CHAPTER 6
MECHANICAL PROPERTIES OF METALS
6-8, 6-19, AND 6-42
6.8 A cylindrical rod of copper (E = 110 GPa, 16  106 psi) having a yield strength of 240 MPa (35,000 psi) is to
be subjected to a load of 6660 N (1500 lb f). If the length of the rod is 380 mm (15.0 in.), what must be the diameter
to allow an elongation of 0.50 mm (0.020 in.)?
Solution
This problem asks us to compute the diameter of a cylindrical specimen of copper in order to allow an
elongation of 0.50 mm. Employing Equations 6.1, 6.2, and 6.5, assuming that deformation is entirely elastic
 =
Or, solving for d0
F
=
A0
F
l
= E
d 2 
l0
  0 
 4 
 

4 l0F
 E l
d0 =
(4) (380  10 3 m) (6660 N)
= 
() (110  10 9 N / m2 )(0.50  10 3 m)

= 7.65  10-3 m = 7.65 mm (0.30 in.)
1
6.19 Consider a cylindrical specimen of some hypothetical metal alloy that has a diameter of 8.0 mm (0.31 in.). A
tensile force of 1000 N (225 lbf) produces an elastic reduction in diameter of 2.8  10-4 mm (1.10  10-5 in.).
Compute the modulus of elasticity for this alloy, given that Poisson's ratio is 0.30.
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
E =

F
=
=
z
A0z
F
2
d 
 z   0 
 2 
=
4F
 z  d02
d
From the definition of Poisson's ratio, (Equation 6.8) and realizing that for the transverse strain, x=

d0

d
z =  x = 

d0 

Therefore, substitution of this expression for z into the above equation yields

E =
=
4F
 z  d02
=
4F 
 d0 d
(4)(1000 N )(0.30)
= 1.705  10 11 Pa = 170.5 GPa (24.7
 (8  10 3 m)(
2.8  10 7 m)

2
 10 6 psi)
6.41 Using the data in Problem 6.28 and Equations 6.15, 6.16, and 6.18a, generate a true stress–true strain plot for
aluminum. Equation 6.18a becomes invalid past the point at which necking begins; therefore, measured diameters
are given below for the last four data points, which should be used in true stress computations.
Load
Length
Diameter
N
lbf
mm
in.
mm
in.
46,100
10,400
56.896
2.240
11.71
0.461
42,400
10,100
57.658
2.270
10.95
0.431
42,600
9,600
58.420
2.300
10.62
0.418
36,400
8,200
59.182
2.330
9.40
0.370
Solution
These true stress-strain data are plotted below.
3
6-42: A tensile test is performed on a metal specimen, and it is found that a true plastic strain of 0.2 is
produced when a true stress of 575 Mpa; for the same metal the value of K is 860Mpa; calculate the
true strain that results from the application of a true stress of 600Mpa
Solution: True stress is given by, 𝜎𝑇 = 𝐾 ∈𝑛𝑇
Taking ln each side of the equation gives, 𝑛 =
So, 𝑛 =
ln 𝜎𝑇 −ln 𝐾
ln ∈𝑇
ln 575−ln 860
ln 0.2
= 0.25
Now the true strain when applied true stress, 𝜎𝑇 = 600 𝑀𝑝𝑎
∈𝑇 = 𝑒
ln 𝜎𝑇 −ln 𝐾
𝑛
4
=𝑒
ln 600−ln 860
0.25
= 0.237