Fundamentals of Materials Science and Engineering: An Integrated Approach, 3e

GTBL042-07 GTBL042-Callister-v2 August 6, 2007 12:43

202 • Chapter 7 / Mechanical Properties

Stress (10

3

psi)

70

60

50

40

30

20

10

0

500

400

300

200

100

(^00) 0.10 0.20 0.30 0.40
Strain
Stress (MPa)
Tensile strength
450 MPa (65,000 psi)
A
Yield strength
250 MPa (36,000 psi)
MPa
200
100
0
40
30
20
10
(^00) 0.005
103 psi
Figure 7.12 The
stress–strain
behavior for the
brass specimen
discussed in
Example Problem
7.3.
Solution
(a)The modulus of elasticity is the slope of the elastic or initial linear portion
of the stress–strain curve. The strain axis has been expanded in the inset,
Figure 7.12, to facilitate this computation. The slope of this linear region is
the rise over the run, or the change in stress divided by the corresponding
change in strain; in mathematical terms,
E=slope=
σ


=

σ 2 −σ 1

 2 − 1

(7.10)

Inasmuch as the line segment passes through the origin, it is convenient to

take bothσ 1 and 1 as zero. Ifσ 2 is arbitrarily taken as 150 MPa, then 2

will have a value of 0.0016. Therefore,

E=

(150−0) MPa

0. 0016 − 0

= 93 .8 GPa (13. 6 × 106 psi)

which is very close to the value of 97 GPa (14× 106 psi) given for brass in

Table 7.1.

(b)The 0.002 strain offset line is constructed as shown in the inset; its intersec-

tion with the stress–strain curve is at approximately 250 MPa (36,000 psi),

which is the yield strength of the brass.

(c)The maximum load that can be sustained by the specimen is calculated by

using Equation 7.1, in whichσ is taken to be the tensile strength, from

Figure 7.12, 450 MPa (65,000 psi). Solving forF, the maximum load, yields

F=σA 0 =σ

(

d 0

2

) 2

π

=(450× 106 N/m^2 )

(

12. 8 × 10 −^3 m

2

) 2

π=57,900 N (13,000 lbf)