for copper and iron. You may want to consult Table 3.1.
Solution
This problem asks that we compute the magnitudes of the Burgers vectors for copper and iron. For Cu,
which has an FCC crystal structure, R = 0.1278 nm (Table 3.1) and, from Equation 3.1
a = 2R 2 = (2)(0.1278 nm)( 2)
= 0.3615 nm
Also, from Equation 7.1a, the Burgers vector for FCC metals is
a
b = 〈110〉
2
Therefore, the values for u, v, and w in Equation 7.11 are 1, 1, and 0, respectively. Hence, the magnitude of the
Burgers vector for Cu is
a
2
b =
=
0.3615 nm
2
u2 + v 2 + w 2
(1 )2 + (1 )2 + (0)2 = 0.2556 nm
For Fe which has a BCC crystal structure, R = 0.1241 nm (Table 3.1) and, from Equation 3.4
a=
4R
3
=
(4)(0.1241 nm)
3
= 0.2866 nm
Also, from Equation 7.1b, the Burgers vector for BCC metals is
a
b = 〈111〉
2
Therefore, the values for u, v, and w in Equation 7.11 are 1, 1, and 1, respectively. Hence, the magnitude of the
Burgers vector for Fe is
b =
0.2866 nm
2
(1)2 + (1)2 + (1)2 = 0.2482 nm
7.6 (a) In the manner of Equations 7.1a to 7.1c, specify the Burgers vector for the simple cubic crystal
structure whose unit cell is shown in Figure 3.3. Also, simple cubic is the crystal structure for the edge dislocation
of Figure 4.4, and for its motion as presented in Figure 7.1. You may also want to consult the answer to Concept
Check 7.1.
(b) On the basis of Equation 7.11, formulate an expression for the magnitude of the Burgers vector, |b|, for
the simple cubic crystal structure.
Solution
(a) This part of the problem asks that we specify the Burgers vector for the simple cubic crystal structure
(and suggests that we consult the answer to Concept Check 7.1). This Concept Check asks that we select the slip
system for simple cubic from four possibilities. The correct answer is {100} 010 . Thus, the Burgers vector will lie
in a 010 -type direction. Also, the unit slip distance is a (i.e., the unit cell edge length, Figures 4.4 and 7.1).
Therefore, the Burgers vector for simple cubic is
b = a 010
Or, equivalently
b = a 100
(b) The magnitude of the Burgers vector, |b|, for simple cubic is
b = a(12 + 02 + 02 )1 / 2 = a
Slip in Single Crystals
7.7 Consider a metal single crystal oriented such that the normal to the slip plane and the slip direction
are at angles of 60° and 35°, respectively, with the tensile axis. If the critical resolved shear stress is 6.2 MPa, will
an applied stress of 12 MPa cause the single crystal to yield? If not, what stress will be necessary?
Solution
This problem calls for us to determine whether or not a metal single crystal having a specific orientation
and of given critical resolved shear stress will yield. We are given that f = 60°, l = 35°, and that the values of the
critical resolved shear stress and applied tensile stress are 6.2 MPa and 12 MPa, respectively. From Equation 7.2
τ R = σ cos φ cos λ = (12 MPa)(cos 60°)(cos 35°) = 4.91 MPa
Since the resolved shear stress (4.91 MPa) is less that the critical resolved shear stress (6.2 MPa), the single crystal
will not yield.
However, from Equation 7.4, the stress at which yielding occurs is
σy =
τ crss
6.2 MPa
=
= 15.1 MPa
cos φ cos λ (cos 60°)(cos 35°)
7.8 Consider a single crystal of nickel oriented such that a tensile stress is applied along a [001] direction.
If slip occurs on a (111) plane and in a [101] direction and is initiated at an applied tensile stress of 13.9 MPa,
compute the critical resolved shear stress.
Solution
This problem asks that we compute the critical resolved shear stress for nickel. In order to do this, we must
employ Equation 7.4, but first it is necessary to solve for the angles l and f which are shown in the sketch below.
The angle l is the angle between the tensile axis—i.e., along the [001] direction—and the slip direction—i.e.,
[101], and may be determined using Equation 7.6 as
⎡
λ = cos−1 ⎢
⎢
⎢⎣
(
⎤
⎥
⎥
u12 + v12 + w12 u22 + v22 + w22 ⎥
⎦
u1u2 + v1v2 + w1w2
)(
)
where (for [001]) u1 = 0, v1 = 0, w1 = 1, and (for [1 01] ) u2 = –1, v2 = 0, w2 = 1. Therefore, l is equal to
⎡
⎤
⎥
⎢
⎥
⎢ ⎡(0)2 + (0)2 + (1)2 ⎤ ⎡(−1)2 + (0)2 + (1)2 ⎤ ⎥
⎣
⎦
⎣
⎦
⎣
⎦
⎢
λ = cos−1
(0)(−1) + (0)(0) + (1)(1)
⎛ 1 ⎞
= cos−1 ⎜
⎟ = 45°
⎝ 2⎠
Furthermore, f is the angle between the tensile axis—the [001] direction—and the normal to the slip
plane—i.e., the (111) plane; for this case this normal is along a [111] direction. Therefore, again using Equation 7.6
⎡
⎤
⎢
⎥
(0)(1)
+
(0)(1)
+
(1)(1)
φ = cos−1 ⎢
⎥
⎢ ⎡(0)2 + (0)2 + (1)2 ⎤ ⎡(1)2 + (1)2 + (1)2 ⎤ ⎥
⎦⎣
⎦⎦
⎣ ⎣
⎛ 1 ⎞
= cos−1 ⎜
⎟ = 54.7°
⎝ 3⎠
And, finally, using Equation 7.4, the critical resolved shear stress is equal to
τ crss = σ y (cos φ cos λ )
= (13.9 MPa) ⎡⎣cos(54.7°)cos(45°) ⎤⎦
⎛ 1 ⎞⎛ 1 ⎞
= (13.9 MPa) ⎜
⎟⎜
⎟ = 5.68 MPa
⎝ 3⎠⎝ 2⎠
7.9 (a) A single crystal of a metal that has the BCC crystal structure is oriented such that a tensile stress is
applied in the [100] direction. If the magnitude of this stress is 4.0 MPa, compute the resolved shear stress in the
[111] direction on each of the (110), (011), and (101) planes.
(b) On the basis of these resolved shear stress values, which slip system(s) is (are) most favorably
oriented?
Solution
In order to solve this problem, it is necessary to employ Equation 7.2, which means that we first need to
solve for the for angles l and f for the three slip systems.
For each of these three slip systems, the l will be the same—i.e., the angle between the direction of the
applied stress, [100] and the slip direction, [1 11] . This angle l may be determined using Equation 7.6
⎡
λ = cos−1 ⎢⎢
⎢⎣
(
⎤
⎥
⎥
2
2
2
2
2
2
u1 + v1 + w1 u2 + v2 + w2 ⎥
⎦
u1u2 + v1v2 + w1w2
)(
)
where (for [100]) u1 = 1, v1 = 0, w1 = 0, and (for [1 11] ) u2 = 1, v2 = –1, w2 = 1. Therefore, l is determined as
⎡
⎤
⎥
⎢
⎥
⎢ ⎡(1)2 + (0)2 + (0)2 ⎤ ⎡(1)2 + (−1)2 + (1)2 ⎤ ⎥
⎣
⎦
⎣
⎦
⎣
⎦
⎢
λ = cos−1
(1)(1) + (0)( − 1) + (0)(1)
⎛ 1 ⎞
= cos−1 ⎜
⎟ = 54.7°
⎝ 3⎠
Let us now determine f for the angle between the direction of the applied tensile stress—i.e., the [100] direction—
and the normal to the (110) slip plane—i.e., the [110] direction. Again, using Equation 7.6 where u1 = 1, v1 = 0, w1
= 0 (for [100]), and u2 = 1, v2 = 1, w2 = 0 (for [110]), f is equal to
⎡
φ[100]−[110]
⎤
⎥
⎢
⎥
⎢ ⎡(1)2 + (0)2 + (0)2 ⎤ ⎡(1)2 + (1)2 + (0)2 ⎤ ⎥
⎦⎣
⎦⎦
⎣ ⎣
⎢
= cos−1
(1)(1) + (0)(1) + (0)(0)
⎛ 1 ⎞
= cos−1 ⎜
⎟ = 45°
⎝ 2⎠
Now, using Equation 7.2
τ R = σ cosφ cos λ
we solve for the resolved shear stress for this slip system as
τ R (110) − [111] = (4.0 MPa) ⎡⎣cos(45°)cos(54.7°) ⎤⎦ = (4.0 MPa)(0.707)(0.578) = 1.63 MPa
Now, we must determine the value of f for the (011)– [1 11] slip system—that is, the angle between the
direction of the applied stress, [100], and the normal to the (011) plane—i.e., the [011] direction. Again, using
Equation 7.6
⎡
λ[100]−[011]
⎤
⎥
⎢
⎥
2
2
2
2
2
2
⎢ ⎡(1) + (0) + (0) ⎤ ⎡(0) + (1) + (1) ⎤ ⎥
⎦⎣
⎦⎦
⎣ ⎣
⎢
= cos−1
(1)(0) + (0)(1) + (0)(1)
= cos−1 (0) = 90°
Thus, the resolved shear stress for this (011)– [111] slip system is
τ R (011) − [111] = = (4.0 MPa) ⎡⎣cos(90°)cos(54.7°) ⎤⎦ = (4.0 MPa)(0)(0.578) = 0 MPa
And, finally, it is necessary to determine the value of f for the (101) – [111] slip system—that is, the angle
between the direction of the applied stress, [100], and the normal to the (101) plane—i.e., the [101] direction.
