14:440:407 Section 02 Fall 2010 SOLUTION OF HOMEWORK 04
Question 6.3:
A specimen of aluminum having a rectangular cross section 10 mm
12.7
mm (0.4 in.
0.5 in.) is pulled in tension with 35,500 N (8000 lb
f
) force, producing only
elastic deformation.
Calculate the resulting strain.
Solution:
This problem calls for us to calculate the elastic strain that results for an aluminum specimen stressed in tension.
The cross-sectional area is just (10 mm)
(12.7 mm) = 127 mm
2
(= 1.27
10
-4
m
2
= 0.20 in.
2
);
also, the elastic
modulus for Al is given in Table 6.1 as 69 GPa (or 69
10
9
N/m
2
).
Combining Equations 6.1 and 6.5 and solving
for the strain yields
=
E
=
F
A
0
E
=
35,500 N
(
1.27
10
4
m
2
)(
69
10
9
N/m
2
)
= 4.1
10
-3
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Question 6.5:
A steel bar 100 mm (4.0 in.) long and having a square cross section 20 mm
(0.8 in.) on an edge is pulled in tension with a load of 89,000 N (20,000 lb
f
), and experiences
an elongation of 0.10 mm (4.0
10
-3
in.).
Assuming that the deformation is entirely elastic,
calculate the elastic modulus of the steel.
Solution:
This problem asks us to compute the elastic modulus of steel.
For a square cross-section,
A
0
=
b
0
2
, where
b
0
is the edge length.
Combining Equations 6.1, 6.2, and 6.5 and solving for
E
, leads to
E
=
=
F
A
0
l
l
0
=
Fl
0
b
0
2
l
=
(89,000 N)
(
100
10
3
m
)
(
20
10
3
m
)
2
(
0.10
10
3
m
)
= 223
10
9
N/m
2
= 223 GPa
(31.3
10
6
psi)

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Question 6.7:
For a bronze alloy, the stress at which plastic deformation begins is 275 MPa
(40,000 psi), and the modulus of elasticity is 115 GPa (16.7
10
6
psi).
(a) What is the maximum load that may be applied to a specimen with a cross-sectional
area of 325 mm
2
(0.5 in.
2
) without plastic deformation?
(b) If the original specimen length is 115 mm (4.5 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 (
F
y
).
Taking the yield strength to be 275 MPa, and employment of Equation 6.1 leads to
F
y
=
y
A
0
=
(
275
10
6
N/m
2
)(
325
10
-6
m
2
)
= 89,375 N
(20,000 lb
f
)
(b)
The maximum length to which the sample may be deformed without plastic deformation is determined
from Equations 6.2 and 6.5 as
l
i
=
l
0
1
E
= (115 mm) 1
275 MPa
115
10
3
MPa
= 115.28 mm (4.51 in.)
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Question 6.13:
In Section 2.6 it was noted that the net bonding energy E
N
between two
isolated positive and negative ions is a function of interionic distance r as follows:
E
N
A
r
B
r
n
(6.25)
where A, B, and n are constants for the particular ion pair. Equation 6.25 is also valid for
the bonding energy between adjacent ions in solid materials. The modulus of elasticity E is
proportional to the slope of the interionic force–separation curve at the equilibrium
interionic separation; that is,