2
Therefore, the mean coil diameter of the spring is
D
1.225
0.105
1.120 in
=

=
The spring index, see Equation (102), page 501, can be written as
D
C =
d
Therefore, the spring index is
1.120
C
10.67
0.105
=
=
The Bergstrasser factor, see Equation (106), page 501, can be written as
4
2
4
3
B
C
K
C
=

Therefore, the Bergstrasser factor is
(4
10.67)
2
1.126
(4
10.67)
3
B
K
×
+
=
=
×

The spring force corresponding to the torsional yield stress, see Example (101), page 509 and see note
below Equation (107), page 502, can be written as
3
sy
B
π
d S
F =
8 K
D
Note that this equation gives the force that will just yield the spring.
It is a matter of user preference
whether there should be a factor of safety used here (Using S
sy
/n
s
in stead of S
sy
) or not.
Since the
problem does not specifically call for a factor of safety, none will be used.
However, the book does
suggest (see Equation (1021), page 510) that as a rule of thumb, the factor of safety at the solid height
should be greater than or equal to 1.2.
To illustrate this recommendation, see the solution to the next
problem in this homework set.
So, the spring force corresponding to the torsional yield stress (with no
factor of safety) is
3
3
(0.105)
125.4
10
45.2 lb
8
1.126
1.120
F
π
×
×
×
=
=
×
×
The spring rate, see Equation (109), page 502, can be written as
4
3
a
d
G
k =
8 D N
where
a
N is the number of active coils.
The modulus of rigidity for music wire with diameter
0.105 in
d
=
, see Table 105, page 508, is
11.75 Mpsi
G
=
Therefore, the spring rate for the helical compression spring is
4
6
3
(0.105)
11.75
10
11.55 lb/in
8
(1.120)
11
k
×
×
=
=
×
×
The deflection caused by the above spring force, see Example (101), page 509, can be written as