Again, using Equation 7.6
⎡
λ[100]-[101]
⎤
⎥
⎢
⎥
⎢ ⎡(1)2 + (0)2 + (0)2 ⎤ ⎡(1)2 + (0)2 + (−1)2 ⎤ ⎥
⎦⎣
⎦⎦
⎣ ⎣
⎢
= cos−1
(1)(1) + (0)(0) + (0)(−1)
⎛ 1 ⎞
= cos−1 ⎜
⎟ = 45°
⎝ 2⎠
Here, as with the (110)– [111] slip system above, the value of f is 45°, which again leads to
τ R (101) − [111] = (4.0 MPa) ⎡⎣cos(45°)cos(54.7°) ⎤⎦ = (4.0 MPa)(0.707)(0.578) = 1.63 MPa
(b) The most favored slip system(s) is (are) the one(s) that has (have) the largest tR value. Both (110)–
[111] and (101) − [111] slip systems are most favored since they have the same tR (1.63 MPa), which is greater
than the tR value for (011) − [111] (viz., 0 MPa)
7.10 Consider a single crystal of some hypothetical metal that has the FCC crystal structure and is
oriented such that a tensile stress is applied along a [112] direction. If slip occurs on a (111) plane and in a [011]
direction, and the crystal yields at a stress of 5.12 MPa compute the critical resolved shear stress.
Solution
To solve this problem, we use Equation 7.4; however, it is first necessary to determine the values of f and
l. These determinations are possible using Equation 7.6. Now, l is the angle between [112] and [011] directions.
Therefore, relative to Equation 7.6 let us take u1 = 1, v1 = 1, and w1 = 2, as well as u2 = 0, v2 = 1, and w2 = 1. This
leads to
⎡
λ = cos−1 ⎢⎢
⎢⎣
⎧
(
⎤
⎥
⎥
2
2
2
2
2
2
u1 + v1 + w1 u2 + v2 + w2 ⎥
⎦
u1u2 + v1v2 + w1w2
)(
)
⎫
⎪
⎨
⎬
⎪ ⎡(1)2 + (1)2 + (2)2 ⎤ ⎡(0)2 + (1)2 + (1)2 ⎤ ⎪
⎦⎣
⎦⎭
⎩ ⎣
⎪
= cos−1
(1)(0) + (1)(1) + (2)(1)
⎛ 3 ⎞
= cos−1 ⎜
⎟ = 30°
⎝ 12 ⎠
Now for the determination of f, the normal to the (111) slip plane is the [111] direction. Again, using Equation 7.6,
where we now take u1 = 1, v1 = 1, w1 = 2 (for [112] ), and u2 = 1, v2 = 1, w2 = 1 (for [111]). Thus,
⎧
⎫
⎪
⎨
⎬
⎪ ⎡(1)2 + (1)2 + (2)2 ⎤ ⎡(1)2 + (1)2 + (1)2 ⎤ ⎪
⎦⎣
⎦⎭
⎩ ⎣
⎪
φ = cos−1
(1)(1) + (1)(1) + (2)(1)
⎛ 4 ⎞
= cos−1 ⎜
⎟ = 19.5°
⎝ 18 ⎠
It is now possible to compute the critical resolved shear stress (using Equation 7.4) as
τ crss = σ y (cosφ cos λ )
⎛ 4 ⎞⎛ 3 ⎞
= (5.12 MPa) ⎜
= 4.18 MPa
⎝ 18 ⎟⎠ ⎜⎝ 12 ⎟⎠
Strengthening by Grain Size Reduction
7.11 Briefly explain why small-angle grain boundaries are not as effective in interfering with the slip
process as are high-angle grain boundaries.
Solution
Small-angle grain boundaries are not as effective in interfering with the slip process as are high-angle grain
boundaries because there is not as much crystallographic misalignment in the grain boundary region for small-angle,
and therefore not as much change in slip direction.
7.12 Briefly explain why HCP metals are typically more brittle than FCC and BCC metals.
Solution
Hexagonal close packed metals are typically more brittle than FCC and BCC metals because there are
fewer slip systems in HCP.
7.13 (a) From the plot of yield strength versus (grain diameter)–1/2 for a 70 Cu–30 Zn cartridge brass in
Figure 7.15, determine values for the constants σ0 and ky in Equation 7.7.
(b) Now predict the yield strength of this alloy when the average grain diameter is 2.0 × 10–3 mm.
Solution
(a) Perhaps the easiest way to solve for s0 and ky in Equation 7.7 is to pick two values each of sy and d-1/2
from Figure 7.15, and then solve two simultaneous equations, which may be set up. For example
d-1/2 (mm)-1/2
sy (MPa)
4
75
12
175
Since Equation 7.7 is of the form
σ y = σ 0 + k y d −1/2
the two equations are as follows:
75 = σ 0 + 4 k y
175 = σ 0 + 12k y
Simultaneous solution of these equations yields the values of
k y = 12.5 MPa(mm)1/2
s0 = 25 MPa
(b) When d = 2.0 ´ 10-3 mm, d-1/2 = 22.4 mm-1/2, and, using Equation 7.7,
σ y = σ 0 + k y d −1/2
1/2 ⎤
⎡
= (25 MPa) + ⎢12.5 MPa(mm) ⎥ (22.4 mm −1/2 ) = 305 MPa
⎣
⎦
7.14 If it is assumed that the plot in Figure 7.15 is for non-cold-worked brass, determine the grain size of
the alloy in Figure 7.19; assume its composition is the same as the alloy in Figure 7.15.
Solution
From Figure 7.19a, the yield strength of brass at 0%CW is approximately 175 MPa. This yield strength
from Figure 7.15 corresponds to a d-1/2 value of approximately 12.0 (mm) -1/2. Thus,
d=
1
⎡12.0 (mm)−1/2 ⎤
⎣
⎦
2
= 6.94 × 10−3 mm
Strain Hardening
7.15 Two previously undeformed cylindrical specimens of an alloy are to be strain hardened by reducing
their cross-sectional areas (while maintaining their circular cross sections). For one specimen, the initial and
deformed radii are 15 and 12 mm, respectively. The second specimen, with an initial radius of 11 mm, must have the
same deformed hardness as the first specimen; compute the second specimen’s radius after deformation.
