Design Guide - Solvay Plastics

length to its diameter) and the degree of fiber orientation.
Perpendicular to the fiber orientation, the fibers act more
as fillers than as reinforcing agents.
When molding polymers, there are instances where melt
fronts meet (commonly known as weld lines) such as
when the plastic melt flows around a core pin. However,
the reinforcement in the plastic, if present, does not
cross the weld line. Thus the weld line does not have
the strength of the reinforced polymer and at times can
even be less than the matrix polymer itself. These factors
must be taken into account when designing parts with
reinforced plastics.
Designing for Equivalent Part Stiffness
Sometimes, a design engineer wants to replace a metal
part with one made of plastic, but still wants to retain
the rigidity of the metal part. There are two fairly simple
ways to maintain the stiffness of a part when substituting
one material with another material — even though the
materials have different moduli of elasticity.
In the first method, the cross sectional thickness is
increased to provide the stiffness. In the second, ribs are
added to achieve greater stiffness. An example of each
approach follows.
Changing Section Thickness
In reviewing the deflection equations in Table 46, the
deflection is always proportional to the load and length
and inversely proportional to the modulus of elasticity
and moment of inertia.
Selecting one case, for example, both ends fixed
with a uniformly distributed load, the deflection is
determined by:
Y
FL
=
3
384EI
Therefore, to equate the stiffness using two different
materials, the deflections are equated as follows:
FL 3
384EI
= Y =
metal
FL 3
384EI
plastic
If the metal part were magnesium having a modulus of
elasticity E of 44.8 GPa (6.5 Mpsi) and the thermoplastic
chosen to replace it was Amodel AS-1145 HS resin
having a modulus of elasticity of 13.8 GPa (2.0 Mpsi),
we need to increase the moment of inertia I of the plastic
version by increasing the part thickness or adding ribs.
Substituting the E values in equation 1:
(44.8 × 10 9 )I metal = (13.8 × 10 9 )I Amodel
3.25 I metal = I Amodel
From Table 47, the moment of inertia for rectangular
sections is:
I =
bd 3
12
where b is the width and d is the thickness of the
section, substituting into our equation to determine the
required thickness yields:
3.25 d 3 metal = d 3 Amodel
If d for the metal part is 2.54 mm (0.10 in.) then the
thickness is Amodel resin is:
d
Amodel
3
= 3.25(2.54) 3 = 3.76 mm (0.148 in.)
or 48% thicker than the magnesium part. However, ribs
can be used to effectively increase the moment of inertia
as discussed in the next section.
Adding Ribs to Maintain Stiffness
In the last section, it was determined that if a metal
part were to be replaced with a part molded of Amodel
AS-1145 HS resin, a part thickness of 3.7 mm (0.148 in.)
would be required to equal the stiffness of a 2.5 mm
(0.100 in.) thick magnesium part.
By incorporating ribs into the Amodel design, the wall
thickness and weight can be reduced very efficiently
and yet be as stiff as the magnesium part.
Since the load and length are to remain the same, the
FL 3 becomes a constant on both sides of the equation
and what remains is:
Equation 1
{EI} metal = {EI} plastic
This, then, is the governing equation for equating
part stiffness.
68 Amodel ® PPA Design Guide
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