8.3 Potential Energy of a Helical Spring
79
Fig. 8.1 For calculation of the potential energy of elastic strain of a round rod
A =
l
2
N 2
EF
+
M 2
u
EI x
+
M 2
k
GI p
.
(8.4)
Taking into account that E = 2(1 + ν)G, formula (8.4) will look as follows:
A =
l
2E
4N 2
πd 2 +
64
πd 4
M
2
u + (1 + ν)M
2
k
.
(8.5)
8.3 Potential Energy of a Helical Spring
Let us consider a cylindrical helical spring shown in Fig. 8.2. Let us dissect the
spring turn by the axial plane. From the condition of equilibrium of the dissected
part of the spring, let us find the moment of internal forces relative to the axis going
through the center of gravity of the turn section in perpendicular to the specified
axial plane:
M = P R,
where R is the average spring radius.
A component of this moment relative to the axis perpendicular to the crosssection plane represents a torque whose magnitude will be
M k = P R cos α,
(8.6)
where α is the inclination level of the helical line of the spring.
Let us calculate the bending moment using the formula:
M u =
M 2 − M 2
k = P R sin α.
(8.7)