Solution
In order for these two cylindrical specimens to have the same deformed hardness, they must be deformed to
the same percent cold work. For the first specimen
⎛ π r2 − π r2 ⎞
⎛ A − Ad ⎞
d
%CW = ⎜ 0
× 100 = ⎜ 0
⎟ × 100
⎟
2
A0 ⎠
π
r
⎝
⎝
⎠
0
in which r0 and rd denote the original and deformed specimen radii, respectively. Substitution of the values for
these two parameters provided in the problem statement leads to the following:
⎛ π r2 − π r2 ⎞
π (15 mm)2 − π (15 mm)2
d
%CW = ⎜ 0
× 100 =
× 100 = 36%CW
⎟
π r02
π (15 mm)2
⎝
⎠
For the second specimen, the deformed radius is computed using the above equation for %CW and solving for rd as
rd = r0 1 −
%CW
100
And insertion of the value of %CW determined above (i.e., 36%CW) and the r0 (11 mm) for the second specimen
leads to
rd = (11 mm) 1 −
36%CW
= 8.80 mm
100
7.16 A cylindrical specimen of cold-worked copper has a ductility (%EL) of 15%. If its cold-worked radius
is 6.4 mm, what was its radius before deformation?
Solution
This problem calls for us to calculate the precold-worked radius of a cylindrical specimen of copper that
has a cold-worked ductility of 15%EL. From Figure 7.19c, copper that has a ductility of 15%EL will have
experienced a deformation of about 20%CW. For a cylindrical specimen, Equation 7.8 becomes
⎛ π r2 − π r2 ⎞
d
%CW = ⎜ 0
⎟ × 100
π r 20
⎝
⎠
Since rd = 6.4 mm, solving for r0 yields
r0 =
rd
%CW
1−
100
=
6.4 mm
20.0
1−
100
= 7.2 mm
7.17 (a) What is the approximate ductility (%EL) of a brass that has a yield strength of 345 MPa?
(b) What is the approximate Brinell hardness of a 1040 steel having a yield strength of 620 MPa?
Solution
In order to solve this part of the problem, it is necessary to consult Figures 7.19a and 7.19c. From Figure
7.19a, a yield strength of 345 MPa for brass corresponds to 20%CW. A brass that has been cold-worked 20% will
have a ductility of about 24%EL (Figure 7.19c).
(b) This portion of the problem asks for the Brinell hardness of a 1040 steel having a yield strength of 620
MPa. From Figure 7.19a, a yield strength of 620 MPa for a 1040 steel corresponds to about 5%CW. A 1040 steel
that has been cold worked 5% will have a tensile strength of about 750 MPa (Figure 7.19b). Finally, using Equation
6.20a
HB =
TS(MPa) 750 MPa
=
= 217
3.45
3.45
Recovery
Recrystallization
Grain Growth
7.18 Explain the differences in grain structure for a metal that has been cold worked and one that has been
cold worked and then recrystallized.
Solution
During cold-working, the grain structure of the metal has been distorted to accommodate the deformation.
Recrystallization produces grains that are equiaxed and smaller than the parent grains.
7.19 (a) What is the driving force for recrystallization?
(b) What is the driving force for grain growth?
Solution
(a) The driving force for recrystallization is the difference in internal energy between the strained and
unstrained material.
(b) The driving force for grain growth is the reduction in grain boundary energy as the total grain boundary
area decreases.
7.20 Consider a hypothetical material that has a grain diameter of 2.1 ´ 10-2 mm. After a heat treatment
at 600°C for 3 h, the grain diameter has increased to 7.2 ´ 10-2 mm. Compute the grain diameter when a specimen
of this same original material (i.e., d0 = 2.1 ´ 10-2 mm) is heated for 1.7 h at 600°C. Assume the n grain diameter
exponent has a value of 2.
Solution
To solve this problem requires that we use Equation 7.9 with n = 2. It is first necessary to solve for the
parameter K in this equation, using values given in the problem statement of d0 (2.1 ´ 10-2 mm) and the grain
diameter after the 3-h heat treatment (7.2 ´ 10-2 mm). The computation for K using a rearranged form of Equation
7.9 in which K becomes the dependent parameter is as follows:
K=
=
d 2 − d02
t
(7.2 × 10−2 mm)2 − (2.1 × 10−2 mm)2
3h
= 1.58 × 10−3 mm2 /h
It is now possible to solve for the value of d after a heat treatment of 1.7 h using a rearranged form of Equation 7.9:
d = d02 + Kt
= (2.1 × 10−2 mm)2 + (1.58 × 10−3 mm2 /h)(1.7 h)
= 5.59 × 10−2 mm
7.21 The average grain diameter for a brass material was measured as a function of time at 650°C, which
is shown in the following table at two different times:
Time (min)
Grain Diameter (mm)
40
5.6 × 10–2
100
8.0 × 10–2
(a) What was the original grain diameter?
(b) What grain diameter would you predict after 200 min at 650°C?
Solution
(a) Using the data given and Equation 7.9 (taking n = 2)—that is
d 2 − d02 = Kt
we may set up two simultaneous equations with d0 and K as unknowns, as follows:
(5.6 × 10−2 mm)2 − d02 = (40 min) K
(8.0 × 10−2 mm)2 − d02 = (100 min) K
Solution of these expressions yields a value for d0, the original grain diameter, of
d0 = 0.031 mm,
and a value for K of
K = 5.44 × 10−5 mm2 /min
(b) At 200 min, the diameter d is computed using a rearranged form of Equation 7.9 (incorporating values
of d0 and K that were just determined) as follows:
d=
=
d02 + Kt
(0.031 mm)2 + (5.44 × 10−5 mm2 /min) (200 min)
= 0.109 mm
7.22
Grain growth is strongly dependent on temperature (i.e., rate of grain growth increases with
increasing temperature), yet temperature is not explicitly included in Equation 7.9.
(a) Into which of the parameters in this expression would you expect temperature to be included?
(b) On the basis of your intuition, cite an explicit expression for this temperature dependence.
Solution
(a) The temperature dependence of grain growth is incorporated into the constant K in Equation 7.9.
(b) The explicit expression for this temperature dependence is of the form
⎛ Q ⎞
K = K0 exp ⎜ −
⎝ RT ⎟⎠
in which K0 is a temperature-independent constant, the parameter Q is an activation energy, and R and T are the gas
constant and absolute temperature, respectively.
7.23 The following yield strength, grain diameter, and heat treatment time (for grain growth) data were
gathered for an iron specimen that was heat treated at 800°C. Using these data compute the yield strength of a
specimen that was heated at 800°C for 3 h. Assume a value of 2 for n, the grain diameter exponent.
Grain diameter (mm)
Yield Strength (MPa)
Heat Treating Time (h)
0.028
300
10
0.010
385
1
Solution
This problem is solved using the following steps:
1. From data given in the problem statement, determine values of d0 and K in Equation 7.9.
2. Using these data, compute the value of d after the heat treatment (800°C for 3 h).
3. From data provided in the problem statement, determine values of s0 and ky in Equation 7.7.
4. Calculate the value of sy using Equation 7.7 incorporating the d value determined in step 2.
Step 1
Using grain diameter-heat treating time data provided in the problem statement, we set up two
simultaneous expressions of Equation 7.9—i.e.,
d 2 − d02 = Kt
as follows:
(0.028 mm)2 − d02 = K(10 h)
(0.010 mm)2 − d02 = K(1 h)
From these expressions it is possible to solve for values of d0 and K:
d0 = 0.0049 mm
K = 7.60 ´ 10-5 mm2/h
Step 2
We now compute the grain size d after the three-hour heat treatment using a rearranged form of Equation
7.9 and the above values for d0 and K as follows:
d = Kt + d02
= (7.60 × 10−5 mm2 /h)(3 h) + (0.0049 mm)2
= 0.0159 mm
Step 3
Using grain diameter-yield strength data provided in the problem statement, we set up two simultaneous
expressions of Equation 7.7—i.e.,
σ y = σ 0 + k y d −1/2
as follows:
300 MPa = σ 0 + k y (0.028 mm)−1/2
385 MPa = σ 0 + k y (0.010 mm)−1/2
from which we determine values for ky and s0; these values are as follows:
ky = 21.25 MPa-mm1/2
s0 = 172.5 MPa
Step 4
Finally, it is possible to calculate the value of sy using Equation 7.7 incorporating the d value determined
in step 2. Thus,
σ y = σ 0 + k y d −1/2
= 172.5 MPa + (21.25 MPa-mm1/2 )(0.0159 mm)−1/2
= 341 MPa
DESIGN PROBLEMS
Strain Hardening
Recrystallization
7.D1 Determine whether it is possible to cold work steel so as to give a minimum Brinell hardness of 240
and at the same time have a ductility of at least 15%EL. Justify your answer.
