General Design Principles for DuPont Engineering Polymers - Module

Example 1
It there are no restrictions on the geometry of the new
cross-ribbed wall design, the curve can be used to
determine the dimension that will satisfy a required
reduction in part weight.
FLAT PLATE RIBBED STRUCTURE
Known: Present wall thickness (tA) = 0.0675 in
Required: Material reduction equals 40% or:
VB = 0.60
VA From Figure 4.08:
(tA) (N) = 0.135, or N = 0.135 × 1 = 2 ribs per in
W 0.0675
tB = 0.437, or: tB = (0.437) (0.0675)
tA = 0.030 in wall thickness
T = 1.875, or: T = (1.875) (0.0675)
tA = 0.127 in rib plus wall thickness
Example 2
If melt flow of the resin limits the redesigned wall
thickness, part geometry can be calculated as follows:
Known: Present wall thickness (tA) = 0.050 in
Required: Minimum wall thickness (tB) = 0.020 in
or tB = 0.020 = 0.4
tA 0.050
From Figure 4.08:
T = 1.95, or: T = (1.95) (0.050) = 0.098 in
tA (tA) (N) = 0.125, or N = 0.125 × 1 = 2.5 ribs per in
W 0.05
VB = 0.55
VA Thus, the 0.02 inch wall design has an overall height
of 0.098 inch, a rib spacing of 2.5 per inch (5 ribs
every 2 inches) and a 45 percent material saving.
Example 3
If the overall wall thickness is the limitation because
of internal or exterior size of the part, other dimensions
can be found on the curve:
27
Known: Present wall thickness (tA) = 0.25 in
Required: Maximum height of ribbed wall
(T) = 0.425 in
or T = 0.425 = 1.7
tA 0.25
From Figure 4.08:
(tA) (N) = 0.165, or N = 0.165 × 1 = 0.66 rib per in
W 0.250
tB = 0.53, or: tB = (0.53) (0.25) = 0.133 in
tA V B = 0.71
V A
The ribbed design provides a material reduction of
29 percent, will use 0.66 ribs per inch (2 ribs every
3 inches) and will have a wall thickness of 0.133 inch.
If thinner ribs are desired for functional or appearance
reasons, the same structure can be obtained by holding
the product of the number of ribs and the rib thickness
constant. In this example, if the rib wall thickness
were cut in half to 0.067 inch, the number of ribs
should be increased from 2 every 3 inches to 4 every
3 inches.
Example 4
If the number of ribs per inch is limited because of
possible interference with internal components of the
product, or by the need to match rib spacing with an
adjoining structure or decorative elements, the designer
can specify the number of ribs and then determine
the other dimensions which will provide a
minimum volume.
Known: Present wall thickness (tA) = 0.0875 in
Required: Ribs per in (N) = 2
Therefore, for a base (W) of unity:
(tA) (N) = (0.0875) (2) = 0.175
W 1
From Figure 4.08:
tB = 0.56, or: TB = (0.56) (0.0875) = 0.049 in
tA T = 1.67, or: T = (1.67) (0.0875) = 0.146 in
t A
V B = 0.75
V A
The resulting design has an overall height of 0.127
inch, a wall thickness of 0.063 inch and a material
saving of 25 percent. (An alternate solution obtained
with a V B/V A value of 0.90 provides a material saving
of only 10 percent. The choice depends on the suitability
of wall thickness and overall height.)