Design Guide - Solvay Plastics

Designing for Stiffness
When a design engineer considers replacing a metal
component with plastic, one consideration is the rigidity
or stiffness of the component. If the application requires
that the maximum deflection under load remain at the
current value, then the plastic component must have
stiffness equivalent to the metal component.
Table 38 provides the deflection equations for a variety
of beams. Selecting the beam with both ends fixed and
a uniformly distributed load, the deflection (Y) is given by
Y =
FL 3
384EI
To arrive at a part design with equivalent stiffness, start
by equating the deflection equations for the plastic and
metal parts as follows:
FL 3
384EI
metal
=
FL 3
384EI
plastic
Assuming a constant load and length, and removing
the factors common to both sides of the equation, the
governing equation becomes:
Equation 1
[EI] metal = [EI] plastic
Knowing that the modulus of elasticity (E) of these
materials are substantially different, it becomes apparent
that the dimensions of the parts will need to be changed
to adjust the moment of inertia (I).
For example, if the metal part is produced from
magnesium with a modulus of 6.5 psi x 10 6 psi
(44.9 GPa) and the replacement plastic material is Udel
GF-130 with a modulus of 1.07 psi x 10 6 psi (7.38 GPa),
then the required increase in the moment of inertia can
be calculated from Equation 2.
Equation 2
(6.5 psi x 10 6 psi)(I magnesium )=(1.07 psi x 10 6 psi)(I Udel )
or (44.9 GPa)(I magnesium )=(7.38 GPa)(I Udel )
I Udel =6.07 I magnesium
Increasing Section Thickness
One way to increase the moment of inertia is to increase
the section thickness.
From Table 39, the formula for the moment of inertia of a
rectangular section is:
Substituting into Equation 2 and eliminating
common factors:
If the section thickness in magnesium was 0.100 inch
(2.54 mm), then
3
d
Udel
= 6.07(.001)
or
d
3
6.07d 3
Udel
d
Udel
=
magnesium
3
d
Udel
= 6.07(16.38)
d
Udel
To achieve equivalent stiffness by simply increasing
section thickness requires an 82% increase.
Adding Ribs to Maintain Stiffness
Another method of achieving stiffness is to increase the
moment of inertia by adding ribs. The use of ribs can
reduce wall thickness and weight, and still provide the
required stiffness.
Because the material is the same as in the previous
example, the moment of inertia of the rib design is equal
to the moment of inertia of the 0.182 inch (4.63 mm)-thick
plate. The moment of inertia of the plate is given by:
I plate
=
=
0.182 inch
4.63 mm
bd 3
12
If we assign 1 inch (25.4 mm) to the width, then I = 5.02
x 10 -4 in 4 (209 mm 4 ).
Calculating the rib and wall thickness that will give
the same moment of inertia is done using the following
equations for the distance from the neutral axis to the
extreme fiber (c), the moment of inertia (I), and the
area (A).
d 2 t + s 2 (b − t)
c = d −
2(bs + ht)
I =
=
tc 3 + b(d − c) 3 − (b − t)(d − c − s) 3
3
I =
bd 3
12
A = bs + ht
where b is the width and d is the thickness of the section.
Design Information
Udel ® Polysulfone Design Guide
45