Solution
This problem calls for us to determine whether it is possible to cold work steel so as to give a minimum
Brinell hardness of 240 and a ductility of at least 15%EL. According to Figure 6.19, a Brinell hardness of 240
corresponds to a tensile strength of 800 MPa. Furthermore, from Figure 7.19b, in order to achieve a tensile strength
of 800 MPa, deformation of at least 13%CW is necessary. Finally, if we cold-work the steel to 13%CW, then the
ductility is 15%EL from Figure 7.19c. Therefore, it is possible to meet both of these criteria by plastically
deforming the steel.
7.D2 A cylindrical specimen of cold-worked steel has a Brinell hardness of 240.
(a) Estimate its ductility in percent elongation.
(b) If the specimen remained cylindrical during deformation and its original radius was 10 mm, determine
its radius after deformation.
Solution
(a) For this portion of the problem we are to determine the ductility of cold-worked steel that has a Brinell
hardness of 240. From Figure 6.19, for steel, a Brinell hardness of 240 corresponds to a tensile strength of 800 MPa,
which, from Figure 7.19b, requires a deformation of 12.5%CW. Furthermore, 12.5%CW yields a ductility of about
15%EL for steel, Figure 7.19c.
(b) We are now asked to determine the radius after deformation if the non-cold-worked radius is 10 mm.
From Equation 7.8 and for a cylindrical specimen
⎛ A − Ad ⎞
%CW = ⎜ 0
⎟ × 100
⎝ A0 ⎠
⎡ π r2 − π r 2 ⎤
d ⎥ × 100
=⎢ 0
⎢⎣
⎥⎦
π r02
Now, solving for rd from this expression, we get
rd = r0 1 −
%CW
100
And, finally, for 15%CW
rd = (10 mm) 1 −
15%CW
= 9.22 mm
100
7.D3 A cylindrical rod of 1040 steel originally 11.4 mm in diameter is to be cold worked by drawing; the
circular cross section will be maintained during deformation. A cold-worked tensile strength in excess of 825 MPa
and a ductility of at least 12%EL are desired. Furthermore, the final diameter must be 8.9 mm. Explain how this
may be accomplished.
Solution
This problem calls for us to explain the procedure by which a cylindrical rod of 1040 steel may be
deformed so as to produce a given final diameter (8.9 mm), as well as a specific minimum tensile strength (825
MPa) and minimum ductility (12%EL). First let us calculate the percent cold work and attendant tensile strength
and ductility if the drawing is carried out without interruption. From Equation 7.8, for a cylindrical specimen having
original and deformed diameters of 11.4 mm and 8.9 mm, respectively,
⎛ A − Ad ⎞
%CW = ⎜ 0
⎟ × 100
⎝ A0 ⎠
2
=
⎛d ⎞
⎛d ⎞
π⎜ 0⎟ − π⎜ d⎟
⎝ 2⎠
⎝ 2⎠
⎛d ⎞
π⎜ 0⎟
⎝ 2⎠
2
=
2
× 100
2
⎛ 11.4 mm ⎞
⎛ 8.9 mm ⎞
π⎜
−π⎜
⎟
2
2 ⎟⎠
⎝
⎠
⎝
⎛ 11.4 mm ⎞
π⎜
⎟⎠
2
⎝
2
2
× 100 = 40%CW
At 40%CW, the steel will have a tensile strength on the order of 900 MPa (Figure 7.19b), which is adequate;
however, the ductility will be less than 9%EL (Figure 7.19c), which is insufficient.
Instead of performing the drawing in a single operation, let us initially draw some fraction of the total
deformation, then anneal to recrystallize, and, finally, cold-work the material a second time in order to achieve the
final diameter, tensile strength, and ductility.
Reference to Figure 7.19b indicates that 17%CW is necessary to yield a tensile strength of 825 MPa.
Similarly, a maximum of 24%CW is possible for 12%EL (Figure 7.19c). The average of these extremes is
20.5%CW. If the final diameter after the first drawing is d0′ , then, from Equation 7.8
2
20.5%CW =
⎛ d′ ⎞
⎛ 8.9 mm ⎞
π⎜ 0 ⎟ − π⎜
⎝ 2 ⎟⎠
⎝ 2⎠
⎛ d′ ⎞
π⎜ 0 ⎟
⎝ 2⎠
2
And, solving for d0′ , yields
d0′ =
8.9 mm
1 −
20.5%CW
100
= 9.98 mm
2
× 100
7.D4 Consider the brass alloy discussed in Problem 7.21. Given the following yield strengths for the two
specimens, compute the heat treatment time required at 650°C to give a yield strength of 90 MPa. Assume a value
of 2 for n, the grain diameter exponent.
Time (min)
Yield Strength (MPa)
40
80
100
70
Solution
This problem is solved using the following steps:
1. From data provided in the problem statement and Problem 7.21, determine values of s0 and ky in Equation 7.7.
2. Calculate the value of d that is required for to give a yield strength of 90 MPa using these values and Equation
7.7.
3. From data given in Problem 7.21, determine values of d0 and K in Equation 7.9.
4. Calculate the heat-treating time required to give the d value determined in step 2, using Equation 7.9.
Step 1
Using grain diameter-yield strength data provided in the problem statement and Problem 7.21, we set up
two simultaneous expressions of Equation 7.7—i.e.,
σ y = σ 0 + k y d −1/2
as follows:
80 MPa = σ 0 + k y (0.056 mm)−1/2
70 MPa = σ 0 + k y (0.080 mm)−1/2
From these expressions it is possible to solve for values of s0 and ky:
ky = 14.49 MPa-mm1/2
s0 = 18.75 MPa
Step 2
We now compute the grain size d required to give a yield strength of 90 MPa using a rearranged from of
Equation 7.7 and the above values of s0 and ky as follows:
⎛ ky ⎞
d=⎜
⎟
⎝ σ y −σ0 ⎠
2
⎛ 14.49 MPa-mm1/2 ⎞
=⎜
⎟
⎝ 90 MPa − 18.75 MPa ⎠
2
= 0.041 mm
Step 3
Using grain diameter-heat treating time data provided in Problem 7.41, we set up two simultaneous expressions of
Equation 7.9—i.e.,
d 2 − d02 = Kt
as follows:
(0.056 mm)2 − d02 = K(40 min)
(0.080 mm)2 − d02 = K(100 min)
From which we determine values for d0 and K as follows:
d0 = 0.031 mm
K = 5.44 ´ 10-5 mm2/min
Step 4
And finally, we calculate the heat-treating time t required to give the d value determined in step 2 (0.041
mm), using a rearranged form of Equation 7.9:
t=
=
d 2 − d02
K
(0.041 mm)2 − (0.031 mm)2
5.44 × 10−5 mm2 /min
= 13.2 min
CHAPTER 8
FAILURE
PROBLEM SOLUTIONS
Principles of Fracture Mechanics
8.1 Estimate the theoretical fracture strength of a brittle material if it is known that fracture occurs by the
propagation of an elliptically shaped surface crack of length 0.5 mm and a tip radius of curvature of 5 × 10–3 mm,
when a stress of 1035 MPa is applied.
Solution
In order to estimate the theoretical fracture strength of this material it is necessary to calculate sm using
Equation 8.1 given that s0 = 1035 MPa, a = 0.5 mm, and rt = 5 ´ 10-3 mm. Thus,
⎛ a⎞
σ m = 2σ 0 ⎜ ⎟
⎝ ρt ⎠
1/2
⎛ 0.5 mm ⎞
= (2)(1035 MPa) ⎜
⎟
⎝ 5 × 10−3 mm ⎠
1/2
= 2.07 × 104 MPa = 20.7 GPa
8.2 If the specific surface energy for aluminum oxide is 0.90 J/m2, then using data in Table 12.5, compute
the critical stress required for the propagation of an internal crack of length 0.40 mm.
