1
ME 452: Machine Design II
Solution to Homework Set 10
Problem 1
(Problem 103 Page 561)
A helical compression spring is wound using 2.5 mm diameter music wire.
The spring has an outside
diameter of 31 mm with plain ground ends, and 14 total coils.
(a)
What should the free length be to ensure that when the spring is compressed solid the torsional stress
does not exceed the yield strength, that is, that it is solidsafe?
(b)
What force is needed to compress this spring to closure?
(c)
Estimate the spring rate.
(d)
Is there a possibility that the spring might buckle in service?
SOLUTION:
(a)
The helical compression spring has 14 total coils; i.e.,
N
t
= 14
The number of active coils for plain and ground ends, from Table 101, page 521, can be written as
1
=

a
t
N
N
Therefore, the number of active coils is
14
1
13
a
N
=

=
The solid length for plain and ground ends, from Table 101, page 521, can be written as
s
t
L
d N
=
Therefore, the solid length of the compression spring is
s
L
= 2.5
14
35 mm
×
=
The ultimate tensile strength for a spring material, see Equation (1014), page 523, can be written as
ut
m
A
S
d
=
For music wire, the intercept and the exponent, see Table 104, page 525, are
m
2211 MPa.mm
A
=
and
0.145
m
=
Therefore, the ultimate tensile strength for the spring material is
0.145
2211
1936 MPa
(2.5)
ut
S
=
=
No setting operation was discussed in the problem statements, so we will assume none was performed
on the spring.
Therefore, the torsional yield strength of the music wire, see Table 106, page 526 and
Example (101), page 527, can be written as