CHAPTER 7
MECHANICAL PROPERTIES
PROBLEM SOLUTIONS
Concepts of Stress and Strain
7.4 For a brass alloy, the stress at which plastic deformation begins is 345 MPa (50,000 psi), and the
modulus of elasticity is 103 GPa (15.0 × 106 psi).
(a) What is the maximum load that may be applied to a specimen with a cross-sectional area of 130 mm2
(0.2 in.2) without plastic deformation?
(b) If the original specimen length is 76 mm (3.0 in.), what is the maximum length to which it may be
stretched without causing plastic deformation?
Solution
(a) This portion of the problem calls for a determination of the maximum load that can be applied without
plastic deformation (Fy). Taking the yield strength to be 345 MPa, and employment of Equation 7.1 leads to
Fy =  y A0 = (345  106 N/m2 )(130  10-6 m2 )
= 44,850 N (10,000 lbf)

(b) The maximum length to which the sample may be deformed without plastic deformation is determined
from Equations 7.2 and 7.5 as

 
li = l0 1   = l0 1  

E 

345 MPa 
= (76
mm)
1


= 76.25 mm (3.01 in.)

 103  103 MPa 
7.12 A cylindrical rod 500 mm (20.0 in.) long, having a diameter of 12.7 mm (0.50 in.), is to be
subjected to a tensile load. If the rod is to experience neither plastic deformation nor an elongation of more than
 the applied load is 29,000 N (6500 lb f), which of the four metals or alloys listed below are
1.3 mm (0.05 in.) when
possible candidates? Justify your choice(s).
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Material
Modulus of
Elasticity (GPa)
Yield Strength
(MPa)
Tensile Strength
(MPa)
Aluminum alloy
70
255
420
Brass alloy
100
345
420
Copper
110
210
210
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 7.1; a material to be used for this application must necessarily have a yield strength greater
than this value. Thus,
 =
F
=
A0
29,000 N
12.7  103 m 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 aluminum, brass, and steel using Equations 7.2 and 7.5
in order to determine whether or not this elongation is less than 1.3 mm. For aluminum
l =
 l0
(230 MPa)(500 mm)
=
= 1.64 mm
E
70  103 MPa
Thus, aluminum is not a candidate.
For brass

l =
 l0
(230 MPa)(500 mm)
=
= 1.15 mm
E
100  103 MPa
Thus, brass is a candidate. And, for steel

l =
 l0
(230 MPa)(500 mm)
=
= 0.56 mm
E
207  103 MPa
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
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Therefore, of these four alloys, only brass and steel satisfy the stipulated criteria.
7.15 A cylindrical specimen of stainless steel having a diameter of 12.8 mm (0.505 in.) and a gauge length
of 50.800 mm (2.000 in.) is pulled in tension. Use the load–elongation characteristics tabulated below to complete
parts (a) through (f).
Load
Length
N
lbf
mm
in.
0
0
50.800
2.000
12,700
2,850
50.825
2.001
25,400
5,710
50.851
2.002
38,100
8,560
50.876
2.003
50,800
11,400
50.902
2.004
76,200
17,100
50.952
2.006
89,100
20,000
51.003
2.008
92,700
20,800
51.054
2.010
102,500
23,000
51.181
2.015
107,800
24,200
51.308
2.020
119,400
26,800
51.562
2.030
128,300
28,800
51.816
2.040
149,700
33,650
52.832
2.080
159,000
35,750
53.848
2.120
160,400
36,000
54.356
2.140
159,500
35,850
54.864
2.160
151,500
34,050
55.880
2.200
124,700
28,000
56.642
2.230
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.
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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 7.10) as
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E =

400 MPa  0 MPa
=
= 200  103 MPa = 200 GPa (29  106 psi)

0.002  0
(c) For the yield strength, the 0.002 strain offset line is drawn dashed. It intersects the stress-strain curve at

approximately 750 MPa (112,000 psi ).
(d) The tensile strength is approximately 1250 MPa (180,000 psi), corresponding to the maximum stress on
the complete stress-strain plot.
(e) The 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 7.14, the modulus of resilience is just
Ur =
 2y
2E
which, using data computed above gives a value of

(750 MPa)2
= 1.40  106 J/m3 (210 in.- lbf /in.3 )
(2) (200  103 MPa)
7.18 A steel alloy to be used for a spring application must have a modulus of resilience of at least
Ur =
2.07 MPa (300 psi). 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 7.14; the value of E for steel given in Table 7.1 is 207 GPa. Solving for y from this expression
yields
y =
2U r E =
(2) (2.07 MPa) (207  103 MPa)
= 925 MPa (134,000 psi)

7.22 Find the toughness (or energy to cause fracture) for a metal that experiences both elastic and plastic
deformation. Assume Equation 7.5 for elastic deformation, that the modulus of elasticity is 103 GPa (15 × 10 6 psi),
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 7.19, in which the values for K and n are 1520 MPa (221,000
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psi) 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). The easiest way to do this
is to integrate both elastic and plastic regions, and then add them together.

Toughness =  d
0.007

=
0.60
E d  +
0

E2
=
2
 Kn d 
0.007
0.007
0.60
K
+
(n 1)
(n  1)
0
=
0.007
6
2
103  109 N / m2
2 + 1520  10 N/ m (0.60)1.15  (0.007)1.15
(0.007
)
 2
(1.0  0.15)


= 7.33  108 J/m3 (1.07  105 in.-lb /in.3)
f

Elastic Recovery After Plastic Deformation
7.24 A steel alloy specimen having a rectangular cross section of dimensions 19 mm × 3.2 mm
( 3 in. 
4
1
8
in.) has the stress–strain behavior shown in Figure 7.33. If this specimen is subjected to a tensile force of 110,000 N
(25,000 lbf) then


(a) Determine the elastic and plastic strain values.
(b) If its original length is 610 mm (24.0 in.), 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
(25,000 lbf) is applied to the steel specimen and then released. First it becomes necessary to determine the applied
stress using Equation 7.1; thus
 =
F
F
=
A0
b0 d0
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Copyright Act without the permission of the copyright owner is unlawful.
where b0 and d0 are cross-sectional width and depth (19 mm and 3.2 mm, respectively). Thus
 =
(19
110,000 N
= 1.810  109 N / m2  1810 MPa (265,000 psi)
 103 m)(3.2  103 m)
From Figure 7.33, this point is in the plastic region so the specimen will be both elastic and plastic strains. The total

strain at this point, t, is about 0.020. We are able to estimate the amount of permanent strain recovery e from
Hooke's law, Equation 7.5 as
e =

E
And, since E = 207 GPa for steel (Table 7.1)

e =
1810 MPa
= 0.0087
207  103 MPa
The value of the plastic strain, p is just the difference between the total and elastic strains; that is

p = t – e = 0.020 – 0.0087 = 0.0113
(b) If the initial length is 610 mm (24.0 in.) then the final specimen length li may be determined from a
rearranged form of Equation 7.2 using the plastic strain value as
li = l0(1 + p) = (610 mm)(1 + 0.0113) = 616.7 mm (24.26 in.)
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