Solution
We may determine the critical stress required for the propagation of an internal crack in aluminum oxide
using Equation 8.3. Taking the value of 393 GPa (Table 12.5) as the modulus of elasticity, and realizing that values
for gs (0.90 J/m2) and 2a (0.40 mm) are given in the problem statement, leads to
⎛ 2E γ s ⎞
σc = ⎜
⎝ π a ⎟⎠
1/2
1/2
⎡
⎤
⎢
⎥
9
2
⎢ (2) (393 × 10 N/m ) (0.90 N/m) ⎥
=⎢
⎥
⎛ 0.4 × 10−3 m ⎞
⎢
⎥
(
π
)
⎜
⎟
⎢
⎥
2
⎝
⎠
⎣
⎦
= 33.6 × 106 N/m2 = 33.6 MPa
8.3 A specimen of a 4340 steel alloy with a plane strain fracture toughness of 54.8 MPa m is exposed to
a stress of 1030 MPa. Will this specimen experience fracture if the largest surface crack is 0.5 mm long? Why or
why not? Assume that the parameter Y has a value of 1.0.
Solution
This problem asks us to determine whether or not the 4340 steel alloy specimen will fracture when exposed
to a stress of 1030 MPa. This requires that we solve for sc from Equation 8.6, given the values of KIc
(54.8 MPa m) , the largest value of a (0.5 mm), and Y (1.0). Thus, the critical stress for fracture is equal to
σc =
=
K Ic
Y πa
54.8 MPa m
(1.0) (π )(0.5 × 10−3 m)
= 1380 MPa
Therefore, fracture will not occur because this specimen will tolerate a stress of 1380 MPa before fracture, which is
greater than the applied stress of 1030 MPa.
8.4 Suppose that a wing component on an aircraft is fabricated from an aluminum alloy that has a planestrain fracture toughness of 26.0 MPa m . It has been determined that fracture results at a stress of 112 MPa when
the maximum internal crack length is 8.6 mm. For this same component and alloy, compute the stress level at which
fracture will occur for a critical internal crack length of 6.0 mm.
Solution
This problem asks us to determine the stress level at which a wing component on an aircraft will fracture
for a given fracture toughness (26 MPa m ) and maximum internal crack length (6.0 mm), given that fracture
occurs for the same component using the same alloy at one stress level (112 MPa) and another internal crack length
(8.6 mm). (Note: Because the cracks are internal, their lengths are equal to 2a.) It first becomes necessary to solve
for the parameter Y for the conditions under which fracture occurred using Equation 8.5. Therefore,
Y=
=
K Ic
σ πa
26 MPa m
⎛ 8.6 × 10−3 m ⎞
(112 MPa) (π ) ⎜
⎟
2
⎝
⎠
= 2.0
Now we will solve for sc (for a crack length of 6 mm) using Equation 8.6 as
σc =
K Ic
Y πa
=
26 MPa m
⎛ 6 × 10−3 m ⎞
(2.0) (π ) ⎜
⎟
2
⎝
⎠
= 134 MPa
8.5
A large plate is fabricated from a steel alloy that has a plane strain fracture toughness of
82.4 MPa m. If the plate is exposed to a tensile stress of 345 MPa during service use, determine the minimum
length of a surface crack that will lead to fracture. Assume a value of 1.0 for Y.
Solution
For this problem, we are given values of KIc (82.4 MPa m) , s (345 MPa), and Y (1.0) for a large plate
and are asked to determine the minimum length of a surface crack that will lead to fracture. All we need do is to
solve for ac using Equation 8.7; therefore
ac =
1 ⎛ K Ic ⎞
π ⎜⎝ σ Y ⎟⎠
2
1 ⎡ 82.4 MPa m ⎤
= ⎢
⎥
π ⎢⎣ (345 MPa)(1.0) ⎥⎦
2
= 0.0182 m = 18.2 mm
8.6 A structural component in the form of a wide plate is to be fabricated from a steel alloy that has a
plane-strain fracture toughness of 98.9 MPa m and a yield strength of 860 MPa. The flaw size resolution limit of
the flaw detection apparatus is 3.0 mm. If the design stress is one-half the yield strength and the value of Y is 1.0,
determine whether a critical flaw for this plate is subject to detection.
Solution
This problem asks that we determine whether or not a critical flaw in a wide plate is subject to detection
given the limit of the flaw detection apparatus (3.0 mm), the value of KIc (98.9 MPa m ) , the design stress (sy/2
in which sy = 860 MPa), and Y = 1.0. We first need to compute the value of ac using Equation 8.7; thus
ac =
1 ⎛ K Ic ⎞
π ⎜⎝ σ Y ⎟⎠
2
⎡
⎤
⎢
⎥
1 ⎢ 98.9 MPa m ⎥
=
⎥
π ⎢ ⎛ 860 MPa ⎞
⎢⎜
⎟⎠ (1.0) ⎥
2
⎝
⎣
⎦
2
= 0.0168 m = 16.8 mm
Therefore, the critical flaw is subject to detection since this value of ac (16.8 mm) is greater than the 3.0 mm
resolution limit.
Fracture Toughness Testing
8.7 The following tabulated data were gathered from a series of Charpy impact tests on a tempered 4340
steel alloy.
Temperature (°C)
Impact Energy (J)
0
105
–25
104
–50
103
–75
97
–100
63
–113
40
–125
34
–150
28
–175
25
–200
24
(a) Plot the data as impact energy versus temperature.
(b) Determine a ductile-to-brittle transition temperature as the temperature corresponding to the average
of the maximum and minimum impact energies.
(c) Determine a ductile-to-brittle transition temperature as the temperature at which the impact energy is
50 J.
Solution
(a) The plot of impact energy versus temperature is shown below.
(b) The average of the maximum and minimum impact energies from the data is
Average =
105 J + 24 J
= 64.5 J
2
As indicated on the plot by the one set of dashed lines, the ductile-to-brittle transition temperature according to this
criterion is about –100°C.
(c) Also, as noted on the plot by the other set of dashed lines, the ductile-to-brittle transition temperature
for an impact energy of 50 J is about –110°C.
8.8 What is the maximum carbon content possible for a plain carbon steel that must have an impact energy
of at least 200 J at -50°C?
Solution
From the curves in Figure 8.17, for only the 0.11 wt%C and 0.01 wt% C steels are impact energies (and
therefore, ductile-to-brittle temperatures) greater than 200 J.
concentration.
Therefore, 0.11 wt% is the maximum carbon
Cyclic Stresses
The S–N Curve
8.9 A fatigue test was conducted in which the mean stress was 70 MPa, and the stress amplitude was 210
MPa.
(a) Compute the maximum and minimum stress levels.
(b) Compute the stress ratio.
(c) Compute the magnitude of the stress range.
Solution
(a) Given the values of sm (70 MPa) and sa (210 MPa) we are asked to compute smax and smin. From
Equation 8.11
σm =
σ max + σ min
= 70 MPa
2
It is the case that
σ max + σ min = 140 MPa
(8.11a)
Furthermore, utilization of Equation 8.13 yields
σa =
σ max − σ min
= 210 MPa
2
Such that
σ max − σ min = 420 MPa
Simultaneously solving Equations 8.11a and 8.13a leads to
σ max = 280 MPa
σ min = − 140 MPa
(b) Using Equation 8.14 the stress ratio R is determined as follows:
R=
σ min −140 MPa
=
= − 0.50
σ max
280 MPa
(c) The magnitude of the stress range sr is determined using Equation 8.12 as
(8.13a)
σ r = σ max − σ min = 280 MPa − ( − 140 MPa)
= 420 MPa
8.10 A cylindrical 4340 steel bar is subjected to reversed rotating-bending stress cycling, which yielded
the test results presented in Figure 8.21. If the maximum applied load is 5,000 N, compute the minimum allowable
bar diameter to ensure that fatigue failure will not occur. Assume a factor of safety of 2.25 and that the distance
between load-bearing points is 55.0 mm.
