EMSE 201 – Introduction to Materials Science & Engineering
Fall 2010
Case Western Reserve University
1 of 4
Case School of Engineering
Solution to Homework #5
Explain
how answers were obtained. Give appropriate
units
for all numerical answers.
1)
(C&R problem 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
×
10
6
psi).
a)
(3 points) What is the maximum load that may be applied to a specimen with a crosssectional area of 130
mm
2
(0.2 in.
2
) without plastic deformation?
b)
(3 points) 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
(from the publisher’s solution manual; used with permission)
a)
This portion of the problem calls for a determination of the maximum load that can be
applied without plastic deformation (
F
y
).
Taking the yield strength to be 345 MPa, and
employment of Equation 7.1 leads to
F
y
=
σ
y
A
0
=
(
345
×
10
6
N/m
2
)(
130
×
10
6
m
2
)
= 44,850 N
(10,000 lb
f
)
(3 pts)
b)
The maximum length to which the sample may be deformed without plastic deformation is
determined from Equations 7.2 and 7.5 as
l
i
=
l
0
1
+
ε
(
)
=
l
0
1
+
σ
E
⎛
⎝
⎜
⎞
⎠
⎟
= (76 mm)
1
+
345
MPa
103
×
10
3
MPa
⎡
⎣
⎢
⎤
⎦
⎥
= 76.25 mm (3.002 in.)
(3 pts)
2)
(5 points) A brass alloy is known to have a yield strength of 240 MPa (35,000 psi), a tensile strength of 310
MPa (45,000 psi), and an elastic modulus of 110 GPa (16.0
×
106 psi).
A cylindrical specimen of this alloy
15.2 mm (0.60 in.) in diameter and 380 mm (15.0 in.) long is stressed in tension and found to elongate 1.9 mm
(0.075 in.).
On the basis of the information given, is it possible to compute the magnitude of the load that is
necessary to produce this change in length?
If so, calculate the load.
If not, explain why (C&R problem 7.10)
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 Spring '09
 Tensile strength, yield strength, MPa, Materials Science & Engineering

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