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ME 452: Machine Design II
Solution to Homework Set 12
Problem 1.
(Problem 822, page 448).
To determine the member stiffness use the approach outlined by Eq. (820), see page 415. Compare your
result with the answer obtained from the finiteelement study, see Eq. (823). To determine the length of
the bolt use Table A17, page 1015. To determine the height of the hexagonal nut use Table A31, page
1035.
Determine the preload using Eqs. (830) and (831), see page 427.
Determine the load factor guarding against overproof loading using Eq. (828), see page 425.
Also determine the load factor guarding against joint separation using Eq. (829), see page 426.
Solution:
The stiffness of the bolt, see Equation (817), page 413 can be written as
dt
b
td
AAE
k
Al
=
+
The area of the unthreaded portion of the bolts, see Table (87), page 412, can be written as
2
4
d
d
A
π
=
Therefore, the area of the unthreaded portion of the bolts is given as
2
2
(12)
113 mm
4
d
A
×
==
The tensile stress area for metric threads, see Table (81), page 398, for a diameter of 12 mm and
assuming coarse pitch
, is given as
2
84.3 mm
t
A
=
The length of the grip, see Figure for Problem (820), page 453, is given as
20
25
45 mm
G
L
=
+=
The height of the hexagonal head of the bolt, see Table (A31), page 1035 , for a 12 mm size bolt and
assuming regular hexagonal nut, is given as
10.8 mm
H
=
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The total length of the bolt is given as
Total length
G
LH
=
+
Therefore, the total length of the bolt is given as
Total length
45 10.8
55.8 mm
=
+=
The length of a standard bolt that is nearest to the value obtained above, see Table (A17), page 1015, is
given as
60 mm
L
=
The thread length, see Equation (814), page 408, for a bolt length of 60 mm and a bolt diameter of 12
mm can be written as
2
6
T
LD
=
+
Therefore, the thread length is given as
(
2
1
2
)
6
3
0
m
m
T
L
=
×+
=
The length of the unthreaded portion in the grip, see Example (84), page 427 and Table 87(a), page
412, can be written as
dT
lL
L
=
−
Therefore, the length of the unthreaded portion in the grip can be written as
6
0
3
0
3
0
m
m
d
l
=
−=
The length of the threaded portion in the grip, see Example (84), page 427 and Table 87(a), page 412,
can be written as
tG
d
lLl
=
−
Therefore, the length of the threaded portion in the grip is given as
45 30
15 mm
t
l
=
The modulus of elasticity for the bolts assuming they are made of steel, see Table (88), page 416, is
given as
207 GPa
E
=
Therefore, the stiffness of the bolts is given as
66
9
6
63
6 3
(113 10 ) (84.3 10 ) (207 10 )
466.8 10 N/m
(113 10
15 10 )
(84.3 10
30 10 )
b
k
−−
− −
×××××
==
×
××
×
+
×
3
Note that in this problem the two members are made of different materials. The cylinder head is made of
steel whereas the cylinder is made of Grade 30 cast iron (see Figure for Problem 820, page 453). The
member stiffness needs to be calculated using the procedure outlined on pages 414 and 415.
Since the members do not have an identical thickness, three regions (frustum’s) need to considered to
calculate the overall member stiffness. These three frustums are shown in the figure below.
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This note was uploaded on 02/09/2012 for the course ME 352 taught by Professor Staff during the Fall '08 term at Purdue University.
 Fall '08
 Staff
 Machine Design

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