Solution
From Figure 8.21, the fatigue limit for this steel is 480 MPa. For rotating-bending tests on a cylindrical
specimen of diameter d0, stress is defined in Equation 8.15 as
σ=
16FL
π d03
When we divide stress by the factor of safety (N), the above equation becomes Equation 8.16—that is
σ 16FL
=
N π d3
0
And solving for d0 realizing that s is the fatigue limit (480 MPa or 480 ´ 106 N/m2) and incorporating values for F
(5,000 N), L (55.0 mm = 55 ´ 10-3 m), and N (2.25) provided in the problem statement, leads to
⎛ 16FLN ⎞
d0 = ⎜
⎝ σπ ⎟⎠
1/3
1/3
⎡ (16)(5,000 N)(55.0 × 10−3 m)(2.25) ⎤
⎢
⎥
(480 × 106 N/m2 )(π )
⎢⎣
⎥⎦
0.0187 m = 18.7 mm
8.11 A cylindrical rod of diameter 6.7 mm fabricated from a 70Cu-30Zn brass alloy is subjected to
rotating-bending load cycling; test results (as S-N behavior) are shown in Figure 8.21. If the maximum and
minimum loads are +120 N and –120 N, respectively, determine its fatigue life. Assume that the separation between
loadbearing points is 67.5 mm.
Solution
In order to solve this problem, we compute the maximum stress using Equation 8.15, and then determine
the fatigue life from the curve in Figure 8.21 for the brass material. In Equation 8.15, F is the maximum applied
load, which for this problem is +120 N. Values for L and d0 are provided in the problem statement—viz. 67.5 mm
(67.5 ´ 10-3 m) and 6.7 mm (6.7 ´ 10-3 m), respectively. Therefore, the maximum stress s is equal to
σ=
=
16FL
π d03
(16)(120 N)(67.5 × 10−3 m)
(π )(6.7 × 10−3 m)3
= 137 × 106 N/m2 = 137 MPa
From Figure 8.21 and the curve for brass, the logarithm of the fatigue life (log Nf) at 137 MPa is about 6.5, which
means that the fatigue life is equal to
N f = 106.5 cycles = 3 × 106 cycles
8.12 The fatigue data for a brass alloy are given as follows:
Stress Amplitude (MPa)
Cycles to Failure
170
3.7 × 104
148
1.0 × 105
130
3.0 × 105
114
1.0 × 106
92
1.0 × 107
80
1.0 × 108
74
1.0 × 109
(a) Make an S–N plot (stress amplitude versus logarithm of cycles to failure) using these data.
(b) Determine the fatigue strength at 4 × 106 cycles.
(c) Determine the fatigue life for 120 MPa.
Solution
(a) The fatigue data for this alloy are plotted below.
(b) As indicated by one set of dashed lines on the plot, the fatigue strength at 4 ´ 106 cycles [log (4 ´ 106)
= 6.6] is about 100 MPa.
(c) As noted by the other set of dashed lines, the fatigue life for 120 MPa is about 6 ´ 105 cycles (i.e., the
log of the lifetime is about 5.8).
8.13 Suppose that the fatigue data for the brass alloy in Problem 8.12 were taken from bending-rotating
tests and that a rod of this alloy is to be used for an automobile axle that rotates at an average rotational velocity of
1800 revolutions per minute. Give the maximum bending stress amplitude possible for each of the following lifetimes
of the rod: (a) 1 year, (b) 1 month, (c) 1 day, and (d) 1 hour.
Solution
We are asked to compute the maximum torsional stress amplitude possible at each of several fatigue
lifetimes for the brass alloy the fatigue behavior of which is given in Problem 8.12. For each lifetime, first compute
the number of cycles, and then read the corresponding fatigue strength from the above plot.
(a) Fatigue lifetime = (1 yr)(365 days/yr)(24 h/day)(60 min/h)(1800 cycles/min) = 9.5 ´ 108 cycles. The
stress amplitude corresponding to this lifetime is about 74 MPa.
(b) Fatigue lifetime = (30 days)(24 h/day)(60 min/h)(1800 cycles/min) = 7.8 ´ 107 cycles. The stress
amplitude corresponding to this lifetime is about 80 MPa.
(c) Fatigue lifetime = (24 h)(60 min/h)(1800 cycles/min) = 2.6 ´ 106 cycles. The stress amplitude
corresponding to this lifetime is about 115 MPa.
(d) Fatigue lifetime = (60 min/h)(1800 cycles/min) = 108,000 cycles. The stress amplitude corresponding
to this lifetime is about 145 MPa.
8.14 Three identical fatigue specimens (denoted A, B, and C) are fabricated from a nonferrous alloy. Each
is subjected to one of the maximum-minimum stress cycles listed in the following table; the frequency is the same for
all three tests.
Specimen
smax (MPa)
smin (MPa)
A
+450
–150
B
+300
–300
C
+500
–200
(a) Rank the fatigue lifetimes of these three specimens from the longest to the shortest.
(b) Now justify this ranking using a schematic S–N plot.
Solution
For this problem we are given, for three identical fatigue specimens of the same material, smax and smin
data, and are asked to rank the lifetimes from the longest to the shortest. In order to do this, it is necessary to
compute both the mean stress and stress amplitude for each specimen. From Equation 8.11, mean stresses are
calculated as follows:
σm =
σ m (A) =
450 MPa + (−150 MPa)
= 150 MPa
2
σ m (B) =
σ m (C) =
σ max + σ min
2
300 MPa + (−300 MPa)
= 0 MPa
2
500 MPa + (−200 MPa)
= 150 MPa
2
Furthermore, using Equation 8.13, stress amplitudes are determined:
σa =
σ max − σ min
2
σ a (A) =
450 MPa − (−150 MPa)
= 300 MPa
2
σ a (B) =
300 MPa − (−300 MPa)
= 300 MPa
2
σ a (C) =
500 MPa − (−200 MPa)
= 350 MPa
2
On the basis of these results, the fatigue lifetime for specimen B will be greater than specimen A which in turn will
be greater than specimen C. This conclusion is based upon the following S-N plot on which curves are plotted for
two sm values.
Crack Initiation and Propagation
Factors That Affect Fatigue Life
8.15 List four measures that may be taken to increase the resistance to fatigue of a metal alloy.
Answer
Four measures that may be taken to increase the fatigue resistance of a metal alloy are as follows:
(1) Polish the surface to remove stress amplification sites.
(2) Reduce the number of internal defects (pores, etc.) by means of altering processing and fabrication
techniques.
(3) Modify the design to eliminate notches and sudden contour changes.
(4) Harden the outer surface of the structure by case hardening (carburizing, nitriding) or shot peening.
Generalized Creep Behavior
8.16 The following creep data were taken on an aluminum alloy at 480°C and a constant stress of 2.75
MPa. Plot the data as strain versus time, then determine the steady-state or minimum creep rate. Note: The initial
and instantaneous strain is not included.
Time (min)
Strain
Time (min)
Strain
0
0.00
18
0.82
2
0.22
20
0.88
4
0.34
22
0.95
6
0.41
24
1.03
8
0.48
26
1.12
10
0.55
28
1.22
12
0.62
30
1.36
14
0.68
32
1.53
16
0.75
34
1.77
Solution
These creep data are plotted below
The steady-state creep rate (De/Dt) is the slope of the linear region (i.e., the straight line that has been superimposed
on the curve). Taking strain values at 0 min and 30 min, leads to the following:
Δε
1.20 − 0.25
=
= 3.2 × 10−2 min−1
Δt 30 min − 0 min
Stress and Temperature Effects
8.17 For a cylindrical S-590 alloy specimen (Figure 8.33) originally 14.5 mm in diameter and 400 mm
long, what tensile load is necessary to produce a total elongation of 52.7 mm after 1150 h at 650°C? Assume that
the sum of instantaneous and primary creep elongations is 4.3 mm.
