General Design Principles for DuPont Engineering Polymers - Module

Cantilever Springs
Tests were conducted on cantilever springs of Delrin ®
acetal resin to obtain accurate load deflection data on
this type of spring geometry commonly used in
molded parts. The measurements were made on a
machine with a constant rate of cross-head movement.
Standard 127 × 12.7 × 3.2 mm (5 × 1 ⁄ 2 × 1 ⁄ 8 in) bars of
Delrin ® acetal resin were placed in a fixture and loads
recorded at spring lengths of 25 to 89 mm (1.00 to
3.50 in) in increments of 13 mm ( 1 ⁄ 2 in). The data were
tabulated for each deflection increment of 2.5 mm
(0.10 in) and plotted in terms of spring rate vs. deflection
as shown in Figure 6.02.
Values of load were corrected to eliminate the frictional
factor and to show the change in spring rate in a
direction perpendicular to the unstrained plane of the
spring.
Figure 6.02 is a plot of the data obtained on springs of
3.2 mm ( 1 ⁄ 8 in) thickness. This information can be used
for thicknesses up to 6.4 mm ( 1 ⁄ 4 in), but beyond this,
accuracy will diminish.
This chart will be of use in the design of any cantilever
type spring in Delrin ® acetal resin that is operated
intermittently at room temperature. As a general rule,
because of creep under continuous load, it is not
desirable to maintain a constant strain on a spring of
Delrin ® acetal resin. In fact, any spring which will be
under a constant load or deflection, or which is
required to store energy, should be made of a lower
creep material such as metal, rather than Delrin ®
acetal resin.
The two sample problems that follow illustrate the use
of the charts.
Problem #1
A spring, integral with a molded part, is to provide
a spring rate (W/y) of 0.88 N/mm (5 lb/in) when
deflected (y) 12.7 mm (0.50 in). The maximum spring
length is limited only by the size of the interior
dimensions of the enclosure for the part which is
50.8 mm (2.0 in). The width (b) can be a maximum
of 12.7 mm (0.50 in). Find the thickness (h) of the
spring required.
37
Solution
(Use Figure 6.02.) Choose a value for b that will
provide clearance within the housing. Therefore, let
b = 6.35 mm (0.25 in) to allow room for modification
if necessary.
Since the chart is designed for a value of b = 25.4 mm
(1.0 in), W/y must be multiplied by 25.4/6.35 to
obtain chart value, or:
Wc/y = (0.88) (25.4/6.35) = 3.5 N/mm (20 lb/in)
Locate the W/y = 3.5 N/mm (20 lb/in) horizontal line
on the chart. There is now a wide choice of y/L and
L/h values on this line that will provide a solution.
Next, assume that L = 50.8 mm (2.00 in), the maximum
length allowable.
Thus y/L = 12.7/50.8 = 0.25
This y/L intersects the Wc/y = 3.5 N/mm (20 lb/in)
line at a point slightly below the line of L/h = 16.
Interpolating between L/h = 16 and L/h = 18 a value
of L/h = 16.5 is found. h can now be found.
L/h = 16.5
h = L/16.5 = 50.8/16.5 = 3.1 mm (0.121 in)
The stress level can be picked off the chart as
approximately 46 MPa (6600 psi). However, were
the spring to operate many times in a short period, say
60 times/min, it would be desirable to use a lower
stress level. Referring to the chart, it can be seen that
the stress can be lowered by decreasing y/L, increasing
L/h or decreasing Wc/y. Since the length is limited to 50.8 mm (2.00 in),
Wc/y would have to be decreased. By doubling b
to 12.7 mm (0.50 in) (maximum width available)
Wc/y can be reduced from 3.5 N/mm (20 lb/in) to
1.8 N/mm (10 lb/in). At a y/L value of 0.25, this new
L/h value can be interpolated as 21.
Hence h = L/21 = 50.8/21 = 2.4 mm (0.095 in) and the
new value of stress would be 42 MPa (6100 psi).