Chapter 6
1. Name the three principal ways in which a load may be applied to a specimen.
tension
compression
shear
The three principal ways in which a load may be applied to a specimen
are tension, compression, and shear.
2. A specimen of some metal having a rectangular cross section 10.8 mm x 12.5 mm is pulled in
tension with a force of 34,300 N, which produces only elastic deformation. Given that the
elastic modulus of this metal is 79 GPa, calculate the resulting strain.
0.00322
Hooke's law is as follows:
Furthermore, the applied stress is equal to
where l and w are the cross-sectional length and width, respectively. Equating to one another
these two expressions for σ leads to
and solving for the strain
which, for this problem yields a value of
= 0.00322
3. Consider a brass alloy the stress-strain behavior of which is shown below.
A cylindrical specimen of this alloy having a length of 63 mm must elongate only 0.11 mm
when a tensile load of 53,500 N is applied. Under these circumstances, what is the radius of
the specimen (in mm)?
10.7 mm
For this problem, strain may be computed as follows:
We next locate the point on the stress-strain curve at this strain value, and then read the
corresponding stress value, which is 150 MPa. From the definition of stress for a specimen of
a circular cross-section
it is possible to solve for the specimen radius, r0 as
= 0.0107 m = 10.7 mm
4. Consider a brass alloy the stress-strain behavior of which is shown below.
A cylindrical specimen of this alloy 20 mm in diameter and 188 mm long is to be pulled in
tension. Calculate the stress (in MPa) necessary to cause a 0.0105 mm reduction in diameter.
Assume a value of 0.34 for Poisson's ratio.
135 MPa
Poisson's ratio is defined as
which is the negative ratio of transverse and longitudinal strains. For this cylindrical specimen
that has an initial diameter d0, the transverse strain is equal to
Inasmuch as we know that the value of ν is 0.34, it is possible to solve for the longitudinal
strain from the definition of Poisson's ratio as
Next, locate the point on the stress-strain curve at this strain value and read the corresponding
stress from the vertical axis; the value of this stress is about 135 MPa.
5. For a bronze alloy, the stress at which plastic deformation begins is 267 MPa and the modulus
of elasticity is 115 GPa.
(a) What is the maximum load (in N) that may be applied to a specimen having a crosssectional area of 300 mm2 without plastic deformation?
(b) If the original specimen length is 137 mm, what is the maximum length (in mm) to which
it may be stretched without causing plastic deformation?
(a) 80,100 N
(b) 137.3 mm
(a) The maximum stress that may be applied without plastic deformation taking place is the
yield strength, σy; thus, from the definition of stress
the load at yielding (or maximum load), Fy, is
= 80,100 N
(b) Combining Hooke's law (at the stress corresponding to the yield strength)
with the definition of strain
as
and solving for the maximum length li gives
which, for this problem is equal to
= 137.3 mm
6. A cylindrical bar of metal having a diameter of 18.8 mm and a length of 198 mm is deformed
elastically in tension with a force of 49,400 N. Given that the elastic modulus and Poisson's
ratio of the metal are 67.1 GPa and 0.34, respectively, determine the following:
(a) The amount by which this specimen will elongate (in mm) in the direction of the applied
stress.
(b) The change in diameter of the specimen (in mm). Indicate an increase in diameter with a
positive number and a decrease with a negative number.
(a) 0.525 mm
(b) - 0.0169 mm
(a) Combining Hooke's law - i.e.,
with the definition of stress for a cylindrical specimen--i.e.,
and also incorporating the expression for longitudinal strain--i.e.,
we get the following:
Then solving for the change in length, Δl, leads to
and, for this problem, the following value for Δl:
= 0.525 x 10-3 m = 0.525 mm
(b) When the definition for Poisson's ratio--i.e.,
is combined with the expression for transverse strain,
and longitudinal strain,
The following equation results:
and, finally solving for the change in diameter, Δd, leads to
which, for this problem, yields the following value for Δd:
= - 0.0169 mm
7. Consider a brass alloy the stress-strain behavior of which is shown below.
A cylindrical specimen of this alloy 12.8 mm in diameter and 109.1 mm long is pulled in
tension; after the load is released the final length is 109.3 mm. Calculate the magnitude of the
load (in N) necessary to cause this elongation.
32,170 N
It is first necessary to calculate the final strain of the specimen after the load release. From the
definition of strain
Next locate this strain value on the bottom horizontal axis of the inset plot, and then draw a
straight line from this strain value parallel to the linear elastic region. This line intersects the
stress-strain curve at a stress of about 250 MPa. From the definition of stress for a cylindrical
specimen of diameter d
solving for the applied load leads to
= 32,170 N
8. An alloy has a yield strength of 805 MPa and an elastic modulus of 107 GPa. Calculate the
modulus of resilience for this alloy [in J/m^3 (which is equivalent to Pa)] given that it
exhibits linear elastic stress-strain behavior.
3.03E6 J/m^3
The modulus of resilience Ur may be calculated using the following equation:
Using values of σy and E provided in the problem statement, the modulus of resilience is
equal to
9. A cylindrical specimen of a metal alloy 48.5 mm long and 9.72 mm in diameter is stressed in
tension. A true stress of 369 MPa causes the specimen to plastically elongate to a length of
54.7 mm. If it is known that the strain-hardening exponent for this alloy is 0.20, calculate the
true stress (in MPa) necessary to plastically elongate a specimen of this same material from a
length of 48.5 mm to a length of 57.2 mm.
393 MPa
True stress and strain are related to one another as follows:
The true strain is defined by
By substitution the equation defining true strain into the first equation and after some
rearrangement, it is possible to solve for the value of K using the first set of l0and li values as
follows:
(The "1" subscript is used to denote the first set of values given in the problem statement.)
Now it is possible to determine the value of σT for the second set of data (σT2) by combining
the first two equations, and by substitution of the above expression for K as follows:
Thus, the value of σT2 for this problem is
= 393 MPa
10. A 10-mm-diameter Brinell hardness indenter produced an indentation 1.55 mm in diameter
in a steel alloy when a load of 500 kg was used. Calculate the Brinell hardness (in HB) of
this material.
263 HB
The Brinell hardness may be calculated using the following equation:
where P, D, and d are, respectively, the applied load (in kg), the ball diameter (in mm), and
the indentation diameter (in mm). Thus,
= 263 HB
11. Using the following figure
estimate the (a) Brinell hardness (in HB), and (b) Rockwell hardness (in HRB) for a brass
alloy having the stress-strain behavior shown below.
(a) 125 HB
(b) 80 HRB
(a) From the stress-strain plot, the tensile strength of this alloy (i.e., the maximum point on
the curve) is 450 MPa. Using the tensile strength-hardness plot (the line for brass) a tensile
strength of 450 MPa corresponds to a Brinell hardness of approximately 125 HB (as read
from the bottom horizontal axis).
(b) For this same tensile strength value, the Rockwell hardness (as read from the top
horizontal axis) is approximately 80 HRB.