Solution
It is first necessary to calculate the steady state creep rate so that we may utilize Figure 8.33 in order to
determine the tensile stress. The steady state elongation, Dls, is just the difference between the total elongation and
the sum of the instantaneous and primary creep elongations; that is,
Δls = 52.7 mm − 4.3 mm = 48.4 mm
Now the steady state creep rate, ε!s , is just
ε!s =
Δε
Δt
Δls 48.4 mm
l
= 0 = 400 mm
Δt
1150 h
= 1.05 × 10−4 h −1
Employing the 650°C line in Figure 8.33, a steady state creep rate of 1.05 ´ 10-4 h-1 corresponds to a stress s of
about 300 MPa. From this we may compute the tensile load (for this cylindrical specimen) using Equation 6.1 as
follows:
⎛d ⎞
F = σ A0 = σπ ⎜ 0 ⎟
⎝ 2⎠
2
Here d0 is the original cross-sectional diameter (14.5 mm = 14.5 ´ 10-3 m). Solving this equation for tensile load,
incorporating the value of stress determined from Figure 8.33 above (300 MPa = 300 ´ 106 N/m2) leads to
⎛d ⎞
F = σπ ⎜ 0 ⎟
⎝ 2⎠
= (300 × 106 N/m2 )(π )
2
⎛ 14.5 × 10−3 m ⎞
⎜
⎟
2
⎝
⎠
= 49,500 N
2
8.18 A cylindrical specimen 13.2 mm in diameter of an S-590 alloy is to be exposed to a tensile load of
27,000 N. At approximately what temperature will the steady-state creep be 10-3 h-1?
Solution
Let us first determine the stress imposed on this specimen using values of cross-section diameter and
applied load; this is possible using Equation 6.1 as
σ=
F
=
A0
F
⎛d ⎞
π⎜ 0⎟
⎝ 2⎠
2
Inasmuch as d0 = 13.2 mm (13.2 ´ 10-3 m) and F = 27,000 N, the stress is equal to
σ=
27,000 N
⎛ 13.2 × 10−3 m ⎞
π⎜
⎟
2
⎝
⎠
2
= 197 × 106 N/m2 = 197 MPa
Or approximately 200 MPa. From Figure 8.33 the point corresponding 10-3 h-1 and 200 MPa lies approximately
midway between 730°C and 815°C lines; therefore, the temperature would be approximately 775°C.
8.19 A cylindrical component constructed from an S-590 alloy (Figure 8.32) has a diameter of 14.5 mm.
Determine the maximum load that may be applied for it to survive 10 h at 925°C.
Solution
We are asked in this problem to determine the maximum load that may be applied to a cylindrical S-590
alloy component that must survive 10 h at 925°C. From Figure 8.32, the stress corresponding to 10 h is about 100
MPa. In order to determine the applied load, we use the expression used to compute stress, as given in Equation 6.1,
which for a cylindrical specimen takes the form:
σ=
F
=
A0
F
⎛d ⎞
π⎜ 0⎟
⎝ 2⎠
2
From which the force F (or load) is equal to
⎛d ⎞
F = σπ ⎜ 0 ⎟
⎝ 2⎠
2
For a stress of 100 MPa (100 ´ 106 N/m2) and a cross-sectional diameter of 14.5 mm (14.5 ´ 10-3 m) the load is
equal to
⎛ 14.5 × 10−3 m ⎞
F = (100 × 106 N/m2 )(π ) ⎜
⎟
2
⎝
⎠
= 16,500 N
2
8.20 From Equation 8.21, if the logarithm of ε!s is plotted versus the logarithm of σ, then a straight line
should result, the slope of which is the stress exponent n. Using Figure 8.33, determine the value of n for the S-590
alloy at 925°C, and for the initial (lower-temperature) straight line segments at each of 650°C, 730°C, and 815°C.
Solution
The slope of the line from a log ε!s versus log s plot yields the value of n in Equation 8.21; that is
n=
Δ log ε!s
Δ log σ
We are asked to determine the values of n for the creep data at the four temperatures in Figure 8.33. This is
accomplished by taking ratios of the differences between two log ε!s and log s values. (Note: Figure 8.33 plots log
s versus log ε!s ; therefore, values of n are equal to the reciprocals of the slopes of the straight-line segments.)
Thus, for the line segment for 925°C
n925 =
Δ log ε!s
log (1) − log (10−4 )
=
Δ log σ log (200 MPa) − log (60 MPa)
=
0 − (−4)
= 7.65
2.301 − 1.778
While for 815°C
n815 =
log (1) − log (10−6 )
Δ log ε!s
=
= 8.2
Δ log σ log (350 MPa) − log (65 MPa)
And at 730°C
n730 =
log (1) − log (10−6 )
Δ log ε!s
=
= 11.1
Δ log σ log (450 MPa) − log (130 MPa)
And, finally, at 650°C
(
)
−1
− log (7 × 10−6 )
Δ log ε!s log 2 × 10
n650 =
=
= 10.2
Δ log σ
log (600 MPa) − log (220 MPa)
8.21 Steady-state creep rate data are given in the following table for a nickel alloy at 538°C (811 K):
ε!s (h-1 )
s (MPa)
10-7
22.0
10-6
36.1
Compute the stress at which the steady-state creep is 10-5 h-1 (also at 538°C).
Solution
This problem gives ε!s values at two different stress levels and 538°C, and asks that we determine the
stress at which the steady-state creep rate is 10-5 h-1 (also at 538°C). It is possible to solve this problem using
Equation 8.21. First of all, it is necessary to determine values of K1 and n. Taking the logarithms of both sides of
Equation 8.21 leads to the following expression:
log ε!s = log K1 + n log σ
(8.21a)
We can now generate two simultaneous equations from the data given in the problem statement in which the two
unknowns are K1 and n. These two equations are as follows:
log(10−7 ) = log K1 + n log(22.0 MPa)
log(10−6 ) = log K1 + n log(36.1 MPa)
Simultaneous solution to these expressions leads to the following:
K1 = 6.12 ´ 10-14
n = 4.63
Using these values, we may determine the stress at which ε!s = 10−5 h −1 . Let us rearrange Equation 8.21a above
such that log s is the dependent variable:
log σ =
log ε!s − log K1
n
Incorporation of values of ε!s , K1, and n yields the following:
log σ =
log(10−5 ) − log(6.12 × 10−14 )
4.63
= 1.774
From which we determine s as follows:
σ = 101.774 = 59.4 MPa
8.22 Steady-state creep data taken for an iron at a stress level of 140 MPa are given here:
ε!s (h–1)
T (K)
6.6 × 10–4
1090
8.8 × 10–2
1200
If it is known that the value of the stress exponent n for this alloy is 8.5, compute the steady-state creep rate at 1300
K and a stress level of 83 MPa.
Solution
This problem gives ε!s values at two different temperatures and 140 MPa, and the value of the stress
exponent n = 8.5, and asks that we determine the steady-state creep rate at a stress of 83 MPa and 1300 K.
Taking natural logarithms of both sides of Equation 8.22 yields
ln ε!s = ln K2 + n ln σ −
Qc
RT
With the given data there are two unknowns in this equation—namely K2 and Qc. Using the data provided in the
problem statement we can set up two independent equations as follows:
ln(6.6 × 10−4 h −1 ) = ln K2 + (8.5) ln(140 MPa) −
Qc
(8.31 J/mol-K)(1090 K)
ln(8.8 × 10−2 h −1 ) = ln K2 + (8.5) ln(140 MPa) −
Qc
(8.31 J/mol-K)(1200 K)
Now, solving simultaneously for K2 and Qc leads to K2 = 57.5 h-1 and Qc = 483,500 J/mol. Thus, it is now possible
to solve for ε!s at 83 MPa and 1300 K using Equation 8.22 as
⎛ Q ⎞
ε!s = K2σ nexp ⎜ − c ⎟
⎝ RT ⎠
⎡
⎤
483,500 J/mol
= (57.5 h −1 ) (83 MPa)8.5 exp ⎢ −
⎥
⎣ (8.31 J/mol-K)(1300 K) ⎦
= 4.31 × 10−2 h −1
8.23 (a) Using Figure 8.32 compute the rupture lifetime for an S-590 alloy that is exposed to a tensile
stress of 400 MPa at 815°C.
(b) Compare this value to the one determined from the Larson-Miller plot of Figure 8.34, which is for this
same S-590 alloy.
Solution
(a) From Figure 8.32 using the 815°C the rupture lifetime at 400 MPa is about 10-2 h.
(b) Using Figure 8.34, at a stress of 400 MPa, the value of the Larson-Miller parameter is about m = 19.7,
which is equal to 103 T(20 + log tr) for T in K and tr in h. We take the temperature to be 815°C + 273 = 1088 K.
Therefore, we may write the following:
⎛ 1 ⎞
19.7 = ⎜
(1088 K)(20 + logtr ) = 1.088(20 + logtr )
⎝ 103 ⎟⎠
Or
19.7
= 18.11 = 20 + logtr
1.088
Which leads to
logtr = 18.11 − 20 = −1.89
And, solving for tr we obtain
tr = 10−1.89 h = 1.3 × 10−2 h
Which value is very close to the 10-2 h that was obtained using Figure 8.32.
DESIGN PROBLEMS
The Fatigue S-N Curve
8.D1 A cylindrical metal bar is to be subjected to reversed and rotating-bending stress cycling. Fatigue
failure is not to occur for at least 107 cycles when the maximum load is 250 N. Possible materials for this
application are the seven alloys having S-N behaviors displayed in Figure 8.21. Rank these alloys from least to
most costly for this application. Assume a factor of safety of 2.0 and that the distance between load-bearing points
is 80.0 mm (0.0800 m). Use cost data found in Appendix C for these alloys as follows:
Alloy designation
(Figure 8.21)
Alloy designation
(Cost data to use—Appendix C)
EQ21A-T6 Mg
AZ31B (extruded) Mg
70Cu-30Zn brass
Alloy C26000
2014-T6 Al
Alloy 2024-T3
Ductile cast iron
Ductile irons (all grades)
1045 steel
Steel alloy 1040 Plate,
cold rolled
Steel alloy 4340 Bar,
normalized
Alloy Ti-5Al-2.5Sn
4340 steel
Ti-5Al-2.5Sn titanium
You may also find useful data that appears in Appendix B.
Solution
The following steps will be used to solve this problem:
1. Determine the fatigue strength or endurance limit at 107cycles for each alloy.
2. Using these data, compute the original cross-sectional diameter for each alloy.
3. Calculate the volume of material required using these d0 values.
4. Using density values from Table B.1, determine the mass of material required for each alloy.
5. Compute the cost using mass data and cost per unit mass data found in Appendix C
Step 1
Below are tabulated the fatigue limits or fatigue strengths at 107cycles as taken from Figure 8.21:
Alloy
s (Fatigue limit or fatigue strength)
EQ21A-T6 Mg
(MPa)
100
70Cu-30Zn brass
115
2014-T6 Al
170
Ductile cast iron
220
1045 steel
310
4340 steel
485
Ti-5Al-2.5Sn titanium
490
Step 2
In order to compute the cross-sectional diameter d0 for each alloy it is necessary to use Equation 8.16—viz.,
σ 16FL
=
N π d3
0
which incorporates the following values for the parameters in this expression:
s = the fatigue strength at 107 cycles or endurance limit (Figure 8.21)
N = the factor of safety (2.0)
F = maximum applied load (250 N)
L = distance between load-bearing points (80.0 mm)
Solving for d0 from Equation 8.16 leads to
⎛ 16FLN ⎞
d0 = ⎜
⎝ πσ ⎟⎠
1/3
The following table lists values of d0 that were calculated using the above equation:
Alloy
d0 (mm)
EQ21A-T6 Mg
12.7
70Cu-30Zn brass
12.1
2014-T6 Al
10.6
Ductile cast iron
9.75
1045 steel
8.69
4340 steel
7.50
Ti-5Al-2.5Sn titanium
7.46
Step 3
Using these d0 values we may compute the cylinder volume V for each alloy using the following equation:
⎛d ⎞
V = πl ⎜ 0 ⎟
⎝ 2⎠
2
Here, l is the cylinder length. For the sake of convenience, let us arbitrarily assume that this length is L, the distance
between load-bearing points from above—i.e., 80.0 mm. (Note: inasmuch as densities are expressed in units of
grams per centimeters cubed, we now choose to express lengths and diameters in centimeters—i.e., l = 8.0 cm and
d0 values in the above table divided by a factor of 10.) The table below presents cylindrical specimen volumes for
these seven alloys.
Alloy
d0 (cm)
V (cm3)
EQ21A-T6 Mg
1.27
10.1
70Cu-30Zn brass
1.21
9.20
2014-T6 Al
1.06
7.06
Ductile cast iron
0.975
5.97
1045 steel
0.869
4.74
4340 steel
0.750
3.53
Ti-5Al-2.5Sn titanium
0.746
3.50
Step 4
The next step is to determine the mass of each alloy in its cylinder. This is possible by multiplying the
cylinder volume (V) by the alloy density (r). Density values are tabulated in Table B.1 of Appendix B. The
following table lists volumes, densities, and cylinder masses for the seven alloys. (Note: Table B.1 does not present
densities for all of the seven alloys. Therefore, in some cases it has been necessary to use approximate density
values.)
Alloy
V (cm3)
r (g/cm3)
Alloy mass (g)
EQ21A-T6 Mg
10.1
1.80
18.2
70Cu-30Zn brass
9.20
8.50
78.2
2014-T6 Al
7.06
2.70
19.1
Ductile cast iron
5.97
7.10
42.4
1045 steel
4.74
7.85
37.2
4340 steel
3.53
7.85
27.7
Ti-5Al-2.5Sn titanium
3.50
4.50
15.8
Step 5
Now, to calculate the cost of each alloy it is necessary to multiply alloy mass (from above converted into kilograms)
by cost data found in Appendix C. (Notes: the problem statement lists alloy cost data that is to be used in these
computations. Also, in Appendix C when cost ranges are cited, average values are used.) These mass and cost data
are tabulated below.
Alloy
Alloy mass (kg)
Cost per unit mass
($US/kg)
Cost
($US)
EQ21A-T6 Mg
[AZ31B (extruded) Mg]
70Cu-30Zn brass
(Alloy C26000)
0.0182
12.10
0.22
0.0782
9.95
0.78
2014-T6 Al
(Alloy 2024-T3)
0.0191
16.00
0.31
Ductile cast iron
[Ductile irons (all grades)]
0.0424
2.60
0.11
1045 steel
(Steel alloy 1040 Plate,
cold rolled)
0.0372
2.20
0.08
4340 steel
(Steel alloy 4340 Bar,
normalized)
0.0277
3.60
0.10
Ti-5Al-2.5Sn titanium
(Alloy Ti-5Al-2.5Sn)
0.0158
115.00
1.82
And, finally, the following lists the ranking of these alloys from least to most costly.
1045 steel
4340 steel
Ductile cast iron
EQ21A-T6 Mg
2014-T6 Al
70Cu-30Zn brass
Ti-5Al-2.5Sn titanium
Data Extrapolation Methods
8.D2 An S-590 iron component (Figure 8.34) must have a creep rupture lifetime of at least 20 days at
650°C (923 K). Compute the maximum allowable stress level.
Solution
This problem asks that we compute the maximum allowable stress level to give a rupture lifetime of 20
days for an S-590 iron component at 923 K. It is first necessary to compute the value of the Larson-Miller
parameter as follows:
{
m = T (20 + log tr ) = (923 K) 20 + log ⎡⎣(20 days)(24 h/day) ⎤⎦
}
= 20.9 × 103
From the curve in Figure 8.34, this value of the Larson-Miller parameter corresponds to a stress level of about 280
MPa.
8.D3 For an 18-8 Mo stainless steel (Figure 8.36), predict the time to rupture for a component that is
subjected to a stress of 100 MPa at 600°C (873 K).
Solution
This problem asks that we determine, for an 18-8 Mo stainless steel, the time to rupture for a component
that is subjected to a stress of 100 MPa at 600°C (873 K). From Figure 8.36, the value of the Larson-Miller
parameter at 100 MPa is about 22.6 ´ 103, for T in K and tr in h. Therefore,
22.6 × 103 = T (20 + log tr )
= 873 (20 + log tr )
Which leads to the following:
25.89 = 20 + log tr
And upon rearrangement
log tr = 5.89
Solving for leads to tr
tr = 105.89 = 7.8 × 105 h = 89 yr